diff --git "a/.ragatouille/colbert/indexes/cbseclass10index/collection.json" "b/.ragatouille/colbert/indexes/cbseclass10index/collection.json" --- "a/.ragatouille/colbert/indexes/cbseclass10index/collection.json" +++ "b/.ragatouille/colbert/indexes/cbseclass10index/collection.json" @@ -1,8 +1,84 @@ [ - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-VERY SHORT QUESTIONS.txt\nVery Short Answer Type Question [1 mark] \nOne Sentence Answer \n1. What is a redox reaction? \n2. What is corrosion? Explain its advantage and disadvantage. \n3. What is rancidity? How can we reduce the problem of rancidity? \n4. How is corrosion different from rusting? \n5. What is meant by endothermic and exothermic reactions? Give suitable example \nfor each. \n6. Define different types of chemical reaction and give examples for each. \n7. Why is photosynthesis considered as an endothermic reaction? \n8. In electrolysis of water, why is the volu me of gas collected over one electrode \ndouble that of the other electrode? \n9. What happens when water is added to solid calcium oxide taken in a container? \nWrite a chemical formula for the same. \n10. Give three types of decomposition reaction. \n11. Name the compound us ed for testing CO2-gas.", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-VERY SHORT QUESTIONS.txt\nDirection (Q 12 to 16): In the following Questions, the Assertion and Reason have \nbeen put forward. Read the statements carefully and choose the correct \nalternative from the following:", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-VERY SHORT QUESTIONS.txt\n(a) Both the Assertion and Reason are correct and the reason is the correct \nexplanation of the Assertion. \n(b) The Assertion and the reason are correct but the Reason is not the correct \nexplanation of the Assertion. \n(c) Assertion is true but the Reason is false. \n(d) The statement of the Assertion is false but the reason is true. \n12. Assertion : AgBr is used on photographic and X -ray film \nReason : AgBr is photosensitive and changes to Ag and bromine in presence \n of sunlight and undergoes decomposition reaction. \n13. Assertion: Magnesium ribbon keeps on burning in atmosphere of nitrogen. \nReason : Magnesium reacts with nitrogen to form magnesium nitride and \n this reaction is combination reaction . \n14. Assertion :Zinc reacts with sulphuric acid to form to form zinc sulphate and \n Hydrogen gas and it is displacement reaction. \nReason : Zinc reacts with oxygen to form Zinc oxide", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-VERY SHORT QUESTIONS.txt\nHydrogen gas and it is displacement reaction. \nReason : Zinc reacts with oxygen to form Zinc oxide \n15. Assertion : MnO 2 + 4 HCl ---\uf0e0 MnCl 2 + Cl 2 + 2 H 2O is r edox reaction. \nReason : MnO 2 oxides HCl to Cl 2 and gets reduced to MnCl 2 \n16. Assertion : lead nitrate on thermal decomposition gives lead oxide , brown \n coloured nitrogen dioxide and oxygen gas . \nReason : Lead nitrate reacts wi th potassium iodide to form yellow ppt of lead \niodide \n And the reaction is double displacement as well as precipitation \nreaction.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nTest items \n \n \n \nLOB: Compare the characteristics of initial & final substances in order to check whether the \nchange is physical or chemical \n \n1) A student poured 100 mL of water in a bottle and added 40 mL vinegar to it. A balloon was filled \nwith 20 g baking soda and was fixed at the mouth of the bottle. Slowly the shape of the balloon \nchanged, as shown. \n \n \n \nThe student claims that a chemical change happened when the two substances were mixed. Is \nthe claim made by the student correct? \n \n(a) Yes, as a new substance was formed in the form of a gas. \n(b) Yes, as the mass remains the same throughout the experiment. \n(c) No, as the formation of bubbles in the mixture shows a physical change. \n(d) No, as the change in the shape and size of the balloon shows a physical change.", + "(c) No, as the formation of bubbles in the mixture shows a physical change. \n(d) No, as the change in the shape and size of the balloon shows a physical change. \nCorrect Answer: Option (a) \n \n2) A student makes a list of some activities he observes one day.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\n2) A student makes a list of some activities he observes one day. \n \n \n \n \nWhich activity can the student classify as a chemical change? \n \n(a) Activity 1, as the properties of the substances in the mixture change. \n(b) Activity 2, as the physical state of the apple pie changes when cut. \n(c) Activity 3, as the shape of the can changes. \n(d) Activity 4, as the shape and size of the wooden log changes. \nCorrect Answer: Option (a) \nLOB : Relate the substances taking part in the chemical reaction & substances formed in the \nchemical reaction in order to classify them as reactants & products \n1) Sodium and chlorine are reacted and as a result, sodium chloride is formed which is also called \ntable salt. What option gives the reactants and products of the reaction?", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\n(a) reactants -sodium; products - chlorine \n(b) reactants -sodium and table salt; products - chlorine \n(c) reactants -tables salt; products - sodium and chlorine \n(d) reactants -sodium and chlorine; products - sodium chloride \nCorrect Answer: Option (d) \n \n \n2) The image shows some chemical reactions. \n \n \n \nWhich option identifies the reactants and products of the reactions? \n \n(a) \n(b) \n(c) \n(d) \nCorrect Answer: Option (c) \nLOB : Use chemical symbols & chemical formulae correctly in order to acquire the skill of writing \nchemical equations \n1) A student performs an experiment to form aluminium chloride from aluminium and chlorine. \nWhich options gives the chemical equation of the reaction?", + "Which options gives the chemical equation of the reaction? \n(a) Al + Cl2 \uf0e0\uf020AlCl 2 \n(b) 2Al + Cl2 \uf0e0\uf0202AlCl \n(c) 2Al + 3Cl 2 \uf0e0\uf0202AlCl 3 \n(d) 3Al + 3Cl 2 \uf0e0\uf0203AlCl 3 \n \nCorrect Answer: Option (c)", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nCorrect Answer: Option (c) \n \n \n2) A researcher adds barium hydroxide to hydrochloric acid to form a white -colored barium \nchloride. Which option gives the balanced chemical equation of the reaction? \n \n(a) HCl + Ba(OH) 2 \uf0e0\uf020BaCl 2 + 2HOH \n \n(b) 2HCl + Ba(OH) 2 \uf0e0\uf020BaCl 2 + 2HOH \n \n(c) 2HCl + Ba(OH) 2 \uf0e0\uf020BaH 2 + 2HCl + O2 \n \n(d) HCl + 2Ba(OH) \uf0e0\uf0202BaCl 2 + 2HOH + O2 \n \nCorrect Answer: Option (b) \n \n \n \n \n \n \n \nLOB : Apply Law of Conservation of Mass in order to balance chemical equations \n1) A student writes a balanced chemical equation.", + "Pb(s) + CuCl 2(aq) \uf0ae\uf020PbCl 2(aq) + Cu(s) \nWhich option gives the number of elements on the LHS and RHS of the chemical equation? \n(a) \n \n \n(b) \n \n \n \n(c) \n \n \n(d) \n \n \n \nCorrect Answer: Option (d) \n \n \n2) The image shows a balanced chemical equation of the reaction between sodium and chlorine to \nform sodium chloride. \n \n \n \nWhich option shows the number of atoms on both sides of the reaction? \n(a)", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nWhich option shows the number of atoms on both sides of the reaction? \n(a) \n \n \n(b) \n \n \n(c) \n \n \n(d) \n \n \nCorrect Answer: Option (b)", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nLOB: Categorize the given reactions as (combination/ decomposition) based on the reactants & \nproducts of a chemical reaction \n1) A student writes a chemical equation of the reaction between carbon monoxide and hydrogen. \nCO 2 + 2H 2 -> CH 3OH \nHow can the reaction be classified? \n(a) The reaction is an example of a combination reaction as a compound separates into two \ncompounds. \n(b) The reaction is an example of a decomposition reaction as a compound dissociates into \ntwo compounds. \n(c) The reaction is an example of a combination reaction as two compounds react to form a \nsingle compound. \n(d) The reaction is an example of a decomposition reaction as two compounds react to form \na single compound.", + "(c) The reaction is an example of a combination reaction as two compounds react to form a \nsingle compound. \n(d) The reaction is an example of a decomposition reaction as two compounds react to form \na single compound. \nCorrect Answer: Option (c) \n2) A student learns that some products are formed as a result of combining two compounds while \nsome compounds are formed as a result of dissociation of two compounds. The image shows", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nsome compounds are formed as a result of dissociation of two compounds. The image shows \ntwo reactions.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nWhich reaction is an example of a combination reaction and a decomposition reaction? \n(a) both the reactions are examples of combination reaction \n(b) both the reactions are examples of a decomposition reaction \n(c) reaction P is an example of a combination reaction while reaction Q is an example of a \ndecomposition reaction \n(d) P is an example of a decomposition reaction while reaction Q is an example of a \ncombination reaction \nCorrect Answer: Option (c)", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nLOB: Classify the given reaction as displacement or double displacement based on the type of \nreactants used & products formed \n1) A student adds lead and silver to two different test tubes containing an equal amount of copper \nsulphate solution. The student observes that the color of the solution in the test tube with lead \nchanges. What explains the change in the colour of the solution? \n(a) A displacement reaction takes place as lead replaces copper from the solution. \n(b) A combination reaction takes place as lead combines with sulpha te in the solution. \n(c) decomposition reaction takes place as copper dissociates from sulphate in the solution. \n(d) A double displacement reaction takes place as copper dissociates from sulphate and \nlead combines with sulphate in the solution.", + "(c) decomposition reaction takes place as copper dissociates from sulphate in the solution. \n(d) A double displacement reaction takes place as copper dissociates from sulphate and \nlead combines with sulphate in the solution. \nCorrect Answer: Option (a) \n2) The chemical reaction between potassium chloride and silver nitrate is given by the chemical", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nlead combines with sulphate in the solution. \nCorrect Answer: Option (a) \n2) The chemical reaction between potassium chloride and silver nitrate is given by the chemical \nequation.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nWhat can be inferred from the chemical equation? \n(a) silver nitrate and potassium undergo a decomposition reaction to form silver chloride \nand potassium nitrate \n(b) silver nitrate and potassium undergo a displacement reaction to form silver chloride \nand potassium nitrate \n(c) silver nitrate and potassium undergo a combination reaction to form silver chloride and \npotassium nitrate \n(d) silver nitrate and potassium undergo double displacement reaction to form silver \nchloride and potassium nitrate \nCorrect Answer: Option (d) \n \n \nLOB: Predict the reaction as Oxidation or Reduction based on the addition/ removal of oxygen/ \nhydrogen/ electrons to the reactants to form products. \n1) The image shows a reaction between zinc and hydrogen. \n Which option shows oxidation? \n(a) Zn \uf0ae\uf020Zn!\" \n(b) 2H! \uf0ae\uf020H2 \n(c) Zn!\"", + "1) The image shows a reaction between zinc and hydrogen. \n Which option shows oxidation? \n(a) Zn \uf0ae\uf020Zn!\" \n(b) 2H! \uf0ae\uf020H2 \n(c) Zn!\" \uf0ae\uf020Zn \n(d) H 2 \uf0ae2H! \nCorrect Answer: Option (a) \n \n \n2) The image shows a reaction between iron oxide and hydrogen.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\n2) The image shows a reaction between iron oxide and hydrogen. \n \n \nWhich option shows the compounds undergoing oxidation and reduction? \n(a) \n \n \n \n(b) \n \n \n(c) \n \n \n(d) \n \n \nCorrect Answer: Option (a) \n \n \n \n \n \nLOB: Observe colour change in iron, copper and silver articles over time in order to outline the \neffects of corrosion in our surroundings (real life situations, stating any two)", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nLOB: Observe colour change in iron, copper and silver articles over time in order to outline the \neffects of corrosion in our surroundings (real life situations, stating any two) \n \n1) A student notices that a new hammer made of iron is shiny while an old one kept in the toolbox \nfor long has a reddish -brown powder deposit over it. What does the change in colour of the \nhammer indicate? \n(a) effect of moisture on metals \n(b) iron hammer turns brown after some time \n(c) effects of kept in a box for a longer duration \n(d) iron changes colour when kept with other tools \nCorrect Answer: Option (a) \n2) A student notices that her silver jewellery turned dull and had a gray -black film over it after \nwearing for a few months. What results in the change in colour of the silver metal?", + "What results in the change in colour of the silver metal? \n(a) dust deposits over the jewellery which changes its colour", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\n(b) the jewellery comes in contact with air, moisture, and acids and corrodes \n(c) the polish over the jewellery was removed after wearing for a few months \n(d) silver breaks due to wear and tear and turns its colour changes due to rusting \nCorrect Answer: Option (b) \n \n \nLOB: Detect changes in smell, colour, taste of food items overtime, in order to explain effects of \noxidation on food items \n \n1) A student learns that food companies fill bags of chips with nitrogen gas. What is the purpose \npacking it with nitrogen? \n(a) it prevents rancidity of chips \n(b) it keeps the mosquitoes away from chips \n(c) it keeps the chips dry if the pack falls in water \n(d) prevents chips from spilling out when the pack is opened \n \nCorrect Answer: Option (a)", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nCorrect Answer: Option (a) \n \n \n \n \n2) A student notices that the bread kept out has a green coloured coating over it after a few days. \nWhat explains the reason for the student\u2019s observation? \n(a) the oils in the bread oxidises and causes rancidity \n(b) bread comes in contact with atmospheric moisture and corrodes \n(c) the oils in the bread reduces and cause the change in the colour of the bread \n(d) comes in contact with the atmospheric nitrogen and a layer deposit over it", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nCorrect Answer: Option (a) \nSuggested Teacher Resources \n1 Activity \nObjective Categorize the given reactions as (combination/ decomposition) based on the \nreactants & products of a chemical reaction. \nPrerequisite Identify the changes that a reaction mixture undergoes during a chemical \nreaction, what are reactants, what are products \nMaterial \nRequired Test tubes, boiling tube, distilled water, burner, sugar, iron fillings, sulphur \npowder \nVocabulary Combination Reaction: A reaction in which a single product is formed from \ntwo or more reactants is known as a combination reaction. \n \nDecomposition Reaction: A reaction in which a single substance decomposes \nto give two or more substances. \n \nProcedure Procedure 1: \n1. Take a small amount of iron fillings and sulphur powder in a boiling \ntube. \n2.", + "Procedure Procedure 1: \n1. Take a small amount of iron fillings and sulphur powder in a boiling \ntube. \n2. Heat the boiling tube strongly over the flame of a burner or spirit lamp \nand observe changes.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nProcedure 2: \n1. Take about 2 g sugar crystals in a dry boiling tube. (Note the colour of \nthe crystals.) \n2. Heat the boiling tube over the flame of a burner or spirit lamp and \nobserve changes. \nReflection \nQuestions \nText to real \nworld \nconnection Propane is a fuel used to provide heat for some homes. It is stored in large \ntanks as shown here. Can you predict the chemical equation for burning of \npropane? \n \n \nBeyond the \nclassroom What type of reaction takes place inside the human and plant cells? \n \n \n \n \nMaterial \nrequired Fresh potato chips, 2 glass jars with lid, Aluminium foil. \nProcedure 1. Wrap the glass jar with aluminium foil. Tape the foil in place so that no \nlight can enter the container. \n2.", + "Procedure 1. Wrap the glass jar with aluminium foil. Tape the foil in place so that no \nlight can enter the container. \n2. Place fresh potato chips in the foil -wrapped jar and in a similar clear jar \nwithout foil around it.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nActivity \n Reaction No. Materials \nused Chemicals \nused Describe the \nreactants \n(number of \nreactants) Procedure Describe \nthe \nproducts \n(number of \nproducts) Type of \nreaction \nCalcium \nOxide + \nwater \nHeating \nSugar \n \n 3. Taste the potato chips and rate their flavour on a 5-point scale, 1 being \nextremely dislike the flavour and 5 being extremely like the flavour. \n(Enter your data in a table.) \n4. Place the two jars on a window sill where they will be exposed to \nsunlight. \n5. Taste potato chips from each jar at intervals of 1-2 days for 1-2 weeks \nand record the taste of potatoes in both the jars. \n6. Make a graph of your data, noting the flavor of the potato chips stored \nthese two ways versus storage time.", + "6. Make a graph of your data, noting the flavor of the potato chips stored \nthese two ways versus storage time. The y-axis should be the flavor \nscore and the x-axis the time in days. \nReflection \nquestions 1. What effect does the aluminium foil have on the taste of potatoes after 2 \nweeks?", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nscore and the x-axis the time in days. \nReflection \nquestions 1. What effect does the aluminium foil have on the taste of potatoes after 2 \nweeks? \n2. Is there anything else that will affect the taste of potatoes over 2 weeks \napart from the sunlight?", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-IMPORTANT NOTES.txt\nFocus Points \n1.Chemical Reaction: During chemical reactions, the chemical composition of \nsubstances changes or new substances are formed. \n2. Chemical Equation: Chemical reactions can be written in chemical equation form \nwhich should always be balanced. \n3. Types of Chemical Reactions: \nCombination reaction: A single product is formed from two or more reactants. \n2Mg + O 2 \u2192 2MgO \nDecomposition reaction: A single reactant breaks down to yield two or more \nproducts. \n\uf0b7 Thermal decomposition: 2Pb(NO 2)2 \u2192 2PbO + 4NO 2 + O 2 \n\uf0b7 Electrolysis: 2H20 \u2192 2H 2 + O 2 \n\uf0b7 Photochemical reaction: 2AgBr \u2192 2Ag + Br 2 \nDisplacement reaction: One element is displaced by another element. \nZn + CuSO 4 \u2192 ZnSO 4 + Cu \nDouble displacement reaction: Exchange of ions between reactants.", + "Zn + CuSO 4 \u2192 ZnSO 4 + Cu \nDouble displacement reaction: Exchange of ions between reactants. \nAgNO 3 + NaCl \u2192 AgCl + NaNO 3 \nRedox reaction: Both oxidation and reduction take place simultaneously. \nCuO + H 2 \u2192 Cu + H 2O", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-IMPORTANT NOTES.txt\nDouble displacement reaction: Exchange of ions between reactants. \nAgNO 3 + NaCl \u2192 AgCl + NaNO 3 \nRedox reaction: Both oxidation and reduction take place simultaneously. \nCuO + H 2 \u2192 Cu + H 2O \nExothermic reaction: A chemical reaction in which heat energy is evolved. \nC + O 2 \u2192 CO 2 (g) + heat \nEndothermic reaction: A chemical reaction in which heat en ergy is absorbed. \nZnCO 3 + Heat \u2192 ZnO + CO 2 \nRedox reaction: Chemical reaction in which both oxidation and reduction take place \nsimultaneously. \n4. Oxidation: Reaction that involves the gain of oxygen or loss of hydrogen. 5. Reduction: Reaction that shows the loss of oxygen or gain of hydrogen. \n ZnO + C \u2192 Zn + CO \n ZnO is reduced to Zn \u2014reduction. \n C is oxidized to CO \u2014Oxidation. \n6.", + "5. Reduction: Reaction that shows the loss of oxygen or gain of hydrogen. \n ZnO + C \u2192 Zn + CO \n ZnO is reduced to Zn \u2014reduction. \n C is oxidized to CO \u2014Oxidation. \n6. Effects of Oxidation Reactions in Our Daily Life: \n\uf0b7 Corrosion: It is an undesirable change that occurs in metals when they are \nattacked by moisture, air, acids and bases.", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-IMPORTANT NOTES.txt\n6. Effects of Oxidation Reactions in Our Daily Life: \n\uf0b7 Corrosion: It is an undesirable change that occurs in metals when they are \nattacked by moisture, air, acids and bases. \nExample, Corrosion (rusting) of Iron: Fe 2O3. nH 2O (Hydrated iron oxide) \n\uf0b7 Rancidity: Undesirable change that takes place in oil containing food items \ndue to the oxidation of fat ty acids. \nPreventive methods of rancidity: Adding antioxidants to the food materials, \nstoring food in the airtight container, flushing out air with nitrogen gas and \nrefrigeration.", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nChemical Reactions and Equations \nChemical Reactions and Equations \nAny process that involves the rearrangement of structure of the substance or conversion of \nreactants into products is defined as Chemical Reaction . \nFor a Chemical Reaction to occur, the change can be observed in the form of - \n\uf0b7 Change in State : Melting of ice into water. \n\uf0b7 Change in Colour : Iron rusting which has colour change from silver to reddish brown.", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\n\uf0b7 Change in Temperature : There are two types of reaction i.e Exothermic and Endothermic \nReaction. \nExothermic Reactions: Those reactions in which energy is released in the form of heat are \ncalled Exothermic Reactions . \nExamples - \n(1) All combustion reactions e.g. \nCH 4+ 2O 2 \u2014> CO 2 + 2H 2O + Heat \n(2) Thermite reactions e.g. \n2A 1 + Fe 2O3 \u2014> 2Fe + Al 2O3 + Heat \nCombinations are generally exothermic in nature. The decomposition of organic matters into \ncompost is an example of exothermic reaction. \nEndothermic Reactions: Those reactions in which energy is absorbed are called Endothermic \nReactions . \nExamples - \n \nalso, the reaction of photosynthesis -", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nalso, the reaction of photosynthesis - \n \n\uf0b7 Evolution of any gas: When Zinc reacts with sulphuric acid it gives hydrogen gas. \nZn + H2 SO 4 \u2192 ZnSO 4 + H2 \nFormation of Precipitate : When a soluble carbonate reacts with Barium, Barium Carbonate \nprecipitate can be observed. \nChange in State \nSome chemical reactions are characterized by a change in state. \n\uf0b7 When wax is burned (in the form of wax candle,) then water and carbon dioxide are formed. \n\uf0b7 Now, wax is a liquid whereas carbon dioxide is a gas. This means that during the combustion \nreaction of wax, the physical state changes from solid to liquid and gas. \nPhysical Change \n\uf0b7 In this change identity of the substance remains same. \n\uf0b7 For Example, Melting, Boiling etc. \nChemical Change \n\uf0b7 The identity of the substances change \n\uf0b7 Reactants are converted into substance due to formation or broken down of older bonds", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nChemical Equation \nThe symbolic representation of chemical reaction using symbols and formulae is known \nas Chemical Equation . For this, reactants are written in left hand side whereas products are \nwritten on the right. \nBalanced Chemical Equation \nA balanced chemical equation is the one where the number of atoms involved in reactants side is \nequal to number of atoms on product side. \n \nEq.1. Example of Balanced Chemical Equation \nSteps to form Balanced Equation \nTo show how to balance the equation, the following equation is used - \nFe + H2O \u2192 Fe 3O4 + H2 \nStep 1: First of all, draw the boxes around each formula as shown below -", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nStep 2: Find out the number of atoms of each element. For Example , on reactant side, 1 for Fe, \n2 H, and 1 O and on product side we have, 3 for Fe, 4 for O and 2 for H. \nStep 3: Start to balance the equation with the compound having maximum number of atoms. While \nbalancing does not alter the formula of the compound. \nStep 4: One by one balance each element on reactant and product side. \n \nStep 5: After balancing number of atoms on both the side of the equation, finally check the \ncorrectness of the balanced equation. \n \nStep 6: then write the symbols of the physical state of reactants and products as shown below - \n3Fe(s) + 4H 2O (g) \u2192 Fe 3O4 (s) + 4H 2 (g) \nThis above equation represents the balanced equation.", + "Balancing a Redox Reaction \nThe basic ionic form of the equation is - \nFe2+ + Cr 2O72- \u2192 Fe3+ + Cr3+ \nOxidation half reaction is - \n \nReduction half reaction is -", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nReduction half reaction is - \n \nUse the reduction half method to balance the equation. Balance the atoms in each half of the \nreaction except H and O atoms. \nCr 2O72- (aq) \u2192 2 Cr3+(aq) \nAdd water molecules as the reaction is taking place in acidic solution. This is to balance the O \natoms and hydrogen ions. \nCr 2O72- (aq) + 14 H+(aq) \u2192 2 Cr3+(aq) + 7H 2O (I) \nThen balance the charges in both half reactions.", + "This is to balance the O \natoms and hydrogen ions. \nCr 2O72- (aq) + 14 H+(aq) \u2192 2 Cr3+(aq) + 7H 2O (I) \nThen balance the charges in both half reactions. \nFe2+(aq) \u2192 Fe3+(aq) + e- \nCr 2O72- (aq) + 14 H+ + 6e- \u2192 2 Cr3+ + 7H 2O \n6 Fe2+(aq) \u2192 6 Fe3+(aq) + 6e- \nTwo half of the equations are added to get the overall reaction \n6Fe2+(aq) + Cr 2O72-(aq) + 14H+(aq) \u2192 6Fe3+(aq) + 2Cr3+(aq) + 7H 2O (I) \nTypes of Chemical Reaction \n\uf0b7 Combination Reaction is reaction when single product is formed from the combination of two \nor more reactants. For Example -", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nEq.2. Example of Combination Reaction \nReactions can be exothermic as well as endothermic. Exothermic reaction release heats and \nraises the temperature of the surroundings. For Example , Respiration is an example of exothermic \nreaction. \n \nEq.3. Example of Exothermic Reaction \nEndothermic reaction involved the absorption of the heat and thus it cools the surrounding. The \ndecomposition of dead organic material is an endothermic reaction. \n\uf0b7 Decomposition Reaction is type of reaction which involves breakdown of single reactant into \nsimpler products. Decomposition of silver chloride into silver and chlorine in presence of \nsunlight is an example of decomposition reaction. \n \nEq.4. Example of Deco mposition Reaction \n\uf0b7 Displacement Reaction is a reaction in which more reactive element will \ndisplaces the less reactive element.", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nEq.4. Example of Deco mposition Reaction \n\uf0b7 Displacement Reaction is a reaction in which more reactive element will \ndisplaces the less reactive element. \n \nEq. 5. Example of Displacement Reaction \n\uf0b7 Double Displacement Reaction is a type of reaction in which cations and anions in the \nreactants switch the places to form new products. \n \nEq. 6. Example of Double Displacement Reaction \n\uf0b7 Redox Reaction is also known as Oxidation -reduction Reaction . In this type of reaction \ntransfer of electrons occurs between the two species. Oxidation is def ined as addition of oxygen \nor removal of hydrogen. Reduction is defined as removal of oxygen or addition of hydrogen. \nOxidizing agent is the one which gains the electrons and is reduced in a chemical reaction. \nReducing agent is oxidized in a chemical react ion and it loses the electrons. Fluorine is the \nstrongest oxidizing agent. Formic acid is a reducing agent", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nEq.7. Example of Redox Reaction \nCorrosion \nMetals are prone to corrosion. It is a slow conversion of metals into some undesirable compounds. \nThis occur may be due to reaction with oxygen, gases, acids etc. When irons reacts with \natmospheric oxygen and moisture, a red layer is formed on the surface of the iron, this process is \nknown as Rusting . \n \nEq. 8. Equation for Iron Rusting \nRancidity \nWhen food containing fats and oils are exposed to the atmosphere, the oxidation of fat and oil \noccurs, this is known as Rancidity . \nMethods to Prevent Rancidity \n\uf0b7 Store cooking oils from direct sunlight. \n\uf0b7 Food should be placed at low temperature. \n\uf0b7 By adding antioxidants food can be protected from rancidity. \n\uf0b7 Packing material should replace the air with nitrogen. \n\uf0b7 Minimize the use of salts in fried foods.", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nLESSON PLAN \nClass: 10 Chapter: 1 ( Chemical Reactions and Equations )", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nTopics to be covered Chemical reactions, writing chemical equations, Balancing \nchemical equations, types of chemical reactions -\ncombination reactions, decomposition reactions, \ndisplacement reactions, double displacement reactions, \nneutralization reactions , precipitation reactions, redox \nreactions \nPrevious Knowledge \u2022 Students know differences between physical \nchange and chemical change \n\u2022 Studied the symbols of elements and chemical \nformulae of molecules. \n\u2022 Also studied law of conservation of mass. \nLearning Objectives \u2022 To understand the chemical reactions in daily life \n\u2022 Able to write the chemical equations from the \nstatements \n\u2022 Understand how to balance chemical reactions \n\u2022 Study different types of chemical reactions \n\u2022 Define the terms reactants, products, precipitate, \nneutralization etc. \nLearning Experiences \u2022 In combination reactions two or more substances \ncombine to form a single compound . \n\u2022 In decomposition reactions a single compound", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nneutralization etc. \nLearning Experiences \u2022 In combination reactions two or more substances \ncombine to form a single compound . \n\u2022 In decomposition reactions a single compound \nbreaks down into two or more substances. \n\u2022 Displacement reactions are the reactions in which a \nmore active element displace a less active element \nfrom its solution. \n\u2022 In double displacement reactions ions are \nexchanged between two compounds. \n\u2022 REDuction + Oxidation = REDOX reaction \nResources NCERT book \nReference book \u2013Prdeeps Publications \nYou tube ,education channels", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nTeaching Aids Usage of ICT( power point) multimedia., Lab activities( \nvirtual lab also) \n \nProcedure Class starts with the introduction \nThe topics are presenting with power point , audiovisual \naids, virtual lab and lab activities \nAt the end homework and assignments are given \n \nAssessment Methods Oral questioning , class tests and monthly tests", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\n1 \n SALWAN PUBLIC SCHOOL , RAJENDRA NAGAR \nCLASS \u2013 X \nLESSON PLAN \nCHAPTER -1: CHEMICAL REACTIONS AND EQUATIONS \nCHAPTER -1: Science :( CHEMICAL REACTIONS AND EQUATIONS ) \nClass Transaction \n \n \n \n \n(10-12 days) PART -II", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\n(10-12 days) PART -II \n \nTotal: 5 periods (approx. 35 min each) \nSub-Topic: Types of chemical reactions -Combination Reaction and different \nactivities. \nSub-Topic: Types of chemical reactions -Decomposition Reaction and different \nactivities. \nSub-Topic: Types of chemical reactions -Displacement Reaction and different \nactivities. \nSub-Topic : Type of chemical reactions \u2013Double displacement reaction and \ndifferent activities. \nSub-Topic :- Oxidation and reduction reaction (Corrosion and rancidity) \nPre-requisite for the \ncourse (KNOWLEDGE) This lesson requires: \n\uf0b7 Knowledge of the basic properties and mechanisms of chemical \nreactions. \n\uf0b7 The nature and the identity of the initial substance have somewhat \nchanged. \n\uf0b7 What is actually meant by a chemical reaction? How do we come to \nknow that a chemical reaction has taken place? \n \nLearning Objectives", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nLearning Objectives \n \n Students will be able to: \n\uf0b7 Classify a chemical reaction as a combination, decomposition, \nsingle replacement, double replacement, exothermic or \nendothermic reaction. \n\uf0b7 Differentiate between the types of chemical reactions. \n\uf0b7 Predict the products of a chemical reaction. \n\uf0b7 Understand the different sets of conditions required such as \ntemperature, light etc in reactions. \n\uf0b7 Know the effects of oxidation reactions in everyday life like \ncorrosion and rancidity.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nTeachers ' activity \n Introduction: Teacher will introduce the topic by performing various \nactivities: \nEvidence -based active -learning instructional strategies : \n\uf0b7 Problem solving (investigating and applying established principles \nto justify an observation) Activity: - \nExample -1-: Formation of calcium hydroxide \nExample -2: Heating of calcium carbonate \nExample -3: Electrolysis of water 2 \n Example -4: Reaction of iron nails with copper sulphate solution \nExample -5: Reaction between Sodium Sulphate and Barium Chloride \nExample -6: Reaction between Cu and O 2; CuO and H 2. \n\uf0b7 Critical thinking and collaboration (Competitive evaluation of given \ninformation) \nRecall Activity : \nYou are given test tube, the solutions of lead (II) nitrate and potassium iodide \nand are instructed to mix together in a test tube. Give answers by activity : \n(i)What was the colour of the precipitate formed? Can you name the \n compound precipitated?", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nand are instructed to mix together in a test tube. Give answers by activity : \n(i)What was the colour of the precipitate formed? Can you name the \n compound precipitated? \n(ii)Write the balanced chemical equation for this reaction. \n(iii)Is this also a double displacement reaction?", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\n\uf0b7 The ability to understand the effects of oxidation reactions in \neveryday life? \nExample - Browning of cut surfaces of fruits. Formation of metal oxide on \nmetal like iron rust -Corrosion, Burning of crackers , Rancidity: \n\uf0b7 Recapitulation (working effectively with others): \n Example -Discuss brain storming questions in class in group. \n \nLearning Outcomes Students will be able to know \n\uf0b7 displacement and double displacement reaction. \n\uf0b7 precipitation reactions which produce insoluble salts. \n\uf0b7 importance of Redox reaction in ever day life. Oxidation is the gain of \noxygen or loss of hydrogen. Reduction is the loss of oxygen or gain of \nhydrogen. \n\uf0b7 molecular interactions and chemical reactions in the body .", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nAssessment \uf0b7 Continuously monitor student progress by asking oral questions. \n\uf0b7 Providing active involvement of students in their own learning by \ngiving daily Practice Problems in the class. \n\uf0b7 Peer assessment in Lab Activities \n\uf0b7 Written test \nSelf-Study, \nAssignments , \nHomework \uf0b7 Self-Study (Independent Practice) -The teacher would ensure that \neach student should bring NCERT Book and tries to be independent \nat the work as well as be under the teacher\u2019s guidance. \n\uf0b7 Guided practice followed by Independent Practice ( Question \nBank): \nOn completion of the chapter, students will be answer the following \nrevision questions: \nQ.1.Give some examples of combination and decomposition reactions. \nQ.2. What happens when Ferrous sulphate crystal is heated? Write the \nequations also. \nQ.3.Dissolve 0.5 gram silver nitrate 10 ml of water in a test tube and place a 3 \n Copper wire in a test tube. \nExplain your observation.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nequations also. \nQ.3.Dissolve 0.5 gram silver nitrate 10 ml of water in a test tube and place a 3 \n Copper wire in a test tube. \nExplain your observation. \nQ.4. What is the difference between displacement and double displacement \nreactions? Write equations for these reactions. \nQ.5.Take about 2 g silver bromide in a china dish and Place this china dish in \nsunlight for some time. \n(a)Observe the colour of the silver bromide after some time \n(b).Name type of the reaction. \nQ.6.An iron knife kept dipped in a blue copper sulphate solution turns the \nblue solution light green. Why? \nHomework: The students, with the help of the teacher, will solve the \nquestions from NCERT back e xercise in their notebook.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\n1 \n SALWAN PUBLIC SCHOOL , RAJENDRA NAGAR \nCLASS \u2013 X \nLESSON PLAN \nCHAPTER -1: CHEMICAL REACTIONS AND EQUATIONS \nCHAPTER -1: Science :( CHEMICAL REACTIONS AND EQUATIONS ) \nClass Transaction \n \n \n \n(9-10 days) PART -I \n \nTotal: 4 periods (approx. 35 min each) \n \nSub-Topic: Introduction regarding situations of daily life -Physical and \n chemical changes \nSub-Topic: Chemical equations(writing equations) \n \nSub-Topic: Chemical equations(writing balancing equations) \n \nSub-Topic : Type of chemical reactions (combina tion and decomposition) \n \nPre-requisite for the \ncourse (KNOWLEDGE) This lesson requires: \n\uf0b7 Basic knowledge of the situations of daily life and think what happens \nwhen physical and chemical changes take place \n\uf0b7 The nature and the identity of the initial substance have somewhat \nchanged. \n\uf0b7 What is actually meant by a chemical reaction?", + "\uf0b7 What is actually meant by a chemical reaction? How do we come to \nknow that a chemical reaction has taken place? \n \nLearning Objectives", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nLearning Objectives \n \n Students will be able to: \n\uf0b7 recognize the Physical and chemical processes associated with \nbiological and industrial processes affecting life and the \nenvironment. \n\uf0b7 understand chemical reaction they encounter every day. \n\uf0b7 represent equations to represent the reactions . \n\uf0b7 apply the principles of conservation of mass to balance chemical \nreactions \n\uf0b7 identify the different types of chemical reaction and explain.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nTeachers ' activity \n Introduction: Teacher will introduce the topic by asking various questions \nfrom previous knowledge : \n\uf0b7 Analytical skills (scrutinizing an observation): To identify physical \nand chemical changes. Example -Paper is torn or Paper is burnt, ice \nis melting or iron nail is kept exposed to moist air. \n Evidence -based active -learning instructional strategies : \n\uf0b7 Problem solving (investigating and applying established principles \nto justify an observation) \n 2 \n \nExample - Activity: -Burning of a magnesium ribbon in air and collection of \nmagnesium oxide in a watch -glass. \nConcept insight: Always remember that magnesium being a reactive metal will \nreact with oxygen if kept in open. So, it has to be cleaned before it is burned in \nair. \nActivity: -Formation of hydrogen gas by the action of dilute hydrochloric acid on \nzinc. \nConcept insight: The key to this answer is to remember that metals react with", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nair. \nActivity: -Formation of hydrogen gas by the action of dilute hydrochloric acid on \nzinc. \nConcept insight: The key to this answer is to remember that metals react with \nhydrochloric acid to form respective metal chlorides an d liberate hydrogen gas. \n\uf0b7 Critical thinking and collaboration (Competitive evaluation of given \ninformation) \nExample - To make a chemical equation more informative, the physical states of \nthe reactants and products are mentioned along with their chemical form ulae \nfrom given information of different activities. \nConcept insight: First convert the word equation into skeletal equation with \nstate and then balance the different atoms on both sides of the equation. \n\uf0b7 The ability to understand complex concepts (Inspect ing and analyzing \nreasons for variation in established principles);Types of chemical \nreactions: \nExample - Combination and decomposition reactions : Chemical reactions", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nreasons for variation in established principles);Types of chemical \nreactions: \nExample - Combination and decomposition reactions : Chemical reactions \ninvolve the breaking and making of bonds between atoms to produce new \nsubstances and balan cing equations with the help of law of conservation of \nmass \nConcept insight: Remember that a chemical equation should be balanced to \nfollow the Law of conservation of mass. \nRecapitulation (working effectively with others): \nExample -Discuss brain storming questions in class in group.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nLearning Outcomes Students will be able to know: \n\uf0b7 A complete chemical equation which represents the reactants, \nproducts and their physical states symbolically. \n\uf0b7 How to write the balanced chemical equation . \n\uf0b7 A combination and d ecomposition reactions. \n\uf0b7 Reactions in which heat is given out along with the products are \ncalled exothermic reactions. \n\uf0b7 Reactions in which energy is absorbed are known as endothermic \nreactions.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nAssessment \uf0b7 Continuously monitor student progress by asking oral questions. \n\uf0b7 Providing active involvement of students in their own learning by \ngiving daily Practice Problems in the class. \n\uf0b7 Peer assessment in Lab Activities \n\uf0b7 Written test \nSelf-Study, \uf0b7 Self-Study (Independent Practice) -The teacher would ensure that 3 \n Assignments , \nHomework each student should bring NCERT Book and tries to be independent \nat the work as well as be under the teacher\u2019s guidance. \n\uf0b7 Guided practice followed by Independent Practice ( Question \nBank): \nOn completion of the chapter, students will be answer the following \nrevision questions: \nQ.1.Why should a magnesium ribbon be cleaned before burning in air? \nQ.2. Write a balanced chemical equation with state symbols for the \nreactions: Sodium hydroxide solution (in water) reacts with hydrochloric \nacid so lution (in water) to produce sodium chloride solution and water.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nreactions: Sodium hydroxide solution (in water) reacts with hydrochloric \nacid so lution (in water) to produce sodium chloride solution and water. \nQ.3.Take a few zinc granules in a conical flask or a test tube and a dd dilute \nhydrochloric acid or sulphuric acid to this. \n(a)Do you observe anything happening aroun d the zinc granules? \n(b)Touch the conical flask or test tube. Is there any change in its \ntemperature? \nQ.4.Take a small amount of calcium oxide or quick lime in a beaker , slowly \nadd water to this and touch the beaker. \n(a)Do you feel any change in temperature? \n(b) Write the balanced chemical equation with physical state. \nQ.5.Take about 2 g silver chloride in a china dish and Place this china dish in \nsunlight for some time. \n(a)Observe the colour of the silver chloride after some time \n(b).Name type of the reaction.", + "Q.5.Take about 2 g silver chloride in a china dish and Place this china dish in \nsunlight for some time. \n(a)Observe the colour of the silver chloride after some time \n(b).Name type of the reaction. \nQ.6.Take about 2 g barium hydroxide in a test tube. Add 1 g of ammonium", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nsunlight for some time. \n(a)Observe the colour of the silver chloride after some time \n(b).Name type of the reaction. \nQ.6.Take about 2 g barium hydroxide in a test tube. Add 1 g of ammonium \nchloride and mix with the help of a glass rod. Touch the bottom of the test \ntube with your palm. \n(a)What do you feel? \n(b)Is this an exothermic or endothermic reaction? \nHomework: The students, with the help of the teacher, will solve the \nquestions from NCERT back e xercise in their notebook.", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\nChemical Reactions and Equations \nCompetency Based Questions", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\n1. A compound 'X' used for drinking, has pH =7. Its acidified solution \nundergoes decomposition in presence of electricity to produce gases ' \nY' and 'Z' The volume of Y is double than Z. Y is highly combustible \nwhereas Z is supporter of combustion. identify X. Y & Z and write th e \nchemical reactions involved. \n2. An aqueous solution of metal nitrate P reacts with sodium bromide \nsolution to form yellow not of compound O which is used in \nphotography. O on exposure to sunlight undergoes decomposition \nreaction to form metal present in P a long with reddish brown gas. \nIdentify P & O. Write the chemical reaction & type of chemical reaction. \n3. Bhawana took a pale green substance A in a test tube, and heated it \nover the flame of a burner. A brown coloured residue B was formed \nalong with evolution of two gases with burning smell of sulphur. Identify \nA & B. Write the chemical reaction involved. \n4.", + "A brown coloured residue B was formed \nalong with evolution of two gases with burning smell of sulphur. Identify \nA & B. Write the chemical reaction involved. \n4. A student took 2 -3 g of a substance X in a glass beaker & poured water", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\nalong with evolution of two gases with burning smell of sulphur. Identify \nA & B. Write the chemical reaction involved. \n4. A student took 2 -3 g of a substance X in a glass beaker & poured water \nover it slowly. He observed bubbles along with hissing noise. The \nbeaker becomes quite hot. Identify X. What type of reaction is it? \n5. A substance X used for coating iron articles is added to a blue solution \nof a reddish brown metal Y. The colour of the solution gets discharged . \nIdentify X and Y & also the type of reaction. \n6. A reddish brown ve ssel developed a green coloured solid X when left \nopen in air for a long time. When reacted with dil H 2S04.", + "Identify X and Y & also the type of reaction. \n6. A reddish brown ve ssel developed a green coloured solid X when left \nopen in air for a long time. When reacted with dil H 2S04. it forms a blue \ncoloured solution along with brisk effervescence due to colourless & \nodourless gas Z. X decomposes to form black coloured oxide Y of a \nreddish brown metal along with gas Z. Identify X. Y. & Z. \n7. A student has mixed the solutions of lead (II) nitrate and potassium \niodide,", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\nreddish brown metal along with gas Z. Identify X. Y. & Z. \n7. A student has mixed the solutions of lead (II) nitrate and potassium \niodide, \n(i) What was the colour of the precipitate fo rmed? Can you name the \ncompound precipitated? \n(ii) Write the balanced chemical equation for this reaction, \n(iii) What type of reaction is it? 8. Observe the following activity & answer the Questions", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\na. Do you observe anything happening around the zinc granules \nb. Is there any change in its temperature? \nc. Why is glass tube not dipp ed in dil H 2SO 4? \nd. How is H 2 gas collected by downward displacement or upward \ndisplacement of water? \ne. Is H 2 gas soluble or insoluble in water? \nf. Is H 2 gas heavier or lighter than air? \n9. A reddish brown metal X when heated in presence of oxygen forms a \nblack compound Y which is ba sic in nature when heated with hydrogen \ngas gives back X. Identify X & Y. Write the chemical reaction between Y \n& H 2 .Identify the substance being oxidized & reduced \n10. Name the type of reaction seen in the diagram below. Write the equation \nfor the reaction.", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\n11. A student burnt a metal A found in the form of ribbon. The ribbon burnt \nwith a dazzling flame & a white powder B is formed which is basic in \nnature. Identify A & B. Write the balanced chemical equation. \n12. A student drop ped few p ieces of marble in dilute H Cl contained in a test \ntube. The gas evolved was passed through lime water. What change \nwould be observed in lime water? Write chemical reactions for both the \nchanges observed. \n13. Astha has been collecting silver coins and Co pper coins. One day she \nobserved a black coating on silver coins and a green coating on Co pper \ncoins. Which chemical phenomenon is responsible for these coatings? \nWrite the chemical name of black and green coatings \n14. Identify the type of chemical reaction \n(i) \n(ii) \n15. A student took 2 -3 g o f a substance X in a glass beaker & poured water \nover it slowly. He observed bubbles along with hissing noise.", + "Identify the type of chemical reaction \n(i) \n(ii) \n15. A student took 2 -3 g o f a substance X in a glass beaker & poured water \nover it slowly. He observed bubbles along with hissing noise. The \nbeaker becomes quite hot. Identify X. What type of reaction is it?", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\nover it slowly. He observed bubbles along with hissing noise. The \nbeaker becomes quite hot. Identify X. What type of reaction is it? \n16. A substance X used for coating iron articles is added to a blue sol ution \nof a reddish brown metal. T he colour of the solution gets discharged \nIdentify X and Y & also the type of reaction. \n17. A solution of a substance \u2018X\u2019 is used for white washing \n i. Name the substance \u2018X\u2019 and writes its formula. \n ii. Write the reaction of the substance \u2018X\u2019 named in (i) above with water \n18. A shiny brown coloured element \u2018X\u2019 on heating in air becomes black in \ncolour. Name the element \u2018X\u2019 and the black coloured compound formed. \n19. An aqueous solution of metal nitrate P reacts with sodium bromide \nsolution to form yellow ppt of compound Q which is used in \nphotography. Q on exposure to sunlight undergoes decomposition \nreaction to form metal present in P along with reddish brown gas.", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\nsolution to form yellow ppt of compound Q which is used in \nphotography. Q on exposure to sunlight undergoes decomposition \nreaction to form metal present in P along with reddish brown gas. \nIdentify P &Q. Write the chemical reaction & type of chemic al reaction. \n20. A reddish brown vessel developed a green coloured solid X When left \nopen in air for a long time. When reacted with dil , it forms a blue \ncoloured solution along with brisk effervescence due to colourless & \nodourless gas Z. X decomposes to for m black coloured oxide Y of a \nreddish brown metal along with gas Z, Identify X, Y, & Z. \n21. A metal is heated with dil H 2SO 4. The gas evolved is collected by the \nmethod shown in the figure: Answer the following", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\n(a) Name the gas. \n(b) Name the method of collection of gas. \n(c) Is the gas soluble or insoluble in water? \n(d) Is the gas lighter or heavier than air?", "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-MCQ.txt\nChemical Reactions and Equations \nMCQ Type Questions \n1. Some crystal s of copper sulphate were dissolved in water. The colour of the \nsolution obtained would be \n(a) Green \n(b) Red \n(c) Blue \n(d) Brown \n2. When dilute HCl is added to zinc pieces taken in a test tube \n(a) No change take place \n(b) The colour of the solution becomes yellow \n(c) A pungent smelling gas gets liberated \n(d) A small bubbles of H2 gas appear on the surface of zinc pieces. \n3. PbS reacts with ozone (O3) and forms PbSO4. As per the balanced equation, \nmolecues of ozone required for e very one molecule of PbS is/are \n(a) 4 \n(b) 3 \n(c) 2 \n(d) 1 \n4. Chemically rust is \n(a) Hydrated ferrous oxide \n(b) Hydrated ferric oxide \n(c) Only ferric oxide \n(d) None of these \n5.", "Chemically rust is \n(a) Hydrated ferrous oxide \n(b) Hydrated ferric oxide \n(c) Only ferric oxide \n(d) None of these \n5. Which of the following reactions is not correct \n(a) Zn + CuSO 4 -----\uf0e0 ZnSO 4 +Cu \n(b) 2 Ag + Cu(NO 3)2 --\uf0e0 2 AgNO 3 + Cu \n(c) Fe + CuSO 4 -\uf0e0 FeSO 4 + Cu \n(d) Mg + 2 HCl ---\uf0e0 MgCl 2 + H 2", "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-MCQ.txt\n5. Which of the following reactions is not correct \n(a) Zn + CuSO 4 -----\uf0e0 ZnSO 4 +Cu \n(b) 2 Ag + Cu(NO 3)2 --\uf0e0 2 AgNO 3 + Cu \n(c) Fe + CuSO 4 -\uf0e0 FeSO 4 + Cu \n(d) Mg + 2 HCl ---\uf0e0 MgCl 2 + H 2 \n6. Copper displaces which of the following metals from its salt solution: \n(a) ZnSO 4 \n(b) FeSO 4 \n(c) AgNO 3 \n(d) NiSO 4 \n7. In an electric cell where electrolysis is carried out , anode has: \n(a) Positive charge \n(b) Negative charge \n(c) Connected to negative terminal of the battery \n(d) None of these is correct \n8. The reaction H 2 + Cl 2 --\uf0e0 2 HCl represents: \n(a) oxidation \n(b) reduction \n(c) decomposition \n(d) combination \n9.", @@ -12,6 +88,53 @@ "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-MCQ.txt\nsolution is an example of \n(a) Decomposition reaction \n(b) Displacement reation \n(c) Double displacement reaction \n(d) Neutralization reaction \n16. What happens when dilute hydrochloric acid is added to iro n filings? \n(a) hydrogen gas and iron chloride are produced \n(b) chlorine gas and iron hydroxides are produced \n(c) no reaction takes place \n(d) iron salt and water are produced. \n17. Which of the following gases can be used for the storage of fresh sample of \nan oil for a longer time? \n(a) Carbon dioxide or oxygen \n(b) Nitrogen or helium \n(c) Helium or oxygen \n(d) Nitrogen or oxygen. \n18. In the decomposition of lead (II) nitrate to give lead (II) oxide, nitrogen dioxide \nand oxygen gas, the coefficient of nitrogen dioxide ( in the balanced equation) \nis \n(a) 1 \n(b) 2 \n(c) 3 (d) 4 \n19.", "In the decomposition of lead (II) nitrate to give lead (II) oxide, nitrogen dioxide \nand oxygen gas, the coefficient of nitrogen dioxide ( in the balanced equation) \nis \n(a) 1 \n(b) 2 \n(c) 3 (d) 4 \n19. We store silver chloride in dark coloured bottles because it is \n(a) A white solid \n(b) Undergoes redox reaction \n(c) To avoid action by sunlight \n(d) None of the above.", "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-MCQ.txt\n(b) 2 \n(c) 3 (d) 4 \n19. We store silver chloride in dark coloured bottles because it is \n(a) A white solid \n(b) Undergoes redox reaction \n(c) To avoid action by sunlight \n(d) None of the above. \n20. Silver articles turn blak when kept in the open for a few days due to formation \nof \n(a) H2S \n(b) AgS \n(c) AgSO 4 \n(d) Ag2S", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-VERY SHORT QUESTIONS.txt\nVery Short Answer Type Question [1 mark] \nOne Sentence Answer \n1. What is a redox reaction? \n2. What is corrosion? Explain its advantage and disadvantage. \n3. What is rancidity? How can we reduce the problem of rancidity? \n4. How is corrosion different from rusting? \n5. What is meant by endothermic and exothermic reactions? Give suitable example \nfor each. \n6. Define different types of chemical reaction and give examples for each. \n7. Why is photosynthesis considered as an endothermic reaction? \n8. In electrolysis of water, why is the volu me of gas collected over one electrode \ndouble that of the other electrode? \n9. What happens when water is added to solid calcium oxide taken in a container? \nWrite a chemical formula for the same. \n10. Give three types of decomposition reaction. \n11. Name the compound us ed for testing CO2-gas.", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-VERY SHORT QUESTIONS.txt\nDirection (Q 12 to 16): In the following Questions, the Assertion and Reason have \nbeen put forward. Read the statements carefully and choose the correct \nalternative from the following:", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-VERY SHORT QUESTIONS.txt\n(a) Both the Assertion and Reason are correct and the reason is the correct \nexplanation of the Assertion. \n(b) The Assertion and the reason are correct but the Reason is not the correct \nexplanation of the Assertion. \n(c) Assertion is true but the Reason is false. \n(d) The statement of the Assertion is false but the reason is true. \n12. Assertion : AgBr is used on photographic and X -ray film \nReason : AgBr is photosensitive and changes to Ag and bromine in presence \n of sunlight and undergoes decomposition reaction. \n13. Assertion: Magnesium ribbon keeps on burning in atmosphere of nitrogen. \nReason : Magnesium reacts with nitrogen to form magnesium nitride and \n this reaction is combination reaction . \n14. Assertion :Zinc reacts with sulphuric acid to form to form zinc sulphate and \n Hydrogen gas and it is displacement reaction. \nReason : Zinc reacts with oxygen to form Zinc oxide", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-VERY SHORT QUESTIONS.txt\nHydrogen gas and it is displacement reaction. \nReason : Zinc reacts with oxygen to form Zinc oxide \n15. Assertion : MnO 2 + 4 HCl ---\uf0e0 MnCl 2 + Cl 2 + 2 H 2O is r edox reaction. \nReason : MnO 2 oxides HCl to Cl 2 and gets reduced to MnCl 2 \n16. Assertion : lead nitrate on thermal decomposition gives lead oxide , brown \n coloured nitrogen dioxide and oxygen gas . \nReason : Lead nitrate reacts wi th potassium iodide to form yellow ppt of lead \niodide \n And the reaction is double displacement as well as precipitation \nreaction.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\nVery Short Answer Type Questions [2 Marks]", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n1. \u201cWe need to balance a skeletal chemical equation.\u201d Give reason to justify the \nstatement. \n2. Giving an example list two information which make a chemical equation \nmore useful (informative). \n3. Name the reducing agent in the following reaction: \n 3MnO 2 + 4Al \u2014\u2014\u2014\u2014 > 3Mn + 2Al 2O3 \nState which is more reactive, Mn or Al and why? \n4. (i) Write a balanced chemical equation for process of photosynthesis. \n (ii)When do desert plants take up carbon dioxide and perform \nphotosynthe sis? \n5. What is observed when a solution of potassium iodide solution is added to \na solution of lead nitrate? Name the type of reaction. Write a balanced \nchemical equation to represent the above chemical reaction. \n6. Write balanced chemical equations for the fol lowing reactions.", + "5. What is observed when a solution of potassium iodide solution is added to \na solution of lead nitrate? Name the type of reaction. Write a balanced \nchemical equation to represent the above chemical reaction. \n6. Write balanced chemical equations for the fol lowing reactions. \n(i) Silver bromide on exposure to sunlight decomposes into silver and \nbromine, \n(ii) Sodium metal reacts with water to form sodium hydroxide and hydrogen \ngas.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n(i) Silver bromide on exposure to sunlight decomposes into silver and \nbromine, \n(ii) Sodium metal reacts with water to form sodium hydroxide and hydrogen \ngas. \n7. Identify the type of reaction(s) in the following equations. \n(i)CH 4 + 2O 2 ----\uf0e0 CO 2 + 2 H2O \n(ii) Pb(NO 3)2 + 2KI \u2014\u2014\u2013>Pbl 2 + 2KNO 3 \n(iii) CaO + H2O \u2014\u2014\u2013> Ca(OH) 2 \n(iv) CuSO 4 + Zn \u2014\u2014\u2013> ZnSO 4 + Cu \n8. Write balanced equation for the reaction between magnesium and \nhydrochloric acid. Name the products obtained, identify the type of reaction. \n9. Describe an activity to observe what happens when quick lime is added to \nwater taken in a beaker. State two important observations and name the type \nof reaction taking place. \n10.", + "Name the products obtained, identify the type of reaction. \n9. Describe an activity to observe what happens when quick lime is added to \nwater taken in a beaker. State two important observations and name the type \nof reaction taking place. \n10. What is the colour of ferrous sulphate crystals? How does this colour change \nafter heating? \n11. Why does the colour of copper sulphate solution change when an iron nail is \ndipped in it? Write two observations.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\nafter heating? \n11. Why does the colour of copper sulphate solution change when an iron nail is \ndipped in it? Write two observations. \n12. Translate the following statement into chemical equation and then balance it : \nBarium chloride reacts with aluminium sulphate to gi ve aluminium chloride \nand a precipitate of barium sulphate. State the two types in which this reaction \ncan be classified. \n13. Why decomposition reactions are called the opposite of combination \nreactions? Write equations for these reactions. \n14. AgN0 3 (aq) + NaCl( aq)\u2014\u2014\u2014\u2014\u2014\u2014 \u2013 > AgCl (s)\u2193 + NaN0 3(aq) \nFeS + H 2S0 4\u2014\u2014\u2014\u2014 - > FeS0 4 + H 2S\u2191 \nConsider the above mentioned two chemical equations with two different \nkinds of arrows (\u2191and \u2193) along with product. What do these two different \narrows indicate? \n15.", + "What do these two different \narrows indicate? \n15. Hydrogen being a highly infla mmable gas and oxygen being a supporter of \ncombustion, yet water which is a compound made up of hydrogen and oxygen", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\narrows indicate? \n15. Hydrogen being a highly infla mmable gas and oxygen being a supporter of \ncombustion, yet water which is a compound made up of hydrogen and oxygen \nis used to extinguish fire. Why? 16. Using a suitable chemical equation, justify that some chemical reactions are \ndetermined by: \n(i) change in c olour, (ii) change in temperature. \n17. (a) A solution of substance \u2018X\u2019 is used for white washing. What is the \nsubstance \u2018X\u2019? State the chemical reaction of \u2018X\u2019 with water. \n(b) Why does the colour of copper sulphate solution change when an iron nail \nis dipped i n it? \n18. Write the balanced equation for the following reaction and identify the type of \nreaction in each case. \n(i) Potassium bromide + Barium iodide \u2014-> Potassium iodide + Barium \nbromide. \n(ii) Hydrogen(g) + Chlorine(g) \u2014-> Hydrogen chloride(g) \n19.", + "(i) Potassium bromide + Barium iodide \u2014-> Potassium iodide + Barium \nbromide. \n(ii) Hydrogen(g) + Chlorine(g) \u2014-> Hydrogen chloride(g) \n19. A zinc plate was put into a solution of copper sulphate kept in a glass \ncontainer. It was found that blue colour of the solution gets fader and fader", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n19. A zinc plate was put into a solution of copper sulphate kept in a glass \ncontainer. It was found that blue colour of the solution gets fader and fader \nwith the passage of time. After few days, when zinc plate was taken out of the \nsolution, a number of holes were observed o n it. \n(i) State the reason for changes observed on the zinc plate. \n(ii) Write the chemical equation for the reaction involved. \n20. A white salt on heating decomposes to give brown fumes and a residue is \nleft behind. \n(i) Name the salt. \n(ii) Write the equation f or the decom -position reaction. \n21. When a solution of potassium iodide is added to a solution of lead nitrate in a \ntest tube, a reaction takes place. \n(a) What type of reaction is this? \n(b) Write a balanced chemical equation to represent the above reaction. \n22. Define combination reaction. Give one example of a combination reaction \nwhich is also exothermic \n23. (a) Classify the following reactions into different types.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n22. Define combination reaction. Give one example of a combination reaction \nwhich is also exothermic \n23. (a) Classify the following reactions into different types. \n (i) Na 2SO 4 + BaCl 2 -----\uf0e0 BaSO 4 + NaCl \n (ii) CaO + H 2O --------\uf0e0 Ca(OH) 2 \n (iii) CaCO 3 ----------\uf0e0 CaO + CO 2 \n(b) Which of the above reaction(s) is/are precipitation reaction(s)? Why a \nreaction is called precipitation reaction?", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n24. Write a balanced chemical equation for the reaction between sodium chloride \nand silver nitrate indicating the physical state of the reactants and the \nproducts. \n25. What is a redox reaction? When a magnesium ribbon burns in air with a \ndazzling flame and forms a white ash, is magnesium oxidized or reduced? \nWhy? \n26. Write any two observations in an activity w hich may suggest that a \nchemical reaction has taken place. Give an example in support of your \nanswer. \n27. When the powder of a common metal is heated in an open china dish, its \ncolour turns black. However, when hydrogen is passed over the hot black \nsubstance s o formed, it regains its original colour. Based on the above \ninformation, answer the following questions. \n(i) What type of chemical reaction takes place in each of the two given steps? (ii) Name the metal initially taken in the powder form. Write balanced chemical \nequations for both reactions.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n(i) What type of chemical reaction takes place in each of the two given steps? (ii) Name the metal initially taken in the powder form. Write balanced chemical \nequations for both reactions. \n28. In electrolysis of water, why is the volume of gas collected over one electrode \ndouble that of gas collected over the other electrode? \n29. Name the products formed on strongly heating ferrous sulphate crystals. What \ntype of chemical reaction occurs in this change? \n30. What is an oxidation reaction? Give an example of oxidation reaction. Is \noxidation an exothermic or an endothermic reaction? \n31. Describe an activity to demonstrate the change that takes place when white \nsilver chlor ide is kept in sunlight. State the type of chemical reaction which \ntakes place. \n32. When magnesium ribbon burns in air or oxygen, a product is formed. State \nthe type of chemical reaction and name the product formed in the reaction. \nWrite balanced chemical equa tion of this reaction.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\nthe type of chemical reaction and name the product formed in the reaction. \nWrite balanced chemical equa tion of this reaction. \n33. Distinguish between a displacement reaction and a double displacement \nreaction. Identify the displacement and the double displacement reaction from \nthe following reactions. \n(i) CuSO 4 + Zn ----\uf0e0 ZnSO 4 +Cu \n(ii) Pb(NO 3)2 + KI -----\uf0e0 PbI 2 + 2 KNO 3", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\nShort Answer Type Questions (II) [3 Marks] \n \n34. Write the chemical equation of the reaction in which the following changes \nhave taken place with an example of each: \n(i) Change in colour \n(ii) Change in temperature \n(iii) Formation of precipitate \n35. State the type of chemical reactions and chemical equations that take place in \nthe following: \n(i) Magnesium wire is burnt in air. \n(ii) Electric current is passed through water. \n(iii) Ammonia and hydrogen chloride gases \u2018are mixed. \n36. (a) Write the essential c ondition for the following reaction to take place: \n 2AgBr \u2014-> 2Ag + Br 2 \nWrite one application of this reaction. \n(b) Complete the following chemical equation of a chemical reaction \n Heat \n 2FeS0 4 \u2014\uf0e0 Fe2O3 +\u2026\u2026. + \u2026\u2026 \n(c) What happens when water is added to quick lime? Write chemical \nequation.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n37. 2g of ferrous sulphate crystals are heated in a dry boiling tube. \n(i) List any two observations. \n(ii) Name the type of chemical reaction taking place. \n(iii) \u2018Write the chemical equation for the reaction. \n38. Write chemical equation reactions taking place when carried out with the help \nof \n(a) Iron reacts with steam (b) Magnesium reacts with dil HCl \n(c) Copper is heated in air. \n39. Which products will be obtained when lead nit rate is heated simply? Write \nbalanced chemical equation for the reaction? State the type of chemical \nreaction that occur in the change. \n40. What is meant by skeletal type chemical equation? What does it represent? \nUsing the equation for electrolytic decomposit ion of water, differentiate \nbetween a skeletal chemical equation and a balanced chemical equation. \n41. What is rancidity? Mention any two ways by which rancidity can be prevented. \n42. Write balanced chemical equation for the reactions that take place during", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n41. What is rancidity? Mention any two ways by which rancidity can be prevented. \n42. Write balanced chemical equation for the reactions that take place during \nrespir ation. Identify the type of combination reaction that takes place during \nthis process and justify the name. Give one more example of this type of \nreaction. \n43. What is redox reaction? Identify the substance oxidised and the substance \nreduced in the following reactions. \n(i)2PbO + C \u2014\u2013> 2Pb + CO 2 \n(ii)MnO 2 + 4HCl \u2014\u2013> MnCl 2 + 2H 20 + Cl 2 \n44. Write the balanced chemical equations for the following reaction and identify \nthe type of reaction. \nThermite reaction, iron (III) oxide reacts with aluminium and gives molten iron \nand aluminium oxide. \n45. A solution of potassium chloride when mixed with silver nitrate solution, an \ninsoluble white substance is formed.", + "Thermite reaction, iron (III) oxide reacts with aluminium and gives molten iron \nand aluminium oxide. \n45. A solution of potassium chloride when mixed with silver nitrate solution, an \ninsoluble white substance is formed. Write the chemical reaction involved and \nalso mention the type of the chemica l reaction?", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\ninsoluble white substance is formed. Write the chemical reaction involved and \nalso mention the type of the chemica l reaction? \n46. Write balanced equations for the following mentioning the type of reaction \ninvolved. \n(i) Aluminium + Bromine \u2014\u2013> Aluminium bromide \n(ii) Calcium carbonate \u2014\u2013> Calcium oxide + Carbon dioxide \n(iii) Silver chloride \u2014\u2013>Silver + Chlorine \n47. (a) Why is res piration considered as an exothermic reaction? \n(b) Define the terms oxidation and reduction. \n(c) Identify the substance that is oxidised and reduced in the following \nreaction \n Zn + CuO ----\uf0e0 ZnO + Cu", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n48. What is meant by \n(i) precipitation reac tion, \n(ii) exothermic reaction, \n(iii) oxidation reaction? \nWrite balanced chemical equations for an example of each . \n49. You might have noted that when copper powder is heated in a china dish, the \nsurface of copper powder becomes coated with a black colour substance. \n(i) How has this black coloured substance formed? \n(ii) What is that black substance? \n(iii) Write the chemical equation of the reaction that takes place. \n50. (a) What happens chemically when quicklime is added to water filled in a \nbucket? \n(b) On what basis is a chemical equation balanced? \n(c ) What change in colour is observed when white silver chloride is left \nexposed to sunlight? State the type of chemical reaction in this change. 51. What happens when an aqueous solution of sodium sulphate reac ts with an \naqueous solution of barium chloride? State the physical conditions of", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\naqueous solution of barium chloride? State the physical conditions of \nreactants in which the reaction between them will not take place. Write the \nbalanced chemical equation for the reaction and name the type of reaction. \n52. When you have mixed the solutions of lead(II) nitrate and potassium iodide, \n(i) what was the colour of the precipitate formed and can you name the \nprecipitate? \n(ii) Write the balanced chemical equation for this reaction. \n(iii) Is this also a double displac ement reaction? \n53. Name the type of reaction represented by the following equation: \n(i) CaO + H 2O ---\uf0e0 Ca(OH) 2 \n(ii) 3BaCl 2 + Al 2(SO 4)3 ---\uf0e0 3 BaSO 4 + 2 AlCl 3 \n heat \n(iii) 2 FeSO 4 ----------------- \uf0e0 Fe2O3 + SO 2 + SO 3 \n54.", + "Write the chemical equation of the reaction in which the following changes \nhave taken place with an example of each: \n(i) Change in colour \n(ii) Change in temperature \n(iii) Formation of precipitate", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\nhave taken place with an example of each: \n(i) Change in colour \n(ii) Change in temperature \n(iii) Formation of precipitate \n55. Balance the following chemical equations. \n(i) BaCl 2 + H 2SO 4 ------\uf0e0 BaSO 4 + HCl \n(ii) Ca(OH) 2 + HNO 3 ----\uf0e0 Ca(NO 3)2 + H 2O \n(iii)Pb(NO 3)2 ------\uf0e0 PbO +NO 2 +O 2 \n(iv) MnO 2 + HCl ----\uf0e0 MnCl 2 + H 2O + Cl 2 \n56. Balance the following chemical reactions and identify the type of reaction.", + "Balance the following chemical reactions and identify the type of reaction. \n(a) Mg (s) + Cl 2 (g) ---\uf0e0 MgCl 2(s) \n(b) HgO (s) --\uf0e0 Hg(l) + O 2(g) \n(c) Na(s) + S (s) ----------\uf0e0 Na2S(s) \n(d) TiCl4 (l) + Mg (s) ----\uf0e0 Ti(s) + MgCl 2(s) \n(e) CaO (s) + SiO 2(s) ---\uf0e0 CaSiO 3(s) \n(f) H2O2(l) --------\uf0e0 H2O(l) + O 2(g) \n57. During the reaction of some metas with dilute hydrochloric acid, following \noberservations were made. \n(a) Silver meta does not show any change \n(b) The temperature of the reaction mixture rises when aluminium (Al) is \nadded. \n(c) The reaction os sodium metal is found to be highly explosive", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n(a) Silver meta does not show any change \n(b) The temperature of the reaction mixture rises when aluminium (Al) is \nadded. \n(c) The reaction os sodium metal is found to be highly explosive \n(d) Some bubbles of a gas are seen when lead (Pb) is reacted with acid. \nExplain these observations givi ng suitable reasons.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-5 MARKS.txt\nLong Answer Type Questions [ 5 Marks ] \n1. (a) Define a balanced chemical equation. Why should an equation be \nbalanced? \n(b) Write the balanced chemical equation for the following reaction: \n(i) Phosphorus burns in presence of chlorine to form phosphorus penta \nchloride. \n(ii) Burning of natural gas. \n(iii) The process of respiration. \n2. (a) Explain two ways by which food industries prevent rancidity. \n(b) Discuss the importance of decomposition reacti on in metal industry with \nthree points. \n3. (a) Write one example for each of decomposion reaction carried out with help \nof \n(i) Electricity (ii) Heat (iii) Light \n(b) Which of the following statements is correct and why copper can displace \nsilver from silver ni trate and silver can displace copper from copper sulphate \nsolution. \n4. What happens when a piece of \n(a) Zinc metal is added to copper suphste solution? \n(b) Aluminium metal is added to dilute hydrochloric acid?", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-5 MARKS.txt\nsolution. \n4. What happens when a piece of \n(a) Zinc metal is added to copper suphste solution? \n(b) Aluminium metal is added to dilute hydrochloric acid? \n(c) Silver metal is added to copper sulphate solution? \nAlso write the balanced chemical equation if the reaction occurs. \n5. On heating blue coloured powder of copper (II) nitrate in a boiling tube , \ncopper oxide (black), oxygen gas and a brown gas X is formed. \n(a) Write a balanced chemical equation of the reaction \n(b) Identif y the brown gas x evolved. \n(c) Identify the type of reaction \n(d) What could be the pH range of aqueous solution of the gas x? \n6. On adding a drop of barium chloride solution to an aqueous solution of \nsodium sulphate, white precipitate is obtained. \n(a) Write the balanced chemical equation of the reaction involved. \n(b) What other name can be given to this precipitation reaction? \n(c) On adding diu te hydrochloric acid to the reaction mixture, white precipitate \ndisappears. Why?", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-5 MARKS.txt\n(b) What other name can be given to this precipitation reaction? \n(c) On adding diu te hydrochloric acid to the reaction mixture, white precipitate \ndisappears. Why? \n7. You are provided with two containers made up of copper and aluminium. You \nare aso provided with solutions of dilute HCl, dilute HNO 3, ZnCl 2 and H 2O. In \nwhich of the above cont ainers of these solutions can be kept?", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nChemical Reactions and Equations\nIntroduction to Chemical Reactions and Equations\nPhysical and chemical changes\nChemical change - one or more new substances with new physical and chemical properties\nare formed.\nExample: Fe(s)\u00a0 + \u00a0CuSO 4(aq) \u2192FeSO 4(aq) +Cu(s)\u00a0\n \u00a0 \u00a0 \u00a0 (Blue)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (Green)\u00a0 \u00a0 \u00a0 \u00a0\nHere, when copper sulphate reacts with iron, two new substances, i.e., ferrous sulphate and\ncopper are formed.\nPhysical change - change in colour or state occurs but no new substance is formed.\nExample: Water changes to steam on boiling but no new substance is formed(Even though\nsteam and water look different when they are made to react with a piece of Na, they react\nthe same way and give the exact same products). This involves only change in state (liquid\nto vapour).\u00a0\nObservations that help determine a chemical reaction\nA chemical reaction can be determined with the help of any of the following observations:\na) Evolution of a gas\nb) Change in temperature", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nObservations that help determine a chemical reaction\nA chemical reaction can be determined with the help of any of the following observations:\na) Evolution of a gas\nb) Change in temperature\nc) Formation of a precipitate\nd) Change in colour\ne) Change of state\nChemical reaction\nChemical reactions are chemical changes in which reactants transform into products by\nmaking or breaking of bonds(or both) between different atoms.\nTypes of chemical reactionsTaking into consideration different factors, chemical reactions are grouped into multiple\ncategories.\nFew examples are:\n\u25cfCombination\n\u25cfDecomposition\n\u25cfSingle Displacement\n\u25cfDouble displacement\n\u25cfRedox\n\u25cfEndothermic\n\u25cfExothermic\n\u25cfPrecipitation\n\u25cfNeutralisation\nChemical Reactions and Equations I\nWord equation\nA\u00a0 word equation is a\u00a0chemical reaction\u00a0expressed in words rather than\u00a0chemical\nformulas.\u00a0It helps identify the reactants and products in a chemical reaction.\nFor example,\u00a0\nSodium + Chlorine \u2192 Sodium chloride", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nformulas.\u00a0It helps identify the reactants and products in a chemical reaction.\nFor example,\u00a0\nSodium + Chlorine \u2192 Sodium chloride\nThe above equation means: \"Sodium reacts with chlorine to form sodium chloride.\"\u00a0\nSymbols of elements and their valencies\nA symbol is the chemical code for an element. Each element has one or two letter atomic\nsymbol, which is the abbreviated form of its name.\nValency is the combining capacity of an\u00a0element. It can be considered as the number of\nelectrons lost, gain or shared by an atom when it combines with another atom to form a\nmolecule.\nWriting chemical equations\nRepresentation of a chemical reaction in terms of symbols and chemical formulae of the\nreactants and products is known as a chemical equation.\nZn(s) +dil.H2SO 4(aq) \u2192ZnSO 4(aq) +H2(\u2191)\n\u00a0(Reactants)\u00a0 \u00a0(Products)\n\u2022 For solids, the symbol is \"(s)\".\n\u2022 For liquids, it is \"(l)\".", + "Zn(s) +dil.H2SO 4(aq) \u2192ZnSO 4(aq) +H2(\u2191)\n\u00a0(Reactants)\u00a0 \u00a0(Products)\n\u2022 For solids, the symbol is \"(s)\".\n\u2022 For liquids, it is \"(l)\".\n\u2022 For gases, it is \"(g)\".\u2022 For aqueous solutions, it is \"(aq)\".", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nZn(s) +dil.H2SO 4(aq) \u2192ZnSO 4(aq) +H2(\u2191)\n\u00a0(Reactants)\u00a0 \u00a0(Products)\n\u2022 For solids, the symbol is \"(s)\".\n\u2022 For liquids, it is \"(l)\".\n\u2022 For gases, it is \"(g)\".\u2022 For aqueous solutions, it is \"(aq)\".\n\u2022 For gas produced in the reaction, it is represented by \"(\u2191)\".\n\u2022 For precipitate formed in the reaction, it is represented by \"(\u2193)\".\nBalancing of a Chemical Reaction\nConservation of mass\nAccording to the law of conservation of mass, no atoms can be created or destroyed in a\nchemical reaction, so the number of atoms for each element in the reactants side has to\nbalance the number of atoms that are present in the products side.\nIn other words, the total mass of the products formed in a chemical reaction is equal to the\ntotal mass of the reactants participated in a chemical reaction.", + "In other words, the total mass of the products formed in a chemical reaction is equal to the\ntotal mass of the reactants participated in a chemical reaction.\nBalanced chemical equation\nThe chemical equation in which the number of atoms of each element\u00a0in the reactants side\nis equal to that of the products side is called a balanced chemical equation.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nBalanced chemical equation\nThe chemical equation in which the number of atoms of each element\u00a0in the reactants side\nis equal to that of the products side is called a balanced chemical equation.\nSteps for balancing chemical equations\nHit and trial method: While balancing the equation, change the coef\ufb01cients (the numbers in\nfront of the compound or molecule) so that the number of atoms of each element is same\non each side of the chemical equation.\u00a0\nShort-cut technique for balancing a chemical equation\nExample:\naCaCO 3+bH3PO 4\u2192cCa 3(PO 4)2+dH2CO 3\nSet up a series of simultaneous equations, one for each element.", + "Short-cut technique for balancing a chemical equation\nExample:\naCaCO 3+bH3PO 4\u2192cCa 3(PO 4)2+dH2CO 3\nSet up a series of simultaneous equations, one for each element.\nCa: a=3c\nC:\u00a0 \u00a0a=d\nO:\u00a0 \u00a03a+4b=8c+3d\nH:\u00a0 \u00a03b=2d\nP:\u00a0 \u00a0 b=2c\nLet's set c=1\nThen a=3 and\nd=a=3\nb=2c=2So a=3; b=2; c=1; d=3\nThe balanced equation is\n3CaCO 3+ 2H 3PO 4\u2192Ca3(PO 4)2+ 3H 2CO 3\nChemical Reactions and Equations II\nTypes of chemical reactions\nTaking into consideration different factors, chemical reactions are grouped into multiple\ncategories.\nFew examples are:\n\u25cfCombination", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nChemical Reactions and Equations II\nTypes of chemical reactions\nTaking into consideration different factors, chemical reactions are grouped into multiple\ncategories.\nFew examples are:\n\u25cfCombination\n\u25cfDecomposition\n\u25cfSingle Displacement\n\u25cfDouble displacement\n\u25cfRedox\n\u25cfEndothermic\n\u25cfExothermic\n\u25cfPrecipitation\n\u25cfNeutralisation\nCombination reaction\nIn a combination reaction, two elements or one element and one compound or two\ncompounds combine\u00a0to give one single product.\nH2+Cl2\u2192 2HCl\nelement + element \u2192 compound\n2CO +O2\u2192 2CO 2\ncompound + element \u2192 compound\nNH 3+HCl \u2192NH 4Cl\ncompound + compound \u2192 compound\nDecomposition reaction\nA single reactant decomposes on the application of heat or light or electricity to give two or\nmore products.\nTypes of decomposition reactions:\na. Decomposition reactions which require heat - thermolytic decomposition or thermolysis.\nThermal decomposition of HgO\nb. Decomposition reactions which require light - photolytic decomposition\u00a0or\u00a0photolysis.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nThermal decomposition of HgO\nb. Decomposition reactions which require light - photolytic decomposition\u00a0or\u00a0photolysis.\nPhotolytic decomposition of H2O2\nc. Decomposition reactions which require electricity - electrolytic decomposition or\nelectrolysis.\nElectrolytic decomposition of H 2O\nDisplacement reaction\nMore reactive element displaces a less reactive element from its compound or solution.i)Zn(s) +CuSO 4(aq) \u2192ZnSO 4(aq) +Cu(s)\nii)Cu(s) + 2AgNO 3(aq) \u2192Cu(NO 3)2(aq) + 2Ag (s)\nDouble displacement reaction\nAn exchange of ions between the reactants takes place to give new products.\nFor example,\u00a0Al 2(SO4)3(aq) + 3Ca( OH)2(aq) \u2192 2Al (OH)3(aq) + 3CaSO 4(s)\nPrecipitation reaction\nAn insoluble compound called precipitate forms when two solutions containing soluble salts\nare combined.", + "For example, Pb( NO 3)2(aq) + 2KI (aq) \u2192 2KNO 3(aq) +PbI 2(\u2193)(s)(yellow )\nRedox reaction\nOxidation and reduction take place simultaneously.\nOxidation: Substance loses electrons or gains oxygen or loses hydrogen.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nRedox reaction\nOxidation and reduction take place simultaneously.\nOxidation: Substance loses electrons or gains oxygen or loses hydrogen.\nReduction: Substance gains electrons or loses oxygen or gains hydrogen.\nOxidising agent - a substance that oxidises another substance and self-gets reduced.\nReducing agent - a substance that reduces another substance and self-gets oxidised.\nExamples:\n1.Fe(s) +CuSO 4(aq) \u2192FeSO 4(aq) +Cu(s)\u00a0 \u00a0 \u00a0 \u00a0(Blue)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (Green)\nFe\u2192Fe+2+ 2e \u2212 \u00a0(oxidation ) ; Fe - reducing agent.\nCu+2+ 2e \u2212 \u2192Cu(s)\u00a0(reduction ) ; Cu - oxidising agent.\n2.ZnO +C\u2192Zn+CO\nZnO reduces to Zn\u00a0\u2192 reduction\nC oxidises to CO \u2192 oxidation\nZnO - Oxidising agent\nC - Reducing agent\nEndothermic and exothermic reaction\nExothermic reaction - heat is evolved during a reaction.", + "Most of the combination reactions\nare exothermic.\nAl+Fe2O3\u2192Al2O3+Fe+heat\nCH 4+ 2O 2\u2192CO 2+ 2H 2O+heat\nEndothermic - Heat is required to carry out the reaction.\n6CO 2+ 6H 2O+Sunlight \u2192C6H12O6+ 6O 2\n\u00a0 \u00a0 \u00a0 \u00a0Glucose", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nare exothermic.\nAl+Fe2O3\u2192Al2O3+Fe+heat\nCH 4+ 2O 2\u2192CO 2+ 2H 2O+heat\nEndothermic - Heat is required to carry out the reaction.\n6CO 2+ 6H 2O+Sunlight \u2192C6H12O6+ 6O 2\n\u00a0 \u00a0 \u00a0 \u00a0Glucose\nMost of the\u00a0decomposition reactions are endothermic.\nCorrosion\nGradual deterioration of a material, usually a metal, by the action of moisture, air or\nchemicals in the surrounding environment.", + "Corrosion\nGradual deterioration of a material, usually a metal, by the action of moisture, air or\nchemicals in the surrounding environment.\nRusting:\n4Fe(s) + 3O 2(from \u00a0air) +xH 2O(moisture ) \u2192 2Fe 2O3.xH 2O(rust)\nCorrosion of copper:\nCu(s) +H2O(moisture ) +CO 2(from \u00a0air) \u2192CuCO 3.Cu(OH)2(green )\nCorrosion of silver:\nAg(s) +H2S(from \u00a0air) \u2192Ag2S(black) +H2(g)\nRancidity\nIt refers to oxidation of fats and oils in food that is kept for a long time. It gives foul smell\nand bad taste to food. Rancid food causes stomach infection on consumption.\nPrevention:\n(i) Use of air-tight containers(ii) Packaging with nitrogen\n(iii) Refrigeration\n(iv) Addition of antioxidants or preservatives", "CHAPTER 1 -CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nChapter 1 \nCHEMICAL REACTIONS AND EQUATIONS \nChem ical Reaction: The tra nsformation of chemical substance into another \nchemical substance is known as Chemical Reaction. \nFor example: Rusting of iron, the setting of milk into curd, digestion of food, \nrespiration, etc. \nIn a chemical reaction, a new substance is formed which is completely different in \nproperties from the original substance, so in a chemical reaction, a chemical change \ntakes place. \nOnly a rearrangement of atoms takes place in a chemical r eaction. \n\u2022 The substances which take part in a chemical reaction are called reactants. \n\u2022 The new substances produced as a result of a chemical reaction are called \nproducts. \nExample: The burning of magnesium in the air to form magnesium oxide is an \nexample of a chemical reaction. \n 2Mg(s) + O 2(g) 2MgO(s) \nBefore burning in air, the magnesium ribbon is cleaned by rubbing with sandpaper.", "CHAPTER 1 -CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nexample of a chemical reaction. \n 2Mg(s) + O 2(g) 2MgO(s) \nBefore burning in air, the magnesium ribbon is cleaned by rubbing with sandpaper. \nThis is done to remove the protective layer of basic magnesium carbonate from the \nsurface of the magnesi um ribbon. \nReactant: Substances which take part in a chemical reaction are called reactants. \nExample: Mg and O 2. \nProduct: New substance formed after a chemical reaction is called a product. \nExample: MgO. \nCharacteristics of Chemical Reactions: \n(i) Evolution of gas: The chemical reaction between zinc and dilute sulphuric acid is \ncharacterised by the evolution of hydrogen gas. \n Zn(s) + H 2SO 4(aq) \u2192 ZnSO 4(aq) + H 2(g) \u2191 \n(ii) Change in Colour: The chemical reacti on between citric acid and purple \ncoloured potassium permanganate solution is characterised by a change in colour \nfrom purple to colourless.", "The chemical reaction between sulphur dioxide gas and acidified potassium", @@ -46,49 +169,7 @@ "CHAPTER 1 -CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nC + O 2 \u2192 CO 2 (g) + heat \nEndothermic reaction: A chemical reaction in which heat energy is absorbed. \nZnCO 3 + Heat \u2192 ZnO + CO 2 \nRedox reaction: Chemical reaction in which both oxidation and reduction take place \nsimultaneously. \n4. Oxidation: Reaction that involves the gain of oxygen or loss of hydrogen. \n5. Reduction: Reaction that shows the loss of oxygen or gain of hydrogen. \nZnO + C \u2192 Zn + CO \nZnO is reduced to Zn \u2014reduction. C is oxidized to CO \u2014Oxidation. \n6. Effects of Oxidation Reactions in Our Daily Life: \n\u2022 Corrosion: It is an undesirable change that occurs in metals when they are \nattacked by moisture, air, acids and bases. \nExample, Corrosion (rusting) of Iron: Fe 2O3.", "Effects of Oxidation Reactions in Our Daily Life: \n\u2022 Corrosion: It is an undesirable change that occurs in metals when they are \nattacked by moisture, air, acids and bases. \nExample, Corrosion (rusting) of Iron: Fe 2O3. nH 2O (Hydrated iron oxide) \u2022 Rancidity: Undesirable change that takes place in oil containing food items \ndue to the oxidation of fatty acids. \nPreventive methods of rancidity: Adding antioxidants to the food materials,", "CHAPTER 1 -CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\ndue to the oxidation of fatty acids. \nPreventive methods of rancidity: Adding antioxidants to the food materials, \nstoring food in the airtight container, flush ing out air with nitrogen gas and \nrefrigeration.", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\nChemical Reactions and Equations \nCompetency Based Questions", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\n1. A compound 'X' used for drinking, has pH =7. Its acidified solution \nundergoes decomposition in presence of electricity to produce gases ' \nY' and 'Z' The volume of Y is double than Z. Y is highly combustible \nwhereas Z is supporter of combustion. identify X. Y & Z and write th e \nchemical reactions involved. \n2. An aqueous solution of metal nitrate P reacts with sodium bromide \nsolution to form yellow not of compound O which is used in \nphotography. O on exposure to sunlight undergoes decomposition \nreaction to form metal present in P a long with reddish brown gas. \nIdentify P & O. Write the chemical reaction & type of chemical reaction. \n3. Bhawana took a pale green substance A in a test tube, and heated it \nover the flame of a burner. A brown coloured residue B was formed \nalong with evolution of two gases with burning smell of sulphur. Identify \nA & B. Write the chemical reaction involved. \n4.", - "A brown coloured residue B was formed \nalong with evolution of two gases with burning smell of sulphur. Identify \nA & B. Write the chemical reaction involved. \n4. A student took 2 -3 g of a substance X in a glass beaker & poured water", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\nalong with evolution of two gases with burning smell of sulphur. Identify \nA & B. Write the chemical reaction involved. \n4. A student took 2 -3 g of a substance X in a glass beaker & poured water \nover it slowly. He observed bubbles along with hissing noise. The \nbeaker becomes quite hot. Identify X. What type of reaction is it? \n5. A substance X used for coating iron articles is added to a blue solution \nof a reddish brown metal Y. The colour of the solution gets discharged . \nIdentify X and Y & also the type of reaction. \n6. A reddish brown ve ssel developed a green coloured solid X when left \nopen in air for a long time. When reacted with dil H 2S04.", - "Identify X and Y & also the type of reaction. \n6. A reddish brown ve ssel developed a green coloured solid X when left \nopen in air for a long time. When reacted with dil H 2S04. it forms a blue \ncoloured solution along with brisk effervescence due to colourless & \nodourless gas Z. X decomposes to form black coloured oxide Y of a \nreddish brown metal along with gas Z. Identify X. Y. & Z. \n7. A student has mixed the solutions of lead (II) nitrate and potassium \niodide,", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\nreddish brown metal along with gas Z. Identify X. Y. & Z. \n7. A student has mixed the solutions of lead (II) nitrate and potassium \niodide, \n(i) What was the colour of the precipitate fo rmed? Can you name the \ncompound precipitated? \n(ii) Write the balanced chemical equation for this reaction, \n(iii) What type of reaction is it? 8. Observe the following activity & answer the Questions", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\na. Do you observe anything happening around the zinc granules \nb. Is there any change in its temperature? \nc. Why is glass tube not dipp ed in dil H 2SO 4? \nd. How is H 2 gas collected by downward displacement or upward \ndisplacement of water? \ne. Is H 2 gas soluble or insoluble in water? \nf. Is H 2 gas heavier or lighter than air? \n9. A reddish brown metal X when heated in presence of oxygen forms a \nblack compound Y which is ba sic in nature when heated with hydrogen \ngas gives back X. Identify X & Y. Write the chemical reaction between Y \n& H 2 .Identify the substance being oxidized & reduced \n10. Name the type of reaction seen in the diagram below. Write the equation \nfor the reaction.", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\n11. A student burnt a metal A found in the form of ribbon. The ribbon burnt \nwith a dazzling flame & a white powder B is formed which is basic in \nnature. Identify A & B. Write the balanced chemical equation. \n12. A student drop ped few p ieces of marble in dilute H Cl contained in a test \ntube. The gas evolved was passed through lime water. What change \nwould be observed in lime water? Write chemical reactions for both the \nchanges observed. \n13. Astha has been collecting silver coins and Co pper coins. One day she \nobserved a black coating on silver coins and a green coating on Co pper \ncoins. Which chemical phenomenon is responsible for these coatings? \nWrite the chemical name of black and green coatings \n14. Identify the type of chemical reaction \n(i) \n(ii) \n15. A student took 2 -3 g o f a substance X in a glass beaker & poured water \nover it slowly. He observed bubbles along with hissing noise.", - "Identify the type of chemical reaction \n(i) \n(ii) \n15. A student took 2 -3 g o f a substance X in a glass beaker & poured water \nover it slowly. He observed bubbles along with hissing noise. The \nbeaker becomes quite hot. Identify X. What type of reaction is it?", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\nover it slowly. He observed bubbles along with hissing noise. The \nbeaker becomes quite hot. Identify X. What type of reaction is it? \n16. A substance X used for coating iron articles is added to a blue sol ution \nof a reddish brown metal. T he colour of the solution gets discharged \nIdentify X and Y & also the type of reaction. \n17. A solution of a substance \u2018X\u2019 is used for white washing \n i. Name the substance \u2018X\u2019 and writes its formula. \n ii. Write the reaction of the substance \u2018X\u2019 named in (i) above with water \n18. A shiny brown coloured element \u2018X\u2019 on heating in air becomes black in \ncolour. Name the element \u2018X\u2019 and the black coloured compound formed. \n19. An aqueous solution of metal nitrate P reacts with sodium bromide \nsolution to form yellow ppt of compound Q which is used in \nphotography. Q on exposure to sunlight undergoes decomposition \nreaction to form metal present in P along with reddish brown gas.", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\nsolution to form yellow ppt of compound Q which is used in \nphotography. Q on exposure to sunlight undergoes decomposition \nreaction to form metal present in P along with reddish brown gas. \nIdentify P &Q. Write the chemical reaction & type of chemic al reaction. \n20. A reddish brown vessel developed a green coloured solid X When left \nopen in air for a long time. When reacted with dil , it forms a blue \ncoloured solution along with brisk effervescence due to colourless & \nodourless gas Z. X decomposes to for m black coloured oxide Y of a \nreddish brown metal along with gas Z, Identify X, Y, & Z. \n21. A metal is heated with dil H 2SO 4. The gas evolved is collected by the \nmethod shown in the figure: Answer the following", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-COMPETENCY BASED QUESTIONS.txt\n(a) Name the gas. \n(b) Name the method of collection of gas. \n(c) Is the gas soluble or insoluble in water? \n(d) Is the gas lighter or heavier than air?", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\n1 \n SALWAN PUBLIC SCHOOL , RAJENDRA NAGAR \nCLASS \u2013 X \nLESSON PLAN \nCHAPTER -1: CHEMICAL REACTIONS AND EQUATIONS \nCHAPTER -1: Science :( CHEMICAL REACTIONS AND EQUATIONS ) \nClass Transaction \n \n \n \n(9-10 days) PART -I \n \nTotal: 4 periods (approx. 35 min each) \n \nSub-Topic: Introduction regarding situations of daily life -Physical and \n chemical changes \nSub-Topic: Chemical equations(writing equations) \n \nSub-Topic: Chemical equations(writing balancing equations) \n \nSub-Topic : Type of chemical reactions (combina tion and decomposition) \n \nPre-requisite for the \ncourse (KNOWLEDGE) This lesson requires: \n\uf0b7 Basic knowledge of the situations of daily life and think what happens \nwhen physical and chemical changes take place \n\uf0b7 The nature and the identity of the initial substance have somewhat \nchanged. \n\uf0b7 What is actually meant by a chemical reaction?", - "\uf0b7 What is actually meant by a chemical reaction? How do we come to \nknow that a chemical reaction has taken place? \n \nLearning Objectives", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nLearning Objectives \n \n Students will be able to: \n\uf0b7 recognize the Physical and chemical processes associated with \nbiological and industrial processes affecting life and the \nenvironment. \n\uf0b7 understand chemical reaction they encounter every day. \n\uf0b7 represent equations to represent the reactions . \n\uf0b7 apply the principles of conservation of mass to balance chemical \nreactions \n\uf0b7 identify the different types of chemical reaction and explain.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nTeachers ' activity \n Introduction: Teacher will introduce the topic by asking various questions \nfrom previous knowledge : \n\uf0b7 Analytical skills (scrutinizing an observation): To identify physical \nand chemical changes. Example -Paper is torn or Paper is burnt, ice \nis melting or iron nail is kept exposed to moist air. \n Evidence -based active -learning instructional strategies : \n\uf0b7 Problem solving (investigating and applying established principles \nto justify an observation) \n 2 \n \nExample - Activity: -Burning of a magnesium ribbon in air and collection of \nmagnesium oxide in a watch -glass. \nConcept insight: Always remember that magnesium being a reactive metal will \nreact with oxygen if kept in open. So, it has to be cleaned before it is burned in \nair. \nActivity: -Formation of hydrogen gas by the action of dilute hydrochloric acid on \nzinc. \nConcept insight: The key to this answer is to remember that metals react with", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nair. \nActivity: -Formation of hydrogen gas by the action of dilute hydrochloric acid on \nzinc. \nConcept insight: The key to this answer is to remember that metals react with \nhydrochloric acid to form respective metal chlorides an d liberate hydrogen gas. \n\uf0b7 Critical thinking and collaboration (Competitive evaluation of given \ninformation) \nExample - To make a chemical equation more informative, the physical states of \nthe reactants and products are mentioned along with their chemical form ulae \nfrom given information of different activities. \nConcept insight: First convert the word equation into skeletal equation with \nstate and then balance the different atoms on both sides of the equation. \n\uf0b7 The ability to understand complex concepts (Inspect ing and analyzing \nreasons for variation in established principles);Types of chemical \nreactions: \nExample - Combination and decomposition reactions : Chemical reactions", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nreasons for variation in established principles);Types of chemical \nreactions: \nExample - Combination and decomposition reactions : Chemical reactions \ninvolve the breaking and making of bonds between atoms to produce new \nsubstances and balan cing equations with the help of law of conservation of \nmass \nConcept insight: Remember that a chemical equation should be balanced to \nfollow the Law of conservation of mass. \nRecapitulation (working effectively with others): \nExample -Discuss brain storming questions in class in group.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nLearning Outcomes Students will be able to know: \n\uf0b7 A complete chemical equation which represents the reactants, \nproducts and their physical states symbolically. \n\uf0b7 How to write the balanced chemical equation . \n\uf0b7 A combination and d ecomposition reactions. \n\uf0b7 Reactions in which heat is given out along with the products are \ncalled exothermic reactions. \n\uf0b7 Reactions in which energy is absorbed are known as endothermic \nreactions.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nAssessment \uf0b7 Continuously monitor student progress by asking oral questions. \n\uf0b7 Providing active involvement of students in their own learning by \ngiving daily Practice Problems in the class. \n\uf0b7 Peer assessment in Lab Activities \n\uf0b7 Written test \nSelf-Study, \uf0b7 Self-Study (Independent Practice) -The teacher would ensure that 3 \n Assignments , \nHomework each student should bring NCERT Book and tries to be independent \nat the work as well as be under the teacher\u2019s guidance. \n\uf0b7 Guided practice followed by Independent Practice ( Question \nBank): \nOn completion of the chapter, students will be answer the following \nrevision questions: \nQ.1.Why should a magnesium ribbon be cleaned before burning in air? \nQ.2. Write a balanced chemical equation with state symbols for the \nreactions: Sodium hydroxide solution (in water) reacts with hydrochloric \nacid so lution (in water) to produce sodium chloride solution and water.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nreactions: Sodium hydroxide solution (in water) reacts with hydrochloric \nacid so lution (in water) to produce sodium chloride solution and water. \nQ.3.Take a few zinc granules in a conical flask or a test tube and a dd dilute \nhydrochloric acid or sulphuric acid to this. \n(a)Do you observe anything happening aroun d the zinc granules? \n(b)Touch the conical flask or test tube. Is there any change in its \ntemperature? \nQ.4.Take a small amount of calcium oxide or quick lime in a beaker , slowly \nadd water to this and touch the beaker. \n(a)Do you feel any change in temperature? \n(b) Write the balanced chemical equation with physical state. \nQ.5.Take about 2 g silver chloride in a china dish and Place this china dish in \nsunlight for some time. \n(a)Observe the colour of the silver chloride after some time \n(b).Name type of the reaction.", - "Q.5.Take about 2 g silver chloride in a china dish and Place this china dish in \nsunlight for some time. \n(a)Observe the colour of the silver chloride after some time \n(b).Name type of the reaction. \nQ.6.Take about 2 g barium hydroxide in a test tube. Add 1 g of ammonium", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN TEACHER.txt\nsunlight for some time. \n(a)Observe the colour of the silver chloride after some time \n(b).Name type of the reaction. \nQ.6.Take about 2 g barium hydroxide in a test tube. Add 1 g of ammonium \nchloride and mix with the help of a glass rod. Touch the bottom of the test \ntube with your palm. \n(a)What do you feel? \n(b)Is this an exothermic or endothermic reaction? \nHomework: The students, with the help of the teacher, will solve the \nquestions from NCERT back e xercise in their notebook.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\n1 \n SALWAN PUBLIC SCHOOL , RAJENDRA NAGAR \nCLASS \u2013 X \nLESSON PLAN \nCHAPTER -1: CHEMICAL REACTIONS AND EQUATIONS \nCHAPTER -1: Science :( CHEMICAL REACTIONS AND EQUATIONS ) \nClass Transaction \n \n \n \n \n(10-12 days) PART -II", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\n(10-12 days) PART -II \n \nTotal: 5 periods (approx. 35 min each) \nSub-Topic: Types of chemical reactions -Combination Reaction and different \nactivities. \nSub-Topic: Types of chemical reactions -Decomposition Reaction and different \nactivities. \nSub-Topic: Types of chemical reactions -Displacement Reaction and different \nactivities. \nSub-Topic : Type of chemical reactions \u2013Double displacement reaction and \ndifferent activities. \nSub-Topic :- Oxidation and reduction reaction (Corrosion and rancidity) \nPre-requisite for the \ncourse (KNOWLEDGE) This lesson requires: \n\uf0b7 Knowledge of the basic properties and mechanisms of chemical \nreactions. \n\uf0b7 The nature and the identity of the initial substance have somewhat \nchanged. \n\uf0b7 What is actually meant by a chemical reaction? How do we come to \nknow that a chemical reaction has taken place? \n \nLearning Objectives", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nLearning Objectives \n \n Students will be able to: \n\uf0b7 Classify a chemical reaction as a combination, decomposition, \nsingle replacement, double replacement, exothermic or \nendothermic reaction. \n\uf0b7 Differentiate between the types of chemical reactions. \n\uf0b7 Predict the products of a chemical reaction. \n\uf0b7 Understand the different sets of conditions required such as \ntemperature, light etc in reactions. \n\uf0b7 Know the effects of oxidation reactions in everyday life like \ncorrosion and rancidity.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nTeachers ' activity \n Introduction: Teacher will introduce the topic by performing various \nactivities: \nEvidence -based active -learning instructional strategies : \n\uf0b7 Problem solving (investigating and applying established principles \nto justify an observation) Activity: - \nExample -1-: Formation of calcium hydroxide \nExample -2: Heating of calcium carbonate \nExample -3: Electrolysis of water 2 \n Example -4: Reaction of iron nails with copper sulphate solution \nExample -5: Reaction between Sodium Sulphate and Barium Chloride \nExample -6: Reaction between Cu and O 2; CuO and H 2. \n\uf0b7 Critical thinking and collaboration (Competitive evaluation of given \ninformation) \nRecall Activity : \nYou are given test tube, the solutions of lead (II) nitrate and potassium iodide \nand are instructed to mix together in a test tube. Give answers by activity : \n(i)What was the colour of the precipitate formed? Can you name the \n compound precipitated?", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nand are instructed to mix together in a test tube. Give answers by activity : \n(i)What was the colour of the precipitate formed? Can you name the \n compound precipitated? \n(ii)Write the balanced chemical equation for this reaction. \n(iii)Is this also a double displacement reaction?", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\n\uf0b7 The ability to understand the effects of oxidation reactions in \neveryday life? \nExample - Browning of cut surfaces of fruits. Formation of metal oxide on \nmetal like iron rust -Corrosion, Burning of crackers , Rancidity: \n\uf0b7 Recapitulation (working effectively with others): \n Example -Discuss brain storming questions in class in group. \n \nLearning Outcomes Students will be able to know \n\uf0b7 displacement and double displacement reaction. \n\uf0b7 precipitation reactions which produce insoluble salts. \n\uf0b7 importance of Redox reaction in ever day life. Oxidation is the gain of \noxygen or loss of hydrogen. Reduction is the loss of oxygen or gain of \nhydrogen. \n\uf0b7 molecular interactions and chemical reactions in the body .", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nAssessment \uf0b7 Continuously monitor student progress by asking oral questions. \n\uf0b7 Providing active involvement of students in their own learning by \ngiving daily Practice Problems in the class. \n\uf0b7 Peer assessment in Lab Activities \n\uf0b7 Written test \nSelf-Study, \nAssignments , \nHomework \uf0b7 Self-Study (Independent Practice) -The teacher would ensure that \neach student should bring NCERT Book and tries to be independent \nat the work as well as be under the teacher\u2019s guidance. \n\uf0b7 Guided practice followed by Independent Practice ( Question \nBank): \nOn completion of the chapter, students will be answer the following \nrevision questions: \nQ.1.Give some examples of combination and decomposition reactions. \nQ.2. What happens when Ferrous sulphate crystal is heated? Write the \nequations also. \nQ.3.Dissolve 0.5 gram silver nitrate 10 ml of water in a test tube and place a 3 \n Copper wire in a test tube. \nExplain your observation.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nequations also. \nQ.3.Dissolve 0.5 gram silver nitrate 10 ml of water in a test tube and place a 3 \n Copper wire in a test tube. \nExplain your observation. \nQ.4. What is the difference between displacement and double displacement \nreactions? Write equations for these reactions. \nQ.5.Take about 2 g silver bromide in a china dish and Place this china dish in \nsunlight for some time. \n(a)Observe the colour of the silver bromide after some time \n(b).Name type of the reaction. \nQ.6.An iron knife kept dipped in a blue copper sulphate solution turns the \nblue solution light green. Why? \nHomework: The students, with the help of the teacher, will solve the \nquestions from NCERT back e xercise in their notebook.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-5 MARKS.txt\nLong Answer Type Questions [ 5 Marks ] \n1. (a) Define a balanced chemical equation. Why should an equation be \nbalanced? \n(b) Write the balanced chemical equation for the following reaction: \n(i) Phosphorus burns in presence of chlorine to form phosphorus penta \nchloride. \n(ii) Burning of natural gas. \n(iii) The process of respiration. \n2. (a) Explain two ways by which food industries prevent rancidity. \n(b) Discuss the importance of decomposition reacti on in metal industry with \nthree points. \n3. (a) Write one example for each of decomposion reaction carried out with help \nof \n(i) Electricity (ii) Heat (iii) Light \n(b) Which of the following statements is correct and why copper can displace \nsilver from silver ni trate and silver can displace copper from copper sulphate \nsolution. \n4. What happens when a piece of \n(a) Zinc metal is added to copper suphste solution? \n(b) Aluminium metal is added to dilute hydrochloric acid?", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-5 MARKS.txt\nsolution. \n4. What happens when a piece of \n(a) Zinc metal is added to copper suphste solution? \n(b) Aluminium metal is added to dilute hydrochloric acid? \n(c) Silver metal is added to copper sulphate solution? \nAlso write the balanced chemical equation if the reaction occurs. \n5. On heating blue coloured powder of copper (II) nitrate in a boiling tube , \ncopper oxide (black), oxygen gas and a brown gas X is formed. \n(a) Write a balanced chemical equation of the reaction \n(b) Identif y the brown gas x evolved. \n(c) Identify the type of reaction \n(d) What could be the pH range of aqueous solution of the gas x? \n6. On adding a drop of barium chloride solution to an aqueous solution of \nsodium sulphate, white precipitate is obtained. \n(a) Write the balanced chemical equation of the reaction involved. \n(b) What other name can be given to this precipitation reaction? \n(c) On adding diu te hydrochloric acid to the reaction mixture, white precipitate \ndisappears. Why?", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-5 MARKS.txt\n(b) What other name can be given to this precipitation reaction? \n(c) On adding diu te hydrochloric acid to the reaction mixture, white precipitate \ndisappears. Why? \n7. You are provided with two containers made up of copper and aluminium. You \nare aso provided with solutions of dilute HCl, dilute HNO 3, ZnCl 2 and H 2O. In \nwhich of the above cont ainers of these solutions can be kept?", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nLESSON PLAN \nClass: 10 Chapter: 1 ( Chemical Reactions and Equations )", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nTopics to be covered Chemical reactions, writing chemical equations, Balancing \nchemical equations, types of chemical reactions -\ncombination reactions, decomposition reactions, \ndisplacement reactions, double displacement reactions, \nneutralization reactions , precipitation reactions, redox \nreactions \nPrevious Knowledge \u2022 Students know differences between physical \nchange and chemical change \n\u2022 Studied the symbols of elements and chemical \nformulae of molecules. \n\u2022 Also studied law of conservation of mass. \nLearning Objectives \u2022 To understand the chemical reactions in daily life \n\u2022 Able to write the chemical equations from the \nstatements \n\u2022 Understand how to balance chemical reactions \n\u2022 Study different types of chemical reactions \n\u2022 Define the terms reactants, products, precipitate, \nneutralization etc. \nLearning Experiences \u2022 In combination reactions two or more substances \ncombine to form a single compound . \n\u2022 In decomposition reactions a single compound", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nneutralization etc. \nLearning Experiences \u2022 In combination reactions two or more substances \ncombine to form a single compound . \n\u2022 In decomposition reactions a single compound \nbreaks down into two or more substances. \n\u2022 Displacement reactions are the reactions in which a \nmore active element displace a less active element \nfrom its solution. \n\u2022 In double displacement reactions ions are \nexchanged between two compounds. \n\u2022 REDuction + Oxidation = REDOX reaction \nResources NCERT book \nReference book \u2013Prdeeps Publications \nYou tube ,education channels", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-LESSON PLAN FOR TEACHERS.txt\nTeaching Aids Usage of ICT( power point) multimedia., Lab activities( \nvirtual lab also) \n \nProcedure Class starts with the introduction \nThe topics are presenting with power point , audiovisual \naids, virtual lab and lab activities \nAt the end homework and assignments are given \n \nAssessment Methods Oral questioning , class tests and monthly tests", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-IMPORTANT NOTES.txt\nFocus Points \n1.Chemical Reaction: During chemical reactions, the chemical composition of \nsubstances changes or new substances are formed. \n2. Chemical Equation: Chemical reactions can be written in chemical equation form \nwhich should always be balanced. \n3. Types of Chemical Reactions: \nCombination reaction: A single product is formed from two or more reactants. \n2Mg + O 2 \u2192 2MgO \nDecomposition reaction: A single reactant breaks down to yield two or more \nproducts. \n\uf0b7 Thermal decomposition: 2Pb(NO 2)2 \u2192 2PbO + 4NO 2 + O 2 \n\uf0b7 Electrolysis: 2H20 \u2192 2H 2 + O 2 \n\uf0b7 Photochemical reaction: 2AgBr \u2192 2Ag + Br 2 \nDisplacement reaction: One element is displaced by another element. \nZn + CuSO 4 \u2192 ZnSO 4 + Cu \nDouble displacement reaction: Exchange of ions between reactants.", - "Zn + CuSO 4 \u2192 ZnSO 4 + Cu \nDouble displacement reaction: Exchange of ions between reactants. \nAgNO 3 + NaCl \u2192 AgCl + NaNO 3 \nRedox reaction: Both oxidation and reduction take place simultaneously. \nCuO + H 2 \u2192 Cu + H 2O", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-IMPORTANT NOTES.txt\nDouble displacement reaction: Exchange of ions between reactants. \nAgNO 3 + NaCl \u2192 AgCl + NaNO 3 \nRedox reaction: Both oxidation and reduction take place simultaneously. \nCuO + H 2 \u2192 Cu + H 2O \nExothermic reaction: A chemical reaction in which heat energy is evolved. \nC + O 2 \u2192 CO 2 (g) + heat \nEndothermic reaction: A chemical reaction in which heat en ergy is absorbed. \nZnCO 3 + Heat \u2192 ZnO + CO 2 \nRedox reaction: Chemical reaction in which both oxidation and reduction take place \nsimultaneously. \n4. Oxidation: Reaction that involves the gain of oxygen or loss of hydrogen. 5. Reduction: Reaction that shows the loss of oxygen or gain of hydrogen. \n ZnO + C \u2192 Zn + CO \n ZnO is reduced to Zn \u2014reduction. \n C is oxidized to CO \u2014Oxidation. \n6.", - "5. Reduction: Reaction that shows the loss of oxygen or gain of hydrogen. \n ZnO + C \u2192 Zn + CO \n ZnO is reduced to Zn \u2014reduction. \n C is oxidized to CO \u2014Oxidation. \n6. Effects of Oxidation Reactions in Our Daily Life: \n\uf0b7 Corrosion: It is an undesirable change that occurs in metals when they are \nattacked by moisture, air, acids and bases.", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS-IMPORTANT NOTES.txt\n6. Effects of Oxidation Reactions in Our Daily Life: \n\uf0b7 Corrosion: It is an undesirable change that occurs in metals when they are \nattacked by moisture, air, acids and bases. \nExample, Corrosion (rusting) of Iron: Fe 2O3. nH 2O (Hydrated iron oxide) \n\uf0b7 Rancidity: Undesirable change that takes place in oil containing food items \ndue to the oxidation of fat ty acids. \nPreventive methods of rancidity: Adding antioxidants to the food materials, \nstoring food in the airtight container, flushing out air with nitrogen gas and \nrefrigeration.", + "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS 2.txt\nChemical Reactions & \nequations \nPower Point Presentation END", "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS -TEXTBOOK NOTES.txt\nChemical Reactions\nand Equations1 CHAPTER\nConsider the following situations of daily life and think what happens\nwhen \u2013\n\u0084milk is left at room temperature during summers.\n\u0084an iron tawa/pan/nail is left exposed to humid atmosphere.\n\u0084grapes get fermented.\n\u0084food is cooked.\n\u0084food gets digested in our body.\n\u0084we respire.\nIn all the above situations, the nature and the identity of the initialsubstance have somewhat changed. We have already learnt about physicaland chemical changes of matter in our previous classes. Whenever a chemicalchange occurs, we can say that a chemical reaction has taken place.\nYou may perhaps be wondering as to what is actually meant by a\nchemical reaction. How do we come to know that a chemical reactionhas taken place? Let us perform some activities to find the answer to\nthese questions.\nFigure 1.1\nBurning of a magnesium ribbon in air and collection of magnesiumoxide in a watch-glassActivity 1.1Activity 1.1Activity 1.1Activity 1.1Activity 1.1", "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS -TEXTBOOK NOTES.txt\nthese questions.\nFigure 1.1\nBurning of a magnesium ribbon in air and collection of magnesiumoxide in a watch-glassActivity 1.1Activity 1.1Activity 1.1Activity 1.1Activity 1.1\nCAUTION: This Activity needs\nthe teacher\u2019s assistance. It\nwould be better if studentswear eye protection.\n\u0084Clean a magnesium ribbonabout 2 cm long by rubbingit with sandpaper.\n\u0084Hold it with a pair of tongs.Burn it using a spirit lamp orburner and collect the ash so\nformed in a watch-glass as\nshown in Fig. 1.1. Burn themagnesium ribbon keeping it\nas far as possible from your\neyes.\n\u0084What do you observe?", "1.1. Burn themagnesium ribbon keeping it\nas far as possible from your\neyes.\n\u0084What do you observe?\n\u201cFacts are not science \u2014 as the dictionary is not literature.\u201d\nMartin H. FischerScience 2Activity 1.2Activity 1.2Activity 1.2Activity 1.2Activity 1.2\nFigure 1.2\nFormation of hydrogengas by the action ofdilute sulphuric acid onzincFrom the above three activities, we can say that any ofthe following observations helps us to determine whethera chemical reaction has taken place \u2013\n\u0084change in state\n\u0084change in colour", @@ -153,85 +234,6114 @@ "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS -TEXTBOOK NOTES.txt\n11. Why are decomposition reactions called the opposite of combination reactions?\nWrite equations for these reactions.Science 16Group Activity\nPerform the following activity.\n\u0084Take four beakers and label them as A, B, C and D.\n\u0084Put 25 mL of water in A, B and C beakers and copper sulphate solution in beaker D.\n\u0084Measure and record the temperature of each liquid contained in the beakers above.\n\u0084Add two spatulas of potassium sulphate, ammonium nitrate, anhydrous copper\nsulphate and fine iron fillings to beakers A, B, C and D respectively and stir.\n\u0084Finally measure and record the temperature of each of the mixture above.\nFind out which reactions are exothermic and which ones are endothermic in nature.12. Write one equation each for decomposition reactions where energy is supplied in\nthe form of heat, light or electricity.\n13. What is the difference between displacement and double displacement reactions?\nWrite equations for these reactions.", "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS -TEXTBOOK NOTES.txt\nthe form of heat, light or electricity.\n13. What is the difference between displacement and double displacement reactions?\nWrite equations for these reactions.\n14. In the refining of silver, the recovery of silver from silver nitrate solution involved\ndisplacement by copper metal. Write down the reaction involved.\n15. What do you mean by a precipitation reaction? Explain by giving examples.16. Explain the following in terms of gain or loss of oxygen with two examples each.\n(a) Oxidation\n(b) Reduction\n17. A shiny brown coloured element \u2018X\u2019 on heating in air becomes black in colour.\nName the element \u2018X\u2019 and the black coloured compound formed.\n18. Why do we apply paint on iron articles?19. Oil and fat containing food items are flushed with nitrogen. Why?20. Explain the following terms with one example each.\n(a) Corrosion\n(b) Rancidity", "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS.txt\nChemical Reactions & \nequations \nPower Point Presentation END", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\nVery Short Answer Type Questions [2 Marks]", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n1. \u201cWe need to balance a skeletal chemical equation.\u201d Give reason to justify the \nstatement. \n2. Giving an example list two information which make a chemical equation \nmore useful (informative). \n3. Name the reducing agent in the following reaction: \n 3MnO 2 + 4Al \u2014\u2014\u2014\u2014 > 3Mn + 2Al 2O3 \nState which is more reactive, Mn or Al and why? \n4. (i) Write a balanced chemical equation for process of photosynthesis. \n (ii)When do desert plants take up carbon dioxide and perform \nphotosynthe sis? \n5. What is observed when a solution of potassium iodide solution is added to \na solution of lead nitrate? Name the type of reaction. Write a balanced \nchemical equation to represent the above chemical reaction. \n6. Write balanced chemical equations for the fol lowing reactions.", - "5. What is observed when a solution of potassium iodide solution is added to \na solution of lead nitrate? Name the type of reaction. Write a balanced \nchemical equation to represent the above chemical reaction. \n6. Write balanced chemical equations for the fol lowing reactions. \n(i) Silver bromide on exposure to sunlight decomposes into silver and \nbromine, \n(ii) Sodium metal reacts with water to form sodium hydroxide and hydrogen \ngas.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n(i) Silver bromide on exposure to sunlight decomposes into silver and \nbromine, \n(ii) Sodium metal reacts with water to form sodium hydroxide and hydrogen \ngas. \n7. Identify the type of reaction(s) in the following equations. \n(i)CH 4 + 2O 2 ----\uf0e0 CO 2 + 2 H2O \n(ii) Pb(NO 3)2 + 2KI \u2014\u2014\u2013>Pbl 2 + 2KNO 3 \n(iii) CaO + H2O \u2014\u2014\u2013> Ca(OH) 2 \n(iv) CuSO 4 + Zn \u2014\u2014\u2013> ZnSO 4 + Cu \n8. Write balanced equation for the reaction between magnesium and \nhydrochloric acid. Name the products obtained, identify the type of reaction. \n9. Describe an activity to observe what happens when quick lime is added to \nwater taken in a beaker. State two important observations and name the type \nof reaction taking place. \n10.", - "Name the products obtained, identify the type of reaction. \n9. Describe an activity to observe what happens when quick lime is added to \nwater taken in a beaker. State two important observations and name the type \nof reaction taking place. \n10. What is the colour of ferrous sulphate crystals? How does this colour change \nafter heating? \n11. Why does the colour of copper sulphate solution change when an iron nail is \ndipped in it? Write two observations.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\nafter heating? \n11. Why does the colour of copper sulphate solution change when an iron nail is \ndipped in it? Write two observations. \n12. Translate the following statement into chemical equation and then balance it : \nBarium chloride reacts with aluminium sulphate to gi ve aluminium chloride \nand a precipitate of barium sulphate. State the two types in which this reaction \ncan be classified. \n13. Why decomposition reactions are called the opposite of combination \nreactions? Write equations for these reactions. \n14. AgN0 3 (aq) + NaCl( aq)\u2014\u2014\u2014\u2014\u2014\u2014 \u2013 > AgCl (s)\u2193 + NaN0 3(aq) \nFeS + H 2S0 4\u2014\u2014\u2014\u2014 - > FeS0 4 + H 2S\u2191 \nConsider the above mentioned two chemical equations with two different \nkinds of arrows (\u2191and \u2193) along with product. What do these two different \narrows indicate? \n15.", - "What do these two different \narrows indicate? \n15. Hydrogen being a highly infla mmable gas and oxygen being a supporter of \ncombustion, yet water which is a compound made up of hydrogen and oxygen", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\narrows indicate? \n15. Hydrogen being a highly infla mmable gas and oxygen being a supporter of \ncombustion, yet water which is a compound made up of hydrogen and oxygen \nis used to extinguish fire. Why? 16. Using a suitable chemical equation, justify that some chemical reactions are \ndetermined by: \n(i) change in c olour, (ii) change in temperature. \n17. (a) A solution of substance \u2018X\u2019 is used for white washing. What is the \nsubstance \u2018X\u2019? State the chemical reaction of \u2018X\u2019 with water. \n(b) Why does the colour of copper sulphate solution change when an iron nail \nis dipped i n it? \n18. Write the balanced equation for the following reaction and identify the type of \nreaction in each case. \n(i) Potassium bromide + Barium iodide \u2014-> Potassium iodide + Barium \nbromide. \n(ii) Hydrogen(g) + Chlorine(g) \u2014-> Hydrogen chloride(g) \n19.", - "(i) Potassium bromide + Barium iodide \u2014-> Potassium iodide + Barium \nbromide. \n(ii) Hydrogen(g) + Chlorine(g) \u2014-> Hydrogen chloride(g) \n19. A zinc plate was put into a solution of copper sulphate kept in a glass \ncontainer. It was found that blue colour of the solution gets fader and fader", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n19. A zinc plate was put into a solution of copper sulphate kept in a glass \ncontainer. It was found that blue colour of the solution gets fader and fader \nwith the passage of time. After few days, when zinc plate was taken out of the \nsolution, a number of holes were observed o n it. \n(i) State the reason for changes observed on the zinc plate. \n(ii) Write the chemical equation for the reaction involved. \n20. A white salt on heating decomposes to give brown fumes and a residue is \nleft behind. \n(i) Name the salt. \n(ii) Write the equation f or the decom -position reaction. \n21. When a solution of potassium iodide is added to a solution of lead nitrate in a \ntest tube, a reaction takes place. \n(a) What type of reaction is this? \n(b) Write a balanced chemical equation to represent the above reaction. \n22. Define combination reaction. Give one example of a combination reaction \nwhich is also exothermic \n23. (a) Classify the following reactions into different types.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n22. Define combination reaction. Give one example of a combination reaction \nwhich is also exothermic \n23. (a) Classify the following reactions into different types. \n (i) Na 2SO 4 + BaCl 2 -----\uf0e0 BaSO 4 + NaCl \n (ii) CaO + H 2O --------\uf0e0 Ca(OH) 2 \n (iii) CaCO 3 ----------\uf0e0 CaO + CO 2 \n(b) Which of the above reaction(s) is/are precipitation reaction(s)? Why a \nreaction is called precipitation reaction?", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n24. Write a balanced chemical equation for the reaction between sodium chloride \nand silver nitrate indicating the physical state of the reactants and the \nproducts. \n25. What is a redox reaction? When a magnesium ribbon burns in air with a \ndazzling flame and forms a white ash, is magnesium oxidized or reduced? \nWhy? \n26. Write any two observations in an activity w hich may suggest that a \nchemical reaction has taken place. Give an example in support of your \nanswer. \n27. When the powder of a common metal is heated in an open china dish, its \ncolour turns black. However, when hydrogen is passed over the hot black \nsubstance s o formed, it regains its original colour. Based on the above \ninformation, answer the following questions. \n(i) What type of chemical reaction takes place in each of the two given steps? (ii) Name the metal initially taken in the powder form. Write balanced chemical \nequations for both reactions.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n(i) What type of chemical reaction takes place in each of the two given steps? (ii) Name the metal initially taken in the powder form. Write balanced chemical \nequations for both reactions. \n28. In electrolysis of water, why is the volume of gas collected over one electrode \ndouble that of gas collected over the other electrode? \n29. Name the products formed on strongly heating ferrous sulphate crystals. What \ntype of chemical reaction occurs in this change? \n30. What is an oxidation reaction? Give an example of oxidation reaction. Is \noxidation an exothermic or an endothermic reaction? \n31. Describe an activity to demonstrate the change that takes place when white \nsilver chlor ide is kept in sunlight. State the type of chemical reaction which \ntakes place. \n32. When magnesium ribbon burns in air or oxygen, a product is formed. State \nthe type of chemical reaction and name the product formed in the reaction. \nWrite balanced chemical equa tion of this reaction.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\nthe type of chemical reaction and name the product formed in the reaction. \nWrite balanced chemical equa tion of this reaction. \n33. Distinguish between a displacement reaction and a double displacement \nreaction. Identify the displacement and the double displacement reaction from \nthe following reactions. \n(i) CuSO 4 + Zn ----\uf0e0 ZnSO 4 +Cu \n(ii) Pb(NO 3)2 + KI -----\uf0e0 PbI 2 + 2 KNO 3", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\nShort Answer Type Questions (II) [3 Marks] \n \n34. Write the chemical equation of the reaction in which the following changes \nhave taken place with an example of each: \n(i) Change in colour \n(ii) Change in temperature \n(iii) Formation of precipitate \n35. State the type of chemical reactions and chemical equations that take place in \nthe following: \n(i) Magnesium wire is burnt in air. \n(ii) Electric current is passed through water. \n(iii) Ammonia and hydrogen chloride gases \u2018are mixed. \n36. (a) Write the essential c ondition for the following reaction to take place: \n 2AgBr \u2014-> 2Ag + Br 2 \nWrite one application of this reaction. \n(b) Complete the following chemical equation of a chemical reaction \n Heat \n 2FeS0 4 \u2014\uf0e0 Fe2O3 +\u2026\u2026. + \u2026\u2026 \n(c) What happens when water is added to quick lime? Write chemical \nequation.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n37. 2g of ferrous sulphate crystals are heated in a dry boiling tube. \n(i) List any two observations. \n(ii) Name the type of chemical reaction taking place. \n(iii) \u2018Write the chemical equation for the reaction. \n38. Write chemical equation reactions taking place when carried out with the help \nof \n(a) Iron reacts with steam (b) Magnesium reacts with dil HCl \n(c) Copper is heated in air. \n39. Which products will be obtained when lead nit rate is heated simply? Write \nbalanced chemical equation for the reaction? State the type of chemical \nreaction that occur in the change. \n40. What is meant by skeletal type chemical equation? What does it represent? \nUsing the equation for electrolytic decomposit ion of water, differentiate \nbetween a skeletal chemical equation and a balanced chemical equation. \n41. What is rancidity? Mention any two ways by which rancidity can be prevented. \n42. Write balanced chemical equation for the reactions that take place during", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n41. What is rancidity? Mention any two ways by which rancidity can be prevented. \n42. Write balanced chemical equation for the reactions that take place during \nrespir ation. Identify the type of combination reaction that takes place during \nthis process and justify the name. Give one more example of this type of \nreaction. \n43. What is redox reaction? Identify the substance oxidised and the substance \nreduced in the following reactions. \n(i)2PbO + C \u2014\u2013> 2Pb + CO 2 \n(ii)MnO 2 + 4HCl \u2014\u2013> MnCl 2 + 2H 20 + Cl 2 \n44. Write the balanced chemical equations for the following reaction and identify \nthe type of reaction. \nThermite reaction, iron (III) oxide reacts with aluminium and gives molten iron \nand aluminium oxide. \n45. A solution of potassium chloride when mixed with silver nitrate solution, an \ninsoluble white substance is formed.", - "Thermite reaction, iron (III) oxide reacts with aluminium and gives molten iron \nand aluminium oxide. \n45. A solution of potassium chloride when mixed with silver nitrate solution, an \ninsoluble white substance is formed. Write the chemical reaction involved and \nalso mention the type of the chemica l reaction?", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\ninsoluble white substance is formed. Write the chemical reaction involved and \nalso mention the type of the chemica l reaction? \n46. Write balanced equations for the following mentioning the type of reaction \ninvolved. \n(i) Aluminium + Bromine \u2014\u2013> Aluminium bromide \n(ii) Calcium carbonate \u2014\u2013> Calcium oxide + Carbon dioxide \n(iii) Silver chloride \u2014\u2013>Silver + Chlorine \n47. (a) Why is res piration considered as an exothermic reaction? \n(b) Define the terms oxidation and reduction. \n(c) Identify the substance that is oxidised and reduced in the following \nreaction \n Zn + CuO ----\uf0e0 ZnO + Cu", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n48. What is meant by \n(i) precipitation reac tion, \n(ii) exothermic reaction, \n(iii) oxidation reaction? \nWrite balanced chemical equations for an example of each . \n49. You might have noted that when copper powder is heated in a china dish, the \nsurface of copper powder becomes coated with a black colour substance. \n(i) How has this black coloured substance formed? \n(ii) What is that black substance? \n(iii) Write the chemical equation of the reaction that takes place. \n50. (a) What happens chemically when quicklime is added to water filled in a \nbucket? \n(b) On what basis is a chemical equation balanced? \n(c ) What change in colour is observed when white silver chloride is left \nexposed to sunlight? State the type of chemical reaction in this change. 51. What happens when an aqueous solution of sodium sulphate reac ts with an \naqueous solution of barium chloride? State the physical conditions of", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\naqueous solution of barium chloride? State the physical conditions of \nreactants in which the reaction between them will not take place. Write the \nbalanced chemical equation for the reaction and name the type of reaction. \n52. When you have mixed the solutions of lead(II) nitrate and potassium iodide, \n(i) what was the colour of the precipitate formed and can you name the \nprecipitate? \n(ii) Write the balanced chemical equation for this reaction. \n(iii) Is this also a double displac ement reaction? \n53. Name the type of reaction represented by the following equation: \n(i) CaO + H 2O ---\uf0e0 Ca(OH) 2 \n(ii) 3BaCl 2 + Al 2(SO 4)3 ---\uf0e0 3 BaSO 4 + 2 AlCl 3 \n heat \n(iii) 2 FeSO 4 ----------------- \uf0e0 Fe2O3 + SO 2 + SO 3 \n54.", - "Write the chemical equation of the reaction in which the following changes \nhave taken place with an example of each: \n(i) Change in colour \n(ii) Change in temperature \n(iii) Formation of precipitate", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\nhave taken place with an example of each: \n(i) Change in colour \n(ii) Change in temperature \n(iii) Formation of precipitate \n55. Balance the following chemical equations. \n(i) BaCl 2 + H 2SO 4 ------\uf0e0 BaSO 4 + HCl \n(ii) Ca(OH) 2 + HNO 3 ----\uf0e0 Ca(NO 3)2 + H 2O \n(iii)Pb(NO 3)2 ------\uf0e0 PbO +NO 2 +O 2 \n(iv) MnO 2 + HCl ----\uf0e0 MnCl 2 + H 2O + Cl 2 \n56. Balance the following chemical reactions and identify the type of reaction.", - "Balance the following chemical reactions and identify the type of reaction. \n(a) Mg (s) + Cl 2 (g) ---\uf0e0 MgCl 2(s) \n(b) HgO (s) --\uf0e0 Hg(l) + O 2(g) \n(c) Na(s) + S (s) ----------\uf0e0 Na2S(s) \n(d) TiCl4 (l) + Mg (s) ----\uf0e0 Ti(s) + MgCl 2(s) \n(e) CaO (s) + SiO 2(s) ---\uf0e0 CaSiO 3(s) \n(f) H2O2(l) --------\uf0e0 H2O(l) + O 2(g) \n57. During the reaction of some metas with dilute hydrochloric acid, following \noberservations were made. \n(a) Silver meta does not show any change \n(b) The temperature of the reaction mixture rises when aluminium (Al) is \nadded. \n(c) The reaction os sodium metal is found to be highly explosive", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-SHORT ANSWER TYPE QUESTIONS.txt\n(a) Silver meta does not show any change \n(b) The temperature of the reaction mixture rises when aluminium (Al) is \nadded. \n(c) The reaction os sodium metal is found to be highly explosive \n(d) Some bubbles of a gas are seen when lead (Pb) is reacted with acid. \nExplain these observations givi ng suitable reasons.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nTest items \n \n \n \nLOB: Compare the characteristics of initial & final substances in order to check whether the \nchange is physical or chemical \n \n1) A student poured 100 mL of water in a bottle and added 40 mL vinegar to it. A balloon was filled \nwith 20 g baking soda and was fixed at the mouth of the bottle. Slowly the shape of the balloon \nchanged, as shown. \n \n \n \nThe student claims that a chemical change happened when the two substances were mixed. Is \nthe claim made by the student correct? \n \n(a) Yes, as a new substance was formed in the form of a gas. \n(b) Yes, as the mass remains the same throughout the experiment. \n(c) No, as the formation of bubbles in the mixture shows a physical change. \n(d) No, as the change in the shape and size of the balloon shows a physical change.", - "(c) No, as the formation of bubbles in the mixture shows a physical change. \n(d) No, as the change in the shape and size of the balloon shows a physical change. \nCorrect Answer: Option (a) \n \n2) A student makes a list of some activities he observes one day.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\n2) A student makes a list of some activities he observes one day. \n \n \n \n \nWhich activity can the student classify as a chemical change? \n \n(a) Activity 1, as the properties of the substances in the mixture change. \n(b) Activity 2, as the physical state of the apple pie changes when cut. \n(c) Activity 3, as the shape of the can changes. \n(d) Activity 4, as the shape and size of the wooden log changes. \nCorrect Answer: Option (a) \nLOB : Relate the substances taking part in the chemical reaction & substances formed in the \nchemical reaction in order to classify them as reactants & products \n1) Sodium and chlorine are reacted and as a result, sodium chloride is formed which is also called \ntable salt. What option gives the reactants and products of the reaction?", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\n(a) reactants -sodium; products - chlorine \n(b) reactants -sodium and table salt; products - chlorine \n(c) reactants -tables salt; products - sodium and chlorine \n(d) reactants -sodium and chlorine; products - sodium chloride \nCorrect Answer: Option (d) \n \n \n2) The image shows some chemical reactions. \n \n \n \nWhich option identifies the reactants and products of the reactions? \n \n(a) \n(b) \n(c) \n(d) \nCorrect Answer: Option (c) \nLOB : Use chemical symbols & chemical formulae correctly in order to acquire the skill of writing \nchemical equations \n1) A student performs an experiment to form aluminium chloride from aluminium and chlorine. \nWhich options gives the chemical equation of the reaction?", - "Which options gives the chemical equation of the reaction? \n(a) Al + Cl2 \uf0e0\uf020AlCl 2 \n(b) 2Al + Cl2 \uf0e0\uf0202AlCl \n(c) 2Al + 3Cl 2 \uf0e0\uf0202AlCl 3 \n(d) 3Al + 3Cl 2 \uf0e0\uf0203AlCl 3 \n \nCorrect Answer: Option (c)", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nCorrect Answer: Option (c) \n \n \n2) A researcher adds barium hydroxide to hydrochloric acid to form a white -colored barium \nchloride. Which option gives the balanced chemical equation of the reaction? \n \n(a) HCl + Ba(OH) 2 \uf0e0\uf020BaCl 2 + 2HOH \n \n(b) 2HCl + Ba(OH) 2 \uf0e0\uf020BaCl 2 + 2HOH \n \n(c) 2HCl + Ba(OH) 2 \uf0e0\uf020BaH 2 + 2HCl + O2 \n \n(d) HCl + 2Ba(OH) \uf0e0\uf0202BaCl 2 + 2HOH + O2 \n \nCorrect Answer: Option (b) \n \n \n \n \n \n \n \nLOB : Apply Law of Conservation of Mass in order to balance chemical equations \n1) A student writes a balanced chemical equation.", - "Pb(s) + CuCl 2(aq) \uf0ae\uf020PbCl 2(aq) + Cu(s) \nWhich option gives the number of elements on the LHS and RHS of the chemical equation? \n(a) \n \n \n(b) \n \n \n \n(c) \n \n \n(d) \n \n \n \nCorrect Answer: Option (d) \n \n \n2) The image shows a balanced chemical equation of the reaction between sodium and chlorine to \nform sodium chloride. \n \n \n \nWhich option shows the number of atoms on both sides of the reaction? \n(a)", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nWhich option shows the number of atoms on both sides of the reaction? \n(a) \n \n \n(b) \n \n \n(c) \n \n \n(d) \n \n \nCorrect Answer: Option (b)", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nLOB: Categorize the given reactions as (combination/ decomposition) based on the reactants & \nproducts of a chemical reaction \n1) A student writes a chemical equation of the reaction between carbon monoxide and hydrogen. \nCO 2 + 2H 2 -> CH 3OH \nHow can the reaction be classified? \n(a) The reaction is an example of a combination reaction as a compound separates into two \ncompounds. \n(b) The reaction is an example of a decomposition reaction as a compound dissociates into \ntwo compounds. \n(c) The reaction is an example of a combination reaction as two compounds react to form a \nsingle compound. \n(d) The reaction is an example of a decomposition reaction as two compounds react to form \na single compound.", - "(c) The reaction is an example of a combination reaction as two compounds react to form a \nsingle compound. \n(d) The reaction is an example of a decomposition reaction as two compounds react to form \na single compound. \nCorrect Answer: Option (c) \n2) A student learns that some products are formed as a result of combining two compounds while \nsome compounds are formed as a result of dissociation of two compounds. The image shows", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nsome compounds are formed as a result of dissociation of two compounds. The image shows \ntwo reactions.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nWhich reaction is an example of a combination reaction and a decomposition reaction? \n(a) both the reactions are examples of combination reaction \n(b) both the reactions are examples of a decomposition reaction \n(c) reaction P is an example of a combination reaction while reaction Q is an example of a \ndecomposition reaction \n(d) P is an example of a decomposition reaction while reaction Q is an example of a \ncombination reaction \nCorrect Answer: Option (c)", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nLOB: Classify the given reaction as displacement or double displacement based on the type of \nreactants used & products formed \n1) A student adds lead and silver to two different test tubes containing an equal amount of copper \nsulphate solution. The student observes that the color of the solution in the test tube with lead \nchanges. What explains the change in the colour of the solution? \n(a) A displacement reaction takes place as lead replaces copper from the solution. \n(b) A combination reaction takes place as lead combines with sulpha te in the solution. \n(c) decomposition reaction takes place as copper dissociates from sulphate in the solution. \n(d) A double displacement reaction takes place as copper dissociates from sulphate and \nlead combines with sulphate in the solution.", - "(c) decomposition reaction takes place as copper dissociates from sulphate in the solution. \n(d) A double displacement reaction takes place as copper dissociates from sulphate and \nlead combines with sulphate in the solution. \nCorrect Answer: Option (a) \n2) The chemical reaction between potassium chloride and silver nitrate is given by the chemical", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nlead combines with sulphate in the solution. \nCorrect Answer: Option (a) \n2) The chemical reaction between potassium chloride and silver nitrate is given by the chemical \nequation.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nWhat can be inferred from the chemical equation? \n(a) silver nitrate and potassium undergo a decomposition reaction to form silver chloride \nand potassium nitrate \n(b) silver nitrate and potassium undergo a displacement reaction to form silver chloride \nand potassium nitrate \n(c) silver nitrate and potassium undergo a combination reaction to form silver chloride and \npotassium nitrate \n(d) silver nitrate and potassium undergo double displacement reaction to form silver \nchloride and potassium nitrate \nCorrect Answer: Option (d) \n \n \nLOB: Predict the reaction as Oxidation or Reduction based on the addition/ removal of oxygen/ \nhydrogen/ electrons to the reactants to form products. \n1) The image shows a reaction between zinc and hydrogen. \n Which option shows oxidation? \n(a) Zn \uf0ae\uf020Zn!\" \n(b) 2H! \uf0ae\uf020H2 \n(c) Zn!\"", - "1) The image shows a reaction between zinc and hydrogen. \n Which option shows oxidation? \n(a) Zn \uf0ae\uf020Zn!\" \n(b) 2H! \uf0ae\uf020H2 \n(c) Zn!\" \uf0ae\uf020Zn \n(d) H 2 \uf0ae2H! \nCorrect Answer: Option (a) \n \n \n2) The image shows a reaction between iron oxide and hydrogen.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\n2) The image shows a reaction between iron oxide and hydrogen. \n \n \nWhich option shows the compounds undergoing oxidation and reduction? \n(a) \n \n \n \n(b) \n \n \n(c) \n \n \n(d) \n \n \nCorrect Answer: Option (a) \n \n \n \n \n \nLOB: Observe colour change in iron, copper and silver articles over time in order to outline the \neffects of corrosion in our surroundings (real life situations, stating any two)", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nLOB: Observe colour change in iron, copper and silver articles over time in order to outline the \neffects of corrosion in our surroundings (real life situations, stating any two) \n \n1) A student notices that a new hammer made of iron is shiny while an old one kept in the toolbox \nfor long has a reddish -brown powder deposit over it. What does the change in colour of the \nhammer indicate? \n(a) effect of moisture on metals \n(b) iron hammer turns brown after some time \n(c) effects of kept in a box for a longer duration \n(d) iron changes colour when kept with other tools \nCorrect Answer: Option (a) \n2) A student notices that her silver jewellery turned dull and had a gray -black film over it after \nwearing for a few months. What results in the change in colour of the silver metal?", - "What results in the change in colour of the silver metal? \n(a) dust deposits over the jewellery which changes its colour", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\n(b) the jewellery comes in contact with air, moisture, and acids and corrodes \n(c) the polish over the jewellery was removed after wearing for a few months \n(d) silver breaks due to wear and tear and turns its colour changes due to rusting \nCorrect Answer: Option (b) \n \n \nLOB: Detect changes in smell, colour, taste of food items overtime, in order to explain effects of \noxidation on food items \n \n1) A student learns that food companies fill bags of chips with nitrogen gas. What is the purpose \npacking it with nitrogen? \n(a) it prevents rancidity of chips \n(b) it keeps the mosquitoes away from chips \n(c) it keeps the chips dry if the pack falls in water \n(d) prevents chips from spilling out when the pack is opened \n \nCorrect Answer: Option (a)", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nCorrect Answer: Option (a) \n \n \n \n \n2) A student notices that the bread kept out has a green coloured coating over it after a few days. \nWhat explains the reason for the student\u2019s observation? \n(a) the oils in the bread oxidises and causes rancidity \n(b) bread comes in contact with atmospheric moisture and corrodes \n(c) the oils in the bread reduces and cause the change in the colour of the bread \n(d) comes in contact with the atmospheric nitrogen and a layer deposit over it", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nCorrect Answer: Option (a) \nSuggested Teacher Resources \n1 Activity \nObjective Categorize the given reactions as (combination/ decomposition) based on the \nreactants & products of a chemical reaction. \nPrerequisite Identify the changes that a reaction mixture undergoes during a chemical \nreaction, what are reactants, what are products \nMaterial \nRequired Test tubes, boiling tube, distilled water, burner, sugar, iron fillings, sulphur \npowder \nVocabulary Combination Reaction: A reaction in which a single product is formed from \ntwo or more reactants is known as a combination reaction. \n \nDecomposition Reaction: A reaction in which a single substance decomposes \nto give two or more substances. \n \nProcedure Procedure 1: \n1. Take a small amount of iron fillings and sulphur powder in a boiling \ntube. \n2.", - "Procedure Procedure 1: \n1. Take a small amount of iron fillings and sulphur powder in a boiling \ntube. \n2. Heat the boiling tube strongly over the flame of a burner or spirit lamp \nand observe changes.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nProcedure 2: \n1. Take about 2 g sugar crystals in a dry boiling tube. (Note the colour of \nthe crystals.) \n2. Heat the boiling tube over the flame of a burner or spirit lamp and \nobserve changes. \nReflection \nQuestions \nText to real \nworld \nconnection Propane is a fuel used to provide heat for some homes. It is stored in large \ntanks as shown here. Can you predict the chemical equation for burning of \npropane? \n \n \nBeyond the \nclassroom What type of reaction takes place inside the human and plant cells? \n \n \n \n \nMaterial \nrequired Fresh potato chips, 2 glass jars with lid, Aluminium foil. \nProcedure 1. Wrap the glass jar with aluminium foil. Tape the foil in place so that no \nlight can enter the container. \n2.", - "Procedure 1. Wrap the glass jar with aluminium foil. Tape the foil in place so that no \nlight can enter the container. \n2. Place fresh potato chips in the foil -wrapped jar and in a similar clear jar \nwithout foil around it.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nActivity \n Reaction No. Materials \nused Chemicals \nused Describe the \nreactants \n(number of \nreactants) Procedure Describe \nthe \nproducts \n(number of \nproducts) Type of \nreaction \nCalcium \nOxide + \nwater \nHeating \nSugar \n \n 3. Taste the potato chips and rate their flavour on a 5-point scale, 1 being \nextremely dislike the flavour and 5 being extremely like the flavour. \n(Enter your data in a table.) \n4. Place the two jars on a window sill where they will be exposed to \nsunlight. \n5. Taste potato chips from each jar at intervals of 1-2 days for 1-2 weeks \nand record the taste of potatoes in both the jars. \n6. Make a graph of your data, noting the flavor of the potato chips stored \nthese two ways versus storage time.", - "6. Make a graph of your data, noting the flavor of the potato chips stored \nthese two ways versus storage time. The y-axis should be the flavor \nscore and the x-axis the time in days. \nReflection \nquestions 1. What effect does the aluminium foil have on the taste of potatoes after 2 \nweeks?", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-TEST QUESTIONS.txt\nscore and the x-axis the time in days. \nReflection \nquestions 1. What effect does the aluminium foil have on the taste of potatoes after 2 \nweeks? \n2. Is there anything else that will affect the taste of potatoes over 2 weeks \napart from the sunlight?", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nChemical Reactions and Equations\nIntroduction to Chemical Reactions and Equations\nPhysical and chemical changes\nChemical change - one or more new substances with new physical and chemical properties\nare formed.\nExample: Fe(s)\u00a0 + \u00a0CuSO 4(aq) \u2192FeSO 4(aq) +Cu(s)\u00a0\n \u00a0 \u00a0 \u00a0 (Blue)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (Green)\u00a0 \u00a0 \u00a0 \u00a0\nHere, when copper sulphate reacts with iron, two new substances, i.e., ferrous sulphate and\ncopper are formed.\nPhysical change - change in colour or state occurs but no new substance is formed.\nExample: Water changes to steam on boiling but no new substance is formed(Even though\nsteam and water look different when they are made to react with a piece of Na, they react\nthe same way and give the exact same products). This involves only change in state (liquid\nto vapour).\u00a0\nObservations that help determine a chemical reaction\nA chemical reaction can be determined with the help of any of the following observations:\na) Evolution of a gas\nb) Change in temperature", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nObservations that help determine a chemical reaction\nA chemical reaction can be determined with the help of any of the following observations:\na) Evolution of a gas\nb) Change in temperature\nc) Formation of a precipitate\nd) Change in colour\ne) Change of state\nChemical reaction\nChemical reactions are chemical changes in which reactants transform into products by\nmaking or breaking of bonds(or both) between different atoms.\nTypes of chemical reactionsTaking into consideration different factors, chemical reactions are grouped into multiple\ncategories.\nFew examples are:\n\u25cfCombination\n\u25cfDecomposition\n\u25cfSingle Displacement\n\u25cfDouble displacement\n\u25cfRedox\n\u25cfEndothermic\n\u25cfExothermic\n\u25cfPrecipitation\n\u25cfNeutralisation\nChemical Reactions and Equations I\nWord equation\nA\u00a0 word equation is a\u00a0chemical reaction\u00a0expressed in words rather than\u00a0chemical\nformulas.\u00a0It helps identify the reactants and products in a chemical reaction.\nFor example,\u00a0\nSodium + Chlorine \u2192 Sodium chloride", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nformulas.\u00a0It helps identify the reactants and products in a chemical reaction.\nFor example,\u00a0\nSodium + Chlorine \u2192 Sodium chloride\nThe above equation means: \"Sodium reacts with chlorine to form sodium chloride.\"\u00a0\nSymbols of elements and their valencies\nA symbol is the chemical code for an element. Each element has one or two letter atomic\nsymbol, which is the abbreviated form of its name.\nValency is the combining capacity of an\u00a0element. It can be considered as the number of\nelectrons lost, gain or shared by an atom when it combines with another atom to form a\nmolecule.\nWriting chemical equations\nRepresentation of a chemical reaction in terms of symbols and chemical formulae of the\nreactants and products is known as a chemical equation.\nZn(s) +dil.H2SO 4(aq) \u2192ZnSO 4(aq) +H2(\u2191)\n\u00a0(Reactants)\u00a0 \u00a0(Products)\n\u2022 For solids, the symbol is \"(s)\".\n\u2022 For liquids, it is \"(l)\".", - "Zn(s) +dil.H2SO 4(aq) \u2192ZnSO 4(aq) +H2(\u2191)\n\u00a0(Reactants)\u00a0 \u00a0(Products)\n\u2022 For solids, the symbol is \"(s)\".\n\u2022 For liquids, it is \"(l)\".\n\u2022 For gases, it is \"(g)\".\u2022 For aqueous solutions, it is \"(aq)\".", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nZn(s) +dil.H2SO 4(aq) \u2192ZnSO 4(aq) +H2(\u2191)\n\u00a0(Reactants)\u00a0 \u00a0(Products)\n\u2022 For solids, the symbol is \"(s)\".\n\u2022 For liquids, it is \"(l)\".\n\u2022 For gases, it is \"(g)\".\u2022 For aqueous solutions, it is \"(aq)\".\n\u2022 For gas produced in the reaction, it is represented by \"(\u2191)\".\n\u2022 For precipitate formed in the reaction, it is represented by \"(\u2193)\".\nBalancing of a Chemical Reaction\nConservation of mass\nAccording to the law of conservation of mass, no atoms can be created or destroyed in a\nchemical reaction, so the number of atoms for each element in the reactants side has to\nbalance the number of atoms that are present in the products side.\nIn other words, the total mass of the products formed in a chemical reaction is equal to the\ntotal mass of the reactants participated in a chemical reaction.", - "In other words, the total mass of the products formed in a chemical reaction is equal to the\ntotal mass of the reactants participated in a chemical reaction.\nBalanced chemical equation\nThe chemical equation in which the number of atoms of each element\u00a0in the reactants side\nis equal to that of the products side is called a balanced chemical equation.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nBalanced chemical equation\nThe chemical equation in which the number of atoms of each element\u00a0in the reactants side\nis equal to that of the products side is called a balanced chemical equation.\nSteps for balancing chemical equations\nHit and trial method: While balancing the equation, change the coef\ufb01cients (the numbers in\nfront of the compound or molecule) so that the number of atoms of each element is same\non each side of the chemical equation.\u00a0\nShort-cut technique for balancing a chemical equation\nExample:\naCaCO 3+bH3PO 4\u2192cCa 3(PO 4)2+dH2CO 3\nSet up a series of simultaneous equations, one for each element.", - "Short-cut technique for balancing a chemical equation\nExample:\naCaCO 3+bH3PO 4\u2192cCa 3(PO 4)2+dH2CO 3\nSet up a series of simultaneous equations, one for each element.\nCa: a=3c\nC:\u00a0 \u00a0a=d\nO:\u00a0 \u00a03a+4b=8c+3d\nH:\u00a0 \u00a03b=2d\nP:\u00a0 \u00a0 b=2c\nLet's set c=1\nThen a=3 and\nd=a=3\nb=2c=2So a=3; b=2; c=1; d=3\nThe balanced equation is\n3CaCO 3+ 2H 3PO 4\u2192Ca3(PO 4)2+ 3H 2CO 3\nChemical Reactions and Equations II\nTypes of chemical reactions\nTaking into consideration different factors, chemical reactions are grouped into multiple\ncategories.\nFew examples are:\n\u25cfCombination", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nChemical Reactions and Equations II\nTypes of chemical reactions\nTaking into consideration different factors, chemical reactions are grouped into multiple\ncategories.\nFew examples are:\n\u25cfCombination\n\u25cfDecomposition\n\u25cfSingle Displacement\n\u25cfDouble displacement\n\u25cfRedox\n\u25cfEndothermic\n\u25cfExothermic\n\u25cfPrecipitation\n\u25cfNeutralisation\nCombination reaction\nIn a combination reaction, two elements or one element and one compound or two\ncompounds combine\u00a0to give one single product.\nH2+Cl2\u2192 2HCl\nelement + element \u2192 compound\n2CO +O2\u2192 2CO 2\ncompound + element \u2192 compound\nNH 3+HCl \u2192NH 4Cl\ncompound + compound \u2192 compound\nDecomposition reaction\nA single reactant decomposes on the application of heat or light or electricity to give two or\nmore products.\nTypes of decomposition reactions:\na. Decomposition reactions which require heat - thermolytic decomposition or thermolysis.\nThermal decomposition of HgO\nb. Decomposition reactions which require light - photolytic decomposition\u00a0or\u00a0photolysis.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nThermal decomposition of HgO\nb. Decomposition reactions which require light - photolytic decomposition\u00a0or\u00a0photolysis.\nPhotolytic decomposition of H2O2\nc. Decomposition reactions which require electricity - electrolytic decomposition or\nelectrolysis.\nElectrolytic decomposition of H 2O\nDisplacement reaction\nMore reactive element displaces a less reactive element from its compound or solution.i)Zn(s) +CuSO 4(aq) \u2192ZnSO 4(aq) +Cu(s)\nii)Cu(s) + 2AgNO 3(aq) \u2192Cu(NO 3)2(aq) + 2Ag (s)\nDouble displacement reaction\nAn exchange of ions between the reactants takes place to give new products.\nFor example,\u00a0Al 2(SO4)3(aq) + 3Ca( OH)2(aq) \u2192 2Al (OH)3(aq) + 3CaSO 4(s)\nPrecipitation reaction\nAn insoluble compound called precipitate forms when two solutions containing soluble salts\nare combined.", - "For example, Pb( NO 3)2(aq) + 2KI (aq) \u2192 2KNO 3(aq) +PbI 2(\u2193)(s)(yellow )\nRedox reaction\nOxidation and reduction take place simultaneously.\nOxidation: Substance loses electrons or gains oxygen or loses hydrogen.", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nRedox reaction\nOxidation and reduction take place simultaneously.\nOxidation: Substance loses electrons or gains oxygen or loses hydrogen.\nReduction: Substance gains electrons or loses oxygen or gains hydrogen.\nOxidising agent - a substance that oxidises another substance and self-gets reduced.\nReducing agent - a substance that reduces another substance and self-gets oxidised.\nExamples:\n1.Fe(s) +CuSO 4(aq) \u2192FeSO 4(aq) +Cu(s)\u00a0 \u00a0 \u00a0 \u00a0(Blue)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (Green)\nFe\u2192Fe+2+ 2e \u2212 \u00a0(oxidation ) ; Fe - reducing agent.\nCu+2+ 2e \u2212 \u2192Cu(s)\u00a0(reduction ) ; Cu - oxidising agent.\n2.ZnO +C\u2192Zn+CO\nZnO reduces to Zn\u00a0\u2192 reduction\nC oxidises to CO \u2192 oxidation\nZnO - Oxidising agent\nC - Reducing agent\nEndothermic and exothermic reaction\nExothermic reaction - heat is evolved during a reaction.", - "Most of the combination reactions\nare exothermic.\nAl+Fe2O3\u2192Al2O3+Fe+heat\nCH 4+ 2O 2\u2192CO 2+ 2H 2O+heat\nEndothermic - Heat is required to carry out the reaction.\n6CO 2+ 6H 2O+Sunlight \u2192C6H12O6+ 6O 2\n\u00a0 \u00a0 \u00a0 \u00a0Glucose", - "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-REVISION NOTES.txt\nare exothermic.\nAl+Fe2O3\u2192Al2O3+Fe+heat\nCH 4+ 2O 2\u2192CO 2+ 2H 2O+heat\nEndothermic - Heat is required to carry out the reaction.\n6CO 2+ 6H 2O+Sunlight \u2192C6H12O6+ 6O 2\n\u00a0 \u00a0 \u00a0 \u00a0Glucose\nMost of the\u00a0decomposition reactions are endothermic.\nCorrosion\nGradual deterioration of a material, usually a metal, by the action of moisture, air or\nchemicals in the surrounding environment.", - "Corrosion\nGradual deterioration of a material, usually a metal, by the action of moisture, air or\nchemicals in the surrounding environment.\nRusting:\n4Fe(s) + 3O 2(from \u00a0air) +xH 2O(moisture ) \u2192 2Fe 2O3.xH 2O(rust)\nCorrosion of copper:\nCu(s) +H2O(moisture ) +CO 2(from \u00a0air) \u2192CuCO 3.Cu(OH)2(green )\nCorrosion of silver:\nAg(s) +H2S(from \u00a0air) \u2192Ag2S(black) +H2(g)\nRancidity\nIt refers to oxidation of fats and oils in food that is kept for a long time. It gives foul smell\nand bad taste to food. Rancid food causes stomach infection on consumption.\nPrevention:\n(i) Use of air-tight containers(ii) Packaging with nitrogen\n(iii) Refrigeration\n(iv) Addition of antioxidants or preservatives", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS 2.txt\nChemical Reactions & \nequations \nPower Point Presentation END", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nChemical Reactions and Equations \nChemical Reactions and Equations \nAny process that involves the rearrangement of structure of the substance or conversion of \nreactants into products is defined as Chemical Reaction . \nFor a Chemical Reaction to occur, the change can be observed in the form of - \n\uf0b7 Change in State : Melting of ice into water. \n\uf0b7 Change in Colour : Iron rusting which has colour change from silver to reddish brown.", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\n\uf0b7 Change in Temperature : There are two types of reaction i.e Exothermic and Endothermic \nReaction. \nExothermic Reactions: Those reactions in which energy is released in the form of heat are \ncalled Exothermic Reactions . \nExamples - \n(1) All combustion reactions e.g. \nCH 4+ 2O 2 \u2014> CO 2 + 2H 2O + Heat \n(2) Thermite reactions e.g. \n2A 1 + Fe 2O3 \u2014> 2Fe + Al 2O3 + Heat \nCombinations are generally exothermic in nature. The decomposition of organic matters into \ncompost is an example of exothermic reaction. \nEndothermic Reactions: Those reactions in which energy is absorbed are called Endothermic \nReactions . \nExamples - \n \nalso, the reaction of photosynthesis -", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nalso, the reaction of photosynthesis - \n \n\uf0b7 Evolution of any gas: When Zinc reacts with sulphuric acid it gives hydrogen gas. \nZn + H2 SO 4 \u2192 ZnSO 4 + H2 \nFormation of Precipitate : When a soluble carbonate reacts with Barium, Barium Carbonate \nprecipitate can be observed. \nChange in State \nSome chemical reactions are characterized by a change in state. \n\uf0b7 When wax is burned (in the form of wax candle,) then water and carbon dioxide are formed. \n\uf0b7 Now, wax is a liquid whereas carbon dioxide is a gas. This means that during the combustion \nreaction of wax, the physical state changes from solid to liquid and gas. \nPhysical Change \n\uf0b7 In this change identity of the substance remains same. \n\uf0b7 For Example, Melting, Boiling etc. \nChemical Change \n\uf0b7 The identity of the substances change \n\uf0b7 Reactants are converted into substance due to formation or broken down of older bonds", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nChemical Equation \nThe symbolic representation of chemical reaction using symbols and formulae is known \nas Chemical Equation . For this, reactants are written in left hand side whereas products are \nwritten on the right. \nBalanced Chemical Equation \nA balanced chemical equation is the one where the number of atoms involved in reactants side is \nequal to number of atoms on product side. \n \nEq.1. Example of Balanced Chemical Equation \nSteps to form Balanced Equation \nTo show how to balance the equation, the following equation is used - \nFe + H2O \u2192 Fe 3O4 + H2 \nStep 1: First of all, draw the boxes around each formula as shown below -", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nStep 2: Find out the number of atoms of each element. For Example , on reactant side, 1 for Fe, \n2 H, and 1 O and on product side we have, 3 for Fe, 4 for O and 2 for H. \nStep 3: Start to balance the equation with the compound having maximum number of atoms. While \nbalancing does not alter the formula of the compound. \nStep 4: One by one balance each element on reactant and product side. \n \nStep 5: After balancing number of atoms on both the side of the equation, finally check the \ncorrectness of the balanced equation. \n \nStep 6: then write the symbols of the physical state of reactants and products as shown below - \n3Fe(s) + 4H 2O (g) \u2192 Fe 3O4 (s) + 4H 2 (g) \nThis above equation represents the balanced equation.", - "Balancing a Redox Reaction \nThe basic ionic form of the equation is - \nFe2+ + Cr 2O72- \u2192 Fe3+ + Cr3+ \nOxidation half reaction is - \n \nReduction half reaction is -", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nReduction half reaction is - \n \nUse the reduction half method to balance the equation. Balance the atoms in each half of the \nreaction except H and O atoms. \nCr 2O72- (aq) \u2192 2 Cr3+(aq) \nAdd water molecules as the reaction is taking place in acidic solution. This is to balance the O \natoms and hydrogen ions. \nCr 2O72- (aq) + 14 H+(aq) \u2192 2 Cr3+(aq) + 7H 2O (I) \nThen balance the charges in both half reactions.", - "This is to balance the O \natoms and hydrogen ions. \nCr 2O72- (aq) + 14 H+(aq) \u2192 2 Cr3+(aq) + 7H 2O (I) \nThen balance the charges in both half reactions. \nFe2+(aq) \u2192 Fe3+(aq) + e- \nCr 2O72- (aq) + 14 H+ + 6e- \u2192 2 Cr3+ + 7H 2O \n6 Fe2+(aq) \u2192 6 Fe3+(aq) + 6e- \nTwo half of the equations are added to get the overall reaction \n6Fe2+(aq) + Cr 2O72-(aq) + 14H+(aq) \u2192 6Fe3+(aq) + 2Cr3+(aq) + 7H 2O (I) \nTypes of Chemical Reaction \n\uf0b7 Combination Reaction is reaction when single product is formed from the combination of two \nor more reactants. For Example -", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nEq.2. Example of Combination Reaction \nReactions can be exothermic as well as endothermic. Exothermic reaction release heats and \nraises the temperature of the surroundings. For Example , Respiration is an example of exothermic \nreaction. \n \nEq.3. Example of Exothermic Reaction \nEndothermic reaction involved the absorption of the heat and thus it cools the surrounding. The \ndecomposition of dead organic material is an endothermic reaction. \n\uf0b7 Decomposition Reaction is type of reaction which involves breakdown of single reactant into \nsimpler products. Decomposition of silver chloride into silver and chlorine in presence of \nsunlight is an example of decomposition reaction. \n \nEq.4. Example of Deco mposition Reaction \n\uf0b7 Displacement Reaction is a reaction in which more reactive element will \ndisplaces the less reactive element.", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nEq.4. Example of Deco mposition Reaction \n\uf0b7 Displacement Reaction is a reaction in which more reactive element will \ndisplaces the less reactive element. \n \nEq. 5. Example of Displacement Reaction \n\uf0b7 Double Displacement Reaction is a type of reaction in which cations and anions in the \nreactants switch the places to form new products. \n \nEq. 6. Example of Double Displacement Reaction \n\uf0b7 Redox Reaction is also known as Oxidation -reduction Reaction . In this type of reaction \ntransfer of electrons occurs between the two species. Oxidation is def ined as addition of oxygen \nor removal of hydrogen. Reduction is defined as removal of oxygen or addition of hydrogen. \nOxidizing agent is the one which gains the electrons and is reduced in a chemical reaction. \nReducing agent is oxidized in a chemical react ion and it loses the electrons. Fluorine is the \nstrongest oxidizing agent. Formic acid is a reducing agent", - "CHAPTER 1 CHEMICAL REACTIONS AND EQUATIONS5.txt\nEq.7. Example of Redox Reaction \nCorrosion \nMetals are prone to corrosion. It is a slow conversion of metals into some undesirable compounds. \nThis occur may be due to reaction with oxygen, gases, acids etc. When irons reacts with \natmospheric oxygen and moisture, a red layer is formed on the surface of the iron, this process is \nknown as Rusting . \n \nEq. 8. Equation for Iron Rusting \nRancidity \nWhen food containing fats and oils are exposed to the atmosphere, the oxidation of fat and oil \noccurs, this is known as Rancidity . \nMethods to Prevent Rancidity \n\uf0b7 Store cooking oils from direct sunlight. \n\uf0b7 Food should be placed at low temperature. \n\uf0b7 By adding antioxidants food can be protected from rancidity. \n\uf0b7 Packing material should replace the air with nitrogen. \n\uf0b7 Minimize the use of salts in fried foods." + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\n1. Why should a magnesium ribbon be cleaned before burning in the air?\nSolution:\nMagnesium ribbon should be cleaned before burning in air because Magnesium metal reacts with the atmospheric oxygen and forms Magnesium Oxide (MgO) layer which is a very stable compound. In order to prevent further reactions with Oxygen, it is therefore necessary to clean the ribbon by to remove the layer of MgO.\n2. Write a balanced equation for the following chemical reactions.\ni) Hydrogen + Chloride \u2014-> Hydrogen chloride\nii) Barium chloride + Aluminium sulphate \u2014-> Barium sulphate + Aluminium chloride\niii) Sodium + Water \u2014-> Sodium hydroxide + Hydrogen\nSolution:\ni)\u00a0H2\u00a0+ Cl2\u00a0\u2192 2HCl\nii)\u00a03BaCl2\u00a0+ Al2(SO4)3\u00a0\u21923BaSO4\u00a0+\u00a02AlCl3\u00a0\niii)\u00a02Na + 2H2O \u2192 2NaOH + H2\n3.", + "Write a balanced chemical equation with state symbols for the following reactions\ni) Solutions of Barium chloride and Sodium sulphate in water react to give insoluble Barium sulphate and solution of Sodium chloride.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\ni) Solutions of Barium chloride and Sodium sulphate in water react to give insoluble Barium sulphate and solution of Sodium chloride.\nii) Sodium hydroxide solution in water reacts with the hydrochloric acid solution to produce Sodium chloride solution and water.\nSolution:\ni)\u00a0BaCl2\u00a0+ Na2SO4\u00a0\u2192 BaSO4\u00a0+ 2NaCl\nii)\u00a0NaOH + HCl \u2192 NaCl + H2O\nIn-text questions set 2 Page number \u2013\u00a010\n1. A solution of a substance, \u2018X,\u2019 is used for whitewashing.\n(i) Name the substance \u2018X\u2019 and write its formula.\n(ii) Write the reaction of the substance \u2018X\u2019 named in (i) above with water.\nSolution:\ni) The substance \u2018X\u2019 which is used in whitewashing is quick lime or\u00a0Calcium Oxide\u00a0and its formula is\u00a0CaO.\nii) CaO + H2O \u2192 Ca(OH)2\n2.", + "Solution:\ni) The substance \u2018X\u2019 which is used in whitewashing is quick lime or\u00a0Calcium Oxide\u00a0and its formula is\u00a0CaO.\nii) CaO + H2O \u2192 Ca(OH)2\n2. Why is the amount of gas collected in one of the test tubes in Activity 1.7 double the amount collected in the other? Name this gas.\nSolution:", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\nii) CaO + H2O \u2192 Ca(OH)2\n2. Why is the amount of gas collected in one of the test tubes in Activity 1.7 double the amount collected in the other? Name this gas.\nSolution:\nIn activity 1.7, gas collected in one of the test tubes is double of the amount collected in the other because water gets hydrolysed to release H2\u00a0and O2\u00a0gas. Here, after electrolysis two molecules of Hydrogen and one molecule of oxygen gas are released; hence, the amount of Hydrogen collected would be double than that of oxygen.\nIn-text questions set 3 Page number \u2013 13\n1. Why does the colour of copper sulphate solution change when an iron nail is dipped in it?\nSolution:\nWhen an iron nail is dipped in the copper sulphate solution, iron displaces copper from the copper sulphate because iron is more reactive than copper. Therefore, the colour of the copper sulphate solution changes. The reaction is\nFe + CuSO4\u00a0\u2192 FeSO4\u00a0+ Cu\n2.", + "Therefore, the colour of the copper sulphate solution changes. The reaction is\nFe + CuSO4\u00a0\u2192 FeSO4\u00a0+ Cu\n2.\u00a0Give an example of a double displacement reaction other than the one given in Activity 1.10.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\nFe + CuSO4\u00a0\u2192 FeSO4\u00a0+ Cu\n2.\u00a0Give an example of a double displacement reaction other than the one given in Activity 1.10.\n\u00a0Solution:\nReaction Between silver nitrate (AgNO3) and Sodium chloride (NaCl) is an example of a double displacement reaction. During the reaction, negative and positive ions trade positions as a result of the formation of white silver chloride precipitate. The chemical reaction is given below.\nAg+\u00a0+ NO3\u2013\u00a0+ Na+\u00a0+ Cl\u2013\u00a0\u2192 AgCl + Na+\u00a0+ NO3\u2013\n3. Identify the substances that are oxidised and that are reduced in the following equation.", + "The chemical reaction is given below.\nAg+\u00a0+ NO3\u2013\u00a0+ Na+\u00a0+ Cl\u2013\u00a0\u2192 AgCl + Na+\u00a0+ NO3\u2013\n3. Identify the substances that are oxidised and that are reduced in the following equation.\ni)\u00a04Na(s) + O2(g) \u2192 2Na2O(s)\nii)\u00a0CuO(s) + H2(g) \u2192 Cu(s) + H2O(l)\nSolution:\nThe Sodium (Na) in the first equation is getting oxidized with the addition of Oxygen (O2), and the Copper (Cu) in the second equation is reduced due to the addition of Hydrogen (H2).\nExercise Questions Page number \u2013 14-16\n1. Which of the statements about the reaction below are incorrect?\n2PbO(s) + C(s) \u2192 2Pb(s) + CO2(g)\n(a) Lead is getting reduced", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\nExercise Questions Page number \u2013 14-16\n1. Which of the statements about the reaction below are incorrect?\n2PbO(s) + C(s) \u2192 2Pb(s) + CO2(g)\n(a) Lead is getting reduced\n(b) Carbon Dioxide is getting oxidised\n(c) Carbon is getting oxidised\n(d) Lead oxide is getting reduced\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (i) (a) and (b)\n\u00a0 \u00a0 \u00a0 \u00a0(ii) (a) and (c)\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (iii) (a), (b) and (c)\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (iv) all the above\nSolution:\n(i) (a) and (b)\nExplanation: (a) Because Oxygen is being removed and (b) Because the removed oxygen from Lead is added to the elemental Carbon.\n\n2. \u00a0Fe2O3\u00a0+ 2Al \u2192 Al2O3\u00a0+ 2Fe\nThe above reaction is an example of a\nCombination reaction\nDouble displacement reaction\nDecomposition reaction\nDisplacement reaction\nSolution:\nThe answer is 4. Displacement reaction.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\nThe above reaction is an example of a\nCombination reaction\nDouble displacement reaction\nDecomposition reaction\nDisplacement reaction\nSolution:\nThe answer is 4. Displacement reaction.\nExplanation: The Oxygen from the Ferrous oxide is getting displaced to the Aluminium metal to form Aluminium Oxide. In this reaction, Aluminum is a more reactive metal than Fe. Therefore, Al will displace Fe from its oxide. This type of chemical reaction, called displacement reaction, is in which one of the elements displaces another. Here, less reactive metal is displaced by more reactive metal. Since one-time displacement is occurring, it is called a single displacement reaction.\n\n452\n3. What happens when dilute hydrochloric acid is added to iron fillings? Tick the correct answer.\nHydrogen gas and Iron chloride are produced.\nChlorine gas and Iron hydroxide are produced.\nNo reaction takes place.\nIron salt and water are produced.\nSolution:\nHydrogen gas and Iron chloride are produced.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\nChlorine gas and Iron hydroxide are produced.\nNo reaction takes place.\nIron salt and water are produced.\nSolution:\nHydrogen gas and Iron chloride are produced.\nExplanation: The Chlorine from Hydrogen chloride is displaced by the Iron fillings to undergo the following reaction.\n2HCl + Fe \u2192 FeCl2\u00a0+ H2\n\n\n4. What is a balanced chemical equation? Why should a chemical equation be balanced?\nSolution:\nA balanced equation is the one in which number of different atoms on both the reactant and product sides are equal. Balancing chemical equations is necessary for the reaction should obey The Law of Conservation of mass. Balancing the chemical equation has no defined method and is purely a trial-and-error attempt.\n\n\n5. \u00a0Translate the following statements into chemical equations and balance them.\n(a) Hydrogen gas combines with nitrogen to form ammonia.\n(b) Hydrogen sulphide gas burns in the air to give water and sulphur dioxide.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\n(a) Hydrogen gas combines with nitrogen to form ammonia.\n(b) Hydrogen sulphide gas burns in the air to give water and sulphur dioxide.\n(c) Barium chloride reacts with aluminium sulphate to give Aluminium chloride and a precipitate of barium sulphate.\n(d) Potassium metal reacts with water to give potassium hydroxide and Hydrogen gas.\nSolution:\n(a) Unbalanced: H2\u00a0+ N2\u00a0\u2192 NH3\nBalanced: 3H2\u00a0+ N2\u00a0\u2192 2NH3\n\n\n(b) Unbalanced: H2S + O2\u00a0\u2192 H2O + SO2\nBalanced: 2H2S + 3O2\u00a0\u2192 2H2O + 2SO2\n\n\n(c) \u00a0Unbalanced:\nBaCl2\u00a0+ Al2(SO4)3\u00a0\u2192 AlCl3\u00a0+ BaSO4\nBalanced: 3BaCl2\u00a0+ Al2(SO4)3\u00a0\u2192 2AlCl3\u00a0+ 3BaSO4", + "(c) \u00a0Unbalanced:\nBaCl2\u00a0+ Al2(SO4)3\u00a0\u2192 AlCl3\u00a0+ BaSO4\nBalanced: 3BaCl2\u00a0+ Al2(SO4)3\u00a0\u2192 2AlCl3\u00a0+ 3BaSO4\n\n\n(d) Unbalanced: K + H2O \u2192 KOH + H2\nBalanced: \u00a02K + 2H2O \u2192 2KOH + H2\n\n\n6. Balance the following chemical equations.\n(a) HNO3\u00a0+ Ca(OH)2\u00a0\u2192 Ca(NO3)2\u00a0+ H2O\n(b) NaOH + H2SO4\u00a0\u2192 Na2SO4\u00a0+ H2O\n(c) NaCl + AgNO3\u00a0\u2192 AgCl + NaNO3\n(d) BaCl2\u00a0+ H2SO4\u00a0\u2192 BaSO4\u00a0+ HCl\nSolution:\n(a) 2HNO3\u00a0+ Ca(OH)2\u00a0\u2192 Ca(NO3)2\u00a0+ 2H2O\n\n\n(b) 2NaOH + H2SO4\u00a0\u2192 Na2SO4\u00a0+ 2H2O", + "(b) 2NaOH + H2SO4\u00a0\u2192 Na2SO4\u00a0+ 2H2O\n\n\n(c) NaCl + AgNO3\u00a0\u2192 AgCl + NaNO3\n\n\n(d) BaCl2\u00a0+ H2SO4\u00a0\u2192 BaSO4\u00a0+ 2HCl", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\n(d) BaCl2\u00a0+ H2SO4\u00a0\u2192 BaSO4\u00a0+ HCl\nSolution:\n(a) 2HNO3\u00a0+ Ca(OH)2\u00a0\u2192 Ca(NO3)2\u00a0+ 2H2O\n\n\n(b) 2NaOH + H2SO4\u00a0\u2192 Na2SO4\u00a0+ 2H2O\n\n\n(c) NaCl + AgNO3\u00a0\u2192 AgCl + NaNO3\n\n\n(d) BaCl2\u00a0+ H2SO4\u00a0\u2192 BaSO4\u00a0+ 2HCl\n\n\n7. Write the balanced chemical equation for the following reactions.\nCalcium hydroxide + Carbon dioxide \u2014-> Calcium carbonate + Water\nZinc + Silver nitrate \u2014-> Zinc nitrate + Silver\nAluminium + Copper chloride \u2014-> Aluminium chloride + Copper\nBarium chloride + Potassium sulphate \u2014-> Barium sulphate + Potassium chloride\nSolution:\n2Ca(OH)2\u00a0+ 2CO2\u00a0\u2192 2CaCO3\u00a0+ 2H2O", + "Zn + 2AgNO3\u00a0\u2192 Zn(NO3)2\u00a0+ 2Ag\n\n\n2Al + 3CuCl2\u00a0\u2192 2AlCl3\u00a0+ 3Cu\n\n\nBaCl2\u00a0+ K2SO4\u00a0\u2192 BaSO4\u00a0+ 2KCl\n\n\n8. Write a balanced chemical equation for the following and identify the type of reaction of each case.\nKBr + BaI2\u00a0\u2192 KI + BaBr2\nZnCO3\u00a0\u2192 ZnO + CO2\nH2\u00a0+ Cl \u2192 HCl\nMg + HCl \u2192 MgCl2\u00a0+ H2\nSolution:\n2KBr + BaI2\u00a0\u2192 2KI + BaBr2\u00a0(Double Displacement Reaction)\n\n\nZnCO3\u00a0\u2192 ZnO + CO2\u00a0(Decomposition Reaction)\n\n\nH2\u00a0+ Cl \u2192 2HCl (Combination Reaction)", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\nH2\u00a0+ Cl \u2192 HCl\nMg + HCl \u2192 MgCl2\u00a0+ H2\nSolution:\n2KBr + BaI2\u00a0\u2192 2KI + BaBr2\u00a0(Double Displacement Reaction)\n\n\nZnCO3\u00a0\u2192 ZnO + CO2\u00a0(Decomposition Reaction)\n\n\nH2\u00a0+ Cl \u2192 2HCl (Combination Reaction)\n\n\nMg + 2HCl \u2192 MgCl2\u00a0+ H2\u00a0(Displacement Reaction)\n\n\n9. What is meant by exothermic and endothermic reactions? Give examples.\nSolution:\nAn endothermic reaction occurs when energy is absorbed from the surroundings in the form of heat (For example, Photosynthesis, melting of ice, evaporation). Conversely, an exothermic reaction is one in which energy is released from the system into the surroundings (Examples: Explosions, concrete setting, nuclear fission and fusion).\n\n10. Why is respiration considered to be an exothermic reaction?\nSolution:", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\n10. Why is respiration considered to be an exothermic reaction?\nSolution:\nFor the survival of life, we require energy. We obtain this energy from the food we eat. The food molecules, through the process of digestion, is broken down into simpler molecule like glucose. These substances come in contact with the Oxygen present in our body cells to form Carbon dioxide and water along with a certain amount of energy (Respiration process). Since the energy is in the form of heat (that maintains our body temperature), respiration is considered to be an exothermic reaction. The reaction taking place is:\nC6H12O6\u00a0+ 6O2\u00a0\u2192 6CO2\u00a0+ 6H2O + Energy\n\n\n11. Why are decomposition reactions called the opposite of Combination reactions? Write equations for decomposition reactions.\nSolution:", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\nC6H12O6\u00a0+ 6O2\u00a0\u2192 6CO2\u00a0+ 6H2O + Energy\n\n\n11. Why are decomposition reactions called the opposite of Combination reactions? Write equations for decomposition reactions.\nSolution:\nCombination reaction is said to be the reaction between two or more molecules to form a larger molecule; whereas the decomposition reaction is defined as the splitting of larger molecules into two or more smaller molecules. This essentially explains that the decomposition reaction is the opposite of the combination reaction.\nIn most cases, the decomposition reaction is endothermic since the heat from the surrounding or induced heat is used to break the bonds of the larger molecule. A few examples of decomposition reactions are\nZnCO3\u00a0\u2192 ZnO + CO2\nCaCO3\u00a0+ Energy \u2192 CaO + CO2\n2HgO \u2192 2Hg + O2\n\n\n12. Write one equation each for decomposition reactions in which energy is supplied in the form of heat, light or electricity.\nSolution:\n(a) Thermal decomposition reaction (Thermolysis)", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\n12. Write one equation each for decomposition reactions in which energy is supplied in the form of heat, light or electricity.\nSolution:\n(a) Thermal decomposition reaction (Thermolysis)\nDecomposition of potassium chlorate: When heated strongly, potassium chlorate decomposes into potassium chloride and oxygen. This reaction is used for the preparation of oxygen.\n2KClO3\u00a0+ Heat \u2192 2KCl + 3O2\n(b) Electrolytic decomposition reaction (Electrolysis)\nDecomposition of sodium chloride: On passing electricity through molten sodium chloride, it decomposes into sodium and chlorine.\n\n(c) Photodecomposition reaction (Photolysis)\nDecomposition of hydrogen peroxide: In the presence of light, hydrogen peroxide decomposes into water and oxygen.\n\n13. \u00a0What is the difference between displacement and double displacement reactions? Write relevant equations for the above.\nSolution:", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\n13. \u00a0What is the difference between displacement and double displacement reactions? Write relevant equations for the above.\nSolution:\nA displacement reaction is one when a more reactive substance displaces a less reactive one from its salt solution, whereas a double displacement reaction is one where a mutual exchange of ions happens between two compounds.\nIn a displacement reaction, only a single displacement takes place, whereas in the double displacement reaction, as the name suggests, two displacement takes place between the molecules.\nExample:\nDisplacement reaction\nMg + 2HCl \u2192 MgCl2\u00a0+ H2\nDouble displacement reaction\n2KBr + BaI2\u00a0\u2192 2KI + BaBr2\n\n\n14. In the refining of Silver, the recovery of silver from Silver nitrate solution involves displacement reaction by Copper metal. Write down the reaction involved.\nSolution:\nCu(s) + 2AgNO3(aq) \u2192 Cu(NO3)2(aq) + 2Ag(s)\n\n\n15. What do you mean by a precipitation reaction? Explain by giving examples.\nSolution:", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\nSolution:\nCu(s) + 2AgNO3(aq) \u2192 Cu(NO3)2(aq) + 2Ag(s)\n\n\n15. What do you mean by a precipitation reaction? Explain by giving examples.\nSolution:\nWhen two solutions containing soluble salts are combined, a double displacement reaction takes place in which the ions are exchanged between the compounds. When one of such compounds formed is in solid form (that is insoluble in aqua), then it settles down at the bottom of the container. This solid is known as the precipitate and the respective reaction is termed as the precipitation reaction. A few examples of precipitation reactions are\nCdSO4(aq) + K2S(aq) \u2192 CdS(s) + K2SO4(aq)\n2NaOH(aq) + MgCl2(aq) \u2192 2NaCl(aq) + Mg(OH)2(s)", + "16. Explain the following in terms of the gain of oxygen with two examples each.\n(a) Oxidation\n(b) Reduction\nSolution:\n(a) In a chemical reaction, when the oxygen is added to the element to form its respective oxide it is the element being oxidised. Example:\n4Na(s) + O2(g) \u2192 2Na2O(s)\nH2S + O2\u00a0\u2192 H2O + SO2", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\nSolution:\n(a) In a chemical reaction, when the oxygen is added to the element to form its respective oxide it is the element being oxidised. Example:\n4Na(s) + O2(g) \u2192 2Na2O(s)\nH2S + O2\u00a0\u2192 H2O + SO2\n\n\n(b) In a chemical reaction, when the oxygen is removed from the compound, then it is said to be reduced. Example:\nCuO(s) + H2(g) \u2192 Cu(s) + H2O(l)\n2HgO \u2192 2Hg + O2\n\n\n17. A shiny brown coloured element \u2018X\u2019 on heating in the air becomes black in colour. Name the element \u2018X\u2019 and the black-coloured compound formed.\n\nSolution:\nThe shiny brown-coloured element is Copper metal (Cu). When the metal is heated in air, it reacts with atmospheric oxygen to form copper oxide. Hence, the black-coloured compound is copper oxide.\n2Cu(s) + O2(g) \u2192 2CuO(s)\n\n\n18) Why do we apply paint on iron articles?\nSolution:", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\n2Cu(s) + O2(g) \u2192 2CuO(s)\n\n\n18) Why do we apply paint on iron articles?\nSolution:\nIron articles are painted to prevent them from rusting. When left unpainted, the metal surface comes in contact with the atmospheric oxygen and in the presence of moisture it from Iron(III) oxide. But if painted, the surface does not come in contact with moisture and air, thus prevents Rusting.\n\n\n19) Oil and Fat containing food items are flushed with Nitrogen. Why?\nSolution:\nThe main purpose of flushing Nitrogen into food packets that contain oil and fat items is to prevent Rancidity which occurs when the oil or fat reacts with the oxygen letting out an unpleasant smell and taste. Therefore, by flushing Nitrogen, an unreactive surrounding is created, thus preventing rancidity.\n\n\n20) Explain the following terms with one example each.\n(a) Corrosion\n(b) Rancidity\nSolution:", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -BOOK BACK QUESTION WITH ANSWERS.txt\n20) Explain the following terms with one example each.\n(a) Corrosion\n(b) Rancidity\nSolution:\n(a) Corrosion is a process where a refined metal is oxidised by atmospheric oxygen to form a more stable compound, such as oxides. The metal gradually degrades during the corrosion process. Rusting of iron is a good example of corrosion where the iron is converted to Iron oxide. Millions of dollars are spent annually in preventing rusting from bridges and other monuments.\n\n(b) The condition produced by the aerial oxidation of the oil and fat present in the food material that produces an unpleasant taste and smell. The rancidity is retarded when the food is kept inside the refrigerator since the low temperature does not promote the oxidation reaction.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\nClass 10 Chapter 1 Chemical Reactions and Equations Important Questions with Answers\nMultiple Choice Type Questions\nQ1.\u00a0Which of the following gases is used to store fat and oil-containing foods for a long time?\nCarbon dioxide\nOxygen\nNitrogen\nNeon\nAnswer:\n(3) Nitrogen gas is used to store fat and oil-containing foods for a long time.\nQ2.\u00a0The chemical reaction between Hydrogen sulphide and iodine to give Hydrogen iodide and sulphur is given below:\nH2S + I2\u00a0\u2192 2HI + S.\nThe reducing and oxidising agents involved in this redox reaction are:\nIodine and sulphur, respectively\nIodine and hydrogen sulphide, respectively\nSulphur and iodine, respectively\nHydrogen sulphide and sulphur, respectively\nAnswer:\n(2) Iodine is an oxidising agent, and hydrogen sulphide is the reducing agent in the reaction mentioned above.\nH2S + I2\u00a0\u2192 2HI + S.\nShort Answer Type Questions\nQ1.", + "H2S + I2\u00a0\u2192 2HI + S.\nShort Answer Type Questions\nQ1.\u00a0Write the balanced chemical equations for the following reactions and identify the type of reaction in each case.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\nH2S + I2\u00a0\u2192 2HI + S.\nShort Answer Type Questions\nQ1.\u00a0Write the balanced chemical equations for the following reactions and identify the type of reaction in each case.\n(a )Nitrogen gas is treated with hydrogen gas in the presence of a catalyst at 773K to form ammonia gas.\n(b )Sodium hydroxide solution is treated with acetic acid to form sodium acetate and water.\n(c ) Ethanol is warmed with ethanoic acid to form ethyl acetate in the presence of concentrated H2SO4.\n(d) Ethene is burnt in the presence of oxygen to form carbon dioxide and water and releases heat and light.\nAnswer:\n(a )\n\ufffd2(\ufffd)+3\ufffd2(\ufffd)\u2192\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd773\ufffd2\ufffd\ufffd3(\ufffd)\nIt is an addition reaction.\n(b ) NaOH (aq) + CH3COOH (aq) \u2192 CH3COONa (aq) + H2O (l)\nIt is a double displacement or a neutralisation reaction.", + "(b ) NaOH (aq) + CH3COOH (aq) \u2192 CH3COONa (aq) + H2O (l)\nIt is a double displacement or a neutralisation reaction.\n(c )\n\ufffd2\ufffd5\ufffd\ufffd(\ufffd)+\ufffd\ufffd3\ufffd\ufffd(\ufffd)\u2192\ufffd+\ufffd\ufffd3\ufffd\ufffd\ufffd\ufffd2\ufffd5(\ufffd)+\ufffd2\ufffd(\ufffd)\nIt is a double displacement or an esterification reaction.\n(d ) C2H4\u00a0(g) + 3 O2\u00a0(g) \u2192 2 CO2\u00a0(g) + 2 H2O (g) + Heat + light\nIt is a redox or a combustion reaction.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\nIt is a double displacement or an esterification reaction.\n(d ) C2H4\u00a0(g) + 3 O2\u00a0(g) \u2192 2 CO2\u00a0(g) + 2 H2O (g) + Heat + light\nIt is a redox or a combustion reaction.\nQ2.\u00a0Write the balanced chemical equations for the following reactions and identify the type of\nreaction in each case.\n(a ) In the thermite reaction, iron (III) oxide reacts with aluminium and gives molten iron and aluminium oxide.\n(b ) Magnesium ribbon is burnt in an atmosphere of nitrogen gas to form solid magnesium nitride.\n(c ) Chlorine gas is passed in an aqueous potassium iodide solution to form potassium\nchloride solution and solid iodine.\n(d ) Ethanol is burnt in the air to form carbon dioxide and water and releases heat.\nAnswer:\n(a ) Fe2O3\u00a0(s) + 2 Al (s) \u2192 Al2O3\u00a0(s) + 2 Fe (l) + Heat\nIt is a displacement or redox reaction.", + "(d ) Ethanol is burnt in the air to form carbon dioxide and water and releases heat.\nAnswer:\n(a ) Fe2O3\u00a0(s) + 2 Al (s) \u2192 Al2O3\u00a0(s) + 2 Fe (l) + Heat\nIt is a displacement or redox reaction.\n(b ) 3 Mg (s) + N2\u00a0(g) \u2192 Mg3N2\u00a0(s)\nIt is a combination reaction.\n(c ) 2 KI (aq) + Cl2\u00a0(g) \u2192 2 KCl (aq) + I2\u00a0(s)", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\nIt is a displacement or redox reaction.\n(b ) 3 Mg (s) + N2\u00a0(g) \u2192 Mg3N2\u00a0(s)\nIt is a combination reaction.\n(c ) 2 KI (aq) + Cl2\u00a0(g) \u2192 2 KCl (aq) + I2\u00a0(s)\nQ3.", + "(b ) 3 Mg (s) + N2\u00a0(g) \u2192 Mg3N2\u00a0(s)\nIt is a combination reaction.\n(c ) 2 KI (aq) + Cl2\u00a0(g) \u2192 2 KCl (aq) + I2\u00a0(s)\nQ3.\u00a0Complete the missing components / variables given as x and y in the following reactions\n(a) Pb(NO3)2\u00a0(aq) + 2 KI (aq) \u2192 Pbl2\u00a0(x) + 2 KNO3\u00a0(y)\n(b) Cu (s) + 2 AgNO3\u00a0(aq) \u2192 Cu(NO3)2\u00a0(aq) + x(s)\n(c) Zn (s) + H2SO4\u00a0(aq) \u2192 ZnSO4\u00a0(x) + H2\u00a0(y)\n(d)\n\ufffd\ufffd\ufffd\ufffd3(\ufffd)\u2192\ufffd\ufffd\ufffd\ufffd(\ufffd)+\ufffd\ufffd2(\ufffd)\nAnswer:\n(a) Pb(NO3)2\u00a0(aq) + 2 KI (aq) \u2192 Pbl2\u00a0(s) + 2 KNO3\u00a0(aq)\n(b) Cu (s) + 2 AgNO3\u00a0(aq) \u2192 Cu(NO3)2\u00a0(aq) + 2 Ag(s)\n(c) Zn (s) + H2SO4\u00a0(aq) \u2192 ZnSO4\u00a0(aq) + H2\u00a0(g)\n(d)\n\ufffd\ufffd\ufffd\ufffd3(\ufffd)\u2192\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd\ufffd(\ufffd)+\ufffd\ufffd2(\ufffd)\nQ4.", + "Which among the following changes are exothermic or endothermic in nature?\n(a) Decomposition of ferrous sulphate\n(b) Dilution of sulphuric acid\n(c) Dissolution of sodium hydroxide in water\n(d) Dissolution of ammonium chloride in water\nAnswer:\n(a ) The decomposition of ferrous sulphate is an example of an endothermic reaction because heat is absorbed during this reaction.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\n(d) Dissolution of ammonium chloride in water\nAnswer:\n(a ) The decomposition of ferrous sulphate is an example of an endothermic reaction because heat is absorbed during this reaction.\n(b ) The dilution of sulphuric acid is an example of an exothermic reaction because heat is released during this reaction.\n(c ) The dissolution of sodium hydroxide in water is an example of an exothermic reaction because heat is released during this reaction.\n(d ) The dissolution of ammonium chloride in water is an example of an endothermic reaction because heat is absorbed during this reaction.\nQ5.\u00a0Identify the reducing agent in the following reactions\n(a ) 4 NH3\u00a0+ 5 O2\u00a0\u2192 4 NO + 6 H2O\n(b ) H2O + F2\u00a0\u2192 HF + HOF\n(c ) Fe2O3\u00a0+ 3 CO \u2192 2 Fe + 3 CO2\n(d ) 2 H2\u00a0+ O2\u00a0\u2192 2 H2O\nAnswer:\n(a ) Here, ammonia (NH3\u00a0) is the reducing agent.", + "(b ) Here, water (H2O) is the reducing agent.\n(c ) Here, carbon monoxide (CO) is the reducing agent.\n(d ) Here, hydrogen (H2\u00a0) is the reducing agent.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\n(a ) Here, ammonia (NH3\u00a0) is the reducing agent.\n(b ) Here, water (H2O) is the reducing agent.\n(c ) Here, carbon monoxide (CO) is the reducing agent.\n(d ) Here, hydrogen (H2\u00a0) is the reducing agent.\nQ6.\u00a0Identify the oxidising agent (oxidant) in the following reactions\n(a ) Pb3O4\u00a0+ 8 HCI \u2192 3 PbCl2\u00a0+ Cl2\u00a0+ 4 H2O\n(b ) 2 Mg + O2\u00a0\u2192 2 MgO\n(c ) CuSO4\u00a0+ Zn \u2192 Cu + ZnSO4\n(d ) V2O5\u00a0+ 5 Ca \u2192 2 V + 5 CaO\n(e ) 3 Fe + 4 H2O \u2192 Fe3O4\u00a0+ 4 H2\n(f ) CuO + H2\u00a0\u2192 Cu + H2O\nAnswer:\n(a ) Pb3O4\u00a0is the oxidising agent here.", + "The oxidation state of Pb in Pb3O4\u00a0reduces from + 6 to + 2 in PbCl2.\u00a0Thus it acts as an oxidising agent.\n(b ) O2\u00a0is the oxidising agent here. The oxidation state of oxygen in elemental form O2\u00a0reduces from 0 to \u2013 2 in MgO.\u00a0Thus it acts as an oxidising agent.\n(c ) CuSO4\u00a0is the oxidising agent here. The oxidation state of Cu in CuSO4\u00a0reduces from + 2 to 0 in Cu.\u00a0Thus it acts as an oxidising agent.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\n(c ) CuSO4\u00a0is the oxidising agent here. The oxidation state of Cu in CuSO4\u00a0reduces from + 2 to 0 in Cu.\u00a0Thus it acts as an oxidising agent.\n(d ) V2O5\u00a0is the oxidising agent here. The oxidation state of V in V2O5\u00a0reduces from + 5 to 0 in V.\u00a0Thus, it acts as an oxidising agent.\n(e ) H2O is the oxidising agent here. The oxidation state of oxygen in H2O reduces from \u2013 2 to \u2013 3 in H2O.\u00a0Thus it acts as an oxidising agent.\n(f ) CuO is the oxidising agent here. The oxidation state of Cu in CuO reduces from + 2 to 0 in Cu.\u00a0Thus, it acts as an oxidising agent.\nQ7.\u00a0Write the balanced chemical equations for the following reactions\n(a ) Sodium carbonate on reaction with hydrochloric acid in equal molar concentrations gives sodium chloride and sodium hydrogen carbonate.", + "Thus, it acts as an oxidising agent.\nQ7.\u00a0Write the balanced chemical equations for the following reactions\n(a ) Sodium carbonate on reaction with hydrochloric acid in equal molar concentrations gives sodium chloride and sodium hydrogen carbonate.\n(b ) Sodium hydrogen carbonate on reaction with hydrochloric acid gives sodium chloride, water and liberates carbon dioxide.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\n(b ) Sodium hydrogen carbonate on reaction with hydrochloric acid gives sodium chloride, water and liberates carbon dioxide.\n(c ) On treatment with potassium iodide, copper sulphate precipitates cuprous iodide (Cu2I2), liberates iodine gas and forms potassium sulphate.\nAnswer:\n(a ) Na2CO3\u00a0+ HCl \u2192 NaCl + NaHCO3\n(b ) NaHCO3\u00a0+ HCl \u2192 NaCl + H2O + CO2\n(c ) 2 CuSO4\u00a0+ 4 KI \u2192 Cu2I2\u00a0+ 2 K2SO4\u00a0+ I2\nQ8.\u00a0A solution of potassium chloride, when mixed with silver nitrate solution, an insoluble white substance is formed. Write the chemical reaction involved and also mention the type of the chemical reaction?\nAnswer:\nChemical reaction: KCl + AgNO3\u00a0\u2192 KNO3\u00a0+ AgCl\nIt is a double displacement reaction.\nQ9.\u00a0Ferrous sulphate decomposes with the evolution of a gas having a characteristic dour of burning sulphur.", + "Answer:\nChemical reaction: KCl + AgNO3\u00a0\u2192 KNO3\u00a0+ AgCl\nIt is a double displacement reaction.\nQ9.\u00a0Ferrous sulphate decomposes with the evolution of a gas having a characteristic dour of burning sulphur. Write the chemical reaction involved and identify the type of reaction.\nAnswer:\nFeSO4\u00a0(s) + Heat \u2192 Fe2O3\u00a0(s) + SO2\u00a0(g) + SO3\u00a0(g)\nIt is a thermal decomposition reaction.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\nAnswer:\nFeSO4\u00a0(s) + Heat \u2192 Fe2O3\u00a0(s) + SO2\u00a0(g) + SO3\u00a0(g)\nIt is a thermal decomposition reaction.\nQ10.\u00a0Why do fireflies glow at night?\nAnswer:\nFireflies glow at night because of a chemical reaction involving light\u2019s emission. Fireflies store a protein (luciferin) that combines with oxygen in the air to form a new substance (oxyluciferin) and the evolution of energy in light.\nQ11.\u00a0Grapes hanging on the plant do not ferment, but after being plucked from the plant can be\nfermented. Under what conditions do these grapes ferment? Is it a chemical or a physical\nchange?\nAnswer:\nWhen attached to the plants, Grapes are living, and therefore, their immune system prevents fermentation. The microbes can grow in the plucked grapes, which can be fermented under anaerobic conditions. This is a chemical change.\nQ12.\u00a0Which among the following are physical or chemical changes?", + "Answer:\nWhen attached to the plants, Grapes are living, and therefore, their immune system prevents fermentation. The microbes can grow in the plucked grapes, which can be fermented under anaerobic conditions. This is a chemical change.\nQ12.\u00a0Which among the following are physical or chemical changes?\n(a ) Evaporation of petrol\n(b ) Burning of Liquefied Petroleum Gas (LPG)\n(c ) Heating of an iron rod to red hot.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\nQ12.\u00a0Which among the following are physical or chemical changes?\n(a ) Evaporation of petrol\n(b ) Burning of Liquefied Petroleum Gas (LPG)\n(c ) Heating of an iron rod to red hot.\n(d ) Curdling of milk\n(e ) Sublimation of solid ammonium chloride\nAnswer:\n(a ) Evaporation of petrol is a physical change as it only gets converted from one physical state to another.\n(b ) Burning of Liquefied Petroleum Gas (LPG) is a chemical change as heating produces carbon dioxide and water.\n(c ) The heating of an iron rod to red hot is a physical change as heating involves only temperature change.\n(d ) The curdling of milk is a chemical change as it affects the chemical composition of the milk.\nQ13.\u00a0We made the following observations during the reaction of some metals with dilute hydrochloric acid.\n(a) Silver metal does not show any change\n(b) The temperature of the reaction mixture rises when aluminium (Al) is added.\n(c) The sodium metal reaction is highly explosive.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\n(a) Silver metal does not show any change\n(b) The temperature of the reaction mixture rises when aluminium (Al) is added.\n(c) The sodium metal reaction is highly explosive.\n(d) Some gas bubbles are seen when lead (Pb) is reacted with the acid.\nExplain these observations giving suitable reasons.\nAnswer:\n(a ) Silver does not show any characteristics change because silver is less reactive than hydrogen. Thus, it cannot displace hydrogen from dilute hydrochloric acid.\n(b ) The reaction between aluminium (Al) and hydrochloric acid is highly exothermic. Thus, the temperature of the reaction mixture rises.\n(c ) Sodium is a highly reactive metal. It reacts with hydrochloric acid, vigorously forming hydrogen gas and a large amount of heat.\n(d ) When lead reacts with hydrochloric acid, the gas bubbles observed are hydrogen gas.\nPb (s) + 2 HCl (aq) \u2192 PbCl2\u00a0(s) + H2\u00a0(g)", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\n(d ) When lead reacts with hydrochloric acid, the gas bubbles observed are hydrogen gas.\nPb (s) + 2 HCl (aq) \u2192 PbCl2\u00a0(s) + H2\u00a0(g)\nQ14.\u00a0A substance X, an oxide of a group 2 element, is used intensively in the cement industry. This element is present in bones also. On treatment with water, it forms a solution which turns red litmus blue. Identify X and also write the chemical reactions involved.\nAnswer:\nHere, X is calcium oxide.\nCalcium oxide is used intensively in the cement industry.\nThe element present in it (in bones also) is calcium.\nOn treatment with water, calcium oxide forms a solution of calcium hydroxide [Ca(OH)2], which is an alkali. Hence, it turns red litmus blue.\nCaO (s) + H2O (l) \u2192 Ca(OH)2\u00a0(aq) + Heat\nQ15.\u00a0Write a balanced chemical equation for each following reaction and classify\nthem.", + "Hence, it turns red litmus blue.\nCaO (s) + H2O (l) \u2192 Ca(OH)2\u00a0(aq) + Heat\nQ15.\u00a0Write a balanced chemical equation for each following reaction and classify\nthem.\n(a ) Lead acetate solution is treated with dilute hydrochloric acid to form lead chloride and acetic acid solution.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\nthem.\n(a ) Lead acetate solution is treated with dilute hydrochloric acid to form lead chloride and acetic acid solution.\n(b ) A piece of sodium metal is added to absolute ethanol to form sodium ethoxide and hydrogen gas.\n(c ) Iron (III) oxide on heating with carbon monoxide gas reacts to form solid iron and liberates carbon dioxide gas.\n(d ) Hydrogen sulphide gas reacts with oxygen gas to form solid sulphur and liquid water\nAnswer:\n(a ) Pb(CH3COO)2\u00a0+ 2 HCl \u2192 PbCl2\u00a0+ 2 CH3COOH\nIt is a double displacement reaction.\n(b ) 2 Na + 2 C2H5OH \u2192 2 C2H5ONa+ H2\nIt is a displacement or a redox reaction.\n(c ) Fe2O3\u00a0+ 3 CO \u2192 2 Fe + 3 CO2\nIt is a redox reaction.\n(d ) 2 H2S + O2\u00a0\u2192 2 S + 2 H2O\nIt is a redox reaction.\nQ16.", + "(c ) Fe2O3\u00a0+ 3 CO \u2192 2 Fe + 3 CO2\nIt is a redox reaction.\n(d ) 2 H2S + O2\u00a0\u2192 2 S + 2 H2O\nIt is a redox reaction.\nQ16.\u00a0Why do we store silver chloride in dark coloured bottles?\nAnswer:\nWe store silver chloride in the dark-coloured bottles because silver chloride decomposes into silver and chlorine gas in sunlight.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\nAnswer:\nWe store silver chloride in the dark-coloured bottles because silver chloride decomposes into silver and chlorine gas in sunlight.\nQ17.\u00a0Balance the following chemical equations and identify the type of chemical reaction.\n(a ) Mg (s) + Cl2\u00a0(g) \u2192 MgCI2\u00a0(s)\n(b ) HgO (s) + Heat \u2192 Hg (l) + O2\u00a0(g)\n(c ) Na (s) + S (s) \u2192 Na2S (s)\n(d ) TlCl4\u00a0(l) + Mg (s) \u2192 Tl (s) + MgCl2\u00a0(s)\n(e ) CaO (s) + SiO2\u00a0(s) \u2192 CaSiO3\u00a0(s)\n(f ) H2O2\u00a0(l) + UV \u2192 H2O (l) + O2\u00a0(g)\nAnswer:\n(a ) Mg (s) + Cl2\u00a0(g) \u2192 MgCI2\u00a0(s)\nIt is a combination reaction.", + "(b ) 2 HgO (s) + Heat \u2192 2 Hg (l) + O2\u00a0(g)\nIt is a thermal decomposition reaction.\n(c ) 2 Na (s) + S (s) \u2192 Na2S (s)\nIt is a combination reaction.\n(d ) TlCl4\u00a0(l) + 2 Mg (s) \u2192 Tl (s) + 2 MgCl2\u00a0(s)\nIt is a displacement reaction.\n(e ) CaO (s) + SiO2\u00a0(s) \u2192 CaSiO3\u00a0(s)\nIt is a combination reaction.\n(f ) 2 H2O2\u00a0(l) + UV \u2192 2 H2O (l) + O2\u00a0(g)\nIt is a decomposition reaction.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\nIt is a displacement reaction.\n(e ) CaO (s) + SiO2\u00a0(s) \u2192 CaSiO3\u00a0(s)\nIt is a combination reaction.\n(f ) 2 H2O2\u00a0(l) + UV \u2192 2 H2O (l) + O2\u00a0(g)\nIt is a decomposition reaction.\nQ18.\u00a0A magnesium ribbon is burnt in oxygen to give a white compound X accompanied by light emission. If the burning ribbon is now placed in an atmosphere of nitrogen, it continues to burn and forms a compound Y.\n(a) Write the chemical formulae of X and Y.\n(b) Write a balanced chemical equation when X is dissolved in water.\nAnswer:\nHere, X is magnesium oxide, and Y is magnesium nitride.\n(a ) The chemical formulae of X are MgO and Y is Mg3N2.\n(b ) When X is dissolved in water following reaction occurs.\nMgO + H2O \u2192 Mg(OH)2\nQ19.\u00a0Zinc liberates hydrogen gas when reacted with dilute hydrochloric acid, whereas copper does not. Explain why?\nAnswer:", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\nMgO + H2O \u2192 Mg(OH)2\nQ19.\u00a0Zinc liberates hydrogen gas when reacted with dilute hydrochloric acid, whereas copper does not. Explain why?\nAnswer:\nZinc is more reactive than copper as Zinc is placed above hydrogen, and copper is placed below hydrogen in the activity series of metals. Thus, zinc liberates hydrogen gas when reacted with dilute hydrochloric acid, whereas copper does not.\nQ20.\u00a0A silver article generally turns black when kept in the open for a few days. The article, when rubbed with toothpaste again, starts shining.\n(a ) Why do silver articles turn black when kept in the open for a few days? Name the phenomenon involved.\n(b ) Name the black substance formed and give its chemical formula.\nAnswer:\n(a ) The silver article turns black when kept in the air because the silver article reacts with sulphur compounds such as hydrogen sulphide (H2S) present in the air to form silver sulphide Ag2S. This phenomenon is called corrosion. It is also known as tarnishing of silver.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\n(b ) The black substance is silver sulphide. Its chemical formula is Ag2S.\nRelated Videos\nBalancing of Chemical Equations\n\n8,642\nLong Answer Type Questions\nQ1.\u00a0On heating blue coloured powder of copper (I) nitrate in a boiling tube, copper oxide\n(black), oxygen gas, and a brown gas X is formed\n(a) Write a balanced chemical equation of the reaction.\n(b) Identity the brown gas X evolved.\n(c) Identify the type of reaction.\n(d) What could be the pH range of the aqueous solution of the gas X?\nAnswer:\n(a ) 2 CuNO3\u00a0(s) + Heat \u2192 2 CuO (s) + 4 NO2\u00a0(g) + O2\u00a0(g)\n(b ) The brown gas is of nitrogen dioxide.\n(c ) It is a thermal decomposition reaction.\n(d ) NO2\u00a0gas reacts with water to produce nitric acid. Thus, its pH range will be less than 7.\nQ2.", + "(c ) It is a thermal decomposition reaction.\n(d ) NO2\u00a0gas reacts with water to produce nitric acid. Thus, its pH range will be less than 7.\nQ2.\u00a0Give the characteristic tests for the following gases\n(a ) CO2\n(b ) SO2\n(c ) O2\n(d ) H2\nAnswer:\nThe characteristics test for\n(a ) CO2: CO2\u00a0turns lime water milky due to the formation of insoluble calcium carbonate.\nCO2\u00a0+ Ca(OH)2\u00a0\u2192 CaCO3\u00a0+ H2O", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\n(a ) CO2\n(b ) SO2\n(c ) O2\n(d ) H2\nAnswer:\nThe characteristics test for\n(a ) CO2: CO2\u00a0turns lime water milky due to the formation of insoluble calcium carbonate.\nCO2\u00a0+ Ca(OH)2\u00a0\u2192 CaCO3\u00a0+ H2O\n(b ) SO2: SO2\u00a0turns purple coloured acidic potassium permanganate solution colourless.\n5 SO2\u00a0+ 2 KMnO4\u00a0+ 2 H2O \u2192 K2SO4\u00a0+ 2 MnSO4\u00a0+ 2 H2SO4\n(c ) O2: We can confirm the evolution of oxygen gas by bringing a burning candle near the mouth of the test tube containing the reaction mixture. The intensity of the flame increases because oxygen supports burning.\n(d ) H2: Hydrogen (H2) gas burns with a pop sound when a burning candle is brought near it.\nQ3.\u00a0What happens when a piece of\n(a) Zinc metal is added to copper sulphate solution?", + "The intensity of the flame increases because oxygen supports burning.\n(d ) H2: Hydrogen (H2) gas burns with a pop sound when a burning candle is brought near it.\nQ3.\u00a0What happens when a piece of\n(a) Zinc metal is added to copper sulphate solution?\n(b) Aluminium metal is added to dilute hydrochloric acid?\n(c) Silver metal is added to copper sulphate solution?\nAlso, write the balanced chemical equation if the reaction occurs\nAnswer:", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\n(b) Aluminium metal is added to dilute hydrochloric acid?\n(c) Silver metal is added to copper sulphate solution?\nAlso, write the balanced chemical equation if the reaction occurs\nAnswer:\n(a ) Zinc metal reacts with copper sulphate solution and forms colourless zinc sulphate and reddish-brown copper metal.\nZn (s) + CuSO4\u00a0(aq) \u2192 ZnSO4\u00a0(aq) + Cu (s)\n(b ) Aluminium metal reacts with dilute hydrochloric acid to form aluminium chloride and hydrogen gas.\n2 Al (s) + 6 HCl (aq) \u2192 2 AlCl3\u00a0(aq) + 3 H2\u00a0(g)\n(c ) Silver is less reactive than copper. Hence, no reaction will occur.\nQ4.\u00a0When zinc granules are treated with a dilute solution of H2SO4, HCI, HNO3, NaCI and NaOH. Write the chemical equations if a reaction occurs.\nAnswer:\nZinc granules react with dilute sulphuric acid to form zinc sulphate and hydrogen gas.", + "When zinc granules are treated with a dilute solution of H2SO4, HCI, HNO3, NaCI and NaOH. Write the chemical equations if a reaction occurs.\nAnswer:\nZinc granules react with dilute sulphuric acid to form zinc sulphate and hydrogen gas.\nZn (s) + H2SO4\u00a0(aq) \u2192 ZnSO4\u00a0(aq) + H2\u00a0(g)\nZinc granules react with dilute hydrochloric acid to form zinc chloride and hydrogen gas.\nZn (s) + H2SO4\u00a0(aq) \u2192 ZnCl2\u00a0(aq) + H2\u00a0(g)", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\nZn (s) + H2SO4\u00a0(aq) \u2192 ZnSO4\u00a0(aq) + H2\u00a0(g)\nZinc granules react with dilute hydrochloric acid to form zinc chloride and hydrogen gas.\nZn (s) + H2SO4\u00a0(aq) \u2192 ZnCl2\u00a0(aq) + H2\u00a0(g)\nZinc granules react with dilute nitric acid to form zinc nitrate, water and dinitrogen gas.\nZn (s) + H2SO4\u00a0(aq) \u2192 Zn(NO3)2\u00a0(aq) + H2O (l) + N2O (g)\nZinc does not react with sodium chloride\nZn (s) + NaCl \u2192 No Reaction.\nZinc granules react with dilute sodium hydroxide to form zinc hydroxide and hydrogen gas.\nZn (s) + NaOH (aq) \u2192 Zn(OH)2\u00a0(aq) + Na (g)\nQ5.", + "Zinc granules react with dilute sodium hydroxide to form zinc hydroxide and hydrogen gas.\nZn (s) + NaOH (aq) \u2192 Zn(OH)2\u00a0(aq) + Na (g)\nQ5.\u00a0A white precipitate is obtained when adding a drop of barium chloride solution to an aqueous sodium sulphite solution.\n(a ) Write a balanced chemical equation of the reaction involved\n(b ) What other name can be given to this precipitation reaction?\n(c ) On adding dilute hydrochloric acid to the reaction mixture, white residue disappears. Why?\nAnswer:\n(a ) BaCl2\u00a0+ Na2SO3\u00a0\u27f6 BaSO3\u00a0+ 2 NaCl\n(b ) It can be assigned as a double displacement reaction.", + "CHAPTER 1-CHEMICAL REACTIONS AND EQUATIONS -IMPORTANT QUESTION WITH ANSWERS.txt\n(c ) On adding dilute hydrochloric acid to the reaction mixture, white residue disappears. Why?\nAnswer:\n(a ) BaCl2\u00a0+ Na2SO3\u00a0\u27f6 BaSO3\u00a0+ 2 NaCl\n(b ) It can be assigned as a double displacement reaction.\n(c ) On adding dilute hydrochloric acid to the reaction mixture, white residue disappears due to the formation of barium chloride.\nBaSO3\u00a0+ 2 HCl \u27f6 BaCl2\u00a0+ SO2\u00a0+ H2O\nQ6.\u00a0You are provided with two containers made up of copper and aluminium. You are also\nprovided with dilute HCI, HNO3, ZnCl2\u00a0and H2O solutions. In which of the above\ncontainers we can keep these solutions?\nAnswer:\nThe solution of dilute HCI, HNO3, ZnCl2\u00a0and H2O can be kept in a container made of copper since copper is a less reactive metal and is placed below the hydrogen in the reactivity series.", + "In which of the above\ncontainers we can keep these solutions?\nAnswer:\nThe solution of dilute HCI, HNO3, ZnCl2\u00a0and H2O can be kept in a container made of copper since copper is a less reactive metal and is placed below the hydrogen in the reactivity series. Hence it does not react with HCI, HNO3, ZnCl2\u00a0and H2O. At the same time, aluminium is a highly reactive metal and can react with these solutions. Thus container made of copper is suitable to keep the given solutions.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nCHAPTER \u2013 2 \nACIDS, BASES AND SALTS \n \nq \nAcids Bases \n\u2013 Sour in taste \u2013 Bitter in taste \n\u2013 Change the blue litmus to red \u2013 Change red litmus to blue \n\u2013 eg. HydrochloricAcid HCl eg.", + "HydrochloricAcid HCl eg. Sodium hydroxide NaOH \n\u2013 Sulphuric Acid H2SO4 Potassium hydroxide KOH \n\u2013 Nitric Acid HNO3 Calcium hydroxide Ca(OH)2 \n\u2013 Acetic Acid CH3 COOH \u2013 Ammonium hydroxide NH4OH \nq Some Naturally occuring acids \nVinegar \u2013 Acetic Acid \nOrange \u2013 Citric Acid \nLemon \u2013 Citric Acid \nTamarind \u2013 Tartaric Acid \nTomato \u2013 Oxalic Acid \nSour milk (Curd) \u2013 Lactic Acid \nAnt and Nettle sting \u2013 Methanoic Acid \nq Acid \u2013 Base Indicators \u2013 Indicate the presence of an acid or base in a \nsolution. \nq Litmus solution \u2013 It is a natural indicator. It is a purple day extracted from \nLichens. Other examples are Red Cabbage and coloured petals of Petunia \nand turmeric. \nq Olfactory indicators \u2013 Show odour changes in acidic or basic media.", + "It is a purple day extracted from \nLichens. Other examples are Red Cabbage and coloured petals of Petunia \nand turmeric. \nq Olfactory indicators \u2013 Show odour changes in acidic or basic media. eg. \nonion and clove.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nQ A cid \u2013 Base Indicators", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nS. No. Name of the \nIndicator Colour Change \nwith Acid Colour Change \nwith Base \nA. Blue litmus solution To red No change \nB. Red litmus solution No change To blue \nC. Turmeric No change To red \nD. Methyl orange To red To yellow \nE. Phenolphthalein (colourless) No change To pink \nq Dilute Acid : Contains only a small amounts of acid and a large amount of \nwater. \nq Concentrated Acid : A concentrated acid contains a large amount of acid and \na small amount of water. \nq Chemical Properties of Acids and Bases \nAcid + Metal \u23af\u23af\u2192 Salt + Hydrogen \n(Refer activity 2.3 on page No.", + "q Chemical Properties of Acids and Bases \nAcid + Metal \u23af\u23af\u2192 Salt + Hydrogen \n(Refer activity 2.3 on page No. 19 of NCERT Book) \n2HCl + Zn \u23af\u23af\u2192 ZnCl2 + H2 \n2HNO3 + Zn \u23af\u23af\u2192 Zn (NO3)2 + H2 \nH2SO4 + Zn \u23af\u23af\u2192 ZnSO4 + H2 \n2CH3COOH + Zn \u23af\u23af\u2192 (CH3COO)2 Zn + H2 \nq Pop test : When a buring candle is brought near a test tube containing \nhydrogen gas it burns with a \u2018Pop\u2019 sound. This test is conducted for examining \nthe presence of hydrogen gas.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nq Pop test : When a buring candle is brought near a test tube containing \nhydrogen gas it burns with a \u2018Pop\u2019 sound. This test is conducted for examining \nthe presence of hydrogen gas. \nq Base + Metal \u23af\u23af\u2192 Salt + Hydrogen \nNaOH + Zn \u23af\u23af\u2192 Na2ZnO2 + H2 \nSodium Zincate", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nq Action of Acids with metal Carbonates and metal bicarbonates \nMetal Carbonate + Acid \u23af\u23af\u2192 Salt + Carbondio xide + Water \nNa2CO3 \n \n(s) + 2HCl(aq) \u23af\u23af\u2192 2NaCl(aq) + H2O( l) + CO2(g) \nMetal bicarbonate + Acid \u23af\u23af\u2192 Salt + Carbondio xide + Water \nNaHCO3 + HCl \u23af\u23af\u2192 NaCl + CO2 + H2O \nq Lime water Test : On passing the CO2 gas evolved through lime water,", + "Ca(OH)2 (aq) + CO2(g) \u23af\u23af\u2192 CaCO3(s) + H2O( l) \nLime water White precipitate \nOn passing excess CO2 the following reaction takes place \nCaCO3(s) + H2O(l) + CO2(g) \u23af\u23af\u2192 Ca(HCO3)2 aq \nSoluble in water q Neutralisation Reactions \nBase + Acid \u23af\u23af\u2192 Salt + Water \nNaOH(aq) + HCl(aq) \u23af\u23af\u2192 NaCl(aq) + H2O(l) \nNeutralisation reacton takes place when the effect of a base is nullified by \nan acid and vice versa to give salt and water. \nq Reactions of metal oxides with acids \nMetal Oxide + Acid \u23af\u23af\u2192 Salt + Water", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nCuO + HCl \u23af\u23af\u2192 CuCl2 + H2O \nCopperoxide Hydrochloric \nacid Copper + Water \nchloride \nNote : Appearance of blue green colour of the solution because of formation \nof CuCl2. \nMetallic oxides are said to be basic oxides because they give salt and water \non reacting with acids. \nq Reaction of Non Metallic Oxide with Base \nNon metallic oxide + Base \u23af\u23af\u2192 Salt + Water \nCa(OH)2 + CO2 \u23af\u23af\u2192 CaCO3 + H2Note : Non Metallic oxides are said to be acidic in nature because on reacting with a \nbase they produce Salt and Water. \nq All acidic solutions conduct electricity \nRefer activity 2.3 on page 22 of NCERT Book \n\u2013 Glowing of bulb indicates that there is a flow of electric current through the \nsolution.", + "q All acidic solutions conduct electricity \nRefer activity 2.3 on page 22 of NCERT Book \n\u2013 Glowing of bulb indicates that there is a flow of electric current through the \nsolution. \nq Acids or bases in a Water Solution \nAcids produce H+ions in the presence of water \nHCl + H2O \u23af\u23af\u2192 H3O+ + Cl\u2013 \nH3O+ \u2013 Hydronium ion. \n\u2013 H+ion cannot exist alone. It exists as H+(aq) or (H3O+) hydronium ion. \nH+ + H2O \u23af\u23af\u2192 H3O+", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nAcids produce H+ions in the presence of water \nHCl + H2O \u23af\u23af\u2192 H3O+ + Cl\u2013 \nH3O+ \u2013 Hydronium ion. \n\u2013 H+ion cannot exist alone. It exists as H+(aq) or (H3O+) hydronium ion. \nH+ + H2O \u23af\u23af\u2192 H3O+ \n\u2013 Bases provide (OH\u2013) ions in the presence of water \nNaOH H2O Na+ + OH\u2013", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nKOH (s) \u23af\u23af\u23af\u23af\u2192 H2O K+ (aq) \n + OH\u2013 (aq) \n(s) \u23af\u23af\u23af\u23af\u2192 (aq) (aq) \nMg(OH) H2O Mg2+(aq) + 2OH\u2013(aq) \n \nq Alkalis 2(s) \u23af\u23af\u23af\u23af\u2192 \nAll bases donot dissolve in water. An alkali is a base that dissolves in water. \nCommon alkalis are \nNaOH Sodium hydroxide \nKOH Potassium hydroxide \nCa(OH)2 Calcium hydroxide \nNH4OH : Ammonium hydroxide \nNote : All alkalis are bases but all bases are not alkalis. \nq Precaution must be taken while mixing acid or base with water. The acid must \nalways be added to wate r with constant stirring as it is highly exothermic \nreaction.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nWhen an acid or a base is mixed with water they become dilute. This results \nin the decrease in the concentration of H3O+ or OH\u2013 per unit volume in acids \nand bases respectively. \nq Strength of an Acid or Base \nStrength of acids and bases depends on the no. of H+ions and OH\u2013ions \nproduced respectively. \nWith the help of a universal indicator we can find the strength of an acid or \nbase. This indicator is called PH scale. \npH = Potenz in German means power. \nThis scale measures from 0 (very acidic) to 14 (very alkaline) 7 Neutral \n(water in Neutral). \npH paper : Is a paper which is used for measuring PH. \nVariation of PH", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\n\u2013 strong Acids give rise to more H+ions. \neg. HCl, H2SO4 and HNO3. \n\u2013 Weak Acids give rise to less H+ ions \neg. CH3 COOH, H2 CO3 (Carbonic acid) \n\u2013 Strong Bases \u2013 Strong bases give rise to more OH\u2013 ions. \neg. NaOH, KOH, Ca(OH)2 \n\u2013 Weak Bases : give rise to less OH\u2013 ions. \neg. NH4OHS. \nNo. PH \nValue Colour of the \npH Paper Nature of \nSolution H+ion \nConc. O H \u2013 i o n \nConc. \n1. 0 Dark red Highly acidic very high very low \n2. 4 Orange or yellow Acidic high low \n3. 7: Green Neutral Equal Equal \n4. 10 Bluish green or blue Alkaline low high \n5. 14 Dark blue or voilet highly basic very low very high \n \n \n \n \n \nq More about Salts \nSalts and their derivation", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nS. No. Name of Salt Formula Derived from Derived from \n1. Potassium Sulphate K2SO4 KOH H2SO4 \n2. Sodium Sulphate Na2SO4 NaOH H2SO4 \n3. Sodium Chloride NaCl NaOH HCl \n4. Ammonium Chloride NH4Cl NH4OH HCl \nNote : NaCl and Na2 SO4 belong to the family of sodium salts as they have \nthe same radicals. Similarly NaCl and KCl belong to the family of chloride \nsalts. \nImportance of pH in our daily life \nq Importance of pH in our digestive system \u2013 pH level of our body regulates \nour digestive system. In case of indigestion our stomach produces acid in a \nvery large quantity because of which we feel pain and irritation in our stomach. \nTo get relief from this pain antacids are used. These antacids neutralises the \nexcess acid and we get relief.", + "In case of indigestion our stomach produces acid in a \nvery large quantity because of which we feel pain and irritation in our stomach. \nTo get relief from this pain antacids are used. These antacids neutralises the \nexcess acid and we get relief. \nq pH of Acid Rain : When pH of rain water is less than 5.6 it is called Acid \nRain.When this acidic rain flows into rivers these also get acidic, which causes \na threat to the survival of aquatic life.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nRain.When this acidic rain flows into rivers these also get acidic, which causes \na threat to the survival of aquatic life. \nq pH of Soil : Plants require a specific range of pH for their healthy growth. If \npH of soil of any particular place is less or more than normal than the farmers \nadd suitable fertilizers to it. \nq Our body functions between the range of 7.0 to 7.8 living organisms can \nsurvive only in the narrow range of pH change. \nq Tooth decay and pH : Bacteria present in the mouth produce acids by \ndegredation of sugar and food particles remaining in the mouth. Using toothpaste \nwhich is generally basic can neutralise the excess acid and prevent tooth \ndecay. \nq Bee sting or Nettle sting contains methanoic acid which causes pain and \nirritation. When we use a weak base like baking soda on it we get relief.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nNeutral Salts : Strong Acid + Strong base \npH value is 7 \neg. NaCl, CaSO4 \nAcidic Salts : Strong Acid + weak base \npH value is less than 7 \neq. NH4Cl, NH4 NO3 \nBasic Salts : Strong base + weak acid \npH value is more than 7 \neg. CaCO3, CH3 COONa \nq Chemicals from Common Salt \n\u2013 Sodium chloride is called as common salt used in our food. It is derived \nfrom seawater. \n\u2013 Rock Salt is the brown coloured large crystals. This s mined like coal. \n\u2013 Common Salt is an important raw material for many materials of daily use \nsuch as. \nSodium hydroxide \nWashing Soda \nBleaching Power. \nq Sodium Hydroxide \nPreparation : Prepared by the method called chlor -alkali \nCalled chlor -alkali because we get chlorine and a base in this.", + "Sodium hydroxide \nWashing Soda \nBleaching Power. \nq Sodium Hydroxide \nPreparation : Prepared by the method called chlor -alkali \nCalled chlor -alkali because we get chlorine and a base in this. \n2NaCl(aq) + 2H2O( l) \u23af\u23af\u2192 2NaOH(aq) + Cl2(g) + H2(g)", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nq Bleaching Power \nPreparation \u23af\u23af\u2192 Ca(OH)2 + Cl2 \u23af\u23af\u2192 CaOCl2 + H2O \ncalcium hydroxide chlorine bleaching water \npower \nuses in textile, factories and laundry, used as disinfectant \nq Baking Soda \n\u2013 Common name \u2013 Sodium Hydrogen Carbonate \n \nPreparation NaCl + H2O + CO2 + NH3 \u23af\u23af\u2192 NH4Cl + NaHCO3 \nSodium Water Carbon Ammonia Sodium hydrogen \nchloride dioxide carbonate", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\nPreparation NaCl + H2O + CO2 + NH3 \u23af\u23af\u2192 NH4Cl + NaHCO3 \nSodium Water Carbon Ammonia Sodium hydrogen \nchloride dioxide carbonate \n \nOn heating NaHCO3 produces : \nNaHCO3 \u23af\u23afHe\u23afat\u23af\u2192 Na2CO3 + H2O + CO2 \nCO2 produced causes dough to rise and make cakes, pastries spongy. \nUses : In household, ingredients of antacid \nIn making baking power \nOn heating baking powder produces \nNaHCO3 + H+ \u23af\u23af\u2192 CO2 + H2O + Sodium Salt of acid \nq Washing Soda \nPreparation : Recrystallisation of sodium carbonate \nNa2CO3 + 10H2O \u23af\u23afHe\u23afat\u23af\u2192 \nUses Na2CO3.", + "10H2O \n\u2013 Used in glass, soap and paper industry \n\u2013 Cleaning agent for domestic purposes. \n\u2013 Removal of hardness of water. \n\u2013 Manufacturere of borax. \nq Water of crystallisation : Fixed no. of water molecules present in one formula \nunit of a salt. \n\u2013 On heating copper sulphate crystals water droplets appear, formula of \nhydrated copper sulphate \u2013 CuSO4. 5H2O.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.txt\n\u2013 gypsum also contains water of crystallisation. \n\u2013 Formula of gypsum \u2013 CaSO4.2H2O \n\u2013 On heating gypsum at 373k it becomes CaSO4.\u00bdH2O is plaster of \nparis. \n\u2013 Plaster of Paris is used as plaster for fractured bones. \n\u2013 When plaster of Paris is mixed with water it changes to gypsum. \nCaSO4.\u00bd H2O + 1\u00bd H2O \u23af\u23af\u2192 CaSO4.2H2O \nUses of plaster of Paris : Making toys, decorative material and smooth surfaces.", + "CHAPTER 2-ACIDS,BASES AND SALTS -REVISION NOTES.txt\nChemical Reactions & \nequations \nPower Point Presentation END", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n\uf041\uf063\uf069\uf064\uf073\uf02c\uf020\uf042\uf061\uf073\uf065\uf073\n\uf061\uf06e\uf064\uf020\uf053\uf061\uf06c\uf074\uf073\uf032 \uf043\uf048\uf041\uf050\uf054\uf045\uf052\nYou have lear nt in your previous classes that the sour and bitter\ntastes of food are due to acids and bases, respectively, present in them.\nIf someone in the family is suffering from a problem of acidity after\novereating, which of the following would you suggest as a remedy\u2013 lemon\njuice, vinegar or baking soda solution?\n\uf06eWhich property did you think of while choosing the remedy?\nSurely you must have used your knowledge about the ability of\nacids and bases to nullify each other\u2019s effect.\n\uf06eRecall how we tested sour and bitter substances without tastingthem.", + "\uf06eWhich property did you think of while choosing the remedy?\nSurely you must have used your knowledge about the ability of\nacids and bases to nullify each other\u2019s effect.\n\uf06eRecall how we tested sour and bitter substances without tastingthem.\nYou alr eady know that acids ar e sour in taste and change the colour\nof blue litmus to red, whereas, bases are bitter and change the colour of\nthe r ed litmus to blue. Litmus is a natural indicator , tur meric is another\nsuch indicator. Have you noticed that a stain of curry on a white cloth", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nthe r ed litmus to blue. Litmus is a natural indicator , tur meric is another\nsuch indicator. Have you noticed that a stain of curry on a white cloth\nbecomes reddish-brown when soap, which is basic in nature, is scrubbedon it? It turns yellow again when the cloth is washed with plenty ofwater . You can also use synthetic indicators such as methyl orange and\nphenolphthalein to test for acids and bases.\nIn this Chapter, we will study the reactions of acids and bases, how\nacids and bases cancel out each other\u2019s effects and many more interestingthings that we use and see in our day-to-day life.\n\uf044\uf06f\uf020\uf059\uf06f\uf075\uf020\uf04b\uf06e\uf06f\uf077\uf03fLitmus solution is a purple dye, which is extracted from lichen, a plant belonging tothe division Thallophyta, and is commonly used as an indicator. When the litmus\nsolution is neither acidic nor basic, its colour is purple. There are many other natural", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nsolution is neither acidic nor basic, its colour is purple. There are many other natural\nmaterials like red cabbage leaves, turmeric, coloured petals of some flowers such asHydrangea , Petunia and Geranium , which indicate the presence of acid or base in a\nsolution. These are called acid-base indicators or sometimes simply indicators.\n2019-20\nScience \uf031\uf038\uf032\uf02e\uf031 \uf032\uf02e\uf031\uf032\uf02e\uf031 \uf032\uf02e\uf031\uf032\uf02e\uf031\uf055\uf04e\uf044\uf045\uf052\uf053\uf054 \uf055\uf04e\uf044\uf045\uf052\uf053\uf054\uf055\uf04e\uf044\uf045\uf052\uf053\uf054 \uf055\uf04e\uf044\uf045\uf052\uf053\uf054\uf055\uf04e\uf044\uf045\uf052\uf053\uf054", + "\uf055\uf04e\uf044\uf045\uf052\uf053\uf054\uf055\uf04e\uf044\uf045\uf052\uf053\uf054 \uf041\uf04e\uf044\uf049\uf04e\uf047\uf020\uf054\uf048\uf045\uf020\uf043\uf048\uf045\uf04d\uf049\uf043\uf041\uf04e\uf044\uf049\uf04e\uf047\uf020\uf054\uf048\uf045\uf020\uf043\uf048\uf045\uf04d\uf049\uf043\uf041\uf04e\uf044\uf049\uf04e\uf047\uf020\uf054\uf048\uf045\uf020\uf043\uf048\uf045\uf04d\uf049\uf043\uf041\uf04e\uf044\uf049\uf04e\uf047\uf020\uf054\uf048\uf045\uf020\uf043\uf048\uf045\uf04d\uf049\uf043\uf041\uf04e\uf044\uf049\uf04e\uf047\uf020\uf054\uf048\uf045\uf020\uf043\uf048\uf045\uf04d\uf049\uf043", + "\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\n\uf041\uf043\uf049\uf044", + "\uf046\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\n\uf041\uf043\uf049\uf044\uf053\uf020\uf041\uf04e\uf044\uf020\uf042\uf041\uf053\uf045\uf053\uf041\uf043\uf049\uf044\uf053\uf020\uf041\uf04e\uf044\uf020\uf042\uf041\uf053\uf045\uf053\uf041\uf043\uf049\uf044\uf053\uf020\uf041\uf04e\uf044\uf020\uf042\uf041\uf053\uf045\uf053\uf041\uf043\uf049\uf044\uf053\uf020\uf041\uf04e\uf044\uf020\uf042\uf041\uf053\uf045\uf053\uf041\uf043\uf049\uf044\uf053\uf020\uf041\uf04e", + "\uf049\uf044\uf053\uf020\uf041\uf04e\uf044\uf020\uf042\uf041\uf053\uf045\uf053\uf041\uf043\uf049\uf044\uf053\uf020\uf041\uf04e\uf044\uf020\uf042\uf041\uf053\uf045\uf053\n\uf032\uf02e\uf031\uf02e\uf031\uf020\uf041\uf063\uf069\uf064\uf073\uf020\uf061\uf06e\uf064\uf020\uf042\uf061\uf073\uf065\uf073\uf020\uf069\uf06e\uf020\uf074\uf068\uf065\uf020\uf04c\uf061\uf062\uf06f\uf072\uf061\uf074\uf06f\uf072", + "\uf061\uf073\uf065\uf073\uf020\uf069\uf06e\uf020\uf074\uf068\uf065\uf020\uf04c\uf061\uf062\uf06f\uf072\uf061\uf074\uf06f\uf072 \uf079\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\nThese indicators tell us whether a substance is", + "indicators tell us whether a substance is acidic or basic by\nchange in colour. Ther e are some substances whose odour changes in\nacidic or basic media. These are called olfactory indicators. Let us try", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nchange in colour. Ther e are some substances whose odour changes in\nacidic or basic media. These are called olfactory indicators. Let us try\nout some of these indicators.\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\n\uf03f1. You have been pr ovided with three test tubes. One of them contains\ndistilled water and the other two contain an acidic solution and a basic\nsolution, r espectively. If you are given only r ed litmus paper , how will\nyou identify the contents of each test tube?\n\uf06eCollect the following solutions from the science laboratory\u2013hydrochloric acid (HCl), sulphuric acid (H\n2SO4), nitric acid (HNO3),\nacetic acid (CH3COOH), sodium hydroxide (NaOH), calcium\nhydroxide [Ca(OH)2], potassium hydroxide (KOH), magnesium\nhydroxide [Mg(OH)2], and ammonium hydroxide (NH4OH).", + "\uf06ePut a drop of each of the above solutions on a watch-glass one by\none and test with a drop of the indicators shown in Table 2.1.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nhydroxide [Mg(OH)2], and ammonium hydroxide (NH4OH).\n\uf06ePut a drop of each of the above solutions on a watch-glass one by\none and test with a drop of the indicators shown in Table 2.1.\n\uf06eWhat change in colour did you observe with red litmus, blue litmus,phenolphthalein and methyl orange solutions for each of the\nsolutions taken?\n\uf06eTabulate your observations in Table 2.1.", + "\uf06eWhat change in colour did you observe with red litmus, blue litmus,phenolphthalein and methyl orange solutions for each of the\nsolutions taken?\n\uf06eTabulate your observations in Table 2.1.\nTable 2.1\nSample Red Blue Phenolph- Methyl\nsolution litmus litmus -thalein orange\nsolution solution solution solution\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf032\n\uf06eTake some finely chopped onions in a plastic bag along with some\nstrips of clean cloth.", + "T ie up the bag tightly and leave over night in\nthe fridge. The cloth strips can now be used to test for acids andbases.\n\uf06eTake two of these cloth strips and check their odour.\n\uf06eKeep them on a clean surface and put a few drops of dilute HCl\nsolution on one strip and a few drops of dilute NaOH solution on\nthe other .\n2019-20", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n\uf06eKeep them on a clean surface and put a few drops of dilute HCl\nsolution on one strip and a few drops of dilute NaOH solution on\nthe other .\n2019-20\nAcids, Bases and Salts \uf031\uf039Which of these \u2013 vanilla, onion and clove, can be used as olfactory\nindicators on the basis of your observations?\nLet us do some more activities to understand the chemical properties\nof acids and bases.", + "2019-20\nAcids, Bases and Salts \uf031\uf039Which of these \u2013 vanilla, onion and clove, can be used as olfactory\nindicators on the basis of your observations?\nLet us do some more activities to understand the chemical properties\nof acids and bases.\n\uf032\uf02e\uf031\uf02e\uf032\uf020\uf048\uf06f\uf077\uf020\uf064\uf06f\uf020\uf041\uf063\uf069\uf064\uf073\uf020\uf061\uf06e\uf064\uf020\uf042\uf061\uf073\uf065\uf073\uf020\uf052\uf065\uf061\uf063\uf074\uf020\uf077\uf069\uf074\uf068\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\uf03f\uf06eRinse both cloth strips with water and again check their odour.\n\uf06eNote your observations.\n\uf06eNow take some dilute vanilla essence and clove oil and check their\nodour .", + "\uf06eNote your observations.\n\uf06eNow take some dilute vanilla essence and clove oil and check their\nodour .\n\uf06eTake some dilute HCl solution in one test tube and dilute NaOHsolution in another. Add a few dr ops of dilute vanilla essence to\nboth test tubes and shake well. Check the odour once again andrecord changes in odour, if any.\n\uf06eSimilarly, test the change in the odour of clove oil with dilute HCland dilute NaOH solutions and record your observations.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n\uf06eSimilarly, test the change in the odour of clove oil with dilute HCland dilute NaOH solutions and record your observations.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf033\nCAUTION : This activity needs the teacher\u2019s assistance.\n\uf06eSet the apparatus as shown in Fig. 2.1.", + "\uf06eSet the apparatus as shown in Fig. 2.1.\n\uf06eTake about 5 mL of dilute sulphuric acid in a test tube and add a\nfew pieces of zinc granules to it.\n\uf06eWhat do you observe on the surface of zinc granules?\n\uf06ePass the gas being evolved through the soap solution.\n\uf06eWhy are bubbles formed in the soap solution?\n\uf06eTake a burning candle near a gas filled bubble.\n\uf06eWhat do you observe?\n\uf06eRepeat this Activity with some more acids like HCl, HNO3 and\nCH3COOH.\n\uf06eAre the observations in all the cases the same or different?\nFigure 2.1Figure 2.1Figure 2.1Figure 2.1Figure 2.1 Reaction of zinc granules with dilute sulphuric acid and testing hydrogen\ngas by burning\n2019-20\nScience \uf032\uf030Note that the metal in the above reactions displaces hydrogen atoms", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\ngas by burning\n2019-20\nScience \uf032\uf030Note that the metal in the above reactions displaces hydrogen atoms\nfrom the acids as hydrogen gas and forms a compound called a salt.\nThus, the reaction of a metal with an acid can be summarised as \u2013\nAcid + Metal \u2192 Salt + Hydrogen gas\nCan you now write the equations for the reactions you have observed?", + "Thus, the reaction of a metal with an acid can be summarised as \u2013\nAcid + Metal \u2192 Salt + Hydrogen gas\nCan you now write the equations for the reactions you have observed?\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf034\nThe reactions occurring in the above Activity are written as \u2013\nTest tube A: Na CO HCl(aq) Cl(aq) H O(l) + CO23", + "reactions occurring in the above Activity are written as \u2013\nTest tube A: Na CO HCl(aq) Cl(aq) H O(l) + CO23 2 2 (s) Na (g) + \u2192 + 2 2\nTest tube B: NaHCO HCl(aq) Cl(aq) H O(l) + CO3 22 (s) Na (g) +\u2192 +\nOn passing the carbon dioxide gas evolved through lime water,\nCa(OH) CO H O(l)2 2 2 (aq) (g) CaCO s3 +\u2192 +()\n(Lime water) (White precipitate)\uf06ePlace a few pieces of granulated zinc metal in a test tube.\n\uf06eAdd 2 mL of sodium hydroxide solution and warm the contents\nof the test tube.\n\uf06eRepeat the rest of the steps as in Activity 2.3 and record yourobservations.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n\uf06eAdd 2 mL of sodium hydroxide solution and warm the contents\nof the test tube.\n\uf06eRepeat the rest of the steps as in Activity 2.3 and record yourobservations.\nThe reaction that takes place can be written as follows.\n2NaOH(aq) + Zn(s) \u2192 Na2ZnO2(s) + H2(g)\n (Sodium zincate)\nYou find again that hydr ogen is formed in the reaction. However,\nsuch reactions are not possible with all metals.\n\uf032\uf02e\uf031\uf02e\uf033\uf020\uf048\uf06f\uf077\uf020\uf064\uf06f\uf020\uf04d\uf065\uf074\uf061\uf06c\uf020\uf043\uf061\uf072\uf062\uf06f\uf06e\uf061\uf074\uf065\uf073\uf020\uf061\uf06e\uf064\uf020\uf04d\uf065\uf074\uf061\uf06c\n\uf048\uf079", + "\uf061\uf072\uf062\uf06f\uf06e\uf061\uf074\uf065\uf073\uf020\uf061\uf06e\uf064\uf020\uf04d\uf065\uf074\uf061\uf06c\n\uf048\uf079\uf064\uf072\uf06f\uf067\uf065\uf06e\uf063\uf061\uf072\uf062\uf06f\uf06e\uf061\uf074\uf065\uf073\uf020\uf052\uf065\uf061\uf063\uf074\uf020\uf077\uf069\uf074\uf068\uf020\uf041\uf063\uf069\uf064\uf073\uf03f\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf035\uf041\uf063\uf074\uf069\uf076", + "\uf020\uf032\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf035\n\uf06eTake two test tubes, label them as A\nand B.\n\uf06eTake about 0.5 g of sodium carbonate(Na\n2CO3) in test tube A and about\n0.5 g of sodium hydrogencarbonate(NaHCO\n3) in test tube B.\n\uf06eAdd about 2 mL of dilute HCl to boththe test tubes.\n\uf06eWhat do you observe?", + "\uf06eAdd about 2 mL of dilute HCl to boththe test tubes.\n\uf06eWhat do you observe?\n\uf06ePass the gas produced in each case\nthrough lime water (calcium\nhydroxide solution) as shown in\nFig. 2.2 and record your observations.Figure 2.2Figure 2.2Figure 2.2Figure 2.2Figure 2.2", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n\uf06ePass the gas produced in each case\nthrough lime water (calcium\nhydroxide solution) as shown in\nFig. 2.2 and record your observations.Figure 2.2Figure 2.2Figure 2.2Figure 2.2Figure 2.2\nPassing carbon dioxide gasthrough calcium hydroxidesolution\n2019-20\nAcids,", + "2.2 and record your observations.Figure 2.2Figure 2.2Figure 2.2Figure 2.2Figure 2.2\nPassing carbon dioxide gasthrough calcium hydroxidesolution\n2019-20\nAcids, Bases and Salts \uf032\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf036\n\uf06eTake about 2 mL of dilute NaOH solution in a test tube and add\ntwo drops of phenolphthalein solution.", + "\uf06eWhat is the colour of the solution?\n\uf06eAdd dilute HCl solution to the above solution drop by drop.\n\uf06eIs there any colour change for the reaction mixture?\n\uf06eWhy did the colour of phenolphthalein change after the addition\nof an acid?\n\uf06eNow add a few drops of NaOH to the above mixture.\n\uf06eDoes the pink colour of phenolphthalein reappear?\n\uf06eWhy do you think this has happened?On passing excess carbon dioxide the following reaction takes place:\nCaCO s Ca(HCO aq3 3 () )( ) + \u2192 H O(l)+ CO (g) 2 2 2\n(Soluble in water)", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n\uf06eWhy do you think this has happened?On passing excess carbon dioxide the following reaction takes place:\nCaCO s Ca(HCO aq3 3 () )( ) + \u2192 H O(l)+ CO (g) 2 2 2\n(Soluble in water)\nLimestone, chalk and marble are different forms of calcium carbonate.\nAll metal carbonates and hydrogencarbonates react with acids to give a\ncorresponding salt, carbon dioxide and water.\nThus, the reaction can be summarised as \u2013\nIn the above Activity, we have observed that the effect of a base is\nnullified by an acid and vice-versa. The reaction taking place is written as \u2013\nNaOH(aq) + HCl(aq) \u2192 NaCl(aq) + H2O(l)\nThe reaction between an acid and a base to give a salt and water is\nknown as a neutralisation reaction. In general, a neutralisation reaction\ncan be written as \u2013\nBase + Acid \u2192 Salt + W ater\n\uf032\uf02e\uf031\uf02e\uf035\uf020", + "In general, a neutralisation reaction\ncan be written as \u2013\nBase + Acid \u2192 Salt + W ater\n\uf032\uf02e\uf031\uf02e\uf035\uf020\uf052\uf065\uf061\uf063\uf074\uf069\uf06f\uf06e\uf020\uf06f\uf066\uf020\uf04d\uf065\uf074\uf061\uf06c\uf06c\uf069\uf063\uf020\uf04f\uf078\uf069\uf064\uf065\uf073\uf020\uf077\uf069\uf074\uf068\uf020\uf041\uf063\uf069\uf064\uf073\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079", + "\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf037\n\uf06eTake a small amount of copper oxide in a beaker and add dilute\nhydrochloric acid slowly while stirring.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf037\n\uf06eTake a small amount of copper oxide in a beaker and add dilute\nhydrochloric acid slowly while stirring.\n\uf06eNote the colour of the solution. What has happened to the copperoxide?", + "\uf06eNote the colour of the solution. What has happened to the copperoxide?\nYou will notice that the colour of the solution becomes blue-gr een\nand the copper oxide dissolves. The blue-green colour of the solution is\ndue to the formation of copper(II) chloride in the reaction.", + "What has happened to the copperoxide?\nYou will notice that the colour of the solution becomes blue-gr een\nand the copper oxide dissolves. The blue-green colour of the solution is\ndue to the formation of copper(II) chloride in the reaction. The general\nreaction between a metal oxide and an acid can be written as \u2013\nMetal oxide + Acid \u2192 Salt + W aterMetal carbonate/Metal hydrogencarbonate + Acid \u2192 Salt + Carbon dioxide + W ater\n\uf032\uf02e\uf031\uf02e\uf034\uf020\uf048\uf06f\uf077\uf020\uf064\uf06f\uf020\uf041\uf063\uf069\uf064\uf073\uf020\uf061\uf06e\uf064\uf020\uf042\uf061\uf073\uf065\uf073\uf020\uf052\uf065\uf061\uf063\uf074\uf020\uf077\uf069\uf074\uf068\uf020\uf065\uf061\uf063\uf068\uf020\uf06f\uf074\uf068\uf065\uf072\uf03f\n2019-20\nScience \uf032\uf032Now write and balance the equation for the above reaction.", + "Since\nmetallic oxides react with acids to give salts and water, similar to the\nreaction of a base with an acid, metallic oxides are said to be basic oxides.\n\uf032\uf02e\uf031\uf02e\uf036\uf020\uf052\uf065\uf061\uf063\uf074\uf069\uf06f\uf06e\uf020\uf06f\uf066\uf020\uf061\uf020\uf04e\uf06f\uf06e\uf02d\uf06d\uf065\uf074\uf061\uf06c\uf06c\uf069\uf063\uf020\uf04f\uf078\uf069\uf064\uf065\uf020\uf077\uf069\uf074\uf068\uf020\uf042\uf061\uf073\uf065", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nmetallic oxides react with acids to give salts and water, similar to the\nreaction of a base with an acid, metallic oxides are said to be basic oxides.\n\uf032\uf02e\uf031\uf02e\uf036\uf020\uf052\uf065\uf061\uf063\uf074\uf069\uf06f\uf06e\uf020\uf06f\uf066\uf020\uf061\uf020\uf04e\uf06f\uf06e\uf02d\uf06d\uf065\uf074\uf061\uf06c\uf06c\uf069\uf063\uf020\uf04f\uf078\uf069\uf064\uf065\uf020\uf077\uf069\uf074\uf068\uf020\uf042\uf061\uf073\uf065\nYou saw the reaction between carbon dioxide and calcium hydroxide\n(lime water) in Activity 2.5. Calcium hydroxide, which is a base, reacts\nwith carbon dioxide to pr oduce a salt and water.", + "Calcium hydroxide, which is a base, reacts\nwith carbon dioxide to pr oduce a salt and water. Since this is similar to\nthe reaction between a base and an acid, we can conclude that non-metallic oxides are acidic in nature.\n\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. Why should curd and sour substances not be kept in brass and copper\nvessels?\n2. Which gas is usually liberated when an acid reacts with a metal?\nIllustrate with an example. How will you test for the presence of\nthis gas?\n3. Metal compound A reacts with dilute hydrochloric acid to produceeffervescence. The gas evolved extinguishes a bur ning candle. W rite a\nbalanced chemical equation for the reaction if one of the compoundsformed is calcium chloride.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nbalanced chemical equation for the reaction if one of the compoundsformed is calcium chloride.\n\uf032\uf02e\uf032\uf020\uf057\uf048\uf041 \uf032\uf02e\uf032\uf020\uf057\uf048\uf041\uf032\uf02e\uf032\uf020\uf057\uf048\uf041 \uf032\uf02e\uf032\uf020\uf057\uf048\uf041\uf032\uf02e\uf032\uf020\uf057\uf048\uf041 \uf054\uf020\uf044\uf04f\uf020\uf041\uf04c\uf04c\uf020\uf041\uf043\uf049\uf044\uf053\uf020\uf041\uf04e\uf044\uf020\uf041\uf04c\uf04c\uf020\uf042\uf041\uf053\uf045\uf053\uf020\uf048\uf041\uf054\uf020\uf044\uf04f\uf020\uf041\uf04c\uf04c", + "\uf020\uf041\uf04c\uf04c\uf020\uf042\uf041\uf053\uf045\uf053\uf020\uf048\uf041\uf054\uf020\uf044\uf04f\uf020\uf041\uf04c\uf04c\uf020\uf041\uf043\uf049\uf044\uf053\uf020\uf041\uf04e\uf044\uf020\uf041\uf04c\uf04c\uf020\uf042\uf041\uf053\uf045\uf053\uf020\uf048\uf041\uf054\uf020\uf044\uf04f\uf020\uf041\uf04c\uf04c\uf020\uf041\uf043\uf049\uf044\uf053\uf020\uf041\uf04e\uf044\uf020\uf041\uf04c\uf04c\uf020\uf042\uf041\uf053\uf045\uf053\uf020\uf048\uf041\uf054\uf020\uf044\uf04f\uf020\uf041\uf04c\uf04c\uf020\uf041", + "\uf04c\uf04c\uf020\uf042\uf041\uf053\uf045\uf053\uf020\uf048\uf041\uf054\uf020\uf044\uf04f\uf020\uf041\uf04c\uf04c\uf020\uf041\uf043\uf049\uf044\uf053\uf020\uf041\uf04e\uf044\uf020\uf041\uf04c\uf04c\uf020\uf042\uf041\uf053\uf045\uf053\uf020\uf048\uf041\uf054\uf020\uf044\uf04f\uf020\uf041\uf04c\uf04c\uf020\uf041\uf043\uf049\uf044\uf053\uf020\uf041\uf04e\uf044\uf020\uf041\uf04c\uf04c\uf020\uf042\uf041\uf053\uf045\uf053\uf020\uf048\uf041 \uf056\uf045\uf020\uf049\uf04e", + "\uf041\uf04e\uf044\uf020\uf041\uf04c\uf04c\uf020\uf042\uf041\uf053\uf045\uf053\uf020\uf048\uf041 \uf056\uf045\uf020\uf049\uf04e \uf056\uf045\uf020\uf049\uf04e\uf056\uf045\uf020\uf049\uf04e \uf056\uf045\uf020\uf049\uf04e\uf056\uf045\uf020\uf049\uf04e\n\uf043\uf04f\uf04d\uf04d\uf04f\uf04e\uf03f\uf043\uf04f\uf04d\uf04d\uf04f\uf04e\uf03f\uf043\uf04f\uf04d\uf04d\uf04f\uf04e\uf03f\uf043\uf04f\uf04d\uf04d\uf04f\uf04e\uf03f\uf043\uf04f\uf04d\uf04d\uf04f\uf04e\uf03f\nIn Section 2.1 we have seen that all acids have similar chemical\nproperties.", + "Section 2.1 we have seen that all acids have similar chemical\nproperties. What leads to this similarity in pr operties? We saw in Activity\n2.3 that all acids generate hydrogen gas on reacting with metals, sohydrogen seems to be common to all acids. Let us perform an Activity to\ninvestigate whether all compounds containing hydrogen are acidic.", + "What leads to this similarity in pr operties? We saw in Activity\n2.3 that all acids generate hydrogen gas on reacting with metals, sohydrogen seems to be common to all acids. Let us perform an Activity to\ninvestigate whether all compounds containing hydrogen are acidic.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf038\n\uf06eTake solutions of glucose, alcohol,\nhydrochloric acid, sulphuric acid, etc.", + "\uf06eFix two nails on a cork, and place the cork ina 100 mL beaker .\n\uf06eConnect the nails to the two terminals of a6 volt battery through a bulb and a switch, as", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nhydrochloric acid, sulphuric acid, etc.\n\uf06eFix two nails on a cork, and place the cork ina 100 mL beaker .\n\uf06eConnect the nails to the two terminals of a6 volt battery through a bulb and a switch, as\nshown in Fig. 2.3.\n\uf06eNow pour some dilute HCl in the beaker andswitch on the current.\n\uf06eRepeat with dilute sulphuric acid.\n\uf06eWhat do you observe?\n\uf06eRepeat the experiment separately withglucose and alcohol solutions. What do you\nobserve now?\n\uf06eDoes the bulb glow in all cases?\nFigure 2.3Figure 2.3Figure 2.3Figure 2.3Figure 2.3\nAcid solution in waterconducts electricity\n2019-20\nAcids, Bases and Salts \uf032\uf033The bulb will start glowing in the case of acids, as shown in Fig. 2.3.\nBut you will observe that glucose and alcohol solutions do not conduct\nelectricity.", + "2.3.\nBut you will observe that glucose and alcohol solutions do not conduct\nelectricity. Glowing of the bulb indicates that there is a flow of electric\ncurrent through the solution. The electric current is carried through the\nacidic solution by ions.\nAcids contain H+ ion as cation and anion such as Cl\u2013 in HCl, NO3\u2013 in\nHNO3, SO2\u2013", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\ncurrent through the solution. The electric current is carried through the\nacidic solution by ions.\nAcids contain H+ ion as cation and anion such as Cl\u2013 in HCl, NO3\u2013 in\nHNO3, SO2\u2013\n4 in H2SO4, CH3COO\u2013 in CH3COOH. Since the cation present in\nacids is H+, this suggests that acids produce hydrogen ions, H+(aq), in\nsolution, which are responsible for their acidic properties.\nRepeat the same Activity using alkalis such as sodium hydroxide, calcium\nhydroxide, etc. What can you conclude from the results of this Activity?", + "Repeat the same Activity using alkalis such as sodium hydroxide, calcium\nhydroxide, etc. What can you conclude from the results of this Activity?\n\uf032\uf02e\uf032\uf02e\uf031\uf020\uf057\uf068\uf061\uf074\uf020\uf048\uf061\uf070\uf070\uf065\uf06e\uf073\uf020\uf074\uf06f\uf020\uf061\uf06e\uf020\uf041\uf063\uf069\uf064\uf020\uf06f\uf072\uf020\uf061\uf020\uf042\uf061\uf073\uf065\uf020\uf069\uf06e\uf020\uf061\uf020\uf057\uf061\uf074\uf065\uf072\uf020\uf053\uf06f\uf06c\uf075\uf074\uf069\uf06f\uf06e\uf03f\nDo acids produce ions only in aqueous solution? Let us test this.", + "Let us test this.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf039\n\uf06eTake about 1g solid NaCl in a clean and\ndry test tube and set up the apparatus as\nshown in Fig. 2.4.\n\uf06eAdd some concentrated sulphuric acid tothe test tube.\n\uf06eWhat do you observe? Is there a gas comingout of the delivery tube?", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\ndry test tube and set up the apparatus as\nshown in Fig. 2.4.\n\uf06eAdd some concentrated sulphuric acid tothe test tube.\n\uf06eWhat do you observe? Is there a gas comingout of the delivery tube?\n\uf06eTest the gas evolved successively with dryand wet blue litmus paper.\n\uf06eIn which case does the litmus paper changecolour?\n\uf06eOn the basis of the above Activity, what doyou infer about the acidic character of:\n(i)dry HCl gas\n(ii) HCl solution?\nFigure 2.4 Figure 2.4 Figure 2.4 Figure 2.4 Figure 2.4 Preparation of HCl gas\nThis experiment suggests that hydrogen ions in HCl are produced\nin the pr esence of water. The separation of H+ ion from HCl molecules\ncannot occur in the absence of water .\nHCl + H2O \u2192 H3O+ + Cl\u2013\nHydrogen ions cannot exist alone, but they exist after combining\nwith water molecules.", + "The separation of H+ ion from HCl molecules\ncannot occur in the absence of water .\nHCl + H2O \u2192 H3O+ + Cl\u2013\nHydrogen ions cannot exist alone, but they exist after combining\nwith water molecules. Thus hydrogen ions must always be shown as\nH+(aq) or hydronium ion (H3O+).\nH+ + H2O \u2192 H3O+\nWe have seen that acids give H3O+ or H+(aq) ion in water . Let us see\nwhat happens when a base is dissolved in water.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nH+(aq) or hydronium ion (H3O+).\nH+ + H2O \u2192 H3O+\nWe have seen that acids give H3O+ or H+(aq) ion in water . Let us see\nwhat happens when a base is dissolved in water.\nNaOH(s)HO2\uf8e7\u2192\uf8e7 \uf8e7 \uf8e7Na+ (aq) + OH\u2013 (aq)Note to teachers: If the climate is very humid, you will have to pass the gas produced\nthrough a guard tube (drying tube) containing calcium chloride to dry the gas.\n2019-20\nScience \uf032\uf034KOH(s) K (aq)+ OH (aq)HO2 +\uf8e7 \u2192\uf8e7 \uf8e7 \uf8e7\u2013\nMg(OH)2(s)HO2\uf8e7 \u2192\uf8e7 \uf8e7 \uf8e7Mg2+(aq)+2OH\u2013(aq)\nBases generate hydr oxide (OH\u2013) ions in water.", + "Bases which are soluble\nin water are called alkalis.\uf044\uf06f\uf020\uf059\uf06f\uf075\n\uf04b\uf06e\uf06f\uf077\uf03fAll bases do not dissolve in water. An alkali is a base that dissolves in water. They\nare soapy to touch, bitter and corrosive. Never taste or touch them as they may\ncause harm. Which of the bases in the Table 2.1 are alkalis?\nNow as we have identified that all acids generate H+(aq) and all\nbases generate OH\u2013(aq), we can view the neutralisation reaction as\nfollows \u2013\nAcid + Base \u2192 Salt + W ater\nH X + M OH \u2192 MX + HOH", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nNow as we have identified that all acids generate H+(aq) and all\nbases generate OH\u2013(aq), we can view the neutralisation reaction as\nfollows \u2013\nAcid + Base \u2192 Salt + W ater\nH X + M OH \u2192 MX + HOH\nH+(aq) + OH\u2013 (aq) \u2192 H2O(l)\nLet us see what is involved when water is mixed with an acid or a base.", + "\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf030\n\uf06eTake 10 mL water in a beaker .\n\uf06eAdd a few drops of concentrated H2SO4 to it and swirl the\nbeaker slowly.\n\uf06eTouch the base of the beaker.\n\uf06eIs there a change in temperature?", + "\uf06eAdd a few drops of concentrated H2SO4 to it and swirl the\nbeaker slowly.\n\uf06eTouch the base of the beaker.\n\uf06eIs there a change in temperature?\n\uf06eIs this an exothermic or endothermic process?\n\uf06eRepeat the above Activity with sodium hydroxide pellets\nand record your observations.Figure 2.5Figure 2.5Figure 2.5Figure 2.5Figure 2.5\nWarning sign displayed\non containers containingconcentrated acids andbases\nThe process of dissolving an acid or a base in water is a highly\nexothermic one. Care must be taken while mixing concentrated nitric\nacid or sulphuric acid with water . The acid must always be added slowly", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nexothermic one. Care must be taken while mixing concentrated nitric\nacid or sulphuric acid with water . The acid must always be added slowly\nto water with constant stirring. If water is added to a concentrated acid,the heat generated may cause the mixture to splash out and cause burns.The glass container may also break due to excessive local heating. Lookout for the warning sign (shown in Fig. 2.5) on the can of concentratedsulphuric acid and on the bottle of sodium hydroxide pellets.\nMixing an acid or base with water results in decrease in the\nconcentration of ions (H\n3O+/OH\u2013) per unit volume. Such a process is\ncalled dilution and the acid or the base is said to be diluted.\n2019-20\nAcids, Bases and Salts \uf032\uf035\uf032\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf053\uf054\uf052\uf04f\uf04e\uf047\uf020\uf041\uf052\uf045", + "and Salts \uf032\uf035\uf032\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf053\uf054\uf052\uf04f\uf04e\uf047\uf020\uf041\uf052\uf045\uf020\uf041\uf043\uf049\uf044\uf020\uf04f\uf052\uf020\uf042\uf041\uf053\uf045\uf020\uf053\uf04f\uf04c\uf055\uf054\uf049\uf04f\uf04e\uf053\uf03f\uf032\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf053\uf054\uf052\uf04f\uf04e\uf047\uf020\uf041\uf052\uf045\uf020\uf041\uf043\uf049\uf044\uf020\uf04f\uf052\uf020\uf042\uf041\uf053\uf045\uf020\uf053\uf04f\uf04c\uf055\uf054\uf049\uf04f\uf04e", + "\uf041\uf043\uf049\uf044\uf020\uf04f\uf052\uf020\uf042\uf041\uf053\uf045\uf020\uf053\uf04f\uf04c\uf055\uf054\uf049\uf04f\uf04e\uf053\uf03f\uf032\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf053\uf054\uf052\uf04f\uf04e\uf047\uf020\uf041\uf052\uf045\uf020\uf041\uf043\uf049\uf044\uf020\uf04f\uf052\uf020\uf042\uf041\uf053\uf045\uf020\uf053\uf04f\uf04c\uf055\uf054\uf049\uf04f\uf04e\uf053\uf03f\uf032\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf053\uf054\uf052\uf04f\uf04e\uf047\uf020\uf041\uf052\uf045\uf020\uf041", + "\uf03f\uf032\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf053\uf054\uf052\uf04f\uf04e\uf047\uf020\uf041\uf052\uf045\uf020\uf041\uf043\uf049\uf044\uf020\uf04f\uf052\uf020\uf042\uf041\uf053\uf045\uf020\uf053\uf04f\uf04c\uf055\uf054\uf049\uf04f\uf04e\uf053\uf03f\uf032\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf053\uf054\uf052\uf04f\uf04e\uf047\uf020\uf041\uf052\uf045\uf020\uf041\uf043\uf049\uf044\uf020\uf04f\uf052\uf020\uf042\uf041\uf053\uf045\uf020\uf053\uf04f\uf04c\uf055\uf054\uf049\uf04f\uf04e\uf053\uf03f", + "\uf049\uf044\uf020\uf04f\uf052\uf020\uf042\uf041\uf053\uf045\uf020\uf053\uf04f\uf04c\uf055\uf054\uf049\uf04f\uf04e\uf053\uf03f\nWe know how acid-base indicators can be used to distinguish between", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nWe know how acid-base indicators can be used to distinguish between\nan acid and a base. We have also lear nt in the previous section about\ndilution and decrease in concentration of H+ or OH\u2013 ions in solutions.\nCan we quantitatively find the amount of these ions present in a solution?Can we judge how strong a given acid or base is?\nWe can do this by making use of a universal indicator, which is a\nmixture of several indicators. The universal indicator shows differentcolours at different concentrations of hydrogen ions in a solution.\nA scale for measuring hydrogen ion concentration in a solution, called\npH scale has been developed. The p in pH stands for \u2018 potenz \u2019 in German,\nmeaning power . On the pH scale we can measure pH generally fr om\n0 (very acidic) to 14 (very alkaline). pH should be thought of simply as anumber which indicates the acidic or basic nature of a solution. Higherthe hydronium ion concentration, lower is the pH value.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nThe pH of a neutral solution is 7. Values less than 7 on the pH scale\nrepresent an acidic solution. As the pH value increases from 7 to 14, it\nrepresents an increase in OH\n\u2013 ion concentration in the solution, that is,\nincrease in the strength of alkali (Fig. 2.6). Generally paper impregnated\nwith the universal indicator is used for measuring pH.\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. Why do HCl, HNO3, etc., show acidic characters in aqueous solutions\nwhile solutions of compounds like alcohol and glucose do not show acidic\ncharacter?\n2. Why does an aqueous solution of an acid conduct electricity?\n3. Why does dry HCl gas not change the colour of the dry litmus paper?\n4. While diluting an acid, why is it recommended that the acid should be\nadded to water and not water to the acid?\n5.", + "Why does an aqueous solution of an acid conduct electricity?\n3. Why does dry HCl gas not change the colour of the dry litmus paper?\n4. While diluting an acid, why is it recommended that the acid should be\nadded to water and not water to the acid?\n5. How is the concentration of hydronium ions (H3O+) affected when a\nsolution of an acid is diluted?\n6. How is the concentration of hydroxide ions (O H\u2013) affected when excess\nbase is dissolved in a solution of sodium hydroxide?", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nsolution of an acid is diluted?\n6. How is the concentration of hydroxide ions (O H\u2013) affected when excess\nbase is dissolved in a solution of sodium hydroxide?\nFigure 2.6Figure 2.6Figure 2.6Figure 2.6Figure 2.", + "6. How is the concentration of hydroxide ions (O H\u2013) affected when excess\nbase is dissolved in a solution of sodium hydroxide?\nFigure 2.6Figure 2.6Figure 2.6Figure 2.6Figure 2.6 Variation of pH with the change in concentration of H+(aq) and OH\u2013(aq) ions\n2019-20\nScience \uf032\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf031\n\uf06eTest the pH values\nof solutions given in\nTable 2.", + "2.\n\uf06eRecord your observations.\n\uf06eWhat is the nature of each\nsubstance on the basis of\nyour observations?\nFigure 2.7 Figure 2.7 Figure 2.7 Figure 2.7 Figure 2.7 pH of some common substances shown on a pH paper (colours are only a rough guide)\nThe strength of acids and bases depends on the number of H+ ions\nand OH\u2013 ions produced, respectively. If we take hydrochloric acid and\nacetic acid of the same concentration, say one molar , then these pr oduce\ndifferent amounts of hydrogen ions. Acids that give rise to more H+ ions", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nacetic acid of the same concentration, say one molar , then these pr oduce\ndifferent amounts of hydrogen ions. Acids that give rise to more H+ ions\nare said to be strong acids, and acids that give less H+ ions are said to be\nweak acids. Can you now say what weak and strong bases are?\n\uf032\uf02e\uf033\uf02e\uf031\uf020\uf049\uf06d\uf070\uf06f\uf072\uf032\uf02e\uf033\uf02e\uf031\uf020\uf049\uf06d\uf070\uf06f\uf072\uf032\uf02e\uf033\uf02e\uf031\uf020\uf049\uf06d\uf070\uf06f\uf072\uf032\uf02e\uf033\uf02e\uf031\uf020\uf049\uf06d\uf070\uf06f\uf072\uf032\uf02e\uf033\uf02e\uf031\uf020\uf049\uf06d\uf070\uf06f\uf072", + "\uf074\uf061\uf06e\uf063\uf065\uf020\uf06f\uf066\uf020\uf070\uf048\uf020\uf069\uf06e\uf020\uf045\uf076\uf065\uf072\uf074\uf061\uf06e\uf063\uf065\uf020\uf06f\uf066\uf020\uf070\uf048\uf020\uf069\uf06e\uf020\uf045\uf076\uf065\uf072\uf074\uf061\uf06e\uf063\uf065\uf020\uf06f\uf066\uf020\uf070\uf048\uf020\uf069\uf06e\uf020\uf045\uf076\uf065\uf072\uf074\uf061\uf06e\uf063\uf065\uf020\uf06f\uf066\uf020\uf070\uf048\uf020\uf069\uf06e\uf020\uf045\uf076\uf065\uf072\uf074\uf061\uf06e\uf063\uf065\uf020\uf06f\uf066\uf020", + "\uf066\uf020\uf070\uf048\uf020\uf069\uf06e\uf020\uf045\uf076\uf065\uf072\uf074\uf061\uf06e\uf063\uf065\uf020\uf06f\uf066\uf020\uf070\uf048\uf020\uf069\uf06e\uf020\uf045\uf076\uf065\uf072 \uf079\uf064\uf061\uf079\uf020\uf04c\uf069\uf066\uf065 \uf079\uf064\uf061\uf079\uf020\uf04c\uf069\uf066\uf065\uf079\uf064\uf061\uf079\uf020\uf04c\uf069\uf066\uf065 \uf079\uf064\uf061\uf079\uf020\uf04c\uf069\uf066\uf065\uf079\uf064\uf061\uf079\uf020\uf04c\uf069\uf066\uf065\nAre plants and animals pH sensitive?\nOur body works within the pH range of 7.0 to 7.8.", + "plants and animals pH sensitive?\nOur body works within the pH range of 7.0 to 7.8. Living organisms cansurvive only in a narrow range of pH change. When pH of rain water is\nless than 5.6, it is called acid rain. When acid rain flows into the rivers, it\nlowers the pH of the river water. The survival of aquatic life in such riversbecomes difficult.Table 2.2\n S. Solution Colour of Approx- Nature of\nNo. pH paper -imate substance\npH value\n1Saliva (before meal)\n2Saliva (after meal)\n3Lemon juice", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nS. Solution Colour of Approx- Nature of\nNo. pH paper -imate substance\npH value\n1Saliva (before meal)\n2Saliva (after meal)\n3Lemon juice\n4Colourless aerated\ndrink\n5Carrot juice\n6Coffee\n7Tomato juice\n8Tap water\n91M NaOH\n10 1M HCl\n2019-20\nAcids, Bases and Salts \uf032\uf037What is the pH of the soil in your backyard?\nPlants require a specific pH range for their healthy growth. To find out\nthe pH required for the healthy growth of a plant, you can collect the soilfrom various places and check the pH in the manner described below inActivity 2.12. Also, you can note down which plants are growing in theregion from which you have collected the soil.Acids in other planets\nThe atmosphere of venus is made up of thick white and yellowish clouds of\nsulphuric acid. Do you think life can exist on this planet?", + "Also, you can note down which plants are growing in theregion from which you have collected the soil.Acids in other planets\nThe atmosphere of venus is made up of thick white and yellowish clouds of\nsulphuric acid. Do you think life can exist on this planet?\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf032\n\uf06ePut about 2 g soil in a test tube and add 5 mL water to it.", + "\uf06eShake the contents of the test tube.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf032\n\uf06ePut about 2 g soil in a test tube and add 5 mL water to it.\n\uf06eShake the contents of the test tube.", + "\uf06eShake the contents of the test tube.\n\uf06eFilter the contents and collect the filtrate in a test tube.\n\uf06eCheck the pH of this filtrate with the help of universal\nindicator paper .\n\uf06eWhat can you conclude about the ideal soil pH for the growth ofplants in your region?\npH in our digestive system\nIt is very interesting to note that our stomach produces hydrochloric\nacid. It helps in the digestion of food without harming the stomach.During indigestion the stomach produces too much acid and this causes\npain and irritation. To get rid of this pain, people use bases called\nantacids. One such remedy must have been suggested by you at the\nbeginning of this Chapter. These antacids neutralise the excess acid.\nMagnesium hydroxide (Milk of magnesia), a mild base, is often used for\nthis purpose.\npH change as the cause of tooth decay", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nbeginning of this Chapter. These antacids neutralise the excess acid.\nMagnesium hydroxide (Milk of magnesia), a mild base, is often used for\nthis purpose.\npH change as the cause of tooth decay\nTooth decay starts when the pH of the mouth is lower than 5.5. Tooth\nenamel, made up of calcium hydroxyapatite (a crystalline form of calcium\nphosphate) is the hardest substance in the body. It does not dissolve in\nwater, but is corroded when the pH in the mouth is below 5.5. Bacteria\npresent in the mouth produce acids by degradation of sugar and food\nparticles remaining in the mouth after eating. The best way to prevent\nthis is to clean the mouth after eating food. Using toothpastes, which are\ngenerally basic, for cleaning the teeth can neutralise the excess acid and\nprevent tooth decay.\nSelf defence by animals and plants through chemical warfare\nHave you ever been stung by a honey-bee? Bee-sting leaves an acid", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nprevent tooth decay.\nSelf defence by animals and plants through chemical warfare\nHave you ever been stung by a honey-bee? Bee-sting leaves an acid\nwhich causes pain and irritation. Use of a mild base like baking sodaon the stung area gives relief. Stinging hair of nettle leaves inject\nmethanoic acid causing burning pain.\uf044\uf06f\uf020\uf059\uf06f\uf075\n\uf04b\uf06e\uf06f\uf077\uf03f\n2019-20\nScience \uf032\uf038\uf032\uf02e\uf034\uf020\uf04d\uf04f\uf052\uf045\uf020\uf041\uf042\uf04f\uf055\uf054\uf020\uf053\uf041\uf04c\uf032\uf02e\uf034\uf020\uf04d\uf04f\uf052\uf045\uf020\uf041\uf042\uf04f\uf055\uf054\uf020\uf053\uf041\uf04c\uf032\uf02e", + "\uf04c\uf032\uf02e\uf034\uf020\uf04d\uf04f\uf052\uf045\uf020\uf041\uf042\uf04f\uf055\uf054\uf020\uf053\uf041\uf04c\uf032\uf02e\uf034\uf020\uf04d\uf04f\uf052\uf045\uf020\uf041\uf042\uf04f\uf055\uf054\uf020\uf053\uf041\uf04c\uf032\uf02e\uf034\uf020\uf04d\uf04f\uf052\uf045\uf020\uf041\uf042\uf04f\uf055\uf054\uf020\uf053\uf041\uf04c\uf032\uf02e\uf034\uf020\uf04d\uf04f\uf052\uf045\uf020\uf041\uf042\uf04f\uf055\uf054\uf020\uf053\uf041\uf04c \uf054\uf053 \uf054\uf053\uf054\uf053 \uf054\uf053\uf054\uf053\nIn the previous sections we have", + "\uf041\uf042\uf04f\uf055\uf054\uf020\uf053\uf041\uf04c \uf054\uf053 \uf054\uf053\uf054\uf053 \uf054\uf053\uf054\uf053\nIn the previous sections we have seen the formation of salts during\nvarious reactions. Let us understand more about their preparation,\nproperties and uses.\n\uf032\uf02e\uf034\uf02e\uf031\uf020\uf046\uf061\uf06d\uf069\uf06c\uf079\uf020\uf06f\uf066\uf020\uf053\uf061\uf06c\uf074\uf073\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079", + "\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf033\n\uf06eWrite the chemical formulae of the salts given below.\nPotassium sulphate, sodium sulphate, calcium sulphate,\nmagnesium sulphate, copper sulphate, sodium chloride, sodium\nnitrate, sodium carbonate and ammonium chloride.Nature provides neutralisation options", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nmagnesium sulphate, copper sulphate, sodium chloride, sodium\nnitrate, sodium carbonate and ammonium chloride.Nature provides neutralisation options\nNettle is a herbaceous plant which grows in the wild. Its leaves have stinging hair,\nwhich cause painful stings when touched accidentally. This is due to the methanoic\nacid secreted by them. A traditional remedy is rubbing the\narea with the leaf of the dock plant, which often grows beside\nthe nettle in the wild. Can you guess the nature of the dock\nplant? So next time you know what to look out for if you\naccidentally touch a nettle plant while trekking. Are you aware\nof any other effective traditional remedies for such stings?", + "Can you guess the nature of the dock\nplant? So next time you know what to look out for if you\naccidentally touch a nettle plant while trekking. Are you aware\nof any other effective traditional remedies for such stings?\nTable 2.3 Some naturally occurring acids\nNatural source Acid Natural source Acid\nVinegar Acetic acid Sour milk (Curd) Lactic acid\nOrange Citric acid Lemon Citric acid\nTamarind Tartaric acid Ant sting Methanoic acid\nTomato Oxalic acid Nettle sting Methanoic acid\n\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nVinegar Acetic acid Sour milk (Curd) Lactic acid\nOrange Citric acid Lemon Citric acid\nTamarind Tartaric acid Ant sting Methanoic acid\nTomato Oxalic acid Nettle sting Methanoic acid\n\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n1. You have two solutions, A and B. The pH of solution A is 6 and pH of\nsolution B is 8. Which solution has more hydrogen ion concentration?\nWhich of this is acidic and which one is basic?\n2. What effect does the concentration of H+(aq) ions have on the nature of the\nsolution?\n3. Do basic solutions also have H+(aq) ions? If yes, then why are these basic?\n4.", + "Which of this is acidic and which one is basic?\n2. What effect does the concentration of H+(aq) ions have on the nature of the\nsolution?\n3. Do basic solutions also have H+(aq) ions? If yes, then why are these basic?\n4. Under what soil condition do you think a farmer would treat the soil of hisfields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or\nchalk (calcium carbonate)?\uf03f\uf044\uf06f\uf020\uf059\uf06f\uf075\uf020\uf04b\uf06e\uf06f\uf077\uf03f\n2019-20\nAcids, Bases and Salts \uf032\uf039\uf06eIdentify the acids and bases from which the above salts may be\nobtained.\n\uf06eSalts having the same positive or negative radicals are said tobelong to a family. For example, NaCl and Na\n2SO4 belong to the", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nobtained.\n\uf06eSalts having the same positive or negative radicals are said tobelong to a family. For example, NaCl and Na\n2SO4 belong to the\nfamily of sodium salts. Similarly, NaCl and KCl belong to the familyof chloride salts. How many families can you identify among the\nsalts given in this Activity?", + "For example, NaCl and Na\n2SO4 belong to the\nfamily of sodium salts. Similarly, NaCl and KCl belong to the familyof chloride salts. How many families can you identify among the\nsalts given in this Activity?\n\uf032\uf02e\uf034\uf02e\uf032\uf020\uf070\uf048\uf020\uf06f\uf066\uf020\uf053\uf061\uf06c\uf074\uf073\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf034\n\uf06eCollect", + "the following salt samples \u2013 sodium chloride, potassium\nnitrate, aluminium chloride, zinc sulphate, copper sulphate,\nsodium acetate, sodium carbonate and sodium hydrogencarbonate\n(some other salts available can also be taken).\n\uf06eCheck their solubility in water (use distilled water only).\n\uf06eCheck the action of these solutions on litmus and find the pH\nusing a pH paper.\n\uf06eWhich of the salts are acidic, basic or neutral?\n\uf06eIdentify the acid or base used to form the salt.\n\uf06eReport your observations in Table 2.4.\nSalts of a strong acid and a strong base\nare neutral with pH value of 7. On the other", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n\uf06eIdentify the acid or base used to form the salt.\n\uf06eReport your observations in Table 2.4.\nSalts of a strong acid and a strong base\nare neutral with pH value of 7. On the other\nhand, salts of a strong acid and weak base\nare acidic with pH value less than 7 and those\nof a strong base and weak acid are basic in\nnature, with pH value more than 7.\n\uf032\uf02e\uf034\uf02e\uf033\uf020\uf043\uf068\uf065\uf06d\uf069\uf063\uf061\uf06c\uf073\uf020\uf066\uf072\uf06f\uf06d\uf020\uf043\uf06f\uf06d\uf06d\uf06f\uf06e\uf020\uf053\uf061\uf06c\uf074\nBy now you have learnt that the salt formedby the combination of hydrochloric acid and\nsodium hydroxide solution is called sodium\nchloride. This is the salt that you use in food.", + "This is the salt that you use in food.\nYou must have observed in the above Activity\nthat it is a neutral salt.\nSeawater contains many salts dissolved\nin it. Sodium chloride is separated from thesesalts. Deposits of solid salt are also found in\nseveral parts of the world. These large crystals\nare often brown due to impurities. This is\ncalled rock salt. Beds of rock salt were formed\nwhen seas of bygone ages dried up. Rock salt\nis mined like coal.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nare often brown due to impurities. This is\ncalled rock salt. Beds of rock salt were formed\nwhen seas of bygone ages dried up. Rock salt\nis mined like coal.\nYou must have heard about Mahatma Gandhi\u2019s Dandi March. Did\nyou know that sodium chloride was such an important symbol in our\nstruggle for freedom?Table 2.4\nSalt pH Acid used Base used\n2019-20\nScience \uf033\uf030Common salt \u2014 A raw material for chemicals\nThe common salt thus obtained is an important raw material for various\nmaterials of daily use, such as sodium hydroxide, baking soda, washing\nsoda, bleaching powder and many more. Let us see how one substance\nis used for making all these different substances.\nSodium hydroxide\nWhen electricity is passed through an aqueous solution of sodiumchloride (called brine), it decomposes to form sodium hydroxide. The\nprocess is called the chlor -alkali process because of the products formed\u2013\nchlor for chlorine and alkali for sodium hydroxide.", + "The\nprocess is called the chlor -alkali process because of the products formed\u2013\nchlor for chlorine and alkali for sodium hydroxide.\n2NaCl(aq) + 2H\n2O(l) \u2192 2NaOH(aq) + Cl2(g) + H2(g)", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nprocess is called the chlor -alkali process because of the products formed\u2013\nchlor for chlorine and alkali for sodium hydroxide.\n2NaCl(aq) + 2H\n2O(l) \u2192 2NaOH(aq) + Cl2(g) + H2(g)\nChlorine gas is given off at the anode, and hydrogen gas at the cathode.\nSodium hydroxide solution is formed near the cathode. The three\nproducts produced in this process are all useful. Figure 2.8 shows the\ndifferent uses of these products.\nBleaching powderYou have alr eady come to know that chlorine is produced during the\nelectrolysis of aqueous sodium chloride (brine). This chlorine gas is usedfor the manufacture of bleaching powder . Bleaching powder is produced\nby the action of chlorine on dry slaked lime [Ca(OH)\n2]. Bleaching powder\nis represented as CaOCl2, though the actual composition is quite\ncomplex.", + "This chlorine gas is usedfor the manufacture of bleaching powder . Bleaching powder is produced\nby the action of chlorine on dry slaked lime [Ca(OH)\n2]. Bleaching powder\nis represented as CaOCl2, though the actual composition is quite\ncomplex.\nCa(OH)2 + Cl2 \u2192 CaOCl2 + H2OFigure 2.8Figure 2.8Figure 2.8Figure 2.8Figure 2.8 Important products from the chlor-alkali process\n2019-20\nAcids, Bases and Salts \uf033\uf031Bleaching powder is used \u2013", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\ncomplex.\nCa(OH)2 + Cl2 \u2192 CaOCl2 + H2OFigure 2.8Figure 2.8Figure 2.8Figure 2.8Figure 2.8 Important products from the chlor-alkali process\n2019-20\nAcids, Bases and Salts \uf033\uf031Bleaching powder is used \u2013\n(i)for bleaching cotton and linen in the textile industry, for bleaching\nwood pulp in paper factories and for bleaching washed clothes\nin laundry;\n(ii)as an oxidising agent in many chemical industries; and\n(iii) to make drinking water free from germs.\nBaking soda\nThe baking soda is commonly used in the kitchen for making tasty crispypakoras, etc. Sometimes it is added for faster cooking. The chemicalname of the compound is sodium hydrogencarbonate (NaHCO\n3). It is\nproduced using sodium chloride as one of the raw materials.", + "Sometimes it is added for faster cooking. The chemicalname of the compound is sodium hydrogencarbonate (NaHCO\n3). It is\nproduced using sodium chloride as one of the raw materials.\nNaCl + H O + CO + NH NH Cl + NaHCO2 23 4 3 \u2192\n(Ammonium (Sodium\nchloride) hydrogencarbonate)\nDid you check the pH of sodium hydrogencarbonate in Activity 2.14?", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nNaCl + H O + CO + NH NH Cl + NaHCO2 23 4 3 \u2192\n(Ammonium (Sodium\nchloride) hydrogencarbonate)\nDid you check the pH of sodium hydrogencarbonate in Activity 2.14?\nCan you correlate why it can be used to neutralise an acid? It is a mildnon-corrosive basic salt. The following reaction takes place when it isheated during cooking \u2013\n2NaHCO Na CO + H O + CO3 23 2 2Heat\uf8e7 \u2192\uf8e7 \uf8e7 \uf8e7\n(Sodium (Sodium\nhydrogencarbonate) carbonate)\nSodium hydrogencarbonate has got various uses in the household.\nUses of Baking soda\n(i)For making baking powder, which is a mixture of baking soda(sodium hydrogencarbonate) and a mild edible acid such astartaric acid.", + "Uses of Baking soda\n(i)For making baking powder, which is a mixture of baking soda(sodium hydrogencarbonate) and a mild edible acid such astartaric acid. When baking powder is heated or mixed in water,the following reaction takes place \u2013\nNaHCO\n3 + H+ \u2192 CO2 + H2O + Sodium salt of acid\n (From any acid)\nCarbon dioxide produced during the reaction can cause bread orcake to rise making them soft and spongy.\n(ii)Sodium hydrogencarbonate is also an ingredient in antacids.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n(From any acid)\nCarbon dioxide produced during the reaction can cause bread orcake to rise making them soft and spongy.\n(ii)Sodium hydrogencarbonate is also an ingredient in antacids.\nBeing alkaline, it neutralises excess acid in the stomach and\nprovides relief.\n(iii) It is also used in soda-acid fire extinguishers.\nWashing sodaAnother chemical that can be obtained from sodium chloride isNa\n2CO3.10H2O (washing soda). Y ou have seen above that sodium\ncarbonate can be obtained by heating baking soda; recrystallisation ofsodium carbonate gives washing soda. It is also a basic salt.\nNa CO H O Na CO H O\nSodium\ncarbonate23 22 32 10 10 +\u2192 .\n(\n)\n2019-20\nScience \uf033\uf032\nWhat does 10H2O signify? Does it make Na2CO3 wet? W e will address\nthis question in the next section.\nSodium carbonate and sodium hydrogencarbonate are useful\nchemicals for many industrial processes as well.", + "Does it make Na2CO3 wet? W e will address\nthis question in the next section.\nSodium carbonate and sodium hydrogencarbonate are useful\nchemicals for many industrial processes as well.\nUses of washing soda\n(i)Sodium carbonate (washing soda) is used in glass, soap and\npaper industries.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nchemicals for many industrial processes as well.\nUses of washing soda\n(i)Sodium carbonate (washing soda) is used in glass, soap and\npaper industries.\n(ii)It is used in the manufacture of sodium compounds such as borax.\n(iii) Sodium carbonate can be used as a cleaning agent for domestic\npurposes.\n(iv) It is used for removing per manent hardness of water .\n\uf032\uf02e\uf034\uf02e\uf034\uf020\uf041\uf072\uf065\uf020\uf074\uf068\uf065\uf020\uf043\uf072\uf079\uf073\uf074\uf061\uf06c\uf073\uf020\uf06f\uf066\uf020\uf053\uf061\uf06c\uf074\uf073\uf020\uf072\uf065\uf061\uf06c\uf06c\uf079\uf020\uf044\uf072", + "\uf079\uf03f\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf032\uf02e\uf031\uf035\n\uf06eHeat a few crystals of copper sulphate\nin a dry boiling tube.\n\uf06eWhat is the colour of the coppersulphate after heating?\n\uf06eDo you notice water droplets in theboiling tube?", + "a few crystals of copper sulphate\nin a dry boiling tube.\n\uf06eWhat is the colour of the coppersulphate after heating?\n\uf06eDo you notice water droplets in theboiling tube? Where have these come\nfrom?\n\uf06eAdd 2-3 drops of water on the sampleof copper sulphate obtained after\nheating.\n\uf06eWhat do you observe? Is the bluecolour of copper sulphate restored?\nFigure 2.9Figure 2.9Figure 2.9Figure 2.9Figure 2.9\nRemoving waterof crystallisationCopper sulphate crystals which seem to be dry contain water of", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nFigure 2.9Figure 2.9Figure 2.9Figure 2.9Figure 2.9\nRemoving waterof crystallisationCopper sulphate crystals which seem to be dry contain water of\ncrystallisation. When we heat the crystals, this water is removed and the\nsalt turns white.\nIf you moisten the crystals again with water, you will find that blue\ncolour of the crystals reappears.\nWater of crystallisation is the fixed number of water molecules pr esent\nin one formula unit of a salt. Five water molecules are present in one\nformula unit of copper sulphate. Chemical formula for hydrated copper\nsulphate is Cu SO4. 5H2O. Now you would be able to answer the question\nwhether the molecule of Na2CO3.10H2O is wet.\nOne other salt, which possesses water of crystallisation is gypsum.\nIt has two water molecules as water of cyrstallisation. It has the chemicalformula CaSO\n4.2H2O. Let us look into the use of this salt.", + "One other salt, which possesses water of crystallisation is gypsum.\nIt has two water molecules as water of cyrstallisation. It has the chemicalformula CaSO\n4.2H2O. Let us look into the use of this salt.\nPlaster of ParisOn heating gypsum at 373 K, it loses water molecules and becomes\ncalcium sulphate hemihydrate (", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n4.2H2O. Let us look into the use of this salt.\nPlaster of ParisOn heating gypsum at 373 K, it loses water molecules and becomes\ncalcium sulphate hemihydrate (\n421CaSO . H O2). This is called Plaster of\n2019-20\nAcids, Bases and Salts \uf033\uf033Paris, the substance which doctors use as plaster for supporting\nfractured bones in the right position. Plaster of Paris is a white powder\nand on mixing with water, it changes to gypsum once again giving a\nhard solid mass.\nCaSO .1\n2H O+11\n2H O CaSO .2H O4 2 2 4 2\n(Plaster of Paris ) Gypsum )\u2192\n(\nNote that only half a water molecule is shown to be attached as water\nof crystallisation. How can you get half a water molecule? It is written in\nthis form because two formula units of CaSO4 share one molecule of\nwater.", + "How can you get half a water molecule? It is written in\nthis form because two formula units of CaSO4 share one molecule of\nwater. Plaster of Paris is used for making toys, materials for decorationand for making sur faces smooth. Try to find out why is calcium sulphate\nhemihydrate called \u2018Plaster of Paris\u2019 ?\n1. What is the common name of the compound CaOCl2?", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nhemihydrate called \u2018Plaster of Paris\u2019 ?\n1. What is the common name of the compound CaOCl2?\n2. Name the substance which on treatment with chlorine yields bleaching\npowder .\n3. Name the sodium compound which is used for softening har d water .\n4. What will happen if a solution of sodium hydrocarbonate is heated?\nGive the equation of the reaction involved.\n5. Write an equation to show the r eaction between Plaster of Paris and\nwater .\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f\n\uf057\uf068\uf061\uf074\uf020\uf079 \uf06f\uf075\uf020\uf068\uf061\uf076\uf065\uf020\uf06c\uf065\uf061\uf072\uf06e\uf074\n\uf06e Acid-base indicators are dyes or mixtures of dyes which are used to indicate the\npresence of acids and bases.", + "\uf06e Acidic nature of a substance is due to the formation of H+(aq) ions in solution.\nFormation of OH\u2013(aq) ions in solution is responsible for the basic nature of a\nsubstance.\n\uf06e When an acid reacts with a metal, hydrogen gas is evolved and a corresponding\nsalt is formed.\n\uf06e When a base reacts with a metal, along with the evolution of hydrogen gas a salt is", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\nsubstance.\n\uf06e When an acid reacts with a metal, hydrogen gas is evolved and a corresponding\nsalt is formed.\n\uf06e When a base reacts with a metal, along with the evolution of hydrogen gas a salt is\nformed which has a negative ion composed of the metal and oxygen.\n\uf06e When an acid reacts with a metal carbonate or metal hydrogencarbonate, it gives\nthe corr esponding salt, carbon dioxide gas and water.\n\uf06e Acidic and basic solutions in water conduct electricity because they produce\nhydrogen and hydroxide ions respectively.\n2019-20\nScience \uf033\uf034\uf06e The strength of an acid or an alkali can be tested by using a scale called the pH\nscale (0-14) which gives the measure of hydrogen ion concentration in a solution.\n\uf06e A neutral solution has a pH of exactly 7, while an acidic solution has a pH less\nthan 7 and a basic solution a pH more than 7.\n\uf06e Living beings carry out their metabolic activities within an optimal pH range.", + "\uf06e A neutral solution has a pH of exactly 7, while an acidic solution has a pH less\nthan 7 and a basic solution a pH more than 7.\n\uf06e Living beings carry out their metabolic activities within an optimal pH range.\n\uf06e Mixing concentrated acids or bases with water is a highly exothermic process.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n\uf06e Living beings carry out their metabolic activities within an optimal pH range.\n\uf06e Mixing concentrated acids or bases with water is a highly exothermic process.\n\uf06e Acids and bases neutralise each other to for m corr esponding salts and water .\n\uf06e Water of crystallisation is the fixed number of water molecules pr esent in one for mula\nunit of a salt.\n\uf06e Salts have various uses in everyday life and in industries.\n\uf045\uf058\uf045\uf052\uf043\uf049\uf053\uf045\uf053\n1. A solution turns red litmus blue, its pH is likely to be\n(a)1 (b) 4 (c) 5 (d) 10\n2.A solution reacts with crushed egg-shells to give a gas that turns lime-water milky.\nThe solution contains\n(a)NaCl (b) HCl (c) LiCl (d) KCl\n3.10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a\ngiven solution of HCl.", + "The solution contains\n(a)NaCl (b) HCl (c) LiCl (d) KCl\n3.10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a\ngiven solution of HCl. If we take 20 mL of the same solution of NaOH, the amount\nHCl solution (the same solution as before) required to neutralise it will be\n(a)4 mL (b) 8 mL (c) 12 mL (d) 16 mL", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\ngiven solution of HCl. If we take 20 mL of the same solution of NaOH, the amount\nHCl solution (the same solution as before) required to neutralise it will be\n(a)4 mL (b) 8 mL (c) 12 mL (d) 16 mL\n4.Which one of the following types of medicines is used for treating indigestion?\n(a)Antibiotic\n(b)Analgesic\n(c)Antacid\n(d)Antiseptic\n5.Write word equations and then balanced equations for the reaction taking\nplace when \u2013\n(a)dilute sulphuric acid reacts with zinc granules.\n(b)dilute hydrochloric acid reacts with magnesium ribbon.\n(c)dilute sulphuric acid reacts with aluminium powder.\n(d)dilute hydrochloric acid reacts with iron filings.\n6.Compounds such as alcohols and glucose also contain hydrogen but are not\ncategorised as acids. Describe an Activity to prove it.\n7.Why does distilled water not conduct electricity, whereas rain water does?", + "(d)dilute hydrochloric acid reacts with iron filings.\n6.Compounds such as alcohols and glucose also contain hydrogen but are not\ncategorised as acids. Describe an Activity to prove it.\n7.Why does distilled water not conduct electricity, whereas rain water does?\n2019-20\nAcids, Bases and Salts \uf033\uf0358.Why do acids not show acidic behaviour in the absence of water?", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n7.Why does distilled water not conduct electricity, whereas rain water does?\n2019-20\nAcids, Bases and Salts \uf033\uf0358.Why do acids not show acidic behaviour in the absence of water?\n9.Five solutions A,B,C,D and E when tested with universal indicator showed pH as\n4,1,11,7 and 9, respectively. Which solution is\n(a)neutral?\n(b)strongly alkaline?\n(c)strongly acidic?\n(d)weakly acidic?\n(e)weakly alkaline?\nArrange the pH in increasing order of hydrogen-ion concentration.\n10. Equal lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloricacid (HCl) is added to test tube A, while acetic acid (CH\n3COOH) is added to test\ntube B. Amount and concentration taken for both the acids are same. In which testtube will the fizzing occur more vigorously and why?\n11. Fresh milk has a pH of 6. How do you think the pH will change as it turns intocurd? Explain your answer.", + "In which testtube will the fizzing occur more vigorously and why?\n11. Fresh milk has a pH of 6. How do you think the pH will change as it turns intocurd? Explain your answer.\n12. A milkman adds a very small amount of baking soda to fresh milk.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n11. Fresh milk has a pH of 6. How do you think the pH will change as it turns intocurd? Explain your answer.\n12. A milkman adds a very small amount of baking soda to fresh milk.\n(a)Why does he shift the pH of the fresh milk from 6 to slightly alkaline?\n(b)Why does this milk take a long time to set as curd?\n13. Plaster of Paris should be stored in a moistur e-proof container. Explain why?\n14. What is a neutralisation reaction? Give two examples.\n15. Give two important uses of washing soda and baking soda.\n(I) Prepare your own indicator\n\uf06e Crush beetr oot in a mortar .\n\uf06e Add sufficient water to obtain the extract.\n\uf06e Filter the extract by the procedure learnt by you in earlier classes.\n\uf06e Collect the filtrate to test the substances you may have tasted earlier .\n\uf06e Arrange four test tubes in a test tube stand and label them as A,B,C and D. Pour2 mL each of lemon juice solution, soda-water, vinegar and baking soda solutionin them respectively.", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n\uf06e Arrange four test tubes in a test tube stand and label them as A,B,C and D. Pour2 mL each of lemon juice solution, soda-water, vinegar and baking soda solutionin them respectively.\n\uf06e Put 2-3 drops of the beetroot extract in each test tube and note the colour changeif any. Write your observation in a Table.\n\uf06e You can pr epar e indicators by using other natural materials like extracts of r ed\ncabbage leaves, coloured petals of some flowers such as Petunia , Hydrangea and\nGeranium .\uf047\uf072\uf06f\uf075\uf070\uf020\uf020\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\n2019-20\nScience \uf033\uf036(II) Preparing a soda-acid fire extinguisher\nThe reaction of acids with metal hydrogencarbonates is used in the fire extinguishers\nwhich produce carbon dioxide.", + "\uf06e Take 20 mL of sodium hydrogencarbonate (NaHCO3) solution in a wash-bottle.\n\uf06e Suspend an ignition tube containing dilute sulphuric acid in the wash-bottle\n(Fig. 2.10) .\n\uf06e Close the mouth of the wash-bottle.\n\uf06e Tilt the wash-bottle so that the acid from the ignition tube mixes with the sodium", + "CHAPTER 2-ACIDS,BASES AND SALTS -text book.txt\n(Fig. 2.10) .\n\uf06e Close the mouth of the wash-bottle.\n\uf06e Tilt the wash-bottle so that the acid from the ignition tube mixes with the sodium\nhydrogencarbonate solution below.\n\uf06e You will notice discharge coming out of the nozzle.\n\uf06e Direct this discharge on a burning candle. What happens?\nFigure 2.10Figure 2.10Figure 2.10Figure 2.10Figure 2.10 (a) Ignition tube containing dilute sulphuric acid suspended in a wash-bottle containing\nsodium hydrogencarbonate, (b) Discharge coming out of the nozzle\n2019-20", + "CHAPTER 2-ACIDS,BASES AND SALTS - LONG ANSWER TYPE QUESTIONS.txt\n1. A white powder \u201cX\u201d is used in making breads and cakes. \nIts main ingradient is used as an antacid. Name X. write \nits chemical name and formula. Write the reaction \noccurring when a solution of X is heated \n2. Blue coloured crystals of a substance \u201cX\u201d on heating loses \nits colour on heating , but regains the colour when water is \nadded into it. Explain why? \n3. Define water of crystallization. How many molecules of \nwater of crystallization is present in a formula unit of \nPlaster o f Paris ? Why? What happens when plaster of \nParis is kept uncovered for a long time. Write the equation \nof reaction occurring \n4. Given below are some of the salts. Magnesium carbonate , \npotassium chloride, aluminum sulphate, sodium nitrate. \nIdentify the acid and base from which it is formed. Predict \nthe nature of the above salts \n5.", + "Given below are some of the salts. Magnesium carbonate , \npotassium chloride, aluminum sulphate, sodium nitrate. \nIdentify the acid and base from which it is formed. Predict \nthe nature of the above salts \n5. Why a) HCl is a strong acid where as acetic aacid is a weak \nacid \nb) Acid is always added to water for diluting the acid", + "CHAPTER 2-ACIDS,BASES AND SALTS - LONG ANSWER TYPE QUESTIONS.txt\nthe nature of the above salts \n5. Why a) HCl is a strong acid where as acetic aacid is a weak \nacid \nb) Acid is always added to water for diluting the acid \nc) Glucose solution do not conduct electricity \nd) Carbon dioxide turns lime water milky \ne) onion is an olfactory indicator \n6. Draw a schematic labelled diagram to show the reaction of \ndilute sulphuric acid with zinc granules. Write the balanced \nchemical equation of reaction occurring. How will you test the gas evolved? \n7.i) Write an activity to show that acids do not show the acidic \nbehavior in the absence of water. \nii) Write the products of chlor alkali process along with their 2 \nuses each \n8.i) Write an activity to show that water of crystallization is \npresent in copper sulphate crystals.How many molecules of \nwater of crustallisation is present in copper sulphate crystals \nIi)Define water of crystallization.iii) Name the common name for \nsodium carbonate deca hydrate", + "CHAPTER 2-ACIDS,BASES AND SALTS - LONG ANSWER TYPE QUESTIONS.txt\nwater of crustallisation is present in copper sulphate crystals \nIi)Define water of crystallization.iii) Name the common name for \nsodium carbonate deca hydrate \n9. A white coloured compound \u201cX\u201d used for making decorative \nitems w as kept open in a vessel for few days. Later it was found \nto be a changed to a very hard mass \u201cY\u201d. Identify X and Y. \nWrite the reaction of conversion of X to Y. Can you convert Y \nback to X? How \n10. Why a) Dry acids do not change colour of dry litmus \nb) Exc ess CO 2 makes milkiness of lime water disappear \nc)An equal strength solution of HCl &NaOH do not change \ncolour of pH paper \nd)Which gas is produced when crushed egg shells react with \nhydrochloric acid? \ne) It is not advisable to keep tamarind in metal contai ner", + "CHAPTER 2-ACIDS,BASES AND SALTS -SHORT ANSWER TYPE QUESTIONS.txt\nShort answer questions \n1. What are olfactory indicators? Give examples \n2. How does sodium carbonate react with hydrochloric acid? \n3. Define neutralization reaction. Give example. \n4. How can we conclude metallic oxides are basic in nature. \nJustify your answer \n5. Should we prefer baking soda over tamarind while cleaning \nan old copper vessel? Justify \n6. The formula of glucose is C 6H12O6. It has a lot of hydrogen \natoms as indicated in the formula. Then why does it not act \nas acid? \n7. Why is the formula of Plaster of Paris written as CaSO 4 \u00bd \nH2O? \n8. Will a solution of alcohol in water conducts electricity? If not \nwhy? \n9. What happens when a wet litmus paper is shown to dry HCl \ngas? Will there be any colour change? Why? \n10. Define the term dilution.", + "8. Will a solution of alcohol in water conducts electricity? If not \nwhy? \n9. What happens when a wet litmus paper is shown to dry HCl \ngas? Will there be any colour change? Why? \n10. Define the term dilution. What happens to the \nconcentration of H+ ion when an acid is diluted? \n11. Does rain water conducts electricity? Why? \n12. What are antacids? Give the name of a house hold item", + "CHAPTER 2-ACIDS,BASES AND SALTS -SHORT ANSWER TYPE QUESTIONS.txt\nconcentration of H+ ion when an acid is diluted? \n11. Does rain water conducts electricity? Why? \n12. What are antacids? Give the name of a house hold item \nwhich can be used as antacid \n13. How does tooth decay occur? Explain in terms of pH \nchange \n14. Name the acids present in the following a) ora nges b) \ntamarind c) tomato d) Nettle sting \n15. What is \u201cchlor alkali\u201d process? Write the equation", + "CHAPTER 2-ACIDS,BASES AND SALTS -VERY SHORT ANSWER TYPE QUESTIONS.txt\nVSA \n \n1. Name an olfactory indicator \n2. Define water of crystallization \n3. Can baking soda be used as antacid? Why? \n4. Why is the formula of plaster of Paris written as CaSO 4 \u00bd \nH2O ? \n5. Which sodium compound is used to remove permanent \nhardness of water? \n6. What is the common name of CaSO 4 2 H 2O ? \n7. Name the acid present in tomato \n8. Which is the hardest substance in our body? \n9. What should be the pH for a rain to be labelled as acid rain?", + "CHAPTER 2-ACIDS,BASES AND SALTS -VERY SHORT ANSWER TYPE QUESTIONS.txt\n10. What happens to H+ ion concentration when an acid \nis diluted in water? \n11.Name the raw materials required for the manufacture \nof bleaching powder \n12. Give another important use of sodium compound \nused for softening hard water \n13. Can curd be used as an antacid. Justify your answer \n14. When CaCO 3 reacts with \u201cX\u201d , it gives CaCl 2, water \nand CO 2. Identify \u201cX\u201d \n15. Which calcium compound has powerful smell of \nchlorine and is also used as disinfecting drinking water? \n16. What is the common name given to a mixture of \nNaHCO 3 and an edible acid like tartaric acid? 17. What is the use of universal indicator? \n18. Name an acid base indicator which is easy to use for \na visually impaired person \n19. Name a natural indicator extracted from lichens \n20. Why a solution of glucose cannot change the colour \nof litmus paper?", + "CHAPTER 2-ACIDS,BASES AND SALTS - COMPETENCY BASED QUESTIONS.txt\n1. Acid \u2013 base indicators are dyes or mixtures of dyes which are \nused to indicate the presence of acids and bases. Examples \nare litmus, phenolphthalein, methyl orange etc. these \nindicators tell us whether a substance is acidic or basic by \nchange in colour . There are some substances whose odour \nchanges in acidic or basic media. These are called olfactory \nindicators. Onion is an example of olfactory indicator. Acidic \nnature of a substance is due to the formation of H+ ions in \nsolution whereas formation of OH- ions in solution is \nresponsible for basic nature of a substance. \na) What is the colour of phenolphthalein in acidic medium? ( \npink, yellow, colourless, blue) \nb) Name an olfactory indicator other than onion. \nc) What happens to the concentration of H+ ions when an \nacidic solution is diluted with water?", + "( \npink, yellow, colourless, blue) \nb) Name an olfactory indicator other than onion. \nc) What happens to the concentration of H+ ions when an \nacidic solution is diluted with water? \nd) Can you distinguish between acetic acid and HCl of same \nconcentration using above mentioned indicators? If not, \nwhat should be used instead?", + "CHAPTER 2-ACIDS,BASES AND SALTS - COMPETENCY BASED QUESTIONS.txt\nacidic solution is diluted with water? \nd) Can you distinguish between acetic acid and HCl of same \nconcentration using above mentioned indicators? If not, \nwhat should be used instead? \n2. A reaction in which an acid and a base react to form salt and \nwater is calle d neutralization reaction. Nature provides \nneutralization options. Nettle plant causes painful stings when \ntouched. This is due to methanoic acid secreted by them. A \nremedy is rubbing this area with the leaf of dock plant which \noften grows beside nettle in the wild. \na) Which acid is present in tomato? (lactic acid, methanoic \nacid, oxalic acid, citric acid) \nb) Under what soil conditions does a farmer treat the soil in \nfields with quick lime? \nc) If you mix 10 mL NaOH and 10mL HCl of equal strengths, \nwhat will be the pH of the resultant solution?", + "c) If you mix 10 mL NaOH and 10mL HCl of equal strengths, \nwhat will be the pH of the resultant solution? ( =7, >7, <7, \n\u22487)", + "CHAPTER 2-ACIDS,BASES AND SALTS - COMPETENCY BASED QUESTIONS.txt\n3. Water of crystallization is the fixed no. of water \nmolecules chemically attached to each formula unit of salt in \nits crystalline form. There are many salts with water of \ncrystallization, blue coloured copper sulphate, washing soda, \nPlaster of Paris, gypsum are some of the examples. When we \nheat these crystals, they become anhydrous and lose some \nof its properties. \n \na) Why is formula of Plaster of Paris written as CaSO 4 \u00bd \nH2O? \nb) Blue coloured copper sulphate crystals loses its colour on \nheating to white. Can this be considered a physical or \nchemical change? Justify your answer. \nc) Write an equation to show the reaction between Plaster of \nParis and water. \nd) What is the chemical name of washing soda? 4. The pH value of any solution is a number which simply \nrepresents the acidity and basicity of that solution.", + "c) Write an equation to show the reaction between Plaster of \nParis and water. \nd) What is the chemical name of washing soda? 4. The pH value of any solution is a number which simply \nrepresents the acidity and basicity of that solution. The pH \nvalue of any solution is numerically equal to the logarithm of \nthe inverse of the hydrogen ion (H+) concentration.", + "CHAPTER 2-ACIDS,BASES AND SALTS - COMPETENCY BASED QUESTIONS.txt\na) What will be the pH value for distilled water? \nb) Using above picture can you predict the colour of pH paper \nwhen dipped in lime juice? \nc) Why do we use basic tooth pastes commonly? \n5. Salt, in chemistry , substance produced by the reaction of an \nacid with a base. A salt consists of the positive ion (cat ion) of \na base and the negative ion (anion) of an acid.", + "c) Why do we use basic tooth pastes commonly? \n5. Salt, in chemistry , substance produced by the reaction of an \nacid with a base. A salt consists of the positive ion (cat ion) of \na base and the negative ion (anion) of an acid. The different \ntype of salts are \nType of Salt Type of \nAcid Type of \nBase Example \n Neutral \n pH = 7 Strong \nAcids \n Examples: \n HCl \n H2SO 4 Strong \nBases \n Examples: \n NaOH \n KOH NaCl \n K2SO 4 \nType of Salt Type of \nAcid Type of \nBase Example \n Acidic \n pH < 7 Strong \nAcids \n Examples: \n HCl \n HNO 3 weak Bases \n Examples: \n NH 4OH \n Mg(OH) 2 NH 4Cl \n Mg(NO 3)2 \n Basic \n pH > 7 Weak Acids \n Examples: \n H2CO 3 \n CH 3COOH Strong \nBases \n Examples: \n NaOH \n KOH Na2CO 3 \n \nCH 3COOK", + "CHAPTER 2-ACIDS,BASES AND SALTS - COMPETENCY BASED QUESTIONS.txt\nNH 4OH \n Mg(OH) 2 NH 4Cl \n Mg(NO 3)2 \n Basic \n pH > 7 Weak Acids \n Examples: \n H2CO 3 \n CH 3COOH Strong \nBases \n Examples: \n NaOH \n KOH Na2CO 3 \n \nCH 3COOK \ni) What will be the nature of magnesium sulphate ? \nii) Give the formula of salt formed by weak acid and weak \nbase? \niii) From which acid and base is potassium hydrogen \ncarbonate is formed? \n6. Here are some ways neutralisation is used: \n\uf0b7 Farmers use lime (calcium oxide) to neutralise acid soils. \n\uf0b7 Your stomach contains hydrochloric acid, and too much of this \ncauses indigestion. Antacid tablets contain bases such as \nmagnesium hydroxide and magnesium carbonate to neutralise \nthe extra acid.", + "\uf0b7 Your stomach contains hydrochloric acid, and too much of this \ncauses indigestion. Antacid tablets contain bases such as \nmagnesium hydroxide and magnesium carbonate to neutralise \nthe extra acid. \uf0b7 Bee stings are acidi c. They can be neutralised using baking \npowder, which contains sodium hydrogen carbonate \ni) Define neutralisation reaction \nii) How does tooth paste helps in neutralisation \niii) Name the plant whose leaves can be used to relieve", + "CHAPTER 2-ACIDS,BASES AND SALTS - COMPETENCY BASED QUESTIONS.txt\npowder, which contains sodium hydrogen carbonate \ni) Define neutralisation reaction \nii) How does tooth paste helps in neutralisation \niii) Name the plant whose leaves can be used to relieve \nthepainful stings of leaves of nettle \n7. When dry crystals of green ferrous sulphate is heated in a dry \ntest tube water can be seen on the cooler sides of test tube. \nWhy? \n8. Zinc granules were taken in a bottle and dil HCl was added \ninto it .A balloon covered the mouth of the bottle to collect the \ngas. When the end was tied the balloon started floating up in \nair. Why? Write the reaction occurring during the formation of \ngas. How can we test this gas? \n9. Can plaster of Paris be converted to Gypsum and vice versa? \nExplain how \n10. You are p rovided with lime wa ter in which few dr ops of \nphenolphthalein is added (forms pink solution) and a straw. \nUsing this how can you change the solution back to \ncolourless?", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nDELHI PUBLIC SCHOOL GHAZIABAD \nCLASS \u2013 X \nLESSON PLAN \nChapter 2: Acids, Bases and Salts \n \n Unit Chapter 2: Acids, Bases and Salts", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nClass Transaction Total: 12 periods (approx. 40 min each) \nPre-requisite for \nthe course This lesson requires previous knowledge of - \n\uf0b7 What is an indicator? \n\uf0b7 What are acids and bases? \nAssessment of \nqualifying \nknowledge 1. Written test \n2. Lab activity \n3. Group Discussion \n4. HW notebook \nObjective Given the content (chapter) the learners will be able to \n\uf0b7 Answer questions based on the chemical properties of acids and \nbases with 90% accuracy. \n\uf0b7 Describe the method of preparation, properties and uses of \ncommon salt, bleaching powder, baking soda, washing soda and \nplaster of paris and write chemical equations for the same in their \nnotebook with 95% \nLearning \nOutcomes KNOWLEDGE - Students would be able to \n1. compare chemical properties of acids and bases, write equations and \ngive illustration \n2. analyze and compare pH of given solutions.and find chemical nature of \ngiven solution", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\n1. compare chemical properties of acids and bases, write equations and \ngive illustration \n2. analyze and compare pH of given solutions.and find chemical nature of \ngiven solution \n3. appreciate the importance of pH in daily life and apply neutralization/ pH \nto solve problems in day to day life . \n4. identify strong and weak acids and bases \n5. describe preparation of different salts and su ggest their uses in day to day \nlife \n. \nSKILLS AND COMPETENCIES - Students would be able to", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\n1. Express the chemical properties through equations and illustration \ndevelop critical thinking and collaboration in the process. \n2. Built character amongst themselves by discussing \n/communicating and applying neutralization/ pH to solve \nproblems in day to day life . \n3. Develop the ability to analyze and compare pH of given solutions. \n4. Carry out practical in lab , develop character and collaboration in \nthe process. \n5. Prepare cluster web for the preparation, properties and uses of salts \ncreatively \n \nTransaction \nMethodology Transaction would proceed in the following manner - \n (The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Brain Storming -The class would start with a discussion on what the \nstudents have already learnt in the previous classes and hence what is it that \nthey would learn now. They would also be told the significance of the topic \nthat they would be studying.", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nActive lea rning - Wipro Technique (KWL - What you Know, What you \nwill Learn , What you have learned). . (Critical thinking) \nStudents will be asked how much they know about acids and bases. \nStudents\u2019 responses will be listed on the blackboard by using the KWL \nchart and handouts throughout the lesson. \n \nGroup learning -. (Communication, Collaboration) \nCircle the Sage ; Think, pair share ; Experi mentation \n \nDifferentiation - (Character ) \nInterest centres or interest group . Independent study \n \nQuestioning - Pose, pause, ponce and bounce, Multiple level question - \nTeacher will prepare a list of question s on different topics and will ask \nthese questions during discussions. ( Critical thinking, Communication) \n \nGuided practice followed by Independent Practice - NCERT questions to \nbe discussed in the classroom.", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nGuided practice followed by Independent Practice - NCERT questions to \nbe discussed in the classroom. \n \nStrategies \nThink - Pair and Share, group work - collaborative, tabulation, Wipro - G.O.s(web \nchart, flow chart and compare - contrast matrix) , Experimentation \nDOL 1, 2, 3, 4", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nTechniques to be used: \nQuiz \nDaily Practice Problem \nMCQ \nPeer Assessment \nStudent -teacher interaction, \nsilent/ loud reading, \ncollaborative learning, \nResources Text Book: \nNCERT text book for science \nReference Book \n1. CBSE Exemplar \n2. Sample papers / last years papers \nSelf Study, Home \nWork, \nAssignments Independent Practice : Students would do the questions in their H.W \nnotebooks as mentioned in the monthly planner by which we can assess \ntheir critical thinking and creativity \n HW notebooks to be marked as per the given plan: \nAssessment Parameters : The total marks for the act ivity is 5 marks \nOn time submission\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nHW notebooks to be marked as per the given plan: \nAssessment Parameters : The total marks for the act ivity is 5 marks \nOn time submission\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nPresentation/ Neatness (creativity) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nContent ( critical thinking and \ncreativity) \u2026\u2026\u2026.\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026......3 marks \nIt is also advised that the students come to the class with proper \nbackground knowledge of the topic under discussion. They can refer to the \nresources stated above. \n \nAssessments 3 UNIT TESTS (20, 30, 30)\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.80 Marks \nMID TERM EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u202680 Marks \nHALF YEARLY EXAMINATION \u2026\u2026\u2026\u2026\u2026\u202680 Marks \nPRE BOARD -I EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026..80 Marks \nPRE BOARD -II EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026. 80 Marks \nANNUAL BOARD EXAMINATION \u2026\u2026\u2026\u2026\u2026.80 Marks", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nPERIODIC TEST \nBy administering pen paper test we assess creativity and critical \nthinking. \nAverage of the best two tests to be taken that will have a weightage of 10 \nmarks. \nBest 2 Tests out of: Units tests, Midterm examination, Half Yearly \nexamination. \nINTERNAL ASSESSMENT \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u202620 Marks \nPeriodical Test 10 Marks \n \nBy administering pen paper test we assess creativity and critical \nthinking.", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nBy administering pen paper test we assess creativity and critical \nthinking. \n \nSubject Enrichment 05 Marks \nWith the help of subject enrichment activities we assess \ncommunication ,collaboration, character building and citizenship \nNote Book Submission 05 Marks \nBy writing the answers of three levels of spiral question we can assess \ntheir knowledge (Level 1 questions) critical thinking ( Level 2 questions) , \ncreativity (Level 3 questions) \nNote Book Submission 05 Marks \nAddressing \nClassroom \nDiversity Due to various social backgrounds and multiple intelligences, the \nclassroom might be a diverse arena. The following techniques can be used \nfor various groups: \n \nFor gifted students: \n\uf0b7 High order thinking questions \n\uf0b7 Encouragement for referring other resources \n For weak students: \n\uf0b7 Asset management classes \n\uf0b7 Buddy help to be provided \n\uf0b7 Provide grade -up classes", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nFor differently abled students: \n\uf0b7 Ignore spelling mistakes and formulae, if not written \n\uf0b7 Call parents at regular intervals \n\uf0b7 Provide grade -up classes", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nMarks The weightage would be given by CBSE. \n Following questions will be asked to check students\u2019 understanding \n1. What happens when dilute hydrochloric acid is added to a test tube \ncontaining \na) Zinc granules \nb) NaOH solution \nc) Sodium bi carbonate \n Write chemical equations for the reactions. \n2. Give reasons: \na) dry HCl gas not change the colour of the dry litmus paper. \nb) Plaster of Paris should be stored in a moisture -proof container. \nc) While diluting an acid, why is it recommended that the acid s hould be \nadded to water and not water to the acid? \n3. A chemical compound having smell of chlorine is used to remove \nyellowness of white clothes in laundries. Name the compound and write \nthe chemical equation involved in its preparation. \n4. When electricity is passed through a common salt solution, sodium hydroxide is \nproduced along with the liberation of two gases \u201eX \u201f and \u201eY\u201f. \u201eX\u201f burns with a", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\n4. When electricity is passed through a common salt solution, sodium hydroxide is \nproduced along with the liberation of two gases \u201eX \u201f and \u201eY\u201f. \u201eX\u201f burns with a \npop sound whereas \u201eY \u201f is used for disinfecting drinking water. \n(i) Identify X and Y. \n (ii) Give the chemical equation for the reaction stated above. \n (iii) State the reaction of Y with dry slaked lime. \n5. A metal carbonate X on reacting with an acid gives a gas which when passed \nthrough a solution Y gives the carbonate back. On 1the other hand, a gas G that is \nobtained at anode during electrolysis of brine is passed on dry Y, it gives a \ncompound Z, used for disinfecting drinking water. Identity X, Y, G and Z.", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\n, \n \n \nTopic DAY 1 - Introduction to indicators \nClass Transaction 40 min each) \nPre-requisite for This topic requires b asic knowledge of indicators which students had the course already learnt in class VII \nAssessment of \nqualifying \nknowledge Knowledge testing by asking examples of indicators and their colour \nchanges by teacher in class \nObjective Students will be able to identify different types of indicators and their use s. \n \nLearning \nOutcomes KNOWLEDGE - Students will know and understand \n\uf0b7 Colour change given by some indicators with acids and bases \n \nSKILLS AND COMPETENCIES - Students would be able to", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nLearning \nOutcomes KNOWLEDGE - Students will know and understand \n\uf0b7 Colour change given by some indicators with acids and bases \n \nSKILLS AND COMPETENCIES - Students would be able to \n \n\uf0b7 Critically analyse the colour changes by various indicators for \nidentification of acid and base. \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner - \nAnticipatory Set: 10 min \nWipro Technique (KWL - What you Know, What you will Learn , What \nyou have learned). \n\uf0b7 Students\u2019 responses will be listed on the blackboard by using the \nKWL chart \n \nDOL 1, 2, 3", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nDOL 1, 2, 3 \n \nDiscussion of topic through Collaborative Learning: 15 min \n (Critical Thinking and Problem Solving) \nThink, pair share Students in pairs will name some indicators and \nclassify them as natural and synthetic. \nStudents in group of 5 students each will carry out activity 2.1and 2.2 \nTeacher will discuss with the students and introduce the term Olfactory \nindicators. \nStudents teacher interaction on the topic \nTeacher will ask questions based on activities \n \nGuided practice: 10 min \nStudents will record their observa tion and draw out the conclusion for the \nsame. \nIndependent Practice: NCERT Questions \nClosure: 5 min \nInvolving students in summarizing by asking them to prepare questions \nbased on the topic .", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nResources Text Book: NCERT text book for Science. \nReference Book -CBSE Exemplar \nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, Independent Practice : NCERT Questions Assignments \nAssessments Assignment \n \nTopic DAY 2 - Chemical Properties of acids \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires b asic knowledge of acids which students had already \nlearnt in classes VII \nAssessment of \nqualifying \nknowledge Knowledge testing by asking stud ents to write names of some acids \nObjective Students will be able to explain different chemical properties of acids. \nLearning \nOutcomes \nSKILLS AND COMPETENCIES - Students would be able to \nWrite chemical equations for reactions shown by acids with metals and \nmetal carbonates/ metal carbonates develop critical thinking and \ncollaboration in the process.", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner - \nAnticipatory Set: 5 min \n *Teacher asks questions based on previous day learning \n, \nDOL 1, 2, 3 \nDiscussion of topic th rough Collaborative Learning: 20 min \n (Critical Thinking and Problem Solving and collaboration ) \nStrategies Demo by teacher supported by PPT/ video \ngroup work - collaborative, tabulation \nStudents in group of 5 students will observe activity 2.3 and 2.4 and note \ntheir observations \nThe students will then draw out the conclusion and record it. \nStudents teacher interaction on the topic \nTeacher will ask questions based on activities \nGuided practice: 10 min . Students will record their observation for \nproperties of acids in tabular form. \nIndependent Practice: NCERT Questions \nClosure: 5 min \nSummarization by students with the muddiest point discussed by the", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nproperties of acids in tabular form. \nIndependent Practice: NCERT Questions \nClosure: 5 min \nSummarization by students with the muddiest point discussed by the \nteacher", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nResources Text Book: \nNCERT text book for Science. \nReference Book \nCBSE Exemplar \nClosure Summarization by students with the muddiest point discussed by the \nteacher \n Self Study, Home \nWork, \nAssignments Independent Pr actice: NCERT Questions \nAssessments Peer assessment \n \nTopic Day 3 - Reaction of metal carbonates and metal bicarbonates with \nacids \nClass Transaction 40 min each) \nPre-requisite for \nthe course This topic requires basic knowledge of writing balanced chemical \nequations \nAssessment of \nqualifying \nknowledge Knowledge testing by asking questions based on previous day learning \nObjective Students will be able to explain different chemical properties of acids. \nLearning \nOutcomes SKILLS AND COMPETENCIES - Students would be able to \ndevelop the ability to analyze reaction of metal carbonates and metal \nbicarbonates with acids \nStudents will be able to identify carbon dioxide gas by carrying out a test \nTransaction \nMethodology", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nbicarbonates with acids \nStudents will be able to identify carbon dioxide gas by carrying out a test \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner - \nAnticipatory Set: 5 min \nRecapitulation by asking questions based on previous day\u2019s learning \n, \nDOL 1, 2, 3 \nDiscussion of topic through Collaborative Learning: 20 min \n (Critical Thinking, collaboration and communication) \nStrategies Demo by teacher supported by PPT/ video \nThink - Pair and Share, group work - collaborative, tabulation \nThink, pair share strategy Students in pairs \nwill observe activity 2.5 note their observations \nThe students will then draw out the conclusion and record it. \nStudents teacher interaction on the topic \nTeacher will ask questions based on activities \n.Guided practice: 10 min", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nThe students will then draw out the conclusion and record it. \nStudents teacher interaction on the topic \nTeacher will ask questions based on activities \n.Guided practice: 10 min \nStudents w ill record their observation for properties of acids in tabular \nform. \nClosure: 5 min \nSummarization by students \nIndependent Practice - NCERT Questions \nResources Text Book: \nNCERT text book for Science. \nReference Book \nCBSE Exemplar \nClosure Involving students in summarizing the topic \nSelf Study, Home \nWork, \nAssignments NCERT Questions Assessments Quiz \n \nTopic DAY 4 - Neutralisation reactio n; reaction of non metallic oxide with \nbase ; reaction of metallic oxides with acid", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires b asic knowledge of writing chemical equation s \nAssessment of \nqualifying \nknowledge Knowledge testing by asking students to write chemical equations for \ngiven reaction \nObjective Students will be able to identify acids on the basis of their chemical \nproperties \nLearning \nOutcomes KNOWLEDGE - Students will be able to suggest products of neutralization \nreactions and write equation for the same. \nSKILLS AND COMPETENCIES - Students would b e able to \ndevelop the ability to analyze reaction of non metallic oxide with base and \nmetallic oxides with acid and identify the chemical nature of metallic \nand non metallic oxides..", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner - \nAnticipatory Set: 5 min \nRecapitulation by asking questions based on previous day\u2019s learning \nDOL 1, 2, 3 \nDiscussion of topic through Collaborative Learning: 20 min \n (Critical Thinking , collaboration , character and communication) \nStrategies \n Experimentation group work - collaborative, tabulation, \nStudents will be divided in groups of 5 students each to study the chemical \nproperties of acids. \n\uf0b7 Students will carry out the reactions of acids with base \n\uf0b7 Students will carry out the reactions to find the chemical nature of \nmetallic oxides and non metallic oxides. \n\uf0b7 Students w ill record their observation . \n\uf0b7 Studen ts teacher interaction on the activities \n\uf0b7 Teacher will reinforce their understanding by asking questions \nGuided practice: 10 min", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\n\uf0b7 Students w ill record their observation . \n\uf0b7 Studen ts teacher interaction on the activities \n\uf0b7 Teacher will reinforce their understanding by asking questions \nGuided practice: 10 min \nStudents will record their observation for chemical properties of acids and \nbases in tabular form and write chemical equations \nClosure: 5 min \nSummarization by students \nIndependent Practice - NCERT Questions \nResources Text Book: \nNCERT text book for Science. \nReference Book CBSE Exemplar \nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments NCERT Questions \nAssessments Questioning", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nTopic DAY 5 - Practical Properties of acids and bases. \nClass Transaction 40 min each) \nPre-requisite for \nthe course This topic requires b asic knowledge of chemical properties of acids and \nbases \nAssessment of \nqualifying \nknowledge Knowledge testing by asking students to write chemical equations for \ngiven reaction \nObjective To study the properties of acids (dilute hydrochloric acid) by their reaction \nwith the followin g: \na) Litmus solution (red/blue) \nb) Zinc metal \nc) Solid sodium carbonate \nLearning \nOutcomes SKILLS AND COMPETENCIES - Students would be able to \nhave hands on experiment and enhance their ability to make correct \nobservations. \n.develop the ability to compare chemical properties of acids and bases. \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner - \nAnticipatory Set: 5 min", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner - \nAnticipatory Set: 5 min \n Recapitulation by asking questions based on previous day\u2019s learning \nStrategies DEAL, Experimentation , group work - collaborative, \ntabulation, \nDOL 1, 2, 3", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nDiscussion of topic through Collaborative Learning: 20 min \n (Critical Thinking and Problem Solving , collaboration and Character \nbuilding: Comes prepared to learn and bringing of material on the day \nof lab visit and working as a team \nTeaching strategy \u2013 Hands on experiment, Discussion and tabulation \nInstructional Input : Teacher will explain the practical, Collaborative \nwork; Students will work in group of three students each \nTeacher\u2019s role as a facilitator", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nGuided practice: 10 min \nStudents will observe and record their observation in tabular form. \nClosure: 5 min \nSummarization by students Independent Practice - NCERT Questions \nResources Text Book: \nNCERT text book for Science. \nReference Book -CBSE Exemplar \nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments Independent Practice : NCERT Questions \nAssessments Peer assessment \n \nTopic DAY 6 - Subtopic - Dilution of acid and base \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires b asic knowledge of concentrated and dilute solutions \nAssessment of \nqualifying \nknowledge Knowledge testing by asking questions related to concentrated and dilute \nsolutions \nObjective Students will be able to understand the correct way of process of dilution \nof acids o r base", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nLearning \nOutcomes SKILLS AND COMPETENCIES - Students would be able to \n\uf0b7 Critically analyse the the process of dilution and suggest correct \nway for dilution \n \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner - \nAnticipatory Set: 5 min \n Recapitulation by asking questions based on previous day\u2019s learning \n \nDiscussion of topic th rough Collaborative Learning: 20 min \n (Critical Thinking and Problem Solving , collaboration, communication \nand character ) \n.Activity Pose, pause, pounce and bounce \n\uf0b7 Students will be divided in groups and read activity 2.10 to \nunderstand dilution of acids and bases. \n\uf0b7 Class discussion \n\uf0b7 Teacher will ask questions. Students will be able to tell the correct \nprocedure to be followed for dilution. \n\uf0b7 Recording of the above on the basis of their understanding,", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nGuided prac tice: 10 min \nStudents will be able to tell the precautions to be taken while carrying out \ndilution. \nIndependent Practice: NCERT Questions Closure: 5 min \nSummarization by students with the muddiest point discussed by the \nteacher. \nResources Text Book: \nNCERT text book for Science. \nReference Book \nCBSE Exemplar \nClosure Summarization by students with the muddiest point discussed by the \nteacher.", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nSelf Study, Home \nWork, \nAssignments NCERT Questions \nAssessments Oral Test \nTopic DAY 7 - Subtopic - pH scale \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires b asic knowledge of use of indicators to find chemical \nnature \nAssessment of \nqualifying \nknowledge Knowledge testing by asking students colour change with indicators \nObjective Students will be able to understand the strength of acidic and basic \nsolution in terms of pH \nLearning \nOutcomes SKILLS AND COMPETENCIES - Students would be able to \n\uf0b7 Critically analyse the colour produced by given solution on pH \npaper and interpre t its chemical nature and compare strength with \nother solutions. \n\uf0b7 Identify given acids as strong acids or weak acids \n\uf0b7 Identify given bases as strong base or weak base \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\n\uf0b7 Identify given bases as strong base or weak base \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner - \nAnticipatory Set: 5 min \n Recapitulation by asking questions based on previous day\u2019s learning", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nDOL 1, 2, 3 \n \nDiscussion of topic th rough Collaborative Learning: 20 min \n (Critical Thinking and Problem Solving , collaboration, communication \nand character ) \nStrategies Group wor k - collaborative, tabulation, \n.Activity \nStudents will be divided in groups and carry out activity 2.11 to know \nthe pH values of different solutions . \n\uf0b7 Students will carry out the activity with different test solutions using \npH paper. \n\uf0b7 Tabulation of the above on the basis of their observations. \n\uf0b7 Students teacher interaction \n\uf0b7 Students in group will prepare questions and ask the other groups \n\uf0b7 Teacher\u2019s role as a facilitator \nGuided practice: 10 min \nStudents will record their observation for different solutions and write their \nnature. \nIndependent Practice: NCERT Questions \nClosure: 5 min \nSummarization by students with the muddiest point discussed by the \nteacher.", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nResources Text Book: \nNCERT text book for Science. \nReference Book \nCBSE Exemplar \nClosure Summarization by students with the muddiest point discussed by the \nteacher. \n \nSelf Study, Home \nWork, \nAssignments NCERT Questions \nAssessments Oral Test \n \n \nTopic Day 8 -Practical \nClass Transaction 40 min each) \nPre-requisite for \nthe course This topic requires basic knowledge of type of reactions \nAssessment of \nqualifying \nknowledge Recording of the observations made in lab \nObjective To find the pH of the following samples by using pH paper / universal \nindicator: \na) Dilute hydrochloric acid solution (A) \nb) Dilute sodium hydroxide solution (B) \nc) Dilute ethanoic acid solution (C) \nd) Lemon juice (D) \ne) Water (E) \nf) Dilute sodium hydrogen carbonate solution (F)", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nLearning \nOutcomes SKILLS AND COMPETENCIES - Students would be able to \nhave hands on experiment and enhance their ability to make correct \nobservations. \nCriticall y analyse the colour produced by given solution on pH paper and \ninterpre t its chemical nature and compare strength with other solutions. \n \nTransaction \nMethodology Transaction would proceed in the following manner - \nAnticipatory Set: 5 min (The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Recapitulation by asking questions based on previous day\u2019s learning \n \nActivity DEAL, , tabulation, \nDOL 1, 2, 3", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nActivity DEAL, , tabulation, \nDOL 1, 2, 3 \n \nDiscussion of topic through Collaborative Learning: 20 min \n (Critical Thinking and Problem Solving , collaboration and Character \nbuilding: Comes prepared to learn and bringing of material on the day \nof lab visit and working as a team \nTeaching strategy \u2013 Hands on experiment, Discussion and tabulation \nInstructional Input : Teacher will explain the practical, Collaborative \nwork; Students will work in group of three students each", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nGuided practice: 10 min \nStudents will observe and record their observation in tabular form. \nClosure: 5 min \nSummarization by students \nIndependent Practice -Students will complete practical in and answer \nrelated questions. \nResources Text Book: \nNCERT text book for Science. \nReference Book \nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments NCERT Questions \nAssessments Observations recorded in the spiral \n \nTopic DAY 9 Subtopic Importance of pH in everyday life \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires b asic knowledge of neutralization reactions \nAssessment of \nqualifying \nknowledge Questions based on neutralization reactions \nObjective Students will be able to explain the importance of pH in daily life", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nLearning \nOutcomes SKILLS AND COMPETENCIES - Students would be able to \n\uf0b7 Creatively make web cluster on Importance of pH in everyday life . \n\uf0b7 Apply neutralization and suggest solutions for the problem due \nto excess acid or base. \nTransaction \nMethodology \n(The teacher can Transaction would proceed in the following manner - \nAnticipatory Set: 5 min \nRecapitulation by asking questions based on previous day\u2019s learning use the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Strategies / Technique Questioning, group work - collaborative, \nDOL 1, 2, 3,4", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nDiscussion of topic through Collaborative Learning: 20 min \n (Creative Thinking, collaboration and communication) \n\uf0b7 Students in the group of 5 will discuss Importance of pH in \neveryday life \n\uf0b7 Groups will give presentation \n\uf0b7 Students teacher interaction \n\uf0b7 Questioning -Teacher will ask ques tions on the basis of uses of pH \nin day to day life \n\uf0b7 children will make web cluster on Importance of pH in everyday \nlife \n\uf0b7 Teacher\u2019s role as a facilitator \nGuided practice: 10 min \n NCERT Questions \nClosure: 5 min \nSummarization by students with the muddiest point discussed by the \nteacher. \nIndependent Practice - NCERT and exemplar Questions \nResources Text Book: \nNCERT text book for Science. \nReference Book \nCBSE Exemplar \nClosure Involving students in summarizing \nSelf Study, Home \nWork, \nAssignments NCERT Questions \nAssessments Quiz \n \nTopic DAY 10 - Subtopic Family of salts \nClass Transaction 40 min each) \nPre-requisite for", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nSelf Study, Home \nWork, \nAssignments NCERT Questions \nAssessments Quiz \n \nTopic DAY 10 - Subtopic Family of salts \nClass Transaction 40 min each) \nPre-requisite for \nthe course This topic requires b asic knowledge of chemical reactions \nAssessment of \nqualifying \nknowledge Knowledge testing by asking students question based on previous \nknowledge \nObjective Students will be able to explain preparation properties and uses different \ntypes of salts used by us. \nLearning \nOutcomes SKILLS AND COMPETENCIES - Students would be able to", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nDescribe the preparation, properties and uses of salts with the help of \ncluster web creatively \n Transaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner - \nAnticipatory Set: 5 min \n Recapitulation by asking questions based on previous day\u2019s learning \nDOL 1, 2, 3,4", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nDiscussion of topic through Collaborative Learning: 25 min \n (Creative Thinking and collaboration , ) \nStrategies / Technique Questioning, group wor k - collaborative, \ntabulation, \nThe students in a group of 5 will discuss types of salts topic from NCERT. \n\uf0b7 They will classify salts on the basis of acid/ base from which have \nbeen produced, record this in their notebooks \n\uf0b7 They will discuss activity 2.14 in their groups and complete table \n2.4 and find pH of given salts \n\uf0b7 Students teacher interaction \n\uf0b7 Teacher will a sk questions to check their understanding \nGuided practice: 5 min \nNCERT Questions \nIndependent Practice : NCERT Questions \nClosure: 5 min \nInvolving students in summarizing by asking them to prepare questions \nbased on the topic \nResources Text Book: NCERT text book for Science. \nReference Book CBSE Exemplar \nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork,", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nReference Book CBSE Exemplar \nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments NCERT Questions \nAssessments Oral Test \n \nTopic DAY 11 - Chemicals obtained from common salt \nClass Transaction 40 min each) \nPre-requisite for \nthe course This topic requires b asic knowledge of chemical reactions \nAssessment of \nqualifying \nknowledge Knowledge testing by asking students question based on previous \nknowledge \nObjective Students will be able to explain preparation properties and uses \ndifferent types of salts used by us. \nLearning \nOutcomes SKILLS AND COMPETENCIES - Students would be able to", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nDescribe the preparation, properties and uses of salts with th e help of \ncluster web creatively \n \nTransaction Transaction would proceed in the following manner - Methodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Anticipatory Set: 5 min \n Recapitulation by asking questions based on previous day\u2019s learning \nDOL 1, 2, 3 \n \nDiscussion of topic through Collaborative Learning: 25 min \n (Creative Thinking, collaboration , character ) \nStrategies / Technique Questioning , group work - collaborative, \ntabulation, Wipro - G.O.s(web chart, flow chart)", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\n\uf0b7 Class will be divided in groups of 5 students each and allotted one \ncompound (caustic soda, bleaching powder and baking soda) \n\uf0b7 The students in group will discuss the preparation, properties and \nuses of allotted salt . Each group will give presentation on the topic \n\uf0b7 Other group s will add more information. \n\uf0b7 Students teacher interaction \n\uf0b7 Teacher will ask questions to check their understanding \n\uf0b7 children make web cluster on preparation, properties and uses of \nsalts \nGuided practice: 5 min \nNCERT Questions \nIndependent Practice : NCERT Questions \nClosure: 5 min \nInvolving students in summarizing by asking them to prepare questions \nbased on the topic \nResources Text Book: \nNCERT text book for Science. \nReference Book \nCBSE Exemplar \nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments NCERT Questions \nAssessments Oral Test", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments NCERT Questions \nAssessments Oral Test \n \nTopic DAY 12 - Water of crystallization in salts , POP \nClass Transaction 40 min each) \nPre-requisite for \nthe course This topic requires b asic knowledge of chemical reactions \nAssessment of \nqualifying \nknowledge Knowledge testing by asking students question based on previous \nknowledge \nObjective Students will be able to explain the terms hydrated salts, anhydrous salts, \nWater of crystallization in salts \nLearning \nOutcomes SKILLS AND COMPETENCIES - Students would be able to \ncompare the prepa ration, properties and uses of hydrated s alts and develop collaboration and character in the process \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner -", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner - \nAnticipatory Set: 5 min \n One minute paper by asking questions based on previous day\u2019s learning \nDiscussion of topic through demo by teacher, group activity , \nQuestioning - Pose, pause, pounce and bounce : 25 min \n\uf0b7 Teacher will carry out activity 2.15 and introduce the terms \nhydrated salts , anhydrous salts, Water of crystallization in salts \n\uf0b7 Class will be divided in groups of 5 students each and allotted one \ncompound (hydrated salts Washing soda and plaster of Paris) \n\uf0b7 The students in group will discuss the preparation, properties and \nuses of allotted salt. Each group will give presentation on the topic \n\uf0b7 Other groups will add more information. \n\uf0b7 Students teacher interaction \n\uf0b7 Teacher will ask questions to check their understanding", + "CHAPTER 2-ACIDS,BASES AND SALTS - LESSON PLAN.txt\n\uf0b7 Other groups will add more information. \n\uf0b7 Students teacher interaction \n\uf0b7 Teacher will ask questions to check their understanding \n\uf0b7 Children will complete web cluster on preparation, properties and \nuses of salts \nGuided practice: 5 min \nStudents will record the observations made on heating hydrated salts. \nIndependent Practice : NCERT Questions \nClosure: 5 min \nInvolving students in summarizing by asking them to prepare questions \nbased on the topic \nResources Text Book: \nNCERT text book for Science. \nReference Book \nCBSE Exemplar \nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments NCERT Questions \nAssessments Oral Test \n \n Closure of chapter by mind mapping and open discussion", + "CHAPTER 2-ACIDS,BASES AND SALTS -FOCUS POINTS.txt\nFocus Spot: \nReaction of Metal Carbonates and Metal \nHydrogencarbonates with Acids. \nWhy: \nStudents are baffled about the reaction of metal \ncarbonates and metal hydrogencarbonates with acids as \nthey do not get how these give the corresponding salt, \ncarbon dioxide and water. \nImpact: \nStudents are unable to construct a correct chemical \nequation of metal oxides and carbonates as they apply the \nsalt product of metal oxides in metal carb onates and vice \nversa. \n \n \nFocus Spot: \nChemicals derived from Common Salt. \nWhy: \nStudents find it difficult to understand chemicals derived \nfrom common salt because they are unable to remember equations that involve the origination of products from \ncommon sal t in the correct form. \nImpact: \nStudents are unable to comprehend the origin of baking \nsoda, washing soda and bleaching powder due to \nimproper retention of chemical reactions involving \ncommon salt.", + "CHAPTER 2-ACIDS,BASES AND SALTS MCQ TYPE QUESTIONS.txt\nMCQ", + "CHAPTER 2-ACIDS,BASES AND SALTS MCQ TYPE QUESTIONS.txt\n1. The colour of phenolphthalein in acidic medium (yellow, pink, \ncolourless, blue) \n2. The gas which burns with a pop sound ( H2, CO 2, CuSO 4, CH 4) \n3. Milkiness of lime water disappear when excess CO 2 is passed \ndue to the formation of ( calcium hydroxide, calcium chloride, \ncalcium bicarbonate, calcium carbonate) \n4. Greenish coating on copper articles can be removed by using \n( lemon, tamarind, vinegar, all of the above) \n5. Rain water usually have pH ( =7, <7, >7, \u22487) \n6. The acid present in tomato is ( lac tic acid, oxalic acid, tartaric \nacid, methanoic acid) \n7. A solution of sodium carbonate has pH around 13.", + "The acid present in tomato is ( lac tic acid, oxalic acid, tartaric \nacid, methanoic acid) \n7. A solution of sodium carbonate has pH around 13. The nature \nof solution is ( strongly acidic, weakly acidic, strongly alkaline, \nweakly alkaline) \n8. The gas which is produced at anode during chlor -alkali \nprocess ( Cl 2, H2, NaOH, O2) \n9. The common name for the substance used as antacid ( \nwashing soda, plaster of Paris, bleaching powder, baking \nsoda)", + "CHAPTER 2-ACIDS,BASES AND SALTS MCQ TYPE QUESTIONS.txt\nprocess ( Cl 2, H2, NaOH, O2) \n9. The common name for the substance used as antacid ( \nwashing soda, plaster of Paris, bleaching powder, baking \nsoda) \n10. The total no. of molecules of water of crystallization in \nwashing soda is (2, \u00bd, 10,5) 11. Among the following, which is NOT a form of calcium \ncarbonate ( chalk, limestone, egg shell, quick lime) \n12. If a solution turns blue litmus to red, what colour will be \nobserved if the same is placed on pH paper ( green, blue, red, \nnone of the above) \n13. Identify a fruit/ veg etable which is basic in nature? \n(lemon, tomato, broccoli, oranges) \n14. Calcium carbonate reacts with hy drochloric acid to form \nX,Y,Z. W hat are X,Y,Z ?", + "Identify a fruit/ veg etable which is basic in nature? \n(lemon, tomato, broccoli, oranges) \n14. Calcium carbonate reacts with hy drochloric acid to form \nX,Y,Z. W hat are X,Y,Z ? \n(a) X: Ca(OH) 2, Y:H 2O, Z:CO 2 \n(b) X:CaCl 2 , Y:H 2O, Z:CO 2 \n(c) X:Ca(OH) 2 , Y:H 2O, Z:H 2 \n(d) X:CaCl 2 , Y:H 2O, Z:H 2", + "CHAPTER 2-ACIDS,BASES AND SALTS MCQ TYPE QUESTIONS.txt\n15. When you clean a metal vessel with tamarind the \nreaction taking place is: \nMetal oxide + X ----------\u2192Salt + W ater. \n What is X ? (acid, base, hydrogen, carbondioxide) \n16. A student wants to make a basic salt. which of the \nfollowing pairs of acid and base should he use. (a) HCl + \nNaOH (b) H 2CO 3 + NaOH (c) HCl+NH 4OH (d) H 2SO 4 + KOH \n17. Ph of milk of magnesia is 10. Among following solutions \nwhich can neutralize milk of magnesia (baking soda, \ncommon salt, vinegar, none of the above) \n18. The acidic salt among following is (ammonium chloride, sodium chloride, potassium chloride, calcium chloride) \n19. The gas which is produced at cathode during chlor -\nalkali process is (hydrogen, chlorine, wa ter, sodium \nhydroxide) \n20.", + "The acidic salt among following is (ammonium chloride, sodium chloride, potassium chloride, calcium chloride) \n19. The gas which is produced at cathode during chlor -\nalkali process is (hydrogen, chlorine, wa ter, sodium \nhydroxide) \n20. The number of water molecules present in white \ncoloured copper sulphate crystal is (0, 1,2,5) \n21. The gas responsible for softening of cakes and bread \nwhich is formed when baking powder is heated is (carbon", + "CHAPTER 2-ACIDS,BASES AND SALTS MCQ TYPE QUESTIONS.txt\ncoloured copper sulphate crystal is (0, 1,2,5) \n21. The gas responsible for softening of cakes and bread \nwhich is formed when baking powder is heated is (carbon \nmonoxide, carbon dioxide, chlorine, tartaric acid) \n22. What happens to hydrogen ions when lime juice is \nadded to vinegar solution? (increases, decreas es, do not \nchange) \n23. Which sodium compound is used in the manuf acture of \nborax, glass, paper (sodium chloride, sodium bicarbonate, \nsodium carbonate, sodium nitrate) \n24. Which among the following is an alkali (magnesium \nhydroxide, calcium hydroxide, sodium hydrox ide, aluminum \nhydroxide. \n25.", + "Which among the following is an alkali (magnesium \nhydroxide, calcium hydroxide, sodium hydrox ide, aluminum \nhydroxide. \n25. Zinc metal react with X to form zinc sulphate along with \na gas Y which burns with pop sound X and Y are (a) X: \nH2SO 4 Y : CO 2 (b) X: H 2SO 4 Y : H 2 (c) X: HCl Y : H 2 (d) \nX: HCl Y : CO 2", + "CHAPTER 3- METALS AND NON METALS-NOTES.txt\nNAVODAYA VIDYALAYA SAMITI \ne content preparation \nCLASS X -SCIENCE \nCHAPTER NO:3 \nPrepared by, \nDEVIKRISHNA K P \nPGT CHEMISTRY \nJNV DAVANGERE Properties of Metals \nProperties \n1.Generally solids \n2.Hard \n3.Malleable and ductile(gold \nis the most ductile metal) \n4.Good conductors of heat \nand electricity (Silver & \ncopper is the best \nconductor) \n5.Metallic Lusture \n6.Sonorous \n7.High m.p and b.p Exceptions \n\u2022Mercury is a liquid \n\u2022Lithium , Sodium are soft \n\u2022Gallium and cesium have very \nlow m.p \u2013melts in the palm Properties of Non-metals \nProperties \n1.Generally solids or gases \n2.Soft\n3.Brittle \n4.Bad conductor of heat and \nElectricity \n5.", + "Sodium are soft \n\u2022Gallium and cesium have very \nlow m.p \u2013melts in the palm Properties of Non-metals \nProperties \n1.Generally solids or gases \n2.Soft\n3.Brittle \n4.Bad conductor of heat and \nElectricity \n5.No Lusture Exceptions \n\u2022Bromine is a liquid \n\u2022Diamond- an allotrope of \ncarbon is the hardest substance \n\u2022Graphite is a good conductor \n\u2022Iodine shows lusture Chemical Properties of Metals Reaction with Air/Oxygen \n\u2756Almost all metals combine with oxygen to \nform metal oxides. \n\u0000Metal + Oxygen \u2192 Metal oxide \n\u2756metal oxides are basic in nature.", + "CHAPTER 3- METALS AND NON METALS-NOTES.txt\n\u2756Almost all metals combine with oxygen to \nform metal oxides. \n\u0000Metal + Oxygen \u2192 Metal oxide \n\u2756metal oxides are basic in nature. \n\u2756But some metal oxides show both acidic as \nwell as basic behaviour-Amphoteric oxides. \n\u0000Such metal oxides react with both acids as well as \nbases to produce salts and water. \n\u2714Eg-Al2O3, ZnO \n\u2756Most metal oxides are insoluble in water \n\u2756But some of the metal oxides dissolve in \nwater to form alkalis \n\u0000Eg-Sodium oxide , Potassium oxide \nRelative reactivity of Metals towards \nOxygen \n\u2756All metals do not react with oxygen at the same rate. \n\u0000potassium and sodium react so vigorously that they catch \nfire if kept in the open. \n\u2714Hence, to protect them and to prevent accidental fires, they are \nkept immersed in kerosene oil. \n\u0000At ordinary temperature, the surfaces magnesium , \naluminium, zinc, lead, etc., are covered with a thin layer of \noxide.", + "\u2714Hence, to protect them and to prevent accidental fires, they are \nkept immersed in kerosene oil. \n\u0000At ordinary temperature, the surfaces magnesium , \naluminium, zinc, lead, etc., are covered with a thin layer of \noxide. \n\u2714The protective oxide layer prevents the metal from further \noxidation.", + "CHAPTER 3- METALS AND NON METALS-NOTES.txt\n\u0000At ordinary temperature, the surfaces magnesium , \naluminium, zinc, lead, etc., are covered with a thin layer of \noxide. \n\u2714The protective oxide layer prevents the metal from further \noxidation. \n\u0000Iron does not burn on heating but iron filings burn \nvigorously when sprinkled in the flame of the burner. \n\u0000 Copper does not burn, but the hot metal is coated with a \nblack coloured layer of copper oxide. \n\u0000Silver and gold do not react with oxygen even at high \ntemperatures Reaction with Water \n\u2756Metals react with water to produce a metal \noxide and hydrogen \n\u2756Metal oxides that are soluble in water \ndissolve to form metal hydroxide. \n\u0000Metal + Water \u2192 Metal oxide + Hydrogen \n\u0000Metal oxide + Water \u2192 Metal hydroxide Relative reactivity of Metals towards water \n\u2756Potassium and Sodium react violently with cold water. \n\u0000the reaction is so violent and exothermic that the evolved \nhydrogen immediately catches fire. \n\u2756The reaction of calcium with water is less violent.", + "CHAPTER 3- METALS AND NON METALS-NOTES.txt\n\u0000the reaction is so violent and exothermic that the evolved \nhydrogen immediately catches fire. \n\u2756The reaction of calcium with water is less violent. \n\u0000The heat evolved is not sufficient for the hydrogen to catch \nfire.\n\u0000Calcium starts floating \n\u2714the bubbles of hydrogen gas formed stick to the surface of the \nmetal. \n\u2756Magnesium does not react with cold water. It reacts \nwith hot water \n\u0000form magnesium hydroxide and hydrogen. \n\u0000 It also starts floating \n\u2756Aluminium, iron and zinc do not react either with cold \nor hot water. \n\u0000they react with steam to form the metal oxide and \nhydrogen. \n\u2756lead, copper, silver and gold do not react with w aterReaction with Acids \n\u2756Metal + Acid \u2192 Salt + Hydrogen \n\u2756Hydrogen gas is not evolved when a metal \nreacts with nitric acid. \n\u0000It is because HNO3 is a strong oxidising \nagent.", + "\u0000It is because HNO3 is a strong oxidising \nagent. \n\u0000It oxidises the H2 produced to water and itself \ngets reduced to any of the nitrogen oxides \n(N2O, NO,NO2). \n\u0000But magnesium (Mg) and manganese (Mn)", + "CHAPTER 3- METALS AND NON METALS-NOTES.txt\nagent. \n\u0000It oxidises the H2 produced to water and itself \ngets reduced to any of the nitrogen oxides \n(N2O, NO,NO2). \n\u0000But magnesium (Mg) and manganese (Mn) \nreact with very dilute HNO3 to evolve H2 gas. Reaction with Other Metal salt \nsolution \n\u2756A more reactive metal displaces less \nreactive metal from its solution \n\u0000Metal A + Salt solution of B \u2192 Salt solution of \nA + Metal B \n\u2756Displacement reaction can be used to \ndetermine the relative reactivity of Metals Reactivity Series \n\u2756Metals arranged in the decreasing order of \ntheir reactivities \nWhy Inert Gases are Inert?", + "\u2756The outer most shell of noble/inert gases \ncontain 8 electrons (except He) \n\u2756No other elements have 8 electrons in \ntheir valence shell \n\u2756The configuration of 8 electrons is very \nstable which makes the noble gases inert \n\u0000The configuration of 8 electron is known as \noctet \n\u2756Atoms of other elements attains stability if \nthey get 8 electrons in their outermost \nshellHow \u2018Na\u2019 attains stability", + "CHAPTER 3- METALS AND NON METALS-NOTES.txt\n\u0000The configuration of 8 electron is known as \noctet \n\u2756Atoms of other elements attains stability if \nthey get 8 electrons in their outermost \nshellHow \u2018Na\u2019 attains stability \n\u2756\u2018Na\u2019 will loose one electron from its outer most \nshell \nHow \u2018Cl\u2019 attains stability \n\u2756\u2018Cl\u2019 will gain one electron to its outermost \nshell \nWhen \u2018Na\u2019 and \u2018Cl\u2019 react\u2026\u2026 \n\u2756Sodium and chloride ions, are oppositely \ncharged, \n\u0000They attract each other \n\u0000They are held by strong electrostatic forces of \nattraction to exist as sodium chloride (NaCl).", + "\u2756sodium chloride does not exist as molecules \nbut aggregates of oppositely charged ions P rop erties of ionic compounds \u2756Physical nature \n\u0000Solids, Hard, Brittle in nature \n\u2756Melting and Boiling Point \n\u0000High melting and Boiling point \n\u2756Solubility \n\u0000Soluble in polar solvents and insoluble in nonpolar \nsolvents \n\u2756Conduction of electricity \n\u0000Conducts electricity through solution.Non \nconductor in solid state https://youtu.be/wh_JyiS-6bE https://youtu.be/RV7Z1HnJJ9A LINKS:", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf04d\uf065\uf074\uf061\uf06c\uf073\uf020\uf061\uf06e\uf064\uf020\uf04e\uf06f\uf06e\uf02d\uf06d\uf065\uf074\uf061\uf06c\uf073\uf033 \uf043\uf048\uf041\uf050\uf054\uf045\uf052\nIn Class IX you have learnt about various elements. You have seen\nthat elements can be classified as metals or non-metals on the basis of\ntheir properties.\n\uf06eThink of some uses of metals and non-metals in your daily life.\n\uf06eWhat properties did you think of while categorising elements\nas metals or non-metals?\n\uf06eHow are these properties related to the uses of these elements?\nLet us look at some of these properties in detail.", + "\uf06eWhat properties did you think of while categorising elements\nas metals or non-metals?\n\uf06eHow are these properties related to the uses of these elements?\nLet us look at some of these properties in detail.\n\uf033\uf02e\uf031\uf020\uf050\uf048\uf059\uf053\uf049\uf043\uf033\uf02e\uf031\uf020\uf050\uf048\uf059\uf053\uf049\uf043\uf033\uf02e\uf031\uf020\uf050\uf048\uf059\uf053\uf049\uf043\uf033\uf02e\uf031\uf020\uf050\uf048\uf059\uf053\uf049\uf043\uf033\uf02e\uf031\uf020\uf050\uf048\uf059\uf053\uf049\uf043", + "\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\n\uf033\uf02e\uf031\uf02e\uf031\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\nThe easiest way to start grouping substances is by comparing theirphysical properties.", + "easiest way to start grouping substances is by comparing theirphysical properties. Let us study this with the help of the following\nactivities. For performing A ctivities 3.1 to 3.6, collect the samples of\nfollowing metals \u2013 iron, copper, aluminium, magnesium, sodium, lead,\nzinc and any other metal that is easily available.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nfollowing metals \u2013 iron, copper, aluminium, magnesium, sodium, lead,\nzinc and any other metal that is easily available.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\n\uf06eTake samples of ir on, copper , aluminium and magnesium. Note\nthe appearance of each sample.", + "Note\nthe appearance of each sample.\n\uf06eClean the surface of each sample by rubbing them with sand paper\nand note their appearance again.\nMetals, in their pure state, have a shining surface. This property is\ncalled metallic lustre.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf032\n\uf06eTake small pieces of ir on, copper, aluminium, and magnesium.", + "Try to cut these metals with a sharp knife and note yourobservations.\n\uf06eHold a piece of sodium metal with a pair of tongs.CAUTION: Always handle sodium metal with care. Dry it by\npressing between the folds of a filter paper.\n\uf06ePut it on a watch-glass and try to cut it with a knife.\n\uf06eWhat do you observe?\n2019-20\nDownload all NCERT books PDF from www.ncert.online", + "CHAPTER 3- METALS AND NON METALS-text book.txt\npressing between the folds of a filter paper.\n\uf06ePut it on a watch-glass and try to cut it with a knife.\n\uf06eWhat do you observe?\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineScience \uf033\uf038You will find that some metals can be beaten into thin sheets. This\nproperty is called malleability.", + "\uf06eWhat do you observe?\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineScience \uf033\uf038You will find that some metals can be beaten into thin sheets. This\nproperty is called malleability. Did you know that gold and silver are the\nmost malleable metals?\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf033\n\uf06eTake pieces of ir on, zinc, lead and copper.", + "\uf06ePlace any one metal on a block of iron and strike it four or five\ntimes with a hammer . What do you observe?\n\uf06eRepeat with other metals.\n\uf06eRecord the change in the shape of these metals.You will find that metals are generally hard. The hardness varies\nfrom metal to metal.", + "What do you observe?\n\uf06eRepeat with other metals.\n\uf06eRecord the change in the shape of these metals.You will find that metals are generally hard. The hardness varies\nfrom metal to metal.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf034\n\uf06eList the metals whose wires you have seen in daily life.\nThe ability of metals to be drawn into thin wires is called ductility.", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf034\n\uf06eList the metals whose wires you have seen in daily life.\nThe ability of metals to be drawn into thin wires is called ductility.\nGold is the most ductile metal.", + "The ability of metals to be drawn into thin wires is called ductility.\nGold is the most ductile metal. You will be surprised to know that a wir e\nof about 2 km length can be drawn from one gram of gold.\nIt is because of their malleability and ductility that metals can be\ngiven different shapes according to our needs.\nCan you name some metals that are used for making cooking vessels?\nDo you know why these metals are used for making vessels? Let us do\nthe following Activity to find out the answer .", + "It is because of their malleability and ductility that metals can be\ngiven different shapes according to our needs.\nCan you name some metals that are used for making cooking vessels?\nDo you know why these metals are used for making vessels? Let us do\nthe following Activity to find out the answer .\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf035\n\uf06eTake an aluminium or copper\nwire.", + "Clamp this wire on a\nstand, as shown in Fig. 3.1.\n\uf06eFix a pin to the free end of thewire using wax.\n\uf06eHeat the wire with a spirit lamp,candle or a burner near the\nplace where it is clamped.\n\uf06eWhat do you observe after sometime?\n\uf06eNote your observations. Doesthe metal wire melt?", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf06eHeat the wire with a spirit lamp,candle or a burner near the\nplace where it is clamped.\n\uf06eWhat do you observe after sometime?\n\uf06eNote your observations. Doesthe metal wire melt?\nThe above activity shows that metals are good conductors of heat\nand have high melting points. The best conductors of heat are silver and\ncopper. Lead and mer cury are comparatively poor conductors of heat.\nDo metals also conduct electricity? Let us find out.Figure 3.1Figure 3.1Figure 3.1Figure 3.1Figure 3.1\nMetals are goodconductors of heat.\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf033\uf039You must have seen that the wir es that carry curr ent\nin your homes have a coating of polyvinylchloride (PVC)\nor a rubber-like material. Why are electric wir es coated\nwith such substances?\nWhat happens when metals strike a hard surface? Do they produce", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nin your homes have a coating of polyvinylchloride (PVC)\nor a rubber-like material. Why are electric wir es coated\nwith such substances?\nWhat happens when metals strike a hard surface? Do they produce\na sound? The metals that produce a sound on striking a hard surfaceare said to be sonorous. Can you now say why school bells are made of\nmetals?\n\uf033\uf02e\uf031\uf02e\uf032\uf020\uf04e\uf06f\uf06e\uf02d\uf06d\uf065\uf074\uf061\uf06c\uf073\nIn the previous Class you have learnt that there are very few non-metalsas compared to metals. Some of the examples of non-metals are carbon,\nsulphur , iodine, oxygen, hydrogen, etc. The non-metals ar e either solids\nor gases except bromine which is a liquid.\nDo non-metals also have physical properties similar to that of metals?", + "Some of the examples of non-metals are carbon,\nsulphur , iodine, oxygen, hydrogen, etc. The non-metals ar e either solids\nor gases except bromine which is a liquid.\nDo non-metals also have physical properties similar to that of metals?\nLet us find out.\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf036\n\uf06eSet up an electric circuit as shown in Fig.", + "3.2.\n\uf06ePlace the metal to be tested in the circuit\nbetween terminals A and B as shown.\n\uf06eDoes the bulb glow? What does this indicate?\nFigure 3.2Figure 3.2Figure 3.2Figure 3.2Figure 3.2", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf06ePlace the metal to be tested in the circuit\nbetween terminals A and B as shown.\n\uf06eDoes the bulb glow? What does this indicate?\nFigure 3.2Figure 3.2Figure 3.2Figure 3.2Figure 3.2\nMetals are goodconductors of electricity.", + "\uf06eDoes the bulb glow? What does this indicate?\nFigure 3.2Figure 3.2Figure 3.2Figure 3.2Figure 3.2\nMetals are goodconductors of electricity.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf037\n\uf06eCollect samples of carbon (coal or graphite), sulphur and iodine.", + "\uf06eCarry out the Activities 3.1 to 3.4 and 3.6 with these non-metals\nand record your observations.\nCompile your observations regarding metals and non-metals in Table 3.1.\nTable 3.1\nElement Symbol Type of Hardness Malleability Ductility Conducts Sonority\nsurface Electricity\nOn the bases of the observations recorded in Table 3.1, discuss the\ngeneral physical properties of metals and non-metals in the class. Y ou\nmust have concluded that we cannot group elements according to their\nphysical properties alone, as there are many exceptions. For example \u2013", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nmust have concluded that we cannot group elements according to their\nphysical properties alone, as there are many exceptions. For example \u2013\n(i)All metals except mercury exist as solids at room temperature.In Activity 3.5, you have observed that metals have high melting\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineScience \uf034\uf030points but gallium and caesium have very low melting points.\nThese two metals will melt if you keep them on your palm.\n(ii)Iodine is a non-metal but it is lustrous.\n(iii) Carbon is a non-metal that can exist in different forms. Each\nform is called an allotrope. Diamond, an allotrope of carbon, is\nthe hardest natural substance known and has a very high melting\nand boiling point. Graphite, another allotrope of carbon, is a\nconductor of electricity.\n(iv) Alkali metals (lithium, sodium, potassium) are so soft that theycan be cut with a knife. They have low densities and low melting\npoints.", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nconductor of electricity.\n(iv) Alkali metals (lithium, sodium, potassium) are so soft that theycan be cut with a knife. They have low densities and low melting\npoints.\nElements can be more clearly classified as metals and non-metals\non the basis of their chemical properties.", + "(iv) Alkali metals (lithium, sodium, potassium) are so soft that theycan be cut with a knife. They have low densities and low melting\npoints.\nElements can be more clearly classified as metals and non-metals\non the basis of their chemical properties.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf038\n\uf06eTake a magnesium ribbon and some sulphur powder.", + "\uf06eBurn the magnesium ribbon. Collect the ashes formed and dissolve\nthem in water .\n\uf06eTest the r esultant solution with both r ed and blue litmus paper .\n\uf06eIs the product formed on burning magnesium acidic or basic?\n\uf06eNow bur n sulphur powder . Place a test tube over the bur ning\nsulphur to collect the fumes produced.\n\uf06eAdd some water to the above test tube and shake.\n\uf06eTest this solution with blue and red litmus paper.\n\uf06eIs the product formed on burning sulphur acidic or basic?\n\uf06eCan you write equations for these reactions?\n\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. Give an example of a metal which\n(i)is a liquid at room temperature.", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf06eIs the product formed on burning sulphur acidic or basic?\n\uf06eCan you write equations for these reactions?\n\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. Give an example of a metal which\n(i)is a liquid at room temperature.\n(ii)can be easily cut with a knife.\n(iii) is the best conductor of heat.\n(iv) is a poor conductor of heat.\n2. Explain the meanings of malleable and ductile.", + "Give an example of a metal which\n(i)is a liquid at room temperature.\n(ii)can be easily cut with a knife.\n(iii) is the best conductor of heat.\n(iv) is a poor conductor of heat.\n2. Explain the meanings of malleable and ductile.\n\uf033\uf02e\uf032\uf020\uf043\uf048\uf045\uf04d\uf049\uf043\uf033\uf02e\uf032\uf020\uf043\uf048\uf045\uf04d\uf049\uf043\uf033\uf02e\uf032\uf020\uf043\uf048\uf045\uf04d\uf049\uf043\uf033\uf02e\uf032\uf020\uf043\uf048\uf045\uf04d\uf049\uf043\uf033\uf02e\uf032\uf020\uf043\uf048\uf045\uf04d\uf049\uf043 \uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020", + "\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\uf020\uf04d\uf045\uf054\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\uf020\uf04d\uf045\uf054\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\uf020\uf04d\uf045\uf054\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\uf020\uf04d\uf045\uf054\uf041\uf04c\uf020\uf050\uf052", + "\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\uf020\uf04d\uf045\uf054\uf041\uf04c\uf020\uf050\uf052\uf04f\uf050\uf045\uf052\uf054\uf049\uf045\uf053\uf020\uf04f\uf046\uf020\uf04d\uf045\uf054 \uf041\uf04c\uf053 \uf041\uf04c\uf053\uf041\uf04c\uf053 \uf041\uf04c\uf053\uf041\uf04c\uf053\nWe will learn about the chemical properties of metals in the following\nSections 3.2.1 to 3.2.4. For this, collect the samples of following metals \u2013\naluminium, copper , iron, lead, magnesium, zinc and sodium. Most non-metals produce acidic oxides when dissolve in water . On\nthe other hand, most metals, give rise to basic oxides.", + "2.4. For this, collect the samples of following metals \u2013\naluminium, copper , iron, lead, magnesium, zinc and sodium. Most non-metals produce acidic oxides when dissolve in water . On\nthe other hand, most metals, give rise to basic oxides. You will be lear ning\nmore about these metal oxides in the next section.\n2019-20\nDownload all NCERT books PDF from www.ncert.online", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nthe other hand, most metals, give rise to basic oxides. You will be lear ning\nmore about these metal oxides in the next section.\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf034\uf031\uf033\uf02e\uf032\uf02e\uf031\uf020\uf057\uf068\uf061\uf074\uf020\uf068\uf061\uf070\uf070\uf065\uf06e\uf073\uf020\uf077\uf068\uf065\uf06e\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\uf020\uf061\uf072\uf065\uf020\uf062\uf075\uf072\uf06e\uf074\uf020\uf069\uf06e\uf020\uf041\uf069\uf072\uf03f\nYou have seen in Activity 3.8 that magnesium bur ns in air with a dazzling\nwhite flame.", + "Do all metals react in the same manner? Let us check by\nperforming the following Activity.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf039\nCAUTION : The following activity needs the teacher\u2019s assistance.\nIt would be better if students wear eye protection.\n\uf06eHold any of the samples taken above with a pair of tongs and try\nburning over a flame.", + "It would be better if students wear eye protection.\n\uf06eHold any of the samples taken above with a pair of tongs and try\nburning over a flame. Repeat with the other metal samples.\n\uf06eCollect the product if formed.\n\uf06eLet the products and the metal surface cool down.\n\uf06eWhich metals burn easily?\n\uf06eWhat flame colour did you observe when the metal burnt?\n\uf06eHow does the metal surface appear after burning?", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf06eLet the products and the metal surface cool down.\n\uf06eWhich metals burn easily?\n\uf06eWhat flame colour did you observe when the metal burnt?\n\uf06eHow does the metal surface appear after burning?\n\uf06eArrange the metals in the decreasing order of their reactivitytowards oxygen.\n\uf06eAre the products soluble in water?\nAlmost all metals combine with oxygen to form metal oxides.\nMetal + Oxygen \u2192 Metal oxide\n For example, when copper is heated in air , it combines with oxygen\nto form copper(II) oxide, a black oxide.\n2Cu + O2 \u2192 2CuO\n(Copper) (Copper(II) oxide)\nSimilarly, aluminium forms aluminium oxide.\n4Al +3O2 \u2192 2Al2O3\n(Aluminium) (Aluminium oxide)\nRecall from Chapter 2, how copper oxide reacts with hydrochloric acid.\nWe have learnt that metal oxides are basic in natur e. But some metal\noxides, such as aluminium oxide, zinc oxide show both acidic as well as\nbasic behaviour.", + "We have learnt that metal oxides are basic in natur e. But some metal\noxides, such as aluminium oxide, zinc oxide show both acidic as well as\nbasic behaviour. Such metal oxides which react with both acids as well as\nbases to produce salts and water are known as amphoteric oxides.", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nbasic behaviour. Such metal oxides which react with both acids as well as\nbases to produce salts and water are known as amphoteric oxides.\nAluminium oxide reacts in the following manner with acids and bases \u2013\nAl2O3 + 6HCl \u21922AlCl3+3H2O\nAl2O3 + 2NaOH \u21922NaAlO2+H2O\n(Sodium\naluminate)\n Most metal oxides are insoluble in water but some of these dissolve\nin water to form alkalis. Sodium oxide and potassium oxide dissolve in\nwater to produce alkalis as follows \u2013\nNa2O(s) +H2O(l) \u2192 2NaOH(aq)\nK2O(s) +H2O(l) \u2192 2KOH(aq)\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineScience \uf034\uf032We have observed in Activity 3.9 that all metals do not react with\noxygen at the same rate. Different metals show different reactivities\ntowards oxygen.", + "Different metals show different reactivities\ntowards oxygen. Metals such as potassium and sodium react so\nvigorously that they catch fire if kept in the open. Hence, to protect them\nand to prevent accidental fires, they are kept immersed in kerosene oil.", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nvigorously that they catch fire if kept in the open. Hence, to protect them\nand to prevent accidental fires, they are kept immersed in kerosene oil.\nAt ordinary temperature, the surfaces of metals such as magnesium,\naluminium, zinc, lead, etc., are covered with a thin layer of oxide. The\nprotective oxide layer prevents the metal from further oxidation. Iron\ndoes not burn on heating but iron filings burn vigorously when sprinkled\nin the flame of the bur ner. Copper does not burn, but the hot metal is\ncoated with a black coloured layer of copper(II) oxide. Silver and gold donot react with oxygen even at high temperatures.\uf044\uf06f\uf020\uf059\uf06f\uf075\uf020\uf04b\uf06e\uf06f\uf077\uf03fAnodising is a process of forming a thick oxide layer of aluminium. Aluminium\ndevelops a thin oxide layer when exposed to air. This aluminium oxide coat makes itresistant to further corrosion. The resistance can be improved further by making theoxide layer thicker .", + "Aluminium\ndevelops a thin oxide layer when exposed to air. This aluminium oxide coat makes itresistant to further corrosion. The resistance can be improved further by making theoxide layer thicker . During anodising, a clean aluminium article is made the anode", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nand is electrolysed with dilute sulphuric acid. The oxygen gas evolved at the anodereacts with aluminium to make a thicker protective oxide layer . This oxide layer can\nbe dyed easily to give aluminium articles an attractive finish.\nAfter performing Activity 3.9, you must have observed that sodium\nis the most reactive of the samples of metals taken here. The reaction of\nmagnesium is less vigorous implying that it is not as reactive as sodium.But burning in oxygen does not help us to decide about the reactivity of\nzinc, iron, copper or lead. Let us see some more reactions to arrive at a\nconclusion about the order of reactivity of these metals.", + "Let us see some more reactions to arrive at a\nconclusion about the order of reactivity of these metals.\n\uf033\uf02e\uf032\uf02e\uf032\uf020\uf057\uf068\uf061\uf074\uf020\uf068\uf061\uf070\uf070\uf065\uf06e\uf073\uf020\uf077\uf068\uf065\uf06e\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\uf020\uf072\uf065\uf061\uf063\uf074\uf020\uf077\uf069\uf074\uf068\uf020\uf057", + "\uf061\uf074\uf065\uf072\uf03f\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf030\nCAUTION: This Activity needs the teacher\u2019s assistance.\n\uf06eCollect the samples of the same metals as in Activity 3.9.", + "This Activity needs the teacher\u2019s assistance.\n\uf06eCollect the samples of the same metals as in Activity 3.9.\n\uf06ePut small pieces of the samples separately in beakers half-filled\nwith cold water .", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf06eCollect the samples of the same metals as in Activity 3.9.\n\uf06ePut small pieces of the samples separately in beakers half-filled\nwith cold water .\n\uf06eWhich metals reacted with cold water? Arrange them in the\nincreasing order of their r eactivity with cold water.\n\uf06eDid any metal produce fire on water?\n\uf06eDoes any metal start floating after some time?\n\uf06ePut the metals that did not react with cold water in beakers\nhalf-filled with hot water.\n\uf06eFor the metals that did not react with hot water , arrange the\napparatus as shown in Fig. 3.3 and observe their reaction with steam.\n\uf06eWhich metals did not react even with steam?\n\uf06eArrange the metals in the decreasing order of reactivity with water .\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf034\uf033Metals react with water and produce a metal oxide and hydrogen\ngas.", + "2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf034\uf033Metals react with water and produce a metal oxide and hydrogen\ngas. Metal oxides that are soluble in water dissolve in it to further form\nmetal hydroxide. But all metals do not react with water .", + "CHAPTER 3- METALS AND NON METALS-text book.txt\ngas. Metal oxides that are soluble in water dissolve in it to further form\nmetal hydroxide. But all metals do not react with water .\nMetal +Water\u2192Metal oxide + Hydrogen\nMetal oxide +Water\u2192Metal hydroxide\nMetals like potassium and sodium r eact violently with cold water . In\ncase of sodium and potassium, the reaction is so violent and exothermic\nthat the evolved hydrogen immediately catches fire.\n2K(s) + 2H2O(l) \u2192 2KOH(aq) + H2(g) + heat energy\n2Na(s) + 2H2O(l) \u2192 2NaOH(aq) + H2(g) + heat energy\nThe reaction of calcium with water is less violent. The heat evolved is\nnot sufficient for the hydrogen to catch fire.\nCa(s) + 2H2O(l) \u2192 Ca(OH)2(aq) + H2(g)\nCalcium starts floating because the bubbles of hydrogen gas formed\nstick to the surface of the metal.\nMagnesium does not r eact with cold water .", + "Ca(s) + 2H2O(l) \u2192 Ca(OH)2(aq) + H2(g)\nCalcium starts floating because the bubbles of hydrogen gas formed\nstick to the surface of the metal.\nMagnesium does not r eact with cold water . It reacts with hot water\nto form magnesium hydroxide and hydrogen. It also starts floating dueto the bubbles of hydrogen gas sticking to its surface.", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nMagnesium does not r eact with cold water . It reacts with hot water\nto form magnesium hydroxide and hydrogen. It also starts floating dueto the bubbles of hydrogen gas sticking to its surface.\nMetals like aluminium, iron and zinc do not react either with cold or\nhot water. But they r eact with steam to for m the metal oxide and hydrogen.\n2Al(s) + 3H\n2O(g) \u2192 Al2O3(s) + 3H2(g)\n3Fe(s) +4H2O(g) \u2192 Fe3O4(s) + 4H2(g)\nMetals such as lead, copper , silver and gold do not r eact with water at all.", + "\uf033\uf02e\uf032\uf02e\uf033\uf020\uf057\uf068\uf061\uf074\uf020\uf068\uf061\uf070\uf070\uf065\uf06e\uf073\uf020\uf077\uf068\uf065\uf06e\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\uf020\uf072\uf065\uf061\uf063\uf074\uf020\uf077\uf069\uf074\uf068\uf020\uf041\uf063\uf069\uf064\uf073\uf03f\nYou have already learnt that metals react with acids to give a salt and\nhydrogen gas.Figure 3.3Figure 3.3Figure 3.3Figure 3.3Figure 3.3 Action of steam on a metal\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineScience \uf034\uf034Metal + Dilute acid \u2192 Salt + Hydrogen\nBut do all metals react in the same manner? Let us find out.", + "Let us find out.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf031", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nwww.cbse.onlineScience \uf034\uf034Metal + Dilute acid \u2192 Salt + Hydrogen\nBut do all metals react in the same manner? Let us find out.", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nwww.cbse.onlineScience \uf034\uf034Metal + Dilute acid \u2192 Salt + Hydrogen\nBut do all metals react in the same manner? Let us find out.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf031\n\uf06eCollect all the metal samples except sodium and potassium again.", + "If the samples ar e tar nished, rub them clean with sand paper .\nCAUTION: Do not take sodium and potassium as they react\nvigor ously even with cold water.\n\uf06ePut the samples separately in test tubes containing dilutehydrochloric acid.\n\uf06eSuspend thermometers in the test tubes, so that their bulbs aredipped in the acid.\n\uf06eObserve the rate of formation of bubbles carefully.\n\uf06eWhich metals reacted vigorously with dilute hydrochloric acid?\n\uf06eWith which metal did you record the highest temperature?\n\uf06eArrange the metals in the decreasing order of reactivity with dilute\nacids.\nWrite equations for the reactions of magnesium, aluminium, zinc\nand iron with dilute hydrochloric acid.", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf06eArrange the metals in the decreasing order of reactivity with dilute\nacids.\nWrite equations for the reactions of magnesium, aluminium, zinc\nand iron with dilute hydrochloric acid.\nHydrogen gas is not evolved when a metal reacts with nitric acid. It is\nbecause HNO3 is a strong oxidising agent. It oxidises the H2 produced to\nwater and itself gets reduced to any of the nitrogen oxides (N2O, NO,\nNO2). But magnesium (Mg) and manganese (Mn) react with very dilute\nHNO3 to evolve H2 gas.\nYou must have observed in Activity 3.11, that the rate of for mation\nof bubbles was the fastest in the case of magnesium. The reaction was\nalso the most exothermic in this case. The reactivity decreases in the\norder Mg > Al > Zn > Fe. In the case of copper , no bubbles wer e seen and\nthe temperature also remained unchanged.", + "The reaction was\nalso the most exothermic in this case. The reactivity decreases in the\norder Mg > Al > Zn > Fe. In the case of copper , no bubbles wer e seen and\nthe temperature also remained unchanged. This shows that copper doesnot react with dilute HCl.\uf044\uf06f\uf020\uf059\uf06f\uf075\uf020\uf04b\uf06e\uf06f\uf077\uf03fAqua regia, (Latin for \u2018royal water\u2019) is a freshly prepared mixture of concentrated", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nthe temperature also remained unchanged. This shows that copper doesnot react with dilute HCl.\uf044\uf06f\uf020\uf059\uf06f\uf075\uf020\uf04b\uf06e\uf06f\uf077\uf03fAqua regia, (Latin for \u2018royal water\u2019) is a freshly prepared mixture of concentrated\nhydrochloric acid and concentrated nitric acid in the ratio of 3:1. It can dissolve\ngold, even though neither of these acids can do so alone. Aqua regia is a highly\ncorrosive, fuming liquid. It is one of the few reagents that is able to dissolve gold andplatinum.\n\uf033\uf02e\uf032\uf02e\uf034\uf020\uf048\uf06f\uf077\uf020\uf064\uf06f\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\uf020\uf072\uf065\uf061\uf063\uf074\uf020\uf077\uf069\uf074\uf068", + "\uf020\uf064\uf06f\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\uf020\uf072\uf065\uf061\uf063\uf074\uf020\uf077\uf069\uf074\uf068\uf020\uf053\uf06f\uf06c\uf075\uf074\uf069\uf06f\uf06e\uf073\uf020\uf06f\uf066\uf020\uf06f\uf074\uf068\uf065\uf072\uf020\uf04d\uf065\uf074\uf061\uf06c\n\uf053\uf061\uf06c\uf074\uf073\uf03f\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069", + "\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf032\n\uf06eTake a clean wire of copper and an iron nail.\n\uf06ePut the copper wire in a solution of iron sulphate and the iron\nnail in a solution of copper sulphate taken in test tubes (Fig. 3.4).\n\uf06eRecord your observations after 20 minutes.", + "ake a clean wire of copper and an iron nail.\n\uf06ePut the copper wire in a solution of iron sulphate and the iron\nnail in a solution of copper sulphate taken in test tubes (Fig. 3.4).\n\uf06eRecord your observations after 20 minutes.\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf034\uf035\uf06eIn which test tube did you find that a reaction has occurred?", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf034\uf035\uf06eIn which test tube did you find that a reaction has occurred?\n\uf06eOn what basis can you say that a reaction has actually taken\nplace?\n\uf06eCan you correlate your observations for the Activities 3.9, 3.10and 3.11?\n\uf06eWrite a balanced chemical equation for the r eaction that has taken\nplace.\n\uf06eName the type of reaction.\nReactive metals can displace less\nreactive metals from their compounds in\nsolution or molten form.\nWe have seen in the previous sections\nthat all metals are not equally reactive. W e\nchecked the reactivity of various metalswith oxygen, water and acids. But all\nmetals do not react with these reagents.\nSo we were not able to put all the metal\nsamples we had collected in decreasing\norder of their reactivity. Displacement\nreactions studied in Chapter 1 give better\nevidence about the reactivity of metals.", + "But all\nmetals do not react with these reagents.\nSo we were not able to put all the metal\nsamples we had collected in decreasing\norder of their reactivity. Displacement\nreactions studied in Chapter 1 give better\nevidence about the reactivity of metals. It\nis simple and easy if metal A displaces", + "CHAPTER 3- METALS AND NON METALS-text book.txt\norder of their reactivity. Displacement\nreactions studied in Chapter 1 give better\nevidence about the reactivity of metals. It\nis simple and easy if metal A displaces\nmetal B from its solution, it is more reactive than B.\nMetal A + Salt solution of B \u2192 Salt solution of A + Metal B\nWhich metal, copper or iron, is more reactive according to your\nobservations in Activity 3.12?\n\uf033\uf02e\uf032\uf02e\uf035\uf020\uf054\uf068\uf065\uf020\uf052\uf065\uf061\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf053\uf065\uf072\uf069\uf065\uf073\nThe reactivity series is a list of metals arranged in the order of theirdecreasing activities. After performing displacement experiments\n(Activities 1.9 and 3.12), the following series, (Table 3.2) known as the\nreactivity or activity series has been developed.", + "After performing displacement experiments\n(Activities 1.9 and 3.12), the following series, (Table 3.2) known as the\nreactivity or activity series has been developed.\nTable 3.2 Activity series : Relative reactivities of metals\nK Potassium Most reactive\nNa Sodium\nCa Calcium\nMg Magnesium\nAl Aluminium\nZn Zinc Reactivity decreases\nFe Iron\nPb Lead\n[H] [Hydrogen]\nCu Copper\nHg Mercury\nAg Silver\nAu Gold Least reactiveFigure 3.4Figure 3.4Figure 3.4Figure 3.4Figure 3.4", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nMg Magnesium\nAl Aluminium\nZn Zinc Reactivity decreases\nFe Iron\nPb Lead\n[H] [Hydrogen]\nCu Copper\nHg Mercury\nAg Silver\nAu Gold Least reactiveFigure 3.4Figure 3.4Figure 3.4Figure 3.4Figure 3.4\nReaction of metals withsalt solutions\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineScience \uf034\uf036\uf03f\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n1. Why is sodium kept immersed in kerosene oil?\n2. Write equations for the r eactions of\n(i)iron with steam\n(ii)calcium and potassium with water\n3. Samples of four metals A, B, C and D were taken and added to the\nfollowing solution one by one. The results obtained have been tabulated\nas follows.", + "Write equations for the r eactions of\n(i)iron with steam\n(ii)calcium and potassium with water\n3. Samples of four metals A, B, C and D were taken and added to the\nfollowing solution one by one. The results obtained have been tabulated\nas follows.\n\uf033\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf044\uf04f\uf020\uf04d\uf045\uf054\uf033\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf044\uf04f\uf020\uf04d\uf045\uf054\uf033\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf044\uf04f\uf020\uf04d\uf045\uf054\uf033\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf044\uf04f\uf020\uf04d\uf045\uf054\uf033\uf02e\uf033\uf020\uf048\uf04f\uf057\uf020\uf044\uf04f\uf020\uf04d\uf045\uf054", + "\uf041\uf04c\uf053\uf020\uf041\uf04e\uf044\uf020\uf04e\uf04f\uf04e\uf02d\uf04d\uf045\uf054\uf041\uf04c\uf053\uf020\uf041\uf04e\uf044\uf020\uf04e\uf04f\uf04e\uf02d\uf04d\uf045\uf054\uf041\uf04c\uf053\uf020\uf041\uf04e\uf044\uf020\uf04e\uf04f\uf04e\uf02d\uf04d\uf045\uf054\uf041\uf04c\uf053\uf020\uf041\uf04e\uf044\uf020\uf04e\uf04f\uf04e\uf02d\uf04d\uf045\uf054\uf041\uf04c\uf053\uf020\uf041\uf04e\uf044\uf020\uf04e\uf04f\uf04e\uf02d\uf04d\uf045\uf054 \uf041\uf04c\uf053\uf020\uf052\uf045\uf041", + "\uf041\uf04c\uf053\uf020\uf052\uf045\uf041 \uf041\uf04c\uf053\uf020\uf052\uf045\uf041\uf041\uf04c\uf053\uf020\uf052\uf045\uf041 \uf041\uf04c\uf053\uf020\uf052\uf045\uf041\uf041\uf04c\uf053\uf020\uf052\uf045\uf041 \uf043\uf054\uf03f \uf043\uf054\uf03f\uf043\uf054\uf03f \uf043\uf054\uf03f\uf043\uf054\uf03f\nIn the above activities, you saw the reactions of metals with a number of\nreagents. Why do metals react in this manner? Let us recall what we", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nIn the above activities, you saw the reactions of metals with a number of\nreagents. Why do metals react in this manner? Let us recall what we\nlearnt about the electronic configuration of elements in Class IX. We\nlearnt that noble gases, which have a completely filled valence shell, showlittle chemical activity. W e, ther efore, explain the reactivity of elements\nas a tendency to attain a completely filled valence shell.\nLet us have a look at the electronic configuration of noble gases and\nsome metals and non-metals.\nWe can see fr om T able 3.3 that a sodium atom has one electr on in its\noutermost shell. If it loses the electron from its M shell then its L shell\nnow becomes the outermost shell and that has a stable octet. The nucleus\nof this atom still has 11 protons but the number of electrons hasbecome 10, so there is a net positive charge giving us a sodium cationNa", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nof this atom still has 11 protons but the number of electrons hasbecome 10, so there is a net positive charge giving us a sodium cationNa\n+. On the other hand chlorine has seven electrons in its outermost shellUse the Table above to answer the following questions about metals\nA, B, C and D.\n(i) Which is the most reactive metal?\n(ii) What would you observe if B is added to a solution of Copper(II)\nsulphate?\n(iii) Arrange the metals A, B, C and D in the order of decreasing\nreactivity.\n4. Which gas is produced when dilute hydrochloric acid is added to areactive metal? W rite the chemical r eaction when ir on reacts with dilute\nH\n2SO4.\n5. What would you observe when zinc is added to a solution of iron(II)sulphate?", + "W rite the chemical r eaction when ir on reacts with dilute\nH\n2SO4.\n5. What would you observe when zinc is added to a solution of iron(II)sulphate? Write the chemical reaction that takes place.Metal Iron(II) sulphate Copper(II) sulphate Zinc sulphate Silver nitrate\nA No reaction Displacement\nB Displacement No reaction\nC No reaction No reaction No reaction Displacement\nD No reaction No reaction No reaction No reaction", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nA No reaction Displacement\nB Displacement No reaction\nC No reaction No reaction No reaction Displacement\nD No reaction No reaction No reaction No reaction\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf034\uf037and it requires one more electron to complete its octet. If sodium and chlorine\nwere to react, the electron lost by sodium could be taken up by chlorine.After gaining an electron, the chlorine atom gets a unit negative charge,because its nucleus has 17 protons and there are 18 electrons in its K, L\nand M shells. This gives us a chloride anion C1\n\u2013. So both these elements\ncan have a give-and-take relation between them as follows (Fig. 3.5).", + "This gives us a chloride anion C1\n\u2013. So both these elements\ncan have a give-and-take relation between them as follows (Fig. 3.5).\nNa Na + e\n2,8,1 2,8+\n(Sodium cation )\u2192\u2013\nCl +e Cl\n2,8,7 2,8,8\n(Chloride anion )\u2013\u2013\u2192\nFigure 3.5Figure 3.5Figure 3.5Figure 3.5Figure 3.5 Formation of sodium chloride\nSodium and chloride ions, being oppositely charged, attract each", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nCl +e Cl\n2,8,7 2,8,8\n(Chloride anion )\u2013\u2013\u2192\nFigure 3.5Figure 3.5Figure 3.5Figure 3.5Figure 3.5 Formation of sodium chloride\nSodium and chloride ions, being oppositely charged, attract each\nother and are held by strong electrostatic forces of attraction to exist as\nsodium chloride (NaCl). It should be noted that sodium chloride does\nnot exist as molecules but aggregates of oppositely charged ions.\nLet us see the formation of one more ionic compound, magnesium\nchloride (Fig.", + "It should be noted that sodium chloride does\nnot exist as molecules but aggregates of oppositely charged ions.\nLet us see the formation of one more ionic compound, magnesium\nchloride (Fig. 3.6) .Table 3.3 Electronic configurations of some elements\nType of Element Atomic Number of\nelement number electrons in shells\nK LM N\nNoble gases Helium (He) 22\nNeon (Ne) 10 28\nArgon (Ar) 18 2 88\nMetals Sodium (Na) 11 2 81\nMagnesium (Mg) 12 2 82\nAluminium (Al) 13 2 83\nPotassium (K) 19 2 88 1\nCalcium (Ca) 20 2 88 2\nNon-metals Nitrogen (N) 7 25\nOxygen (O) 8 26\nFluorine (F) 9 27\nPhosphorus (P) 15 2 85\nSulphur (S) 16 2 86\nChlorine (Cl) 17 2 87\n2019-20\nDownload all NCERT books PDF from www.ncert.online", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nNon-metals Nitrogen (N) 7 25\nOxygen (O) 8 26\nFluorine (F) 9 27\nPhosphorus (P) 15 2 85\nSulphur (S) 16 2 86\nChlorine (Cl) 17 2 87\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineScience \uf034\uf038Mg Mg e2+\n(Magnesium cation )\uf8e7 \u2192\uf8e7 + 2\n282 28\u2013\n,, ,\nCl +e Cl\n2,8,7 2,8,8\n(Chloride anion)\u2013 \u2013\uf8e7 \u2192\uf8e7\nFigure 3.6Figure 3.6Figure 3.6Figure 3.6Figure 3.6 Formation of magnesium chloride\nThe compounds formed in this manner by the transfer of electrons\nfrom a metal to a non-metal are known as ionic compounds or\nelectrovalent compounds.", + "Can you name the cation and anion presentin MgCl\n2?\n\uf033\uf02e\uf033\uf02e\uf031\uf020\uf050\uf072\uf06f\uf070\uf065\uf072 \uf074\uf069\uf065\uf073\uf020\uf06f\uf066\uf020\uf049\uf06f\uf06e\uf069\uf063\uf020\uf043\uf06f\uf06d\uf070\uf06f\uf075\uf06e\uf064\uf073\nTo learn about the properties of ionic compounds,", + "let us perform the\nfollowing Activity:\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf033\n\uf06eTake samples of sodium chloride, potassium iodide, barium\nchloride or any other salt from the science laboratory.\n\uf06eWhat is the physical state of these salts?", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf06eTake samples of sodium chloride, potassium iodide, barium\nchloride or any other salt from the science laboratory.\n\uf06eWhat is the physical state of these salts?\n\uf06eTake a small amount of a sample on a metal spatula and\nheat directly on the flame (Fig. 3.7). Repeat with other samples.\n\uf06eWhat did you observe? Did the samples impart any colourto the flame? Do these compounds melt?\n\uf06eTry to dissolve the samples in water , petr ol and kerosene.\nAre they soluble?\n\uf06eMake a circuit as shown in Fig. 3.8 and insert the electrodesinto a solution of one salt. What did you observe? Test the\nother salt samples too in this manner.\n\uf06eWhat is your inference about the nature of thesecompounds?", + "Are they soluble?\n\uf06eMake a circuit as shown in Fig. 3.8 and insert the electrodesinto a solution of one salt. What did you observe? Test the\nother salt samples too in this manner.\n\uf06eWhat is your inference about the nature of thesecompounds?\nFigure 3.7Figure 3.7Figure 3.7Figure 3.7Figure 3.7\nHeating a salt sample on aspatula\nFigure 3.8Figure 3.8Figure 3.8Figure 3.8Figure 3.8Testing the conductivity ofa salt solutionTable 3.4 Melting and boiling points of some ionic compounds\nIonic Melting point Boiling point\ncompound (K) (K)\nNaCl 1074 1686\nLiCl 887 1600", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nIonic Melting point Boiling point\ncompound (K) (K)\nNaCl 1074 1686\nLiCl 887 1600\nCaCl21045 1900\nCaO 2850 3120\nMgCl2981 1685\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf034\uf039You may have observed the following general pr operties for ionic\ncompounds\u2014\n(i)Physical nature : Ionic compounds are solids and are somewhat\nhard because of the strong force of attraction between the positive\nand negative ions. These compounds are generally brittle andbreak into pieces when pressure is applied.\n(ii)Melting and Boiling points : Ionic compounds have high melting\nand boiling points (see Table 3.4). This is because a considerableamount of ener gy is r equired to break the strong inter -ionic\nattraction.\n(iii) Solubility : Electrovalent compounds are generally soluble in\nwater and insoluble in solvents such as kerosene, petrol, etc.", + "This is because a considerableamount of ener gy is r equired to break the strong inter -ionic\nattraction.\n(iii) Solubility : Electrovalent compounds are generally soluble in\nwater and insoluble in solvents such as kerosene, petrol, etc.\n(iv)Conduction of Electricity : The conduction of electricity through", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nwater and insoluble in solvents such as kerosene, petrol, etc.\n(iv)Conduction of Electricity : The conduction of electricity through\na solution involves the movement of charged particles. A solution\nof an ionic compound in water contains ions, which move to the\nopposite electrodes when electricity is passed through thesolution. Ionic compounds in the solid state do not conductelectricity because movement of ions in the solid is not possible\ndue to their rigid structure. But ionic compounds conduct\nelectricity in the molten state. This is possible in the molten statesince the elecrostatic forces of attraction between the oppositelycharged ions are overcome due to the heat. Thus, the ions move\nfreely and conduct electricity.\n\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. (i) Write the electron-dot structur es for sodium, oxygen and\nmagnesium.\n(ii) Show the formation of Na2O and MgO by the transfer of electrons.\n(iii) What are the ions present in these compounds?", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nmagnesium.\n(ii) Show the formation of Na2O and MgO by the transfer of electrons.\n(iii) What are the ions present in these compounds?\n2. Why do ionic compounds have high melting points?\n\uf033\uf02e\uf034 \uf033\uf02e\uf034\uf033\uf02e\uf034 \uf033\uf02e\uf034\uf033\uf02e\uf034\uf04f\uf043\uf043\uf055\uf052\uf052\uf045\uf04e\uf043\uf045\uf020\uf04f\uf046\uf020\uf04d\uf045\uf054\uf04f\uf043\uf043\uf055\uf052\uf052\uf045\uf04e\uf043\uf045\uf020\uf04f\uf046\uf020\uf04d\uf045\uf054\uf04f\uf043\uf043\uf055\uf052\uf052\uf045\uf04e\uf043\uf045\uf020\uf04f\uf046\uf020\uf04d", + "\uf04f\uf046\uf020\uf04d\uf045\uf054\uf04f\uf043\uf043\uf055\uf052\uf052\uf045\uf04e\uf043\uf045\uf020\uf04f\uf046\uf020\uf04d\uf045\uf054\uf04f\uf043\uf043\uf055\uf052\uf052\uf045\uf04e\uf043\uf045\uf020\uf04f\uf046\uf020\uf04d\uf045\uf054\uf04f\uf043\uf043\uf055\uf052\uf052\uf045\uf04e\uf043\uf045\uf020\uf04f\uf046\uf020\uf04d\uf045\uf054 \uf041\uf04c\uf053 \uf041\uf04c\uf053\uf041\uf04c\uf053 \uf041\uf04c\uf053\uf041\uf04c\uf053\nThe earth\u2019s crust is the major source of metals. Seawater also contains\nsome soluble salts such as sodium chloride, magnesium chloride, etc.", + "\uf041\uf04c\uf053\uf041\uf04c\uf053\nThe earth\u2019s crust is the major source of metals. Seawater also contains\nsome soluble salts such as sodium chloride, magnesium chloride, etc.\nThe elements or compounds, which occur naturally in the earth\u2019s crust,\nare known as minerals. At some places, minerals contain a very high\npercentage of a particular metal and the metal can be profitably extracted\nfrom it. These minerals are called ores.\n\uf033\uf02e\uf034\uf02e\uf031\uf020\uf045\uf078\uf074\uf072\uf061\uf063\uf074\uf069\uf06f\uf06e\uf020\uf06f\uf066\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\nYou have lear nt about the r eactivity series of metals. Having this\nknowledge, you can easily understand how a metal is extracted from its", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf033\uf02e\uf034\uf02e\uf031\uf020\uf045\uf078\uf074\uf072\uf061\uf063\uf074\uf069\uf06f\uf06e\uf020\uf06f\uf066\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\nYou have lear nt about the r eactivity series of metals. Having this\nknowledge, you can easily understand how a metal is extracted from its\nore. Some metals are found in the earth\u2019s crust in the free state. Someare found in the form of their compounds. The metals at the bottom ofthe activity series are the least reactive. They are often found in a free\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineScience \uf035\uf030state. For example, gold, silver , platinum and copper ar e found in the\nfree state. Copper and silver are also found in the combined state as\ntheir sulphide or oxide ores.", + "For example, gold, silver , platinum and copper ar e found in the\nfree state. Copper and silver are also found in the combined state as\ntheir sulphide or oxide ores. The metals at the top of the activity series(K, Na, Ca, Mg and Al) are so reactive that they are never found in\nnature as free elements. The metals in the middle of the activity series\n(Zn, Fe, Pb, etc.) are moderately reactive. They are found in the earth\u2019scrust mainly as oxides, sulphides or carbonates. Y ou will find that", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n(Zn, Fe, Pb, etc.) are moderately reactive. They are found in the earth\u2019scrust mainly as oxides, sulphides or carbonates. Y ou will find that\nthe ores of many metals are oxides. This is because oxygen is a veryreactive element and is very abundant on the earth.\nThus on th e basis of reactivity, we can group the metals into the\nfollowing three categories (Fig. 3.9) \u2013 (i) Metals of low reactivity; (ii) Metalsof medium reactivity; (iii) Metals of high reactivity.", + "Thus on th e basis of reactivity, we can group the metals into the\nfollowing three categories (Fig. 3.9) \u2013 (i) Metals of low reactivity; (ii) Metalsof medium reactivity; (iii) Metals of high reactivity. Different techniquesare to be used for obtaining the metals falling in each category.K\nNaCa\nMg\nAl\nZn\nFe\nPbCu\nAg\nAuReduction using\ncarbon\nFound in nativestateElectrolysis\nFigure 3.9Figure 3.9Figure 3.9Figure 3.9Figure 3.9\nActivity series andrelated metallurgy\nFigure 3.10Figure 3.10Figure 3.10Figure 3.10Figure 3.10 Steps involved in the extraction of metals from ores\n\uf033\uf02e\uf034\uf02e\uf032\uf045\uf06e\uf072\uf069\uf063\uf068\uf06d\uf065\uf06e\uf074\uf020\uf06f\uf066\uf020\uf04f\uf072\uf065\uf073\nOres mined from the earth are usually contaminated with large amounts", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf033\uf02e\uf034\uf02e\uf032\uf045\uf06e\uf072\uf069\uf063\uf068\uf06d\uf065\uf06e\uf074\uf020\uf06f\uf066\uf020\uf04f\uf072\uf065\uf073\nOres mined from the earth are usually contaminated with large amounts\nof impurities such as soil, sand, etc., called gangue. The impurities must\nbe removed from the ore prior to the extraction of the metal. The processesSeveral steps are involved in the extraction of pure metal from\nores. A summary of these steps is given in Fig.3.10. Each step is\nexplained in detail in the following sections.\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf035\uf031used for removing the gangue from the ore are based on the differences\nbetween the physical or chemical properties of the gangue and the ore.\nDifferent separation techniques are accordingly employed.", + "Different separation techniques are accordingly employed.\n\uf033\uf02e\uf034\uf02e\uf033\uf045\uf078\uf074\uf072\uf061\uf063\uf074\uf069\uf06e\uf067\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\uf020\uf04c\uf06f\uf077\uf020\uf069\uf06e\uf020\uf074\uf068\uf065\uf020\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf053\uf065\uf072\uf069\uf065\uf073\nMetals low in the activity series are very unreactive. The oxides of thesemetals can be reduced to metals by heating alone. For example, cinnabar\n(HgS) is an ore of mer cury. When it is heated in air , it is first converted", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n(HgS) is an ore of mer cury. When it is heated in air , it is first converted\ninto mercuric oxide (HgO). Mercuric oxide is then reduced to mercuryon further heating.\n2HgS(s) + 3O (g) 2HgO(s) + 2SO (g)2 2Heat\uf8e7\u2192\uf8e7 \uf8e7 \uf8e7\n2HgO(s) 2Hg(l) + O (g)2Heat\uf8e7\u2192\uf8e7 \uf8e7 \uf8e7\nSimilarly, copper which is found as Cu2S in nature can be obtained\nfrom its or e by just heating in air.\n2Cu S + 3O (g) 2Cu O(s) + 2SO (g)\n2Cu O + Cu S 22 22\n22Heat\nHeat\uf8e7\u2192\uf8e7 \uf8e7 \uf8e7\n\uf8e7\u2192\uf8e7 \uf8e7 \uf8e7 \uf8e76Cu(s) + SO", + "2Cu O(s) + 2SO (g)\n2Cu O + Cu S 22 22\n22Heat\nHeat\uf8e7\u2192\uf8e7 \uf8e7 \uf8e7\n\uf8e7\u2192\uf8e7 \uf8e7 \uf8e7 \uf8e76Cu(s) + SO (g)2\n\uf033\uf02e\uf034\uf02e\uf034\uf020\uf045\uf078\uf074\uf072\uf061\uf063\uf074\uf069\uf06e\uf067\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\uf020\uf069\uf06e\uf020\uf074\uf068\uf065\uf020\uf04d\uf069\uf064\uf064\uf06c\uf065\uf020\uf06f\uf066\uf020\uf074\uf068\uf065\uf020\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf053\uf065\uf072\uf069\uf065\uf073\nThe metals in the middle of", + "metals in the middle of the activity series such as iron, zinc, lead,\ncopper , are moderately reactive. These are usually present as sulphides\nor carbonates in nature. It is easier to obtain a metal from its oxide, ascompared to its sulphides and carbonates. Therefore, prior to reduction,\nthe metal sulphides and carbonates must be converted into metal\noxides. The sulphide ores are converted into oxides by heating strongly", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nthe metal sulphides and carbonates must be converted into metal\noxides. The sulphide ores are converted into oxides by heating strongly\nin the pr esence of excess air. This process is known as roasting. The\ncarbonate ores are changed into oxides by heating strongly in limitedair. This process is known as calcination. The chemical reaction that\ntakes place during roasting and calcination of zinc ores can be shownas follows \u2013\nRoasting\n2ZnS(s) + 3O (g) 2ZnO(s) + 2SO (g)2 2Heat\uf8e7\u2192\uf8e7 \uf8e7\uf8e7\nCalcination\nZnCO (s) ZnO(s) + CO (g)3 2Heat\uf8e7\u2192\uf8e7 \uf8e7 \uf8e7\nThe metal oxides are then reduced to the corresponding metals by\nusing suitable reducing agents such as carbon. For example, when zincoxide is heated with carbon, it is reduced to metallic zinc.", + "For example, when zincoxide is heated with carbon, it is reduced to metallic zinc.\nZnO(s) + C(s) \u2192 Zn(s) + CO(g)\nYou ar e already familiar with the pr ocess of oxidation and r eduction\nexplained in the first Chapter . Obtaining metals fr om their compounds\nis also a reduction process.\nBesides using carbon (coke) to reduce metal oxides to metals,", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nexplained in the first Chapter . Obtaining metals fr om their compounds\nis also a reduction process.\nBesides using carbon (coke) to reduce metal oxides to metals,\nsometimes displacement reactions can also be used. The highly reactive\nmetals such as sodium, calcium, aluminium, etc., are used as reducing\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineScience \uf035\uf032agents because they can displace metals of lower reactivity from their\ncompounds. For example, when manganese dioxide is heated with\naluminium powder , the following reaction takes place \u2013\n3MnO2(s) + 4Al(s) \u2192 3Mn(l) + 2Al2O3(s) + Heat\nCan you identify the substances that are getting oxidised\nand reduced?\nThese displacement reactions are highly exothermic. The\namount of heat evolved is so large that the metals are producedin the molten state. In fact, the reaction of iron(III) oxide (Fe\n2O3)", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nThese displacement reactions are highly exothermic. The\namount of heat evolved is so large that the metals are producedin the molten state. In fact, the reaction of iron(III) oxide (Fe\n2O3)\nwith aluminium is used to join railway tracks or crackedmachine parts. This reaction is known as the thermit reaction.", + "The\namount of heat evolved is so large that the metals are producedin the molten state. In fact, the reaction of iron(III) oxide (Fe\n2O3)\nwith aluminium is used to join railway tracks or crackedmachine parts. This reaction is known as the thermit reaction.\nFe\n2O3(s) + 2Al(s) \u2192 2Fe(l) + Al2O3(s) + Heat\n\uf033\uf02e\uf034\uf02e\uf035\uf020\uf045\uf078\uf074\uf072\uf061\uf063\uf074\uf069\uf06e\uf067\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\uf020\uf074\uf06f\uf077\uf061\uf072\uf064\uf073\uf020\uf074\uf068\uf065\uf020\uf054 \uf06f\uf070\uf020\uf06f\uf066\uf020\uf074\uf068\uf065\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf053\uf065\uf072\uf069\uf065\uf073\nThe metals high up in the reactivity series are very reactive.", + "They cannotbe obtained from their compounds by heating with carbon. For example,\ncarbon cannot reduce the oxides of sodium, magnesium, calcium,\naluminium, etc., to the respective metals. This is because these metals\nhave more affinity for oxygen than carbon. These metals are obtained\nby electrolytic reduction. For example, sodium, magnesium and calcium\nare obtained by the electrolysis of their molten chlorides. The metals\nare deposited at the cathode (the negatively charged electrode), whereas,", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nare obtained by the electrolysis of their molten chlorides. The metals\nare deposited at the cathode (the negatively charged electrode), whereas,\nchlorine is liberated at the anode (the positively charged electrode). The\nreactions are \u2013\nAt cathode Na+ + e\u2013\u2192Na\nAt anode 2Cl \u2013\u2192Cl\n2 + 2e \u2013\nSimilarly, aluminium is obtained by the electrolytic reduction of\naluminium oxide.\n\uf033\uf02e\uf034\uf02e\uf036\uf020\uf052\uf065\uf066\uf069\uf06e\uf069\uf06e\uf067\uf020\uf06f\uf066\uf020\uf04d\uf065\uf074\uf061\uf06c\uf073\nThe metals produced by various reduction processesdescribed above are not very pure. They contain\nimpurities, which must be removed to obtain pure metals.\nThe most widely used method for refining impure metals\nis electrolytic refining.", + "They contain\nimpurities, which must be removed to obtain pure metals.\nThe most widely used method for refining impure metals\nis electrolytic refining.\nElectrolytic Refining: Electrolytic Refining: Electrolytic Refining: Electrolytic Refining: Electrolytic Refining: Many metals, such as copper , zinc,\ntin, nickel, silver, gold, etc., ar e refined electr olytically. In\nthis process, the impure metal is made the anode and athin strip of pure metal is made the cathode. A solution of", + "CHAPTER 3- METALS AND NON METALS-text book.txt\ntin, nickel, silver, gold, etc., ar e refined electr olytically. In\nthis process, the impure metal is made the anode and athin strip of pure metal is made the cathode. A solution of\nthe metal salt is used as an electrolyte. The apparatus is\nset up as shown in Fig. 3.12. On passing the current\nthrough the electrolyte, the pure metal from the anode\ndissolves into the electrolyte. An equivalent amount of pure\nFigure 3.11Figure 3.11Figure 3.11Figure 3.11Figure 3.11\nThermit process forjoining railway tracks\nFigure 3.12Figure 3.12Figure 3.12Figure 3.12Figure 3.12Electr olytic r efining of copper . The\nelectrolyte is a solution of acidified coppersulphate. The anode is impur e copper ,\nwhereas, the cathode is a strip of purecopper .", + "The\nelectrolyte is a solution of acidified coppersulphate. The anode is impur e copper ,\nwhereas, the cathode is a strip of purecopper . On passing electric curr ent, pur e\ncopper is deposited on the cathode.\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf035\uf033metal from the electrolyte is deposited on the cathode. The soluble", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf035\uf033metal from the electrolyte is deposited on the cathode. The soluble\nimpurities go into the solution, whereas, the insoluble impurities settle\ndown at the bottom of the anode and are known as anode mud.\n1. Define the following terms.\n(i)Mineral (ii) Ore (iii) Gangue\n2. Name two metals which are found in nature in the free state.\n3. What chemical process is used for obtaining a metal from its oxide?", + "1. Define the following terms.\n(i)Mineral (ii) Ore (iii) Gangue\n2. Name two metals which are found in nature in the free state.\n3. What chemical process is used for obtaining a metal from its oxide?\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf033\uf02e\uf031\uf034\n\uf06eTake three test tubes and place clean iron nails\nin each of them.", + "\uf06eLabel these test tubes A, B and C. Pour somewater in test tube A and cork it.\n\uf06ePour boiled distilled water in test tube B, addabout 1 mL of oil and cork it. The oil will float on\nwater and prevent the air from dissolving in the\nwater .\n\uf06ePut some anhydrous calcium chloride intest tube C and cork it. Anhydrous calcium", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nwater and prevent the air from dissolving in the\nwater .\n\uf06ePut some anhydrous calcium chloride intest tube C and cork it. Anhydrous calcium\nchloride will absorb the moisture, if any, from\nthe air . Leave these test tubes for a few days and\nthen observe (Fig. 3.13).\nFigure 3.13Figure 3.13Figure 3.13Figure 3.13Figure 3.13\nInvestigating the conditions under which ironrusts. In tube A, both air and water arepresent. In tube B, there is no air dissolvedin the water . In tube C, the air is dry .You will observe that iron nails rust in test tube A,\nbut they do not rust in test tubes B and C. In the test\ntube A, the nails ar e exposed to both air and water . In\nthe test tube B, the nails are exposed to only water , and\nthe nails in test tube C are exposed to dry air . What\ndoes this tell us about the conditions under which ironarticles", + "In\nthe test tube B, the nails are exposed to only water , and\nthe nails in test tube C are exposed to dry air . What\ndoes this tell us about the conditions under which ironarticles rust?\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f\n\uf033\uf02e\uf035\uf020\uf043\uf04f\uf052\uf052\uf04f\uf053\uf049\uf04f\uf04e\uf033\uf02e\uf035\uf020\uf043\uf04f\uf052\uf052\uf04f\uf053\uf049\uf04f\uf04e\uf033\uf02e\uf035\uf020\uf043\uf04f\uf052\uf052\uf04f\uf053\uf049\uf04f\uf04e\uf033\uf02e\uf035\uf020\uf043\uf04f\uf052\uf052\uf04f\uf053\uf049\uf04f\uf04e\uf033\uf02e\uf035\uf020\uf043\uf04f\uf052\uf052\uf04f\uf053\uf049\uf04f\uf04e\nYou", + "hav e learnt the following about corrosion in Chapter 1 \u2013", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf03f\n\uf033\uf02e\uf035\uf020\uf043\uf04f\uf052\uf052\uf04f\uf053\uf049\uf04f\uf04e\uf033\uf02e\uf035\uf020\uf043\uf04f\uf052\uf052\uf04f\uf053\uf049\uf04f\uf04e\uf033\uf02e\uf035\uf020\uf043\uf04f\uf052\uf052\uf04f\uf053\uf049\uf04f\uf04e\uf033\uf02e\uf035\uf020\uf043\uf04f\uf052\uf052\uf04f\uf053\uf049\uf04f\uf04e\uf033\uf02e\uf035\uf020\uf043\uf04f\uf052\uf052\uf04f\uf053\uf049\uf04f\uf04e\nYou hav e learnt the following about corrosion in Chapter 1 \u2013\n\uf06eSilver articles become black after some time when exposed to air .", + "This is because it reacts with sulphur in the air to form a coating\nof silver sulphide.\n\uf06eCopper reacts with moist carbon dioxide in the air and slowly losesits shiny brown surface and gains a green coat. This green\nsubstance is basic copper carbonate.\n\uf06eIron when exposed to moist air for a long time acquires a coatingof a brown flaky substance called rust.Let us find ou\nt the conditions under which iron rusts.\nA B C\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineScience \uf035\uf034 \uf044\uf06f\uf020\uf059\uf06f\uf075\uf020\uf04b\uf06e\uf06f\uf077\uf03fPure gold, known as 24 carat gold, is very soft. It is, therefore, not suitable for making\njewellery. It is alloyed with either silver or copper to make it hard. Generally, in India,\n22 carat gold is used for making ornaments. It means that 22 parts of pure gold is", + "CHAPTER 3- METALS AND NON METALS-text book.txt\njewellery. It is alloyed with either silver or copper to make it hard. Generally, in India,\n22 carat gold is used for making ornaments. It means that 22 parts of pure gold is\nalloyed with 2 parts of either copper or silver.\nIf one of the metals is mercury, then the alloy is known as an\namalgam. The electrical conductivity and melting point of an alloy is\nless than that of pure metals. For example, brass, an alloy of copper andzinc (Cu and Zn), and bronze, an alloy of copper and tin (Cu and Sn), arenot good conductors of electricity whereas copper is used for making\nelectrical circuits. Solder, an alloy of lead and tin (Pb and Sn), has a low\nmelting point and is used for welding electrical wir es together .", + "Solder, an alloy of lead and tin (Pb and Sn), has a low\nmelting point and is used for welding electrical wir es together .\n\uf033\uf02e\uf035\uf02e\uf031\uf020\uf050\uf072\uf065\uf076\uf065\uf06e\uf074\uf069\uf06f\uf06e\uf020\uf06f\uf066\uf020\uf043\uf06f\uf072\uf072\uf06f\uf073\uf069\uf06f\uf06e\nThe rusting of iron can be prevented by painting, oiling, greasing,galvanising, chrome plating, anodising or making alloys.\nGalvanisation is a method of protecting steel and iron from rusting", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nThe rusting of iron can be prevented by painting, oiling, greasing,galvanising, chrome plating, anodising or making alloys.\nGalvanisation is a method of protecting steel and iron from rusting\nby coating them with a thin layer of zinc. The galvanised article is\nprotected against rusting even if the zinc coating is broken. Can you\nreason this out?\nAlloying is a very good method of improving the properties of a metal.\nWe can get the desir ed pr operties by this method. For example, ir on is\nthe most widely used metal. But it is never used in its pure state. This isbecause pure iron is very soft and stretches easily when hot. But, if it is\nmixed with a small amount of carbon (about 0.05 %), it becomes hardand strong. When iron is mixed with nickel and chromium, we getstainless steel, which is hard and does not rust. Thus, if iron is mixed\nwith some other substance, its properties change. In fact, the properties", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nwith some other substance, its properties change. In fact, the properties\nof any metal can be changed if it is mixed with some other substance.The substance added may be a metal or a non-metal. An alloy is ahomogeneous mixture of two or more metals, or a metal and a non-\nmetal. It is prepared by first melting the primary metal, and then,\ndissolving the other elements in it in definite proportions. It is then cooledto room temperature.\nIron pillar at DelhiThe wonder of ancient Indian metallurgy\nThe iron pillar near the Qutub Minar in Delhi was built more\nthan 1600 years ago by the iron workers of India. They had\ndeveloped a process which prevented iron from rusting. Forits quality of rust resistance it has been examined by scientistsfrom all parts of the world.", + "They had\ndeveloped a process which prevented iron from rusting. Forits quality of rust resistance it has been examined by scientistsfrom all parts of the world. The iron pillar is 8 m high and\nweighs 6 tonnes (6000 kg).\uf04d\uf06f\uf072\uf065\uf020\uf074\uf06f\uf020\uf04b\uf06e\uf06f\uf077\uf021\n2019-20\nDownload all NCERT books PDF from www.ncert.online", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nweighs 6 tonnes (6000 kg).\uf04d\uf06f\uf072\uf065\uf020\uf074\uf06f\uf020\uf04b\uf06e\uf06f\uf077\uf021\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf035\uf035In which cases will you find displacement reactions taking place?\n2. Which metals do not corrode easily?\n3. What are alloys?\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. Metallic oxides of zinc, magnesium and copper were heated with the\nfollowing metals.", + "3. What are alloys?\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. Metallic oxides of zinc, magnesium and copper were heated with the\nfollowing metals.\nMetal Zinc Magnesium Copper\nZinc oxide\nMagnesium oxide\nCopper oxide\n\uf057\uf068\uf061\uf074\uf020\uf079 \uf06f\uf075\uf020\uf068\uf061\uf076\uf065\uf020\uf06c\uf065\uf061\uf072\uf06e\uf074\n\uf06eElements can be classified as metals and non-metals.\n\uf06eMetals are lustrous, malleable, ductile and are good conductors of heat and\nelectricity. They are solids at room temperature, except mercury which is a liquid.\n\uf06eMetals can form positive ions by losing electrons to non-metals.\n\uf06eMetals combine with oxygen to form basic oxides. Aluminium oxide and zinc oxide\nshow the properties of both basic as well as acidic oxides.", + "\uf06eMetals can form positive ions by losing electrons to non-metals.\n\uf06eMetals combine with oxygen to form basic oxides. Aluminium oxide and zinc oxide\nshow the properties of both basic as well as acidic oxides. These oxides are knownas amphoteric oxides.\n\uf06eDifferent metals have different reactivities with water and dilute acids.", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nshow the properties of both basic as well as acidic oxides. These oxides are knownas amphoteric oxides.\n\uf06eDifferent metals have different reactivities with water and dilute acids.\n\uf06eA list of common metals arranged in order of their decreasing reactivity is known\nas an activity series.\n\uf06eMetals above hydrogen in the Activity series can displace hydrogen from dilute\nacids.\n\uf06eA more reactive metal displaces a less reactive metal from its salt solution.\n\uf06eMetals occur in nature as free elements or in the form of their compounds.\n\uf06eThe extraction of metals from their ores and then refining them for use is knownas metallurgy.\n\uf06eAn alloy is a homogeneous mixture of two or more metals, or a metal and a\nnon-metal.\n\uf06eThe surface of some metals, such as iron, is corroded when they are exposed tomoist air for a long period of time. This phenomenon is known as corrosion.\n\uf06eNon-metals have properties opposite to that of metals. They are neither malleable", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n\uf06eNon-metals have properties opposite to that of metals. They are neither malleable\nnor ductile. They are bad conductors of heat and electricity, except for graphite,\nwhich conducts electricity.\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineScience \uf035\uf036\uf06eNon-metals form negatively charged ions by gaining electrons when reacting with\nmetals.\n\uf06eNon-metals form oxides which are either acidic or neutral.\n\uf06eNon-metals do not displace hydrogen from dilute acids. They react with hydrogen\nto form hydrides.\n\uf045\uf058\uf045\uf052\uf043\uf049\uf053\uf045\uf053\n1. Which of the following pairs will give displacement reactions?\n(a) NaCl solution and copper metal\n(b) MgCl2 solution and aluminium metal\n(c) FeSO4 solution and silver metal\n(d) AgNO3 solution and copper metal.\n2.Which of the following methods is suitable for preventing an iron frying pan from\nrusting?", + "(a) NaCl solution and copper metal\n(b) MgCl2 solution and aluminium metal\n(c) FeSO4 solution and silver metal\n(d) AgNO3 solution and copper metal.\n2.Which of the following methods is suitable for preventing an iron frying pan from\nrusting?\n(a) Applying grease\n(b) Applying paint\n(c) Applying a coating of zinc\n(d) All of the above.", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n2.Which of the following methods is suitable for preventing an iron frying pan from\nrusting?\n(a) Applying grease\n(b) Applying paint\n(c) Applying a coating of zinc\n(d) All of the above.\n3.An element reacts with oxygen to give a compound with a high melting point. Thiscompound is also soluble in water . The element is likely to be\n(a) calcium\n(b) carbon\n(c) silicon\n(d) iron.\n4.Food cans are coated with tin and not with zinc because\n(a) zinc is costlier than tin.\n(b) zinc has a higher melting point than tin.\n(c) zinc is more reactive than tin.\n(d) zinc is less reactive than tin.\n5.You ar e given a hammer, a battery, a bulb, wires and a switch.\n(a) How could you use them to distinguish between samples of metals and\nnon-metals?\n(b) Assess the usefulness of these tests in distinguishing between metals and\nnon-metals.\n6.What are amphoteric oxides? Give two examples of amphoteric oxides.", + "(a) How could you use them to distinguish between samples of metals and\nnon-metals?\n(b) Assess the usefulness of these tests in distinguishing between metals and\nnon-metals.\n6.What are amphoteric oxides? Give two examples of amphoteric oxides.\n7.Name two metals which will displace hydrogen from dilute acids, and two metals\nwhich will not.\n2019-20", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nnon-metals.\n6.What are amphoteric oxides? Give two examples of amphoteric oxides.\n7.Name two metals which will displace hydrogen from dilute acids, and two metals\nwhich will not.\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.onlineMetals and Non-metals \uf035\uf0378.In the electrolytic refining of a metal M, what would you take as the anode, the\ncathode and the electrolyte?\n9.Pratyush took sulphur powder on a spatula and heated it. He collected the gas\nevolved by inverting a test tube over it, as shown in figure below.\n(a) What will be the action of gas on\n(i)dry litmus paper?\n(ii) moist litmus paper?\n(b) Write a balanced chemicalequation for the reaction\ntaking place.\n10. State two ways to prevent the\nrusting of iron.\n11. What type of oxides are formed when\nnon-metals combine with oxygen?\n12.", + "(ii) moist litmus paper?\n(b) Write a balanced chemicalequation for the reaction\ntaking place.\n10. State two ways to prevent the\nrusting of iron.\n11. What type of oxides are formed when\nnon-metals combine with oxygen?\n12. Give reasons\n(a) Platinum, gold and silver areused to make jewellery.\n(b) Sodium, potassium andlithium are stored under oil.", + "CHAPTER 3- METALS AND NON METALS-text book.txt\nnon-metals combine with oxygen?\n12. Give reasons\n(a) Platinum, gold and silver areused to make jewellery.\n(b) Sodium, potassium andlithium are stored under oil.\n(c) Aluminium is a highly reactivemetal, yet it is used to make\nutensils for cooking.\n(d) Carbonate and sulphide oresare usually converted into\noxides during the process of\nextraction.\n13. You must have seen tarnished copper vessels being cleaned with lemon or tamarind\njuice. Explain why these sour substances are effective in cleaning the vessels.\n14. Differentiate between metal and non-metal on the basis of their chemical properties.\n15. A man went door to door posing as a goldsmith. He promised to bring back the", + "CHAPTER 3- METALS AND NON METALS-text book.txt\n14. Differentiate between metal and non-metal on the basis of their chemical properties.\n15. A man went door to door posing as a goldsmith. He promised to bring back the\nglitter of old and dull gold ornaments. An unsuspecting lady gave a set of goldbangles to him which he dipped in a particular solution. The bangles sparkledlike new but their weight was reduced drastically. The lady was upset but after afutile argument the man beat a hasty retreat. Can you play the detective to findout the nature of the solution he had used?\n16. Give reasons why copper is used to make hot water tanks and not steel (an alloyof iron).Collection of gas\n2019-20\nDownload all NCERT books PDF from www.ncert.online\nwww.cbse.online", + "CHAPTER 3- METALS AND NON METALS-Reading NOTES.txt\nCLASS X \nCHAPTER 3 \nMETALS AND NON -METALS -PROPERTIES\nPART -I\nPHYSICAL PROPERTIES OF METALS\n\uf075Metals are solids. (except Mercury)\n\uf075Metals are hard. (except Lithium, Potassium, Sodium)\n\uf075Metals have metallic lustre. (shine)\n\uf075Metals are malleable. (can be beaten into thin sheets)\n\uf075Metals are ductile. (can be drawn into long wires)\n\uf075Metals have high melting points. (Gallium and Caesium have low melting \npoints. They melt in the palm of the hand)\n\uf075Metals are good conductors of heat. (Best conductors are Silver and \nCopper. Poor conductors are Lead and Mercury)\n\uf075Metals are good conductors of electricity. (Best conductors are Silver and \nCopper)\n\uf075Metals are sonorous.", + "(Best conductors are Silver and \nCopper. Poor conductors are Lead and Mercury)\n\uf075Metals are good conductors of electricity. (Best conductors are Silver and \nCopper)\n\uf075Metals are sonorous. (produce sound when beaten)\nPHYSICAL PROPERTIES OF NON METALS\n\uf075Non metals may be solids, liquids or gases. (Solids \u2013Carbon, Sulphur, \nPhosphorus etc. Liquid \u2013Bromine, Gases \u2013Oxygen, Hydrogen, Nitrogen \netc.)\n\uf075Non metals are soft. ( except Diamond which is the hardest natural \nsubstance)", + "CHAPTER 3- METALS AND NON METALS-Reading NOTES.txt\nPhosphorus etc. Liquid \u2013Bromine, Gases \u2013Oxygen, Hydrogen, Nitrogen \netc.)\n\uf075Non metals are soft. ( except Diamond which is the hardest natural \nsubstance)\n\uf075Non metals do not have lustre.( except Iodine)\n\uf075Non metals are not malleable.\n\uf075Non metals are not ductile.\n\uf075Non metals have low melting points and low boiling points.\n\uf075Non metals are bad conductors of heat.\n\uf075Non metals are bad conductors of electricity. ( except Graphite)\n\uf075Non metals are not sonorous.\nCHEMICAL PROPERTIES OF METALS\n1. Metals combine with oxygen to form metal oxides.", + "\uf075Non metals are bad conductors of heat.\n\uf075Non metals are bad conductors of electricity. ( except Graphite)\n\uf075Non metals are not sonorous.\nCHEMICAL PROPERTIES OF METALS\n1. Metals combine with oxygen to form metal oxides.\n\uf075Metal + Oxygen \u2192Metal oxide\n\uf0752Cu + O2\u21922CuO\n(Copper(II) oxide) \n\uf0752Mg+O2\u21922MgO\n(Magnesium oxide)\n\uf0754Al + 3O2\u21922Al2O3 Burning of Magnesium ribbon in air\n(Aluminium oxide)\nNATURE OF METAL OXIDES :\n\uf075Some metal oxides are basic oxides because they react with water \nto form bases .\n\uf075 4Na + O2\u21922Na2O\n\uf075 Na2O + H2O \u21922NaOH \n\uf075Some metal oxides show both acidic and basic properties.", + "CHAPTER 3- METALS AND NON METALS-Reading NOTES.txt\nto form bases .\n\uf075 4Na + O2\u21922Na2O\n\uf075 Na2O + H2O \u21922NaOH \n\uf075Some metal oxides show both acidic and basic properties. \nThey are called Amphoteric oxides.\nEg:-Aluminium oxide, Zinc oxide etc.", + "They are called Amphoteric oxides.\nEg:-Aluminium oxide, Zinc oxide etc.\n\uf075 Al2O3+ 6HCl \u2192AlCl3+ 3H2O\n\uf075 Al2O3+ 2NaOH \u21922NaAlO2+ H2O\n(Sodium aluminate)CHEMICAL PROPERTIES OF METALS\nCHEMICAL PROPERTIES OF METALS\n2.Metal + Water \u2192Metal oxide + Hydrogen\n\uf075Metal oxide + Water \u2192Metal hydroxide\n\uf0752K(s) + 2H2O(l) \u21922KOH(aq) + H2(g) + heat energy\n\uf0752Na(s) + 2H2O(l) \u21922NaOH(aq) + H2(g) + heat energy\n\uf075Metals like potassium and sodium react violently\nwith cold water.\n\uf075Ca(s) + 2H2O(l) \u2192Ca(OH)2(aq) + H2(g)\n\uf075Calcium starts floating because the bubbles of \nhydrogen gas formed stick to the surface of the metal.", + "\uf075Ca(s) + 2H2O(l) \u2192Ca(OH)2(aq) + H2(g)\n\uf075Calcium starts floating because the bubbles of \nhydrogen gas formed stick to the surface of the metal. \nCHEMICAL PROPERTIES OF METALS\n\uf075Metals like Aluminium, Iron and Zinc do \nnot react either with cold or hot water. \nBut they react with steam to form the \nmetal oxide and hydrogen.\n\uf0752Al(s) + 3H2O(g) \u2192Al2O3(s) + 3H2(g)", + "CHAPTER 3- METALS AND NON METALS-Reading NOTES.txt\n\uf075Metals like Aluminium, Iron and Zinc do \nnot react either with cold or hot water. \nBut they react with steam to form the \nmetal oxide and hydrogen.\n\uf0752Al(s) + 3H2O(g) \u2192Al2O3(s) + 3H2(g)\n\uf0753Fe(s) + 4H2O(g) \u2192Fe3O4(s) + 4H2(g)\n\uf075Metals such as lead, copper, silver and \ngold do not react with water at all.\nCHEMICAL PROPERTIES OF METALS\n3.Metal + Dilute acid \u2192Salt + Hydrogen\n\uf075Mg + 2HCl \u2192MgCl2 + H2\n\uf0752Al + 6HCl \u21922AlCl3+ 3H2\n\uf075Zn + 2HCl \u2192ZnCl2+ H2\n\uf075Copper, silver and gold do not react with dilute HCl.", + "\uf075Hydrogen gas is not evolved when metals react with \nnitric acid (HNO3) because it is a strong oxidising agent \nand it oxidises the H2produced to water and is itself \nreduced to oxides of nitrogen.\nCHEMICAL PROPERTIES OF METALS\n4. DISPLACEMENT REACTIONS\n\uf075Metal A + Salt solution of B \u2192Salt solution of A + Metal B\n(MORE REACTIVE) (LESS REACTIVE)\n\uf075Fe + CuSO4\u2192FeSO4+ Cu\nREACTIVITY \nSERIES OF \nMETALS\uf075K Potassium (Most reactive)\n\uf075Na Sodium", + "CHAPTER 3- METALS AND NON METALS-Reading NOTES.txt\n(MORE REACTIVE) (LESS REACTIVE)\n\uf075Fe + CuSO4\u2192FeSO4+ Cu\nREACTIVITY \nSERIES OF \nMETALS\uf075K Potassium (Most reactive)\n\uf075Na Sodium\n\uf075Ca Calcium\n\uf075Mg Magnesium\n\uf075Al Aluminium\n\uf075Zn Zinc Reactivity decreases\n\uf075Fe Iron\n\uf075Pb Lead\n\uf075H Hydrogen\n\uf075Cu Copper\n\uf075Hg Mercury\n\uf075Ag Silver\n\uf075Au Gold (Least reactive)\nHOW DO METALS AND NON -METALS \nREACT?\n\uf075Metals :-lose electrons and become positive ions. So, they \nare electropositive in nature .\n\uf075Non metals :-gain electrons and become negative ions. So, \nthey are electronegative in nature.\nEXAMPLE -FORMATION OF SODIUM CHLORIDE :\n\uf075The atomic number of Na is 11, its electronic configuration is \n2,8,1.", + "\uf075Non metals :-gain electrons and become negative ions. So, \nthey are electronegative in nature.\nEXAMPLE -FORMATION OF SODIUM CHLORIDE :\n\uf075The atomic number of Na is 11, its electronic configuration is \n2,8,1.\n\uf075It has 1 valence electron, so it loses 1 electron to form Na+ion. \n\uf075The atomic number of Cl is 17, its electronic configuration is 2,8,7.\n\uf075It has 7 valence electrons, so it gains 1 electron to form Cl-ion. \n\uf075Then, the attraction between the Na+ion and Cl-ion results in the", + "CHAPTER 3- METALS AND NON METALS-Reading NOTES.txt\n\uf075It has 7 valence electrons, so it gains 1 electron to form Cl-ion. \n\uf075Then, the attraction between the Na+ion and Cl-ion results in the \nformation of sodium chloride molecule, NaCl.\nProperties of Ionic Compounds\n\uf075They are formed by the transfer of electrons and \nare made up of ions.\n\uf075They are hard crystalline solids.\n\uf075They have high melting points and boiling points.\n\uf075They are soluble in water but insoluble in organic \nsolvents (like petrol, kerosene etc.)\n\uf075They conduct electricity in molten state or in \nsolution.", + "CHAPTER 3- METALS AND NON METALS-NOTES FOR CHILDREN.txt\nCLASS X \nCHAPTER 3 \nMETALS AND NON -METALS -\nMETALLURGY and CORROSION\nPART -II\nMETALLURGY\n\uf075The earth \u2019s crust is the major source of metals. \n\uf075Minerals -The elements or compounds, which occur \nnaturally in the earth \u2019s crust, are known as minerals. \n\uf075Ores -The minerals from which a metal can be \nprofitably and conveniently extracted are called ores. \n\uf075Gangue -The impurities present in the ore like rock \nparticles, sand particles, clay particles etc.METALLURGICAL STEPS\n\uf075Metals are extracted from their ores.", + "\uf075Gangue -The impurities present in the ore like rock \nparticles, sand particles, clay particles etc.METALLURGICAL STEPS\n\uf075Metals are extracted from their ores. Main steps \ninvolved are : -\n\uf075Concentration of the ore ( Ore -Enrichment )\n\uf075Conversion of concentrated ore into metal oxide\n( ROASTING/CALCINATION )\n\uf075Reduction to the metal oxide to metal\n\uf075Refining ( Purification of the metal )\u2022Extraction of metals low in the reactivity series \nEg:-Mercury is obtained from its ore Cinnabar \n(HgS) by heating in the presence of oxygen. \nWhen it is heated in the presence of oxygen it is \nfirst converted into mercuric oxide ( HgO) and on", + "CHAPTER 3- METALS AND NON METALS-NOTES FOR CHILDREN.txt\nEg:-Mercury is obtained from its ore Cinnabar \n(HgS) by heating in the presence of oxygen. \nWhen it is heated in the presence of oxygen it is \nfirst converted into mercuric oxide ( HgO) and on \nfurther heating it is reduced to mercury.\n\uf0d8 2HgS + 3O2\u21922HgO + 2SO2\n\uf0d8 2HgO \u21922Hg + O2\u2022Extraction of metals medium in the reactivity series \n\uf075Metals in the middle of the activity series like Zn, Fe, Pb, are found as \noxide, sulphide or carbonate ores. It is easier to obtain metals from their \noxides than from their sulphides or carbonates. \n\uf075Roasting is heating of concentrated ore in the presence of excess oxygen. \nIt is used to convert sulphide ore into metal oxide.", + "It is easier to obtain metals from their \noxides than from their sulphides or carbonates. \n\uf075Roasting is heating of concentrated ore in the presence of excess oxygen. \nIt is used to convert sulphide ore into metal oxide.\n\uf075 2ZnS + 3O2 \u21922ZnO + 2SO2\n\uf075Calcination is heating of concentrated ore in the limited supply of oxygen. \nIt is used to convert carbonate ore into metal oxide.\n\uf075 ZnCO3 \u2192ZnO + CO2\n\uf075The oxide ore is then reduced to the metal by heating with a reducing \nagent. The most common reducing agent is coke (carbon).", + "CHAPTER 3- METALS AND NON METALS-NOTES FOR CHILDREN.txt\n\uf075 ZnCO3 \u2192ZnO + CO2\n\uf075The oxide ore is then reduced to the metal by heating with a reducing \nagent. The most common reducing agent is coke (carbon).\n\uf075 ZnO + C \u2192Zn + CO\u2022Extraction of metals medium in the reactivity \nseries (THERMITE REACTION)\n\uf0d8Aluminium as a reducing agent\n\uf075The reaction between metal oxides and aluminium \nis highly exothermic and the metals are obtained in \nthe molten state. Such reactions are called \nthermite reactions .\n\uf0753MnO2+ 4Al \u21923Mn + 2Al2O3+ Heat\n\uf075The reaction between iron oxide and aluminium \nproduces molten iron. This reaction is used to join \nrail tracks, broken machine parts etc.", + "\uf0753MnO2+ 4Al \u21923Mn + 2Al2O3+ Heat\n\uf075The reaction between iron oxide and aluminium \nproduces molten iron. This reaction is used to join \nrail tracks, broken machine parts etc.\n\uf075Fe2O3+ 2Al \u2192Al2O3+ 2Fe + Heat\n\u2022Extraction of metals high in the reactivity series \n\uf075Metals atthetopoftheactivity series likeK,Na,\nCa,Aletc.cannot beobtained from their ores by\nsimple heating orbyheating with reducing agents .\nThey are obtained byElectrolytic Reduction of\ntheir molten chlorides .\n\uf075 Eg:-When electric current ispassed through", + "CHAPTER 3- METALS AND NON METALS-NOTES FOR CHILDREN.txt\nsimple heating orbyheating with reducing agents .\nThey are obtained byElectrolytic Reduction of\ntheir molten chlorides .\n\uf075 Eg:-When electric current ispassed through\nmolten sodium chloride, sodium metal isdeposited\natthecathode andchlorine gasisdeposited atthe\nanode .\n\uf075 At Cathode : Na ++ e -\u2192Na (Sodium metal)\n\uf075 At Anode : 2Cl -\u2192Cl2+ 2e -(Chlorine gas)ELECTROLYTIC REFINING OF COPPER\n\uf075The removal of impurities from the metal \nto obtain the pure metal is called refining \nof metals. The most common method for \nrefining of metals is Electrolytic Refining . \n\uf075A thick block of impure copper is made the \nanode and a thin sheet of pure copper is made \nthe cathode . The electrolyte is acidified copper \nsulphate solution .", + "The most common method for \nrefining of metals is Electrolytic Refining . \n\uf075A thick block of impure copper is made the \nanode and a thin sheet of pure copper is made \nthe cathode . The electrolyte is acidified copper \nsulphate solution .\n\uf075When electric current is passed through the \nelectrolyte, pure copper from the anode is \ndeposited at the cathode and the impurities \nsettle down as anode mud . \nCORROSION OF METALS\n\uf075CORROSION -The slow attack of atmospheric gases, moisture", + "CHAPTER 3- METALS AND NON METALS-NOTES FOR CHILDREN.txt\nelectrolyte, pure copper from the anode is \ndeposited at the cathode and the impurities \nsettle down as anode mud . \nCORROSION OF METALS\n\uf075CORROSION -The slow attack of atmospheric gases, moisture \nand acids on the surface of metals which results in the \nformation of a new layer of sulphide, carbonate, oxide etcto \nbe formed on the surface. \nCorrosion of metals can be prevented by :\n\uf075Applying oil or grease.\n\uf075Applying paint.\n\uf075By galvanization. (Coating with zinc)\n\uf075By tinning. (Coating with tin)\n\uf075By electroplating. (Coating a less reactive metal like chromium)\n\uf075By alloying. (Making alloys)To show that both air and moisture are necessary \nfor the RUSTING OF IRON \n\uf075PROCEDURE : Take three dry test tubes marked \n1,2,3 and put iron nails in each of them.", + "(Making alloys)To show that both air and moisture are necessary \nfor the RUSTING OF IRON \n\uf075PROCEDURE : Take three dry test tubes marked \n1,2,3 and put iron nails in each of them. Put \nsome anhydrous calcium chloride in test tube 1 to \nabsorb moisture. Pour some boiled distilled water \nin test tube 2 and pour some oil over it to \nprevent air into the test tube. Pour some water in", + "CHAPTER 3- METALS AND NON METALS-NOTES FOR CHILDREN.txt\nsome anhydrous calcium chloride in test tube 1 to \nabsorb moisture. Pour some boiled distilled water \nin test tube 2 and pour some oil over it to \nprevent air into the test tube. Pour some water in \ntest tube 3. Cork the test tubes and leave them \naside for a few days.\n\uf075OBSERVATIONS : The nails in test tube 1 does not \nrust because it had only air and no water. The nails in \ntest tube 2 does not rust because it had only water \nand no air. The nails in test tube 3 gets rusted because \nit had air and water. \nALLOYS\n\uf075An alloy is a homogeneous mixture of two or more metals or a metal \nand a non metal.\n\uf075The electrical conductivity and melting point of an alloy is less than pure \nmetals .", + "ALLOYS\n\uf075An alloy is a homogeneous mixture of two or more metals or a metal \nand a non metal.\n\uf075The electrical conductivity and melting point of an alloy is less than pure \nmetals .\nEXAMPLES:\n\uf075Steel\u2013Iron, Carbon ( hard, tough and strong)\n\uf075Stainless steel \u2013Iron, Chromium, Nickel( hard, does not rust)\n\uf075Brass\u2013Copper, Zinc (malleable, strong, resistant to corrosion)\n\uf075Bronze\u2013Copper, Tin ( very strong, highly resistant to corrosion)", + "CHAPTER 3- METALS AND NON METALS-NOTES FOR CHILDREN.txt\n\uf075Stainless steel \u2013Iron, Chromium, Nickel( hard, does not rust)\n\uf075Brass\u2013Copper, Zinc (malleable, strong, resistant to corrosion)\n\uf075Bronze\u2013Copper, Tin ( very strong, highly resistant to corrosion)\n\uf075Solder \u2013Lead, Tin (low melting point ,used for welding electrical \nwires together)\n\uf075If one of the metals in an alloy is mercury, it is called an amalgam .", + "CHAPTER 3- METALS AND NON METALS- LONG ANSWER.txt\nLONG ANSWER TYPE \n \nQuestion 1 . What are acidic, basic and amphoteric oxides? Give two examples of each. \n \nQuestion 2. Name two metals which will displace hydrogen from dilute acids, and two metals \nwhich will not. Also give the chemical equations for those metals which displace hydrogen \nfrom acid. \n \nQuestion 3 . Describe the electrolytic refining of impure copper metal and given the di agram of \nthe electrolytic cell and the reactions involved. \n \nQuestion 4 . Pratyush took sulphur powder on a spatula and heated it. He collected the gas \nevolved by inverting a test tube over it, as shown in figure below. \n(a) What will be the a ction of gas on \n(i) dry litmus paper? \n(ii) Moist litmus paper? \n(b) Write a balanced chemical equation for the reaction taking place. \n(c) How does this gas react with Hydrogen sulphide gas? Give equ ation. \n \nQuestion 5. Describe the steps involved in the extraction of metal from its ore.", + "CHAPTER 3- METALS AND NON METALS- LONG ANSWER.txt\nQuestion 5. Describe the steps involved in the extraction of metal from its ore. \n \n \nQuestion 6. Give reasons for the following: \n(a) Platin um, gold and silver are used to make jewellery. \n(b) Sodium, potassium and lithium are stored under oil. \n(c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking. \n(d) Carbonate and sulphide ores are usually converted into oxides during the process of \nextraction. \n(e) Zinc metal decolorizes the blue copper sulphate solution. \n \nQuestion 7. \n \ni) You must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. \nExplain why these sour substances are effective in cleaning the vessels. \n \nii) Differentiate between metal and non -metal o n the basis of their chemical properties.", + "CHAPTER 3- METALS AND NON METALS- LONG ANSWER.txt\nii) Differentiate between metal and non -metal o n the basis of their chemical properties. \n \niii) A man went door to door posing as a goldsmith. He promised to bring back the glitter of old \nand dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he \ndipped in a particular solution. The bangles sparkled like new but their weight was reduced \ndrastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can \nyou play the detective to find out the nature of the solution he had used? \n \niv) Give reasons why copper is used to make hot water tanks and not steel (an alloy of iron) is \nnot.", + "CHAPTER 3- METALS AND NON METALS-SHORT ANSWER.txt\nSHORT ANSWER TYPE \nQuestion 1. What happens when iron is heated to a high temperature? Give chemical equation. \n \nQuestion 2. What happens when copper is heated to a very high temperature . Give chemical \nequation. \n \nQuestion 3. Write an experiment to show that copper does not react with dilute HCI and H 2So4. \n \nQuestion 4. You are given a hammer, a battery, a bulb, wires and a switch. How could you use \nthem to distinguish between samples of metals and non -metals? Assess the usefulness of these \ntests in distinguishing between metals and non -metals. \n \nQuestion 5. What happens when Copper react with dilute and concentrated nitric acid? \n \nQuestion 6. How does metal displace Hydrogen from acid. Give Chemical equation. \n \nQuestion 7.Write one important use of Caustic soda and Plaster of Paris.", + "CHAPTER 3- METALS AND NON METALS-SHORT ANSWER.txt\nQuestion 6. How does metal displace Hydrogen from acid. Give Chemical equation. \n \nQuestion 7.Write one important use of Caustic soda and Plaster of Paris. \n \nQuestion 8. What do you mean by roasting and calcination? Illustrate with an appropriate \nexample each. \nQuestion 9. Explain briefly the froth flotation process for the concentration of sulphide ore. \nQuestion 10. Give chemical equations involved when Zinc Oxide and Ferric oxide are strongly \nheated with coke.", + "CHAPTER 3- METALS AND NON METALS-VERY SHORT ANSWER TYPE QUESTIONS.txt\nVERY SHORT ANSWER TYPE \n \nQuestion -1 \nName one metal which displaces hydrogen from acid . \nQuestion -2 \nName a metal which is stored in kerosene o il. \nQuestion -3 \nMetals are electropositive in nature. Why? \nQuestion -4 \nGive equation for the reaction of sodium with water. \nQuestion -5 \nWhy is sodium kept in kerosene ? \nQuestion -6 \nName the element or compound which is associated with Hall\u2019s process . \nQuestion -7 \nWhich metal is used in nuclear reactors and aerospace projects ? \nQuestion -8 \nName an ore of mercury ? \nQuestion -9 \nWrite the chemical formula of washing soda. \nQuestion -10. \nWrite on important use of bleaching powder. \n \nQuestion -11 \nWhat type of ores are concentrated by froth floatation process ? \nQuestion -12 \nWhat type of metals is extracted by electrolytic reduction ? \nQuestion -13 \nWhy does common salt appear dump ?.", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\nDownload all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 16Chap 3 : Metals and Non Metals www.cbse.online \nCHAPTER CHAPTER 33\nMetals and Non Metals\n1. OBJECTIVE QUESTIONS\n1. Which of the following metal has highest melting \npoint?(a) Copper (b) Silver\n(c) Sodium (d) Tungsten\nAns : (d) Tungsten\nT\nungsten has the highest melting point among the \nmetals.\n2. The composition of aqua-regia is\n(a) Dil.HCl : Conc. HNO 3 :: 3 : 1\n(b) Conc. HCl : Dil. HNO 3 :: 3 : 1\n(c) Conc. HCl : Conc. HNO 3 :: 3 : 1\n(d) Dil.HCl : Dil. HNO 3 :: 3 : 1\nAns : (c) Conc. HCl : Conc. HNO 3 :: 3 : 1\nConc.", + "HNO 3 :: 3 : 1\n(d) Dil.HCl : Dil. HNO 3 :: 3 : 1\nAns : (c) Conc. HCl : Conc. HNO 3 :: 3 : 1\nConc. HCl and conc. HNO 3 in 3 : 1 ratio form aqua-\nregia. Aqua-regia is a highly corrosive, fuming liquid. It can dissolve all metals even gold and platinum also.\nNO NEED TO PURCHASE ANY BOOKS\nFor session 2019-2020 free pdf will be available at www.cbse.online for1. Previous 15 Years Exams Chapter-wise Question \nBank", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\nNO NEED TO PURCHASE ANY BOOKS\nFor session 2019-2020 free pdf will be available at www.cbse.online for1. Previous 15 Years Exams Chapter-wise Question \nBank\n2. Previous Ten Years Exam Paper (Paper-wise).3. 20 Model Paper (All Solved).4. NCERT Solutions\nAll material will be solved and free pdf. It will be \nprovided by 30 September and will be updated regularly.\nDisclaimer : www.cbse.online is not affiliated to Central Board of Secondary Education, New Delhi in any manner. www.cbse.online is a private organization which provide free \nstudy material pdfs to students. At www.cbse.online CBSE stands for Canny Books \nFor School Education\n3. Food cans are coated with tin and not with zinc \nbecause(a) zinc is costlier than tin.\n(b) zinc has a higher melting point than tin.(c) zinc is more reactive than tin.(d) zinc is less reactive than tin.\nAns : (c) zinc is more reactive than tin.\n4. An element reacts with oxygen to give a compound", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\nAns : (c) zinc is more reactive than tin.\n4. An element reacts with oxygen to give a compound \nwith a high melting point. This compound is soluble in water. The element is likely to be-(a) calcium (b) carbon\n(c) silicon (d) iron\nAns : (a) calcium Ca O2\n2+ CaO$ (ionic compound)\n CaO H O 2+ ()Ca OH 2 $\n5. Which of the following is a characteristic of metals?\n(a) They have one to three valence electrons\n(b) They have 4 to 8 valence electrons(c) They are brittle(d) They are capable to form anions easily\nAns : (a) They have one to three valence electrons\nMetal can easily giv\nen up their electrons and form \nelectropositive ions. They have one to three valence \nelectrons. They are not brittle and do not form anions.\n6. A reactive metal ()M is treated with HS O24(dil).", + "They have one to three valence \nelectrons. They are not brittle and do not form anions.\n6. A reactive metal ()M is treated with HS O24(dil). The \ngas is evolved and is collected over the water as shown in the figure.\nThe correct conclusion drawn is/are(a) the gas is hydrogen\n(b) the gas is lighter than air(c) the gas is SO\n2 and is lighter than air", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\nThe correct conclusion drawn is/are(a) the gas is hydrogen\n(b) the gas is lighter than air(c) the gas is SO\n2 and is lighter than air\n(d) Both (a) and (b)\nAns : (d) Both (a) and (b)\nWhen an\ny reactive metal ()M reacts with the acid \nHS O24(dil), it evolves hydrogen gas ()H2. It is lighter \nthan air.\n() ( . )HS O d i lMs M 24 $ + -sulphate ()Hg2+\n7.", + "It is lighter \nthan air.\n() ( . )HS O d i lMs M 24 $ + -sulphate ()Hg2+\n7. An alloy is\n(a) an element\n(b) a compound(c) a homogeneous mixture(d) a heterogeneous mixture\nAns : (c) a homogeneous mixture\nAn allo\ny is a homogeneous mixture of different metals \nor a metal and a non-metal.File Revision Date : 7 July 2019\nCBSE Objective Questions Exam 2019-2020\nCLASS : 10th \nSUB : Science \nFor 15 Years Exams Chapter-wise Question Bankvisit www.cbse.online or whatsapp at 8905629969Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 17Chap 3 : Metals and Non Metals www.cbse.online \n8. When iron filings are heated in a steam of dry hydrogen", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\n8. When iron filings are heated in a steam of dry hydrogen \nchloride, the compound formed is FeCl x where x is-\n(a) 1 (b) 2\n(c) 3 (d) 4\nAns : (b) 2\n9. Silver articles become black on prolonged exposure to \nair. This is due to the formation of(a) Ag N\n3 (b) Ag O3\n(c) Ag S3 (d) Ag S3 and Ag N3\nAns : (c) Ag S3\nSilver aricle become black because silver reacts with \nHS2 gas present in air to form black coating of .Ag S2\nThe reaction is\nAg H S Ag S H\n()Silver Hydrogen\nsulphideSilver\nsulphide\nblack22 2$ ++\n10. The best malleable metal is-\n(a) aluminium (b) silver\n(c) gold (d) lead\nAns : (c) gold\n11. Which of the following only contain non-metals?", + "The best malleable metal is-\n(a) aluminium (b) silver\n(c) gold (d) lead\nAns : (c) gold\n11. Which of the following only contain non-metals?\n(a) Carbohydrates (b) Proteins\n(c) Alloys (d) Both (a) and (b)\nAns : (d) Both (a) and (b)\nCarb\nohydrates contain carbon (C), hydrogen (H) \nand oxygen (O) as their components, while proteins", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\n(a) Carbohydrates (b) Proteins\n(c) Alloys (d) Both (a) and (b)\nAns : (d) Both (a) and (b)\nCarb\nohydrates contain carbon (C), hydrogen (H) \nand oxygen (O) as their components, while proteins \ncontain carbon (C), nitrogen (N), hydrogen (H) and oxygen (O) but alloys are mixture of metals and may be some non-metals. Hence, option (d) is the correct answer.\n12. Which of the following is not a property of non-metals?(a) They are neither malleable nor ductile\n(b) They are brittle(c) They are sonorous(d) They are poor conductor of heat and electricity \n(except graphite)\nAns : (c) They are sonorous\nAlmost all the non-metals pro\nduce no metallic sound \non hitting. Thus, they are not sonorous.\n13. Which of the following metal will not give ()Hg2 with \nHO2?", + "Thus, they are not sonorous.\n13. Which of the following metal will not give ()Hg2 with \nHO2?\n(a) ()Na s H O 2 2$ + (b) ()Mg s H O 22$ +\n(c) ()Zn s H O 2$ + (d) Cu H O 2$ +\nAns : (d) Cu H O2$ +\nMetals placed below the hydrogen in reactivity series, \nwill not give ()Hg2 with water ()HO2. Decreasing \norder of reactivity of metals is", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\n(c) ()Zn s H O 2$ + (d) Cu H O 2$ +\nAns : (d) Cu H O2$ +\nMetals placed below the hydrogen in reactivity series, \nwill not give ()Hg2 with water ()HO2. Decreasing \norder of reactivity of metals is\nNa Mg Zn Cu>> >\n14. Metals are refined by using different methods. Which of the following metals are refined by electrolytic refining?1. Au 2. Cu\n3. Na 4. K(a) 1 and 2 (b) 1 and 3(c) 2 and 3 (d) 2 and 4\nAns : (a) 1 and 2\nElectrolytic refining is used for metals lik\ne Cu, Zn, \nAg, Au etc.\nThe method to be used for refining an impure \nmetal depends on the nature of the metal as well as \non the nature of impurities present in it.\n15.", + "The method to be used for refining an impure \nmetal depends on the nature of the metal as well as \non the nature of impurities present in it.\n15. Beakers ,AB and C contain zinc sulphate, silver \nnitrate and iron (II) sulphate solutions respectively. Copper pieces are added to each beaker. Blue colour will appear in case of(a) beaker A (b) beaker B\n(c) beaker C (d) all the beakers\nAns : (b) beaker B\nCopp\ner is more reactive than silver thus, displaces \nsilver from its salt solution.", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\n(c) beaker C (d) all the beakers\nAns : (b) beaker B\nCopp\ner is more reactive than silver thus, displaces \nsilver from its salt solution.\n16. Galvanisation is a method of protecting iron from rusting by coating it with a thin layer of(a) gallium (b) aluminium\n(c) zinc (d) silver\nAns : (c) zinc\nGalv\nanisation is a method of protecting iron from \nrusting by coating it with a thin layer of zinc (Zn) \nmetal.\nNO NEED TO PURCHASE ANY BOOKS\nFor session 2019-2020 free pdf will be available at www.cbse.online for1. Previous 15 Years Exams Chapter-wise Question \nBank\n2. Previous Ten Years Exam Paper (Paper-wise).3. 20 Model Paper (All Solved).4. NCERT Solutions\nAll material will be solved and free pdf. It will be \nprovided by 30 September and will be updated regularly.", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\nAll material will be solved and free pdf. It will be \nprovided by 30 September and will be updated regularly.\nDisclaimer : www.cbse.online is not affiliated to Central Board of Secondary Education, New Delhi in any manner. www.cbse.online is a private organization which provide free study material pdfs to students. At www.cbse.online CBSE stands for Canny Books For School Education\n17. Aluminium does not oxidise readily in air because-(a) it is high in the electrochemical series\n(b) it is low in the electrochemical series(c) the metal does not combine with oxygen(d) the metal is covered with a layer of oxide which \ndoes not rub off\nAns : (d) the metal is cover4ed with a layer of oxide \nwhic\nh does not rub off\n18. In each test tubes ,,AB C and D, 2mL of solution of \n()Al SO24 3 in water was filled.", + "In each test tubes ,,AB C and D, 2mL of solution of \n()Al SO24 3 in water was filled. Clean pieces of zinc was \nplaced in test tube A, clean iron nail was put in test \ntub\ne B, silver (Ag) was placed in test tube C and a \nclean copp\ner wire was placed in test tube D.\nWhic", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\nplaced in test tube A, clean iron nail was put in test \ntub\ne B, silver (Ag) was placed in test tube C and a \nclean copp\ner wire was placed in test tube D.\nWhic\nh of the following option (s) is/are correct about \nabove experiment?(a) Zinc is more reactive than aluminium\n(b) Copper is more reactive than aluminium(c) Zinc is more reactive than copperDownload all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 18Chap 3 : Metals and Non Metals www.cbse.online \n(d) Zinc, iron, silver and copper are less reactive than \naluminium\nAns : (d) Zinc, iron, silver and copper are less reactive \nthan aluminium\nZn, Fe, Ag and Cu are less reactive than aluminium. \nAluminium resides at the top of the activity series. While Zn, Fe, Ag and Cu lies below aluminium in the activity series. Thus, being less reactive than aluminium, they cannot displace Al from its salt solution i.e. ()Al\nSO24 3 solution.", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\nSO24 3 solution.\n19. On the basis of the sequence of the given reactions identify the most and least reactive elements:\n XY A+ XA Y$ + ...(1)\n XY B+ XB Y$ + ...(2)\n ZX A+ ZA X$ + ...(3)\n(a) X and Z (b) Y and Z\n(c) Z and X (d) Z and Y\nAns : (d) Z and Y\n\u2018Z\u2019 is the more reactive element and Y is the least \nreactiv \ne element Z easily displaces X, while X easily \ndisplaces Y.\n20. A metal M has electronic configuration 2, 8, 3 and \no\nccurs in earth\u2019s crust and its oxide MO23. It is more \nreactive than zinc. Which of the following options (s) \nis/are correct?(a) The metal M is iron\n(b) The \nmetal M is lead\n(c) \nThe ore form which metal M is extracted in \nhaematite.", + "It is more \nreactive than zinc. Which of the following options (s) \nis/are correct?(a) The metal M is iron\n(b) The \nmetal M is lead\n(c) \nThe ore form which metal M is extracted in \nhaematite.\n(d) \nThe ore form which metal M is extracted is \nbauxite.\nAns : (d) The ore form which metal M is extracted \nis bauxite.\nElectronic configuration of ,, M28 3=Hence, the outer orbital consist of 3 electrons. It is \nmore reactiv\ne than zinc. So, it should be aluminium", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\nis bauxite.\nElectronic configuration of ,, M28 3=Hence, the outer orbital consist of 3 electrons. It is \nmore reactiv\ne than zinc. So, it should be aluminium \n(atomic number 13=). The ore from which metal M \nis extracted is bauxite.\n21. Metal M reacts with oxygen to form metallic oxide \nOM. This oxide reacts with moisture and carbon \ndio\nxide of the atmosphere to form a basic carbonate \nmetal M. The metal \u2018 M\u2019 is\n(a) Cu \n(b) Fe\n(c) Zn (d) Cr\nAns : (a) Cu\nSince, the metal forms an o\nxide OM, the metal is \ndiv\nalent. Since it forms a basic carbonate when \nexposed to moisture and carbon dioxide, therefore it \nmust be copper.\n() Cu CO O H O CuCO Cu OH 2\n()From moist airBasic copper carbonate222 3 2 $ $ ++ +12344444 4 4 4\n22.", + "() Cu CO O H O CuCO Cu OH 2\n()From moist airBasic copper carbonate222 3 2 $ $ ++ +12344444 4 4 4\n22. Which of the following methods is suitable for preventing an iron frying pan from rusting?(a) applying grease\n(b) applying paint(c) applying a coating of zinc(d) all of the above.\nAns : (c) applying a coating of zinc", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\n(b) applying paint(c) applying a coating of zinc(d) all of the above.\nAns : (c) applying a coating of zinc\n23. A student mistakenly used a wet gas jar to collect \nsulphur dioxide. Which one of the following tests of the gas is likely to fail?(a) Odour\n(b) Effect on acidified KC rO\n22 7 solution\n(c) Solubility test(d) None of these\nAns : (d) None of these\n24. A student puts one big iron nail each in four test \ntubes containig solutions of zinc sulphate, aluminium sulphate, copper sulphate and iron sulphate. A reddish brown coating was observed only on the surface of iron nail which was put in the solution of(a) zinc sulphate (b) iron sulphate\n(c) copper sulphate (d) aluminium sulphate\nAns : (c) copper sulphate\nF\ne displaces copper from copper sulphate ()CuSO 4 \nsolution.\nCuSO Fe FeSO Cu\nReddish brown44$ ++\n25.", + "CuSO Fe FeSO Cu\nReddish brown44$ ++\n25. Which of the following is not a characteristics of \nmetal?(a) Malleable\n(b) Electro-positive nature(c) Ductile(d) None of these\nAns : (d) None of these\nAll are c", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\nReddish brown44$ ++\n25. Which of the following is not a characteristics of \nmetal?(a) Malleable\n(b) Electro-positive nature(c) Ductile(d) None of these\nAns : (d) None of these\nAll are c\nharacteristics of metal.\n26. Which of teh following reactions not occur?", + "Which of the following is not a characteristics of \nmetal?(a) Malleable\n(b) Electro-positive nature(c) Ductile(d) None of these\nAns : (d) None of these\nAll are c\nharacteristics of metal.\n26. Which of teh following reactions not occur?\n(a) () ( ) ( ) () ( ) AgNO aq Zn s Zn NO aq Ag s22 33 2 $ ++\n(b) ( ) () ( ) () CuSO aq Zn s ZnSO aq Cu s44 $ ++Download all GUIDE and Sample Paper pdfs from www.cbse.online or www.rava.org.in Page 19Chap 3 : Metals and Non Metals www.cbse.online \n(c) () ( ) ( ) () ( ) AgNO aq Fe s Fe NO aq Ag s22 33 2 $ ++\n(d) () ( ) ( ) () CuSO aq Ag s Cu s Ag SO aq 2 42 4 $ ++\nAns : (d) () ( ) ( ) () CuSO aq Ag s Cu s Ag SO aq 2 42 4 $ ++\nSilver is less reactive than copper, hence cannot \ndisplace copper from its salt solution.", + "27. Pure gold is-(a) 24 carats (b) 22 carats\n(c) 20 carats (d) 18 carats\nAns : (a) 24 carats\n28. When a metal is added to dilute HCl solution, there is \nno evolution of gas. Metal is-(a) K (b) Na\n(c) Ag (d) Zn", + "CHAPTER 3- METALS AND NON METALS- MCQs.txt\n(c) 20 carats (d) 18 carats\nAns : (a) 24 carats\n28. When a metal is added to dilute HCl solution, there is \nno evolution of gas. Metal is-(a) K (b) Na\n(c) Ag (d) Zn\nAns : (c) Ag\nAg is b\nelow hydrogen in reactivity series.\n29. The correct order of increasing chemical reactivity is-\n(a) Zndouble bond>single bond \n- This is to signify that the energy required to break three bonds is higher \nthan that for two bonds or a single bond. \n \n Bond length", + "CHAPTER 4-CARBON AND ITS COMPOUNDS-STUDY MATERIAL.txt\nBond length \n \n- Bond length is determined by the distance between nuclei of the two \natoms in a bond. \n- The order of bond length for multiple bonds is: Triple bondspeedofsoundinliquid>speedofsoundingas\nNocreditforanyotheranswer\nV.SCATTERINGOFLIGHT\nWhetherthecoloroftheoceanandthecoloroftheskyarerelated?Isthisthesamereason?But\ntheyoccurindependentlyofeachother.Inboththecases,thepreferentialabsorptionoflong\nwavelengthlightgivesrisetoblue.TheOceanlooksblue,thebluewavelengthreturned\npredominantlythanlongerwavelength.Notethatthiseffectonlyworksifthewaterisverypure.\nPerhapsthemosthelpfulresponsecamefromMichaelKrugerofthedepartmentofphysicsat\nuniversityofMissouri.Theskyisbluenotbecausetheatmosphereabsorbstheothercolors,but", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\nPerhapsthemosthelpfulresponsecamefromMichaelKrugerofthedepartmentofphysicsat\nuniversityofMissouri.Theskyisbluenotbecausetheatmosphereabsorbstheothercolors,but\ntheatmospheretendstoscattershorterwavelengthlighttoagreaterextentthanlongerwavelengthlight.Whenyoulookupdaytimeskyyouseebluenomatterwhereyoulook.This\nscatteringiscalledRayleighscattering.Theamountoflightscatteringgoesasthefrequencyof\nthelighttofourthpower.NitrogenandOxygenmoleculesaresmallerthanwavelengthoflight\ncannotcausescattering.Samewaywecanthinkaboutwhitenessofmilk.\n1.Whichphenomenonoflightisthereasonforbluecolorofsky?\n2.Theintensityofscatteredlightmorewillbe____wavelength\nA.Longer\nB.Shorter\nC.Bothequally.", + "1.Whichphenomenonoflightisthereasonforbluecolorofsky?\n2.Theintensityofscatteredlightmorewillbe____wavelength\nA.Longer\nB.Shorter\nC.Bothequally.\n3SayYesorNoforthefollowing\n(I)Redcolorofsunduringsunsetisduetoscattering\n(ii)Scatteringissameforallcolors\n(iii)Twinklingofstarsareduetoscatteringoflight\n.\nCorrectresponses\n1.Scattering\n2.Shorter\n3.Yes,No,Nointhatorder\nVI.ATMOSPHERICREFRACTIONSomefriendsweredoingboatinginalake.Theysawabeautifulfishinthelake.Oneofthem", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\n.\nCorrectresponses\n1.Scattering\n2.Shorter\n3.Yes,No,Nointhatorder\nVI.ATMOSPHERICREFRACTIONSomefriendsweredoingboatinginalake.Theysawabeautifulfishinthelake.Oneofthem\ntriedtocatchit,bythinkingthatitisveryclosetotheboat.Otherfriendstoldhimthatthefishes\naredeepinwaterandheshouldnottrytocatchthem.\n1Whydidfishappearclosertotheboatwhenitactualwasdeepinwater?\n2Nametwodailylifesituationwherethisphenomenoninvolved.\n3Ifthespeedoflightinwateris2.25X108m/s,calculatetherefractiveindexofthewater.\nCorrectresponses\n1.Refraction.\n2.Advancesunriseandtwinklingofstars.", + "3Ifthespeedoflightinwateris2.25X108m/s,calculatetherefractiveindexofthewater.\nCorrectresponses\n1.Refraction.\n2.Advancesunriseandtwinklingofstars.\n3.n=c/v=3x108/2.25x108=1.23.VII.THUNDERSTORMANDWHITESKY\nMr.Kamaleshwastravelinginhiscaroneday,whileitwasrainingheavilywiththunder\nandlightning.Fromthenewsontheradio,hecametoknowthatthewhetherdepartment\nhadannouncedofstormwithlightning.Suddenlythevisibilitybecameverylowand\nthereforehepulledoverhiscar.Ashestoppedhesawtwochildrenwhohadtakenshelter\nunderatree.Thechildrenweredrenchedintherain.Hemadethechildrentakeshelterin", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\nthereforehepulledoverhiscar.Ashestoppedhesawtwochildrenwhohadtakenshelter\nunderatree.Thechildrenweredrenchedintherain.Hemadethechildrentakeshelterin\nhiscar.Aftertherain,whentheylookedatthesky,theskywhiteincolour.\n1.Answeryesornotothefollowingquestions:\na.Itissafetostandunderanisolatedtreeinthunderstorm.\nb.Itissafetositinsideavehicleduringthunderstorm.\n2.State,whataccordingtoyou,aretheknowledgeofscienceheapplied.\n3.Whyskyappeartobewhite?\nCorrectresponses\n1.a.No\nb.Yes.\n2.Electrostaticshielding.\n3.Ifthesizeofthemoleculespresentintheatmospherearelargercomparedtothewave\nlengthoflight,noselectivescattering.Soallcoloursoflightscatterequallyandhencesky\nappeartobewhite.\nVIIITotalinternalreflectionFig.", + "3.Ifthesizeofthemoleculespresentintheatmospherearelargercomparedtothewave\nlengthoflight,noselectivescattering.Soallcoloursoflightscatterequallyandhencesky\nappeartobewhite.\nVIIITotalinternalreflectionFig.\u202f1:Underwaterplantsinanaquarium,andtheirinvertedimages(top)formedbytotal\ninternalreflectioninthewater-airsurface.\nTotalinternalreflection(TIR)istheopticalphenomenoninwhichthesurfaceofthewaterina\nfish-tank(forexample)whenviewedfrombelowthewaterlevel,reflectstheunderwaterscene", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\nTotalinternalreflection(TIR)istheopticalphenomenoninwhichthesurfaceofthewaterina\nfish-tank(forexample)whenviewedfrombelowthewaterlevel,reflectstheunderwaterscene\nlikeamirror,withnolossofbrightness(Fig.\u202f1).Ingeneral,TIRoccurswhenwavesinone\nmediumreachtheboundarywithanothermediumatasufficientlyslantingangle,providedthat\nthesecond(\"external\")mediumistransparenttothewavesandallowsthemtotravelfasterthan\ninthefirst(\"internal\")medium.TIRoccursnotonlywithelectromagneticwavessuch\naslightandmicrowaves,butalsowithothertypesofwaves,includingsoundandwaterwaves.\nInthecaseofanarrowtrainofwaves,suchasalaserbeam(Fig.", + "Inthecaseofanarrowtrainofwaves,suchasalaserbeam(Fig.\u202f2),wetendtodescribethe\nreflectionintermsof\"rays\"ratherthanwaves.Inamediumwhosepropertiesareindependentof\ndirection,suchasair,water,orglass,each\"ray\"isperpendiculartothe\nassociatedwavefronts.[importance?]Fig.\u202f2:Repeatedtotalinternalreflectionofa405\u202fnmlaserbeambetweenthefrontandback\nsurfacesofaglasspane.Thecolorofthelaserlightitselfisdeepviolet;butitswavelengthis", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\nassociatedwavefronts.[importance?]Fig.\u202f2:Repeatedtotalinternalreflectionofa405\u202fnmlaserbeambetweenthefrontandback\nsurfacesofaglasspane.Thecolorofthelaserlightitselfisdeepviolet;butitswavelengthis\nshortenoughtocausefluorescenceintheglass,whichre-radiatesgreenishlightinall\ndirections,renderingthezigzagbeamvisible.", + "Refractionisgenerallyaccompaniedbypartialreflection.Whenwavesarerefractedfroma\nmediumoflowerpropagationspeedtoamediumofhigherpropagationspeed(e.g.,fromwater\ntoair),theangleofrefraction(betweentherefractedrayandthelineperpendiculartothe\nrefractingsurface)isgreaterthantheangleofincidence(betweentheincidentrayandthe\nperpendicular).Astheangleofincidenceapproachesacertainlimit,calledthecriticalangle,the\nangleofrefractionapproaches90\u00b0,atwhichtherefractedraybecomesparalleltothesurface.As\ntheangleofincidenceincreasesbeyondthecriticalangle,theconditionsofrefractioncanno\nlongerbesatisfied;sothereisnorefractedray,andthepartialreflectionbecomestotal.For", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\ntheangleofincidenceincreasesbeyondthecriticalangle,theconditionsofrefractioncanno\nlongerbesatisfied;sothereisnorefractedray,andthepartialreflectionbecomestotal.For\nvisiblelight,thecriticalangleisabout49\u00b0forincidenceatthewater-to-airboundary,andabout\n42\u00b0forincidenceatthecommonglass-to-airboundary.", + "DetailsofthemechanismofTIRgiverisetomoresubtlephenomena.Whiletotalreflection,by\ndefinition,involvesnocontinuingflowofpoweracrosstheinterfacebetweenthetwomedia,the\nexternalmediumcarriesaso-calledevanescentwave,whichtravelsalongtheinterfacewithan\namplitudethatfallsoffexponentiallywithdistancefromtheinterface.The\"total\"reflectionis\nindeedtotaliftheexternalmediumislossless(perfectlytransparent),continuous,andofinfinite\nextent,butcanbeconspicuouslylessthantotaliftheevanescentwaveisabsorbedbyalossy\nexternalmedium(\"attenuatedtotalreflectance\"),ordivertedbytheouterboundaryofthe\nexternalmediumorbyobjectsembeddedinthatmedium(\"frustrated\"TIR).", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\nexternalmedium(\"attenuatedtotalreflectance\"),ordivertedbytheouterboundaryofthe\nexternalmediumorbyobjectsembeddedinthatmedium(\"frustrated\"TIR).\nUnlikepartialreflectionbetweentransparentmedia,totalinternalreflectionisaccompaniedbya\nnon-trivialphaseshift(notjustzeroor180\u00b0)foreachcomponentofpolarization(perpendicular\norparalleltotheplaneofincidence),andtheshiftsvarywiththeangleofincidence.The\nexplanationofthiseffectbyAugustin-JeanFresnel,in1823,addedtotheevidenceinfavorof\nthewavetheoryoflight.\n1.Whicharetheconditionsfortotalinternalreflection?", + "1.Whicharetheconditionsfortotalinternalreflection?\n2.Theangleofreflectionatcriticalangleis-------\n3.Thecriticalangleforcertainmediumis300.Findtherefractiveindexofthatmedium.Responses\n1.Lightshouldtravelfromdensertorarermediumandtheangleofincidenceshouldbe\ngreaterthancriticalangle\n2.90o\n3.n=1/sinC=2\nIXFormationofrainbow\nOnedayAnjithandfamilyweregoingtotheirvillageinacar.Duringthejourneyitwasraining.\nChildrenwereenjoyingtravellinginrain.Itwasahighrangeroute.Aftersometimerainstopped", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\n3.n=1/sinC=2\nIXFormationofrainbow\nOnedayAnjithandfamilyweregoingtotheirvillageinacar.Duringthejourneyitwasraining.\nChildrenwereenjoyingtravellinginrain.Itwasahighrangeroute.Aftersometimerainstopped\nandstartedtogetbrightsunlight.Itwasatthattime,hisyoungersonnoticedarainbowinthe\nsky.Hiscuriousmindcouldnotstophimtoaskhowitforms?Anjithexplainedhischildren\nhowarainbowform.\n1.Whicharethephenomenoninvolvedintheformationofrainbow?\n2.Whichactsasprismintheformationofrainbow?\n3.Sayyesornotothefollowingstatements(a)Rainbowisformedonasuunydaywithoutrain\n(b)Rainbowcannotbeseenfromthesurfaceofmoon.", + "2.Whichactsasprismintheformationofrainbow?\n3.Sayyesornotothefollowingstatements(a)Rainbowisformedonasuunydaywithoutrain\n(b)Rainbowcannotbeseenfromthesurfaceofmoon.\nCorrectresponses\n1.Refraction,dispersionandinternalreflection\n2.Waterdropletspresentintheatmosphere\n3.No,Yesinthatorder\nXEYEDEFECTANDX-RAYS\nPriyanka\u2019sgrandfatheroncefellfromthestairs.Hewasnotwearingspectacles.Hegot\nfracturesonhisarmandleg.Priyankawasworriedabouttheconditiononhis\ngrandfatherasherparentswereabroadandshewasalonetotakecareofhergrandfather.With", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\nfracturesonhisarmandleg.Priyankawasworriedabouttheconditiononhis\ngrandfatherasherparentswereabroadandshewasalonetotakecareofhergrandfather.With\nthehelpofherneighbour,shetookhergrandfathertoahospitalweremultipleX-raysweredone\nthosesheknewthatthemultipleX-rayexposureismorebeneficialthantheharmitcauses,\nbecausethiswouldenablethedoctorstoknowtheactualconditionofhergrandfather\u2019sfractures.\nInthehospitalshemetherfriedwhocamewithherfatherforanendoscopyforherfather.\na.Whichtypesoflenspeopleuseintheiroldage?\nb.WriteanytwouseofX-rays.\nc.Nametheprincipleusedinendoscopy.Correctresponses\na.Bifocallens.", + "a.Whichtypesoflenspeopleuseintheiroldage?\nb.WriteanytwouseofX-rays.\nc.Nametheprincipleusedinendoscopy.Correctresponses\na.Bifocallens.\nb.X-raysareusedradiotherapytocureskindisease,cancerandtumors.Theyareusedby\ndetectiveagenciestodetectthegold,silveranddiamondsetc.Concealedinbagsorthebody\nofaperson.\nc.Totalinternalreflection.\nXI.SCIENCEFAIRANDTWINKLINGOFSTARS\nTherewasasciencefairinatown.KushalandChiragvisitedthefairinthe\nevening.Inthefair,theysawamirrorinwhichtheygotafunnyimage.Theupper", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\nc.Totalinternalreflection.\nXI.SCIENCEFAIRANDTWINKLINGOFSTARS\nTherewasasciencefairinatown.KushalandChiragvisitedthefairinthe\nevening.Inthefair,theysawamirrorinwhichtheygotafunnyimage.Theupper\npartofthebodywasbiginsize,themiddlepartwasofnormalsizeandthelowerpartofthe\nbodyshowedverysmallsize.Itwasgettingdarkwhentheywerereturninghome.\nButonthewayKushalwasthinkingaboutthecauseoftheimage.Chiragexplaindthe\nreasonbehindit.Happiliytheywerecomingtothehome.Atthattimetheycould\nseethebeatifulskyandstars.Withacuriousmind,Kushalaskedaboutthestars.", + "Chiragnarratedaboutthecauseofformationofstars..1.Givereasonforthetwinklingofstars\n2.Theupper,middleandlowerpartofthemirrorusedinthefairare-----,------and-------\n3.SayYesorNotothefollowing\n(a)Convexmirroralwysformsrealimage.\n(b)Concavemirroralwaysformsrealimage.\n(c)Planemirroralwaysformserectedimage\nCorrectresponses\n1.Atmospericrefraction\n2.Concavemirror,planemirrorandconvexmirrorinthatorder\n3.No,No,Yesinthatorder\nXII Diamonds&FiberOptics", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\n(c)Planemirroralwaysformserectedimage\nCorrectresponses\n1.Atmospericrefraction\n2.Concavemirror,planemirrorandconvexmirrorinthatorder\n3.No,No,Yesinthatorder\nXII Diamonds&FiberOptics\nWhenlightpassesfromonemediumintoasecond,lessdensemedium,thelightbendsaway\nfromthenormal.Ataparticularincidentangle,theangleofrefractionwillbe90degrees,andthe\nrefractedraywouldtravelalongtheboundarybetweenthetwomedia.", + "Theincidentangleatwhichthisoccursiscalledthecriticalangle.Thisanglecanbecalculated\nbyusingSnell'sLaw:\nsin(criticleangle)=(n2/n1)(sin90degrees)=(n2/n1)\nWheren1=indexofrefractionin1stmedium\nn2=indexofrefractionin2ndmedium\nIftheangleofincidenceisincreasedbeyondthecriticalangle,thelightrayswillbetotally\nreflectedbackintotheincidentmedium.Thiseffectiscalledtotalinternalreflection.Butnote\nthattotalinternalreflectioncannotoccuriflightistravelingfromalessdensemediumtoa\ndenserone.Diamondsachievetheirbrilliancepartiallyfromtotalinternalreflection.Becausediamondshave", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\nthattotalinternalreflectioncannotoccuriflightistravelingfromalessdensemediumtoa\ndenserone.Diamondsachievetheirbrilliancepartiallyfromtotalinternalreflection.Becausediamondshave\nahighindexofrefraction(about2.3),thecriticalangleforthetotalinternalreflectionisonly\nabout25degrees.Incidentlightthereforestrikesmanyoftheinternalsurfacesbeforeitstrikes\nonelessthan25degreesandemerges.Aftermanysuchreflections,thecolorsinthelightare\nseparated,andseenindividually.", + "Inthelate20thcentury,anew,practicalapplicationoftotalreflectionwasfoundintheuseof\nfiberoptics.Iflightentersasolidglassorplastictubeobliquely,thelightcanbetotallyreflected\nattheboundaryofthetubeand,afteranumberofsuccessivetotalreflections,emergefromthe\notherend.Glassfiberscanbedrawntoaverysmalldiameter,coatedwithamaterialoflowerrefractiveindex,andthenassembledintoflexiblebundlesorfusedintoplatesoffibersusedto\ntransmitimages,telephonecalls,videosignals,andcomputerdata.Thereflectivepropertiesof", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\ntransmitimages,telephonecalls,videosignals,andcomputerdata.Thereflectivepropertiesof\nfiberopticstrandsmakefiberopticcablesmoreefficientthancablesthattransmitsignals\nelectrically.Theflexiblebundles,whichcanbeusedtoprovideilluminationaswellasto\ntransmitimages,arevaluableinmedicalexamination,astheycanbeinsertedintovarious\nopenings.\n1.Whatiscriticalangle?\n2.Iftheincidentrayismakinganangle30owiththeinterfacewhatistheangleofincidence?\n4.Sayyesornotothefollowingstatement\n(a)Totalinternalreflectionalwaystakesplacewhenlighttravelsfromglasstoair\n(b)Opticalfibresareusedinmedicalfield\n(c)Totalinternalreflectiontakesplacewhenlighttravelsfromwatertoglass\nCorrectresponses\n1.Theangleofinthedenserdensermediumatwhichangleofrefractionbecome90o\n2.60o\n4.No,Yes,NointhatorderXIIIFiberOptics:EndoscopestoTelephones\n.", + "Fiberopticsisoneapplicationoftotalinternalreflectionthatisinwideuse.Incommunications,\nitisusedtotransmittelephone,internet,andcableTVsignals.Fiberopticsemploysthe", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\n.\nFiberopticsisoneapplicationoftotalinternalreflectionthatisinwideuse.Incommunications,\nitisusedtotransmittelephone,internet,andcableTVsignals.Fiberopticsemploysthe\ntransmissionoflightdownfibersofplasticorglass.Becausethefibersarethin,lightentering\noneislikelytostriketheinsidesurfaceatananglegreaterthanthecriticalangleand,thus,be\ntotallyreflectedTheindexofrefractionoutsidethefibermustbesmallerthaninside,a\nconditionthatiseasilysatisfiedbycoatingtheoutsideofthefiberwithamaterialhavingan\nappropriaterefractiveindex.Infact,mostfibershaveavaryingrefractiveindextoallowmore\nlighttobeguidedalongthefiberthroughtotalinternalrefraction.Raysarereflectedaround\ncornersasshown,makingthefibersintotinylightpipes.", + "Bundlesoffiberscanbeusedtotransmitanimagewithoutalens.Theoutputofadevicecalled\nanendoscope.Endoscopesareusedtoexplorethebodythroughvariousorificesorminor\nincisions.Lightistransmitteddownonefiberbundletoilluminateinternalparts,andthe", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\nanendoscope.Endoscopesareusedtoexplorethebodythroughvariousorificesorminor\nincisions.Lightistransmitteddownonefiberbundletoilluminateinternalparts,andthe\nreflectedlightistransmittedbackoutthroughanothertobeobserved.Surgerycanbeperformed,\nsuchasarthroscopicsurgeryonthekneejoint,employingcuttingtoolsattachedtoandobserved\nwiththeendoscope.Samplescanalsobeobtained,suchasbylassoinganintestinalpolypfor\nexternalexamination.", + "Fiberopticshasrevolutionizedsurgicaltechniquesandobservationswithinthebody.There\nareahostofmedicaldiagnosticandtherapeuticuses.Theflexibilityofthefiberopticbundle\nallowsittonavigatearounddifficultandsmallregionsinthebody,suchastheintestines,the\nheart,bloodvessels,andjoints.Transmissionofanintenselaserbeamtoburnaway\nobstructingplaquesinmajorarteriesaswellasdeliveringlighttoactivatechemotherapydrugs\narebecomingcommonplace.Opticalfibershaveinfactenabledmicrosurgeryandremote\nsurgerywheretheincisionsaresmallandthesurgeon\u2019sfingersdonotneedtotouchthe\ndiseasedtissue.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\narebecomingcommonplace.Opticalfibershaveinfactenabledmicrosurgeryandremote\nsurgerywheretheincisionsaresmallandthesurgeon\u2019sfingersdonotneedtotouchthe\ndiseasedtissue.\nFibersinbundlesaresurroundedbyacladdingmaterialthathasalowerindexofrefractionthan\nthecore.Thecladdingpreventslightfrombeingtransmittedbetweenfibersinabundle.\nWithoutcladding,lightcouldpassbetweenfibersincontact,sincetheirindicesofrefractionare\nidentical.Sincenolightgetsintothecladding(thereistotalinternalreflectionbackintothecore),\nnonecanbetransmittedbetweencladfibersthatareincontactwithoneanother.Thecladding\npreventslightfromescapingoutofthefiber;insteadmostofthelightispropagatedalongthe\nlengthofthefiber,minimizingthelossofsignalandensuringthataqualityimageisformedat\ntheotherend.Thecladdingandanadditionalprotectivelayermakeopticalfibersflexibleand\ndurable.", + "1.Nametheprincipleofopticalfibres\n2.Amongcoreandcladding,whichhasmorerefractiveindex?\n5.Sayyesornotothefollowingstatement", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\ndurable.\n1.Nametheprincipleofopticalfibres\n2.Amongcoreandcladding,whichhasmorerefractiveindex?\n5.Sayyesornotothefollowingstatement\n(d)Totalinternalreflectionalwaystakesplacewhenlighttravelsfromglasstoair\n(e)Opticalfibresareusedinmedicalfield\n(f)Totalinternalreflectiontakesplacewhenlighttravelsfromwatertoglass\nCorrectresponses\n1.Totalinternalreflection\n2.Core\n3.No,Yes,NointhatorderXIVEYEDONATION\nNewsspreadinavillagethatavillagerhasexpiredduetoheartattach.Buthehas\ndonatedbeautifuleyestooneofhisfriends.Allthemembersofthevillagefeltverysadforhis\nuntimelydeath,butontheotherhandtheywereoverwhelmedonhearingthedonationofhis\neyestohisfriendwhowoundnowbeabletoseethisbeautifulnature.\na.Namethepartoftheeyethatisusedduringeyeplant.", + "a.Namethepartoftheeyethatisusedduringeyeplant.\nb.Namethedefectthatcanbecorrectedbythistransplant.\nc.Whatotherorganscanbedonatedafterdeath.\nCorrectresponses\na.Cornea.\nb.Defectcausedduetocornea.\nc.Kidney,heart,leveretc.XVDEFECTSOFVISION\nSumaandSujaarebestfriendsandstudyinclassX.Recently,Sumahasbeenfacingdifficulty", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Competency based questions.txt\nCorrectresponses\na.Cornea.\nb.Defectcausedduetocornea.\nc.Kidney,heart,leveretc.XVDEFECTSOFVISION\nSumaandSujaarebestfriendsandstudyinclassX.Recently,Sumahasbeenfacingdifficulty\ninreadingtheblackboardtextfromthelastdesk.Sujaislittleuncomfortableandwonderswhy\nSumacomplaintssittingonthelastdesk.LateronSumatoldtoherparentsregardingth\neproblem.Theywenttoaneyespecialist.Heprescribedalensofpower-1.25Dforcorrection\nofhervision.\n1.Namethedefect\n2.Calculatethefocallengthofthelens?\n3.Whatisthenatureofthelens?\nCorrectresponses\n1.Myopia\n2.F=-0.8m\n3.Diverginglensorconcavelens", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -IMPORTANT POINT.txt\nStudentsfindmiopiadifficulttoumderstandastheydo\nnotgetthedefinitionofmiopiaandthelensesusedto\nounteracttheproblemalongwithothervisionary\ndefect.\nImpact:\nStudentsareunabletocommentonthevisionarydefects\nthatcausessomeonetolosefarsightedness.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Long answer questions.txt\n1.Howdoesrefractiontakesplaceintheatmosphere?Whydostarstwinklebut\nnottheplanets?\nAns:Thedensityofdifferentlayersofatmosphereatdifferentheightis\ndifferent.Higherthedensity,higherwillbetherefractiveindex.So,underthe\nnormalcondition,therefractiveindexofair,goesondecreasingaswemove\ntowardstheoutmostregionofatmosphere.Whenastarlightentersintothe\natmosphereobliquely,itrefractedcontinuouslyandmoveintotheregionofhigher\nrefractiveindex.Soitbendstowardsthenormal.Thusrefractionoflighttakes\nplaceintheatmosphere.", + "Twinklingofstars:Theapparentpositionofastarisslightlydifferentfrom\ntheactualpositionduetoatmosphericrefraction.Further,thisapparentposition\nisnotstationarybutkeepsonchangingduetothechangeinatmospheric\nconditionslikedensity,temperature,etc.Thepathoftheraysoflightcoming\nfromthestargoesonvaryingslightly.Theamountoflightenteringoureyes\nfromastarincreasesordecreasesrandomlywithtime.Sometimes,thestar\nappearsbrightandothertimesitappearsfainter.Thisgivesrisetothetwinkling\neffectofthestar.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Long answer questions.txt\nfromastarincreasesordecreasesrandomlywithtime.Sometimes,thestar\nappearsbrightandothertimesitappearsfainter.Thisgivesrisetothetwinkling\neffectofthestar.\nTheplanetsdonotshowtwinklingeffect:Astheplanetsaremuch\nclosertotheearth,theamountoflightreceivedfromthemismuchgreaterand\nfluctuationscostintheamountoflightduetoatmosphericrefractionare\nnegligibleascomparedtotheamountoflightreceivedfromthem.\n2Givereasonforthefollowing\n(a)Dangersignalsareredincolour\n(b)Sunisvisibletwominutesbeforeactualsunrise\n\u00a9Skyappearstobeblue\n(d)Starsappearstotwinkle\n(e)Skyappearstobedarkfromthesurfaceofthemoon\nAns:(a)Duetoitslongerwavelengthredisleastscatteredandcanbeseen\nfromfarplaces.\n(b)Becauseofatmosphericrefraction.", + "(b)Becauseofatmosphericrefraction.\n(c)Duetotheselectivescatteringofthebluecomponentpresentinthe\nsunlightbythe moleculespresentintheatmosphere.\n(d)Starsappearstotwinkleduetoatmosphericrefractionofstarlight\nandphysicalconditionsoftheearth\u2019satmosphereisnotbeing\nstationary.\n(e)Skyappearstobedarkfromthesurfaceofthemoonbecausethereare", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Long answer questions.txt\n(d)Starsappearstotwinkleduetoatmosphericrefractionofstarlight\nandphysicalconditionsoftheearth\u2019satmosphereisnotbeing\nstationary.\n(e)Skyappearstobedarkfromthesurfaceofthemoonbecausethereare\nno atmosphereparticlestoscattersunlight.\n3Explaintherefractionoflightthroughatriangularglassprismusingalabelled\nraydiagram.Hencedefinetheangleofdeviation.Ans:ArayoflightPEisincidentatthethefirstsurfaceABofprismand\nmovefromrarertodenser.Onrefractionitbendstowardsthenormalandfollow\nthepathEFasrefractedrayinsidetheprism.AtthesecondsurfaceAC,the\nrefractedrayEF,getsrefractedagainandbendsawayfromthenormalintheform\nofemergentrayFS.\nAngleofdeviation:Theangleformedbetweentheincidentrayproducedinthe\nforwarddirectionandemergentrayproducedinthebackwarddirectionthrough\ntheprismiscalledangleofdeviation.", + "Angleofdeviation:Theangleformedbetweentheincidentrayproducedinthe\nforwarddirectionandemergentrayproducedinthebackwarddirectionthrough\ntheprismiscalledangleofdeviation.\n4.Astudentcannotseeclearlyacharthangingonawallplacedatadistance3\nmetrefromhiseye.\n(i)Namethedefectofvisionheissufferingfrom.Listtwocauses.\n(ii)Drawadiagramtoillustratethisdefect.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Long answer questions.txt\n4.Astudentcannotseeclearlyacharthangingonawallplacedatadistance3\nmetrefromhiseye.\n(i)Namethedefectofvisionheissufferingfrom.Listtwocauses.\n(ii)Drawadiagramtoillustratethisdefect.\n(iii)Withthehelpofdiagramshowhowthisdefectcanbecorrected?\nAns:Myopiaorshortsightedness.\nCausesofMyopia:Elongationofeyeball,decreaseinfocallengthoftheeye\nlens.5.(a)Whatisdispersionofwhitelight?Whatisthecauseofsuchdispersion?\nDrawadiagramtoshowthedispersionofwhitelightbyaglassprism.\n(b)Aglassprismisabletoproduceaspectrumwhenwhitelightpassesthrough\nitbutaglassslabdoesnotproduceanyspectrum.Explainwhyisitso?\n.Ans:\n(a)Thesplittingupofwhitelightintoitsconstituentcoloursiscalleddispersion.", + ".Ans:\n(a)Thesplittingupofwhitelightintoitsconstituentcoloursiscalleddispersion.\nThecoloursequenceisgivenbytheacronymVIBGYOR\u2013Violet,Indigo,Blue,\nGreen,Yellow,OrangeandRed.Thiscolourpatterniscalledaspectrum.\nDispersiontakesplacebecausethespeedoflightofdifferentcoloursthroughaglass\nprismisdifferentandso,refractiveindex,therefore,eachcolourbends(refracts)", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Long answer questions.txt\nDispersiontakesplacebecausethespeedoflightofdifferentcoloursthroughaglass\nprismisdifferentandso,refractiveindex,therefore,eachcolourbends(refracts)\nthroughdifferentangleswithrespecttoincidentrayastheypassthroughaprism.\nTheredcolourhasmaximumspeedinglassprism.So,itisleastdeviatedwhilethe\nvioletcolourhasminimumspeedsoitsdeviationismaximum.Thus,therayofeach\ncolouremergesalongdifferentpathsandbecomesdistinct.", + "Theredcolourhasmaximumspeedinglassprism.So,itisleastdeviatedwhilethe\nvioletcolourhasminimumspeedsoitsdeviationismaximum.Thus,therayofeach\ncolouremergesalongdifferentpathsandbecomesdistinct.\n(b)Fordispersion,thetworefractingsurfacesmustbeinclinedtoeachotherasincaseofprism.Inrectangularglassslab,therefractingsurfacesareparalleltoeach\nother.So,dispersioncannotoccur.Thisisduetofactthattherectangularglassslab\ncanbeconsideredasequivalentoftwoidenticalprismsininvertedpositionplacedin\naninvertedpositionwithrespecttoeachother.Thedeviationanddispersion\nproducedbythesecondinvertedprismisequalandoppositetothatproducedbythe\nfirstprism.\nTherefore,therewillneitherbedispersion,nordeviation,Le.secondinvertedprism\nrecombinesthe,colourtogiveawhitelightparalleltotheincidentrayagainand", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Long answer questions.txt\nfirstprism.\nTherefore,therewillneitherbedispersion,nordeviation,Le.secondinvertedprism\nrecombinesthe,colourtogiveawhitelightparalleltotheincidentrayagainand\nwillundergoonlylateraldisplacement.Hence,rectangularglassslabcannotproduce\nanyspectrum.\n6.(a)Explainthefollowingtermsusedinrelationtodefectsinvisionand\ncorrectionprovidedbythem:\n(i)Myopia(ii)Bifocallenses(iii)Far-sightedness.\n(b)Whyisthenormaleyeunabletofocusonanobjectplacedwithin10cm\nfromtheeye?\nAns:(a)(i)Myopia:Short-sightednessiscausedduetoexcessivecurvatureincornea\norelongationofeyeball.Imageisformedbeforeoftheretina.So,aconcavelensis\nusedtocorrectit.\n(ii)Bifocallenses:Theseareusedtocorrectpresbyopiceye.Thesecontainlenses\nwithupperconcaveandlowerconvexsurface.Presbyopiaariseswithage.", + "(ii)Bifocallenses:Theseareusedtocorrectpresbyopiceye.Thesecontainlenses\nwithupperconcaveandlowerconvexsurface.Presbyopiaariseswithage.\n(iii)Far-sightedness:Hypermetropiaorfar-sightednessiscausedduetogreaterfocal\nlengthofeyelensand/orwheneyeballbecomessmaller.Imageisformedbeyondthe\nretinaandcanbecorrectedusingaconvexlens.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Long answer questions.txt\n(iii)Far-sightedness:Hypermetropiaorfar-sightednessiscausedduetogreaterfocal\nlengthofeyelensand/orwheneyeballbecomessmaller.Imageisformedbeyondthe\nretinaandcanbecorrectedusingaconvexlens.\n(b)Thefocallengthofthelenscannotbechangeduptoanextentthatobjectsnearer\nthan25cmcanbeviewed(itsnearpointis25cm).Soitisnotabletofocusfor10\ncm.\n7WhatisTyndalleffect?What\u00a1sitscause?Explaintwophenomenaobservedin\ndailylifewhicharebasedonTyndalleffect\nAns:\nTyndalleffect:whenabeamoflightispassedthroughacolloidalsolution,placedin\nadarkroom,thepathofthebeambecomesilluminatedwhenobservedthrougha\nmicroscopeperpendiculartothepathoflight.ThisisknownasTyndalleffectThus.", + "Tyndalleffectisduetoscatteringoflightbythecolloidalparticlesandthecolloidal\nparticlesareseentobemovingaspointsoflightmovingagainstadarkbackground.\nDailylifephenomena\n(i)Whensunlightpassesthroughacanopyofadenseforest,thetinywaterdroplets\ninthemistscatterlightandbecomevisible.(ii)Whenafinebeamofsunlightentersasmokefilledroomthroughasmallhole,", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Long answer questions.txt\nDailylifephenomena\n(i)Whensunlightpassesthroughacanopyofadenseforest,thetinywaterdroplets\ninthemistscatterlightandbecomevisible.(ii)Whenafinebeamofsunlightentersasmokefilledroomthroughasmallhole,\nthesmokeparticlesbecomevisibleduetothescatteringoflight.\n8Writethefunctionofthepartsgivenbelow\n(i)Iris(ii)ciliarymuscles(iii)crystallinelens(iv)pupil(v)retina\nAns:\nIris:Itcontrolsthesizeofthepupil\nCiliarymuscles:Ithelpstheeyelenstofocustheimageoftheobjectonthe\nretinaby increasingordecreasingthecurvatureofeye\nlensandholdsthelensin position\nCrystallinelens:Itformsarealandinvertedimageoftheobjectonthe\nretina.\nPupil:Itregulatesandcontrolstheamountoflightenteringtheeye.", + "Pupil:Itregulatesandcontrolstheamountoflightenteringtheeye.\nRetina:Itisthescreenwhereimageisformedanditconvertslightinto\nelectrical signalsthataresenttothebrainthroughopticnerves\n9(a)Writetheimportanceofciliarymusclesinthehumaneye.Namethedefectof\nvisionthatarisesduetothegradualweakeningoftheciliarymusclesinoldage.\nWhattypeoflensesarerequiredbythepersonssufferingfromthisdefecttosee", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD -Long answer questions.txt\nvisionthatarisesduetothegradualweakeningoftheciliarymusclesinoldage.\nWhattypeoflensesarerequiredbythepersonssufferingfromthisdefecttosee\ntheobjectsclearly?\n(b)Nametheeyedefectinwhicheyelensbecomescloudyandmilkyandhowit\ncanbecorrected?\nAns:(a)Functionofciliarymuscles\nIthelpstheeyelenstofocustheimageoftheobjectontheretinaby\nincreasingordecreasingthecurvatureofeyelensandholdsthelensin\nposition\nDefectofvision:Presbyopia\nCorrection:Byusingbifocallenses\n(b)Cataractanditcanbecorrectedbycataractsurgery\n10Whatishypermetropia?Listtwocauses.Drawadiagramtoillustratethis\ndefect.Alsoshowadiagramhowthisdefectcanbecorrectedusingalens.\nAns:Theinabilityofaneyeinviewingthenearbyobject.", + "Ans:Theinabilityofaneyeinviewingthenearbyobject.\nReason:(i)Theverysmallsizeoftheeyeball.\n(ii)Focallengthoftheeyelensislarge.", + "CHAPTER 11-ELECTRCITY-READING MATERIAL.txt\nChapter 12 - Electricity. \nElectricity is a form of energy which is very handy,mean s it can be easily converted to \nother forms. \nElectric current and circuit : Electric current in a circuit is defined as the rate of flow \nof electric charge through any section of the circuit. That is I = Q / t . The S I unit of \nelectric current is ampere. \nOne am pere is 1 Coulomb per second . 1 A = 1 C / 1 s. \nAmmeter is the device used to measure electric current in a circuit. It is always \nconn ected in series. \nElectric circuit is a closed and continuous path of electric current. \nPotential difference : Potential difference between two points in a circuit is maintained \nby a cell. Potential difference between two points is defined as the work done( energy \nspend ) by the cell to move 1 C charge between the points. That is V = W / Q. \nS I unit of potential difference is volt ( V). 1 V = 1 J / 1C", + "CHAPTER 11-ELECTRCITY-READING MATERIAL.txt\nspend ) by the cell to move 1 C charge between the points. That is V = W / Q. \nS I unit of potential difference is volt ( V). 1 V = 1 J / 1C \nPotential difference is measured by using a voltmeter. It is connected in parallel to the \ndevice. \nOhm\u2019s law : states that the current flowing through a conductor is proportional to the \npotential difference across it under constant temperature of the conductor. That is \nV \u03b1 I or V = I R where R is known as the resistance of the conductor. The graph \nconnectin g V and I for a conductor will be a straight line.", + "CHAPTER 11-ELECTRCITY-READING MATERIAL.txt\nResistance of a conductor is defined as the ratio of potential difference to the current. \nR = V / I . S I unit of R is Ohm ( \u2126 ). \nFactors affecting resistance of a conductor : Resistance of a conductor is \n(i) proportional to its length and (ii) inversely proportional to its cross sectional area. \nR \u03b1 L / A or R = \u03c1 L / A where \u03c1 is called resistivity of the material of the conductor. \nResistivity of the material of the condu ctor is the resistance offered by a conductor of \nunit length and unit cross sectional area. Its S I unit is Ohm metre (\u2126 m ) \nResistivity varies with temperature of the conductor. It increases with temperature. \nGood conductors have low resist ivity where a s insulators have high resistivity. \nAlloys have higher resistivity than pure metals. Alloys are used as heating elements. \nCombinations of resistors : Resistances are connected in series as well as parallel.", + "CHAPTER 11-ELECTRCITY-READING MATERIAL.txt\nAlloys have higher resistivity than pure metals. Alloys are used as heating elements. \nCombinations of resistors : Resistances are connected in series as well as parallel. \nIn series connection, current through each resistor is the same, potential differences are \ndifferent. Total resistance of the combination is always more than the individual \nresistance values. The effective resistance is calculated by R = R 1 + R 2 + R 3", + "CHAPTER 11-ELECTRCITY-READING MATERIAL.txt\nIn parallel combination, the voltage across each resistor is same and is equal to the \nsource voltage. Current through each resistor is different. The net resistance of the \ncombination is always less than the least resistance available in the combination. T he \nexpression to calculate effective resistance is 1/ R = 1/ R 1 +1/ R 2 + 1 / R 3", + "CHAPTER 11-ELECTRCITY-READING MATERIAL.txt\nParallel combination is used in house hold circuit. \nHeating effect of electric current( Joule\u2019s law of heating) : When electric current \nflows through a resistor, electrical energy is converted in to heat energy . The amount of \nheat energy generated in a resistor is \n(i) directly proportional to the resistance of the resistor \n(ii) directly proportional to square of electric current \n(iii) directly proportional to the time of pa ssage of electric current. That is \n H = I 2 R t .Known as Joule\u2019s law of heating. \nThe heat generated in a resistor per unit time is known as its electrical power. \nThat is power , P = I 2 R = V2 / R = V I . \nS I unit of power is Watt (W) \nCommercial unit of energy is kW h also called as \u2018Unit\u2019 \n1 kW h = 3.6 x 106 J", + "CHAPTER 11-ELECTRCITY-m c q type of questions.txt\n1. A piece of wire of resistance R is cut in to five equal parts. These parts are then connected in \nparallel. If the equivalent resistance of this combination is R\u2019 , then the ratio R / R \u2018 is - \n (a) 1/ 25 (b) 1/5 (c) 6 (d) 25", + "CHAPTER 11-ELECTRCITY-m c q type of questions.txt\n2.. Which of the following terms does not represent electrical power in a circuit? \n(a) I2R (b) I R2 (c) V I (d) V2 / R \n3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V , the power \nconsumed will be \n(a) 100 W (b) 75 W (c) 50 W (d) 25 W \n4. Two conducting wires of the same material and of equal lengths and equal diameters are \nfirst connected in series and then in parallel in a circuit across the same potential difference. \nThe ratio of heat produced in series and parallel combination woul d be\u2026 \n(a) 1 : 2 (b) 2 : 1 (c) 1: 4 (d) 4 : 1 \n5. A cylindrical conductor of length \u2018l\u2019 and uniform cross section \u2018A\u2019 has resistance \u2018R\u2019 .", + "A cylindrical conductor of length \u2018l\u2019 and uniform cross section \u2018A\u2019 has resistance \u2018R\u2019 . The \narea of cross section of another conductor of same material and same resistance but of \nlength \u20182l\u2019 is \n( a) A/2 (b) 3 A / 2 (c) 2 A (d) 3 A \n6. The maximum resistance which can be made using four resistors each of resistance \u00bd \u2126 is \n( a) 2 \u2126 (b) 1 \u2126 (c) 2.5 \u2126 (d) 8 \u2126", + "CHAPTER 11-ELECTRCITY-m c q type of questions.txt\n( a) A/2 (b) 3 A / 2 (c) 2 A (d) 3 A \n6. The maximum resistance which can be made using four resistors each of resistance \u00bd \u2126 is \n( a) 2 \u2126 (b) 1 \u2126 (c) 2.5 \u2126 (d) 8 \u2126 \n7. In house hold circuit , we employ ....... combination of resistances \n (a) seri es (b) parallel ( c) both series and parallel ( d ) none of these. \n8. The correct form of Ohm\u2019s law is \n (a) V = I /R ( b) I = V / R (c) R = I / V (d) none of these. \n9. The effective resistance when two identical resistors of resistances \u2018R\u2019 each when \nconnected in parallel is \n (a) 2 R (b) R / 2 (c) R (d) 4 R \n10.", + "9. The effective resistance when two identical resistors of resistances \u2018R\u2019 each when \nconnected in parallel is \n (a) 2 R (b) R / 2 (c) R (d) 4 R \n10. Working principle of electrical fuse is ...... \n (a) heating effect of electric current \n (b) chemical effect of electric current. \n (c) magnetic effect of electric cu rrent \n (d) None of these. \n 11. The correct relation connecting current and charge in a conductor is:", + "CHAPTER 11-ELECTRCITY-m c q type of questions.txt\n(c) magnetic effect of electric cu rrent \n (d) None of these. \n 11. The correct relation connecting current and charge in a conductor is: \n (a) q = I t (b) q = I / t (c) q = t / I (d) None of these \n12.The S I unit of charge is \n (a) ampere (b) Coulomb (c) Volt (d) Ohm \n13. A wire of resistance R is cut into two equal halves. The resistance of the half is \n ( a) R (b) R / 2 (c) 2 R ( d) none of these. \n14. When the temperature of a conductor increases ,its resistance\u2026\u2026. \n (a) increases (b) decreases (c) No change (d) initially increases then decreses. \n15. The alloys have \u2026\u2026\u2026.", + "14. When the temperature of a conductor increases ,its resistance\u2026\u2026. \n (a) increases (b) decreases (c) No change (d) initially increases then decreses. \n15. The alloys have \u2026\u2026\u2026. resistances than pure metals \n (a) lower (b) higher (c) same (d) none of these \n16. The working principle of fuse wire is : \n (a) Archimedes\u2019 principle (b) Jo ules law of heating (c) Newton\u2019s laws of \nmotion (d) none of these \n17. Fuse wire should have \u2026\u2026\u2026\u2026 melting point", + "CHAPTER 11-ELECTRCITY-m c q type of questions.txt\n(a) Archimedes\u2019 principle (b) Jo ules law of heating (c) Newton\u2019s laws of \nmotion (d) none of these \n17. Fuse wire should have \u2026\u2026\u2026\u2026 melting point \n(a) low (b) high (c) any value (d) none of these \n18. When resistors are combined in series . \u2026\u2026.. through each resistor is same \n (a) resistance (b) potential difference (c) current (d) heat generated \n19. The equivalent resistance when three resistors of 3 ohms connected in series is\u2026 \n (a) 3 ohms (b) 9 ohms (c) 1 ohm ( d) none of these \n20. Three identical resistances o f 6 ohms connected in parallel. The equivalent resistance of \nthe combination is \u2026. \n (a) 6 ohms (b) 2 ohms (c) 18 ohms (d) none of these. \n21.", + "Three identical resistances o f 6 ohms connected in parallel. The equivalent resistance of \nthe combination is \u2026. \n (a) 6 ohms (b) 2 ohms (c) 18 ohms (d) none of these. \n21. In house hold circuits , we always use \u2026\u2026\u2026. Combination \n (a) series (b) parallel (c) both (d) none of these \n22. In a parallel combination of resistances, \u2026\u2026\u2026 is same for all resistors \n (a) current (b) resistance (c) potential difference (d) heat", + "CHAPTER 11-ELECTRCITY-m c q type of questions.txt\n22. In a parallel combination of resistances, \u2026\u2026\u2026 is same for all resistors \n (a) current (b) resistance (c) potential difference (d) heat \n23. The slope of V \u2013 I graph gives \u2026\u2026.. \n (a) current (b) resistance (c) potential difference (d) heat \n24. Ammeter is always connected in \u2026\u2026\u2026 with the circuit \n (a) parallel (b) series (c) both (d) none of these 25. Voltmeter is always connected \u2026\u2026\u2026 in a circuit \n (a) parallel (b) series (c) both (d) none of these", + "CHAPTER 11-ELECTRCITY-STUDY MATERIAL.txt\nNAVODAYA VIDYALAYA SAMITI \nE \u2013CONTENT FOR \nCHAPTER 12 \nELECTRICITY \n \n \n PREPARED BY \n REKHA ASHOK \n VICE PRINCIPAL \nINSTRUCTIONAL OBJECTIVES. \n\u2022 Student comes to know the basic terminology in electricity \n\u2022 Student understand s the concept of electric potential and \npotential difference. \n\u2022 Student understands Ohm\u2019 s law \n\u2022. Student understand s laws of combination of resistances \n\u2022Student improves drawing skills. \n\u2022Student develops interpretation skills from graphs. \n\u2022To develop Scientific attitude among children. \n\u2022To develop the competency level of students \n\u2022To develop scientific Values .", + "CHAPTER 11-ELECTRCITY-STUDY MATERIAL.txt\nFLOW OF CHARGES (Q) \nELECTRIC CURRENT (I) \nELECTRIC POTENTIAL (V) \nPOTENTIAL DIFFERENCE \n(V) COULOMB \nAMPERE \nVOLT \nVOLT \nRESISTANCE (R) OHM \nWHETHER ALL MATERIALS WILL ALLOW \nCURRENT TO PASS THROUGH IT? \nCONDUCTORS \n\u2022Substances which allow electric current to pass \nthrough it .eg.copper,silver,aluminium,graphite \nINSULATORS \n\u2022Substances which do not allow electric current \nto pass through it. eg. Rubber \n,ceramic,glass,mica BY PROVIDING \nPOTENTIAL DIFFERENCE \nHow? \nBY CONNECTING CELL, \nBATTERY OR POWER SUPPLY How\n? \nMAKE A BULB \nGLOW \nHow\n? \nBY PASSING CURRENT \nTHROUGH IT Electric current is \nrate of flow of \ncharges What? \nWhat?", + "BY CONNECTING CELL, \nBATTERY OR POWER SUPPLY How\n? \nMAKE A BULB \nGLOW \nHow\n? \nBY PASSING CURRENT \nTHROUGH IT Electric current is \nrate of flow of \ncharges What? \nWhat? Difference in \nelectric potential \nenergy \nElectric potential difference is \nthe amount of work done in \nbringing a unit positive charge \nfrom one point to another Electric current is the rate of flow of charge .", + "CHAPTER 11-ELECTRCITY-STUDY MATERIAL.txt\nThe SI unit of current is Ampere (A) \n 1 ampere = 1 coulomb/ 1 sec \nOne ampere is that current flowing through a material if one \ncoulomb of charge passes through it in one second. \n The device that can measure current is named as AMMETER. \nAmmeter is connected in series in a circuit POTENTIAL DIFFERENCE \n Potential difference is defined as the amount of work done in \nbringing a unit positive charge from one point to another point.(V) \n \n V= W/Q W - work done \n Q- Charge \n 1volt=1joule/1coulomb \nThe potential difference between any two points is said to be \n1volt if 1joule of work is to be done in moving 1couloumb of \ncharge between the points. \nIn an electric circuit current flowing is directly proportional to the", + "CHAPTER 11-ELECTRCITY-STUDY MATERIAL.txt\n1volt if 1joule of work is to be done in moving 1couloumb of \ncharge between the points. \nIn an electric circuit current flowing is directly proportional to the \npotential difference applied under constant physical conditions \n V \u03b1 I \n V = RI \n Where R is the constant of proportionality and called as the \nresistance of the material. \n Resistance is the property of a material by which it opposes \nthe flow of current through it. \n Resistance varies from material to material. Good conductors \nwill have low resistance and bad conductors and insulators will \nhave more resistance OHM\u2019S LAW \n \n v \n I R V=IR \nI=V/R \nR=V/I REMEMBER A - AMMETER \nV - VOLTMETER \nR - RESISTANCE \nK - KEY \nRh - Rheostat A PRACTICAL OHM\u2019LAW CIRCUIT V \u2013I GRAPH \nFactors affecting resistance of a material \n\ud835\udc45=\ud835\udf0c\ud835\udc59\n\ud835\udc34", + "CHAPTER 11-ELECTRCITY-STUDY MATERIAL.txt\n\ud835\udf0c=\ud835\udc45\ud835\udc34\n\ud835\udc59 \n \n\ud835\udf0c=\ud835\udc45 \ud835\udc64\u210e\ud835\udc52\ud835\udc5b \ud835\udc34=1 \ud835\udc4e\ud835\udc5b\ud835\udc51 \ud835\udc59=1 \n GROUPING OF RESISTORS \n SERIES \n \nV=V1+V2+V3 \n \nV=IR OHM\u2019S LAW \n \nTherefore IRs = IR1+IR2+ IR3 \n \nSo Rs = R1+R2+R3 PARALLEL \n \n \n I=I1+I2+I3 \n \nI=V/R OHM\u2019S LAW \n \nTherefore \nV/RP=V/R1+V/R2+V/R3", + "CHAPTER 11-ELECTRCITY-STUDY MATERIAL.txt\nSo 1/RP=1/R1+1/R2+1/R3 HEATING EFECT OF ELECTRIC \nCURRENT \nHeat energy will be produced by a material when current passes \nthrough it .According to Joule\u2019s law of heating effect \n H= work done=VXQ= VIt OR H=(V2/R)t \n OR H=I2 Rt \nTherefore the heat energy produced : - \n1.Directly depends on the resistance when current is constant. \nand \n2.Inversely proportional to resistance when potential difference remains \nconstant. \n \n Heat energy is measured in JOULES(J) \n ELECTRIC POWER \nRATE AT WHICH ELECTRICAL ENERGY IS CONSUMED \n P=E/t \n I= Q/t \n Q=It \n V=W/Q= W/It \n W=VIt \n Also Power = W/t = VIt/t = VI \u2026\u2026\u2026\u2026.1 \nHence P = VI P = VI \n = IR x I or = V x V/R", + "CHAPTER 11-ELECTRCITY-STUDY MATERIAL.txt\nAlso Power = W/t = VIt/t = VI \u2026\u2026\u2026\u2026.1 \nHence P = VI P = VI \n = IR x I or = V x V/R \n = I2R \u2026\u2026\u2026\u2026\u20262 = V2/R \u2026\u2026\u2026\u2026.3 \n Electric power is measured in Watts (W)", + "CHAPTER 11-ELECTRCITY-STUDY MATERIAL.txt\nCOMMERCIAL UNIT OF ELECTRICAL \nENERGY \nElectrical energy is measured commercially in a unit named \nkilowatt hour (kWhr) \n1kWhr=1000Wx 3600s \n =36x105 ws \n=36x105 J ELECTRICAL ENERGY CONVERTED \nTO HEAT ENERGY IN HEATING COILS \nHeating coils will have high resistance and high melting point. \nThey are made of alloys such as nichrome. Alloys will not get \noxidised when they get in contact with air. \nElectric bulbs \n \nBulb filaments are made of \ntungsten because it has \n 1.High melting point \n2 .High tensile strength \n3.Low vapour pressure Electric fuse \nMaterials used for making electric fuse should \nhave \n \n1.Low melting point \n \n2.Low resistivity \n E-LINKS \n\u2022https:// www.khanacademy.org/science/in -in-class10th -physics \n\u2022https:// www.topperlearning.com/cbse -class -10-physics \n\u2022https:// diksha.gov.in/explore", + "CHAPTER 11-ELECTRCITY-SHORT ANSWER TYPE QUESTIONS.txt\nSHORT ANSWER TYPE QUESTIONS. \n1. A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10 -6\u2126 m. What will be the length \nof this wire to make its resistance 10 \u2126 ? How much does the resistance change if the diameter is \ndoubled? \n2. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the \ncircuit. Find the value of resistance of the resistor. \n3. A battery of 9 V is connected in series with resistors 0.2 \u2126 , 0.3\u2126, 0.4\u2126, 0.5 \u2126 and 12 \u2126, \nrespe ctively. How much current would flow through the 12\u2126 resistor? \n4. How many 176 \u2126 resistors in parallel are required to carry 5 A on a 220 V line?", + "How much current would flow through the 12\u2126 resistor? \n4. How many 176 \u2126 resistors in parallel are required to carry 5 A on a 220 V line? \n5..Show how you would connect three resistors , each of resistance 6 \u2126, so that the combination has \na resista nce of (i) 9 \u2126 and (ii) 4 \u2126 \n6. Several electric bulbs designed to be used on a 220 V line, are rated 10 W. How many lamps can \nbe connected in parallel with each other across the two wires of 220 V line if the maximum", + "CHAPTER 11-ELECTRCITY-SHORT ANSWER TYPE QUESTIONS.txt\n6. Several electric bulbs designed to be used on a 220 V line, are rated 10 W. How many lamps can \nbe connected in parallel with each other across the two wires of 220 V line if the maximum \nallowable current is 5 A. \n7. A hot plat e of electric oven connected to a 220 V line has two resistance coils A and B, each of \n24 \u2126 resistance, which may be used separately, in series ,or in parallel. What are the currents in the \nthree cases? \n8. Compute the heat generated while transferring 960 00 C of charge in one hour through a potential \ndifference of 50 V. \n9. An electrtic bulb is connected to a 220 V generator. The current is 0.5 A. What is the power of the \nbulb? \n10. An electric refrigerator rated 400 W operates 8 hour per day . What is the cost of the energy to \noperate iot for 30 days at Rs 3.00 per kW h? \n11.", + "10. An electric refrigerator rated 400 W operates 8 hour per day . What is the cost of the energy to \noperate iot for 30 days at Rs 3.00 per kW h? \n11. Compare the power used in the 2 \u2126 resistor in each of the following circuits: \n (i) a 6 V battery in series with 1 \u2126 and 2 \u2126 resistors and", + "CHAPTER 11-ELECTRCITY-SHORT ANSWER TYPE QUESTIONS.txt\n11. Compare the power used in the 2 \u2126 resistor in each of the following circuits: \n (i) a 6 V battery in series with 1 \u2126 and 2 \u2126 resistors and \n (ii) a 4 V battery in parall el with 12\u2126 and 2 \u2126 resistors . \n12Two lamps ,one rated 100 W at 220 V and the other 60 W at 220 V, are connected in parallel to \nelectric mains supply. What current is drawn from the line if the supply voltage is 220 V? \n13. Which uses more energy , a 250 W T V set in I hr, or 1200 W toaster in 110 minutes? \n14. An electric heater of resistance 8 \u2126 draws 15 A from the service mains 2 hours. Calculate the rate \nat which heat is developed in the heater. \n15. (a) Define the S I Unit of current. \n (b) Name the device used for measuring electric current. How it is connected in a circuit?", + "Calculate the rate \nat which heat is developed in the heater. \n15. (a) Define the S I Unit of current. \n (b) Name the device used for measuring electric current. How it is connected in a circuit? \n ( c) The filament of an electric lamp , which draws a current of 0.2 A, is used for 5 hours . \n Calculate the amount of charge flowing through the circuit. \n 16. (i) D efine 1 V", + "CHAPTER 11-ELECTRCITY-SHORT ANSWER TYPE QUESTIONS.txt\n( c) The filament of an electric lamp , which draws a current of 0.2 A, is used for 5 hours . \n Calculate the amount of charge flowing through the circuit. \n 16. (i) D efine 1 V \n (ii) Name the device used for measuring potential difference. How it is connected in a \ncircuit? \n (iii) Calculate the work done to move 2 C of charge between two points having a potential \ndifference of 6 V \n17. How the current through a conduc tor change when the potential difference across the \nconductor is doubled? What happens to the resistance of the conductor in this case? \n18. Draw the circuit diagram to study Ohm\u201fs law. Explain the function of the rheostat in the \ncircuit. \n19. What happens to the resistance and resistivity of a wire in the following cases \n (i) wire is cut into two equal halves \n (ii) wire is stretched to double its length, \n20.", + "Explain the function of the rheostat in the \ncircuit. \n19. What happens to the resistance and resistivity of a wire in the following cases \n (i) wire is cut into two equal halves \n (ii) wire is stretched to double its length, \n20. Two resistors of resistances 2 \u2126 and 4 \u2126 are connected in series with a bat tery of 6 V", + "CHAPTER 11-ELECTRCITY-SHORT ANSWER TYPE QUESTIONS.txt\n(i) wire is cut into two equal halves \n (ii) wire is stretched to double its length, \n20. Two resistors of resistances 2 \u2126 and 4 \u2126 are connected in series with a bat tery of 6 V \nwith an ammeter to measure the current through 4 \u2126 resistor and a voltmeter to measure the \nvoltage across 2 \u2126 resistor with a closed plug key. \n (i) Draw the circuit diagram \n (ii) Calculate the readings of ammeter and voltmeter. \n21. Five resistors of identical resistances are connected (i) in series and (ii) in parallel. \n Calculate the ratio of equivalent resistance of series with that of parallel. \n22. (i) Define electric power. Write its S I unit. \n (ii) A bulb has a power of 10W. Calculate the current it draws when connected across a \nbattery of 12 V. \n23. (a) How does the resistivity of alloys compare with those of pure metals from which they \nmay have beenformed?", + "(ii) A bulb has a power of 10W. Calculate the current it draws when connected across a \nbattery of 12 V. \n23. (a) How does the resistivity of alloys compare with those of pure metals from which they \nmay have beenformed? \n(b) Calculate the resistance of a 1 km long copper wire of area of cr oss section 2 x 10-2 cm2.", + "CHAPTER 11-ELECTRCITY-SHORT ANSWER TYPE QUESTIONS.txt\nmay have beenformed? \n(b) Calculate the resistance of a 1 km long copper wire of area of cr oss section 2 x 10-2 cm2. \nThe resistivity of copper is 1.62 x 10-8 \u2126 m \n24. (a) Distinguish between resistance and resistivity . \n (b) A wire is stretched so that its length becomes 6/5 times of its original length. If the \noriginal resistance is 25 \u2126 , find its new resistance. \n25. Write three advantages of connecting the electrical appliances in parallel with the mains \nthan in series. \n26. How many 40 W,220 V lamps can be safely connected to a 220 V, 5 A line? \n27 . How much current will an electric iro n draw from a 220 V source if the resistance of its \nelement when hot is 55 \u2126 .Also calculate the wattage of the iron when it is operated on 220 V 28.", + "27 . How much current will an electric iro n draw from a 220 V source if the resistance of its \nelement when hot is 55 \u2126 .Also calculate the wattage of the iron when it is operated on 220 V 28. A current of 5 A is passed through a conductor of 12 \u2126 for 2 minutes .Calculate the \namount of heat produc ed. \n29. The V \u2013 I graph for the two wires A and B are shown in the figure. If we connect both the", + "CHAPTER 11-ELECTRCITY-SHORT ANSWER TYPE QUESTIONS.txt\namount of heat produc ed. \n29. The V \u2013 I graph for the two wires A and B are shown in the figure. If we connect both the \nwires one by one to the same battery , which of the two will produce more heat per unit \ntime?Justify your answer.", + "CHAPTER 11-ELECTRCITY-SHORT ANSWER TYPE QUESTIONS.txt\nA \n V B", + "CHAPTER 11-ELECTRCITY-SHORT ANSWER TYPE QUESTIONS.txt\nI \n30. Three resistors of resistances R 1 , R2 and R 3 are connected (i) in series , and (ii) in \nparallel. Write the expressions for the equivalent resistance of the combinations for the \nequivalent resistance of the combination in each case. \n31. Two resistors of 12 \u2126 each are connected to a battery of 3 V. Calculate the ratio of the \npower consumed by the resulting combinations with minimum resistance and maximum \nresistance. \n32. Write the relation between resistance and resistivity of the material of a conductor in the \nshape of a cylinder of length \u201el\u201f and area of cross section \u201eA\u201f, Hence derive the S I unit of \nelectrical resistivity. \n33. Resistance of a metal wire of length 5 m is 100 \u2126. If the area of cross section of the wire \nis 3 x 10 -7 m 2, calculate the resistivity of the metal. \n34. An electric bulb is rated 200 V, 100 W. What is its resistance? Calculate the energy", + "CHAPTER 11-ELECTRCITY-SHORT ANSWER TYPE QUESTIONS.txt\nis 3 x 10 -7 m 2, calculate the resistivity of the metal. \n34. An electric bulb is rated 200 V, 100 W. What is its resistance? Calculate the energy \nconsumed by three such bulbs if they glow continuously for 10 hours for complete month of \nNovember. Calculate the total cost if the rate is Rs 6.50 per unit \n35. What is meant by the statement \u201c the resistance of a conductor is one ohm\u201d? \n36.Define electric power. Writ e an expression relating electric power , potential difference \nand resistance. \n37. How many 132 \u2126 resistors in parallel are required to carry 5 A on a 220 V line? \n38. A torch bulb is rated 5 V and 500 mA. Calculate its (i ) power (ii) resistance (iii) energy \nconsumed when it is lighted for 2 \u00bd hours.", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nElectricity12 CHAPTER\nElectricity has an important place in modern society. It is a controllable\nand convenient form of energy for a variety of uses in homes, schools,\nhospitals, industries and so on. What constitutes electricity? How does\nit flow in an electric circuit? What are the factors that control or regulate\nthe curr ent thr ough an electric cir cuit? In this Chapter , we shall attempt\nto answer such questions. W e shall also discuss the heating ef fect of\nelectric current and its applications.\n12.1 ELECTRIC CURRENT AND CIRCUIT12.1 ELECTRIC CURRENT AND CIRCUIT12.1 ELECTRIC CURRENT AND CIRCUIT12.1 ELECTRIC CURRENT AND CIRCUIT12.1 ELECTRIC CURRENT AND CIRCUIT\nWe are familiar with air curr ent and water curr ent. W e know that flowing\nwater constitute water current in rivers. Similarly, if the electric charge\nflows through a conductor (for example, through a metallic wire), we\nsay that ther e is an electric curr ent in the conductor . In a tor ch, we", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nflows through a conductor (for example, through a metallic wire), we\nsay that ther e is an electric curr ent in the conductor . In a tor ch, we\nknow that the cells (or a battery, when placed in proper order) provide\nflow of char ges or an electric curr ent thr ough the tor ch bulb to glow. W e\nhave also seen that the torch gives light only when its switch is on. What\ndoes a switch do? A switch makes a conducting link between the cell and\nthe bulb. A continuous and closed path of an electric current is called an\nelectric circuit. Now, if the circuit is broken anywhere (or the switch of the\ntorch is turned off), the current stops flowing and the bulb does not glow.\nHow do we express electric current? Electric current is expressed by\nthe amount of charge flowing through a particular area in unit time. In\nother words, it is the rate of flow of electric charges. In circuits using\nmetallic wir es, electr ons constitute the flow of char ges. However , electr ons", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nother words, it is the rate of flow of electric charges. In circuits using\nmetallic wir es, electr ons constitute the flow of char ges. However , electr ons\nwere not known at the time when the phenomenon of electricity was first\nobserved. So, electric current was considered to be the flow of positive\ncharges and the direction of flow of positive charges was taken to be the\ndirection of electric current. Conventionally, in an electric circuit the\ndirection of electric current is taken as opposite to the direction of the\nflow of electrons, which are negative charges.\n2020-21\nScience 200If a net charge Q, flows across any cross-section of a conductor in\ntime t , then the current I, through the cross-section is\nIQ\nt= (12.1)\nThe SI unit of electric charge is coulomb (C), which is equivalent to\nthe char ge contained in nearly 6 \u00d7 1018 electr ons. (W e know that an\nelectron possesses a negative charge of 1.6 \u00d7 10\u201319 C.) The electric", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nthe char ge contained in nearly 6 \u00d7 1018 electr ons. (W e know that an\nelectron possesses a negative charge of 1.6 \u00d7 10\u201319 C.) The electric\ncurrent is expressed by a unit called ampere (A), named after the\nFrench scientist, Andre-Marie Ampere (1775\u20131836). One ampere is\nconstituted by the flow of one coulomb of charge per second, that is,\n1 A = 1 C/1 s. Small quantities of current are expressed in milliampere\n(1 mA = 10\u20133 A) or in microampere (1 \u00b5A = 10\u20136 A).\nAn instrument called ammeter measures electric\ncurrent in a circuit. It is always connected in series\nin a circuit through which the current is to be\nmeasured. Figure 12.1 shows the schematic\ndiagram of a typical electric circuit comprising a\ncell, an electric bulb, an ammeter and a plug key.", + "It is always connected in series\nin a circuit through which the current is to be\nmeasured. Figure 12.1 shows the schematic\ndiagram of a typical electric circuit comprising a\ncell, an electric bulb, an ammeter and a plug key.\nNote that the electric current flows in the circuit\nfrom the positive ter minal of the cell to the negative\nterminal of the cell thr ough the bulb and ammeter .\nFigure 12.1Figure 12.1Figure 12.1Figure 12.1Figure 12.1", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nfrom the positive ter minal of the cell to the negative\nterminal of the cell thr ough the bulb and ammeter .\nFigure 12.1Figure 12.1Figure 12.1Figure 12.1Figure 12.1\nA schematic diagram of an electric circuit\ncomprising \u2013 cell, electric bulb, ammeter and\nplug key\nQUESTIONS\n?Example 12.1\nA current of 0.5 A is drawn by a filament of an electric bulb for 10\nminutes. Find the amount of electric charge that flows through the\ncircuit.\nSolution\nWe are given, I = 0.5 A; t = 10 min = 600 s.\nFrom Eq. (12.1), we have\nQ=It\n=0.5 A \u00d7 600 s\n=300 C\n1. What does an electric circuit mean?\n2. Define the unit of current.\n3. Calculate the number of electrons constituting one coulomb of charge.", + "(12.1), we have\nQ=It\n=0.5 A \u00d7 600 s\n=300 C\n1. What does an electric circuit mean?\n2. Define the unit of current.\n3. Calculate the number of electrons constituting one coulomb of charge.\n2020-21\nElectricity 20112.2 ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE12.2 ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE12.2 ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE12.2 ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE12.2 ELECTRIC POTENTIAL AND POTENTIAL DIFFERENCE", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nWhat makes the electric charge to flow? Let us consider the analogy of\nflow of water . Char ges do not flow in a copper wir e by themselves, just as\nwater in a perfectly horizontal tube does not flow. If one end of the tube\nis connected to a tank of water kept at a higher level, such that there is a\npressure difference between the two ends of the tube, water flows out of\nthe other end of the tube. For flow of charges in a conducting metallic\nwire, the gravity, of course, has no role to play; the electrons move only\nif there is a difference of electric pressure \u2013 called the potential differenc e \u2013\nalong the conductor . This dif ference of potential may be pr oduced by a\nbattery, consisting of one or more electric cells. The chemical action within\na cell generates the potential difference across the terminals of the cell,\neven when no current is drawn from it. When the cell is connected to a\nconducting circuit element, the potential difference sets the charges in", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\neven when no current is drawn from it. When the cell is connected to a\nconducting circuit element, the potential difference sets the charges in\nmotion in the conductor and produces an electric current. In order to\nmaintain the current in a given electric circuit, the cell has to expend its\nchemical energy stored in it.\nWe define the electric potential dif ference between two points in an\nelectric circuit carrying some current as the work done to move a unit\ncharge from one point to the other \u2013\nPotential dif ference ( V) between two points = Work done ( W)/Char ge (Q)\nV=W/Q (12.2)\nThe SI unit of electric potential difference is volt (V), named after\nAlessandr o Volta (174 5\u20131827), an Italian physicist. One volt is the\u2018Flow\u2019 of charges inside a wire\nHow does a metal conduct electricity? Y ou would think that a low-ener\ngy electr on\nwould have great dif ficulty passing through a solid conductor . Inside the solid, the", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nHow does a metal conduct electricity? Y ou would think that a low-ener\ngy electr on\nwould have great dif ficulty passing through a solid conductor . Inside the solid, the\natoms are packed together with very little spacing between them. But it turns out\nthat the electrons are able to \u2018travel\u2019 through a perfect solid crystal smoothly and\neasily, almost as if they were in a vacuum. The \u2018motion\u2019 of electrons in a conductor ,\nhowever , is very dif ferent fr om that of char ges in empty space. When a steady curr ent\nflows thr ough a conductor , the electr ons in it move with a certain average \u2018drift speed\u2019.\nOne can calculate this drift speed of electrons for a typical copper wire carrying a\nsmall current, and it is found to be actually very small, of the order of 1 mm s-1. How\nis it then that an electric bulb lights up as soon as we turn the switch on? It cannot be\nthat a current starts only when an electron from one terminal of the electric supply", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nis it then that an electric bulb lights up as soon as we turn the switch on? It cannot be\nthat a current starts only when an electron from one terminal of the electric supply\nphysically reaches the other terminal through the bulb, because the physical drift of\nelectrons in the conducting wires is a very slow process. The exact mechanism of the\ncurrent flow, which takes place with a speed close to the speed of light, is fascinating,\nbut it is beyond the scope of this book. Do you feel like probing this question at an\nadvanced level?More to Know!\n2020-21\nScience 202potential difference between two points in a current carrying conductor\nwhen 1 joule of work is done to move a charge of 1 coulomb from one\npoint to the other .\nTherefore, 1 volt = 1 joule\n1 coulomb(12.3)\n1 V = 1 J C\u20131\nThe potential difference is measured by means of an instrument called\nthe voltmeter . The voltmeter is always connected in parallel acr oss the\npoints between which the potential difference is to be measured.", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nthe voltmeter . The voltmeter is always connected in parallel acr oss the\npoints between which the potential difference is to be measured.\nExample 12.2\nHow much work is done in moving a charge of 2 C across two points\nhaving a potential difference 12 V?\nSolution\nThe amount of charge Q, that flows between two points at potential\ndifference V (= 12 V) is 2 C. Thus, the amount of work W, done in\nmoving the charge [from Eq. (12.2)] is\nW= VQ\n= 12 V \u00d7\n 2 C\n= 24 J.\nQUESTIONS\n?\n12.3 CIRCUIT DIAGRAM12.3 CIRCUIT DIAGRAM12.3 CIRCUIT DIAGRAM12.3 CIRCUIT DIAGRAM12.3 CIRCUIT DIAGRAM\nWe know that an electric cir cuit, as shown in Fig. 12.1, comprises a cell\n(or a battery), a plug key, electrical component(s), and connecting wires.", + "12.1, comprises a cell\n(or a battery), a plug key, electrical component(s), and connecting wires.\nIt is often convenient to draw a schematic diagram, in which different\ncomponents of the circuit are represented by the symbols conveniently\nused. Conventional symbols used to represent some of the most", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\ncomponents of the circuit are represented by the symbols conveniently\nused. Conventional symbols used to represent some of the most\ncommonly used electrical components are given in Table 12.1.1. Name a device that helps to maintain a potential difference across a\nconductor .\n2. What is meant by saying that the potential difference between two points\nis 1 V?\n3. How much energy is given to each coulomb of charge passing through a\n6 V battery?\n2020-21\nElectricity 203Table 12.1 Symbols of some commonly used components in circuit diagrams\nSl. Components Symbols\nNo.", + "3. How much energy is given to each coulomb of charge passing through a\n6 V battery?\n2020-21\nElectricity 203Table 12.1 Symbols of some commonly used components in circuit diagrams\nSl. Components Symbols\nNo.\n1 An electric cell\n2 A battery or a combination of cells\n3 Plug key or switch (open)\n4 Plug key or switch (closed)\n5 A wire joint\n6 Wires crossing without joining\n7 Electric bulb or \n8 A resistor of resistance R\n9 Variable r esistance or r heostat or \n10 Ammeter\n11 Voltmeter\n12.4 OHM\u2019S LA12.4 OHM\u2019S LA12.4 OHM\u2019S LA12.4 OHM\u2019S LA12.4 OHM\u2019S LA W WW WW\nIs there a relationship between the potential difference across a", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\n10 Ammeter\n11 Voltmeter\n12.4 OHM\u2019S LA12.4 OHM\u2019S LA12.4 OHM\u2019S LA12.4 OHM\u2019S LA12.4 OHM\u2019S LA W WW WW\nIs there a relationship between the potential difference across a\nconductor and the current through it? Let us explore with an Activity.\nActivity 12.1Activity 12.1Activity 12.1Activity 12.1Activity 12.1\n/square6Set up a circuit as shown in Fig. 12.2, consisting of a nichrome wire XY of length, say 0.5 m,\nan ammeter , a voltmeter and four cells of 1.5 V each. (Nichr ome is an alloy of nickel, chr omium,\nmanganese, and iron metals.)\n2020-21\nScience 204Figure 12.3Figure 12.3Figure 12.3Figure 12.3Figure 12.3\nV\u2013I graph for a nichrome wire.", + "2020-21\nScience 204Figure 12.3Figure 12.3Figure 12.3Figure 12.3Figure 12.3\nV\u2013I graph for a nichrome wire. A\nstraight line plot shows that as the\ncurrent through a wire increases, the\npotential difference across the wire\nincreases linearly \u2013 this is Ohm\u2019s law.S. Number of cells Current through Potential difference V/I\nNo. used in the the nichrome across the (volt/ampere)\ncircuit wire, I nichrome\n(ampere) wire, V (volt)\n1 1\n2 2\n3 3\n4 4", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nNo. used in the the nichrome across the (volt/ampere)\ncircuit wire, I nichrome\n(ampere) wire, V (volt)\n1 1\n2 2\n3 3\n4 4\nIn this Activity, you will find that approximately the\nsame value for V/I is obtained in each case. Thus the V\u2013I\ngraph is a straight line that passes through the origin of\nthe graph, as shown in Fig. 12.3. Thus, V/I is a constant\nratio.\nIn 1827, a German physicist Georg Simon Ohm\n(1787\u20131854) found out the relationship between the current\nI, flowing in a metallic wire and the potential difference across\nits terminals. The potential difference, V, across the ends of\na given metallic wire in an electric circuit is directly\nproportional to the current flowing through it, provided its\ntemperature remains the same. This is called Ohm\u2019s law.", + "The potential difference, V, across the ends of\na given metallic wire in an electric circuit is directly\nproportional to the current flowing through it, provided its\ntemperature remains the same. This is called Ohm\u2019s law. In\nother words \u2013\nV \u221d I (12.4)\nor V/I =constant\n=R\nor V =IR (12.5)\nIn Eq. (12.4), R is a constant for the given metallic wire\nat a given temperature and is called its resistance. It is", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nother words \u2013\nV \u221d I (12.4)\nor V/I =constant\n=R\nor V =IR (12.5)\nIn Eq. (12.4), R is a constant for the given metallic wire\nat a given temperature and is called its resistance. It is\nthe property of a conductor to resist the flow of chargesFigure 12.2Figure 12.2Figure 12.2Figure 12.2Figure 12.2 Electric circuit for studying Ohm\u2019s law/square6First use only one cell as the source in the\ncircuit. Note the reading in the ammeter I,\nfor the current and reading of the voltmeter\nV for the potential difference across the\nnichrome wire XY in the circuit. Tabulate\nthem in the Table given.\n/square6Next connect two cells in the circuit and note\nthe respective readings of the ammeter and\nvoltmeter for the values of current through\nthe nichrome wire and potential difference\nacross the nichrome wire.\n/square6Repeat the above steps using three cells and\nthen four cells in the circuit separately.", + "/square6Repeat the above steps using three cells and\nthen four cells in the circuit separately.\n/square6Calculate the ratio of V to I for each pair of\npotential difference V and current I.", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\n/square6Repeat the above steps using three cells and\nthen four cells in the circuit separately.\n/square6Calculate the ratio of V to I for each pair of\npotential difference V and current I.\n/square6Plot a graph between V and I, and observe the nature of the graph.\n2020-21\nElectricity 205through it. Its SI unit is ohm, r epresented by the Gr eek letter \u2126. Accor ding\nto Ohm\u2019s law,\nR = V/I (12.6)\nIf the potential difference across the two ends of a conductor is 1 V\nand the current through it is 1 A, then the resistance R, of the conductor\nis 1 \u2126. That is, 1 ohm = 1 volt\n1 ampere\nAlso from Eq. (12.5) we get\nI = V/R (12.7)\nIt is obvious from Eq. (12.7) that the current through a resistor is\ninversely proportional to its resistance. If the resistance is doubled the\ncurrent gets halved.", + "(12.5) we get\nI = V/R (12.7)\nIt is obvious from Eq. (12.7) that the current through a resistor is\ninversely proportional to its resistance. If the resistance is doubled the\ncurrent gets halved. In many practical cases it is necessary to increase\nor decrease the current in an electric circuit. A component used to\nregulate current without changing the voltage source is called variable", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nor decrease the current in an electric circuit. A component used to\nregulate current without changing the voltage source is called variable\nresistance. In an electric circuit, a device called rheostat is often used\nto change the r esistance in the cir cuit. W e will now study about electrical\nresistance of a conductor with the help of following Activity.\nActivity 12.2Activity 12.2Activity 12.2Activity 12.2Activity 12.2\n/square6Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0 \u2013 5 A range), a plug\nkey and some connecting wires.\n/square6Set up the circuit by connecting four dry cells of 1.5 V each in series with the ammeter\nleaving a gap XY in the circuit, as shown in Fig. 12.4.\nFigure 12.4Figure 12.4Figure 12.4Figure 12.4Figure 12.4\n/square6Complete the cir cuit by connecting the nichr ome wir e in the gap XY . Plug the key.", + "12.4.\nFigure 12.4Figure 12.4Figure 12.4Figure 12.4Figure 12.4\n/square6Complete the cir cuit by connecting the nichr ome wir e in the gap XY . Plug the key. Note\ndown the ammeter reading. Take out the key from the plug. [Note: Always take out the key", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\n/square6Complete the cir cuit by connecting the nichr ome wir e in the gap XY . Plug the key. Note\ndown the ammeter reading. Take out the key from the plug. [Note: Always take out the key\nfrom the plug after measuring the current through the circuit. ]\n/square6Replace the nichrome wire with the torch bulb in the circuit and find the current through it by\nmeasuring the r eading of the ammeter .\n/square6Now r epeat the above step with the 10 W bulb in the gap XY .\n/square6Are the ammeter readings different for different components connected in the gap XY?\nWhat do the above observations indicate?\n/square6You may r epeat this Activity by keeping any material component in the gap. Observe the\nammeter readings in each case. Analyse the observations.\nIn this Activity we observe that the current is different for different\ncomponents. Why do they differ? Certain components offer an easy path\nfor the flow of electric curr ent while the others r esist the flow. W e know\n2020-21", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\ncomponents. Why do they differ? Certain components offer an easy path\nfor the flow of electric curr ent while the others r esist the flow. W e know\n2020-21\nScience 206that motion of electrons in an electric circuit constitutes an electric\ncurrent. The electr ons, however , are not completely fr ee to move within a\nconductor . They ar e restrained by the attraction of the atoms among\nwhich they move. Thus, motion of electrons through a conductor is\nretarded by its resistance. A component of a given size that offers a low\nresistance is a good conductor . A conductor having some appreciable\nresistance is called a r esistor . A component of identical size that of fers a\nhigher r esistance is a poor conductor . An insulator of the same size of fers\neven higher resistance.", + "A conductor having some appreciable\nresistance is called a r esistor . A component of identical size that of fers a\nhigher r esistance is a poor conductor . An insulator of the same size of fers\neven higher resistance.\n12.5 FA 12.5 FA12.5 FA 12.5 FA12.5 FA CTORS ON WHICH THE RESISTCTORS ON WHICH THE RESISTCTORS ON WHICH THE RESISTCTORS ON WHICH THE RESISTCTORS ON WHICH THE RESIST ANCE OF AANCE OF AANCE OF AANCE OF AANCE OF A", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nCONDUCTOR DEPENDSCONDUCTOR DEPENDSCONDUCTOR DEPENDSCONDUCTOR DEPENDSCONDUCTOR DEPENDS\nActivity 12.3Activity 12.3Activity 12.3Activity 12.3Activity 12.3\n/square6Complete an electric cir cuit consisting of a cell, an ammeter , a nichrome wir e of length l\n[say, marked (1)] and a plug key, as shown in Fig. 12.5.\nFigure 12.5 Figure 12.5 Figure 12.5 Figure 12.5 Figure 12.5 Electric circuit to study the factors on which the resistance of conducting wires depends\n/square6Now, plug the key. Note the curr ent in the ammeter .\n/square6Replace the nichrome wire by another nichrome wire of same thickness but twice the\nlength, that is 2 l [marked (2) in the Fig. 12.5].\n/square6Note the ammeter reading.\n/square6Now replace the wire by a thicker nichrome wire, of the same length l [marked (3)].", + "12.5].\n/square6Note the ammeter reading.\n/square6Now replace the wire by a thicker nichrome wire, of the same length l [marked (3)]. A\nthicker wire has a larger cross-sectional area. Again note down the current through the\ncircuit.\n/square6Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig. 12.5] in the circuit.", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\ncircuit.\n/square6Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig. 12.5] in the circuit.\nLet the wire be of the same length and same area of cross-section as that of the first nichrome\nwire [marked (1)]. Note the value of the current.\n/square6Notice the difference in the current in all cases.\n/square6Does the current depend on the length of the conductor?\n/square6Does the current depend on the area of cross-section of the wire used?\nIt is observed that the ammeter reading decreases to one-half when\nthe length of the wire is doubled. The ammeter reading is increased when\na thicker wire of the same material and of the same length is used in the\ncircuit. A change in ammeter reading is observed when a wire of different\nmaterial of the same length and the same area of cross-section is used.\nOn applying Ohm\u2019s law [Eqs.", + "A change in ammeter reading is observed when a wire of different\nmaterial of the same length and the same area of cross-section is used.\nOn applying Ohm\u2019s law [Eqs. (12.5) \u2013 (12.7)], we observe that the\n2020-21\nElectricity 207resistance of the conductor depends (i) on its length, (ii) on its area of", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nOn applying Ohm\u2019s law [Eqs. (12.5) \u2013 (12.7)], we observe that the\n2020-21\nElectricity 207resistance of the conductor depends (i) on its length, (ii) on its area of\ncross-section, and (iii) on the nature of its material. Precise measurements\nhave shown that resistance of a uniform metallic conductor is directly\nproportional to its length (l ) and inversely proportional to the area of\ncross-section ( A). That is,\nR \u221d l (12.8)\nand R \u221d 1/A (12.9)\nCombining Eqs. (12.8) and (12.9) we get\nR \u221d l\nA\nor, R = \u03c1l\nA(12.10)\nwhere \u03c1 (rho) is a constant of proportionality and is called the electrical\nresistivity of the material of the conductor . The SI unit of r esistivity is\n\u2126 m. It is a characteristic property of the material.", + "The SI unit of r esistivity is\n\u2126 m. It is a characteristic property of the material. The metals and alloys\nhave very low resistivity in the range of 10\u20138 \u2126 m to 10\u20136 \u2126 m. They are\ngood conductors of electricity. Insulators like rubber and glass have\nresistivity of the order of 1012 to 1017 \u2126 m. Both the resistance and", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\ngood conductors of electricity. Insulators like rubber and glass have\nresistivity of the order of 1012 to 1017 \u2126 m. Both the resistance and\nresistivity of a material vary with temperature.\nTable 12.2 reveals that the resistivity of an alloy is generally higher\nthan that of its constituent metals. Alloys do not oxidise (bur n) readily\nat high temperatures. For this reason, they are commonly used in\nelectrical heating devices, like electric iron, toasters etc. Tungsten is used\nalmost exclusively for filaments of electric bulbs, whereas copper and\naluminium are generally used for electrical transmission lines.", + "For this reason, they are commonly used in\nelectrical heating devices, like electric iron, toasters etc. Tungsten is used\nalmost exclusively for filaments of electric bulbs, whereas copper and\naluminium are generally used for electrical transmission lines.\nTable 12.2 Electrical resistivity * of some substances at 20\u00b0C\nMaterial Resistivity ( \u2126\u2126\u2126\u2126\u2126 m)\nConductors Silver 1.60 \u00d7 10\u20138\nCopper 1.62 \u00d7 10\u20138\nAluminium 2.63 \u00d7 10\u20138\nTungsten 5.20 \u00d7 10\u20138\nNickel 6.84 \u00d7 10\u20138\nIron 10.0 \u00d7 10\u20138\nChromium 12.9 \u00d7 10\u20138\nMercury 94.0 \u00d7 10\u20138\nManganese 1.84 \u00d7 10\u20136\nAlloys Constantan 49 \u00d7 10\u20136\n(alloy of Cu and Ni)\nManganin 44 \u00d7 10\u20136\n(alloy of Cu, Mn and Ni)", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nNickel 6.84 \u00d7 10\u20138\nIron 10.0 \u00d7 10\u20138\nChromium 12.9 \u00d7 10\u20138\nMercury 94.0 \u00d7 10\u20138\nManganese 1.84 \u00d7 10\u20136\nAlloys Constantan 49 \u00d7 10\u20136\n(alloy of Cu and Ni)\nManganin 44 \u00d7 10\u20136\n(alloy of Cu, Mn and Ni)\nNichrome 100 \u00d7 10\u20136\n(alloy of Ni, Cr\n, Mn and Fe)\nInsulators Glass 1010 \u2013 1014\nHard rubber 1013 \u2013 1016\nEbonite 1015 \u2013 1017\nDiamond 1012 - 1013\nPaper (dry) 1012\n* You need not memorise these values. Y ou can use these values for solving numerical\nproblems.\n2020-21\nScience 208Example 12.3\n(a)How much current will an electric bulb draw from a 220 V source,\nif the resistance of the bulb filament is 1200 \u2126 ?", + "Y ou can use these values for solving numerical\nproblems.\n2020-21\nScience 208Example 12.3\n(a)How much current will an electric bulb draw from a 220 V source,\nif the resistance of the bulb filament is 1200 \u2126 ? (b) How much\ncurrent will an electric heater coil draw from a 220 V source, if\nthe resistance of the heater coil is 100 \u2126?\nSolution\n(a)We are given V = 220 V ; R = 1200 \u2126.\nFrom Eq. (12.6), we have the current I = 220 V/1200 \u2126 = 0.18 A.\n(b)We are given, V = 220 V , R = 100 \u2126.\nFrom Eq. (12.6), we have the current I = 220 V/100 \u2126 = 2.2 A.", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nFrom Eq. (12.6), we have the current I = 220 V/1200 \u2126 = 0.18 A.\n(b)We are given, V = 220 V , R = 100 \u2126.\nFrom Eq. (12.6), we have the current I = 220 V/100 \u2126 = 2.2 A.\nNote the difference of current drawn by an electric bulb and electric\nheater from the same 220 V source!\nExample 12.4\nThe potential difference between the terminals of an electric heater\nis 60 V when it draws a current of 4 A from the source. What\ncurrent will the heater draw if the potential difference is increased\nto 120 V?", + "Example 12.4\nThe potential difference between the terminals of an electric heater\nis 60 V when it draws a current of 4 A from the source. What\ncurrent will the heater draw if the potential difference is increased\nto 120 V?\nSolution\nWe are given, potential differ ence V = 60 V , curr ent I = 4 A.\nAccording to Ohm\u2019s law, 60 V= = 154 AVRI= \u2126.\nWhen the potential difference is increased to 120 V the current is\ngiven by\ncurrent = 120 V= = 8 A15 V\nR \u2126.\nThe current through the heater becomes 8 A.\nExample 12.5\nResistance of a metal wire of length 1 m is 26 \u2126 at 20\u00b0C. If the\ndiameter of the wire is 0.3 mm, what will be the resistivity of the\nmetal at that temperature? Using Table 12.2, predict the material\nof the wire.", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\ndiameter of the wire is 0.3 mm, what will be the resistivity of the\nmetal at that temperature? Using Table 12.2, predict the material\nof the wire.\nSolution\nWe are given the resistance R of the wire = 26 \u2126, the diameter\nd = 0.3 mm = 3 \u00d7\n 10-4 m, and the length l of the wire = 1 m.\nTherefore, from Eq. (12.10), the resistivity of the given metallic wire is\n\u03c1= (RA/l) = (R \u03c0d2/4l)\nSubstitution of values in this gives\n\u03c1= 1.84 \u00d7 10\u20136 \u2126 m\nThe resistivity of the metal at 20\u00b0C is 1.84 \u00d7 10\u20136 \u2126 m. From\nTable 12.2, we see that this is the resistivity of manganese.", + "2020-21\nElectricity 209Example 12.6\nA wire of given material having length l and area of cross-section A\nhas a resistance of 4 \u2126. What would be the resistance of another wire\nof the same material having length l/2 and area of cross-section 2 A?\nSolution\nFor first wire\n R1\u03c1=l\nA= 4\u2126\nNow for second wire\nR2\u03c1/2=2l\nA \u03c1 =1\n4l\nA\nR21=4R1\nR2= 1\u2126\nThe r esistance of the new wir e is 1 \u2126.\nQUESTIONS", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nSolution\nFor first wire\n R1\u03c1=l\nA= 4\u2126\nNow for second wire\nR2\u03c1/2=2l\nA \u03c1 =1\n4l\nA\nR21=4R1\nR2= 1\u2126\nThe r esistance of the new wir e is 1 \u2126.\nQUESTIONS\n?1. On what factors does the resistance of a conductor depend?\n2. Will current flow more easily through a thick wire or a thin wire of the\nsame material, when connected to the same source? Why?\n3. Let the resistance of an electrical component remains constant while\nthe potential difference across the two ends of the component decreases\nto half of its former value. What change will occur in the current through\nit?\n4. Why are coils of electric toasters and electric irons made of an alloy\nrather than a pure metal?\n5. Use the data in Table 12.2 to answer the following \u2013\n(a)Which among iron and mercury is a better conductor?\n(b)Which material is the best conductor?", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nrather than a pure metal?\n5. Use the data in Table 12.2 to answer the following \u2013\n(a)Which among iron and mercury is a better conductor?\n(b)Which material is the best conductor?\n12.6 RESIST12.6 RESIST12.6 RESIST12.6 RESIST12.6 RESIST ANCE OF A SYSTEM OF RESISTORSANCE OF A SYSTEM OF RESISTORSANCE OF A SYSTEM OF RESISTORSANCE OF A SYSTEM OF RESISTORSANCE OF A SYSTEM OF RESISTORS\nIn preceding sections, we lear nt about some simple electric cir cuits. W e\nhave noticed how the current through a conductor depends upon its\nresistance and the potential difference across its ends. In various electrical\ngadgets, we often use resistors in various combinations. W e now ther efore\nintend to see how Ohm\u2019s law can be applied to combinations of resistors.\nThere ar e two methods of joining the r esistors together .", + "In various electrical\ngadgets, we often use resistors in various combinations. W e now ther efore\nintend to see how Ohm\u2019s law can be applied to combinations of resistors.\nThere ar e two methods of joining the r esistors together . Figur e 12.6\nshows an electric circuit in which three resistors having resistances R1,\nR2 and R3, respectively, are joined end to end. Here the resistors are said\nto be connected in series.", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nshows an electric circuit in which three resistors having resistances R1,\nR2 and R3, respectively, are joined end to end. Here the resistors are said\nto be connected in series.\n2020-21\nScience 210Figure 12.7Figure 12.7Figure 12.7Figure 12.7Figure 12.7 Resistors in parallelFigure 12.6Figure 12.6Figure 12.6Figure 12.6Figure 12.6 Resistors in series\nFigure 12.7 shows a combination of resistors in which three resistors\nare connected together between points X and Y . Here, the r esistors ar e\nsaid to be connected in parallel.\nActivity 12.4Activity 12.4Activity 12.4Activity 12.4Activity 12.4\n/square6Join three resistors of different values in series. Connect them\nwith a battery, an ammeter and a plug key, as shown in Fig. 12.6.", + "Connect them\nwith a battery, an ammeter and a plug key, as shown in Fig. 12.6.\nYou may use the r esistors of values like 1 \u2126, 2 \u2126, 3 \u2126 etc., and a\nbattery of 6 V for performing this Activity.\n/square6Plug the key. Note the ammeter reading.\n/square6Change the position of ammeter to anywhere in between the", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nbattery of 6 V for performing this Activity.\n/square6Plug the key. Note the ammeter reading.\n/square6Change the position of ammeter to anywhere in between the\nresistors. Note the ammeter reading each time.\n/square6Do you find any change in the value of current through the\nammeter?12.6.1 Resistors in Series\nWhat happens to the value of current when a number of resistors are\nconnected in series in a circuit? What would be their equivalent\nresistance? Let us try to understand these with the help of the following\nactivities.\n2020-21\nElectricity 211You will observe that the value of the current in the ammeter is the\nsame, independent of its position in the electric circuit. It means that in\na series combination of resistors the current is the same in every part of\nthe cir cuit or the same curr ent thr ough each r esistor .", + "It means that in\na series combination of resistors the current is the same in every part of\nthe cir cuit or the same curr ent thr ough each r esistor .\nActivity 12.5Activity 12.5Activity 12.5Activity 12.5Activity 12.5\n/square6In Activity 12.4, insert a voltmeter across the ends X and Y of the", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nActivity 12.5Activity 12.5Activity 12.5Activity 12.5Activity 12.5\n/square6In Activity 12.4, insert a voltmeter across the ends X and Y of the\nseries combination of three resistors, as shown in Fig. 12.6.\n/square6Plug the key in the circuit and note the voltmeter reading. It\ngives the potential difference across the series combination of\nresistors. Let it be V. Now measur e the potential dif ference acr oss\nthe two terminals of the battery. Compare the two values.\n/square6Take out the plug key and disconnect the voltmeter . Now insert\nthe voltmeter acr oss the ends X and P of the first r esistor , as\nshown in Fig. 12.8.\nFigure 12.8Figure 12.8Figure 12.8Figure 12.8Figure 12.8\n/square6Plug the key and measure the potential difference across the first\nresistor . Let it be V1.", + "12.8.\nFigure 12.8Figure 12.8Figure 12.8Figure 12.8Figure 12.8\n/square6Plug the key and measure the potential difference across the first\nresistor . Let it be V1.\n/square6Similarly, measure the potential difference across the other two\nresistors, separately. Let these values be V2 and V3, respectively.\n/square6Deduce a relationship between V, V1, V2 and V3.", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\n/square6Similarly, measure the potential difference across the other two\nresistors, separately. Let these values be V2 and V3, respectively.\n/square6Deduce a relationship between V, V1, V2 and V3.\nYou will observe that the potential dif ference V is equal to the sum of\npotential differences V1, V2, and V3. That is the total potential difference\nacross a combination of resistors in series is equal to the sum of potential\ndifference across the individual resistors. That is,\nV = V1 + V2 + V3(12.11)\nIn the electric circuit shown in Fig. 12.8, let I be the current through\nthe circuit. The current through each resistor is also I . It is possible to\nreplace the three resistors joined in series by an equivalent single resistor\nof resistance R , such that the potential difference V across it, and the\ncurrent I through the circuit remains the same. Applying the Ohm\u2019s law\nto the entire circuit, we have\nV = I R (12.12)\n2020-21", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nof resistance R , such that the potential difference V across it, and the\ncurrent I through the circuit remains the same. Applying the Ohm\u2019s law\nto the entire circuit, we have\nV = I R (12.12)\n2020-21\nScience 212On applying Ohm\u2019s law to the three resistors separately, we further\nhave\nV1=I R1[12.13(a)]\nV2=I R2[12.13(b)]\nand V3=I R3[12.13(c)]\nFrom Eq. (12.11),\nI R= I R1 + I R2 + I R3\nor\nRs= R1 +R2 + R3(12.14)\nWe can conclude that when several r esistors ar e joined in series, the\nresistance of the combination Rs equals the sum of their individual\nresistances, R1, R2, R3, and is thus greater than any individual resistance.", + "Example 12.7\nAn electric lamp, whose resistance is 20 \u2126, and a conductor of 4 \u2126\nresistance are connected to a 6 V battery (Fig. 12.9). Calculate (a) the\ntotal resistance of the circuit, (b) the current through the circuit, and\n(c) the potential dif ference acr oss the electric lamp and conductor .\nSolution\nThe resistance of electric lamp, R1 = 20 \u2126 ,", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\ntotal resistance of the circuit, (b) the current through the circuit, and\n(c) the potential dif ference acr oss the electric lamp and conductor .\nSolution\nThe resistance of electric lamp, R1 = 20 \u2126 ,\nThe r esistance of the conductor connected in series, R2 = 4 \u2126.\nThen the total resistance in the circuit\nR= R1 + R2\nRs=20 \u2126 + 4 \u2126 = 24 \u2126.\nThe total potential difference across the two terminals of the battery\nV = 6 V .", + "Now by Ohm\u2019s law, the current through the circuit is given by\nI=V/Rs\n=6 V/24 \u2126\n=0.25 A.Figure 12.9Figure 12.9Figure 12.9Figure 12.9Figure 12.9 An electric lamp connected in series with\na resistor of 4 \u2126 to a 6 V battery\n2020-21\nElectricity 213Applying Ohm\u2019s law to the electric lamp and conductor separately,\nwe get potential difference across the electric lamp,\nV1=20 \u2126 \u00d7 0.25 A\n=5 V;\nand,\nthat acr oss the conductor , V2 = 4 \u2126 \u00d7 0.25 A\n = 1 V .\nSuppose that we like to replace the series combination of electric", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nV1=20 \u2126 \u00d7 0.25 A\n=5 V;\nand,\nthat acr oss the conductor , V2 = 4 \u2126 \u00d7 0.25 A\n = 1 V .\nSuppose that we like to replace the series combination of electric\nlamp and conductor by a single and equivalent r esistor . Its resistance\nmust be such that a potential difference of 6 V across the battery\nterminals will cause a current of 0.25 A in the circuit. The resistance\nR of this equivalent resistor would be\nR=V/I\n=6 V/ 0.25 A\n=24 \u2126.\nThis is the total resistance of the series circuit; it is equal to the sum\nof the two resistances.\nQUESTIONS\n?1. Draw a schematic diagram of a circuit consisting of a battery of three\ncells of 2 V each, a 5 \u2126 resistor , an 8 \u2126 resistor , and a 12 \u2126 resistor , and\na plug key, all connected in series.\n2.", + "QUESTIONS\n?1. Draw a schematic diagram of a circuit consisting of a battery of three\ncells of 2 V each, a 5 \u2126 resistor , an 8 \u2126 resistor , and a 12 \u2126 resistor , and\na plug key, all connected in series.\n2. Redraw the circuit of Question 1, putting in an ammeter to measure\nthe current through the resistors and a voltmeter to measure the\npotential dif ference acr oss the 12 \u2126 resistor . What would be the r eadings\nin the ammeter and the voltmeter?", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nthe current through the resistors and a voltmeter to measure the\npotential dif ference acr oss the 12 \u2126 resistor . What would be the r eadings\nin the ammeter and the voltmeter?\n12.6.2 Resistors in Parallel\nNow, let us consider the arrangement of three resistors joined in parallel\nwith a combination of cells (or a battery), as shown in Fig.12.7.\nActivity 12.6Activity 12.6Activity 12.6Activity 12.6Activity 12.6\n/square6Make a parallel combination, XY, of three\nresistors having resistances R1, R2, and R3,\nrespectively. Connect it with a battery, a\nplug key and an ammeter , as shown in\nFig. 12.10. Also connect a voltmeter in\nparallel with the combination of resistors.\n/square6Plug the key and note the ammeter reading.\nLet the current be I. Also take the voltmeter\nreading. It gives the potential dif ference V,\nacross the combination.", + "Also connect a voltmeter in\nparallel with the combination of resistors.\n/square6Plug the key and note the ammeter reading.\nLet the current be I. Also take the voltmeter\nreading. It gives the potential dif ference V,\nacross the combination. The potential\ndifference acr oss each r esistor is also V. This\ncan be checked by connecting the voltmeter\nacross each individual resistor (see", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nacross the combination. The potential\ndifference acr oss each r esistor is also V. This\ncan be checked by connecting the voltmeter\nacross each individual resistor (see\nFig. 12.11).Figure 12.10Figure 12.10Figure 12.10Figure 12.10Figure 12.10\n2020-21\nScience 214/square6Take out the plug from the key. Remove the ammeter and voltmeter from the circuit.\nInsert the ammeter in series with the resistor R1, as shown in Fig. 12.11. Note the ammeter\nreading, I1.\nFigure 12.11Figure 12.11Figure 12.11Figure 12.11Figure 12.11\n/square6Similarly, measure the currents through R2 and R3. Let these be I2 and I3, respectively.\nWhat is the relationship between I, I1, I2 and I3?\nIt is observed that the total current I , is equal to the sum of the\nseparate currents through each branch of the combination.", + "Let these be I2 and I3, respectively.\nWhat is the relationship between I, I1, I2 and I3?\nIt is observed that the total current I , is equal to the sum of the\nseparate currents through each branch of the combination.\nI = I1 + I2 + I3(12.15)\nLet Rp be the equivalent resistance of the parallel combination of\nresistors. By applying Ohm\u2019s law to the parallel combination of resistors,\nwe have\nI = V/Rp(12.16)", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nI = I1 + I2 + I3(12.15)\nLet Rp be the equivalent resistance of the parallel combination of\nresistors. By applying Ohm\u2019s law to the parallel combination of resistors,\nwe have\nI = V/Rp(12.16)\nOn applying Ohm\u2019s law to each resistor , we have\nI1 = V /R1; I2 = V /R2; and I3 = V /R3(12.17)\nFrom Eqs. (12.15) to (12.17), we have\n V/Rp = V/R1 + V/R2 + V/R3\nor\n 1/Rp = 1/R1 + 1/R2 + 1/R3(12.18)\nThus, we may conclude that the reciprocal of the equivalent resistance\nof a group of resistances joined in parallel is equal to the sum of the\nreciprocals of the individual resistances.\nExample 12.8\nIn the circuit diagram given in Fig.", + "Example 12.8\nIn the circuit diagram given in Fig. 12.10, suppose the resistors R1,\nR2 and R3 have the values 5 \u2126, 10 \u2126, 30 \u2126, respectively, which have\nbeen connected to a battery of 12 V . Calculate (a) the curr ent thr ough\neach r esistor , (b) the total curr ent in the cir cuit, and (c) the total cir cuit\nresistance.\nSolution\nR1 = 5 \u2126, R2 = 10 \u2126, and R3 = 30 \u2126.\nPotential dif ference acr oss the battery, V = 12 V .", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\neach r esistor , (b) the total curr ent in the cir cuit, and (c) the total cir cuit\nresistance.\nSolution\nR1 = 5 \u2126, R2 = 10 \u2126, and R3 = 30 \u2126.\nPotential dif ference acr oss the battery, V = 12 V .\nThis is also the potential difference across each of the individual\nresistor; therefore, to calculate the current in the resistors, we use\nOhm\u2019s law.\nThe current I1, through R1 = V/ R1\nI1=12 V/5 \u2126\n = 2.4 A.", + "The current I1, through R1 = V/ R1\nI1=12 V/5 \u2126\n = 2.4 A.\n2020-21\nElectricity 215The current I2, through R2=V/ R2\nI2=12 V/10 \u2126=1.2 A.\nThe current I3, through R3=V/R3\nI3=12 V/30 \u2126=0.4 A.\nThe total current in the circuit,\nI= I1 + I2 + I3\n=(2.4 + 1.2 + 0.4) A\n=4 A\nThe total resistance Rp, is given by [Eq. (12.18)]\n1 1 1 1 1\n5 10 30 3pR= + + =\nThus, Rp = 3 \u2126.\nExample 12.9\nIf in Fig.", + "(12.18)]\n1 1 1 1 1\n5 10 30 3pR= + + =\nThus, Rp = 3 \u2126.\nExample 12.9\nIf in Fig. 12.12, R1 = 10 \u2126, R2 = 40 \u2126, R3 = 30 \u2126, R4 = 20 \u2126, R5 = 60 \u2126,\nand a 12 V battery is connected to the arrangement. Calculate\n(a) the total resistance in the circuit, and (b) the total current flowing\nin the circuit.\nSolution", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nand a 12 V battery is connected to the arrangement. Calculate\n(a) the total resistance in the circuit, and (b) the total current flowing\nin the circuit.\nSolution\nSuppose we replace the parallel resistors R1 and R2 by an\nequivalent resistor of resistance, R \u2032. Similarly we replace\nthe parallel resistors R3, R4 and R5 by an equivalent single\nresistor of resistance R\u2033. Then using Eq. (12.18), we have\n1/ R\u2032 = 1/10 + 1/40 =5/40; that is R\u2032 = 8 \u2126 .", + "Then using Eq. (12.18), we have\n1/ R\u2032 = 1/10 + 1/40 =5/40; that is R\u2032 = 8 \u2126 .\nSimilarly, 1/ R\u2033=1/30 + 1/20 + 1/60 = 6/60;\nthat is, R\u2033 = 10 \u2126.\nThus, the total resistance, R = R\u2032 + R\u2033 = 18 \u2126.\nTo calculate the current, we use Ohm\u2019s law, and get\n I = V/R = 12 V/18 \u2126 = 0.67 A.\nWe have seen that in a series cir cuit the curr ent is constant thr oughout\nthe electric circuit. Thus it is obviously impracticable to connect an electric\nbulb and an electric heater in series, because they need currents of widely\ndifferent values to operate properly (see Example 12.3). Another major", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nbulb and an electric heater in series, because they need currents of widely\ndifferent values to operate properly (see Example 12.3). Another major\ndisadvantage of a series circuit is that when one component fails the circuit is\nbroken and none of the components works. If you have used \u2018fairy lights\u2019 to\ndecorate buildings on festivals, on marriage celebrations etc., you might have\nseen the electrician spending lot of time in trouble-locating and replacing the\n\u2018dead\u2019 bulb \u2013 each has to be tested to find which has fused or gone. On the\nother hand, a parallel circuit divides the current through the electrical gadgets.\nThe total resistance in a parallel circuit is decreased as per Eq. (12.18). This\nis helpful particularly when each gadget has different resistance and requires\ndifferent current to operate properly.Figure 12.12Figure 12.12Figure 12.12Figure 12.12Figure 12.12\nAn electric circuit showing\nthe combination of series\nand parallel resistors\n2020-21", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\ndifferent current to operate properly.Figure 12.12Figure 12.12Figure 12.12Figure 12.12Figure 12.12\nAn electric circuit showing\nthe combination of series\nand parallel resistors\n2020-21\nScience 21612.7 HEA 12.7 HEA12.7 HEA 12.7 HEA12.7 HEA TING EFFECT OF ELECTRIC CURRENTTING EFFECT OF ELECTRIC CURRENTTING EFFECT OF ELECTRIC CURRENTTING EFFECT OF ELECTRIC CURRENTTING EFFECT OF ELECTRIC CURRENT\nWe know that a battery or a cell is a sour ce of electrical ener gy. The\nchemical reaction within the cell generates the potential difference between\nits two terminals that sets the electrons in motion to flow the current\nthrough a r esistor or a system of r esistors connected to the battery. W e\nhave also seen, in Section 12.2, that to maintain the current, the source\nhas to keep expending its energy. Where does this energy go?", + "W e\nhave also seen, in Section 12.2, that to maintain the current, the source\nhas to keep expending its energy. Where does this energy go? A part of\nthe source energy in maintaining the current may be consumed into\nuseful work (like in rotating the blades of an electric fan). Rest of the", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nthe source energy in maintaining the current may be consumed into\nuseful work (like in rotating the blades of an electric fan). Rest of the\nsource energy may be expended in heat to raise the temperature of\ngadget. W e often observe this in our everyday life. For example, an electric\nfan becomes warm if used continuously for longer time etc. On the other\nhand, if the electric circuit is purely resistive, that is, a configuration of\nresistors only connected to a battery; the source energy continually gets\ndissipated entirely in the form of heat. This is known as the heating\neffect of electric current. This effect is utilised in devices such as electric\nheater , electric ir on etc.\nConsider a current I flowing through a resistor of resistance R. Let\nthe potential difference across it be V (Fig. 12.13). Let t be the time during\nwhich a charge Q flows across. The work done in moving the charge \nQ\nthrough a potential difference V is VQ. Therefore, the source must supply", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nwhich a charge Q flows across. The work done in moving the charge \nQ\nthrough a potential difference V is VQ. Therefore, the source must supply\nenergy equal to VQ in time t. Hence the power input to the circuit by the\nsource is\nP V VI =Q\nt= (12.19)\nOr the energy supplied to the circuit by the source in time t is P \u00d7 t,\nthat is, VIt. What happens to this energy expended by the source? This\nenergy gets dissipated in the resistor as heat. Thus for a steady\ncurrent I, the amount of heat H\n produced in time t is\nH = VIt (12.20)QUESTIONS\n?1. Judge the equivalent resistance when the following are connected in\nparallel \u2013 (a) 1 \u2126 and 106 \u2126, (b) 1 \u2126 and 103 \u2126, and 106 \u2126.\n2.", + "Judge the equivalent resistance when the following are connected in\nparallel \u2013 (a) 1 \u2126 and 106 \u2126, (b) 1 \u2126 and 103 \u2126, and 106 \u2126.\n2. An electric lamp of 100 \u2126, a toaster of resistance 50 \u2126, and a water\nfilter of resistance 500 \u2126 are connected in parallel to a 220 V source.\nWhat is the resistance of an electric iron connected to the same source\nthat takes as much current as all three appliances, and what is the\ncurrent through it?", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nWhat is the resistance of an electric iron connected to the same source\nthat takes as much current as all three appliances, and what is the\ncurrent through it?\n3. What are the advantages of connecting electrical devices in parallel\nwith the battery instead of connecting them in series?\n4. How can three resistors of resistances 2 \u2126, 3 \u2126, and 6 \u2126 be connected\nto give a total resistance of (a) 4 \u2126, (b) 1 \u2126?\n5. What is (a) the highest, (b) the lowest total resistance that can be secured\nby combinations of four coils of resistance 4 \u2126, 8 \u2126, 12 \u2126, 24 \u2126?\n2020-21\nElectricity 217Applying Ohm\u2019s law [Eq. (12.5)], we get\nH = I2 Rt (12.21)\nThis is known as Joule\u2019s law of heating.", + "2020-21\nElectricity 217Applying Ohm\u2019s law [Eq. (12.5)], we get\nH = I2 Rt (12.21)\nThis is known as Joule\u2019s law of heating. The\nlaw implies that heat produced in a resistor is\n(i) directly proportional to the square of current\nfor a given resistance, (ii) directly proportional to\nresistance for a given current, and (iii) directly\nproportional to the time for which the current flows\nthrough the r esistor . In practical situations, when", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nresistance for a given current, and (iii) directly\nproportional to the time for which the current flows\nthrough the r esistor . In practical situations, when\nan electric appliance is connected to a known\nvoltage source, Eq. (12.21) is used after\ncalculating the current through it, using the\nrelation I = V/R .\nExample 12.10\nAn electric iron consumes energy at a rate of 840 W when heating\nis at the maximum rate and 360 W when the heating is at the\nminimum. The voltage is 220 V . What ar e the curr ent and the\nresistance in each case?\nSolution\nFrom Eq. (12.19), we know that the power input is\nP = V I\nThus the current I = P/V\n(a)When heating is at the maximum rate,\nI = 840 W/220 V = 3.82 A;\nand the resistance of the electric iron is\nR = V/I = 220 V/3.82 A = 57.60 \u2126.", + "(b)When heating is at the minimum rate,\nI = 360 W/220 V = 1.64 A;\nand the resistance of the electric iron is\nR = V/I = 220 V/1.64 A = 134.15 \u2126.\nExample 12.11\n100 J of heat is produced each second in a 4 \u2126 resistance. Find the", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nI = 360 W/220 V = 1.64 A;\nand the resistance of the electric iron is\nR = V/I = 220 V/1.64 A = 134.15 \u2126.\nExample 12.11\n100 J of heat is produced each second in a 4 \u2126 resistance. Find the\npotential dif ference acr oss the r esistor .\nSolution\nH = 100 J, R = 4 \u2126, t = 1 s, V = ?\nFrom Eq. (12.21) we have the current through the resistor as\nI = \u221a(H/Rt )\n= \u221a[100 J/(4 \u2126 \u00d7 1 s)]\n=5 A\nThus the potential dif fer\nence acr oss the r esistor , V [from Eq.", + "(12.21) we have the current through the resistor as\nI = \u221a(H/Rt )\n= \u221a[100 J/(4 \u2126 \u00d7 1 s)]\n=5 A\nThus the potential dif fer\nence acr oss the r esistor , V [from Eq. (12.5)] is\nV=IR\n=5 A \u00d7 4 \u2126\n=20 V.Figure 12.13Figure 12.13Figure 12.13Figure 12.13Figure 12.13\nA steady current in a purely resistive electric circuit\n2020-21\nScience 21812.7.1 Practical Applications of Heating Effect of\nElectric Current\nThe generation of heat in a conductor is an inevitable consequence of\nelectric current. In many cases, it is undesirable as it converts useful electrical\nenergy into heat. In electric circuits, the unavoidable heating can increase\nthe temperatur e of the components and alter their pr operties. However ,", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nenergy into heat. In electric circuits, the unavoidable heating can increase\nthe temperatur e of the components and alter their pr operties. However ,\nheating effect of electric current has many useful applications. The electric\nlaundry ir on, electric toaster , electric oven, electric kettle and electric heater\nare some of the familiar devices based on Joule\u2019s heating.\nThe electric heating is also used to produce light, as in an electric\nbulb. Here, the filament must retain as much of the heat generated as is\npossible, so that it gets very hot and emits light. It must not melt at such\nhigh temperature. A strong metal with high melting point such as\ntungsten (melting point 3380\u00b0C) is used for making bulb filaments. The\nfilament should be ther mally isolated as much as possible, using\ninsulating support, etc. The bulbs are usually filled with chemically\ninactive nitrogen and argon gases to prolong the life of filament. Most of", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\ninsulating support, etc. The bulbs are usually filled with chemically\ninactive nitrogen and argon gases to prolong the life of filament. Most of\nthe power consumed by the filament appears as heat, but a small part\nof it is in the form of light radiated.\nAnother common application of Joule\u2019s heating is the fuse used in\nelectric circuits. It protects circuits and appliances by stopping the flow\nof any unduly high electric current. The fuse is placed in series with\nthe device. It consists of a piece of wire made of a metal or an alloy of\nappropriate melting point, for example aluminium, copper , iron, lead\netc. If a current larger than the specified value flows through the circuit,\nthe temperature of the fuse wire increases. This melts the fuse wire and\nbreaks the circuit. The fuse wire is usually encased in a cartridge of\nporcelain or similar material with metal ends. The fuses used for domestic\npurposes are rated as 1 A, 2 A, 3 A, 5 A, 10 A, etc. For an electric iron", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nporcelain or similar material with metal ends. The fuses used for domestic\npurposes are rated as 1 A, 2 A, 3 A, 5 A, 10 A, etc. For an electric iron\nwhich consumes 1 kW electric power when operated at 220 V, a current\nof (1000/220) A, that is, 4.54 A will flow in the circuit. In this case, a 5 A\nfuse must be used.QUESTIONS\n?1. Why does the cord of an electric heater not glow while the heating element\ndoes?\n2. Compute the heat generated while transferring 96000 coulomb of charge\nin one hour thr ough a potential dif ference of 50 V .\n3. An electric iron of resistance 20 \u2126 takes a current of 5 A. Calculate the\nheat developed in 30 s.", + "2. Compute the heat generated while transferring 96000 coulomb of charge\nin one hour thr ough a potential dif ference of 50 V .\n3. An electric iron of resistance 20 \u2126 takes a current of 5 A. Calculate the\nheat developed in 30 s.\n2020-21\nElectricity 21912.8 ELECTRIC POWER12.8 ELECTRIC POWER12.8 ELECTRIC POWER12.8 ELECTRIC POWER12.8 ELECTRIC POWER\nYou have studied in your earlier Class that the rate of doing work is\npower . This is also the rate of consumption of ener gy.\nEquation (12.21) gives the rate at which electric energy is dissipated", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nYou have studied in your earlier Class that the rate of doing work is\npower . This is also the rate of consumption of ener gy.\nEquation (12.21) gives the rate at which electric energy is dissipated\nor consumed in an electric cir cuit. This is also ter med as electric power .\nThe power P is given by\nP = VI\nOr P = I2R = V2/R (12.22)\nThe SI unit of electric power is watt (W). It is the power consumed by\na device that carries 1 A of current when operated at a potential difference\nof 1 V . Thus,\n 1 W = 1 volt \u00d7 1 ampere = 1 V A (12.23)\nThe unit \u2018watt\u2019 is very small. Therefore, in actual practice we use a\nmuch larger unit called \u2018kilowatt\u2019. It is equal to 1000 watts. Since electrical\nenergy is the product of power and time, the unit of electric energy is,\ntherefore, watt hour (W h).", + "Therefore, in actual practice we use a\nmuch larger unit called \u2018kilowatt\u2019. It is equal to 1000 watts. Since electrical\nenergy is the product of power and time, the unit of electric energy is,\ntherefore, watt hour (W h). One watt hour is the energy consumed when\n1 watt of power is used for 1 hour . The commer cial unit of electric ener gy\nis kilowatt hour (kW h), commonly known as \u2018unit\u2019.", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\ntherefore, watt hour (W h). One watt hour is the energy consumed when\n1 watt of power is used for 1 hour . The commer cial unit of electric ener gy\nis kilowatt hour (kW h), commonly known as \u2018unit\u2019.\n1 kW h =1000 watt \u00d7 3600 second\n=3.6 \u00d7 106 watt second\n=3.6 \u00d7 106 joule (J)More to Know!Many people think that electrons are consumed in an electric circuit. This is wrong!\nWe pay the electricity board or electric company to provide ener gy to move electr ons\nthrough the electric gadgets like electric bulb, fan and engines. W e pay for the ener gy\nthat we use.\nExample 12.12\nAn electric bulb is connected to a 220 V generator . The curr ent is\n0.50 A. What is the power of the bulb?\nSolution\nP=VI\n=220 V \u00d7 0.50 A\n=110 J/s\n=110 W .\nExample 12.13\nAn electric refrigerator rated 400 W operates 8 hour/day.", + "The curr ent is\n0.50 A. What is the power of the bulb?\nSolution\nP=VI\n=220 V \u00d7 0.50 A\n=110 J/s\n=110 W .\nExample 12.13\nAn electric refrigerator rated 400 W operates 8 hour/day. What is\nthe cost of the energy to operate it for 30 days at Rs 3.00 per kW h?\n2020-21\nScience 220QUESTIONS\n?1. What determines the rate at which energy is delivered by a current?", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nthe cost of the energy to operate it for 30 days at Rs 3.00 per kW h?\n2020-21\nScience 220QUESTIONS\n?1. What determines the rate at which energy is delivered by a current?\n2. An electric motor takes 5 A from a 220 V line. Determine the power of\nthe motor and the energy consumed in 2 h.\nWhat y ou hav e learnt\n/square6A stream of electrons moving through a conductor constitutes an electric current.\nConventionally, the direction of current is taken opposite to the direction of flow of\nelectrons.\n/square6The SI unit of electric current is ampere.\n/square6To set the electrons in motion in an electric circuit, we use a cell or a battery. A cell\ngenerates a potential difference across its terminals. It is measured in volts (V).\n/square6Resistance is a pr operty that r esists the flow of electr ons in a conductor . It contr ols\nthe magnitude of the current. The SI unit of resistance is ohm (\u2126 ).\n/square6Ohm\u2019s law: The potential difference across the ends of a resistor is directly", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nthe magnitude of the current. The SI unit of resistance is ohm (\u2126 ).\n/square6Ohm\u2019s law: The potential difference across the ends of a resistor is directly\nproportional to the current through it, provided its temperature remains the same.\n/square6The resistance of a conductor depends directly on its length, inversely on its area of\ncross-section, and also on the material of the conductor .\n/square6The equivalent resistance of several resistors in series is equal to the sum of\ntheir individual resistances.\n/square6A set of resistors connected in parallel has an equivalent resistance Rp given by\n1 2 31 1 1 1...\npR R R R= + + +\n/square6The electrical energy dissipated in a resistor is given by\nW = V \u00d7 I \u00d7 t\n/square6The unit of power is watt (W). One watt of power is consumed when 1 A of current\nflows at a potential dif ference of 1 V .\n/square6The commercial unit of electrical energy is kilowatt hour (kWh).", + "One watt of power is consumed when 1 A of current\nflows at a potential dif ference of 1 V .\n/square6The commercial unit of electrical energy is kilowatt hour (kWh).\n1 kW h = 3,600,000 J = 3.6 \u00d7 106 J.Solution", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nflows at a potential dif ference of 1 V .\n/square6The commercial unit of electrical energy is kilowatt hour (kWh).\n1 kW h = 3,600,000 J = 3.6 \u00d7 106 J.Solution\nThe total energy consumed by the refrigerator in 30 days would be\n400 W \u00d7 8.0 hour/day \u00d7\n 30 days = 96000 W h\n= 96 kW h\nThus the cost of energy to operate the refrigerator for 30 days is\n96 kW h \u00d7 Rs 3.00 per kW h = Rs 288.00\n2020-21\nElectricity 221EXERCISES\n1.A piece of wire of resistance R is cut into five equal parts. These parts are then\nconnected in parallel. If the equivalent resistance of this combination is R\u2032, then the\nratio R/R \u2032 is \u2013\n(a)1/25 (b) 1/5 (c) 5 (d) 25\n2.Which of the following terms does not represent electrical power in a circuit?", + "If the equivalent resistance of this combination is R\u2032, then the\nratio R/R \u2032 is \u2013\n(a)1/25 (b) 1/5 (c) 5 (d) 25\n2.Which of the following terms does not represent electrical power in a circuit?\n(a)I2R (b) IR2(c) VI (d) V2/R\n3.An electric bulb is rated 220 V and 100 W . When it is operated on 110 V , the\npower consumed will be \u2013\n(a)100 W (b) 75 W (c) 50 W (d) 25 W\n4.Two conducting wires of the same material and of equal lengths and equal diameters", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\npower consumed will be \u2013\n(a)100 W (b) 75 W (c) 50 W (d) 25 W\n4.Two conducting wires of the same material and of equal lengths and equal diameters\nare first connected in series and then parallel in a circuit across the same potential\ndifference. The ratio of heat produced in series and parallel combinations would be \u2013\n(a)1:2 (b) 2:1 (c) 1:4 (d) 4:1\n5.How is a voltmeter connected in the circuit to measure the potential difference between\ntwo points?\n6.A copper wire has diameter 0.5 mm and resistivity of 1.6 \u00d7 10\u20138 \u2126 m. What will be\nthe length of this wire to make its resistance 10 \u2126? How much does the resistance\nchange if the diameter is doubled?", + "6.A copper wire has diameter 0.5 mm and resistivity of 1.6 \u00d7 10\u20138 \u2126 m. What will be\nthe length of this wire to make its resistance 10 \u2126? How much does the resistance\nchange if the diameter is doubled?\n7.The values of current I flowing in a given resistor for the corresponding values of\npotential difference V across the resistor are given below \u2013\nI (amperes) 0.5 1.0 2.0 3.0 4.0\nV (volts) 1.6 3.4 6.7 10.2 13.2\nPlot a graph between V and I and calculate the resistance of that resistor .", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nI (amperes) 0.5 1.0 2.0 3.0 4.0\nV (volts) 1.6 3.4 6.7 10.2 13.2\nPlot a graph between V and I and calculate the resistance of that resistor .\n8.When a 12 V battery is connected acr oss an unknown r esistor , there is a curr ent\nof 2.5 mA in the circuit. Find the value of the resistance of the r esistor .\n9.A battery of 9 V is connected in series with r esistors of 0.2 \u2126 , 0.3 \u2126 , 0.4 \u2126 , 0.5 \u2126\nand 12 \u2126, respectively. How much current would flow through the 12 \u2126 resistor?\n10.How many 176 \u2126 resistors (in parallel) are required to carry 5 A on a 220 V line?", + "How much current would flow through the 12 \u2126 resistor?\n10.How many 176 \u2126 resistors (in parallel) are required to carry 5 A on a 220 V line?\n11.Show how you would connect three resistors, each of resistance 6 \u2126 , so that the\ncombination has a r esistance of (i) 9 \u2126 , (ii) 4 \u2126 .\n12.Several electric bulbs designed to be used on a 220 V electric supply line, are\nrated 10 W . How many lamps can be connected in parallel with each other acr oss\nthe two wires of 220 V line if the maximum allowable current is 5 A?", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nrated 10 W . How many lamps can be connected in parallel with each other acr oss\nthe two wires of 220 V line if the maximum allowable current is 5 A?\n13.A hot plate of an electric oven connected to a 220 V line has two resistance coils\nA and B, each of 24 \u2126 resistance, which may be used separately, in series, or in\nparallel. What are the currents in the three cases?\n14.Compare the power used in the 2 \u2126 resistor in each of the following circuits:\n(i) a 6 V battery in series with 1 \u2126 and 2 \u2126 resistors, and (ii) a 4 V battery in parallel\nwith 12 \u2126 and 2 \u2126 resistors.\n2020-21\nScience 22215.Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected\nin parallel to electric mains supply. What current is drawn from the line if the supply\nvoltage is 220 V?", + "2020-21\nScience 22215.Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected\nin parallel to electric mains supply. What current is drawn from the line if the supply\nvoltage is 220 V?\n16.Which uses mor e ener gy, a 250 W TV set in 1 hr , or a 1200 W toaster in 10 minutes?\n17.An electric heater of resistance 8 \u2126 draws 15 A from the service mains 2 hours.", + "CHAPTER 11-ELECTRCITY-TEXTBOOK.txt\nvoltage is 220 V?\n16.Which uses mor e ener gy, a 250 W TV set in 1 hr , or a 1200 W toaster in 10 minutes?\n17.An electric heater of resistance 8 \u2126 draws 15 A from the service mains 2 hours.\nCalculate the rate at which heat is developed in the heater .\n18.Explain the following.\n(a) Why is the tungsten used almost exclusively for filament of electric lamps?\n(b) Why are the conductors of electric heating devices, such as bread-toasters\nand electric irons, made of an alloy rather than a pure metal?\n(c) Why is the series arrangement not used for domestic circuits?\n(d) How does the resistance of a wire vary with its area of cross-section?\n(e) Why are copper and aluminium wires usually employed for electricity\ntransmission?\n2020-21", + "CHAPTER 11-ELECTRCITY-lesson_plan_2.txt\nCLASS - X \nCHAPTER 12: ELECTRICITY \nLESSON PLAN - 2 \n \nTOPICS FACTORS ON WHICH RESISTANCE OF A CONDUCTOR DEPENDS \nCOMBINATION OF RESISTORS \n \nDURATION \n \n 4 PERIODS (Approx. 40 minutes each) \nPeriod 1: Electrical resistance \n Period 2: Rheostat \nPeriod 3: Resistances in Series Combination \nPeriod 4: Resistances in Parallel Combination \n \nPRE-REQUISITE \nFOR THE COURSE This topic requires knowledge of : \n\u2022 Some materials are good conductors while some are poor conductors of \nelectric current. \nLEARNING \nOBJECTIVES Students will be able to: \n\u2022 Understand how resistance affects the flow of current through the conductor \n\u2022 List the factors on which resistance of a conductor depends \n\u2022 Identify the combination of resistors as series or parallel \n\u2022 Solve numerical problems based on combination of resistors \nTEACHING \nMETHODOLOGY \nRecapitulation: 5 min \nQuick revision of the previous topic learnt would be taken up.", + "CHAPTER 11-ELECTRCITY-lesson_plan_2.txt\nTransaction of the Lesson Plan : Activity , Think Pair Share \n \nThe facilitator begins the session with an activity 12.2 mentioned in NCERT. The \nstudents observe the current in the Ammeter when (i) Nichrome wire, (ii) 10 W bulb \nand (iii) Torch bulb, is connected in a circuit. \n \nThe facilitator explains the difference o bserved in the reading of ammeter in the three \ncases and hence explain the cause of resistance. \n \nThink Pair Share: Discussion of topic through Collaborative Learning: 15 min (Critical \nThinking, Collaboration) \nUsing the knowledge of cause of resistance, t he facilitator asks students to predict \nhow would the resistance of a wire (made of one type of material) change: \n\u2022 If it is folded such that its length becomes half ? \n\u2022 If wire of same length but different area of section is used? \n\u2022 If a different material wire of same length and same area of cross -section is \nused?", + "CHAPTER 11-ELECTRCITY-lesson_plan_2.txt\nAfter the discussion, t he facilitator then explains why the resistance of the wire in the \ncircuit: (i) is directly proportional to its length , (ii) is inversely proportional to its area \nof cross -section and (iii) depends on the nature of its material ? \n \nDiscussion of topic and Guided Practice 20 min Inquiry based learning (Critical \nThinking and Problem Solving) Using the table 12.2 of NCERT - Electrical resistivity of some substances at 200C, the \nstudents are asked to find out the following: \n\u2022 Which material is the best conductor of electricity? \n\u2022 Arrange the resistivity of conductors, insulators and alloys in increasing order. \n\u2022 Which materials are use d in electrical heating devices and why?", + "CHAPTER 11-ELECTRCITY-lesson_plan_2.txt\nTo study the system (or combination) of resistors, the facilitators ask students to \npredict and observe what happens if i n a simple electrical circuit consisting of a bulb \nB1 connected to a switch and a battery , \ni. bulb B2 is connected in series to B1? \nii. bulb B2 is connected in parallel to B1? \nWhich arrangement is used in domestic circuits \u2013 series or parallel? Why? \n \nGuided practice followed by Independent Practice - NCERT questions to be discussed \nin the classroom. \n \nValue Points learned : - \n\u2022 Resistance of a conductor depends on its length, area of cross -section and its \nmaterial. \n\u2022 Resistivity (or conductivity) of a material is its characteristic property. \n\u2022 In a series combination of resistors, same current flows through all the \nresistors. \n\u2022 In parallel combination of resistors, potential difference across each resistor is \nsame ,", + "CHAPTER 11-ELECTRCITY-lesson_plan_2.txt\nClosu re: \n5 min A short oral test would be taken to check proper assimilation of the topic \ndiscussed. \nLearning \nOutcomes: Students would be able to: \n\u2022 Identify the materials as good conductors or insulators on the basis of their \nresistivity. \n\u2022 To explain the cause of resistance and the factors on which it depends. \n\u2022 Analyse that current remains same in series combination of resistors and \npotential difference remains same in parallel combination of resistors. \n\u2022 Solve numerical problems based on series and parallel combination of \nresistor s.", + "CHAPTER 11-ELECTRCITY-lesson_plan_2.txt\nSELF -STUDY, \nHOME WORK, \nASSIGNMENTS Students would do the given questions in their H.W. notebooks: \n\u2022 Will current flow more easily through a thick wire or a thin wire of the same \nmaterial, when connected to the same source and why? \n\u2022 Why are coils of electric toasters and electric irons made of an alloy rather \nthan a pure metal? \n\u2022 Why is tungsten used exclusively for filament of electric lamps? \n\u2022 Why is the series arrangement not used for domestic circuits?", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nCBSE Class 10 Science Notes Chapter 12 Electricity\nElectricity: Electric current, electric circuit, voltage or electric potential, resistance and (Ohm\u2019s law).\nElectric Current: The flow of electric charge is known as Electric Current, Electric current is carried by moving electrons through a conductor.\nBy convention, electric current flows in the opposite direction to the movement of electrons.\nElectric Circuit: Electric circuit is a continuous and closed path of electric current.\nExpression of Electric Current: Electric current is denoted by the letter \u2018I\u2019. Electric current is expressed by the rate of flow of electric charges. Rate of flow means, the amount of charge flowing through a particular area in unit time.\n\nIf a net electric charge (Q) flows through a cross-section of a conductor in time t, then,\n\nWhere I is electric current, Q is a net charge and t is a time in second.\nS.I. Unit of Electric Charge and Current: S.I. unit of electric charge is coulomb (C).", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nWhere I is electric current, Q is a net charge and t is a time in second.\nS.I. Unit of Electric Charge and Current: S.I. unit of electric charge is coulomb (C).\nOne coulomb is nearly equal to 6 \u00d7 1018 electrons. S.I. unit of electric current is ampere (A). Ampere is the flow of electric charge through a surface at the rate of one coulomb per second. This means, if 1 coulomb of electric charge flows through a cross section for 1 second, it would be equal to 1 ampere.\nTherefore, 1 A = 1 C/1 s\nSmall Quantity of Electric Current: Small quantity of electric current is expressed in milliampere and microampere. Milliampere is written as mA and microampere as pA.\n1 mA (milliampere) = 10-3 A\n1 pA (microampere) = 10-6 A\nAmmeter: An apparatus to measure electric current in a circuit.,\nCharge: Like mass, the charge is the fundamental property of matter.", + "1 mA (milliampere) = 10-3 A\n1 pA (microampere) = 10-6 A\nAmmeter: An apparatus to measure electric current in a circuit.,\nCharge: Like mass, the charge is the fundamental property of matter. There are two types of charge\n(i) Positive charge.\n(ii) Negative charge.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nCharge: Like mass, the charge is the fundamental property of matter. There are two types of charge\n(i) Positive charge.\n(ii) Negative charge.\nPositive and Negative Charge: The charge acquired by a glass rod when rubbed with silk is called a positive charge and the charge acquired by an ebonite rod when rubbed with wool is called negative charge.\nProperties of Electric Charge:\n(i) Unlike charges attract each other and like charges repel each other.\n(ii) The.force between two charges varies directly as the product of two charges and inversely as the square of the distance (r) between both charges (q1 and q2).\n\nS.I. unit of charge is coulomb (C).\n1 coulomb = 1 ampere \u00d7 1 second.\n1C = 1A \u00d7 1s\nThus, the quantity of charge which flows through a circuit when one ampere of current flows through it in one second is known as a 1-coulomb charge.\nElectric Potential and Potential Difference\nElectric Potential: The amount of electric potential energy at a point is called electric potential.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nElectric Potential and Potential Difference\nElectric Potential: The amount of electric potential energy at a point is called electric potential.\nPotential Difference: The difference in the amount of electric potential energy between two points in an electric circuit is called electric potential difference.\nElectric potential difference is known as voltage, which is equal to the amount of work done to move the unit charge between two points against static electric field.\n\nTherefore, Voltage = \\frac { Workdone }{ Charge }\n\nVoltage or electric potential difference is denoted by V\u2019. Therefore, V = \\frac { W }{ Q }\nWhere, W = Work done and Q = Charge\nS.I. Unit of Electric Potential Difference (Voltage)\nS.I. unit of electric potential difference is volt and denoted by \u2018V\u2019 This is named in honour of Italian Physicist Alessandro Volta.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nS.I. Unit of Electric Potential Difference (Voltage)\nS.I. unit of electric potential difference is volt and denoted by \u2018V\u2019 This is named in honour of Italian Physicist Alessandro Volta.\nSince joule is the unit of work and Coulomb is the unit of charge, 1 volt of electric potential difference is equal to the 1 joule of work to be done to move a charge of 1 coulomb from one point to another in an electric circuit. Therefore\n1V = 1Joule/1Coulomb = 1J/1C\n1V = 1JC-1\nVoltmeter: An apparatus to measure the potential difference or electric potential difference between two points in an electric circuit.\nGalvanometer: It is a device to detect current in an electric circuit.\nOhm\u2019s Law: Ohm\u2019s Law states that the potential difference between two points is directly proportional to the electric current, at a constant temperature.\nThis means potential difference V varies as electric current.\nV \u221d I\nV = RI\nI = \\frac { V }{ R }\nR = \\frac { V }{ I }", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nThis means potential difference V varies as electric current.\nV \u221d I\nV = RI\nI = \\frac { V }{ R }\nR = \\frac { V }{ I }\nWhere, R is constant for the given conductor at a given temperature and is called resistance.\nResistance: Resistance is the property of conductor which resists the flow of electric current through it.\nS.I. unit of resistance is ohm. Ohm is denoted by Greek letter \u2018Q\u2019\n1 Ohm: 1 ohm (Q) of resistance (R) is equal to the flow 1A of current through a conductor between two points having a potential difference equal to 1V.\nThis means; 1\u2126 = \\frac { 1V }{ 1A }\nFrom the expression of Ohm\u2019s Law, it is obvious that electric current through a resistor is inversely proportional to resistance. This means electric current will decrease with an increase in resistance and vice versa. The graph of V (potential difference) versus I (electric current) is always a straight line.\n\nGraph of Potential Difference (V) Vs Electric Current (I)\nVoltage, i.e.", + "This means electric current will decrease with an increase in resistance and vice versa. The graph of V (potential difference) versus I (electric current) is always a straight line.\n\nGraph of Potential Difference (V) Vs Electric Current (I)\nVoltage, i.e. Potential diffrence (V) = ?", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nGraph of Potential Difference (V) Vs Electric Current (I)\nVoltage, i.e. Potential diffrence (V) = ?\nWe know, from Ohm\u2019s Law that,\nR = \\frac { V }{ I }\n15 \u2126 = \\frac { V }{ 15A }\nV = 225V\nResistance: Resistance is a property of conductor due to which it resists the flow of electric current through it. A component that is used to resist the flow of electric current in a circuit is called a resistor.\nIn practical application, resistors are used to increase or decrease the electric current.\nVariable Resistance: The component of an electric circuit which is used to regulate the current, without changing the voltage from the source, is called variable resistance.\nRheostat: This is a device which is used in a circuit to provide variable resistance.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nRheostat: This is a device which is used in a circuit to provide variable resistance.\nCause of Resistance in a Conductor: Flow of electrons in a conductor is electric current. The positive particles of conductor create hindrance to flow of electrons, because of attraction between them, this hindrance is the cause of resistance in the flow of electricity.\nFactors on Which Resistance of a Conductor Depends: Resistance in a conductor depends on nature, length and area of cross section of the conductor.\n(i) Nature of Material: Some materials create least hindrance and hence, are called good conductors. Silver is the best conductor of electricity. While some other materials create more hindrance in the flow of electric current, i.e. flow of electrons through them. Such materials are called bad conductors. Bad conductor are also known as insulators. Hard plastic is the one of the best insulators of electricity.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\n(ii) Length of Conductor: Resistance (R) is directly proportional to the length of the conductor. This means, resistance increases with increase in length of the conductor. This is the cause that long electric wires create more resistance to the electric current. Thus, Resistance (R) \u221d length of conductor (l)\nor, R \u221d l \u2026(i)\n(iii) Area of Cross Section: Resistance R is inversely proportional to the area of cross section (A) of the conductor. This means R will decrease with an increase in the area of conductor and vice versa. More area of conductor facilitates the flow of electric current through more area and thus, decreases the resistance. This is the cause that thick copper wire creates less resistance to the electric current.\nThus, resistance (R) \u221d 1/Area of cross section of conductor (A)\nor, R \u221d \\frac { l }{ A } \u2026.(ii)\nFrom equations (i) and (ii)\nR \u221d \\frac { l }{ A }\nR = \u03c1 \\frac { l }{ A }", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nThus, resistance (R) \u221d 1/Area of cross section of conductor (A)\nor, R \u221d \\frac { l }{ A } \u2026.(ii)\nFrom equations (i) and (ii)\nR \u221d \\frac { l }{ A }\nR = \u03c1 \\frac { l }{ A }\nWhere, \u03c1 (rho) is the proportionality constant. It is called the electrical resistivity of the material of conductor.\nFrom equation (iii) RA = \u03c1l \u21d2 \u03c1 = \\frac { RA }{ l } ..(iv)\nThe S.I. of Resistivity: Since, the S.I. unit of R is Q, S.I. unit of area is m2 and S.I. unit of length is m. Hence, unit of resistivity (\u03c1) = \\frac { \\Omega \\times { { m }^{ 2 } } }{ m } = \u2126m\nThus, S.I. unit of resistivity (\u03c1) is \u2126m.", + "unit of length is m. Hence, unit of resistivity (\u03c1) = \\frac { \\Omega \\times { { m }^{ 2 } } }{ m } = \u2126m\nThus, S.I. unit of resistivity (\u03c1) is \u2126m.\nResistivity: It is defined as the resistance offered by a cube of a material of side 1m when current flows perpendicular to its opposite faces. It\u2019s S.I. unit is ohm-meter (\u2126m).\nResistivity, \u03c1 = \\frac { RA }{ l }\nResistivity is also known as specific resistance.\nResistivity depends on the nature of the material of the conductor.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nResistivity, \u03c1 = \\frac { RA }{ l }\nResistivity is also known as specific resistance.\nResistivity depends on the nature of the material of the conductor.\nMaterials having a resistivity in the range of 10-8 \u2126m to 10-6 \u2126m are considered as very good conductors. Silver has resistivity equal to 1.60 \u00d7 10-8 \u2126m and copper has resistivity equal to 1.62 \u00d7 10-8 \u2126m.\nRubber and glass are very good insulators. They have a resistivity in the order of 10-12 \u2126m to 10-8 \u2126m.\nThe resistivity of materials varies with temperature.\nCombination of resistors (Series and Parallel combination), the heating effect of electric current and electric power.\nCombination of Resistors\n(i) Series combination\n(ii) Parallel combination.\n1. Resistors in Series: When resistors are joined from end to end, it is called in series. In this case, the total resistance of the system is equal to the sum of the resistance of all the resistors in the system.", + "1. Resistors in Series: When resistors are joined from end to end, it is called in series. In this case, the total resistance of the system is equal to the sum of the resistance of all the resistors in the system.\n\nLet, three resistors R1, R2, and R3 get connected in series.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nLet, three resistors R1, R2, and R3 get connected in series.\nPotential difference across A and B = V\nPotential difference across R1, R2 and R3 = V1, V2 and V3\nCurrent flowing through the combination = I\nWe, know that\nV= V1 + V2 + V3 \u2026. (i)\nAccording to Ohm\u2019s Law :\nV1 = IR1, V2 = IR2 and V3 = IR3 \u2026.. (ii)\nLet, total resistance = Rs\nThen, V = IRs \u2026(iii)\nFrom equations (i) and (ii) and (iii)\nIRs = IR1 + IR2 + IR3\nRs = R1 + R2 + R3\nWhen the resistors are connected in series, the current flowing through each resistor is the same and is equal to the total current.\n2. Resistors in Parallel: When resistors are joined in parallel, the reciprocal of the total resistance of the system is equal to the sum of reciprocal of the resistance of resistors.\n\nLet three resistors R1, R2 and R3 connected in parallel.", + "2. Resistors in Parallel: When resistors are joined in parallel, the reciprocal of the total resistance of the system is equal to the sum of reciprocal of the resistance of resistors.\n\nLet three resistors R1, R2 and R3 connected in parallel.\nPotential difference across point A and B = V\nTotal current flowing between point A and B = I\nCurrents flowing through resistors R1, R2 and R3 = I1, I2 and I3 respectively.\nWe, know that,", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nPotential difference across point A and B = V\nTotal current flowing between point A and B = I\nCurrents flowing through resistors R1, R2 and R3 = I1, I2 and I3 respectively.\nWe, know that,\nI = I1 + I2 + I3 \u2026\u2026.(i)\nSince, the potential difference across R1, R2, and R3 is the same = V\nAccording to Ohm\u2019s Law,\n\nIn parallel combination, the potential difference across each resistor is the same and is equal to the total potential difference.\nThe total current through the circuit can be calculated by adding the electric current through individual resistors.\nItotal = 6A + 48A + 30A + 12A + 24A = 120A", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nThe total current through the circuit can be calculated by adding the electric current through individual resistors.\nItotal = 6A + 48A + 30A + 12A + 24A = 120A\nHeating Effect of Electric Current: When electric current is supplied to a purely resistive conductor, the energy of electric current is dissipated entirely in the form of heat and as a result, resistor gets heated. The heating of resistor because of dissipation of electrical energy is commonly known as Heating Effect of Electric Current. Some examples are as follows : When electric energy is supplied to an electric bulb, the filament gets heated because of which, it gives light. The heating of electric bulb happens because of heating effect of electric current.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nCause of Heating Effect of Electric Current: Electric current generates heat to overcome the resistance offered by the conductor through which it passes. Higher the resistance, the electric current will generate higher amount of heat. Thus, generation of heat by electric current while passing through a conductor is an inevitable consequence. This heating effect is used in many appliances, such as electric iron, electric heater, electric geyser, etc.\nJoule\u2019s Law Of Heating: Let, an electric current, I is flowing through a resistor having resistance = R.\nThe potential difference through the resistor is = V.\nThe charge, Q flows through the circuit for the time, t\nThus, work done in moving of charge (Q) of potential difference (V),\nW = V \u00d7 Q\nSince this charge, Q flows through the circuit for time t\nTherefore, power input (P) to the circuit can be given by the following equation :\nP = \\frac { W }{ T }\nP = V \u00d7 \\frac { Q }{ t } \u2026..(i)\nWe know, electric current, I = \\frac { Q }{ t }", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nTherefore, power input (P) to the circuit can be given by the following equation :\nP = \\frac { W }{ T }\nP = V \u00d7 \\frac { Q }{ t } \u2026.(i)\nWe know, electric current, I = \\frac { Q }{ t }\nSubstituting \\frac { Q }{ t } = I in equation (i), we get,\nP = VI \u2026(ii)\ni.e. P = VI\nSince, the electric energy is supplied for time ?, thus, after multiplying both sides of equation (ii) by time t, we get,\nP \u00d7 t = VI \u00d7 t = VIt \u2026\u2026(iii)\ni.e. P = VIt\nThus, for steady current I, the heat produced (H) in time t is equal to VIt\nH = VIt i.e. H = VIt\nWe know, according to Ohm\u2019s Law,\nV = IR\nBy substituting this value of V in equation (iii), we get,", + "for steady current I, the heat produced (H) in time t is equal to VIt\nH = VIt i.e. H = VIt\nWe know, according to Ohm\u2019s Law,\nV = IR\nBy substituting this value of V in equation (iii), we get,\nH = IR \u00d7 It\nH = I2Rt \u2026\u2026(iv)\nThe expression (iv) is known as Joule\u2019s Law of Heating, which states that heat produced in a resistor is directly proportional to the square of current given to the resistor, directly proportional to the resistance for a given current and directly proportional to the time for which the current is flowing through the resistor.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nElectric Bulb: In an electric bulb, the filament of bulb gives light because of the heating effect of electricity. The filament of bulb is generally, made of tungsten metal, having melting point equal to 3380\u00b0C.\nElectric Iron: The element of electric iron is made of alloys having high melting poir^ Electric heater and geyser work on the same mechanism.\nElectric Fuse: Electric fuse is used to protect the electric appliances from high voltage if any. Electric fuse is made of metal or alloy of metals, such as aluminum, copper, iron, lead, etc. In the case of flow of higher voltage than specified, fuse wire melts and protect the electric appliances.\nFuse of 1A, 2A, 3A, 5A, 10A, etc., used for domestic purpose.", + "In the case of flow of higher voltage than specified, fuse wire melts and protect the electric appliances.\nFuse of 1A, 2A, 3A, 5A, 10A, etc., used for domestic purpose.\nSuppose, if an electric heater consumes 1000W at 220 V.\nThen electric current in circuit\nI = \\frac { P }{ V }\nI = \\frac { 1000W }{ 220V } = 4.5 A\nThus, in this case of 5A should be used to protect the electric heater in the flow of higher voltage.\nElectric Power", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nI = \\frac { P }{ V }\nI = \\frac { 1000W }{ 220V } = 4.5 A\nThus, in this case of 5A should be used to protect the electric heater in the flow of higher voltage.\nElectric Power\nS.I. unit of electric power is watt (W).\n1W = 1 volt \u00d7 1 ampere = 1V \u00d7 1A\nI kilowatt or 1kW = 1000 W\nConsumption of electricity (electric energy) is generally measured in kilowatt.\nUnit of electric energy is kilowatt-hour (kWh).\n1 kWh = 1000 watt \u00d7 1 hour = 1 unit = 1000 W \u00d7 3600 s\n1 kWh = 3.6 x 106 watt second = 3.6 \u00d7 106 J\nConductor: The material which can allow the flow of electrons through itself is called the conductor. It has a large number of free electrons. It offers low opposition in the flow of current.\nInsulator: The material which does not allow the flow of electrons through itself is called insulator.", + "It has a large number of free electrons. It offers low opposition in the flow of current.\nInsulator: The material which does not allow the flow of electrons through itself is called insulator. It has less or no free electrons. It offers high opposition in the flow of current.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nInsulator: The material which does not allow the flow of electrons through itself is called insulator. It has less or no free electrons. It offers high opposition in the flow of current.\nElectric Current: The amount of flow charge through any cross-sectional area of a conductor in unity time is called Electric Current.\nIt is represented by \u2018I\u2019\nI = \\frac { Q }{ T }\nUnit of Electric Current: It is CS-1 (coulomb per second) or Ampere (A). Electric Current is a scalar quantity. It is measured by an ammeter.\nDirection: The direction of conventional current (or practical current) is opposite to the flow of electrons.\nElectric potential: Electric Potential at any point in the electric field is defined as the amount of work done to bring the unit positive charge from infinity (from outside the electric field) to that point.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nV =\\frac { W }{ Q }, S.I. unit of Electric Potential is JC-1 or volt (V). It is a scalar quantity. The +ve charge flows from higher to lower potential. The -ve charge flows from lower to a higher potential. The difference of electric potential between any two points in the electric field is called Electric Potential difference. It is known as a voltage which is equal to the work done per unit charge between two points against the static electric field.\nVAB = VA \u2013 VB = \\frac { { W }_{ AB } }{ Q }\nElectric Potential difference is measured by a voltmeter.\nOhm\u2019s Law: According to this law \u201cUnder the constant physical condition the potential difference across the conductor is directly proportional to the current flowing through the conductor.\u201d\nV \u221d I\nV = IR \u2026[Where R is proportionality constant called resistance of conductor]\n\u21d2 I = \\frac { V }{ R }\nR depends upon nature, geometry and physical condition of the conductor.", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nV \u221d I\nV = IR \u2026[Where R is proportionality constant called resistance of conductor]\n\u21d2 I = \\frac { V }{ R }\nR depends upon nature, geometry and physical condition of the conductor.\nThe heat generated by electric current: The potential difference between two points in an electrical field is equal to the work done in moving a unit charge from one point to another.\nThen, work is done, W = VQ and Q = I \u00d7 t\nW = V \u00d7 I \u00d7 t\nFrom Ohm\u2019s Law, we know that\nV = IR\nW = IR \u00d7 I \u00d7 t = I2.Rt\nSince heat produced by the electric current is equal to work done, W\nH = W\n\u21d2 H (heat) = I2Rt Joule.\nResistance: Ratio of the applied voltage to the current flowing in the conductor is called resistance of the conductor.\n\u21d2 R = \\frac { V }{ I }\nS.I. Unit of resistance is VA-1 or ohm (\u2126).\nResistance is the opposition offered by the conductor in the flow of current.", + "\u21d2 R = \\frac { V }{ I }\nS.I. Unit of resistance is VA-1 or ohm (\u2126).\nResistance is the opposition offered by the conductor in the flow of current.\nPractically it is\nR \u221d L (L is the length of a conductor)\nR \u221d 1/A (A is the area of a conductor)\nSo, R \u221d L/A", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nResistance is the opposition offered by the conductor in the flow of current.\nPractically it is\nR \u221d L (L is the length of a conductor)\nR \u221d 1/A (A is the area of a conductor)\nSo, R \u221d L/A\nR = \u03c1L/A \u2026[Where p is proportionality constant called specific resistance of conductor\nIt only depend upon nature (material) and temperature of conductor.\nSpecific resistance or Resistivity = \u03c1 = RA /L\nIt\u2019s S.I. Unit is Qm\nCombination of resistance:\nIn this combination the current across every component is same but potential across every component is different.\nIf resistance R1, R2 and R3 are connected in series with a battery of Potential V, then equivalence resistance of the combination\nR = R1 + R2 + R3\nThe parallel combination of resistance:\nIn this combination the current across every component is different. But potential across every component is the same.\nIf resistance R1, R2 and R3 are connected in parallel with a battery of Potential V, then equivalence resistance of combination", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nIf resistance R1, R2 and R3 are connected in parallel with a battery of Potential V, then equivalence resistance of combination\n\\frac { 1 }{ R } =\\frac { 1 }{ { R }_{ 1 } } +\\frac { 1 }{ { R }_{ 2 } } +\\frac { 1 }{ { R }_{ 3 } }\nElectric Energy is amount of work done to maintain the continuous flow of electric current in the circuit.\nIts S.I. unit is joule (J).\nElectric power (P): The electric work done per unit time is called electric power.\nElectric Power = \\frac { Electric\\quad work\\quad done }{ Time\\quad taken }\nor P = \\frac { W }{ t }\nElectric power is also defined as the electric energy consumed per unit time.\nP = \\frac { E }{ t }\nS.I. unit of electric power is Watt. When one joule of energy is used for one second, electric power is equal to one watt.", + "P = \\frac { E }{ t }\nS.I. unit of electric power is Watt. When one joule of energy is used for one second, electric power is equal to one watt.\nDerivation of formula for electric power:\nWe know that electric work done, W = V \u00d7 I \u00d7 t or P = \\frac { VIt }{ t }\nP = VI\nElectric power in watts = Volts \u00d7 ampere\nAlso V = IR \u2026[According to Ohm\u2019s Law]\nSo P = IR \u00d7 I\nP = I2R", + "CHAPTER 11-ELECTRCITY-NOTES.txt\nWe know that electric work done, W = V \u00d7 I \u00d7 t or P = \\frac { VIt }{ t }\nP = VI\nElectric power in watts = Volts \u00d7 ampere\nAlso V = IR \u2026[According to Ohm\u2019s Law]\nSo P = IR \u00d7 I\nP = I2R\nWe know that I = \\frac { V }{ R }\nP = (\\frac { V }{ R })2 \u00d7 R = \\frac { { V }^{ 2 } }{ R } Watt\nThe maximum value of electric current that can pass through an electric appliance without damaging electric appliance is called current rating of electric appliance.", + "CHAPTER 11-ELECTRCITY-ppt.txt\nELECTRICITY\n, Chapter 12What is a Charge?\n\u2022It is the fundamental property of matter \nbecause of which it experiences force.\n\u2022Charges are of two type \u2013positive charge \nand negative charge.\n\u2022Since charge of an electron (e = 1.6x10-19C) \nis very verysmall .\n\u2022Therefore a more practical unit of charge is \nCoulomb (C) and one coulomb is equal to \ncombined charge of 6.25x1018electrons .\nWhat is the basic difference between \nconductors and insulators?\nThink before moving to the next slideGeneral response of Students\n\u2022Conductors are substances which allow current to pass through \nthem. For example Iron, Silver, Copper, Aluminum etc.\n\u2022Insulators are substances which do not allow current to pass \nthrough them. For example Plastic, Wood, Glass, Rubber etc.\n\u2022These responses are OK upto 8thStandard. \n\u2022Since students have learnt atomic model or structure of an \natom in 9thstandard, this knowledge must be included to give \nthem better understanding of concept of conductors and", + "CHAPTER 11-ELECTRCITY-ppt.txt\n\u2022Since students have learnt atomic model or structure of an \natom in 9thstandard, this knowledge must be included to give \nthem better understanding of concept of conductors and \ninsulators.Concept of free electron\n\u2022Matter is made up of atoms and each atom has a \npositively charged nucleus at the center with \nnegatively charged electron revolve around it.\n\u2022Negatively charge electron is bound to positively \ncharge nucleus because of its charge.\n\u2022If we provide energy to electron, it will not \nremain bound to the nucleus and just come out of \nan atom.\n\u2022This type of electron is known as free electron \nbecause it is not bound to any particular nucleus \nand therefore free to move within the material.\nAdvance definition of Conductors and \ninsulators\n\u2022For some materials, energy required to free an electron is less \nand it can be acquired from the surrounding at room \ntemperature.\n\u2022These free electrons contribute to the electric current. Other", + "CHAPTER 11-ELECTRCITY-ppt.txt\n\u2022For some materials, energy required to free an electron is less \nand it can be acquired from the surrounding at room \ntemperature.\n\u2022These free electrons contribute to the electric current. Other \nelectrons would remain bound to the nucleus and do not \ncontribute to electric current.\n\u2022The materials which can produce free electron at room \ntemperature are known as conductors (because of less energy \nrequired to free an electron)\n\u2022The materials which can not produce free electron at room \ntemperature are known as insulators (because of large amount \nof energy required to free an electron) \nElectric Current\n\u2022Flow of free electrons through a conductor in a given direction is \nknown as electric current .\n\u2022SI unit of current is Ampere (A)\n\u2022Electric current is measured as I = Q/t , where Q is charge in \nCoulombs (C) and t is time in seconds (s )\nAn activity for understanding flow of charge \nIn order understand the flow of charge through a conductor, let us consider a", + "CHAPTER 11-ELECTRCITY-ppt.txt\nCoulombs (C) and t is time in seconds (s )\nAn activity for understanding flow of charge \nIn order understand the flow of charge through a conductor, let us consider a \nhollow pipe containing 7 balls. If we try to push one more ball into the pipe \nfrom one end, a ball would come out from the other end of the pipe. It means \nthat at any given time, there cannot be more than 7 balls in the pipe because \nthere is no space for an extra ball .\nMechanism of flow of charge\nGoing by the same logic that we have understood from the activity, a \nconductor may have billions or trillions of free electrons, but there is no \nspace for an additional electron. It means that when we push an additional \nelectron into the conductor from one end, an electron would come out from \nthe other end. This happens instantly, it means that there is no time gap \nbetween pushing an electron from one end and jumping out of electron from \nthe other end, either the length of the conductor is I meter or 1 kilometer.", + "CHAPTER 11-ELECTRCITY-ppt.txt\nbetween pushing an electron from one end and jumping out of electron from \nthe other end, either the length of the conductor is I meter or 1 kilometer. \nIs there any charge on conductor, when \ncurrent is passing through it?\nThere is a general misconception amongst the students, that during \nflow of current through the conductor, additional charge enters the \nconductor and because of this it will no longer remain neutral. As we \nhave understand from the activity, there is no space for an additional \nelectron in the conductor. Therefore the number of electrons in the \nconductor remains constant even during the passage of current \nthrough it. So there is no charge on the conductor, ever though the \ncurrent is passing through it.\n\u2022In order to measure to the flow of electrons through a conductor, let us consider we are \npushing 10 electrons into the conductor in one second. 10 electrons would come out from the", + "CHAPTER 11-ELECTRCITY-ppt.txt\n\u2022In order to measure to the flow of electrons through a conductor, let us consider we are \npushing 10 electrons into the conductor in one second. 10 electrons would come out from the \nother side of the conductor during the same period. So the flow of electron becomes 10 \nelectrons per second. If we push 1 million electrons into the conductor in one second, the \nflow of current becomes 1 million electrons per second.\n\u2022Going by the same logic, When one coulomb of charge (One Coulomb = 6.25x1018electrons ) \npasses through a conductor in one second, the flow of charge is said to be one ampereWhat is one Ampere?\nWhat is an ammeter?\n\u2022It is a device used to measure the \ncurrent in the circuit.\n\u2022It is always connected in series with the \nresistor.\n\u2022It has very low internal resistance.\n\u2022Positive terminal of Ammeter is \nconnected to the positive side of the \ncircuit and negative to the negative \nside.\nWhat is the direction of conventional \ncurrent?\nSince the free electrons are not bound to any", + "CHAPTER 11-ELECTRCITY-ppt.txt\nconnected to the positive side of the \ncircuit and negative to the negative \nside.\nWhat is the direction of conventional \ncurrent?\nSince the free electrons are not bound to any \nparticular nucleus, they can move freely \nwithin the conductor. But by convention, the \nflow of positive charge is considered to be \nthe direction of current through the \nconductor . The direction of conventional \ncurrent is opposite to the flow of free \nelectrons through the conductor.\nIn order to pass the charge through a \nconductor, we have to push additional charge \nfrom one end and at the same time, withdraw \nsame amount of charge from the other end. \nThis can be achieved with the help of device \nknown as Battery or an Electric Cell .How to make the flow of charge happen \nthrough a conductor?\nWhat is a Battery?\n\u2022When two or more electric cells are \nconnected in series, the system thus obtained \nis known as Battery.\n\u2022Battery or electric cell has two terminals \u2013\npositive terminal and negative terminal.", + "CHAPTER 11-ELECTRCITY-ppt.txt\n\u2022When two or more electric cells are \nconnected in series, the system thus obtained \nis known as Battery.\n\u2022Battery or electric cell has two terminals \u2013\npositive terminal and negative terminal.\n\u2022Positive terminal is known as high potential \nbecause of its high potential energy, while \nnegative terminal is known as low potential \non account of its low energy.\nElectric Potential and Potential difference\n\u2022The ability of a battery\u2019s terminal to push (or \npull) a unit charge into the conductor is known \nas Electric potential .\n\u2022The difference of electric potential between the \npositive and negative terminals of a battery \ndefines battery\u2019s power to push charge into the \nconductor or more specifically it is known as \npotential difference .\nWhat is voltage?\n\u2022Potential difference is also known as Voltage .\n\u2022Potential difference or voltage between any two point is defined as \nthe work done in moving a unit charge from one point to another.", + "CHAPTER 11-ELECTRCITY-ppt.txt\n\u2022Potential difference is also known as Voltage .\n\u2022Potential difference or voltage between any two point is defined as \nthe work done in moving a unit charge from one point to another.\n\u2022It is given as V = W/Q , where W is work done in joules and Q is \ncharge in Coulombs.\n\u2022SI unit of voltage is Volts (V).\n\u2022Voltage between two points is said to be one volt, when one joule of \nwork is done in moving one coulomb of charge between these points.\nWhat is a Voltmeter?\n\u2022It is a device used to measure the voltage \nacross the resistance in the circuit.\n\u2022It is always connected in parallel to the \nresistance.\n\u2022It has very high internal resistance.\n\u2022Positive terminal of the Voltmeter is \nconnected to the positive side of the \ncircuit and negative to the negative side.\nNow putting everything together to \nunderstand how a bulb glows?\nThink before moving to next slideGlowing of a bulb\n\u2022When the circuit is turned on, the positive terminal( rich in", + "CHAPTER 11-ELECTRCITY-ppt.txt\nNow putting everything together to \nunderstand how a bulb glows?\nThink before moving to next slideGlowing of a bulb\n\u2022When the circuit is turned on, the positive terminal( rich in \npositive charge ) of the battery which is at high potential \npushes positive charge into the bulb through connecting wires.\n\u2022At the same time, negative terminal( deficiency of positive \ncharge ) of the battery which is at low potential pulls positive \ncharge from the bulb because of its polarity through \nconnecting wire.\n\u2022This act of pushing charge through the positive terminal and at \nthe same time, withdrawing the same amount of charge \nthrough negative terminal is known as completing the circuit . \n\u2022As the circuit becomes complete, the bulb starts glowing. The \nbulb will not glow, if the circuit is not complete.\n\u2022It means that we cannot make a bulb glow by connecting it \nonly to high potential or low potential terminals.\n\u2022We have to create a potential difference by connecting the", + "CHAPTER 11-ELECTRCITY-ppt.txt\n\u2022It means that we cannot make a bulb glow by connecting it \nonly to high potential or low potential terminals.\n\u2022We have to create a potential difference by connecting the \nconductor between high potential and low potential terminals \nof the battery, to pass the current through it.\nConnecting wireConnecting wireOhm\u2019s law\n\u2022We have to create a potential difference across a conductor by \nconnecting it to battery, if current is to flow through the conductor.\n\u2022The amount of current passing through the conductor depends upon \nthe potential difference. This is known as Ohm\u2019s law.\n\u2022It states that the current through a conductor or wire is directly \nproportional to potential difference across the conductor provided \nthe temperature remains constant. Mathematically V = IR , \nwhere V is potential difference, I current and R resistance of the \nconductor\nGraphical Representation of Ohm\u2019s law\nThe slope of the graph \nrepresents the resistance \nof the conductor\nVoltage (V)Current (I)", + "CHAPTER 11-ELECTRCITY-ppt.txt\nconductor\nGraphical Representation of Ohm\u2019s law\nThe slope of the graph \nrepresents the resistance \nof the conductor\nVoltage (V)Current (I)\nWhat is Resistance of a conductor?\n\u2022It\u2019s the fundamental property of substances \nbecause of which they obstruct the flow of \ncharge through them.\n\u2022SI unit of resistance is ohm ( \u03a9).\n\u2022Conductors have very low resistance while \ninsulators have infinite resistance.\n\u2022Materials with zero or negligible resistance are \nknown as Superconductors .\nFactors affecting the resistance of wire or \nconductor\n\u2022Resistance (R) is directly proportional to the length of the wire \n(l). It means that longer wire has more resistance as compared to \nshorter wire.\n\u2022Resistance (R) is inversely proportional to area of cross section of \nwire (A). It means that thin wires are more resistive than a thick \nwire.\n\u2022Mathematically R = \u03c1l/A, where \u03c1 is the resistivity of the wire.\nWhat is resistivity?\n\u2022It is the resistance offered by a unit \nlength and unit area of cross section of", + "CHAPTER 11-ELECTRCITY-ppt.txt\nwire.\n\u2022Mathematically R = \u03c1l/A, where \u03c1 is the resistivity of the wire.\nWhat is resistivity?\n\u2022It is the resistance offered by a unit \nlength and unit area of cross section of \nwire.\n\u2022Its SI unit is \u03a9m.\n\u2022Its value depends upon the nature of \nmaterial. It means that the resistivity \nof copper is different from resistivity \nof aluminum.\n\u2022Value of resistivity of few materials are \ngiven in the table.\nMaterial Resistivity ( \u03a9m)\nSilver 1.60x10-8\nCopper 1.62x10-8\nAluminium 2.63x10-8\nIron 10.0x10-8\nNichrome 100x10-6\nGlass 1010\u20131014\nRubber 1013\u20131016\nCombination of Resistors\nSeries and Parallel combinations.Series combination of resistors\n\u2022Current in series combination remains \nconstant through all the resistors. \n\u2022Voltage is divided amongst different \nresistors in series combination, V 1across R 1\nand V 2across R 2.", + "\u2022Voltage is divided amongst different \nresistors in series combination, V 1across R 1\nand V 2across R 2.\n\u2022Voltage across a resistor depends upon the \nvalue of its resistance .\n\u2022Equivalent resistance of two given resistors \nin series is calculated as\nR = R 1+ R2", + "CHAPTER 11-ELECTRCITY-ppt.txt\nand V 2across R 2.\n\u2022Voltage across a resistor depends upon the \nvalue of its resistance .\n\u2022Equivalent resistance of two given resistors \nin series is calculated as\nR = R 1+ R2\nParallel combination of resistors\n\u2022Voltage remains constant across all the \nresistors connected in parallel.\n\u2022Current is divided amongst different branches \nof resistors, I 1through R 1and I 2through R 2\n\u2022Equivalent resistance of two resistors \nconnected in parallel is given as\n1\nR=1\nR1+1\nR2\nWhat is heating effect of current?\n\u2022Heat produced by a conductor or resistance when current passes \nthrough it is known as Heating effect of current . Heat produced is \nmeasured in joules.\n\u2022Voltage across a resistor is some time called as voltage drop because \nthe resistor converts electrical energy into heat or some other form of \nenergy.\n\u2022Factors effecting the heat produced (H) is \n\uf071directly proportional to current through the conductor(I)\n\uf071directly proportional to resistance of the conductor(R)", + "CHAPTER 11-ELECTRCITY-ppt.txt\nenergy.\n\u2022Factors effecting the heat produced (H) is \n\uf071directly proportional to current through the conductor(I)\n\uf071directly proportional to resistance of the conductor(R)\n\uf071directly proportional to time for which current passes through the \nconductor (t)\n\u2022Mathematically H = I2Rt\nElectric Power\n\u2022Heat produced by a device or circuit over a \nperiod of time is known as Electric Power .\n\u2022SI unit of Power \u2013Watts\n\u2022Power of a device is said to be one watt if it \nconsumes or produces one joule of energy in \none second.\n\u2022Commercial unit of energy \u2013Unit or kWh.\n\u20221kWh or unit = 3.6x106joules\n\u2022Electric power is given as \nP= W/t, P = I2R, P = V2/R, P = VI\nThe End", + "CHAPTER 11-ELECTRCITY-NOTES OF LESSON.txt\nLESSON PLAN \nNam e of the teacher: RAGHAVAN K P \nDesignation: P G T Physics \nClass : X \nSchedule June \nLesson: Electricity \n \nMajor topics: electric charge, electric potential, electric current, Ohm\u2019s law, resistance, \nresistivity \n \nInstructional Objectives: \n Student comes to know the basic terminology in electricity \nStudent understand s the concept of electric potential and p.d. \nStudent verifies that condition for p.d is satisfied in all situations of current flow. \nStudent identifies electrical components in a circuit \nStudent understands Ohm\u2019 s law \nStudent applies Ohm\u2019s law to find resistance/ current \nStudent recognize s the general applicability of resistivity and specific nature of \nresistance \nStudent improves drawing skills. \nStudent develops interpretation sk ills from graphs . \n \nTeaching Aids: Tester , Ammeter, voltmeter, Rheostat, RB, multimeter etc", + "CHAPTER 11-ELECTRCITY-NOTES OF LESSON.txt\nTeaching Aids: Tester , Ammeter, voltmeter, Rheostat, RB, multimeter etc \n \n \n \nActivities: To identify electrical components \n To verify Ohm\u2019s law \n \n \n \nEvaluation : Identification of circuit symbols, identifying the potential di fference, problems \nfrom Ohm\u2019s law and resistivity. \n \n \n \n \n \n \n \n \nLESSON PLAN \nName of the teacher: RAGHAVAN K P \nDesignation: P G T Physics \nClass : X \nSchedule July 15 \nLesson: Electricity \n \nMajor topics: Combination of resistances -effective resistance , electrical networks , heating \neffect of electricity, Joule\u2019s law", + "CHAPTER 11-ELECTRCITY-NOTES OF LESSON.txt\nMajor topics: Combination of resistances -effective resistance , electrical networks , heating \neffect of electricity, Joule\u2019s law \n \nInstructional Objectives: \nStudent understand s laws of combination of resistances \nStudent interpret networks as series / parallel circuits \nStudent simplifies complex circuits . \nStudent identifies the energy conversion in various electrical devices . \nStudent applies Ohm\u2019s law to find resistance / current . \nStudent cites examples for heating effect of electric current \nStudent develops drawing skills \nPrevious Knowledge: \nElectrical devi ces, circuits -open and closed, symbols \n \nTeaching aids : resistances, , symbol chart, ammeter, voltmeter , heating coil \nActivities : \n \nVerification of laws of combination of resistances, combining resistances to get \ndesired resistance , observing heating ef fect of current", + "CHAPTER 11-ELECTRCITY-NOTES OF LESSON.txt\nVerification of laws of combination of resistances, combining resistances to get \ndesired resistance , observing heating ef fect of current \n \nEvaluation : \nTo cite examples for heating effect of electric current , problems on combinations, \nJoule\u2019s law", + "CHAPTER 11-ELECTRCITY-IMPORTAT POINTS.txt\nFOCAL POINTS \nFocus Spot: \nResistors in Series. \nWhy: \nStudents find it difficult to understand the significance behind the connections of \nresistors as they do not know the indications for different materials that comprises a \ncircuit. Students are unaware of the indication of circuits, especially complex circuits. \nImpact: \nStudents draw inaccurate circuit diagram as they are unable to visualise effective \nresistance and current flow among resistors that are meant to be in seri es connection. \nFocus Spot: \nResistors connected in parallel. \nWhy: \nIt is tough for students to understand the concept of resistors connected in parallel as \nthey do not get the connection of resistors in parallel and tend to visualise it same as that \nof resis tors in series and draw inaccurate circuit diagram. \nImpact: \nStudents will miscalculate the solutions of parallel resistance due to improper circuit \ndiagram leading to incorrect answers.", + "CHAPTER 11-ELECTRCITY-Competency based questions.txt\nCompetency based questions . \n1. On the basis of your understanding of the following paragraph and the related studies \nconcept. \n The Tehri dam is the highest dam in India and one of the highest in the World. The \nTehri dam withholds a reservoir of capacity 4.0 km2 and surface area 52 km2. It is used for \nirrigation , municipal water supply and the generation of 1000 MW of hydroelectricity. \nThe Tehridam has been the object of protests . Environment activist Shri. Sunder \nLalBahuguna led \u2018Anti Tehri dam movement\u2019 from 1980s to 2014. The protest was against \nthe displacement of town inhabitants and environmental consequences of the weak eco \nsystem. The relocation of more than 1,00,000 people from the area has led to protract ed legal \nbattles over resettlement rights and ultimately resulted in the delayed completion of the \nproject. \n(a) How is hydropower harnessed? \n(b) Define 1 MW \n( c ) Mention two disadvantages of constructing Tehri dam.", + "CHAPTER 11-ELECTRCITY-Competency based questions.txt\nproject. \n(a) How is hydropower harnessed? \n(b) Define 1 MW \n( c ) Mention two disadvantages of constructing Tehri dam. \n(d) What happens when water from great heights is made to fall on blades of turbine? \n2. For the given question two statements are given, one labelled as Assertion (A) and the \nother labelled as Reason ( R). Select the correct answer to these questions from the codes (a), \n(b), (c) and (d) as giv en below. \n(a) Both (A) and ( R ) are true and ( R ) is correct explanation of Assertion \n(b) Both (A) and ( R ) are true but ( R ) is not correct explanation of Assertion \n( c) ( A) is true but (R ) is false \n(d) (A) is false but (R ) is true \nASSERTION ( A ) :Alloys are commonly used in electrical heating devices like electric iron \nand heater.", + "REASON (R) : Resistivity of an alloy is generally higher than that of its constituent metals \nbut the alloys have low melting points than their constituent metals.", + "CHAPTER 11-ELECTRCITY-Competency based questions.txt\nand heater. \nREASON (R) : Resistivity of an alloy is generally higher than that of its constituent metals \nbut the alloys have low melting points than their constituent metals. \n3. An electric geyser is known to consume 2.2 \u2018units\u2019 of electrical energy per hour of its use. It is \ndesigned to wor k on the mains voltage of 220V. \na) What is the \u2018power -rating\u2019 of this device? \nb) What is the current flowing through this device when it is co nnected across the \u2018mains\u2019? \nc) What is t he \u2018resistance\u2019 of this device? d) Does the resistance of this device remain constan t during its operation/working? \ne) Which of the two - a 100W, 220V lamp, or a 10W, 220V night lamp \u2013 has a higher resistance?", + "CHAPTER 11-ELECTRCITY-Competency based questions.txt\n4) Read the given information carefully and an swer the questions that follow: \n\uf0b7 The resistivity of copper is less than that of aluminium which, in turn i s less than that of \nconstantan. \n\uf0b7 There are nine wires, labelled as A, B, C, D, E, F, G, H, I, that have been des igned as per the \ntabular details given below: \n \nAnswer the following questions: \na) Arrange the three values R1, R2, R3 in increasing order. \nb) Arrange the three values R7, R8, R9 in decreasing order. \nc) Which of the two (i) R4 and R6 (ii) R1 and R8 has a lower value? \nd) Will the graph between R1, R2, R3 (on the y -axis) and their corresponding length values (on \nthe x -axis) be a straight line? \ne) What is the likely value of the ratio R4/R6? \n5) Observe the given circuit diagram carefully and answer the questions that fol low:", + "CHAPTER 11-ELECTRCITY-Competency based questions.txt\na) What are the currents drawn by the least and the h ighest of the three resistors? \nb) What is the curr ent drawn by the 6 \u03a9 resistor? \nc) What would be the reading of the ammeter? \nd) How much current does the least of the three resistors , draw from the given bat tery? \ne) Is the equivalent resistance of the parallel combination sho wn, more or less than the least (= \n4 \u2013 2) \u03a9, of the individual resistances? \n \n6) Observe the given circuit carefully and answer the questions that follow : \n \na) What is the total equival ent resist ance of the circuit? \nb) What would be the reading of the ammeter? \nc) What would be the reading of a voltmeter connected between the points Q and R? \nd) What would be the reading of the ammeter if only the highest of the three resistors (= 5\u03a9) \nwere present alone in the circuit? \ne) What would be the reading of a voltmeter conne cted between the points P and S", + "CHAPTER 11-ELECTRCITY-Competency based questions.txt\n7) The figure below shows three cylindrical copper conductors along with their face areas \nlengths. Compare the resistance and the resistivity of the three conduct ors. Justify your \nanswer. \n L 3L L/3 \n A A/3 \n 3A \n \n8) The values of current(I) flowing through a given resistor( R), for the corresponding values of \npotential difference (V) across the resistor are as given below: \nV (volts) 0.5 1.0 1.5 2.0 2.5 3.0 4.0 5.0 \nI (amperes 0.1 0.2 0.3 0.4 0.5 0.6 0.8 1.0 \nPlot a graph between current (I) and the potential difference (V) and determine the resistance (R) of \nthe resistor.", + "CHAPTER 11-ELECTRCITY-Competency based questions.txt\n9) a) Explain how does a cell maintain current in a circuit. \n b) In the circuit given below the resistance of the path xTy = 2 \u03a9 and that xZy = 6 \u03a9. \n i) Find the equivalent resistance between x and y. \n ii) Find the current in the main circ uit. \n iii) Calculate the current that flows through the path xTy and xZy. \n 15\u03a9 \n x \n Z T \n Y 6V \n Y \n10) Study the following circuit an d answer the following questions that follow : \n \n \n \n 10\u03a9 \n 15\u03a9 \na) State the type of combination of the two resisto rs in the circuit. \nb) What will be the potential difference across the above combination if applied potential is 6 V? \nc) What will be the ammeter reading?", + "CHAPTER 11-ELECTRCITY-Very short answer type questions.txt\nONE MARK TYPE QUSETIONS \n1. Write the S I unit of charge, \n2. Calculate the number of electron in 1 C of charge, \n3. Define Electric current. \n4. Define 1 ampere. \n5. Name the electrical device used for measuring current. \n6. How is an ammeter is included in a circuit? \n7. A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount \nof electric charge that flows through the circuit. \n8. Define an electric circuit. \n9. Define 1 V of potential difference. \n10. Calculate the work done to move a charge of 2 C between two points of potential difference \n6 V. \n11. Name the device used for measuring potential difference. How it is included in the circuit? \n12. How much energy is given to each coulomb of charge passing through a 6 V battery? \n13. Draw the circuit symbols of (i) closed plug key (ii) A wire joint. \n14.", + "How it is included in the circuit? \n12. How much energy is given to each coulomb of charge passing through a 6 V battery? \n13. Draw the circuit symbols of (i) closed plug key (ii) A wire joint. \n14. State Ohm\u2019s law. \n15. Draw the graph between V and I for a wire \n16. Write the mathematical form of Ohm\u2019s law.", + "CHAPTER 11-ELECTRCITY-Very short answer type questions.txt\n13. Draw the circuit symbols of (i) closed plug key (ii) A wire joint. \n14. State Ohm\u2019s law. \n15. Draw the graph between V and I for a wire \n16. Write the mathematical form of Ohm\u2019s law. \n17. Define resistance. Write its S I unit \n18. Define 1 ohm. \n19. Mention the factors on which resistance of a conductor depends on its shape. \n20. Define resistivity of a conductor. How the resistivity of a conductor vary with length and \ncross sectional area of the conductor. \n21. Write the S I unit of resistivity. \n22. How the resistivity of a conductor varies with temperature of the conductor? \n23. Which has higher resistivity: an alloy or a pure metal? \n24. Why alloys are used in heating electrical devices? \n25. What are the advantages of parallel connection of resistors than series connections? \n26.", + "23. Which has higher resistivity: an alloy or a pure metal? \n24. Why alloys are used in heating electrical devices? \n25. What are the advantages of parallel connection of resistors than series connections? \n26. How can three resistors of resistances 2 \u2126 , 3\u2126 and 6 \u2126 be connect ed to give a total \nresistance of (a) 4\u2126 and (b) 1 \u2126 \n27. State Joule\u2019s law of heating.", + "CHAPTER 11-ELECTRCITY-Very short answer type questions.txt\n26. How can three resistors of resistances 2 \u2126 , 3\u2126 and 6 \u2126 be connect ed to give a total \nresistance of (a) 4\u2126 and (b) 1 \u2126 \n27. State Joule\u2019s law of heating. \n28. Mention any two practical uses of heating effect of electric current. \n29. Write the properties of a fuse wire. \n30. What are the material generally used for making a fuse wire? \n31. Define electric power. Write its S I unit. \n32. Write the commercial unit of energy. \n33. Convert I kW h in to joules. \n34. What determines the rate at which energy is delivered by a current? \n35 ..A piece of wire of resistance R is cut in to five equal parts. These part s are then connected in \nparallel. If the equivalent resistance of this combination is R\u2019 , then the ratio R / R \u2018 is - \n36. Which of the following terms does not represent electrical power in a circuit?", + "These part s are then connected in \nparallel. If the equivalent resistance of this combination is R\u2019 , then the ratio R / R \u2018 is - \n36. Which of the following terms does not represent electrical power in a circuit? \n(a) I2R (b) I R2 (c) V I (d) V2 / R \n37. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V , the power", + "CHAPTER 11-ELECTRCITY-Very short answer type questions.txt\n(a) I2R (b) I R2 (c) V I (d) V2 / R \n37. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V , the power \nconsumed will be (a) 100 W (b) 75 W (c) 50 W (d) 25 W \n38. Two conducting wires of the same material and of equal lengths and equal diameters \nare first connected in se ries and then in parallel in a circuit across the same potential \ndifference. The ratio of heat produced in series and parallel combination would be\u2026 \n(a) 1 : 2 (b) 2 : 1 (c) 1: 4 (d) 4 : 1 \n39. Why does the cord of an electric heater not glow while heat ing element does? \n40. Why tungsten is used almost exclusively for filament of electric lamps? \n41. Why is the series arrangement is not used for domestic circuits? \n42.", + "Why does the cord of an electric heater not glow while heat ing element does? \n40. Why tungsten is used almost exclusively for filament of electric lamps? \n41. Why is the series arrangement is not used for domestic circuits? \n42. How does the resistance of a wire vary with its area of cross \u2013 section? \n43. Why are copp er and aluminium wires usually employed for electricity transmission?", + "CHAPTER 11-ELECTRCITY-Very short answer type questions.txt\n42. How does the resistance of a wire vary with its area of cross \u2013 section? \n43. Why are copp er and aluminium wires usually employed for electricity transmission? \n44. A cylindrical conductor of length \u2018l\u2019 and uniform cross section \u2018A\u2019 has resistance \u2018R\u2019 . The \narea of cross section of another conductor of same material and same resistance but of \nlength \u20182l\u2019 is \n( a) A/2 (b) 3 A / 2 (c) 2 A (d) 3 A \n45. The maximum resistance which can be made using four resistors each of resistance \u00bd \u2126 \nis \n( a) 2 \u2126 (b) 1 \u2126 (c) 2.5 \u2126 (d) 8 \u2126 \n46. Define 1 kW h", + "CHAPTER 11-ELECTRCITY-LESSON PLAN.txt\nCLASS - X \nCHAPTER 12: ELECTRICITY \nLESSON PLAN - 1 \n \nTOPIC S ELECTRIC CHARGE, ELECTRIC CURRENT AND ELECTRIC CIRCUIT \nELECTRIC POT ENTIAL, POTENTIAL DIFFERENCE \n OHM\u2019S LAW \n \nDURATION \n \n 4 PERIODS (Approx. 40 minutes each) \nPeriod 1: Electric current and circuit \nPeriod 2: Electric potential and Potential difference \nPeriod 3: Components used in circuit diagram and Ohm\u2019s Law \nPeriod 4: Questions based on the topics covered \n \nPRE-REQUISITE \nFOR THE COURSE This topic requires knowledge of : \n\u2022 Types of charges and interaction between them. \n\u2022 Knowledge about simple electric circuits and their components. \nLEARNING \nOBJECTIVES Students will be able to: \n\u2022 Understand that free electrons are the charge carriers in conductors \n\u2022 Define electric current \n\u2022 List the factors that control or regulate the flow of current in an electric \ncircuit. \n\u2022 Explain the function of components", + "CHAPTER 11-ELECTRCITY-LESSON PLAN.txt\nSKILLS and COMPETENCIES : \n1) Critical thinking and problem solving (working effectively with others), \n2) Character building (Taking tasks fearlessly and achieves goals.) \n3) Communication (evaluating the information critically and competently ). \n4) Collaboration (responding open mindedly to different ideas) \n5) Creativity (developing, implementing and communicating new ideas) \n \nTEACHING \nMETHODOLOGY Transaction of the Lesson Plan : Recall , question, connect and summarize \n \nThe facilitator helps the students in recalling the activity done by them in previous \nclasses of glowing a torch bulb on connecting it with a cell. \n \nBrainstorming Question s: - \nWhat is the source of current in a circuit? \nWhat makes the charges flow in a c ircuit? \nWhat type of charge carriers constitute current?", + "CHAPTER 11-ELECTRCITY-LESSON PLAN.txt\nBrainstorming Question s: - \nWhat is the source of current in a circuit? \nWhat makes the charges flow in a c ircuit? \nWhat type of charge carriers constitute current? \n \nConnect : \nActivity : The facilitator will divide the students into groups of 4 and the students \nperform in groups to make an electric circuit using potatoes (or lemon), connecting \nwires, LED bulb and a switch. \nThe students conclude that fruits and vegetables conduct electricity . \n \nExplain and Summarize : \nThe facilitator then explains and define the following terms : \nelectric current, unit of electric current, potential difference and unit of potential \ndifference. Chemical reaction in a cell leads to excess of negative (or positive) charges on one \nelectrode and deficiency of the same charge on the other electr ode. This results in \ndevelopment of electric potential difference between the two electrodes of a cell.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN.txt\nWhen a cell/battery is joined in a closed electric circuit, the potential difference \nbetween its terminals causes an electric current in the circuit (o r electric current starts \nflowing through it!). There are many types of electrical conductors such as metallic \nwires, ionic conductors, electrolytes etc. \n \nActivity to study the relation between potential difference through a conductor and \nthe current through it : Ohm\u2019s Law \nStudents are asked to set -up an electric circuit based on the given circuit diagram. This \ncircuit may comprise a cell (or a battery), a resistor, an ammeter, a voltmeter and a \nkey. \n \n \n\u27a2 Students should be made to realise that it is immaterial where in the circuit \nthe ammeter and key are connected as long as they are connected in series \nwith the resistor/cell. \n \n\u27a2 The facilitator may also clarify that the Ohm\u2019s Law is valid only for conductors \nprovided their temperature do not change.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN.txt\n\u27a2 The facilitator may also clarify that the Ohm\u2019s Law is valid only for conductors \nprovided their temperature do not change. \n \nGuided practice foll owed by Independent Practice - NCERT questions to be discussed \nin the classroom. \n \nValue Points learned: - \n\u2022 Flow of charges in a definite direction constitutes an electric current. \n\u2022 An electric current between two points is due to potential difference between \nthem. \n\u2022 An electric circuit provides a closed path for flow of charges, that is, the \nelectric current. \n\u2022 Electric potential difference between two points in a circuit is equivalent to the \nwork done in carrying a unit charge from one point to another. \n\u2022 Current through a metallic conductor is directly proportional to the potential \ndifference across its ends. \n \nClosure: \n5 min A short oral test would be taken to check proper assimilation of the topic \ndiscussed.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN.txt\nClosure: \n5 min A short oral test would be taken to check proper assimilation of the topic \ndiscussed. \n \nLearning \nOutcomes Students would be able to: \n\u2022 Explain the direction of flow of current \n\u2022 Define current and its SI unit \n\u2022 Define electric potential \n\u2022 State Ohm\u2019s law and apply it in circuits \nSELF -STUDY, \nHOME WORK, \nASSIGNMENTS Students would do the given questions in their H.W. notebooks : \n\u2022 Define electric current and its SI unit. \n\u2022 Calculate the number of electrons constituting one coulomb of charge. \n\u2022 State the function of Ammeter in a circuit. \n\u2022 How is an ammeter connected in a circuit? \n\u2022 How is a voltmeter connected in a circuit?", + "CHAPTER 11-ELECTRCITY-long answer type questions.txt\nChapter No.12 : ELECTRICITY \nLong answer Type questions \n1. (a) A current \u2018I\u2019 is flowing through a resistor \u2018R\u2019 having potential difference \u2018V\u2019 across its \nends and in time \u2018t\u2019 charge \u2018Q\u2019 flows. Derive an expression for heat produced. \n(b) What will happen to heat produced if the current passing through an electric heater has \nbeen halved? \n2. (a) Express electrical power in terms of current and potential difference. Define its S I Unit \n (b) The power rating of an appliance is 100W -220 V. What does it indicate?. If this bulb is \nconnected across 110 V, Calculate its power. \n3. Two similar resistances are connected in series and in parallel, in two different circuits one \nby one. The same current is passed through both the combinations. In which case more heat \nwill be produced? Give suitable reason for your answer with diagram. \n4.(a) State and explain Ohm\u2019s law. Express it mathematically and plot a graph between V and \nI for a conductor.", + "CHAPTER 11-ELECTRCITY-long answer type questions.txt\nwill be produced? Give suitable reason for your answer with diagram. \n4.(a) State and explain Ohm\u2019s law. Express it mathematically and plot a graph between V and \nI for a conductor. \n (b) Define resistance of a resistor. Also define 1 O hm. \n5. With a neat diagram explain how three resistors are connected in series across a battery. \nAlso derive an expression for the equivalent resistance of the combination. \n6.. With a neat diagram explain how three resistors are connected in parallel acros s a battery. \nAlso derive an expression for the equivalent resistance of the combination. \n7. (i) What are the factors on which resistance of a conductor depends? \n (ii) Define resistivity of a conductor. Write its S I unit. \n (iii) What happens to the resistance of a conductor when (a) its length is doubled (b) \nradius is halved? \n8. (a) State and explain Joule\u2019s law of heating. \n (b) Mention any two applications of Joule\u2019s law of heating", + "CHAPTER 11-ELECTRCITY-long answer type questions.txt\nradius is halved? \n8. (a) State and explain Joule\u2019s law of heating. \n (b) Mention any two applications of Joule\u2019s law of heating \n( c) Why alloys are used as the heating elements in electrical appliances?E \n9. (a) Establish the relationship between kWh and Joules. \n (b) A torch bulb is rated 6 V and 750 mA. Calculate the energy consumed by the bulb \nin 4 hours. \n 10. The resistivity values of some substances are given below. \nMaterials A B C C E \nResistivity \n ( \u2126 m) 1.6 x 10-8 6.4 x 10-8 10 x 10-8 96 x 10-8 100 x 10-6 \n Answer the following questions in relation to them giving justification for each \n(i) Which material is best for making connecting cords? \n(ii) Which material do you suggest to be u sed in heating devices? \n(iii) You have two wires of same length and same thickness.", + "(ii) Which material do you suggest to be u sed in heating devices? \n(iii) You have two wires of same length and same thickness. One is made of material A and \nanother of material D. If the resistance of wire made of A is 2 \u2126, what is the resistance of the \nother wire? \n11.", + "CHAPTER 11-ELECTRCITY-Lesson_plan_3.txt\nSALWAN PUBLIC SCHOOL AFTERNOON, \n RAJENDRA NAGAR \nCLASS \u2013 X SCIENCE \nLESSON PLAN \nCHAPTER 12 : ELECTRICITY \n \nTOPIC HEATING EFFECT OF CURRENT AND ITS APPLICATIONS \nELECTRIC POWER", + "CHAPTER 11-ELECTRCITY-Lesson_plan_3.txt\nDURATION \n 2 PERIODS (Approx. 40 minutes each) \nPeriod 1: Heating effect of current \nPeriod 2: Electric power \nPRE - REQUISITE \nFOR THE COURSE This topic requires knowledge of \n\uf0b7 Electric current , Electric potential , Resistance etc. \n\uf0b7 Series and Parallel combination of resistances \nLEARNING \nOBJECTIVE Students will be able to: \n\uf0b7 State and explain Heating effect of current. \n\uf0b7 List applications of the effect. \n\uf0b7 Identify and explain factors responsible for production of heat by a \nconductor. \n\uf0b7 Solve numerical based on heating effect of current. \n\uf0b7 State and explain electric power. \n\uf0b7 Explain unit of power and commercial unit of energy \n\uf0b7 Solve numerical related to electric power. \n \nSKILLS and COMPETENCIES : \n1) Critical thinking and problem solving (working effectively with others), \n2) Character building (Taking tasks fearlessly and achieves goals.) \n3) Communication (evaluating the information critically and competently ).", + "CHAPTER 11-ELECTRCITY-Lesson_plan_3.txt\n2) Character building (Taking tasks fearlessly and achieves goals.) \n3) Communication (evaluating the information critically and competently ). \n4) Collaboration (responding open mindedly to different ideas) \n5) Creativity (developing, implementing and communicating new ideas) \nTRANSACTION \nMETHODOLOGY \n Transaction of the Lesson Plan : Activity , Think Pair Share", + "CHAPTER 11-ELECTRCITY-Lesson_plan_3.txt\nThe facilitator begins the session with discuss on the question \nWhy do certain bodies become hot when we pass current through them, like \nelectric iron, Electric bulb etc. \nAfter the discuss ion, facilitator introduces the concept of heating effect of \ncurrent and explains it to the student. Facilitator will ask the students to \nidentify other applications of the effect.", + "CHAPTER 11-ELECTRCITY-Lesson_plan_3.txt\nThe facilitator discusses with the students the factors affecting the heating produce by a conductor. Heat produce is \n\uf0fc directly proportional to the square of current through the conductor. \n\uf0fc directly proportional to the resistance of the wire \n\uf0fc directly proportional to the time for which current passes through the \nconductor. \n \nThink Pair Share: Discussion of topic through Collaborative Learning: 15 min \n(Critical Thinking, Collaboration) \nUsing the knowledge of Heating effect by a conductor , the facilitator asks \nstudents to predict what would happen to the heat produce by a wire: \n\uf0b7 If it is folded such that its length becomes half ? \n\uf0b7 If current passing through the wire is doubled ? \n\uf0b7 If time for which the current flowing through the wire is reduce to 1/3 \nof its initial value?", + "CHAPTER 11-ELECTRCITY-Lesson_plan_3.txt\nExplain and summarize \nThe facilitator then explains and define the following terms : \nElectric power, SI unit of power, different relations to calculate power \ncommercial unit of energy and its relation in term of joules . \nDemonstrates to the students how to convert power of a device in watts to \ncommercial units. Discuss with the student ways to calculate the energy \nconsumed by house hold device like tubelights, bulbs, TV, AC etc. \n \nDiscussion of topic and Guided Practice 20 min Inquiry based learning \n(Critical Thinking and Problem Solving) \nAfter the discussion, t he facilitator demonstrates to the students how to \nsolve numerical problems related to heating effect of current and electric \npower \n \nGuided practice followed by Independent Practice - NCERT questions to be \ndiscussed in the classroom.", + "CHAPTER 11-ELECTRCITY-Lesson_plan_3.txt\nGuided practice followed by Independent Practice - NCERT questions to be \ndiscussed in the classroom. \n \nValue Points learned : - \n\uf0b7 Heat produced by a conductor depends upon the current passing \nthrough the conductor . \n\uf0b7 Heat produced is directly proportional to the resistance of a \nconductor . \n\uf0b7 Heat produced depends on time interval for which current passing \nthrough the co nductor . \n\uf0b7 Electric power and its SI unit \n\uf0b7 Commerci al unit of energy and its relationship with energy in joules. \n\uf0b7 To solve numerical using formula of electric power and heat produced \nby a conductor Closure: \n5 min A short oral test would be taken to check proper assimilation of the \ntopic discussed \n \nThe faci litator helps the students in recalling the factors related to heat \nproduced by a conductor, applications of the effect, commercial unit of \nenergy and different relation to to calculate the power of circuit or \nresistance .", + "CHAPTER 11-ELECTRCITY-Lesson_plan_3.txt\nRevision Question s: - \nHow does the heat produce by a conductor depend upon its resistance? \nHow much the heat produce by a conductor would be affect, if the current \npassing through the conductor is reduce to half of its initial value? \nDefine one watt. \nWhat is the commercial unit of ener gy? \nWhat are the different formulas for calculating power? \nLEARNING \nOUTCOMES Students would be able to: \n\uf0b7 Explain the heating effect of current \n\uf0b7 Explain different factors on which the heat produced by a conductor \ndepends . \n\uf0b7 Explain electric power. \n\uf0b7 Explain SI unit of power and commercial unit of energy. \n\uf0b7 Solve numerical base on heating effect of current and electric power. \nSELF STUDY, \nHOME WORK, \nASSIGNMENTS Students would do the given questions in their H.W. notebooks : \n\uf0b7 Define electric power and its SI unit \n\uf0b7 Explain commercial unit of energy and its relation in terms of joules. \n\uf0b7 Define heating effect of current.", + "CHAPTER 11-ELECTRCITY-Lesson_plan_3.txt\n\uf0b7 Define electric power and its SI unit \n\uf0b7 Explain commercial unit of energy and its relation in terms of joules. \n\uf0b7 Define heating effect of current. \n\uf0b7 List applications of heating effect of current. \n\uf0b7 Solve numerical problems related to heating effect of current and \nelectric power.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nCLASS \u2013 X \nCOMPETENCY BASED LESSON PLAN \nCHAPTER-ELECTRICITY \n \n Unit ELECTRICITY \nClass Transaction Total: 10 periods (approx. 40 min each) \nPre-requisite for \nthe course Students will be able to recall \n1. importance of electricity in day to day life \n2. materials which are good and bad conductor of elect ricity. \nAssessment of \nqualifying \nknowledge 1. Written test \n2. Lab activity \n3. Group Discussion \n4. HW notebook \n \nObjective To develop scientific temper and inquisit iveness in students. Also students \nwill be able to apply and relate learned concepts i n their daily lives. \nDOL D1, D2, D3 \nLearning \nOutcomes KNOWLEDGE- Students will be able to", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\n1. Define electric current, potential difference, resi stance, resistivity \nand power. \n2. Deduce Ohm's law and verify it experimentally. \n3. Understand the meaning of \u2018series and parallel comb ination\u2019 of \nresistors, learn the way of connecting a given numb er of resistors \nReasons for arrangement of devices, \u2018in parallel\u2019 w ith each other, in \ndomestic circuits. \n4. Describe Heating effect of electric current and der ive Joule's law of \nheating. \n5. Find an expression for electric power and derive co mmercial unit of \nelectrical energy is KWh \n \nSKILLS AND COMPETENCIES- Students would be able to", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\n1. Critically solve numerical based on current, potential differ ence, \nresistance, resistivity and power. \n2. Analyse change in resistance by the change in length and a rea of \nwire and series and parallel combination. \n3. Collaborate to understand the meaning of \u2018series and parallel \ncombination\u2019 of resistors, learn the way of connect ing a given \nnumber of resistors Reasons for arrangement of devi ces, \u2018in \nparallel\u2019 with each other, in domestic circuits \n4. Communicate about applications of Heating Effect of Electric \nCurrent in our daily life. \nTransaction \nMethodology \n(The teacher can Transaction would proceed in the following manner- \nSet induction and Brain Storming -The class would start with a discussion use the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) on what the students have already learnt in the pre vious classes and hence \nwhat is it that they would learn now. They would al so be told the", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nwherever \napplicable, and can \nuse any other too.) on what the students have already learnt in the pre vious classes and hence \nwhat is it that they would learn now. They would al so be told the \nsignificance of the topic that they would be studyi ng. \nIntroduction of the topic - ppt and digital content would be shared", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nGroup learning- The class will be divided in the gr oup of 5 students. \nCircle the Sage - The question will be asked by the teacher about t he \ndifferent physical quantities and components of ele ctric circuit and the \nstudents who knows the answer will stand at the var ious corners of the \nroom. (Communication, Collaboration) \n \nDifferentiation- The students will draw the circuit diagrams and solve \nnumericals. ( Critical thinking) \n \nQuestioning- Multiple level question - Teacher will prepare a list of \nquestion about the resistance in series and paralle l combination and will \nask these questions during discussions. Students wi ll answer ( Analytically) \n \nBrain Storming -The class would start with a discussion on what th e \nstudents have already learnt in the previous classe s and hence what is it that \nthey would learn now. They would also be told the s ignificance of the topic \nthat they would be studying.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nGuided practice followed by Independent Practice- NCERT, Exemplar, \nspiral questions to be discussed in the classroom. \n \nTechniques to be used: \nSet induction \nClass discussion' \nQuiz \nDaily Practice Problem \nMCQ \nPeer Assessment \nCase Studies \nLab Activities \nGraphic Organizers for different terms \nLesson progress using Smart Class \nThink Pair Share for solving numerical \nMind Mapping for lesson summarization \nMuddiest Point and One Minute Paper for Closure \nAny Other \n \nResources Text Book: \nNCERT text book for Science \nNCERT Exemplar Self Study, Home \nWork, \nAssignments Independent Practice : Students would do the questions in their H.W \nnotebooks as mentioned in the monthly planner by wh ich we can assess \ntheir critical and creative thinking", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nHW notebooks to be marked as per the given plan: \nAssessment Parameters : The total marks for the activity is 5 marks \nOn time submission\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nPresentation/ Neatness (creativity) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nContent (analytical and critical thinking) \u2026\u2026\u2026\u2026\u2026\u2026\u2026......3 marks \n \nIt is also advised that the students come to the cl ass with proper \nbackground knowledge of the topic under discussion. They can refer to the \nresources stated above. \n \nAssessment 3 UNIT TESTS (20, 30, 30)\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.80 Marks \nMID TERM EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u202680 Marks \nHALF YEARLY EXAMINATION \u2026\u2026\u2026\u2026\u2026\u202680 Marks \nPRE BOARD-I EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026.80 Marks \nPRE BOARD-II EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026.80 Marks \nANNUAL BOARD EXAMINATION \u2026\u2026\u2026\u2026\u2026.80 Marks \n \nPERIODIC TEST \nBy administering pen paper test we assess creativit y and critical \nthinking. \nAverage of the best two tests to be taken that will have a weightage of 10 \nmarks.", + "Average of the best two tests to be taken that will have a weightage of 10 \nmarks. \nBest 2 Tests out of: Units tests, Midterm examinat ion, Half Yearly \nexamination.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nINTERNAL ASSESSMENT \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u202620 Marks \nPeriodical Test 10 Marks \n \nBy administering pen paper test we assess creativit y and critical \nthinking. \n \nSubject Enrichment 05 M arks \nWith the help of subject enrichment activities we a ssess \ncommunication, collaboration, character building an d citizenship. The \nstudents in a group of two will be performing hands on practical of \nverification of Ohm\u2019s law and relate various terms used in daily life \nNote Book Submission 05 Ma rks \nBy writing the answers of different levels of question we can assess \ntheir knowledge, critical thinking and creativity. \n \n \n Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used for various groups: \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Initially simple questions to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes and formulae, if not writt en \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nMarks The weightage would be given by CBSE. \nQuestions 1. State in brief the meaning of an electric circuit. \n2. What is meant by potential difference between two p oints? \n3. Name the instrument used for measuring: (i) Potenti al difference \n(ii) Current \n4. Name the property by virtue of which two conductors having \nidentical structures offer different resistances to the flow of current \nwhen connected to the same source of electric curre nt. \n5. In an electric circuit state the relationship betwe en the direction of \nconventional current and the direction of flow of e lectrons. \n6. State Ohm\u2019s law. Express it mathematically and appl y it to define \nthe unit of resistance. \n7. A given length of a wire is doubled on itself and t his process is \nrepeated once again. By what factor does the resist ance of the wire \nchanges? \n8.", + "Express it mathematically and appl y it to define \nthe unit of resistance. \n7. A given length of a wire is doubled on itself and t his process is \nrepeated once again. By what factor does the resist ance of the wire \nchanges? \n8. Determine the ratio of the (a) highest (b) lowest r esistance that can \nbe secured by combining four coils of resistance 4 \u2126, 8 \u2126, 12 \u2126,", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nchanges? \n8. Determine the ratio of the (a) highest (b) lowest r esistance that can \nbe secured by combining four coils of resistance 4 \u2126, 8 \u2126, 12 \u2126, \nand 24 \u2126 in an electric circuit. \n9. Why are the heating elements of electric toaster an d electric iron \nmade of an alloy rather than pure metal? An electri c iron of \nresistance 20 \u2126 takes a current of 5 A. Calculate the heat develop ed \nin 30 seconds. \n10. Which is having more resistance a 220V, 100 W bulb or a 220V, \n60 W bulb?", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nTopic DAY 1- concept of Electric Circuit, Electric Current and Potential \ndifference \n \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires basic knowledge of electrons an d their flow and cells \nwhich is already learnt in class VII and VIII \nAssessment of \nqualifying \nknowledge Facilitator asks the students to make down list of resources of current \nwhich they use in their own home and compare their list with other \nstudents in groups \nObjective Student will be able to understand that w hat is conventional current and its \ndirection, formulae and unit of current and potenti al \n \nLearning \nOutcomes KNOWLEDGE- Students will know and understand \n\u2022 Discuss the importance of definitions and units \n\u2022 Appreciate the formulae. \n \nSKILLS AND COMPETENCIES- Students would be able to \n \n\u2022 Critically analyse the numerical and solve \n\u2022 Collaborate to find1ampere is equal to how many coulomb", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nSKILLS AND COMPETENCIES- Students would be able to \n \n\u2022 Critically analyse the numerical and solve \n\u2022 Collaborate to find1ampere is equal to how many coulomb \n \nAttitudes and beliefs- Students will be able to \nDevelop citizenship by visualizing and observing the quantities in day to \nday lives and sensitizing others \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \n \nDifferentiation- The students will define SI units of current and potential \nand solve numericals ( Character, Citizenship) \nGuided practice followed by Independent Practice- NCERT questions to be \ndiscussed in the classroom.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nTechniques to be used: \nQuiz \nDaily Practice Problem \nMCQ \nPeer Assessment \nStudent - teacher interaction, \nWipro- G.O.s(web chart, flow chart and differentiat ion table,compare- \ncontrast matrix), silent/ loud reading, \ncollaborative learning, \nResearch work/surveys", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nResources Text Book: \nNCERT text book for Science. \nReference Book \nNCERT Exemplar \nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments Independent Practice : Spiral level 1Questions \nAssessments Assessment of skit presented by various groups on the basis of content, \ncommunication, collaboration, time management and m essage given by \neach group to sensitize others. \nAddressing \nClassroom \nDiversity Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups: \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Initially simple to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nFor weak students: \n\u2022 Initially simple to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes. \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes \n \n \n \n \nUnit Day 2 -Circuit diagram \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires basic knowledge of components o f the circuit which is \nalready learnt in class VII and VIII \nAssessment of \nqualifying \nknowledge Written test \nLab activity \nGroup Discussion \nHW notebook \nObjective Students will be able to identify the sym bols of cell, battery resistance \nswitch and other components and draw circuit diagrams . \nDOL D1, D2, D3 \nLearning \nOutcomes KNOWLEDGE- Students will be able to \nStudents will gain the skill to draw circuit diagra ms. \nStudents will recognize the function of every compo nent in the \ncircuit", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nSKILLS AND COMPETENCIES- Students would be able to \nCommunicate about different symbols used in circuit diagram and \ntheir importance \n \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nSet induction and Brain Storming -The class would start with a discussion \non what the students have already learnt in the pre vious classes and hence \nwhat is it that they would learn now. They would al so be told the \nsignificance of the topic that they would be studyi ng. \nIntroduction of the topic - ppt and digital content would be shared \n \n \nDifferentiation- The students will draw the circuit diagrams. ( Critical \nthinking) \n \nQuestioning- Multiple level question - Teacher will prepare a list of \nquestion about the resistance in series and paralle l combination and will \nask these questions during discussions. Students wi ll answer ( Analytically)", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nGuided practice followed by Independent Practice- NCERT, Exemplar, \nspiral questions to be discussed in the classroom. \n \nTechniques to be used: \nSet induction \nClass discussion' \nQuiz \nDaily Practice Problem \nMCQ \nPeer Assessment \nCase Studies \nLab Activities \nGraphic Organizers for different terms \nLesson progress using Smart Class \nThink Pair Share for solving numerical Mind Mapping for lesson summarization \nMuddiest Point and One Minute Paper for Closure \nAny Other \n \nResources Text Book: \nNCERT text book for Science \nReference Book \nNCERT Exemplar \n \nSelf Study, Home \nWork, \nAssignments Independent Practice : Students would do the questions in their H.W \nnotebooks as mentioned in the monthly planner by wh ich we can assess \ntheir critical and creative thinking", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nHW notebooks to be marked as per the given plan: \nAssessment Parameters : The total marks for the activity is 5 marks \nOn time submission\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nPresentation/ Neatness (creativity) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nContent (analytical and critical thinking) \u2026\u2026\u2026\u2026\u2026\u2026\u2026......3 marks \n \nIt is also advised that the students come to the cl ass with proper \nbackground knowledge of the topic under discussion. They can refer to the \nresources stated above. \n \nAssessment The students in a group of two will be p erforming activity of connecting an \nelectric circuit. \nBy writing the answers of three different difficulty l evel of question we \ncan assess their knowledge (remembering type questions) critical \nthinking ( analytical questions) , creativity (HOT questions) \n \n \nAddressing \nClassroom Diversity Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups:", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Initally simple to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes and formulae, if not writt en \n\u2022 Call parents at regular intervals \n\u2022 Provide grade -up classes \n \nMarks The weightage would be given by CBSE. \n \n \nUnit Day 3 -Resistance \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires basic knowledge of current and potential difference \nwhich is already learnt \nAssessment of \nqualifying \nknowledge Written test \nLab activity \nGroup Discussion \nHW notebook", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nObjective Students will be able to relate that pote ntial difference is directly \nproportional to current and plot V -I graph to calculate Resistance. \nDOL D1, D2, D3 \nLearning \nOutcomes KNOWLEDGE- Students will be able to \n\u2022 Students will understand concept of resistance. \n\u2022 Students will gain the skill to draw circuit diagra ms. \n \nSKILLS AND COMPETENCIES- Students would be able to \n \nCritically solve numerical based on current, potential differ ence, \nand Resistance.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nSKILLS AND COMPETENCIES- Students would be able to \n \nCritically solve numerical based on current, potential differ ence, \nand Resistance. \n \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nSet induction and Brain Storming -The class would start with a discussion \non what the students have already learnt in the pre vious classes and hence \nwhat is it that they would learn now. They would al so be told the \nsignificance of the topic that they would be studyi ng. \nIntroduction of the topic - ppt and digital content would be shared \n \n \nDifferentiation- The students will able to calculat e resistance by graphical \nand mathematical method. ( Critical thinking) \n \nGuided practice followed by Independent Practice- NCERT, Exemplar, \nspiral questions to be discussed in the classroom.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nGuided practice followed by Independent Practice- NCERT, Exemplar, \nspiral questions to be discussed in the classroom. \n \nTechniques to be used: \nSet induction \nClass discussion' \nQuiz Daily Practice Problem \nMCQ \nPeer Assessment \nCase Studies \nLab Activities \nGraphic Organizers for different terms \nLesson progress using Smart Class \nThink Pair Share for solving numerical \nMind Mapping for lesson summarization \nMuddiest Point and One Minute Paper for Closure \nAny Other \n \nResources Text Book: \nNCERT text book for Science \nReference Book \nCBSE Exemplar \nDPSG Spiral \n \nSelf Study, Home \nWork, \nAssignments Independent Practice : Students would do the questions in their H.W \nnotebooks as mentioned in the monthly planner by wh ich we can assess \ntheir critical and creative thinking", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nHW notebooks to be marked as per the given plan: \nAssessment Parameters : The total marks for the activity is 5 marks \nOn time submission\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nPresentation/ Neatness (creativity) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nContent (analytical and critical thinking) \u2026\u2026\u2026\u2026\u2026\u2026\u2026......3 marks \n \nIt is also advised that the students come to the cl ass with proper \nbackground knowledge of the topic under discussion. They can refer to the \nresources stated above. \n \nAssessment By writing the answers of three different difficulty l evel question we \ncan assess their knowledge (remembering type questions) critical \nthinking (analytical questions) , creativity (HOT questions) \n \n \nAddressing \nClassroom Diversity Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups: \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Initially simple to be completed \u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes and formulae, if not writt en \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes \n \nMarks The weightage would be given by CBSE. \n \n \nUnit Day 4 -Resistivity \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires basic knowledge of current, pot ential difference and \nresistance which is already learnt . \nAssessment of \nqualifying \nknowledge Written test \nLab activity \nGroup Discussion \nHW notebook", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nObjective To develop scientific temper and inquisit iveness in students. Also, students \nwill be able to apply and relate learned concepts i n their daily lives. \nDOL D1, D2, D3 \nLearning \nOutcomes KNOWLEDGE- Students will be able to \nStudents will be able to define and solve numerical problems based \non resistivity. \n \nSKILLS AND COMPETENCIES- Students would be able to \n \nCritically solve numerical based on Resistivity. \nAnalyse change in resistance by the change in length and a rea of \nwire.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nSKILLS AND COMPETENCIES- Students would be able to \n \nCritically solve numerical based on Resistivity. \nAnalyse change in resistance by the change in length and a rea of \nwire. \n \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nSet induction and Brain Storming -The class would start with a discussion \non what the students have already learnt in the pre vious classes and hence \nwhat is it that they would learn now. They would al so be told the \nsignificance of the topic that they would be studyi ng. \nIntroduction of the topic - ppt and digital content would be shared \n \n \nDifferentiation- The students will be able to solve numerical based on \nresistivity. ( Critical thinking) \n Guided practice followed by Independent Practice- NCERT, Exemplar, \nother questions to be discussed in the classroom.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nTechniques to be used: \nSet induction \nClass discussion' \nQuiz \nDaily Practice Problem \nMCQ \nPeer Assessment \nCase Studies \nLab Activities \nGraphic Organizers for different terms \nLesson progress using Smart Class \nThink Pair Share for solving numerical \nMind Mapping for lesson summarization \nMuddiest Point and One Minute Paper for Closure \nAny Other \n \nResources Text Book: \nNCERT text book for Science \n NCERT Exemplar \nSelf Study, Home \nWork, \nAssignments Independent Practice : Students would do the questions in their H.W \nnotebooks as mentioned in the monthly planner by wh ich we can assess \ntheir critical and creative thinking \n \nHW notebooks to be marked as per the given plan: \nAssessment Parameters : The total marks for the activity is 5 marks \nOn time submission\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nPresentation/ Neatness (creativity) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nContent (analytical and critical thinking) \u2026\u2026\u2026\u2026\u2026\u2026\u2026......3 marks", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nIt is also advised that the students come to the cl ass with proper \nbackground knowledge of the topic under discussion. They can refer to the \nresources stated above. \n \nAssessment \nThe students in a group of two will be performing h ands on practical of \nverification of Ohm\u2019s law and relate various terms used in daily life \nNote Book Submission 05 Ma rks \nBy writing the answers of three different levels of qu estion we can \nassess their knowledge (simple knowledge based questions) critical \nthinking (application based questions) , creativity (HOT questions) \nAddressing \nClassroom Diversity \n Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups: \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nFor weak students: \n\u2022 Simple knowledge based questions to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes and formulae, if not writt en \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes \n \n \nMarks The weightage would be given by CBSE. \n \n \n \nUnit Day 5-series combination of resistance. \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires basic knowledge of resistance w hich is already learnt \nin previous class. \nAssessment of \nqualifying \nknowledge Written test \nLab activity \nGroup Discussion \nHW notebook", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nObjective Students will be able to make a circuit o f series combination, derive \nrelation of equivalent resistance in series combina tion and solve numerical \nbased on series combination in their notebook with 90% accuracy \nDOL D1, D2, D3 \nLearning \nOutcomes KNOWLEDGE- Students will be able to \n\u2022 The students will be demonstrated hands on activity 12.4 and 12.5 \nof NCERT of making simple electric circuit for seri es combination \nof resistors. \n\u2022 The students will make simple electric circuit for series \ncombination of resistors resistors. \n\u2022 The students will derive relation equivalent resist ance for series \ncombination of resistors. \n\u2022 The students will solve numericals based on resista nce combination \nand circuit analysis using think pair square. \n \nSKILLS AND COMPETENCIES - Students would be able to \nCritically solve numerical based on series combination of \nresistance.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nSKILLS AND COMPETENCIES - Students would be able to \nCritically solve numerical based on series combination of \nresistance. \n \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nSet induction and Brain Storming -The class would start with a discussion \non what the students have already learnt in the pre vious classes and hence \nwhat is it that they would learn now. They would al so be told the \nsignificance of the topic that they would be studyi ng. \nIntroduction of the topic - ppt and digital content would be shared \n \nQuestioning- Multiple level question - Teacher will prepare a list of \nquestion about the resistance in combination and wi ll ask these questions \nduring discussions. Students will answer ( Analytically) \n \n \nGuided practice followed by Independent Practice- NCERT, Exemplar, \nspiral questions to be discussed in the classroom.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nGuided practice followed by Independent Practice- NCERT, Exemplar, \nspiral questions to be discussed in the classroom. \n \nTechniques to be used: \nSet induction \nClass discussion' \nQuiz \nDaily Practice Problem \nMCQ \nPeer Assessment \nCase Studies \nLab Activities \nGraphic Organizers for different terms \nLesson progress using Smart Class \nThink Pair Share for solving numerical \nMind Mapping for lesson summarization \nMuddiest Point and One Minute Paper for Closure \nAny Other \n \nResources Text Book: \nNCERT text book for Science \n NCERT Exemplar \n \nSelf Study, Home \nWork, \nAssignments Independent Practice : Students would do the questions in their H.W \nnotebooks as mentioned in the monthly planner by wh ich we can assess \ntheir critical and creative thinking", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nHW notebooks to be marked as per the given plan: Assessment Parameters : The total marks for the activity is 5 marks \nOn time submission\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nPresentation/ Neatness (creativity) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nContent (analytical and critical thinking) \u2026\u2026\u2026\u2026\u2026\u2026\u2026......3 marks \n \nIt is also advised that the students come to the cl ass with proper \nbackground knowledge of the topic under discussion. They can refer to the \nresources stated above. \n \nAssessment With the help of subject enrichment activities we a ssess \ncommunication ,collaboration, character building an d citizenship. The \nstudents in a group of two will be performing hands on practical of \nverification of series combination. \n \nAddressing \nClassroom Diversity \n Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups:", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Initially Knowledge based questions to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes and formulae, if not writt en \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes \n \n \nMarks The weightage would be given by CBSE. \n \nUnit Day 7- Parallel combination of resistance. \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires basic knowledge of resistance w hich is already learnt in \nclass . \nAssessment of \nqualifying \nknowledge Written test \nLab activity \nGroup Discussion \nHW notebook", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nObjective Students will be able to make simple circ uit of parallel combination, derive \nrelation of equiv alent resistance in parallel combination and solve numerical based on parallel combination in their no tebook with 90% \naccuracy \nDOL D1, D2, D3 \nLearning \nOutcomes KNOWLEDGE- Students will be able to \n\u2022 The students will be demonstrated hands on activity 12.5 of \nNCERT of making simple electric circuit for paralle l combination \nof \n\u2022 The students will make simple electric circuit for parallel \ncombination of resistors. \n\u2022 The students will derive relation equivalent resist ance for parallel \ncombination of resistors. \n\u2022 The students will solve numericals based on resista nce combination \nand circuit analysis using think pair square. \n \nSKILLS AND COMPETENCIES- Students would be able to", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nSKILLS AND COMPETENCIES- Students would be able to \n \n1. Critically solve numerical based on Resistance. \n2. Analyse change in resistance by the change in length and a rea of \nwire and series and parallel combination. \n3. Communicate about importance of parallel combination in \ndomestic circuit. \n \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nSet induction and Brain Storming -The class would start with a discussion \non what the students have already learnt in the pre vious classes and hence \nwhat is it that they would learn now. They would al so be told the \nsignificance of the topic that they would be studyi ng. \nIntroduction of the topic - ppt and digital content would be shared \n \n \nDifferentiation- The students will draw the circuit diagrams. ( Critical \nthinking)", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nDifferentiation- The students will draw the circuit diagrams. ( Critical \nthinking) \n \nQuestioning- Multiple level question - Teacher will prepare a list of \nquestion about the resistance in series and paralle l combination and will \nask these questions during discussions. Students wi ll answer ( Analytically) \n \n \nGuided practice followed by Independent Practice- NCERT, Exemplar, \nspiral questions to be discussed in the classroom. \n \nTechniques to be used: \nSet induction \nClass discussion' Quiz \nDaily Practice Problem \nMCQ \nPeer Assessment \nCase Studies \nLab Activities \nGraphic Organizers for different terms \nLesson progress using Smart Class \nThink Pair Share for solving numerical \nMind Mapping for lesson summarization \nMuddiest Point and One Minute Paper for Closure \nAny Other \n \nResources Text Book: \nNCERT text book for Science \nNCERT Exemplar \nDPSG Spiral", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nResources Text Book: \nNCERT text book for Science \nNCERT Exemplar \nDPSG Spiral \n \nSelf Study, Home \nWork, \nAssignments Independent Practice : Students would do the questions in their H.W \nnotebooks as mentioned in the monthly planner by wh ich we can assess \ntheir critical and creative thinking \n \nHW notebooks to be marked as per the given plan: \nAssessment Parameters : The total marks for the activity is 5 marks \nOn time submission\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nPresentation/ Neatness (creativity) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nContent (analytical and critical thinking) \u2026\u2026\u2026\u2026\u2026\u2026\u2026......3 marks", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nIt is also advised that the students come to the cl ass with proper \nbackground knowledge of the topic under discussion. They can refer to the \nresources stated above. \n \nAssessment 3 UNIT TESTS (20, 30, 30) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.80 Marks \nMID TERM EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u202680 Marks \nHALF YEARLY EXAMINATION \u2026\u2026\u2026\u2026\u2026\u202680 Marks \nPRE-BOARD-I EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026.80 Marks \nPRE-BOARD-II EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026.80 Marks \nANNUAL BOARD EXAMINATION \u2026\u2026\u2026\u2026\u2026.80 Marks \n \nPERIODIC TEST \nBy administering pen paper test, we assess creativi ty and critical \nthinking. \nAverage of the best two tests to be taken that will have a weightage of 10 \nmarks. \nBest 2 Tests out of: Units tests, Midterm examinat ion, Half Yearly \nexamination. \nINTERNAL ASSESSMENT \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u202620 Marks \nPeriodical Test 10 Marks \n \nBy administering pen paper test we assess creativit y and critical \nthinking.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nBy administering pen paper test we assess creativit y and critical \nthinking. \n \nSubject Enrichment 05 M arks \nWith the help of subject enrichment activities we a ssess \ncommunication ,collaboration, character building an d citizenship. The \nstudents in a group of two will be performing hand s on practical of \nparallel combination and relate to various terms us ed in daily life \nNote Book Submission 05 Ma rks \nBy writing the answers of three different levels of qu estions we can \nassess their knowledge (remembering type questions) critical thinking ( \nanalytical questions) , creativity (HOT questions) \n \n \n Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups: \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Initially simple remembering type questions to be c ompleted \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes and formulae, if not writt en \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes \n \n \nMarks The weightage would be given by CBSE. \n \nUnit Day 7- Joule\u2019s law of heating \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires basic knowledge of components o f the circuit which is \nalready learnt in class VII and VIII \nAssessment of \nqualifying \nknowledge Written test \nLab activity \nGroup Discussion \nHW notebook \nObjective The students will be able to establish th e relation between, H, I, R and t, \ndiscuss practical problem due to heating and solve numericals based on \nheat produced. \nDOL D1, D2, D3 \nLearning \nOutcomes KNOWLEDGE-", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\n\u2022 The students will derive Joule\u2019s law of heating. \n\u2022 The students will solve numericals based on heating effect of \nelectric current. \n\u2022 The students will discuss different applications of heating effect of \nelectric current. \n\u2022 The students will solve pratical problems and numer icals based on \nheating effect of electric current. \n \nSKILLS AND COMPETENCIES- Students would be able to \n \nCritically solve numerical based on heating effect of current . \nCommunicate about different components of domestic circuit and \ntheir importance", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nCritically solve numerical based on heating effect of current . \nCommunicate about different components of domestic circuit and \ntheir importance \n \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nSet induction and Brain Storming -The class would start with a discussion \non what the students have already learnt in the pre vious classes and hence \nwhat is it that they would learn now. They would al so be told the \nsignificance of the topic that they would be studyi ng. \nIntroduction of the topic - ppt and digital content would be shared \n \n \nDifferentiation- The students will solve numericals . ( Critical thinking) \n \n \nGuided practice followed by Independent Practice- NCERT, Exemplar \nquestions to be discussed in the classroom.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nDifferentiation- The students will solve numericals . ( Critical thinking) \n \n \nGuided practice followed by Independent Practice- NCERT, Exemplar \nquestions to be discussed in the classroom. \n \nTechniques to be used: \nSet induction \nClass discussion' \nQuiz \nDaily Practice Problem \nMCQ \nPeer Assessment \nCase Studies \nLab Activities Graphic Organizers for different terms \nLesson progress using Smart Class \nThink Pair Share for solving numerical \nMind Mapping for lesson summarization \nMuddiest Point and One Minute Paper for Closure \nAny Other \n \nResources Text Book: \nNCERT text book for Science \n NCERT Exemplar \n \nSelf Study, Home \nWork, \nAssignments Independent Practice : Students would do the questions in their H.W \nnotebooks as mentioned in the monthly planner by wh ich we can assess \ntheir critical and creative thinking", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nHW notebooks to be marked as per the given plan: \nAssessment Parameters : The total marks for the activity is 5 marks \nOn time submission\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nPresentation/ Neatness (creativity) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nContent (analytical and critical thinking) \u2026\u2026\u2026\u2026\u2026\u2026\u2026......3 marks \n \nIt is also advised that the students come to the cl ass with proper \nbackground knowledge of the topic under discussion. They can refer to the \nresources stated above. \n \nAssessment With the help of subject enrichment activities we a ssess \ncommunication ,collaboration, character building an d citizenship. The \nstudents in a group of two will be performing hands on activity of verifying \nJoule\u2019s law \nBy writing the answers of three different types of que stion we can \nassess their knowledge (remembering questions) critical thinking \n( analytical questions) , creativity (HOT questions)", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nAddressing \nClassroom Diversity Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups: \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Simple remembering type to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n For differently abled students: \n\u2022 Ignore spelling mistakes and formulae, if not writt en \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes \n \n \nMarks The weightage would be given by CBSE. \n \n \nUnit Day 8- Electrical Power and commercial unit of electrical energy \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires basic knowledge of components o f the circuit which is \nalready learnt in class VII and VIII \nAssessment of \nqualifying \nknowledge Written test \nLab activity \nGroup Discussion \nHW notebook", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nObjective To students will be able to apply formula of power and establish the \nrelation between electrical unit and commercial unit of ener gy. \nDOL D1, D2, D3 \nLearning \nOutcomes KNOWLEDGE- \nThe students will be able to define electric power and commercial \nunit of electric energy \n \nSKILLS AND COMPETENCIES- Students would be able to \n \nCritically solve numerical based on power. \nCommunicate about different components of domestic circuit and \ntheir importance", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nSKILLS AND COMPETENCIES- Students would be able to \n \nCritically solve numerical based on power. \nCommunicate about different components of domestic circuit and \ntheir importance \n \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nSet induction and Brain Storming -The class would start with a discussion \non what the students have already learnt in the pre vious classes and hence \nwhat is it that they would learn now. They would al so be told the \nsignificance of the topic that they would be studyi ng. \nIntroduction of the topic - ppt and digital content would be shared \n \n \nDifferentiation- The students will able to apply fo rmula of power to solve \nnumerical. ( Critical thinking)", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nDifferentiation- The students will able to apply fo rmula of power to solve \nnumerical. ( Critical thinking) \n \nGuided practice followed by Independent Practice- NCERT, Exemplar \nquestions to be discussed in the classroom. \nTechniques to be used: \nSet induction \nClass discussion' \nQuiz \nDaily Practice Problem \nMCQ \nPeer Assessment \nCase Studies \nLab Activities \nGraphic Organizers for different terms \nLesson progress using Smart Class \nThink Pair Share for solving numerical \nMind Mapping for lesson summarization \nMuddiest Point and One Minute Paper for Closure \nAny Other \n \nResources Text Book: \nNCERT text book for Science \nNCERT Exemplar \n \nSelf Study, Home \nWork, \nAssignments Independent Practice : Students would do the questions in their H.W \nnotebooks as mentioned in the monthly planner by wh ich we can assess \ntheir critical and creative thinking", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nHW notebooks to be marked as per the given plan: \nAssessment Parameters : The total marks for the activity is 5 marks \nOn time submission\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nPresentation/ Neatness (creativity) \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nContent (analytical and critical thinking) \u2026\u2026\u2026\u2026\u2026\u2026\u2026......3 marks \n \nIt is also advised that the students come to the cl ass with proper \nbackground knowledge of the topic under discussion. They can refer to the \nresources stated above. \n \nAssessment By writing the answers of three different levels of qu estion we can \nassess their knowledge (simple remembering type questions) critical \nthinking (analytical questions) , creativity (HOT questions) \n \n \nAddressing \nClassroom Diversity Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups:", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \nFor weak students: \n\u2022 Simple remembering type questions to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes and formulae, if not writt en \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes \n \n \nMarks The weightage would be given by CBSE. \nDay: 9", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nAim: To study the dependence of the potential differenc e across a resistor on the current \n through it and to determine its resistance and to v erify the Ohm\u2019s law. \nMaterial required: A resistor of about 5 \u2126, an ammeter ( 0 - 3 A), a voltmeter (0 - 10 V), 4 dry \ncells of 1.5 V each a battery eliminator, a plug ke y, connecting wires, sand paper. \nTheory: According to the Ohm\u2019s law, the potential differenc e ( V) across the ends of a resistor is \ndirectly proportional to the current ( I) through it provided its temperature remains the s ame. \nV = RI \n \n \n R is a constant for the given resistor at a given tem perature and is called its resistance. \nA graph between the potential difference across the two ends of a resistor and the current \nthrough it is a straight line passing through the o rigin. The slope of this graph gives the \nresistance R of the resistor. \nProcedure:", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nthrough it is a straight line passing through the o rigin. The slope of this graph gives the \nresistance R of the resistor. \nProcedure: \n1. Note the range and least count of the given am meter and the voltmeter. \n2. Set up the circuit by connecting different com ponents with the help of connecting wires. \n3. Initially connect only one cell in the circuit . In case a battery eliminator is used, keep the \n rating of the eliminator at the minimum \n4. Insert the key in the plug and note the readin gs of ammeter and voltmeter and record them. \n5. Now instead of using one cell in the circuit, connect two cells in the circuit. \n6. Note and record the voltmeter and ammeter read ings. \n7. Repeat the experiment by connecting three and four cells in the circuit. \nObservation and calculation: \n(i) Range of the ammeter = ___ - ___ A. \n(ii) Least count of the ammeter = ___ A.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nObservation and calculation: \n(i) Range of the ammeter = ___ - ___ A. \n(ii) Least count of the ammeter = ___ A. \n(iii) Range of the voltmeter = ___ -___ V. \n(iv) Least count of the voltmeter = ___ V. \nMean value of resistance R of the resistor = _____ \u2126 \nS.No Current \nthrough the \nresistor I Potential difference across \nthe end of resistor V Resistance of the \nresistor R = V/I \n (A) (V) ( \u2126) \n1 \n2 \n3 \n4", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nGraph: Plot the graph with I and V along the x- and y-axes respectively on the graph paper. \nFind the slope of this graph by choosing two points P and Q on it. \n Slope = QM/MP \n \nGraph between V and I", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nResult: Resistance obtained from the calculations = \u2126 \n Resistance obtained from the graph = \u2126 \n\u2022 The value of resistance R of resistor for all values of current through it re mains the same (or \nalmost same). The graph between V and I is a straight line and passes through the origin. T his \nverifies the Ohm\u2019s law. \nPrecautions: \n1. The connecting wires should be thick copper wires a nd the insulation of their ends should be \n removed using the sand paper. \n2. Connections should be tight otherwise some ext ernal resistance may introduce in the circuit. \n3. The ammeter should be connected in series with the resistor such that the current enters at \n the positive terminal and leaves at the negat ive terminal of the ammeter. \n4. Voltmeter should always be connected in parall el to resistor. \n5. The pointers of the ammeter and voltmeter shou ld be at zero mark when no current through \n the circuit.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nDay: 10 \nEXPERIMENT No. 2 \nAim: To determine the equivalent resistance of two resistors connected in series \ncombination. \nMaterial required: Two resistors of (each of 2 \u2126 resistance), an ammeter (range 0 \u2013 5 A), a \nvoltmeter (range 0 \u2013 5 V), three dry cells of 1.5 V each a plug key, connecting wires. \nTheory: When two resistors of resistance R1 and R2 respectively are connected in a series \ncombination. then their equivalent resistance Rs i s given by \n Rs = R1 + R2 ------------------------ eq. 1 \n \n \nProcedure: \n1. Note the range and least count of the given am meter and the voltmeter. \n2. Place the given resistors one after the other and join the ends. \n3. Set up the circuit by connecting different com ponents with the help of connecting wires \nas \n shown in the circuit diagram. \n4.", + "2. Place the given resistors one after the other and join the ends. \n3. Set up the circuit by connecting different com ponents with the help of connecting wires \nas \n shown in the circuit diagram. \n4. Make sure that the positive and negative termi nals of the ammeter and voltmeter are \n correctly connected in the circuit.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nas \n shown in the circuit diagram. \n4. Make sure that the positive and negative termi nals of the ammeter and voltmeter are \n correctly connected in the circuit. \n5. Insert the key in the plug and note the readin gs of ammeter and voltmeter and record \nthem. \n6. Repeat the activity for three different values of current through the circuit and record \nthe \n readings of the ammeter and voltmeter in each case. \nObservation and calculation: Range of the ammeter = ___ - ___ A. Least count of the ammeter = ___ A. \nRange of the voltmeter = ___ - ___ V. \nLeast count of the voltmeter = ___ V. \nResistance of first resistor R1 = ___ \u2126 , Resistance of second resistor, R2 = ___ \u2126", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nS. \nNo. Current through \nthe series \ncombination, I s Potential \ndifference across \nthe series V s Equivalent Resistance \nof combination, Rs= \nVs / I s Average value \nof R s \n (A) (V) (\u2126) (\u2126) \n1 \n2 \n3 \n4 \n \nR1 = ____ \u2126, R2 = _____ \u2126 \n \nEquivalent resistance = R1 + R2 = ___ \u2126 \nResult: Resistance of the series combination (from observation table) :_____________ \u2126 \nResistance of the series combination (calculated by using Eq. 1 ) :_____________ \u2126", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nPrecaution: \n1. The connecting wires should be thick copper wi res and the insulation of their ends should \n be removed using the sand paper. \n2. Connections should be tight otherwise some ext ernal resistance may introduce in the circuit. \n3. The ammeter should be connected in series with the combination of resistors such that the \n current enters at the positive terminal and l eaves at the negative terminal of the ammeter. \n4. Voltmeter should always be connected in parall el to the combination of resistors. \n5. The pointers of the ammeter and voltmeter shou ld be at zero mark when no current flows \n through the circuit.. \n6. Current should be passed through the circuit f or a short time while taking observations; \n otherwise current would cause unnecessary hea ting in the circuit. Heating may change the \n resistance of resistors.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nEXPERIMENT No. 3 \nAim: To determine the equivalent resistance of two resistors connected in parallel combination. \nMaterial required: Two resistors of (each of 2 \u2126 resistance), an ammeter (range 0 \u2013 5 A), \na voltmeter (range 0 \u2013 5 V), three dry cells of 1.5 V each , a plug key, connecting wires, \nTheory: When two resistors of resistance R1 and R2 respectively are connected in a parallel \ncombination then their equivalent resistance RP is given by \nEquivalent resistance(R P) = R 1R2 / (R 1+ R 2) \n \nProcedure: \n1. Note the range and least count of the given ammeter and the voltmeter. \n2. Place the given resistors side by side and joi n end A with the end C, and end B with end D. \n3. Set up the circuit by connecting different com ponents with the help of connecting wires as \n shown in the circuit diagram . \n4.", + "2. Place the given resistors side by side and joi n end A with the end C, and end B with end D. \n3. Set up the circuit by connecting different com ponents with the help of connecting wires as \n shown in the circuit diagram . \n4. Note the readings of the ammeter and voltmeter and record them.", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nshown in the circuit diagram . \n4. Note the readings of the ammeter and voltmeter and record them. \n5. Repeat the activity for three different values of current through the circuit and record the \n readings of the ammeter and voltmeter in each case. \nObservation and calculation: \n(i) Range of the ammeter = ____ - ____ A. \n(ii) Least count of the ammeter = ____ A. \n(iii) Range of the voltmeter = ____ - ____ V. \n(iv) Least count of the voltmeter = ____ V \n(vi) Resistance of first resistor, R1 = ____ \u2126, Resistance of second resistor, R2 = ____ \u2126", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nS.No. Current through the \nparallel combination, \nIP Potential difference \nacross the parallel \ncombination, V P Equivalent \nResistance of \ncombination, R P \nRP = V P / I P Average \nvalue of R P \n (A) (V) (\u2126) (\u2126) \n \n \n \n \n \nR1 = ____ \u2126, R2 = _____ \u2126", + "CHAPTER 11-ELECTRCITY-LESSON PLAN (2).txt\nR1 = ____ \u2126, R2 = _____ \u2126 \n \n Equivalent resistance = R 1R2 / (R 1+ R 2) \nResult and discussion: Resistance of the parallel combination (from observ ation table):______ \nResistance of the parallel combination (calculated by using Eq. 1):________________ \nPrecautions: \n1. The connecting wires should be thick copper wi res and the insulation of their ends should \n be removed using the sand paper. \n2. Connections should be tight otherwise some ext ernal resistance may introduce in the circuit. \n3. The ammeter should be connected in series with the combination of resistors such that the \n current enters at the positive terminal and l eaves at the negative terminal of the ammeter. \n4. Voltmeter should always be connected in parall el to the combination of resistors. \n5. The pointers of the ammeter and voltmeter shou ld be at zero mark when no current flows \n through the circuit..", + "CHAPTER 11-ELECTRCITY-ppt2.txt\nELECTRICITY\n1.Electric Current\n2.Conventional Current\n3.Electric Circuit\n4.Electric Potential and Potential Difference\n5.Symbols used in Circuit Diagram\n6. Ohm\u2019s Law\n7. Resistance and Resistivity \n8. Conductance and Conductivity\n9.Series and Parallel combination of resistors\n10.Heating Effect of Current\n11.Joule\u2019s Law of Heating\n12.Electric Power and Electric Energy\n13.Electric FuseElectric Current\nThe electric current is defined as the charge flowing through any section \nof the conductor in one second .\nDifferent types of current:\nI\nt0abc\nd) Alternating current whose \nmagnitude varies continuously \nand direction changes \nperiodicallya)Steady current which does not \nvary with time \nb)&c) Varying current whose \nmagnitude varies with timed(if the rate of flow of charge is steady) I=Q\ntSI unit of electric charge is coulomb (C) and SI unit of time is second (s).\nSI unit of electric current is coulomb / second or C/s.\nOrSI unit of electric current is ampere (A).", + "SI unit of electric current is coulomb / second or C/s.\nOrSI unit of electric current is ampere (A).(From the formula)", + "CHAPTER 11-ELECTRCITY-ppt2.txt\ntSI unit of electric charge is coulomb (C) and SI unit of time is second (s).\nSI unit of electric current is coulomb / second or C/s.\nOrSI unit of electric current is ampere (A).(From the formula)\n(Named after the French \nscientist, Andre -Marie Ampere\n(1775 \u20131836))\nNote:\n\u27a2Q = ne ( Q \u2013charge, n \u2013no. of electrons and e \u2013electronic charge)\n\u27a2Charge of one electron is 1.6 x 10-19coulomb.\n\u27a21 coulomb is equivalent to the charge of 6.25 x 1018electrons.\n\u27a21 coulomb, in fact, is very high charge.\n\u27a2Therefore, charge is expressed in smaller units like milli coulomb (mc) or\nmicro coulomb ( \u00b5C).", + "\u27a21 coulomb, in fact, is very high charge.\n\u27a2Therefore, charge is expressed in smaller units like milli coulomb (mc) or\nmicro coulomb ( \u00b5C).\n\u27a2Similarly current is expressed in milli ampere (mA) or micro ampere (\u00b5A).Conventional Current\nConventional current is the current \nwhose direction is along the direction of \nthe motion of positive charge under the \naction of electric field.+\n+\n+\n+-\n-\n-\n-\nConventional current due to motion of \nelectrons is in the direction opposite to \nthat of motion of electrons.+ + +\nI\n+\n+\n+\n+\n-\n-\n-\n-", + "CHAPTER 11-ELECTRCITY-ppt2.txt\naction of electric field.+\n+\n+\n+-\n-\n-\n-\nConventional current due to motion of \nelectrons is in the direction opposite to \nthat of motion of electrons.+ + +\nI\n+\n+\n+\n+\n-\n-\n-\n-\nI- - -Electric Circuit\nA continuous and closed path of an electric current is called an electric \ncircuit.\nA simple electric circuit consists of a cell (or battery), switch (or key) and \nload (or resistance).\nAn ammeter (current measuring instrument) is always connected in series\nin an electric circuit.\nA voltmeter (potential measuring instrument) is always connected in \nparallel (shunt) across the load (or resistance).\nA Simple Electric CircuitEI I\nAn Electric Circuit with \nAmmeter and VoltmeterEAV\nII\nR\nK KElectric Potential\nElectric potential is a physical quantity which determines the flow of charges \nfrom one body to another.\nIt is a physical quantity that determines the degree of electrification of a body.\nElectric Potential at a point in the electric field is defined as the work done in", + "CHAPTER 11-ELECTRCITY-ppt2.txt\nfrom one body to another.\nIt is a physical quantity that determines the degree of electrification of a body.\nElectric Potential at a point in the electric field is defined as the work done in \nmoving (without any acceleration) a unit positive charge from infinity to that \npoint against the electrostatic force irrespective of the path followed.\nSI unit of electric potential is volt (V) or J C-1or Nm C-1.\n(named after Alessandro Volta (1745 -1827), an Italian Physicist)\nElectric potential at a point is one volt if one joule of work is done in moving \none coulomb charge from infinity to that point in the electric field.V =W\nQElectric Potential Difference between any two points in the electric field is \ndefined as the work done in moving (without any acceleration) a unit positive \ncharge from one point to the other against the electrostatic force irrespective \nof the path followed.Electric Potential Difference\n1.Electric potential and potential difference are scalar quantities.", + "CHAPTER 11-ELECTRCITY-ppt2.txt\nof the path followed.Electric Potential Difference\n1.Electric potential and potential difference are scalar quantities.\n2.Electric potential at infinity is zero.\n3.Electric potential near an isolated positive charge (q > 0) is positive and that \nnear an isolated negative charge (q < 0) is negative.\n4.cgs unit of electric potential is stat volt . 1 stat volt = 1 erg / stat coulombWAB\nQVB-VA= \u2206V =Ohm\u2019s Law\nThe electric current flowing through a conductor is directly \nproportional to the potential difference across the two ends of the \nconductor when physical conditions such as temperature, mechanical \nstrain, etc. remain the same.\nI\nI \u03b1VorV \u03b1IorV = R I\nVI\n0EOhm\u2019s Law Experiment\nGraph is linear (straight line)V\nR\nEA\nKIIResistance\nThe resistance of conductor is the opposition offered by the \nconductor to the flow of electric current through it.\nResistance in terms of physical features of the conductor\nAR = \u03c1l where \u03c1is proportionality constant called \u2018 resistivity \u2019", + "CHAPTER 11-ELECTRCITY-ppt2.txt\nconductor to the flow of electric current through it.\nResistance in terms of physical features of the conductor\nAR = \u03c1l where \u03c1is proportionality constant called \u2018 resistivity \u2019 \nor \u2018specific resistance \u2019, l is the length of the conductor \nand A is the area of cross section of the conductor.\nResistance is directly proportional to length and inversely proportional \nto cross -sectional area of the conductor and depends on nature of \nmaterial.\nResistivity depends upon nature of material and noton the geometrical \ndimensions of the conductor. R =V\nI\nBoth the resistance and resistivity vary with temperature.SI unit of resistivitiy is \u2018ohm -m\u2019 or \u03a9m.SI unit of resistance is \u2018ohm \u2019 and its symbol is \u2018 \u03a9\u2019.\nIf the potential difference across the two ends of a conductor is 1 Vand \nthe current through it is 1 A, then the resistance of the conductor is \nsaid to be 1 ohm .\nConductance and conductivity\nConductance is the reciprocal of resistance. Its S.I unit is mho .", + "CHAPTER 11-ELECTRCITY-ppt2.txt\nthe current through it is 1 A, then the resistance of the conductor is \nsaid to be 1 ohm .\nConductance and conductivity\nConductance is the reciprocal of resistance. Its S.I unit is mho .\nConductivity is the reciprocal of resistivity. Its S.I unit is mho / m.Resistance of a system of resistors\nResistors are said to be \nin series combination \nwhen they are connected \nend to end.Resistors in series\nA- +R1 R2 R3\nVI II I\nV1 V2 V3\nIn series combination, \n1.Current through each \nresistor is same.\nV = V1+ V2+ V3\nIR = IR1+ IR2+ IR3\nOr R = R1+ R2+ R3\nThe Effective or Equivalent Resistance of the resistors in a series \ncombination is the sum of the resistances of the individual resistors.K\n2. Potential difference across resistors of unequal resistance is different. \nHowever, the total p.d. is the sum of the p.d.", + "Potential difference across resistors of unequal resistance is different. \nHowever, the total p.d. is the sum of the p.d. across the individual resistors.\nNote:\nThe Effective or Equivalent Resistance of the resistors in a series", + "CHAPTER 11-ELECTRCITY-ppt2.txt\nHowever, the total p.d. is the sum of the p.d. across the individual resistors.\nNote:\nThe Effective or Equivalent Resistance of the resistors in a series \ncombination is greater than the greatest of them.I IResistors are said to be in \nparallel combination when \nthey are connected \ntogether between two \npoints.Resistors in parallel\nA- +\nVI IV\nIn parallel combination, \n1.Potential difference \nacross each resistor is \nsame.\nI = I1+ I2+ I3The reciprocal of Effective or Equivalent \nResistance of the resistors in a parallel \ncombination is the sum of the reciprocals \nof resistances of the individual resistors.R1\nR3R2 I I\nKI3I1\nI2\n=V\nR+V\nR1V\nR3+V\nR2\n=1\nR+1\nR11\nR3+1\nR2Or2. Current through resistors of unequal resistance is different. However, the \ntotal current supplied by the battery is the sum of the currents through \nthe individual resistors.", + "Current through resistors of unequal resistance is different. However, the \ntotal current supplied by the battery is the sum of the currents through \nthe individual resistors.\nNote:\nThe Effective or Equivalent Resistance of the \nresistors in a parallel combination is smaller", + "CHAPTER 11-ELECTRCITY-ppt2.txt\nthe individual resistors.\nNote:\nThe Effective or Equivalent Resistance of the \nresistors in a parallel combination is smaller \nthan the smallest of them.I IJoule\u2019s Law:\nWhen an electric current is passed through a conducting wire, the electrical \nenergy is converted into heat energy. This effect is called \u2018heating effect of \nelectric current\u2019.\nIt is also called \u2018Joule Heating\u2019.\nJoule\u2019s Law givesQ = I2R t Q is in Joule, I in Amp, R in ohm and t in sec.\nI2R t\nJQ =Q is in Calorie, I in Amp, R in ohm and t in sec.\nCause of Heating Effect of Current:\nWhen a battery is connected to the ends of a conductor, electric field is set \nup. The free electrons move in the direction opposite to the field. They \nundergo collision with the positive ions and atoms of the conductor. The \naverage kinetic energy of vibration of the ions and the atoms increases and \nhence the temperature increases.\nNote: The course of electron through resistor is like that of the stone falling", + "CHAPTER 11-ELECTRCITY-ppt2.txt\naverage kinetic energy of vibration of the ions and the atoms increases and \nhence the temperature increases.\nNote: The course of electron through resistor is like that of the stone falling \nin water.HEATING EFFECT OF CURRENTElectric Power\nElectric power is the rate at which work is done by an electric current.\nP = V I P = I2R P = V2/ R\nSI unit of power is \u2018watt\u2019.\nOther units are \u2018 kW\u2019 and \u2018 hp\u2019.\n1 kW = 1000 W 1 hp = 746 W\nElectric Energy\nElectric energy is the total work done by an electric current in a given time.\nE = V I t E = I2R t E = V2t / R\nCommercial Unit of Electric Energy is kWh or B.O.T.U .\n1 kWh = 1000 w x 3600 s\n= 3600000 w -s or joule\n= 3.6 x 106jouleElectric Heating Appliances use Nichrome wire (alloy of Ni and Cr).", + "1 kWh = 1000 w x 3600 s\n= 3600000 w -s or joule\n= 3.6 x 106jouleElectric Heating Appliances use Nichrome wire (alloy of Ni and Cr).\nIt is used because:\ni)It has high specific resistance ii) It has high melting point\niii)It has high malleability iv) It is not easily oxidised.\nFor given V, P \u03b1I P \u03b11 / R", + "CHAPTER 11-ELECTRCITY-ppt2.txt\nIt is used because:\ni)It has high specific resistance ii) It has high melting point\niii)It has high malleability iv) It is not easily oxidised.\nFor given V, P \u03b1I P \u03b11 / R\ni.e. i) Higher the power of the appliance, more is the current drawn\nii) Higher the power of the appliance, less is the resistance.Electric Fuse\nElectric fuse is a protective device used in series with an electric circuit or \nan electric appliance to save it from damage due to overheating produced \nby strong current in the circuit or appliance.\nFuse is generally made of alloy of 63% tin and 37% lead . \nIt has high resistance and low melting point .\nMore of Electricity in Higher Class\u2026", + "CHAPTER 11-ELECTRCITY-IMPORTANT NOTESS.txt\nFOCUS POINTS . \nChapter : 12 ELECTRICITY \n1. The relation between current and potential difference across a resistor needs to be understood \nproperly by students. The expression V / I = R , where independence of R with V and I has to \nbe stressed upon . More numerical and conceptual problems have to be done. \nThe interpretation of V \u2013 I graph its slope determination gives R to be explained. \nThe V \u2013 I graph with temperature of wire may also be explained. \n2. The expression for calc ulation of resistance with length and cross sectional area , \nnumerical problems have to be solved. The variation of resistance with length while cutting \nand stretching a wire has to be focussed upon. \n3. In the combination of resistances , the properties of current, voltage and equivalent \nresistance have to be discussed in detail. More problems with combination of resistances to \nbe given as practice. Most students has confusion of calculating effective resi stance in", + "CHAPTER 11-ELECTRCITY-IMPORTANT NOTESS.txt\nresistance have to be discussed in detail. More problems with combination of resistances to \nbe given as practice. Most students has confusion of calculating effective resi stance in \nparallel combination. They take R = 1/R 1 + 1/R 2 instead 1/ R = 1/R 1 + 1/R 2 \n4. Joule\u2019s law of heating expression H = I2 R t = ( V2 / R ) t = V I t . Dependence of H on R \nunder constant current and constant voltage needs to be stressed. \n5. Electric power P = V I = I2 R = V2 / R . More numericals have to be done . \n6. In numericals students should be practised to take all the quantities in S I units only. \n7. Practice of drawing circuit diagrams with neat labelling has to be given.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-QUESTIONS.txt\nFr. A gnel school", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-QUESTIONS.txt\nDescription: \nElectricity and magnetism are ultimately inextricably linked. Electricity is \nthe term given to a group of physical phenomena involving electric \ncharges, their motions, and their effects. Magnetism is the force \ngenerated by the passage of an electric curren t through matter. \nThe magnetic effect of electric current - one of the important utility of \nelectric current, demonstrates that electricity and magnetism are linked \nto each other. \n Moving or changing magnetic field generates electric field and moving \nor ch anging electric field generates magnetic field. \n \nCurrent carrying conductor has magnetic field around it and a conductor \nmoving in the magnetic field can generate electricity in it. \nLearning Objective: \n\u00b7 Differentiate between the physical situations in which this rule and \nthe situations, in which Fleming\u2019s left hand rule, are to be used. \n\u00b7 Realize that when a ny two of the three (mutually perpendicular)", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-QUESTIONS.txt\nthe situations, in which Fleming\u2019s left hand rule, are to be used. \n\u00b7 Realize that when a ny two of the three (mutually perpendicular) \ndirections are given we can use this rule for finding the third unknown \ndirection. \n (a) direction of movement of the conductor \n (b) direction of magnetic field The students will be able to : \n\u00b7 Describe mag netic field and field lines. \n\u00b7 Explain magnetic field due to a straight current carrying \nconductor. \n\u00b7 Summarise the factors on which strength and direction of \nmagnetic field around a straight conductor. \n\u00b7 State the right hand thumb rule. \n\u00b7 Apply the right hand thumb rule. \n\u00b7 Demonstrate magnetic field due to a current through a circular \nloop \n\u00b7 Analyse the magnetic field pattern around a solenoid carrying \ncurrent \n\u00b7 Express force on a current carrying conductor in a magnetic \nfield. \n\u00b7 State Fleming\u2019s right hand rule.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-QUESTIONS.txt\ncurrent \n\u00b7 Express force on a current carrying conductor in a magnetic \nfield. \n\u00b7 State Fleming\u2019s right hand rule. \n\u00b7 Illustrate the use of the Fleming\u2019s right hand rule in some \nsimple situations \n (c) direction of induced current/emf \n\u00b7 Define electromagnetic induction . \n\u00b7 State the principle of an (a) electric motor (b) electric generator. \u00b7 Discuss the importance of fuse and earthing in domestic circuit \n\u00b7 Explain how does the electric fuse and earthing protect the \ndomestic circuit \n \n\u00b7 Demonstrate the understanding of terms (a) overloading (b) short \ncircuit \nPerception: \nUnderstanding: \u00b7 Make an electromagnet and explain its use in daily life \n\u00b7 Understand the working of an domestic electric circuits \n\u00b7 Draw the schematic diagram of common domestic circuit. \nChapter will be introduced by presenting a situation .", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-QUESTIONS.txt\n\u00b7 Understand the working of an domestic electric circuits \n\u00b7 Draw the schematic diagram of common domestic circuit. \nChapter will be introduced by presenting a situation . \nSuspending a bar magnet and observing the direction in which it is \ncoming to rest . \nNote the deflection when a compass is brought near to it. . \nThen setting up an electric circuit and bringing a compass near it \nshows the same deflection. \nQuestions to be asked : \nWhat is the analogy in the activities ? \nWhat do you infer from this ? \n \nLink to be used \n//ctsp:class contents/class X/physics/magnetism.ppt \n \n Tutorials from EXTRA -MARKS WILL BE SHOWN TO THE \nSTUDENTS \nQuestions to be raised to students \nHow does a magnetic compass shows the N -S direction of a place ? \nWhy does the compass deflect when placed near a current carrying \nwire ? \nWhy does the needle deflection changes with the reversal of current \nflow ? \nThere by deducing the right hand thumb rule.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-QUESTIONS.txt\nWhy does the compass deflect when placed near a current carrying \nwire ? \nWhy does the needle deflection changes with the reversal of current \nflow ? \nThere by deducing the right hand thumb rule. \nDemonstration of marking of field lines around a bar magnet and \ndiscussing the properties of magnetic lines of force. \nAt the end of the lesson students will be able to : \n Describe the following terms related to magnetism (a) \nmagnetic field (b) magnetic lines of force \n State the SI unit of magnetic field \n Explain the characteristics of magnetic lines of fo rce \n Draw the magnetic field lines around (a) bar magnet (b) \nCurrent carrying straight conductor (c) Current carrying circular \nloop (d)Current carrying solenoid Analysis: \n Draw the magnetic field lines around a current carrying (a) \nstraight conductor (b) circular loop (c) solenoid. \nTo experimentally investigate the magnetic field of a straight \nconductor or a solenoid.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-QUESTIONS.txt\nstraight conductor (b) circular loop (c) solenoid. \nTo experimentally investigate the magnetic field of a straight \nconductor or a solenoid. \n\u2022 Suspend a magnet fr om a thread into glycerin, honey, or some \nother clear, viscous fluid. \n\u2022 Gently sprinkle iron filings on the liquid. \n\u2022 The iron filings remain suspended, arranging themselves three -\ndimensionally in the magnetic field around the magnet. \n\u2022 Draw a representation of the magnetic field pattern. \n\u2022 Trying to remove the iron filings from the magnet may be difficult, \nbut less so if you wrap the magnet in plastic sandwich wrap \nbeforehand. \nIf an electromagnet can be set up for this demonstration, it not only \nhelps to remove the iron filings, but it can also be used to illustrate the \ncollapse of the magnetic field when the current to the electromagnet is \ndiscontinued \nSynthesis: State Fleming\u2019s right hand thumb rule and to use this rule to \nfind the direction of magnetic field.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-QUESTIONS.txt\ndiscontinued \nSynthesis: State Fleming\u2019s right hand thumb rule and to use this rule to \nfind the direction of magnetic field. \nhttp://www.khanacademy.org/science/physics/v/magnetism -6--\nmagnetic -field-due-to-current \nExperimentation \nAllow a definite am ount of current to flow through a flexible \nconducting wire. Change the shape of the wire and measure the field \nstrength at a fixed distance, by changing the shape of the wire. Use a \nmagnetic compass to compare the field strength. Analyse whether \nwith the c hange of shape of conductor the magnetic field strength \nvaries. \nQuestion \n Does a definite amount of current always exhibits the same \nvalue of magnetism ? \n Determine the factors on which field strength depends with a \n(a) straight conductor (b) circular loop (c) solenoid. \nDiscussion of a situation \nIn a television screen the charged ions striking at the screen \nilluminates different points .", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-QUESTIONS.txt\n(a) straight conductor (b) circular loop (c) solenoid. \nDiscussion of a situation \nIn a television screen the charged ions striking at the screen \nilluminates different points . \n What make the charged particles to strike at varies points? \n Magnetic force is one of the fundamental type of force , can it \nbe deflecting charged particles. Similarly current carrying Evaluation: \nQuestions raised to students \nElectric energy can be changed to mechanical energy, can the reverse \nenergy transformation possible , think about the possible circuit set up \nCreativity: \nDesign analog electric meters like ammeter, voltmeter, \ngalvanometer which can be operated on motor principle.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-QUESTIONS.txt\nhttp://www.thesecond school.com/agnelone/noida/m/lesson_plans/view/\nMagnetic -Effect -of-electric -current \n conductor can be pulled by a magnet or in turn magnet exert \nforce on a conductor placed near by. \n What is the magnitude and direction of force experienced by a \ncurrent carrying conductor in a magnetic field ? \n Magnet can exert force on a current carrying element, how can \nit bring rotational motion ? What should be the direction of \nmagnetic field? \n How can this rotational motion kept in continuation ? \n Design an electric motor ? What are the essential components \nof an electric motor ? \n How is an ac motor different from a dc motor ? \nWith just a C cell, paperclips, a small magnet, rubber band and a coil \nof wire, students can build an operating electric motor that illustrates \nthe role of electricity and magnetism in a simple, but fundamental, \ntechnology.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES (2).txt\n13. MAGNETIC EFFECT OF ELECTRIC OF ELECTRIC CURRENT \nSUB: SCIENCE (PHYSICS) CLASS: X", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES (2).txt\nB. SHORT ANSWER TYPE: - \n1. Write any three magnetic properties \n2. Write any three properties of magnetic field lines. \n3. State \u201cRight Hand thumb rule\u201d. Draw necessary diagram to explain it. \n4. With the help of diagram explain how a current carrying coil will behave like \nmagnetic \u2018North pole\u2019 and \u2018South pole\u2019. \n5. State the any one similarity and dissimilarity between solenoid and bar magnet. \n6. Explain Fleming\u2019s Left -hand rule. \n7. Explain Fleming\u2019s right hand rule. \n8. What is fuse? Which are the materials used to prepare it? \n9. What are the differences between mo tor and generator? \n10. What is the role of fuse connected in series? Why we have to use suitable \nrating fuse and should not replace with another rating material? \n11. What is electromagnet? How soft iron can be converted into electromagnet? \n12. State one main difference between A.C and D.C. Why A.C is preferred over \nD.C for long range transmission of electric power? Name one source each of D.C", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES (2).txt\n12. State one main difference between A.C and D.C. Why A.C is preferred over \nD.C for long range transmission of electric power? Name one source each of D.C \nand A.C. \n13. State the co nsequences that can lead to a short circuit. Or \nOne of the major cause of fire in office building is short circuiting. List three factors \nwhich may lead to the short circuit.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-PEDAGOGY.txt\n13. MAGNETIC EFFECT OF ELECTRIC CURRENT", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nMagnetic Effects of\nElectric Current13 CHAPTER\nIn the previous Chapter on \u2018Electricity\u2019 we learnt about the heating\neffects of electric current. What could be the other effects of electric\ncurrent? W e know that an electric curr ent-carrying wir e behaves like a\nmagnet. Let us perform the following Activity to reinforce it.\nActivity 13.1Activity 13.1Activity 13.1Activity 13.1Activity 13.1\n/square6Take a straight thick copper wire and place it\nbetween the points X and Y in an electric circuit,\nas shown in Fig. 13.1. The wire XY is kept\nperpendicular to the plane of paper .\n/square6Horizontally place a small compass near to this\ncopper wire. See the position of its needle.\n/square6Pass the current through the circuit by\ninserting the key into the plug.", + "13.1. The wire XY is kept\nperpendicular to the plane of paper .\n/square6Horizontally place a small compass near to this\ncopper wire. See the position of its needle.\n/square6Pass the current through the circuit by\ninserting the key into the plug.\n/square6Observe the change in the position of the\ncompass needle.Figure 13.1\nCompass needle is deflected on passing an electric\ncurrent through a metallic conductor\nWe see that the needle is deflected. What does it mean? It means that", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\ncompass needle.Figure 13.1\nCompass needle is deflected on passing an electric\ncurrent through a metallic conductor\nWe see that the needle is deflected. What does it mean? It means that\nthe electric current through the copper wire has produced a magnetic\neffect. Thus we can say that electricity and magnetism are linked to each\nother . Then, what about the r everse possibility of an electric ef fect of\nmoving magnets? In this Chapter we will study magnetic fields and such\nelectr omagnetic ef fects. W e shall also study about electr omagnets and\nelectric motors which involve the magnetic effect of electric current, and\nelectric generators which involve the electric effect of moving magnets.\nHans Christian Oersted (1777\u20131851)\nHans Christian Oersted, one of the leading scientists of the 19th\ncentury, played a crucial role in understanding electromagnetism . In\n1820 he accidentally discovered that a compass needle got deflected", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\ncentury, played a crucial role in understanding electromagnetism . In\n1820 he accidentally discovered that a compass needle got deflected\nwhen an electric current passed through a metallic wire placed nearby.\nThrough this observation Oersted showed that electricity and\nmagnetism were related phenomena. His research later created\ntechnologies such as the radio, television and fiber optics. The unit of\nmagnetic field str ength is named the oersted in his honor .Resistor\nLong straight\nconductor\n2020-21\nScience 22413.1 MAGNETIC FIELD AND FIELD LINES13.1 MAGNETIC FIELD AND FIELD LINES13.1 MAGNETIC FIELD AND FIELD LINES13.1 MAGNETIC FIELD AND FIELD LINES13.1 MAGNETIC FIELD AND FIELD LINES\nWe are familiar with the fact that a compass needle gets deflected when\nbrought near a bar magnet. A compass needle is, in fact, a small bar\nmagnet. The ends of the compass needle point approximately towards\nnorth and south directions. The end pointing towards north is called north", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nmagnet. The ends of the compass needle point approximately towards\nnorth and south directions. The end pointing towards north is called north\nseeking or north pole. The other end that points towards south is called\nsouth seeking or south pole. Through various activities we have observed\nthat like poles r epel, while unlike poles of magnets attract each other .\nQUESTION\n?1. Why does a compass needle get deflected when brought near\na bar magnet?\nActivity 13.2Activity 13.2Activity 13.2Activity 13.2Activity 13.2\n/square6Fix a sheet of white paper on a drawing\nboard using some adhesive material.\n/square6Place a bar magnet in the centre of it.\n/square6Sprinkle some iron filings uniformly\naround the bar magnet (Fig. 13.2). A\nsalt-sprinkler may be used for this\npurpose.\n/square6Now tap the board gently.\n/square6What do you observe?", + "/square6Sprinkle some iron filings uniformly\naround the bar magnet (Fig. 13.2). A\nsalt-sprinkler may be used for this\npurpose.\n/square6Now tap the board gently.\n/square6What do you observe?\nFigure 13.2Figure 13.2Figure 13.2Figure 13.2Figure 13.2\nIron filings near the bar magnet align\nthemselves along the field lines.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\n/square6Now tap the board gently.\n/square6What do you observe?\nFigure 13.2Figure 13.2Figure 13.2Figure 13.2Figure 13.2\nIron filings near the bar magnet align\nthemselves along the field lines.\nThe iron filings arrange themselves in a pattern as shown\nFig. 13.2. Why do the iron filings arrange in such a pattern? What does\nthis pattern demonstrate? The magnet exerts its influence in the region\nsurrounding it. Therefore the iron filings experience a force. The force\nthus exerted makes iron filings to arrange in a pattern. The region\nsurrounding a magnet, in which the force of the magnet can be detected,\nis said to have a magnetic field. The lines along which the iron filings\nalign themselves represent magnetic field lines.\nAre there other ways of obtaining magnetic field lines around a bar\nmagnet? Y es, you can yourself draw the field lines of a bar magnet.", + "The lines along which the iron filings\nalign themselves represent magnetic field lines.\nAre there other ways of obtaining magnetic field lines around a bar\nmagnet? Y es, you can yourself draw the field lines of a bar magnet.\nActivity 13.3Activity 13.3Activity 13.3Activity 13.3Activity 13.3\n/square6Take a small compass and a bar magnet.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nmagnet? Y es, you can yourself draw the field lines of a bar magnet.\nActivity 13.3Activity 13.3Activity 13.3Activity 13.3Activity 13.3\n/square6Take a small compass and a bar magnet.\n/square6Place the magnet on a sheet of white paper fixed on a drawing\nboard, using some adhesive material.\n/square6Mark the boundary of the magnet.\n/square6Place the compass near the north pole of the magnet. How does\nit behave? The south pole of the needle points towards the north\npole of the magnet. The north pole of the compass is directed\naway from the north pole of the magnet.\n2020-21\nMagnetic Effects of Electric Current 225Magnetic field is a quantity that has both direction and magnitude.\nThe direction of the magnetic field is taken to be the direction in which a\nnorth pole of the compass needle moves inside it. Therefore it is taken\nby convention that the field lines emerge from north pole and merge at\nthe south pole (note the arrows marked on the field lines in Fig. 13.4).", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nby convention that the field lines emerge from north pole and merge at\nthe south pole (note the arrows marked on the field lines in Fig. 13.4).\nInside the magnet, the direction of field lines is from its south pole to its\nnorth pole. Thus the magnetic field lines are closed curves.\nThe relative strength of the magnetic field is shown by the degree of\ncloseness of the field lines. The field is str onger , that is, the for ce acting\non the pole of another magnet placed is greater where the field lines are\ncrowded (see Fig. 13.4).\nNo two field-lines ar e found to cr oss each other . If they did, it would\nmean that at the point of intersection, the compass needle would point\ntowards two directions, which is not possible.", + "13.4).\nNo two field-lines ar e found to cr oss each other . If they did, it would\nmean that at the point of intersection, the compass needle would point\ntowards two directions, which is not possible.\n13.2 13.213.2 13.213.2 MA MAMAMAMAGNETIC FIELD DUE TO A CURRENTGNETIC FIELD DUE TO A CURRENTGNETIC FIELD DUE TO A CURRENTGNETIC FIELD DUE TO A CURRENTGNETIC FIELD DUE TO A CURRENT -\n-- --C CC CCARRYINGARRYINGARRYING ARRYINGARRYING", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\n-- --C CC CCARRYINGARRYINGARRYING ARRYINGARRYING\nCONDUCTORCONDUCTORCONDUCTORCONDUCTORCONDUCTOR\nIn Activity 13.1, we have seen that an electric current through a\nmetallic conductor produces a magnetic field around it. In order to\nfind the direction of the field produced let us repeat the activity in the\nfollowing way \u2013Figure 13.3Figure 13.3Figure 13.3Figure 13.3Figure 13.3\nDrawing a magnetic field line with the help of a\ncompass needle/square6Mark the position of two ends of the needle.\n/square6Now move the needle to a new position\nsuch that its south pole occupies the\nposition previously occupied by its north\npole.\n/square6In this way, proceed step by step till you\nreach the south pole of the magnet as\nshown in Fig. 13.3.\n/square6Join the points marked on the paper by a\nsmooth curve. This curve represents\na field line.", + "/square6In this way, proceed step by step till you\nreach the south pole of the magnet as\nshown in Fig. 13.3.\n/square6Join the points marked on the paper by a\nsmooth curve. This curve represents\na field line.\n/square6Repeat the above procedure and draw as\nmany lines as you can. Y ou will get a\npattern shown in Fig. 13.4. These lines\nrepresent the magnetic field around the", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\na field line.\n/square6Repeat the above procedure and draw as\nmany lines as you can. Y ou will get a\npattern shown in Fig. 13.4. These lines\nrepresent the magnetic field around the\nmagnet. These are known as magnetic\nfield lines.\n/square6Observe the deflection in the compass\nneedle as you move it along a field line.\nThe deflection increases as the needle is\nmoved towards the poles.\nFigure 13.4Figure 13.4Figure 13.4Figure 13.4Figure 13.4\nField lines around a bar magnet\n2020-21\nScience 22613.2.1 Magnetic Field due to a Current through a Straight\nConductor\nWhat determines the pattern of the magnetic field generated by a current\nthrough a conductor? Does the pattern depend on the shape of the\nconductor? W e shall investigate this with an activity.\nWe shall first consider the patter n of the magnetic field ar ound a\nstraight conductor carrying current.Activity 13.4Activity 13.4Activity 13.4Activity 13.4Activity 13.4", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nWe shall first consider the patter n of the magnetic field ar ound a\nstraight conductor carrying current.Activity 13.4Activity 13.4Activity 13.4Activity 13.4Activity 13.4\n/square6Take a long straight copper wire, two or three cells of 1.5 V each, and a plug key. Connect\nall of them in series as shown in Fig. 13.5 (a).\n/square6Place the straight wire parallel to and over a compass needle.\n/square6Plug the key in the circuit.\n/square6Observe the direction of deflection of the north pole of the needle. If the current flows from\nnorth to south, as shown in Fig. 13.5 (a), the north pole of the compass needle would move\ntowards the east.\n/square6Replace the cell connections in the circuit as shown in Fig. 13.5 (b). This would result in\nthe change of the direction of current through the copper wire, that is, from south to\nnorth.", + "/square6Replace the cell connections in the circuit as shown in Fig. 13.5 (b). This would result in\nthe change of the direction of current through the copper wire, that is, from south to\nnorth.\n/square6Observe the change in the dir ection of deflection of the needle. Y ou will see that now the", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nthe change of the direction of current through the copper wire, that is, from south to\nnorth.\n/square6Observe the change in the dir ection of deflection of the needle. Y ou will see that now the\nneedle moves in opposite direction, that is, towards the west [Fig. 13.5 (b)]. It means that\nthe direction of magnetic field produced by the electric current is also reversed.\nFigure 13.5Figure 13.5Figure 13.5Figure 13.5Figure 13.5 A simple electric circuit in which a straight copper wire is placed parallel to and over a compass\nneedle. The deflection in the needle becomes opposite when the direction of the current is reversed.(a) (b)\nActivity 13.5Activity 13.5Activity 13.5Activity 13.5Activity 13.5\n/square6Take a battery (12 V), a variable resistance (or a rheostat), an\nammeter (0\u20135 A), a plug key, connecting wires and a long straight\nthick copper wire.", + "/square6Insert the thick wire through the centre, normal to the plane of a\nrectangular cardboard. Take care that the cardboard is fixed and", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nthick copper wire.\n/square6Insert the thick wire through the centre, normal to the plane of a\nrectangular cardboard. Take care that the cardboard is fixed and\ndoes not slide up or down.\n2020-21\nMagnetic Effects of Electric Current 227What happens to the deflection of the compass needle placed at a\ngiven point if the current in the copper wire is changed? To see this, vary\nthe curr ent in the wire. W e find that the deflection in the needle also\nchanges. In fact, if the current is increased, the deflection also increases.\nIt indicates that the magnitude of the magnetic field produced at a given\npoint increases as the current through the wire increases.\nWhat happens to the deflection of the needle if the compass is moved\naway from the copper wire but the current through the wire remains the\nsame? To see this, now place the compass at a farther point from the\nconducting wire (say at point Q). What change do you observe? W e see", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nsame? To see this, now place the compass at a farther point from the\nconducting wire (say at point Q). What change do you observe? W e see\nthat the deflection in the needle decreases. Thus the magnetic field\nproduced by a given current in the conductor decreases as the distance\nfrom it increases. From Fig. 13.6, it can be noticed that the concentric\ncircles representing the magnetic field around a current-carrying straight\nwire become larger and larger as we move away from it.\n13.2.2 Right-Hand Thumb Rule\nA convenient way of finding the direction of magnetic field associated\nwith a current-carrying conductor is given in Fig. 13.7./square6Connect the copper wire vertically between the\npoints X and Y, as shown in Fig. 13.6 (a), in\nseries with the battery, a plug and key.\n/square6Sprinkle some iron filings uniformly on the\ncardboar d. (You may use a salt sprinkler for this\npurpose.)", + "13.6 (a), in\nseries with the battery, a plug and key.\n/square6Sprinkle some iron filings uniformly on the\ncardboar d. (You may use a salt sprinkler for this\npurpose.)\n/square6Keep the variable of the rheostat at a fixed\nposition and note the current through the\nammeter .", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\ncardboar d. (You may use a salt sprinkler for this\npurpose.)\n/square6Keep the variable of the rheostat at a fixed\nposition and note the current through the\nammeter .\n/square6Close the key so that a current flows through\nthe wire. Ensure that the copper wire placed\nbetween the points X and Y remains vertically\nstraight.\n/square6Gently tap the cardboard a few times. Observe\nthe patter n of the ir on filings. You would find\nthat the iron filings align themselves showing\na pattern of concentric circles around the\ncopper wire (Fig. 13.6).\n/square6What do these concentric circles represent?\nThey represent the magnetic field lines.\n/square6How can the direction of the magnetic field be\nfound? Place a compass at a point (say P) over\na circle. Observe the direction of the needle. The\ndirection of the north pole of the compass\nneedle would give the direction of the field lines\nproduced by the electric current through the\nstraight wir e at point P . Show the dir ection by\nan arrow.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\ndirection of the north pole of the compass\nneedle would give the direction of the field lines\nproduced by the electric current through the\nstraight wir e at point P . Show the dir ection by\nan arrow.\n/square6Does the direction of magnetic field lines get\nreversed if the direction of current through the\nstraight copper wire is reversed? Check it.\nFigure 13.6Figure 13.6Figure 13.6Figure 13.6Figure 13.6\n(a) A pattern of concentric circles indicating\nthe field lines of a magnetic field around a\nstraight conducting wire. The arrows in the\ncircles show the direction of the field lines.\n(b) A close up of the pattern obtained.(a)\n(b)Variable\nresistance\n2020-21\nScience 228Imagine that you are holding a current-carrying straight\nconductor in your right hand such that the thumb points towards\nthe direction of current. Then your fingers will wrap around the\nconductor in the direction of the field lines of the magnetic field, as\nshown in Fig. 13.7.", + "Then your fingers will wrap around the\nconductor in the direction of the field lines of the magnetic field, as\nshown in Fig. 13.7. This is known as the right-hand thumb rule *.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nconductor in the direction of the field lines of the magnetic field, as\nshown in Fig. 13.7. This is known as the right-hand thumb rule *.\nExample 13.1\nA current through a horizontal power line flows in east to west\ndirection. What is the direction of magnetic field at a point directly\nbelow it and at a point directly above it?\nSolution\nThe current is in the east-west direction. Applying the right-hand\nthumb rule, we get that the magnetic field (at any point below or\nabove the wire) turns clockwise in a plane perpendicular to the wire,\nwhen viewed from the east end, and anti-clockwise, when viewed\nfrom the west end.Figure 13.7Figure 13.7Figure 13.7Figure 13.7Figure 13.7\nRight-hand thumb rule\n*This rule is also called Maxwell\u2019s corkscrew rule.", + "If we consider ourselves driving a\ncorkscrew in the direction of the current, then the direction of the rotation of\ncorkscrew is the direction of the magnetic field.13.2.3 Magnetic Field due to a Current through a\nCircular Loop", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\ncorkscrew in the direction of the current, then the direction of the rotation of\ncorkscrew is the direction of the magnetic field.13.2.3 Magnetic Field due to a Current through a\nCircular Loop\nWe have so far observed the patter n of the magnetic field lines\nproduced around a current-carrying straight wire. Suppose\nthis straight wire is bent in the form of a circular loop and a\ncurrent is passed through it. How would the magnetic field\nlines look like? W e know that the magnetic field produced\nby a current-carrying straight wire depends inversely on the\ndistance from it. Similarly at every point of a current-carrying\ncircular loop, the concentric circles representing the magnetic\nfield around it would become larger and larger as we move\naway from the wire (Fig. 13.8). By the time we reach at the\ncentre of the circular loop, the arcs of these big circles would\nappear as straight lines. Every point on the wire carrying\ncurrent would give rise to the magnetic field appearing as", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\ncentre of the circular loop, the arcs of these big circles would\nappear as straight lines. Every point on the wire carrying\ncurrent would give rise to the magnetic field appearing as\nstraight lines at the center of the loop. By applying the right\nhand rule, it is easy to check that every section of the wire\ncontributes to the magnetic field lines in the same direction\nwithin the loop.Figure 13.8Figure 13.8Figure 13.8Figure 13.8Figure 13.8\nMagnetic field lines of the field\nproduced by a current-carrying\ncircular loop?QUESTIONS\n1. Draw magnetic field lines around a bar magnet.\n2. List the properties of magnetic field lines.\n3. Why don\u2019t two magnetic field lines intersect each other?\n2020-21\nMagnetic Effects of Electric Current 229We know that the magnetic field produced by a curr ent-carrying\nwire at a given point depends directly on the current passing through it.\nTherefore, if there is a circular coil having n turns, the field produced is", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nwire at a given point depends directly on the current passing through it.\nTherefore, if there is a circular coil having n turns, the field produced is\nn times as large as that produced by a single turn. This is because the\ncurrent in each circular turn has the same direction, and the field due to\neach turn then just adds up.\nActivity 13.6Activity 13.6Activity 13.6Activity 13.6Activity 13.6\n/square6Take a rectangular cardboard having two holes.\nInsert a circular coil having large number of turns\nthrough them, normal to the plane of the cardboard.\n/square6Connect the ends of the coil in series with a battery,\na key and a rheostat, as shown in Fig. 13.9.\n/square6Sprinkle iron filings uniformly on the cardboard.\n/square6Plug the key.\n/square6Tap the cardboard gently a few times.", + "/square6Connect the ends of the coil in series with a battery,\na key and a rheostat, as shown in Fig. 13.9.\n/square6Sprinkle iron filings uniformly on the cardboard.\n/square6Plug the key.\n/square6Tap the cardboard gently a few times. Note the\npattern of the iron filings that emerges on the\ncardboard.Figure 13.9Figure 13.9Figure 13.9Figure 13.9Figure 13.9\nMagnetic field produced by a current-\ncarrying circular coil.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\npattern of the iron filings that emerges on the\ncardboard.Figure 13.9Figure 13.9Figure 13.9Figure 13.9Figure 13.9\nMagnetic field produced by a current-\ncarrying circular coil.\n13.2.4 Magnetic Field due to a Current in a Solenoid\nA coil of many circular turns of insulated copper wire wrapped\nclosely in the shape of a cylinder is called a solenoid. The pattern\nof the magnetic field lines around a current-carrying solenoid is\nshown in Fig. 13.10. Compare the pattern of the field with the\nmagnetic field around a bar magnet (Fig. 13.4). Do they look\nsimilar? Y es, they are similar . In fact, one end of the solenoid\nbehaves as a magnetic north pole, while the other behaves as the\nsouth pole. The field lines inside the solenoid are in the form of\nparallel straight lines. This indicates that the magnetic field is\nthe same at all points inside the solenoid.", + "The field lines inside the solenoid are in the form of\nparallel straight lines. This indicates that the magnetic field is\nthe same at all points inside the solenoid. That is, the field is\nuniform inside the solenoid.\nA strong magnetic field produced inside a solenoid can be", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nthe same at all points inside the solenoid. That is, the field is\nuniform inside the solenoid.\nA strong magnetic field produced inside a solenoid can be\nused to magnetise a piece of magnetic material, like soft iron,\nwhen placed inside the coil (Fig. 13.11). The magnet so formed is\ncalled an electromagnet.Figure 13.10Figure 13.10Figure 13.10Figure 13.10Figure 13.10\nField lines of the magnetic field\nthrough and around a current\ncarrying solenoid.\nFigure 13.11Figure 13.11Figure 13.11Figure 13.11Figure 13.11\nA current-carrying solenoid coil\nis used to magnetise steel rod\ninside it \u2013 an electromagnet.QUESTIONS\n?1. Consider a circular loop of wire lying in\nthe plane of the table. Let the current\npass through the loop clockwise. Apply\nthe right-hand rule to find out the\ndirection of the magnetic field inside\nand outside the loop.\n2.", + "Consider a circular loop of wire lying in\nthe plane of the table. Let the current\npass through the loop clockwise. Apply\nthe right-hand rule to find out the\ndirection of the magnetic field inside\nand outside the loop.\n2. The magnetic field in a given region is\nuniform. Draw a diagram to represent it.\n2020-21", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nthe right-hand rule to find out the\ndirection of the magnetic field inside\nand outside the loop.\n2. The magnetic field in a given region is\nuniform. Draw a diagram to represent it.\n2020-21\nScience 23013.3 13.313.3 13.313.3 FORCE ON A CURRENTFORCE ON A CURRENTFORCE ON A CURRENTFORCE ON A CURRENTFORCE ON A CURRENT - ----C\nCC CCARRYING CONDUCTORARRYING CONDUCTORARRYING CONDUCTORARRYING CONDUCTORARRYING CONDUCTOR\nIN A MAGNETIC FIELDIN A MAGNETIC FIELDIN A MAGNETIC FIELDIN A MAGNETIC FIELDIN A MAGNETIC FIELD\nWe have lear nt that an electric curr ent flowing thr\nough a conductor\nproduces a magnetic field. The field so produced exerts a force on a\nmagnet placed in the vicinity of the conductor . French scientist Andre\nMarie Ampere (1775\u20131836) suggested that the magnet must also exert\nan equal and opposite for ce on the curr ent-carrying conductor .", + "The field so produced exerts a force on a\nmagnet placed in the vicinity of the conductor . French scientist Andre\nMarie Ampere (1775\u20131836) suggested that the magnet must also exert\nan equal and opposite for ce on the curr ent-carrying conductor . The for ce\ndue to a magnetic field acting on a current-carrying conductor can be", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nan equal and opposite for ce on the curr ent-carrying conductor . The for ce\ndue to a magnetic field acting on a current-carrying conductor can be\ndemonstrated through the following activity.3. Choose the correct option.\nThe magnetic field inside a long straight solenoid-carrying current\n(a) is zero.\n(b) decreases as we move towards its end.\n(c) increases as we move towards its end.\n(d) is the same at all points.\nActivity 13.7Activity 13.7Activity 13.7Activity 13.7Activity 13.7\n/square6Take a small aluminium rod AB (of about 5 cm). Using\ntwo connecting wires suspend it horizontally from a\nstand, as shown in Fig. 13.12.\n/square6Place a strong horse-shoe magnet in such a way that\nthe rod lies between the two poles with the magnetic\nfield directed upwards. For this put the north pole of\nthe magnet vertically below and south pole vertically\nabove the aluminium rod (Fig. 13.12).", + "/square6Place a strong horse-shoe magnet in such a way that\nthe rod lies between the two poles with the magnetic\nfield directed upwards. For this put the north pole of\nthe magnet vertically below and south pole vertically\nabove the aluminium rod (Fig. 13.12).\n/square6Connect the aluminium rod in series with a battery,\na key and a rheostat.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nthe magnet vertically below and south pole vertically\nabove the aluminium rod (Fig. 13.12).\n/square6Connect the aluminium rod in series with a battery,\na key and a rheostat.\n/square6Now pass a current through the aluminium rod from\nend B to end A.\n/square6What do you observe? It is observed that the rod is\ndisplaced towar ds the left. Y ou will notice that the r od\ngets displaced.\n/square6Reverse the direction of current flowing through the\nrod and observe the direction of its displacement. It is\nnow towards the right.\nWhy does the rod get displaced?Figure 13.12Figure 13.12Figure 13.12Figure 13.12Figure 13.12\nA current-carrying rod, AB, experiences\na force perpendicular to its length and\nthe magnetic field. Support for the\nmagnet is not shown here, for simplicity.\nThe displacement of the rod in the above activity suggests that a\nforce is exerted on the current-carrying aluminium rod when it is placed\nin a magnetic field.", + "Support for the\nmagnet is not shown here, for simplicity.\nThe displacement of the rod in the above activity suggests that a\nforce is exerted on the current-carrying aluminium rod when it is placed\nin a magnetic field. It also suggests that the direction of force is also", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nforce is exerted on the current-carrying aluminium rod when it is placed\nin a magnetic field. It also suggests that the direction of force is also\nreversed when the direction of current through the conductor is reversed.\nNow change the direction of field to vertically downwards by\ninterchanging the two poles of the magnet. It is once again observed that\n2020-21\nMagnetic Effects of Electric Current 231the direction of force acting on the current-carrying rod gets reversed. It\nshows that the direction of the force on the conductor depends upon the\ndirection of current and the direction of the magnetic field. Experiments\nhave shown that the displacement of the rod is largest (or the magnitude\nof the force is the highest) when the direction of current is at right angles\nto the direction of the magnetic field. In such a condition we can use a\nsimple rule to find the dir ection of the for ce on the conductor .\nIn Activity 13.7, we considered the", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nto the direction of the magnetic field. In such a condition we can use a\nsimple rule to find the dir ection of the for ce on the conductor .\nIn Activity 13.7, we considered the\ndirection of the current and that of the magnetic\nfield perpendicular to each other and found\nthat the force is perpendicular to both of them.\nThe three directions can be illustrated through\na simple rule, called Fleming\u2019s left-hand rule.\nAccording to this rule, stretch the thumb,\nforefinger and middle finger of your left hand\nsuch that they are mutually perpendicular\n(Fig. 13.13). If the first finger points in the\ndirection of magnetic field and the second\nfinger in the direction of current, then the\nthumb will point in the direction of motion or\nthe for ce acting on the conductor .\nDevices that use current-carrying conductors and magnetic fields\ninclude electric motor , electric generator , loudspeakers, microphones\nand measuring instruments. In the next few sections we shall study", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\ninclude electric motor , electric generator , loudspeakers, microphones\nand measuring instruments. In the next few sections we shall study\nabout electric motors and generators.\nExample 13.2\nAn electron enters a magnetic field at right angles to it, as shown\nin Fig. 13.14. The direction of force acting on the electron will be\n(a) to the right.\n(b)to the left.\n(c)out of the page.\n(d)into the page.\nSolution\nAnswer is option (d). The direction of force is perpendicular to the\ndirection of magnetic field and current as given by Fleming\u2019s left hand\nrule. Recall that the direction of current is taken opposite to the direction\nof motion of electrons. The force is therefore directed into the page.Figure 13.13Figure 13.13Figure 13.13Figure 13.13Figure 13.13\nFleming\u2019s left-hand rule\nFigure 13.14Figure 13.14Figure 13.14Figure 13.14Figure 13.14\nQUESTIONS\n?1. Which of the following property of a proton can change while it moves", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nFleming\u2019s left-hand rule\nFigure 13.14Figure 13.14Figure 13.14Figure 13.14Figure 13.14\nQUESTIONS\n?1. Which of the following property of a proton can change while it moves\nfreely in a magnetic field? (Ther e may be mor e than one corr ect answer .)\n(a) mass (b) speed\n(c) velocity (d) momentum\n2020-21\nScience 2322. In Activity 13.7, how do we think the displacement of rod AB will be\naffected if (i) current in rod AB is increased; (ii) a stronger horse-shoe\nmagnet is used; and (iii) length of the rod AB is increased?\n3. A positively-charged particle (alpha-particle) projected towards west is\ndeflected towards north by a magnetic field. The direction of magnetic\nfield is\n(a) towards south (b) towards east\n(c) downward (d) upward\nMagnetism in medicine\nAn electric current always produces a magnetic field.", + "The direction of magnetic\nfield is\n(a) towards south (b) towards east\n(c) downward (d) upward\nMagnetism in medicine\nAn electric current always produces a magnetic field. Even weak ion currents that\ntravel along the nerve cells in our body produce magnetic fields. When we touch", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nMagnetism in medicine\nAn electric current always produces a magnetic field. Even weak ion currents that\ntravel along the nerve cells in our body produce magnetic fields. When we touch\nsomething, our nerves carry an electric impulse to the muscles we need to use. This\nimpulse produces a temporary magnetic field. These fields are very weak and are about\none-billionth of the earth\u2019s magnetic field. Two main organs in the human body where\nthe magnetic field produced is significant , are the heart and the brain. The magnetic field\ninside the body forms the basis of obtaining the images of different body parts. This is\ndone using a technique called Magnetic Resonance Imaging (MRI). Analysis of these\nimages helps in medical diagnosis. Magnetism has, thus, got important uses in medicine.\n13.4 ELECTRIC MOTOR13.4 ELECTRIC MOTOR13.4 ELECTRIC MOTOR13.4 ELECTRIC MOTOR13.4 ELECTRIC MOTOR\nAn electric motor is a rotating device that converts electrical energy to", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\n13.4 ELECTRIC MOTOR13.4 ELECTRIC MOTOR13.4 ELECTRIC MOTOR13.4 ELECTRIC MOTOR13.4 ELECTRIC MOTOR\nAn electric motor is a rotating device that converts electrical energy to\nmechanical energy. Electric motor is used as an important component\nin electric fans, refrigerators, mixers, washing machines, computers, MP3\nplayers etc. Do you know how an electric motor works?\nAn electric motor , as shown in Fig. 13.15, consists of a r ectangular\ncoil ABCD of insulated copper wire. The coil is placed between the two\npoles of a magnetic field such that the arm AB and CD are perpendicular\nto the direction of the magnetic field. The\nends of the coil are connected to the two\nhalves P and Q of a split ring. The inner sides\nof these halves are insulated and attached\nto an axle. The external conducting edges of\nP and Q touch two conducting stationary\nbrushes X and Y, respectively, as shown in\nthe Fig. 13.15.\nCurrent in the coil ABCD enters from the\nsource battery through conducting brush", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nP and Q touch two conducting stationary\nbrushes X and Y, respectively, as shown in\nthe Fig. 13.15.\nCurrent in the coil ABCD enters from the\nsource battery through conducting brush\nX and flows back to the battery through\nbrush Y . Notice that the curr ent in ar m AB\nof the coil flows from A to B. In arm CD it\nflows from C to D, that is, opposite to the\ndirection of current through arm AB. On\napplying Fleming\u2019s left hand rule for the\ndirection of force on a current-carryingFigure 13.15\nA simple electric motor\nMore to Know!\nN\n2020-21\nMagnetic Effects of Electric Current 233conductor in a magnetic field (see Fig. 13.13). W e find that the for ce\nacting on arm AB pushes it downwards while the force acting on arm\nCD pushes it upwards. Thus the coil and the axle O, mounted free to\nturn about an axis, rotate anti-clockwise. At half rotation, Q makes\ncontact with the brush X and P with brush Y .", + "Thus the coil and the axle O, mounted free to\nturn about an axis, rotate anti-clockwise. At half rotation, Q makes\ncontact with the brush X and P with brush Y . Therefor e the current in\nthe coil gets reversed and flows along the path DCBA. A device that", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\ncontact with the brush X and P with brush Y . Therefor e the current in\nthe coil gets reversed and flows along the path DCBA. A device that\nreverses the direction of flow of current through a circuit is called a\ncommutator . In electric motors, the split ring acts as a commutator . The\nreversal of current also reverses the direction of force acting on the two\narms AB and CD. Thus the arm AB of the coil that was earlier pushed\ndown is now pushed up and the arm CD previously pushed up is now\npushed down. Therefore the coil and the axle rotate half a turn more in\nthe same direction. The reversing of the current is repeated at each half\nrotation, giving rise to a continuous rotation of the coil and to the axle.\nThe commercial motors use (i) an electromagnet in place of permanent\nmagnet; (ii) large number of turns of the conducting wire in the current-\ncarrying coil; and (iii) a soft iron core on which the coil is wound. The", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nmagnet; (ii) large number of turns of the conducting wire in the current-\ncarrying coil; and (iii) a soft iron core on which the coil is wound. The\nsoft iron core, on which the coil is wound, plus the coils, is called an\narmatur e. This enhances the power of the motor .\nActivity 13.8Activity 13.8Activity 13.8Activity 13.8Activity 13.8QUESTIONS\n?1. State Fleming\u2019s left-hand rule.\n2. What is the principle of an electric motor?\n3. What is the role of the split ring in an electric motor?\n13.5 ELECTROMAGNETIC INDUCTION13.5 ELECTROMAGNETIC INDUCTION13.5 ELECTROMAGNETIC INDUCTION13.5 ELECTROMAGNETIC INDUCTION13.5 ELECTROMAGNETIC INDUCTION\nWe have studied that when a current-carrying conductor is placed in a\nmagnetic field such that the direction of current is perpendicular to the\nmagnetic field, it experiences a force.", + "This force causes the conductor to\nmove. Now let us imagine a situation in which a conductor is moving\ninside a magnetic field or a magnetic field is changing around a fixed", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nmove. Now let us imagine a situation in which a conductor is moving\ninside a magnetic field or a magnetic field is changing around a fixed\nconductor . What will happen? This was first studied by English physicist\nMichael Faraday. In 1831, Faraday made an important breakthrough\nby discovering how a moving magnet can be used to generate electric\ncurrents. To observe this effect, let us perform the following activity.\n/square6Take a coil of wire AB having a large number of turns.\n/square6Connect the ends of the coil to a galvanometer as shown in\nFig. 13.16.\n/square6Take a strong bar magnet and move its north pole towards the\nend B of the coil. Do you find any change in the galvanometer\nneedle?\n2020-21\nScience 234/square6There is a momentary deflection in the needle of\nthe galvanometer , say to the right. This indicates\nthe presence of a current in the coil AB. The\ndeflection becomes zero the moment the motion\nof the magnet stops.\n/square6Now withdraw the north pole of the magnet away", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nthe presence of a current in the coil AB. The\ndeflection becomes zero the moment the motion\nof the magnet stops.\n/square6Now withdraw the north pole of the magnet away\nfrom the coil. Now the galvanometer is deflected\ntoward the left, showing that the current is now\nset up in the direction opposite to the first.\n/square6Place the magnet stationary at a point near to\nthe coil, keeping its north pole towards the end\nB of the coil. W e see that the galvanometer needle\ndeflects toward the right when the coil is moved\ntowards the north pole of the magnet. Similarly\nthe needle moves toward left when the coil is\nmoved away.\n/square6When the coil is kept stationary with respect to\nthe magnet, the deflection of the galvanometer\ndrops to zero.", + "Similarly\nthe needle moves toward left when the coil is\nmoved away.\n/square6When the coil is kept stationary with respect to\nthe magnet, the deflection of the galvanometer\ndrops to zero. What do you conclude from this\nactivity?Figure 13.16Figure 13.16Figure 13.16Figure 13.16Figure 13.16\nMoving a magnet towards a coil sets up a\ncurrent in the coil circuit, as indicated by\ndeflection in the galvanometer needle.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nactivity?Figure 13.16Figure 13.16Figure 13.16Figure 13.16Figure 13.16\nMoving a magnet towards a coil sets up a\ncurrent in the coil circuit, as indicated by\ndeflection in the galvanometer needle.\nA galvanometer is an instrument that can detect\nthe presence of a current in a circuit. The pointer\nremains at zero (the centre of the scale) for zero\ncurrent flowing through it. It can deflect either\nto the left or to the right of the zero mark\ndepending on the direction of current.\nYou can also check that if you had moved south pole of the magnet\ntowards the end B of the coil, the deflections in the galvanometer would\njust be opposite to the previous case. When the coil and the magnet are\nboth stationary, ther e is no deflection in the galvanometer . It is, thus,\nclear from this activity that motion of a magnet with respect to the coil\nproduces an induced potential difference, which sets up an induced\nelectric current in the circuit.\nMichael Faraday (1791\u20131867)", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nproduces an induced potential difference, which sets up an induced\nelectric current in the circuit.\nMichael Faraday (1791\u20131867)\nMichael Faraday was an experimental physicist. He had no formal\neducation. He worked in a book-binding shop during his early years.\nHe used to read books that came for binding. This way Faraday\ndeveloped his interest in science. He got an opportunity to listen to\nsome public lectures by Humphrey Davy of Royal Institute. He made\ncareful notes of Davy\u2019s lectures and sent them to Davy. Soon he was\nmade an assistant in Davy\u2019s laboratory at the Royal Institute. Faraday\nmade several path-breaking discoveries that include electromagnetic\ninduction and the laws of electrolysis. Several universities conferred\non him the honorary degrees but he turned down such honours. Faraday loved his\nscience work mor e than any honour .\n2020-21\nMagnetic Effects of Electric Current 235Let us now perform a variation of Activity 13.8 in which the moving", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nscience work mor e than any honour .\n2020-21\nMagnetic Effects of Electric Current 235Let us now perform a variation of Activity 13.8 in which the moving\nmagnet is replaced by a current-carrying coil and the current in the coil\ncan be varied.\nActivity 13.9Activity 13.9Activity 13.9Activity 13.9Activity 13.9\n/square6Take two different coils of copper wire having\nlarge number of turns (say 50 and 100 turns\nrespectively). Insert them over a non-conducting\ncylindrical r oll, as shown in Fig. 13.17. (Y ou may\nuse a thick paper roll for this purpose.)\n/square6Connect the coil-1, having larger number of\nturns, in series with a battery and a plug key.\nAlso connect the other coil-2 with a\ngalvanometer as shown.\n/square6Plug in the key. Observe the galvanometer . Is\nthere a deflection in its needle?", + "Also connect the other coil-2 with a\ngalvanometer as shown.\n/square6Plug in the key. Observe the galvanometer . Is\nthere a deflection in its needle? Y ou will observe\nthat the needle of the galvanometer instantly\njumps to one side and just as quickly returns to\nzero, indicating a momentary current in coil-2.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nthere a deflection in its needle? Y ou will observe\nthat the needle of the galvanometer instantly\njumps to one side and just as quickly returns to\nzero, indicating a momentary current in coil-2.\n/square6Disconnect coil-1 fr om the battery. Y ou will\nobserve that the needle momentarily moves, but\nto the opposite side. It means that now the\ncurrent flows in the opposite direction in coil-2.Figure 13.17Figure 13.17Figure 13.17Figure 13.17Figure 13.17\nCurrent is induced in coil-2 when current\nin coil-1 is changed\nIn this activity we observe that as soon as the current in coil-1 reaches\neither a steady value or zero, the galvanometer in coil-2 shows no\ndeflection.\nFrom these observations, we conclude that a potential difference is\ninduced in the coil-2 whenever the electric current through the coil\u20131\nis changing (starting or stopping). Coil-1 is called the primary coil and\ncoil-2 is called the secondary coil. As the current in the first coil changes,", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nis changing (starting or stopping). Coil-1 is called the primary coil and\ncoil-2 is called the secondary coil. As the current in the first coil changes,\nthe magnetic field associated with it also changes. Thus the magnetic\nfield lines around the secondary coil also change. Hence the change in\nmagnetic field lines associated with the secondary coil is the cause of\ninduced electric current in it. This process, by which a changing\nmagnetic field in a conductor induces a curr ent in another conductor ,\nis called electromagnetic induction. In practice we can induce current\nin a coil either by moving it in a magnetic field or by changing the\nmagnetic field around it. It is convenient in most situations to move\nthe coil in a magnetic field.\nThe induced current is found to be the highest when\nthe direction of motion of the coil is at right angles to the\nmagnetic field. In this situation, we can use a simple rule\nto know the direction of the induced current. Stretch the", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nthe direction of motion of the coil is at right angles to the\nmagnetic field. In this situation, we can use a simple rule\nto know the direction of the induced current. Stretch the\nthumb, forefinger and middle finger of right hand so that\nthey ar e perpendicular to each other , as shown in\nFig. 13.18. If the forefinger indicates the direction of the\nmagnetic field and the thumb shows the direction of motion\nof conductor , then the middle finger will show the dir ection\nof induced current. This simple rule is called Fleming\u2019s\nright-hand rule.Figure 13.18Figure 13.18Figure 13.18Figure 13.18Figure 13.18\nFleming\u2019s right-hand rule\n2020-21\nScience 23613.6 ELECTRIC GENER13.6 ELECTRIC GENER13.6 ELECTRIC GENER13.6 ELECTRIC GENER13.6 ELECTRIC GENER A AAAATORTORTORTORTOR\nBased on the phenomenon of electromagnetic induction, the experiments\nstudied above generate induced current, which is usually very small.", + "This principle is also employed to produce large currents for use in homes", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nstudied above generate induced current, which is usually very small.\nThis principle is also employed to produce large currents for use in homes\nand industry. In an electric generator , mechanical ener gy is used to r otate\na conductor in a magnetic field to produce electricity.\nAn electric generator , as shown in Fig. 13.19, consists of a r otating\nrectangular coil ABCD placed between the two poles of a permanent\nmagnet. The two ends of this coil are connected to\nthe two rings R1 and R2. The inner side of these\nrings are made insulated. The two conducting\nstationary brushes B1 and B2 are kept pressed\nseparately on the rings R1 and R2, respectively. The\ntwo rings R1 and R2 are internally attached to an\naxle. The axle may be mechanically rotated from\noutside to rotate the coil inside the magnetic field.\nOuter ends of the two brushes are connected to\nthe galvanometer to show the flow of current in the\ngiven external circuit.\nWhen the axle attached to the two rings is", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nOuter ends of the two brushes are connected to\nthe galvanometer to show the flow of current in the\ngiven external circuit.\nWhen the axle attached to the two rings is\nrotated such that the arm AB moves up (and the\narm CD moves down) in the magnetic field produced\nby the permanent magnet. Let us say the coil ABCD\nis rotated clockwise in the arrangement shown in Fig. 13.19. By applying\nFleming\u2019s right-hand rule, the induced currents are set up in these arms\nalong the directions AB and CD. Thus an induced current flows in the\ndirection ABCD. If there are larger numbers of turns in the coil, the current\ngenerated in each turn adds up to give a large current through the coil.\nThis means that the current in the external circuit flows from B2 to B1.\nAfter half a rotation, arm CD starts moving up and AB moving down.\nAs a result, the directions of the induced currents in both the arms\nchange, giving rise to the net induced current in the direction DCBA.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nAs a result, the directions of the induced currents in both the arms\nchange, giving rise to the net induced current in the direction DCBA.\nThe current in the external circuit now flows from B1 to B2. Thus after\nevery half rotation the polarity of the current in the respective arms\nchanges. Such a current, which changes direction after equal intervals\nof time, is called an alternating current (abbreviated as AC). This device\nis called an AC generator .\nTo get a direct current (DC, which does not change its direction with\ntime), a split-ring type commutator must be used. With this arrangement,\none brush is at all times in contact with the arm moving up in the field,\nwhile the other is in contact with the ar m moving down. W e have seen\nthe working of a split ring commutator in the case of an electric motorQUESTION\n?1. Explain different ways to induce current in a coil.\nFigure 13.19Figure 13.19Figure 13.19Figure 13.19Figure 13.19\nIllustration of the\nprinciple of electric\ngenerator", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\n?1. Explain different ways to induce current in a coil.\nFigure 13.19Figure 13.19Figure 13.19Figure 13.19Figure 13.19\nIllustration of the\nprinciple of electric\ngenerator\n2020-21\nMagnetic Effects of Electric Current 237(see Fig. 13.15). Thus a unidirectional current is produced. The generator\nis thus called a DC generator .\nThe difference between the direct and alternating currents is that the\ndirect current always flows in one direction, whereas the alternating\ncurrent reverses its direction periodically. Most power stations\nconstructed these days produce AC. In India, the AC changes direction\nafter every 1/100 second, that is, the frequency of AC is 50 Hz. An\nimportant advantage of AC over DC is that electric power can be\ntransmitted over long distances without much loss of energy.\nQUESTIONS\n?1. State the principle of an electric generator .\n2. Name some sources of direct current.\n3. Which sources produce alternating current?\n4. Choose the correct option.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nQUESTIONS\n?1. State the principle of an electric generator .\n2. Name some sources of direct current.\n3. Which sources produce alternating current?\n4. Choose the correct option.\nA rectangular coil of copper wires is rotated in a magnetic field. The\ndirection of the induced current changes once in each\n(a) two revolutions (b) one revolution\n(c) half revolution (d) one-fourth revolution\n13.7 DOMESTIC ELECTRIC CIRCUITS13.7 DOMESTIC ELECTRIC CIRCUITS13.7 DOMESTIC ELECTRIC CIRCUITS13.7 DOMESTIC ELECTRIC CIRCUITS13.7 DOMESTIC ELECTRIC CIRCUITS\nIn our homes, we receive supply of electric power through a main supply\n(also called mains), either supported through overhead electric poles or\nby underground cables. One of the wires in this supply, usually with\nred insulation cover , is called live wir e (or positive). Another wir e, with\nblack insulation, is called neutral wire (or negative). In our country, the\npotential dif ference between the two is 220 V .", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nblack insulation, is called neutral wire (or negative). In our country, the\npotential dif ference between the two is 220 V .\nAt the meter -boar d in the house, these wir es pass into an electricity\nmeter through a main fuse. Through the main switch they are connected\nto the line wires in the house. These wires supply electricity to separate\ncircuits within the house. Often, two separate circuits are used, one of\n15 A current rating for appliances with higher power ratings such as\ngeysers, air coolers, etc. The other circuit is of 5 A current rating for\nbulbs, fans, etc. The earth wir e, which has insulation of gr een colour , is\nusually connected to a metal plate deep in the earth near the house.\nThis is used as a safety measure, especially for those appliances that\nhave a metallic body, for example, electric pr ess, toaster , table fan,\nrefrigerator , etc. The metallic body is connected to the earth wir e, which\nprovides a low-resistance conducting path for the current. Thus, it", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nrefrigerator , etc. The metallic body is connected to the earth wir e, which\nprovides a low-resistance conducting path for the current. Thus, it\nensures that any leakage of current to the metallic body of the appliance\nkeeps its potential to that of the earth, and the user may not get a severe\nelectric shock.\n2020-21\nScience 238Figure 13.20 gives a schematic diagram of one of the common\ndomestic circuits. In each separate circuit, different appliances can be\nconnected across the live and neutral wires. Each appliance has a\nseparate switch to \u2018ON\u2019/\u2018OFF\u2019 the flow of current through it. In order\nthat each appliance has equal potential difference, they are connected\nparallel to each other .\nElectric fuse is an important component of all domestic cir cuits. W e\nhave already studied the principle and working of a fuse in the previous\nchapter (see Section 12.7). A fuse in a circuit prevents damage to the\nappliances and the circuit due to overloading. Overloading can occur", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nchapter (see Section 12.7). A fuse in a circuit prevents damage to the\nappliances and the circuit due to overloading. Overloading can occur\nwhen the live wire and the neutral wire come into direct contact. (This\noccurs when the insulation of wires is damaged or there is a fault in the\nappliance.) In such a situation, the current in the circuit abruptly\nincreases. This is called short-circuiting. The use of an electric fuse\nprevents the electric circuit and the appliance from a possible damage\nby stopping the flow of unduly high electric current. The Joule heating\nthat takes place in the fuse melts it to break the electric circuit.\nOverloading can also occur due to an accidental hike in the supply\nvoltage. Sometimes overloading is caused by connecting too many\nappliances to a single socket.Figure 13.20Figure 13.20Figure 13.20Figure 13.20Figure 13.20 A schematic diagram of one of the common domestic circuits\nQUESTIONS\n?1. Name two safety measures commonly used in electric circuits and", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nQUESTIONS\n?1. Name two safety measures commonly used in electric circuits and\nappliances.\n2. An electric oven of 2 kW power rating is operated in a domestic electric\ncircuit (220 V) that has a current rating of 5 A. What result do you\nexpect? Explain.\n3. What precaution should be taken to avoid the overloading of domestic\nelectric circuits?\n2020-21\nMagnetic Effects of Electric Current 239What y ou hav e learnt\n/square6A compass needle is a small magnet. Its one end, which points towards north, is\ncalled a north pole, and the other end, which points towards south, is called a\nsouth pole.\n/square6A magnetic field exists in the region surrounding a magnet, in which the force of\nthe magnet can be detected.\n/square6Field lines are used to represent a magnetic field. A field line is the path along\nwhich a hypothetical free north pole would tend to move. The direction of the\nmagnetic field at a point is given by the direction that a north pole placed at that", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nwhich a hypothetical free north pole would tend to move. The direction of the\nmagnetic field at a point is given by the direction that a north pole placed at that\npoint would take. Field lines are shown closer together where the magnetic field is\ngreater.\n/square6A metallic wire carrying an electric current has associated with it a magnetic field.\nThe field lines about the wire consist of a series of concentric circles whose direction\nis given by the right-hand rule.\n/square6The pattern of the magnetic field around a conductor due to an electric current\nflowing thr ough it depends on the shape of the conductor . The magnetic field of a\nsolenoid carrying a current is similar to that of a bar magnet.\n/square6An electromagnet consists of a core of soft iron wrapped around with a coil of\ninsulated copper wire.\n/square6A current-carrying conductor when placed in a magnetic field experiences a force.\nIf the direction of the field and that of the current are mutually perpendicular to", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\ninsulated copper wire.\n/square6A current-carrying conductor when placed in a magnetic field experiences a force.\nIf the direction of the field and that of the current are mutually perpendicular to\neach other , then the for ce acting on the conductor will be perpendicular to both\nand will be given by Fleming\u2019s left-hand rule. This is the basis of an electric motor .\nAn electric motor is a device that converts electric energy into mechanical energy.\n/square6The phenomenon of electromagnetic induction is the production of induced current\nin a coil placed in a region where the magnetic field changes with time. The magnetic\nfield may change due to a relative motion between the coil and a magnet placed\nnear to the coil. If the coil is placed near to a current-carrying conductor , the\nmagnetic field may change either due to a change in the current through the\nconductor or due to the relative motion between the coil and conductor . The", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nmagnetic field may change either due to a change in the current through the\nconductor or due to the relative motion between the coil and conductor . The\ndirection of the induced current is given by the Fleming\u2019s right-hand rule.\n/square6A generator converts mechanical energy into electrical energy. It works on the basis\nof electromagnetic induction.\n/square6In our houses we receive AC electric power of 220 V with a frequency of 50 Hz. One\nof the wires in this supply is with red insulation, called live wire. The other one is of\nblack insulation, which is a neutral wire. The potential difference between the two\nis 220 V . The thir d is the earth wir e that has gr een insulation and this is connected\nto a metallic body deep inside earth. It is used as a safety measure to ensure that\nany leakage of curr ent to a metallic body does not give any sever e shock to a user .\n/square6Fuse is the most important safety device, used for protecting the circuits due to", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nany leakage of curr ent to a metallic body does not give any sever e shock to a user .\n/square6Fuse is the most important safety device, used for protecting the circuits due to\nshort-circuiting or overloading of the circuits.\n2020-21\nScience 240EXERCISES\n1.Which of the following correctly describes the magnetic field near a long\nstraight wire?\n(a) The field consists of straight lines perpendicular to the wire.\n(b) The field consists of straight lines parallel to the wire.\n(c) The field consists of radial lines originating from the wire.\n(d) The field consists of concentric circles centred on the wire.\n2.The phenomenon of electromagnetic induction is\n(a) the process of charging a body.\n(b) the process of generating magnetic field due to a current passing\nthrough a coil.\n(c) producing induced current in a coil due to relative motion between a\nmagnet and the coil.\n(d) the pr ocess of r otating a coil of an electric motor .\n3.The device used for producing electric current is called a", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nmagnet and the coil.\n(d) the pr ocess of r otating a coil of an electric motor .\n3.The device used for producing electric current is called a\n(a) generator .\n(b) galvanometer .\n(c) ammeter .\n(d) motor .\n4.The essential difference between an AC generator and a DC generator is that\n(a) AC generator has an electromagnet while a DC generator has\npermanent magnet.\n(b) DC generator will generate a higher voltage.\n(c) AC generator will generate a higher voltage.\n(d) AC generator has slip rings while the DC generator has a commutator .\n5.At the time of short circuit, the current in the circuit\n(a) reduces substantially.\n(b) does not change.\n(c) increases heavily.\n(d) vary continuously.\n6.State whether the following statements are true or false.\n(a) An electric motor converts mechanical energy into electrical energy.\n(b) An electric generator works on the principle of electromagnetic induction.\n(c) The field at the centre of a long circular coil carrying current will be", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\n(b) An electric generator works on the principle of electromagnetic induction.\n(c) The field at the centre of a long circular coil carrying current will be\nparallel straight lines.\n(d) A wire with a green insulation is usually the live wire of an electric supply.\n7.List two methods of producing magnetic fields.\n2020-21\nMagnetic Effects of Electric Current 2418.How does a solenoid behave like a magnet? Can you determine the north and\nsouth poles of a current \u2013carrying solenoid with the help of a bar magnet? Explain.\n9.When is the force experienced by a curren t\u2013carrying conductor placed in a magnetic\nfield largest?\n10.Imagine that you are sitting in a chamber with your back to one wall. An electron\nbeam, moving horizontally from back wall towards the front wall, is deflected by a\nstrong magnetic field to your right side. What is the direction of magnetic field?\n11.Draw a labelled diagram of an electric motor . Explain its principle and working.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nstrong magnetic field to your right side. What is the direction of magnetic field?\n11.Draw a labelled diagram of an electric motor . Explain its principle and working.\nWhat is the function of a split ring in an electric motor?\n12.Name some devices in which electric motors are used.\n13.A coil of insulated copper wir e is connected to a galvanometer . What will happen if\na bar magnet is (i) pushed into the coil, (ii) withdrawn from inside the coil, (iii) held\nstationary inside the coil?\n14.Two cir cular coils A and B ar e placed closed to each other . If the curr ent in the coil\nA is changed, will some current be induced in the coil B? Give reason.\n15.State the rule to determine the direction of a (i) magnetic field produced around a\nstraight conductor -carrying curr ent, (ii) force experienced by a curr ent-carrying\nstraight conductor placed in a magnetic field which is perpendicular to it, and\n(iii) current induced in a coil due to its rotation in a magnetic field.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-textbook.txt\nstraight conductor placed in a magnetic field which is perpendicular to it, and\n(iii) current induced in a coil due to its rotation in a magnetic field.\n16.Explain the underlying principle and working of an electric generator by drawing\na labelled diagram. What is the function of brushes?\n17.When does an electric short circuit occur?\n18.What is the function of an earth wire? Why is it necessary to earth metallic\nappliances?\n2020-21", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\nPHYSICS POWER POINT PRESENTATIO N\nBY:\nSHIVAMURTHY M\nPGT PHYSICS \nJNV CHIKKAMAGALURMAGNETIC EFFECT OF CURRENT\n1.Introduction: Magnetic properties\n2.Magnetic Effect of Current \u2013Oersted\u2019s Experiment\n3.Ampere\u2019s Swimming Rule\n4.Maxwell\u2019s Cork Screw Rule\n5.Right Hand Thumb Rule\n6.Magnetic Field due to Infinitely Long Straight Current \u2013carrying Conductor\n7.Magnetic Field due to a Circular Loop carrying current\n8.Magnetic Field due to a Solenoid\n9.Fleming\u2019s left hand rule\n10.Electric motor(AC/DC)\n11.Electromagnetic Induction and Fleming\u2019s right hand rule\n12.Electric generator (AC/ DC): Dynamo\n13.Domestic electric circuitTopics coveredMAGNETIC FIELD AND FIELD LINES\nProperties of Magnets : \n1.They have two permanent poles namely, North and South and poles cannot be isolated.\n2.Like poles repel each other and unlike poles attract each other.", + "2.Like poles repel each other and unlike poles attract each other. \n3.When magnet is cut into pieces they will behave like separate magnets with same \nproperty but lesser magnetic strength compared to whole magnet.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\n3.When magnet is cut into pieces they will behave like separate magnets with same \nproperty but lesser magnetic strength compared to whole magnet. \n4.The freely suspended bar magnet will align in geographic N \u2013S direction. \n5.When magnet is subjected to mechanical damages (falling on a hard surface, heating, \npiling, hammering etc.) they lose their magnetic properties\n6.Depending on application we have, i) dumbbell magnet, ii) ring magnet, iii) bar \nmagnet, iv) horse shoe magnet, v) compass needle etc. \n7.Note: There is a magnet without poles is called toroid, whose north and south poles are \njoined. \n8.Magnets are naturally available (Earth) and can be prepared artificially for use. Introduction: MAGNETIC FIELD AND FIELD LINES\nDefinition: Magnetic field is space or region surrounding the magnet (or current carrying \nconductor) in which another magnet will experience the force of magnetostatics. It is \ndenoted by B SI unit T (tesla) and a vector .", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\nconductor) in which another magnet will experience the force of magnetostatics. It is \ndenoted by B SI unit T (tesla) and a vector . \nDefinition: Magnetic field lines are the imaginary path in which an isolated north pole can \ntravel from one pole to another in space. \nProperties: \n1.Magnetic field lines or Magnetic lines of force, travel from north pole and reach south \npole in its vicinity or travel up to infinity.\n2.M.F.L. are continuous closed curves and can pass through conductors\n3.Note: Superconductors does not allow MFL to enter inside hence B = 0 (inside).\n4.The tangent drawn at any point on the MFL gives direction of MF at that point. \n5.No two MFL can intersect each other since two value of MF doesn\u2019t exist at a point. \n6.MFL are crowded at strong MF (poles ) but diverged at weak MF region. Introduction: MAGNETIC FIELD AND FIELD LINES OF A BAR MAGNET\nExperimental image Illustrative diagram \nNMagnetic Effect of electric Current:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\nExperimental image Illustrative diagram \nNMagnetic Effect of electric Current:\nAn electric current (i.e. flow of electric charge) produces magnetic effect in \nthe space around the conductor called strength of Magnetic field or simply \nMagnetic field.\nHans Christian Oersted\u2019s Experiment:\nWhen current is passing through the wire over \nthe compass needle in anti -clockwise direction, \nthe needle was found to deflect towards West \ndirection from its normal position.E\nKI\nN\nE\nK\nI\nWhen current direction is reversed through the wire \nover the compass needle, the needle was found to \ndeflect towards East direction from its normal \ndirection.B B\nRules to determine the direction of magnetic field: \nAmpere\u2019s Swimming Rule or\nSNOW Rule :\nImagining a man who swims in the \ndirection of current from south to north \nfacing a magnetic needle kept under \nhim such that current enters his feet \nthen the North pole of the needle will \ndeflect towards his left hand, i.e. \ntowards West.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\nfacing a magnetic needle kept under \nhim such that current enters his feet \nthen the North pole of the needle will \ndeflect towards his left hand, i.e. \ntowards West.\nMaxwell\u2019s Cork Screw Rule or Right \nHand Screw Rule:\nIf the forward motion of an imaginary \nright handed screw is in the direction of \nthe current through a linear conductor, \nthen the direction of rotation of the \nscrew gives the direction of the \nmagnetic lines of force around the \nconductor.SI\nI\n I\nRight Hand Thumb Rule or Curl Rule:\nIf a current carrying conductor is imagined to be held in the right \nhand such that the thumb points in the direction of the current, \nthen the tips of the fingers encircling the conductor will give the \ndirection of the magnetic lines of force .\nMagnetic Field Lines due to current carrying conductor:\nDirection of Bis determined by Right Hand Screw Rule . \nIt is perpendicular to the plane of the diagram and entering into the plane at as shown in diagram .\nI\nB\nBII", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\nDirection of Bis determined by Right Hand Screw Rule . \nIt is perpendicular to the plane of the diagram and entering into the plane at as shown in diagram .\nI\nB\nBII\nBI IDifferent views of direction of current and magnetic field due to circular loop of \na current carrying coil:\nB\nII\nB B\nCurrent moving in clock wise direction \nrepresenting the face towards us (Inwards) as \nSouth pole of bar magnet Current moving in anti -clock wise direction \nrepresenting the face towards right (outwards) as \nNorth pole of bar magnet \nMagnetic Field due to a current carrying Solenoid:\nI Ix x x x x x x\nTIP:\nWhen we look at any end of the coil carrying current, if the current is in anti-clockwise\ndirection then that end of coil behaves like North Pole and if the current is in clockwise\ndirection then that end of the coil behaves like South Pole . Hence it behave like bar magnet . BForce acting on a current carrying conductor: \n1.A current carrying conductor is freely suspended from a stand", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\n1.A current carrying conductor is freely suspended from a stand \nwith the help of thread and connecting wires as shown in fig.\n2.A horse shoe magnet is placed over the conductor at centre \nsuch that, north pole on top and south pole at bottom and \ndirection of motion of conductor is noted.\n3.The direction of poles are now reversed and the direction of \nmotion of conductor is noted.\n4.The polarity of cell is reversed and above steps is repeated \nand the direction of motion of conductor is noted in each \ncase.\n5. The conclusion is drawn from the observations. \nConclusion: i) The direction of motion of conductor is towards \neast, when the current is outward plane perpendicular with \ndirection of magnetic field from N \u2013S and current moving into \nthe plane and direction of magnetic field is S \u2013N.\nii) The direction of motion of conductor is towards west \nfor other two directions of current and MF.ACTIVITY: \n3 D view: https://youtu.be/rtjGH0B -vVAFLEMING\u2019S LEFT HAND RULE:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\nii) The direction of motion of conductor is towards west \nfor other two directions of current and MF.ACTIVITY: \n3 D view: https://youtu.be/rtjGH0B -vVAFLEMING\u2019S LEFT HAND RULE:\nIStretch first three fingers of your left hand \nmutually perpendicular to each other such that, \nfore (Index) finger represents direction of \nmagnetic field, middle finger represent direction \nof current, then thumb finger show the direction \nof force (motion) of the conductor. \nConclusion: i) The direction of motion of \nconductor is towards east, when the current is \noutward plane perpendicular with direction of \nmagnetic field from N \u2013S and current moving into \nthe plane and direction of magnetic field is S \u2013N.\nii) The direction of motion of conductor is \ntowards west for other two directions of current \nand MF.\nELECTRIC MOTOR:\nAC MOTOR: DC MOTOR: \nIt is a device which converts electrical energy (AC/DC current) into mechanical \nenergy.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\ntowards west for other two directions of current \nand MF.\nELECTRIC MOTOR:\nAC MOTOR: DC MOTOR: \nIt is a device which converts electrical energy (AC/DC current) into mechanical \nenergy. \nPrinciple : When a current carrying conductor placed between magnetic field it will \nundergo rotation based on Fleming\u2019s left hand rule. \nConstruction: As shown in diagram all the components are placed and necessary \nAC/DC current is given. \nWorking: When current is passed through armature in clock wise direction, \naccording to Fleming\u2019s left hand rule, left part (towards North) of armature \nwill experience upward motion and right part will experience downward \nmotion and it starts rotating, due to commutators the direction of rotation is \nsuch that it will be according AC/DC cycle which are in contact with metallic \n(carbon) brush and hence to the source.\nConclusion: Electrical energy is converted into mechanical energy.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\nsuch that it will be according AC/DC cycle which are in contact with metallic \n(carbon) brush and hence to the source.\nConclusion: Electrical energy is converted into mechanical energy. \nApplication : Fan, mixer, blender, washing machine etc. AC/DC MOTOR: The relative motion between closed conducting wire and magnet will induce \nelectric current in the coil without any electrochemical reaction.\nFaraday\u2019s Experiments:\nFig. having solenoid and magnet produce current whenever there is relative motion \nbetween coil and magnet which is noticed in deflection of galvanometer needle. \nAlso notice that the direction of current is reversed when there is change in the pole \nof the bar magnet and its relative motion with coil as we notice that, current moving \nin anticlockwise represents North pole of magnet and in clockwise it will behave as \nSouth pole of the magnet. Faraday\u2019s Electromagnetic Induction: \nCurrent \ndirection\nStretch first three fingers of your right hand", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\nSouth pole of the magnet. Faraday\u2019s Electromagnetic Induction: \nCurrent \ndirection\nStretch first three fingers of your right hand \nmutually perpendicular to each other such \nthat, fore (Index) finger represents \ndirection of magnetic field, middle finger \nrepresent direction of current induced, then \nthumb finger show the direction of force \n(motion) of the conductor. Fleming\u2019s right hand rule: \nAlternating current (AC) is time varying \ncurrent which varies sinusoidal or \ncosinusoidal with time. \nDirect Current (DC) is having fixed magnitude \nand won\u2019t change with time. AC and DC current: ELECTRIC GENERATOR: It is a device which converts mechanical energy into \nelectrical energy (AC/DC current). \nPrinciple: When a coil is rotated between magnetic \nfield there will be induced current/emf based on \nElectromagnetic Induction. \nConstruction: As shown in diagram all the components \nare placed and necessary AC/DC current is drawn \nacross load resistor.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\nElectromagnetic Induction. \nConstruction: As shown in diagram all the components \nare placed and necessary AC/DC current is drawn \nacross load resistor. \nWorking: When armature is rotated between strong \nmagnetic field according to Electromagnetic Induction \nand Fleming\u2019s right hand rule there will be induced \ncurrent/emf in the coil and it is drawn across resistor, \nwhich are in contact with metallic (carbon) brush and \nhence to the battery.\nConclusion: Hence the mechanical energy is converted \ninto electrical energy. \nLoad \nResistor\nADVANTAGES OF AC OVER DC\n1.The cost of producing AC much lesser than that of DC.\n2.AC can be easily converted into DC using rectifier. \n3.AC can be easily transmitted for long distances compared to DC.\n4.AC can be controlled much easily compared to DC without much loss of energy.\nDISADVANTAGES OF AC OVER DC\n1.AC cannot be used in electrolysis as that of DC.\n2.AC is more dangerous than DC.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\n4.AC can be controlled much easily compared to DC without much loss of energy.\nDISADVANTAGES OF AC OVER DC\n1.AC cannot be used in electrolysis as that of DC.\n2.AC is more dangerous than DC.\n3.AC will have more magnitude (Positive value) compared to DC magnitude. The electric circuit designed for the purpose of domestic use (house, factory, SSI etc) is \ncalled domestic electric circuit. DOMESTIC ELECTRIC CIRCUIT: \nThe electric circuit is having \nLive wire ( + ve): \nNeutral wire( -ve): \nEarth wire (0 V): \nInsulated colour wires, \nTwo circuits one of 15 A \nrating fuse for Heavy \nappliances such as Heater, \noven etc. and 5 A fuse for \nbulb, fan etc.\nThe applied voltage across \nmains is, 220 \u2013240 V in India \nwith 50 Hz frequency.S1S2 S3\n1.Live wire : It is kept as positive with 220 V supply from mains.\n2.Neutral wire : It is kept at 0 V w.r.t.", + "The applied voltage across \nmains is, 220 \u2013240 V in India \nwith 50 Hz frequency.S1S2 S3\n1.Live wire : It is kept as positive with 220 V supply from mains.\n2.Neutral wire : It is kept at 0 V w.r.t. live wire for completing the circuit \nconnection. \n3.Earth wire : It is connected to earth through metallic wire which is at 0 V and", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\n2.Neutral wire : It is kept at 0 V w.r.t. live wire for completing the circuit \nconnection. \n3.Earth wire : It is connected to earth through metallic wire which is at 0 V and \ncan allow the excess charges produced easily whenever electrical hazard \ntakes place.\n4. Switch ( S1, S2\u2026) : It is controlling component which is connected in series \nwith live wire to operate appliances whenever needed. \n5. Fuse : It is safety device which is connected in series with appliances with \nsuitable rating for easy melting and disconnection whenever there is short \ncircuit or overloading in circuit. It is made up of alloys of Al, Fe, Cu, Sn, Pb etc. \n6. MCB (Miniature Circuit Breaker): Designed to break circuit during electrical \nhazards based on electromagnetic induction or eddy current production. 7. Short Circuit : When live wire coincide with neutral wire directly (accidentally) \nthere will be sudden flow of charges from high potential to low potential", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\nthere will be sudden flow of charges from high potential to low potential \ncausing enormous heat according to Joule\u2019s heating effect in the circuit .\n8. Over loading : When there are more number of appliances are connected in \nparallel at a time the effective resistance will reduce too small giving the \ncondition of short circuiting called as overloading. \n9. Electrical Hazard : It is electric shock which produce enormous heat which is \ncapable of burning whole domestic electric circuit due to short circuit or \noverloading or failure in the earthing of appliances. \nSUMMARY:\n1.Magnetic field and magnetic field line properties \n2.Current carrying conductor and coil possess magnetic properties\n3.Various laws and rules explaining the direction of magnetic field produced due to \ncurrent carrying conductor, such as right hand thumb rule .\n4.When current appear to travel in anti clockwise direction in a closed coil it behave", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\ncurrent carrying conductor, such as right hand thumb rule .\n4.When current appear to travel in anti clockwise direction in a closed coil it behave \nlike North pole , if it is clockwise it resembles South pole of a magnet\n5.The combination of such closed loops form solenoid and when current is passed \nthrough it behaves like bar magnet. \n6.The direction of force experienced by current carrying conductor in uniform \nmagnetic field \u2013Fleming\u2019s left hand rule, its application in electric motors such as \ngrinder, fan, washing machine etc. \n7.Production of induced current/ emf due to relative motion between coil and \nmagnet \u2013Electromagnetic induction . 8. Current is induced in a coil when subjected rotation in a uniform magnetic field \u2013\nFleming\u2019s right rule \n9.Application of EMI in producing current due to various sources \u2013AC generation in", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-reading notes.txt\nFleming\u2019s right rule \n9.Application of EMI in producing current due to various sources \u2013AC generation in \nHydroelectric power plants, wind mill etc. \n10.Purposeful use of electricity using domestic electric circuit with proper safety \ndevices and precautions", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-MATERIAL.txt\n13. MAGNETIC EFFECT OF ELECTRIC OF ELECTRIC CURRENT \nSUB: SCIENCE (PHYSICS) CLASS: X \n \n \nD. MULTIPLE CHOICE QUESTIONS (MCQ) : - \n1. The switches are kept commonly in electrical appliances \na) Live wire b) Earth wire c) Neutral wire d) Any one of above \n \n2. Name the component which reverses current in DC motor. \na) Carbon brushes b) A battery c) Magnets d) Commutator \n \n3. A 3 - pin ma ins plug is fitted to cable for 2kW electric heater to be used for \n250V ac supply. Which of the following statement is not correct? \na) A fuse of 20 A is most appropriate value to use \nb) The neutral wire is coloured black \nc) A fuse of 15A should be fitte d in live wire \nd) The green wire should be connected to earth pin \n \n4.", + "a) A fuse of 20 A is most appropriate value to use \nb) The neutral wire is coloured black \nc) A fuse of 15A should be fitte d in live wire \nd) The green wire should be connected to earth pin \n \n4. The connection of appliances is made in ________ \na) Parallel b) Series c) both a) or b) d) neither a) nor b)", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-MATERIAL.txt\n4. The connection of appliances is made in ________ \na) Parallel b) Series c) both a) or b) d) neither a) nor b) \n \n5. When a small magnetic needle placed in E \u2013 W direction kept under a \nstrong current flowing in N \u2013 S carrying conductor, its north pole will direct \ntowards, \na) West \u2013 East b) South \u2013 West c) North \u2013 West d) North \u2013 East \n \n6. Main fuse is connected to the ______ wire. \na) Earth wire b) Live wire c) Neutral wire d) any of these. \n \n7. The instrument which convert electrical energy into mechanical energy is, \na) Motor b) Generator c) Iron d) oven \n \n8. The instrument which convert mechanical energy into electrical energy is, \na) Motor b) Generator c) Iron d) oven \n \n9.", + "The instrument which convert electrical energy into mechanical energy is, \na) Motor b) Generator c) Iron d) oven \n \n8. The instrument which convert mechanical energy into electrical energy is, \na) Motor b) Generator c) Iron d) oven \n \n9. Colour of wire in which switches need to be connected is, \na) Red (Brown) b) Blue c) Black d) Green \n10. The current carrying straight wire will exhibit magnetic around it in, a) Circular and endless b) Oval and endless c) Straight line d) All the above", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-MATERIAL.txt\n11. The instrument which does not work on Joule\u2019s heating effect in domestic \ncircuit is, a) Iron b) Oven c) Fuse d) Heater", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-study material.txt\n13. MAGNETIC EFFECT OF ELECTRIC OF ELECTRIC CURRENT \nSUB: SCIENCE (PHYSICS) CLASS: X", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-study material.txt\nA. LONG ANSWER TYPE: - \n1. Explain properties of magnet and magnetic field lines. \n2. Explain Oersted\u2019s experiment. What is the conclusion we draw from it? \n3. Draw magnetic field line due to i) Bar magnet ii) Straight current carrying \nconductor iii) Current carrying loop in a) clock wise current and b) anti clock wise \ncurrent direction iv) current carrying solenoid. \n4. State Fleming\u2019s left -hand rule. Explain the activity with the help of neat labelled \ndiagram. \n5. What is the principle of electric motor? Explain its working using labelled \ndiagram. \n6. State and explain Faraday\u2019s law of \u201celectromagnetic induction\u201d. \n7. State principle, construction and working of DC generator. \n8. Why parallel combination is preferred over series combination in \u201cDomestic \nelectric circuit\u201d? \n9. Explain importance of the following: a) Short circuit b) overl oading c) Fuse d) \nJoule\u2019s heating effect and e) Earth wire.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-study material.txt\nelectric circuit\u201d? \n9. Explain importance of the following: a) Short circuit b) overl oading c) Fuse d) \nJoule\u2019s heating effect and e) Earth wire. \n10. Write the advantages and disadvantages of AC and DC current in their \nrespective purposes. \n11. a. Mention effect of electric current on which the working of an electric fuse is \nbased on? \nb. Draw schematic labelled diagram of Domestic electric circuit consisting of \nmain fuse, meter, distribution box and electrical points with necessary wires and \nearth. \nc. Explain the terms overloading and short circuit. \n12. a) Describe an activity to demonst rate the pattern of magnetic field lines \naround a straight current carrying conductor . \nb) State the rule to find the direction of magnetic field associated with \ncurrent carrying conductors and \nc) What is the shape of the magnetic field which produce bar m agnet \npattern? 13.(a) Describe an activity to demonstrate the pattern of magnetic field lines", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-study material.txt\ncurrent carrying conductors and \nc) What is the shape of the magnetic field which produce bar m agnet \npattern? 13.(a) Describe an activity to demonstrate the pattern of magnetic field lines \naround a straight conductor carrying current. \n(b) State the rule to find the direction of magnetic field associated with a current \ncarrying conductor. \n(c) What is the shape of a current carrying conductor whose magnetic field \npattern resembles that of a bar -magnet ? \n14.(a) Explain why there are two separate circuits one for high power rating \nappliances and other for low power rating appliances. \n(b) A domestic circuit has 5A fuse. How many bulbs of rating 100W, 220V can be \nsafely used in this circuit? Justify your answer. \n15. (i) Design an activity with the help of two nails, very thin aluminium st rip, a 12 \nV Battery and a key to illustrates. \n(ii) Cable of a microwave oven has three wires inside it which have insulation of", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-study material.txt\nV Battery and a key to illustrates. \n(ii) Cable of a microwave oven has three wires inside it which have insulation of \ndifferent colours black, green and red. Mention the significance of the three \ncolours and potential difference between red and b lack one.", + "MAGNETIC EFFECT OF ELECTRIC CURRENT LESSON PLAN.txt\nCLASS \u2013 X \nPHYSICS LESSON PLAN \nCHAPTER \u2013 13 MAGNETIC EFFECT OF ELECTRIC CURRENT.", + "MAGNETIC EFFECT OF ELECTRIC CURRENT LESSON PLAN.txt\nUNIT 13. MAGNETIC EFFECT OF ELECTRIC CURRENT \nClass \ntransaction Total: 15 periods of 40 mins each \nPre- requisite \nfor the course This lesson requires basic knowledge of electric current \nand magnetic properties \nAssessment of \nqualifying \nKnowledge 1. Pen and pa per test \n2. Assignment \n3. Oral test or quiz \n4. Lab activity or field work \n5. Home work \n6. Group Discussion \n7. Projects \nObjective The learners are able compare and use the magnetic \nproperties in association with electric properties in daily \nlife and conversion of one form of energy into another to \ntheir capacity. \nLearning \nOutcomes KNOWLEDGE - \n\u2022 Students will know and understand concepts \n Students will be able to draw diagrams systematically . \n\u2022 Students will be able to design electric circuit using \nvarious \ncomponents from real life. \n\u2022 Students will be able to compare and contrast t he effects \nof electric current and magn etic field .", + "MAGNETIC EFFECT OF ELECTRIC CURRENT LESSON PLAN.txt\nvarious \ncomponents from real life. \n\u2022 Students will be able to compare and contrast t he effects \nof electric current and magn etic field . \n\u2022 Students will be able to interpret working of electric \nmotor and generator . \n\u2022 Students will be able to experiment magnetic effect of \nelectric \ncurrent. \n\u2022 Students will be able to apply magnetic effect of electric \ncurrent in \nreal life.", + "MAGNETIC EFFECT OF ELECTRIC CURRENT LESSON PLAN.txt\nSKILL S AND COMPETENCIES - \n Students would be able to \n\u2022 Critically analyse a) the magnetic effect electric current \nand \nb) production of electric current from magnetic field . \n\u2022 Appreciate importance of magnetic effect of electric \ncurrent and production of electric current Collaboratively. \n\u2022 Collaborate in groups and experiment how current will \nproduce magnetic effect in an electric circui t. \n\u2022 Design Creative methods to construct electric motor \nand generator. \n\u2022 Built Character amongst them selves by discussing \nimportance of electricity and magnetic properties . \nAnswer knowledge, understanding, application and \ndiagrammatic \nskill based questions based on the topic. \nATTITUDES AND BELIEFS \u2013 \nStudents will be able to \nDevelop citizenship by visualizing and observing day to \nday phenomena and demonstrate it to sensitize others. \nTransaction \nMethodology \n(The teacher \ncan \nuse the \nmentioned \ntechniques, \nwherever \napplicable, and", + "MAGNETIC EFFECT OF ELECTRIC CURRENT LESSON PLAN.txt\nday phenomena and demonstrate it to sensitize others. \nTransaction \nMethodology \n(The teacher \ncan \nuse the \nmentioned \ntechniques, \nwherever \napplicable, and \ncan use any \nother too.) Transaction would proceed in the following manner - \nActive Learning - Frayer Model on magnetic effect of \nElectric Current and electromagnetic induction. \n(Critical Thinking) \nGroup Learning - The class will be divided into eight \ngroups. Jigsaw \nExpert Group: The group will be given worksheet about \nthe magnetic properties of electric current and EMI . \nLearning Group: Students will teach each other the \nmagnetic effect of electric cu rrent and electromagnetic \ninduction. \n(Communication, Collaboration) \nDifferentiation - The students in groups of 8 will be \ndemonstrating the magnetic effect of electric current and \nEMI. \n (Character, Communication) \nQuestioning - Multiple level questions - Teac her will \nprepare a list of question about magnetic effect of electric", + "MAGNETIC EFFECT OF ELECTRIC CURRENT LESSON PLAN.txt\nEMI. \n (Character, Communication) \nQuestioning - Multiple level questions - Teac her will \nprepare a list of question about magnetic effect of electric \ncurrent and EMI and will ask these questions \nafter discussion. \n(Critical Thinking, Communication) \nBrain Storming -The class would start with a discussion \non what the students have already learnt in the previous \nclasses and hence what is it that they would learn now. \nThey would also be told the significance of the topic \nthat they would be studying. Introduction of the topic - PPT and Digital Content would \nbe shared. \nFacilitation of co ntent \nThe facilitator will enable the learners to understand the \nfundamental concept of magnetic effect of electric current . \nHe/ She will compare and contrast differen ce between \nmagnetic and electric field. \nApproaches \nThe facilitator should apply some or a ll the approaches \nwhile administering the teaching learning process. \n\u2022 Didactic Approach \n\u2022 Summarizing Approach", + "MAGNETIC EFFECT OF ELECTRIC CURRENT LESSON PLAN.txt\nApproaches \nThe facilitator should apply some or a ll the approaches \nwhile administering the teaching learning process. \n\u2022 Didactic Approach \n\u2022 Summarizing Approach \n\u2022 Feedback Approach \nTechniques to be used: \n\u2022 Quiz \n\u2022 MCQ \n\u2022 PPT \n\u2022 Computer aided simulation/3D pictures \n\u2022 Skit/Role playing \n\u2022 Group discussion \n\u2022 HOTS questions \n\u2022 Peer Assessment \n\u2022 Student -teacher interaction, Wipro - G.O.s(web chart, \nflow chart and differentiation table, compare - contrast \nmatrix), silent/ loud reading, collaborative learning, \n\u2022 Lab Activities \nGuided practice to be followed by Independent Practice - \nNCERT questions to be discussed in the classroom.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-LESSON PLAN.txt\nLesson Plan \n \nTeacher: _____________________________________ \n \n \n \n1. Lesson Objectives: \nDefine magnetic field and field line \nExplain how sound is produced \n \n \n2. a) Pre-requisites \nCompass needle and it\u2019s use \nMagnetic effects of electric current \n b) Pre-assessment : \nWhat is a compass needle and what are the uses of it \nWhat effects can be observed when we pass current \nthrough a conductor \n \n \n \n3. a) Content : \nMagnetic Effects of Electric Current \n Magnetic field due to a current carrying conductor \n \n \n \n \n \nb) Key Terms \n \nMagnetic Field, Field lines, Compass Needle, Deflection 4. Integrated Links: \na) CCL \n \nMagnetic Resonance Imaging(MRI) \n \n \nb) RL \n \nHardware Disease ,MRI, Earth\u2019s Magnetism \n \n \nc) Value integration \n \n5. Resources: \nEduBrisk website \n 6. Response to Previous Lesson Assessment: \n \n \n \n7. Motivator / Opening activity :", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-LESSON PLAN.txt\nHardware Disease ,MRI, Earth\u2019s Magnetism \n \n \nc) Value integration \n \n5. Resources: \nEduBrisk website \n 6. Response to Previous Lesson Assessment: \n \n \n \n7. Motivator / Opening activity : \n \nShow a video on the Magnetic Field lines and with the help of a bar magnet and compass needle magnetic field lines \nare plotted in a plane sheet. Grade: 10 Section: A & B \nDate: ________ _______ Subject: Physics \nChapt : Magnetic effect of Current car rying \nconductor \nTopic: Magnetic field due to a cu rrent carrying \nconductor \nDuration: 40 Mins", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-LESSON PLAN.txt\n8. Action Plan \nNo. Teacher Action Student Action \n1 \nHelps the students in performing the activity \nregarding mapping of magnetic field using bar \nmagnet and compass needle Perform the activity on Mapping Magnetic field lines \n2 Shows Magnetic Field on a current carrying conductor \nWith the help of diagrams Students learn that a magnetic field will exists around \na current carrying conductor \n3 Shows how to find out the direction of magnetic field \nusing Right hand thumb rule Students understands Right hand thumb rule and \ntries to find out magnetic field direction in the \navailable figures. \n4 \nShows Magnetic Field around a current carrying loop \nwith the help of diagram \nStudents learn that a magnetic field will exists around \na current carrying loop \n5 Teacher shows a video regarding Magnetic field due \nto a current in a circular coil Students understands the production of magnetic \nfield around a current carrying conductor", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-LESSON PLAN.txt\n5 Teacher shows a video regarding Magnetic field due \nto a current in a circular coil Students understands the production of magnetic \nfield around a current carrying conductor \n6 Discuss Magnetic field due to a current carrying \nsolenoid with the help of learned concepts Students understands the magnetic field inside a \nsolenoid", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-LESSON PLAN.txt\n9. Differential I nstruction s: \nSEND Basic Intermediate AGT \nWhy does a compass \nneedle get deflected when \na bar magnet is brought \nnear to it? \nWhat are the properties of \nmagnetic field lines? \nState right hand thumb \nrule? What are the different ways \nto induce current in a coil? \nA current through a \nhorizontal power line flows in \neast to west direction .What \nis the direction of magnetic \nfield at appoint directly above \nit and at a point directly \nbelow it ? Which all are the factors \non which the magnetic \nfield due to a current \ncarrying conductor \ndepends? \nWhy to magnetic lines of \nforce do not intersects \neach other Electricity and magnetism \nare linked to each other. \nDo you agree with this \nstatement; Justify your \nanswer \n \n \n \n \n \n \n \n \n \n \n10. Plenary \na) Closure: \nStudents summarize the topics learnt in the lesson.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-LESSON PLAN.txt\n10. Plenary \na) Closure: \nStudents summarize the topics learnt in the lesson. \n \n \nb) Lesson A ssessment: \n1. What is magnetic field? \n2. How can we say that there is magnetic field around a current carrying conductor? \n3. Wha t is the method followed to find out the direction of magnetic field? \n4. There is a force of attraction between two current carr ying conductors when the current in them is \u2026\u2026.. \n \n11. Reinforcement / Homework / Project / Assignment: \n1.Plot Magnetic Field lines of different bar magnets", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT- competency based questions.txt\n13. MAGNETIC EFFECT OF ELECTRIC OF ELECTRIC CURRENT \nSUB: SCIENCE (PHYSICS) CLASS: X", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT- competency based questions.txt\n1. What happens to a magnetic needle is brough t close to current carrying \nconductor? When, \ni) Towards north of the conductor when current is moving from west to \neast. \nii) Towards south of the conductor when current is moving from west to \neast \niii) Towards north of the conductor when current is moving from east to \nwest \niv) Towards south of the conductor when current is moving from east to \nwest. \nv) Above the conductor when current is moving from east to west \nvi) Below the conductor when current is moving from east to west \n2. What is the direction of motion of c urrent carrying conductor ? When, \ni) Current is m oving from s outh to no rth and magnetic fi eld is \nperpendicular into the plane o f the paper.", + "What is the direction of motion of c urrent carrying conductor ? When, \ni) Current is m oving from s outh to no rth and magnetic fi eld is \nperpendicular into the plane o f the paper. \nii) Current is m oving from s outh to no rth and magnetic fi eld is \nperpen dicular out of the plane o f the paper \niii) Current is m oving from North to South and magnetic fi eld is \nperpendicular into the plane o f the paper .", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT- competency based questions.txt\nperpen dicular out of the plane o f the paper \niii) Current is m oving from North to South and magnetic fi eld is \nperpendicular into the plane o f the paper . \niv) Current is m oving from North to South and magnetic fi eld is \nperpendicular outwards the plane o f the paper ? \n3. State and explain Fleming \u2019s left hand rule. What are its applications ? \n4. State and explain Faraday \u2019s laws of electromagnetic induction. Describe \nthe activit ies to explain induced current with the help of galvanometer \nwhen, \ni) North pole of bar magnet facing tow ards solenoid. \nii) South pole of bar magnet facing tow ards solenoid. \niii) Bar magnet is kept inside solenoid without motion . \n5. Plot the graph between induced voltage/current versus length, illustrating \ninduced current when bar magnet is allow ed to pass through the solen oid \nof length \u2018l\u2019 completely . \n6.", + "iii) Bar magnet is kept inside solenoid without motion . \n5. Plot the graph between induced voltage/current versus length, illustrating \ninduced current when bar magnet is allow ed to pass through the solen oid \nof length \u2018l\u2019 completely . \n6. Explain the principle, construc tion, working and applications of electric AC \nmotor.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT- competency based questions.txt\ninduced current when bar magnet is allow ed to pass through the solen oid \nof length \u2018l\u2019 completely . \n6. Explain the principle, construc tion, working and applications of electric AC \nmotor. \n7. Explain the principle, construc tion, working and applications of electric DC \nmotor. 8. Explain the principle, construc tion, working and applications of electric AC \ngenerato r. \n9. Explain the principle, construc tion, working and applications of electric DC \ngenerat or. \n10. Explain the pr inciple and working of safety devices in domestic \nelectric circuit , \ni) Fuse, \nii) MCB and \niii) Earthing wire. \n11. What is i) short circuit and ii) Overloading? \n12. How to protect from electrical hazards in domestic electric circuit? \n13. A house catchup fire due to electric al short circuit , what are the \nmeasures to be adopted for avoid ing such electrical hazar ds? \n14.", + "12. How to protect from electrical hazards in domestic electric circuit? \n13. A house catchup fire due to electric al short circuit , what are the \nmeasures to be adopted for avoid ing such electrical hazar ds? \n14. Why 5 \uf057 resistance fuse wire should not be used in circuit having \nheavy electrical appliances?", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT- competency based questions.txt\nmeasures to be adopted for avoid ing such electrical hazar ds? \n14. Why 5 \uf057 resistance fuse wire should not be used in circuit having \nheavy electrical appliances? \n15. Why 25 \uf057 resistance fuse wire should not be used in electrical \ncircuit having low load electrical appliances? \n16. Why fuse wire is made of allo ys but not with elem ental metal wires? \n17. It is advi sed to switch of f or plug out from the mains of electric circuit \nfor electrical appliances such as, TV, Rad io or such gadgets during \nthunder and lightning. W hy? \n18. What is t he function of step down transformer in controlling the \nsupply of voltage for the electrical appl iances?", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nClass 10 Science Revision Notes \nChapter 13 Magnetic Effects of Electric Current \n \nIn this chapter, we will study the effects of electric current : Moving charges or electric \ncurrent generates a magnetic field. \n1. Hans Christian Oersted (1777 -1851) \nOersted showed that electricity and magnetism are related to each other. His \nresearch later used in radio, television etc. \nThe unit of magnetic field strength is named Oersted in his honour. \n2. Oersted Experiment", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nOn passing the current through the copper wire XY in the circuit, the compass needle \nwhich is placed near the conductor gets deflected. If we reverse the direction of \ncurrent, the compass needle deflects in reverse direction. If we stop the flow of \ncurrent, the needle comes at rest. \nHence, it can be concluded that electricity and magnetism are linked to each other. It \nshows that whenever the current will flow through the conductor, then a magnetic \nfield will develop. \n3. Magnetic Field : It is the region surrounding a magnet, in which forc e of magnet \ncan be detected. It is a vector quantity, having both direction& magnitude. \n4. Compass Needle : It is a small bar magnet, whose north end is pointing towards \nnorth pole and south end is pointing towards south pole of earth. \n5. Magnetic Field Lines : The tangent to the magnetic field line at a point gives the \ndirection of magnetic field at that point.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nHence, magnetic field line is a path along which a hypothetical free north pole tend \nto move towards south pole. \n6. Characteristics of Magnetic field lines : \n(a) The direction of magnetic field lines outside the magnet is always from north pole \nto south pole of bar magnet and are indicated by an arrow.Inside the magnet, the \ndirection of field lines is from its south pol e to north pole. Thus magnetic field lines \nare closed curves. \n(b) The strength of magnetic field is expressed by the closeness of magnetic field \nlines. Closer the lines, more will be the strength and farther the lines, less will be the \nmagnetic field stren gth. \n(c) No two field lines will intersect each other.If they intersects, then at point of \nintersection the compass needle will show two directions of magnetic field which is \nnot possible. \n \n7. Magnetic field due to Current Carrying Conductor", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\n7. Magnetic field due to Current Carrying Conductor \n \nThe above e lectric circuit in which a copper wire is placed parallel to a compass \nneedle, shows the deflection in needle gets reversed, when the direction of current \nreversed. Hence electricity and magnetism are related to each other. \n8. Right Hand Thumb Rule \nIt is a convenient way of finding the direction of magnetic field associated with \ncurrent carrying conductor. Hold the straight wire carrying current in your right hand \nsuch that thumb points towards the direction of current, then your folded fingers \naround the c onductor will show the direction of magnetic field. \n \nThis rule also called Maxwell\u2019s corkscrew rule. \n9. Magnetic Field due to Current through a Straight Conductor", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nThis rule also called Maxwell\u2019s corkscrew rule. \n9. Magnetic Field due to Current through a Straight Conductor \n \nLet a current carrying conductor be suspended vertically and the electric current is \nflowi ng from south to north. In this case, the direction of magnetic field will be \nanticlockwise. If the current is flowing from north to south, the direction of magnetic \nfield will be clockwise. \nA current carrying straight conductor has magnetic field in the f orm of concentric \ncircles; around it. Magnetic field of current carrying straight conductor can be shown \nby magnetic field lines. \n \n10. Magnetic Field due to Current through a circular Loop", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\n10. Magnetic Field due to Current through a circular Loop \n \nEvery point on the wire carrying current give rise to the magnet ic field,appearing as a \nstraight line at the centre of loop. By applying Right hand Thumb rule, we can find \nthe direction of magnetic field at every section of the wire. \n11. Solenoid :A Coil of many circular turns of insulated copper wire wrapped closely \nin the shape of a cylinder is called solenoid. \n12. Magnetic field due to a current in a solenoid :", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\n\u2022 Using R.H. Thumb Rule, we can draw the pattern of magnetic field lines \naround a current carrying \u2018Solenoid\u2019. \n\u2022 One end of the solenoid behaves as a magnetic north pole,while the other \nend behave as the South Pole. \n\u2022 The filed lines inside the solenoid are in form of parallel straight lines, that \nimplies that magnetic field inside the solenoid is same at all points i.e. Field is \nuniform. \nThe strength of the magnetic field produced depends upon \n(a) the number of turns \n(b) Strength of current in the solenoid used in making solenoid. \n13. Electromagnet : Strong magnetic field inside the solenoid can be used to \nmagnetise a magnetic material for example soft iron, when it is placed inside the coil. \nThe magnet so formed is called electromagnet.It is a temporary magnet. \nProperties of Magnetic Field: \n\u2022 The magnitude; of magnetic field increases with increase in electric current \nand decreases with decrease in e lectric current.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nProperties of Magnetic Field: \n\u2022 The magnitude; of magnetic field increases with increase in electric current \nand decreases with decrease in e lectric current. \n\u2022 The magnitude of magnetic field; produced by electric current; decreases with \nincrease in distance and vice -versa. The size of concentric circles of magnetic \nfield lines increases with distance from the conductor, which shows that \nmagnetic field decreases with distance. \n\u2022 Magnetic field lines are always parallel to each other. \n\u2022 No two field lines cross each other. \n14. Force on a current carrying conductor in a magnetic field. \nAndre Marie Ampere (1775 \u20131836) suggested that the magnet also exert an equal \nand opposite force on the current carrying conductor.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nWe will observe that the rod will displace i.e. the rod will experience a force, when it \nis placed in magnetic field, in a perpendicular direction to its length. \n\u2022 The direction of the exerted force will be reversed if the direction of current \nthrough the conductor is reversed. \n\u2022 If we change the direction of field by inter changing the two poles of the \nmagnet, again the direction of exert force will change. \n\u2022 Therefore the direction of exerted force depends on \n(a) direction of current \n(b) direction of magnetic field lines. \n15. Left Hand Fleming Rule \n \nOR", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\n\u2022 According to this rule, stretch thumb , forefinger , and middle finge r of \nyour left hand such that they are mutually perpendicular to each other. \nIf fore finger represents direction of magnetic field & middle finger represent \ndirection of current, then thumb will point in the direction of motion or force acting \non the conductor. \n\u2013 ELECTRIC MOTOR: \nElectrical energy is converted into mechanical energy by using an electric motor. \nElectric motor works on the basis of rule suggested by Marie Ampere and Fleming\u2019s \nLeft Hand Rule. \nIn an electric motor, a rectangular coil is suspended between the two poles of a \nmagnetic field. The electric su pply to the coil is connected with a commutator. \nCommutator is a device which reverses the direction of flow of electric current \nthrough a circuit. \nWhen electric current is supplied to the coil of electric motor, it gets deflected \nbecause of magnetic field . As it reaches the half way, the split ring which acts as", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nthrough a circuit. \nWhen electric current is supplied to the coil of electric motor, it gets deflected \nbecause of magnetic field . As it reaches the half way, the split ring which acts as \ncommutator reverses the direction of flow of electric current. Reversal of direction of \ncurrent reverses the direction of forces acting on the coil. The change in direction of \nforce pushes the coil ; and it moves another half turn. Thus, the coil completes one \nrotation around the axle. Continuation of this process keeps the motor in rotation. \nIn commercial motor, electromagnet; instead of permanent magnet; and armature is \nused. Armature is a soft iro n core with large number of conducting wire turns over it. \nLarge number of turns of conducting wire enhances the magnetic field produced by \narmature.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\n16. Michael Faraday : Gave the law of Electro magnetic Induction : When a \nconductor is set to move insid e a magnetic field or a magnetic field is set to be \nchanging around a conductor, electric current is induced in the conductor. \n17. Galvanometer : It is an instrument that can detect the presence of a current in a \ncircuit. If pointer is at zero (the centre of scale) then there will be no flow of current. \nIf the pointer deflect on either side right or left, this will show the direction of current. \nRepresented by", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nELECTRIC GENERATOR : \nThe structure of electric generator is similar to that of an electric motor . In case of an \nelectric generator a rectangular armature is placed within the magnetic field of a \npermanent magnet. The armature is attached to wire and is positioned in way that it \ncan move around an axle. \nWhen the armature moves within the magnetic field an electric current is induced. \nThe direction of induced current changes, when the armature crosses the halfway \nmark of its rotation. Thus, the direction of current changes once in every rotation. \nDue to this, the electric generator usually produces alternate current, i.e. AC. To \nconvert an AC generator into a DC generator, a split ring commutator is used. This \nhelps in producing direct current. \n \n18. Electromagnetic Induction: Can be explained by two experiments \n(a) First Experiment \u201cSelf Induction \u201d", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nIn this experiment, when the north pole of bar magnet is brought closer to the coil or \naway from the coil, we see momentary deflection in the needle of galvanometer on \neither side of null point. First right and then left. \nSimilarly, if we keep the magnet stationary and coil is made to move towards or away \nfrom the north pole of magnet. Again we will observe deflection in the needle of \ngalvanometer. \nIf both bar magnet and coil are kept stationary, there will be no deflection in \ngalvanometer . This experiment can also be done with the south pole of magnet, we \nwill observe the deflection in galvanometer, but it would be in opposite direction to \nthe previous case. \n It concludes that motion of magnet with respect to coil or vice -versa, changes \nthe magnetic field. Due to this change in magnetic field lines, potential difference is \ninduced in the same coil, which set up an induced current in the circuit. \n(b) Second Experiment : Mutual Induction", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nIn this experiment plug in the key that connects coi l with battery and observe the \ndeflection in galvanometer. Now plug out the key that disconnect the coil -1 from \nbattery and observe the deflection in galvanometer, which will be in reverse \ndirection. \nHence, we conclude that potential difference is induced in secondary coil (coil -2), \nwhenever there is a change in current, in primary coil(coil -1) (by on and off of key). \nThis is because, whenever there is change in current in primary coil \n \nMagnetic field associated with it also changes", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nMagnetic field associated with it also changes \n \nNow, magnetic field l ines around the secondary coil (coil -2) will change and induces \nthe electric current in it (observed by the deflection of needle of Galvanometer in \nsecondary circuit) \nThis process, by which changing of strength of current in primary coil, induces a \ncurrent in secondary coil is called Electromagnetic Induction\u201d \nThe induced current is found to be highest when the direction of motion of coil is at \nright angles to the magnetic field. \n19. Fleming\u2019s Right Hand Rule \n \nOR", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nOR \n \nRule can be defined as : \nStretch, thumb, forefinger, and middle finger of right hand, so that they are \nperpendicular to each other. The forefinger indicates direction of magnetic field, \nthumb shows the direction of motion of conductor, then the middle finger will shows \nthe direction of ind uced current. \nElectrical generator is based on the principle of electromagnetic induction. It \nconvert mechanical energy into electrical energy.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\n21. Advantages of Alternate Current (AC) over Direct Current (DC) \nElectric power can be transmitted to longe r distances without much loss of energy. \nTherefore cost of transmission is low. \nIn India the frequency of AC is 50Hz. It means after every 1/100 second it changes its \ndirection. \n22. Domestic Electric Circuits : \nIn our homes, the electric power supplied is of potential difference V = 220V and \nfrequency 50Hz. \nIt consist of three wires : \u2013 \n(1) Wire with red insulation cover \u2013 LIVE WIRE (POSITIVE) Live wire is at high potential \nof 220V \n(2) Wire with black insulation cover \u2013 NEUTRAL WIRE(NEGATIVE) Neutral wire is at \nzero potential Therefore, the potential difference between the two is 220V. \n(3) Wire with Green insulation cover \u2013 EARTH WIRE \nIt is connected to a copper plate deep in the earth near house. \nThe metallic body of the appliances is connected with the eart h wire as a safety \nmeasure.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\nEarth wire provide a low resistance to the current hence any leakage of current to the \nmetallic body of the appliances, keep its potential equal to that of earth. That means \nzero potential and the user is saved from severe ele ctric shock. \nPoint to be noted in Domestic Circuit \n(a) Each appliance has a separate switch of ON/OFF \n(b) In order to provide equal potential difference to each appliance, they should be \nconnected parallel to each other. So that they can be operated at any time.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-RESOURCE MATERIAL.txt\n24. Short Circuiting \nDue to fault in the appliances or damage in the insulation of two wires, the circuit will \noffer zero or negligible resistance to the flow of current. Due to low resistance, large \namount of current will flow. \nAccording to Joule \u2019s law of heating effect , heat is produced in live wire and produces \nspark, damaging the device and wiring. \n25. Overloading \nOverloading can be caused by (1) Connecting too many appliances to a single socket \nor (2) accidental rise in supply voltage if the total current drawn by the appliances at \na particular time exceeds the bearing capacity of that wire, it will get heated up. This \nis known as overloading. Fuse a safety device can prevent the circuit from \noverloading and short circuiting. \n \n \nCopied from MYCBSEGUIDE.COM.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-TEACHING METHODS.txt\nFr. A gnel school", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-TEACHING METHODS.txt\nDescription: \nElectricity and magnetism are ultimately inextricably linked. Electricity is \nthe term given to a group of physical phenomena involving electric \ncharges, their motions, and their effects. Magnetism is the force \ngenerated by the passage of an electric curren t through matter. \nThe magnetic effect of electric current - one of the important utility of \nelectric current, demonstrates that electricity and magnetism are linked \nto each other. \n Moving or changing magnetic field generates electric field and moving \nor ch anging electric field generates magnetic field. \n \nCurrent carrying conductor has magnetic field around it and a conductor \nmoving in the magnetic field can generate electricity in it. \nLearning Objective: \n\u00b7 Differentiate between the physical situations in which this rule and \nthe situations, in which Fleming\u2019s left hand rule, are to be used. \n\u00b7 Realize that when a ny two of the three (mutually perpendicular)", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-TEACHING METHODS.txt\nthe situations, in which Fleming\u2019s left hand rule, are to be used. \n\u00b7 Realize that when a ny two of the three (mutually perpendicular) \ndirections are given we can use this rule for finding the third unknown \ndirection. \n (a) direction of movement of the conductor \n (b) direction of magnetic field The students will be able to : \n\u00b7 Describe mag netic field and field lines. \n\u00b7 Explain magnetic field due to a straight current carrying \nconductor. \n\u00b7 Summarise the factors on which strength and direction of \nmagnetic field around a straight conductor. \n\u00b7 State the right hand thumb rule. \n\u00b7 Apply the right hand thumb rule. \n\u00b7 Demonstrate magnetic field due to a current through a circular \nloop \n\u00b7 Analyse the magnetic field pattern around a solenoid carrying \ncurrent \n\u00b7 Express force on a current carrying conductor in a magnetic \nfield. \n\u00b7 State Fleming\u2019s right hand rule.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-TEACHING METHODS.txt\ncurrent \n\u00b7 Express force on a current carrying conductor in a magnetic \nfield. \n\u00b7 State Fleming\u2019s right hand rule. \n\u00b7 Illustrate the use of the Fleming\u2019s right hand rule in some \nsimple situations \n (c) direction of induced current/emf \n\u00b7 Define electromagnetic induction . \n\u00b7 State the principle of an (a) electric motor (b) electric generator. \u00b7 Discuss the importance of fuse and earthing in domestic circuit \n\u00b7 Explain how does the electric fuse and earthing protect the \ndomestic circuit \n \n\u00b7 Demonstrate the understanding of terms (a) overloading (b) short \ncircuit \nPerception: \nUnderstanding: \u00b7 Make an electromagnet and explain its use in daily life \n\u00b7 Understand the working of an domestic electric circuits \n\u00b7 Draw the schematic diagram of common domestic circuit. \nChapter will be introduced by presenting a situation .", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-TEACHING METHODS.txt\n\u00b7 Understand the working of an domestic electric circuits \n\u00b7 Draw the schematic diagram of common domestic circuit. \nChapter will be introduced by presenting a situation . \nSuspending a bar magnet and observing the direction in which it is \ncoming to rest . \nNote the deflection when a compass is brought near to it. . \nThen setting up an electric circuit and bringing a compass near it \nshows the same deflection. \nQuestions to be asked : \nWhat is the analogy in the activities ? \nWhat do you infer from this ? \n \nLink to be used \n//ctsp:class contents/class X/physics/magnetism.ppt \n \n Tutorials from EXTRA -MARKS WILL BE SHOWN TO THE \nSTUDENTS \nQuestions to be raised to students \nHow does a magnetic compass shows the N -S direction of a place ? \nWhy does the compass deflect when placed near a current carrying \nwire ? \nWhy does the needle deflection changes with the reversal of current \nflow ? \nThere by deducing the right hand thumb rule.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-TEACHING METHODS.txt\nWhy does the compass deflect when placed near a current carrying \nwire ? \nWhy does the needle deflection changes with the reversal of current \nflow ? \nThere by deducing the right hand thumb rule. \nDemonstration of marking of field lines around a bar magnet and \ndiscussing the properties of magnetic lines of force. \nAt the end of the lesson students will be able to : \n Describe the following terms related to magnetism (a) \nmagnetic field (b) magnetic lines of force \n State the SI unit of magnetic field \n Explain the characteristics of magnetic lines of fo rce \n Draw the magnetic field lines around (a) bar magnet (b) \nCurrent carrying straight conductor (c) Current carrying circular \nloop (d)Current carrying solenoid Analysis: \n Draw the magnetic field lines around a current carrying (a) \nstraight conductor (b) circular loop (c) solenoid. \nTo experimentally investigate the magnetic field of a straight \nconductor or a solenoid.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-TEACHING METHODS.txt\nstraight conductor (b) circular loop (c) solenoid. \nTo experimentally investigate the magnetic field of a straight \nconductor or a solenoid. \n\u2022 Suspend a magnet fr om a thread into glycerin, honey, or some \nother clear, viscous fluid. \n\u2022 Gently sprinkle iron filings on the liquid. \n\u2022 The iron filings remain suspended, arranging themselves three -\ndimensionally in the magnetic field around the magnet. \n\u2022 Draw a representation of the magnetic field pattern. \n\u2022 Trying to remove the iron filings from the magnet may be difficult, \nbut less so if you wrap the magnet in plastic sandwich wrap \nbeforehand. \nIf an electromagnet can be set up for this demonstration, it not only \nhelps to remove the iron filings, but it can also be used to illustrate the \ncollapse of the magnetic field when the current to the electromagnet is \ndiscontinued \nSynthesis: State Fleming\u2019s right hand thumb rule and to use this rule to \nfind the direction of magnetic field.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-TEACHING METHODS.txt\ndiscontinued \nSynthesis: State Fleming\u2019s right hand thumb rule and to use this rule to \nfind the direction of magnetic field. \nhttp://www.khanacademy.org/science/physics/v/magnetism -6--\nmagnetic -field-due-to-current \nExperimentation \nAllow a definite am ount of current to flow through a flexible \nconducting wire. Change the shape of the wire and measure the field \nstrength at a fixed distance, by changing the shape of the wire. Use a \nmagnetic compass to compare the field strength. Analyse whether \nwith the c hange of shape of conductor the magnetic field strength \nvaries. \nQuestion \n Does a definite amount of current always exhibits the same \nvalue of magnetism ? \n Determine the factors on which field strength depends with a \n(a) straight conductor (b) circular loop (c) solenoid. \nDiscussion of a situation \nIn a television screen the charged ions striking at the screen \nilluminates different points .", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-TEACHING METHODS.txt\n(a) straight conductor (b) circular loop (c) solenoid. \nDiscussion of a situation \nIn a television screen the charged ions striking at the screen \nilluminates different points . \n What make the charged particles to strike at varies points? \n Magnetic force is one of the fundamental type of force , can it \nbe deflecting charged particles. Similarly current carrying Evaluation: \nQuestions raised to students \nElectric energy can be changed to mechanical energy, can the reverse \nenergy transformation possible , think about the possible circuit set up \nCreativity: \nDesign analog electric meters like ammeter, voltmeter, \ngalvanometer which can be operated on motor principle.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-TEACHING METHODS.txt\nhttp://www.thesecond school.com/agnelone/noida/m/lesson_plans/view/\nMagnetic -Effect -of-electric -current \n conductor can be pulled by a magnet or in turn magnet exert \nforce on a conductor placed near by. \n What is the magnitude and direction of force experienced by a \ncurrent carrying conductor in a magnetic field ? \n Magnet can exert force on a current carrying element, how can \nit bring rotational motion ? What should be the direction of \nmagnetic field? \n How can this rotational motion kept in continuation ? \n Design an electric motor ? What are the essential components \nof an electric motor ? \n How is an ac motor different from a dc motor ? \nWith just a C cell, paperclips, a small magnet, rubber band and a coil \nof wire, students can build an operating electric motor that illustrates \nthe role of electricity and magnetism in a simple, but fundamental, \ntechnology.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-notes.txt\n13. MAGNETIC EFFECT OF ELECTRIC OF ELECTRIC CURRENT \nSUB: SCIENCE (PHYSICS) CLASS: X", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-notes.txt\nC. VERY SHORT ANSWER TYPE: - \n1. What does an electric current carrying wire behaves like? \n2. How do you define field lines? \n3. What do you mean by solid? \n4. What do you understand by electromagnet? \n5. Draw magnetic field lines due to current through circular loop. \n6. State the properties of magnetic field lines. \n7. Name the type of current used in house and dry cell. \n8. Name the physical quantities of electromagnetic induction. \n9. Give one applications of EMI \n10. Name any two domestic appliances which work based on Joule\u2019s law of heating \neffect. \n11. What constitutes magnet field? \n12.Wh y and when does a current carrying conductor kept in a magnetic field \nexperience force? List the factors on which direction of this force depends? \n13.How is the strength of magnetic field near a straight current -conductor \n(i) related to the strength of cur rent in the conductor? \n(ii) is affected by changing the direction of flow of current in the conductor?", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-notes.txt\n(i) related to the strength of cur rent in the conductor? \n(ii) is affected by changing the direction of flow of current in the conductor? \n14.What is meant by the term \u2018frequency of an alternating current\u2019? What is its \nvalue in India? Why is an alternating current considered to be advantageo us over \ndirect current for long range transmission of electric energy? \n15. The given magnet is divided into three parts A, B, and C .", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-notes.txt\nName the parts where the strength of the magnetic field is: (i) maximum \n(ii) minimum . \nHow will the density of magnetic f ield lines differ at these parts? \n16. A compass needle is placed near a current -carrying wire. State your \nobservation for the following cases, and give reason for the same in each case. \n(a) Magnitude of electric current in the wire is increased. \n(b) The compass needle is displaced away from the wire. \n \n17. (a) Two magnets are lying side by side as shown below. \nDraw magnetic field line between poles P and Q. \n \n(b) What does the degree of closeness of magnetic field lines near the poles \nsignify? \n \n18. Magnetic field lines of two magnets are shown in fig. A and fig. B. \n \nSelect the figure that represents the co rrect pattern of field lines. Give reasons for \nyour answer. Also name the poles of the magnets facing each other.", + "CHAPTER 13-OUR ENVIRONMENT-competancy based questions.txt\nCOMPETANCY LEVEL \nI.A village pond is found fully covered with algae and scum. The people in the \nvillage use the water from this pond for their house hold purposes. It is also \nfound that many children in that village is suffering from diarrhea and muscle \ncramps. As a precaution the doctor advice them not to drink water from the \npond, and if at all they use it ,it should be boiled and cooled. \nQ1. As a science student tell the reason f or the over growth of algae in the pond. \nQ2. What change will come to the quality of water when it is boiled and cooled? \nQ3. What are the common causes of diarrhea? \nII. Our experience says that the forest cover in India is declining day by day and \nalso the re is disappearance of more number of wild species of plants and \nanimals. As a part of the country\u2019s development many multi - purpose dams are", + "CHAPTER 13-OUR ENVIRONMENT-competancy based questions.txt\nalso the re is disappearance of more number of wild species of plants and \nanimals. As a part of the country\u2019s development many multi - purpose dams are \nalso found to come up in various parts of the country. Correspondingly we are \nexperiencing change in rain fall pat tern and unexpected cyclones and floods. \nQ1.What is the required forest cover in the hilly region and plain area in india? \nQ2.What are the various uses of multipurpose dams. \nQ3. Do you found any relation between the decline in forest cover and change in \nrainfall pattern, if so write the effect \nIII. Presently there is legal ban on running of hazardous industries in residential areas and near \nschools. The running of such industries in these areas are found causing a lot of sound pollution, air \npollution and w ater pollution . Due to such pollutions the general health of residents are affected by \nthe development of nervous disorders, psychological problems, respiratory diseases etc. The impact", + "CHAPTER 13-OUR ENVIRONMENT-competancy based questions.txt\nthe development of nervous disorders, psychological problems, respiratory diseases etc. The impact \nof the running of industries are not only limited to human life, but also found have it\u2019s impact on \nNatural ecosystems also. We can read that many important species of plants and animals are \ndisappeared in the recent past due to habitat destructuction , fragmentation and even due to bio \nmagnification. We know that many indu stries releases hot water to natural water bodies which lead \nto alteration of the temperature of water, killing many aquatic life specially fishes. \nQ 1. Why do you think that the change in temperature of water bodies affect fish life? \nQ 2. Write few air po llutants released from industries. \nIV. Over recent decades one of the commonest characteristics manifest in the developing nations \nhas been the disparity between rapid urban population growth and sanitation infrastructure", + "CHAPTER 13-OUR ENVIRONMENT-competancy based questions.txt\nIV. Over recent decades one of the commonest characteristics manifest in the developing nations \nhas been the disparity between rapid urban population growth and sanitation infrastructure \nprovision. This disparity is being worsened by the challenges of poor waste management practices \nimpacting on the deteriorating ecosystems of the rapidly transforming cities in these countries. The \nproduct of this mismatch, described as \u2018urbanisation without health\u2019, is the catalogue of \novercrowding, growth in illegal settlements, uncollected household waste, and the absence of \nwater, sanitation and other basic facilities which are typical of many urban centres in Africa, Asia and South America.As a result many millions of the urban poor li ve in neighbor hoods typically hazardous \nto their everyday health and general well -being \nQ1. According to the content what worsened the disparity between rapid urban \nPopulation growth and sanitation infra structure provision?", + "CHAPTER 13-OUR ENVIRONMENT-competancy based questions.txt\nto their everyday health and general well -being \nQ1. According to the content what worsened the disparity between rapid urban \nPopulation growth and sanitation infra structure provision? \nQ2. Which of the following is the product of mismatch between rapid \nUrbanization and sanitation infrastructure provision. \na) Urbanization without health \nb) Health without sanitation \nc) Health without nutrition \nd) Health with out infrastructure.", + "CHAPTER 13-OUR ENVIRONMENT- Innovative paedagogy-1.txt\nINNOVATIVE PAEDAGOGY \n1.Students can be directed to present environment related articles from news \n papers and magzines and can have discussion in the class. \n2. Videos of environmental issues can be shown to students. And can ask them to \n write the synopsis of it. \n3. Student teams can develop skit on the inter relationship between various \n components of ecosystem and cane presented in the class as well as part of \n CCA activities. \n4. Feild trips can be conducted for students to important gardens and parks to \n gather firsthand knowledge about ecosystems. \n5. Visits to Garbage treatment units can be arranged. \n6. Students can be asked to prepare PPt on ozone depletion and it\u2019s impact \n7. Projects can be assigned to students. \n8. Brainstorming sessions can be arranged in the class. \n9. Cross word puzzles can be designed and administered to students. \n10. seminars.", + "CHAPTER 13-OUR ENVIRONMENT-Lesson Plan.txt\nLESSON PLAN \nCHAPTER 15 OUR ENVIRONMENT \nTOPIC Our Environment \nClass Transaction Total 3 -4 Periods \n Sub Topic \n 1.Ecosystem \n 2.Trophic levels \n 3.Food chain and food web \n 4.Energy flow through Fooc chain \n 5.Ozone Depletion \n 6.Managing garbage \nPre requisite for the Topic Previos Knowledge of \n 1.Importance of organisms in the environment \n 2.Role of organisms in the environment \n 3.Types of organisms in the environment \nLearning objectives *Explain the relationship between organisms in \nthe \n environment \n *Explain the function and position of various \norganisms \n in the environment and their inter relations. \n * Enable the students to distinguish between \nproducers , \n consumers of varioys levels and decoposers. \n *Explain the concept of food chain and food web \n *Explain the flow of energy through various \ntrophic levels \n * To make aware of the students about the \nimpact of \n various human activites on environment", + "CHAPTER 13-OUR ENVIRONMENT-Lesson Plan.txt\n*Explain the flow of energy through various \ntrophic levels \n * To make aware of the students about the \nimpact of \n various human activites on environment \n *Explain ozone depletion and it's impact on the \nlife of \n organisms etc. \nMethodology Brain storming can start with discussion on what \nalready \n students know about environment and hence \nwhat is it \n they would learn now. \n They would also be told the significance of the \ntopic \n Lecture cum discussion, demonstration \n Explanation: Teacher will explain the type of \norganisms \n present in the environment by giving \nexamples.Teacher \n can also ask for different types of habitats present on earth \nLearning Out come 1.Students know to define \nenvironment,ecosystem, \n food chain , food web, trophic level, biological \nmagnification \n biodegradable, non biodegrdable \nsubstances,ozone hole \n 2.They are able to identify producers, consumers \nof variuos \n levels and decomposers in the environment.", + "CHAPTER 13-OUR ENVIRONMENT-Lesson Plan.txt\nmagnification \n biodegradable, non biodegrdable \nsubstances,ozone hole \n 2.They are able to identify producers, consumers \nof variuos \n levels and decomposers in the environment. \n 3.They are able to understand the concepts of \nfood chain \n food web, trophic level, Energy flow through \nfood chain \n 4.They know the features of energy flow \nthroughecosystem \n and it's significance. \n 5.They are able to construct food chains and food \nwebs in \n various ecosystems. \n 6.They understand the significance of ozone layer \n 7.They understand the impact of garbage in our \nlife and \n know how to handle it \n Students are able to apply the knowledge gained \nto protect \n our Environment etc. \nAssignment and home work 1.Make a list of biodegradable and \nnonbiodegrdable wastes \n 2.Make a list of natural and man made \necosystems. \n 3.Prepare models of food chains and food webs. \n 4.Explain different effective ways for the disposal \nof garbage", + "CHAPTER 13-OUR ENVIRONMENT-Lesson Plan.txt\nnonbiodegrdable wastes \n 2.Make a list of natural and man made \necosystems. \n 3.Prepare models of food chains and food webs. \n 4.Explain different effective ways for the disposal \nof garbage \n 5.Identify the various sources of biological \nmagnifications \n taking place in your village. \n*****", + "CHAPTER 13-OUR ENVIRONMENT-Long answer questions.txt\nLONG ANSWER TYPE \n1. Explain the phenomenon of \u201cbiological magnification\u201d How does it affect \norganisms belonging to different trophic levels particularly the tertiary \nconsumers? \n2. a) What are biodegradable and non - biodegradable substances? Select two \nbiodegradable pollutants from the following: Agricultural waste, glass, plastic, \nsewage, DDT. \nb) Consider the following food chain which occurs in a forest: Grass -> Deer -> \nLion \nIf 10000 J o f solar energy is available to the grass, how much energy would be \nAvailable to the deer to transfer it to the lion? \n3.Indicate the flow of energy in an ecosystem. Why is it unidirectional? Justify. \n4. a)What are decomposers? What will be the consequence o f their absence in an \necosystem? \nb)Name the wastes which are generated in your house daily. What measures \nwould you take for their disposal?", + "CHAPTER 13-OUR ENVIRONMENT-Long answer questions.txt\necosystem? \nb)Name the wastes which are generated in your house daily. What measures \nwould you take for their disposal? \n5. Explain some harmful effects of agricultural practices on the environment. \n6. Explain the phenomenon of \u201cbiologi cal magnification\u201d How does it affect organisms belonging to \ndifferent trophic levels particularly the tertiary consumers? \n7.\u201cDamage to the ozone layer is a cause for concern.\u201d Justify this statement. Suggest any two steps \nto limit this damage. \n8. Explain the effect of human activities on our environment. \n9. a)Decomposers are very much essential for the sustenance of ecosystems, \nExplain \nb) Explain 10% law of energy transfer through food chain with a suitable \nexample \n10. a) Why food chains cannot prolong ind efinitely? \nb) Which are the different levels of consumers in a food chain, On what basis \nthey are categorized", + "CHAPTER 13-OUR ENVIRONMENT- Reading material (2).txt\nCHAPTER \u201315\nOUR ENVIRONMENTIntroduction \nWe have the word \nenvironment often being \nused on the television, in \nnewspapers and by people \naround us. Environment (F. \nenviron \u2013about ) is the \nsum total of all external \nfactors, substances, livingbeings, and conditions that \nsurround an organism and \ninfluence the same without \nbecoming its part. It \nconsists of both living and \nnon-living components, \neach of which has several \nsub-components which are \ninterlinked andinterdependent on each \nother. In this chapter, we \nshall be studying how \nvarious factors in the \nenvironment interact with \neach other and how we \nimpact the environment. Sub-Concept 15.1 \nWhat happens when we \nadd our waste to the \nenvironment?Effect of adding waste to \nthe environment \n\uf0a7Human activities produce \na lot of waste materials \nwhich are thrown away \ninto the environment. \nThese wastes cause pollution of air, water \nand soil. The waste \nmaterials produced are \nof two main types. They \nare biodegradable", + "CHAPTER 13-OUR ENVIRONMENT- Reading material (2).txt\nwhich are thrown away \ninto the environment. \nThese wastes cause pollution of air, water \nand soil. The waste \nmaterials produced are \nof two main types. They \nare biodegradable \nwastes and non \nbiodegradable wastes.i)Biodegradable wastes: are \nwastes which are \ndecomposed into harmless \nsubstances by \nmicroorganisms. E. g. \nvegetables, fruits, pulses, \ncereals, cotton, jute, wool, \nwood, leather, paper, \nanimal dung, animal bones \netc.ii) Non biodegradable \nwastes :-are wastes which \nare not decomposed by \nmicroorganisms. E. g. \npolythene bags, plastics, \nsynthetic fibres, glass, \nmetals, synthetic rubber, \ninsecticides, pesticides \netc. Sub-Concept 15.2\nEco-system \u2013What \nare its components?Ecosystem :An ecosystem \nconsists of all the living \norganisms in an area along \nwith the non living \ncomponents and their \ninteraction.\nThere are different types of \necosystems.", + "Sub-Concept 15.2\nEco-system \u2013What \nare its components?Ecosystem :An ecosystem \nconsists of all the living \norganisms in an area along \nwith the non living \ncomponents and their \ninteraction.\nThere are different types of \necosystems. They are : -\ni)Natural ecosystems :-like \nforests, deserts, grass lands, \nmountains, ponds, lakes,", + "CHAPTER 13-OUR ENVIRONMENT- Reading material (2).txt\nwith the non living \ncomponents and their \ninteraction.\nThere are different types of \necosystems. They are : -\ni)Natural ecosystems :-like \nforests, deserts, grass lands, \nmountains, ponds, lakes, \nrivers, oceans etc. ii) Artificial ecosystems : -\nlike gardens, parks, crop \nfields, aquarium, zoo etc.\nComponents of an \necosystem:\nAn ecosystem consists of \ntwo main components. They \nare biotic and abiotic \ncomponents.i) Biotic components :-are \nthe living components like \nplants, animals \nandmicroorganisms. They \nconsist of producers, \nconsumers and decomposers.\n\uf0a7Producers :-are green \nplants which produce food by \nphotosynthesis.\uf0a7Consumers :-are \nherbivores which get their \nfood directly from plants, \ncarnivores which get their \nfood indirectly from plants \nand omnivores which get \ntheir food directly or \nindirectly from plants.\uf0a7Decomposers :-are \nmicroorganisms which \ndecompose dead plants and \nanimals. They decompose \ncomplex organic substances \ninto simple inorganic", + "CHAPTER 13-OUR ENVIRONMENT- Reading material (2).txt\ntheir food directly or \nindirectly from plants.\uf0a7Decomposers :-are \nmicroorganisms which \ndecompose dead plants and \nanimals. They decompose \ncomplex organic substances \ninto simple inorganic \nsubstances in the soil which \nare again used by plants.ii) Abiotic components :-are \nthe non living components \nlike air, water, soil, \nminerals, sunlight , \ntemperature, wind etc.Food chain :\nA food chain is the flow of food \nenergy from one organism to \nthe next and to the next and so \non. They usually start with a \nproducer (plants) and end with \na carnivore. In a food chain an \norganism gets food from one \ngroup of organisms.Eg:Grass Deer Lion\n(producer) (1 consumer) (2 consumer)\nGrass Insects Frog Snakes\n(producer) (1 consumer) (2 consumer) (3 consumer)\nGrass Moth Frog Snake HawkFood web :\nFood web is a group of \nseveral interconnected food \nchains.", + "In a food web an \norganism gets food from \nmore than one group of \norganisms.\n\uf0a7\nTrophic levels : -\n\uf0a7Each step in a food chain", + "CHAPTER 13-OUR ENVIRONMENT- Reading material (2).txt\nFood web is a group of \nseveral interconnected food \nchains. In a food web an \norganism gets food from \nmore than one group of \norganisms.\n\uf0a7\nTrophic levels : -\n\uf0a7Each step in a food chain \nwhere transfer of food \nenergy takes place is called \ntrophic level.\n\uf0a7The first trophic level \nconsists of producers.\uf0a7The second trophic level \nconsists of primary \nconsumers.\n\uf0a7The third trophic level \nconsists of secondary \nconsumers. \uf0a7The fourth trophic level \nconsists of tertiary \nconsumers.", + "\uf0a7The third trophic level \nconsists of secondary \nconsumers. \uf0a7The fourth trophic level \nconsists of tertiary \nconsumers.\nSince the transfer of food \nenergy decreases at every \ntrophic level, the number \nof trophic levels are limited \nand do not exceed four or \nfive.Energy flow in trophic\nlevels \nGreen plants (producers) \nabsorb about 1% of solar \nenergy falling on the \nleaves and stores it as \nfood energy during \nphotosynthesis.During the transfer of food \nenergy from one trophic\nlevel to the next, 90% of the \nenergy is lost to the \nenvironment and only 10% \nis transferred to the next \ntrophic level. So there is a decrease in", + "CHAPTER 13-OUR ENVIRONMENT- Reading material (2).txt\nenergy from one trophic\nlevel to the next, 90% of the \nenergy is lost to the \nenvironment and only 10% \nis transferred to the next \ntrophic level. So there is a decrease in \nthe amount of food energy \ntransferred at every trophic\nlevel by 10%. This is \nknown as the 10% law. Biological magnification \n(Biomagnification)\nHarmful chemicals like \ninsecticides and pesticides \nwhich are used to protect \ncrops from insects and pests \nare absorbed by plants and \nenter the food chain. Since \nthese chemicals are nonbiodegradable, they get \naccumulated at every \ntrophic level and their \nconcentration increases. \nSince human beings occupy \nthe highest trophic level, \nthe concentration of these \nharmful chemicals is \nmaximum in our bodies.The increase in \nconcentration of harmful \nchemicals in the bodies of \norganisms at higher \ntrophic levels is called \nbiological magnification.Sub-Concept: 15.3\nHow do our activities \naffect the \nenvironment?Depletion of ozone layer in \nthe atmosphere:", + "CHAPTER 13-OUR ENVIRONMENT- Reading material (2).txt\norganisms at higher \ntrophic levels is called \nbiological magnification.Sub-Concept: 15.3\nHow do our activities \naffect the \nenvironment?Depletion of ozone layer in \nthe atmosphere: \nOzone molecule contains \nthree oxygen atoms (O3). At \nhigher levels in the \natmosphere the UV radiation \nsplits some oxygen \nmolecules (O2) into free \noxygen atoms which \ncombineswith oxygen molecules (O2) \nto form ozone. It is highly \npoisonous.\nUV radiation \nO2 O + O \nO2+ O O3The ozone layer present in \nthe higher layer of the \natmosphere protects the \nearth from the harmful UV \nradiation from the sun. UV \nradiation causes skin cancer \nin humans. The ozone layer is being \ndamaged by the use of \nchemicals like chloro fluoro\ncarbons (CFCs) used in \nrefrigerators and fire \nextinguishers. So the use of \nCFCs is now being reduced \nto protect the ozone layer.", + "The ozone layer is being \ndamaged by the use of \nchemicals like chloro fluoro\ncarbons (CFCs) used in \nrefrigerators and fire \nextinguishers. So the use of \nCFCs is now being reduced \nto protect the ozone layer. Managing the garbage we \nproduce :\nThe household waste is \ncalled garbage. Some of \nthe garbage is \nbiodegradable and some \nare non biodegradable.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material (2).txt\nto protect the ozone layer. Managing the garbage we \nproduce :\nThe household waste is \ncalled garbage. Some of \nthe garbage is \nbiodegradable and some \nare non biodegradable. \nGarbage causes pollution\uf0a7of air, water and soil. So \nit should be disposed \nproperly.\nSome of the methods of \ngarbage disposal are : -\ni) Land fills\nii) Recycling \niii) Production of biogas \nand manureiv) Preparation \nof compost\nv) Incineration\nvi) Sewage \ntreatment \nSewage treatmentSewage treatmentSummary\n\uf0a7The various components \nof an ecosystems are \ninterdependent.\n\uf0a7The produces make the \nenergy from the sunlight \navailable to rest of the \necosystem.\uf0a7There is a loss of energy \nas we go from one tropic \nlevel to the next this \nlimits the number of \ntropic levels in a food \nchain. \n\uf0a7Human actvisists have an \nimpact on the \uf0a7The use of chemicals like \nCFCs has endangered the \nozone layer.", + "\uf0a7Human actvisists have an \nimpact on the \uf0a7The use of chemicals like \nCFCs has endangered the \nozone layer.\n\uf0a7The disposal of waste we \ngenerate is causing \nserious environmental \nproblems.", + "CHAPTER 13-OUR ENVIRONMENT-short answer type questions.txt\nSHORT ANSWER TYPE \n1.How is ozone formed in the upper atmosphere? Why is the damage of ozone \nlayer a cause of concern to us? State a cause of this damage. \n2. Construct an aquatic food chain showing four trophic levels. \n3. Explain \u2018biological magnification\u2019 with the help of an example. \n4. Describe how decomposers facilitate recycling of matter in order to maintain \nbalance in the ecosystem. \n5. \u201cDamage to the ozone layer is a cause for concern.\u201d Justify this statement. \nSuggest any two steps to limit this damage. \u2018 \n6. Why did United Nations act to control the production of chlorofluorocarbons \n(CFCs) used in refrigerators? \n7. What is the important function of presence of ozone in earth\u2019s atmosphere ? \n8. Give an example to illustrate that indiscriminate use of pesticides may result -in \nthe degradation of the environment. \n9. Differentiate between natural and artificial ecosystem.", + "CHAPTER 13-OUR ENVIRONMENT-short answer type questions.txt\n8. Give an example to illustrate that indiscriminate use of pesticides may result -in \nthe degradation of the environment. \n9. Differentiate between natural and artificial ecosystem. \n10. Why are pesticides considered as pollutants despite being useful to th e \nfarmers? \n11.What is an ecosystem? Mention its components \n12. Differentiate between natural and artificial ecosystem. \n13. What is the role of consumers in the food chain? \n14. Write the long term environmental impact of landfills. \n15. list the ill effects of depletion of ozone layer on our health.", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nScience 256Our Environment15 CHAPTER\nWe have heard the word \u2018environment\u2019 often being used on the\ntelevision, in newspapers and by people around us. Our elders tell\nus that the \u2018environment\u2019 is not what it used to be earlier; others say thatwe should work in a healthy \u2018environment\u2019; and global summits involvingthe developed and developing countries are regularly held to discuss\n\u2018environmental\u2019 issues. In this chapter, we shall be studying how various\nfactors in the environment interact with each other and how we impactthe environment. In Class IX, we saw how different materials are cycledin the environment in separate biogeochemical cycles. In these cycles,\nessential nutrients like nitrogen, carbon, oxygen and water are changed\nfrom one form to another. We shall now see how human activities affectthese cycles.", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nessential nutrients like nitrogen, carbon, oxygen and water are changed\nfrom one form to another. We shall now see how human activities affectthese cycles.\n15.115.115.115.115.1 WHAWHAWHAWHAWHA T HAPPENS WHEN WE ADD OUR WT HAPPENS WHEN WE ADD OUR WT HAPPENS WHEN WE ADD OUR WT HAPPENS WHEN WE ADD OUR WT HAPPENS WHEN WE ADD OUR W ASTE TOASTE TOASTE TOASTE TOASTE TO\nTHE ENVIRONMENT?THE ENVIRONMENT?THE ENVIRONMENT?THE ENVIRONMENT?THE ENVIRONMENT?\nIn our daily activities, we generate a lot of material that are thrown\naway. What are some of these waste materials? What happens after wethrow them away? Let us perform an activity to find answers to thesequestions.\nActivity 15.1Activity 15.1Activity 15.1Activity 15.1Activity 15.1\n\u0084Collect waste material from your homes.", + "What happens after wethrow them away? Let us perform an activity to find answers to thesequestions.\nActivity 15.1Activity 15.1Activity 15.1Activity 15.1Activity 15.1\n\u0084Collect waste material from your homes. This could include all\nthe waste generated during a day, like kitchen waste (spoilt food,\nvegetable peels, used tea leaves, milk packets and empty cartons),\nwaste paper , empty medicine bottles/strips/bubble packs, old and", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nthe waste generated during a day, like kitchen waste (spoilt food,\nvegetable peels, used tea leaves, milk packets and empty cartons),\nwaste paper , empty medicine bottles/strips/bubble packs, old and\ntorn clothes and broken footwear.\n\u0084Bury this material in a pit in the school garden or if there is nospace available, you can collect the material in an old bucket/\nflower pot and cover with at least 15 cm of soil.\n\u0084Keep this material moist and observe at 15-day intervals.\n\u0084What are the materials that remain unchanged over long periods\nof time?\n\u0084What are the materials which change their form and structureover time?\n\u0084Of these materials that are changed, which ones change thefastest?\nOur Environment 257We have seen in the chapter on \u2018Life Processes\u2019 that the food we eat is\ndigested by various enzymes in our body. Have you ever wondered why\nthe same enzyme does not break-down everything we eat? Enzymes are\nspecific in their action, specific enzymes are needed for the break-down", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nthe same enzyme does not break-down everything we eat? Enzymes are\nspecific in their action, specific enzymes are needed for the break-down\nof a particular substance. That is why we will not get any energy if we\ntry to eat coal! Because of this, many human-made materials like plastics\nwill not be broken down by the action of bacteria or other saprophytes.\nThese materials will be acted upon by physical processes like heat and\npressure, but under the ambient conditions found in our environment,\nthese persist for a long time.\nSubstances that are broken down by biological processes are said\nto be biodegradable. How many of the substances you buried were\nbiodegradable? Substances that are not broken down in this manner\nare said to be non-biodegradable. These substances may be inert and\nsimply persist in the environment for a long time or may harm the various\nmembers of the eco-system.\nActivity 15.2Activity 15.2Activity 15.2Activity 15.2Activity 15.2", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nsimply persist in the environment for a long time or may harm the various\nmembers of the eco-system.\nActivity 15.2Activity 15.2Activity 15.2Activity 15.2Activity 15.2\n\u0084Use the library or internet to find out more about biodegradable\nand non-biodegradable substances.\n\u0084How long are various non-biodegradable substances expected tolast in our environment?\n\u0084These days, new types of plastics which are said to be biodegradableare available. Find out more about such materials and whether\nthey do or do not harm the environment.\nQUESTIONSQUESTIONSQUESTIONSQUESTIONSQUESTIONS\n?1. Why are some substances biodegradable and some non-biodegradable?\n2. Give any two ways in which biodegradable substances would affect the\nenvironment.\n3. Give any two ways in which non-biodegradable substances would affect\nthe environment.", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\n2. Give any two ways in which biodegradable substances would affect the\nenvironment.\n3. Give any two ways in which non-biodegradable substances would affect\nthe environment.\n15.2 ECO15.2 ECO15.2 ECO15.2 ECO15.2 ECO -SYSTEM \u2014 WHA-SYSTEM \u2014 WHA-SYSTEM \u2014 WHA-SYSTEM \u2014 WHA-SYSTEM \u2014 WHA T ARE ITS COMPONENTS?T ARE ITS COMPONENTS?T ARE ITS COMPONENTS?T ARE ITS COMPONENTS?T ARE ITS COMPONENTS?\nAll organisms such as plants, animals, microorganisms and human\nbeings as well as the physical surroundings interact with each other\nand maintain a balance in nature. All the interacting organisms in an\narea together with the non-living constituents of the environment form\nan ecosystem. Thus, an ecosystem consists of biotic components\ncomprising living organisms and abiotic components comprising\nphysical factors like temperature, rainfall, wind, soil and minerals.\nFor example, if you visit a garden you will find different plants, such", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nphysical factors like temperature, rainfall, wind, soil and minerals.\nFor example, if you visit a garden you will find different plants, such\nas grasses, trees; flower bearing plants like rose, jasmine, sunflower;Science 258and animals like frogs, insects and birds. All these living organisms\ninteract with each other and their growth, reproduction and otheractivities are affected by the abiotic components of ecosystem. So a gardenis an ecosystem. Other types of ecosystems are forests, ponds and lakes.These are natural ecosystems while gardens and crop-fields are human-made (artificial) ecosystems.\nWe have seen in earlier classes that organisms can be grouped as", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nWe have seen in earlier classes that organisms can be grouped as\nproducers, consumers and decomposers according to the manner inwhich they obtain their sustenance from the environment. Let us recallwhat we have learnt through the self sustaining ecosystem created byus above. Which organisms can make organic compounds like sugarand starch from inorganic substances using the radiant energy of theSun in the presence of chlorophyll? All green plants and certain blue-green algae which can produce food by photosynthesis come under thiscategory and are called the producers.\nOrganisms depend on the producers either directly or indirectly for\ntheir sustenance? These organisms which consume the food produced,either directly from producers or indirectly by feeding on other consumersare the consumers. Consumers can be classed variously as herbivores,carnivores, omnivores and parasites. Can you give examples for each ofthese categories of consumers?", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\n\u0084Imagine the situation where you do not clean the aquarium andsome fish and plants have died. Have you ever thought whathappens when an organism dies? The microorganisms, comprisingbacteria and fungi, break-down the dead remains and wasteproducts of organisms. These microorganisms are the decomposersas they break-down the complex organic substances into simple\ninorganic substances that go into the soil and are used up once\nmore by the plants. What will happen to the garbage, and deadanimals and plants in their absence? Will the natural replenishmentof the soil take place, even if decomposers are not there?Activity 15.3Activity 15.3Activity 15.3Activity 15.3Activity 15.3\n\u0084You might have seen an aquarium. Let us try to design one.\n\u0084What are the things that we need to keep in mind when we create\nan aquarium? The fish would need a free space for swimming (it\ncould be a large jar), water , oxygen and food.\n\u0084We can provide oxygen through an oxygen pump (aerator) and", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nan aquarium? The fish would need a free space for swimming (it\ncould be a large jar), water , oxygen and food.\n\u0084We can provide oxygen through an oxygen pump (aerator) and\nfish food which is available in the market.\n\u0084If we add a few aquatic plants and animals it can become a self-\nsustaining system. Can you think how this happens? An aquarium\nis an example of a human-made ecosystem.\n\u0084Can we leave the aquarium as such after we set it up? Why does\nit have to be cleaned once in a while? Do we have to clean ponds\nor lakes in the same manner? Why or why not?Our Environment 25915.2.1 Food Chains and Webs\nIn Activity 15.4 we have formed a series of organisms feeding\non one another. This series or organisms taking part at variousbiotic levels form a food chain (Fig. 15.1).\nEach step or level of the food chain forms a trophic level.\nThe autotrophs or the producers are at the first trophic level.They fix up the solar energy and make it available for", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nEach step or level of the food chain forms a trophic level.\nThe autotrophs or the producers are at the first trophic level.They fix up the solar energy and make it available for\nheterotrophs or the consumers. The herbivores or the primary\nconsumers come at the second, small carnivores or thesecondary consumers at the third and larger carnivores or\nthe tertiary consumers form the fourth trophic level (Fig. 15.2).\nWe know that the food we eat acts as a fuel to provide us\nenergy to do work. Thus the interactions among various\ncomponents of the environment involves flow of energy from\none component of the system to another. As we have studied,the autotrophs capture the energy present in sunlight and\nconvert it into chemical energy. This energy supports all the\nactivities of the living world. From autotrophs, the energy goesto the heterotrophs and decomposers. However , as we saw in\nthe previous Chapter on \u2018Sources of Energy\u2019, when one form", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nactivities of the living world. From autotrophs, the energy goesto the heterotrophs and decomposers. However , as we saw in\nthe previous Chapter on \u2018Sources of Energy\u2019, when one form\nof energy is changed to another, some energy is lost to theenvironment in forms which cannot be used again. The flow\nof energy between various components of the environment\nhas been extensively studied and it has been found that \u2013\n\u0084The green plants in a terrestrial ecosystem capture about\n1% of the energy of sunlight that falls on their leaves and\nconvert it into food energy.\n\u0084When green plants are eaten by primary consumers, a great\ndeal of energy is lost as heat to the environment, some amount\ngoes into digestion and in doing work and the rest goestowards growth and reproduction. An average of 10% of the\nfood eaten is turned into its own body and made available for\nthe next level of consumers.\n\u0084Therefore, 10% can be taken as the average value for the", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nfood eaten is turned into its own body and made available for\nthe next level of consumers.\n\u0084Therefore, 10% can be taken as the average value for the\namount of organic matter that is present at each step and\nreaches the next level of consumers.Activity 15.4Activity 15.4Activity 15.4Activity 15.4Activity 15.4\n\u0084While creating an aquarium did you take care not to put an aquatic\nanimal which would eat others? What would have happened\notherwise?\n\u0084Make groups and discuss how each of the above groups oforganisms are dependent on each other.\n\u0084Write the aquatic organisms in order of who eats whom and form\na chain of at least three steps. \u2192 \u2192 \n\u0084Would you consider any one group of organisms to be of primary\nimportance? Why or why not?", + "\u0084Write the aquatic organisms in order of who eats whom and form\na chain of at least three steps. \u2192 \u2192 \n\u0084Would you consider any one group of organisms to be of primary\nimportance? Why or why not?\nFigure 15.1\nFood chain in nature\n(a) in forest, (b) in\ngrassland and (c) in a\npond\nFigure 15.2\nTrophic levelsScience 260\n\u0084Since so little energy is available for the\nnext level of consumers, food chains\ngenerally consist of only three or four", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\ngrassland and (c) in a\npond\nFigure 15.2\nTrophic levelsScience 260\n\u0084Since so little energy is available for the\nnext level of consumers, food chains\ngenerally consist of only three or four\nsteps. The loss of energy at each step is\nso great that very little usable energy\nremains after four trophic levels.\n\u0084There are generally a greater numberof individuals at the lower trophic levels\nof an ecosystem, the greatest number\nis of the producers.\n\u0084The length and complexity of foodchains vary greatly. Each organism is\ngenerally eaten by two or more other\nkinds of organisms which in turn are\neaten by several other organisms. So\ninstead of a straight line food chain, the\nrelationship can be shown as a series\nof branching lines called a food web\n(Fig. 15.3).\nFrom the energy flow diagram (Fig. 15.4),\ntwo things become clear. Firstly, the flow of\nenergy is unidirectional. The energy that is\ncaptured by the autotrophs does not revert\nback to the solar input and the energy which", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\ntwo things become clear. Firstly, the flow of\nenergy is unidirectional. The energy that is\ncaptured by the autotrophs does not revert\nback to the solar input and the energy which\npasses to the herbivores does not come back\nto autotrophs. As it moves progressively\nthrough the various trophic levels it is no\nlonger available to the previous level.Figure 15.3\nFood web, consisting of many food chains\nFigure 15.4 Diagram showing flow of energy in an ecosystemOur Environment 261Another interesting aspect of food chain is how unknowingly some\nharmful chemicals enter our bodies through the food chain. You have\nread in Class IX how water gets polluted. One of the reasons is the use ofseveral pesticides and other chemicals to protect our crops from diseasesand pests. These chemicals are either washed down into the soil or intothe water bodies. From the soil, these are absorbed by the plants alongwith water and minerals, and from the water bodies these are taken up", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nby aquatic plants and animals. This is one of the ways in which they\nenter the food chain. As these chemicals are not degradable, these getaccumulated progressively at each trophic level. As human beings occupythe top level in any food chain, the maximum concentration of thesechemicals get accumulated in our bodies. This phenomenon is knownas biological magnification. This is the reason why our food grains such\nas wheat and rice, vegetables and fruits, and even meat, contain varying\namounts of pesticide residues. They cannot always be removed bywashing or other means.\nActivity 15.5Activity 15.5Activity 15.5Activity 15.5Activity 15.5\n\u0084Newspaper reports about pesticide levels in ready-made food items\nare often seen these days and some states have banned these\nproducts. Debate in groups the need for such bans.\n\u0084What do you think would be the source of pesticides in these fooditems? Could pesticides get into our bodies from this source through\nother food products too?", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\n\u0084What do you think would be the source of pesticides in these fooditems? Could pesticides get into our bodies from this source through\nother food products too?\n\u0084Discuss what methods could be applied to reduce our intake ofpesticides.\nQUESTIONSQUESTIONSQUESTIONSQUESTIONSQUESTIONS\n?1. What are trophic levels? Give an example of a food chain and state the\ndifferent trophic levels in it.\n2. What is the role of decomposers in the ecosystem?\n15.315.315.315.315.3 HOW DO OUR ACTIVITIES AFFECT THEHOW DO OUR ACTIVITIES AFFECT THEHOW DO OUR ACTIVITIES AFFECT THEHOW DO OUR ACTIVITIES AFFECT THEHOW DO OUR ACTIVITIES AFFECT THE\nENVIRONMENT?ENVIRONMENT?ENVIRONMENT?ENVIRONMENT?ENVIRONMENT?\nWe are an integral part of the environment. Changes in the environment", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nENVIRONMENT?ENVIRONMENT?ENVIRONMENT?ENVIRONMENT?ENVIRONMENT?\nWe are an integral part of the environment. Changes in the environment\naffect us and our activities change the environment around us. We havealready seen in Class IX how our activities pollute the environment. Inthis chapter, we shall be looking at two of the environmental problemsin detail, that is, depletion of the ozone layer and waste disposal.Science 26215.3.1 Ozone Layer and How it is Getting Depleted\nOzone (O3) is a molecule formed by three atoms of oxygen. While O2,\nwhich we normally refer to as oxygen, is essential for all aerobic forms oflife. Ozone, is a deadly poison. However, at the higher levels of the\natmosphere, ozone performs an essential function. It shields the surface\nof the earth from ultraviolet (UV) radiation from the Sun. This radiationis highly damaging to organisms, for example, it is known to cause skincancer in human beings.\nOzone at the higher levels of the atmosphere is a product of UV", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nOzone at the higher levels of the atmosphere is a product of UV\nradiation acting on oxygen (O\n2) molecule. The higher energy UV\nradiations split apart some moleculer oxygen (O2) into free oxygen (O)\natoms. These atoms then combine with the molecular oxygen to form\nozone as shown\u2014\nOO + O2UV\u23af\u2192\u23af\u23af\nOO O23\nOzone )+\u2192\n(\nThe amount of ozone in the atmosphere began to drop sharply in\nthe 1980s. This decrease has been linked to synthetic chemicals like\nchlorofluorocarbons (CFCs) which are used as refrigerants and in fire\nextinguishers. In 1987, the United Nations Environment Programme\n(UNEP) succeeded in forging an agreement to freeze CFC production at\n1986 levels.\nActivity 15.6Activity 15.6Activity 15.6Activity 15.6Activity 15.6\n\u0084Find out from the library, internet or newspaper reports, which\nchemicals are responsible for the depletion of the ozone layer.", + "Activity 15.6Activity 15.6Activity 15.6Activity 15.6Activity 15.6\n\u0084Find out from the library, internet or newspaper reports, which\nchemicals are responsible for the depletion of the ozone layer.\n\u0084Find out if the regulations put in place to control the emission ofthese chemicals have succeeded in reducing the damage to the", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nchemicals are responsible for the depletion of the ozone layer.\n\u0084Find out if the regulations put in place to control the emission ofthese chemicals have succeeded in reducing the damage to the\nozone layer. Has the size of the hole in the ozone layer changed in\nrecent years?\nActivity 15.7Activity 15.7Activity 15.7Activity 15.7Activity 15.7\n\u0084Find out what happens to the waste generated at home. Is there a\nsystem in place to collect this waste?\n\u0084Find out how the local body ( panchayat , municipal corporation,\nresident welfare association) deals with the waste. Are there\nmechanisms in place to treat the biodegradable and non-\nbiodegradable wastes separately?15.3.2 Managing the Garbage we Produce\nVisit any town or city, and we are sure to find heaps of garbage all over\nthe place. Visit any place of tourist interest and we are sure to find the\nplace littered with empty food wrappers. In the earlier classes we have\ntalked about this problem of dealing with the garbage that we generate.", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nplace littered with empty food wrappers. In the earlier classes we have\ntalked about this problem of dealing with the garbage that we generate.\nLet us now look at the problem a bit more deeply.Our Environment 263Improvements in our life-style have resulted in greater amounts of\nwaste material generation. Changes in attitude also have a role to play,with more and more things we use becoming disposable. Changes inpackaging have resulted in much of our waste becoming non-biodegradable. What do you think will be the impact of these on our\nenvironment?Activity 15.8Activity 15.8Activity 15.8Activity 15.8Activity 15.8\n\u0084Calculate how much waste is generated at home in a day.\n\u0084How much of this waste is biodegradable?\n\u0084Calculate how much waste is generated in the classroom\nin a day.\n\u0084How much of this waste is biodegradable?\n\u0084Suggest ways of dealing with this waste.", + "\u0084How much of this waste is biodegradable?\n\u0084Calculate how much waste is generated in the classroom\nin a day.\n\u0084How much of this waste is biodegradable?\n\u0084Suggest ways of dealing with this waste.\nActivity 15.9Activity 15.9Activity 15.9Activity 15.9Activity 15.9\n\u0084Find out how the sewage in your locality is treated. Are there", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\n\u0084Suggest ways of dealing with this waste.\nActivity 15.9Activity 15.9Activity 15.9Activity 15.9Activity 15.9\n\u0084Find out how the sewage in your locality is treated. Are there\nmechanisms in place to ensure that local water bodies are not\npolluted by untreated sewage.\n\u0084Find out how the local industries in your locality treat their wastes.\nAre there mechanisms in place to ensure that the soil and water\nare not polluted by this waste?\nActivity 15.10Activity 15.10Activity 15.10Activity 15.10Activity 15.10\n\u0084Search the internet or library to find out what hazardous materials\nhave to be dealt with while disposing of electronic items. How would\nthese materials affect the environment?\n\u0084Find out how plastics are recycled. Does the recycling processhave any impact on the environment?Disposable cups in trains\nIf you ask your parents, they will probably remember a time when tea in trains was", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\nIf you ask your parents, they will probably remember a time when tea in trains was\nserved in plastic glasses which had to be returned to the vendor. The introduction ofdisposable cups was hailed as a step forward for reasons of hygiene. No one at that timeperhaps thought about the impact caused by the disposal of millions of these cups on adaily basis. Some time back, \nkulhads , that is, disposable cups made of clay, were\nsuggested as an alternative. But a little thought showed that making these kulhads on\na large scale would result in the loss of the fertile top-soil. Now disposable paper-cups\nare being used. What do you think are the advantages of disposable paper-cups over\ndisposable plastic cups?Think it overScience 264What you have learnt\n\u0084The various components of an ecosystem are interdependent.\n\u0084The producers make the energy from sunlight available to the rest of the ecosystem.\n\u0084There is a loss of energy as we go from one trophic level to the next, this limits the", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\n\u0084The producers make the energy from sunlight available to the rest of the ecosystem.\n\u0084There is a loss of energy as we go from one trophic level to the next, this limits the\nnumber of trophic levels in a food-chain.\n\u0084Human activities have an impact on the environment.\n\u0084The use of chemicals like CFCs has endangered the ozone layer. Since the ozonelayer protects against the ultraviolet radiation from the Sun, this could damagethe environment.\n\u0084The waste we generate may be biodegradable or non-biodegradable.\n\u0084The disposal of the waste we generate is causing serious environmental problems.\nEXERCISES\n1. Which of the following groups contain only biodegradable items?\n(a) Grass, flowers and leather\n(b) Grass, wood and plastic\n(c) Fruit-peels, cake and lime-juice\n(d) Cake, wood and grass\n2. Which of the following constitute a food-chain?\n(a) Grass, wheat and mango\n(b) Grass, goat and human\n(c) Goat, cow and elephant\n(d) Grass, fish and goat", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\n(d) Cake, wood and grass\n2. Which of the following constitute a food-chain?\n(a) Grass, wheat and mango\n(b) Grass, goat and human\n(c) Goat, cow and elephant\n(d) Grass, fish and goat\n3. Which of the following are environment-friendly practices?\n(a) Carrying cloth-bags to put purchases in while shopping\n(b) Switching off unnecessary lights and fans\n(c) Walking to school instead of getting your mother to drop you on her scooter\n(d) All of the aboveQUESTIONSQUESTIONSQUESTIONSQUESTIONSQUESTIONS\n?1. What is ozone and how does it affect any ecosystem?\n2. How can you help in reducing the problem of waste disposal? Give any\ntwo methods.Our Environment 2654. What will happen if we kill all the organisms in one trophic level?\n5. Will the impact of removing all the organisms in a trophic level be different for\ndifferent trophic levels? Can the organisms of any trophic level be removed withoutcausing any damage to the ecosystem?", + "CHAPTER 13-OUR ENVIRONMENT-textbook.txt\ndifferent trophic levels? Can the organisms of any trophic level be removed withoutcausing any damage to the ecosystem?\n6. What is biological magnification? Will the levels of this magnification be different at\ndifferent levels of the ecosystem?\n7. What are the problems caused by the non-biodegradable wastes that we generate?\n8. If all the waste we generate is biodegradable, will this have no impact on the\nenvironment?\n9. Why is damage to the ozone layer a cause for concern? What steps are being taken\nto limit this damage?", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nOur Environment Class 10 Notes Science Chapter 15 \nBiodegradable and Non -biodegradable Wastes, Ecosystem, Components of Ecosystem. The environment \nincludes our physical surroundings like air (or atmosphere), water bodies, soil (land and all the \norganisms such as plants, animals, human beings and micro -organisms like bacteria and fungi (called \ndecomposers). The waste materials produced by the various activities of man \u00a3nd animals are poisonous \nto some extent and can be divided into two main groups \n1. Biodeg radable Wastes: Substances that are broken down by the biological processes are said to be \nbiodegradable. These substances are decomposed through the actions of fungi, bacteria, and other \nliving organisms. Temperature and sunlight also play an important ro le in the decomposition of \nbiodegradable substances. \nFor Examples: Food waste, trees leaves, urine and fecal matter, sewage agricultural residue, paper, \nwood, cloth, cow -dung etc.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nbiodegradable substances. \nFor Examples: Food waste, trees leaves, urine and fecal matter, sewage agricultural residue, paper, \nwood, cloth, cow -dung etc. \n2. Non -Biodegradable Wastes: Substances that are not broken down by biologic al \nprocesses. These substances may be in solid, liquid or gaseous form. These substances are inert and \nsimply persist in the environment for a long time or may harm the various members of the ecosystem. \nFor Examples: These includes DDT (Di -chloro -di phenyl trichloro ethane -in-pheneyle the cheoro ethane), \ninsecticides, pesticides, mercury, lead, arsenic aluminum, plastics, polythene bags, glass, radioactive \nwastes. These non -biodegradable wastes are major pollutants of the environment. \nHarmful effects of bio degradable and Non -Biodegradable Substance \n1. The waste destroys the natural beauty and our surroundings become dirty. \n2. Decomposition of these wastes results in the production of foul smell, which spreads to the", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\n1. The waste destroys the natural beauty and our surroundings become dirty. \n2. Decomposition of these wastes results in the production of foul smell, which spreads to the \nsurrounding areas. \n3. These wastes may also block t he drains creating pools of waste, which becomes the breeding \nsites of mosquitoes. The latter is carriers of diseases like malaria and dengue. \nDifference between Biodegradable and Non -Biodegradable wastes \nBiodegradable wastes Non -Biodegradable wastes \n1. The wastes that are broken down naturally by microbial \naction. 1. The wastes that are not broken down by the microbes. \n2. Biodegradation forms harmless and non - poisonous \nproducts. 2. No such action is possible. 3. They release raw materials back to nature. 3. They do not release raw materials. \n4. They pollute the environment only when they are produced \nin quantity beyond the capacity of the environment to \ndegrade them. 4. Non -biodegradable wastes pollute the environment even in", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\n4. They pollute the environment only when they are produced \nin quantity beyond the capacity of the environment to \ndegrade them. 4. Non -biodegradable wastes pollute the environment even in \nsmall quantity. \n5. Bioconcentration does not occur. 5. Bioconcentration or biomagnifications occurs when wastes \nenter food chains. \n6. Recycling is possible both naturally or through human \nefforts. 6. Recycling is possible only through human efforts. \nEcosystem : An ecosystem is a self -contained unit of living things (plants, animals and \ndecomposers), and their non -living environment (soil, air and water). For example; a forest, a pond, a \nlake, a green land etc. \nIn an ecosystem, energy and matter are continuously exchanged betw een living and non\u00acliving \ncomponents. \nAn ecosystem can be both natural or man -made. Some examples of natural ecosystems are grass land, \nforest, sea, river, desert, mountain, pond, lake etc.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\ncomponents. \nAn ecosystem can be both natural or man -made. Some examples of natural ecosystems are grass land, \nforest, sea, river, desert, mountain, pond, lake etc. \nThe desert, grass land and mountains represent the terrestrial ecosystem (land -based ecosystem). \nThe ponds, rivers, lakes and sea represent the aquatic ecosystem (water -based ecosystem). Man -\nmade artificial ecosystems are garden, crop fields, park, aquarium, etc.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nComponents of Ecosystem : There are two components of an ecosystem : (i) biotic component and (ii) \nabiotic component. \n1. Biotic component: It includes three types of organisms : \n(a) Producers: All green plants, blue green algae can produce their food (Sugar and starch) from \ninorganic substance using light energy (Photosynthesis). Therefore, all green plants are called producers. \nThey are also called autotrophs. \nPlanktons are very minute or microscopic organisms freely floating on the surface of water in a pond, \nlake, river or ocean. Planktons are of two typ es : Phytoplanktons and Zooplanktons. \nThe microscopic aquatic plants freely floating on the surface of water are called phytoplanktons. \nThe microscopic aquatic animals freely floating on water are called zooplanktons. The freely floating \nprotozoa are an ex ample of zooplankton. \n(b) Consumers: They are organisms which consume other organisms or their products as their food. All", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nprotozoa are an ex ample of zooplankton. \n(b) Consumers: They are organisms which consume other organisms or their products as their food. All \nanimals belong to this category. The consumers depend upon producers for their food directly or \nindirectly. They get their food by ea ting other organisms or their products. For example, man, goat, \ndeer, fish, lion, cow, buffalo, etc., are common consumers. \nThe consumers can be classified into the following types : \n\uf0b7 Herbivores. \n\uf0b7 Carnivores. \n\uf0b7 Parasite. \n\uf0b7 Omnivores.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\n(i) Herbivores: These are organisms (animals) which get their food by eating the producers (or plant) \ndirectly. Herbivores are also called first order consumers. Some common examples of herbivores are : \ndeer, rabbit, rat, squirrel, goat, cattle, etc. \n(ii) Carnivores: These are orga nisms (animals) which consume other animals. Therefore, carnivores feed \non the flesh of herbivores. These are also called primary carnivores or second order consumers. Some \ncommon examples are snake, wild cat, jackal, frog, some birds, fishes, etc. \nThere a re animals which prey upon primary carnivores. They are called second order consumers or third \norder consumers. For example, owl, peacock, tiger, lion, etc., are some second order carnivores and may \nbe eaten by third order carnivores. The carnivores which are not preyed upon further are called top \ncarnivores. For example, lion is a top carnivore.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nbe eaten by third order carnivores. The carnivores which are not preyed upon further are called top \ncarnivores. For example, lion is a top carnivore. \n(iii) Omnivores: The organisms which feed on both plants and animals are called omnivores. Human \nbeings are common example of omnivores because they eat both plant s (For example; pulses, grams, \noilseeds, fruit, etc.) and animal products (milk, meat, egg, etc.). \n(c) Decomposers: Fungi and bacteria which break down (decompose) the dead plants, animals complex \ncompounds into the simpler one. The decomposers help in the replenishment of natural resources. \nThese are also known as microorganism or saprotrophs. These are also called reducers. \nImportance of Decomposers \n\uf0b7 Decomposers help in disposing of the wastes and dead bodies of plants and animals. Therefore, \nthey clean th e environment and create space for a living of newer generations of organisms.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\n\uf0b7 Decomposers help in disposing of the wastes and dead bodies of plants and animals. Therefore, \nthey clean th e environment and create space for a living of newer generations of organisms. \n\uf0b7 The decomposers release minerals and other raw materials trapped in organic matter. These are \npicked up by plants. This also helps to maintain the fertility of soil. \n\uf0b7 The decompo sers produce some acids which are useful in solubilization of some minerals. \n\uf0b7 Decomposers help in recycling the materials in the biosphere so that, the process of life may go \non and on like an unending chain. \n2. Abiotic Components: These are non -living comp onents of an ecosystem. These include the physical \nenvironment. \n\uf0b7 Edaphic factors like soil texture, topography, water, and air. \n\uf0b7 Inorganic substances like carbon dioxide, nitrogen, oxygen, water, phosphorus, sodium, \npotassium, and calcium. These are involved in the cyclic of materials in the ecosystem.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\n\uf0b7 Inorganic substances like carbon dioxide, nitrogen, oxygen, water, phosphorus, sodium, \npotassium, and calcium. These are involved in the cyclic of materials in the ecosystem. \n\uf0b7 Organic compounds like proteins, carbohydrates, and lipids. These largely form the living body \nand link the abiotic and biotic components. \nClimatic factors: These are sunlight temperature, pressure humidity, mo isture, rainfall, etc. these factors \naffect the distribution of the organisms. \nFunctions of an Ecosystem \n\uf0b7 Ecosystem indicates available solar energy and the efficiency of an ecosystem to trap the same. \n\uf0b7 It gives information about the available essential mine rals and their recycling periods. \n\uf0b7 It provides knowledge about the web of interactions and inter -relationship among the various \npopulation as well as between the population and the abiotic environment. \n\uf0b7 It helps human beings to know about conservation of resources, protection from pollution and", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\npopulation as well as between the population and the abiotic environment. \n\uf0b7 It helps human beings to know about conservation of resources, protection from pollution and \ninputs required for maximizing productivity. \n\uf0b7 In the ecosystem, two processes of energy flow and biogeochemical cycles (nutrients \nmovement) proceed side by side. The energy flow is unidirectional while the movement o f \nnutrients is cyclic. Food chain, Food web, Trophic levels. Flow of energy ten percent law, Depletion of the ozone layer, \nBiological magnification. Mode of waste disposal. \nFood Chain: The sequence of living organisms in a community in which one organism c onsumes another \norganism to transfer food energy, is called a food chain. \nA food chain is unidirection where transfer of energy takes place in only one direction. \nOR \nFood chain is sequential process which represents \u201cwho eats whom\u201d. \nOR", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nA food chain is unidirection where transfer of energy takes place in only one direction. \nOR \nFood chain is sequential process which represents \u201cwho eats whom\u201d. \nOR \nFood chain refers to an arrangement of different biotic groups in a sequence of energy transfer. These \nbiotic groups are producer herbivores, carnivores. \nFor example, T1(Grass) \u2192 T2(Deer) \u2192 T3(Lion) \nExamples of Food Chains: Simple food chain operating in a grass land or fores t \nGrass(Producer) \u2192 Deer(Herbivore) \u2192 Lion(Carnivore) \nIn this food chain, grass represent the producers (first tropic level). Grass synthesize their own food by \nthe process of photosynthesis. Grass is eaten up by deer, which represents the herbivores or th e \nprimary consumers. Deer in turn is consumed by lion, the carnivores or the secondary consumers.", + "Grass synthesize their own food by \nthe process of photosynthesis. Grass is eaten up by deer, which represents the herbivores or th e \nprimary consumers. Deer in turn is consumed by lion, the carnivores or the secondary consumers. \nA food chain in grassland which has four steps is : \nGrass(Producers) \u2192 Insect(Herbivores) \u2192 Frog(Carnivores) \u2192 Eagle(Secondary Carnivore) \nSignificance of Food Chains", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nA food chain in grassland which has four steps is : \nGrass(Producers) \u2192 Insect(Herbivores) \u2192 Frog(Carnivores) \u2192 Eagle(Secondary Carnivore) \nSignificance of Food Chains \n\uf0b7 The study of food chains helps in understanding food relationships and interactions among the \nvarious organisms in an ecosystem. The food chains, transfer energy and materials between \nvarious living components of an ecosystem. \n\uf0b7 The food chains trans fer energy and materials between various living components in an \necosystem or biosphere. \n\uf0b7 The food chains give dynamicity to an ecosystem or biosphere. \n\uf0b7 The movement of toxic substances like pesticides, weedicides, etc., through food chains, can \nprove very harmful. \nFood Web: The inter -connected food chains operating in an ecosystem which establish a network of \nrelationship between various species, are called a food web.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nprove very harmful. \nFood Web: The inter -connected food chains operating in an ecosystem which establish a network of \nrelationship between various species, are called a food web. \nIn a food web, one organism may occupy a position in more than one food chain. An organis m can obtain its food from different sources and in turn, may be eaten up by different types of organisms.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nTrophic Levels: The various steps in the food chain at which the transfer of food (or energy) takes place \nis called trophic levels. \nThere is a grad ual decrease in the amount of energy transfer from one trophic level to the next trophic \nlevel in a food chain.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nSo only 10% of energy is transferred to next trophic level while 90% of energy is used by present trophic \nlevel in its life processes. \nThe var ious trophic levels are given below : \n\uf0b7 The plant or the producers constitute the first trophic level. \n\uf0b7 The herbivores or primary consumers form the second trophic level. \n\uf0b7 Carnivores or secondary consumers make up the third trophic level. \n\uf0b7 Large carnivores or t he tertiary consumers which feed upon the small carnivores constitute the \nfourth trophic level. \nFlow Open Energy \nEnergy is used and conveyed from one trophic level to another in a food chain. This is called flow of \nenergy. Green plants capture about 1% of the solar energy incident on the Earth through the \nbiochemical process of photosynthesis. A part of this trapped energy is used by plants in performing \ntheir metabolic activities and some energy is released as heat into the atmosphere. The remaining", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\ntheir metabolic activities and some energy is released as heat into the atmosphere. The remaining \nenergy is chemical energy stored in the plants as \u2018carbohydrates\u2019. When plants are eaten up by \nherbivores, the chemical energy stored in the plants is transferred to these animals. These animals \n(herbivores) utilize some of this energy for metabolic activities, some energy is \u201creleased as heat and the \nremaining energy is stored. The process of energy transferred is similarly repeated with carnivores and \nso on. \nTen percent law: Ten percent law states that only 10 percent of the energy entering a particular trophic \nlevel of organisms is available for transfer to the next higher trophic level.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nFor example, Suppose 1000 J of solar energy is received by green plants, then only 1% of solar energy \navailable on earth is utilized by plants. So only 10 J (1% of 1000 J) is trapped by plants and the rest 990 J \nof energy is lost to the environment. So, plants utilizes only 10 J of energy. Next, only 10% of the 10 J \nenergy of plant, that is, 1 J, is available to the herbivore animal while 9 J is lost to the environment. \nAgain, just 10% of the 1 J of energy of herbivore animals is utilized by carnivore animals. Thus, carnivore \nanimals have only 0.1 J of energy while 0.9 J is lost to the environment. \nEnvironmental Problems: Changes in the environment affect us and our activities change the \nenvironment around us. This led to the slow degradation of the environment that arose many \nenvironmental problems. For Example; depletion of the Ozone Layer and waste disposal.", + "Environmental Problems: Changes in the environment affect us and our activities change the \nenvironment around us. This led to the slow degradation of the environment that arose many \nenvironmental problems. For Example; depletion of the Ozone Layer and waste disposal. \nDepletion of Ozone Layer: Ozone (O3) layer is largely found in the s tratosphere which is a part of our", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nenvironmental problems. For Example; depletion of the Ozone Layer and waste disposal. \nDepletion of Ozone Layer: Ozone (O3) layer is largely found in the s tratosphere which is a part of our \natmosphere from 12 km -50 km above sea level. This region is called ozonosphere. Ozone is deadly \npoisonous at the ground level. \nOzone is formed as a result of the following photochemical reaction.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nOzone layer is a prote ctive blanket around earth which absorbs most of the harmful U.V. (Ultraviolet) \nradiation of the Sun, thus, protecting the living beings of the Earth from health hazards like skin cancer, \ncataract in eyes, weaken immune system, destruction of plants etc. T he decline of Ozone layer thickness \nin Antarctica was first discovered in 1985 and was termed as OZONE HOLE. \nSteps taken to limit damage of ozone layer: Excessive use of CFCs (Chloro Flouro Carbon) a synthetic, \ninert chemical. For example; Freon which are used as refrigerants and also in fire extinguishers caused \nOzone depletion in the upper atmosphere. A single chlorine atom can destroys 1,00,000 Ozone \nmolecules. U.N.E.P. (United Nation Environment Programme) did an excellent job in forging an \nagreement to freeze CFC production at 1986 levels (KYOTO Protocol) by all countries. \nBiological Magnification: The increase in concentration of harmful chemical substances like pesticides in", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nagreement to freeze CFC production at 1986 levels (KYOTO Protocol) by all countries. \nBiological Magnification: The increase in concentration of harmful chemical substances like pesticides in \nthe body of living organisms at each trophic level of a food chain is called biological magnification. \nExample:", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nMaximum concentration of such chemicals gets accumulated in human bodies. \nGarbage Disposal: Industrialization and rise in demand of consumer goods have created a major \nproblem in the form of wastes/garbage accumulation and its disposal especially in urban areas. \nThe disposal of waste should be done in a scientific way. There are different methods of waste disposal. \nThe method to be used depends on the nature of the waste. Some of the important modes of waste \ndisposal are : \n\uf0b7 Incineration: Burning of waste on high temperature to form ash is called incineration. This \nprocess is carried out in an incinerator. Incineration is used to destroy household, chemical and \nbiological wastes. \n\uf0b7 Open dumping: A conventional method in which solid waste are dumped in selected areas of a \ntown. It actually cause pollution \n\uf0b7 Land fillings: Wastes are dumped in low living areas and are compacted by rolling with \nbulldozers", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\ntown. It actually cause pollution \n\uf0b7 Land fillings: Wastes are dumped in low living areas and are compacted by rolling with \nbulldozers \n\uf0b7 Composting: Organic wastes are filled into a compost pit (2m \u00d7 1m \u00d7 1m). It i s then covered with \na thin layer of soil. After about three months the same garbage filled inside the pit changes into \norganic manure. \n\uf0b7 Recycling: The solid wastes is broken down into its constituent simpler materials. These \nmaterials are then used to make new items. Even non -bio degradable solid wastes like plastic, \nmetal can be recycled. \n\uf0b7 Reuse: A very simple conventional technique of using an item again and again. For example; \npaper can be reused for making envelops, etc\u2026 \nEnvironment: The combination of al l the physical and biological conditions affecting the responses of \nliving organisms is called environment. \nBiodegradable wastes: The wastes which are broken down by the activity of micro\u00acorganisms and enter", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nliving organisms is called environment. \nBiodegradable wastes: The wastes which are broken down by the activity of micro\u00acorganisms and enter \ninto the biogeochemical cycle are known as biode gradable wastes. \nNon -biodegradable wastes: The wastes which cannot be broken down by the enzymes produced by \nmicroorganisms into simpler and harmless products in nature are called non - biodegradable wastes. \nGarbage: Domestic wastes including the kitchen wa ste are termed as garbage. \nIncineration: Destruction of waste materials by burning at high temperature is called incineration. \nBiotic Community: A group of various populations of organisms living in a region is called the biotic \ncommunity. \nEcosystems: The self-contained and distinct functional unit capable of independent existence made by \nthe interaction of living and non -living components is called an ecosystem. \nEcosystem component consists of two components - Abiotic and biotic", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nthe interaction of living and non -living components is called an ecosystem. \nEcosystem component consists of two components - Abiotic and biotic \n\uf0b7 Abiotic: Components consist of inorganic and organic substances and climatic factors. \n\uf0b7 Biotic: Components consist of a living organism. \nAutotrophs: Those organisms which can produce their own food are called autotrophs or producers. All \ngreen plants are producers. \nConsumers: Those org anisms which are unable to synthesise their food themselves and consume the \nfood produced by producers or eat other organisms as food, are known as consumers. \nDecomposers: Bacteria and fungi which break down the complex organic compounds present in the \ndead plants and animals and their products into simpler substances are known as decomposers. \nFood Chain: A series of organisms in a community in which one organism consumes another organism to \ntransfer food energy is called a food chain. \nCharacteristics:", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nFood Chain: A series of organisms in a community in which one organism consumes another organism to \ntransfer food energy is called a food chain. \nCharacteristics: \n\uf0b7 A food chain helps in understanding the food relationship and interactions among various \norganisms in an ecosystem. \n\uf0b7 There is a progressive decline in the amount of energy available as we move from one trophic \nlevel to another in a food chain. \nTrophic levels: Each step of the food chain is known as a trophic level. 10% of food taken by one trophic \nlevel is available for the next trophic level. \nFood Web: The web formed by the interconnection of food -chains of the various trophic levels is called \na food web. \nBioma gnification: The increase in the concentration of the harmful chemicals in the body of an \norganism per unit its mass at each successive trophic level in a food chain is known as biomagnification.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\norganism per unit its mass at each successive trophic level in a food chain is known as biomagnification. \nOzone Depletion: The thinning of the ozone layer is called o zone depletion. Ozone -depleting substances: Chlorofluorocarbons (CFCs), oxides of nitrogen, methane, carbon \ntetrachloride and chlorine are the ozone -depleting substances. \nFour practices which can help in the protection of our environment: \n\uf0b7 Disposal of wastes after separating them into biodegradable and non -biodegradable waste \nmaterial. \n\uf0b7 Use of unleaded petrol and alternate sources of energy, and keeping the engine properly tuned \nand serviced and the tyres inflated to the right pressure so that the vehicl e runs efficiently. \n\uf0b7 The use of gunny bags and paper bags in place of polythene/plastic bags. \n\uf0b7 Activities such as gardening, rain -water harvesting and use of compost in place of fertilizers will \nhelp protect our environment from further damage.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\n\uf0b7 Activities such as gardening, rain -water harvesting and use of compost in place of fertilizers will \nhelp protect our environment from further damage. \nHarmful effec ts of agricultural practices on the environment. \n\uf0b7 Excessive use of fertilisers changes the chemistry of soil and kills useful microbes. \n\uf0b7 Excessive use of non -biodegradable chemical pesticides leads to biological magnification. \n\uf0b7 Extensive cropping causes loss of soil fertility. \n\uf0b7 Excess use of groundwater for agriculture lowers the water table. \n\uf0b7 Agricultural practices lead to some amount of damage to the natural ecosystem/habitat. \n1. Environment: The physical, chemical and \u2018 biological conditions of the region in which an organism \nlives is called its Environment. It includes air, light, soil, temperature, water and the presence or absence \nof other organisms, i.e., the conditions for development or growth. \nThe environment has three main components, viz :", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nof other organisms, i.e., the conditions for development or growth. \nThe environment has three main components, viz : \n\uf0b7 Physical su rroundings [soil, air and water bodies] \n\uf0b7 Living organisms [plants, animals, decomposers (bacteria and fungi)] \n\uf0b7 Meteorological factors (or climatic factors) . [sunlight, temperature, rainfall, humidity, pressure \nand wind speed]. \n2. Physical environment: It is also called a abiotic or non -living environment. It includes : \n\uf0b7 Soil, water bodies and air on the surface of the earth. \n\uf0b7 Meteorological factors. \nThe physical environment is essential for : \n\uf0b7 Supply of nutrient elements to the living beings. \uf0b7 Providing sp ace to the organisms for living. \n\uf0b7 Controlling weather of a place. \n3. Biotic (or biological) environment: It includes : \n\uf0b7 Plants. \n\uf0b7 Animals (including human beings). \n\uf0b7 Decomposers (bacteria and.fungi). \nOther important constituents of the biotic environment includes Kites and vultures as they feed on dead", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\n\uf0b7 Plants. \n\uf0b7 Animals (including human beings). \n\uf0b7 Decomposers (bacteria and.fungi). \nOther important constituents of the biotic environment includes Kites and vultures as they feed on dead \norganisms and act as scavengers (cleansing agents) of the environment. \n4. Ecosystem: A community of organisms, interacting with each other, plus the environment in which \nthey \nlive and with which they also in teract. The examples of the ecosystem are a pond; a desert; a forest; a \nlake; a river; a mountain; the sea. \nAll the above ecosystems are made up of two main components.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\n5. Autotrophs (Producers) and Heterotrophs (consumers):", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\n6. Food chain: The sequence of living organisms in a . community in which one organism.e^ts other and \nis \nitself eaten by another organism to transfer energy is called a food chain. It is also defined as, \u201cchain of \norganisms, existing in any natural community, through which energy is transferred\u201d. \n7. Ozone layer : Ozone (O 3) is a molecule formed by three atoms of oxygen unlike oxygen which is \nrequired for respiration by aerobic forms, ozone is a deadly poison. However, at the higher levels of the \natmosphere, \nozone performs an essentia l function. It shields the surface of the earth from ultraviolet (UV) radiation \nfrom the Sun. This radiation is highly damaging to organisms, for example, it is known to cause skin \ncancer in human beings. \nOzone at the higher levels of the atmosphere is a p roduct of UV radiation, acting on oxygen (O2) \nmolecule. The higher energy UV radiations split apart some molecular", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\ncancer in human beings. \nOzone at the higher levels of the atmosphere is a p roduct of UV radiation, acting on oxygen (O2) \nmolecule. The higher energy UV radiations split apart some molecular \noxygen (O2) into free oxygen (O) atoms. These atoms then combine with the molecular oxygen to form \nozone as shown:", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\nDepletion of ozone layer : Ozone layer gets depleted \u2013 due to the use of chemicals called aerosol, spray \npropellants like chlorofluorocarbons. Depletion of \u2018 ozone layer would cause skin cancer in men and \nanimals and severe damage to the plants. \n8. Biological magnification: It mea ns accumulation of non -biodegradable chemicals (like pesticides) in \nthe living organisms (like plants, animals, including man) in a food chain. \u201cThe increase in the \nconcentration of harmful chemicals in the body of living organisms at each trophic level of a food chain \nis called biological magnification\u201d. \n9. Biodegradable wastes and Non -biodegradable wastes: \n \n10. A generalised food chain: \n \n11. Food web: A food web is a network of food chains which establish a network of relationships \nbetween various speci es. Food web showing 8 interlinked food chains.", + "CHAPTER 13-OUR ENVIRONMENT- Reading material.txt\n11. Food web: A food web is a network of food chains which establish a network of relationships \nbetween various speci es. Food web showing 8 interlinked food chains. \n \n \n13. The flow of energy between various components of the environment: \n\uf0b7 Green plants capture about 1 % of the energy and convert it into food energy. \n\uf0b7 About f % of the food eaten is turned into the body of a n organism and made available for the \nnext level of consumers. \n\uf0b7 About 10% of organic matter is present at each step and reaches the next level of consumers. \n\uf0b7 Since so little energy is available for the next level of consumers, food chains generally consist o f \nonly three or four steps. The loss of energy at each step is so great that very little usable energy \nremains after four trophic levels. \n\uf0b7 There are generally a greater number of individuals at the lower trophic levels of an ecosystem, \nthe greatest number is of the producers.", + "CHAPTER 13-OUR ENVIRONMENT-MCQ.txt\nMCQ \n1.Which of the following is a non -biodegradable thing? \na) banana peel b)tin can c)cotton d)leather \n2. Which of the following components of sunlight is responsible for the formation of ozone in the \nstratosphere? \na) infra red rays b)UV Rays c)Visible light d)red light \n3. Garbage can be disposed more environment friendly by \na) incineration b)Land filling c)Recycling d)composting. \n4. Which of the following is an alternative name for primary consumers? \na)Carnivore b)herbivore c)s aprophyte d)Omnivore \n5. Which of the following statements remain true with regard to the flow of energy in the \necosystem? \na) Energy remain steady b) Go on increasing from lower to higher trophic levels \nc) Converted from one form to another.d) No energy rea ches at higher trophic levels \n6.", + "a) Energy remain steady b) Go on increasing from lower to higher trophic levels \nc) Converted from one form to another.d) No energy rea ches at higher trophic levels \n6. Decomposition is mainly carried out by \na)Worms b) bacteria c)virus d) flies. \n7. Which of the following may be a conclusion of the excessive exposure of \nhumans to sun\u2019s ultraviolet rays?", + "CHAPTER 13-OUR ENVIRONMENT-MCQ.txt\n6. Decomposition is mainly carried out by \na)Worms b) bacteria c)virus d) flies. \n7. Which of the following may be a conclusion of the excessive exposure of \nhumans to sun\u2019s ultraviolet rays? \ni. Peptic ulcers \nii. Eye disease like cataract \niii. Damage to lungs \niv. Skin cancer \n(a) (i) and (iv) \n(b) (ii), (iii) and (iv) \n(c) (ii) and (iv) \n(d) Only (iv) \n8. If 100 J energy is available at the producer level in a food chain then the energy \navailable to the secondary consumer will be: \n(a) 10 J \n(b) 0.1 J \n(c) 1 J \n(d) 0.01 J 9. What will happen if deer is missing in the food chain given below?", + "What will happen if deer is missing in the food chain given below? \nGrass \u2192 Deer \u2192 Tiger \n(a) The population of tiger increases \n(b) The population of grass decreases \n(c) Tiger will start eating grass \n(d) The population of tiger decreases and the population of grass increases \n10. Which of the following substances will not be converted to compost when \nadded in a composting pit? \n(a) Waste paper \n(b) Fruit and vegetable peels", + "CHAPTER 13-OUR ENVIRONMENT-MCQ.txt\n10. Which of the following substances will not be converted to compost when \nadded in a composting pit? \n(a) Waste paper \n(b) Fruit and vegetable peels \n(c) Human and animal excreta \n(d) Plastic bags \n11. Th e gas released by CFC when it comes in contact with UV rays is \na) hydrogen b) nitrogen c) ozone d) chlorine. \n12. Which of the following statements about a producer is right? \na) They do not take up carbon dioxide \nb) Do not release Oxygen \nc) Absorb energy fr om sun. \nd) Depend other organisms for food? \n13. Every food chain in the ecosystem begins with\u2026\u2026\u2026. which are the original \nsource of food. \na) Saprophytes \nb) Parasites \nc) Producers \nd) consumers \n14.We should reduce the use of the plastic bags, bottles etc. bec ause:a) They are not durable \nb) They are non -biodegradable \nc) They are made of toxic materials \nd) They react with the atmospheric gases \n15.", + "bec ause:a) They are not durable \nb) They are non -biodegradable \nc) They are made of toxic materials \nd) They react with the atmospheric gases \n15. Among the following choose the correct option which contains only biodegradable items?", + "CHAPTER 13-OUR ENVIRONMENT-MCQ.txt\nc) They are made of toxic materials \nd) They react with the atmospheric gases \n15. Among the following choose the correct option which contains only biodegradable items? \ni. Wood, paper, PVC ii. Paper, seeds, detergent, \niii. Paper, animal excreta, wood \niv. Wool, leaves, paper \n(a) (i), (ii) and (iii) \n(b) (i) and (iii) \n(c) (ii), (iii) and (iv) \n(d) (iii) and (iv) \n16. Which among the following statements is incorrect in view of the plants? \n(a) They co nvert the solar energy into mechanical energy \n(b) They prepare their food from organic compounds \n(c) They are also called producers \n(d) They are the initial source of energy in a food chain \n17. In a food chain the second trophic level is occupied by: \n(a) C arnivores \n(b) Autotrophs \n(c) Herbivores \n(d) Producers \n18.", + "In a food chain the second trophic level is occupied by: \n(a) C arnivores \n(b) Autotrophs \n(c) Herbivores \n(d) Producers \n18. Green plants utilize \u2026\u2026\u2026 percent of sun\u2019s energy to prepare their food by the process of \nphotosynthesis? \n(a) 1 percent \n(b) 10 percent \n(c) 20 percent \n(d) 99 percent", + "CHAPTER 13-OUR ENVIRONMENT-MCQ.txt\n(d) Producers \n18. Green plants utilize \u2026\u2026\u2026 percent of sun\u2019s energy to prepare their food by the process of \nphotosynthesis? \n(a) 1 percent \n(b) 10 percent \n(c) 20 percent \n(d) 99 percent \n19. The process of accumulation of harmful chemical substances like pesticides, in the body of living \norganisms at each trophic level of a food chain is known as: \n(a) Biological magnification \n(b) Biological accumulation \n(c) Chemical magnification \n(d) Chemical accumulation \n20. . Which of the following may be a conclusion of the excessive exposure of humans to sun\u2019s \nultraviolet rays? \ni. Peptic ulcers ii. Eye disease like cataract \niii. Damage to lungs \niv. Skin cancer \n(a) (i) and (iv) \n(b) (ii), (iii) and (iv) \n(c) (ii) and (iv) \n(d) O nly (iv) \n21. Which among the following is a correct full form for DDT?", + "Skin cancer \n(a) (i) and (iv) \n(b) (ii), (iii) and (iv) \n(c) (ii) and (iv) \n(d) O nly (iv) \n21. Which among the following is a correct full form for DDT? \n(a) Dichlorodiphenyltrichloroethane \n(b) Dichlorodiphenyltetrachloroethane \n(c) Dichlorodecaphenyltrichloroethane \n(d) Dichlorodiethyltrichloroethane", + "CHAPTER 13-OUR ENVIRONMENT-MCQ.txt\n(a) Dichlorodiphenyltrichloroethane \n(b) Dichlorodiphenyltetrachloroethane \n(c) Dichlorodecaphenyltrichloroethane \n(d) Dichlorodiethyltrichloroethane \n22. What will happen if deer is missin g in the food chain given below? \nGrass \u2192 Deer \u2192 Tiger \n(a) The population of tiger increases \n(b) The population of grass decreases \n(c) Tiger will start eating grass \n(d) The population of tiger decreases and the population of grass increases \n23. Global warmi ng is a phenomenon related to: \n(a) Evaporation \n(b) Ecological balance \n(c) Greenhouse effect \n(d) Desertification", + "CHAPTER 13-OUR ENVIRONMENT- PPT.txt\nCHAPTER -15\nOUR \nENVIRONMENTEcosystem and its components\nA) ECOSYSTEM AN ECOSYSTEM CONSISTS OF ALL THE L IVING \nORGANISMS I N AN AREA ALONG WITH THE NON L IVING COMPONENTS.\nB) COMPONENTS OF AN ECOSYSTEM\nAN ECOSYSTEM CONSISTS OF TWO MAIN COMPONENTS. THEY ARE BIOTIC \nAND ABIOTIC COMPONENTS.\nI) BIOTIC COMPONENTS ARE THE L I VING COMPONENTS L IKE PL ANTS, \nANIMALS AND MI CROORGANISMS. THEY CONSIST OF PRODUCERS, \nCONSUMERS AND DECOMPOSERS.\n. II) ABIOTIC COMPONENTS ARE THE NON L IVING COMPONENTS L IKE AI R, \nWATER, S OIL, H UMIDITY, MINERALS, S UNLIGHT , T EMPERATURE, WI ND E TC.Food chain\nA FOOD CHAIN I S THE S ERIES THAT S HOWS WHO EATS WHOM I N AN \nECOSYSTEM.", + "THEY US UALLY START WITH A PRODUCER ( PLANTS) AND END \nWITH A CARNIVORE. \nFood web\nFOOD WEB IS A NETWORK OF SEVERAL INTERCONNECTED FOOD \nCHAINS.. IN AN ECOSYSTEM ONE ORGANISM CAN BE EATEN BY \nSEVERAL OTHER ORGANISMS\nTrophic levels\nEACH STEP OR L EVEL I N A FOOD CHAIN I S CALLED TROPHIC L EVEL.", + "CHAPTER 13-OUR ENVIRONMENT- PPT.txt\nCHAINS.. IN AN ECOSYSTEM ONE ORGANISM CAN BE EATEN BY \nSEVERAL OTHER ORGANISMS\nTrophic levels\nEACH STEP OR L EVEL I N A FOOD CHAIN I S CALLED TROPHIC L EVEL.\nTHE FIRST TROPHIC L EVEL CONSISTS OF PRODUCERS.\nTHE S ECOND TROPHIC L EVEL CONSISTS OF PR IMARY CONSUMERS.\nTHE THIRD TROPHIC L EVEL CONSISTS OF S ECONDARY CONSUMERS. \nTHE FOURTH TROPHIC L EVEL CONSISTS OF T ERTIARY CONSUMERS.\nEnergy flow in trophic levels\nGREEN PLANTS (PRODUCERS) ABSORB ABOUT 1% OF SOLAR ENERGY \nFALLING ON THE LEAVES AND STORES IT AS FOOD ENERGY DURING \nPHOTOSYNTHESIS.", + "Energy flow in trophic levels\nGREEN PLANTS (PRODUCERS) ABSORB ABOUT 1% OF SOLAR ENERGY \nFALLING ON THE LEAVES AND STORES IT AS FOOD ENERGY DURING \nPHOTOSYNTHESIS.\nDURING THE TRANSFER OF FOOD ENERGY FROM ONE TROPHIC \nLEVEL TO THE NEXT, A GREAT DEAL OF ENERGY IS LOST AS HEAT TO \nTHE ENVIRONMENT, SOME AMOUNT GOES INTO DIGESTION AND IN \nDOING WORK AND THE REST GOES TOWARDS GROWTH AND \nREPRODUCTION. AN AVERAGE OF 10% OF THE FOOD EATEN IS TURNED \nINTO ITS OWN BODY AND MADE AVAILABLE FOR THE NEXT LEVEL OF \nCONSUMERS. \nBiological magnification \nHARMFUL CHEMICALS LIKE INSECTICIDES AND PESTICIDES WHICH ARE", + "CHAPTER 13-OUR ENVIRONMENT- PPT.txt\nINTO ITS OWN BODY AND MADE AVAILABLE FOR THE NEXT LEVEL OF \nCONSUMERS. \nBiological magnification \nHARMFUL CHEMICALS LIKE INSECTICIDES AND PESTICIDES WHICH ARE \nUSED TO PROTECT CROPS FROM INSECTS AND PESTS ARE ABSORBED BY \nPLANTS AND ENTER THE FOOD CHAIN. SINCE THESE CHEMICALS ARE \nNON BIODEGRADABLE, THEY GET ACCUMULATED AT EVERY TROPHIC \nLEVEL AND THEIR CONCENTRATION INCREASES PROGRESSIVELY WITH \nEACH TROPHIC LEVEL. SINCE HUMAN BEINGS OCCUPY THE HIGHEST \nTROPHIC LEVEL, THE CONCENTRATION OF THESE HARMFUL CHEMICALS \nIS MAXIMUM IN OUR BODIES.", + "SINCE HUMAN BEINGS OCCUPY THE HIGHEST \nTROPHIC LEVEL, THE CONCENTRATION OF THESE HARMFUL CHEMICALS \nIS MAXIMUM IN OUR BODIES.\nTHE INCREASE IN CONCENTRATION OF HARMFUL CHEMICALS IN THE \nBODIES OF ORGANISMS PROGRESSIVELY WITH EACH TROPHIC LEVEL IS \nCALLED BIOLOGICAL MAGNIFICATION.Human activities affect the environment\nA) DEPLETION OF OZONE LAYER IN THE ATMOSPHERE\nOZONE MOLECULE CONTAINS THREE OXYGEN ATOMS (O3). AT \nHIGHER LEVELS IN THE ATMOSPHERE THE UV RADIATION SPLITS SOME \nOXYGEN MOLECULES (O2) INTO FREE OXYGEN ATOMS WHICH", + "CHAPTER 13-OUR ENVIRONMENT- PPT.txt\nOZONE MOLECULE CONTAINS THREE OXYGEN ATOMS (O3). AT \nHIGHER LEVELS IN THE ATMOSPHERE THE UV RADIATION SPLITS SOME \nOXYGEN MOLECULES (O2) INTO FREE OXYGEN ATOMS WHICH \nCOMBINES WITH OXYGEN MOLECULES (O2) TO FORM OZONE. IT IS \nHIGHLY POISONOUS .\nTHE OZONE LAYER PRESENT IN THE HIGHER LAYER OF THE \nATMOSPHERE PROTECTS THE EARTH FROM THE HARMFUL UV \nRADIATION FROM THE SUN. UV RADIATION CAUSES SKIN CANCER IN \nHUMANS. \nTHE OZONE LAYER IS BEING DAMAGED BY THE USE OF CHEMICALS \nLIKE CHLORO FLUORO CARBONS (CFCS) USED IN REFRIGERATORS AND \nFIRE EXTINGUISHERS. SO THE USE OF CFCS IS NOW BEING REDUCED TO \nPROTECT THE OZONE LAYER.", + "SO THE USE OF CFCS IS NOW BEING REDUCED TO \nPROTECT THE OZONE LAYER. \nManaging the waste we produce\nTHE HOUSEHOLD WASTE IS CALLED GARBAGE. SOME OF THE \nGARBAGE IS BIODEGRADABLE AND SOME ARE NON BIODEGRADABLE. \nGARBAGE CAUSES POLLUTION OF AIR, WATER AND SOIL. SO IT SHOULD \nBE DISPOSED PROPERLY.\nSOME OF THE METHODS OF MANAGEMENT OF GARBAGE DISPOSAL \nARE\nI) REUSE\nII) RECYCLING \nIII) PRODUCTION OF BIOGAS AND MANURE\nIV) PREPARATION OF COMPOST", + "CHAPTER 13-OUR ENVIRONMENT-very short answer questions.txt\nVery short answer type Questions \n1. Which are the major groups of organisms acting as decomposers? \n2. What is the mode of nutrition of fungi? \n3. Mention any one way non -biodegradable waste affect our health. \n4. Name the chemicals mainly depleting ozone layer. \n5. Give any two examples for man - made ecosystem. \n6. What amount of energy is passed from one trophic level to the successive \ntrophic levels of a food chain? \n7. Name another group of organisms that can pr oduce their food other than \nplants. \n8. Which group of biotic components of an ecosystem plays a significant role in \nthe recycling of nutrients.? \n9. How much percent of sunlight falling on leaf surface can be used by plants for \nphotosynthesis? \n10.What is oz one hole? \n11. Name the law which governs the flow of energy through food chain. \n12. Write any one way the complete removal of decomposers affect the life of \nproducers.", + "CHAPTER 13-OUR ENVIRONMENT-very short answer questions.txt\n10.What is oz one hole? \n11. Name the law which governs the flow of energy through food chain. \n12. Write any one way the complete removal of decomposers affect the life of \nproducers. \n13. Write the basic reason for the formation of food web. \n14. Write any two examples for biodegradable garbage. \n15. What are trophic levels?. \n16. Which gas is released by CFC , when it comes in contact with UV rays? \n17. How does ozone layer beneficial for organisms? \n18. In which way incineration useful for the disposal of garbage? \n19. What ar e biodegradable waste? \n20. Write any two measures you may adapt in your life to reduce the problems \ndue to garbage. \nMCQ \n1.Which of the following is a non -biodegradable thing? \na) banana", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nLESSON PLAN NO.1 \nGeneral Aims : (1) To develop the scientific attitude among the students. \n(2) To develop attitude towards environmental education among the students. \n(3) To develop clear thinking and environmental awareness among the students. \n(4) To develop attitude towards environmental pollution among the students. \nSpecific Aim : To provide knowledge to students about \u201cEcosystem and its types\u201d \nP.K. Assumed : It is assumed that the students will be aware of the terms environment , biotic \nand abiotic components of environment and ecology. \nP.K. Testing : \nQ1. Define environment . \nQ2. What are the two main components of environment ? \nQ3. Name the branch of study that deals with the interaction between living organism and \ntheir environment? \nQ4. Define ecosystem and what are the various types of ecosystem? \nAnnouncement of the Topic : \nWell studen ts, today we will study about \u201cE cosystem and its types\u201d", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\ntheir environment? \nQ4. Define ecosystem and what are the various types of ecosystem? \nAnnouncement of the Topic : \nWell studen ts, today we will study about \u201cE cosystem and its types\u201d \nPresentation : Lesson will be developed by the lecture method. \nSubject Matter PT\u2019s Activity Pupil\u2019s Activit y BB Summary \nThe living \norganisms in the \npond, the water in \nit, the stones and \nthe mud at the \nbottom make up \nan ecosystem. An \necosystem may \neither be natural or \nartificial. They \nagain can be \ndivided broadly \ninto terrestrial and \naquatic ecosystem \nTerrestrial PT will explain \nthe definition \nverbally. \nPT will perform \nan activity. \nPT will ask one \nof the students to \ncome and ask \nhim/her to put \nsome H 2O in \nbeaker than \ncome the mud \nand some leaves \nin the beaker. \nThen PT will Students will \nlisten carefully \nto the definition \nand observe the \nactivity. Two types of Ecosystem: \nNatural and Artificial", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nAquatic Ecosystem and \nTerrestrial Ecosystem \n \nTerrestrial Ecosystem \nincludes \nForest, Grassland and \nDesserts", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nEcosystem include \nForest \nEcosystems , \nGrasslands and \nDesserts. show it to whole \nclass and the n \nask them \necosystem look \nlike this. \n \nForest ecosystem: \nTropical rainforest \nhave the highest \nrainfall of any of \nthe ecosystem and \nthe greatest \ndiversity in plants \nand animals. Trees \ngrow tall and \nmany other plants \ngrow around them. PT will show the \nchart of forest \necosystem and \nexplain the type \nof plants and \nanimals in \ntropical \nrainforest Pupil will \nobserve carefully \nthe chart and \nalso listen \ncarefully. Forest Ecosystem: Tropical \nRainforest ha ve the highest \nrainfall and greatest diversity \nof plants & animals \n(b) Grasslands: are \ncharacterized by \nlow growing \nplants and a few \nscattered trees.", + "Forest Ecosystem: Tropical \nRainforest ha ve the highest \nrainfall and greatest diversity \nof plants & animals \n(b) Grasslands: are \ncharacterized by \nlow growing \nplants and a few \nscattered trees. \nTree growth is \nlimited by regular \nperiods of drought, \ngrazing and \noccasional fires \neg: Tropical \nSavannah in \nAfrica & Prairies \nin North America PT will show the \nchart of \nGrassland \necosystem and \nexplain the \ncauses for the \nlimited growth", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\ngrazing and \noccasional fires \neg: Tropical \nSavannah in \nAfrica & Prairies \nin North America PT will show the \nchart of \nGrassland \necosystem and \nexplain the \ncauses for the \nlimited growth \nof the plants. Pupil will \nobserve the chart \ncarefully Grasslands: Tropical \nSavannah in Africa in North \nAmerica \nDesert Ecosys tem: \nDeserts are the \ndriest of all the \necosystem, they \nreceive less than \n30cm rainfall per \nyear. Desert plants \nhave adaptations \nlike leaves reduced \nto spines cactus. PT will explain \nthe desert \necosystem using \nblack board. Students will \nnote down from \nblack board Desert Ecosystem: Less \nAverage Rainfall - 30cm per \nyear \nII.", + "PT will explain \nthe desert \necosystem using \nblack board. Students will \nnote down from \nblack board Desert Ecosystem: Less \nAverage Rainfall - 30cm per \nyear \nII. Aquatic \nEcosystem \nincluding \nFreshwater and \nMarine water \nEcosystem & \nEstuarine water PT will explain \nthe Aquatic \necosystem \nverbally Students will \nlisten carefully Aquatic Ecosystem : \nFreshwater , Marine & \nEstuarine water \n(a) Freshwater \nEcosystem They \nfall into two \ncategories lotic", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nthe Aquatic \necosystem \nverbally Students will \nlisten carefully Aquatic Ecosystem : \nFreshwater , Marine & \nEstuarine water \n(a) Freshwater \nEcosystem They \nfall into two \ncategories lotic \nand lentic PT will explain \nfresh water \necosystem \nverbally. Students will \nlisten carefully Freshwater Ecosystem :lotic \nand lentic \ni) Lotic ecosystem \nhas flowing water \nenvironments. \nThey have \nunidirectional flow \nof water. Lotic ecosystem has flowing \nwater \nii) Lentic \necosystems have \nstanding water \nenvironments and \ncontain stag nant \nwater it has three \nbasic life zones \nlittoral, limnetic \nand profoundal. PT will explain \nthree zones of \nlentic ecosystem \nusing black \nboard Student will note \nin copies and \nlisten carefully Lentic ecosystems have \nstanding water . \nThree basic life zones littoral, \nlimnetic and profoundal. \nb) Saltwater \nEcosystem or \nMari ne Ecosystem \ninclude Sea.", + "Three basic life zones littoral, \nlimnetic and profoundal. \nb) Saltwater \nEcosystem or \nMari ne Ecosystem \ninclude Sea. It \ncovers 70% of \nearth\u2019s surface and \nis characterised by \nits high", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nThree basic life zones littoral, \nlimnetic and profoundal. \nb) Saltwater \nEcosystem or \nMari ne Ecosystem \ninclude Sea. It \ncovers 70% of \nearth\u2019s surface and \nis characterised by \nits high \nconcentration of \nsalt and mineral \nions. It is divided \ninto intertidal, \nneritic and oceanic \nzone. Interdial \nzone in place \nwhere land and \nwater meets. \nNeritic zone \nextends from \nintertidal zone to \ncontinental shelf. \nOceanic zone is \nopen ocean. It \nfurther include \nPelagic or open \nwater and Benthic \nor Ocean Bottom \nZones. Deepest \npart of ocean is \nknown as abyssal PT will explain \nthe ecosystem \nusing blackboard", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \noceanic zone \nusing blackboard \n \n \n \n \n \n \n Pupil s will listen \ncarefully \n \n \n \n \n \n \n \n \nPupil s will note \ndown carefully \n Saltwater Ecosystem or \nMarine Ecosystem This \necosystem is divided into 3 \nzones \nIntertidal, neritic and oceanic \nzone \n \nOceanic zone \nPelagic or open water and \nBenthic or Ocean Bottom \nZones. \nDeepest part of ocean is \nknown as abyssal zone", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nOceanic zone \nPelagic or open water and \nBenthic or Ocean Bottom \nZones. \nDeepest part of ocean is \nknown as abyssal zone \n \n \nPhotic zone :whic h receives \nlight and Aphotic zone is \nwhere there is no light at all. zone. Photic zone \nis the zone of \nocean which \nreceives light a nd \naphotic zone is \nwhere there is no \nlight at all. \nTypes of Marine \nLife: Marine \nhabitats contain \nthree main \necological groups \nof organisms.", + "zone. Photic zone \nis the zone of \nocean which \nreceives light a nd \naphotic zone is \nwhere there is no \nlight at all. \nTypes of Marine \nLife: Marine \nhabitats contain \nthree main \necological groups \nof organisms. \n(1) Plankton \n(2) Nekton \n(3) Benthos \n(1) Plankton: \nFree floating \norganisms \nincapable of \nswimming they \ninclude \nPhytoplankton \nplant plankton, \nphotosynthetic \nblue green \n(cyanobacteria) \nand green algae \nZooplankton \u2013 \nanimal plankton \u2013 \nthe non \nphotosynthetic \nspecies PT will explain \nthe types of \nmarine life using \nflash cards \n \n \n \n \n \nPT will explain 3 \ntypes one by one \nwith examples of \neach types and \nwrite on black \nboard Pupil s will \nobserve and \nlisten carefully", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain 3 \ntypes one by one \nwith examples of \neach types and \nwrite on black \nboard Pupil s will \nobserve and \nlisten carefully \n \n \n \n \n \n \n \nPupil s will listen \ncarefully and \nnote down Types of Marine Life: \n \n \n \n \n(1) Plankton \n(2) Nekton \n(3) Benthos \n \nPlankton: Free floating \norganisms incapable of \nswimming Phytoplankton : \nphotosynthetic blue green \nalgae (cyanobacteria) and \nZooplankton : non \nphotosynthetic \n \n \n \n \n \n \n \n(2) Nekton: are \nstronger \nswimming species \ne.g. fish, turtles, \nwhales PT will ask the \nquestion name \nthe marine \nspecies which \ncan swim? and \nwrite on black \nboard Pupil s will \nanswer fishes, \nwhales, starfish, \nturtle and note \ndown. Nekton: are stronger \nswimming species e.g.", + "and \nwrite on black \nboard Pupil s will \nanswer fishes, \nwhales, starfish, \nturtle and note \ndown. Nekton: are stronger \nswimming species e.g. fish, \nturtles, whales \n(3) Benthos the \nplants and animals \nthat live on the \nbottom in cluding \nseaweeds, \nbarnacles, lob sters, \nworms, \ncrustaceans", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nbenthos and \nwrite on black \nboard Pupil s will listen \ncarefully and \nnote down Benthos : plants and animals \nthat live on the bottom eg. \nseaweeds, barnacles, lobsters, \nworms, crustaceans C) Estuarine/ \nBrackish Coastal \nWater Ecosystems. \nIt includes \nshallow, partially \nenclosed areas \nwhere freshwater \nand sea water mix. \nIt has highly \nvariable \nenvironments. \nMangrove tree is \nwell ad apted to \nswamp\u2019s salty \nenvironm ent. PT will explain \nthe Estuarine/ \nBrackish coastal \nwater ecosystem \nusing black \nboard \n \n \n \n \n \n \nPT will show the \nflash card of \nmangrove tree. Pupil will listen \ncarefully and \nnote down . \n \n \n \n \n \n \n \n \nStudents will \nobserve \ncarefully. Estuarine/ Brackish Coastal \nwater Ecosystems. It includes \nshallow, partially enclosed \nareas where freshwater and \nsea water meet and mix \nExample mangrove tree", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nComponents of \nEcosystem: \nA number of \nfactors affect the \ndistribution of \norganisms in an \necosystem. These \nphysical factors are \nknown as abiotic. \nBiotic factors \nwhich involve the \neffects of other \nliving organisms, \nincluding humans \non the distribution \nand behaviours of \nspecies in the \necosystem. PT will explain \nthe components \nof ecosystem \nusing \nblackboard. \n Pupil s will listen \nand note down \nin copies. Components of Ecosystem \n \nPhysical factors are known as \nabiotic \n \n \n \n \nBiotic factors : involve the \neffects of other living \norganisms \nThe physical \nfacto rs are divided \ninto: \n1) Climatic it \nincludes \ntemperature light, \nwind and water \navailability. \n2) Soil -referred as \nEdaphic factors. \n3) Topographic \ninclude attitude, \naspect (whether \nnorth -facing or \nsouth -facing) and \ninclination \n(steepness of PT will explain \nphysical facto rs \nusing \nblackboard .", + "3) Topographic \ninclude attitude, \naspect (whether \nnorth -facing or \nsouth -facing) and \ninclination \n(steepness of PT will explain \nphysical facto rs \nusing \nblackboard . \n \n \n \n \n \n \n \n \n \n \n \n \n Pupil s will listen \nand note down \nin copies.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPupil s will listen \nand note down \nin copies. \n \n \n \n \n \n \n \n \n \n \n \n \n \n Physical factors: \n \n1)Climatic: temperature light, \nwind and water availability. \n2) Soil or Edaphic factors. \n3) Topographic :attitude, \naspect (whether north -facing \nor south -facing) and \ninclination (steepness of \nslope) latitude etc. \n4)Salinity, wave action, \noxygen concentration \n \n \n \n \n slope) latitude etc. \n4) Others \u2013 such \nsalinity, wave \naction, oxygen \nconcentration etc. \nwhich are relevant \nin specific \nsituations \ncatastrophic events \nlike fire, flood etc. \nPlants that can \ntolerate high levels \nof salts are known \nas halophytes \n \n \n \n \n \n \nPT will explain \nand write the \ndefinition of \nhalophytes on \nblackboard. \n \n \n \n \n \n \nStudent will \nlisten and write \non copies", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nand write the \ndefinition of \nhalophytes on \nblackboard. \n \n \n \n \n \n \nStudent will \nlisten and write \non copies \n \n \n \n \n \n \nHalophytes \nPlants that can tolerate high \nlevel of salts are known as \nhalophytes \nBiotic Component \ninclude \nrelationships \nbetween living \norganisms . Biotic \ncomponents \ninclude \n1) Autotrophs: \nThey make their \nown food using an \nexternal energy \nsource and a \nsimple inorganic \nsupply of carbon \ndioxide. \nPhoto autotrophs: \nuse sunlight as \ntheir energy source \nand included \nphotosynthetic \ngreen plants, algae \nand bacteria. \nChemoautotrophs: \nutilize energy \nderived solely \nfrom chemical \nreactions and \ninclude only \nbacteria , occurring \nin the nitrogen \ncycle and in deep \nsea hydro - thermal \nvent communities. \nHeterotrophs: They \ncannot make their \nown food, they PT will explain \nthe biotic \ncomponent \nverbally.", + "Heterotrophs: They \ncannot make their \nown food, they PT will explain \nthe biotic \ncomponent \nverbally. \nPT will ask , \nname ty pes of \norganism s on \nthe basis of \npreparation of \nfood.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nusing \nblackboard \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n Pupil will listen \ncarefully. \n \n \nPupil s will \nanswer \nAutotrophs and \nHetero trophs. \n \n \n \nPupil s will listen \ncarefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nBiotic Component s \nAutotrophs and Hetero trophs \n \n \n \n \n \nAutotrophs: They make their \nown food . \n \nPhoto autotrophs \n \n \n \n \n \nChemoautotrophs", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nGeneralisation :- So,student s today we have studied about the types of ecosystem i.e. \nterrestrial and aquatic ecosystem and also the components of the ecosystem biotic and abiotic \nand interaction between these components. \nFinal Recapitulation \nQues: Define Terrestrial ecosystem? \nQues: Name the t ypes of Aquatic ecosystem? \nQues: What do you mean by B iotic components of ecosystem ? \nQues: What do you mean by Abiotic components of ecosystem ? \nHome work \nPT will ask the students to draw the flow chart of type of ecosystem and components of \necosystem. \n must take in food \nmolecules from \ntheir surroundings. PT will explain \nusing \nblackboard Pupil s will listen \ncarefully \n Heterotrophs: They cannot \nmake their own food \nThe Decomposers: \nThese are mainly \nbacteria and fungi. \nThey break down \nthe bodies of dead \nanimals and plants, \nusually by extra \ncellular digestion \nand absorb the \nsoluble products. \nInteraction \nbetween Biotic and \nAbiotic \nComponents:", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nThey break down \nthe bodies of dead \nanimals and plants, \nusually by extra \ncellular digestion \nand absorb the \nsoluble products. \nInteraction \nbetween Biotic and \nAbiotic \nComponents: \nAll living \norganisms are part \nof an ever \nchanging system \nthe ecosystem. \nWithin an \necosystem plants \nand animals grow, \ninteract, reproduce \nand die. PT will explain \nDecomposers \nverbally", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nthe interaction \nbetween Biotic \nand Abiotic \ncomponent s \nverbally. Pupil s will listen \nattentively \n \n \n \n \n \n \n \n \n \nPupil s will listen \nattentively \n \n \n \n \n \n \n Decomposer : break down the \nbodies of dead animals and \nplants \n \n \n \n \n \n \n \n \nInteraction between Biotic \nand Abiotic Components : \nclosely related in ecosystem", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nLESSON PLAN NO.2 \nGeneral Aims : (1) To develop the scientific attitude among the students . \n(2) To develop positive and healthy attitude towards environmental education among the \nstudents . \n(3) To develop clear thinking and environmental awareness among the students . \n(4) To develop positive and healthy attitude towards environmen tal pollution among the \nstudents . \nSpecific Aim : To provide knowledge to students about \u201c Energy flow and nutrient recycling \u201d. \nP.K. Assumed : It is ass umed that the students will be aware of the terms environment , biotic \nand abiotic components of environment , ecology and types of ecosystem . \nP.K. Testing : \nQ1. Define environment . \nQ2. What are the two main components of environment ? \nQ3. Name the branch of study that deals with the interaction between living organism and \ntheir environment . \nQ4. Explain the mechanism of energy flow in the ecosystem . \nAnnouncement of the Topic :", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\ntheir environment . \nQ4. Explain the mechanism of energy flow in the ecosystem . \nAnnouncement of the Topic : \nWell students, today we will study about \u201c Energy flow and nutrient recycling \u201d \nPrese ntation: Lesson will be developed by the lecture method. \nSubject Matter PT\u2019s Activity Pupil\u2019s Activity BB Summary \nEnergy flows \nfrom one \ntrophic level to \nanother. Trophic \nlevels are the \nsteps in the food \nchain \ncomprising of \ndifferent types \nof organisms. \nAutotrophs PT will explain \nthe concept of \ntrophic level \nverbally", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nautotrophs with Students will \nlisten carefully . \n \n \n \n \n \n \nStudents will \nobserve and \nlisten carefully . Trophic level \nSteps in food chain \n \n \n \n \n \n \n prepare their \nown food using \nan external \nenergy source. \nCarnivores are \nflesh eating \nanimals. They \nfeed on \nherbivores or \nplant eating \nanimals. \nSecondary \ncarnivores like \nsharks feed on \nfish which is a \nprimary \ncarnivore. \nLion is an \nexample of \nCarni vorous \nAnimal. \nDecomposers \nbreak down the \nbodies of dead \nanimals and \nplants. Process \nof \ndecomposition \nallows mineral \nnutrients to be \nrecycled in the \necosystem. \nFungi and \nbacteria are \nexamples of \ndecomposers. \nEarthworm is an \nexample of \nDecomposer \nDifferent \norganisms of an \necosystem \nlinked together \nby their \nnutritional \nrequirements \nform a food \nchain.", + "Fungi and \nbacteria are \nexamples of \ndecomposers. \nEarthworm is an \nexample of \nDecomposer \nDifferent \norganisms of an \necosystem \nlinked together \nby their \nnutritional \nrequirements \nform a food \nchain. \nNetworks of a \nnumber of food the help of flash \ncards \n \nPT will explain \ncarnivores with \nthe help of flash \ncards", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \ncarnivores with \nthe help of flash \ncards \n \n \n \n \n \n \n \n \n \nPT will explain \ndecomposers \nwith the help of \nflash cards \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nPT will explain \nthe conce pt of \nfoodchain using \nflash cards \n \n \n \n \n \n \n \nPT will explain \nfoodweb with \nthe help of flash \ncards \n \nStudents will \nobserve and \nlisten carefully \n \n \n \n \n \n \n \n \n \nPupils will \nobserve the flash \ncards carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nPupils will note \ndown carefully \n \n \n \n \n \n \n \n \n \nPupils will \nobserve the flash \ncards carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nFood chain \n \n \n \n \n \n \n \n \nchains existing \nin an ecosystem \nform a food web \nSteps in a food \nchain at which \ntransfer of food \nenergy takes \nplace are known \nas tro phic \nlevels. 10% \nenergy is \ntransferred \n from one \ntrophic level to \nother Trophic \nlevel.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nWater cycle \nMaximum water \nis found in \noceans. By \nevaporation and \ntranspiration it \nis taken up into \nclouds. It \nreturns to \noceans, land and \nlakes by \nrainfall. Water \nthat enters the \nsoil returns to \nthe oceans \nthrough run off. \nCarbon cycle \nMost carbon is \nfound as carbon \ndioxide in the \natmosphere and \ntaken in by \ngreen plants in \nphotosynthesis \nthen eaten by \nanimals. It is \nrecycled back to \natmosphere by \nrespiration of \nplants, animals \nand microbes.", + "It is \nrecycled back to \natmosphere by \nrespiration of \nplants, animals \nand microbes. \nNitrogen cycle \nMost of \n \n \n \n \n \n \n \n \nPT will explain \nwater cycle with \nthe help of chart \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nPT will explain \ncarbon cycle \nwith the help of \nchart \n \n \n \n \n \n \n \n \n \n \n \n \nPT will explain \nnitrogen cycle \nwith the help of \nchart \n \n \n \n \n \n \n \n \n \nPupils will \nobserve the \nchart carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nPupils will \nobserve the \nchart carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \nPupils will \nobserve the \nchart carefully \n Food web \n \n \n \n \nWater cycle", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPupils will \nobserve the \nchart carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \nPupils will \nobserve the \nchart carefully \n Food web \n \n \n \n \nWater cycle \n \n \n \n \n \n \n \n \n \n \n \n \nCarbon cycle \n \n \n \n \n \n \n \n \n \n \n \natmospheric \nnitrogen is \nunavailable for \nprotein \nsynthesis. It is \ntaken by plants \nin the form of \nnitrates from the \nsoil and \nincorporated \ninto protein. Nitrogen cycle", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\natmospheric \nnitrogen is \nunavailable for \nprotein \nsynthesis. It is \ntaken by plants \nin the form of \nnitrates from the \nsoil and \nincorporated \ninto protein. Nitrogen cycle \n \n \nGeneralisation : So, student s today we have studied about trophic levels that are the steps in \nthe food chain comprising of different types of organisms. Autotrophs prepare their own \nfood. Carnivores are flesh eating animals. Decomposers break down the bodies of dead \nanimals and plants. Networks of a number of food chai ns existing in an ecosystem form a \nfood web. Water,carbon and nitrogen cycle. \nFinal Recapitulation : \nQues: Define food chain. \nQues: Give an example of autotro phs. \nQues: Define food web. \nQues: What do you mean by Decomposers ? \nHome work \nPT will ask the stude nts to draw food web .", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nLESSON PLAN NO.3 \nGeneral Aims : (1) To develop the scientific attitude among the students . \n(2) To develop attitude towards environmental education among the students . \n(3) To develop clear thinking and environmental awareness among the students . \n(4) To develop attitude towards environmental pollution among the students . \nSpecific Aim : To provide knowledge to students about \u201c Renewable sources of energy \u201d \nPk Assumed : It is assumed that the students will be aware of the components and types of \necosystem and nutrient recycling . \nPk Testing : \nQ1.What are the various types of ecosystem? \nQ2.Give an example of terrestrial ecosystem. \nQ3.How is carbon recycled in nature? \nQ4.What are renewable sources of energy? \nAnnouncement of the Topic: \nWell students, today we will study about \u201c Renewable sources of energy \u201d \nPresentation: Lesson will be developed by the lecture method.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nSubject Matter PT\u2019s Activity Pupil\u2019s \nActivity BB Summary \nResources are \ncategorised as \nrenewable and \nnonrenewable \nEnergy generated by \nusing wind, tides, \nsolar, geothermal \nheat, and biomass \nincluding farm and \nanimal waste as well \nas human excreta is \nknown as non -\nconventional energy. \nAll these sources are PT will \nexplain \nrenewable \nand non \nrenewable \nsources of \nenergy with \nthe help of \nchart \n \n \n \n \n Pupils will \nobserve the \nchart carefully", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPupils will \nobserve the \nchart carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nrenewable or \ninexhaustible and do \nnot cause \nenvironm ental \npollution. More over \nthey do not require \nheavy expenditure \nNon conventional/ \nRenewable sources \nof \n energy \nWind Energy: \nWind power is \nharnessed by setting \nup a windmill which \nis used for pumping \nwater, grinding \ngrain and generating \nelectricity. \nTidal Energy: \nSea water keeps on \nrising and falling \nalternatively twice a \nday under the \ninfluence of \ngravitational pull of \nmoon and sun. This \nphenomenon is \nknown as tides. The \nGulf of Kuchchh is \nbest suited for tidal \nenergy. \nSolar Energy: \nSun is the source o f \nall energy on the \nearth. It is most \nabundant, \ninexhaustible and \nuniversal source of \nenergy.", + "The \nGulf of Kuchchh is \nbest suited for tidal \nenergy. \nSolar Energy: \nSun is the source o f \nall energy on the \nearth. It is most \nabundant, \ninexhaustible and \nuniversal source of \nenergy. India has \ndeveloped \ntechnology to use \nsolar energy for \ncooking, water \nheating, water \ndissimilation, space \nheating, crop drying \netc.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will \nexplain \nrenewable \nsources of \nenergy with \nthe help of \nchart \n \n \n \n \n \nPT will \nexplain wind \nenergy with \nthe help of \nchart \n \n \n \n \nPT will \nexplain tidal \nenergy \nverbally \n \n \n \n \n \n \n \n \nPT will \nexplain solar \nenergy with \nthe help of \nblackboard \n \n \n \n \n \n \n \n \n \n \n \n Pupils will \nobserve the \nchart carefully \n \n \n \n \n \n \n \n \n \nPupils will \nobserve the \nchart carefully \n \n \n \n \n \n \nStudents will \nlisten \ncarefully \n \n \n \n \n \n \n \n \n \nStudents will \nlisten \ncarefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nRenewable sources of energy \n \n \n \n \nWind Energy: \nWind power is harnessed by setting \nup a windmill which is used for \npumping water, grinding grain and \ngenerating electricity \n \n \nTidal energy: by sea tides \n \n \n \n \n \n \n \nSolar energy: by sun, used for \nfor cooking, water heating, water \ndissimilation, space heating, crop \ndrying etc.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nTidal energy: by sea tides \n \n \n \n \n \n \n \nSolar energy: by sun, used for \nfor cooking, water heating, water \ndissimilation, space heating, crop \ndrying etc. \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n4. Geo -Thermal \nEnergy: \nGeo-thermal energy \nis the heat of the \nearth's interior. This \nenergy is manifested \nin the hot springs. \nIndia is not very rich \nin Geothermal \nenergy source. \nImportance of non -\nconventional \nsources of energy: \n1. The non -\nconventional \nsources of energy \nare abundant in \nnature. \n2. These are \nrenewable resources. \nThe non -\nconventional \nsources of energy \ncan be renewed with \nminimum effort and \nmoney. \n3.Non conventional \nsources of energy \nare pollution -free \nand eco -friendly \n \n PT will \nexplain \ngeothermal \nenergy \nverbally \n \n \n \n \n \nPT will \nexplain \nImportance \nof non -\nconventional \nsources of \nenergy \nverbally \n \n \n \n \n \n \n \n \n \n \n \n \n Students will \nlisten \ncarefully", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will \nexplain \nImportance \nof non -\nconventional \nsources of \nenergy \nverbally \n \n \n \n \n \n \n \n \n \n \n \n \n Students will \nlisten \ncarefully \n \n \n \n \n \n \n \nStudents will \nlisten \ncarefully Geothermal Energy: Heat enrgy of \nearth\u2019s interior \n \n \n \n \n \n \n \n \nImportance of non -conventional \nsources of energy: \n*Abundant \n*Renewed freely \n*Pollution free", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nGeneralisation :- So, student s today we have studied about the renewable sources of energy. \nEnergy generated by using wind, tides, solar, geothermal heat, and biomass including farm \nand animal waste as well as human excreta is known as non -conventional ener gy. Non \nconventional sources of energy are pollution -free and eco -friendly . \nFinal Recapitulation \nQues: Define renewable sources of energy. \nQues: Name renewable sources of energy ? \nQues: How can wind energy be used? \nQues: What do you mean by solar energy? \nHome work \nPT will ask the students to draw the flow chart of type s of renewable sources of energy. LESSON PLAN NO.4 \nGeneral Aims : (1) To develop the scientific attitude among the students . \n(2) To develop attitude towards environmental education among the students . \n(3) To develop clear thinking and environmental awareness among the students . \n(4) To develop attitude towards environmental pollution among the students .", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\n(3) To develop clear thinking and environmental awareness among the students . \n(4) To develop attitude towards environmental pollution among the students . \nSpecific Aim : To provide knowledge to students about \u201c Destruction of ecosystem & \nDepletion of Resources\u201d \nPk Assumed :It is assumed that the students will be aware of renewable and non renewable \nsources of energy and types of ecosystem. \nPk Testing : \nQ1. What are the various types of ecosystems? \nQ2. Define renewable and non renewable sources of ene rgy. \nQ3. How are the resources depleted? \nQ4. What are the causes of Destruction of ecosystem ? \nAnnouncement of the Topic: \nWell students, today we will study about \u201c Destruction of ecosystem & Depletion of \nResources \u201d \nPresentation: Lesson w ill be developed by the lecture method. \nSubject Matter PT\u2019s Activity Pupil Activity BB Summary \nPollution is addition of \ntoxic substances in \nenvironment as a result of \nhuman action and natural \ndisasters.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nSubject Matter PT\u2019s Activity Pupil Activity BB Summary \nPollution is addition of \ntoxic substances in \nenvironment as a result of \nhuman action and natural \ndisasters. \nA pollutant is defined \nsimply as any waste \nmaterial that pollutes water, \nair or soil. \nAir pollution is the \nintroduction of chemicals, \nparticulate matter, or PT will explain \npollution using \nblackboard.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \npollutant using \nblackboard \n \n \n \n Students will \nlisten carefully \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n Pollution is addition of toxic \nsubstances in environment \nas a result of human action \nand natural disasters \n \nA pollutant is defined \nsimply as any waste material \nthat pollutes water, air or \nsoil.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nbiological materials that \ncause harm or discomfort \nto humans or other living \norganisms, or cause \ndamage to the natural \nenvironment or built \nenvironment, into the \natmosphere. \nAir pollution leads to \nsevere breathing \nproblems. There is a \n\"natural\" greenhouse \neffect that keeps the \nEarth's climate warm and \nhabitable. There is also the \n\"man -made\" greenhouse \neffect, which is the \nenhancement of Earth's \nnatural greenhouse e ffect \nby the addition of \ngreenhouse gases from the \nburning of fossil fuels \nmainly petroleum and \nnatural gas. Greenhouse \ngases reduce the rate at \nwhich the Earth's surface \nloses infrared radiation to \nouter space ,this makes the \nEarth's surface and lower \natmo sphere warmer than \nthey would otherwise be . \nThis leads to global \nwarming .The phenomenon \nof unequal rise in \ntemperature of earth\u2019s \nsurface due to human \nactivities like pollution \nand deforestation is called \nas global warming. \nWater pollution is the \ncontamina tion of water", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nof unequal rise in \ntemperature of earth\u2019s \nsurface due to human \nactivities like pollution \nand deforestation is called \nas global warming. \nWater pollution is the \ncontamina tion of water \nbodies (e.g. lakes, rivers, \noceans, aquifers and \ngroundwater). Water \npollution occurs when \npollutants are discharged \ndirectly or indirectly into \nwater bodies without \nadequate treatment to \nremove harmful", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nair pollution \nusing \nblackboard \n \n \n \n \n \n \n \n \nPT will explain \ngreen house \neffect using \nblackboard \n \n \n \n \n \n \n \n \n \n \n \n \n \nPT will explain \nglobal warming \nusing \nblackboard \n \n \n \n \n \nPT will explain \nwater pollution \nusing \nblackboard \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n Air pollution is the \nintroduction of chemicals, \nparticulate matter, or \nbiological materials that \ncause harm to the \natmosphere. \n \n \n \n \n \n \n \nGreenhouse effect keeps \nthe Earth's climate warm \nand habitable. \nBurning of fossil fuels \nmainly petroleum and \nnatural gas. Greenhouse \ngases reduce the rate at \nwhich the Earth's surface \nloses infrared radiation to \nouter space , this makes the \nEarth's surface and lower \natmosphere warmer .", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nThe phenomenon of \nunequal rise in temperature \nof earth\u2019s surface due to \nhuman activities like \npollution and deforestation \nis called as global warming \n \n \n \nWater pollution is the \ncontamination of water \nbodies", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\ncompounds. Water \npollution leads to the \ndecrease or even \nextinction of water \ndwelling species. \nEutrophication is the \nresult of water pollution. It \nis the ecosystem response \nby growth of algae on \nsurface of water due to \nthe addition of artificial or \nnatural substances, such as \nnitrates and phosphates, \nthrough fertilizers or \nsewage, to an aquatic \nsystem. \nLand pollution is the \ndemolition of Earth's land \nsurfaces often caused by \nhuman activities and their \nmisuse of land re sources. \nUrbanization and \nindustrialization are major \ncauses of land pollution. \nGrowth of human \npopulation leads to habitat \ndeterioration. \nCauses of ecosystem \ndestruction: \nNatural forest fires and \nmanual cutting of trees \ndestroy the ecosystem. \nClearing the forests, forest \nfires lead to destruction of \nthe ecosystem \nThe growth of industries \nbesides human \ndevelopment has lead to \npollution. Acid rain is one \nof the results of \nIndustrialisation \nAnother problem", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nfires lead to destruction of \nthe ecosystem \nThe growth of industries \nbesides human \ndevelopment has lead to \npollution. Acid rain is one \nof the results of \nIndustrialisation \nAnother problem \nassociated with excessive \nirrigation on poorly \ndraine d soils is \nwaterlogging. This occurs \n(as is common for \nsalinization) in poorly \ndrained soils where water \ncan't penetrate deeply", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \neutrophication \nusing \nblackboard \n \n \n \n \n \n \n \n \nPT will explain \nland pollution \nusing \nblackboard \n \n \n \n \n \n \nPT will explain \ncause of \necosystem \ndestruction \nblackboard \n \n \n \n \n \n \n \n \n \nPT will explain \ncause of \necosystem \ndestruction \nwater logging \nblackboard \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \n \n \n \nEutrophication is the result \nof water pollution. It is the \necosystem response by \ngrowth of algae on surface \nof water due to the \naddition of unwanted \nsubstances \n \n \n \n \n \nLand pollution is the \ndemolition of Earth's land \nsurfaces \n \n \n \n \n \n \n \nCauses of ecosystem \ndestruction \nNatural forest fires and \nmanual cutting of trees \n \n \n \n \n \n \n \n \n \n \nExcessive irrigation on \npoorly drained soils is \nwaterlogging.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nOverexploitation of \nnatural resources due to \nhuman activity lead to \npressures and intruding on \nnatural habitat. This is \ntermed a s encroachment. \nIn Shifting cultivation \nknown as slash and burn \ncultivation ,a small patch \nof native forest is cleared \nby felling trees and then \nby burning them nutrients \nare released. Seeds are \nplanted on soil ash \nmixture. Plants like maize, \nbanana are pl anted. It \nleads to reduced fertility \nof the soil. \nConstruction of dams for \nhydroelectricity \ngeneration, irrigation lead \nto permanent \nenvironmental change \nleading to loss of \nresources by flooding an \narea not previously \ncovered with water \nExtraction of miner als \nfrom earth also lead to \npollution. For example \nStrip mining of coal leads \nto extensive disruption of \nland surface. Extraction of \noil by offshore drilling \nand transportation may \nlead to massive spill and \nthus causing harm to \nmarine environment. \nWars ca use vast \ndevastation on human life", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nland surface. Extraction of \noil by offshore drilling \nand transportation may \nlead to massive spill and \nthus causing harm to \nmarine environment. \nWars ca use vast \ndevastation on human life \nand civilisation along with \nirreversible changes in \nenvironment and \nenvironmental pollution. \nThe atom bombs dropped \nin Hiroshima and \nNagasaki are examples of \nenvironmental pollution \ncaused by radiations.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nshifting \ncultivation \nusing \nblackboard \n \n \n \n \n \n \n \nPT will explain \nconstruction of \ndams, extraction \nof minerals, \nwars as cause of \necosystem \ndestruction \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nPT will explain \nwars as cause of \necosystem \ndestruction \nusing \nblackboard \n \n \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nShifting cultivation known \nas slash and burn \ncultivation ,a small patch of \nnative forest is cleared by \nfelling trees and then by \nburning them nutrients are \nreleased. Seeds are planted \non soil ash mixture. \n \n \n \n \nCauses of ecosystem \ndestruction: \nconstruction of dams, \nextraction of minerals, wars", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nCauses of ecosystem \ndestruction: \nconstruction of dams, \nextraction of minerals, wars \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nWars cause vast \ndevastation on human life \nThe atom bombs dropped \nin Hiroshima and Nagasaki \nare examples of \nenvironmental pollution \ncaused by radiations \n \n \n \n In silviculture natural \nhabitat such as forests and \nwetlands are cleared and \ndrained for agricultural \nuse or sometimes for the \ncultivation of trees. \nSilviculture leads to loss \nof habitat of living \nspecies. \n PT will explain \nsilviculture \nusing \nblackboard \n \n \n \n Students will \nlisten carefully", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nStudents will \nlisten carefully \n \n \n \n \n \n \n \n Silviculture : natural habitat \nsuch as forests and \nwetlands are cleared and \ndrained for agricultural use \n \nGeneralisation : So, student s today we have studied about pollution and its types and various \ncauses of ecosystem destruction. Natural forest fires , Clearing the forests, Industrialisation, \nWaterlogging, Overexploitation of natural resources , Shifting cultivation , Silviculture. \nFinal Recapitulation : \nQues: Define Waterlogging. \nQues: Explain Shifting cultivation. \nQues: Define Silviculture. \nHome work \nPT will ask the students to make a collage on the causes of ecosystem destruction .", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nLESSON PLAN NO.5 \nGeneral Aims : (1) To develop the scientific attitude among the students . \n(2) To develop attitude towards environmental education among the students . \n(3) To develop clear thinking and environmental awareness among the students . \n(4) To develop a ttitude towards environmental pollution among the students . \nSpecific Aim : To provide knowledge to students about \u201c Conservation of resources \u201d. \nPk Assumed :- It is assumed that the students will be aware of renewable and non renewable \nresources, depletion of resources and destruction of ecosystem. \nPk Testing : \nQ1. Define renewable resources. \nQ2. What are non renewable resources. \nQ3. What are the causes of depletion of resources? \nQ4. How can we conserve resources? \nAnnouncement of the Topic: \nWell students, t oday we will study about \u201c Conservation of resources \u201d \nPresentation: Lesson will be developed by t he lecture method. \nSubject Matter PT s Activity Pupil Activity BB Summary", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nWell students, t oday we will study about \u201c Conservation of resources \u201d \nPresentation: Lesson will be developed by t he lecture method. \nSubject Matter PT s Activity Pupil Activity BB Summary \nScientific management \nof man\u2019s surroundings \nto prevent its \nexploitation or \ndestruction is termed as \nconservation. \nReforestation is an \nexample of \nconservation. \nManagement of \nresources to meet \nchanging human needs \nand improving the \nquality of environment \nis termed as sustainable \ndevelopment. PT will explain the \nconcept of \nconservat ion \nverbally.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nsustainable \ndevelopment \nverbally \n Students will \nlisten carefully \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n Scientific management of \nman\u2019s surroundings to \nprevent its exploitation or \ndestruction is termed as \nconservation \nUsing the available \nenergy efficiently is \ntermed as energy \nconservation \nManagement of \nresources to meet \nchanging human needs \nand improving the \nquality of environment is \ntermed as sustainable \ndevelopment", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nEnvironmental, social \nand economic are three \nspheres of sustainability. \nSteps taken to preserve \nthe wild life to maintain \nbalance is termed as \nconservation of wildlife. \nAnimals should not be \nkilled to get their skin, \nantlers, bones, feather, \nivory etc. The species \nnear extinction should be \nlooked after carefully. \nNational parks and \nsanctuaries are steps \ntaken to conserve \nwildlife. \nIUCN -International \nUnion for conservation \nof nature and natural \nresources. \nWWF -World wildlife \nfund focuses on \nconservation of wildlife. \nCONSERVATION OF \nAGRICULTURE \nGrowing two or more \ncrops on the same piece \nof land alternatively so \nas to replenish the \nfertility of soil is called \ncrop rotation. It is an \nexample of conservation \nof agriculture. \nCrop rotation means to \nplant crops in turns such \nthat nutrient balance of \nthe soil is maintained. \nCrop rotation means to \nplant crops in turns such \nthat nutrient balance of \nthe soil is maintained.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nplant crops in turns such \nthat nutrient balance of \nthe soil is maintained. \nCrop rotation means to \nplant crops in turns such \nthat nutrient balance of \nthe soil is maintained. \nPlanting two or more \ncrops at the same time \nside by side to maintain \nthe fertility of soil is", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nconservation of \nwild life verbally. \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nPT will explain \nconservation of \nagriculture, crop \nrotation with the \nhelp of chart \n \n \n \n \n \n \n \n \n \n \n \n \n \nPT will explain \nconservation of \nagriculture, mixed \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nStudents will \nobserve and \nlisten carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nStudents will \nobserve and \nlisten carefully \n \n \n \nConservation of Agriculture \nCrop rotation : growing two \ncrops alternatively", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nMixed cropping \ncalled mixed cropping. \nOnly green manure s are \nused in organic farming. \nBiofertilisers are used \nand biological control of \npest occur in organic \nfarming \nVermicomposting is the \nprocess of using worms \nand micro -organisms to \nturn kitchen waste into a \nblack, earthy -smelling, \nnutrient -rich humus. \nEarthworms are \ncommonly used for the \nprocess of converting \nkitchen waste into \nbiomanure this is \n called as \nVermicomposting \nBiomass refers to all \nplant material and \nanimal excreta when \nconsidered as an energy \nsource. Some important \nkinds of biomass are \ninferior wood, urban \nwaste, bagasse, farm \nanimal and human \nwaste. \nCompressed Natural Gas \nis a domestically \navailable, economical, \nclean burning, \nalternative fuel source \nfor vehicles .", + "Some important \nkinds of biomass are \ninferior wood, urban \nwaste, bagasse, farm \nanimal and human \nwaste. \nCompressed Natural Gas \nis a domestically \navailable, economical, \nclean burning, \nalternative fuel source \nfor vehicles . It has \nhigher initial investment, \nlower running cost, \nlesser power delivery, \nlesser tune -ups required, \nbulkier and heavier \nstorage tanks, limited \navailability, no carbon \ndeposits, fewer oil", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nhigher initial investment, \nlower running cost, \nlesser power delivery, \nlesser tune -ups required, \nbulkier and heavier \nstorage tanks, limited \navailability, no carbon \ndeposits, fewer oil \nchanges, safer, lighter, \nhigh ignition \ntemperature, popular \nwith commercial \nvehicles. \n cropping with a \nchart", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nconservation of \nagriculture, \nvermicomposting \nverbally \n \n \n \n \n \n \n \n \n \nPT will explain \nbiogas formation \nverbally \n \n \n \n \n \n \n \nPT will explain \nCNG using BB \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \nStudents will \nlisten carefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nPaddy n brassica planted at \nthe same time it is an \nexample of mixed cropping. \n \n \nVermicomposting is the \nprocess of using worms and \nmicro -organisms to turn \nkitchen waste into a black, \nearthy -smelling, nutrient -\nrich humus. \n \n \n \n \nBiogas plant \n \n \n \n \n \n \n \nCompressed Natural Gas is \na domestically available, \neconomical, clean burning, \nalternative fuel source for \nvehicles .", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nBiogas plant \n \n \n \n \n \n \n \nCompressed Natural Gas is \na domestically available, \neconomical, clean burning, \nalternative fuel source for \nvehicles . \n \n \n \n \n \nLPG - Liquified \nPetroleum Gas is a \nhydrocarbon fuel, lig hter \nthan petroleum. Can be \nused as automotive fuel \nIt has lower initial \ninvestment, higher \nrunning cost, increased \npower delivery, more \nrefined, required smaller \nand lighter storage \ntanks, far better \navailability, cleaner \nemissions, heavier, low \nignition temperature, \npopular with private \nvehicles. PT will explain \nLPG using BB \n \n \n \n Students will \nlisten carefully LPG - Liquified Petroleum \nGas is a hydrocarbon fuel, \nlighter than petroleum. Can \nbe used as automotive fuel", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nStudents will \nlisten carefully LPG - Liquified Petroleum \nGas is a hydrocarbon fuel, \nlighter than petroleum. Can \nbe used as automotive fuel \n \nGeneralisation :- So, student s today we have studie d about conservation, sustainable \ndevelopment, crop rotation, mixed cropping, vermicomposting, LPG,CNG. \nFinal Recapitulation \nQues: Define sustainable development. \nQues: Explain crop rotation. \nQues: Define mixed cropping. \nQues: What is vermicomposting? \nHome work \nPT will ask the students to write a short note on various techniques of conservation.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nLESSON PLAN NO.6 \nGeneral Aims : (1) To develop the scientific attitude among the students . \n(2) To develop attitude towards environmental education among t he students . \n(3) To develop clear thinking and environmental awareness among the students . \n(4) To develop attitude towards environmental pollution among the students . \nSpecific Aim : To provide knowledge to students about \u201c Waste generation and management \u201d. \nPk Assumed :- It is assumed that the students will be aware of types of pollution . \nPk Testing : \nQ1. Define Pollution. \nQ2. What are the various types of pollution? \nQ3. What is the main cause of land pollution? \nQ4. How can waste be managed? \nAnnouncement of the Topic: \nWell students, today we will study about \u201c Waste generation and management \u201d \nPresentation: Lesson w ill be developed by the lecture method. \nSubject Matter PT\u2019s Activity Pupil\u2019s Activity BB Summary \nAccording to the states of \nmatter, wastes are of three", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPresentation: Lesson w ill be developed by the lecture method. \nSubject Matter PT\u2019s Activity Pupil\u2019s Activity BB Summary \nAccording to the states of \nmatter, wastes are of three \ntypes solid, liquid and \ngaseous. \nWaste management is \nessential to reduce the \nlevel of air, water, soil \npollution. To improve \nhuman health, to preserve \nand recycle resources. \nWaste management \ninvolves: \nCollection, segregation \nand recycling. \nCollection of wastes: \nMunicipal corporation \ncleans the roads and \nsewers. Each city has PT will ex plain \nwaste \nmanagement", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nStudents will listen \ncarefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n Waste management \ninvolves: \nCollection, \nsegregation and \nrecycling", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\ncommunal sites for waste \ndisposal. Sweepers collect \nwaste door to door. \nSegregation of wastes: \nWaste is segregated as \norganic, inorganic, \nhazardous, non hazardous, \nbiodegradable, non \nbiodegradable, industrial, \ndomestic, recyclable, non \nrecyclable. Different \ncoloured dustbins are \nused. \n1.Green coloured :All \nrecyclable material is \nthrown in this dustbin e.g. \nleftover chapatti, bread, \nfruit and vegetable \npeelings, paper, wood \nshaving s, leaf plates, \ncardboard, newspaper etc. \n2.Blue coloured: Non \nbiological wastes are \ncollected in these eg. \nplastics, glass, polythene, \nmetal, tin boxes, hospital \nwastes. \n3.Red coloured :Infectious \nwastes, radioactive wastes, \nmedical wastes are \ncollected i n these. \nRecycling: Mechanical \nprocessing: Collected \nwaste is passed through \nprocessing plant. High \nspeed wind blown through \nprocessor.", + "3.Red coloured :Infectious \nwastes, radioactive wastes, \nmedical wastes are \ncollected i n these. \nRecycling: Mechanical \nprocessing: Collected \nwaste is passed through \nprocessing plant. High \nspeed wind blown through \nprocessor. Heavy and light \nwaste and magnetic and \nnon magnetic waste is \nseparated. Recycling of \npaper ,steel and metals is", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nprocessing plant. High \nspeed wind blown through \nprocessor. Heavy and light \nwaste and magnetic and \nnon magnetic waste is \nseparated. Recycling of \npaper ,steel and metals is \ndone by machines. \nLiquid waste treatment: \nFollowing steps are \napplied: \nPrimary treatment: Large \nsuspended and floating \nmatter eg. plastic, wood, \npaper removed by sieving. \nSuspended matter of small", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nsegregation of \nwastes \n \n \n \n \n \n \n \n \n \n \n \nPT will explain \nuse of different \ncoloured \ndustbins using \nblackboard \n \n \n \n \n \n \n \n \n \n \n \nPT will explain \nrecycling using \nblackboard \n \n \n \n \n \n \n \nPT will explain \nliquid waste \ntreatment using \nblackboard \n \n \n \n \n \nStudents will listen \ncarefully \n \n \n \n \n \n \n \n \n \n \n \n \nStudents will listen \ncarefully \n \n \n \n \n \n \n \n \n \n \n \n \n \n \nStudents will listen \ncarefully \n \n \n \n \n \n \n \n \nStudents will listen \ncarefully \n \n \n \n \n \n \n \nSegregation of \nwastes: Waste is \nsegregated as \norganic, inorganic, \nhazardous, non \nhazardous, \nbiodegradable, non \nbiodegradable, \nindustrial, domestic, \nrecyclable, non \nrecyclable. \n \n \n \nDifferent coloured \ndustbins are used.", + "Different coloured \ndustbins are used. \nGreen coloured :All \nrecyclable material \n \n \nBlue coloured: Non \nbiological wastes \n \n \nRed coloured \n:Infectious wastes, \nradioactive wastes, \nmedical wastes \n \n \nRecycling: \nMechanical \nprocessing \n \n \n \n \n \n \nLiquid waste \ntreatment", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nBlue coloured: Non \nbiological wastes \n \n \nRed coloured \n:Infectious wastes, \nradioactive wastes, \nmedical wastes \n \n \nRecycling: \nMechanical \nprocessing \n \n \n \n \n \n \nLiquid waste \ntreatment \n \nPrimary treatment: \nLarge suspended and \nfloating matter e g. \nplastic, wood, paper \nremoved by sieving size is removed. The \nsuspended materials settle \ndown in the sedimentation \ntanks. Alum or lime are \nadded to the chambers. \nThese chemical coagulants \nsettle down the suspended \nmaterials. \nSecondary treatment: The \noutgoing water from the \nfirst chamber is lead i nto \nthis chamber. The total \nmass is created and mixed \nwith microbes to produce \na biomass which settles \ndown and is free from \niron. \nTertiary treatment: The \nwaste water is now passed \nthrough clear quartz tubes \nexposed to intense \nultraviolet light. \n \n \n \n \n \n \nPT will explain \nsecondary \ntreatment using \nblackboard \n \n \n \n \n \n \nPT will explain \ntertiary \ntreatment using \nblackboard", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nsecondary \ntreatment using \nblackboard \n \n \n \n \n \n \nPT will explain \ntertiary \ntreatment using \nblackboard \n \n \n \n \n \n \n \nStudents will listen \ncarefully \n \n \n \n \n \n \n \n \nStudents will listen \ncarefully \n \n \n \n \n \n \nSecondary treatment \n \nThe total mass is \ncreated and mixed \nwith microbes to \nproduce a biomass \nwhich settles down \n \n \nTertiary treatment: \nThe waste water is \nnow passed through \nclear quartz tubes \nexposed to intense \nultraviolet light", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nTertiary treatment: \nThe waste water is \nnow passed through \nclear quartz tubes \nexposed to intense \nultraviolet light \n \nGeneralisation :- So, student today we have studied about waste generation ,segregation of \nwastes, use of coloured dustbins to segregate wastes, recycling, primary, secondary and \ntertiary treatment. \nFinal Recapitulation \nQues: How is waste segregated? \nQues: Which coloured bin should be used to throw medical waste? \nQues: Define recycling . \nQues: What are the three processes of liquid waste treatment? \nHome work \nPT will ask the students to write down the process of waste management.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nLESSON PLAN NO.7 \nGeneral Aims : (1) To develop the scientific attitude among the students . \n(2) To develop attitude towards environmental education among the students . \n(3) To develop clear thinking and environmental awareness among the students . \n(4) To develop attitude towards environmental pollution among the students . \nSpecific Aim : To provide knowledge to students about \u201c Environmental Values and Ethics \u201d. \nPk Assumed :- It is assumed that the students will be aware of environment, ecosystem, types \nof po llution, waste management. \nPk Testing : \nQ1. Define Pollution . \nQ2. What are the various types of pollution? \nQ3. Define Environment. \nQ4. What are environmental laws? \nAnnouncement of the Topic: \nWell students, today we will study about \u201c Environmental Values an d Ethics \u201d \nPresentation: Lesson w ill be developed by the lecture method.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nSubject Matter PT s Activity Pupil\u2019s \nActivity BB Summary \nHuman rights are \ncommonly understood as \n\"inalienable fundamental \nrights to which a person is \ninherently entitled simply \nbecause she or he is a \nhuman being.\u201c \nHuman rights are thus \nconceived as universal \n(applicable everywhere) \nand egalitarian (the same \nfor everyone). T hese \nPT will explain \nhuman rights \nusing \nblackboard \n \n \n \n \n \n \n \nStudents will \nlisten \ncarefully \n \n \n \n \n \n \n \n Human rights are \nfundamental rights to \nwhich a person is \ninherently entitled simply \nbecause she or he is a \nhuman being .", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nrights may exist as natural \nrights or as legal rights, in \nboth national and \ninternational law. \nGuidelines/rules \nformulated by the \ngovernment to protect our \nenvironment are called \nenvironmental laws. \nNorms that have been \nformed by the government \nto pro tect environment are \ncalled Environmental \nLaws \nFWPCA Law prevents the \nindustries from \ndischarging waste \neffluents into water \nbodies . \nFundamental duties: \nArticle 51 A of the \nconstitution states that \u201dIt \nshall be the duty of every \ncitizen of India to protec t \nand improve the natural \nenvironment including \nforests, lakes, rivers and \nwildlife and have \ncompassion for living \ncreatures.\u201d \nChild welfare: Child \nlabour below 14 years \nshas been banned by the \ngovernment. \nChild labour refers to the \nemployment of children at \nregular and sustained \nlabour. This practice is \nconsidered exploitative by \nmany international \norganizations and is illegal \nin many countries.", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nPT will explain \nenvironmental \nlaws using \nblackboard \n \n \n \n \n \n \n \n \n \nPT will explain \nfundamental \nduties using \nblackboard \n \n \n \n \n \n \n \n \n \n \nPT will explain \nchild welfare \nand child \nlabour using \nblackboard \n \n \n \n \n \nStudents will \nlisten \ncarefully \n \n \n \n \n \n \n \n \n \n \nStudents will \nlisten \ncarefully \n \n \n \n \n \n \n \n \n \n \n \nStudents will \nlisten \ncarefully \n \n \n \n \n \nGuidelines/rules \nformulated by the \ngovernment to protect our \nenvironment are called \nenvironmental laws. \n \nFWPCA Law prevents the \nindustries from \ndischarging waste \neffluents into water bodies . \n \n \n \nFundamental duties: \nArticle 51 A \n \n \n \n \n \n \n \n \n \n \n \n \nChild welfare:Child labour \nhas been banned by the \ngovernment", + "CHAPTER 13-OUR ENVIRONMENT-notes.txt\nFundamental duties: \nArticle 51 A \n \n \n \n \n \n \n \n \n \n \n \n \nChild welfare:Child labour \nhas been banned by the \ngovernment \n \nGeneralisation :- So student s today we have studied about environmental laws ,human \nrights, fundamental duties and child labour. \n \nFinal Recapitulation \nQues: Define Environmental laws. \nQues: Which law prevents disposing waste effluents into water bodies ? \nQues: Which artic le deals with environment protection? \nQues: What is child labour? \nHome work \nPT will ask the students to write a short note on environmental laws.", + "CHAPTER 13-OUR ENVIRONMENT- Innovative paedagogy.txt\n1. Hands on Learning: \nThis is the best teaching method invented so far that involves the active participation of students to \nexperience scientific concepts than to just have an audience view. \n2. Role Play \nThis innovative method is becoming an integral part of science education as students can \nintellectually and physically involve through activities while learning a new concept. \n3. Visual clues \nUsing visual clues easily supplements auditory information and students can easily connect better \nwith ideas. \n4. Science Text Cards \nThis is an innovative teaching method to convey the science facts in an easy and organized way. In \nthis activity, statements related to science concepts are written on index cards \nWord Games \n5. This is a creative strategy to help eleme ntary students to experience the language of science. \nHangman, Pictionary, Dingbats, Bingo, Scrabble, Odd One Out, Charades, Trivial Pursuit etc. are the \ncommon games. \n6. Graphic Organizers", + "CHAPTER 13-OUR ENVIRONMENT- Innovative paedagogy.txt\nHangman, Pictionary, Dingbats, Bingo, Scrabble, Odd One Out, Charades, Trivial Pursuit etc. are the \ncommon games. \n6. Graphic Organizers \nThis teaching approach is helpful for students to interact with sc ience in a more organized and \nstructured way. Teachers can use different types of templates to represent the data according to the \ntopic that is being handled. \n7. Social media \nThe different social media platforms can be used wisely to teach science to make the science \nclassrooms more interesting and engaging. For instance, students can be asked to follow scientists in \nTwitter and share his/her new thoughts and findings in class or to use Feedly for improving the \ncontent of their research projects. \n8. Virtua l science labs \nThere are many virtual science labs available online for free and therefore, this approach almost \ngives hands on experience of learning the subject without much expense. \n9. Word walls", + "CHAPTER 13-OUR ENVIRONMENT- Innovative paedagogy.txt\nThere are many virtual science labs available online for free and therefore, this approach almost \ngives hands on experience of learning the subject without much expense. \n9. Word walls \nScience word walls in classrooms can stimulate the intere st of students in the subject and an \nopportunity for them to illustrate different concepts. In an advanced technique, a more in -depth \nunderstanding of different scientific terms can be made possible with pictures that accompany the \nwords. \n10. Crossover Lea rning \nIn this teaching method, students are given opportunity to learn in multiple settings such as inside \nthe classroom, museums, field trips or clubs. \n Teachers can use this method effectively by proposing a question in class, taking students to \ninformal settings to find answers and then, going back to classrooms to discuss and share the \nfindings. This way, learners can record, link, recall and share their diverse learning events to explore", + "CHAPTER 13-OUR ENVIRONMENT- Innovative paedagogy.txt\nfindings. This way, learners can record, link, recall and share their diverse learning events to explore \nthe topic and get a deeper level understanding. \n11. Science museu ms \nGive opportunity to students to visit a science museum as part of the learning process. This gives \nthem access to innovative resources and they can visualize data they learned in class. \n12. Projects \nThis can be an individual activity or group activity w hich helps students to showcase the application \nside of what they learnt through theory.", + "CHAPTER 13-OUR ENVIRONMENT- Innovative paedagogy.txt\nThis method involves choosing the idea, building a plan, executing the plan and finally evaluating it. \nWhen students pass through these stages, they can improve their skills to express ideas, problem \nsolving, overcoming the challenges, team work and self assessment. \n13. Multimedia Approach \nThis method is a blend of text, audio, animation, video, still images or interactivity content forms to \nteach diverse difficult to understand concepts in science. \n \nThe educator can convey vast information using advanced media, devices and techniques, and \ninvolve a wide range of activities to provide a meaningful learning experience. \n14. Science kit \nThis is one of the innovative method s of teaching science to kids and adults alike. Let it be physical \nscience, life science, or earth science, teachers are now making use of science kits to help students \nhave an engaging educational experience.", + "CHAPTER 13-OUR ENVIRONMENT- Innovative paedagogy.txt\nThey can conduct small experiments with guide d instructions to identify rock specimens, to watch \nout for insect metamorphosis or demonstrate chemistry reactions. \n15. Peer -to-Peer Teaching \nStudents take the role of teaching each other that promotes excitement learning science. In Peer -to-\nPeer Teaching approach, they are really engaged in the content by discussing scientific topics, \ngenerating questions and working in teams to explore new information. \n16.Field trips \nIn the middle of boring classroom sessions, take your students out for field trips to ex perience \nscience while learning. Go to an aquarium, a nature center, a scientist\u2019s home or visit a pet store. \n \nYou can also spend some time with the students in the bird park or simply go on a nature walk to \nexperience the science around while learning new concepts.", + "CHAPTER 13-OUR ENVIRONMENT-important notes.txt\nOur Environment Focus points \n \n1. Mode of nutrition in animals and plants \n2. Ecosystem : trophic levels , food chains . \n3. Pyramid of trophic levels. \n4. Law of conservation of energy and energy flow \n5. Human activities and pollution \n6. Ozone layer and it\u2019s depletion \n7. Garbage and it\u2019s management. \n8. Types of consumers. \n9. Bio degradable and non - biodegradable waste.", + "CHAPTER 13-OUR ENVIRONMENT-Focus Points.txt\nFOCUS POINT \n1. Classification of the components of ecosystem \n2. Different types of Consumers \n3. Food chain, Food webs and their significance \n4. Energy flow and 10% law of energy transfer \n5. Human activities and their impact on environment. \n6. Methods of disposal of garbage. \n7. Biological magnification. \n8. Ozone layer, depletion of ozone layer and it\u2019s impact", + "sample question papers 8.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 28 Page 1\nhttps://qrbook.page.link/app\nSample Paper 28\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho\nwever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv\ne Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac\nh. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each.", + "5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 8.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac\nh with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. In each test tub\nes A, B, C and D, ml2 of solution of ()AlSO24 3 in water was filled. Clean pieces of zinc was \nplaced in test tube A, clean iron nail was put in test tube B, silver ()Ag was placed in test tube C and a clean \ncopper wire was placed in test tube D.\nWhich of the following option(s) is/are correct about above experiment?\n(a) Copper is more reactiv\ne than aluminium.\n(b) Zinc is more reactive than aluminium.\n(c) Zinc, iron, silver and copp\ner are less reactive than aluminium.", + "(a) Copper is more reactiv\ne than aluminium.\n(b) Zinc is more reactive than aluminium.\n(c) Zinc, iron, silver and copp\ner are less reactive than aluminium.\n(d) Zinc is more reactive than copp\ner.\n2. A light bar magnet is susp", + "sample question papers 8.txt\ne than aluminium.\n(b) Zinc is more reactive than aluminium.\n(c) Zinc, iron, silver and copp\ner are less reactive than aluminium.\n(d) Zinc is more reactive than copp\ner.\n2. A light bar magnet is susp\nended near a current-carrying wire as shown below. What will happen to the light bar \nmagnet?\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 28 CBSE Science Class 10\nhttps://qrbook.page.link/app\n(a) It will rotate an ti-clockwise.\n(b)\n It will rotate clo\nckwise.\n(c)\n It will mo\nve to the left.\n(d)\n It will mo\nve to the right.\n3.\n Nastic \nmovements are non-directional responses to the stimuli. These movements are independent of the direction \nof the stimulus. On the other hand, growth movements which occur in the direction of the stimulus are called \ntropic movements.Which of the following movements is a nastic movement?", + "3.\n Nastic \nmovements are non-directional responses to the stimuli. These movements are independent of the direction \nof the stimulus. On the other hand, growth movements which occur in the direction of the stimulus are called \ntropic movements.Which of the following movements is a nastic movement?(a)\n Closing up of lea\nves of a sensitive plant on being touched with an object.\n(b)\n Bending of sho", + "sample question papers 8.txt\ntropic movements.Which of the following movements is a nastic movement?(a)\n Closing up of lea\nves of a sensitive plant on being touched with an object.\n(b)\n Bending of sho\not of a plant in response to light\n(c)\n Mo\nvement of root of a plant towards a source of water.\n(d)\n Clim\nbing up of a plant on an object by using tendrils. \n4.\n The diagram sho\nws part of the human respiratory system.\nWhat are W, X, Y and Z ?\nS.No. Bronchus Bronchiole Larynx Rings of cartilage\na. W X Z Y\nb. X Z Y W\nc. Y W X Z\nd. Z Y W X\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 28 P age 3\nhttps://qrbook.page.link/app\n5. The diagram shows the arrangement of cells inside the leaf of a green plant. (No cell contents are shown). Which \ncells normally contain chloroplasts?", + "The diagram shows the arrangement of cells inside the leaf of a green plant. (No cell contents are shown). Which \ncells normally contain chloroplasts?\n(a) 1 and 4\n(b)\n 1 and 2\n(c)\n 2 and 4\n(d)\n 2 and 3\n6.\n The table sho\nws the changes that occur in girls during puberty.", + "sample question papers 8.txt\ncells normally contain chloroplasts?\n(a) 1 and 4\n(b)\n 1 and 2\n(c)\n 2 and 4\n(d)\n 2 and 3\n6.\n The table sho\nws the changes that occur in girls during puberty.\n(i) Increases in the size of the breasts.\n(ii) Beginning of menstruation.(iii) Darkening of skin around the nipples.\nWhat is the likely significance of these changes?\n(a)\n Sexual maturation\n(b)\n Aging of the b\nody\n(c)\n Abnormal division of the cells\n(d)\n Pro\nduction of germ cells\n7.\n Clean small pieces of magnesium, zinc, aluminium, iron and copp\ner by rubbing them with a piece of sand paper. \nTake them in separate test tubes. Add about ml10 of dilute hydrochloric acid to each of them.\nIdentify the correct statement(s).(i)\n The rate of ev\nolution of hydrogen gas bubbles is not same in all the test tubes.\n(ii)\n The rate of formation of bubbles is the fastest in the case of magnesium.", + "Add about ml10 of dilute hydrochloric acid to each of them.\nIdentify the correct statement(s).(i)\n The rate of ev\nolution of hydrogen gas bubbles is not same in all the test tubes.\n(ii)\n The rate of formation of bubbles is the fastest in the case of magnesium.\n(iii)\n The reactivit\ny decreases in the order:\n Mg Zn Al Fe Cu 22 22\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 8.txt\n(iii)\n The reactivit\ny decreases in the order:\n Mg Zn Al Fe Cu 22 22\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 28 CBSE Science Class 10\nhttps://qrbook.page.link/app\n(iv) In the case of copper, no bubbles are seen and the temperature also remains unchanged. This shows that \ncopper does not react with dilute HCl.\n(a)\n (iii) and (iv)\n(b)\n (ii) and (iv)\n(c)\n (i), (ii) and (iv)\n(d)\n Only (iii)\n8.\n Whic\nh of the following statements about the given reaction are correct?", + "(a)\n (iii) and (iv)\n(b)\n (ii) and (iv)\n(c)\n (i), (ii) and (iv)\n(d)\n Only (iii)\n8.\n Whic\nh of the following statements about the given reaction are correct?\n() () () () FesH OF eO sH gg 34 4 23 42$ ++\n(i) Iron metal is getting o xidised\n(ii)\n W\nater is getting reduced\n(iii)\n W\nater is acting as reducing agent\n(iv)\n W\nater is acting as oxidising agent\n(a)\n (iii) and (iv)\n(b)\n (i),(ii) and (iv)\n(c)\n (ii) and (iv)\n(d)\n (i),(ii) and (iii)\n9.\n F\nour students (A), (B), (C) and (D) separately measured the pH values of each one of the given samples of distilled", + "sample question papers 8.txt\n(b)\n (i),(ii) and (iv)\n(c)\n (ii) and (iv)\n(d)\n (i),(ii) and (iii)\n9.\n F\nour students (A), (B), (C) and (D) separately measured the pH values of each one of the given samples of distilled \nwater, oxalic acid, dilute hydrochloric acid and a solution of sodium hydroxide using pH papers.\nStudent Distilled water Oxalic acid Dil. Hydrochloric Sodium Hydroxide\nA. 7 1 1 1\nB. 7 3 1 1\nC. 7 1 1 13\nD. 7 3 1 13\nWhich one of the following represents correct pH value?\n(a)\n B\n(b) A\n(c) D\n(d) C\n10. The table sho ws the formulae of three organic compounds that belong to the same homologous series.", + "7 3 1 13\nWhich one of the following represents correct pH value?\n(a)\n B\n(b) A\n(c) D\n(d) C\n10. The table sho ws the formulae of three organic compounds that belong to the same homologous series.\nFirst member of the homologous series CH OC H 33--\nSecond member of the homologous series CH CH OC H 32 3 --\nThird member of the homologous series CH CH CH OC H 32 23--\nWhat is the general formula of this series?(a)\n CH O nn22+\n(b) CH O nn2\n(c) CH OH nn22+\n(d) CH OH nn22+\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 8.txt\nWhat is the general formula of this series?(a)\n CH O nn22+\n(b) CH O nn2\n(c) CH OH nn22+\n(d) CH OH nn22+\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 28 P age 5\nhttps://qrbook.page.link/app\n11. An acid ()A with sodium hydrogen carbonate is used in making the cakes fluffy and spongy. It is due to the release \nof gas ()B in the reaction. Here, ()A and ()B are:\n(a)\n ()A : Tartaric acid, ()B, O2\n(b) ()A : Oxalic acid, ()B, CO2 \n(c) ()A : Tartaric acid, ()B. CO2\n(d) ()A : Succinic acid, ()B. H2 \n12. \nThe above lens has a focal length of 10 cm. The object of height 2 mm is placed at a distance of 5 cm from the \npole.", + "The above lens has a focal length of 10 cm. The object of height 2 mm is placed at a distance of 5 cm from the \npole. Find the height of the image.(a)\n 6.67 mm\n(b)\n 4 cm\n(c)\n 3.33 mm\n(d)\n 4 mm\n13.\n Rah\nul is a skilled painter. He mixed a white coloured powder, compound X with water. The compound X reacted", + "sample question papers 8.txt\npole. Find the height of the image.(a)\n 6.67 mm\n(b)\n 4 cm\n(c)\n 3.33 mm\n(d)\n 4 mm\n13.\n Rah\nul is a skilled painter. He mixed a white coloured powder, compound X with water. The compound X reacted \nvigorously with water to produce a compound Y and a large amount of heat. Then, Rahul used the compound Y \nfor white washing the walls.\nName the compound Y that Rahul got after mixing X with water.\n(a)\n Calcium o\nxide\n(b)", + "sample question papers 8.txt\nCalcium\n(c)\n Calcium h\nydroxide\n(d)\n Calcium carb\nonate\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 28 CBSE Science Class 10\nhttps://qrbook.page.link/app\n14. Iden tify P, Q R and S in the case of normal human from given flow chart and select the correct option.\nP Q R S\na. XX XY XY XX\nb. XY XX XX XY\nc. XX XY XX XY\nd. XY XX XY XX\n15.\n Figure (A), (B), (C) and (D) resp\nectively correspond to\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "Figure (A), (B), (C) and (D) resp\nectively correspond to\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 28 P age 7\nhttps://qrbook.page.link/app\n(a) the short-sighted eye, the correction of long sightedness, the long-sighted eye and the correction of short-\nsightedness\n(b)\n the short-sigh\nted eye, the correction of short sightedness, . the long-sighted eye and the correction of long-\nsightedness\n(c)\n the", + "sample question papers 8.txt\nsightedness\n(b)\n the short-sigh\nted eye, the correction of short sightedness, . the long-sighted eye and the correction of long-\nsightedness\n(c)\n the \nlong-sighted eye, the correction of short-sightedness, the short-sighted eye and the correction of long-\nsightedness\n(d)\n None of the ab\nove\n16.", + "sample question papers 8.txt\nFor the diagram shown, according to the new Cartesian sign convention the magnification of the image formed \nwill have the following specifications:(a)\n Sign \u2014 P\nositive, Value \u2014 More than 1\n(b)\n Sign \u2014 P\nositive, Value \u2014 Less than 1\n(c)\n Sign \u2014 Negativ\ne, Value \u2014 More than 1\n(d)\n Sign \u2014 Negativ\ne, Value \u2014 Less than 1\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17.\n Assertion (A):\n Mendel selected pea plants for this experiments.\nReason (R): Pea plant is self pollinating with short life cycle and bears visible contrasting characters.\n(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n18.", + "(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n18.\n Assertion (A):\n When the length of a wire is doubled, then its resistance also gets doubled.", + "sample question papers 8.txt\nAssertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n18.\n Assertion (A):\n When the length of a wire is doubled, then its resistance also gets doubled.\nReason (R): The resistance of a wire is directly proportional to its length.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n19.\n Assertion (A):\n Transpiration is a necessary evil.\nReason (R): It causes water loss but helps in absorption and upward movement of water and minerals.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)", + "sample question papers 8.txt\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 28 CBSE Science Class 10\nhttps://qrbook.page.link/app\n20. Assertion (A): Following are the structural isomers of butane.\nReason (R): Structural isomers have the same molecular formula, but they differ in their structures.\n(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21.\n F", + "sample question papers 8.txt\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21.\n F\nor the same angle of incidence in media A, B and C, the angles of refraction are 20\u00b0, 30\u00b0 and 40\u00b0 respectively. In \nwhich medium will the velocity of light be maximum? Give reason in support of your answer.\n\u0001o\u0001\nWhy is the colour of the clear sky blue?\n22.\n (i)\n What is mean\nt by garbage? List two classes into which garbage is classified.\n(ii)\n What do w\ne actually mean when we say that the \u201cenzymes are specific in their action\u201d?\n23.\n What are the strategies of plan\nts to get rid of their wastes?\n24.\n What is the adv\nantage of having four chambered heart?\n25.\n A studen\nt performs the following four experiments.\nBased on the above experiments:\n(i)\n In whic\nh test tube(s) no reaction occurred? Give reason.", + "24.\n What is the adv\nantage of having four chambered heart?\n25.\n A studen\nt performs the following four experiments.\nBased on the above experiments:\n(i)\n In whic\nh test tube(s) no reaction occurred? Give reason.\n(ii)\n Arrange the giv\nen metal samples in the increasing order of reactivity.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne", + "sample question papers 8.txt\n(i)\n In whic\nh test tube(s) no reaction occurred? Give reason.\n(ii)\n Arrange the giv\nen metal samples in the increasing order of reactivity.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 28 P age 9\nhttps://qrbook.page.link/app\n\u0001o\u0001\nThe following reaction takes place when aluminium powder is heated with MnO 2:\n() () () () MnOs AlsM nA lO ll 34 32 22 3 \" ++ + Heat\n(i)\n Is aluminium getting reduced?\n(ii)", + "sample question papers 8.txt\nIs MnO 2 getting oxidised?\n26.\n What is feedbac\nk mechanism of harmonic regulation? Take the example of insulin to explain this phenomenon.\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27.\n Dra\nw a neat and well labelled diagram for electrolytic refining of copper. Write the reactions involved.\n28.\n W\nrite the essential function performed by ozone at the higher levels of the Earth\u2019s atmosphere? How is it \nproduced? Name the synthetic chemicals mainly responsible for the drop of amount of ozone in the atmosphere. \nHow can the use of these chemicals be reduced?\n29.\n (i)\n A circuit con\ntains a battery, a variable resistor and a solenoid. The figure below show the magnetic field \npattern produced by the current in the solenoid.\n(a) State ho w the magnetic field pattern indicates regions where the magnetic field is stronger?\n(b)\n What happ\nens to the magnetic field when the current in the circuit is reversed?\n(ii)\n State the direction of magnetic field in the follo", + "sample question papers 8.txt\n(b)\n What happ\nens to the magnetic field when the current in the circuit is reversed?\n(ii)\n State the direction of magnetic field in the follo\nwing case.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 28 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001o\u0001\nName, state and explain with an example the rule used to determine the direction of force experienced by a current \ncarrying conductor placed in a uniform magnetic field.\n30.\n An organic comp\nound X on heating with conc. HSO24 forms a compound Y which on addition of one molecule \nof hydrogen in the presence of nickel forms a compound Z. One molecule of compound Z on combustion forms \ntwo molecules of CO 2 and three molecules of HO2. Identify giving reasons the compounds X, \u2018Y and Z. Write \nthe chemical equations for all the chemical reactions involved.\n31.\n (i)\n With the help of lab", + "sample question papers 8.txt\nthe chemical equations for all the chemical reactions involved.\n31.\n (i)\n With the help of lab\nelled ray diagram, show the path followed by a narrow beam of monochromatic light \nwhen it passes through a glass prism.\n(ii)\n What w\nould happen if this beam is replaced by a narrow beam of white light?\n32.\n W\nrite one example of each of the following tropic movements:\n(i)\n P\nositive phototropism.\n(ii)\n Negativ\ne phototropism.\n(iii)\n P\nositive geotropism.\n(iv)\n Negativ\ne geotropism.\n(v)", + "sample question papers 8.txt\nHydrotropism.\n(vi)\n \nChemotropism.\n\u0001o\u0001\n(i)\n Name the part of h\numan brain which controls.\n(a)\n v\noluntary actions and\n(b)\n in\nvoluntary actions\n(ii)\n W\nrite the function of peripheral nervous system. Name the components of this system stating their origin.\n33.", + "sample question papers 8.txt\nIn the above circuit, if the current reading in the ammeter A is 2A, what would be the value of R1?\nSECTION-D \n\u0001Question no. 34 to 36 are Long answer questions.\n34.\n (i)\n Define reflex arc. Dra\nw a flowchart showing the sequence of events which occur during sneezing.\n(ii)\n List four plan\nt hormones. Write one function of each.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 28 P age 11\nhttps://qrbook.page.link/app\n\u0001o\u0001\n(i)\n Iden\ntify the endocrine glands A, B, C, D, E and F in the given diagram.\n(ii)\n List the functions of eac\nh part.\n35. (i) Dra w magnetic field lines produced around a current carrying straight conductor passing through a \ncardboard. Name, state and apply the rule to mark the direction of these field lines.", + "(ii)\n List the functions of eac\nh part.\n35. (i) Dra w magnetic field lines produced around a current carrying straight conductor passing through a \ncardboard. Name, state and apply the rule to mark the direction of these field lines.\n(ii)\n Ho\nw will the strength of the magnetic field change when the point where magnetic field is to be determined", + "sample question papers 8.txt\n(ii)\n Ho\nw will the strength of the magnetic field change when the point where magnetic field is to be determined \nis moved away from the straight wire carrying constant current? Justify your answer.\n36.\n A metal nitrate A on heating gives yellowish brown coloured metal oxide along with brown gas B and a colourless \ngas C. Aqueous solution of A on reaction with potassium iodide forms a yellow precipitate of compound D. \nIdentify A, B, C and D. Also identify the types of both the reactions. Metal present in A is used in alloy which \nis used for soldering purposes.\n\u0001o\u0001\n(i)\n Crystals \nof a substance changed their colour on heating in a closed test tube but regained it after sometime \nwhen they were allowed to cool down. Name the substance, write its formula and explain the phenomenon \ninvolved.\n(ii)\n Name the comp\nound whose one formula unit is associated with 10 water molecules. How is it prepared? Give \nequations of related reactions. Give two uses of the compound.\nSECTION-E", + "sample question papers 8.txt\ninvolved.\n(ii)\n Name the comp\nound whose one formula unit is associated with 10 water molecules. How is it prepared? Give \nequations of related reactions. Give two uses of the compound.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.\n37.\n A studen\nt wants to project the image of a candle flame on the walls of the school laboratory by using a mirror.\n(i)\n Whic\nh type of mirror should he use and why?\n(ii)\n A\nt what distance, in terms of focal length of the mirror, should he place the candle flame to get the \nmagnified image on the wall?\n\u0001o\u0001\n(iii)\n Dra\nw a ray diagram to show the formation of the image in this case.\n(iv)\n Can he use this mirror to pro\nject a diminished image of the candle flame on the same wall? State \u2018how\u2019 if \nyour answer is \u2018yes\u2019 and \u2018why not\u2019 if your answer is \u2018no\u2019.", + "(iv)\n Can he use this mirror to pro\nject a diminished image of the candle flame on the same wall? State \u2018how\u2019 if \nyour answer is \u2018yes\u2019 and \u2018why not\u2019 if your answer is \u2018no\u2019.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 8.txt\nyour answer is \u2018yes\u2019 and \u2018why not\u2019 if your answer is \u2018no\u2019.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 12 Sample P aper 28 CBSE Science Class 10\nhttps://qrbook.page.link/app\n38. Mrs. Tomar uses a compound of sodium X to make pakoras crispy. It is a mild non-corrosive basic salt, also used \nas an ingredient in antacids. It is produced using sodium chloride as one of the raw materials.\n(i)\n Iden\ntify the compound of sodium X. Is the pH value of X solution lower than or higher than 7?\n(ii)\n W\nrite the chemical equation of preparation of X. Write the chemical reaction involved when Xis heated.\n\u0001o\u0001\nState the chemical properties on which the following uses of X are based :\n(i)\n as an an\ntacid.\n(ii)\n as a so\nda fire extinguisher.\n39.\n An organism A which cannot move from one place to another, makes a simple food B from the substances C", + "sample question papers 8.txt\n(i)\n as an an\ntacid.\n(ii)\n as a so\nda fire extinguisher.\n39.\n An organism A which cannot move from one place to another, makes a simple food B from the substances C \nand D available in the environment. This food is made in the presence of a green coloured substance E present \nin organs F in the presence of light energy in a process called G. Some of the simple food B also gets converted \ninto a complex food H for storage purposes. The food H gives a blue-black colour with dilute iodine solution.\n(i)\n What is (a) organism A (b) food B, and (c) food H?\n(ii)\n What are C and D?\n(iii)\n In eac\nh of the following situations what happens to the rate of process G?\n(a)\n Cloudy da\nys.\n(b)\n Stomata get blo\ncked due to dust.\n\u0001o\u0001\n(iv)\n Explain the mec\nhanism of process G.\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 6.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 30 Page 1\nhttps://qrbook.page.link/app\nSample Paper 30\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho\nwever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv\ne Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac\nh. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each.", + "5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 6.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac\nh with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Which among the follo\nwing is/are double displacement reaction(s)?\n(i) Pb Cucl Pbcl Cu 22$++\n(ii) NaSO BaCl BaSO NaCl 2 24 24$ ++\n(iii) CO CO 22$+\n(iv) CH OC OH O 22 42 22$++\n(a) Only (ii)\n(b) (i) and (iv)\n(c) (iii) and (iv)\n(d) (i) and (ii)\n2. \nExamine the above figure and state which of the following option is correct? (one small box in the figure is equal \nto 1 cm).", + "Examine the above figure and state which of the following option is correct? (one small box in the figure is equal \nto 1 cm).(a)\n The mirror has a focal length of -3 cm and will pro\nduce an image of magnification -1.\n(b) The mirror has a focal length of -6 cm and will pro\nduce an image of magnification +1.", + "sample question papers 6.txt\nto 1 cm).(a)\n The mirror has a focal length of -3 cm and will pro\nduce an image of magnification -1.\n(b) The mirror has a focal length of -6 cm and will pro\nduce an image of magnification +1.\n(c) The mirror has a focal length of -6 cm and will pro\nduce an image of magnification -1.\n(d) The mirror has a focal length of -3 cm and will pro\nduce an image of magnification +1.Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 30 CBSE Science Class 10\nhttps://qrbook.page.link/app\n3. The image sho ws a bud developing on a Hydra.\nHow does the bud develop in the Hydra?\n(a)\n Bud dev\nelops due to repetitive cell division at a specific site\n(b)\n Bud dev\nelops due to separation of body parts of Hydra\n(c)\n Bud dev\nelops due to attachment of another Hydra at a specific site\n(d)\n Bud dev\nelops due to change in the environmental conditions\n4.", + "Differen\nt organs of human eye are labelled as A to F.\nThe Structure of Human Eye", + "sample question papers 6.txt\n(d)\n Bud dev\nelops due to change in the environmental conditions\n4.\n Differen\nt organs of human eye are labelled as A to F.\nThe Structure of Human Eye\nWhen light rays enter the eye, most of the refraction occurs at the:(a)", + "sample question papers 6.txt\npart B\n(b) part D\n(c) part E\n(d) outer surface of part F\n5. The giv en figure represents a single nephron from a mammalian kidney. Identify the labelled parts, match them \nwith the options (I-IV) and select the correct answer.(i)\n The site of Ultrafiltration\n(ii)\n Collect the urine an mak\ne it more concentrated \n(iii)\n The main site of reabsorption of glucose an amino acids.\n(iv)\n Largely resp\nonsible for the maintenance of blood pH\n(a)\n (i)-A, (ii)-B, (iii)-C, (iv)-D\n(b)\n (i)-A, (ii)-E, (iii)-C, (iv)-D\n(c)\n (i)-E, (ii)-E, (iii)-D, (iv)-A\n(d)\n (i)-B, (ii)-A, (iii)-C, (iv)-E\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.", + "Click Here To InstallCBSE Science Class 10 Sample P aper 30 P age 3\nhttps://qrbook.page.link/app\n6. The table giv en below shows the reaction of a few elements with acids and bases to evolve Hydrogen gas.\nElement Acid Base\nA. # #\nB. { {\nC. { #\nD. { {\nWhich of these elements form amphoteric oxides ?\n(a)\n B and D\n(b)\n A and D\n(c)", + "sample question papers 6.txt\nElement Acid Base\nA. # #\nB. { {\nC. { #\nD. { {\nWhich of these elements form amphoteric oxides ?\n(a)\n B and D\n(b)\n A and D\n(c)\n C and D\n(d)\n A and C\n7.", + "sample question papers 6.txt\nThe angle of incidence from air to glass at the point O on the hemispherical glass slab is :\n(a)\n 0c\n(b) 45c\n(c) 180c\n(d) 90c\n8. What happ ens when a solution of an acid is mixed with a solution of a base in a test tube?\n(i)\n The temp\nerature of the solution increases.\n(ii)\n The temp\nerature of the solution decreases.\n(iii)\n The temp\nerature of the solution remains the same.\n(iv)\n Salt formation tak\nes place.\n(a)\n (i) and (ii)\n(b)\n Only (i)\n(c)\n (i) and (iv)\n(d)\n (ii) and (iii)\n9.\n A studen\nt obtains a blurred image of a distant object on a screen using a convex lens. To obtain a distinct image \non the screen, he should move the lens:(a)\n to\nwards the screen.\n(b)\n a\nway from the screen\n(c)\n either to\nwards or away from the screen depending upon the position of the object.\n(d)\n to a p\nosition very far away from the screen.", + "(b)\n a\nway from the screen\n(c)\n either to\nwards or away from the screen depending upon the position of the object.\n(d)\n to a p\nosition very far away from the screen.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 30 CBSE Science Class 10", + "sample question papers 6.txt\n(d)\n to a p\nosition very far away from the screen.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 30 CBSE Science Class 10\nhttps://qrbook.page.link/app\n10. The molecular form ulae of three organic compounds are shown below. Choose the correct option.\nOrganic Compound Molecular Compound\nP CH38\nQ CH51 0\nR CH46\nIdentify the incorrect statement about these three hydrocarbons.\n(a)\n P, Q both differ by - CH 2 unit.\n(b)\n ALL ha\nve different general formula.\n(c)\n Q is an alkene.\n(d)\n P is an alkane.\n11.\n pH of differen\nt solutions are given in the table below:\nSolution pH\nP 2.2-2.4\nQ 13.8-14.0\nR 6.5-7.5\nS 8.0-9.0\nArrange these solutions in the increasing order of H+ ion concentration.", + "(a)\n SRQP11 1\n(b) PRSQ11 1\n(c) RSQP11 1\n(d) QSRP11 1\n12. Whic h of the following is the correct electronic arrangement of sodium oxide?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 6.txt\n(b) PRSQ11 1\n(c) RSQP11 1\n(d) QSRP11 1\n12. Whic h of the following is the correct electronic arrangement of sodium oxide?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 30 P age 5\nhttps://qrbook.page.link/app\n13. Select the correct statemen t regarding p, q, r and s.\n(a) q represents pulmonary artery that carries oxygenated blood from lungs to heart.\n(b)\n p represents pulmonary veins that carries deoxygenated blood from heart to lungs \n(c)\n Exc\nhange of gases takes place in r and oxygenated blood is carried back through aorta.\n(d)\n Exc\nhange of gases and substances takes place in s and deoxygenated blood is then carried back to the lungs \nthrough vena cava.\n14.\n Whic\nh of the following statements are true about the endocrine glands?", + "(d)\n Exc\nhange of gases and substances takes place in s and deoxygenated blood is then carried back to the lungs \nthrough vena cava.\n14.\n Whic\nh of the following statements are true about the endocrine glands?\n(i)\n They are ductless glands\n(ii)\n They release their hormones in\nto a duct\n(iii)\n They pro\nduce chemical messengers called hormones\n(iv)", + "sample question papers 6.txt\n(i)\n They are ductless glands\n(ii)\n They release their hormones in\nto a duct\n(iii)\n They pro\nduce chemical messengers called hormones\n(iv)\n They release their hormones directly in\nto the bloodstream\n(a)\n (i), (iii) and (iv)\n(b)\n (i) and (iii)\n(c)\n (i) and (iv)\n(d)\n (i), (ii) and (iii)\n15.\n Tw\no pea plants one with round green seeds (RRyy) and another with wrinkled yellow (rrYY) seeds produce \nF1 progeny that have round, yellow (RrYy) seeds. When F1 plants are selfed, the F2 progeny will have new \ncombination of characters.", + "When F1 plants are selfed, the F2 progeny will have new \ncombination of characters. Choose the new combination from the following:\n(i)\n Round, y\nellow\n(ii)\n Round, green\n(iii)\n W\nrinkled, yellow\n(iv)\n W\nrinkled, green\n(a)\n (i) and (iv)\n(b)\n (i) and (ii)\n(c)\n (i) and (iii)\n(d)\n (ii) and (iii)\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 30 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers 6.txt\n(d)\n (ii) and (iii)\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 30 CBSE Science Class 10\nhttps://qrbook.page.link/app\n16. Quic k lime combines vigorously with water to form (A) which reacts slowly with the carbon dioxide in air to form \n(B).Identify the compounds (A) and (B).\n(A) (B)\na. Calcium carbonate Calcium hydroxide\nb. Calcium hydroxide Calcium carbonatec. Calcium Calcium bicarbonate\nd. Calcium bicarbonate Calcium\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17.\n Assertion (A):\n The planets twinkle while the stars do not.\nReason (R): The planets are much closer to the earth than the stars.\n(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).", + "17.\n Assertion (A):\n The planets twinkle while the stars do not.\nReason (R): The planets are much closer to the earth than the stars.\n(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)", + "sample question papers 6.txt\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n18.\n Assertion (A):\n In human beings, the sex of the individual is largely genetically determined.\nReason (R): In snails, sex is not genetically determined.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n19.\n Assertion (A):\n Amoeba always produces two daughter amoebae while Plasmodium divides into many daughter \ncells.Reason (R): Amoeba undergoes binary fission while Plasmodium undergoes multiple fission.(a)", + "sample question papers 6.txt\nAmoeba always produces two daughter amoebae while Plasmodium divides into many daughter \ncells.Reason (R): Amoeba undergoes binary fission while Plasmodium undergoes multiple fission.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n20.\n Assertion (A):\n Burning of natural gas is an endothermic process.\nReason (R): Methane gas combines with oxygen to produce carbon dioxide and water.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.", + "(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n\u0001Cont\no\nnueCo", + "sample question papers 6.txt\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 30 P age 7\nhttps://qrbook.page.link/app\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21.\n (i)\n List the parts of the h\numan eye that control the amount of light entering into it. Explain how they perform \nthis function.\n(ii)\n W\nrite the function of retina in human eye.\n\u0001o\u0001\nExplain using ray diagrams how the defect associated with hypermetropic eye can be corrected.\n22.\n Giv\ne reasons:\n(i)\n Placen\nta is extremely essential for foetal development.\n(ii)\n Uterine lining b\necomes thick and spongy after fertilisation.\n23.\n Kulhads \n(disposable cups made of clay) and disposable paper cups both are used as an alternative for disposable", + "sample question papers 6.txt\n(ii)\n Uterine lining b\necomes thick and spongy after fertilisation.\n23.\n Kulhads \n(disposable cups made of clay) and disposable paper cups both are used as an alternative for disposable \nplastic cups. Which one of these two can be considered as a better alternative to plastic cups and why?\n24.\n (i)\n Name the repro\nductive and non-reproductive parts of bread mould (Rhizopus).\n(ii)\n List an\ny two advantages of vegetative propagation.\n25.\n Observ\ne the given figure and answer the questions that follow:\n(i) Iden tify the gas X.\n(ii)\n W\nrite the chemical reaction involved.\n(iii)\n Whic\nh type of chemical reaction is taking place?\n(iv)\n is it an exothermic reaction or an endothermic reaction?\n\u0001o\u0001\nWhat are strong and w\neak acids? In the following list of acids, separate strong acids from weak acids. Hydrochloric \nacid, citric acid, acetic acid, nitric acid, formic acid, sulphuric acid.\n26.", + "\u0001o\u0001\nWhat are strong and w\neak acids? In the following list of acids, separate strong acids from weak acids. Hydrochloric \nacid, citric acid, acetic acid, nitric acid, formic acid, sulphuric acid.\n26.\n Ho\nw do auxins promote the growth of a tendril around a support ?\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne", + "sample question papers 6.txt\nacid, citric acid, acetic acid, nitric acid, formic acid, sulphuric acid.\n26.\n Ho\nw do auxins promote the growth of a tendril around a support ?\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 30 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27.\n Differen\ntiate between a glass slab and a glass prism. What happens when a narrow beam of (i) a monochromatic \nlight, and (ii) white light passes through (a) glass slab and (b) glass prism?\n28.\n In the follo\nwing food chain, only 2J of energy was available to the peacocks. How much energy would have been \npresent in Grass? Justify your answer. Grass \" Grass Hopper \" Frog \" Snake \" Peacock.\n29.", + "28.\n In the follo\nwing food chain, only 2J of energy was available to the peacocks. How much energy would have been \npresent in Grass? Justify your answer. Grass \" Grass Hopper \" Frog \" Snake \" Peacock.\n29.\n Tw\no resistors with resistance 10 \u2126 and 15 \u2126 are to be connected to a battery of emf 12 V so as to obtain:\n(i)\n minim\num current\n(ii)\n maxim\num current", + "sample question papers 6.txt\n29.\n Tw\no resistors with resistance 10 \u2126 and 15 \u2126 are to be connected to a battery of emf 12 V so as to obtain:\n(i)\n minim\num current\n(ii)\n maxim\num current\nDescribe the mode of connecting the resistances in each case. Calculate the strength of the total current in the \ncircuit in each case.\n\u0001o\u0001\n(i)\n State the relation correlating the electric curren\nt flowing in a conductor and the voltage applied across it. \nAlso draw a graph to show this relationship.\n(ii)\n Find the resistance of a conductor if the electric curren\nt flowing through it is 0.35 A when the potential \ndifference across it is 1.4 V.\n30.\n Analyse \nthe following observation table showing variation of image distance ()v with object distance u in case of \na convex lens and answer the questions that follow without doing any calculations:\nS.No. Object distance u ()cm Image distance v ()cm\n1. -100 +25\n2. -60 +30\n3. -40 +40\n4. -30 +60\n5.", + "Object distance u ()cm Image distance v ()cm\n1. -100 +25\n2. -60 +30\n3. -40 +40\n4. -30 +60\n5. -25 +100\n6. -15 +120\n(i)\n What is the fo", + "sample question papers 6.txt\nS.No. Object distance u ()cm Image distance v ()cm\n1. -100 +25\n2. -60 +30\n3. -40 +40\n4. -30 +60\n5. -25 +100\n6. -15 +120\n(i)\n What is the fo\ncal length of the convex Lens? Give reason to justify your answer.\n(ii)\n W\nrite the serial number of the observation which is not correct. On what basis have you arrived at this \nconclusion?\n(iii)\n Select an appropriate scale and dra\nw a ray diagram for the observation at S.No.2. Also find the approximate \nvalue of magnification.\n31.\n During electrolysis of brine, a gas G is liberated at anode. When this gas G is passed through slaked lime, a \ncompound C is formed, which is used for disinfecting drinking water.\n(i)\n W\nrite formula of G and C.\n(ii)\n State the c\nhemical equation involved.\n(iii)\n What is common name of comp\nound C ? Give its chemical name.\n32.", + "(i)\n W\nrite formula of G and C.\n(ii)\n State the c\nhemical equation involved.\n(iii)\n What is common name of comp\nound C ? Give its chemical name.\n32.\n Nerv\nous and hormonal systems together perform the function of control and coordination in human beings. Justify \nthis statement with the help of an example.\n\u0001o\u0001", + "sample question papers 6.txt\nound C ? Give its chemical name.\n32.\n Nerv\nous and hormonal systems together perform the function of control and coordination in human beings. Justify \nthis statement with the help of an example.\n\u0001o\u0001\nList in tabular form three distinguishing features between cerebrum and cerebellum.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 30 P age 9\nhttps://qrbook.page.link/app\n33. The general form ula of three compounds A, B and C is CH B nn2+ has highest boiling point and C has lowest \nboiling point.\n(i)\n Name the homologous series to whic\nh A, B and C belongs.\n(ii)\n Whic\nh of these have minimum number of carbon atoms.?\n(iii)\n W\nrite the name and molecular formula of 4th member of this homologous series.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34.\n (i)\n Dra\nw a diagram of human excretory system and label on it the following parts:\n(a)", + "sample question papers 6.txt\nKidney\n(b)\n \nUreter\n(c)\n Urinary bladder\n(d)", + "sample question papers 6.txt\nUrethra\n(ii)\n W\nrite one main function each of the labelled parts.\n\u0001o\u0001\n(i)\n Dra\nw a schematic representation of transport and exchange of oxygen and carbon dioxide during \ntransportation of blood in human beings and label on it: Lung capillaries, Pulmonary artery to lungs, Aorta \nto body, Pulmonary veins from lungs.\n(ii)\n What is the adv\nantage of separate channels in mammals and birds for oxygenated and deoxygenated blood?\n35.\n State reason for the follo\nwing statements:\n(i)\n T\nap water conducts electricity whereas distilled water does not.\n(ii)\n Dry h\nydrogen chloride gas does not turn blue litmus red whereas dil. HCl does.\n(iii)\n During summer season, a milkman usually adds a v\nery small amount of baking soda to fresh milk.\n(iv)\n F\nor dilution of acid, acid is added to water and not water into acid.\n(v)\n Ammonia is a base but do\nes not contain hydroxyl group.\n\u0001o\u0001\n(i)\n W\nrite the chemical formula of hydrated copper sulphate and anhydrous copper sulphate. Giving an activity", + "sample question papers 6.txt\n(v)\n Ammonia is a base but do\nes not contain hydroxyl group.\n\u0001o\u0001\n(i)\n W\nrite the chemical formula of hydrated copper sulphate and anhydrous copper sulphate. Giving an activity \nillustrate how these are inter-convertible?\n(ii)\n W\nrite chemical names and formula of plaster of Paris and gypsum.\n36.\n (i)\n State the rule to determine the direction of a\n(a)\n magnetic field pro\nduced around a straight conductor-carrying current.\n(b)\n force \nexperienced by a current-carrying straight conductor placed in a magnetic field which is \nperpendicular to it, and\n(ii)\n Magnetic field lines of t\nwo magnets are shown in fig. (a) and (b).\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 30 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSelect the figure that represent the correct pattern of field lines. Give reason for your answer. Also name the poles \nof the magnet facing each other.\nSECTION-E", + "sample question papers 6.txt\nhttps://qrbook.page.link/app\nSelect the figure that represent the correct pattern of field lines. Give reason for your answer. Also name the poles \nof the magnet facing each other.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.\n37.\n The \nfemale reproductive system includes the ovaries, fallopian tubes, uterus, vagina and mammary glands. These \norgans are involved in the production and transportation of gametes and the production of sex hormones. The \nfemale reproductive system also facilitates the fertilisation of ova by sperm and supports the development of offspring during pregnancy and infancy.\n(i) In whic h part does:\n(a)\n fertilisation tak\ne place\n(b)\n fo\netus develop\n(ii)\n Whic\nh structures in human female are equivalent to the following structures in the male ?\n(a)", + "sample question papers 6.txt\nTestes\n(b)\n V\nas deferenes\nIn each case say in what respect the structures are equivalent ?\n\u0001o\u0001\n(iii)\n W\nrite the number of immature eggs present in the ovaries of a newly born baby girl. Mention what happen \nto these immature eggs when the girl attains puberty ?\n38.\n A studen\nt was asked to perform an experiment to study the force on a current carrying conductor in a magnetic \nfield. He took a small aluminium rod AB, a strong horse-shoe magnet, some connecting wires, a battery and a \nswitch and connected them as shown. He observed that on passing current, the rod gets displaced. On reversing \nthe direction of current, the direction of displacement also gets reversed. On the basis of your understanding of this phenomenon, answer the following questions:\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 30 P age 11\nhttps://qrbook.page.link/app", + "sample question papers 6.txt\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 30 P age 11\nhttps://qrbook.page.link/app\n(i) Wh y does the rod get displaced on passing current through it?\n(ii)\n State the rule that determines the direction of the force on the conductor AB.\n(iii)", + "sample question papers 6.txt\n(a)\n If the U shap\ned magnet is held vertically and the aluminium rod is suspended horizontally with its \nend B towards due north, then on passing current through the rod from B to A as shown, in which \ndirection will the rod be displaced?\n(b)\n Name an\ny two devices that use current carrying conductors and magnetic field.\n\u0001o\u0001\n(iv)\n Dra\nw the pattern of magnetic field lines produced around a current carrying straight conductor held \nvertically on a horizontal cardboard. Indicate the direction of the field lines as well as the direction of \ncurrent flowing through the conductor.\n39.\n Ethanol, \ncommonly knowns alcohol, is an active ingredient of all alcoholic drinks. It is also used in medicines \nsuch as tincture iodine, cough syrups and many tonics. Ethanol\u2019s molecular formula is CHO6 2 which means it", + "sample question papers 6.txt\nsuch as tincture iodine, cough syrups and many tonics. Ethanol\u2019s molecular formula is CHO6 2 which means it \nhas two carbon atoms and one oxygen atom. Inspite of its many benefits, its impact on social behaviour has been questioned as consumption of even a small quantity of ethanol can cause drunkenness.(i)\n What happ\nens when a small piece of sodium is dropped into ethanol?\n(ii)\n Name the comp\nound formed when ethanol is warmed with ethanoic acid in the presence of few drops of \nconc. HSO24.\n(iii)\n What is the role of conc. HSO24 in making ethane from ethanol?\n\u0001o\u0001\n(iv)\n Name t\nwo oxidising agents that are used to convert ethanol to ethanoic acids.\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers21.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 22 Page 1\nhttps://qrbook.page.link/app\nSample Paper 22\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho\nwever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv\ne Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac\nh. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each.", + "5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers21.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac\nh with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. The image shows the repro\nductive organ in females.\nWhich event will likely occur in the ovaries of females after attaining puberty?\n(a) Synthesis of eggs\n(b) F\nertilisation\n(c) Growth and dev\nelopment of embryo\n(d) Production of eggs\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 22 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2. Ho w will you protect yourself from the heat generated while diluting a concentrated acid?", + "Click Here To InstallPage 2 Sample P aper 22 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2. Ho w will you protect yourself from the heat generated while diluting a concentrated acid?\n(a)\n By adding w\nater to acid with constant stirring\n(b)\n By adding acid to w", + "sample question papers21.txt\n2. Ho w will you protect yourself from the heat generated while diluting a concentrated acid?\n(a)\n By adding w\nater to acid with constant stirring\n(b)\n By adding acid to w\nater with constant stirring\n(c)\n By adding base to acid with constan\nt stirring\n(d)\n By adding w\nater to acid followed by base\n3.\n 50.0 mL of tap w\nater was taken in a beaker. Hydrochloric acid was added drop by drop to water. The temperature \nand pH of the solution was noted. The following graph was obtained. Choose the correct statements related to \nthis activity.\nVolume of HCl added (ml).(i)\n The pro\ncess of dissolving an acid in water is highly endothermic.\n(ii)\n The pH of the solution increases rapidly on addition of acid.\n(iii)\n The pH of the solution decreases rapidly on addition of acid.", + "Volume of HCl added (ml).(i)\n The pro\ncess of dissolving an acid in water is highly endothermic.\n(ii)\n The pH of the solution increases rapidly on addition of acid.\n(iii)\n The pH of the solution decreases rapidly on addition of acid.\n(iv)\n The pH of tap w\nater was around 7.0\n(a)\n (i) and (iii)\n(b)\n (i) and (ii)\n(c)\n (ii) and (iv)\n(d)\n (iii) and (iv)\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers21.txt\n(iv)\n The pH of tap w\nater was around 7.0\n(a)\n (i) and (iii)\n(b)\n (i) and (ii)\n(c)\n (ii) and (iv)\n(d)\n (iii) and (iv)\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 22 P age 3\nhttps://qrbook.page.link/app\n4. Observ e the given diagram and identify the correct statements.\n(i) A t anode, oxygen gas is evolved.\n(ii)\n In the test tub\ne covering the anode, the amount of gas collected is double than that of the gas collected in \nthe test tube covering the cathode.\n(iii)\n A\nt cathode, hydrogen gas is evolved.\n(iv)\n It is a decomp\nosition reaction.\n(a)\n (i), (iii) and (iv)\n(b)\n (i), (ii) and (iii) \n(c)\n All the statemen\nts are correct.", + "(iii)\n A\nt cathode, hydrogen gas is evolved.\n(iv)\n It is a decomp\nosition reaction.\n(a)\n (i), (iii) and (iv)\n(b)\n (i), (ii) and (iii) \n(c)\n All the statemen\nts are correct.\n(d)\n (iii) and (iv) \n5.", + "sample question papers21.txt\nIdentify the correct option from the given table which represents the type of reactions occurring in step 1 and step 2.\nOption Endothermic Exothermic\na. \u00fb {\nb. { \u00fb\nc. { {\nd. \u00fb \u00fb\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 22 CBSE Science Class 10\nhttps://qrbook.page.link/app\n6. Whic h one of the following structures correctly depicts the compound CaCl2?\n7. Carefully study the diagram of the h uman respiratory system with labels (i), (ii), (iii) and (iv). Select the option \nwhich gives correct identification and main function and/or characteristic.\n(a) (i) T rachea: It is supported by bony rings for conducting inspired air.\n(b)\n (ii) Ribs: When w\ne breathe out, ribs are lifted.\n(c)\n (iii) Alv\neoli: Thin-walled sac like structures for exchange of gases.", + "(a) (i) T rachea: It is supported by bony rings for conducting inspired air.\n(b)\n (ii) Ribs: When w\ne breathe out, ribs are lifted.\n(c)\n (iii) Alv\neoli: Thin-walled sac like structures for exchange of gases.\n(d)\n (iv) Diaphragm: It is pulled up when w\ne breathe in.\n8.\n When a fuse is rated at 8 A, it means:\n(a)\n it will burn if curren\nt exceeds 8 A\n(b)\n it will w", + "sample question papers21.txt\n(d)\n (iv) Diaphragm: It is pulled up when w\ne breathe in.\n8.\n When a fuse is rated at 8 A, it means:\n(a)\n it will burn if curren\nt exceeds 8 A\n(b)\n it will w\nork only if current is 8 A\n(c)\n it has a resistance of 8\u2126 \n(d) it will not w ork if current is less than 8 A \n9.\n It \nis important to balance the chemical equations to satisfy the law of conservation of mass. Which of the following \nstatements of the law is incorrect?(a)\n The n\number of atoms of each elements remains the same, before and after a chemical reaction.\n(b)\n The total mass of the elemen\nts present in the reactants is equal to the total mass of the elements presents \nin the products. \n(c)\n Mass can neither b\ne created nor can it be destroyed in a chemical reaction.\n(d)\n The c\nhemical composition of the reactants is the same before and after the reaction.", + "(c)\n Mass can neither b\ne created nor can it be destroyed in a chemical reaction.\n(d)\n The c\nhemical composition of the reactants is the same before and after the reaction.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 22 P age 5", + "sample question papers21.txt\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 22 P age 5\nhttps://qrbook.page.link/app\n10. Observ e the image of a single nephron.\nThe amount of liquid passing through in the form of glomerular filtrate is approximately 150-180 litres per day \nwhereas the amount of urine flowing out of all the nephrons is only 1.5 to 1.8 litres per day.Water is getting reabsorbed.In which part of the nephron could the water be getting reabsorbed?(a)\n In the long tubular part\n(b)\n In the Bo\nwman\u2019s cup\n(c)\n In the glomerulus\n(d)\n In the collecting duct\n11.", + "(a)\n In the long tubular part\n(b)\n In the Bo\nwman\u2019s cup\n(c)\n In the glomerulus\n(d)\n In the collecting duct\n11.\n Iden\ntify X, Y and Z respectively in the given reaction\n(a) CH CH 22= , CH3COOH, CH3COOCH2CH3\n(b) CH3COOH, CH CH 22= , CH3COOCH3 \n(c)", + "sample question papers21.txt\nCH3CH3, HCHO, CH3COOH\n(d)\n HCHO, CH3CH3, CH3CH2COOH\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 22 CBSE Science Class 10\nhttps://qrbook.page.link/app\n12. If the k ey in the arrangement is taken out (the circuit is made open) and magnetic field lines are drawn over the \nhorizontal plane ABCD, the lines are:\n(a) elliptical in shap e\n(b)\n concen\ntric circles\n(c)\n concen\ntric circles near the point O but of elliptical shapes as we go away from it\n(d)\n straigh\nt lines parallel to each other\n13.", + "The t\nwo versions of a trait (character) which are brought in by the male and female gametes are situated on:\n(a)\n t\nwo different chromosomes\n(b)\n copies of the same c\nhromosome\n(c)\n an\ny chromosome\n(d)\n sex c\nhromosomes\n14.\n The giv\nen figures show movement seen in Mimosa pudica plant when it is touched.\nSelect the incorrect option regarding this.\n(a)\n The mo\nvement is non-directional that occurs due to turgor changes.", + "sample question papers21.txt\nThe giv\nen figures show movement seen in Mimosa pudica plant when it is touched.\nSelect the incorrect option regarding this.\n(a)\n The mo\nvement is non-directional that occurs due to turgor changes.\n(b)\n The mo\nvement is directional that involves growth. \n(c)\n The mo\nvement is in response to touch and is called thigmonasty.\n(d)\n The mo\nvement is immediate in response to stimulus. \n15.\n The resistance of a resistor is reduced to half of its initial v\nalue. In doing so, if other parameters of the circuit \nremain unchanged, the heating effects in the resistor will become:(a)", + "sample question papers21.txt\nhalf\n(b)\n t\nwo times\n(c)\n four times\n(d)", + "sample question papers21.txt\none-fourthInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 22 P age 7\nhttps://qrbook.page.link/app\n16. A studen t plots V-I graphs for three samples of nichrome wire with resistances R1, R2 and R3. Choose from the \nfollowing statement that holds true for this graph:\n(a) RR R 12 3 22\n(b) RR R 12 3 ==\n(c) RR R 21 3 22\n(d) RR R 32 1 22\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17.\n Assertion (A):\n In many reptiles, sex determination rely entirely on environmental factors.\nReason (R): The temperature at which fertilised eggs are kept determines whether the animal developing in the \neggs will be a male or female.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).", + "Reason (R): The temperature at which fertilised eggs are kept determines whether the animal developing in the \neggs will be a male or female.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)", + "sample question papers21.txt\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n18.\n Assertion (A):\n A compass needle is placed near a current carrying wire. The deflection of the compass needle \ndecreases when the magnitude of electric current in the wire increases.Reason (R): The magnitude of the magnetic field produced at a point increases as the current through the wire increases.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n19.\n Assertion (A):\n Hydrochloric acid helps in the digestion of food in the stomach.", + "sample question papers21.txt\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n19.\n Assertion (A):\n Hydrochloric acid helps in the digestion of food in the stomach.\nReason (R): Hydrochloric acid creates an acidic medium to activate protein digesting enzymes.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 22 CBSE Science Class 10\nhttps://qrbook.page.link/app\n20.", + "Click Here To InstallPage 8 Sample P aper 22 CBSE Science Class 10\nhttps://qrbook.page.link/app\n20. Assertion (A): () () 2( )( ) HS gO gS sH Ol 22 22 2$ ++ is a redox reaction.\nReason (R): In this reaction, oxidation of H2S to S and reduction of O2 to H2O takes place.\n(a)", + "sample question papers21.txt\n20. Assertion (A): () () 2( )( ) HS gO gS sH Ol 22 22 2$ ++ is a redox reaction.\nReason (R): In this reaction, oxidation of H2S to S and reduction of O2 to H2O takes place.\n(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21.\n What is a rain\nbow? Draw a labelled diagram to show the formation of a rainbow.\n\u0001o\u0001\nA beam of white light falling on a glass prism gets split up into seven colours marked 1 to 7 as shown in the \ndiagram.", + "21.\n What is a rain\nbow? Draw a labelled diagram to show the formation of a rainbow.\n\u0001o\u0001\nA beam of white light falling on a glass prism gets split up into seven colours marked 1 to 7 as shown in the \ndiagram.\nWhich two positions correspond closely to the colour of(i)\n a solution of p\notassium permanganate\n(ii)\n danger or stop signal ligh\nts?\n22.\n What is haemoglobin? State the consequences of deficiency of haemoglobin in our b", + "sample question papers21.txt\na solution of p\notassium permanganate\n(ii)\n danger or stop signal ligh\nts?\n22.\n What is haemoglobin? State the consequences of deficiency of haemoglobin in our b\nodies.\n23.\n What is the difference b\netween the organisms belonging to the first and the third trophic levels? Give one example \neach of the organisms belonging to these two trophic levels.\n24.\n Differen\ntiate between an autotroph and a heterotroph.\n25.\n W\nrite the balanced chemical equations for the following reactions and identify the type of reaction in each case.\n(i)\n Nitrogen gas is treated with h\nydrogen gas in the presence of a catalyst at 773K to form ammonia gas.\n(ii)\n Ethene is burn\nt in the presence of oxygen to form carbon dioxide, water and releases heat and light.\n\u0001o\u0001\nBlue litmus solution is added to two test tubes A and B containing dilute HCl and NaOH solution respectively. In \nwhich test tube a colour change will be observed? State the colour change and give its reason.\n26.\n T", + "sample question papers21.txt\nwhich test tube a colour change will be observed? State the colour change and give its reason.\n26.\n T\nrace the sequence of events which occur when a bright light is focused on your eyes.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 22 P age 9\nhttps://qrbook.page.link/app\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27.\n The nature, size and p\nosition of image of an object produced by a lens or mirror are as shown below. Identify the \nlens/mirror (X) used in each case and draw the corresponding complete ray diagram, (size of the object about \nhalf of the image).\n28. (i) State Snell\u2019s la w of refraction of light.", + "Identify the \nlens/mirror (X) used in each case and draw the corresponding complete ray diagram, (size of the object about \nhalf of the image).\n28. (i) State Snell\u2019s la w of refraction of light. \n(ii)\n When a ra\ny of light travelling in air enters obliquely into a glass slab, it is observed that the, Light ray \nemerges parallel to the incident ray but it is shifted sideways slightly. Draw a labelled ray diagram to illustrate it.\n29.\n (i)\n Dra", + "sample question papers21.txt\nemerges parallel to the incident ray but it is shifted sideways slightly. Draw a labelled ray diagram to illustrate it.\n29.\n (i)\n Dra\nw the pattern of magnetic field lines due to a magnetic field through and around a current carrying \ncircular loop.\n(ii)\n Name and state the rule to find out the direction of magnetic field inside and around the lo\nop.\n\u0001o\u0001\nWhat is a solenoid? Draw the pattern of magnetic field lines of\n(i)\n a curren\nt carrying solenoid and a bar magnet.\n(ii)\n List t\nwo distinguishing features between the two fields.\n30.\n Y\nou are given a white solid which is calcium carbonate, CaCO3.\n(i) What happ ens to the white solid after heating?\n(ii)\n What will y\nou observe in the lime water during the heating process?\n(iii)\n W\nrite the reactions involved.Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 22 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers21.txt\n(iii)\n W\nrite the reactions involved.Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 22 CBSE Science Class 10\nhttps://qrbook.page.link/app\n31. The graph sho ws how the pH of the soil in a farmer\u2019s field changed over a period of time\n(i) A t which point A, B, C or D did the farmer apply lime to the field.\n(ii)\n What is the imp\nortance of pH in our daily life?\n(iii)\n Giv\ne two examples showing importance of neutralisation in our daily life.\n32.\n (i)\n State the role pla\nyed by the following in the process of digestion:\n(a)\n Enzyme trypsin\n(b)\n Enzyme lipase\n(ii)\n List t\nwo functions of finger like projection present in the small intestine.\n\u0001o\u0001\n(i)\n W\nrite the correct sequence of steps followed during journey of oxygen rich blood from lungs to various \norgans of human body.\n(ii)\n What happ\nens when the system of blood vessels develop a leak?\n33.", + "\u0001o\u0001\n(i)\n W\nrite the correct sequence of steps followed during journey of oxygen rich blood from lungs to various \norgans of human body.\n(ii)\n What happ\nens when the system of blood vessels develop a leak?\n33.\n Ho\nw can we help in reducing the problem of waste disposal? Suggest any three methods.\nSECTION-D", + "sample question papers21.txt\norgans of human body.\n(ii)\n What happ\nens when the system of blood vessels develop a leak?\n33.\n Ho\nw can we help in reducing the problem of waste disposal? Suggest any three methods.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34.\n Differen\ntiate between the following:\n(i)\n P\nollen tube and style\n(ii)\n fission in Amo\neba and Plasmodium\n(iii)\n F\nragmentation and regeneration\n(iv)\n Bud of Hydra and bud of Bry\nophyllum\n(v)\n V\negetative propagation and spore formation \n\u0001o\u0001\n(i)\n What is pub\nerty?\n(ii)\n Describ\ne in brief the functions of the following parts in the human male reproductive system:\n(a)", + "sample question papers21.txt\nTestes\n(b)\n Seminal v\nesicle\n(c)\n V\nas deferens\n(d)", + "sample question papers21.txt\nUrethra\n(iii)\n Wh\ny are testes located outside the abdominal cavity?\n(iv)\n State ho\nw sperms move towards the female germ cell.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 22 P age 11\nhttps://qrbook.page.link/app\n35. With the help of a diagram of exp erimental setup, describe an activity to show that the force acting on a current \ncarrying conductor placed in a magnetic field increases with increase in field strength.\n36.\n The comp\nounds methanal, ethanal, propanal and butanal belong to the homologous series called aldehydes. The \ntable shows some information on these four aldehydes.\nName Formula Boiling Point/\u00b0C Solubility in water\nMethanal HCHO 21- very soluble\nEthanal CH3CHO 21 very soluble\nPropanal CH3CH2CHO 49 soluble\nButanal CH3CH2CH2CHO 76 slightly soluble\n(i)\n List t\nwo characteristics of this homologous series using the information shown in the table.\n(ii)\n Compare the molecular form", + "sample question papers21.txt\nButanal CH3CH2CH2CHO 76 slightly soluble\n(i)\n List t\nwo characteristics of this homologous series using the information shown in the table.\n(ii)\n Compare the molecular form\nulae of the four aldehydes listed in the table. What conclusion can you make?\n(iii)\n Dra\nw the electronic structure of ethanal. You need to show only the outer shell electrons.\n\u0001o\u0001\n(i)\n Compare soaps and detergen\nts on the basis of their composition and cleansing action in hard water.\n(ii)\n What happ\nens when ethanol is treated with sodium metal? State the behaviour of ethanol in this reaction.\n(iii)\n Dra\nw the structure of cyclohexane.\n(iv)\n Name the follo\nwing compound\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37.\n Mano\nj wanted to fix the rear-view mirror of his bike. He knows that rear-view mirror is an essential safety device", + "sample question papers21.txt\none of these sub-parts.\n37.\n Mano\nj wanted to fix the rear-view mirror of his bike. He knows that rear-view mirror is an essential safety device \nin the vehicle and allows him to see objects at the backside of his vehicle.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 12 Sample P aper 22 CBSE Science Class 10\nhttps://qrbook.page.link/app\nHe bought two mirrors M1 and M2, out of which M1 is curved inwards and M2 is curved outwards.\n(i)\n Based on the giv\nen situation, which mirror should Manoj need to fix as his rear-view mirror and why?\n(ii)\n An ob\nject is placed at the centre of curvature of M1. Find the distance between its image and pole.\n(iii)\n Mano\nj did some preliminary experiment with mirror M1 and found that magnification of the real image of \nan object placed at 10 cm in front of it is 3, at what distance is the image located?\n\u0001o\u0001\n(iv)\n An ob", + "sample question papers21.txt\nan object placed at 10 cm in front of it is 3, at what distance is the image located?\n\u0001o\u0001\n(iv)\n An ob\nject is placed 60 cm in front of M2. The image formed by the mirror is located 30 cm behind the \nmirror. What is the object\u2019s magnification?\n38.\n When a silv\nery grey powder of a solid (A) is mixed with a powder of solid (B) no reaction occurs. But if the \nmixture is ignited and lighted using magnesium ribbon a reaction occurs with evolution of large amount of heat \nforming product (C) which settles down as liquid metal and the solid product (D) formed floats on the liquid (C) in solid form reacts with moisture to form rust.The amount of heat generated during the reaction is so high that the reaction is used in welding of electric conductors, joints in railway tracks.(i)\n Iden\ntify (A), (B), (C) and (D). Write the balanced chemical equation for the reaction. Name the type of \nreaction.\n(ii)\n If (A) reacts with air on heating that will b\ne the nature of oxide formed?", + "(i)\n Iden\ntify (A), (B), (C) and (D). Write the balanced chemical equation for the reaction. Name the type of \nreaction.\n(ii)\n If (A) reacts with air on heating that will b\ne the nature of oxide formed?\n\u0001o\u0001\n(iv)\n Do", + "sample question papers21.txt\nreaction.\n(ii)\n If (A) reacts with air on heating that will b\ne the nature of oxide formed?\n\u0001o\u0001\n(iv)\n Do\nes oxide of (A) react with aqueous NaOH and/or HCl. Give balanced chemical equations.\n39.\n Sex determination is the metho\nd by which distinction between males and females is established in a species. The \nsex of an individual is determined by specific chromosomes. These chromosomes are called sex chromosomes. X \nand Y chromosomes are called sex chromosomes. The normal chromosomes other than the sex chromosomes of an individual are known as autosomes.\n(i) A normal baby girl receives her X chromosome from whom: mother, father, both mother and father or either \nfrom mother or father?\n(ii)\n Whic\nh vital function is not controlled by autosomes?\n(iii)\n A couple has six daugh\nters. What is the possibility of them having a girl next time?\n\u0001o\u0001\n(iv)\n Do genetic com\nbination of mothers play a significant role in determining the sex of a new born?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 14.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 4 Page 1\nhttps://qrbook.page.link/app\nSample Paper 4\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 14.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. The schematic diagram is given below :\nWhich of the following is an incorrect statement?\n(a) A and E are chemically same.\n(b) A and D are chemically same.\n(c) D and E are chemically same.\n(d) C and E are chemically same.\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 4 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2. What happens when dilute hydrochloric acid is added to iron fillings as shown in the figure?", + "Click Here To InstallPage 2 Sample P aper 4 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2. What happens when dilute hydrochloric acid is added to iron fillings as shown in the figure?\n(a) Hydrogen gas and iron c hloride are produced.\n(b) Chlorine gas and iron h ydroxide are produced.", + "sample question papers 14.txt\n2. What happens when dilute hydrochloric acid is added to iron fillings as shown in the figure?\n(a) Hydrogen gas and iron c hloride are produced.\n(b) Chlorine gas and iron h ydroxide are produced.\n(c) No reaction tak es place.\n(d) Iron salt and w ater are produced.\n3. When hydrogen sulphide gas is passed through a blue solution of copper sulphate, a black precipitate of copper \nsulphide is obtained and the sulphuric acid so formed remains in the solution. The reaction is an example of:\n(a) com bination reaction\n(b) displacemen t reaction\n(c) decomp osition reaction\n(d) double displacemen t reaction\n4. Which of the following pair is incorrect in the given table?", + "The reaction is an example of:\n(a) com bination reaction\n(b) displacemen t reaction\n(c) decomp osition reaction\n(d) double displacemen t reaction\n4. Which of the following pair is incorrect in the given table?\nReaction Reaction Name\n(a)CH OC O 2 42 2\"+ HO22+ Combustion reaction and oxidation reaction\n(b) ()PbNO KI 2 32+ PbIK NO 2 23\"+ Double displacement and precipitation reaction\n(c)CaOH O2+ ()Ca OH 2\" Combination reaction\n(d) CuSO Zn 4+ ZnSO Cu 4\" + Combination reaction", + "sample question papers 14.txt\n(b) ()PbNO KI 2 32+ PbIK NO 2 23\"+ Double displacement and precipitation reaction\n(c)CaOH O2+ ()Ca OH 2\" Combination reaction\n(d) CuSO Zn 4+ ZnSO Cu 4\" + Combination reaction\n5. Which of the following metal displace hydrogen from dilute acid?\n(a) Zinc\n(b) Magnesium\n(c) Copper\n(d) Sodium\n6. Which of the following statements is not correct?\n(a) All metal carb onates react with acid to give a salt, water and carbon dioxide.\n(b) All metal o xides react with water to give salt and acid.\n(c) Some metals react with acids to giv e salt and hydrogen.\n(d) Some non-metal o xides react with water to form acid.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 4 P age 3\nhttps://qrbook.page.link/app\n7. Which of the following is not a straight chain hydrocarbon ?", + "Click Here To InstallCBSE Science Class 10 Sample P aper 4 P age 3\nhttps://qrbook.page.link/app\n7. Which of the following is not a straight chain hydrocarbon ?\n(a) HC CH CH CH CH\n|\nCH3 222 2\n3----\n(b) HC CH CH CH CH CH 3 22223-----\n8. The given diagram is the structure of a/an-\n(a) Alimen tary canal", + "sample question papers 14.txt\n(a) HC CH CH CH CH\n|\nCH3 222 2\n3----\n(b) HC CH CH CH CH CH 3 22223-----\n8. The given diagram is the structure of a/an-\n(a) Alimen tary canal\n(b) Respiratory tract\n(c) Nephron\n(d) Small in testine\n9. For the start of respiration, a living cell requires?\n(a) Glucose\n(b) Glucose + 2O\n(c) 2O\n(d) Glucose + A TPInstall NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 4 CBSE Science Class 10\nhttps://qrbook.page.link/app\n10. Human males all the chromosomes are paired perfectly except one. This/these unpaired chromosomes is/are\n(i) large c hromosomes\n(ii) small c hromosomes\n(iii) Y-chromosome\n(iv) X-chromosome\n(a) (i) and (ii)\n(b) (iii) only\n(c) (iii) and (iv)\n(d) (ii) and (iv)\n11.", + "The growth of tendril in pea plants is due to\n(a) effect of ligh t\n(b) effect of gra vity\n(c) rapid cell divisions in tendrillar cells that are a way from the support\n(d) rapid cell divisions in tendrillar cells in con tact with the support", + "sample question papers 14.txt\n(a) effect of ligh t\n(b) effect of gra vity\n(c) rapid cell divisions in tendrillar cells that are a way from the support\n(d) rapid cell divisions in tendrillar cells in con tact with the support\n12. In Spirogyra, asexual reproduction takes place by\n(a) breaking up of filamen ts into smaller bits\n(b) division of a cell in to two cells\n(c) division of a cell in to many cells\n(d) formation of y oung cells from older cells\n13. An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be \nused for it?\n(a) 1 A\n(b) 2 A\n(c) 4 A\n(d) 5 A\n14. A constant current flows in a horizontal wire in the plane of the paper from east to west as shown in Figure. The direction of magnetic field at a point will be North to South.\n(a) directly ab ove the wire\n(b) directly b elow the wire\n(c) at a p oint located in the plane of the paper, on the north side of the wire", + "sample question papers 14.txt\n(a) directly ab ove the wire\n(b) directly b elow the wire\n(c) at a p oint located in the plane of the paper, on the north side of the wire\n(d) at a p oint located in the plane of the paper, on the south side of the wire. Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 4 P age 5\nhttps://qrbook.page.link/app\n15. In the following circuits (Figure), heat produced in the resistor or combination of resistors connected to a V 12 \nbattery will be\n(a) same in all the cases\n(b) minim um in case (i)\n(c) maxim um in case (ii)\n(d) maxim um in case (iii)\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 4 CBSE Science Class 10\nhttps://qrbook.page.link/app\n16.", + "Click Here To InstallPage 6 Sample P aper 4 CBSE Science Class 10\nhttps://qrbook.page.link/app\n16. \nChoose the incorrect statement from the following regarding magnetic lines of field\n(a) The direction of magnetic field at a p oint is taken to be the direction in which the north pole of a magnetic \ncompass needle points.", + "sample question papers 14.txt\n(a) The direction of magnetic field at a p oint is taken to be the direction in which the north pole of a magnetic \ncompass needle points.\n(b) Magnetic field lines are closed curv es.\n(c) If magnetic field lines are parallel and equidistan t, they represent zero field strength.\n(d) Relativ e strength of magnetic field is shown by the degree of closeness of the field lines.\n\u0001Question no. 17 to 20 are Assertion - Reasoning based questions.\n17. Assertion : Silver chloride turns grey is sunlight.\nReason : Silver is one of the least reactive metals.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.", + "sample question papers 14.txt\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : Mendel chose a number of varieties of garden pea as plant material for his experiments.Reason : Garden pea has well defined characters and was bisexual.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : In plants, water is transported through phloem.\nReason : It is because sieve tubes are absent in phloem.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.", + "sample question papers 14.txt\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Both Assertion and Reason are false.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 4 P age 7\nhttps://qrbook.page.link/app\n20. Assertion : In a conductor, free electrons keep on moving but no magnetic force acts on a conductor in a magnetic \nfield.Reason : Force on free electrons due to magnetic field always acts perpendicular to its direction of motion.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.", + "sample question papers 14.txt\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. Why are cooking vessels and water boilers generally made of copper and aluminium as shown in figure ?\n\u0001o\u0001\nHydrogen is a non-metal, still it is given a place in the reactivity series. Why ?\n22. Explain the significance of peristaltic movement that occurs all along the gut during digestion.\n23. How are the breathing movements controlled ?\n24. Urine passed during summer is usually less in quantity and is some what thicker. Why is it so ?\n25. Which component of white light deviates (i) the least and (ii) the most while passing through a glass prism? State the reason of this difference in deviation.\n\u0001o\u0001\nName the part of eye where images formed in a normal human eye. State how the image position changes in \nmyopia and hypermetropia.", + "sample question papers 14.txt\n\u0001o\u0001\nName the part of eye where images formed in a normal human eye. State how the image position changes in \nmyopia and hypermetropia.\n26. Classify the following ecosystems into natural and artificial ecosystem :Forest ecosystem, aquarium, Marine ecosystem and crop land ecosystem.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 4 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. Identify the substances that are oxidised and that are reduced in the following reactions :\n(i) ZnOC ZnCO$++\n(ii) CuOH CuHO 22$++\n(iii) MnOH Cl MnCl HO Cl 42 22 22 $++ +\n28. State which of the following reaction will take place or not and why ?", + "State which of the following reaction will take place or not and why ?\n(i) () () () () ZnsC uSOa qZ nSOa qC us 44$ ++\n(ii) () () () () FesZ nSOa qF eSOa qZ ns 44$ ++\n(iii) () () () () ZnsF eSOa qZ nSOa qF es 44$ ++\n29. (i) What is the role of m utes in stomach ?", + "sample question papers 14.txt\n(i) () () () () ZnsC uSOa qZ nSOa qC us 44$ ++\n(ii) () () () () FesZ nSOa qF eSOa qZ ns 44$ ++\n(iii) () () () () ZnsF eSOa qZ nSOa qF es 44$ ++\n29. (i) What is the role of m utes in stomach ?\n(ii) Ho w exit of food from the stomach is regulated ?\n(iii) Where do es food enter from stomach ?\n\u0001o\u0001\nWrite one function of each of the following components of the transport system in human beings\n(a) Blo od vessels\n(b) Lymph\n(c) Heart\n30. (a) Whic h mirror do we use as a rear-view mirror in vehicles ?\n(b) Dra w a ray diagram to illustrate the formation of an image when an object is placed anywhere in front of \nthe mirror on its principal axis. State the nature and position of the image formed.\n31. (a) Explain in brief, con vex lens is converging in nature.", + "(b) Dra w a ray diagram to illustrate the formation of an image when an object is placed anywhere in front of \nthe mirror on its principal axis. State the nature and position of the image formed.\n31. (a) Explain in brief, con vex lens is converging in nature.\n(b) A con vex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle \nplaced in front of convex lens if the image is equal to the size of the object ? Also find the power of the lens.", + "sample question papers 14.txt\nplaced in front of convex lens if the image is equal to the size of the object ? Also find the power of the lens.\n32. Answer the following questions :\n(i) What is the direction of magnetic field lines outside a bar-magnet ?\n(ii) The magnetic field lines in a giv en region are getting crowded. What does it indicate ?\n(iii) State one adv antage of AC over DC.\n\u0001o\u0001\n(a) Giv e the significance of the following in a domestic circuit:\n(i) electric meter,\n(ii) earthing.\n(b) List t wo precautions that should be taken to avoid overloading.\n33. Explain some harmful effects of agricultural practices on the environment.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 4 P age 9\nhttps://qrbook.page.link/app\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. Write chemical equations to represent what happens when :\n(a) Ethanol burns in air.\n(b) Ethanol reacts with so dium metal.", + "sample question papers 14.txt\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. Write chemical equations to represent what happens when :\n(a) Ethanol burns in air.\n(b) Ethanol reacts with so dium metal.\n(c) Ethanol is heated with alk aline KMnO 4.\n(d) Ethanol is heated with ethanoic acid in presence of few drops of concen trated sulphuric acid.\n(e) Ethanol is heated at 443 K with excess concen trated HSO24.\n\u0001o\u0001\n(a) Giv e a chemical test to distinguish between saturated and unsaturated hydrocarbons.\n(b) Name the pro ducts formed when ethane burns in air. Write the balanced chemical equation for the reaction \nshowing the types of energies liberated.\n(c) Wh y is reaction between methane and chlorine in the presence of sunlight considered a substitution reaction.\n35. (a) \u201cUse of a condom is b eneficial for both the sexes involved in a sexual act.\u201d Justify this statement giving two \nreasons.\n(b) Ho w do oral contraceptive help in avoiding pregnancies ?", + "sample question papers 14.txt\n35. (a) \u201cUse of a condom is b eneficial for both the sexes involved in a sexual act.\u201d Justify this statement giving two \nreasons.\n(b) Ho w do oral contraceptive help in avoiding pregnancies ?\n(c) What is sex selectiv e abortion? How does it affect a healthy society ? (State any one consequence)\n\u0001o\u0001\n(i) Describ e the various steps involved in the process of binary fission with the help of a diagram.\n(ii) Wh y do multicellular organisms use complex way of reproduction ?\n36. When a current is passed through the circular loop of wire, a magnetic field lines near the coil are nearly circular \nand concentric. At the centre of the circular loop, the magnetic field lines are straight.The strength of the magnetic field produced by a current-carrying circular coil (or circular wire) depends on :\n(i) curren t flowing through the coil.\n(ii) radius of the circular coil.\n(iii) n umber of turns of wire in the circular coil.", + "sample question papers 14.txt\n(i) curren t flowing through the coil.\n(ii) radius of the circular coil.\n(iii) n umber of turns of wire in the circular coil.\nThe direction of the field lines can be found by applying Right-Hand Thumb Rule.\n(i) State Righ t-hand Thumb rule.\n(ii) A long horizontal power line is carrying a current of 100 A in the east-west direction. What is the direction \nof magnetic field at a point 1.0 m below it ?\n(iii) What t ype of curve we get, between magnetic field and distance along the axis of a current carrying circular \ncoil ?\n(iv) If a curren t carrying straight conductor is placed in east-west direction, then find the direction of the force \nexperienced by the conductor due to earth\u2019s magnetic field.Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 4 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data -based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.", + "sample question papers 14.txt\nhttps://qrbook.page.link/app\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data -based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37. Refining is the process of purification of metals. One of the important method of refining is electrolysis. In electrolysis, electrical energy is used to bring about a non-spontaneous redox reaction. This is done by passing an electric current through a liquid containing ions, known as an electrolyte. In contrast to metals, the current in electrolytes is carried by the movement of ions rather than the movement of electrons. The solid conductors inserted into the liquid are called electrodes, the one with a positive charge is called the anode (because it attracts anions) and the one with the negative charge is called the cathode.A diagrammatic representation of electrolysis of copper is shown below :\n(i) Name the electrolyte used in refining of copp er.\n(ii) a. Cu Cu e 22$ ++\u2212\n b. Cu eC u 22$++\u2212", + "sample question papers 14.txt\n(i) Name the electrolyte used in refining of copp er.\n(ii) a. Cu Cu e 22$ ++\u2212\n b. Cu eC u 22$++\u2212\n Whic h of these two reactions occur at cathode and anode?\n\u0001o\u0001\n(iii) What is ano de mud ? Name two metals which can be refined by electrolytic method.\n38. The growing size of the human population is a cause of concern for all people. The rate of birth and death in a \ngiven population will determine its size. Reproduction is the process by which organisms increase their population. The process of sexual maturation for reproduction is gradual and takes place while general body growth is still going on. Some degree of sexual maturation does not necessarily mean that the mind or body is ready for sexual acts or for having and bringing up children. Various contraceptive devices are being used by human beings to control the size of population.\n(i) List t wo common signs of sexual maturation in boys and girls.\n(ii) What is the result of rec kless female foeticide ?", + "sample question papers 14.txt\n(i) List t wo common signs of sexual maturation in boys and girls.\n(ii) What is the result of rec kless female foeticide ?\n(iii) Whic h contraceptive method changes the hormonal balance of the body ?\n\u0001o\u0001\n(iv) W rite two factors that determine the size of a population.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 4 P age 11\nhttps://qrbook.page.link/app\n39. When light ray goes from one transparent medium to another transparent medium, it suffers a change in direction, \ninto second medium. The extent of the change in direction suffered by the phenomenon of change in the path of light rays when going from one medium to another medium is known as refraction. Ray is a given pair of media can be expressed in terms of refractive index. The refractive index is related to an important physical quantity in the relative speed of light in different media.", + "sample question papers 14.txt\n(i) A ra y of light enters into the glass from air. Does it bend towards normal ?\n(ii) What is the unit of refractiv e index ?\n(iii) Ligh t enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed \nof light in vacuum is ms 31 081#-.\n\u0001o\u0001\n(iii) When ligh t goes from one medium to another, which of the three parameters, frequency, wavelength, \nvelocity change ?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 28.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 11 Page 1\nhttps://qrbook.page.link/app\nSample Paper 11\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 28.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Which of the following is (are) true when ()HClg is passed through water?\n1. It does not ionise in the solution as it is a covalent compound.\n2. It ionizes in the solution.\n3. It gives b oth hydrogen and hydroxyl ion in the solution.\n4. It forms hydronium ion in the solution due to the com bination of hydrogen ion with water molecule.\n(a) Only 1\n(b) Only 3\n(c) 2 and 4\n(d) 3 and 4Install NODIA App to See the Solutions.", + "4. It forms hydronium ion in the solution due to the com bination of hydrogen ion with water molecule.\n(a) Only 1\n(b) Only 3\n(c) 2 and 4\n(d) 3 and 4Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 11 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2. Which metal have maximum reactivity ?\n(a) K\n(b) Na\n(c) Au\n(d) Pt", + "sample question papers 28.txt\nClick Here To InstallPage 2 Sample P aper 11 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2. Which metal have maximum reactivity ?\n(a) K\n(b) Na\n(c) Au\n(d) Pt\n3. Solid calcium oxide reacts vigorously with water to form calcium hydroxide accompanied by liberation of heat. \nThis process is called slaking of lime. Calcium hydroxide dissolves in water to form its solution called lime water. Which among the following is (are) true about slaking of lime and the solution formed?\n1. It is an endothermic reaction.\n2. It is an exothermic reaction.\n3. The pH of the resulting solution will b e more than seven.\n4. The pH of the resulting solution will b e less than seven.\n(a) 1 and 2\n(b) 2 and 3\n(c) 1 and 4\n(d) 3 and 4\n4. Which of the following are correctly matched in the given table?\n1.Combination reaction Formation of single product.\n2.Decomposition reaction Break down of single, entity.", + "Which of the following are correctly matched in the given table?\n1.Combination reaction Formation of single product.\n2.Decomposition reaction Break down of single, entity.\n3.Thermal decomposition Heat is used.\n4.Displacement reaction Based on reactivity series.", + "sample question papers 28.txt\n1.Combination reaction Formation of single product.\n2.Decomposition reaction Break down of single, entity.\n3.Thermal decomposition Heat is used.\n4.Displacement reaction Based on reactivity series.\n(a) 1, 2 and 3\n(b) 1, 2 and 4\n(c) 1, 3 and 4\n(d) 1, 2, 3 and 4\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 11 P age 3\nhttps://qrbook.page.link/app\n5. A metal rod ()M was dipped in a coloured solution Y(). After some time it was observed that the metal rod starts \ndissolving in the solution and the solution starts fading in colour. However, a coloured precipitate Z() was seen at \nthe bottom of the beaker.", + "A metal rod ()M was dipped in a coloured solution Y(). After some time it was observed that the metal rod starts \ndissolving in the solution and the solution starts fading in colour. However, a coloured precipitate Z() was seen at \nthe bottom of the beaker. ()M, Y() and Z() could be\n(a)M Zn= Y FeSO 4= Z Fe=\n(b)M Cu= Y ()AlSO24 3 = Z Al=\n(c)M Ag= Y CuSO 4= Z Cu=\n(d)M Fe= Y ZnSO 4= Z Zn=\n6. A student adds a few drops of the universal indicator to a dilute solution of sodium bicarbonate taken in a test \ntube.", + "sample question papers 28.txt\n(c)M Ag= Y CuSO 4= Z Cu=\n(d)M Fe= Y ZnSO 4= Z Zn=\n6. A student adds a few drops of the universal indicator to a dilute solution of sodium bicarbonate taken in a test \ntube.\nThe colour of mixture of universal indicator and dilute sodium bicarbonate is:\n(a) Green\n(b) Yellow\n(c) Violet\n(d) Blue\n7. Which of the following represents saponification reaction ?\n(a) CHCOONa NaOHC HN aCOCaO\n34 23 ++\n(b) CH COOH CH OH CH COOC HH OHSO\n32 53 25 224++\n(c) CH COOH Na CH COONaH22 2 33 2 $++\n(d) CH COOC H NaOHC HCOONa CH OH 32 53 25 \" ++Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 11 CBSE Science Class 10\nhttps://qrbook.page.link/app\n8.", + "Click Here To InstallPage 4 Sample P aper 11 CBSE Science Class 10\nhttps://qrbook.page.link/app\n8. The rate of breathing in aquatic organisms is much faster than that seen in terrestrial organisms, because-\n(a) the amoun t of dissolved oxygen in water is fairly high as compared to the amount of oxygen in the air.\n(b) the amoun t of dissolved oxygen in water is fairly low as compared to the amount of oxygen in the air.", + "sample question papers 28.txt\n(b) the amoun t of dissolved oxygen in water is fairly low as compared to the amount of oxygen in the air.\n(c) aquatic organisms need more o xygen to breath.\n(d) aquatic organisms do not ha ve proper organs for breathing.\n9. The process of movement of solvent particles from region of less solute concentration to region of high solute \nconcentration through semi permeable membrane is known as\n(a) Diffusion\n(b) Osmosis\n(c) Transpiration\n(d) Translocation\n10. The maleness of a child is determined by\n(a) the X c hromosome in the zygote\n(b) the Y c hromosome in zygote\n(c) the cytoplasm of germ cell whic h determines the sex\n(d) sex is determined b y chance\n11. In a synapse, chemical signal is transmitted from\n(a) dendritic end of one neuron to axonal end of another neuron.\n(b) axon to cell b ody the same neuron.\n(c) cell b ody to axonal end of the same neuron.\n(d) axonal end of one neuron to dendritic end of another neuron.", + "sample question papers 28.txt\n(b) axon to cell b ody the same neuron.\n(c) cell b ody to axonal end of the same neuron.\n(d) axonal end of one neuron to dendritic end of another neuron.\n12. The correct sequence of reproductive stages seen in flowering plants is \n(a) gametes, zygote, em bryo, seedling\n(b) zygote, gametes, em bryo, seedling\n(c) seedling, em bryo, zygote, gametes\n(d) gametes, em bryo, zygote, seedling\n13. The maximum resistance which can be made using four resistors each of resistance 21\u2126 is :\n(a) 2\u2126 (b) 1\u2126 \n(c) .25\u2126 (d) 8\u2126 \n14. Light rays are deviated by a prism as shown in the figure.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "Light rays are deviated by a prism as shown in the figure.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 11 P age 5\nhttps://qrbook.page.link/app\nThe deviation angle \u03b4 is measured for light rays of different frequency, including blue light and red light. Which \nof the following graph is correct?", + "sample question papers 28.txt\nhttps://qrbook.page.link/app\nThe deviation angle \u03b4 is measured for light rays of different frequency, including blue light and red light. Which \nof the following graph is correct?\n15. Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. The current \nthrough the 40 W bulb will be :\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 11 CBSE Science Class 10\nhttps://qrbook.page.link/app\n(a) 0.4 A\n(b) 0.6 A\n(c) 0.8 A\n(d) 1 A\n16. Which of following source gives monochromatic light?\n(a) So dium lamp\n(b) Mercury lamp\n(c) Spark lamp\n(d) All of the ab ove\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : Chips manufacturers usually flush bags of chips with oxygen gas.\nReason : It adds taste to chips.", + "17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : Chips manufacturers usually flush bags of chips with oxygen gas.\nReason : It adds taste to chips.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.", + "sample question papers 28.txt\nReason : It adds taste to chips.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : Plants lack the nervous system, but they do coordinate.Reason : It is so because of hormones.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : in the daytime, CO 2 generated during respiration is used up for photosynthesis.\nReason : There is no CO 2 release during day.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.", + "sample question papers 28.txt\nReason : There is no CO 2 release during day.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Assertion is false but Reason is true.\n20. Assertion : A small coil carrying current, in equilibrium, is perpendicular to the direction of the uniform magnetic field.Reason : Torque is maximum when plane of coil and direction of the magnetic field are parallel to each other.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 28.txt\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 11 P age 7\nhttps://qrbook.page.link/app\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. Out of the two\u2013hydrochloric acid and acetic acid, which one is considered a strong acid and why ? Write the name/\nmolecular formula of one more strong acid.\n\u0001o\u0001\nName the acid produced in our stomach. What happens if there is an excess of acid in the stomach? How can it \nbe cured ? Name the antacid.\n22. A graph was plotted to show the energy output of two types of respiration. Identify the type of respiration denoted by curves \nA and B.\n23. Why are the chances of variation more in sexually developing organisms ?\n24.", + "Name the antacid.\n22. A graph was plotted to show the energy output of two types of respiration. Identify the type of respiration denoted by curves \nA and B.\n23. Why are the chances of variation more in sexually developing organisms ?\n24. You soak seeds of bean and observe them after 2-3 days. What will be your observations?", + "sample question papers 28.txt\nA and B.\n23. Why are the chances of variation more in sexually developing organisms ?\n24. You soak seeds of bean and observe them after 2-3 days. What will be your observations?\n25. Calculate the equivalent resistance from the following combination of resistors.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 11 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001o\u0001\nWhat would be the readings of ammeter and voltmeter in the given circuit ?\n26. Name the sources from where the green plants obtain C, H and O.\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. Write an experiment to show that copper does not react with dilute HCl.\n28. What is a thermochemical equation ? Give two examples.\n29. How do Mendel\u2019s experiments show that traits may be dominant or recessive?\n\u0001o\u0001\nWith the help of two suitable examples explain why certain experiences and traits earned by people during their", + "sample question papers 28.txt\n29. How do Mendel\u2019s experiments show that traits may be dominant or recessive?\n\u0001o\u0001\nWith the help of two suitable examples explain why certain experiences and traits earned by people during their \nlifetime are not passed onto their next generations. When such traits be passed on?\n30. (a) Ra vi kept a book at a distance of 10 cm from the eyes of his friend Hari. Hari is not able to read anything \nwritten on the book. Explain why ?\n(b) A lens of focal length 5.0 cm is being used by a student in the laboratory as a magnifying glass. His least \ndistance of distinct vision is 25 cm. What magnification is the student getting ?\n31. \n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 11 P age 9\nhttps://qrbook.page.link/app\n(a) A ra y of light is incident at an angle of 45\u00b0 at the interface of medium (1) and medium (2) as shown in the", + "sample question papers 28.txt\nhttps://qrbook.page.link/app\n(a) A ra y of light is incident at an angle of 45\u00b0 at the interface of medium (1) and medium (2) as shown in the \nabove diagram. Redraw this diagram in the answer book and complete it. If the angle of refraction is 30\u00b0 \nfind the refractive index of medium (2) with respect to medium (1).\n (Giv en that sin4 521c= and sin3 0 21c=)\n(b) If second medium is w ater in place of medium (2), will the angle of refraction increase or decrease ? Why ? \n(Refractive index of water /43= )\n32. State Joules law of heating. List two special characteristics of a heating element wire.An electric iron consumes energy at the rate of 880W when heating is at the maximum rate and 440W when the heating is at the minimum rate. The applied voltage is 220 V. Calculate the current and resistance in each case.\n\u0001o\u0001\nStudy the given electric circuit and calculate :\n(i) the curren t flowing through the 4\u2126 resister and", + "sample question papers 28.txt\n\u0001o\u0001\nStudy the given electric circuit and calculate :\n(i) the curren t flowing through the 4\u2126 resister and\n(ii) p otential difference across the combination of two resistor of 8\u2126 each.\n33. (a) \u201cImpro vements in our lifestyle have resulted in greater amounts of waste generation.\u201d Give two examples to \nsupport the given statement. Suggest one change that we can incorporate in our lifestyle in order to reduce \nnon-biodegradable waste.\n(b) The follo wing organisms form a food chain.\n Insect, Ha wk, Grass, Snake, Frog\nWhich of these will have highest concentration of non-biodegradable chemicals? Name the phenomenon.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.", + "sample question papers 28.txt\nInsect, Ha wk, Grass, Snake, Frog\nWhich of these will have highest concentration of non-biodegradable chemicals? Name the phenomenon.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. A student dropped few pieces of marble in dilute hydrochloric acid, contained in a test tube. The evolved gas was then passed through lime water. What change would be observed in lime water ? What will happen if excess of gas is passed through lime water ? Write balanced chemical equations for all the changes observed.\n\u0001o\u0001\n(i) What are strong acids and w eak acids? Give an example for each.\n(ii) A dry pellet of a common base B when kept in open absorbs moisture and turns sticky. The compound is \nalso formed by chlor-alkali process. Identify B. What type of reaction occurs when B is treated with dilute \nhydrochloric acid ? Write the chemical equation.Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 11 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers 28.txt\nhydrochloric acid ? Write the chemical equation.Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 11 CBSE Science Class 10\nhttps://qrbook.page.link/app\n35. Give reasons :\n(a) V entricles have thicker muscular walls than atria.\n(b) T ransport system in plants is slow.\n(c) Circulation of blo od in aquatic vertebrates differs from that in terrestrial vertebrates.\n(d) During the da ytime, water and minerals travel faster through xylem as compared to the night.\n(e) V eins have valves whereas arteries do not.\n\u0001o\u0001\nHow would you classify a green plant, an animal, a fungus and a roundworm based on their modes of nutrition ? \nExplain very briefly the manner how they get their nutrition.\n36. (i) With the help of an activit y, explain the method of inducing electric current in a coil with a moving magnet. \nState the rule used to find the direction of electric current thus generated in the coil.", + "sample question papers 28.txt\nState the rule used to find the direction of electric current thus generated in the coil.\n(ii) Tw o circular coils-1 and coil-2 are kept close to each other as shown in the diagram. Coil-1 is connected to \na battery and key and coil-2 with a galvanometer. State your observation in the galvanometer:\n(a) When k ey K closed;\n(b) When k ey K is opened;\nGive reason for your observations.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.\n37. A student prepared oxygen in a lab by catalytic decomposition of potassium chlorate KClO 3 ^h as shown in the \nfigure. Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen O2^h. The following \nreaction takes place :\n() () () KClO sK ClsO g 32$ +\n(i) Ho w many moles of KClO 3 are required to produce 2.4 moles of O2?", + "The following \nreaction takes place :\n() () () KClO sK ClsO g 32$ +\n(i) Ho w many moles of KClO 3 are required to produce 2.4 moles of O2?\n(ii) Name the elemen t which is reduced in the given reaction.", + "sample question papers 28.txt\nreaction takes place :\n() () () KClO sK ClsO g 32$ +\n(i) Ho w many moles of KClO 3 are required to produce 2.4 moles of O2?\n(ii) Name the elemen t which is reduced in the given reaction.\n(iii) Ho w many moles of KClO 3 give 3 moles of oxygen?\n\u0001o\u0001\n(iv) What is the o xidation state of chlorine in potassium chlorate?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 11 P age 11\nhttps://qrbook.page.link/app\n38. The sexual act always has the potential to lead to pregnancy will make major demands on the body and the mind \nof the woman and if she is not ready for it, her health will be adversely affected. Therefore, many ways have been devised to avoid pregnancy.\n(i) Name an y two bacterial diseases that are caused due to unprotected sex.\n(ii) In what a pill helps in prev enting pregnancy?", + "Therefore, many ways have been devised to avoid pregnancy.\n(i) Name an y two bacterial diseases that are caused due to unprotected sex.\n(ii) In what a pill helps in prev enting pregnancy?\n(iii) What is v asectomy?\n\u0001o\u0001\n(iv) What are the common side-effects of using con traceptive pills?", + "sample question papers 28.txt\n(ii) In what a pill helps in prev enting pregnancy?\n(iii) What is v asectomy?\n\u0001o\u0001\n(iv) What are the common side-effects of using con traceptive pills?\n39. The ciliary muscle muscles of eye control the curvature of the lens in the eye and hence can alter the effective", + "sample question papers 28.txt\n\nfocal length of the system. When the muscles are fully relaxed, the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curvature decreases) and focal length decreases. For a clear vision the image must be on retina. The image distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. It is about 2.5 cm for a grown-up person.A person can theoretically have clear vision of objects situated at any large distance from the eye. The smallest distance at which a person can clearly see is related to minimum possible focal length, The ciliary muscles are most strained in this position. For an average grown-up person minimum distance of object should be around 25 cm.A person suffering for eye defects uses spectacles (Eye glass). The function of lens of spectacles is to form the image of the objects within the range in which person can see clearly.", + "For an average grown-up person minimum distance of object should be around 25 cm.A person suffering for eye defects uses spectacles (Eye glass). The function of lens of spectacles is to form the image of the objects within the range in which person can see clearly. The image of the spectacle-lens becomes object for eye-lens and whose image is formed on retina.The number of spectacle-lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens is equal to the numerical value of the power of lens with sign. For example power of lens required is +3D (converging lens of focal length 100/3 cm) then number of lens will be +3.", + "sample question papers 28.txt\nFor all the calculations required you can use the lens formula and lens maker\u2019s formula. Assume that the eye lens \nis equiconvex lens. Neglect the distance between eye lens and the spectacle lens.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 12 Sample P aper 11 CBSE Science Class 10\nhttps://qrbook.page.link/app\n(i) What do y ou mean by the ciliary muscles ?\n(ii) What is the minim um focal length of eye lens of a normal person ?\n(iii) What is the maxim um focal length of eye lens of normal person ?\n\u0001o\u0001\n(iv) A near-sigh ted man can clearly see object only up-to a distance of 100 cm and not beyond this. What is the \nnumber of the spectacles lens necessary for the remedy of this defect ?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 25.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 7 Page 1\nhttps://qrbook.page.link/app\nSample Paper 7\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 25.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1\u201320.\n1. Which of the following is/are correct for diluting acid?\n1. Adding acid to water b y stirring.\n2. Adding water to acid b y stirring.\n(a) Only 1\n(b) Only 2\n(c) Both 1 and 2\n(d) Neither 1 nor 2\n2. A solution of substance X is used for white washing. Here X is:\n(a) CaO (b) CaO 2\n(c) NaCl (d) KCl\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.", + "A solution of substance X is used for white washing. Here X is:\n(a) CaO (b) CaO 2\n(c) NaCl (d) KCl\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 7 CBSE Science Class 10\nhttps://qrbook.page.link/app\n3. When hydrogen sulphide gas is passed through a blue solution of copper sulphate, a black precipitate of copper", + "sample question papers 25.txt\nhttps://qrbook.page.link/app\n3. When hydrogen sulphide gas is passed through a blue solution of copper sulphate, a black precipitate of copper \nsulphide is obtained and the sulphuric acid so formed remains in the solution. The reaction is an example of-\n(a) a com bination reaction\n(b) a displacemen t reaction\n(c) a decomp osition reaction\n(d) a double decomp osition reaction\n4. Which of the following are exothermic processes?\n1. Reaction of w ater with quick lime.\n2. Dilution of an acid.\n3. Ev aporation of water.\n4. Sublimation of camphor (crystals).\n(a) 1 and 2\n(b) 2 and 3\n(c) 1 and 4\n(d) 3 and 4\n5.", + "3. Ev aporation of water.\n4. Sublimation of camphor (crystals).\n(a) 1 and 2\n(b) 2 and 3\n(c) 1 and 4\n(d) 3 and 4\n5. 2 mL each of concentrated HCl, HNO 3 and a mixture of concentrated HCl and concentrated HNO 3 in the ratio \nof 3 : 1 were taken in test tubes labelled as A, B and C. A small piece of metal was put in each test tube. No \nchange occurred in test tubes A and B but the metal got dissolved in test tube C respectively. The metal could be\n(a) Al\n(b) Au\n(c) Cu\n(d) Pt", + "sample question papers 25.txt\nchange occurred in test tubes A and B but the metal got dissolved in test tube C respectively. The metal could be\n(a) Al\n(b) Au\n(c) Cu\n(d) Pt\n6. Which of the following are present in a dilute aqueous solution of hydrochloric acid?\n(a) HO Cl 3++\u2212\n(b) HO OH 3++\u2212\n(c) Cl OH+\u2212\u2212\n(d) Unionised HCl\n7. Which among the following are unsaturated hydrocarbons ?\n(i) HC CH CH CH 32 23 ---\n(ii) HC CC CH 33/--\n(iii) \n|HC CH CH\nCH33\n3--\n(iv) \n|HC CC H\nCH32\n3\u2212=\n(a) (i) and (iii)\n(b) (ii) and (iii)\n(c) (ii) and (iv)\n(d) (iii) and (iv)\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "Click Here To InstallCBSE Science Class 10 Sample P aper 7 P age 3\nhttps://qrbook.page.link/app\n8. The respiratory route of air in the respiratory tract of human is :\n(a) nostrils \" pharynx \" larynx \" trachea \" alveoli.\n(b) alveoli \" pharynx \" larynx \" trachea \" nostrils.\n(c) alveoli \" larynx \" trachea \" pharynx \" nostrils.\n(d) nostrils \" trachea \" pharynx \" larynx \" alveoli.", + "sample question papers 25.txt\n(b) alveoli \" pharynx \" larynx \" trachea \" nostrils.\n(c) alveoli \" larynx \" trachea \" pharynx \" nostrils.\n(d) nostrils \" trachea \" pharynx \" larynx \" alveoli.\n9. Since the environment is not under the control of the individual organism, the outside source of energy is quite:\n(a) Varied\n(b) Same\n(c) Differ from case to case\n(d) Under the con trol of organism\n10. From the list given below, select the character which can be acquired but not inherited\n(a) colour of ey e\n(b) colour of skin\n(c) size of b ody\n(d) nature of hair\n11. Which of the following statements is correct about receptors ?\n(a) Gustatory receptors detect taste while olfactory receptors detect smell.\n(b) Both gustatory and olfactory receptors detect smell.\n(c) Auditory receptors detect smell and olfactory receptors detect taste.\n(d) Olfactory receptors detect taste and gustatory receptors detect smell.", + "(a) Gustatory receptors detect taste while olfactory receptors detect smell.\n(b) Both gustatory and olfactory receptors detect smell.\n(c) Auditory receptors detect smell and olfactory receptors detect taste.\n(d) Olfactory receptors detect taste and gustatory receptors detect smell.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 25.txt\n(d) Olfactory receptors detect taste and gustatory receptors detect smell.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 7 CBSE Science Class 10\nhttps://qrbook.page.link/app\n12. Which of the following statements are true for flowers?\n(i) Flo wers are always bisexual\n(ii) They are the sexual repro ductive organs\n(iii) They are pro duced in all groups of plants\n(iv) After fertilisation they giv e rise to fruits\n(a) (i) and (iv)\n(b) (ii) and (iii)\n(c) (i) and (iii)\n(d) (ii) and (iv)\n13. \nTwo resistors are shown in the above figure when it connected to a battery will have\n(a) same curren t flowing through them when connected in parallel\n(b) same curren t flowing through them when connected in series\n(c) same p otential difference across them when connected in series\n(d) differen t potential difference across them when connected in parallel", + "sample question papers 25.txt\n(c) same p otential difference across them when connected in series\n(d) differen t potential difference across them when connected in parallel \n14. For a current in a long straight solenoid N and S-poles are created at the two ends. Among the following \nstatements, the incorrect statement is\n(a) The field lines inside the solenoid are in the form of straigh t lines which indicates that the magnetic field is \nthe same at all points inside the solenoid.\n(b) The strong magnetic field pro duced inside the solenoid can be used to magnetise a piece of magnetic \nmaterial like soft iron, when placed inside the coil.\n(c) The pattern of the magnetic field asso ciated with the solenoid is different from the pattern of the magnetic \nfield around a bar magnet.\n(d) The N and S-poles exchange position when the direction of current through the solenoid is reversed.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 25.txt\n(d) The N and S-poles exchange position when the direction of current through the solenoid is reversed.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 7 P age 5\nhttps://qrbook.page.link/app\n15. \nA cylindrical conductor of length l and uniform area of cross-section A has resistance R. Another conductor of \nlength 21 and resistance R of the same material has area of cross section\n(a) A/2 (b) 3A/2\n(c) 2 A (d) 3 A\n16. \nCommercial electric motors do not use\n(a) an electromagnet to rotate the armature\n(b) effectiv ely large number of turns of conducting wire in the current carrying coil\n(c) a p ermanent magnet to rotate the armature\n(d) a soft iron core on whic h the coil is wound\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : A chemical equation should be balanced.", + "17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : A chemical equation should be balanced.\nReason : Number of atoms of each element should be same on reactants as well as products side.", + "sample question papers 25.txt\n17. Assertion : A chemical equation should be balanced.\nReason : Number of atoms of each element should be same on reactants as well as products side.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 7 CBSE Science Class 10\nhttps://qrbook.page.link/app\n18. Assertion : DNA finger printing is a method in which polymerase chain reaction followed by DNA probe is used.\nReason : A DNA finger print is inherited and therefore, resembles that of parents.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).", + "sample question papers 25.txt\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : Carbon monoxide is injurious to the health of the individual.Reason : Carbon monoxide has very strong affinity for the blood.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Both Assertion and Reason are false.\n20. Assertion : The magnetic field is stronger at a point which is nearer to the conductor and goes on decreasing on moving away from the conductor.Reason : The magnetic field \nB produced by a straight current carrying wire is inversely proportional to the \ndistance from the wire.", + "sample question papers 25.txt\nB produced by a straight current carrying wire is inversely proportional to the \ndistance from the wire.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. A metal A, which is used in thermit process, when heated with oxygen gives an oxide B, which is amphoteric in \nnature. Identify A and B. Write down the reactions of oxide B with HCl and NaOH.\n\u0001o\u0001\nIn one method of rust prevention, the iron is not coated with anything. Name the method and define it.\n22. Why does herbivores have longer small intestine than carnivores ?\n23. Do the fresh water animals reabsorb water through their excretory system like marine animal ? Justify your \nanswer.", + "sample question papers 25.txt\n22. Why does herbivores have longer small intestine than carnivores ?\n23. Do the fresh water animals reabsorb water through their excretory system like marine animal ? Justify your \nanswer.\n24. What is translocation ? Why is it essential for plants?\n25. What would the sky look if the earth had no atmosphere ? Why ?\n\u0001o\u0001\nHow do we see colours ?\n26. What is ten percent law ? Explain by an example.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 7 P age 7\nhttps://qrbook.page.link/app\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. Lead nitrate solution is added to a test tube containing potassium iodide solution.\n(a) W rite the name and colour of the compound precipitated.\n(b) W rite the balanced chemical equation for the reaction involved.\n(c) Name the t ype of this reaction justifying your answer.\n28. Give the reaction involved during extraction of zinc from its ore by\n(a) roasting of zinc ore.", + "sample question papers 25.txt\n(c) Name the t ype of this reaction justifying your answer.\n28. Give the reaction involved during extraction of zinc from its ore by\n(a) roasting of zinc ore.\n(b) calcination of zinc ore.\n(c) reduction of zinc o xide.\n29. (a) Wh y is vegetative propagation practised for growing some types of plants ?\n(b) Name the differen t parts of a flower that has germ cells.\n(c) List an y two agents of pollination.\n\u0001o\u0001\n(a) List three distinguishing features b etween sexual and asexual types of reproduction.\n(b) Explain wh y variations are observed in the offsprings of sexually reproducing organisms ?\n30. The image formed by a spherical mirror is real, inverted and is of magnification 2-. If the image is at a distance \nof 30 cm from the mirror, where is the object placed ? Find the focal length of the mirror. List two characteristics \nof the image formed if the object is moved 10 cm towards the mirror.\n31. (i) Define optical cen tre of a spherical lens.", + "sample question papers 25.txt\nof the image formed if the object is moved 10 cm towards the mirror.\n31. (i) Define optical cen tre of a spherical lens.\n(ii) A div ergent lens has a focal length of 20 cm. At what distance should an object of height 4 cm from the \noptical centre of the lens be placed so that its image is formed 10 cm away from the lens. Find the size of the image also.\n(iii) Dra w a ray diagram to show the formation of image in above situation.\n32. The diagram below shows a coil connected to a center zero galvanometer G. The galvanometer shows a deflection \nto the right when the N pole of a powerful magnet is moved to the right as shown.\n(i) Explain wh y the deflection occurs in the galvanometer.\n(ii) Do es the direction of current in the coil appear clockwise or anti-clockwise when viewed from end A?\n(iii) State the observ ation in G when the coil is moved away from N.\n(iv) State the observ ation in G when both coil and the magnet, are moved to the right at the same speed.\n\u0001o\u0001", + "sample question papers 25.txt\n(iii) State the observ ation in G when the coil is moved away from N.\n(iv) State the observ ation in G when both coil and the magnet, are moved to the right at the same speed.\n\u0001o\u0001\n(a) Dra w the magnetic field lines of the field produced due to a current carrying circular loop.\n(b) State the la w used to find the direction of magnetic field around a straight current carrying conductor.Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 7 CBSE Science Class 10\nhttps://qrbook.page.link/app\n33. What do you mean by focal chain ? Give the characteristics of food chain.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. (I) Complete the follo wing reactions :\n(a) CHO HC HCOOH HO. Conc HSO\n25 3224++\n(b) ..... CH Cl HClSunlight\n42++\n(c) ..... CHCH OH HO.", + "34 to 36 are Long answer questions.\n34. (I) Complete the follo wing reactions :\n(a) CHO HC HCOOH HO. Conc HSO\n25 3224++\n(b) ..... CH Cl HClSunlight\n42++\n(c) ..... CHCH OH HO.\nHSOHotconc\n32 2\n24+\n(II) State t wo properties of carbon which lead to huge number of carbon compounds we see around us.\n\u0001o\u0001\nAnswer the following questions :", + "sample question papers 25.txt\n42++\n(c) ..... CHCH OH HO.\nHSOHotconc\n32 2\n24+\n(II) State t wo properties of carbon which lead to huge number of carbon compounds we see around us.\n\u0001o\u0001\nAnswer the following questions :\n(a) Describ e a chemical test to distinguish between ethanol and ethanoic acid.\n(b) Giv e reason for the following :\n(i) Ethanol is used in the preparation of tincture io dine.\n(ii) Ethanoic acid is used in the preserv ation of pickles.\n35. (a) Suggest an y two categories of contraceptive methods to control the size of human population which is \nessential for the prosperity of a country. Also explain about each method briefly.\n(b) Name t wo bacterial and two viral infections each that can get sexually transmitted.\n(c) List t wo advantages of using condom during sexual act.\n\u0001o\u0001\n(i) Describ e the role of prostate gland, seminal vesicle and testes in the human male reproductive system.\n(ii) Ho w is the surgical removal of unwanted pregnancies misused?", + "sample question papers 25.txt\n\u0001o\u0001\n(i) Describ e the role of prostate gland, seminal vesicle and testes in the human male reproductive system.\n(ii) Ho w is the surgical removal of unwanted pregnancies misused?\n(iii) Explain the role of oral con traceptive pills in preventing conception.\n36. Find out the following in the electric circuit given in figure :\n(a) Effectiv e resistance of two 8\u2126 resistors in the combination.\n(b) Curren t flowing through 4\u2126 resistor.\n(c) P otential difference across 4\u2126 resistance.\n(d) P ower dissipated in 4\u2126 resistor.\n(e) Difference in ammeter readings, if an y.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 7 P age 9\nhttps://qrbook.page.link/app\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.", + "sample question papers 25.txt\nhttps://qrbook.page.link/app\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37. The arrangement of metals in a vertical column in the decreasing order of their re-activities is called the reactivity series or activity series of metals. The most reactive metal is at the top position of the reactivity series. The least reactive metal is at the bottom of the reactivity series.Hydrogen, though a non-metal, has been included in the activity series of metals only for comparison. Apart from it, the hydrogen atom also has tendency to lose its valence electron and form cation like the behaviour shown by metals. Thus,\n H He$ ++\u2212\n(i) An elemen t \u2018X\u2019 after reacting with acids liberate hydrogen gas and can displace lead and tin from their salt \nsolution. Write down the Name of X metal.\n(ii) Whic h metal can be displaced by copper from its salt solution ?\n\u0001o\u0001", + "sample question papers 25.txt\nsolution. Write down the Name of X metal.\n(ii) Whic h metal can be displaced by copper from its salt solution ?\n\u0001o\u0001\n(ii) W rite down the name of most reactive metal and which metal does not liberate hydrogen gas after reacting \nwith acid ?\n38. The human brain is the command centre for the human nervous system. It receives signals from the body\u2019s sensory", + "sample question papers 25.txt\nwith acid ?\n38. The human brain is the command centre for the human nervous system. It receives signals from the body\u2019s sensory \norgans and outputs information to the muscles. The human brain has the same basic structure as other mammal brains but is larger in relation to body size than the brains of many other mammals, such as dolphins, whales and elephants.The human brain weighs about 3 lbs. (1.4 kilograms) and makes up about 2% of a humans body weight. On average, male brains are about 10% larger than female brains, according to North-western Medicine in Illinois. The average male has a brain volume of nearly 78 cubic inches (1,274 cubic centimetres), while the average female brain has a volume of 69 cubic inches (1,131 cubic cm). The cerebrum, which is the main part of the brain located in the front area of the skull, makes up 85% of the brain\u2019s weight.\n(i) Name the giv en figure and identify the labelled part Q and R.", + "sample question papers 25.txt\n(i) Name the giv en figure and identify the labelled part Q and R.\n(ii) Whic h region is responding for pain and conscious association ?\n(iii) Giv e two functions of the part \u2018P\u2019.\n\u0001o\u0001\n(iv) F acial muscular activities and auditory reception are respectively controlled by\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 7 CBSE Science Class 10\nhttps://qrbook.page.link/app\n39. Is there a relationship between the radius of curvature R, and focal length f, of a spherical mirror ? For spherical \nmirrors of small apertures, the radius of curvature is found to be equal to twice the focal length. We put this as \nRf2=. This implies that the principal focus of a spherical mirror lies midway between the pole and centre of \ncurvature.\n(i) W rite relation between radius of curvature and focal length.\n(ii) F or which type of mirrors above relation is verified?\n(iii) What should b e size of the aperture ?\n\u0001o\u0001", + "sample question papers 25.txt\ncurvature.\n(i) W rite relation between radius of curvature and focal length.\n(ii) F or which type of mirrors above relation is verified?\n(iii) What should b e size of the aperture ?\n\u0001o\u0001\n(iv) Where is the principle fo cus of a spherical mirror lies?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 12.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 20 Page 1\nhttps://qrbook.page.link/app\nSample Paper 20\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 12.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Consider the following table :\nSubstance pH\nLemon 2.3\nBattery acid x\nSea water 8.5\nApple 3.1\nThe value of x in above table is:\n(a) 0\n(b) 1.3\n(c) 2.5\n(d) 1.9\n2. Ethane ()CH26 on complete combustion gave CO 2 and water. It shows that the results are in accordance with the \nlaw of conservation of mass.", + "Ethane ()CH26 on complete combustion gave CO 2 and water. It shows that the results are in accordance with the \nlaw of conservation of mass. Then, the coefficient of oxygen is equal to\n(a) 7/2\n(b) 3/2\n(c) 5/2\n(d) 9/2\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 20 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers 12.txt\n(b) 3/2\n(c) 5/2\n(d) 9/2\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 20 CBSE Science Class 10\nhttps://qrbook.page.link/app\n3. Magnesium ribbon is rubbed with sand paper before making it to burn. The reason of rubbing the ribbon is to:\n(a) remo ve moisture condensed over the surface of ribbon.\n(b) generate heat due to exothermic reaction.\n(c) remo ve magnesium oxide formed over the surface of magnesium.\n(d) mix silicon from sand pap er (silicon dioxide) with magnesium for lowering ignition temperature of the \nribbon.\n4. Exposure of silver chloride to sunlight for a long duration turns grey due to\nWhich among the following statement(s) is(are) true? \n1. the formation of silv er by decomposition of silver chloride.\n2. sublimation of silv er chloride.\n3. decomp osition of chlorine gas from silver chloride.\n4. o xidation of silver chloride.", + "1. the formation of silv er by decomposition of silver chloride.\n2. sublimation of silv er chloride.\n3. decomp osition of chlorine gas from silver chloride.\n4. o xidation of silver chloride.\n(a) Only 1\n(b) 1 and 3\n(c) 2 and 3\n(d) Only 4", + "sample question papers 12.txt\n2. sublimation of silv er chloride.\n3. decomp osition of chlorine gas from silver chloride.\n4. o xidation of silver chloride.\n(a) Only 1\n(b) 1 and 3\n(c) 2 and 3\n(d) Only 4\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 20 P age 3\nhttps://qrbook.page.link/app\n5. Which among the following statements is incorrect for magnesium metal?\n(a) It burns in o xygen with a dazzling white flame.\n(b) It reacts with cold w ater to form magnesium oxide and evolves hydrogen gas.\n(c) It reacts with hot w ater to form magnesium hydroxide and evolves hydrogen gas.\n(d) It reacts with steam to form magnesium h ydroxide and evolves hydrogen gas.\n6.", + "(b) It reacts with cold w ater to form magnesium oxide and evolves hydrogen gas.\n(c) It reacts with hot w ater to form magnesium hydroxide and evolves hydrogen gas.\n(d) It reacts with steam to form magnesium h ydroxide and evolves hydrogen gas.\n6. The reaction that differs from the rest of the reaction given is-\n(a) formation of calcium o xide from limestone\n(b) formation of aluminium from aluminium o xide\n(c) formation of so dium carbonate from sodium hydrogen carbonate\n(d) formation of mercury from mercuric o xide", + "sample question papers 12.txt\n(b) formation of aluminium from aluminium o xide\n(c) formation of so dium carbonate from sodium hydrogen carbonate\n(d) formation of mercury from mercuric o xide\n7. Mineral acids are stronger acids than carboxylic acids because\n(i) mineral acids are completely ionized.\n(ii) carb oxylic acids are completely ionized\n(iii) mineral acids are partially ionized\n(iv) carb oxylic acids are partially ionized\n(a) (i) and (iv)\n(b) (ii) and (iii)\n(c) (i) and (ii)\n(d) (iii) and (iv)\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 20 CBSE Science Class 10\nhttps://qrbook.page.link/app\n8. Which of the following structures is involved in gaseous exchange in woody stem of a plant as shown in the figure?\n(a) Stomata\n(b) Guard cell\n(c) Lenticel\n(d) Epidermis\n9.", + "Which of the following structures is involved in gaseous exchange in woody stem of a plant as shown in the figure?\n(a) Stomata\n(b) Guard cell\n(c) Lenticel\n(d) Epidermis\n9. Human body stores energy in form of:\n(a) Glucose\n(b) Insulin\n(c) glycogen\n(d) Fructose\n10. What must be preserved in an ecosystem, if the system needs to be maintained?", + "sample question papers 12.txt\n(c) Lenticel\n(d) Epidermis\n9. Human body stores energy in form of:\n(a) Glucose\n(b) Insulin\n(c) glycogen\n(d) Fructose\n10. What must be preserved in an ecosystem, if the system needs to be maintained?\n(a) pro ducers and carnivores\n(b) pro ducers and decomposers\n(c) Carniv ores and decomposers\n(d) Herbiv ores and carnivores\n11. Posture and balance of the body is controlled by\n(a) cerebrum\n(b) cerebellum\n(c) medulla\n(d) pons\n12. A feature of reproduction that is common to Amoeba, Spirogyra and yeast is that\n(a) they repro duce asexually\n(b) they are all unicellular\n(c) they repro duce only sexually\n(d) they are all m ulticellular\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "Click Here To InstallCBSE Science Class 10 Sample P aper 20 P age 5\nhttps://qrbook.page.link/app\n13. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be-\n(a) Plane\n(b) Concave\n(c) Convex\n(d) Either plane or con vex", + "sample question papers 12.txt\nhttps://qrbook.page.link/app\n13. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be-\n(a) Plane\n(b) Concave\n(c) Convex\n(d) Either plane or con vex\n14. When white light passes through the achromatic combination of prisms, then what is observed ?\n(a) Deviation\n(b) Dispersion\n(c) Both deviation and disp ersion\n(d) A tmospheric refraction\n15. A student determines the focal length of a device X, by focusing the image of a far off object on the screen \npositioned as shown in figure The device X is a\n(a) Con vex lens\n(b) Conca ve lens\n(c) Con vex mirror\n(d) Conca ve mirror\n16. A student traces the path of a ray of light through a glass prism for different angles of incidence. He analysis each \ndiagram and draws the following conclusion:\nI. On en tering prism, the light ray bends towards its base.\nII. Ligh t ray suffers refraction at the point of incidence and point of emergence while passing through the prism.", + "sample question papers 12.txt\nI. On en tering prism, the light ray bends towards its base.\nII. Ligh t ray suffers refraction at the point of incidence and point of emergence while passing through the prism.\nIII. Emergen t ray bends at certain angle to the direction of the incident ray.\nIV. While emerging from the prism, the ligh t ray bends towards the vertex of the prism.\nOut of the above inferences, the correct ones are:\n(a) I, I I and III\n(b) I, I II and IV\n(c) I I, III and IV\n(d) I and IV\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 20 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : Photosynthesis is considered as an endothermic reaction.\nReason : Energy gets released in the process of photosynthesis.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.", + "sample question papers 12.txt\nReason : Energy gets released in the process of photosynthesis.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : Our body maintains blood sugar level.Reason : Pancreas secretes insulin which helps to regulate blood sugar levels in the body.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : Artificial kidney is a device used to remove nitrogenous waste products from the blood through dialysis.\nReason : Reabsorption does not occur in artificial kidney.", + "sample question papers 12.txt\n19. Assertion : Artificial kidney is a device used to remove nitrogenous waste products from the blood through dialysis.\nReason : Reabsorption does not occur in artificial kidney.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Assertion is false but Reason is true.\n20. Assertion : The product of resistivity and conductivity of a conductor depends on the material of the conductor.Reason : Because each of resistivity and conductivity depends on the material of the conductor.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\nSECTION-B", + "sample question papers 12.txt\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. What prevents the metals such as magnesium, aluminium, zinc and lead from oxidation at ordinary temperature ?\n\u0001o\u0001\nExplain why sodium hydroxide solution cannot be kept in aluminium containers ? Write equation for the reaction \nthat may take for the same.\n22. If you keep the potted plant horizontally for 2-3 days, what type of movements would be shown by the shoot and root after two or three days. Why ?\n23. What are the rules of inheritance ?\n24. What is meant by pollination? Name and differentiate between the two modes of pollination in flowering plants.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 20 P age 7\nhttps://qrbook.page.link/app\n25. State two positions in which a concave mirror produces a magnified image of a given object. List two differences", + "sample question papers 12.txt\nhttps://qrbook.page.link/app\n25. State two positions in which a concave mirror produces a magnified image of a given object. List two differences \nbetween the two images.\n\u0001o\u0001\nWhat is the difference between virtual images produced by concave, plane and convex mirror ?\n26. Give two examples each of producers, consumers and decomposer.\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. An organic compound with molecular formula CHO38 reacts with sodium metal to produce hydrogen gas. Deduce \nthe possible structure of the compound. Write the balanced chemical equation of the reaction.\n28. Explain the following chemical changes, giving one example in each case :\n(i) Displacemen t or substitution,\n(ii) Dissociation,\n(iii) Isomerisation reaction.\n29. Our government launches campaigns to provide information about AIDS prevention, testing and treatment by", + "sample question papers 12.txt\n(i) Displacemen t or substitution,\n(ii) Dissociation,\n(iii) Isomerisation reaction.\n29. Our government launches campaigns to provide information about AIDS prevention, testing and treatment by \nputting posters, conducting radio shows and using other agencies of advertisements.To which category of diseases AIDS belongs ? Name and explain. What is its causative organism ? Also give two more examples of such diseases.\n\u0001o\u0001\nDistinguish between pollination and fertilisation. Mention the site and the product of fertilisation in a flower.\n30. Why does a ray of light passing through the centre of curvature of a concave mirror after reflection, is reflected \nback along the same path ?\n31. (a) A comp ound lens is made of two lenses in contact having powers .D12 5+ and .D25- . Find the focal \nlength and power of the combination.\n(b) The magnification pro duced by a mirror is 1+. What does this mean ?\n32. In the given circuit, find :\n(a) T otal resistance of the network of resistors", + "sample question papers 12.txt\nlength and power of the combination.\n(b) The magnification pro duced by a mirror is 1+. What does this mean ?\n32. In the given circuit, find :\n(a) T otal resistance of the network of resistors\n(b) Curren t through ammeter A\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 20 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001o\u0001\nThe values of current I flowing in a given resistor for the corresponding values of potential difference V across \nthe resistor are given below:\nI (ampere) 0.5 1.0 2.0 3.0 4.0\nV (volt) 1.6 3.4 6.7 10.2 13.2\nPlot a graph between V and I and calculate the resistance of the resistor.\n33. (a) Ho w many eggs are produced every month by either of the ovaries in a human female ?", + "33. (a) Ho w many eggs are produced every month by either of the ovaries in a human female ? Where does \nfertilization take place in the female reproductive system ?\n(b) What happ ens in case the eggs released by the ovary are not fertilized?\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.", + "sample question papers 12.txt\nfertilization take place in the female reproductive system ?\n(b) What happ ens in case the eggs released by the ovary are not fertilized?\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. Discuss the physical properties of non-metals.\n\u0001o\u0001\nDiscuss the exceptions in the properties of metals and non-metals.\n35. Suggest three contraceptive methods to control the size of human population. Mention two factors that determine \nthe size of population.\n\u0001o\u0001\nHow do the following organisms reproduce by asexual methods ?\n(a) Euglena\n(b) Spirogyra\n(c) Ginger\n(d) Chrysanthemum\n(e) Strawberry\n(f) Mango\n36. A household uses the following electric appliances :\n(i) refrigerator of rating 400 W for 10 hours eac h day.\n(ii) t wo electric fans of rating 80 W each for 6 hours daily.\n(iii) six electric tub es of rating 18 W each for 6 hours daily.\nCalculate the electricity bill for the household for month of June, if cost of electrical energy is .<300 per unit.", + "sample question papers 12.txt\n(iii) six electric tub es of rating 18 W each for 6 hours daily.\nCalculate the electricity bill for the household for month of June, if cost of electrical energy is .<300 per unit.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 20 P age 9\nhttps://qrbook.page.link/app\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.", + "sample question papers 12.txt\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37. Acids, bases and salts are three main categories of chemical compounds. These have certain definite properties which distinguish one class from the other.The acids are sour in taste while bases are bitter in taste. Tasting a substance is not a good way of finding out if it is an acid or a base! Acids and bases can be better distinguished with the help of indicators. Indicators are substances that undergo a change of colour with a change of acidic, neutral or basic medium. Many of these indicators are derived from natural substances such as extracts from flower petals and barrier. Litmus, a purple dye is extracted from the lichen plant. Some indicators are prepared artificially. For example, methyl orange and phenolphthalein. Given below is a table of indicators and their colour change in acidic and basic medium.", + "sample question papers 12.txt\nIndicator Colour in Acid Colour in Alkali\nLitmus Red Blue\nMethyl Pinkish red Yellow\nPhenolphthalein Colourless Pink\n(i) Giv e two examples each of natural and artificial indicators.\n(ii) An aqueous solution turns red litm us solution blue. Excess addition of which solution would reverse the \nchange-ammonium hydroxide solution or hydrochloric acid?\n(iii) What will b e the change in colour when a few drops of phenolphthalein is added to a solution having pH \n8.5.\n\u0001o\u0001\n(iv) What is univ ersal indicator?\n38. Questions are based on the two table given below.", + "(iii) What will b e the change in colour when a few drops of phenolphthalein is added to a solution having pH \n8.5.\n\u0001o\u0001\n(iv) What is univ ersal indicator?\n38. Questions are based on the two table given below. Study these tables related to blood pressure level and answer \nthe question that follow :\nTable-A\nBLOOD PRESSURE CATEGORY SYSTOLIC mm Hg\n(Upper number)DIASTOLIC mm Hg\n(Lower number)\nNormal 120 80\nElevated 120\u2013129 Less than 80\nHigh Blood Pressure (Hypertension)Stage 1130\u2013139 80\u201390\nHigh Blood Pressure (Hypertension)Stage 2140 or higher 90 or higher\nHypertensive crisis (consult your doctor immediately)Higher than 180 Higher than 120", + "sample question papers 12.txt\nHigh Blood Pressure (Hypertension)Stage 2140 or higher 90 or higher\nHypertensive crisis (consult your doctor immediately)Higher than 180 Higher than 120\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 20 CBSE Science Class 10\nhttps://qrbook.page.link/app\nTable-B\nTime of Measurement Blood Pressure\nPatient\u2013X Patient\u2013Y\nMorning 75\u2013115 85\u2013125\nAfternoon 79\u2013122 80\u2013120\nEvening 82\u2013132 75\u2013110\n(i) In the table B, at whic h time patent\u2013Y have ideal normal blood pressure ?\n(ii) Iden tify the patient, which have hypertension stage-1 blood pressure ?\n(iii) Whic h Diet is the best for high blood pressure patient ?\n\u0001o\u0001\n(iv) What is the ideal blo od pressure measurement of a human ?\n39. After coming from playground, Tanu feels very hungry. But still some more time was required by her mother", + "sample question papers 12.txt\n\u0001o\u0001\n(iv) What is the ideal blo od pressure measurement of a human ?\n39. After coming from playground, Tanu feels very hungry. But still some more time was required by her mother \nto cook food. While waiting on dining table Tanu was playing with her spoon. All of sudden she observed two different orientations of her face when she looked her face from both sides of spoon.She was confused why the orientation of her face changed in two cases. She was curious to know why her reflected image appears upside down in the one surface of a spoon but the correct way up in the opposite surface.\n(i) Whic h type of image is formed on the both surface of spoon?\n(ii) As tan u move concave surface of spoon towards her face, again she find that there comes a point (provided \nthe spoon is big enough) where her image flips from inverted to upright. State the condition under which it happens ? Is this image real or virtual?", + "sample question papers 12.txt\nthe spoon is big enough) where her image flips from inverted to upright. State the condition under which it happens ? Is this image real or virtual?\n(iii) The giv en ray diagram depict the correct explanation of the image formed by one surface of the spoon. \nName the surface which can form the image as depicted in given ray diagram?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 20 P age 11\nhttps://qrbook.page.link/app\n\u0001o\u0001\n(iv) T anu was trying to form image using a concave mirror. She got an inverted and real image of same size of \nthe object. Given figure shows four possible positions of the image formed. Figure out the correct position \nand justify it.\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers19.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 19 Page 1\nhttps://qrbook.page.link/app\nSample Paper 19\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers19.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Which of the following are correctly matched?\n1.Common salt formed by sodium hydroxide and hydrochloric acid.\n2.Brine aqueous solution of sodium chloride.\n3.Chlor-alkali process formation of sodium chloride\n(a) 1 and 2\n(b) 2 and 3\n(c) 1 and 3\n(d) 1, 2 and 3\n2. When the gases sulphur dioxide and hydrogen sulphide mix in the presence of water, the reaction is \nSO HS HO S 22 3 22 2\"++ .", + "When the gases sulphur dioxide and hydrogen sulphide mix in the presence of water, the reaction is \nSO HS HO S 22 3 22 2\"++ .\nHere hydrogen sulphide is acting as:\n(a) an oxidising agent\n(b) a reducing agent\n(c) a dehydrating agen t\n(d) a catalyst\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers19.txt\nHere hydrogen sulphide is acting as:\n(a) an oxidising agent\n(b) a reducing agent\n(c) a dehydrating agen t\n(d) a catalyst\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 19 CBSE Science Class 10\nhttps://qrbook.page.link/app\n3. Barium chloride on reacting with ammonium sulphate forms barium sulphate and ammonium chloride. Which of \nthe following correctly represents the type of the reaction involved?\n1. Displacemen t reaction\n2. Precipitation reaction\n3. Com bination reaction\n4. Double displacemen t reaction\n(a) Only 1\n(b) Only 2\n(c) Only 4\n(d) 2 and 4\n4. Which of the following is not a chemical change?\n(a) Burning of a candle.\n(b) Co oking a food\n(c) Sublimation\n(d) Germination of seeds\n5. An element A is soft and can be cut with a knife.", + "Which of the following is not a chemical change?\n(a) Burning of a candle.\n(b) Co oking a food\n(c) Sublimation\n(d) Germination of seeds\n5. An element A is soft and can be cut with a knife. This is very reactive to air and cannot be kept open in air. It \nreacts vigorously with water. Identify the element from the following :\n(a) Mg\n(b) Na\n(c) P\n(d) Ca", + "sample question papers19.txt\nreacts vigorously with water. Identify the element from the following :\n(a) Mg\n(b) Na\n(c) P\n(d) Ca\n6. Which of the following are correctly matched in the given table?\n1.Plants and animals pH range is 7.0 to 7.8\n2.Rain water pH is 7.6\n3.Tooth decay pH less than 5.5\n(a) 1 and 2\n(b) 2 and 3\n(c) 1 and 3\n(d) 1, 2 and 3Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 19 P age 3\nhttps://qrbook.page.link/app\n7. Vinegar is a solution of\n(a) 50%-60% acetic acid in alcohol\n(b) 5%-8% acetic acid in alcohol\n(c) 5%-8% acetic acid in w ater\n(d) 50%-60% acetic acid in w ater\n8. What happens if a person has one kidney removed?", + "What happens if a person has one kidney removed?\n(a) They will accum ulate excess urea\n(b) They will die\n(c) They will con tinue as normal\n(d) They will stop making urine\n9. The function of valves present in auricles and ventricles is-\n(a) It ensures that the blo od flows only in one direction.\n(b) Helps in coagulation of blo od\n(c) Destro y the worn out blood cells", + "sample question papers19.txt\n9. The function of valves present in auricles and ventricles is-\n(a) It ensures that the blo od flows only in one direction.\n(b) Helps in coagulation of blo od\n(c) Destro y the worn out blood cells\n(d) Measure pressure of b ody fluids\n10. Grasshopper in grassland is a\n(a) pro ducer (b) herbiv ore\n(c) carniv ore (d) none of the ab ove\n11. The substance that triggers the fall of mature leaves and fruits from plants is due to\n(a) auxins (b) gibb erellin\n(c) abscisic acid (d) cytokinin\n12. In a flower, the parts that produce male and female gametes (germ cells) are\n(a) stamen and an ther\n(b) filamen t and stigma\n(c) an ther and ovary\n(d) stamen and st yle\n13.", + "In a flower, the parts that produce male and female gametes (germ cells) are\n(a) stamen and an ther\n(b) filamen t and stigma\n(c) an ther and ovary\n(d) stamen and st yle\n13. A thin layer of water is transparent but a very thick layer of water is:\n(a) translucent\n(b) opaque\n(c) most transparen t\n(d) none of theseInstall NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 19 CBSE Science Class 10\nhttps://qrbook.page.link/app\n14.", + "sample question papers19.txt\n(b) opaque\n(c) most transparen t\n(d) none of theseInstall NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 19 CBSE Science Class 10\nhttps://qrbook.page.link/app\n14. \nWhich of the following angles are correctly marked in the above diagram?\n(a) A+, r+ and D+\n(b) A+ and e+\n(c) i+, A+ and D+\n(d) A+, r+ and e+\n15. Which of the following ray diagrams is correct for the ray of light incident on a concave mirror as shown in Figure?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 19 P age 5\nhttps://qrbook.page.link/app\n16. The bluish colour of water in deep sea is due to\n(a) the presence of algae and other plan ts found in water\n(b) reflection of sky in w ater\n(c) scattering of ligh t\n(d) absorption of ligh t by the sea.", + "The bluish colour of water in deep sea is due to\n(a) the presence of algae and other plan ts found in water\n(b) reflection of sky in w ater\n(c) scattering of ligh t\n(d) absorption of ligh t by the sea.\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : FeOA lA lO Fe 22 23 23$++", + "sample question papers19.txt\n(b) reflection of sky in w ater\n(c) scattering of ligh t\n(d) absorption of ligh t by the sea.\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : FeOA lA lO Fe 22 23 23$++\nThe above chemical equation is an example of displacement reaction.\nReason : Aluminium being more reactive than iron, displaces Fe from its oxide.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : Nerve impulse is a one way conduction.\nReason : Nerve impulse is transmitted from dendrite to axon terminals.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).", + "sample question papers19.txt\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 19 CBSE Science Class 10\nhttps://qrbook.page.link/app\n19. Assertion : The main organ of human excretory system is kidney.\nReason : Kidneys perform the function of adding water and nitrogenous wastes from the body.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Assertion is false but Reason is true.\n20. Assertion : If 1\u03c1 and 2\u03c1 be the resistivity of the materials of two resistors of resistances R1 and R2 respectively \nand RR>12 .\nReason : The resistance RAl\u03c1 = & >12rr if RR>12", + "sample question papers19.txt\n20. Assertion : If 1\u03c1 and 2\u03c1 be the resistivity of the materials of two resistors of resistances R1 and R2 respectively \nand RR>12 .\nReason : The resistance RAl\u03c1 = & >12rr if RR>12\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. An element reacts with air (oxygen) to form its oxide. When dissolved in water the solution turns red litmus blue. Is it a metal or a non-metal ? Justify your answer.\n\u0001o\u0001\nWhat properties do you think of while categorizing elements as metals and non-metals.\n22. What is a receptor ? Name the receptors for light, sound and smell.", + "sample question papers19.txt\n\u0001o\u0001\nWhat properties do you think of while categorizing elements as metals and non-metals.\n22. What is a receptor ? Name the receptors for light, sound and smell.\n23. What is the effect of high temperature on the sex-determination in turtle (Chrysema picta) and the lizard (Agama \nagama)?\n24. The organisms formed by asexual reproduction are considered as clones. Why? State the advantage of sexual reproduction over asexual reproduction.\n25. An object is placed at a distance of 30 cm from a concave lens of focal lengths 15 cm. List four characteristics (nature, position, etc.) of the image formed by the lens.\n\u0001o\u0001\nAn object is placed at a distance of 15 cm from concave lens of focal length 30 cm. List four characteristics (nature, \nposition, etc.) of the image formed by the lens.\n26. Write differences between food chain and food web.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 19 P age 7", + "sample question papers19.txt\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 19 P age 7\nhttps://qrbook.page.link/app\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. Name the compound formed when ethanol is heated in excess of conc. sulphuric acid at 443 K. Also write the \nchemical equation of the reaction stating the role of conc. sulphuric acid in it. What would happen if hydrogen is added to the product of this reaction in the presence of catalysts such as palladium or nickel ?\n28. Design an activity to show a decomposition reaction in which light is used to decompose a reactant. Write chemical equation for the reaction and state its one use.\n29. Name the process of reproduction observed in yeast. Design an activity to observe this mode of reproduction in a school laboratory. Name one more organism which reproduces by this mode.\n\u0001o\u0001\nWhat is carpel? Write the function of its various parts.", + "sample question papers19.txt\n\u0001o\u0001\nWhat is carpel? Write the function of its various parts.\n30. What is dispersion of white light ? State it cause. Draw a ray diagram to show the dispersion of white light by a \nglass prism.\n31. (a) State t wo main causes of developing far-sightedness.\n(b) Ho w can this defect of vision be corrected ?\n32. For the circuit diagram given below, calculate :\n(a) the v alue of current through each resistor.\n(b) the total curren t in the circuit.\n(c) the total effectiv e resistance of the circuit.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 19 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001o\u0001\nV-I graph for a conductor is as shown in figure.\n(i) What do y ou infer from this graph ?\n(ii) State the la w expressed here\n(iii) Name the ph ysical quantity represented by the slope of this graph and its unit.\n33.", + "(i) What do y ou infer from this graph ?\n(ii) State the la w expressed here\n(iii) Name the ph ysical quantity represented by the slope of this graph and its unit.\n33. (a) Ho w will an organism be benefited if it reproduces through spores ?", + "sample question papers19.txt\n(ii) State the la w expressed here\n(iii) Name the ph ysical quantity represented by the slope of this graph and its unit.\n33. (a) Ho w will an organism be benefited if it reproduces through spores ?\n(b) Ho w is regeneration different from fragmentation?\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. Write balanced chemical equation for the reactions taking place when :\n(i) Zinc carb onate is calcinated.\n(ii) Zinc sulphide is roasted.\n(iii) Zinc o xide is reduced to zinc.\n(iv) Cinnabar is heated in the air.\n(v) Manganese dio xide is heated with aluminium powder.\n\u0001o\u0001\nA metal M found in nature as sulphide ore ()MS2 is one of the good conductor of heat and electricity and used \nin making electric wires :\n(i) Iden tify the metal M.\n(ii) W rite the balanced chemical equations involved in the process of extraction of the metal.\n(iii) Dra w a labelled diagram of electrolytic refining of the metal.", + "sample question papers19.txt\n(i) Iden tify the metal M.\n(ii) W rite the balanced chemical equations involved in the process of extraction of the metal.\n(iii) Dra w a labelled diagram of electrolytic refining of the metal.\n35. State in brief the function of the following organs in the human female reproductive system :\n(a) Ovary\n(b) F allopian tube\n(c) Uterus\n\u0001o\u0001\n(i) What is v egetative propagation ?\n(ii) What metho ds you will use for growing jasmine and rose plants ?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 19 P age 9\nhttps://qrbook.page.link/app\n36. (a) With the help of a suitable circuit diagram pro ve that the reciprocal of the equivalent resistance of a group \nof resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances. \n(b) In an electric circuit t wo resistors of 12\u2126 each are joined in parallel to a 6 V battery.", + "(b) In an electric circuit t wo resistors of 12\u2126 each are joined in parallel to a 6 V battery. Find the current drawn \nfrom the battery.\nSECTION-E", + "sample question papers19.txt\n(b) In an electric circuit t wo resistors of 12\u2126 each are joined in parallel to a 6 V battery. Find the current drawn \nfrom the battery.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37. Study the table related to colour change with indicators and answer the questions that follow.\nSolutions Colour change with phenolphthalein indicator Colour change with methyl orange indicator\nP Pink Yellow\nQ Colourless Orange\nR Colourless Red\n(i) Name the solution whic h is acidic.\n(ii) Arrange the giv en solutions in increasing order of their pH value.\n(iii) What is the name of solution P and Q ?\n\u0001o\u0001\n(iv) When solution P added to the china rose indicator, what is the colour of solution P ?\n38. A scientist removed some cells from the growing point of a plant and placed it in a suitable medium leading to", + "sample question papers19.txt\n38. A scientist removed some cells from the growing point of a plant and placed it in a suitable medium leading to \nthe formation of a shapeless lump of mass X. X is then transferred to another medium which stimulates it to \ndevelop roots. When X with developed roots is placed in a yet another medium, then it developed shoots to form \ntiny plantlets. These plantlets can then be transplanted in pots or soil where they can grow to form mature plants.\n(i) What is the shap eless lump of mass X known as?\n(ii) What name is giv en to this method of producing new plants?\n(iii) The gro wth medium used in this method contains plant nutrients in the form of a \u2018jelly\u2019 Name this jelly.\n\u0001o\u0001\n(iv) What is the general name of c hemicals used to stimulate the growth of plant cells and development of roots \nand shoots?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 19 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers19.txt\nand shoots?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 19 CBSE Science Class 10\nhttps://qrbook.page.link/app\n39. Three students Jahnvi, Snigdha and Aarati of class X brought three big plane mirrors in their class room for \nscience fest . They fixed the three mirrors: one at the ceiling and the other two on the adjacent wall of the room. All three students was able to see more than three images of herself .Students of other classes also came to see this and felt happy.\n(i) Ho w many images of a single student were formed?\n(ii) Giv e reason for number of images obtained in Q. no. (i).\n(iii) Another studen t of class X takes a mirror which is depressed at the centre and mounts it on a mirror stand \nas shown in diagram. Now he focused the image of a candle flame on a white screen by placing the flame at various distances from the mirror.", + "(i).\n(iii) Another studen t of class X takes a mirror which is depressed at the centre and mounts it on a mirror stand \nas shown in diagram. Now he focused the image of a candle flame on a white screen by placing the flame at various distances from the mirror.\n He noted his observ ations as given below\nSet 1 2 3 4 5 6", + "sample question papers19.txt\nas shown in diagram. Now he focused the image of a candle flame on a white screen by placing the flame at various distances from the mirror.\n He noted his observ ations as given below\nSet 1 2 3 4 5 6\nDistance of the flame from the mirror (cm) 20 25 30 40 60 75\nDistance of the screen from the mirror (cm) 60 37.5 30 42 20 18.72\n Name the mirror and find its fo cal length.\n\u0001o\u0001\n(iv) One set of these observ ations in the above table is incorrect. Identify this set of observation and give reason \nfor your choice.\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 29.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 14 Page 1\nhttps://qrbook.page.link/app\nSample Paper 14\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 29.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Which of the following statements is correct about an aqueous solution of an acid and of base?\n1. Higher the pH, stronger the acid\n2. Higher the pH, weak er the acid\n3. Lower the pH, stronger the base\n4. Lower the pH, weaker the base\n(a) 1 and 3\n(b) 2 and 3\n(c) 1 and 4\n(d) 2 and 4\n2. Which one of the following is the example of precipitation reaction according to the given diagram?", + "Lower the pH, weaker the base\n(a) 1 and 3\n(b) 2 and 3\n(c) 1 and 4\n(d) 2 and 4\n2. Which one of the following is the example of precipitation reaction according to the given diagram?\n(a) () () () () () Cu s AgNO aq Cu NO aq Ag s 22 33 2$ ++\n(b) () () () CuCO sC uOsC OgHeat\n32 +\n(c) () () ()Ng Hg Hg 32 22 3$ +", + "sample question papers 29.txt\n(a) () () () () () Cu s AgNO aq Cu NO aq Ag s 22 33 2$ ++\n(b) () () () CuCO sC uOsC OgHeat\n32 +\n(c) () () ()Ng Hg Hg 32 22 3$ +\n(d) () () () () AgNO aq NaCla q AgCls NaNO aq 33 $ ++Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 14 CBSE Science Class 10\nhttps://qrbook.page.link/app\n3. Which of the following statements about the reaction given below are incorrect?\n() () () () PbOs Cs PbsC Og 22 2 $++\n1. Lead is getting reduced.\n2. Carb on dioxide is getting oxidised.\n3. Carb on is getting oxidised.\n4. Lead o xide is getting reduced.\n(a) 1 and 2\n(b) 3 and 4\n(c) 1 and 3\n(d) 2 and 4\n4.", + "3. Carb on is getting oxidised.\n4. Lead o xide is getting reduced.\n(a) 1 and 2\n(b) 3 and 4\n(c) 1 and 3\n(d) 2 and 4\n4. When ()Ca NO 32 is heated, it gives CaO, ()NO g 2 and ()Og2. The correct number of moles of ()Ca NO 32, CaO, \n()NO g 2 and ()Og2 are present in the reaction are respectively\n(a) 1, 2, 4, 1\n(b) 2, 2, 4, 1\n(c) 2, 1, 3, 2\n(d) 2, 2, 2, 1\n5. Which of the following oxide(s) of iron would be obtained on prolonged reaction of iron with steam?\n(a) FeO\n(b) Fe O23\n(c) Fe O34", + "sample question papers 29.txt\n(a) 1, 2, 4, 1\n(b) 2, 2, 4, 1\n(c) 2, 1, 3, 2\n(d) 2, 2, 2, 1\n5. Which of the following oxide(s) of iron would be obtained on prolonged reaction of iron with steam?\n(a) FeO\n(b) Fe O23\n(c) Fe O34\n(d) Fe O23 and Fe O34\n6. If a few drops of a concentrated acid accidentally spills over the hand of a student, what should be done?\n(a) W ash the hand with saline solution.\n(b) W ash the hand immediately with plenty of water and apply a paste of sodium hydrogen carbonate.\n(c) After w ashing with plenty of water apply solution of sodium hydroxide on the hand.\n(d) Neutralize the acid with a strong alk ali.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "(c) After w ashing with plenty of water apply solution of sodium hydroxide on the hand.\n(d) Neutralize the acid with a strong alk ali.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 14 P age 3\nhttps://qrbook.page.link/app\n7. Oils on treating with hydrogen in the presence of palladium or nickel catalyst form fats as shown in the figure. \nThis is an example of\n(a) Addition reaction\n(b) Substitution reaction\n(c) Displacemen t reaction", + "sample question papers 29.txt\nThis is an example of\n(a) Addition reaction\n(b) Substitution reaction\n(c) Displacemen t reaction\n(d) Oxidation reaction \n8. Only two of the following statements accurately describe what happens in the mouth.\n1. Am ylase breaks down large starch molecules into smaller maltose molecules.\n2. Chewing increases the surface area of fo od for digestion.\n3. Saliv a emulsifies fats into smaller droplets.\n4. T eeth breakup large insoluble molecules into smaller soluble molecules.\nWhich statements are correct?\n(a) 1 and 2\n(b) 2 and 3\n(c) 3 and 4\n(d) 1 and 4\n9. During the process of respiration in plants, the direction of diffusion of oxygen and carbon dioxide depends upon\n(a) the en vironmental conditions\n(b) the requiremen ts of the plant\n(c) b oth (a) and (b)\n(d) none of these\n10. A trait in an organism is shown in the figure influenced by\n(a) paternal DNA only\n(b) maternal DNA only\n(c) b oth maternal and paternal DNA", + "sample question papers 29.txt\n(c) b oth (a) and (b)\n(d) none of these\n10. A trait in an organism is shown in the figure influenced by\n(a) paternal DNA only\n(b) maternal DNA only\n(c) b oth maternal and paternal DNA\n(d) neither b y paternal nor by maternalInstall NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 14 CBSE Science Class 10\nhttps://qrbook.page.link/app\n11. In humans, the life processes are controlled and regulated by\n(a) repro ductive and endocrine systems\n(b) respiratory and nerv ous systems\n(c) endo crine and digestive systems\n(d) nerv ous and endocrine systems\n12. Reproduction is essential for living organisms in order to\n(a) k eep the individual organism alive\n(b) fulfil their energy requiremen t\n(c) main tain growth\n(d) con tinue the species generation after generation\n13.", + "Reproduction is essential for living organisms in order to\n(a) k eep the individual organism alive\n(b) fulfil their energy requiremen t\n(c) main tain growth\n(d) con tinue the species generation after generation\n13. A full length of a distant tall building can definitely be seen by using\n(a) a conca ve mirror\n(b) a con vex mirror\n(c) a plane mirror\n(d) b oth concave as well as plane mirror", + "sample question papers 29.txt\n13. A full length of a distant tall building can definitely be seen by using\n(a) a conca ve mirror\n(b) a con vex mirror\n(c) a plane mirror\n(d) b oth concave as well as plane mirror\n14. A light ray is incident perpendicularly to one face of a 90c prism and is totally internally reflected at the glass-air \ninterface. If the angle of reflection is 45c, we conclude that the refractive index\n(a) n21>\n(b) n 2>\n(c) n21<\n(d) n 212, where will the strength of the magnetic field be larger)\n(a) Inwards, Outwards\n(b) Outwards, Inwards\n(c) Inwards, Inwards,\n(d) Outwards, Outwards\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 9 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : During digestion, carbohydrates are broken down to form glucose.\nReason : Glucose is necessary for breathing.", + "17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : During digestion, carbohydrates are broken down to form glucose.\nReason : Glucose is necessary for breathing.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.", + "sample question papers 26.txt\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : In humans, males play an important role in determining the sex of the child.Reason : Males have two X chromosomes.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : Failure of the kidneys leads to death of the person and there is no way he can survive.Reason : Transplant of kidneys in humans is not possible.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.", + "sample question papers 26.txt\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Both Assertion and Reason are false.\n20. Assertion : There is no change in the energy of a charged particle moving in a magnetic field although a magnetic force is acting on it.Reason : Work done by centripetal force is always zero.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. Give reasons for the following :\n(a) So dium chloride has a high melting point.\n(b) Non-metals do not displace h ydrogen from dilute acids.\n\u0001o\u0001", + "sample question papers 26.txt\n21. Give reasons for the following :\n(a) So dium chloride has a high melting point.\n(b) Non-metals do not displace h ydrogen from dilute acids.\n\u0001o\u0001\nA substance X, which is an oxide of a metal is used intensively in the cement industry. This element is present \nin bones also. On treatment with water it forms a solution which turns red litmus blue. Identify X and also write \nthe chemical reactions involved.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 9 P age 7\nhttps://qrbook.page.link/app\n22. What is parasitism ? Give two examples of parasites one from animals and one from plants.\n23. What is phloem ? Name its two main components (elements) which help in the conduction of food.\n24. Name the parts of a nephron in their proper sequence starting from the point of entry of blood into it upto the \npoint of pouring out of the urine from the nephron.", + "sample question papers 26.txt\n24. Name the parts of a nephron in their proper sequence starting from the point of entry of blood into it upto the \npoint of pouring out of the urine from the nephron.\n25. Why does it take some time to see the objects in a dim-lit room when we enter the room from bright sunlight outside ?\n\u0001o\u0001\nGive reason :\n(a) Danger signals are red.\n(b) W e cannot see an object clearly if it is placed very close to the eyes.\n26. Consider the food chain : \nWhat will happen it lions are removed from the above food chain ?\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. Give chemical explanation for evolution and absorption of heat in a chemical reaction.\n28. (a) Wh y metals are not found in their free state generally ?\n(b) If a strip of aluminium with scratc hed clean surface is dipped into an aqueous solution of copper sulphate \nfor little time, the surface of the strip becomes brownish. What is the reason for this ? Write the balanced \nchemical equation for the reaction.", + "sample question papers 26.txt\nfor little time, the surface of the strip becomes brownish. What is the reason for this ? Write the balanced \nchemical equation for the reaction.\n29. Define pollination. Explain the different types of pollination.\u2019 List two agents of pollination? How does suitable pollination lead to fertilization?\n\u0001o\u0001\nA student wants to germinate dicot seeds. Write the four steps in correct sequence that will help him to perform \nthe experiment in the right way.\n30. (a) State la ws of refraction.\n(b) A ra y of light is incident normally to the surface of a glass slab placed in air. Find the angle of incidence \nand angle of refraction in this case.\n31. A 3 cm tall object is placed 18 cm in front of a concave mirror of focal length 12 cm. At what distance from the mirror should a screen be placed to see a sharp image of the object on the screen. Also calculate the height of the image formed.Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 9 CBSE Science Class 10", + "sample question papers 26.txt\nClick Here To InstallPage 8 Sample P aper 9 CBSE Science Class 10\nhttps://qrbook.page.link/app\n32. Demonstrate that due to motion of a magnet near a solenoid coil an induced current is set up in the coil.\n\u0001o\u0001\nThe flow of current in a circular loop of wire creates a magnetic field at its center. How may existence of this field \nbe detected ? State the rule which helps to predict the direction of this magnetic field.\n33. Give reason to justify the following :\n(i) The existence of decomp oser is essential in a biosphere.\n(ii) Flo w of energy in a food chain is unidirectional.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. The formula of four organic compounds are given below:\n CH CH COOH CHOHCHAB C D\n24 32 52 6\n(i) Whic h one of these compounds A, B, C or D is a saturated hydrocarbon?\n(ii) Iden tify the organic acid and give its structural formula.", + "(ii) Iden tify the organic acid and give its structural formula.\n(iii) Whic h of the above compounds when heated at 443K in the presence of concentrated H2SO4 forms ethene as", + "sample question papers 26.txt\n(ii) Iden tify the organic acid and give its structural formula.\n(iii) Whic h of the above compounds when heated at 443K in the presence of concentrated H2SO4 forms ethene as \nthe major product ? What is the role played by concentrated H2SO4 in this reaction? Also write the chemical \nequation involved.\n(iv) Giv e a chemical equation when B and C react with each other in presence of concentrated H2SO4. Name \nthe major product formed and mention one of its important use.\n\u0001o\u0001\n(a) Carry out the follo wing conversions giving complete conditions for the reaction to take place in each case :\n(i) Ethanoic acid from Ethanol\n(ii) Ethane from Ethene\n(iii) Ester from Ethanoic acid and ethanol\n Also state the names giv en to all the above conversions.\n(b) Detergen ts are preferred over soaps. Why ? (Give one reason)Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 9 P age 9\nhttps://qrbook.page.link/app", + "sample question papers 26.txt\nClick Here To InstallCBSE Science Class 10 Sample P aper 9 P age 9\nhttps://qrbook.page.link/app\n35. (a) Whic h device prevents implantation by irritating the lining of uterus ?\n(b) What could b e the possible reason for declining female : male sex ratio in our country? Suggest two measures \nto achieve 1:1 ratio.\n(c) Name those parts of a flo wer which serve the same function as the following do in animals :\n(i) Testis\n(ii) Ovary\n(iii) Eggs\n(iv) Sperms\n\u0001o\u0001\n(a) With the help of diagram sho w asexual reproduction in Rhizopus.\n(b) Ho w is this method advantageous for Rhizopus?\n(c) Ho w is mode of reproduction in unicellular organisms differ from multicellular organisms?\n36. In our daily life we use two types of electric current whose current time graphs are given below :\n(i) Name the t ype of current in two cases.\n(ii) Iden tify any one source for each type of current.\n(iii) What is the frequency of curren t in case (b) in our country ?", + "sample question papers 26.txt\n(i) Name the t ype of current in two cases.\n(ii) Iden tify any one source for each type of current.\n(iii) What is the frequency of curren t in case (b) in our country ?\n(iv) On the basis of these graphs list t wo differences between the two currents.\n(v) Out of the t wo which one is used in transmitting electric power over long distances and why ?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 9 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37.", + "37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37. Electronic configuration of some of the elements are given :\nType of element Element Atomic number Number of electrons in shell\nMetals Sodium (Na)Magnesium (Mg)Aluminium (Al)1112132 8 12 8 22 8 3\nNon-metals Oxygen (O)Sulphur (S)Chlorine (Cl)816172 62 8 62 8 7", + "sample question papers 26.txt\nType of element Element Atomic number Number of electrons in shell\nMetals Sodium (Na)Magnesium (Mg)Aluminium (Al)1112132 8 12 8 22 8 3\nNon-metals Oxygen (O)Sulphur (S)Chlorine (Cl)816172 62 8 62 8 7\n(i) State one ph ysical property to distinguish between metals and non-metals.\n(ii) What is the nature of the b ond formed when magnesium reacts with chlorine? Write the formula of the \ncompound.\n\u0001o\u0001\n(ii) What is common b etween oxygen and sulphur? Draw the electron dot structure of O2 molecule.\n38. Plant hormones affect gene expression and transcription levels, cellular division and growth. They are naturally", + "sample question papers 26.txt\n\nproduced within plants, but very similar chemicals are produced by fungi and bacteria that can also affect plant growth. A large number of related chemical compounds are synthesized by humans. They are used to regulate the growth of cultivated plants, weeds and in vitro-grown plants and plant cells; these man-made compounds are called plant growth regulators or PGRs for short. Plant hormones are not nutrients, but chemicals that in small amounts promote and influence the growth, development and differentiation of cells and tissues. The biosynthesis of plant hormones within plant tissues is often diffused and not always localized. Plants lack glands to produce and store hormones, because, unlike animals which have two circulatory systems (lymphatic and cardiovascular) powered by a heart that moves fluids around the body. Plants use more passive means to move chemicals around their bodies. Plants utilize simple chemicals as hormones, which move more easily through their tissues. They are often produced and used on a local basis within the plant body. Plant cells produce hormones that affect different regions of the cell producing the hormone.Different hormones can be sorted into different classes, depending on their chemical structures.", + "Plants utilize simple chemicals as hormones, which move more easily through their tissues. They are often produced and used on a local basis within the plant body. Plant cells produce hormones that affect different regions of the cell producing the hormone.Different hormones can be sorted into different classes, depending on their chemical structures. Within each class of hormone the exact structures vary, but they have similar physiological effects. Initial research into plant hormones identified five major classes : abscisic acid, auxin, cytokinins, ethylene and gibberellins. This list was later expanded and brassinosteroids, jasmonates, salicylic acid and strigolactones are now considered as major plant hormones.", + "sample question papers 26.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 9 P age 11\nhttps://qrbook.page.link/app\n(i) What are the factors affected b y the plant hormones ?\n(ii) What do es PGR stands for ?\n(iii) Whic h class does plant hormones fall into ?\n\u0001o\u0001\n(iv) What w ere the five major plant hormones discovered in the initial research ?\n39. A person is suffering from hypermetropia (long sightedness). It is a defect in which a human eye can see far off \nobject clearly, but is unable to see nearby object distinctly. The near point of the person is ..m15 Assume that \nthe near point of the normal eye is 25 cm.\n(i) What t ype of lens should be used in his spectacles?\n(ii) What should b e the focal length of the lens he used ?\n(iii) What will b e the power of the lens ?\n\u0001o\u0001\n(iv) W rite one possible cause of this defect.\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 22.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 13 Page 1\nhttps://qrbook.page.link/app\nSample Paper 13\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 22.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Common salt besides being used in kitchen can also be used as the raw material for making\n1. washing soda\n2. bleaching powder\n3. baking soda\n4. slaked lime\n(a) 1 and 2\n(b) 1, 2 and 4\n(c) 1 and 3\n(d) 1, 3 and 4\n2. Which reaction is used in photography?", + "Which reaction is used in photography?\n(a) () CaOH OC aOHH eat 22$++\n(b) FeSO FeOS OS O 2Heat\n42 32 3 ++\n(c) Cu OC uO 22 2$+\n(d) AgB rA gB r 22sunlight+\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 13 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers 22.txt\n(d) AgB rA gB r 22sunlight+\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 13 CBSE Science Class 10\nhttps://qrbook.page.link/app\n3. Which of the following is/are correct for a balanced chemical equation?\n1. It is based on la w of conservation of mass.\n2. The ph ysical states makes the chemical reaction less informative.\n(a) Only 1\n(b) Only 2\n(c) Both 1 and 2\n(d) Neither 1 nor 2\n4. \nThe following reaction is used for the preparation of oxygen gas in the laboratory\n() () () KClO sK ClsO g 22 3CatalystHeat\n32 +\nWhich of the following statement(s) is(are) correct about the reaction?\n(a) It is a decomp osition reaction and endothermic in nature.\n(b) It is a com bination reaction.\n(c) It is a decomp osition reaction and accompanied by release of heat.", + "(a) It is a decomp osition reaction and endothermic in nature.\n(b) It is a com bination reaction.\n(c) It is a decomp osition reaction and accompanied by release of heat.\n(d) It is a photo chemical decomposition reaction and exothermic in nature.\n5. Which one of the following metals do not react with cold as well as hot water?\n(a) Na\n(b) Ca", + "sample question papers 22.txt\n(d) It is a photo chemical decomposition reaction and exothermic in nature.\n5. Which one of the following metals do not react with cold as well as hot water?\n(a) Na\n(b) Ca\n(c) Mg\n(d) Fe\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 13 P age 3\nhttps://qrbook.page.link/app\n6. An aqueous solution turns red litmus solution blue. \nExcess addition of which of the following solution would reverse the change?\n(a) Baking p owder\n(b) Lime\n(c) Ammonium h ydroxide solution\n(d) Hydro chloric acid\n7. Which of the following statements are usually correct for carbon compounds? These\n(i) are go od conductors of electricity\n(ii) are p oor conductors of electricity\n(iii) ha ve strong forces of attraction between their molecules\n(iv) do not ha ve strong forces of attraction between their molecules.", + "Which of the following statements are usually correct for carbon compounds? These\n(i) are go od conductors of electricity\n(ii) are p oor conductors of electricity\n(iii) ha ve strong forces of attraction between their molecules\n(iv) do not ha ve strong forces of attraction between their molecules.\n(a) (i) and (iii) (b) (ii) and (iii)\n(c) (i) and (iv) (d) (ii) and (iv)", + "sample question papers 22.txt\n(iv) do not ha ve strong forces of attraction between their molecules.\n(a) (i) and (iii) (b) (ii) and (iii)\n(c) (i) and (iv) (d) (ii) and (iv)\n8. Choose the forms in which most plants absorb nitrogen:\n1. A tmospheric nitrogen\n2. Proteins\n3. Nitrates and nitrites\n4. Urea\nChoose the correct option.\n(a) 1 and 4\n(b) 3 and 4\n(c) 1 and 2\n(d) 2 and 3\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 13 CBSE Science Class 10\nhttps://qrbook.page.link/app\n9. During the process of photosynthesis, absorption of light energy is done by\n(a) Leaf\n(b) Midrib\n(c) Vein\n(d) Chlorophyll\n10.", + "During the process of photosynthesis, absorption of light energy is done by\n(a) Leaf\n(b) Midrib\n(c) Vein\n(d) Chlorophyll\n10. The correct sequence of organs in the male reproductive system for transport of sperms is\n(a) testis \" vas deferens \" urethra\n(b) testis \" ureter \" urethra\n(c) testis \" urethra \" ureter\n(d) testis \" vas deferens \" ureter\n11. Which is the correct sequence of the components of a reflex arc ?", + "sample question papers 22.txt\n(b) testis \" ureter \" urethra\n(c) testis \" urethra \" ureter\n(d) testis \" vas deferens \" ureter\n11. Which is the correct sequence of the components of a reflex arc ?\n(a) Receptors \" Muscles \" Sensory neuron \" Motor neuron \" Spinal cord\n(b) Receptors \" Motor neuron \" Spinal cord \" Sensory neuron \" Muscle\n(c) Receptors \" Spinal cord \" Sensory neuron \" Motor neuron \" Muscle\n(d) Receptors \" Sensory neuron \" Spinal cord \" Motor neuron \" Muscle.\n12. Which among the following statements are true for unisexual flowers?\n(i) They p ossess both stamen and pistil\n(ii) They p ossess either stamen or pistil\n(iii) They exhibit cross p ollination\n(iv) Unisexually flo wers possessing only stamens cannot produce fruits\n(a) (i) and (iv) (b) (ii), (iii) and (iv)\n(c) (iii) and (iv) (d) (i), (iii) and (iv)\n13.", + "The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should \nbe the position of the object?", + "sample question papers 22.txt\n(c) (iii) and (iv) (d) (i), (iii) and (iv)\n13. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should \nbe the position of the object?\n(a) Bet ween the principal focus and the centre of curvature\n(b) A t the centre of curvature\n(c) Bey ond the centre of curvature\n(d) Bet ween the pole of the mirror and its principal focus.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 13 P age 5\nhttps://qrbook.page.link/app\n14. In which of the following diagrams is the path of a ray of light passing through a glass prism shown correctly?\n(a) I\n(b) II\n(c) III\n(d) IV\n15. The path of a ray of light coming from air passing through a rectangular glass slab traced by four students are \nshown in figure. Which one of them is correct?Install NODIA App to See the Solutions.", + "(a) I\n(b) II\n(c) III\n(d) IV\n15. The path of a ray of light coming from air passing through a rectangular glass slab traced by four students are \nshown in figure. Which one of them is correct?Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 13 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers 22.txt\nshown in figure. Which one of them is correct?Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 13 CBSE Science Class 10\nhttps://qrbook.page.link/app\n16. Angle of deviation is depends on:\n(a) Angle of prism\n(b) Nature of material of prism\n(c) Angle of incidence on the prism\n(d) All of the ab ove\n\u0001Question no. 17 to 20 are Assertion - Reasoning based questions.\n17. Assertion : Corrosion of iron is commonly known as rusting.\nReason : Corrosion of iron occurs in presence of water and air.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : Cytokinins are present in highest concentration in fruits and seeds.Reason : Cytokinins are responsible for promoting cell division.", + "sample question papers 22.txt\n(d) Both Assertion and Reason are F alse.\n18. Assertion : Cytokinins are present in highest concentration in fruits and seeds.Reason : Cytokinins are responsible for promoting cell division.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : Liver is known as the largest gland of the body.Reason : It secretes salivary amylase.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Assertion is false but Reason is true.", + "sample question papers 22.txt\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Assertion is false but Reason is true.\n20. Assertion : When the length of a wire is doubled, then its resistance also gets doubled.Reason : The resistance of a wire is directly proportional to its length.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. A metal compound A reacts with dilute sulphuric acid to produce a gas which extinguishes a burning candle. \nIdentify the compound A and the gas produced. Write a balanced chemical equation for the reaction if one of the", + "sample question papers 22.txt\nIdentify the compound A and the gas produced. Write a balanced chemical equation for the reaction if one of the \ncompounds formed in the reaction is sodium sulphate.\n\u0001o\u0001\nHow does the enamel of the teeth undergo damage due to the eating of chocolates and sweets ? What should be \ndone to prevent it.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 13 P age 7\nhttps://qrbook.page.link/app\n22. Differentiate between motor neuron and sensory neuron.\n23. Is it correct to say that if fertilization of the egg occurs on a full moon night, the child produced will be a male ? \nGive reason.\n24. Why is DNA copying necessary during reproduction?\n25. Name the type of mirror used in a solar furnace. How is high temperature achieved by this device ?\n\u0001o\u0001\nWhat is meant by refractive index ? If the speed of light in a medium is rd32 of the speed of light in vacuum, find \nthe refractive index of that medium.", + "sample question papers 22.txt\n\u0001o\u0001\nWhat is meant by refractive index ? If the speed of light in a medium is rd32 of the speed of light in vacuum, find \nthe refractive index of that medium.\n26. What will happen to grass-lands if all the grazers/herbivores are removed from there ?\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. What is meant by refining of metals? Describe the electrolytic refining of copper with a neat labelled diagram.\n28. Name the plant Mendel used for his experiment. What type of progeny was obtained by Mendel in F1 and F2 \ngenerations when he crossed the tall and short plants? Write the ratio he obtained in F2 generation plants.\n29. Distinguish between the acquired traits and the inherited traits in tabular form, giving one example for each.\n\u0001o\u0001\nHow did Mendel\u2019s experiments show that different traits are inherited independently? Explain.\n30. (i) Least distance of distinct vision of a long-sigh ted person is 40 cm. He wishes to reduce it to 25 cm by using", + "sample question papers 22.txt\n30. (i) Least distance of distinct vision of a long-sigh ted person is 40 cm. He wishes to reduce it to 25 cm by using \nspectacles. Find the power and nature of the lens used by him.\n(ii) Dra w a ray diagram to show the correction of the defect by using an appropriate lens.\n31. The near point of a person suffering from hypermetropia is 75 cm. Calculate the focal length and power of the lens \nrequired to enable him to read the newspaper which is kept at 25 cm from the eye.\n32. (a) Tw o conductors A and B of resistances 5\u2126 and 10\u2126 respectively can be arranged in parallel and later on \nin series. In each arrangement, the total voltage applied across it is 20 volts. In which arrangement will the voltage across \nA and B be the same and in which case will the current flowing through A and B be the \nsame ?\n(b) Calculate the total resistance for eac h arrangement.\n\u0001o\u0001\nA metallic coil, connected to a 220 V supply, has a resistance of 110 ohm (while hot). How long will it take for this", + "sample question papers 22.txt\nsame ?\n(b) Calculate the total resistance for eac h arrangement.\n\u0001o\u0001\nA metallic coil, connected to a 220 V supply, has a resistance of 110 ohm (while hot). How long will it take for this \ncoil to heat 1 kg of water from 20\u00b0C to 70\u00b0C ? Assume that whole of the heat produced by the coil is taken up by water. (Specific heat of water \n/JkgC 4186 c = )\n33. (i) What is the heigh t of ozone from the equator ?\n(ii) Name the ra ys against which ozone layer provides protection.\n(iii) Name one effect of depletion of ozone.Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 13 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. What is plaster of paris? How is plaster of paris prepared ? Write equation of the reaction involved. What happens \nwhen water is added to plaster of paris? Write an equation to show the reaction between plaster of paris and water.\n\u0001o\u0001", + "sample question papers 22.txt\nwhen water is added to plaster of paris? Write an equation to show the reaction between plaster of paris and water.\n\u0001o\u0001\nGive suitable reasons for the following statements :\n(a) Rain w ater conducts electricity but distilled water does not.\n(b) W e feel burning sensation in the stomach when we overeat.\n(c) A tarnished copp er vessel regains its shine when rubbed with lemon.\n(d) The crystals of w ashing soda change to white powder on exposure to air.\n(e) An aqueous solution of so dium chloride is neutral but and aqueous solution of sodium is basic.\n35. Describe the structure of a sieve tube. Name the cells which are present along sieve tubes in the phloem tissue. \nDraw a labelled diagram of phloem tissue.\n\u0001o\u0001\nWhat is translocation ? Name the cells involved in the transport of food in plants. What are the steps involved in \nthe translocation of food in plants ?\n36. (a) What is a solenoid ? Dra w the pattern of magnetic field lines around a current carrying solenoid.", + "sample question papers 22.txt\nthe translocation of food in plants ?\n36. (a) What is a solenoid ? Dra w the pattern of magnetic field lines around a current carrying solenoid.\n(b) What is the pattern of field lines inside a solenoid? What do they indicate ?\n(c) Ho w is the magnetic field produced in a solenoid used ?\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data -based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.", + "sample question papers 22.txt\n37. Corrosion is the phenomenon of deterioration of surface of metal in presence of air and moisture. It is a natural process and in the presence of a moist atmosphere, chemically active metals get corroded. This is oxidation reaction. Rusting is the process where iron corrodes due to exposure to the atmosphere. The main circumstance of corrosion occurs with iron because it is a structural material in construction, bridges, buildings, rail transport, ships, etc. Aluminium is also an important structural metal, but even aluminium undergoes oxidation reactions. However, aluminium doesn\u2019t corrode or oxidize as rapidly as its reactivity suggests. An alloy of aluminium or any other metal like magnesium can make aluminium stronger and harder.Copper (Cu) corrodes and forms a basic green carbonate and lead corrodes to form a white lead oxide or carbonate.\n(i) What is rusting?\n(ii) Whic h two metals do not corrode easily?\n(iii) List t wo properties of alloys.\n\u0001o\u0001", + "sample question papers 22.txt\n(i) What is rusting?\n(ii) Whic h two metals do not corrode easily?\n(iii) List t wo properties of alloys.\n\u0001o\u0001\n(iv) What is the effect of corrosion on electrical conductivit y ?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 13 P age 9\nhttps://qrbook.page.link/app\n38. Carpel is present at the centre of a flower and is the female reproductive part. It is made of three parts. The", + "sample question papers 22.txt\nswollen bottom part is the ovary, the middle elongated part is the style and the terminal part which may be sticky is the stigma. The ovary contains ovules and each ovule ha an egg cell. The male germ-cell produced by pollen grain fuses with the female gamete present in the ovule. This fusion of the germ-cells or fertilization gives us the zygote which is capable of growing into a new plant. Thus, the pollen needs to be transferred from the stamen to stigma. If this transfer of pollen occurs in the same flower, it is referred to as self-pollination. On the other hand, if the pollen is transferred from one flower to another, it is known as cross-pollination. This transfer of pollen from one flower to another is achieved by agents like wind, water or animals. After the pollen lands on a suitable stigma, it has to reach the female germ-cells which are in the ovary. For this, a tube grows out of the pollen grain and travels through the style to reach the ovary.", + "sample question papers 22.txt\n(i) What is presen t at the centre of the flower?\n(ii) Ho w many parts does the carpel have?\n(iii) What do y ou mean by cross-pollination?\n\u0001o\u0001\n(iv) What do y ou mean by self-pollination?\n39. Convex mirror is used as a rear view mirror in vehicles. Since the image of the object formed is small in size, the \nfield of view is increased. Convex mirror is also used in street lights to diverge light over a large area.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 13 CBSE Science Class 10\nhttps://qrbook.page.link/app\n(i) In driv er\u2019s mirror, what type of image is formed behind the vehicle ?\n(ii) What can y ou say about field of view of a convex mirror ?\n\u0001o\u0001\n(iii) A con vex mirror is used to form the image of an object. What is the nature of formed image ?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 26 Page 1\nhttps://qrbook.page.link/app\nSample Paper 26\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho\nwever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv\ne Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac\nh. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each.", + "5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac\nh with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Study the given figure depicting lo\ncations of several endocrine glands in a human female identify the labelled \nglands and select the option with any four correct labels.\n(a) p-Pituitary, t-Th yroid, u-Adrenal, w-Ovary\n(b) u-pancreas, w-Ovary\n, p-Hypothalamus, q-Pineal\n(c) q-Pituitary, t-adrenal, p-Hyp\nothalamus, v-Pancreas\n(d) s-Parath\nyroid, q-Pineal, v-Ovary, u-Thyroid\n2.", + "A solenoid\u2019s magnetic field will not increase when:\n(a) a soft iron core is inserted inside the solenoid\n(b) the magnitude of current will increase\n(c) the conductor is kept closed", + "sample question papers.txt\n2. A solenoid\u2019s magnetic field will not increase when:\n(a) a soft iron core is inserted inside the solenoid\n(b) the magnitude of current will increase\n(c) the conductor is kept closed\n(d) a soft aluminium core is inserted inside the solenoidInstall NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 26 CBSE Science Class 10\nhttps://qrbook.page.link/app\n3. The image sho ws the process of vegetative propagation in a plant.\nThe shoot of the parent plant is pushed below the soil that results in growth of a new plant. What is the advantage \nof this process?(a)\n This helps gro\nw plants without adding extra manure\n(b)\n This result in plan\nt of different flowers\n(c)\n This allo\nws growth of plants with new genetic composition\n(d)\n This eliminates the need of pro\nducing plant using seeds\n4.\n The graph giv\nen below depicts a neutralisation reaction (acid + alkali $ salt + water). The pH of a solution \nchanges as we add excess of acid to an alkali.\nVolume of acid added", + "sample question papers.txt\n4.\n The graph giv\nen below depicts a neutralisation reaction (acid + alkali $ salt + water). The pH of a solution \nchanges as we add excess of acid to an alkali.\nVolume of acid added\nWhich letter denotes the area of the graph where both acid and salt are present?(a)\n B\n(b) A\n(c) D\n(d) C\n5. () () () () CH OO CO HO l aq aq aq 66 6 61 26 22 2 $ ++\nThe above reaction is a/an:(a)\n endothermic reaction\n(b)\n displacemen\nt reaction\n(c)\n neutralisation reaction\n(d)\n exothermic reaction\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 26 P age 3\nhttps://qrbook.page.link/app\n6. A studen t learns that magnetic field strength around a bar magnet is different at every point.", + "Click Here To InstallCBSE Science Class 10 Sample P aper 26 P age 3\nhttps://qrbook.page.link/app\n6. A studen t learns that magnetic field strength around a bar magnet is different at every point. Which diagram \nshows the correct magnetic field lines around a bar magnet?\n7. Whic h of the given options correctly represents the parent acid and base of calcium carbonate?\nOption Parent acid Parent base\n(a) HCl NaOH\n(b) H2CO3Ca(OH)2", + "sample question papers.txt\n7. Whic h of the given options correctly represents the parent acid and base of calcium carbonate?\nOption Parent acid Parent base\n(a) HCl NaOH\n(b) H2CO3Ca(OH)2\n(c) H3PO3CaSO4\n(d) H2SO4CaSO4\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 26 CBSE Science Class 10\nhttps://qrbook.page.link/app\n8. The temp erature of a conductor is increased. The graph best showing the variation of its resistance is:\n9. The diagram sho ws the reaction between metal and dil. hydrochloric acid.\nWhat is the reason for different behaviour of Mg in test tube B?\n(a)\n Mg reacts with dil. HCl to pro\nduce H2 gas which helps in floating\n(b)\n Mg is ligh\nter element than dil. HCl\n(c)\n Mg reacts with dil. HCl to pro\nduce CO2 gas which helps in floating\n(d)\n Mg reacts with di.", + "HCl to pro\nduce H2 gas which helps in floating\n(b)\n Mg is ligh\nter element than dil. HCl\n(c)\n Mg reacts with dil. HCl to pro\nduce CO2 gas which helps in floating\n(d)\n Mg reacts with di. HCl to pro\nduce N2 gas which helps in floating\n10.\n During the preparation of h\nydrogen chloride gas on a humid day, the gas is usually passed through the guard tube", + "sample question papers.txt\n(d)\n Mg reacts with di. HCl to pro\nduce N2 gas which helps in floating\n10.\n During the preparation of h\nydrogen chloride gas on a humid day, the gas is usually passed through the guard tube \ncontaining calcium chloride. The role of calcium chloride taken in the guard tube is to:(a)\n moisten the gas\n(b)\n absorb the ev\nolved gas\n(c)", + "sample question papers.txt\nabsorb Cl- ion from the evolved gas\n(d)\n absorb moisture from the gas\n11.\n A plan\nt is kept in the dark for two days. A leaf is used in an experiment to investigate the effect of two factors on \nphotosynthesis as shown in the diagram.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 26 P age 5\nhttps://qrbook.page.link/app\nWhat are the colours of Q and R, when the leaf is tested for starch, using iodine solution?\nS. No. Q R\n(a) Blue/black Brown\n(b) Brown Brown\n(c) Blue/black Blue/black\n(d) Brown Blue/black\n12.\n Refer to the giv\nen figure showing a cross between a tall and a short variety of a plant species.\nWhat could be genotype of X, Y and Z in the given figure?", + "Refer to the giv\nen figure showing a cross between a tall and a short variety of a plant species.\nWhat could be genotype of X, Y and Z in the given figure?\nX Y Z\n(a) TT tt Tt\n(b) Tt Tt Tt\n(c) Tt Tt TT\n(d) TT tt TT\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 26 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers.txt\n(d) TT tt TT\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 26 CBSE Science Class 10\nhttps://qrbook.page.link/app\n13. A diagram is giv en below:\nThe statement which defines the diagram and its parts is:\n(a)\n This is an ideal setting of micelle formation with A = hydrophobic end. B = hydrophilic end and C = oil \ndroplet.\n(b)\n This is an ideal setting of micelle formation with A = Hydrophilic end, B = oil droplet and C = Hydrophobic \nend. \n(c)\n This is an ideal setting of micelle formation with A = oil droplet, B = hydrophobic end and C = hydrophilic \nend.\n(d)\n This is an ideal setting of micelle formation with A = oil droplet, B = hydrophilic end and C = hydrophobic \nend. \n14.", + "(d)\n This is an ideal setting of micelle formation with A = oil droplet, B = hydrophilic end and C = hydrophobic \nend. \n14.\n The resistance of a wire of length 300 m and cross-sectional area 1.0 mm2, made of material of resistivity \n.m10 107# \u2126 \u2212 is:\n(a)", + "sample question papers.txt\n3 \u2126 \n(b) 2 \u2126 \n(c) 30 \u2126 \n(d) 20 \u2126 \n15. The diagram sho ws a vertical section through the heart.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 26 P age 7\nhttps://qrbook.page.link/app\nWhat are the functions of the numbered blood vessels?\nS. No. Carries blood to body Carries blood to lungs Carries blood from lungs Carries blood from body\n(a) 1 2 3 4\n(b) 1 3 4 2\n(c) 2 4 3 1\n(d) 3 1 4 2\n16.\n In the reaction of iron with copp\ner sulphate solution:\nCuSO Fe Cu FeSO 44$++\nWhich option in the given table correctly represents the substance oxidised and the reducing agent?", + "In the reaction of iron with copp\ner sulphate solution:\nCuSO Fe Cu FeSO 44$++\nWhich option in the given table correctly represents the substance oxidised and the reducing agent?\nOption Substance oxidised Reducing agent\n(a) Fe Fe\n(b) Fe FeSO4\n(c) Cu Fe\n(d) CuSO4Fe\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17.\n Assertion (A):\n Cerebellum controls the coordination of body movement and posture.\nReason (R): Medulla oblongata controls and regulates the centre for coughing, sneezing and vomiting.", + "sample question papers.txt\n17.\n Assertion (A):\n Cerebellum controls the coordination of body movement and posture.\nReason (R): Medulla oblongata controls and regulates the centre for coughing, sneezing and vomiting.\n(a)\n Both Assertion (A) and Reason are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n18.\n Assertion (A):\n At high temperatures, metal wires have a greater chance of short circuiting.\nReason (R): Both resistance and resistivity of a material vary with temperature.(a)\n Both Assertion (A) and Reason are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)", + "sample question papers.txt\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n19.\n Assertion (A):\n The transfer of pollen grains from the anther of a stamen to the stigma of a carpel is called \npollination.Reason (R): Pollination is done by insects, birds, wind and water.(a)\n Both Assertion (A) and Reason are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.", + "(d)\n Assertion (A) is false but Reason (R) is true.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 26 CBSE Science Class 10\nhttps://qrbook.page.link/app\n20. Assertion (A): Quicklime reacts vigorously with water releasing a large amount of heat.", + "sample question papers.txt\nClick Here To InstallPage 8 Sample P aper 26 CBSE Science Class 10\nhttps://qrbook.page.link/app\n20. Assertion (A): Quicklime reacts vigorously with water releasing a large amount of heat.\nReason (R): A solution of quicklime is used for whitewashing walls.\n(a)\n Both Assertion (A) and Reason are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21.\n A \nperson is unable to see distinctly the objects closer than 1m. Name the defect of vision he is suffering from. Draw \nray diagrams to illustrate the cause of the defect and its correction by suitable lens.\n\u0001o\u0001\nExplain why the planets do not twinkle?\n22.\n What is biological magnification? Will the lev", + "sample question papers.txt\nray diagrams to illustrate the cause of the defect and its correction by suitable lens.\n\u0001o\u0001\nExplain why the planets do not twinkle?\n22.\n What is biological magnification? Will the lev\nels of this magnification be different at different Levels of the \necosystem?\n23.\n Name the parts A, B, C and D of h\numan brain.\n24. Ho w is the amount of urine produced regulated?\n25.\n Iden\ntify the reducing agent in the following reactions:\n(i)\n HO FH FH OF 22$++\n(ii) FeOC OF eC O 32 3 23 2$++\n\u0001o\u0001\nWhich among the following changes are exothermic or endothermic in nature?\n(i)\n Decomp\nosition of ferrous sulphate.\n(ii)\n Dilution of sulph\nuric acid.\n(iii)\n Dissolution of so\ndium hydroxide in water.\n(iv)\n Dissolution of ammonium c\nhloride in water.\n26.\n W\nrite two different ways in which glucose is oxidised to provide energy in human body.", + "(iii)\n Dissolution of so\ndium hydroxide in water.\n(iv)\n Dissolution of ammonium c\nhloride in water.\n26.\n W\nrite two different ways in which glucose is oxidised to provide energy in human body. Write the products formed \nin each case.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 26 P age 9", + "sample question papers.txt\nin each case.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 26 P age 9\nhttps://qrbook.page.link/app\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27.\n (i)\n Create a fo\nod chain of the following organisms: Insect, Hawk, Grass, Snake, Frog?\n(ii)\n Name the organism at the third trophic lev\nel of the created food chain.\n(iii)\n Whic\nh organism of this food chain will have the highest concentration of non-biodegradable chemicals?\n(iv)\n Name the phenomenon asso\nciated with it.\n(v)\n If 10,000 joules of energy is a\nvailable to frogs, how much energy will be available to snakes in this food \nchain?\n28.\n If the image formed b\ny a lens for all positions of an object placed in front of it is always erect and diminished, \nwhat is the nature of this lens? Draw a ray diagram to justify your answer.", + "28.\n If the image formed b\ny a lens for all positions of an object placed in front of it is always erect and diminished, \nwhat is the nature of this lens? Draw a ray diagram to justify your answer. If the numerical value of the power \nof this lens is 10 D, what is its focal length in the Cartesian system?\n29.\n The follo", + "sample question papers.txt\nof this lens is 10 D, what is its focal length in the Cartesian system?\n29.\n The follo\nwing diagram shows two parallel straight conductors carrying same current. Copy the diagram and draw \nthe pattern of the magnetic field lines around them showing their directions. What is the magnitude of magnetic field at a point \u2018X\u2019 which is equidistant from the conductors? Give justification for your answer.\n\u0001o\u0001\nGive reasons for the following:\n(i)\n There is either a con\nvergence or a divergence of magnetic field lines near the ends of a current carrying \nstraight solenoid.\n(ii)\n The curren\nt carrying solenoid when suspended freely rests along a particular direction.\n(iii)\n The burn\nt out fuse should be replaced by another fuse of identical rating.\n30.\n Iden\ntify the compound X on the basis of the reactions given below. Also, write the name and chemical formulae \nof A, B and C.Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 26 CBSE Science Class 10", + "sample question papers.txt\nof A, B and C.Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 26 CBSE Science Class 10\nhttps://qrbook.page.link/app\n31. The image of an ob ject formed by a mirror is real, inverted and is of magnification 1-. If the image is at the \ndistance of 30 cm from the mirror, where is the object placed? Find the position of the image if the object is now \nmoved 20 cm towards the mirror. What is the nature of the image obtained? Justify your answer with the help of ray diagram.\n32.\n List in tabular form three distinguishing features b\netween autotrophic nutrition and heterotrophic nutrition.\n\u0001o\u0001\nDesign an activity to show that CO2 is produced during breathing. Name the intermediate and the end products \nof glucose breakdown in aerobic respiration.\n33.\n What is mean\nt by skeletal type chemical equation? What does it represents? Using the equation for electrolytic", + "sample question papers.txt\nof glucose breakdown in aerobic respiration.\n33.\n What is mean\nt by skeletal type chemical equation? What does it represents? Using the equation for electrolytic \ndecomposition of water, differentiate between a skeletal chemical equation and a balanced chemical equation.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34.\n (i)\n What are dominan\nt and recessive traits?\n(ii)\n Is it p\nossible that a trait is inherited but may not be expressed in the next generation? Give a suitable \nexample to justify this statement.\n\u0001o\u0001\n(i)\n Wh\ny did Mendel choose garden pea for his experiments? Write two reasons.\n(ii)\n What are mono-h\nybrid and dihybrid cross?\n(iii)\n Ho\nw Mendel proved that tallness is the dominant trait and dwarfness is recessive in a pea plant? Explain \nwith the help of a mono-hybrid cross.\n35.\n PQ is a current carrying conductor in the plane of the paper as shown in the figure below.", + "sample question papers.txt\nwith the help of a mono-hybrid cross.\n35.\n PQ is a current carrying conductor in the plane of the paper as shown in the figure below.\n(i) Find the directions of the magnetic fields pro duced by it at points R and S? Given r1 2 r2, where will the \nstrength of the magnetic field be larger? Give reasons.\n(ii)\n Field strength at whic\nh point will be greater?\n(iii)\n If the \npolarity of the battery connected to the wire is reversed, how would the direction of the magnetic field \nbe changed?\n(iv)\n Explain the rule that is used to find the direction of the magnetic field for a straigh\nt current carrying \nconductor.\n36.\n Y\nou are given balls and stick model of six carbon atoms and fourteen hydrogen atoms and sufficient number of \nsticks. In how many ways one can join the models of six carbon atoms and fourteen hydrogen atoms to form \ndifferent molecules of C6H14?\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers.txt\ndifferent molecules of C6H14?\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 26 P age 11\nhttps://qrbook.page.link/app\n\u0001o\u0001\nWrite the chemical formula and name of the compound which is the active ingredient of all alcoholic drinks. List \nits two uses. Write chemical equation and name of the product formed when this compound reacts with:(i)\n so\ndium metal\n(ii)\n hot concen\ntrated sulphuric acid.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.\n37.\n A comp\nound microscope is an instrument which consists of two lenses L1 and L2. The lens L1 called objective, \nforms a real, inverted and magnified image of the given object. This serves as the object for the second lens L2, \nthe eye piece.", + "37.\n A comp\nound microscope is an instrument which consists of two lenses L1 and L2. The lens L1 called objective, \nforms a real, inverted and magnified image of the given object. This serves as the object for the second lens L2, \nthe eye piece. The eye piece functions like a simple microscope or magnifier. It produces the final, image, which is", + "sample question papers.txt\nthe eye piece. The eye piece functions like a simple microscope or magnifier. It produces the final, image, which is \ninverted with respect to the original object, enlarged and virtual.(i)\n What t\nypes of lenses must be L1 and L2?\n(ii)\n If on applying Cartesian sign con\nvention for spherical lenses, the image distance obtained is negative, state \nthe significance of the negative sign.\n(iii)\n If \npower of the eyepiece L2 is 5 dioptres and it forms an image at a distance of 80 cm from its optical centre, \nat what distance should the object be?\n\u0001o\u0001\n(iv)\n What is the v\nalue and sign of magnification (according to the new cartesian sign convention) of the image \nformed by L1 and L2?\n38.\n The gro\nwing size of the human population is a cause of concern for all people. The rate of birth and death in a \ngiven population will determine its size. Reproduction is the process by which organisms increase their population.", + "sample question papers.txt\ngiven population will determine its size. Reproduction is the process by which organisms increase their population. \nThe process of sexual maturation for reproduction is gradual and takes place while general body growth is still going on. Some degree of sexual maturation does not necessarily mean that the mind or body is ready for sexual acts or for having and bringing up children. Various contraceptive devices are being used by human beings to control the size of population.(i)\n What is con\ntraception?\n(ii)\n Whic\nh contraceptive method changes the hormonal balance of the body?\n(iii)\n List t\nwo common signs of sexual maturation in boys and girls.\n\u0001o\u0001\n(iv)\n List an\ny two reasons why the Government has banned prenatal sex determination by law?\n39.\n Rishabh w\nanted to study displacement reactions. He knows that he needs a metal and a salt solution of a different \nmetal. So, he takes two tubes T1 and T2, out of which in T1, he placed a copper wire in iron sulphate solution and", + "sample question papers.txt\nmetal. So, he takes two tubes T1 and T2, out of which in T1, he placed a copper wire in iron sulphate solution and \nin T2, he placed an iron nail in copper sulphate solution as shown below:\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 12 Sample P aper 26 CBSE Science Class 10\nhttps://qrbook.page.link/app\n(i) Based on the giv en information, which test tube will undergo displacement reaction? Also, write the chemical \nreaction occurring in that test tube.\n(ii)\n State the c\nhanges observed in T1 and T2. \n\u0001o\u0001\n(iii)\n What will happ\nen (a) when zinc wire is used in place of copper wire in T1 and (b) when silver nitrate is used \nin place of iron sulphate in T1?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 5.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 3 Page 1\nhttps://qrbook.page.link/app\nSample Paper 3\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 5.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Equal volumes of hydrochloric acid and sodium hydroxide solutions of same concentration are mixed and the pH \nof the resulting solution is checked with a pH paper. What would be the colour obtained? \n(a) Red\n(b) Yellow\n(c) Yellowish green\n(d) Blue\n2. HClZ nClH Y2 22 $++ . In the above reaction, Y is:\n(a) Aluminium\n(b) Copper\n(c) Sodium\n(d) Zinc\n3. Which one of the following pair is correct?", + "HClZ nClH Y2 22 $++ . In the above reaction, Y is:\n(a) Aluminium\n(b) Copper\n(c) Sodium\n(d) Zinc\n3. Which one of the following pair is correct?\nReaction Reaction Type\n(a) () KNOs2 3 () KNOs2 2\" ()Og2+ Displacement reaction\n(b) () () sa q Zn AgNO 2 3 + ()ZnNO 32 \" ()Ag s 2+ Combination reaction", + "sample question papers 5.txt\nReaction Reaction Type\n(a) () KNOs2 3 () KNOs2 2\" ()Og2+ Displacement reaction\n(b) () () sa q Zn AgNO 2 3 + ()ZnNO 32 \" ()Ag s 2+ Combination reaction\n(c) () () Ni NO aq 32 NaOH 2+ ()Ni OH 2 \". () NaNO aq2 3 + Double displacement reaction and precipitation \nreaction\n(d) () ()Ng Hg 3 22+ ()NH g 2 3\" Decomposition reactionInstall NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 3 CBSE Science Class 10\nhttps://qrbook.page.link/app\n4. Which of the following is correct for a physical change?\n1. Only ph ysical properties change.\n2. Large amoun t of heat is absorbed or evolved.\nWhich of the above statements is/are correct?\n(a) Only 1\n(b) Only 2\n(c) Both 1 and 2\n(d) Neither 1 and 2\n5.", + "Only ph ysical properties change.\n2. Large amoun t of heat is absorbed or evolved.\nWhich of the above statements is/are correct?\n(a) Only 1\n(b) Only 2\n(c) Both 1 and 2\n(d) Neither 1 and 2\n5. An element X has electronic configuration 2, 8, 1 and another element Y has electronic configuration 2, 8, 7. They \nform a compound Z. The property that is not exhibited by Z is \n(a) It has high melting p oint.\n(b) It is a go od conductor of electricity in its pure solid state.", + "sample question papers 5.txt\nform a compound Z. The property that is not exhibited by Z is \n(a) It has high melting p oint.\n(b) It is a go od conductor of electricity in its pure solid state.\n(c) It breaks in to pieces when beaten with hammer.\n(d) It is soluble in w ater\n6. Which of the following are used as an antacid to reduce acidity in stomach?\n(a) So dium carbonate and magnesium hydroxide\n(b) Magnesium h ydroxide and sodium hydroxide\n(c) So dium bicarbonate and calcium hydroxide\n(d) So dium bicarbonate and magnesium hydroxide\n7. Structural formula of benzene is\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 3 P age 3\nhttps://qrbook.page.link/app\n8. In the given activity, the lime water of which test tube will get milky faster?", + "Click Here To InstallCBSE Science Class 10 Sample P aper 3 P age 3\nhttps://qrbook.page.link/app\n8. In the given activity, the lime water of which test tube will get milky faster?\n(a) T est tube (a)\n(b) T est tube (b)\n(c) Both test tub e will take same time\n(d) Can\u2019t sa y", + "sample question papers 5.txt\n8. In the given activity, the lime water of which test tube will get milky faster?\n(a) T est tube (a)\n(b) T est tube (b)\n(c) Both test tub e will take same time\n(d) Can\u2019t sa y\n9. When a person eats some egg white, proteins and water enter the stomach. Which substances are found leaving \nthe stomach and leaving the small intestine?\nLeaving the Stomach Leaving the Small Intestine\n(a) Protein, amino acids and water Water\n(b) Amino acids and water Amino acids and water\n(c) Fatty acids, glycerol and water Fatty acids, glycerol and water\n(d) Protein and water Fatty acids and glycerol\n10. Exchange of genetic material takes place in\n(a) v egetative reproduction\n(b) asexual repro duction\n(c) sexual repro duction\n(d) budding\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "Exchange of genetic material takes place in\n(a) v egetative reproduction\n(b) asexual repro duction\n(c) sexual repro duction\n(d) budding\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 3 CBSE Science Class 10\nhttps://qrbook.page.link/app\n11. When a person is suffering from severe cold, he or she cannot", + "sample question papers 5.txt\nClick Here To InstallPage 4 Sample P aper 3 CBSE Science Class 10\nhttps://qrbook.page.link/app\n11. When a person is suffering from severe cold, he or she cannot\n(a) differen tiate the taste of an apple from that of an ice cream.\n(b) differen tiate the smell of a perfume from that of an agarbatti.\n(c) differen tiate red light from green light\n(d) differen tiate a hot object from a cold object.\n12. Which among the following statements are true for sexual reproduction in flowering plants?\n(i) It requires t wo types of gametes\n(ii) F ertilisation is a compulsory event\n(iii) It alw ays results in formation of zygote\n(iv) Offsprings formed are clones\n(a) (i) and (iv) (b) (i), (ii) and (iv)\n(c) (i), (ii) and (iii) (d) (i), (iii) and (iv)\n13.", + "The resistivity does not change if\n(a) the material is c hanged\n(b) the temp erature is changed\n(c) the shap e of the resistor is changed\n(d) b oth material and temperature are changed", + "sample question papers 5.txt\n13. The resistivity does not change if\n(a) the material is c hanged\n(b) the temp erature is changed\n(c) the shap e of the resistor is changed\n(d) b oth material and temperature are changed\n14. A uniform magnetic field exists in the plane of paper pointing from left to right as shown in Figure. In the field \nan electron and a proton move as shown. The electron and the proton experience.\n(a) forces b oth pointing into the plane of paper\n(b) forces b oth pointing out of the plane of paper\n(c) forces p ointing into the plane of paper and out of the plane of paper, respectively.\n(d) force p ointing opposite and along the direction of the uniform magnetic field respectively.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 3 P age 5\nhttps://qrbook.page.link/app\n15. A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure.", + "Click Here To InstallCBSE Science Class 10 Sample P aper 3 P age 5\nhttps://qrbook.page.link/app\n15. A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure. The current recorded \nin the ammeter will be", + "sample question papers 5.txt\nhttps://qrbook.page.link/app\n15. A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure. The current recorded \nin the ammeter will be\n(a) maxim um in (i)\n(b) maxim um in (ii)\n(c) maxim um in (iii)\n(d) the same in all the cases\n16. The most important safety method used for protecting home appliances from short circuiting or overloading is\n(a) earthing\n(b) use of fuse\n(c) use of stabilizers\n(d) use of electric meter.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 3 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001Question no. 17 to 20 are Assertion - Reasoning based questions.\n17. Assertion : When water is added to calcium oxide, a large amount of heat is produced.\nReason : It is an endothermic reaction.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.", + "sample question papers 5.txt\nReason : It is an endothermic reaction.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : The genetic complement of an organism is called genotype.Reason : Genotype is the type of hereditary properties of an organism.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : During the night the effect of root pressure in transport of water is more important.Reason : Stomata is open during day, transpiration takes place which help in transport of water.", + "sample question papers 5.txt\n19. Assertion : During the night the effect of root pressure in transport of water is more important.Reason : Stomata is open during day, transpiration takes place which help in transport of water.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Both Assertion and Reason are false.\n20. Assertion : A solenoid tends to expand, when a current passes through it.Reason : Two straight parallel metallic wires carrying current in same direction attract each other.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\nSECTION-B", + "sample question papers 5.txt\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. Why do the articles made of aluminium not corrode?\n\u0001o\u0001\n(a) Fe CuSO FeSO Cu 44$ ++\n(b) Cu FeSO CuSO Fe 44$ ++\nWhich of the above two reactions will take place and why ?\n22. Stomata of desert plants remain closed during day time. How do they take up carbon dioxide and perform \nphotosynthesis ?\n23. Which is the largest digestive gland present in human body ? What is the name and function of its secretion?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 3 P age 7\nhttps://qrbook.page.link/app\n24. What are the end products formed during fermentation in yeast ? Under what condition a similar process takes \nplace in our body that leads to muscle cramps ?", + "What are the end products formed during fermentation in yeast ? Under what condition a similar process takes \nplace in our body that leads to muscle cramps ?\n25. Why there is no dispersion of light refracted through a rectangular glass slab.", + "sample question papers 5.txt\nplace in our body that leads to muscle cramps ?\n25. Why there is no dispersion of light refracted through a rectangular glass slab.\n\u0001o\u0001\nWhat is meant by near point and far point of an eye? State their values of the normal human eye.\n26. In a food chain comprising frogs, insects, birds and grass, which one of the organisms is likely to have maximum \nconcentration of harmful non-biodegradable chemicals in its body ?\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. Name the type of chemical reaction represented by the following equations :\n(i) () () () CaCO sC aOsC OgHeat\n32 +\n(ii) () () () () CaOs HO lC aOHa q 22$ +\n(iii) () () () () ZnsH SO aq ZnSO aq Hg 24 42$ ++\n28. (a) A non-metal X exists in two different forms Y and Z. Y is the hardest natural substance whereas Z is \na good conductor of electricity. Identify X, Y, Z.", + "(a) A non-metal X exists in two different forms Y and Z. Y is the hardest natural substance whereas Z is \na good conductor of electricity. Identify X, Y, Z.\n(b) An elemen t X on reaction with oxygen forms an oxide OX 2. The oxide when dissolved in water turns blue", + "sample question papers 5.txt\na good conductor of electricity. Identify X, Y, Z.\n(b) An elemen t X on reaction with oxygen forms an oxide OX 2. The oxide when dissolved in water turns blue \nlitmus red. State whether element X is a metal or non-metal.\n(c) Name the metal whic h is alloyed with copper to make bronze.\n29. Explain the process of assimilation of proteins in human digestive system.\n\u0001o\u0001\nWrite three events which occur during the process of photosynthesis.\n30. An object 4 cm in height is placed at 15 cm in front of a concave mirror of focal length 10 cm. At what distance \nfrom the mirror should a screen be placed to obtain a sharp image of the object. Calculate the height of the image.\n31. (a) Name the kind of lens that can form;\n(i) an in verted magnified image.\n(ii) an erect diminished image.\n Dra w ray diagrams to illustrate your answer in each case.\n(b) Dra w a ray diagram to show the image formed of an object placed between f and f2 distances from a \nconvex lens.", + "sample question papers 5.txt\nDra w ray diagrams to illustrate your answer in each case.\n(b) Dra w a ray diagram to show the image formed of an object placed between f and f2 distances from a \nconvex lens.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 3 CBSE Science Class 10\nhttps://qrbook.page.link/app\n32. What would be the reading of ammeter and voltmeter in the given circuit?\n\u0001o\u0001\nTwo resistors with resistances 10\u2126 and 15\u2126 are to be connected to emf 12 V so as to obtain : (i) minimum \ncurrent (ii) maximum current. How will you connect the resistance in each case ? Calculate the strength of the \ntotal current in the circuit in the two cases.\n33. It is said that, there is a need to put a complete ban on the products containing aerosols. What are aerosols ? Why is there a demand to put a ban on them.\nSECTION-D\n\u0001Question no.", + "33. It is said that, there is a need to put a complete ban on the products containing aerosols. What are aerosols ? Why is there a demand to put a ban on them.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. (a) Dra w the structure of ethanoic acid.", + "sample question papers 5.txt\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. (a) Dra w the structure of ethanoic acid.\n(b) Name the comp ound formed when ethanol is heated with ethanoic acid in the presence of conc. HSO24.\n(c) Complete the follo wing equations :\n CH ClSunlight\n42+\n CHO HOCombustion\n25 2+\n\u0001o\u0001\n(a) A comp ound X having formula CHO 24 2 when treated with ethanol and a few drops of conc. HSO24 forms \na sweet smelling substance Y. Name X and Y. Write the equation of the reaction leading to the formation \nof Y from X. What is the function of conc. HSO24 in the above reaction?\n(b) Wh y do soaps form scum instead of lather in hard water ?\n35. (a) Differen tiate between pollen grain and ovule.", + "HSO24 in the above reaction?\n(b) Wh y do soaps form scum instead of lather in hard water ?\n35. (a) Differen tiate between pollen grain and ovule.\n(b) State in brief the functions of the follo wing parts of the human female reproductive system :\n(i) Ovary\n(ii) F allopian tube\n(iii) Uterus\n\u0001o\u0001\n(a) What is v ariation? How is variation created in a population? How does the creation of variation in a species \npromote survival?", + "sample question papers 5.txt\n(i) Ovary\n(ii) F allopian tube\n(iii) Uterus\n\u0001o\u0001\n(a) What is v ariation? How is variation created in a population? How does the creation of variation in a species \npromote survival?\n(b) Explain ho w, offspring and parents of organisms reproducing sexually have the same number of chromosomes.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 3 P age 9\nhttps://qrbook.page.link/app\n36. An electric lamp of resistance 20\u2126 and a conductor of resistance 4\u2126 are connected to a 6 V battery as shown in \nthe circuit. Calculate :\n(a) the total resistance of the circuit,\n(b) the curren t through the circuit,\n(c) the p otential difference across the (i) electric lamp and (ii) conductor, and\n(d) p ower of the lamp.\nSECTION-E\n\u0001Question no.", + "Calculate :\n(a) the total resistance of the circuit,\n(b) the curren t through the circuit,\n(c) the p otential difference across the (i) electric lamp and (ii) conductor, and\n(d) p ower of the lamp.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data -based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.", + "sample question papers 5.txt\n(d) p ower of the lamp.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data -based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37. The earth\u2019s crust is the major source of metals-seawater contains some soluble salts such as sodium chloride, magnesium chloride, etc. The elements or compounds, which occur naturally in the earth\u2019s crust are known as minerals. At some places, minerals contain a very high percentage of a particular metal and the metal can be profitably extracted from it. These minerals are called ores.\n(i) Name the c hief ore of mercury and zinc.\n(ii) W rite equations for the extraction of copper from its sulphide ore.\n\u0001o\u0001\n(ii) Define the pro cess used to convert carbonate ores into metal oxide.\n38. Chronic Kidney Disease (CKD) is a condition characterized by a gradual loss of kidney function over time. CKD", + "sample question papers 5.txt\n\u0001o\u0001\n(ii) Define the pro cess used to convert carbonate ores into metal oxide.\n38. Chronic Kidney Disease (CKD) is a condition characterized by a gradual loss of kidney function over time. CKD \nis also known as chronic renal disease. With increasing life expectancy and prevalence of life style diseases, US has seen a 30% increase in prevalence of Chronic Kidney Disease (CKD) in the last decade. Unfortunately, from India there is no longitudinal study and limited data on the prevalence of CKD.In western countries, diabetes and hypertension account for over 2/3\nrd of the cases of CKD. In India too, diabetes", + "sample question papers 5.txt\nrd of the cases of CKD. In India too, diabetes \nand hypertension today account for 40-60% cases of CKD. As per recent Indian Council of Medical Research data, prevalence of diabetes in Indian adult population has risen to 7.1%, (varying from 5.8% in Jharkhand to 13.5% in Chandigarh) and in urban population (over the age of 40 years) the prevalence is as high as 28%. Likewise, the reported prevalence of hypertension in the adult population today is 17% (14.8% from rural and 21.4% from urban belt). A similar prevalence of 17.4% has been reported by Panesar et al. (in the age group of 20-59 years) even from slum-resettlement colony of Delhi. With rising prevalence of these diseases in India, prevalence of CKD is expected to rise and obviously, this is the key target population to address.A study published in this issue is from a rural belt of Karnataka. The population had a mean age of 39.88 \n!", + "sample question papers 5.txt\n! \n15.87 years with 3.82% prevalence of diabetes and 33.62% of hypertension. Authors found 6.3% prevalence of CKD stage 3; which is the highest reported till date by any Indian worker. It is disturbing to note, the high prevalence of hypertension in a rural setting where over 75% population had normal or low body mass index. \n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 3 CBSE Science Class 10\nhttps://qrbook.page.link/app\nIn comparison to most other published studies from India, the present study population is younger and even the \nprevalence of diabetes is low but surprisingly despite that prevalence of stage 3 CKD is reported to be higher (6.3%). It is disturbing to see the rising prevalence of hypertension and CKD in rural belts. Possibly, with shifting population the difference between urban and rural areas is getting blurred. Undoubtedly, we need more Indian data to validate these findings.", + "sample question papers 5.txt\n(i) What is CKD ?\n(ii) What are the ma jor causes of CKD ?\n(iii) In whic h segment of society is CKD more prevalent?\n\u0001o\u0001\n(iv) What is the highest p ercentage of CKD reported?\n39. A concave lens is thick at the edges and thin at the centre, while a convex lens is thick at the centre and thin at \nthe edges. We can distinguish between a concave lens and a convex lens without touching them. For this keep a book close to a lens and observe the image of the text of the book through the lens. If the letters appear enlarged, then it is a convex lens and if the letters appear diminished then it is a concave lens.\nConvex lens converges light rays and hence known as converging lens. Similarly, concave lens diverges light rays \nand is known as diverging lens. Linear magnification produced by a lens is equal to the ratio of the image distance to the object distance. Power of a lens is defined as the reciprocal of its focal length.\n(i) What t ype of image is always made by a concave lens ?", + "sample question papers 5.txt\n(i) What t ype of image is always made by a concave lens ?\n(ii) If magnification pro duced by a spherical lens is +0.75, then what is the nature of the lens ?\n(iii) What is the p ower of a convex lens with focal length 80 cm ?\n\u0001o\u0001\n(iii) What kind of lens is presen t in human eye ?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question paper 1.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Paper 1 Page 1\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. nstall NODIA App to See the Solutions\nClick Here To In stall Click Here To InstallSample Paper 1\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.2. All questions are compulsory. However, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objective Type questions carrying 1 mark each.4. Section B consists of 6 Very Short questions carrying 02 marks each. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer type questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.", + "sample question paper 1.txt\nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer type questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer type questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks each with sub-parts.\nSECTION-A\nSelect and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. In an attempt to demonstrate electrical conductivity through an electrolyte, the following apparatus (figure) was \nset up.\nWhich among the following statement (s) is (are) correct?1. Bulb will not glow because electrolyte is not acidic.2. Bulb will glow because NaOH is a strong base and furnishes ions for conduction.3. Bulb will not glow because circuit is incomplete.4. Bulb will not glow because it depends upon the type of electrolytic solution.", + "sample question paper 1.txt\nContinue on next page......Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample Paper 1 CBSE Science Class 10\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. Install NODIA App to See the Solution\nClick Here To In stall Click Here To Install(a) 1 and 3\n(b) 2 and 4\n(c) Only 2\n(d) Only 4\n2. The following reaction is an example of a\n() () () ()NH g O g NO g H O g45 4 4 32 2 $ ++\n1. displacement reaction\n2. combination reaction3. redox reaction4. neutralisation reaction(a) 1 and 4\n(b) 2 and 3(c) 1 and 3(d) 3 and 4\n3. Calcium oxide reacts vigorously with water.\nWhich of the following is the incorrect observation of the reaction shown in the above set up?\n(a) It is an endothermic reaction.\n(b) Slaked lime is produced.(c) It is an exothermic reaction.", + "Calcium oxide reacts vigorously with water.\nWhich of the following is the incorrect observation of the reaction shown in the above set up?\n(a) It is an endothermic reaction.\n(b) Slaked lime is produced.(c) It is an exothermic reaction.(d) It is a combination reaction.\n4. NH N H32\n22 3$+\nWith the reference of above reaction which one of the option in the table is correct?\nReactants Products\n(a) N2, H2 NH 3\n(b) NH 3 N2, H2", + "sample question paper 1.txt\n4. NH N H32\n22 3$+\nWith the reference of above reaction which one of the option in the table is correct?\nReactants Products\n(a) N2, H2 NH 3\n(b) NH 3 N2, H2\n(c) N2 H2, NH 3\n(d) NN H23 H2\n5. Which one of the following properties is not general exhibited by ionic compounds?\n(a) Solubility in water\n(b) Electrical conductivity in solid state\n(c) High melting and boiling points(d) Electrical conductivity in molten stateInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Paper 1 Page 3\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. nstall NODIA App to See the Solutions\nClick Here To In stall Click Here To Install6. A student require hard water for an experiment in his laboratory which is not available in the neighbouring area.", + "sample question paper 1.txt\nClick Here To In stall Click Here To Install6. A student require hard water for an experiment in his laboratory which is not available in the neighbouring area. \nIn the laboratory there are some salts, which when dissolved in distilled water can convert it into hard water. Select from the following groups of salts, a group, each salt of which when dissolved in distilled water will make it hard.(a) Sodium chloride, Potassium chloride\n(b) Sodium sulphate, Potassium sulphate\n(c) Sodium sulphate, Calcium sulphate\n(d) Calcium sulphate, Calcium chloride\n7. The correct structural formula of butanoic acid is\n8. The diagram shows the arrangement of cells inside the leaf of a green plant. (No cell contents are shown).\nWhich of the following cells normally contain chloroplasts?\n(a) 2 and 4\n(b) 2 and 3(c) 1 and 2(d) 1 and 4Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample Paper 1 CBSE Science Class 10", + "sample question paper 1.txt\n(a) 2 and 4\n(b) 2 and 3(c) 1 and 2(d) 1 and 4Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample Paper 1 CBSE Science Class 10\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. Install NODIA App to See the Solution\nClick Here To In stall Click Here To Install9. Identify the micro-organism whose nutrition type is shown below :\n(a) Food bacteria\n(b) Yeast(c) Fungus(d) Amoeba\n10. Structure present in a cell which is responsible for determination of the sex of a baby is :\n(a) cytoplasm (b) cell membrane\n(c) nucleus (d) chromosome\n11. What is the correct direction of flow of electrical impulses ?\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Paper 1 Page 5\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions.", + "What is the correct direction of flow of electrical impulses ?\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Paper 1 Page 5\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. nstall NODIA App to See the Solutions\nClick Here To In stall Click Here To Install12. In following diagram the parts A, B and C are sequentially\n(a) cotyledon, plumule and radicle", + "sample question paper 1.txt\nClick Here To In stall Click Here To Install12. In following diagram the parts A, B and C are sequentially\n(a) cotyledon, plumule and radicle\n(b) plumule, radicle and cotyledon(c) plumule, cotyledon and radicle\n(d) radicle, cotyledon and plumule\n13. At the time of short circuit, the electric current in the circuit :\n(a) vary continuously\n(b) does not change(c) reduces substantially\n(d) increases heavily\n14. If the key in the arrangement taken out (the circuit is made open) and magnetic field lines are drawn over the \nhorizontal plane ABCD , the lines are\n(a) concentric circles\n(b) elliptical in shape(c) straight lines parallel to each other(d) concentric circles near the point O but of elliptical shapes as we go away from it.\nContinue on next page......Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample Paper 1 CBSE Science Class 10\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. Install NODIA App to See the Solution", + "sample question paper 1.txt\nClick Here To InstallPage 6 Sample Paper 1 CBSE Science Class 10\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. Install NODIA App to See the Solution\nClick Here To In stall Click Here To Install15. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances \n,RR12 and R3 respectively (Figure). Which of the following is true ?\n(a) RRR123== (b) RRR12322\n(c) RRR32122 (d) RRR23122\n16. Which of the following statement is not correct about the magnetic field?\n(a) Magnetic field lines form a continuous closed curve.(b) Magnetic field line do not interest each other.(c) Direction of tangent at any point on the magnetic field line curve gives the direction of magnetic field at \nthat point.\n(d) Outside the magnet, magnetic field lines go from South to North pole of the magnet.\nQuestion no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : Corrosion of iron is a serious problem.", + "sample question paper 1.txt\nQuestion no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : Corrosion of iron is a serious problem.\nReason : Every year an enormous amount of money is spent to replace damaged iron.\n(a) Both Assertion and Reason are True and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are True but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is True but the Reason is False.\n(d) Both Assertion and Reason are False.\n18. Assertion : Dominant allele is an allele whose phenotype expresses even in the presence of another allele of that \ngene.\nReason : It is represented by a capital letter, e.g. T.(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.", + "sample question paper 1.txt\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : All the plants possess autotrophic mode of nutrition.\nReason : Due to the presence of green coloured pigment chlorophyll in them.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.(d) Both Assertion and Reason are false.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Paper 1 Page 7\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. nstall NODIA App to See the Solutions\nClick Here To In stall Click Here To Install20. Assertion : The magnetic field produced by a current carrying solenoid is independent of its length and cross-\nsection area.", + "sample question paper 1.txt\nClick Here To In stall Click Here To Install20. Assertion : The magnetic field produced by a current carrying solenoid is independent of its length and cross-\nsection area.\nReason : The magnetic field inside the solenoid is uniform.(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).(c) Assertion (A) is true but reason (R) is false.(d) Assertion (A) is false but reason (R) is true.\nSECTION-B\nQuestion no. 21 to 26 are very short answer questions.\n21. A metal is treated with dil HS O24, the gas evolved is collected by the method shown in the figure. Answer the \nfollowing :\n(i) Name the gas.(ii) Name the method of collection of the gas.\nor\nList any two observations when a highly reactive metal is dropped in water.\n22. What do you mean by diffusion ?", + "sample question paper 1.txt\nfollowing :\n(i) Name the gas.(ii) Name the method of collection of the gas.\nor\nList any two observations when a highly reactive metal is dropped in water.\n22. What do you mean by diffusion ?\n23. Which are the first simple molecules of food produced during photosynthesis ? What happens to these simple molecules in the leaves later ?\n24. What is the meaning of the term \u201cassimilation\u201d ?\n25. Draw a neat diagram to show the refraction of a light ray through a glass prism and label on it the angle of incidence and angle of deviation.\nor\nWhat is the scattering of light ? Explain with the help of an example.\n26. What will happen to the garbage and dead animals and plants in absence of microorganisms ?\nContinue on next page......Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample Paper 1 CBSE Science Class 10\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. Install NODIA App to See the Solution\nClick Here To In stall Click Here To InstallSECTION-C", + "sample question paper 1.txt\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. Install NODIA App to See the Solution\nClick Here To In stall Click Here To InstallSECTION-C\nQuestion no. 27 to 33 are short answer questions.\n27. State which of the following chemical reactions will take place or not, giving suitable reason for each :\n(i) () ( ) ( ) ()Zn s CuSO aq ZnSO aq Cu s 44$ ++\n(ii) ( ) () () ( )Fe s ZnSO aq FeSO aq Zn s 44$ ++\n(iii) ( ) () () ( )Zn s FeSO aq ZnSO aq Fe s 44$ ++\n28. Give reasons for the following :(i) Shining surfaces of metals become dull on exposure to air and moisture.(ii) Aluminium is extracted from its ore by electrolysis of molten ore.(iii) Gold is available in the native state.\n29. Mention the three kinds of cells present in blood. Write one function of each.\nor\nWith the help of diagram explain how exchange of gases occurs in leaf of a plant.", + "sample question paper 1.txt\n29. Mention the three kinds of cells present in blood. Write one function of each.\nor\nWith the help of diagram explain how exchange of gases occurs in leaf of a plant.\n30. Manju is uses a concave mirror for image formation for different positions of an object. What inferences can be drawn about the following when an object is placed at a distance of 10 cm from the pole of a concave mirror of focal length 15 cm ?(a) Position of the image(b) Size of the image(c) Nature of the imageDraw a labelled ray diagram to justify your inferences.\n31. (a) Define optical centre of a spherical lens.(b) You are given a convex lens of focal length .cm30 Where would you place an object to get a real, inverted \nand highly enlarged image of the object ? Draw a ray diagram showing the image formation.\n(c) A concave lens has a focal length of 20 cm. At what distance should an object be placed so that it forms an \nimage at 15 cm away from the lens ?", + "sample question paper 1.txt\n(c) A concave lens has a focal length of 20 cm. At what distance should an object be placed so that it forms an \nimage at 15 cm away from the lens ?\n32. Pawan is connected a galvanometer with a coil of insulated copper wire .What would happen if a bar magnet is :(i) Pushed into the coil ?(ii) Withdrawn from inside the coil ?(iii) Held stationary inside the coil ?\nor\nYou are given two identical looking iron bars. Just using these two bars how will you identify whether any or both of these bars is/are a magnet ?\n33. Explain the phenomenon of Biological Magnification. How does it affect organisms belonging to different trophic levels particularly the tertiary consumers ?\nSECTION-D\nQuestion no. 34 to 36 are Long answer questions.\n34. An organic compound A is widely used as a preservative in pickles and has a molecular formula CHO242 . This \ncompound reacts with ethanol to form a sweet smelling compound B.Install NODIA App to See the Solutions.", + "sample question paper 1.txt\ncompound reacts with ethanol to form a sweet smelling compound B.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Paper 1 Page 9\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. nstall NODIA App to See the Solutions\nClick Here To In stall Click Here To Install(a) Identify the compound A.\n(b) Write the chemical equation for the reaction with ethanol to form compound B.\n(c) How can we get compound A from B?\n(d) Name the process and write corresponding chemical equation.\n(e) Which gas is produced when compound A reacts with washing soda ? Write the chemical equation.\nor\n(a) The formula of an ester is CH COOC H 32 5 . Write the structural formulae of the corresponding alcohol and \nthe acid.\n(b) \n(i) Mention the experimental conditions involved in obtaining ethene from ethanol.(ii) Write the chemical equation for the above reaction.\n35. (a) Write the function of following parts in human female reproductive system :", + "sample question paper 1.txt\n35. (a) Write the function of following parts in human female reproductive system :\n(i) Ovary (ii) Oviduct (iii) Uterus\n(b) Describe in brief the structure and function of placenta.\nor\nDefine the terms :(i) Syngamy (ii) Triple fusion(iii) Implantation(iv) Placenta(v) Gestation.\n36. The mobile phone is an excellent communication device. Mobile phones uses electromagnetic radiation in the microwave range. Part of the radio wave emitted by the mobile phone handset will be absorbed by the head.", + "sample question paper 1.txt\nHead is in the \u2018near field\u2019 of radiation, so that most of the heating effect occurs in the head.Temperature in the internal ear, brain increases by 1 degree or more. This adversely affect the functioning of these organs since these have fluid filled cavities. But prolonged heating effect can alter brain functions and hearing ability also. Other harmful effects such as Premature Cataract, Confusion and loss of memory may also be possible. Following figure shows that how mobile phone radiation penetrates the brain.\nHow mobile phone radiation penetrates the brain\nSource: Institute of Electrical and Electronic Engineer\u2019s journal on Micro wave Theory and \nTechniques\nContinue on next page......Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample Paper 1 CBSE Science Class 10\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. Install NODIA App to See the Solution", + "sample question paper 1.txt\nClick Here To InstallPage 10 Sample Paper 1 CBSE Science Class 10\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. Install NODIA App to See the Solution\nClick Here To In stall Click Here To Install(i) What precautions should be taken while using mobile phones ?\n(ii) Which radiations are used in mobile phones ?(iii) How does prolonged heating effect due to mobile radiations can effect adversely ?(iv) In which part of our body, most of the heating effect occurs due to use of mobiles ?\nSECTION-E\nQuestion no. 37 to 39 are case-based/data -based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37. The reactivity series is a list of metals arranged in the order of their decreasing activities. The metal at the top \nof the reactivity series is the most reactive and metal at the bottom is the least reactive. The more reactive metal displaces less reactive metal from its salt solution.\nK Potassium More reactive\nNa Sodium", + "sample question paper 1.txt\nK Potassium More reactive\nNa Sodium\nCa Calcium\nMg Magnesium\nAl Aluminium\nZn Zinc Reactivity decreases.\nFe Iron\nPb Lead\n[H] [Hydrogen]\nCu Copper\nHg Mercury\nAg Silver\nAu Gold Least reactive\n(i) Name the metals which react with steam but not with hot water.\n(ii) What happen when calcium react with nitric acid and which method is used to extract metal present at the \ntop of the reactivity series?\nor\n(ii) Which of the following metals exist in their native states in nature?\nI. Cu\nII. AuIII. ZnIV. Ag\n38. Question numbers i - iv are based on the table given below. Study the table and answer the following questions.\nTable-A\nCharacters Males Females\n1. Total no. of chromosomes 23 pairs 23 pairs\n2. No. of autosome 22 pairs 22 pairs\n3. No. of sex chromosome 1 pair 1 pair\n(i) What is sex determination?\n(ii) What are the sex chromosomes in the males?Install NODIA App to See the Solutions.", + "No. of autosome 22 pairs 22 pairs\n3. No. of sex chromosome 1 pair 1 pair\n(i) What is sex determination?\n(ii) What are the sex chromosomes in the males?Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Paper 1 Page 11", + "sample question paper 1.txt\n(i) What is sex determination?\n(ii) What are the sex chromosomes in the males?Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Paper 1 Page 11\nhttps://qrbook.page.link/appInstall N ODIA A pp to See the Solutions. nstall NODIA App to See the Solutions\nClick Here To In stall Click Here To Install(iii) What are the sex chromosomes in the females?\nor\n(iv) Is the father responsible for the sex of the child?\n39. Lenses are objects made of transparent materials such as glass or clear plastic that has curved surfaces. Diverging", + "sample question paper 1.txt\nor\n(iv) Is the father responsible for the sex of the child?\n39. Lenses are objects made of transparent materials such as glass or clear plastic that has curved surfaces. Diverging \nlenses are thicker at their edges than at their centres and makes light rays passing through them spread out. Converging lenses are thicker in their middle than at this edges and make light rays passing through them focus at a point. These are used in spectacles to help people with poor vision see better. The converging lenses magnify by bending the rays of light that pass through them to meet at a point called focus. Thicker the converging lens is at its centre, the more its magnifies and closer the focus is to the lens.(i) Ravi uses two lenses A and B of same size and same material as shown. P\n1 and P2 are the powers of A and \nB. An object is kept at the same distance from the lens between F and 2F of each lens on the principal axis in turn. Let I", + "sample question paper 1.txt\n1 and P2 are the powers of A and \nB. An object is kept at the same distance from the lens between F and 2F of each lens on the principal axis in turn. Let I\n1 and I2 be the image formed by two lenses respectively. What is the relation of image distances \nof both lens ?\n(ii) Write down the relation between the power of lens of both lenses ?(iii) Meenakshi uses above two lenses A and B along with another two lenses C and D, as shown :\nShe is able to see the subject matter on the black board while sitting in the front row in the classroom but is unable to see the same matter while sitting in the last row.Which of the above four lenses will she require to correct the defect in her vision? Why ?\nor\n(iv) Natasha places an object on the principal axis of above given lens A. One end of this object coincides with \nthe focus F and the other end with 2F. What will be the nature of the image formed by the lens on the other side ?\n\uf02a\uf02a\uf02a\uf02a\uf02a\uf02a\uf02a\uf020", + "sample question papers 3.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 16 Page 1\nhttps://qrbook.page.link/app\nSample Paper 16\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 3.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Which of the following are correctly matched?\n1.Bleaching powder oxidising agent in chemical industries.\n2.Baking powder a mixture of sodium hydrogen carbonate and a mild edible acid.\n3.Washing soda remove permanent hardness of water.\n(a) 1 and 2\n(b) 2 and 3\n(c) 1 and 3\n(d) 1, 2 and 3\n2.", + "2.Baking powder a mixture of sodium hydrogen carbonate and a mild edible acid.\n3.Washing soda remove permanent hardness of water.\n(a) 1 and 2\n(b) 2 and 3\n(c) 1 and 3\n(d) 1, 2 and 3\n2. FeOA lA lO Fe 22 23 23\"++\nThe above reaction is an example of a-\n(a) combination reaction\n(b) double displacement reaction\n(c) decomposition reaction\n(d) displacement reaction\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 3.txt\n(a) combination reaction\n(b) double displacement reaction\n(c) decomposition reaction\n(d) displacement reaction\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 16 CBSE Science Class 10\nhttps://qrbook.page.link/app\n3. The necessary conditions for combustion process to occur are\n1. a vailability of air/oxygen.\n2. a vailability or air/oxygen and fuel.\n3. temp erature of fuel below ignition temperature.\n4. temp erature of fuel above ignition temperature.\nSelect the correct alternative.\n(a) 1 and 2\n(b) 2 and 4\n(c) 3 and 1\n(d) 4 and 1\n4. Which of the following is not a chemical reaction?\n(a) Souring of milk\n(b) Dissolution of sugar in w ater\n(c) Rusting of iron\n(d) Digestion of fo od in the body\n5.", + "Which of the following is not a chemical reaction?\n(a) Souring of milk\n(b) Dissolution of sugar in w ater\n(c) Rusting of iron\n(d) Digestion of fo od in the body\n5. The ability of metals to be drawn into thin wire is known as\n(a) Ductility\n(b) Malleability\n(c) Sonority\n(d) Conductivity\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 3.txt\n(a) Ductility\n(b) Malleability\n(c) Sonority\n(d) Conductivity\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 16 P age 3\nhttps://qrbook.page.link/app\n6. Which of the following statements is/are correct for litmus?\n1. Litm us solution is a purple dye.\n2. It is extracted from lic hen.\n3. In neutral solution. it remains colourless.\n(a) 1 and 2\n(b) 2 and 3\n(c) 1 and 3\n(d) 1, 2 and 3\n7. A soap micelle is shown in th figure. In the soap micelles\n(a) The ionic end of soap is on the surface of the cluster while the carbon chain is in the interior of the cluster.\n(b) Ionic end of soap is in the in terior of the cluster and carbon chain is out of the cluster.", + "In the soap micelles\n(a) The ionic end of soap is on the surface of the cluster while the carbon chain is in the interior of the cluster.\n(b) Ionic end of soap is in the in terior of the cluster and carbon chain is out of the cluster.\n(c) Both ionic end and carb on chain are in the interior of the cluster.\n(d) Both ionic end and carb on chain are on the exterior of the cluster.\n8. Four chambered heart is characteristics feature of?\n(a) Fishes\n(b) Amphibians\n(c) Reptiles\n(d) Mammals", + "sample question papers 3.txt\n(d) Both ionic end and carb on chain are on the exterior of the cluster.\n8. Four chambered heart is characteristics feature of?\n(a) Fishes\n(b) Amphibians\n(c) Reptiles\n(d) Mammals\n9. Translocation is the process in which plants deliver:\n(a) minerals from lea ves to other parts of the plant\n(b) plan t growth hormones from leaves to other parts of the plant\n(c) w ater and organic substance from leaves to other parts of the plant\n(d) all of the ab ove\n10. The two versions of a trait (character) which are brought in by the male and female gametes are situated on\n(a) copies of the same c hromosome\n(b) t wo different chromosomes\n(c) sex c hromosomes\n(d) an y chromosomeInstall NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 16 CBSE Science Class 10\nhttps://qrbook.page.link/app\n11. Choose the incorrect statement about insulin which is shown in the figure:.\n(a) It is pro duced from pancreas.", + "Click Here To InstallPage 4 Sample P aper 16 CBSE Science Class 10\nhttps://qrbook.page.link/app\n11. Choose the incorrect statement about insulin which is shown in the figure:.\n(a) It is pro duced from pancreas.\n(b) It regulates gro wth and development of the body.", + "sample question papers 3.txt\nhttps://qrbook.page.link/app\n11. Choose the incorrect statement about insulin which is shown in the figure:.\n(a) It is pro duced from pancreas.\n(b) It regulates gro wth and development of the body.\n(c) It regulates blo od sugar level.\n(d) Insufficien t secretion of insulin will cause diabetes.\n12. Which of the following is the correct sequence of events of sexual reproduction in a flower?\n(a) p ollination, fertilisation, seedling, embryo\n(b) seedling, em bryo, fertilisation, pollination\n(c) p ollination, fertilisation, embryo, seedling\n(d) em bryo, seedling, pollination, fertilisation.\n13. As light travels from a rarer to a denser medium it will have\n(a) increased v elocity\n(b) decreased v elocity\n(c) decreased w avelength\n(d) b oth (b) and (c)\n14. \nDispersion of light by glass prism is shown in the above figure. Here x and y indicates .......... and .......... colour \nrespectively.", + "Dispersion of light by glass prism is shown in the above figure. Here x and y indicates .......... and .......... colour \nrespectively.\n(a) red, blue\n(b) red, indigo\n(c) red, y ellow\n(d) violet, greenInstall NODIA App to See the Solutions.", + "sample question papers 3.txt\nrespectively.\n(a) red, blue\n(b) red, indigo\n(c) red, y ellow\n(d) violet, greenInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 16 P age 5\nhttps://qrbook.page.link/app\n15. The following figures show the path of light rays through three lenses marked L1, L2 and L3 and their focal points \nF1, F2 and F3 respectively.\nWhich of the following diagram shows the concave lens properties?\n(a) (i)\n(b) (ii)\n(c) (iii)\n(d) (i), (ii)\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 16 CBSE Science Class 10\nhttps://qrbook.page.link/app\n16. Which of the following phenomena contributes significantly to the reddish appearance of sun at sunrise or sunset?", + "Click Here To InstallPage 6 Sample P aper 16 CBSE Science Class 10\nhttps://qrbook.page.link/app\n16. Which of the following phenomena contributes significantly to the reddish appearance of sun at sunrise or sunset?\n(a) Disp ersion of light\n(b) Scattering of ligh t\n(c) T otal internal reflection of light\n(d) Reflection of ligh t from the earth\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.", + "sample question papers 3.txt\n(a) Disp ersion of light\n(b) Scattering of ligh t\n(c) T otal internal reflection of light\n(d) Reflection of ligh t from the earth\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : Carbon dioxide turns lime water milky.\nReason : Carbon dioxide sullies the water.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : Phototropism is caused by auxin.Reason : When light is coming from one side of the plant, auxin diffuses towards the shady side of the shoot.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).", + "sample question papers 3.txt\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : Haemoglobin is not the respiratory pigment in human beings.Reason : It transports oxygen in the human body.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Assertion is false but Reason is true.\n20. Assertion : A fuse wire is always connected in parallel with the mainline.Reason : If a current larger than the specified value flows through the circuit, fuse wire melts.\n(a) Both Assertion and Reason are true and Reason is correct explanation of the assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of the assertion.", + "sample question papers 3.txt\n(a) Both Assertion and Reason are true and Reason is correct explanation of the assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of the assertion.\n(c) Assertion is true but Reason is false.\n(d) Assertion is false but Reason is true.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 16 P age 7\nhttps://qrbook.page.link/app\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. When a metal X is treated with cold water, it gives a base Y with molecular formula XOH (Molecular mass \n40=) and liberates a gas Z which easily catches fire. Identify X, Y and Z and also write the reaction involved.\n\u0001o\u0001\nWhat are the constituents of solder alloy ? Which property of solder makes it suitable for welding electrical wires ?\n22. On touching a hot plate, you suddenly withdraw your hand. Which category of neurons became active first and \nwhich one next ?", + "sample question papers 3.txt\n22. On touching a hot plate, you suddenly withdraw your hand. Which category of neurons became active first and \nwhich one next ?\n23. During crossing why do new features which are not present in the parents appear in the offspring.\n24. Are the two cells formed in reproduction are identical?\n25. What are the properties of the image formed by a plane mirror ?\n\u0001o\u0001\nHow do we distinguish a medium to be rarer or denser?\n26. What would happen if all the decomposers were eliminated from the earth ? Explain.\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. Write the chemical formula and name of the compound which is the active ingredient of all alcoholic drinks. List \nits two uses. Write chemical equation and name of the product formed when this compound reacts with- \n(i) so dium metal\n(ii) hot concen trated sulphuric acid\n28. Study the figure given below and answer the following questions :\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 3.txt\n(ii) hot concen trated sulphuric acid\n28. Study the figure given below and answer the following questions :\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 16 CBSE Science Class 10\nhttps://qrbook.page.link/app\n(a) Name the pro cess depicted in the diagram.\n(b) W rite the composition of the anode and the cathode.\n(c) W rite the balanced chemical equation of the reaction taking place in this case.\n29. (a) What is the role of autosomes ?\n(b) Wh y is it that offspring receives traits from both the parents.\n\u0001o\u0001\nIn one of his experiments with pea plants, Mendel observed that when a pure tall pea plant is crossed with a pure \ndwarf pea plant in the first generation, F1 only tall plants appear.\n(a) What happ ens to the traits of the dwarf plants in this case?\n(b) When the F1 generation plants were self-fertilised, he observed that in the plants of second generation, F2", + "sample question papers 3.txt\n(a) What happ ens to the traits of the dwarf plants in this case?\n(b) When the F1 generation plants were self-fertilised, he observed that in the plants of second generation, F2 \nboth tall plants and dwarf plants were present. Why it happened? Explain briefly.\n30. The apparent altitude of stars appears to be generally more than their true altitudes. Explain, how.\n31. What are the causes of the following defects of vision and how can they be corrected ?\n(a) Cataract\n(b) Presbyopia.\n32. Calculate the total cost of running the following electrical devices in the month of September, if the rate of 1 unit of electricity is ` 6.00.\n(i) Electric heater of 1000 W for 5 hours daily .\n(ii) Electric refrigerator of 400 W for 10 hours daily .\n\u0001o\u0001\n(a) List the factors on whic h the resistance of a conductor in the shape of wire depends.\n(b) Wh y are metals good conductors of electricity whereas glass is a bad conductor of electricity? Give reason.", + "sample question papers 3.txt\n(b) Wh y are metals good conductors of electricity whereas glass is a bad conductor of electricity? Give reason.\n(c) Wh y are alloys commonly used in electrical heating devices? Give reason.\n33. Write one main difference between asexual and sexual mode of reproduction. Which species is likely to have \ncomparatively better chances of survival\u2014the one reproducing asexually or the one reproducing sexually? Give reasons to justify your answer.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 16 P age 9\nhttps://qrbook.page.link/app\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. (a) All ores are minerals but all minerals are not ores. Justify the statemen t.\n(b) What is galv anisation ?\n(c) Explain roasting with the help of a reaction.\n(d) What do y ou mean by amalgam ?\n\u0001o\u0001\nWhat are alloys ? How are they made ? Name the constituents and uses of brass, bronze and solder.", + "sample question papers 3.txt\n(c) Explain roasting with the help of a reaction.\n(d) What do y ou mean by amalgam ?\n\u0001o\u0001\nWhat are alloys ? How are they made ? Name the constituents and uses of brass, bronze and solder.\n35. What is \u2018phototropism\u2019? How does it occur in plants? Describe an activity to demonstrate phototropism.\n\u0001o\u0001\nName the parts C to G on the diagram of a sensory neuron given here. State two ways in which this neuron \ndiffers from a motor neuron.\n36. (a) Deriv e the relation for the equivalent resistance when three resistors of resistances R1, R2 and R3 are \nconnected in parallel.\n(b) Find the minim um resistance that can be made using four resistors, each of 20\u2126 .\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 16 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-E\n\u0001Question no.", + "\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 16 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.", + "sample question papers 3.txt\nhttps://qrbook.page.link/app\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37. Oxidation is the process of gaining of oxygen or losing of hydrogen. Reduction is the process of losing of oxygen or gaining of hydrogen. The substance which undergoes oxidation is the reducing agent while the substance which undergoes reduction is known as the oxidising agent. Oxidation and reduction always take place together and these type of reactions are known as redox reactions. Some of the examples of redox reactions are given below :\n(a) PbOH Cl PbCl Cl HO 83 4 34 22 2 $++ +\n(b) Mg OM gO 22 2$+\n(c) CuSO Zn Cu ZnSO 44$++\n(d) VO Ca VC aO 52 5 25$++\n(e) Fe HO FeOH 34 4 23 42$++\n(f) CuOH CuHO 22$++\n(i) Giv e two examples of oxidation reaction from your everyday life.", + "(ii) W rite the oxidising agent in the reaction a and b.\n\u0001o\u0001", + "sample question papers 3.txt\n(e) Fe HO FeOH 34 4 23 42$++\n(f) CuOH CuHO 22$++\n(i) Giv e two examples of oxidation reaction from your everyday life.\n(ii) W rite the oxidising agent in the reaction a and b.\n\u0001o\u0001\n(iii) Out of o xidation and reduction, which reaction takes place at anode?\n38. Answer given questions on the basis of your understanding of the following paragraph and the related studies \nconcepts.To make a bread dough, a baker mixes flour, sugar and baking powder (mixture of baking soda and tartaric acid). After mixing all the ingredients, the dough is placed in a container for a few hours (in an oven). On heating, the mixture releases carbon dioxide gas leaving bubbles behind. This increases the size of the bread and makes it soft and spongy. Tartaric acid helps in removing bitter taste.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "This increases the size of the bread and makes it soft and spongy. Tartaric acid helps in removing bitter taste.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 16 P age 11\nhttps://qrbook.page.link/app\n(i) Wh y does the bread dough rise?", + "sample question papers 3.txt\nClick Here To InstallCBSE Science Class 10 Sample P aper 16 P age 11\nhttps://qrbook.page.link/app\n(i) Wh y does the bread dough rise?\n(ii) \u2018Y east can be used in place of baking powder for making bread dough\u2019. What is yeast?\n(iii) What w ould you use to measure pH of baking powder?\n\u0001o\u0001\n(iv) Based on the graph represen ted alongside, answer the following questions:\n A bak ery shop started using baking soda instead of baking powder for baking cakes. What could be the \nreason for the decrease in the sale of cakes?\n39. White light is a mixture of seven colours i.e., violet, indigo, blue, green, yellow, orange and red. Every colour has \nits own characteristic wavelength. Different colours with their wavelengths are given below in the table.\nS. No. Colour Wavelength\n1. Red 7900 Ac\n2. Orange 6000 Ac\n3. Yellow 5800 Ac\n4. Green 5400 Ac\n5. Blue 4800 Ac\n6. Indigo 4500 Ac\n7. Violet 4000 Ac", + "sample question papers 3.txt\nS. No. Colour Wavelength\n1. Red 7900 Ac\n2. Orange 6000 Ac\n3. Yellow 5800 Ac\n4. Green 5400 Ac\n5. Blue 4800 Ac\n6. Indigo 4500 Ac\n7. Violet 4000 Ac\nThe phenomenon of splitting white light into seven colours when it passes through a glass prism is called dispersion \nof white light.\n(i) Name the phenomenon o ccurring in nature due to dispersion of light.\n(ii) What is mono chromatic light?\n(iii) Ligh t of two colours \u2018A\u2019 and \u2018B\u2019 pass through a glass prism. \u2018A\u2019 deviates more than \u2018B\u2019 from its path of \nincidence. Which colour has a higher speed in the prism ?\n\u0001o\u0001\n(iv) On whic h factor speed of light depends ?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 30.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 25 Page 1\nhttps://qrbook.page.link/app\nSample Paper 25\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho\nwever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv\ne Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac\nh. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each.", + "5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 30.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac\nh with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Which of the follo\nwing statements are correct in reference to the role of A (shown in the given diagram) during a \nbreathing cycle in human beings?\n(i) It helps to decrease the residual volume of air in lungs.\n(ii) It flattens as we inhale.\n(iii) It gets raised as we inhale.\n(iv) It helps the chest ca\nvity to become larger.\n(a) (iii) and (iv)\n(b) (ii) and (iv)\n(c) (i), (ii) and (iv)\n(d) (i) and (ii)Install NODIA App to See the Solutions.", + "(iv) It helps the chest ca\nvity to become larger.\n(a) (iii) and (iv)\n(b) (ii) and (iv)\n(c) (i), (ii) and (iv)\n(d) (i) and (ii)Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 25 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers 30.txt\n(b) (ii) and (iv)\n(c) (i), (ii) and (iv)\n(d) (i) and (ii)Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 25 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2. The direction of magnetic field around a straigh t conductor carrying current can be determined by:\n(a)\n Fleming\u2019s righ\nt hand rule\n(b)\n Righ\nt hand thumb rule\n(c)\n Lenz\u2019s la\nw\n(d)\n Fleming\u2019s left hand rule\n3.\n What is the correct direction of flo\nw of electrical impulses?\n4. F ollowing observations are observed when calcium oxide reacts vigorously with water.\nIdentify the incorrect observations.\n(i)\n It is an endothermic reaction.\n(ii)\n Slak\ned lime is produced.\n(iii)\n Quic\nk lime is produced.\n(iv)\n It is an exothermic reaction.\n(v)\n It is a com\nbination reaction.", + "Identify the incorrect observations.\n(i)\n It is an endothermic reaction.\n(ii)\n Slak\ned lime is produced.\n(iii)\n Quic\nk lime is produced.\n(iv)\n It is an exothermic reaction.\n(v)\n It is a com\nbination reaction.\n(a)\n (iii) and (iv)\n(b)\n (i) and (ii) \n(c)\n (ii), (iv) and (v)\n(d)\n (i) and (iii)\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 25 P age 3", + "sample question papers 30.txt\n(c)\n (ii), (iv) and (v)\n(d)\n (i) and (iii)\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 25 P age 3\nhttps://qrbook.page.link/app\n5. Tw o bulbs have the following ratings :\n1.\n 40 W, 220 V.\n2\n 20 W, 100 V.\nThe ratio of their resistance is :\n(a)\n 2 : 1\n(b)\n 1 : 2\n(c)\n 1 : 3\n(d)\n 1 : 1\n6.\n In whic\nh of the following setups would the bulb glow?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "In whic\nh of the following setups would the bulb glow?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 25 CBSE Science Class 10\nhttps://qrbook.page.link/app\n(a) (i) and (iv)\n(b)\n (i) and (ii)\n(c)\n (i), (ii) and (iv)\n(d)\n (ii), (iii) and (iv)\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 25 P age 5\nhttps://qrbook.page.link/app\n7. \nWhich of the following two combinations are correct?\nMetal Gas Evolved\n(i) Copper Yes\n(ii) Iron Yes\n(iii) Magnesium No", + "sample question papers 30.txt\nhttps://qrbook.page.link/app\n7. \nWhich of the following two combinations are correct?\nMetal Gas Evolved\n(i) Copper Yes\n(ii) Iron Yes\n(iii) Magnesium No\n(iv) Zinc Yes\n(a)\n i and iii\n(b)\n i and iv\n(c)\n ii and iii\n(d)\n ii and iv\n8.\n If t\nwo parents have the genotypes AA # aa, the probability of having an aa genotype in the F1 generation is:\n(a)\n 50 p\ner cent\n(b)\n 25 p\ner cent\n(c)\n None of these\n(d)\n 75 p\ner cent\n9.\n Ankit observ\ned that the stain of curry on a white shirt becomes reddish-brown when soap is scrubbed on it,.but it \nturns yellow again when the shirt is washed with plenty of water. What might be the reason for his observation?\n(i)\n Soap is acidic in nature.\n(ii)\n Soap is basic in nature.\n(iii)\n T\nurmeric is a natural indicator which gives reddish tinge in bases.\n(iv)\n T\nurmeric is a natural indicator which gives reddish tinge in acids.", + "What might be the reason for his observation?\n(i)\n Soap is acidic in nature.\n(ii)\n Soap is basic in nature.\n(iii)\n T\nurmeric is a natural indicator which gives reddish tinge in bases.\n(iv)\n T\nurmeric is a natural indicator which gives reddish tinge in acids.\n(a)\n (ii) and (iii)\n(b)\n (i) and (ii)\n(c)\n (ii) and (iv)\n(d)\n (i) and (iv)\n10.\n Galv", + "sample question papers 30.txt\n(iv)\n T\nurmeric is a natural indicator which gives reddish tinge in acids.\n(a)\n (ii) and (iii)\n(b)\n (i) and (ii)\n(c)\n (ii) and (iv)\n(d)\n (i) and (iv)\n10.\n Galv\nanisation is a method of protecting iron from rusting by coating with a thin layer of:\n(a)", + "sample question papers 30.txt\naluminium\n(b)\n \ngallium\n(c)\n \nsilver\n(d)", + "sample question papers 30.txt\nzincInstall NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 25 CBSE Science Class 10\nhttps://qrbook.page.link/app\n11. Whic h option correctly lists the changes that occur in females during puberty?\n(a)\n repro\nductive organs enlarge, size of the breasts increases, thick hairs grow on the body\n(b)\n thic\nk hairs grow on face, cracking of voice, enlargement of reproductive organ\n(c)\n size of the breasts increases, b\neginning of menstruation, thick hairs grow on the body\n(d)\n thin hairs gro\nwth occurs on the body, size of the breasts increases, pitch of the voice increases\n12.\n Whic\nh of the following hydrocarbons represent the isomer of Butene?\n13. Observ e the diagram of an activity given below. What does it help to conclude, when the person exhales into the \ntest-tube?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "13. Observ e the diagram of an activity given below. What does it help to conclude, when the person exhales into the \ntest-tube?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 25 P age 7\nhttps://qrbook.page.link/app", + "sample question papers 30.txt\ntest-tube?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 25 P age 7\nhttps://qrbook.page.link/app\n(a) F ermentation occurs in the presence of oxygen\n(b)\n P\nercentage of carbon dioxide is more in inhaled air. \n(c)\n F\nermentation occurs in the presence of carbon dioxide.\n(d)\n P\nercentage of carbon dioxide is more in the exhaled air. \n14.\n Equal \nvolumes of hydrochloric acid and sodium hydroxide solutions of same concentration are mixed and the pH \nof the resulting solution is checked with a pH paper. What would be the colour obtained?\n(a)", + "sample question papers 30.txt\nRed\n(b)\n \nBlue\n(c)", + "sample question papers 30.txt\nOrange\n(d)\n Y\nellowish green\n15.\n The magnetic field lines of solenoid are similar to the magnetic field lines of bar magnet. Whic\nh image correctly \nshows the solenoid as a bar magnet?\n16. The curren t flowing through a wire of resistance 2\u2126 varies with time as shown in figure below. The amount of \nheat produced (in J) in 3 s would be: \n(a) 18 J\n(b)\n 2 J\n(c)\n 10 J\n(d)\n 28 J\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 25 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17.\n Assertion (A):\n Accumulation of variation in a species increases the chances of its survival in changing environment.", + "17 to 20 are Assertion-Reasoning based questions.\n17.\n Assertion (A):\n Accumulation of variation in a species increases the chances of its survival in changing environment.\nReason (R): Accumulation of heat resistance in some bacteria ensure their survival even when temperature in \nenvironment rises too much.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)", + "sample question papers 30.txt\nenvironment rises too much.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n18.\n Assertion (A):\n A solenoid tends to expand, when current passes through it.\nReason (R): Two straight parallel metallic wires carrying current in same direction repel each other.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n19.\n Assertion (A):\n It is necessary to separate oxygenated and de-oxygenated blood in mammals and birds.", + "sample question papers 30.txt\n(d)\n Assertion (A) is false but Reason (R) is true.\n19.\n Assertion (A):\n It is necessary to separate oxygenated and de-oxygenated blood in mammals and birds.\nReason (R): Mammals and birds are warm blooded animals and they depend on environment for their body temperature regulation.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n20.\n Assertion (A):\n Decomposition of vegetable matter into compost is an endothermic reaction.\nReason (R): Decomposition reaction involves breakdown of a single reactant into simpler products.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)", + "sample question papers 30.txt\nBoth Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21.\n Ho\nw can change of size of eyeball be one of the reason for:\n(i)", + "sample question papers 30.txt\nMyopia\n(ii)", + "sample question papers 30.txt\nHypermetropia\nCompare the size of eyeball with that of a normal eye in each case. How does this change of size affect the position \nof image in each case?\n\u0001o\u0001\nState the cause of dispersion of white light by a glass prism. Draw a labelled diagram to illustrate the recombination \nof the spectrum of white light. Why is it essential that the two prisms used for the purpose should be identical and placed in an inverted position with respect to each other?Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 25 P age 9\nhttps://qrbook.page.link/app\n22. Ho w is O2 and CO2 transported in human beings?\n23.\n DDT w\nas sprayed in a lake to regulate breeding of mosquitoes. How would it affect the trophic levels in the \nfollowing food chain associated with a lake? Justify your answer.\nHAWK\n.\nLARGE FISH\n.\nSMALL FISH\n.\nPLANKTON\n.\nAQUATIC ENVIRONMENT\n24.\n List t\nwo different functions performed by pancreas in our body.", + "Justify your answer.\nHAWK\n.\nLARGE FISH\n.\nSMALL FISH\n.\nPLANKTON\n.\nAQUATIC ENVIRONMENT\n24.\n List t\nwo different functions performed by pancreas in our body.\n25.\n A green salt on heating decomp", + "sample question papers 30.txt\nHAWK\n.\nLARGE FISH\n.\nSMALL FISH\n.\nPLANKTON\n.\nAQUATIC ENVIRONMENT\n24.\n List t\nwo different functions performed by pancreas in our body.\n25.\n A green salt on heating decomp\noses to produce a colourless suffocating gas and Leaves behind a reddish brown \nresidue. Name the salt and write the decomposition reaction.\n\u0001o\u0001\nGive suitable reason for the following statements:\n(i)\n W\ne feel burning sensation in the stomach when we overeat.\n(ii)\n The crystals of w\nashing soda change to white powder on exposure to air.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 25 CBSE Science Class 10\nhttps://qrbook.page.link/app\n26. Lab el the parts of a neuron in the given figure:\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27.", + "Lab el the parts of a neuron in the given figure:\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27.\n (i)\n Dra\nw ray diagram to show the principal focus of \n(a)\n a conca\nve mirror, and\n(b)\n a con\nvex mirror.\n(ii)\n In the follo", + "sample question papers 30.txt\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27.\n (i)\n Dra\nw ray diagram to show the principal focus of \n(a)\n a conca\nve mirror, and\n(b)\n a con\nvex mirror.\n(ii)\n In the follo\nwing diagram, MM is a concave mirror and AB is an object. Draw on your answer sheet a ray \ndiagram to show the formation of image of this object.\n28. (i) W rite the essential condition for the following reaction to take place:\n AgBrA gB r 22 2 $ +\n W rite one application of this reaction.\n(ii)\n Complete the follo\nwing chemical equation of a chemical reaction:\n .......... .......... FeSO Fe O 2Heat\n42 3++\n(iii) What happ ens when water is added to quick lime? Write chemical equation.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.", + "Write chemical equation.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 25 P age 11\nhttps://qrbook.page.link/app\n29. (i) Name the p oles P, Q, R and S of the magnets in the following figures a and b.", + "sample question papers 30.txt\nClick Here To InstallCBSE Science Class 10 Sample P aper 25 P age 11\nhttps://qrbook.page.link/app\n29. (i) Name the p oles P, Q, R and S of the magnets in the following figures a and b.\n(ii) State the inference dra wn about the direction of the magnetic field lines on the basis of these diagrams.\n\u0001o\u0001\nState right-hand thumb rule to determine the direction of magnetic field around a current carrying conductor. \nApply this rule to find the direction of magnetic field inside and outside a circular loop of wire lying in the plane of a table and current is flowing through it clockwise.\n30.\n The table b\nelow shows the colour of universal indicator paper (UI paper) at different pH values.\n(i) UI pap er turns purple in oven cleaner solution. What is the pH of oven cleaner solution?\n(ii)\n Suggest the substance in o\nven cleaner solution that turns UI paper purple.\n(iii)\n UI pap\ner turns yellowish-green in milk. What is the pH of milk?\n(iv)\n The milk w", + "sample question papers 30.txt\n(ii)\n Suggest the substance in o\nven cleaner solution that turns UI paper purple.\n(iii)\n UI pap\ner turns yellowish-green in milk. What is the pH of milk?\n(iv)\n The milk w\nas left outside for five days. When the milk was re-tested with UI paper, the paper turned orange. \nWhat has happened to the milk?\n31.\n Ho\nw can we help in reducing the problem of waste disposal? Suggest any three methods.\n32.\n Describ\ne the structure and function of the basic filtering unit of kidney.\n\u0001o\u0001\nHow does nutrition take place in Amoeba? How is it different in Paramoecium?\n33.\n A studen\nt wants to project the image of a candle flame on a screen 80 cm in front of a mirror by keeping the candle \nflame at a distance of 20 cm from its pole.\n(i)\n Whic\nh type of mirror should the student use?\n(ii)\n Find the magnification of the image pro\nduced.\n(iii)\n Find the distance b\netween the object and its image.", + "(i)\n Whic\nh type of mirror should the student use?\n(ii)\n Find the magnification of the image pro\nduced.\n(iii)\n Find the distance b\netween the object and its image.\n(iv)\n Dra\nw a ray diagram to show the image formation in this case and mark the distance between the object and", + "sample question papers 30.txt\nduced.\n(iii)\n Find the distance b\netween the object and its image.\n(iv)\n Dra\nw a ray diagram to show the image formation in this case and mark the distance between the object and \nits image.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 12 Sample P aper 25 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34.\n Based on the giv\nen diagram, answer the questions given below:\n(i) Lab el the parts A, B, C and D.\n(ii)\n Name the hormones secreted b\ny testis and mention its role.\n(iii)\n State the functions of B and C in the pro\ncess of reproduction.\n\u0001o\u0001\n(i)\n Dra\nw a diagram showing germination of pollen on stigma of a flower and mark on it the following organs/\nparts:\n(a)\n P\nollen grain\n(b)\n P\nollen tube\n(c)", + "sample question papers 30.txt\nStigma\n(d)\n F\nemale germ-cell\n(ii)\n State the significance of p\nollen tube.\n(iii)\n Name the parts of flo\nwer that develop after fertilisation into:\n(a)\n \nSeed\n(b)", + "sample question papers 30.txt\nFruit\n35.\n (i)\n Ho\nw will you infer with the help of an experiment that the same current flows through every part of the \ncircuit containing three resistors R1, R2 and R3 in series connected to a battery of V volts?\n(ii)\n Study the follo\nwing circuit and find out:\n(a)\n Curren\nt in 12\u2126 resistor.\n(b)\n Difference in the readings of Al and A2, if any.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 25 P age 13\nhttps://qrbook.page.link/app\n36. Soaps and detergen ts are both types of salts. State the difference between the two. Write the mechanism of the \ncleansing action of soaps. Why do soaps not form lather (foam) with hard water? Mention any two problems that \narise due to the use of detergents instead of soaps.", + "State the difference between the two. Write the mechanism of the \ncleansing action of soaps. Why do soaps not form lather (foam) with hard water? Mention any two problems that \narise due to the use of detergents instead of soaps.\n\u0001o\u0001\nA compound A (C2H4O2) reacts with Na metal to form a compound \u2018B\u2019 and evolves a gas which burns with a pop", + "sample question papers 30.txt\narise due to the use of detergents instead of soaps.\n\u0001o\u0001\nA compound A (C2H4O2) reacts with Na metal to form a compound \u2018B\u2019 and evolves a gas which burns with a pop \nsound. Compound \u2018A\u2019 on treatment with an alcohol \u2018C\u2019 in presence of an acid forms a sweet smelling compound \u2018D\u2019 \n(C4H8O2). On addition of NaOH to \u2018D\u2019 gives back B and C. Identify A, B, C and D. Write the reactions involved.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.\n37.\n The mo\ndes by which various organisms reproduce depend on the body design of the organisms. In asexual \nreproduction, a single individual parent produces offsprings without the involvement of gametes. This method is \na common means of increasing the offsprings rapidly under favourable conditions. Asexual reproduction occurs mostly in unicellular organisms, some plants and certain simple multicellular animals.(i)", + "sample question papers 30.txt\na common means of increasing the offsprings rapidly under favourable conditions. Asexual reproduction occurs mostly in unicellular organisms, some plants and certain simple multicellular animals.(i)\n State the name of the organism in whic\nh binary fission takes place in a definite orientation. Also name the \ndisease caused by this organism.\n(ii)\n Lea\nves of \u2018Bryophyllum\u2019 when they fall on the soil develop into new plants whereas a banana leaf will not \nbe able to do so. Why?\n(iii)\n Explain the pro\ncess of budding in Hydra.\n\u0001o\u0001\n(iv)\n What happ\nens when:\n(a)\n a spirogyra filamen\nt matures and attains a considerable length and\n(b)\n a sp\norangia in Rhizopus bursts on maturation?\n38.\n Electrolysis of w\nater is a popular method used for different applications in various industries. The electrolysis of \nwater is mainly carried out to yield pure hydrogen and oxygen gases. It involves passing an electric current through", + "sample question papers 30.txt\nwater is mainly carried out to yield pure hydrogen and oxygen gases. It involves passing an electric current through \nthe water which results in decomposition of water into hydrogen and oxygen.Pure water is a poor conductor of electricity. Sulphuric acid is added to the water so that the conductance of water increases which makes the reaction faster. The setup for electrolysis of water is given below:\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 14 Sample P aper 25 CBSE Science Class 10\nhttps://qrbook.page.link/app\nThe number of hydrogen molecules produced in electrolysis is twice the number of oxygen molecules. Also, \nhydrogen is double in volume than oxygen.(i)\n Name the gases ev\nolved at cathode and anode respectively. Why is volume of one gas collected at one \nelectrode is double of anode?\n(ii)\n Ho\nw will you test the gas evolved at cathode and at anode?\n\u0001o\u0001\n(iv)\n W\nrite the chemical equation for electrolysis of water.", + "sample question papers 30.txt\nelectrode is double of anode?\n(ii)\n Ho\nw will you test the gas evolved at cathode and at anode?\n\u0001o\u0001\n(iv)\n W\nrite the chemical equation for electrolysis of water.\n Wh\ny are few drops of H2SO4 added to pure water?\n39.\n Adit\nya and his friend Manoj placed a candle flame in front of a convex lens at various distances from it and \nobtained the image of the candle flame on a white screen.\nHe noted down the position of the candle, screen and the lens as underPosition of candle \n20= cm \nPosition of convex lens 50= cm \nPosition of the screen 80= cm\n(i) What is the p osition of the image formed from the convex Lens?\n(ii)\n What is the fo\ncal length of the convex tens? \n\u0001o\u0001\nWhere will the image be formed, if he shifts the candle towards the lens at a position of 35 cm?\n(iii)\n What is the nature of the image formed if Adit\nya shifts the candle towards the lens to 36m?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 24.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 12 Page 1\nhttps://qrbook.page.link/app\nSample Paper 12\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 24.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Sodium hydrogen carbonate when added to acetic acid evolves a gas. Which of the following statements are true \nabout the gas evolved?\n1. It turns lime water milky .\n2. It extinguishes a burnings splinter.\n3. it dissolves in a solution of so dium hydroxide.\n4. It has a pungent o dour.\n(a) 1 and 2\n(b) 1, 2 and 3\n(c) 2, 3 and 4\n(d) 1 and 4Install NODIA App to See the Solutions.", + "4. It has a pungent o dour.\n(a) 1 and 2\n(b) 1, 2 and 3\n(c) 2, 3 and 4\n(d) 1 and 4Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 12 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2. \nA shiny brown coloured element X on heating in air becomes black in colour as shown in the figure. Here X is:", + "sample question papers 24.txt\nhttps://qrbook.page.link/app\n2. \nA shiny brown coloured element X on heating in air becomes black in colour as shown in the figure. Here X is:\n(a) Copper\n(b) Silver\n(c) Aluminium\n(d) Mercury\n3. Which of the following statements about the given reaction are correct?\n() () () () FesH Og FeOs Hg 34 4 23 42$ ++\n1. Iron metal is getting o xidised.\n2. W ater is getting reduced.\n3. W ater is acting as reducing agent.\n4. W ater is acting as oxidising agent.\n(a) 1, 2 and 3\n(b) 3 and 4\n(c) 1, 2 and 4\n(d) 2 and 4\n4. Which one of the following involve a chemical reaction?", + "4. W ater is acting as oxidising agent.\n(a) 1, 2 and 3\n(b) 3 and 4\n(c) 1, 2 and 4\n(d) 2 and 4\n4. Which one of the following involve a chemical reaction?\n(a) Heating magnesium wire in the presence of air at high temp erature\n(b) Ev aporation of water\n(c) Storing on nitrogen gas under pressure\n(d) Keeping p etrol in a China dish in open\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 12 P age 3\nhttps://qrbook.page.link/app\n5.", + "sample question papers 24.txt\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 12 P age 3\nhttps://qrbook.page.link/app\n5. \nIf copper is kept open in air, it slowly loses its shining brown surface and gains a green coating as shown in the \nfigure. It is due to the formation of\n(a) CuSO 4\n(b) CuCO 3\n(c) ()Cu NO 32\n(d) CuO\n6. \nCalcium phosphate is present in tooth enamel. Its nature is\n(a) basic\n(b) acidic\n(c) neutral\n(d) amphoteric\n7. Carbon exists in the atmosphere in the form of\n(a) Carb on monoxide only\n(b) Carb on monoxide in traces and carbon dioxide\n(c) Carb on dioxide only\n(d) Coal\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "Click Here To InstallPage 4 Sample P aper 12 CBSE Science Class 10\nhttps://qrbook.page.link/app\n8. Which of the following is the correct route for blood flow in a human?\n(a) Righ t atrium \" Right ventricle \" Lungs \" Left atrium \" Left ventricle", + "sample question papers 24.txt\nhttps://qrbook.page.link/app\n8. Which of the following is the correct route for blood flow in a human?\n(a) Righ t atrium \" Right ventricle \" Lungs \" Left atrium \" Left ventricle\n(b) Righ t atrium \" Right ventricle \" Left ventricle \" Left atrium \" Lungs\n(c) Left atrium \" Left ventricle \" Right ventricle \" Right atrium \" Lungs\n(d) Left atrium \" Left ventricle \" Lungs \" Right ventricle \" Right atrium\n9. Just as CO 2 is removed from the blood in the lungs, nitrogenous waste such as urea or uric acid are removed from \nblood in the -\n(a) Kidney\n(b) Urinary bladder \n(c) Urethra\n(d) Ureters\n10. A zygote which has an X-chromosome inherited from the father will develop into a\n(a) boy\n(b) girl\n(c) X-c hromosome does not determine the sex of a child\n(d) either b oy or girl\n11.", + "A zygote which has an X-chromosome inherited from the father will develop into a\n(a) boy\n(b) girl\n(c) X-c hromosome does not determine the sex of a child\n(d) either b oy or girl\n11. Dramatic changes of body features associated with puberty are mainly because of secretion of\n(a) estrogen from testes and testosterone from o vary\n(b) estrogen from adrenal gland and testosterone from pituitary gland", + "sample question papers 24.txt\n(a) estrogen from testes and testosterone from o vary\n(b) estrogen from adrenal gland and testosterone from pituitary gland\n(c) testosterone from testes and estrogen from o vary\n(d) testosterone from th yroid gland and estrogen from pituitary gland\n12. Offspring formed as a result of sexual reproduction exhibits more variations because\n(a) sexual repro duction is a lengthy process\n(b) genetic material comes from t wo parents of the same species\n(c) genetic material comes from t wo parents of different species\n(d) genetic material comes from man y parents\n13. Two thin lenses of power .D35+ and .D25- are placed in contact. The power of the lens combination is-\n(a) D 1+\n(b) .D15+\n(c) .D25+\n(d) D 2+\n14. In given figure, a light ray AB is incident normally on one face PQ of an equilateral glass prism.\nThe angles at faces PR is:\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "In given figure, a light ray AB is incident normally on one face PQ of an equilateral glass prism.\nThe angles at faces PR is:\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 12 P age 5", + "sample question papers 24.txt\nThe angles at faces PR is:\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 12 P age 5\nhttps://qrbook.page.link/app\n(a) 60c\n(b) 30c\n(c) 45c\n(d) 90c\n15. A ray of light is refracted as per the following diagram. Which of the following medium is optically denser?\n(a) Medium A\n(b) Medium B\n(c) Cannot b e identify\n(d) Both medium are denser\n16. Which of the following phenomena of light are involved in the formation of a rainbow?\n(a) Reflection, refraction and disp ersion.\n(b) Reflection, disp ersion and total internal reflection.\n(c) Refraction, disp ersion and internal reflection.\n(d) Disp ersion, scattering and total internal reflection.\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : Decomposition reactions are similar to combination reactions.\nReason : Both reactions need a catalyst to occur.", + "sample question papers 24.txt\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : Decomposition reactions are similar to combination reactions.\nReason : Both reactions need a catalyst to occur.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : Reflex actions are automatic and repid responses to stimuli.Reason : These actions are controlled by brain.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.Install NODIA App to See the Solutions.", + "sample question papers 24.txt\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 12 CBSE Science Class 10\nhttps://qrbook.page.link/app\n19. Assertion : Amoeba is not an omnivore organism.\nReason : Lion is a carnivore organism.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Assertion is false but Reason is true.\n20. Assertion : The connecting wires are made of copper.Reason : The electrical conductivity of copper is high.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.", + "sample question papers 24.txt\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. Tooth enamel is one of the hardest substances in our body. How does it undergo damage due to eating chocolates and sweets ? How do tooth pastes prevent this damage ?\n\u0001o\u0001\nThree solutions \nA, B and C have pH values of 6, 2 and 10 respectively. Which of the solutions is highly acidic? \nWhich solution will turn red litmus blue ?\n22. Where does cerebrospinal fluid occur in our body ? Mention its function.\n23. What is a sex chromosome ?\n24. What is the basic event in reproduction?\n25. Identify the nature of the mirror and mention two characteristics of the image formed when magnification \n()m 6=+.", + "Mention its function.\n23. What is a sex chromosome ?\n24. What is the basic event in reproduction?\n25. Identify the nature of the mirror and mention two characteristics of the image formed when magnification \n()m 6=+.\n\u0001o\u0001\nWhere should an object be placed in front of a concave mirror of focal length 20 cm so as to obtain a two times", + "sample question papers 24.txt\n()m 6=+.\n\u0001o\u0001\nWhere should an object be placed in front of a concave mirror of focal length 20 cm so as to obtain a two times \nmagnified real image ?\n26. In a certain study conducted on occurrence of DDT along food chains in an ecosystem, the concentration of DDT in grass was found to be 0.5 ppm (parts per million), in sheep it was 2 ppm and in man it was 10 ppm. Why was the concentration of DDT maximum in case of man ?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 12 P age 7\nhttps://qrbook.page.link/app\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. How does the activity series of metals help us in predicting the relative reactivity of metals ?\n28. Name three chemical reactions in which heat is evolved and three chemical reactions in which heat is absorbed.\n29.", + "27 to 33 are short answer questions.\n27. How does the activity series of metals help us in predicting the relative reactivity of metals ?\n28. Name three chemical reactions in which heat is evolved and three chemical reactions in which heat is absorbed.\n29. Explain the feedback mechanism to regulate the action of the hormones with the help of one suitable example.\n\u0001o\u0001", + "sample question papers 24.txt\n29. Explain the feedback mechanism to regulate the action of the hormones with the help of one suitable example.\n\u0001o\u0001\nNervous and hormonal system together per the function of control and coordination in human beings. Justify the \nstatement.\n30. State the cause of dispersion of white light by a glass prism. How did Newton using two identical glass prism, show that white light is made of seven colours ? Draw a ray diagram to show the path of a narrow beam of while light through a combination of two identical prisms arranged together in inverted position with respect to each other when it is allowed to fall obliquely on one of the faces of the first prism of the combination ?\n31. Due to gradual weakening of ciliary muscles and diminishing flexibility of the eye lens a certain defect of vision arises. Write the name of this defect. Name the type of lens required by such persons to improve the vision. Explain the structure and function of such a lens.", + "sample question papers 24.txt\n32. A hot plate of an electric oven, connected to a 200 V line. It has two resistance coils A and B each of the 30\u2126 \nwhich may be used separately, in series or in parallel. Find the value of the current required in each of the three cases.\n\u0001o\u0001\nState the formula co-relating the electric current flowing in a conductor and the voltage applied across it. Also \nshow this relationship by drawing a graph.What would be the resistance of a conductor if the current flowing through it is 0.35 ampere when the potential difference across it is 1.4 volt ?\n33. (a) What are decomp osers ?\n(b) State in brief the role of decomp osers in the environment.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. Account for the following\n(a) State the relation b etween hydrogen ion concentration of an aqueous solution and its pH.\n(b) An aqueous solution has a pH v alue of 7.0. Is this solution acidic, basic or neutral.\n(c) Whic h has a higher pH value, 1 MHCl or 1 M NaOH solution ?", + "sample question papers 24.txt\n(b) An aqueous solution has a pH v alue of 7.0. Is this solution acidic, basic or neutral.\n(c) Whic h has a higher pH value, 1 MHCl or 1 M NaOH solution ?\n(d) T ooth enamel is one of the hardest substances in our body. How does it undergo damage due to eating \nchocolates and sweets ? What should we do to prevent it.\n(e) Ho w doe []H+ ions exist in water ?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 12 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001o\u0001\n(a) A liquid has a pH less than 7 whic h represents an acidic solution.\n(i) State the nature of solution, if its pH increases from 7 to 14.\n(ii) Men tion the ion whose concentration increases with the increase in pH value.\n(iii) Suggest a metho d that is generally used for measuring pH value.", + "(i) State the nature of solution, if its pH increases from 7 to 14.\n(ii) Men tion the ion whose concentration increases with the increase in pH value.\n(iii) Suggest a metho d that is generally used for measuring pH value.\n(b) Giv e reason for the following :\n(i) T ooth decay starts when the pH of the mouth is lower than 5.5.\n(ii) An tacids are used for treatment of indigestion.", + "sample question papers 24.txt\n(b) Giv e reason for the following :\n(i) T ooth decay starts when the pH of the mouth is lower than 5.5.\n(ii) An tacids are used for treatment of indigestion.\n35. Explain with the help of diagram. Also, indicate what happens to the rate of photosynthesis if stomata get blocked \ndue to dust.\n\u0001o\u0001\nWhy is respiration important for all living organisms?\n36. Briefly explain an activity to plot the magnetic field lines around a bar magnet. Sketch the field pattern for the \nsame specifying field directions.A region \nA has magnetic field lines relatively closer than another region B. Which region has stronger magnetic \nfield ? Give reason to support your answer.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data -based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.", + "sample question papers 24.txt\nfield ? Give reason to support your answer.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data -based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.\n37. When two or more substances react and form some new substance, it is called a chemical reaction. Chemical equation is represented in terms of symbols, molecular formulas, moles, states, etc. As we know, all chemical reaction obeys law of chemical combination. Therefore, chemical reactions need to be balanced. It is done by hit and trial method. The chemical reactions can be classified into different types such as combination reaction, decomposition reaction, displacement reaction, double displacement reaction. The reactions take place in solution is precipitation reactions and neutralisation reactions.\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 12 P age 9\nhttps://qrbook.page.link/app\n(i) Define a c hemical equation.", + "sample question papers 24.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 12 P age 9\nhttps://qrbook.page.link/app\n(i) Define a c hemical equation.\n(ii) Whic h law is followed by all chemical reactions?\n(iii) Name four t ypes of chemical reactions.\n\u0001o\u0001\n(iv) Giv e example of precipitation reactions.\n38. Reproduction in human beings is by sexual reproduction where both the male and female gametes fertilise to give \nrise to an embryo. The fertilization of the human embryo occurs inside the body of the female.\n(i) Name the part of the male repro ductive system where the formation of sperms takes place.\n(ii) What is the placen ta?\n(iii) What is the other name of the o viduct?\n\u0001o\u0001\n(iv) Define the term implan tation.\n39. A near sighted person wears eye glass with power of .50- D for distant vision. Soon, he started having difficulties \nin viewing nearby objects also. His doctor prescribes a correction of .15+ D in near vision section of his bi-focal,", + "sample question papers 24.txt\nin viewing nearby objects also. His doctor prescribes a correction of .15+ D in near vision section of his bi-focal, \nwhich is measured relative to main part of the lens.\n(i) Find the fo cal length of his distant viewing part of lens.\n(ii) Find the fo cal length of near vision section of the lens.\n(iii) What t ype of lens is to be used in this spectacles for near vision ?\n\u0001o\u0001\n(iv) What is the reason of h ypermetropia ?Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 12 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers17.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 24 Page 1\nhttps://qrbook.page.link/app\nSample Paper 24\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho\nwever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv\ne Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac\nh. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each.", + "5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers17.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac\nh with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Magnetic field is produced b\ny the flow of current in a straight wire. The phenomenon was discovered by:\n(a) Fleming\n(b) F\naraday\n(c) Oersted\n(d) Maxw\nell\n2. Why is it imp\nortant to balance a skeletal chemical equation?\n(a) To v\nerify the law of constant proportion.\n(b) To v\nerify law of conservation of energy. \n(c) To v\nerify the law of conservation of momentum.\n(d) To v\nerify the law of conservation of mass. \n3.", + "(a) To v\nerify the law of constant proportion.\n(b) To v\nerify law of conservation of energy. \n(c) To v\nerify the law of conservation of momentum.\n(d) To v\nerify the law of conservation of mass. \n3. Respiratory structures of tw\no different animals a fish and a human being are as shown.\nObserve (a) and (b) and select one characteristic that holds true for both of them.", + "sample question papers17.txt\n3. Respiratory structures of tw\no different animals a fish and a human being are as shown.\nObserve (a) and (b) and select one characteristic that holds true for both of them.\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 24 CBSE Science Class 10\nhttps://qrbook.page.link/app\n(a) Both ha ve thin and moist surface for gaseous exchange.\n(b)\n Both are placed in\nternally in the body of animal. \n(c)\n In b\noth the blood returns to the heart after being oxygenated.\n(d)\n Both are p\noorly supplied with blood vessels to conserve energy. \n4.\n A cross b\netween a tall plant (TT) and short pea plant (tt) resulted in progeny that were all tall plants because:\n(a)\n shortness is the dominan\nt trait.\n(b)\n tallness is the dominan\nt trait. \n(c)\n heigh\nt of pea plant is not governed by gene T or t.", + "(b)\n tallness is the dominan\nt trait. \n(c)\n heigh\nt of pea plant is not governed by gene T or t.\n(d)\n tallness is the recessiv\ne trait. \n5.\n Marble statues are corro", + "sample question papers17.txt\nt trait.\n(b)\n tallness is the dominan\nt trait. \n(c)\n heigh\nt of pea plant is not governed by gene T or t.\n(d)\n tallness is the recessiv\ne trait. \n5.\n Marble statues are corro\nded or stained when they repeatedly come into contact with polluted rain water. Identify \nthe main reason.\n(a) P olluted water is basic in nature hence it reacts with calcium carbonate.\n(b)\n Decomp\nosition of calcium carbonate to calcium oxide. \n(c)\n Calcium carb\nonate dissolves in water to give calcium hydroxide.\n(d)\n P\nolluted water is acidic in nature hence it reacts with calcium carbonate. \n6.\n The table pro\nvides the pH value of four solutions P, Q, R and S :\nSolution pH value\nP 2\nQ 9\nR 5\nS 11\nWhich of the following correctly represents the solutions in increasing order of their hydronium ion concentration?\n(a)", + "sample question papers17.txt\nP>S>Q>R\n(b)\n \nP>Q>R>S\n(c)\n \nS^h and when travels \nfrom a denser medium to a rarer one. it bends away from the normal ir<^h .\nWhere, i = Angle of incidence", + "sample question papers 15.txt\nfrom a denser medium to a rarer one. it bends away from the normal ir<^h .\nWhere, i = Angle of incidence\nand r = Angle of refraction\nWe can see refraction in our daily life, some of the examples are given below :\nThe bottom of a tank or pond containing water appears to be raised due to refraction of light which takes place when light rays pass from the pool of water into the air. The letters appear to be raised when viewed through a glass slab placed over the document because of refraction of light.When a light ray enters in a glass slab, then the emergent ray is parallel to the incident ray but it is shifted sideward slightly.In this case, refraction takes place twice, first when ray enters glass slab from air and second when exits from glass slab to air.\n(i) What do y ou mean by optically rarer and denser medium ?\n(ii) What is the cause of refraction ?\n(iii) Dra w a ray diagram showing refraction through a glass slab.\n\u0001o\u0001", + "sample question papers 15.txt\n(i) What do y ou mean by optically rarer and denser medium ?\n(ii) What is the cause of refraction ?\n(iii) Dra w a ray diagram showing refraction through a glass slab.\n\u0001o\u0001\n(iv) Giv e one example of refraction from our daily life experience other than the two examples given above.Install NODIA App to See the Solutions.\nClick Here To InstallPage 12 Sample P aper 5 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 24 (2).txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 6 Page 1\nhttps://qrbook.page.link/app\nSample Paper 6\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each.", + "5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 24 (2).txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Hard water is not available for an experiment in the school and its vicinity. However, some salts as given below \nare available in the school laboratory.\n1. Sodium Chloride\n2. Sodium Sulphate\n3. Calcium Chloride\n4. Calcium Sulphate\n5. Potassium Chloride\n6. Magnesium Sulphate\nSelect form the following a group of these salts, each member of which may be dissolved in water to make it hard.", + "Sodium Sulphate\n3. Calcium Chloride\n4. Calcium Sulphate\n5. Potassium Chloride\n6. Magnesium Sulphate\nSelect form the following a group of these salts, each member of which may be dissolved in water to make it hard.\n(a) 1, 2, 5\n(b) 1, 3, 5\n(c) 3, 4, 6\n(d) 2, 4, 6\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 6 CBSE Science Class 10", + "sample question papers 24 (2).txt\n(a) 1, 2, 5\n(b) 1, 3, 5\n(c) 3, 4, 6\n(d) 2, 4, 6\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 6 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2. \nWhich of the following gases formed at the anode and the cathode?\n(a) Chlorine, Hydrogen\n(b) Hydrogen, Chlorine\n(c) Hydrogen, Hydrogen\n(d) Chlorine, Chlorine\n3. In the double displacement reaction between aqueous potassium iodide and aqueous lead nitrate, a yellow \nprecipitate of lead iodide is formed. While performing the activity if lead nitrate is not available, which of the following can be used in place of lead nitrate?\n(a) Lead sulphate (insoluble)\n(b) Lead acetate\n(c) Ammonium nitrate\n(d) P otassium sulphate\n4.", + "While performing the activity if lead nitrate is not available, which of the following can be used in place of lead nitrate?\n(a) Lead sulphate (insoluble)\n(b) Lead acetate\n(c) Ammonium nitrate\n(d) P otassium sulphate\n4. Which of the following is the observations of the chemical reaction?\n1. Change in state\n2. Ev olution of a gas\n3. Change in colour\n4. Change in temp erature\n(a) 1, 2 and 3\n(b) 1, 2 and 4\n(c) 1, 3 and 4\n(d) 1, 2, 3 and 4", + "sample question papers 24 (2).txt\n1. Change in state\n2. Ev olution of a gas\n3. Change in colour\n4. Change in temp erature\n(a) 1, 2 and 3\n(b) 1, 2 and 4\n(c) 1, 3 and 4\n(d) 1, 2, 3 and 4\n5. A metal M of moderate reactivity is present as its sulphide X. On heating in air, X converts into is oxide Y and \na gas evolves. On heating Y and X together, the metal M is produced. X and Y respectively are\n(a) X cuprous sulphide, Y cuprous oxide\n(b) X cuprous sulphide, Y cupric oxide\n(c) X sodium sulphide, Y sodium oxide\n(d) X calcium sulphide, Y calcium oxide\n6. Which of the following statements is true for acids?", + "Which of the following statements is true for acids?\n(a) Bitter and c hange red litmus to blue\n(b) Sour and c hange red litmus to blue\n(c) Sour and c hange blue litmus to red\n(d) Bitter and c hange blue litmus to redInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 6 P age 3\nhttps://qrbook.page.link/app\n7. Which of the following is the correct representation of electron dot structure of nitrogen ?", + "sample question papers 24 (2).txt\nClick Here To InstallCBSE Science Class 10 Sample P aper 6 P age 3\nhttps://qrbook.page.link/app\n7. Which of the following is the correct representation of electron dot structure of nitrogen ?\n8. A graph was plotted to show the energy output of two types of respiration. Identify the type of respiration denoted \nby curves A and B.\n(a) Anaerobic respiration , Aerobic respiration\n(b) Aerobic respiration, Anaerobic respiration\n(c) Aerobic respiration, Aerobic respiration\n(d) Anaerobic respiration, Anaerobic respiration\n9. Transpiration helps :\n(a) in the absorption\n(b) in the up ward movement of water minerals dissolved in it from roots to the leaves\n(c) in temp erature regulation\n(d) all of the ab ove\n10. Which of the following the father of genetics ?\n(a) Mendel (b) Ho ok\n(c) F araday (d) Newton\n11.", + "Which of the following the father of genetics ?\n(a) Mendel (b) Ho ok\n(c) F araday (d) Newton\n11. Involuntary actions in the body are controlled by\n(a) medulla in fore brain\n(b) medulla in mid brain\n(c) medulla in hind brain\n(d) medulla in spinal cord", + "sample question papers 24 (2).txt\n(c) F araday (d) Newton\n11. Involuntary actions in the body are controlled by\n(a) medulla in fore brain\n(b) medulla in mid brain\n(c) medulla in hind brain\n(d) medulla in spinal cord\n12. Vegetative propagation refers to formation of new plants from\n(a) stem, ro ots and flowers\n(b) stem, ro ots and leaves\n(c) stem, flo wers and fruits\n(d) stem, lea ves and flowers Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 6 CBSE Science Class 10\nhttps://qrbook.page.link/app\n13. What is the maximum resistance which can be made using five resistors each of /15\u2126 ?\n(a) /15\u2126 (b) 10\u2126 \n(c) 5\u2126 (d) 1\u2126 \n14. The strength of magnetic field around a current carrying conductor is\n(a) in versely proportional to the current but directly proportional to the square of the distance from wire.", + "The strength of magnetic field around a current carrying conductor is\n(a) in versely proportional to the current but directly proportional to the square of the distance from wire.\n(b) directly prop ortional to the current and inversely proportional to the distance from wire.\n(c) directly prop ortional to the distance and inversely proportional to the current", + "sample question papers 24 (2).txt\n(b) directly prop ortional to the current and inversely proportional to the distance from wire.\n(c) directly prop ortional to the distance and inversely proportional to the current\n(d) directly prop ortional to the current but inversely proportional the square of the distance from wire.\n15. V-I graph for the two wires A and B are shown in the figure. If we connect both the wires one by one to the same \nbattery which of the two will produce more heat per unit time ?\n(a) A\n(b) B\n(c) Both A and B\n(d) None of these\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 6 P age 5\nhttps://qrbook.page.link/app\n16. The strength of magnetic field inside a long current carrying straight solenoid is\n(a) more at the ends than at the cen tre\n(b) minim um in the middle\n(c) same at all p oints\n(d) found to increase from one end to the other\n\u0001Question no.", + "The strength of magnetic field inside a long current carrying straight solenoid is\n(a) more at the ends than at the cen tre\n(b) minim um in the middle\n(c) same at all p oints\n(d) found to increase from one end to the other\n\u0001Question no. 17 to 20 are Assertion - Reasoning based questions.", + "sample question papers 24 (2).txt\n(b) minim um in the middle\n(c) same at all p oints\n(d) found to increase from one end to the other\n\u0001Question no. 17 to 20 are Assertion - Reasoning based questions.\n17. Assertion : Changing of colour of copper from reddish brown to black is an example of reduction.\nReason : Hydrogen is removed.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : The sex of the children will be determined by chromosome received from the father.Reason : A human male has one \nX and one Y-chromosome.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).", + "sample question papers 24 (2).txt\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : Egestion in amoeba takes place through a permanent membrane present in them.Reason : Cilia is absent in amoeba.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Both Assertion and Reason are false.\n20. Assertion : Force experienced by moving charge will be maximum if direction of velocity of charge is perpendicular to applied magnetic field.Reason : Force on moving charge is independent of direction of applied magnetic field.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).", + "sample question papers 24 (2).txt\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 6 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. What is meant by galvanisation ? Why is it done ?\n\u0001o\u0001\nWhy do ionic compounds conduct electricity in molten state ?\n22. State the function of Bowman\u2019s Capsule and glomerulus.\n23. State the necessary conditions of autotrophic nutrition and name the by product. Mention the source of this by \nproduct.\n24. In winter the frequency of urination is more. Why ?\n25.", + "22. State the function of Bowman\u2019s Capsule and glomerulus.\n23. State the necessary conditions of autotrophic nutrition and name the by product. Mention the source of this by \nproduct.\n24. In winter the frequency of urination is more. Why ?\n25. Why does the colour of the sky appear blue ? Explain in brief.\n\u0001o\u0001", + "sample question papers 24 (2).txt\nproduct.\n24. In winter the frequency of urination is more. Why ?\n25. Why does the colour of the sky appear blue ? Explain in brief.\n\u0001o\u0001\nWhat type of spectacles should be worn by a person having the defects of myopia as well as hypermetropia? How \ndoes it help ?\n26. If a lake is contaminated with pesticides, which one of the following organisms will contain in its body the maximum concentration of pesticides and why ?Small fish, Pelicans, Zooplanktons, Phytoplanktons, Big fish.\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. When hydrogen sulphide gas is passed through a blue solution of copper sulphate, the colour of the solution fades and a black precipitate is obtained.\n(a) Name the t ype of reaction mentioned above.\n(b) Wh y does the colour of the solution fade away ?\n(c) W rite the chemical name of the black precipitate formed.\n28. (a) Arrange the metals Zn, Mg, Al, Cu and F e in decreasing order of reactivity.\n(b) What w ould you observe when you put", + "sample question papers 24 (2).txt\n(c) W rite the chemical name of the black precipitate formed.\n28. (a) Arrange the metals Zn, Mg, Al, Cu and F e in decreasing order of reactivity.\n(b) What w ould you observe when you put\n(i) Some zinc pieces in to blue copper sulphate solution ?\n(ii) Some copp er pieces into green ferrous sulphate solution.\n(c) Name a metal whic h combines with hydrogen gas. Name the compound formed.\n29. (a) Blo od pressure is high in the arteries and low in the veins. Give the possible reason for such difference.\n(b) What is the ma jor cause of high blood pressure ?\n(c) What ma y happen if a person is having a very high blood pressure ?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 6 P age 7\nhttps://qrbook.page.link/app\n\u0001o\u0001\n(i) Name the follo wing :\n(a) The three carb on molecule that is formed due to break-down of glucose during respiration.", + "sample question papers 24 (2).txt\nhttps://qrbook.page.link/app\n\u0001o\u0001\n(i) Name the follo wing :\n(a) The three carb on molecule that is formed due to break-down of glucose during respiration.\n(b) The nitrogenous w aste that is removed from the blood in our Kidneys.\n(ii) Ho w do unicellular organisms generally remove waste ?\n30. A mirror is fitted in a wall of the AGRA FORT. When you stand at a proper location, a full-size image of the Taj \nMahal can be seen in this mirror.\n(a) What kind of mirror is it ?\n(b) Dra w a ray diagram for such a mirror when the object is at infinity.\n31. You are given a convex lens of focal length 10 cm. Where will you place an object to get a real, inverted and highly enlarged image of the object. Draw a ray diagram.\n32. Answer the following questions :\n(i) What is the direction of magnetic field lines outside a bar-magnet ?\n(ii) What is the SI unit of magnetic field ?\n(iii) What do es crowding of magnetic field lines indicate ?\n\u0001o\u0001", + "sample question papers 24 (2).txt\n(i) What is the direction of magnetic field lines outside a bar-magnet ?\n(ii) What is the SI unit of magnetic field ?\n(iii) What do es crowding of magnetic field lines indicate ?\n\u0001o\u0001\nStudy the diagram given below and answer the questions that follow :\n(a) Wh y do the iron filings arrange in such a pattern?\n(b) What do es this pattern demonstrate ?\n(c) Wh y do the iron filings near the bar magnet seem to align in the shape of closed curves ?\n33. How will you create an artificial aquatic ecosystem, which is self-sustainable?\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. (a) W rite the molecular formula of an organic compound having its name suffixed with \u2018 ol-\u2019 and having two \ncarbon atoms in the molecule.\n With the help of balanced chemical equation indicate what happens when it is heated with excess of \nconcentrated HSO24.\n(b) W rite names of the following compounds :\n(i) HCOOH\n(ii) CH COCH CH 22 3.\n(c) Explain wh y carbon generally forms compounds by covalent bonds.", + "sample question papers 24 (2).txt\nconcentrated HSO24.\n(b) W rite names of the following compounds :\n(i) HCOOH\n(ii) CH COCH CH 22 3.\n(c) Explain wh y carbon generally forms compounds by covalent bonds.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 6 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001o\u0001\n(a) Define catenation. Wh y no other element exhibits the properties of catenation to the extent seen in carbon \ncompounds ?\n(b) Name the t ype of compound formed by the reaction of an organic acid and an alcohol. Write the chemical \nequation for the reaction involved.\n35. (a) Dra w the diagram of female reproductive system and match and mark the part(s) :\n(i) Where blo ck is created surgically to prevent fertilization.\n(ii) Where Copp er-T is inserted?\n(iii) Inside whic h condom can be placed.\n(b) Wh y do more and more people prefer to use condoms?", + "(ii) Where Copp er-T is inserted?\n(iii) Inside whic h condom can be placed.\n(b) Wh y do more and more people prefer to use condoms? What is the principle behind use of condoms?\n\u0001o\u0001", + "sample question papers 24 (2).txt\n(ii) Where Copp er-T is inserted?\n(iii) Inside whic h condom can be placed.\n(b) Wh y do more and more people prefer to use condoms? What is the principle behind use of condoms?\n\u0001o\u0001\nWhat are the three categories of contraceptive methods? Write briefly about each.\n36. (a) What is mean t by the statement, \u201cThe resistance of a conductor is one ohm\u201d?\n(b) Define electric p ower. Write an expression relating electric power, potential difference and resistance.\n(c) Ho w many 132\u2126 resistors in parallel are required to carry 5 A on a 220 V line?\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37. Activity series : Relative reactivities of metals\nPotassiumSodiumCalciumMagnesiumAluminiumZincIronLeadHydrogenCopperMercurySilverGold Most reactive\nReactivity decreases\nLeast reactive\n(i) What happ ens when iron nail is added to copper sulphate solution? What is the colour change ?", + "sample question papers 24 (2).txt\nReactivity decreases\nLeast reactive\n(i) What happ ens when iron nail is added to copper sulphate solution? What is the colour change ?\n(ii) Iden tify the metal which reacts with very dilute nitric acid to evolve hydrogen gas. Name one more metal \nnot given in the above series which reacts in the same way with dilute nitric acid.\n(iii) Name one imp ortant ore of copper with its chemical formula.\n(iv) Whic h method is used to extract sodium from molten sodium chloride ?\n\u0001o\u0001\n(v) Whic h metal is used in the galvanization of iron ?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 6 P age 9\nhttps://qrbook.page.link/app\n38. Nastic movements in plants are not directional movements. They are not dependent on the stimulus and are", + "sample question papers 24 (2).txt\nhttps://qrbook.page.link/app\n38. Nastic movements in plants are not directional movements. They are not dependent on the stimulus and are \ngrowth independent. For example, the leaves of a touch me not plant (Mimosa pudica), fold up immediately when touched. These kinds of changes occur due to the changes in the amount of water in the leaves. Depending on the quantity, they either swell up or shrink. Plant hormones or phytohormones are responsible for the control and coordination of plants. There are different types of hormones, which affect the growth of a plant. Phytohormones are chemical compounds which are released by stimulated cells. These hormones are diffused around the plant cells. They have a role in the cell division, cell enlargement, cell differentiation, fruit growth, falling of leaves, ripening of fruits, ageing of plants etc.\n(i) Name the phenomenon called for the mo vement in growth of plants.\n(ii) What do y ou mean by nastic movement ?", + "sample question papers 24 (2).txt\n(i) Name the phenomenon called for the mo vement in growth of plants.\n(ii) What do y ou mean by nastic movement ?\n(iii) What are the differen t types of harmonies of plants ?\n\u0001o\u0001\n(iv) The plant harmone help in the cell growth at the shoot tips by elongating the cells and help in the growth \nprocess is :\n39. While dealing with the reflection of light by spherical mirrors, we shall follow a set of sign conventions called the \nNew Cartesian Sign Convention. In this convention, the pole (P) of the mirror is taken as the origin. The principal axis of the mirror is take as the \nx-axis of the coordinate system. In a spherical mirror, the distance of the object \nfrom its pole is called the object distance u^h. The distance of the image from the pole of the mirror is called the \nimage distance v^h. Magnification produced by a spherical mirror gives the relative extent to which the image of", + "sample question papers 24 (2).txt\nimage distance v^h. Magnification produced by a spherical mirror gives the relative extent to which the image of \nan object is magnified with respect to the object size. It is expressed as the ratio of the height of the image to the height of the object. It is usually represented by the letter \nm^h.\n(i) Ho w can you calculate the magnification of a spherical mirror ?\n(ii) What do es a negative sign in the value of magnification indicates?\n(iii) Find the fo cal length of a convex mirror whose radius of curvature is 32 cm.\n\u0001o\u0001\n(iv) Wh y does the height of the object is taken to be positive?Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 6 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 18.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 21 Page 1\nhttps://qrbook.page.link/app\nSample Paper 21\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho\nwever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv\ne Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac\nh. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each.", + "5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 18.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac\nh with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. A student places some iron filings around a magnet. The iron filings arrange themselv\nes as shown in image:\nThe student labelled four different regions around the magnet. Where would the magnetic effect be the strongest?\n(a) Q\n(b) P\n(c) S\n(d) R\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 21 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2.", + "Click Here To InstallPage 2 Sample P aper 21 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2. The factors on whic h magnetic field strength produced by current carrying solenoid depends are:\n(a)\n n\number of turns\n(b)\n magnitude of curren\nt\n(c)\n All of the ab\nove\n(d)", + "sample question papers 18.txt\n2. The factors on whic h magnetic field strength produced by current carrying solenoid depends are:\n(a)\n n\number of turns\n(b)\n magnitude of curren\nt\n(c)\n All of the ab\nove\n(d)\n nature of core material\n3.\n Study the exp\nerimental set up shown in given figure and choose the correct option from the following:\nP Q Change observed in calcium hydroxide solution\na. KCO23 Cl2 No change\nb. KHCO 3 CO 2 No change\nc. KHCO 3 H2 Turns milky\nd. KCO23 CO 2 Turns milky\n4.\n Num\nber of electrons shared between carbon-carbon atoms in ethene is:\n(a) 4\n(b)", + "sample question papers 18.txt\n2\n(c)\n \n8\n(d)", + "sample question papers 18.txt\n6\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 21 P age 3\nhttps://qrbook.page.link/app\n5. Whic h among the following is/are double displacement reaction(s)? \n(i)\n Pb CuCl PbCl Cu 22\" ++\n(ii) NaSO BaCl BaSO NaCl 2 24 24\" ++\n(iii) CO CO 22\"+\n(iv) CH OC OH O 22 42 22\"++\n(a) (i) and (ii)\n(b)\n (i) and (iv)\n(c)\n (iii) and (iv)\n(d)\n only (ii)\n6.\n Tw\no pink coloured flowers on crossing resulted in 1 red, 2 pink and 1 white flower progeny. The nature of the \ncross will be :\n(a)\n self p\nollination\n(b)\n double fertilisation\n(c)\n no fertilisation\n(d)\n cross fertilisation\n7.", + "The nature of the \ncross will be :\n(a)\n self p\nollination\n(b)\n double fertilisation\n(c)\n no fertilisation\n(d)\n cross fertilisation\n7.\n When lead nitrate crystals are heated as sho\nwn in the figure:\n(a) a bro wn gas is evolved and a white residue is left behind\n(b)\n a colourless gas is ev\nolved and a yellow residue is left behind\n(c)\n a bro\nwn gas is evolved and a yellow residue is left behind.\n(d)\n a greenish y\nellow gas is evolved and a brown residue is left behind", + "sample question papers 18.txt\nolved and a yellow residue is left behind\n(c)\n a bro\nwn gas is evolved and a yellow residue is left behind.\n(d)\n a greenish y\nellow gas is evolved and a brown residue is left behind\n8.\n Study the follo\nwing table and choose the correct option.\nSalt Parent Acid Parent base Nature of salt\na. Sodium Chloride HCl NaOH Basic\nb. Sodium Carbonate H2CO3NaOH Neutral\nc. Sodium Sulphate HSO24 NaOH Acidic\nd. Sodium Acetate CH3COOH NaOH BasicInstall NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 21 CBSE Science Class 10\nhttps://qrbook.page.link/app\n9. An aqueous solution A turns phenolphthalein solution pink. On addition of an aqueous solution B to A, the pink \ncolour disappears. Which of the following statement is true for solution A and B?\n(a)\n A is strongly acidic and B is a w\neak acid.\n(b)\n A is strongly basic and B is a w\neak base.", + "On addition of an aqueous solution B to A, the pink \ncolour disappears. Which of the following statement is true for solution A and B?\n(a)\n A is strongly acidic and B is a w\neak acid.\n(b)\n A is strongly basic and B is a w\neak base. \n(c)\n A has pH less than 7 and B has pH greater than 7.\n(d)\n A has pH greater than 7 and B has pH less than 7. \n10.\n Whic", + "sample question papers 18.txt\neak acid.\n(b)\n A is strongly basic and B is a w\neak base. \n(c)\n A has pH less than 7 and B has pH greater than 7.\n(d)\n A has pH greater than 7 and B has pH less than 7. \n10.\n Whic\nh one of the following conditions is true for the state of stomata of a green leaf shown in the given diagram?\n(a) Gaseous exc hange is occurring in large amount\n(b)\n Large amoun\nt of water flows into the guard cells\n(c)\n Large amoun\nt of sugar collects in the guard cells\n(d)\n Large amoun\nt of water flows out from the guard cells \n11.\n Iden\ntify the option that indicates the correct enzyme that is secreted in location (i), (ii) and (iii).", + "Iden\ntify the option that indicates the correct enzyme that is secreted in location (i), (ii) and (iii).\n(a)\n (i)-am\nylase, (ii)-pepsin, (iii)-trypsin\n(b)\n (i)-lipase, (ii)-trypsin, (iii)-p\nepsin\n(c)\n (i)-p\nermease, (ii)-carboxylase, (iii)-oxidase\n(d)\n (i)-trypsin, (ii)-am\nylase, (iii)-carboxylase\n12.\n Iden\ntify labels p, q and r and select the correct option.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 18.txt\n(d)\n (i)-trypsin, (ii)-am\nylase, (iii)-carboxylase\n12.\n Iden\ntify labels p, q and r and select the correct option.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 21 P age 5\nhttps://qrbook.page.link/app\np q r\n(a) Motor neuron Sensory neuron Relay neuron\n(b) Sensory neuron Motor neuron Relay neuron\n(c) Relay neuron Motor neuron Sensory neuron\n(d) Sensory neuron Relay neuron Motor neuron\n13.\n The image sho\nws the division in Spirogyra.\nWhat can be concluded about the Spirogyra from this division?\n(a)\n It is a unicellular organism that giv\nes rise to two new equal sized individuals.\n(b)\n It is a m\nulticellular organism gives rise to two new equal sized individuals.\n(c)\n It is a m\nulticellular organism that breaks into pieces that grows into new individuals.", + "(a)\n It is a unicellular organism that giv\nes rise to two new equal sized individuals.\n(b)\n It is a m\nulticellular organism gives rise to two new equal sized individuals.\n(c)\n It is a m\nulticellular organism that breaks into pieces that grows into new individuals.\n(d)\n It is a unicellular organism that breaks in\nto pieces that grows into new individuals.\n14.", + "sample question papers 18.txt\n(c)\n It is a m\nulticellular organism that breaks into pieces that grows into new individuals.\n(d)\n It is a unicellular organism that breaks in\nto pieces that grows into new individuals.\n14.\n Three resistances of 2, 3 and 5\u2126 are connected in parallel to a 10 V battery of negligible internal resistance. The \npotential difference across the 3\u2126 resistance will be:\n(a)\n 3 V\n(b)\n 2 V\n(c)\n 10 V\n(d)\n 5 V\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 21 CBSE Science Class 10\nhttps://qrbook.page.link/app\n15. Whic h arrangement of 3\u2126 resistors will give a total resistance of 7\u2126 ?\n16.", + "Click Here To InstallPage 6 Sample P aper 21 CBSE Science Class 10\nhttps://qrbook.page.link/app\n15. Whic h arrangement of 3\u2126 resistors will give a total resistance of 7\u2126 ?\n16. Iden tify the correct order of reactivity of metals among the following:\n(a)\n Fe Zn Cu Na Al 11 11\n(b) Cu Fe Zn Al Na 11 11 \n(c)\n Cu Zn Al Na Fe 11 11\n(d) Zn Cu Fe Al Na11 11 \n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17.\n Assertion (A):", + "sample question papers 18.txt\n(a)\n Fe Zn Cu Na Al 11 11\n(b) Cu Fe Zn Al Na 11 11 \n(c)\n Cu Zn Al Na Fe 11 11\n(d) Zn Cu Fe Al Na11 11 \n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17.\n Assertion (A):\n The sex of a child in human beings will be determined by the type of chromosome he/she inherits \nfrom the father.\nReason (R): A child who inherits \u2018X\u2019 chromosome from his father would be a girl (XX), while a child who inherits a \u2018Y\u2019 chromosome from the father would be a boy (XY).(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n18.", + "(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n18.\n Assertion (A): In Fleming\u2019s left-hand rule, the direction of magnetic field, force and current are mutually", + "sample question papers 18.txt\n(d)\n Assertion (A) is false but Reason (R) is true.\n18.\n Assertion (A): In Fleming\u2019s left-hand rule, the direction of magnetic field, force and current are mutually \nperpendicular.Reason (R): Fleming\u2019s left-hand rule is applied to measure the induced current.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\n19.\n Assertion (A):\n Resins and gums are stored in old xylem tissue in plants.\nReason (R): Resins and gums facilitate transport of water molecules.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).", + "(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.", + "sample question papers 18.txt\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 21 P age 7\nhttps://qrbook.page.link/app\n20. Assertion (A): The following chemical equation,\nCH OC OH O 22743 26 22 2 $++\nis a balanced chemical equation.\nReason (R): In a balanced chemical equation, the total number of atoms of each element are equal on both sides of the equation.(a)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.", + "(b)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21.\n Wh", + "sample question papers 18.txt\n(c)\n Assertion (A) is true but Reason (R) is false.\n(d)\n Assertion (A) is false but Reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21.\n Wh\ny does absorption of digested food occur mainly in the small intestine?\n22.\n Iden\ntify the oxidising agent in the following reactions:\n(i)\n 3 PbOH Cl PbCl Cl HO 84 34 22 2 $++ +\n(ii) Mg OM gO 22 2$+\n\u0001o\u0001\nComplete the following table:\nPlaster of Paris Bleaching Powder\nChemical equation for its preparation (i) .......... (ii) ..........\nUse (iii) .......... (iv) ..........\n23.\n State with reason an\ny two possible consequences of the elimination of decomposers from the earth.\n24.\n List in tabular form t\nwo differences between reflex action and walking.\n25.\n Name the four parts lab\nelled as a, b, c and d in given diagram and write their functions.\n\u0001o\u0001\nList two causes of presbyopia.", + "24.\n List in tabular form t\nwo differences between reflex action and walking.\n25.\n Name the four parts lab\nelled as a, b, c and d in given diagram and write their functions.\n\u0001o\u0001\nList two causes of presbyopia. Draw labelled diagram of a lens used for the correction of this defect of vision.Install NODIA App to See the Solutions.", + "sample question papers 18.txt\n\u0001o\u0001\nList two causes of presbyopia. Draw labelled diagram of a lens used for the correction of this defect of vision.Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 21 CBSE Science Class 10\nhttps://qrbook.page.link/app\n26. (i) W rite a balanced equation for photosynthesis.\n(ii)\n Whic\nh digestive secretion does not contain any enzyme but is important? Discuss.\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27.\n Name the t\nype of mirror which facilitates\n(i)", + "sample question papers 18.txt\nShaving,\n(ii)\n Observing large images of the teeth of a patien\nt, and\n(iii)\n Observing the rear view in v\nehicles.\nGive reason to justify your answer in each case.\n28.\n The magnification of an image formed b\ny a lens is 1-. If the distance of the image from the optical centre of the \nlens is 25 cm, where is the object placed? Find the nature and focal length of the lens. If the object is displaced \n15 cm towards the optical centre of the lens, where would the image be formed? Draw a ray diagram to justify your answer. \n29.\n What is transpiration? List its t\nwo functions.\n\u0001o\u0001\nList four functions of the human heart. Why is double circulation necessary in the human body?\n30.\n A studen\nt takes three beakers A, B and C filled with aqueous solutions of glucose, alcohol and hydrochloric acid \nrespectively as shown in the following figures :\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "A studen\nt takes three beakers A, B and C filled with aqueous solutions of glucose, alcohol and hydrochloric acid \nrespectively as shown in the following figures :\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 21 P age 9", + "sample question papers 18.txt\nrespectively as shown in the following figures :\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 21 P age 9\nhttps://qrbook.page.link/app\n(i) State y our observation in terms of glowing of bulb when the switch is on.\n(ii)\n Justify y\nour observations by giving reason in each case.\n(iii)\n Men\ntion the change noticed with appropriate reason if the content of beaker B is replaced by sodium \nhydroxide solution.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 21 CBSE Science Class 10\nhttps://qrbook.page.link/app\n31. Consider the follo wing diagram :\n(i) Iden tify \u2018X\u2019 and \u2018Y\u2019.\n(ii)\n What t\nype of reaction is it?\n(iii)\n W\nrite a balanced chemical equation.", + "Consider the follo wing diagram :\n(i) Iden tify \u2018X\u2019 and \u2018Y\u2019.\n(ii)\n What t\nype of reaction is it?\n(iii)\n W\nrite a balanced chemical equation.\n(iv)\n Explain wh\ny is it also called a double displacement reaction?\n32.\n What are magnetic field lines? Justify the follo\nwing statements:\n(i)\n Tw", + "sample question papers 18.txt\n(iii)\n W\nrite a balanced chemical equation.\n(iv)\n Explain wh\ny is it also called a double displacement reaction?\n32.\n What are magnetic field lines? Justify the follo\nwing statements:\n(i)\n Tw\no magnetic field lines never intersect each other.\n(ii)\n Magnetic field lines are closed curv\nes.\n\u0001o\u0001\n(i)\n What is the function of earth wire in electrical instrumen\nts?\n(ii)\n Explain what is short circuiting of an electric supply?\n(iii)\n What is the usual curren\nt rating of the fuse wire in the line to feed:\n(a)\n ligh\nts and fans?\n(b)\n appliances of 2 kW or more p\nower?\n33.\n What is a fo\nod chain? Why is the flow of energy in an ecosystem unidirectional? Explain briefly.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34.\n (i)\n Name the mo\nde of reproduction of the following organisms and state the important feature of each mode:\n(a)", + "sample question papers 18.txt\nPlanaria\n(b)\n \nHydra\n(c)", + "sample question papers 18.txt\nRhizopus\n(ii)\n W\ne can develop new plants from the leaves of Bryophyllum. Comment.\n(iii)\n List t\nwo advantages of vegetative propagation over other modes of reproduction.\n\u0001o\u0001\n(i)\n \u2018\u2018Use \nof a condom is beneficial for both the sexes involved in a sexual act:\u2019\u2019 Justify this statement giving two \nreasons.\n(ii)\n Ho\nw do oral contraceptives help in avoiding pregnancies?\n(ii)\n What is sex selectiv\ne abortion? How does it affect a healthy society? (State any one consequence.)Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 21 P age 11\nhttps://qrbook.page.link/app\n35. In our daily life, w e use two types of electric current whose current-time graphs are given below:\n(i) Iden tify the types of current in each case.\n(ii)\n Iden\ntify any one source of each type of current.\n(iii)\n What is frequency of curren\nt used in domestic supply in India?", + "(ii)\n Iden\ntify any one source of each type of current.\n(iii)\n What is frequency of curren\nt used in domestic supply in India?\n(iv)\n On the basis of graphs, write difference b\netween the two currents.\n(v)\n Out of t", + "sample question papers 18.txt\n(iii)\n What is frequency of curren\nt used in domestic supply in India?\n(iv)\n On the basis of graphs, write difference b\netween the two currents.\n(v)\n Out of t\nwo, which one is used in transmitting electric power over Long distance and why?\n36.\n (i)\n Carry out follo\nwing conversions:\n(a)\n Ethanol to ethene\n(b)\n Ethanol to Ethanoic acid\n(ii)\n Differen\ntiate between addition reaction and substitution reaction. Give one example of each.\n\u0001o\u0001\nWhat are esters? How are esters prepared? Write the chemical equation for the reaction involved. What happens \nwhen an ester reacts with sodium hydroxide? Write the chemical equation for the reaction and also state the name and use of this reaction.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.\n37.\n Refractiv\ne index refers to the measure of the bending of a ray of light when it passes from one medium to another", + "sample question papers 18.txt\n37.\n Refractiv\ne index refers to the measure of the bending of a ray of light when it passes from one medium to another \nmedium. It can also be defined as the ratio of the velocity of a Light ray in vacuum or air to the velocity of Light Install NODIA App to See the Solutions.\nClick Here To InstallPage 12 Sample P aper 21 CBSE Science Class 10\nhttps://qrbook.page.link/app\nin the medium. It has no unit.\n(i)\n The refractiv\ne index of diamond is 2.42. What is the meaning of this statement?\n(ii)\n The diagram b\nelow represents the path of a light ray from three different media A, B and C.\n In whic h of the following media is the speed of light maximum?\n(iii)\n Refractiv\ne index of diamond with respect to glass is 1.6. If the absolute refractive index of glass is 1.5, then \nwhat is the absolute refractive index of diamond?", + "(iii)\n Refractiv\ne index of diamond with respect to glass is 1.6. If the absolute refractive index of glass is 1.5, then \nwhat is the absolute refractive index of diamond?\n\u0001o\u0001\n(iv)\n Refractiv\ne index of turpentine oil, kerosene and alcohol are 1.47, 1.44 and 1.36 respectively. On the basis of", + "sample question papers 18.txt\nwhat is the absolute refractive index of diamond?\n\u0001o\u0001\n(iv)\n Refractiv\ne index of turpentine oil, kerosene and alcohol are 1.47, 1.44 and 1.36 respectively. On the basis of \nthis information, complete the following ray diagram to show path of ray of light through each medium.\n38. Sahil p erformed an experiment to study the inheritance pattern of genes. He crossed tall pea plants (T T) with \nshort pea plants (tt) and obtained all tall plants in F1 generation.\n(i)\n What will b\ne set of genes present in the F1 generation?\n(ii)\n Giv\nen reason why only tall plants are observed in F1 progeny.\n(iii)\n When F1 plants were self-pollinated, a total of 800 plants were produced. How many of these would be tall, \nmedium height or short plant? Give the genotype of F2 generation.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.", + "How many of these would be tall, \nmedium height or short plant? Give the genotype of F2 generation.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 21 P age 13\nhttps://qrbook.page.link/app\n\u0001o\u0001\n(iv)", + "sample question papers 18.txt\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 21 P age 13\nhttps://qrbook.page.link/app\n\u0001o\u0001\n(iv)\n When F1 plants were cross-pollinated with plants having tt genes, a total of 800 plants were produced. How \nmany of these would be tall, medium height or short plants? Give the genotype of F2 generation.\n39.\n A studen\nt took the samples of four metals A, B, C and D and added following solution one by one. The results \nobtained have been tabulated as follows:\nMetal Iron (II) \nsulphateCopper (II) \nsulphateZinc sulphate Silver nitrate\nA No reaction Displacement \u2014 \u2014\nB Displacement \u2014 No reaction \u2014\nC No reaction No reaction No reaction Displacement\nD No reaction No reaction No reaction No reaction\n(i) Whic h is the least reactive and most reactive metal? Give reasons.", + "Give reasons.\n(ii)\n What \nwould be observed, if \u2018B\u2019 is added to a solution of copper (II) sulphate and why? Also, arrange the", + "sample question papers 18.txt\n(i) Whic h is the least reactive and most reactive metal? Give reasons.\n(ii)\n What \nwould be observed, if \u2018B\u2019 is added to a solution of copper (II) sulphate and why? Also, arrange the \nmetals A, B, C and D in order of increasing reactivity.\n\u0001o\u0001\n(iii)\n In whic\nh test tube will the reaction take place? Write your observations for both the test tubes.\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 20.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 18 Page 1\nhttps://qrbook.page.link/app\nSample Paper 18\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 20.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Which of the following are correctly matched in the given table?\n1.Acid + salt metal + hydrogen\n2.Acid + metal carbonate salt + carbon dioxide + water\n3.Metal oxide + acid salt + water\n(a) 1 and 2\n(b) 2 and 3\n(c) 1 and 3\n(d) 1, 2 and 3\n2. The reaction Na Cl NaCl22 2$+ is an example of\n(a) combination reaction\n(b) decomposition reaction\n(c) displacement reaction\n(d) double displacement reaction\n3. Which among the following is (are) double displacement reaction(s)?\n1.", + "The reaction Na Cl NaCl22 2$+ is an example of\n(a) combination reaction\n(b) decomposition reaction\n(c) displacement reaction\n(d) double displacement reaction\n3. Which among the following is (are) double displacement reaction(s)?\n1. Pb CuCl PbCl Cu 22$ ++\n2. NaSO BaCl BaSO NaCl 2 24 24$ ++\n3. CO CO 22$+\n4. CH OC OH O 22 42 22$++\n(a) 1 and 4", + "sample question papers 20.txt\n3. Which among the following is (are) double displacement reaction(s)?\n1. Pb CuCl PbCl Cu 22$ ++\n2. NaSO BaCl BaSO NaCl 2 24 24$ ++\n3. CO CO 22$+\n4. CH OC OH O 22 42 22$++\n(a) 1 and 4\n(b) Only 2\n(c) 1 and 2\n(d) 3 and 4Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 18 CBSE Science Class 10\nhttps://qrbook.page.link/app\n4. Which one of the following is the example of oxidation?\n(a) () () () MgsO gM gO s 22Burnin g\n2+\n(b) () () () () CuOs Hg CusH OgHeat\n22++\n(c) () () () () FeOs AlsA lO sF es 22 23 23$ ++\n(d) None of these\n5. Generally metals react with acids to give salt and hydrogen gas.", + "Generally metals react with acids to give salt and hydrogen gas.\nWhich of the following acids does not give hydrogen gas on reacting with metals (except Mn and Mg)?\n(a) HSO24\n(b) HCl\n(c) HNO 3\n(d) All of these\n6. A solution turns red litmus blue, its pH is likely to be\n(a) 1\n(b) 4\n(c) 5\n(d) 10\n7. The soap molecule has a\n(a) h ydrophilic head and a hydrophobic tail\n(b) h ydrophobic head and a hydrophilic tail", + "sample question papers 20.txt\n(a) 1\n(b) 4\n(c) 5\n(d) 10\n7. The soap molecule has a\n(a) h ydrophilic head and a hydrophobic tail\n(b) h ydrophobic head and a hydrophilic tail\n(c) h ydrophobic head and a hydrophobic tail\n(d) h ydrophilic head and a hydrophilic tail\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 18 P age 3\nhttps://qrbook.page.link/app\n8. The table shows the characteristics of blood in one blood vessel of the body.\nOxygen concentration Carbon dioxide concentration Pressure\nHigh Low High\nWhich blood vessel contains blood with these characteristics?\n(a) V ena cava\n(b) Pulmonary v ein\n(c) Aorta\n(d) Pulmonary artery\n9.", + "Oxygen concentration Carbon dioxide concentration Pressure\nHigh Low High\nWhich blood vessel contains blood with these characteristics?\n(a) V ena cava\n(b) Pulmonary v ein\n(c) Aorta\n(d) Pulmonary artery\n9. The maintenance functions of living organisms must go on even when they are not doing\n(a) An ything particular\n(b) Sleeping\n(c) Mo ving at constant speed\n(d) Hibernation\n10. Which of the following statement for chromosomal theory of inheritance is incorrect?", + "sample question papers 20.txt\n(a) An ything particular\n(b) Sleeping\n(c) Mo ving at constant speed\n(d) Hibernation\n10. Which of the following statement for chromosomal theory of inheritance is incorrect?\n(a) P airing and separation of a pair of the chromosomes would lead to segregation of a factor they carried. \n(b) Beha viour of chromosomes is parallel to the behaviour of genes.\n(c) The t wo alleles of a gene pair are located on homologous sites on homologous chromosomes.\n(d) Chromosomes as w ell as genes occur in pair\n11. In a neuron, conversion of electrical signal to a chemical signal occurs at in\n(a) cell b ody\n(b) axonal end\n(c) dendritic end\n(d) axon\n12.", + "(d) Chromosomes as w ell as genes occur in pair\n11. In a neuron, conversion of electrical signal to a chemical signal occurs at in\n(a) cell b ody\n(b) axonal end\n(c) dendritic end\n(d) axon\n12. In the list of organisms given below, those that reproduce by the asexual method are\n(i) Banana\n(ii) Dog\n(iii) Yeast\n(iv) Amoeba\n(a) (ii) and (iv) (b) (i), (iii) and (iv)\n(c) (i) and (iv) (d) (ii), (iii) and (iv)Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 18 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers 20.txt\n(c) (i) and (iv) (d) (ii), (iii) and (iv)Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 18 CBSE Science Class 10\nhttps://qrbook.page.link/app\n13. A child standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body \nof the same size and that of the legs smaller. The following is the order of combinations for the magic mirror from the top.\n(a) Plane, con vex and concave\n(b) Con vex, concave and plane\n(c) Conca ve, plane and convex\n(d) Con vex, plane and concave\n14. \nHere X and Y indicates:\n(a) Green colour, violet colour\n(b) Red colour, violet colour\n(c) Violet colour, red colour\n(d) Green colour, red colour\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "Here X and Y indicates:\n(a) Green colour, violet colour\n(b) Red colour, violet colour\n(c) Violet colour, red colour\n(d) Green colour, red colour\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 18 P age 5\nhttps://qrbook.page.link/app\n15. Which of the following ray diagrams is correct for the ray of light incident on a lens shown in Figure?", + "sample question papers 20.txt\nhttps://qrbook.page.link/app\n15. Which of the following ray diagrams is correct for the ray of light incident on a lens shown in Figure?\n16. Advanced sunrise and delayed sunset are explained on the basis of\n(a) T yndall effect\n(b) scattering of ligh t\n(c) disp ersion of light\n(d) atmospheric refraction\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 18 CBSE Science Class 10\nhttps://qrbook.page.link/app\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : The following chemical equation,\nCH OC OH O 27 46 66 22 2$++\nis a balanced chemical equation.\nReason : In a balanced chemical equation, the total number of atoms of each element may or may not equal on \nboth side of the equation.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.", + "sample question papers 20.txt\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : Units which make up the nervous system are called neurons.Reason : Nerve impulses are carried by dendrites towards the cell body.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : Excretion is the biological process by which harmful wastes are removed from an organism\u2019s body.Reason : The mode of excretion is completely same in both unicellular and multicellular organisms.", + "sample question papers 20.txt\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Assertion is false but Reason is true.\n20. Assertion : All electric devices shown in the circuit are ideal. The reading of each of ammeter (a) and voltmeter \n()V is zero.\nReason : An ideal voltmeter draws almost no current due to very large resistance, and hence ()V and ()a will read \nzero.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 18 P age 7", + "sample question papers 20.txt\n(d) Assertion (A) is false but reason (R) is true.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 18 P age 7\nhttps://qrbook.page.link/app\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. Metal oxides are basic in nature. But some metal oxides show both acidic as well as basic behaviour. What are \nthese oxides called ? Name one such oxide and write the reaction with an acid and a base.\n\u0001o\u0001\nAluminium occurs in combined state whereas gold is found in free state. Why ?\n22. What name is given to the movement of a plant in the direction of stimulus ? Give any three examples.\n23. How do proteins control the characteristics that are inherited ? Explain with the help of an example.\n24. What are sexually transmitted diseases? Name few of them.\n25. An object is placed at a distance of 40 cm infront of a convex mirror of radius of curvature 40 cm. List four", + "sample question papers 20.txt\n24. What are sexually transmitted diseases? Name few of them.\n25. An object is placed at a distance of 40 cm infront of a convex mirror of radius of curvature 40 cm. List four \ncharacteristics of the image formed by the mirror.\n\u0001o\u0001\nAn object is 2 m away from a lens, which forms an erect image one-fourth the size of the object. Determine the \nfocal length of the lens. What type of lens is this ?\n26. Observe the food chain :\n(a) If autotrophs o ccupying the first trophic level are called producers, what are herbivores called as ?\n(b) Ho w much energy does the lion get in the above food chain ?\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. What is an oxidizing agent ? What happens when an oxidizing agent is added to propanol ? Explain with the help of a chemical equation.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "27. What is an oxidizing agent ? What happens when an oxidizing agent is added to propanol ? Explain with the help of a chemical equation.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 18 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers 20.txt\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 18 CBSE Science Class 10\nhttps://qrbook.page.link/app\n28. In an experiment 20 g of zinc (A) was reacted with g 20 of iodine (B) in a 250 ml beaker containing 50 ml water. \nReaction proceeded with evolution of heat and at the end, reaction mixture did not give any colour on adding \nstarch solution, but some zinc was found settling at the bottom.\n() () ()ZnsI sZ nIsh eat 22$++\n(i) Is the reaction endothermic or exothermic ?\n(ii) Out of the t wo reactants, Zn(s) and ()Is2, which one is left unreacted ? Why?\n(iii) Ho w much of (A) and (B) should he taken so that no reactant is left out at the end of the reaction ?\n29. Name the types of sex chromosomes present in\n(i) h uman male and\n(ii) h uman female.", + "Why?\n(iii) Ho w much of (A) and (B) should he taken so that no reactant is left out at the end of the reaction ?\n29. Name the types of sex chromosomes present in\n(i) h uman male and\n(ii) h uman female.\nWhat will be the sex of the child produced if a sperm carrying Y-chromosome fertilizes the egg ? Name an insect in which similar type of sex determination takes place.\n\u0001o\u0001", + "sample question papers 20.txt\n(ii) h uman female.\nWhat will be the sex of the child produced if a sperm carrying Y-chromosome fertilizes the egg ? Name an insect in which similar type of sex determination takes place.\n\u0001o\u0001\nIn humans, genetically the sex of a child is determined by the father and not by the mother. Explain.\n30. A person cannot see clearly objects beyond a distance of 1.2 m. Name the defect of vision he is suffering from. \nWhat would be the power of correcting lens used to restore proper vision ?\n31. A child is able to read his book comfortably but is unable to read the matter written on the blackboard at certain distance.\n(a) Name the defect of vision he is suffering from.\n(b) With the help of lab elled ray diagram show :\n(i) The ab ove mentioned defect of vision.\n(ii) Correction of the ab ove mentioned defect using a suitable lens.\n32. While studying the dependence of potential difference V across a resistor on the current I passing through it,", + "sample question papers 20.txt\n(ii) Correction of the ab ove mentioned defect using a suitable lens.\n32. While studying the dependence of potential difference V across a resistor on the current I passing through it, \nin order to determine the resistance of the resistor, a student took 5 readings for different values of current and plotted a graph between \nV and I. He got a straight line graph passing through the origin. What does the straight \nline signify? Write the method of determining resistance of the resistor using this graph.\n\u0001o\u0001\nA wire of resistance 6\u2126 is bent to form a closed square. What is the resistance across a diagonal of the square ?\n33. (a) What is the difference b etween self-pollination and cross-pollination ?\n(b) What happ ens to the pollen which falls on a suitable stigma ? Explain.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 18 P age 9\nhttps://qrbook.page.link/app\nSECTION-D", + "sample question papers 20.txt\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 18 P age 9\nhttps://qrbook.page.link/app\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. (a) Compare soaps and detergen ts on the basis of their composition and cleansing action in hard water.\n(b) What happ ens when ethanol is treated with sodium metal? State the behaviour of ethanol in this reaction.\n(c) Dra w the structure of cyclohexane.\n(d) Name the follo wing compound.\n\u0001o\u0001\n(i) Design an activit y to show the conditions needed for iron nails to rust.\n(ii) Wh y do we apply paint on iron articles ?\n35. Differentiate between :\n(a) Plum ule and radicle\n(b) P ollination and fertilization\n\u0001o\u0001\nAnswer the following questions :\n(a) What happ ens if an egg is not fertilized?\n(b) Wh y do we need to adopt contraceptive measures?\n(c) Name one bacterial and one viral sexually transmitted disease.", + "sample question papers 20.txt\nAnswer the following questions :\n(a) What happ ens if an egg is not fertilized?\n(b) Wh y do we need to adopt contraceptive measures?\n(c) Name one bacterial and one viral sexually transmitted disease.\n36. (a) List t wo disadvantages of using a series circuit in homes.\n(b) Calculate the effectiv e resistance between A and B in the circuit given below:\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 18 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37. A metal carbonate X on heating with an acid gives a gas which when passed through a solution Y gives the \ncarbonate back. On the other hand, a gas G that is obtained at anode during electrolysis of brine is passed on dry \nY, it gives a compound Z, used for disinfecting drinking water.", + "On the other hand, a gas G that is obtained at anode during electrolysis of brine is passed on dry \nY, it gives a compound Z, used for disinfecting drinking water.\n(i) Identify X, Y, G and Z.", + "sample question papers 20.txt\nY, it gives a compound Z, used for disinfecting drinking water.\n(i) Identify X, Y, G and Z.\n(ii) What is the nature of the gas ev olved when X is heated ?\n(iii) W rite the reaction involved in the formation of ?G\n\u0001o\u0001\n(iv) W rite the reaction involved when G reacts with Y.", + "sample question papers 20.txt\n\n38. The thyroid is a small, butterfly-shaped gland located at the base of your neck just below the Adam\u2019s apple. It is part of an intricate network of glands called the endocrine system. The endocrine system is responsible for coordinating many of your body\u2019s activities. The thyroid gland manufactures hormones that regulate your body\u2019s metabolism.Several different disorders can arise when your thyroid produces too much hormone (hyperthyroidism) or not enough (hypothyroidism). Four common disorders of the thyroid are Hashimoto\u2019s disease, Graves\u2019 disease, goitre and thyroid nodules.In hyperthyroidism, the thyroid gland is overactive. It produces too much of its hormone. Hyperthyroidism affects about 1 percent of women. It\u2019s less common in men.Graves\u2019 disease is the most common cause of hyperthyroidism, affecting about 70 percent of people with an overactive thyroid.", + "It produces too much of its hormone. Hyperthyroidism affects about 1 percent of women. It\u2019s less common in men.Graves\u2019 disease is the most common cause of hyperthyroidism, affecting about 70 percent of people with an overactive thyroid. Nodules on the thyroid \u2013 a condition called toxic nodular goitre or multinodular goitre can also cause the gland to overproduce its hormones.Excessive thyroid hormone production leads to symptoms such as : restlessness nervousness, racing heart, irritability, increased sweating, shaking, anxiety, trouble sleeping, thin skin, brittle hair and nails, muscle weakness, weight loss, bulging eyes (in Graves\u2019 disease).", + "sample question papers 20.txt\n(i) What is th yroid gland ?\n(ii) What is the function of the th yroid gland ?\n(iii) Name some common disorders of the th yroid.\n\u0001o\u0001\n(iv) Giv e some symptoms of hyperthyroidism.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 18 P age 11\nhttps://qrbook.page.link/app\n39. Rajesh\u2019s father Mr. Jayesh runs a cosmetics and perfumes shop in a crowded market place. Mr. Jayesh i usually \ncomplains at home that there is lot of \u2018shop-lifting\u2019 in his shop which was causing loss to him. Rajesh used to hear such complaints of his father. One day Rajesh went to the market and purchased one big mirror of a special kind. He then went to his father\u2019s shop and fixed the mirror at strategic positions inside the shop as shown in the figure. Mr. Jayesh found that after the installation of the mirror, the shop-lifting almost stopped.", + "He then went to his father\u2019s shop and fixed the mirror at strategic positions inside the shop as shown in the figure. Mr. Jayesh found that after the installation of the mirror, the shop-lifting almost stopped. He was very happy and thanked Rajesh for making this possible.", + "sample question papers 20.txt\n(i) What t ype of mirror was fixed by Rajesh in the shop?\n(ii) Place three tic ks in the table for image formed by such mirror due to which it helps in preventing shop-\nlifting.\nDiminished\nInverted\nReal\nEnlarged\nSame size\nUprightVirtual\n(iii) What sp ecial name/names is/are given to such mirror which help in preventing shop-lifting?\n\u0001o\u0001\n(iv) After few da ys that mirror becomes dirty. During cleaning, Rajesh held the mirror inside the water. So, what \nshould be the change in the focal length of the mirror?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 27.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 10 Page 1\nhttps://qrbook.page.link/app\nSample Paper 10\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 27.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. Some of the substances used in making of a modern safety match box are listed below :\n1. Antimony trisulfide\n2. Glass po wder\n3. Potassium chlorate\n4. Red phosphorus\nThe head of modern safety match stick contains :\n(a) 1 and 4\n(b) 2 and 3\n(c) 3 and 4\n(d) 3 and 1\n2. Which of the following is an example of displacement reaction?", + "Red phosphorus\nThe head of modern safety match stick contains :\n(a) 1 and 4\n(b) 2 and 3\n(c) 3 and 4\n(d) 3 and 1\n2. Which of the following is an example of displacement reaction?\n(a) NaOH HNO NaNO HO 33 2 \" ++\n(b) () Cu AgNO Cu NO Ag 22 33 2\" ++\n(c) Hg O HgO22 2\"+\n(d) () FeCl NaOH NaClF eOH 23 33\" ++Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 10 CBSE Science Class 10\nhttps://qrbook.page.link/app", + "sample question papers 27.txt\n(c) Hg O HgO22 2\"+\n(d) () FeCl NaOH NaClF eOH 23 33\" ++Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 10 CBSE Science Class 10\nhttps://qrbook.page.link/app\n3. Rusting of iron can be prevented by:\n1. Painting\n2. Galvanisation\n3. Electrolytic refining\n4. Alloying\nWhich of the above are correct?\n(a) 1, 2 and 3\n(b) 1, 2 and 4\n(c) 2, 3 and 4\n(d) 1, 2, 3 and 4\n4. Which of the following is a feasible reaction?", + "(a) 1, 2 and 3\n(b) 1, 2 and 4\n(c) 2, 3 and 4\n(d) 1, 2, 3 and 4\n4. Which of the following is a feasible reaction?\n(a) () () () BasK SO aq BaSO aq 24 4$ + ()Ks2+\n(b) () () () () Zns AgNO aq ZnNO aq 2 33 2 $ + ()Ag s 2+\n(c) () () () MgsN aSOa qM gSOa q 24 4$ + ()Na s 2+\n(d) () () () CusM gSOa qC uSOa q 44$ + ()Mg s+\n5. Sodium kept immersed in kerosene oil because-\n(a) So dium is most reactive metal.\n(b) So dium is less reactive metal.\n(c) So dium is not a reactive metal.\n(d) None of these.\n6. A sample of soil is mixed with water and allowed to settle. The clear supernatant solution turns the pH paper", + "sample question papers 27.txt\n(c) So dium is not a reactive metal.\n(d) None of these.\n6. A sample of soil is mixed with water and allowed to settle. The clear supernatant solution turns the pH paper \nyellowish-orange. Which of the following would change the colour of this pH paper to greenish-blue?\n(a) Lemon Juice\n(b) Vinegar\n(c) Common salt\n(d) An an tacid\n7. Identify the unsaturated compounds from the following :\n(i) Propane\n(ii) Propene\n(iii) Propyne\n(iv) Chloropropane\n(a) (i) and (ii) (b) (ii) and (iv)\n(c) (iii) and (iv) (d) (ii) and (iii)Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 10 P age 3\nhttps://qrbook.page.link/app\n8. In the given figure the various trophic levels are shown in a pyramid. At which trophic level is maximum energy \navailable ?", + "Click Here To InstallCBSE Science Class 10 Sample P aper 10 P age 3\nhttps://qrbook.page.link/app\n8. In the given figure the various trophic levels are shown in a pyramid. At which trophic level is maximum energy \navailable ?\n(a) T4\n(b) T2\n(c) T1\n(d) T3\n9. Process of conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen in plants is known as\n(a) Photosynthesis\n(b) Photoperiodism", + "sample question papers 27.txt\n(b) T2\n(c) T1\n(d) T3\n9. Process of conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen in plants is known as\n(a) Photosynthesis\n(b) Photoperiodism\n(c) Plan t nutrition\n(d) Plan t hormone functions\n10. A round, green seeded pea plant (RR yy) is crossed with wrinkled, yellow seeded pea plant, (rr YY) the seeds produced in F\n1 generation are\n(a) round and y ellow\n(b) round and green \n(c) wrinkled and green\n(d) wrinkled and y ellow \n11. Electrical impulse travels in a neuron from\n(a) Dendrite \" axon \" axonal end \" cell body\n(b) Cell b ody \" dendrite \" axon \" axonal end\n(c) Dendrite \" cell body \" axon \" axonal end\n(d) Axonal end \" axon \" cell body \" dendrite\n12. Which among the following is not the function of testes at puberty?", + "Which among the following is not the function of testes at puberty?\n(a) formation of germ cells\n(b) secretion of testosterone\n(c) dev elopment of placenta\n(d) secretion of estrogen\n(a) (i) and (ii)\n(b) (ii) and (iii)\n(c) (iii) and (iv)\n(d) (i) and (iv)", + "sample question papers 27.txt\n(a) formation of germ cells\n(b) secretion of testosterone\n(c) dev elopment of placenta\n(d) secretion of estrogen\n(a) (i) and (ii)\n(b) (ii) and (iii)\n(c) (iii) and (iv)\n(d) (i) and (iv)\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 10 CBSE Science Class 10\nhttps://qrbook.page.link/app\n13. \nIn an electrical circuit two resistors of 2\u2126 and 4\u2126 respectively are connected in series to a 6 V battery as shown \nin the figure. The heat dissipated by the 4\u2126 resistor in 5 s will be\n(a) 5 J\n(b) 10 J\n(c) 20 J\n(d) 30 J\n14. A prism ABC (with BC as base) is placed in different orientations. A narrow beam of white light is incident \non the prism as shown in figure.", + "A prism ABC (with BC as base) is placed in different orientations. A narrow beam of white light is incident \non the prism as shown in figure. In which of the following cases, after dispersion, the third colour from the top \ncorresponds to the colour of the sky?\n(a) (i)\n(b) (ii)\n(c) (iii)\n(d) (iv)\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 27.txt\ncorresponds to the colour of the sky?\n(a) (i)\n(b) (ii)\n(c) (iii)\n(d) (iv)\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 10 P age 5\nhttps://qrbook.page.link/app\n15. When a 4 V battery is connected across an unknown resistor, there is a current of 100 mA in the circuit. The value \nof the resistance of the resister is\n(a) 4\u2126 (b) 40\u2126 \n(c) 400\u2126 (d) .04\u2126 \n16. Which one of the following statements is correct?\n(a) The rain bow is produced by the reflection of white sun light by water drops in the atmosphere.\n(b) The blue colour of the sky is due to scattering of ligh t.\n(c) The stars app ear higher in the sky than actually are,due to scattering of light.\n(d) The planets t winkle at night due to atmospheric refraction of light.", + "(b) The blue colour of the sky is due to scattering of ligh t.\n(c) The stars app ear higher in the sky than actually are,due to scattering of light.\n(d) The planets t winkle at night due to atmospheric refraction of light.\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.", + "sample question papers 27.txt\n(d) The planets t winkle at night due to atmospheric refraction of light.\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : Chemical equations can be made more informative.Reason : We can write physical state of reactants and products, temperature and pressure, name of catalyst used etc.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : Traits like eye colour or height are inherited traits.Reason : Inherited traits are not transferred from parents to young ones.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).", + "sample question papers 27.txt\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : Aerobic animals are not truly aerobic.Reason : Anaerobically they produce lactic acid.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Both Assertion and Reason are false.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 10 CBSE Science Class 10\nhttps://qrbook.page.link/app\n20. Assertion : Electric appliances with metallic body have three connections, whereas an electric bulb has two pin \nconnections.Reason : Three pin connections reduce heating of connecting wires.", + "sample question papers 27.txt\n20. Assertion : Electric appliances with metallic body have three connections, whereas an electric bulb has two pin \nconnections.Reason : Three pin connections reduce heating of connecting wires.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. Name two metals that start floating after sometime when immersed in water and why ?\n\u0001o\u0001\nExplain how mercury is extracted from its sulphide ore (Cinnabar). Give equations of the reactions involved.\n22. A certain tissue in a green plant somehow got blocked and the leaves wilted. What was the tissue that got blocked?\n23. Write one function each of the salivary glands, liver and pancreas.", + "sample question papers 27.txt\n22. A certain tissue in a green plant somehow got blocked and the leaves wilted. What was the tissue that got blocked?\n23. Write one function each of the salivary glands, liver and pancreas.\n24. Why and how does water enter continuously into the root xylem ?\n25. The current flowing through a resistor connected in a circuit and the potential difference developed across its ends \nare as shown in the diagram by milliammeter and voltmeter readings respectively :\n(a) What are the least coun ts of these meters?\n(b) What is the resistance of the resistor?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 10 P age 7\nhttps://qrbook.page.link/app\n\u0001o\u0001\nA V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit \ndiagram to obtain such a graph.\n26. Name the four types of teeth present in adult human and mention their main functions.\nSECTION-C", + "sample question papers 27.txt\ndiagram to obtain such a graph.\n26. Name the four types of teeth present in adult human and mention their main functions.\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. What does a balanced chemical equation convey ?\n28. (a) An elemen t X on reacting with oxygen forms an oxide OX2. The oxide dissolves in water and turns blue \nlitmus red. Predict the nature of the element whether metal or non-metal.\n(b) A solution of copp er sulphate was kept in an iron pot. After few days, the pot developed some holes in it. \nHow will you account for this ?\n29. Define the two main methods of reproduction in living organisms.\n\u0001o\u0001\nList six specific characteristics of sexual reproduction.\n30. Ram placed an object in front of a convex lens of focal length 15 cm. The image formed is three times the size of \nthe object. Calculate the two possible distances of the object from the lens.", + "sample question papers 27.txt\n30. Ram placed an object in front of a convex lens of focal length 15 cm. The image formed is three times the size of \nthe object. Calculate the two possible distances of the object from the lens.\n31. The image of a candle flame placed at a distance of 30 cm from a mirror is formed on a screen placed infront of the mirror at a distance of 60 cm from its pole. What is the nature of the mirror ? Find its focal length, if the height of the flame is 2.4 cm, find the \nh eight of its image. State whether the image formed is erect or inverted.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 10 CBSE Science Class 10\nhttps://qrbook.page.link/app\n32. Diagram below shows a circuit containing a coil wound over a long and thin hollow cardboard tube. Copy the \ndiagram.\n(i) Sho w the polarity acquired by each face of the solenoid.\n(ii) Dra w the magnetic field lines of force inside the coil and also show their direction.", + "sample question papers 27.txt\ndiagram.\n(i) Sho w the polarity acquired by each face of the solenoid.\n(ii) Dra w the magnetic field lines of force inside the coil and also show their direction.\n(iii) Men tion two methods to increase the strength of the magnetic field inside the coil.\n\u0001o\u0001\nState the factors on which the resistance of a cylindrical conductor depends. How will resistance of a conductor \nchange if it is stretched so that its length is doubled ?\n33. (a) What is full form of (i) UNEP (ii) CF Cs.\n(b) On what basis are organisms group ed as producers, consumers and decomposer ?\n(c) W rite two problems that would arise if there were no decomposer in are ecosystem.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. (a) Dra w electron dot structure of methane molecule.\n(b) Iden tify the functional groups present in the following compounds :\n(i) CHO 26\n(ii) CHO 24\n(c) A mixture of oxygen and ethyne is burnt for welding. Why do you think a mixture of ethyne and air is not \nused for welding ?\n\u0001o\u0001", + "sample question papers 27.txt\n(i) CHO 26\n(ii) CHO 24\n(c) A mixture of oxygen and ethyne is burnt for welding. Why do you think a mixture of ethyne and air is not \nused for welding ?\n\u0001o\u0001\n(a) Explain why carbon forms covalent bond? Give two reasons for carbon forming a large number of compounds.\n(b) Explain the formation of ammonia molecule.\n35. What is vegetative propagation? Briefly describe various methods of vegetative propagation.\n\u0001o\u0001\n(a) Dra w diagram of human alimentary canal and label the following :\n(i) P art in which starch digestion starts.\n(ii) P art in which bile is stored.\n(iii) P art in which nutrients are absorbed.\n(iv) P art in which water is absorbed.\n(b) Men tion the role of hydrochloric acid in the stomach.\n(c) What function is serv ed by the following :\n(i) Gastric sphincter\n(ii) Anal sphincterInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 10 P age 9\nhttps://qrbook.page.link/app\n36.", + "Click Here To InstallCBSE Science Class 10 Sample P aper 10 P age 9\nhttps://qrbook.page.link/app\n36. (a) What is an electromagnet ? List an y two uses.", + "sample question papers 27.txt\nClick Here To InstallCBSE Science Class 10 Sample P aper 10 P age 9\nhttps://qrbook.page.link/app\n36. (a) What is an electromagnet ? List an y two uses.\n(b) Dra w a labelled diagram to show how an electromagnet is made.\n(c) State the purp ose of soft iron core used in making an electromagnet.\n(d) List t wo ways of increasing the strength of an electromagnet if the material of the electromagnet is fixed.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data -based questions with 2 to 3 short sub-parts. Internal choice is provided in \none of these sub-parts.\n37. When a metal is attacked by substances around it such as moisture, acids, etc., it is said to corrode, and this process is called corrosion. The black coating on silver, green coating on copper and reddish-brown powder on iron surface are some examples of corrosion.\n(i) What is the name giv en to the corrosion of iron?\n(ii) What is the form ula of green colour coating on copper?", + "sample question papers 27.txt\n(i) What is the name giv en to the corrosion of iron?\n(ii) What is the form ula of green colour coating on copper?\n(iii) Name t wo methods to prevent corrosion of iron.\n\u0001o\u0001\n(iv) Is corrosion a redo x reaction?\n38. The human brain is the central organ of the human nervous system and with the spinal cord makes up the central", + "sample question papers 27.txt\n\nnervous system. The brain consists of the cerebrum, the brain-stem and the cerebellum. It controls most of the activities of the body, processing, integrating and coordinating the information it receives from the sense organs and making decisions as to the instructions sent to the rest of the body. The brain is contained in and protected by, the skull bones of the head. The cerebrum is the largest part of the human brain. It is divided into two cerebral hemispheres. The cerebral cortex is an outer layer of grey matter, covering the core of white matter. The cortex is split into the neocortex and the much smaller allocortex. The neocortex is made up of six neuronal layers, while the allocortex has three or four. Each hemisphere is conventionally divided into four lobes \u2013 the frontal, temporal, parietal and occipital lobes. The frontal lobe is associated with executive functions including self-control, planning, reasoning and abstract thought, while the occipital lobe is dedicated to vision.", + "Each hemisphere is conventionally divided into four lobes \u2013 the frontal, temporal, parietal and occipital lobes. The frontal lobe is associated with executive functions including self-control, planning, reasoning and abstract thought, while the occipital lobe is dedicated to vision. The brain is protected by the skull, suspended in cerebrospinal fluid and isolated from the bloodstream by the blood brain barrier. However, the brain is still susceptible to damage, disease and infection. Damage can be caused by trauma, or a loss of blood supply known as a stroke. The brain is susceptible to degenerative disorders, such as Parkinson\u2019s disease, dementias including Alzheimer\u2019s disease and multiple sclerosis. Psychiatric condition, including schizophrenia and clinical depression, are thought to be associated with brain dysfunctions. The brain can also be the site of tumours, both Install NODIA App to See the Solutions.", + "sample question papers 27.txt\nClick Here To InstallPage 10 Sample P aper 10 CBSE Science Class 10\nhttps://qrbook.page.link/app\nbenign and malignant; these mostly originate from other sites in the body. The study of the anatomy of the brain \nis neuroanatomy, while the study of its function is neuroscience.\n(i) Whic h is the central part of the nervous system?\n(ii) What is the largest part of the h uman brain?\n(iii) What are the functions of the brain?\n\u0001o\u0001\n(iv) What is the branc h which studies the anatomy of brain?\n39. The human eye is like a camera. Its lens system forms an image on a light-sensitive screen called the retina. Light", + "sample question papers 27.txt\n\u0001o\u0001\n(iv) What is the branc h which studies the anatomy of brain?\n39. The human eye is like a camera. Its lens system forms an image on a light-sensitive screen called the retina. Light \nenters the eye through a thin membrane called the cornea. It forms the transparent bulge on the front surface of the eyeball as shown in the figure. The crystalline lens merely provides the finer adjustment of focal length required to focus objects at different distances on the retina. We find a structure called iris behind the cornea. Iris is a dark muscular diaphragm that controls the size of the pupil. The pupil regulates and controls the amount of light entering the eye.\nThe Human Eye\nThere are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) hypermetropia or far-sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.\n(i) What is the function of pu pil in the human eye ?", + "sample question papers 27.txt\n(i) What is the function of pu pil in the human eye ?\n(ii) What is the far p oint and near point of human eye with normal vision ?\n(iii) A student has difficulty reading the blackboard while sitting in the last row. What could be the defect the \nchild is suffering from ?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 10 P age 11\nhttps://qrbook.page.link/app\n\u0001o\u0001\n(iv) What is the function of iris in h uman eye ?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 2.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 15 Page 1\nhttps://qrbook.page.link/app\nSample Paper 15\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho wever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv e Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac h. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t ype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 2.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t ype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac h with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. What happens when a solution of an acid is mixed with a solution of a base in a jar as shown in the figure?\n1. Salt formation takes place.\n2. The temperature of the solution remains the same.\n3. The temperature of the solution decreases.\n4. The temperature of the solution increases.\n(a) Only 1\n(b) 1 and 2\n(c) 2 and 4\n(d) 1 and 4\n2. Which of the following reaction is characterised by the yellow colour of product?", + "4. The temperature of the solution increases.\n(a) Only 1\n(b) 1 and 2\n(c) 2 and 4\n(d) 1 and 4\n2. Which of the following reaction is characterised by the yellow colour of product?\n(a) () () PbOP bOsO g 26 34 2+T\n(b) ()Zn HSOZ nSOH s 24 42$ ++\n(c) Na CO HSON aSOH OC O 23 24 24 22 $ ++ +", + "sample question papers 2.txt\n2. Which of the following reaction is characterised by the yellow colour of product?\n(a) () () PbOP bOsO g 26 34 2+T\n(b) ()Zn HSOZ nSOH s 24 42$ ++\n(c) Na CO HSON aSOH OC O 23 24 24 22 $ ++ +\n(d) () KClO KC lO g 22 3 32 +TInstall NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 15 CBSE Science Class 10\nhttps://qrbook.page.link/app\n3. The statement of law of conservation of mass is:\n(a) mass can neither b e created nor destroy.\n(b) mass can b e created nor destroy.\n(c) mass of the b ody cannot be remain same.\n(d) none of these.\n4. The reaction between carbon and oxygen can be represented as\n() () () Cs Og CO gH eat 22++T\nIn which of the following type(s), the above reaction can be classified?\n1. Com bustion reaction\n2.", + "(d) none of these.\n4. The reaction between carbon and oxygen can be represented as\n() () () Cs Og CO gH eat 22++T\nIn which of the following type(s), the above reaction can be classified?\n1. Com bustion reaction\n2. Displacemen t reaction\n3. Endothermic reaction\n4. Com bination reaction\n(a) 1 and 3\n(b) 1, 3 and 4\n(c) 1 and 4\n(d) 1 Only\n5. Silver articles become black on prolonged exposure to air as shown in the figure. This is due to the formation of", + "sample question papers 2.txt\n4. Com bination reaction\n(a) 1 and 3\n(b) 1, 3 and 4\n(c) 1 and 4\n(d) 1 Only\n5. Silver articles become black on prolonged exposure to air as shown in the figure. This is due to the formation of\n(a) Ag N3\n(b) Ag O2\n(c) Ag S 2\n(d) Ag S 2 and Ag N3\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 15 P age 3\nhttps://qrbook.page.link/app\n6. Which of the following is used for dissolution of gold?\n(a) Hydro chloric acid\n(b) Sulph uric acid\n(c) Nitric acid\n(d) Aqua regia\n7. A molecule of ammonia (NH3) has\n(a) only single b onds\n(b) only double b onds\n(c) only triple b onds\n(d) t wo double bonds and one single bond\n8.", + "A molecule of ammonia (NH3) has\n(a) only single b onds\n(b) only double b onds\n(c) only triple b onds\n(d) t wo double bonds and one single bond\n8. Choose the correct statement that describe the arteries?\n(a) They ha ve thick elastic walls, blood flows under high pressure, collect blood from different organs and bring \nit back to the heart.", + "sample question papers 2.txt\n8. Choose the correct statement that describe the arteries?\n(a) They ha ve thick elastic walls, blood flows under high pressure, collect blood from different organs and bring \nit back to the heart.\n(b) They ha ve thin walls with valves inside, blood flows under low pressure and carry blood away from the heart \nto various organs of the body.\n(c) They ha ve thick elastic walls, blood flows under low pressure, carry blood from the heart to various organs \nof the body.\n(d) They ha ve thick elastic walls without valves inside, blood flows under high pressure and carry blood away \nfrom the heart to different parts of the body.\n9. Inorganic nitrates or nitrites helps the plants for the synthesis of:\n(a) Carbohydrates\n(b) Nitrogen\n(c) Carb on dioxide\n(d) Protein\n10. Select the statements that describe characteristics of genes\n(i) genes are sp ecific sequence of bases in a DNA molecule\n(ii) a gene do es not code for proteins", + "sample question papers 2.txt\n(c) Carb on dioxide\n(d) Protein\n10. Select the statements that describe characteristics of genes\n(i) genes are sp ecific sequence of bases in a DNA molecule\n(ii) a gene do es not code for proteins\n(iii) in individuals of a giv en species, a specific gene is located on a particular chromosome\n(iv) eac h chromosome has only one gene\n(a) (i) and (ii) (b) (i) and (iii)\n(c) (i) and (iv) (d) (ii) and (iv)\n11. A doctor advised a person to take an injection of insulin because\n(a) his blo od pressure was low\n(b) his heart w as beating slowly\n(c) he w as suffering from goitre\n(d) his sugar lev el in blood was high\n12.", + "A doctor advised a person to take an injection of insulin because\n(a) his blo od pressure was low\n(b) his heart w as beating slowly\n(c) he w as suffering from goitre\n(d) his sugar lev el in blood was high\n12. The number of chromosomes in parents and offsprings of a particular species remains constant due to\n(a) doubling of c hromosomes after zygote formation\n(b) halving of c hromosomes during gamete formation\n(c) doubling of c hromosomes after gamete formation\n(d) halving of c hromosomes after gamete formationInstall NODIA App to See the Solutions.", + "sample question papers 2.txt\n(b) halving of c hromosomes during gamete formation\n(c) doubling of c hromosomes after gamete formation\n(d) halving of c hromosomes after gamete formationInstall NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 15 CBSE Science Class 10\nhttps://qrbook.page.link/app\n13. An object is immersed in a fluid. In order that the object becomes invisible, it should\n(a) Beha ve as a perfect reflector\n(b) Absorb all ligh t falling on it\n(c) Ha ve refractive index one\n(d) Ha ve refractive index exactly matching with that of the surrounding fluid\n14. Study the following ray diagram:\nIn this diagram, the angle of incidence, the angle of emergence and the angle of deviation respectively have been \nrepresented by\n(a) ,,yp z\n(b) ,,xq z\n(c) ,,py z\n(d) ,,pz y\n15. Which of the following correctly represents graphical relation between angle of incidence ()i and angle of reflection \n()r ?", + "Which of the following correctly represents graphical relation between angle of incidence ()i and angle of reflection \n()r ?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "sample question papers 2.txt\n()r ?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 15 P age 5\nhttps://qrbook.page.link/app\n16. Twinkling of stars is due to atmospheric\n(a) disp ersion of light by water droplets\n(b) refraction of ligh t by different layers of varying refractive indices\n(c) scattering of ligh t by dust particles\n(d) in ternal reflection of light by clouds\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17. Assertion : The balancing of chemical equations is based on law of conservation of mass.\nReason : Total mass of reactants is equal to total mass of products.\n(a) Both Assertion and Reason are T rue and Reason is the correct explanation of the Assertion.\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.", + "sample question papers 2.txt\n(b) Both Assertion and Reason are T rue but Reason is not the Correct explanation of the Assertion.\n(c) Assertion is T rue but the Reason is False.\n(d) Both Assertion and Reason are F alse.\n18. Assertion : Plant hormones arc growth regulator.Reason : Growth regulators promote or inhibit the growth.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n19. Assertion : In woody plants, gaseous exchange occurs through lenticels.Reason : Lenticels are specialised cells found along with stomata on the stem of woody plants.\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.", + "sample question papers 2.txt\n(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.\n(b) Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.\n(c) Assertion is true but Reason is false.\n(d) Assertion is false but Reason is true.\n20. Assertion : Heater wire must have high resistance will be melting point.Reason : If resistance is high, the electric conductivity will be less.\n(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).\n(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).\n(c) Assertion (A) is true but reason (R) is false.\n(d) Assertion (A) is false but reason (R) is true.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 15 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.", + "sample question papers 2.txt\nClick Here To InstallPage 6 Sample P aper 15 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21. Name the gas which is usually produced when dil. sulphuric acid reacts with a metal. Illustrate it with an example. \nHow will you test the evolution of this gas.\n\u0001o\u0001\nThe soil in a field is highly acidic. List any two materials which can be added to this soil to reduce its acidity. \nGive the reason of your choice.\n22. What are the functions of the forebrain ?\n23. \u201cThe sex of the children is determined by what they inherit from their father and not their mother\u201d. Justify.\n24. Variations are important for the survival of species overtime. Justify this statement with reasons.\n25. State the two laws of reflection of light.\n\u0001o\u0001\nA concave mirror is known as a converging mirror while a convex mirror is known as diverging mirror? Explain \nwhy ?", + "sample question papers 2.txt\n25. State the two laws of reflection of light.\n\u0001o\u0001\nA concave mirror is known as a converging mirror while a convex mirror is known as diverging mirror? Explain \nwhy ?\n26. In the following food chain, 5 J of energy is available to man. How much energy was available at producer level ?\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27. How does rusting take place ? Write the conditions necessary for rusting to take place.\n28. Anhydrous copper sulphate ()CuSO 4 was dissolved in one beaker and hydrated copper sulphate ()CuSO HO542$ \nwas dissolved in another beaker. What heat changes do you expect in these beakers and why ?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 15 P age 7\nhttps://qrbook.page.link/app\n29. Explain Mendel\u2019s experiment with peas on inheritance of characters considering only one visible contrasting \ncharacter.\n\u0001o\u0001", + "sample question papers 2.txt\nhttps://qrbook.page.link/app\n29. Explain Mendel\u2019s experiment with peas on inheritance of characters considering only one visible contrasting \ncharacter.\n\u0001o\u0001\nAn angiosperm plant having red coloured flowers when crossed with the other having the same colour produced 40 \nprogenies out of which 30 plants were with red coloured flowers 10 plants were with white colour flowers.Finds out :\n(a) What is the p ossible genotype of parent plants ?\n(b) Whic h trait is dominated and recessive?\n(c) What is this cross called as and what is its phenot yping ratio ?\n30. A person suffering from short-sightedness can see clearly only upon a distance of 2 metres. Find the nature and power of the lens required to correct his vision.\n31. (a) An ob ject is placed at the focus of a convex lens. Draw a ray diagram to locate the position of the image \nformed, if any state its position and nature.\n(b) An ob ject placed 50 cm from a lens produces a virtual image at a distance of 10 cm from the lens. Find the", + "sample question papers 2.txt\nformed, if any state its position and nature.\n(b) An ob ject placed 50 cm from a lens produces a virtual image at a distance of 10 cm from the lens. Find the \nfocal length of the lens and also state the type of the lens used.\n32. (a) W rite the mathematical expression for Joule\u2019s law of heating.\n(b) Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential \ndifference of 40 V.\n\u0001o\u0001\nTwo electric lamps rated 100 W, 220 V and 25 W, 220 V are connected in parallel. Calculate the total electric \ncurrent in the circuit.\n33. Define an ecosystem. Draw a block diagram to show the flow of energy in an ecosystem.\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34. A metal E is stored under kerosene. When a small piece of it is left open in the air, it catches fire. When the \nproduct formed is dissolved in water it turns red litmus to blue.\n(i) Name the metal E.", + "sample question papers 2.txt\nproduct formed is dissolved in water it turns red litmus to blue.\n(i) Name the metal E.\n(ii) W rite the chemical equation for the reaction when it is exposed to air and when the product is dissolved in \nwater.\n(iii) Explain the pro cess by which the metal is obtained from its molten chloride.\n\u0001o\u0001\n(a) What is an ionic b ond ?\n(b) Ho w is an ionic bond formed ?\n(c) W rits the formation of magnesium chloride.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 15 CBSE Science Class 10\nhttps://qrbook.page.link/app\n35. (a) Dra w the diagram of a flower to show its male and female reproductive parts. Label the following parts in \nit :\n(i) Ovary\n(ii) Anther\n(iii) Filament\n(iv) Stigma\n(b) Ho w does fusion of male and female gametes take place in plants?\n\u0001o\u0001\nDifferentiate between blood and lymph.\n36.", + "Label the following parts in \nit :\n(i) Ovary\n(ii) Anther\n(iii) Filament\n(iv) Stigma\n(b) Ho w does fusion of male and female gametes take place in plants?\n\u0001o\u0001\nDifferentiate between blood and lymph.\n36. What is meant by magnetic force ? Name and explain the rule to determine the direction of force experienced by", + "sample question papers 2.txt\n\u0001o\u0001\nDifferentiate between blood and lymph.\n36. What is meant by magnetic force ? Name and explain the rule to determine the direction of force experienced by \na current carrying conductor in a magnetic field. How does this force gets affected on :\n(i) doubling the magnitude of curren t.\n(ii) rev ersing the direction of current flow and\n(iii) rev ersing the direction of magnetic field?\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.\n37. There are different types of chemical reactions occurring around us or being carried out for the benefit of mandkind, e.g. combination reactions, decomposition reactions, displacement reactions, precipitation reactions, reduction-oxidation (redox) reactions, photochemical reactions etc.Now, answer the following questions :\n(i) Com bustion of coke is a combination reaction. CO 2 is not a pollutant. Then, why is combustion of coke \nharmful?", + "sample question papers 2.txt\n(i) Com bustion of coke is a combination reaction. CO 2 is not a pollutant. Then, why is combustion of coke \nharmful?\n(ii) Whic h reaction followed by two combination reactions are involved in white wash of walls?\n(iii) Giv e one use of tin plating in daily life.\n\u0001o\u0001\n(iv) Ho w photochemical reactions have played an important in photography?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 15 P age 9\nhttps://qrbook.page.link/app\n38. Reproduction (or procreation or breeding) is the biological process by which new individual organisms-\u201coffspring\u201d", + "sample question papers 2.txt\n\nare produced from their \u201cparents\u201d. Reproduction is a fundamental feature of all known life; each individual organism exists as the result of reproduction. There are two forms of reproduction: asexual and sexual. In asexual reproduction, an organism can reproduce without the involvement of another organism. Asexual reproduction is not limited to single-celled organisms. The cloning of an organism is a form of asexual reproduction. By asexual reproduction, an organism creates a genetically similar or identical copy of itself. The evolution of sexual reproduction is a major puzzle for biologists. Sexual reproduction typically requires the sexual interaction of two specialized organisms, called gametes, which contain half the number of chromosomes of normal cells and are created by meiosis, with a male typically fertilizing a female of the same species to create a fertilized zygote. This produces offspring organisms whose genetic characteristics are derived from those of the two parental organisms. Asexual reproduction is a process by which organisms create genetically similar or identical copies of themselves without the contribution of genetic material from another organism. Bacteria divide asexually via binary fission.", + "This produces offspring organisms whose genetic characteristics are derived from those of the two parental organisms. Asexual reproduction is a process by which organisms create genetically similar or identical copies of themselves without the contribution of genetic material from another organism. Bacteria divide asexually via binary fission. Sexual reproduction is a biological process that creates a new organism by combining the genetic material of two organisms in a process that starts with meiosis, a specialized type of cell division.", + "sample question papers 2.txt\n(i) What do y ou mean by reproduction?\n(ii) Ho w many forms of reproduction are there? What are they?\n(iii) What do you mean by asexual reproduction?\n\u0001o\u0001\n(iv) What do y ou mean by sexual reproduction?\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 15 CBSE Science Class 10\nhttps://qrbook.page.link/app\n39. Mr. Rakesh was helping her daughter Aruna at home understanding about the basics of reflection of light. He \nfound the simplest way to convey the role of curved mirrors using stainless steel teaspoon. Aruna got surprised to know that virtual image is that which can be seen but cannot be obtained on the screen. Mr. Rakesh then explained the formation of image using Ray diagrams.\n(i) What do y ou mean by reflection of light?\n(ii) The angle b etween incident ray and reflected ray is 60\u00b0. What is the value of angle of incidence?", + "sample question papers 2.txt\n(i) What do y ou mean by reflection of light?\n(ii) The angle b etween incident ray and reflected ray is 60\u00b0. What is the value of angle of incidence?\n(iii) A ra y of light is incident on a plane mirror at an angle of 30\u00b0.what is the angle of reflection?\n\u0001o\u0001\n(iv) What happ ens to a light ray that is incident normally on a surface?\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "sample question papers 9.txt\nInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample Pap er 27 Page 1\nhttps://qrbook.page.link/app\nSample Paper 27\nClass X 2023-24\nScience (086)\nTime: 3 Hours Max. Marks: 80\nGeneral Instructions:\n1. This question paper consists of 39 questions in 5 sections.\n2. All questions are compulsory. Ho\nwever, an internal choice is provided in some questions. A student is expected \nto attempt only one of these questions.\n3. Section A consists of 20 Objectiv\ne Type questions carrying 1 mark each.\n4. Section B consists of 6 Very Short questions carrying 02 marks eac\nh. Answers to these questions should in the \nrange of 30 to 50 words.\n5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each.", + "5. Section C consists of 7 Short Answer t\nype questions carrying 03 marks each. Answers to these questions should \nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.", + "sample question papers 9.txt\nin the range of 50 to 80 words.\n6. Section D consists of 3 Long Answer t\nype questions carrying 05 marks each. Answer to these questions should \nbe in the range of 80 to 120 words.\n7. Section E consists of 3 source-based/case-based units of assessment of 04 marks eac\nh with sub-parts.\nSECTION-A\n\u0001Select and write one most appropriate option out of the four options given for each of the questions 1 \u2013 20.\n1. The diagram shows the h\numan gut. Which numbered structures secrete digestive enzymes?\n(a) 1, 2, 3 and 4\n(b) 2, 3, 5, and 6\n(c) 1, 2, 3 and 6\n(d) 2, 3, 4 and 5\n\u0001Cont onueCo oue en paueeeeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 2 Sample P aper 27 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2.", + "Click Here To InstallPage 2 Sample P aper 27 CBSE Science Class 10\nhttps://qrbook.page.link/app\n2. Whic h of the following relation is correct?\n(a) ll l 2312 3==\n(b) ll l 3212 3==\n(c) lll23 11 3 ==\n(d) ll l 43 12 3 ==\n3. F our metals P, Q, R and S are tested with water, steam and dilute hydrochloric acid. The table given below", + "sample question papers 9.txt\n(a) ll l 2312 3==\n(b) ll l 3212 3==\n(c) lll23 11 3 ==\n(d) ll l 43 12 3 ==\n3. F our metals P, Q, R and S are tested with water, steam and dilute hydrochloric acid. The table given below \nshows the results of the experiment.\nMetals Reaction with water Reaction with steam Reaction with dil. HCl\nP # { {\nQ # # {\nR { { {\nS # # #\nBetween which two metals should hydrogen be placed in the series?\n(a)\n P and R\n(b)\n P and Q\n(c)\n R and S\n(d)\n Q and S\n4.\n The figure giv\nen below represents the experiment carried out between conc. sulphuric acid and sodium chloride, \nwhich react with each other to form HCl gas.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.", + "The figure giv\nen below represents the experiment carried out between conc. sulphuric acid and sodium chloride, \nwhich react with each other to form HCl gas.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 27 P age 3\nhttps://qrbook.page.link/app\nBlue litmus paper is brought near the mouth of the Delivery tube to check the presence of HCl acid but no change \nis observed in the colour of litmus paper because:\n(a)\n The litm\nus paper used is moist.\n(b)\n The litm", + "sample question papers 9.txt\nis observed in the colour of litmus paper because:\n(a)\n The litm\nus paper used is moist.\n(b)\n The litm\nus paper used is dry.\n(c)\n The litm\nus paper is kept very dose to the mouth of the delivery tube.\n(d)\n Blue litm\nus paper does not change its colour with an acid.\n5.\n In \none of the industrial processes used for manufacture of sodium hydroxide, a gas X is formed as by-product. \nThe gas reacts with lime water to give a compound Y which is used as a bleaching agent in chemical industry. \nThe compound X and Y could be:\n(a)\n CO 2 and CaOCl 2 respectively \n(b)\n H2 and NaHCO 3 respectively\n(c)\n Cl2 and NaHCO 3 respectively \n(d)\n Cl2 and CaOCl 2 respectively\n6.\n The maxim\num resistance which can be made using four resistors each of resistance (1/2) \u2126 is:\n(a)", + "sample question papers 9.txt\n1 \u2126 \n(b) 2 \u2126 \n(c) 8 \u2126 \n(d) 2.5 \u2126 \n7. Whic h of the following is the correct representation of electron dot structure of nitrogen?\n8. Whic h of the following measures can be adopted to prevent or slow down rancidity?\n(i)\n F\nood materials should be packed in air tight container.\n(ii)\n F\nood should be refrigerated.\n(iii)\n F\nood materials and cooked food should be kept away from direct sunlight.\n(a)\n I and I\nII\n(b)\n I\nI and III\n(c)\n I\nI and ID\n(d)\n I, I\nI and III\n9.\n Matc\nh the terms of Column (I) with those of Column (II).", + "(a)\n I and I\nII\n(b)\n I\nI and III\n(c)\n I\nI and ID\n(d)\n I, I\nI and III\n9.\n Matc\nh the terms of Column (I) with those of Column (II).\nColumn(I) Column(II)\n(A) Olfactory receptors (i) Tongue\n(B) Thermoreceptors (temperature receptors) (ii) Eye\n(C) Gustatoreceptors (iii) Nose\n(D) Photoreceptors (iv) Skin\n(a)\n (A)-(iii), (B)-(iv), (C)-(i), (D)-(ii)\n(b)\n (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)\n(c)\n (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)\n(d)\n (A)-(iii), (B)-(ii), (C)-(iv), (D)-(i)Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 27 CBSE Science Class 10", + "sample question papers 9.txt\n(c)\n (A)-(ii), (B)-(iv), (C)-(i), (D)-(iii)\n(d)\n (A)-(iii), (B)-(ii), (C)-(iv), (D)-(i)Install NODIA App to See the Solutions.\nClick Here To InstallPage 4 Sample P aper 27 CBSE Science Class 10\nhttps://qrbook.page.link/app\n10. Iron nail dipp ed in copper sulphate solution for about half an hour\nWhich of the following is the correct observation of the reaction shown in the above set up?\n(a)\n Blue colour of copp\ner sulphate solution is obtained\n(b)\n Copp\ner displaces iron\n(c)\n Reaction is exothermic\n(d)\n No reaction tak\nes place\n11.\n If a round, \ngreen seeded pea plant (RRyy) is crossed with wrinkled, yellow seeded pea plant (rrYY), the seeds \nproduced in F1, generation are:\n(a)\n wrinkled and y\nellow\n(b)\n wrinkled and green\n(c)\n round and y\nellow\n(d)\n round and green\n12.\n Select the incorrect option.", + "Select the incorrect option.\nIndicator Colour in acidic medium Colour in Basic medium\nA. Litmus (purple) Red Blue\nB. Flower of hydrangea plant (Blue) Blue Pink\nC. Red cabbage juice (Purple) Red or Pink Green", + "sample question papers 9.txt\nIndicator Colour in acidic medium Colour in Basic medium\nA. Litmus (purple) Red Blue\nB. Flower of hydrangea plant (Blue) Blue Pink\nC. Red cabbage juice (Purple) Red or Pink Green\nD. Turmeric Juice (Yellow) Colourless Yellow\n(a)", + "sample question papers 9.txt\nA\n(b)\n \nB\n(c)\n \nC\n(d)\n \nD\n13.\n A\nt the centre of a bar magnet, the magnetism is:\n(a)\n \nminimum\n(b)\n \nmaximum\n(c)\n same as at the p\noles\n \n(d)\n \nzero\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 27 P age 5\nhttps://qrbook.page.link/app\n14. The image sho ws the magnetic field lines around a straight current carrying conductor.\nIf the direction of the current in the straight wire is changed, what change in the magnetic field line will be \nobserved?\n(a)", + "sample question papers 9.txt\n(b) \n(c) \n(d) \n15. The figure giv en below represents a single nephron from a mammalian kidney. Identify the labelled parts by \nselecting the most appropriate option.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 6 Sample P aper 27 CBSE Science Class 10\nhttps://qrbook.page.link/app\nA B C\n(a) Renal artery Collecting duct Bowman\u2019s capsule\n(b) Glomerulus Bowman\u2019s capsule Collecting duct(c) Bowman\u2019s capsule Glomerulus Collecting duct\n(d) Collecting duct Glomerulus Bowman\u2019s capsule\n16.\n The image sho\nws a surgical method in females to prevent pregnancy.\nWhich event will be likely prevented from this method?\n(a)\n Pro\nduction of eggs\n(b)\n Maturation of eggs\n(c)\n En\ntry of sperm into the uterus\n(d)\n En\ntry of eggs into the uterusInstall NODIA App to See the Solutions.", + "Which event will be likely prevented from this method?\n(a)\n Pro\nduction of eggs\n(b)\n Maturation of eggs\n(c)\n En\ntry of sperm into the uterus\n(d)\n En\ntry of eggs into the uterusInstall NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 27 P age 7\nhttps://qrbook.page.link/app\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17.", + "sample question papers 9.txt\nClick Here To InstallCBSE Science Class 10 Sample P aper 27 P age 7\nhttps://qrbook.page.link/app\n\u0001Question no. 17 to 20 are Assertion-Reasoning based questions.\n17.\n Assertion (A):\n Monohybrid cross deals with inheritance of one pair of contrasting characters.\nReason (R): Dihybrid cross deals with inheritance of two pairs of contrasting characters.\n(a)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(c)\n Assertion (A) is false but Reason (R) is true.\n(d)\n Assertion (A) is true but Reason (R) is false.\n18.\n Assertion (A):\n A fuse in a circuit prevents damage to the appliances and the circuit due to overloading.\nReason (R): Overloading occurs when the live wire and the neutral wire come into direct contact.", + "(d)\n Assertion (A) is true but Reason (R) is false.\n18.\n Assertion (A):\n A fuse in a circuit prevents damage to the appliances and the circuit due to overloading.\nReason (R): Overloading occurs when the live wire and the neutral wire come into direct contact.(a)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(b)", + "sample question papers 9.txt\nBoth Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(c)\n Assertion (A) is false but Reason (R) is true.\n(d)\n Assertion (A) is true but Reason (R) is false.\n19.\n Assertion (A):\n Leaves are the main site of photosynthesis.\nReason (R): Chloroplast are present in leaves.(a)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(c)\n Assertion (A) is false but Reason (R) is true.\n(d)\n Assertion (A) is true but Reason (R) is false. \n20.", + "(b)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(c)\n Assertion (A) is false but Reason (R) is true.\n(d)\n Assertion (A) is true but Reason (R) is false. \n20.\n Assertion (A):\n Following are the members of a homologous series:\nCH OH 3, CH CH OH 32 ,CH CH CH OH 32 2\nReason (R): A series of compounds with same functional group, but differing by \u2014 CH 2\u2014 unit is called a \nhomologous series.(a)", + "sample question papers 9.txt\nCH OH 3, CH CH OH 32 ,CH CH CH OH 32 2\nReason (R): A series of compounds with same functional group, but differing by \u2014 CH 2\u2014 unit is called a \nhomologous series.(a)\n Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).\n(b)\n Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).\n(c)\n Assertion (A) is false but Reason (R) is true.\n(d)\n Assertion (A) is true but Reason (R) is false.\nSECTION-B\n\u0001Question no. 21 to 26 are very short answer questions.\n21.\n Define the term p\nower of accommodation. Write the modification in the curvature of the eye lens which enables \nus to see the nearby objects clearly?\n\u0001o\u0001\nA glass prism is able to produce a spectrum when white Light passes through it but a rectangular block of same \ntransparent glass does not produce any spectrum. Why?\n22.", + "Write the modification in the curvature of the eye lens which enables \nus to see the nearby objects clearly?\n\u0001o\u0001\nA glass prism is able to produce a spectrum when white Light passes through it but a rectangular block of same \ntransparent glass does not produce any spectrum. Why?\n22.\n Tw\no green plants are kept separately in oxygen free containers, one in the dark and the other in continuous light.", + "sample question papers 9.txt\ntransparent glass does not produce any spectrum. Why?\n22.\n Tw\no green plants are kept separately in oxygen free containers, one in the dark and the other in continuous light. \nWhich one will live longer? Give reasons.\n23.\n What are the problems caused b\ny the non-biodegradable wastes that we generate?Install NODIA App to See the Solutions.\nClick Here To InstallPage 8 Sample P aper 27 CBSE Science Class 10\nhttps://qrbook.page.link/app\n24. In the follo wing figure showing a germinating gram seed, name the parts labelled as A, B, and C:\nWhy is part B considered to be important during germination?\n25.\n Discuss the role of pH in:\n(i)\n Digestiv\ne system, and\n(ii)\n Causes of to\noth decay.\n\u0001o\u0001\n(i)\n Giv\ne the constituents of baking powder.\n(ii)\n Wh\ny does cake or bread swell on adding baking powder? Write chemical equation.\n26.\n (i)\n Name one gustatory receptor and one olfactory receptor presen\nt in human beings.\n(ii)\n W", + "sample question papers 9.txt\n(ii)\n Wh\ny does cake or bread swell on adding baking powder? Write chemical equation.\n26.\n (i)\n Name one gustatory receptor and one olfactory receptor presen\nt in human beings.\n(ii)\n W\nrite (a) and (b) in the given flow chart of neuron through which information travels as an electrical \nimpulse.\nSECTION-C\n\u0001Question no. 27 to 33 are short answer questions.\n27.\n (i)\n W\ne do not clean ponds or takes, but an aquarium needs to be cleaned regularly. Why?\n(ii)\n Wh\ny is ozone layer getting depleted at the higher levels of the atmosphere? Mention one harmful effect \ncaused by its depletion.\n28.\n A studen\nt uses spectacles of focal length - .m25.\n(i)\n Name the defect of vision he is suffering from.\n(ii)\n Whic\nh lens is used for the correction of this defect?\n(iii)\n List t\nwo main causes of developing this defect.\n(iv)\n Compute the p\nower of this lens.\n29.\n What is mean\nt by solenoid? How does a current carrying solenoid behave? Give its main use.\n\u0001o\u0001", + "sample question papers 9.txt\n(iii)\n List t\nwo main causes of developing this defect.\n(iv)\n Compute the p\nower of this lens.\n29.\n What is mean\nt by solenoid? How does a current carrying solenoid behave? Give its main use.\n\u0001o\u0001\nMagnetic field lines are shown in the given diagram. A student makes a statement that the magnetic field at A is \nstronger than at B. Justify this statement. Also redraw the diagram and mark the direction of magnetic field lines.\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 27 P age 9\nhttps://qrbook.page.link/app\n30. Observ e the given figure and answer the following questions:\n(i) Whic h gas is evolved?\n(ii)\n Ho\nw will you test for the presence of gas evolved?\n(iii)\n What will happ\nen to the lime water?\n(iv)\n What happ\nens when the gas is passed for a longer time?\n31.", + "(ii)\n Ho\nw will you test for the presence of gas evolved?\n(iii)\n What will happ\nen to the lime water?\n(iv)\n What happ\nens when the gas is passed for a longer time?\n31.\n (i)\n W\nater has refractive index 1.33 and alcohol has refractive index 1.36. Which of the two mediums is optically", + "sample question papers 9.txt\n(iv)\n What happ\nens when the gas is passed for a longer time?\n31.\n (i)\n W\nater has refractive index 1.33 and alcohol has refractive index 1.36. Which of the two mediums is optically \ndenser? Give reason for your answer.\n(ii)\n Dra\nw a ray diagram to show the path of a ray of light passing obliquely from water to alcohol.\n(iii)\n State the relationship b\netween angle of incidence and angle of refraction in the above case.\n32.\n State the functions of plan\nt hormones. Name four different types of plant hormones.\n\u0001o\u0001\nA squirrel is in a scary situation. Its body has to prepare for either fighting or running away. State the immediate \nchanges that take place in its body so that the squirrel is able to either fight or run.\n33.\n (i)\n Sho\nw diagrammatically the electrons between the atoms in the formation of MgO. Write symbols of cation \nand anion present in MgO.\n(ii)\n Name the solv\nent in which ionic compounds are generally soluble.\n(iii)\n Wh", + "sample question papers 9.txt\nand anion present in MgO.\n(ii)\n Name the solv\nent in which ionic compounds are generally soluble.\n(iii)\n Wh\ny are aqueous solution of ionic\u2019 compounds able to conduct electricity?\n\u0001Cont\no\nnueCo\noue\ne\nn pa\nue\ne\neeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallPage 10 Sample P aper 27 CBSE Science Class 10\nhttps://qrbook.page.link/app\nSECTION-D\n\u0001Question no. 34 to 36 are Long answer questions.\n34.\n (i)\n \u201cBlo\nod circulation in fishes is different from the blood circulation in human beings.\u201d Justify the statement.\n(ii)\n Describ\ne \u201cblood circulation\u201d in human beings.\n\u0001o\u0001\nDesign an activity to show that chlorophyll is essential for photosynthesis.\n35.\n (i)\n Deduce the expression for the equiv\nalent resistance of the parallel combination of three resistors R1, R2 and \nR3.\n(ii)\n Consider the follo\nwing electric circuit.\n(a)\n Whic\nh two resistors are connected in series?", + "35.\n (i)\n Deduce the expression for the equiv\nalent resistance of the parallel combination of three resistors R1, R2 and \nR3.\n(ii)\n Consider the follo\nwing electric circuit.\n(a)\n Whic\nh two resistors are connected in series?\n(b)\n Whic\nh two resistors are connected in parallel?\n(c)\n If ev", + "sample question papers 9.txt\nR3.\n(ii)\n Consider the follo\nwing electric circuit.\n(a)\n Whic\nh two resistors are connected in series?\n(b)\n Whic\nh two resistors are connected in parallel?\n(c)\n If ev\nery resistors of the circuit is of 2 Q, what current will flow in the circuit?.\n36.\n (i)\n Dry p\nellets of a base X when kept in open absorbs moisture and turns sticky. The compound is also formed \nby chlor-alkali process. Write chemical name and formula of X. Describe chlor-alkali process with balanced \nchemical equation. Name the type of reaction that occurs when X is treated with dil. HCl. Write the \nchemical equation.\n(ii)\n While diluting an acid, wh\ny is it recommended that the acid should be added to water and not water to the \nacid?\n\u0001o\u0001\nWrite the main difference between an acid and a base. With the help of suitable examples explain the term \nneutralisation and the formation of:(i)", + "sample question papers 9.txt\nacidic,\n(ii)\n basic and\n(iii)\n neutral salts.\nSECTION-E\n\u0001Question no. 37 to 39 are case-based/data-based questions with 2 to 3 short sub-parts. Internal choice is provided in one of these sub-parts.\n37.\n No\nor, a young student, was trying to demonstrate some properties of light in her science project work. She kept \u2018X\u2019 \ninside the box (as shown in the figure) and with the help of a laser pointer made light rays pass through the holes \non one side of the box. She had a small butter-paper screen to see the spots of light being cast as they emerged.\n\u0001Cont onueCo oue e n pa ue e eeeee \u0001Install NODIA App to See the Solutions.\nClick Here To InstallCBSE Science Class 10 Sample P aper 27 P age 11\nhttps://qrbook.page.link/app\n(i) What could b e the X that she placed inside the box to make the rays behave as shown?\n(ii)\n A \nray of light falls normally on a face of X What are the values of angle of incidence and angle of refraction \nof this ray?\n(iii)", + "sample question papers 9.txt\n(ii)\n A \nray of light falls normally on a face of X What are the values of angle of incidence and angle of refraction \nof this ray?\n(iii)\n She measured the angles of incidence for b\noth the rays on the left side of the box to be 48.6\u00b0. She knew the \nrefractive index of the material X inside the box was 1.5. What will be the approximate value of angle of \nrefraction? (Use the value .. sin4 86 07 5. )\n\u0001o\u0001\n(iv)\n In an exp\neriment with X, a student observed that a ray of light incident at an angle of 55\u00b0 with the normal \non one face of the X, after refraction strikes the opposite face of the X before emerging out into air making \nan angle of 40\u00b0 with the normal. Draw a labelled diagram to show the path of this ray. What value would \nyou assign to the angle of refraction and angle of emergence?\n38.\n The partially digested fo\nod coming from the stomach of a person enters a long and narrow organ A in his body.", + "sample question papers 9.txt\nyou assign to the angle of refraction and angle of emergence?\n38.\n The partially digested fo\nod coming from the stomach of a person enters a long and narrow organ A in his body. \nThe organ A receives the secretions of two glands: liver and pancreas. Liver secretes a greenish-yellow liquid B. \nPancreas secretes pancreatic juice which contains two digestive enzymes C and D. The intestinal juice completes \nthe process of digestion of food. The inner wall of organ A has millions of tiny finger-like projections E which \nhelp in the rapid absorption of digested food into blood stream. The undigested part of food then passes into wider tube \nF which absorbs most of the water from undigested food. The last part of tube F called G stores this \nundigested food (or waste) for some time. The undigested food is then passed out though opening H as faeces in \nthe process known as I.\n(i)\n Enlist the site of syn\nthesis and storage of liquid B.\n(ii)\n What are the digestiv\ne enzymes C and D?\n(iii)\n Wh", + "sample question papers 9.txt\nthe process known as I.\n(i)\n Enlist the site of syn\nthesis and storage of liquid B.\n(ii)\n What are the digestiv\ne enzymes C and D?\n(iii)\n Wh\ny is organ A in herbivores longer than in carnivores?\n\u0001o\u0001\n(iv)\n Name (a) tub\ne F (b) part G (c) opening H and (d)process I.\n39.\n An organic comp\nound A of molecular formula CH24 on reduction gives another compound B of molecular formula \nCH26B$.on reaction with chlorine in the presence of sunlight gives C of molecular formula CHC l 25 .\n(i)\n Name the comp\nounds A and B. Write the chemical equation for the conversion of A to B.\n(ii)\n State the condition required for this reaction to tak\ne place.\n\u0001o\u0001\n(iii)\n Name the comp\nound C and write the chemical equation for the conversion of B to C.\n\u0001\u0001\u0001\u0001\u0001\u0001\u0001\u0001", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nClass 10 Metals and Non Metals NCERT Book Page Number: 40\nQuestion 1\nGive an example of a metal which :\n(i) is a liquid at room temperature.\n(ii) can be easily cut with a knife.\n(iii) is the best conductor of heat.\n(iv) is a poor conductor of heat.\nAnswer:\n(i) Mercury\n(ii) Sodium\n(iii) Silver\n(iv) Lead\nQuestion 2\nExplain the meanings of malleable and ductile.\nAnswer:\nMalleable : A metal that can be beaten into thin sheets on hammering is called malleable.\nDuctile : A metal which can be drawn into thin wires is called ductile.\n\nClass 10 Metals and Non Metals NCERT Book Page Number: 46\nQuestion 1\nWhy is sodium kept immersed in kerosene oil ?\nAnswer:\nSodium is highly reactive. So it is kept immersed in kerosene oil to prevent its reaction with oxygen, moisture and carbon dioxide of air to prevent accidental fires.\nQuestion 2\nWrite equations for the reactions of\n(i) iron with steam.\n(ii) calcium and potassium with water.", + "Answer:\nSodium is highly reactive. So it is kept immersed in kerosene oil to prevent its reaction with oxygen, moisture and carbon dioxide of air to prevent accidental fires.\nQuestion 2\nWrite equations for the reactions of\n(i) iron with steam.\n(ii) calcium and potassium with water.\nAnswer:\n\nQuestion 3", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 2\nWrite equations for the reactions of\n(i) iron with steam.\n(ii) calcium and potassium with water.\nAnswer:\n\nQuestion 3\nSamples of four metals A, B, C and D were taken and added to the following solution one by one.\nThe results obtained have been tabulated as follows :\nUse the Table above to answer the following questions about metals A, B, C and D.\n(i) Which is the most reactive metal ?\n(ii) What would you observe if B is added to a solution of copper (II) sulphate?\n(iii) Arrange the metals A, B, C and D in the order of decreasing reactivity.\nAnswer:\n(i) B is the most reactive metal because it gives displacement reaction with iron (II) sulphate.\n(ii) When metal B is added to copper (II) sulphate solution, a displacement reaction will take place due to which the blue colour of copper (II) sulphate solution will fade and a red-brown deposit of copper will be formed on metal B.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(iii) Metal B is the most reactive because it displaces iron from its salt solution. Metal A is less reactive because it displaces copper from its salt solution. Metal C is still less reactive because it can displace only silver from its salt solution and metal D is the least reactive because it cannot displace any metal from its salt solution. Hence, the decreasing order of reactivity of the metals is B > A > C > D.\nQuestion 4\nWhich gas is produced when dilute hydrochloric acid is added to a reactive metal ? Write the chemical reaction when iron reacts with dilute H2SO4.\nAnswer:\nHydrogen gas is produced when dilute hydrochloric acid is added to a reactive metal.\nChemical reaction when iron reacts with dilute H2SO4 :\nFe(s) + H2SO4(aq) \u2192 FeSO4(aq) + H2(g)\nQuestion 5\nWhat would you observe when zinc is added to a solution of iron (II) sulphate ? Write the chemical reaction that takes place.\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nFe(s) + H2SO4(aq) \u2192 FeSO4(aq) + H2(g)\nQuestion 5\nWhat would you observe when zinc is added to a solution of iron (II) sulphate ? Write the chemical reaction that takes place.\nAnswer:\nZinc is more reactive than iron. Therefore, when zinc is added to a solution of iron (II) sulphate, then the greenish colour of iron (II) sulphate solution fades gradually due to the formation of colourless zinc sulphate solution and iron metal is deposited on zinc.\n\nClass 10 Metals and Non Metals NCERT Book Page Number: 49\nQuestion 1\n(i) Write the electron dot structures for sodium, oxygen and magnesium.\n(ii) Show the formation of Na2O and MgO by the transfer of electrons.\n(iii) What are ions present in these compounds?\nAnswer:\n\n(ii) Formation of Na2O and MgO\n\n(iii) In Na2O, ions present are Na+ and O2-.\nIn MgO, ions present are Mg2+ and O2-.", + "(iii) What are ions present in these compounds?\nAnswer:\n\n(ii) Formation of Na2O and MgO\n\n(iii) In Na2O, ions present are Na+ and O2-.\nIn MgO, ions present are Mg2+ and O2-.\nQuestion 2\nWhy do ionic compounds have high melting points ?\n(iii) What are ions present in these compounds?\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nIn MgO, ions present are Mg2+ and O2-.\nQuestion 2\nWhy do ionic compounds have high melting points ?\n(iii) What are ions present in these compounds?\nAnswer:\nThe ionic compounds are made up of positive and negative ions. There is a strong force of attraction between the oppositely charged ions, so a lot of heat energy is required to break this force of attraction and melt the ionic compound. Due to this, ionic compounds have high melting points.\nClass 10 Metals and Non Metals NCERT Book Page Number: 53\nQuestion 1\nDefine the following terms : (i) Mineral, (ii) Ore and (iii) Gangue.\nAnswer:\n(i) Mineral : The natural materials in which the metals or their compounds are found in earth are called minerals.\n(ii) Ore : Those minerals from which the metals can be extracted conveniently and profitably are called ores.\n(iii) Gangue : The unwanted impurities like sand, rocky material, earth particles, lime stone, mica, etc in an ore are called gangue.\nQuestion 2", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(iii) Gangue : The unwanted impurities like sand, rocky material, earth particles, lime stone, mica, etc in an ore are called gangue.\nQuestion 2\nName two metals which are found in nature in the free state.\nAnswer:\nGold and platinum\nQuestion 3\nWhat chemical process is used for obtaining a metal from its oxide.\nAnswer:\nReduction process is used for obtaining a metal from its oxide.\nFor example, zinc oxide is reduced to metallic zinc by heating with carbon.\nZnO(s) + C(s) \u2192 Zn(s) + CO(g)\nBesides carbon, highly reactive metals like sodium, calcium, aluminium etc. are used as reducing agents. These displace metals of low reactivity from their oxides.\nFor example,\nFe2O3(s) + 2Al(s) \u2192 2Fe(l) + Al2O3(s) + Heat\nGold is Metal or Nonmetal ?", + "are used as reducing agents. These displace metals of low reactivity from their oxides.\nFor example,\nFe2O3(s) + 2Al(s) \u2192 2Fe(l) + Al2O3(s) + Heat\nGold is Metal or Nonmetal ?\nGold is a metal found in nature in the free state\nClass 10 Metals and Non Metals NCERT Book Page Number: 55\nQuestion 1\nMetallic oxides of zinc, magnesium and copper were heated with the following metals :", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nClass 10 Metals and Non Metals NCERT Book Page Number: 55\nQuestion 1\nMetallic oxides of zinc, magnesium and copper were heated with the following metals :\nIn which cases will you find displacement reactions taking place ?\nAnswer:\nA more reactive metal can displace a less reactive metal from its oxide. But out of zinc, magnesium, and copper metals, magnesium is the most reactive, zinc is less reactive whereas copper is the least reactive metal.\nThe displacement will take place in the following cases :\nQuestion 2\nWhich metals do not corrode easily ?\nAnswer:\nGold and Platinum.\nQuestion 3\nWhat are alloys ?\nAnswer:\nAn alloy is a homogeneous mixture of two or more metals, or a metal and a non-metal. For example, bronze is an alloy of copper and tin.\nNCERT Solutions for Class 10 Science Chapter 3 Textbook Chapter End Questions\nMetals and Nonmetals Class 10 Question 1.\nWhich of the following pairs will give displacement reactions ?\n(a) NaCl solution and copper metal.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nMetals and Nonmetals Class 10 Question 1.\nWhich of the following pairs will give displacement reactions ?\n(a) NaCl solution and copper metal.\n(b) MgCl2 solution and aluminium metal.\n(c) FeSO4 solution and silver metal.\n(d) AgNO3 solution and copper metal.\nAnswer:\n(d) AgNO3 solution and copper metal.\nQuestion 2.\nWhich of the following methods is suitable for preventing an iron frying pan from rusting ?\n(a) Applying grease\n(b) Applying paint.\n(c) Applying a coating of zinc\n(d) All the above.\nAnswer:\n(c) Applying a coating of zinc.\nQuestion 3.\nAn element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be\n(a) calcium\n(b) carbon\n(c) silicon\n(d) iron\nAnswer:\n(a) Calcium.\nQuestion 4.", + "Question 3.\nAn element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be\n(a) calcium\n(b) carbon\n(c) silicon\n(d) iron\nAnswer:\n(a) Calcium.\nQuestion 4.\nFood cans are coated with tin and not with zinc because\n(a) zinc is costlier than tin\n(b) zinc has a higher melting point than tin\n(c) zinc is less reactive than tin\n(d) zinc is more reactive than tin.\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(a) zinc is costlier than tin\n(b) zinc has a higher melting point than tin\n(c) zinc is less reactive than tin\n(d) zinc is more reactive than tin.\nAnswer:\n(d) Zinc is more reactive than tin.\nMetals and Non metals Class 10 Question 5.\nYou are given a hammer, a battery, a bulb, wires and a switch.\n(a) How could you use them to distinguish between samples of metals and non-metals?\n(b) Assess the usefulness of these tests in distinguishing between metals and non-metals.\nAnswer:\n(a) Metals can be beaten into thin sheets with a hammer without breaking. Non-metals cannot be beaten with a hammer to form thin sheets. Non-metals break into pieces when hammered. Metals are malleable, while non-metals are non-melleable. When metals are connected into circuit using a battery, bulb, wires and switch, current passes through the circuit and the bulb glows. When non-metals (like sulphur) are connected, the bulb does not light up at all. Metals are good conductors of electricity.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(b) Because of malleability, metals can be casted into sheets. Metals are good conductors of electricity so these can be used for electrical cables.\nQuestion 6.\nWhat are amphoteric oxides ? Give two examples of amphoteric oxides ?\nOR\nWrite chemical equations that show aluminium oxide reacts with acid as well as base. [CBSE2011]\nAnswer:\nThose metal oxides which show basic as well as acidic behaviour are known as amphoteric oxides. In other words, metal oxides that react wtih both acids and bases to form salt and water are called amphoteric oxides. Aluminium oxide and zinc oxide are amphoteric in nature.\n\nQuestion 7.\nName two metals which will displace hydrogen from dilute acids and two metals which will not.\nAnswer:\n(i) Metals above hydrogen in the activity series like sodium and magnesium displace hydrogen from dilute acids.\n(ii) Metals below hydrogen in the activity series like copper, silver do not displace hydrogen from dilute acids.\nQuestion 8.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(ii) Metals below hydrogen in the activity series like copper, silver do not displace hydrogen from dilute acids.\nQuestion 8.\nIn the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte ?\nAnswer:\nCathode \u2013 Pure metal\nAnode \u2013 Impure metal\nElectrolyte \u2013 Metal salt solution\nQuestion 9.\nPratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it, as shown in the figure.\n(a) What will be the action of gas on\n(i) dry litmus paper ?\n(ii) moist litmus paper ?\n(b) Write a balanced chemical equation for the reaction taking place.\n\nAnswer:\n(i) Dry litmus paper \u2013 no action.\n(ii) Moist litmus paper \u2013 becomes red.\n\nQuestion 10.\nState two ways to prevent the rusting of iron.\nAnswer:\nWays to prevent rusting of iron are :\n(a) By painting\n(b) By galvanizing\nQuestion 11.", + "(ii) Moist litmus paper \u2013 becomes red.\n\nQuestion 10.\nState two ways to prevent the rusting of iron.\nAnswer:\nWays to prevent rusting of iron are :\n(a) By painting\n(b) By galvanizing\nQuestion 11.\nWhat type of oxides are formed when non-metals combine with oxygen ?", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAnswer:\nWays to prevent rusting of iron are :\n(a) By painting\n(b) By galvanizing\nQuestion 11.\nWhat type of oxides are formed when non-metals combine with oxygen ?\nAnswer:\nNon-metals combine with oxygen to form acidic oxides or neutral oxides.\n\nQuestion 12.\nGive reasons :\n(a) Platinum, gold and silver are used to make jewellery.\n(b) Sodium, potassium and lithium are stored under oil.\n(c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking.\n(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction.\nAnswer:\n(a) Platinum, gold and silver are used to make jewellery because these are malleable and ductile. These are highly resistant to corrosion.\n(b) Sodium, potassium and lithium are very reactive and catch fire when exposed to air. This is due to their low ignition temperature and high reactivity.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(b) Sodium, potassium and lithium are very reactive and catch fire when exposed to air. This is due to their low ignition temperature and high reactivity.\n(c) Aluminium forms a non-reactive layer of aluminium oxide on its surface. This layer prevents aluminium to react with other substances. That\u2019s why aluminium is used to make cooking utensils.\n(d) It is easier to reduce a metal oxide into free metal. Since it is easier to obtain metals from their oxides than from their carbonates or sulphides directly, therefore, the carbonate and sulphide ores are first converted to oxides for extracting the metals.\nQuestion 13.\nYou must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 13.\nYou must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.\nAnswer:\nThe sour substances such as lemon or tamarind juice contain acids. These acids dissolve the coating of copper oxide or basic copper carbonate present on the surface of tarnished copper vessels and makes them shining red-brown again.\nQuestion 14.\nDifferentiate between metal and non-metal on the basis of their chemical properties. [CBSE 2017 (Delhi)]\nAnswer:\nDifference between metals and non-metals\nQuestion 15.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 14.\nDifferentiate between metal and non-metal on the basis of their chemical properties. [CBSE 2017 (Delhi)]\nAnswer:\nDifference between metals and non-metals\nQuestion 15.\nA man went door-to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty repeat. Can you play the detective to find out the nature of the solution he has used ?\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAnswer:\nThe dishonest goldsmith dipped the gold bangles in aqua-regia (which contains 1 part of concentrated nitric acid and 3 parts of concentrated hydrochloric acid, by volume). Aqua-regia dissolved a considerable amount of gold from gold bangles and hence reduced their weight drastically. The dishonest goldsmith can recover the dissolved gold from aqua-regia by a suitable treatment.\nQuestion 16.\nGive reasons why copper is used to make hot water tanks and not steel (analloy of iron).\nAnswer:\n(i) Copper is a better conductor of heat than steel.\n(ii) Copper does not corrode easily. But steel corrodes easily.\n(iii) Copper does not react with water at any temperature, whereas iron reacts with water on heating.\nNCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals\n.\nQuestion 1\nWhat are amphoteric oxides? Give two examples of amphoteric oxides.\nSolution:\nAmphoteric oxides are the oxides, which react with both acids and bases to form salt and water. E.g. ZnO and Al2O3.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nWhat are amphoteric oxides? Give two examples of amphoteric oxides.\nSolution:\nAmphoteric oxides are the oxides, which react with both acids and bases to form salt and water. E.g. ZnO and Al2O3.\nQuestion 2\nName two metals, which will displace hydrogen from dilute acids, and two metals which will not.\nSolution:\nVery reactive metals like Zn and Mg displace hydrogen from dilute acids. On the other hand less reactive metals like Cu, Ag, etc. do not displace hydrogen from dilute acids.\nQuestion 3\nIn the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?\nSolution:\nAnode is impure, thick block of metal M.\nCathode is a thin strip/wire of pure metal M.\nElectrolyte is a suitable salt solution of metal M.\nMetals and nonmetals Class 10 PDF\nQuestion 4\nState two ways to prevent the rusting of iron.\nSolution:\nBy coating the surface of iron by rust proof paints.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nMetals and nonmetals Class 10 PDF\nQuestion 4\nState two ways to prevent the rusting of iron.\nSolution:\nBy coating the surface of iron by rust proof paints.\nBy applying oil or grease to the surface of iron objects so that supply of air consisting of moisture is cut off form the surface.\nQuestion 5\nWhat types of oxides are formed when non-metals combine with oxygen?\nSolution:\nWhen non-metals combine with oxygen it forms either neutral or acidic oxides. CO is a neutral oxide; N2O5 or N2O3 is an acidic oxide.\nextraction of metals from ores class 10 Question 6\nGive reason\ni. Metals replace hydrogen from dilute acids, where as non-metals do not.\nii. Carbonate and sulphide ores are usually converted into oxides during the process of extraction.\nSolution:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nGive reason\ni. Metals replace hydrogen from dilute acids, where as non-metals do not.\nii. Carbonate and sulphide ores are usually converted into oxides during the process of extraction.\nSolution:\ni. Metals are electropositive in nature. They readily lose electrons. These electrons reduce the protons liberated from the acid to liberate hydrogen gas, where as non-metals possess a tendency to gain electrons and hence they do not furnish electrons to protons liberated from acids. Hence H2 gas is not liberated.\nii. As it is easier to reduce metal oxides to metal, prior to reduction, metal sulphides and carbonates must be converted to oxides.\nQuestion 7\nDifferentiate between metals and non-metals on the basis of their chemical properties.\nSolution:\n\nQuestion 8\nExplain why the surface of some metals acquires a dull appearance when exposed to air for a long time.\nSolution:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nSolution:\n\nQuestion 8\nExplain why the surface of some metals acquires a dull appearance when exposed to air for a long time.\nSolution:\nThis is due to the surface oxidation of metals when exposed to moist air. For e.g. copper turns green on its surface due to the formation of basic copper carbonate Cu(OH) 2. CuCO3. Similarly silver becomes black due to the formation of black Ag2S and Aluminium forms a white coating of Al2O3 on its surface.\nQuestion 9\nState which of the following metals would give hydrogen when added to dilute hydrochloric acid. i. Iron, ii. Copper iii. Magnesium\nCopper does not react with dilute hydrochloric acid at all. This shows that copper is even less reactive than iron.\n\nQuestion 10\nName a non-metallic element, which conducts electricity.\nSolution:\nCarbon in the form of graphite conducts electricity, as there is a free electron in each carbon atom, which moves freely in between the hexagonal layers.\nQuestion 11\nWhich metals do not corrode easily?\nSolution:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 11\nWhich metals do not corrode easily?\nSolution:\nGold and platinum and other noble metals do not corrode in air.\nQuestion 12\nWhat are alloys?\nSolution:\nAlloys are homogeneous mixtures of two or more metals, or a metal and a non-metal.E.g. steel, brass, bronze, etc.\nQuestion 13\nDefine the following terms.\n(i) Minerals\n(ii) Ores\n(iii) Gangue\nSolution:\n(i) Minerals\nAll compounds or elements, which occur naturally in the earth\u2019s crust, are called minerals. Example: Alums, K2SO4.Al2(SO4)3 . 24 H2O, Bauxite Al2O3.2H2O\n(ii) Ores\nThose minerals from which a metal can be profitably extracted are called ores. Bauxite (Al2O3.2H2O) is the ore of Al, copper pyrite CuFeS2. All minerals are not ores but all ores are minerals.", + "Bauxite (Al2O3.2H2O) is the ore of Al, copper pyrite CuFeS2. All minerals are not ores but all ores are minerals.\n(iii) Gangue\nWhen an ore is mined from the earth, it is always found to be contaminated with sand rocky materials. The impurity of sand and rock materials present in the ore is known as gangue.\nQuestion 14\nName two metals that are found in nature in the free state.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 14\nName two metals that are found in nature in the free state.\nSolution:\nGold and platinum are found in the free state in nature.\nQuestion 15\nWhat is chemical process used for obtaining a metal from its oxide?\n\nQuestion 16\nName two metals, which can form hydrides with metals.\nSolution:\nSodium and calcium form stable hydrides on reacting with hydrogen.\nQuestion 17\nDoes every mineral have a definite and a fixed composition? Explain.\nSolution:\nYes, every mineral has a definite and a fixed composition. Minerals are widely distributed in the earth\u2019s crust in the form of oxides, carbonates, sulphides, sulphates, nitrates, etc. These minerals are formed as a result of chemical changes taking place during the formation of earth.\nClass 10 metals and nonmetals Question 18\nExplain the meaning of malleable and ductile.\nSolution:\nMalleable is being able to be beaten/hammered into thin sheets.\nDuctile is being able to be drawn into thin wires.\nQuestion 19", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nExplain the meaning of malleable and ductile.\nSolution:\nMalleable is being able to be beaten/hammered into thin sheets.\nDuctile is being able to be drawn into thin wires.\nQuestion 19\ni. Write the electron dot structures for sodium, oxygen and magnesium.\nii. Show the formation of MgO and Na2O by the transfer of electrons.\niii. What are the ions present in these compounds?\nSolution:\n\nii. Formation of Magnesium oxide\nWhen magnesium reacts with oxygen, the magnesium atom transfers its two outermost electrons to an oxygen atom. By losing 2 elections, the magnesium atoms form a magnesium ion (Mg2+) and by gaining 2 electrons, the oxygen atom forms an oxide ion (O2-).\n\nFormation of Sodium oxide\nTwo sodium atoms transfer their 2 outermost electrons to an oxygen atom. By losing two electrons, the two sodium atoms form two sodiumions (2Na+). And by gaining two electrons, the oxygen atom forms an oxide ion (O2-.)", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\niii. The ions present in sodium oxide compound (Na20) aie sodium ions (2Na+ and oxide ions (O2-).\nThe ions present in Magnesium oxide compound (MgO) are magnesiumions Mg2+ and oxide ions (O2-).\nQuestion 20\nYou must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.\nSolution:\nThe sour substances such as lemon (or tamarind juice) contain acids. These acids dissolve the coating of copper oxide or basic copper carbonate present on the surface of tarnished copper vessels and make them shining red-brown again.\nQuestion 21\nGive an example of a metal which\ni. is a liquid at room temperature.\nii. can be easily cut with a knife.\niii. is the best conductor of heat.\niv. is a poor conductor of heat.\nSolution:\ni. Mercury is in liquid state at room temperature.\nii. Sodium and potassium are soft metals which can be easily cut with a knife.\niii. Silver is the best conductor of electricity.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nSolution:\ni. Mercury is in liquid state at room temperature.\nii. Sodium and potassium are soft metals which can be easily cut with a knife.\niii. Silver is the best conductor of electricity.\niv. Mercury is a poor conductor of heat.\nQuestion 22\nWhy is sodium kept immersed in kerosene?\nSolution:\nSodium metal is kept immersed in kerosene to prevent their reaction with oxygen, moisture and carbon dioxide of air.\nQuestion 23\nWhy do ionic compounds have high melting points?\nSolution:\nThese compounds are made up of positive and negative ions. There is a strong force of attraction between the oppositively charged ions, so a lot of heat energy is required to break this force of attraction and melt the ionic compounds. This is why ionic compounds have high melting points.\nQuestion 24", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 24\nA man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?\nSolution:\nAqua regia (By volume, this contains 3 parts of concentrated hydrochloric acid and 1 part of concentrated nitric acid) is the solution, which is used to sparkle the bangles like new, but their weight will be reduced drastically.\nQuestion 25\nWrite equations for the reactions of\n(i) iron with water\n(ii) calcium and potassium with water\nSolution:\n\nQuestion 26\nWhat would you observe when zinc is added to a sodium of iron(II) sulphate? Write the chemical reaction that takes place?\nSolution:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(ii) calcium and potassium with water\nSolution:\n\nQuestion 26\nWhat would you observe when zinc is added to a sodium of iron(II) sulphate? Write the chemical reaction that takes place?\nSolution:\nZinc is more reactive (more electro positive) than iron. Therefore it displaces iron from its salt solution. The colour of ferrous sulphate is pale green which becomes colourless.\n\nMetals and nonmetals class 10 Question 27\nPratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test-tube over the burning sulphur.\nWhat will be the action of this gas on:\nDry litmus paper?\nMoist litmus paper?\nWrite a balanced chemical equation for the reaction taking place.\nSolution:\na) When sulphur is brunt in air then sulphur dioxide gas is formed.\n(i) Sulphur dioxide gas has no action on dry litmus paper.\n(ii) Sulphur dioxide gas turns moist blue litmus paper to red.", + "Write a balanced chemical equation for the reaction taking place.\nSolution:\na) When sulphur is brunt in air then sulphur dioxide gas is formed.\n(i) Sulphur dioxide gas has no action on dry litmus paper.\n(ii) Sulphur dioxide gas turns moist blue litmus paper to red.\n(b) S(s) + O2(g) \u2014> SO2(g)\nMultiple Choice Questions (MCQs) [1 Mark each]", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(i) Sulphur dioxide gas has no action on dry litmus paper.\n(ii) Sulphur dioxide gas turns moist blue litmus paper to red.\n(b) S(s) + O2(g) \u2014> SO2(g)\nMultiple Choice Questions (MCQs) [1 Mark each]\nMetals and nonmetals class 10 Question 1.\nWhat is the colour of aqueous solution of CuSO4 and FeSO4 as observed in the laboratory?\n(a) CuSO4 \u2013 blue; FeSO4 \u2013 light green\n(b) CuSO4 \u2013 blue; FeSO4 \u2013 dark green\n(c) CuSO4 \u2013 green; FeSO4 \u2013 blue\n(d) CuSO4 \u2013 green; FeSO4 \u2013 colourless\nAnswer:\n(a) Colour of CuSO4 solution is blue and FeSO4 solution is light green.\nMetals and nonmetals class 10 Question 2.\nA student took four test tubes I, II, III and IV containing aluminium sulphate, copper sulphate? ferrous sulphate and zinc sulphate solutions respectively. He placed an iron strip in each of them.", + "Metals and nonmetals class 10 Question 2.\nA student took four test tubes I, II, III and IV containing aluminium sulphate, copper sulphate? ferrous sulphate and zinc sulphate solutions respectively. He placed an iron strip in each of them.\n\nIn which test tube, he found a brown deposit?\n(a) I\n(b) II\n(c) III\n(d) IV\nAnswer:\n(b) In test tube II, because Fe is more reactive than copper but less reactive than Al arid Zn.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nIn which test tube, he found a brown deposit?\n(a) I\n(b) II\n(c) III\n(d) IV\nAnswer:\n(b) In test tube II, because Fe is more reactive than copper but less reactive than Al arid Zn.\nMetals and nonmetals class 10 Question 3.\nAluminium sulphate and copper sulphate solutions were taken in two test tubes I and II respectively. A few pieces of iron filings were then added to both the solutions. The four students A, B, C and D recorded their observations in the form of a table as given below:\nWhich student has recorded the correct observation?\n(a) D\n(b) C\n(c) B\n(d) A\nAnswer:\n(c) Student B\nIron does not react with Al2(SO4)3 solution because iron is less reactive than aluminium. But Fe being more reactive than Cu displaces Cu from CuSO4 solution.\n\nMetals and nonmetals class 10 Question 4.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nMetals and nonmetals class 10 Question 4.\nAqueous solutions of zinc sulphate and iron sulphate were taken in test tubes I and II by four students A, B, C and D. Metal pieces of iron and zinc were dropped in the two solutions and observations made after several hours were recorded in the form of table as given below:\nWhich student has given the correct report?\n(a) B\n(b) D\n(c) A\n(d) C\nAnswer:\n(d) Student C\n(i) Fe is less reactive than zinc. So,\n\n(ii) Zn is more reactive than Fe, so it displaces iron as follows:\n\nMetals and nonmetals class 10 Question 5.\n2 mL each of cone. HCl, cone. HNO3 and a mixture of cone. HCl and cone. HNO3 in the ratio of 3 : 1 were taken in test tubes labelled as A, B and C. A small piece of metal was put in each test tube.", + "HCl, cone. HNO3 and a mixture of cone. HCl and cone. HNO3 in the ratio of 3 : 1 were taken in test tubes labelled as A, B and C. A small piece of metal was put in each test tube. No change occurred in test tubes A\u2019and Bbut the metal got dissolved in test tube C. The metal could be [NCERT Exemplar]\n(a) Al\n(b) Au\n(c) Cu\n(d) Pt\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(a) Al\n(b) Au\n(c) Cu\n(d) Pt\nAnswer:\n(b, d) A mixture of cone. HCl and cone. HNO3 in the ratio of 3 : 1 is known as aqua-regia. Gold (Au) and platinum (Pt) dissolve only in aqua-regia as these metals are very less reactive.\nMetals and nonmetals class 10 Question 6.\nWhen an aluminium strip is kept (a) Green solution of FeSO4 slowly turns brown\n(b) Green solution of FeSO4 rapidly turns brown\n(c) No change in colour of FeSO4\n(d) Green solution of FeSO4 slowly turns colourless\nAnswer:\n(a) The green solution of ferrous sulphate slowly turns brown. As aluminium is more reactive than iron, it displaces iron from ferrous sulphate solution.\n\nMetals and nonmetals class 10 Question 7.\nAluminium is used for making cooking utensils. Which of the following properties of aluminium are responsible for the same?", + "As aluminium is more reactive than iron, it displaces iron from ferrous sulphate solution.\n\nMetals and nonmetals class 10 Question 7.\nAluminium is used for making cooking utensils. Which of the following properties of aluminium are responsible for the same?\n(i) Good thermal conductivity\n(ii) Good electrical conductivity\n(iii) Ductility\n(iv) Fligh melting point [NCERT Exemplar]\n(a) (i) and (ii)\n(b) (i) and (iii)\n(c) (ii) and (iii)\n(d) (i) and (iv)", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(i) Good thermal conductivity\n(ii) Good electrical conductivity\n(iii) Ductility\n(iv) Fligh melting point [NCERT Exemplar]\n(a) (i) and (ii)\n(b) (i) and (iii)\n(c) (ii) and (iii)\n(d) (i) and (iv)\nAnswer:\n(d) Good thermal conductivity, malleability, light weight and high melting point are the properties/of aluminium due to which it-is used for making cooking utensils.\nMetals and nonmetals class 10 Question 8.\nIf copper is kept open in air, it slowly loses its shining brown surface and gains a green coating.", + "Metals and nonmetals class 10 Question 8.\nIf copper is kept open in air, it slowly loses its shining brown surface and gains a green coating. It is due to the formation of [NCERT Exemplar]\n(a) CuSO4\n(b) CuCO3\n(c) CU(NO3)2\n(d) CuO\nAnswer:\n(b) Copper reacts with CO2 present in air and forms a green coating on its surface due to the formation of basic copper carbonate [CuCO3.Cu(OH)2] as:\n\nNCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals (Hindi Medium)\nMetals and Non-Metals\nThere 92 well known naturally occurring minerals of which 70 are metals and rest 20 are the non-metals.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nMetals and Non-Metals\nThere 92 well known naturally occurring minerals of which 70 are metals and rest 20 are the non-metals.\nPhysical Properties of Metals & Non-Metals\nReaction between Metals and Non Metals\nReactivity of an element can be explained as tendency to attain a completely filled outermost shell.\nMetals have 1, 2 or 3 e- in outermost shell and thus it is easier for them to loss e- rather than to gain. They loss e- & gains positive charge & are tenned as cation.\nIn contrast, non-metals have 4-8 e- in outermost shell & thus they gain e- to achieve their octet. They gain e- as well as negative charge & tenned as anion.\nCations & anions attract each other & are held by strong electrostatic force of attraction.\nThe compounds fonned by the transfer of electrons from metal to non-non-metal are known as ionic compounds or electrovalent compound.\n\nOccurrence of Metals\nThe elements or compounds, which occur naturally in the earth\u2019s crust, are known as minerals.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nOccurrence of Metals\nThe elements or compounds, which occur naturally in the earth\u2019s crust, are known as minerals.\nAt some places, minerals contain a very high percentage of a particular metal and the metal can be profitably extracted from it. These minerals arc called ores.\nCorrosion\nCorrosion is the deterioration of materials by chemical interaction with their environment for e.g. darkening of silver articles when exposed to air, gaining of green coat on copper, rusting of iron.\nPrevention: The rusting of iron can be prevented by painting, oiling, greasing, galvanising, chrome plating, anodising or making alloys.\n\u2022 Galvanisation is a method of protecting steel and iron from rusting by coating them with a thin layer of zinc.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n\u2022 Galvanisation is a method of protecting steel and iron from rusting by coating them with a thin layer of zinc.\n\u2022 Alloy is a homogeneous mixture of two or more metals, or a metal & nonmetal. For e.g. stainless steel (alloy of Fe. Ni, & Cr), amalgam (alloy of Hg), brass (alloy of Cu & Zn) etc. The electrical conductivity & melting point of an alloy is less than that of pure metals.\nEnrichment of Ores\nOres mined from the earth are usually contaminated with large amounts of impurities such as soil, sand, etc., called gangue. The impurities must be removed from the ore prior to the extraction of the metal.\nThe processes used for removing the gangue from the ore are based on the differences between physical or chemical properties of the gangue and the ore.\nExtraction of Metals\n\nExtracting Metals towards the Top of the Activity Series\nThese metals are highly reactive & are obtained by electrolytic reduction. For e.g. Na, Mg, & Ca are obtained by the electrolysis of their molten chlorides.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nThese metals are highly reactive & are obtained by electrolytic reduction. For e.g. Na, Mg, & Ca are obtained by the electrolysis of their molten chlorides.\nThe metals are deposited at the cathode whereas chlorine is liberated at anode.\nAt cathode Na+ + e\u2013 \u2192Na\nAt anode 2Cl\u2013 \u2192 Cl2 + 2e\u2013\nSimilarly, aluminium is obtained by the electrolytic reduction of aluminium oxide.\n\nExtracting Metals Low in the Activity Series\nThese metals are the least reactive & are often found in a free state for e.g. An, Ag. Pt & Cu are found in the free state.\nHowever, Cu & Ag are also found in the combined state as their sulphide or oxide ores.\nThe oxides of these metals can be reduced to metals by heating alone. For e.g. cinnabar (HgS), ore of mercury it is heated in air to converted it in mercuric oxide (HgO) which is then reduced to mercury by further heating.", + "The oxides of these metals can be reduced to metals by heating alone. For e.g. cinnabar (HgS), ore of mercury it is heated in air to converted it in mercuric oxide (HgO) which is then reduced to mercury by further heating.\n2HgS(s) + 3O2(g) Heat 2HgO(s) + 2SO2(g)\n2HgO(s)Heat 2Hg(l) + O(g)\nAnother instance is reduction of Cu2S (ore of copper) to copper by heating.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n2HgS(s) + 3O2(g) Heat 2HgO(s) + 2SO2(g)\n2HgO(s)Heat 2Hg(l) + O(g)\nAnother instance is reduction of Cu2S (ore of copper) to copper by heating.\n2Cu2S + 3O2(g)Heat 2Cu2O(s) + 2SO2(g)\n2CU2O + Cu2S Heat 6Cu(s) + SO2\nExtracting Metals Middle in the Activity Series\nThese metals such as Fe, Zn, Pb, Cu, etc are moderately reactive & are usually present as sulphides or carbonates in nature.\nThe sulphide ores are converted into oxides by heating strongly in the presence of excess air which is known as roasting.\nThe carbonate ores are changed into oxides by heating strongly in limited air which is known as calcination.\nThe metal oxides are then reduced to the corresponding metals by using suitable reducing agents such as carbon.\nFor e.g.", + "The carbonate ores are changed into oxides by heating strongly in limited air which is known as calcination.\nThe metal oxides are then reduced to the corresponding metals by using suitable reducing agents such as carbon.\nFor e.g. extraction of Zn\nRoasting: 2ZnS(s) + 3O2(g) Heat . 2ZnO(s) + 2SO2(g)\nCalcination: ZnCO3(s) Heat ZnO(s) + CO2(g)\nReduction: ZnO(s) + C(s) \u2192 Zn(s) + CO(g)", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nFor e.g. extraction of Zn\nRoasting: 2ZnS(s) + 3O2(g) Heat . 2ZnO(s) + 2SO2(g)\nCalcination: ZnCO3(s) Heat ZnO(s) + CO2(g)\nReduction: ZnO(s) + C(s) \u2192 Zn(s) + CO(g)\nSometimes displacement reactions can also be used in place of reduction & highly reactive metals such as Na, Ca, Al, etc., are used as reducing agents.\nFor e.g. 3MnO2(s) + 4Al(s) \u2192 3Mn(l) + 2Al2O3(s) + Heat\nFe2O3(s) + 2Al(s) \u2192 2Fe(l)+Al2O3(s) + Heat\nThis reaction is used to join railway tracks or cracked machine parts and is known as the thermit reaction.\n\nimportant Questions of Metals and Non-Metals Class 10 Science Chapter 3\nQuestion 1.", + "important Questions of Metals and Non-Metals Class 10 Science Chapter 3\nQuestion 1.\nReverse of the following chemical reaction is not possible:\nZn(s)\u00a0+ CuSO4(aq)\u00a0\u2192 ZnSO4(aq)\u00a0+ Cu(s)\nJustify this statement with reason. (Board Term I, 2016)\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 1.\nReverse of the following chemical reaction is not possible:\nZn(s)\u00a0+ CuSO4(aq)\u00a0\u2192 ZnSO4(aq)\u00a0+ Cu(s)\nJustify this statement with reason. (Board Term I, 2016)\nAnswer:\nIf a strip of zinc metal is put in copper sulphate solution, then the blue colour of copper sulphate fades gradually due to the formation of colourless zinc sulphate solution and reddish-brown copper metal is deposited on zinc strip.\n\nIn this reaction, zinc metal being more reactive than copper displaces copper from copper sulphate solution. If however, a strip of copper metal is placed in zinc sulphate solution, then no reaction occurs. This is because copper metal is less reactive than zinc metal and hence, cannot displace zinc from its salt solution.\n\nQuestion 2.\nName a metal which:\n(a) is the best conductor of heat.\n(b) has a very low melting point.\n(c) does not react with oxygen even at high temperature.\n(d) is most ductile. (Board Term I, 2015)\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nName a metal which:\n(a) is the best conductor of heat.\n(b) has a very low melting point.\n(c) does not react with oxygen even at high temperature.\n(d) is most ductile. (Board Term I, 2015)\nAnswer:\n(a) Metal which is the best conductor of heat is silver.\n(b) Gallium has a very low melting point.\n(c) Silver and gold do not react with oxygen even at high temperature.\n(d) Gold is the most ductile metal.\nQuestion 3.\nWhat is meant by amphoteric oxides? Choose the amphoteric oxides from the following :\nNa2O, ZnO, CO2, Al2O3, H2O (Board Term I, 2014)\nAnswer:\nAmphoteric oxides are those which show acidic as well as basic character, i.e., they react with bases as well as acids. ZnO and Al2O3\u00a0are amphoteric oxides.\nQuestion 4.", + "ZnO and Al2O3\u00a0are amphoteric oxides.\nQuestion 4.\nComplete the following equation with balancing: (Board Term I, 2013)\n(i) Al + HCl \u2192\n(ii) Mg + HNO3\u00a0\u2192\nAnswer:\n\nQuestion 5.\nCompare in tabular form the reactivities of the following metals with cold and hot water: (2020)\n(a) Sodium\n(b) Calcium\n(c) Magnesium\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(i) Al + HCl \u2192\n(ii) Mg + HNO3\u00a0\u2192\nAnswer:\n\nQuestion 5.\nCompare in tabular form the reactivities of the following metals with cold and hot water: (2020)\n(a) Sodium\n(b) Calcium\n(c) Magnesium\nAnswer:\n\nQuestion 6.\nGive reason for the following:\n(i) Hydrogen gas is not evolved when most of the metals react with nitric acid.\n(ii) Zinc oxide is considered as an amphoteric oxide.\n(iii) Metals conduct electricity. (Board Term I, 2016)\nAnswer:\n(i) Hydrogen gas is not evolved when most metals react with nitric acid. It is because HNO3\u00a0is a strong oxidising agent. It oxidises the H2\u00a0produced to water and itself gets reduced to any of the nitrogen oxides (N2O, NO, NO2).\n(ii) ZnO reacts both with acids as well as bases to form salt and water. Thus, ZnO is an amphoteric oxide.\n\n(iii) Metals conduct electricity due to the flow of free electrons present in them.\nQuestion 7.", + "(ii) ZnO reacts both with acids as well as bases to form salt and water. Thus, ZnO is an amphoteric oxide.\n\n(iii) Metals conduct electricity due to the flow of free electrons present in them.\nQuestion 7.\n(a) Why does calcium start floating when it reacts with water? Write the balanced chemical equation of the reaction.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 7.\n(a) Why does calcium start floating when it reacts with water? Write the balanced chemical equation of the reaction.\n(b) Name two metals which do not react with water. (Board Term I, 2015)\nAnswer:\n(a) Calcium reacts with cold water to form calcium hydroxide and hydrogen gas.\n\nThe bubbles of hydrogen gas produced stick to the surface of calcium and hence, it starts floating on the surface of water.\n(b) Gold and silver do not react with water.\nQuestion 8.\nState what would happen if:\n(i) some zinc pieces are placed in blue copper sulphate solution.\n(ii) some\u2019copper pieces are placed in green ferrous sulphate solution.\n(iii) an iron nail is dipped in a solution of copper sulphate for some time. (Board Term I, 2014)\nAnswer:\n(i) Refer to answer 1.\n(ii) Cu + FeSO4\u00a0No Reaction\nCu is less reactive than Fe, thus, it cannot displace Fe from FeSO4\u00a0solution.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAnswer:\n(i) Refer to answer 1.\n(ii) Cu + FeSO4\u00a0No Reaction\nCu is less reactive than Fe, thus, it cannot displace Fe from FeSO4\u00a0solution.\n(iii) When an iron nail is dipped in copper sulphate solution, then the blue colour of copper sulphate fades gradually and a reddish brown coating is formed on the iron nail.\n\nAs iron is more reactive than copper, it displaces copper from copper sulphate solution.\nQuestion 9.\nGive reason:\n(a) Aluminium is a reactive metal but is still used for packing food articles.\n(b) Calcium starts floating when water is added to it. (Board Term I, 2014)\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 9.\nGive reason:\n(a) Aluminium is a reactive metal but is still used for packing food articles.\n(b) Calcium starts floating when water is added to it. (Board Term I, 2014)\nAnswer:\n(a) Aluminium is a strong and cheap metal. It is also a good conductor of heat. But it is highly reactive. When it is exposed to moist air, its surface is covered with a thin impervious layer of aluminium oxide (Al2O3). This layer does not allow moist air to come in contact with the fresh metal and hence, protects the metal underneath from further damage or corrosion. Thus, after the formation of this protective layer of Al2O3, aluminium becomes resistant to corrosion. It is because of this reason that although aluminium is a highly reactive metal, it is still used in food packaging.\n(b) Refer to answer 7(a).\nQuestion 10.\n(a) Complete and balance the following chemical equations:\n(i) Al2O3\u00a0+ HCl \u2192\n(ii) K2O + H2O \u2192\n(iii) Fe + H2O \u2192", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(b) Refer to answer 7(a).\nQuestion 10.\n(a) Complete and balance the following chemical equations:\n(i) Al2O3\u00a0+ HCl \u2192\n(ii) K2O + H2O \u2192\n(iii) Fe + H2O \u2192\n(b) An element \u2018X\u2019 displaces iron from the aqueous solution of iron sulphate. List your observations if the element \u2018X\u2019 is treated with the aqueous solutions of copper sulphate, zinc sulphate and silver nitrate. Based on the observations arrange X, Zn, Cu and Ag in increasing order of their reactivities. (2020)\nAnswer:\n(a) (i) Al2O3\u00a0+ 6HCl \u2192 2AlCl3\u00a0+ 3H2O\n(ii) K2O + H2O \u2192 2KOH\n(iii) 3Fe + 4H2O \u2192 Fe3O4\u00a0+ 4H2\n(b) As X displaces iron from its salt solution hence X is more reactive than iron.", + "It will also displace copper from copper sulphate and silver from silver nitrate as both are less reactive than iron. As zinc is more reactive than iron hence, X can be more or less reactive than zinc. Then the order of their reactivities can be\nAg < Cu < Fe < Zn < X or Ag < Cu < Fe < X < Zn.\nQuestion 11.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAg < Cu < Fe < Zn < X or Ag < Cu < Fe < X < Zn.\nQuestion 11.\nA metal \u2018X\u2019 combines with a non-metal \u2018Y\u2019 by the transfer of electrons to form a compound Z.\n(i) State the type of bond in compound Z.\n(ii) What can you say about the melting point and boiling point of compound Z?\n(iii) Will this compound dissolve in kerosene or petrol?\n(iv) Will this compound be a good conductor of electricity? (Board Term I, 2017)\nAnswer:\nX being a metal loses electrons and Y being a non-metal gains electrons to form Z.\n(i) The chemical bond formed by the transfer of electrons from one atom to another is known as an ionic bond. Hence, Z is an ionic compound.\n(ii) Compound Z is an ionic compound thus, it has high melting and boiling points.\n(iii) Ionic compounds are insoluble in non-polar solvents such as kerosene or petrol.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(ii) Compound Z is an ionic compound thus, it has high melting and boiling points.\n(iii) Ionic compounds are insoluble in non-polar solvents such as kerosene or petrol.\n(iv) As Z is an ionic compound, it does not conduct electricity in the solid state because movement of ions in the solid is not possible due to their rigid structure. But it conducts electricity in the molten state or in aqueous solution due to the movement of ions freely.\nQuestion 12.\n(i) By the transfer of electrons, illustrate the formation of bond in magnesium chloride and identify the ions present in this compound.\n(ii) Ionic compounds are solids. Give reasons.\n(iii) With the help of a labelled diagram show the experimental set up of action of steam on a metal. (2020)\nAnswer:\n\n(ii) Ionic compounds are solids because the particles which make up ionic compounds are held together by strong electrostatic bonds.\n\nQuestion 13.\n(a) (i) Write two properties of gold which make it the most suitable metal for ornaments.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 13.\n(a) (i) Write two properties of gold which make it the most suitable metal for ornaments.\n(ii) Name two metals which are the best conductors of heat.\n(iii) Name two metals which melt when you keep them on your palm.\n(b) Explain the formation of ionic compound CaO with electron-dot structure. Atomic numbers of calcium and oxygen are 20 and 8 respectively. (2020)\nAnswer:\n(a) (i) The malleability and ductility properties of gold make it suitable for ornaments.\n(ii) Silver and gold.\n(iii) Gallium and caesium have so low melting points that they melt even on keeping them on palm.\n\nQuestion 14.\n(i) Write down the electronic configuration\nof magnesium and oxygen.\n(ii) Give two general properties of the compound formed by combination of magnesium and oxygen.\n(iii) Show the formation of this compound by the transfer of electrons. (Board Term 1,2014)\nAnswer:\n(i) Atomic number of magnesium (Mg) = 12\n\u2234 Its electronic configuration = 2, 8, 2\nAtomic number of oxygen = 8", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAnswer:\n(i) Atomic number of magnesium (Mg) = 12\n\u2234 Its electronic configuration = 2, 8, 2\nAtomic number of oxygen = 8\nElectronic configuration of oxygen = 2, 6\n(ii) Magnesium (Mg) reacts with oxygen (O2) to form magnesium oxide (MgO).\n2Mg + O2\u00a0\u2192 2MgO\nProperties of MgO are :\n(a) It involves ionic bonding.\n(b) It has high melting point due to strong electrostatic forces of attraction between Mg2+\u00a0and O2-\u00a0ions.\n(iii) In the formation of magnesium oxide, two electrons are transferred from magnesium atom to oxygen atom as represented :\n\nQuestion 15.\nAn ore on treatment with dilute hydrochloric acid produces brisk effervescence. Name the type of ore with one example. What steps will be required to obtain metal from the enriched ore? Also write the chemical equations for the reactions involved in the process. (AI 2019)\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAnswer:\nThe ore on treatment with dilute hydrochloric acid produces brisk effervescence hence, it must be a carbonate ore. Calamine (ZnCO3) is an important carbonate ore of zinc.\nSteps required to obtain metal from the enriched carbonate Are:\n(a) Conversion of the carbonate ore into metal oxide : This is done by calcination (for carbonate ores).\nCalcination is the process of heating the ore strongly in the absence or limited supply of air. The zinc carbonate on heating decomposes to form zinc oxide as shown :\n\n(b) Reduction of the metal oxide to metal : As zinc is moderately reactive, zinc oxide cannot be reduced by heating alone. Hence, it is reduced to zinc by using a reducing agent such as carbon.\n\nThe reduction of metal oxides by heating with coke is called smelting.\nQuestion 16.\n(i) Carbonate of metal \u20182T is abundant in earth crust and its hydroxide is used in \u2018white washing\u2019. Identify metal \u2018XI", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 16.\n(i) Carbonate of metal \u20182T is abundant in earth crust and its hydroxide is used in \u2018white washing\u2019. Identify metal \u2018XI\n(ii) How will you convert this carbonate into its oxide? Name the process and write its equation. (Board Term I, 2014)\nAnswer:\n(i) Calcium carbonate is abundant in earth\u2019s crust and calcium hydroxide is used in white washing. Hence, metal X is calcium (Ca).\n(ii) CaCO3\u00a0is strongly heated in the absence of air to get the metal oxide. This process is called calcination.\n\nQuestion 17.\nZinc is a metal found in the middle of the activity series of metals. In nature, it is found as a carbonate ore, ZnCO3. Mention the steps carried out for its extraction from the ore. Support with equations. (Board Term 1,2013)\nAnswer:\nRefer to answer 15.\nQuestion 18.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAnswer:\nRefer to answer 15.\nQuestion 18.\nCarbon cannot reduce the oxides of sodium, magnesium and aluminium to their respective metals. Why? Where are these metals placed in the reactivity series? How are these metals obtained form their ores? Take an example to explain the process of extraction along with chemical equations. (2020)\nAnswer:\nSodium, magnesium and aluminium have higher affinity towards oxygen than that of carbon because these are highly reactive metals. Hence, carbon cannot reduce the oxides of sodium, magnesium and aluminium to their respective metals. These metals are placed at the top of the reactivity series. The highly reactive metals like Na, Mg, Al, etc. are extracted by electrolytic reduction of their molten chlorides or oxides. Electrolytic reduction is brought about by passing electric current through the molten state. Metal gets deposited at the cathode.\nNaCl \u21cc Na+\u00a0+ Cl\u2013\nAt cathode : Na+\u00a0+ e\u2013\u00a0\u2192 Na\nAt anode : 2Cl\u2013\u00a0\u2192 Cl2\u00a0+ 2e\u2013\nQuestion 19.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nNaCl \u21cc Na+\u00a0+ Cl\u2013\nAt cathode : Na+\u00a0+ e\u2013\u00a0\u2192 Na\nAt anode : 2Cl\u2013\u00a0\u2192 Cl2\u00a0+ 2e\u2013\nQuestion 19.\nWrite balanced chemical equations to explain what happens, when\n(i) Mercuric oxide is heated.\n(ii) Mixture of cuprous oxide and cuprous sulphide is heated.\n(iii) Aluminium is reacted with manganese dioxide.\n(iv) Ferric oxide is reduced with aluminium.\n(v) Zinc carbonate undergoes calcination. (2020)\nAnswer:\n(i) On heating, mercuric oxide decomposes to give mercury .and oxygen.\n\n(ii) On heating mixture of cuprous oxide and cuprous sulphide, copper and sulphur dioxide are produced.\n\n(iii) When aluminium is heated with manganese dioxide, manganese and aluminium oxide are formed.\n\n(iv) Ferric oxide reacts with aluminium to produce aluminium oxide and iron.\n\n(v) On calcination, zinc carbonate produces zinc oxide and carbon dioxide.\n\nQuestion 20.", + "(iii) When aluminium is heated with manganese dioxide, manganese and aluminium oxide are formed.\n\n(iv) Ferric oxide reacts with aluminium to produce aluminium oxide and iron.\n\n(v) On calcination, zinc carbonate produces zinc oxide and carbon dioxide.\n\nQuestion 20.\n(a) List in tabular form three chemical properties on the basis of which we can differentiate between a metal and a non\u00acmetal.\n(b) Give reasons for the following :", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 20.\n(a) List in tabular form three chemical properties on the basis of which we can differentiate between a metal and a non\u00acmetal.\n(b) Give reasons for the following :\n(i) Most metals conduct electricity well.\n(ii) The reaction of iron (III) oxide [Fe2O3] with heated aluminium is used to join cracked machine parts. (Delhi 2019)\nAnswer:\n(b) (i) Refer to answer 6 (iii).\n(ii) The reaction of iron (III) oxide, Fe2O3\u00a0with aluminium is highly exothermic and the iron produced melts. This molten iron is used to join cracked iron parts of machines and railway tracks.\n\nQuestion 21.\n(a) Write the steps involved in the extraction of pure metals in the middle of the activity series from their carbonate ores.\n(b) How is copper extracted from its sulphide ore? Explain the various steps supported by chemical equations. Draw labelled diagram for the electrolytic refining of copper. (2018)\nAnswer:\n(a) Extraction of metals of medium reactivity:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAnswer:\n(a) Extraction of metals of medium reactivity:\nThe metals in the middle of the reactivity series are zinc, iron, lead, etc. The carbonate ores first need to get converted to oxides as it is easier to get metal from their oxides.\nRefer to answer 15.\n(b) Copper glance (Cu2S) when heated in air gets partially oxidised to copper oxide which further reacts with the remaining copper glance to give copper metal.\n\nQuestion 22.\nDraw a schematic diagram of the various steps involved in the extraction of metals from ores for metals of medium reactivity and for metals of low reactivity. (Board Term I, 2018)\nAnswer:\nVarious steps involved in the extraction of a metal from its ore followed by refining of the metal is called \u2018metallurgy\u2019. The steps involved are summarised as follows :\n\nQuestion 23.\n(a) Describe an activity to show that metals are good conductors of electricity.\n(b) Account for the following :\n(i) Hydrogen gas is not evolved when a metal reacts with nitric acid.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 23.\n(a) Describe an activity to show that metals are good conductors of electricity.\n(b) Account for the following :\n(i) Hydrogen gas is not evolved when a metal reacts with nitric acid.\n(ii) For storing sodium metal, it is kept immersed in kerosene.\n(iii) The reaction of iron (III) oxide with aluminium is used to join cracked iron parts of machines. (Board Term I, 2016)\nAnswer:\n(a) Activity : (i) Set up an electric circuit as shown in the figure.\n(ii) Place the metal to be tested (Cu, Al, Ag, Fe, etc.) in the circuit, between the terminal A and B.\n(iii) Switch on the battery.\n\nObservations : Bulb begins to glow.\nConclusion : This indicates that the current is flowing through the metal wire. Hence, metals are good conductors of electricity.\n(b) (i) Refer to answer 6(i).\n(ii) Sodium reacts vigorously with air and catches fire. Also, sodium reacts with water and the hydrogen gas is evolved which catches fire. Therefore, sodium is kept under kerosene.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(ii) Sodium reacts vigorously with air and catches fire. Also, sodium reacts with water and the hydrogen gas is evolved which catches fire. Therefore, sodium is kept under kerosene.\n(iii) Refer to answer 20(b) (ii).\nQuestion 24.\nHow is copper obtained from its ore (Cu2S)?\nWrite only the chemical equations. How is copper thus obtained refined? Name and explain the process along with a labelled diagram. (Board Term I, 2015)\nAnswer:\nRefer to answer 21(b).\nCopper obtained is refined by electrolytic refining.\nElectrolytic refining of crude copper :\nThick block of impure metal acts as anode and a thin strip of pure copper metal acts as cathode. The electrolyse used is aqueous solution of copper sulphate containing a small amount of sulphuric acid. On passing electric current through the electrolyte, the metal from the anode dissolves into the electrolyte. An equivalent amount of copper metal from copper sulphate solution gets deposited on cathode.\n\nQuestion 25.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nQuestion 25.\n(a) Copper produced by heating the ore in air is not very pure. Describe the method used for refining impure copper. Draw labelled diagram of the process.\n(b) Write chemical equations for the reactions taking place when :\n(i) zinc sulphide is heated in air.\n(ii) zinc carbonate is calcined. (Board Term I, 2014)\nAnswer:\nRefer to answer 24.\n\nQuestion 26.\nAssertion (A) : The metals and alloys are good conductors of electricity.\nReason (R) : Bronze is an alloy of copper and tin and it is not a good conductor of electricity.\n(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true.\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true.\nAnswer:\n(c) : Metals and alloys have free electrons in them which can move freely inside them, so they conduct electricity very easily. Bronze is an alloy of copper and tin and it is a very good conductor of electricity.\nQuestion 27.\nName first discovered alloy. Give its composition also. (Board Term I, 2014)\nAnswer:\nBronze is the first discovered alloy. Its composition is copper (90%) and tin (10%).\nQuestion 28.\nList three differentiating features between the processes of galvanisation and alloying.\nAnswer:\nQuestion 29.\nDescribe an activity to find out the conditions under which iron rusts. (Board Term I, 2017)\nAnswer:\nActivity:\n(i) Take three test tubes and put clean nails in each of the three tubes. Label them as A, B and C.", + "Answer:\nQuestion 29.\nDescribe an activity to find out the conditions under which iron rusts. (Board Term I, 2017)\nAnswer:\nActivity:\n(i) Take three test tubes and put clean nails in each of the three tubes. Label them as A, B and C.\n(ii) Pour some water in test tube A and cork it.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAnswer:\nActivity:\n(i) Take three test tubes and put clean nails in each of the three tubes. Label them as A, B and C.\n(ii) Pour some water in test tube A and cork it.\n(iii) In tube B, pour some boiled distilled water along with some turpentine oil and cork it.\n(iv) In test tube C, add some anhydrous calcium chloride and cork it.\n(v) Look these test tubes properly and keep them undisturbed for a few days.\nObservation : Only in test tube A, iron nails get rusted since the nails in this test tube are exposed to both air and water.\nConclusion : Both air and water are required for rusting of iron.\n\nQuestion 30.\nWhy some metal surfaces acquire a dull appearance when they are exposed to moist air? Write colour acquired by the surfaces of copper and silver in such situation and also write the chemical names of the substances due to which it happens. (Board Term I, 2016)\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAnswer:\nWhen a metal has been kept exposed to air for a long time, then it gets a dull appearance. The metals lose their shine or brightness due to the formation of a thin layer of oxide, carbonate or sulphide on their surface and thus, the metal surface gets corroded. The surface of copper gets coated with a green layer in moist air due to the formation of basic copper carbonate, silver articles acquire a blackish tinge due to the formation of silver sulphide.\nQuestion 31.\n(a) Name the following :\n(i) Metal that can be cut by knife\n(ii) Lustrous non-metal\n(iii) Metal that exists in liquid state at room temperature\n(iv) Most malleable and ductile metal\n(v) Metal that is best conductor of electricity\n(vi) Non-metal that can exist in different forms\n(b) How are alloys better than metals? Give composition of solder and amalgam.\nAnswer:\n(a) (i) Sodium\n(ii) Iodine\n(iii) Mercury\n(iv) Gold\n(v) Silver\n(vi) Carbon", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(b) How are alloys better than metals? Give composition of solder and amalgam.\nAnswer:\n(a) (i) Sodium\n(ii) Iodine\n(iii) Mercury\n(iv) Gold\n(v) Silver\n(vi) Carbon\n(b) Alloys are stronger than the metals from which they are made, more resistant to corrosion, have lower melting point, have lower electrical conductivity. Solder is an alloy of lead and tin. An amalgam is an alloy of mercury with another metal.\nQuestion 32.\n(a) Define corrosion.\n(b) What is corrosion of iron called?\n(c) How will you recognise the corrosion of silver?\n(d) Why corrosion of iron is a serious problem?\n(e) How can we prevent corrosion of iron? (Board Term I, 2017)\nAnswer:\n(a) The process of slowly eating up of metals due to their conversion into oxides, carbonates, sulphides, etc., by the action of atmospheric gases and moisture is called corrosion.\n(b) The corrosion of iron is called rusting.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(b) The corrosion of iron is called rusting.\n(c) Silver articles become black after sometime when exposed to air. This is due to formation of a coating of black silver sulphide (Ag2S) on its surface by the action of H2S gas present in the air.\n(d) Corrosion of iron is a serious problem. Every year large amount of money is spent to replace damaged iron articles. Corrosion causes damage to car bodies, bridges and iron railings, ships and to all objects made of metals specially those of iron.\n(e) Corrosion of iron is prevented by coating it with a layer of oil. The reason being that the layer of oil does not allow air and water to react the surface of iron. Corrosion of iron can also be prevented by painting, greasing, galvanising, anodising, electroplating or making alloys.\nQuestion 33.\nGive reason for the following :\n(a) Ionic compounds have higher melting point and higher boiling point.\n(b) Sodium is kept immersed in kerosene.\n(c) Reaction of calcium with water is less violent.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nGive reason for the following :\n(a) Ionic compounds have higher melting point and higher boiling point.\n(b) Sodium is kept immersed in kerosene.\n(c) Reaction of calcium with water is less violent.\n(d) Silver articles become black after some time when exposed to air.\n(e) Prior to reduction the metal sulphides and carbonates must be converted into metal oxides for extracting metals. (Board Term I, 2015)\nAnswer:\n(a) Due to strong forces of attraction, the ions are bound to each other very firmly. As a result, the electrovalent or ionic solids have high melting and boiling points.\n(b) Refer to answer 23(b) (ii).\n(c) Calcium reacts with cold water but the reaction is less violent. The heat evolved is not sufficient for the hydrogen to catch fire.\n\n(d) Refer to answer 32(c).\n(e) The reduction of metal oxides to metal is easier than the reduction of metal sulphides and metal carbonates. Hence, these are first reduced to their corresponding metal oxides.\nQuestion 34.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(e) The reduction of metal oxides to metal is easier than the reduction of metal sulphides and metal carbonates. Hence, these are first reduced to their corresponding metal oxides.\nQuestion 34.\n(a) Metals like iron, silver and copper get corroded on exposure to air. Write the chemical name of the substance deposited on their surface respectively with it\u2019s colour, in each case.\n(b) List four ways by which rusting can be prevented. (Board Term I, 2013)\nAnswer:\n(a) Iron gets corroded and forms ferric oxide which is rust, it is reddish brown in colour. For silver and copper, refer to answer 30.\n(b) The various methods used for preventing the rusting of iron are given below:\n(i) By applying paint : Materials like railings, iron gates, iron bridges, bodies of cars, buses and trucks, etc. are all painted to protect them from rusting. Painting the metal surface does not allow them to come in contact with the moist air and thus, prevents rusting.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(ii) Greasing and oiling : When some grease or oil is applied on the surface of an iron object, then moisture and air cannot come in contact with it and hence, rusting is prevented.\n(iii) Galvanization : It is a method of protecting iron from rusting by coating them with a thin layer of zinc. The iron coated with zinc is called galvanized iron.\n(iv) Electroplating : It is another technique used to prevent articles from rusting. In this process, metals like tin, nickel and chromium which do not corrode are electroplated on iron.\nShort Answer Type Questions[l] [2 Marks]-Year 2015\n35.Write one example of each of\n(i) a metal which is so soft that, it can be cut with knife and a non-metal which is the hardest sustance.\n(ii) a metal and a non-metal which exist as liquid at room temperature.\nAnswer.\n(i) Sodium, carbon (diamond).\n(ii)Mercury is liquid metal, bromine is liquid non-metal.\n36.Mention the names of the metals for the following:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAnswer.\n(i) Sodium, carbon (diamond).\n(ii)Mercury is liquid metal, bromine is liquid non-metal.\n36.Mention the names of the metals for the following:\n(i) Two metals which are alloyed with iron to make stainless steel.\n(ii) Two metals which are used to make jewellary.\nAnswer.\n(i) Nickel and chromium.\n(ii) Gold and platinum.\nCBSE Class 10 Science \u2013 More Resources\nShort Answer Type Question[l] [2 Marks]\u00a0-Year 2013\n37.Give reason for the following:\n(a) School bells are made up of metals.\n(b) Electric wires are made up of copper.\nAnswer.\n(a) It is because metals are sonorous, i.e. they produce sound when struk with a hard substance.\n(b) It-is because copper is good conductor of electricity.\nShort Answer Type Question[ll] [3 Marks]\u00a0-Year 2013\n38. Suggest a method of reduction for the following metals during their metallurgical processes:\n(i) metal \u2018A\u2019 which is one of the last, second or third position in the reactivity.", + "Short Answer Type Question[ll] [3 Marks]\u00a0-Year 2013\n38. Suggest a method of reduction for the following metals during their metallurgical processes:\n(i) metal \u2018A\u2019 which is one of the last, second or third position in the reactivity.\n(ii) metal \u2018B\u2019 which gives vigorous reaction even with water and air.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(i) metal \u2018A\u2019 which is one of the last, second or third position in the reactivity.\n(ii) metal \u2018B\u2019 which gives vigorous reaction even with water and air.\n(iii) metal \u2018C\u2019 which is kept in the middle df activity series.\nAnswer.\n(i) \u2018A\u2019 can be obtained by chemical reduction using carbon or carbon monoxide as reducing agent.\n(ii) \u2018B\u2019 can be obtained by electrolytic reduction.\n(iii) \u2018C\u2019 can be reduced by reducing agent like \u2018Al\u2019.\nVery Short Answer Type Question [1 Mark] -Year 2012\n39.A green layer is gradually formed on a copper plate left exposed to air for a week in a bathroom. What could this green substance be?\nAnswer. It is due to the formation of basic copper carbonate\u00a0[CuC03.Cu(0H)2].\nShort Answer Type Questions[l] [2 Marks]\u00a0-Year 2012\n40.Name the following:\n(a) A metal, which is preserved in kerosene.\n(b) A lustrous coloured non-metal.\n(c) A metal, which can melt while kept on palm.", + "Short Answer Type Questions[l] [2 Marks]\u00a0-Year 2012\n40.Name the following:\n(a) A metal, which is preserved in kerosene.\n(b) A lustrous coloured non-metal.\n(c) A metal, which can melt while kept on palm.\n(d) A metal, which is a poor conductor of heat.\nAnswer.\n(a) Sodium is preserved in kerosene.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(b) A lustrous coloured non-metal.\n(c) A metal, which can melt while kept on palm.\n(d) A metal, which is a poor conductor of heat.\nAnswer.\n(a) Sodium is preserved in kerosene.\n(b) Iodine is lustrous coloured non-metal.\n(c) Gallium. \u2018\n(d) Lead.\n41. Give reason for the following:\n(a) Aluminium oxide is considered as an amphoteric oxide.\n(b) Ionic compounds conduct electricity in molten state.\nAnswer.\n(a) It is because it reacts with acids as well as bases to produce salts and water.\u2019Al\u2019 is less electropositive metal. So, it forms amphoteric oxide which can react with acid as well as base.\n(b) Ionic compounds can conduct electricity in molten state because ions \u2019 become free to move in molten state.\nShort Answer Type Questions[ll] [3 Marks]\u00a0-Year 2012\n42. A metal \u2018X\u2019 acquires a green colour coating on its surface on exposure to air.\n(i) Identify the metal \u2018X\u2019 and name the process responsible for this change.", + "Short Answer Type Questions[ll] [3 Marks]\u00a0-Year 2012\n42. A metal \u2018X\u2019 acquires a green colour coating on its surface on exposure to air.\n(i) Identify the metal \u2018X\u2019 and name the process responsible for this change.\n(ii)Name and write chemical formula of the green coating formed on the metal.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(i) Identify the metal \u2018X\u2019 and name the process responsible for this change.\n(ii)Name and write chemical formula of the green coating formed on the metal.\n(iii) List two important methods to prevent the process.\nAnswer.\n(i) Metal is copper. The process is corrosion.\n(ii)Basic copper carbonate [CuCO3.Cu(0H)2].\n(iii)\n\u2022 It should be coated with tin\n\u2022 It should be mixed with other metals to form alloys.\n43.What are amphoteric oxides? Choose the amphoteric oxides from amongst the following oxides:\nNa2O, ZnO, Al2O3, CO2, H2O\nAnswer.Those oxides which reacts with acids as well as bases to produce salts and water are called amphoteric oxides, e.g. Na2O, ZnO, are amphoteric oxides among given oxides.\n44.Define the terms:\n(i) mineral\n(ii) ore, and\n(iii) gangue.\nAnswer.", + "Na2O, ZnO, are amphoteric oxides among given oxides.\n44.Define the terms:\n(i) mineral\n(ii) ore, and\n(iii) gangue.\nAnswer.\n(i) Mineral: It is a naturally occurring substance from which metal may or may not be extracted profitably or economically, e.g. A1 cannot be extracted profitably from mica.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(iii) gangue.\nAnswer.\n(i) Mineral: It is a naturally occurring substance from which metal may or may not be extracted profitably or economically, e.g. A1 cannot be extracted profitably from mica.\n(ii)Ore: It is a rocky material which contains sufficient quantity of mineral so that metal can be extracted profitably, e.g. zinc blende is an ore of zinc from which zinc can be extracted profitably.\n(iii) Gangue: It is a rocky material which is present along with the mineral in the ore, e.g. FeO is gangue in extraction of copper.\nLong Answer Type Questions [5 Marks] -Year 2012\n45. (a) Write the chemical name of the coating that forms on silver and copper articles when these are left exposed to moist air.\n(b) Explain what is galvanisation. What purpose is served by it?\n(c) Define an alloy. How are alloys prepared? How do the properties of iron change when:\n(i) small quantity of carbon,\n(ii) nickel and chromium are mixed with it.\nAnswer.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(c) Define an alloy. How are alloys prepared? How do the properties of iron change when:\n(i) small quantity of carbon,\n(ii) nickel and chromium are mixed with it.\nAnswer.\n(a) Ag2S (silver sulphide) is formed on silver, basic copper carbonate CuCO3. CU(OH)2\u00a0is formed on copper.\n(b) The process of coating zinc over iron is called galvanisation. It is used to prevent rusting of iron.\n(c) Alloy is a homogeneous mixture of two or more metals. One of them can be non-metal. Alloys are prepared by melting two or more metals together.\n(?) Iron does not rust on adding small,quantity of carbon.\n(ii) When we form alloy of iron with nickel and chromium, we get stainless steel which is malleable and does not get rusted.\n46.(a) Carbon cannot be used as reducing agent to obtain Mg from MgO. Why?\n(b) How is sodium obtained from molten sodium chloride? Give equation of the reactions.\n(c) How is copper obtained from its sulphide ore? Give equations of the reactions.\nAnswer.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(b) How is sodium obtained from molten sodium chloride? Give equation of the reactions.\n(c) How is copper obtained from its sulphide ore? Give equations of the reactions.\nAnswer.\n(a) It is because \u2018Mg\u2019 is stronger reducing agent than carbon.\n(b) Sodium is obtained from molten NaCl by electrolysis.\n\nBlister Copper is purified by electrolytic refining.\n47.Give reasons for the following:\n(i) Silver and copper lose their shine when they are exposed to air. Name the substance formed on their surface in each case.\n(ii) Tarnished copper vessels are cleaned with tamarind juice.\n(iii) Aluminium is more reactive than iron yet there is less corrosion of aluminium as compared to iron when both are exposed to air.\nAnswer.\n(i) These metals get corroded. Silver forms black Ag2S (silver sulphide) and copper form greenish layer of basic copper carbonate CuCO3. CU(OH)2\u00a0.\n(ii) Tamarind contains acid which reacts with basic copper carbonate and product gets dissolved and removed from copper vessel.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(ii) Tamarind contains acid which reacts with basic copper carbonate and product gets dissolved and removed from copper vessel.\n(in) Aluminium forms oxide layer on its surface which does not further react with air.\n48. What are alloys? How are they made? Name the constituents and uses of brass, bronze and solder.\nAnswer. Alloys are homogeneous mixtures of two or more metals. One of them can be a non-metal also. They are made by melting a metal which is in large amount first and then adding the other metal. ,\nBrass contains copper and zinc. It is used for making decorative articles. Bronze contains copper and tin. It is used for making statues and medals. Solder contains lead and tin. It is used for soldering purposes.\nVery Short Answer Type Questions [1 Mark] -Year 2011\n49. A non-metal X exists in two different forms Y and Z. Y is the hardest natural substance, whereas Z is a good conductor of electricity. Identify X, Y and Z.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n49. A non-metal X exists in two different forms Y and Z. Y is the hardest natural substance, whereas Z is a good conductor of electricity. Identify X, Y and Z.\nAnswer.\u00a0\u2018X\u2019 is carbon, \u2018Y\u2019 is diamond as it is the hardest natural substance and \u2018Z\u2019 is graphite as it is good conductor of electricity.\n50. Why does calcium float in water?\nAnswer.\u00a0It is because hydrogen gas is formed which sticks to surface of calcium, therefore it floats. \u2013\n51.Name a non-metal which is lustrous and a metal which is non-lustrous. Iodine is a non-metal which is lustrous,\nAnswer.lead is a non-lustrous metal.\n52.Which gas is liberated when a metal reacts with an acid? How will you test the presence of this gas?\nAnswer. Hydrogen gas is formed. Bring a burning matchstick near to it, H2 will burn explosively with \u2018pop\u2019 sound.\n53.Name the metal which reacts with a very dilute HNOs to evolve hydrogen gas.\nAnswer.", + "How will you test the presence of this gas?\nAnswer. Hydrogen gas is formed. Bring a burning matchstick near to it, H2 will burn explosively with \u2018pop\u2019 sound.\n53.Name the metal which reacts with a very dilute HNOs to evolve hydrogen gas.\nAnswer. Magnesium\n54.Name two metals which are found in nature in the free state.\nAnswer.(i) Gold (ii) Silver", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n53.Name the metal which reacts with a very dilute HNOs to evolve hydrogen gas.\nAnswer. Magnesium\n54.Name two metals which are found in nature in the free state.\nAnswer.(i) Gold (ii) Silver\nAnswer Type Questions[l] [2 Marks] -Year 2011\n55.The way, metals like sodium, magnesium and iron react with air and water is an indication of their relative positions in the \u2018reactivity series\u2019. Is this statement true ? Justify your answer with examples.\nAnswer.Yes, sodium reacts explosively even with cold water, it is most reactive. Magnesium reacts with hot water, it is less reactive than Na. Iron reacts only with steam which shows it is least reactive among the three.\n56. X + YSO4\u2014\u2014-> XSO4+ Y\nY + XSO4\u00a0\u2014\u2014\u2013> No reaction\nOut of the two elements, \u2018X\u2019 and \u2018Y\u2019, which is more reactive and why?\nAnswer.\u00a0\u2018X\u2019 is more reactive than \u2018Y\u2019 because it displaces \u2018Y\u2019 from its salt solution.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nY + XSO4\u00a0\u2014\u2014\u2013> No reaction\nOut of the two elements, \u2018X\u2019 and \u2018Y\u2019, which is more reactive and why?\nAnswer.\u00a0\u2018X\u2019 is more reactive than \u2018Y\u2019 because it displaces \u2018Y\u2019 from its salt solution.\n57.What is an alloy? State the constituents of solder. Which property of solder makes it suitable for welding electrical wires?\nAnswer.\u00a0Alloy is a homogeneous mixture of two or more metals. One of them can be a non-metal also. Solder consists of lead and tin. It has low melting point which makes it suitable for welding electrical wires.\n58. Write chemical equations that shows aluminium oxide reacts with acid as well as base.\nAnswer.\n\nLong Answer Type Questions [5 Marks] -Year 2011\n59.(a) How can the metals at the top of the reactivity series be extracted from their ores? Explain with an example.\n(b) Name any one alloy made from\n(i) a metal and a non-metal, and\n(ii) two metals.\nAnswer.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(b) Name any one alloy made from\n(i) a metal and a non-metal, and\n(ii) two metals.\nAnswer.\n(a) These metals are extracted by electrolytic reduction, e.g. aluminium is obtained from bauxite by electrolytic reduction.\n(b) (i) Steel is made up of iron and carbon.\n(ii) Brass is made up of copper and zinc.\nVery Short Answer Type Questions [1 Mark] -Year 2010\n60.What is the valency of silicon with atomic number 14?\nAnswer. Its valency is equal to 4.\n61.What is the valency of phosphorus with atomic number 15?\nAnswer. Phosphorus has valency 3.\n62.What is the valency of an element with atomic number 35?\nAnswer. Its valency is 1.\nShort Answer Type Question[I] [2 Marks] -Year 2010\n63.Elements magnesium and oxygen respectively belong to group 2 and group 16 of the Modern Periodic Table.", + "62.What is the valency of an element with atomic number 35?\nAnswer. Its valency is 1.\nShort Answer Type Question[I] [2 Marks] -Year 2010\n63.Elements magnesium and oxygen respectively belong to group 2 and group 16 of the Modern Periodic Table. If the atomic numbers of magnesium and oxygen are 12 and 8 respectively, draw their electronic configurations and show the process of formation of their compound by transfer of electrons.\nAnswer.(Mg)=2,8,2", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nAnswer.(Mg)=2,8,2\n\nVery Short Answer Type Question [1 Mark] -Year 2009\n64. Arrange the following metals in the decreasing order of reactivity: Na, K, Cu, Ag.\nAnswer. K > Na > Cu > Ag\nShort Answer Type Questions[ll] [3 Marks] -Year 2009\n65. Give reasons for the following observations:\n(i) Ionic compounds in general have high melting and boiling points.\n(ii) Highly reactive metals cannot be obtained from their oxides by heating\nthem with carbon.\n(iii) Copper vessels get a green coat when left exposed to air in the rainy season.\nAnswer.\u00a0(i) Ionic compounds have high melting and boiling points due to strong force of attraction between oppositely charged ions.\n(ii) It is because these metals themselves are strong reducing agents. Therefore, cannot be reduced by reducing agent like carbon.\n(iii) Copper vessels react with CO2, O2\u00a0and moisture to form green-coloured basic copper carbonate\n[CuCO3.Cu(OH)2].\n66. State reasons for the following observations:", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n(iii) Copper vessels react with CO2, O2\u00a0and moisture to form green-coloured basic copper carbonate\n[CuCO3.Cu(OH)2].\n66. State reasons for the following observations:\n(i) The shining surface of some metals becomes dull when exposed to air for a long time.\n(ii) Zinc fails to evolve hydrogen gas on reacting with dilute nitric acid.\n(iii) Metal sulphides occur mainly in rocks but metal halides occur mostly in sea and lake waters.\nAnswer.\n(i) It is because metal reacts with substances present in atmosphere to form surface compounds which make it dull.\n(ii) It is because dil. HNOs is an oxidising agent therefore zinc gives NO and notH2\u00a0with dil. HNOs.\n(iii) It is because sea water contains sodium chloride due to which metal halides are formed, whereas sulphur is found below rocks. Therefore, metal \u2013 sulphides are formed in rocks.\n67. State reasons for the following:\n(i) Electric wires are covered with rubber like material.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n67. State reasons for the following:\n(i) Electric wires are covered with rubber like material.\n(ii)From dilute hydrochloric acid, zinc can liberate hydrogen gas but copper cannot.\n(iii) Sulphide ore of a metal is first converted to its oxide to extract the metal from it.\nAnswer.\n(i) It is because rubber is an insulator and does not allow current to flow through it.\n(ii) Zinc is more reactive than hydrogen. Therefore, it can displace hydrogen from dilute HCl whereas copper cannot, because it is less reactive than hydrogen. ,\n(iii) It is because it is easier to reduce oxide ore as compared to sulphide ore.\nLong Answer Type Questions [5 Marks] -Year 2009\n68.(a) What is meant by corrosion? Name any two methods used for the prevention\nof corrosion.\n(b) Suppose you have to extract metal M from its enriched sulphide ore. If M is in the middle of the reactivity series, write various steps used in extracting this metal.", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\nof corrosion.\n(b) Suppose you have to extract metal M from its enriched sulphide ore. If M is in the middle of the reactivity series, write various steps used in extracting this metal.\nAnswer. (a) Corrosion is a process in which metal reacts with substances present in the environment to form surface compounds.\nPrevention:\n(i) Galvanisation is a process to prevent corrosion of iron.\n(ii)Electroplating is also used to prevent corrosion.\n(b)(i) Concentration of ores: Sulphide ore will be concentrated by froth- floatation process. Sulphide ore will be collected in froth whereas gangue will be left behind.\n(ii) Roasslng: Sulphide ore is heated strongly in the presence of O2\u00a0to form metal oxide and sulphur dioxide.\n2MS + 3O2\u00a0\u2014\u2014\u2014\u2014- \u25ba 2MO + 2SO2\n(iii) Reduction: MO reacts with carbon (acts as reducing agent) to form metal and CO.\nMO + C \u2014-> M + CO", + "CHAPTER 3- METALS AND NON METALS- QUESTION AND ANSWERS.txt\n2MS + 3O2\u00a0\u2014\u2014\u2014\u2014- \u25ba 2MO + 2SO2\n(iii) Reduction: MO reacts with carbon (acts as reducing agent) to form metal and CO.\nMO + C \u2014-> M + CO\n(iv) Electrolytic refining: Impure metal \u2018M\u2019 is purified by electrolytic refining. Impure metal is taken as anode, pure metal is taken as cathode, soluble salt of metal is taken as electrolyte. Impure metal forms metal ions which gain electrons and form pure metal at cathode.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nCBSE Class 10 Science Notes Chapter 2 Acids Bases and Salts\nIndicators: Indicators are substances which indicate the acidic or basic nature of the solution by the colour change.\nTypes of Indicator: There are many types of indicators. Some common types of indicators are:\n1. Natural Indicators: Indicators obtained from natural sources are called Natural Indicators. Litmus, turmeric, red cabbage, China rose, etc., are some common natural indicators used widely to show the acidic or basic character of substances.\nLitmus: Litmus is obtained from lichens. The solution of litmus is purple in colour. Litmus paper comes in two colours- blue and red.\nAn acid turns blue litmus paper red.\nA base turns red litmus paper blue.\nTurmeric: Turmeric is another natural indicator. Turmeric is yellow in colour. Turmeric solution or paper turns reddish brown with base. Turmeric does not change colour with acid.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nTurmeric: Turmeric is another natural indicator. Turmeric is yellow in colour. Turmeric solution or paper turns reddish brown with base. Turmeric does not change colour with acid.\nRed Cabbage: The juice of red cabbage is originally purple in colour. Juice of red cabbage turns reddish with acid and turns greenish with base.\n2. Olfactory Indicator: Substances which change their smell when mixed with acid or base are known as Olfactory Indicators. For example; Onion, vanilla etc.\nOnion: Paste or juice of onion loses its smell when added with base. It does not change its smell with acid.\nVanilla: The smell of vanilla vanishes with base, but its smell does not vanish with an acid.\nOlfactory Indicators are used to ensure the participation of visually impaired students in the laboratory.\n3. Synthetic Indicator: Indicators that are synthesized in the laboratory are known as Synthetic Indicators. For example; Phenolphthalein, methyl orange, etc.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\n3. Synthetic Indicator: Indicators that are synthesized in the laboratory are known as Synthetic Indicators. For example; Phenolphthalein, methyl orange, etc.\nPhenolphthalein is a colourless liquid. It remains colourless with acid but turns into pink with a base.\nMethyl orange is originally orange in colour. It turns into the red with acid and turns into yellow with base.\nAcids: Acids are sour in taste, turn blue litmus red, and dissolve in water to release H+ ions.\nExample: Sulphuric acid (H2SO4), Acetic Acid (CH3COOH), Nitric Acid (HNO3) etc.\nProperties of Acids:\nAcids have a sour taste.\nTurns blue litmus red.\nAcid solution conducts electricity.\nRelease H+ ions in aqueous solution.\nTypes of Acids: Acids are divided into two types on the basis of their occurrence i.e., Natural acids and Mineral acids.\n(i) Natural Acids: Acids which are obtained from natural sources are called Natural Acids or Organic Acids.", + "Release H+ ions in aqueous solution.\nTypes of Acids: Acids are divided into two types on the basis of their occurrence i.e., Natural acids and Mineral acids.\n(i) Natural Acids: Acids which are obtained from natural sources are called Natural Acids or Organic Acids.\nExamples:\nMethanoic acid (HCOOH)\nAcetic acid (CH3COOH)", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\n(i) Natural Acids: Acids which are obtained from natural sources are called Natural Acids or Organic Acids.\nExamples:\nMethanoic acid (HCOOH)\nAcetic acid (CH3COOH)\nOxalic acid (C2H2O4) etc.\n(ii) Mineral Acids: Acids that are prepared from minerals are known as Mineral Acids Example; Inorganic acids, man-made acids or synthetic acid are also known as Mineral Acids.\nExample:\nHydrochloric acid (HCl)\nSulphuric acid (H2SO4)\nNitric acid (HNO3)\nCarbonic acid (H2CO3)\nPhosphoric acid (H3PO4) etc.\nChemical Properties of Acid:\n(i) Reaction of acids with metal: Acids give hydrogen gas along with respective salt when they react with a metal.\nMetal + Acid \u2192 Salt + Hydrogen\nExamples:\nHydrogen gas and zinc chloride are formed when hydrochloric acid reacts with zinc metal.\n\nHydrogen gas and sodium sulphate are formed when sulphuric acid reacts with sodium metal.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nExamples:\nHydrogen gas and zinc chloride are formed when hydrochloric acid reacts with zinc metal.\n\nHydrogen gas and sodium sulphate are formed when sulphuric acid reacts with sodium metal.\n\nTest For Hydrogen Gas: The gas evolved after reaction of acid with metal can be tested by bringing a lighted candle near it. If the gas bums with a pop sound, then it confirms the evolution of hydrogen gas. Burning with pop sound is the characteristic test for hydrogen gas.\n(ii) Reaction of acids with metal carbonate: Acids give carbon dioxide gas and respective salts along with water when they react with metal carbonates.\nMetal carbonate + Acid \u2192 Salt + Carbon dioxide + Water\nExamples:\nHydrochloric acid gives carbon dioxide gas, sodium chloride along with water when reacts with sodium carbonate.\n\nSulphuric acid gives calcium sulphate, carbon dioxide gas, calcium sulphate and water when it reacts with calcium carbonate", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nSulphuric acid gives calcium sulphate, carbon dioxide gas, calcium sulphate and water when it reacts with calcium carbonate\n\nNitric acid gives sodium nitrate, water and carbon dioxide gas when it reacts with sodium carbonate.\n\n(iii) Reaction of acid with hydrogen carbonates (bicarbonates): Acids give carbon dioxide gas, respective salt and water when they react with metal hydrogen carbonate.\nAcid + Metal hydrogen carbonate \u2192 Salt + Carbon dioxide + Water\nExample:\nSulphuric acid gives sodium sulphate, Carbon dioxide gas and water when it reacts with sodium bicarbonate.\n\nTest For Evolution of Carbon Dioxide Gas: Carbon dioxide turns lime water milky when passed through it. This is the characteristic test for carbon dioxide gas.\nThe gas evolved because of reaction of the acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas. This happens because of the formation of a white precipitate of calcium carbonate.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nBut when excess of carbon dioxide is passed through lime water, it makes milky colour of lime water disappear. This happens because of formation of calcium hydrogen carbonate. As calcium hydrogen carbonate is soluble in water, thus, the milky colour of solution mixture disappears.\nCommon in Acids: Acids give hydrogen gas when they react with metal. This shows that all acids contains hydrogen. For example; Hydrochloric acid (HCl), sulphuric acid (H2SO4), nitric acid (HNO3), etc.\nWhen an acid is dissolved in water, it dissociates hydrogen. The dissociation of hydrogen ion in aqueous solution is the common property in all acids. Because of the dissociation of hydrogen ion in aqueous solution, an acid shows acidic behaviour.\nExamples:\nHydrochloric acid (HCl) gives hydrogen ion (H+) and chloride ion (Cl\u2013) when it is dissolved in water.\n\nAcetic acid (CH3COOH) gives acetate ion (CH3COO\u2013) and hydrogen ion (H+).\n\nAcids\nStrong Acids", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nHydrochloric acid (HCl) gives hydrogen ion (H+) and chloride ion (Cl\u2013) when it is dissolved in water.\n\nAcetic acid (CH3COOH) gives acetate ion (CH3COO\u2013) and hydrogen ion (H+).\n\nAcids\nStrong Acids\nAn acid which is completely ionised in water and produces (H+) is called Strong Acid.\nExamples: Hydrochloric acid (HCl), Sulphuric acid (H2SO4), Nitric acid (HNO3)\nWeak Acids\nAn acid which is partially ionised in water and thus produces a small amount of hydrogen ions (H+) is called a Weak Acid.\nExample: Acetic acid (CH3COOH), Carbonic acid (H2CO3)\nWhen a concentrated solution of acid is diluted by mixing water, then the concentration of Hydrogen ions (H+) or hydronium ion (H3O\u2013) per unit volume decreases.\nBases: Bases are bitter in taste, have soapy touch, turn red litmus blue and give hydroxide ions (OH\u2013) in aqueous solution.", + "Bases: Bases are bitter in taste, have soapy touch, turn red litmus blue and give hydroxide ions (OH\u2013) in aqueous solution.\nExamples: Sodium hydroxide (caustic soda) \u2013 NaOH\nCalcium hydroxide \u2013 Ca(OH)2\nPotassium hydroxide (caustic potash) \u2013 (KOH)\nProperties of Bases:\nHave a bitter taste.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nExamples: Sodium hydroxide (caustic soda) \u2013 NaOH\nCalcium hydroxide \u2013 Ca(OH)2\nPotassium hydroxide (caustic potash) \u2013 (KOH)\nProperties of Bases:\nHave a bitter taste.\nSoapy to touch.\nTurns red litmus blue.\nConducts electricity in solution.\nRelease OH\u2013 ions in Aqueous Solution\nTypes of bases: Bases can be divided in two types \u2013 Water soluble and Water-insoluble.\nThe hydroxide of alkali and alkaline earth metals are soluble in water. These are also known as alkali.\nFor example; sodium hydroxide, magnesium hydroxide, calcium hydroxide, etc. Alkali is considered a strong base.\nChemical properties of bases:\n(i) Reaction of Base with Metals: When alkali (base) reacts with metal, it produces salt and hydrogen gas.\nAlkali + Metal \u2192 Salt + Hydrogen\nExamples: Sodium hydroxide gives hydrogen gas and sodium zincate when reacts with zinc metal.", + "Chemical properties of bases:\n(i) Reaction of Base with Metals: When alkali (base) reacts with metal, it produces salt and hydrogen gas.\nAlkali + Metal \u2192 Salt + Hydrogen\nExamples: Sodium hydroxide gives hydrogen gas and sodium zincate when reacts with zinc metal.\n\nSodium aluminate and hydrogen gas are formed when sodium hydroxide reacts with aluminium metal.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nExamples: Sodium hydroxide gives hydrogen gas and sodium zincate when reacts with zinc metal.\n\nSodium aluminate and hydrogen gas are formed when sodium hydroxide reacts with aluminium metal.\n\n(ii) Reaction of Base with Oxides of Non-metals: Non-metal oxides are acidic in nature. For example; carbon dioxide is a non-metal oxide. When carbon dioxide is dissolved in water it produces carbonic acid.\nTherefore, when a base reacts with non-metal oxide, both neutralize each other resulting respective salt and water.\nBase + Non-metal oxide \u2192 Salt + Water\n(Non-metal oxides are acidic in nature)\nExamples:\nSodium hydroxide gives sodium carbonate and water when it reacts with carbon dioxide.\n\nCalcium hydroxide gives calcium carbonate and water when it reacts with carbon dioxide.\n\n(iii) Neutralisation Reaction: An acid neutralizes a base when they react with each other and respective salt and water are formed.\nAcid + Base \u2192 Salt + Water", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\n(iii) Neutralisation Reaction: An acid neutralizes a base when they react with each other and respective salt and water are formed.\nAcid + Base \u2192 Salt + Water\nSince, the reaction between acid and base both neutralize each other, hence, it is also known as Neutralization Reaction.\nExamples: Sodium chloride and water are formed when hydrochloric acid reacts with sodium hydroxide (a strong base).\n\nIn a similar way, calcium chloride is formed along with water when hydrochloric acid reacts with calcium hydroxide (a base).\n\n(iv) Reaction of Acid with Metal Oxides: Metal oxides are basic in nature. Thus, when an acid reacts with a metal oxide both neutralize each other. In this reaction, the respective salt and water are formed.\nAcid + Metal Oxide \u2192 Salt + Water\n(Metal oxides are basic in nature)\nExamples:", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nAcid + Metal Oxide \u2192 Salt + Water\n(Metal oxides are basic in nature)\nExamples:\nCalcium is a metal, thus, calcium oxide is a metallic oxide which is basic in nature. When an acid, such as hydrochloric acid, reacts with calcium oxide, neutralization reaction takes place and calcium chloride, along with water is formed.\n\nSimilarly, when sulphuric acid reacts with zinc oxide, zinc sulphate and water are formed.\n\nCommon in all bases: A base dissociates hydroxide ion in water, which is responsible for the basic behaviour of a compound.\nExample: When sodium hydroxide is dissolved in water, it dissociates hydroxide ion and sodium ion.\n\nSimilarly, when potassium hydroxide is dissolved in water, it dissociates hydroxide ion and potassium ion.\n\nThus, the base shows its basic character because of dissociation of hydroxide ion.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nSimilarly, when potassium hydroxide is dissolved in water, it dissociates hydroxide ion and potassium ion.\n\nThus, the base shows its basic character because of dissociation of hydroxide ion.\nNeutralisation Reaction: When an acid reacts with a base, the hydrogen ion of acid combines with the hydroxide ion of base and forms water. As these ions combine together and form water instead of remaining free, thus, both neutralize each other.\n\nExample: When sodium hydroxide (a base) reacts with hydrochloric acid, sodium hydroxide breaks into a sodium ion and hydroxide ion and hydrochloric acid breaks into hydrogen ion and chloride ion.\nHydrogen ion and hydroxide ion combine together and form water, while sodium ion and chloride ion combine together and form sodium chloride.\n\nDilution of Acid and Base: The concentration of hydrogen ion in an acid and hydroxide ion in a base, per unit volume, shows the concentration of acid or base.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nDilution of Acid and Base: The concentration of hydrogen ion in an acid and hydroxide ion in a base, per unit volume, shows the concentration of acid or base.\nBy mixing of acid to water, the concentration of hydrogen ion per unit volume decreases. Similarly, by addition of base to water, the concentration of hydroxide ion per unit volume decreases. This process of addition of acid or base to water is called Dilution and the acid or base is called Diluted.\nThe dilution of acid or base is exothermic. Thus, acid or base is always added to water and water is never added to acid or base. If water is added to a concentrated acid or base, a lot of heat is generated, which may cause splashing out of acid or base and may cause severe damage as concentrated acid and base are highly corrosive.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nStrength of Acid and Base: Acids in which complete dissociation of hydrogen ion takes place are called Strong Acids. Similarly, bases in which complete dissociation of hydroxide ion takes place are called Strong Bases.\nIn mineral acid, such as hydrochloric acid, sulphuric acid, nitric acid, etc. hydrogen ion dissociates completely and hence, they are considered as strong acids. Since inorganic acids hydrogen ions do not dissociate completely, so they are weak acids.\n\nFor water or neutral solutions : pH = 7\nFor acidic solutions : pH < 7\nFor basic solution : pH > 7\nUniversal Indicator: Using a litmus paper, phenolphthalein, methyl orange, etc. only the acidic or basic character of a solution can be determined, but the use of these indicators does not give the idea about the strength of acid or base. So, to get the strength as well as acidic and basic nature of a given solution universal indicator is used.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nUniversal indicator shows different colour over the range of pH value from 1 to 14 for a given solution. Universal indicator is available both in the form of strips and solution. Universal indicator is the combination of many indicators, such as water, propanol, phenolphthalein, sodium salt, sodium hydroxide, methyl red, bromothymol blue monosodium salt, and thymol blue monosodium salt. The colour matching chart is supplied with a universal indicator which shows the different colours for different values of pH.\n\npH value shown by different colours role of pH everyday life:\n(i) pH in our digestive system: Dilute HCl (Hydrochloric acid) helps in digestion of food (proteins) in our stomach. Excess acid in stomach causes acidity (indigestion). Antacids like magnesium hydroxide [Mg(OH)2] also known as milk of magnesia and sodium hydrogen carbonate (baking soda) are used to neutralize excess acid.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\n(ii) Tooth decay caused by acids: The bacteria present in our mouth converts the sugar into acids. When the pH of acid formed in the mouth falls below 5.5, tooth-decaying starts. The excess acid has to be removed by cleaning the teeth with a good quality toothpaste because these kinds of toothpaste are alkaline in nature.\n(iii) Soil of pH and plant growth: Most of the plants have a healthy growth when the soil has a specific pH (close to 7) range which should be neither alkaline nor highly acidic. Therefore,\nCompound \u2018X\u2019 is Sodium hydroxide (NaOH).\nCompound \u2018A\u2019 is Zinc sulphate (ZnSO4).\nCompound \u2018B\u2019 is Sodium chloride (NaCl).\nCompound \u2018C\u2019 is Sodium acetate (CH3COONa)", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nCompound \u2018X\u2019 is Sodium hydroxide (NaOH).\nCompound \u2018A\u2019 is Zinc sulphate (ZnSO4).\nCompound \u2018B\u2019 is Sodium chloride (NaCl).\nCompound \u2018C\u2019 is Sodium acetate (CH3COONa)\nSalts: Salts are the ionic compounds which are produced after the neutralization reaction between acid and base. Salts are electrically neutral. There are number of salts but sodium chloride is the most common among them. Sodium chloride is also known as table salt or common salt. Sodium chloride is used to enhance the taste of food.\nCharacteristics of salt:\nMost of the salts are crystalline soild.\nSalts may be transparent or opaque.\nMost of the salts are soluble in water.\nSolution of the salts conducts electricity in their molten state also.\nThe salt may be salty, sour, sweet, bitter and umami (savoury).\nNeutral salts are odourless.\nSalts can be colourless or coloured.\nFamily of Salt: Salts having common acidic or basic radicals are said to belong to the same family.\nExample:", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nNeutral salts are odourless.\nSalts can be colourless or coloured.\nFamily of Salt: Salts having common acidic or basic radicals are said to belong to the same family.\nExample:\n(i) Sodium chloride (NaCl) and Calcium chloride (CaCl2) belongs to chloride family.\n(ii) Calcium chloride (CaCl2) and Calcium sulphate (CaSO4) belongs to calcium family.\n(iii) Zinc chloride (ZnCl2) and Zinc sulphate (ZnSO4) belongs to the zinc family.\nNeutral, Acidic and Basic Salts:\n(i) Neutral Salt: Salts produced because of reaction between a strong acid and strong base are neutral in nature. The pH value of such salts is equal to 7, i.e. neutral.\nExample : Sodium chloride, Sodium sulphate. Postassium chloride, etc.\nSodium chloride (NaCl): It is formed after the reaction between hydrochloric acid (a strong acid) and sodium hydroxide (a strong base).", + "neutral.\nExample : Sodium chloride, Sodium sulphate. Postassium chloride, etc.\nSodium chloride (NaCl): It is formed after the reaction between hydrochloric acid (a strong acid) and sodium hydroxide (a strong base).\n\nSodium Sulphate (Na2SO4): It is formed after the reaction between sodium hydroxide (a strong base) and sulphuric acid ( a strong acid).", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nSodium Sulphate (Na2SO4): It is formed after the reaction between sodium hydroxide (a strong base) and sulphuric acid ( a strong acid).\n\nPotassium Chloride (KCl): It is formed after the reaction between potassium hydroxide (a strong base) and hydrochloric acid (a strong acid).\n\n(ii) Acidic Salts: Salts which are formed after the reaction between a strong acid and weak base are called Acidic salts. The pH value of acidic salt is lower than 7. For example Ammonium sulphate, Ammonium chloride, etc.\nAmmonium chloride is formed after reaction between hydrochloric acid (a strong acid) and ammonium hydroxide (a weak base).\n\nAmmonium sulphate is formed after reaction between ammonium hydroxide (a weak base) and sulphuric acid (a strong acid).\n\n(iii) Basic Salts: Salts which are formed after the reaction between a weak acid and strong base are called Basic Salts. For example; Sodium carbonate, Sodium acetate, etc.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\n(iii) Basic Salts: Salts which are formed after the reaction between a weak acid and strong base are called Basic Salts. For example; Sodium carbonate, Sodium acetate, etc.\nSodium carbonate is formed after the reaction between sodium hydroxide (a strong base) and carbonic acid (a weak acid).\n\nSodium acetate is formed after the reaction between a strong base, sodium hydroxide (a strong base) and acetic acid, (a weak acid).\n\nCause of formation of acidic, basic and neutral salts:\nWhen a strong acid reacts with a weak base, the base is unable to fully neutralize the acid. Due to this, an acidic salt is formed.\nWhen a strong base reacts with a weak acid, the acid is unable to fully neutralize the base. Due to this, a basic salt is formed.\nWhen equally strong acid and a base react, they fully neutralize each other. Due to this, a neutral salt is formed.\npH value of salt:\nNeutral salt: The pH value of a neutral salt is almost equal to 7.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\npH value of salt:\nNeutral salt: The pH value of a neutral salt is almost equal to 7.\nAcidic salt: The pH value of an acidic salt is less than 7.\nBasic salt: The pH value of a basic salt is more than 7.\nSome Important Chemical Compounds\n1. Common Salt (Sodium Chloride): Sodium chloride (NaCl) is also known as Common or Table Salt. It is formed after the reaction between sodium hydroxide and hydrochloric acid. It is a neutral salt. The pH value of sodium chloride is about 7. Sodium chloride is used to enhance the taste of food. Sodium chloride is used in the manufacturing of many chemicals.\nImportant chemical from sodium chloride", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nImportant chemical from sodium chloride\nSodium Hydroxide (NaOH): Sodium hydroxide is a strong base. It is also known as caustic soda. It is obtained by the electrolytic decomposition of solution of sodium chloride (brine). In the process of electrolytic decomposition of brine (aqueous solution of sodium chloride), brine decomposes to form sodium hydroxide. In this process, chlorine is obtained at anode and hydrogen gas is obtained at cathode as by products. This whole process is known as Chlor \u2013 Alkali process.\n\nUse of products after the electrolysis of brine:\nHydrogen gas is used as fuel, margarine, in making of ammonia for fertilizer, etc.\nChlorine gas is used in water treatment, manufacturing of PVC, disinfectants, CFC, pesticides. It is also used in the manufacturing of bleaching powder and hydrochloric acid.\nSodium hydroxide is used for degreasing of metals, manufacturing of paper, soap, detergents, artificial fibres, bleach, etc.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nSodium hydroxide is used for degreasing of metals, manufacturing of paper, soap, detergents, artificial fibres, bleach, etc.\n2. Bleaching Powder (CaOCl2): Bleaching powder is also known as chloride of lime. It is a solid and yellowish white in colour. Bleaching powder can be easily identified by the strong smell of chlorine.\nWhen calcium hydroxide (slaked lime) reacts with chlorine, it gives calcium oxychloride (bleaching powder) and water is formed.\n\nAqueous solution of bleaching powder is basic in nature. The term bleach means removal of colour. Bleaching powder is often used as bleaching agent. It works because of oxidation. Chlorine in the bleaching powder is responsible for bleaching effect.\nUse of Bleaching Powder:\nBleaching powder is used as disinfectant to clean water, moss remover, weed killers, etc.\nBleaching powder is used for bleaching of cotton in textile industry, bleaching of wood pulp in paper industry.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nBleaching powder is used for bleaching of cotton in textile industry, bleaching of wood pulp in paper industry.\nBleaching powder is used as oxidizing agent in many industries, such as textiles industry, paper industry, etc.\n3. Baking Soda (NaHCO3): Baking soda is another important product which can be obtained using byproducts of chlor \u2013 alkali process. The chemical name of baking soda is sodium hydrogen carbonate (NaHCO3) or sodium bicarbonate. Bread soda, cooking soda, bicarbonate of soda, sodium bicarb, bicarb of soda or simply bicarb, etc. are some other names of baking soda.\nPreparation Method: Baking soda is obtained by the reaction of brine with carbon dioxide and ammonia. This is known as Solvay process.\n\nIn this process, calcium carbonate is used as the source of CO2 and the resultant calcium oxide is used to recover ammonia from ammonium chloride.\nProperties of Sodium Bicarbonate:\nSodium bicarbonate is white crystalline solid, but it appears as fine powder.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nProperties of Sodium Bicarbonate:\nSodium bicarbonate is white crystalline solid, but it appears as fine powder.\nSodium hydrogen carbonate is amphoteric in nature.\nSodium hydrogen carbonate is sparingly soluble in water.\nThermal decomposition of sodium hydrogen carbonate (baking soda).\nWhen baking soda is heated, it decomposes into sodium carbonate, carbon dioxide and water.\n2NaHCO3 + heat \u2192 Na2CO3 + CO2 + H2O\nSodium carbonate formed after thermal decomposition of sodium hydrogen carbonate decomposes into sodium oxide and carbon dioxide on further heating.\nNa2CO3 \u2192 Na2O + CO2\nThis reaction is known as Dehydration reaction.\nUse of Baking Soda:\nBaking soda is used in making of baking powder, which is used in cooking as it produces carbon dioxide which makes the batter soft and spongy.\nBaking soda is used as an antacid.\nBaking soda is used in toothpaste which makes the teeth white and plaque free.\nBaking soda is used in cleansing of ornaments made of silver.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nBaking soda is used as an antacid.\nBaking soda is used in toothpaste which makes the teeth white and plaque free.\nBaking soda is used in cleansing of ornaments made of silver.\nSince sodium hydrogen carbonate gives carbon dioxide and sodium oxide on strong heating, thus, it, is used as a fire extinguisher.\nBaking Powder: Baking powder produces carbon dioxide on heating, so it is used in cooking to make the batter spongy. Although, baking soda also produces carbon dioxide on heating, but it is not used in cooking because on heating, baking soda produces sodium carbonate along with carbon dioxide. The sodium carbonate, thus, produced, makes the taste bitter.\n\nBaking powder is the mixture of baking soda and a mild edible acid. Generally, tartaric acid is mixed with baking soda to make baking powder.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nBaking powder is the mixture of baking soda and a mild edible acid. Generally, tartaric acid is mixed with baking soda to make baking powder.\n\nWhen baking powder is heated, sodium hydrogen carbonate (NaHCO3) decomposes to give CO2 and sodium carbonate (Na2CO3). CO2 causes bread and cake fluffy. Tartaric acid helps to remove bitter taste due to formation of Na2CO3.\n4. Washing Soda (Sodium Carbonate)\nPreparation Method: Sodium carbonate is manufactured by the thermal decomposition of sodium hydrogen carbonate obtained by Solvay process.\n\nThe sodium carbonate obtained in this process is dry. It is called Soda ash or Anhydrous sodium carbonate. Washing soda is obtained by rehydration of anhydrous sodium carbonate.\n\nSince there are 10 water molecules in washing soda, hence, it is known as Sodium Bicarbonate Decahydrate.\nSodium carbonate is a crystalline solid and it is soluble in water when most of the carbonates are insoluble in water.\nUse of sodium carbonate:", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nSodium carbonate is a crystalline solid and it is soluble in water when most of the carbonates are insoluble in water.\nUse of sodium carbonate:\nIt is used in the cleaning of cloths, especially in rural areas.\nIn the making of detergent cake and powder.\nIn removing the permanent hardness of water.\nIt is used in glass and paper industries.\nThe water of Crystallization: Many salts contain water molecule and are known as Hydrated Salts. The water molecule present in salt is known as Water of crystallization.\nExamples:\nCopper sulphate pentahydrate (CuSO4.5H2O): Blue colour of copper sulphate is due to presence of 5 molecules of water. When copper sulphate is heated, it loses water molecules and turns: into grey \u2013 white colour, which is known as anhydrous copper sulphate. After adding water, anhydrous copper sulphate becomes blue again.\n\nAcids: Substances which turn blue litmus solution red are called acids. Acids are sour in taste.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nAcids: Substances which turn blue litmus solution red are called acids. Acids are sour in taste.\nBases: Substances which change red litmus solution blue are called bases. They are bitter in taste.\nMineral Acids: Acids which are obtained from minerals like sulphates, nitrates, chlorides etc. are called mineral acids, example, H2SO4 (Sulphuric acid), HNO3 (Nitric acid) and HCl (Hydrochloric acid).\nOrganic Acids: Acids which are obtained from plants and animals are called organic acids. Example citric acid, ascorbic acid, tartaric acid, lactic acid, acetic acid.\nHydronium Ions: They are formed by the reaction of H+ (from acid) and H2O. It is because H+ is unstable.\nUniversal Indicator: A universal indicator is a mixture of indicators which shows a gradual but well-marked series of colour changes over a very wide range of change in concentration of H+ ions.", + "It is because H+ is unstable.\nUniversal Indicator: A universal indicator is a mixture of indicators which shows a gradual but well-marked series of colour changes over a very wide range of change in concentration of H+ ions.\nStrong Acids: Acids which dissociate into ions completely are called strong acids. Example, H2SO4, HCl.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nStrong Acids: Acids which dissociate into ions completely are called strong acids. Example, H2SO4, HCl.\nWeak Acids: Acids which do not dissociate into ions completely are called weak acids. Example, citric acid, acetic acid.\nChemical Properties of Acids:\nAcids react with active metals to give salt and hydrogen gas.\nAcids react with metal carbonates and metal hydrogen carbonates to give salt, water and carbon dioxide.\nAcids react with bases to give salt and water. This reaction is called a neutralization reaction.\nAcids react with metal oxides to give salt and water.\nChemical Properties of Bases:\nReaction with metals: Certain metals such as zinc, aluminium and tin react with alkali solutions on heating and hydrogen gas is evolved.\nReaction with acids: Bases react with acids to form salt and water.\nIndicators: Indicators are substances which indicate the acidic or basic nature of the solution by their colour change.", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nReaction with acids: Bases react with acids to form salt and water.\nIndicators: Indicators are substances which indicate the acidic or basic nature of the solution by their colour change.\npH Scale: A scale for measuring hydrogen ion concentration in a solution.\nThe pH of a solution is defined as the negative logarithm of hydrogen ion concentration in moles per litre.\npH = -log [H+]\npH = -log [H3O+]\nwhere [H+] or [H3O+] represents concentrations of hydrogen ions in a solution.\nThe pH of a neutral solution is 7.\nThe pH of an acidic solution is < 7.\nThe pH of a basic solution is > 7.", + "The pH of a neutral solution is 7.\nThe pH of an acidic solution is < 7.\nThe pH of a basic solution is > 7.\nSome Important Compounds and their Uses:\nEquations of Acids, Bases and Salts:\nAcid + Metal \u2192 Salt + Hydrogen gas\nH2SO4 + Zn \u2192 ZnSO4 + H2\nBase + Metal \u2192 Salt + Hydrogen gas\n2NaOH + Zn \u2192 Na2ZnO2 (Sodium zincate) + H2\nBase + Acid \u2192 Salt + Water\nNaOH (aq) + HCl (aq) \u2192 NaCl (aq) + H2O (l)\nAcids give hydronium ions in water\nHCl + H2O \u2192 H3O+ + Cl\u2013\nBases generate OH- ions in water", + "CHAPTER 2-ACIDS,BASES AND SALTS -NOTES.pdf.txt\nBase + Acid \u2192 Salt + Water\nNaOH (aq) + HCl (aq) \u2192 NaCl (aq) + H2O (l)\nAcids give hydronium ions in water\nHCl + H2O \u2192 H3O+ + Cl\u2013\nBases generate OH- ions in water\nNaOH (aq) + H2O \u2192 Na+ (aq) + O\u2013 (aq)\nReactions Of Important Chemical Compounds:\nPreparation of Bleaching powder: By the action of chlorine on dry slaked lime\nCa(OH)2 + Cl2 \u2192 CaOCl2 + H2O\nOn heating, baking soda liberates CO2\n\nPreparation of Plaster of Paris:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 15 Intext Questions\nQuestion 1\nWhat are trophic levels ? Give an example of food chain and state the different trophic levels in it.\nAnswer:\nTrophic Levels : The various steps in a food chain at which the transfer of food (or energy) takes place are called trophic levels.\nExample : A food chain operating in a grassland is given below :\nGrass \u2192 Insects \u2192 Frog \u2192 Birds\n\nIn this food chain\nGrass represents first trophic level.\nGrasshopper represents second trophic level.\nFrog represents third trophic level.\nEagle represents fourth tropic level.\nQuestion 2\nWhat is the role of decomposers in the ecosystem ?\nAnswer:\n(i) Decomposers help in decomposing the dead bodies of plants and animals and hence act as cleansing agents of the environment.\n(ii) Decomposers also help in putting back the various elements of which dead plants and animals are made, back into the soil, air and water for reuse by the producers like crop plants.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\n(ii) Decomposers also help in putting back the various elements of which dead plants and animals are made, back into the soil, air and water for reuse by the producers like crop plants.\n(iii) They help in recycling of the nutrients.\n(iv) They decompose dead remains thereby providing space for new life to settle in the biosphere.\nPage Number: 262\nQuestion 1\nWhy are some substances biodegradable and some non-biodegradable ?\nAnswer:\nThe microorganism like bacteria and other decomposer organisms (called saprophytes) present in our environment are specific in their action. They break down the materials or products made from natural materials (say, paper) but do not break down man-made materials such as plastics. So, it is due to the property of decomposer organisms of being specific in their action that some waste materials are biodegradable, whereas others are non-biodegradable.\nQuestion 2\nGive any two ways in which biodegradable substances would affect the environment.\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nQuestion 2\nGive any two ways in which biodegradable substances would affect the environment.\nAnswer:\n(i) Biodegradable substances are decomposed by the action of microorganisms and decomposed materials are recycled through geo-chemical cycle.\n(ii) These substances keep the environment clean.\nQuestion 3\nGive any two ways in which non-biodegradable substances would effect the environment.\nAnswer:\n(i) They cause air, water and soil pollution.\n(ii) They may cause bio-magnification in the food chain and end up in humans.\nPage Number: 264\nQuestion 1\nWhat is ozone and how does it affect any ecosystem ?\nAnswer:\nOzone (O3) is an isotope of oxygen, i.e., it is a molecule formed by three atoms of oxygen.\nAt the higher levels of the atmosphere, ozone performs an essential function. It shields the surface of the earth from ultraviolet (UV) radiations from the sun. These radiations are highly damaging to organisms. Ultraviolet rays can cause skin cancer.\nQuestion 2", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nQuestion 2\nHow can you help in reducing the problem of waste disposal ? Give any two methods.\nAnswer:\n(i) Recycling : The solid wastes like paper, plastics and metals, etc. are recycled.\n(ii) Preparation of Compost: Biodegradable domestic wastes such as left over food, fruit and vegetable peels and leaves of potted plants, etc. can be converted into compost by burying in a pit dug into ground.\nNCERT Solutions for Class 10 Science Chapter 15 Textbook Chapter End Questions\nQuestion 1\nWhich of the following groups contain only biodegradable item ?\n(a) Grass, flowers and leather\n(b) Grass, wood and plastic\n(c) Fruit peels, cake and lime juice\n(d) Cake, wood and grass\nAnswer:\n(a) Grass, flowers and leather.\nQuestion 2\nWhich of the following constitutes a food-chain ?\n(a) Grass, wheat and mango\n(b) Grass, goat and human\n(c) Goat, cow and elephant\n(d) Grass, fish and goat\nAnswer:\n(b) Grass, goat and human.\nQuestion 3\nWhich of the following are environment friendly practices ?", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\n(b) Grass, goat and human\n(c) Goat, cow and elephant\n(d) Grass, fish and goat\nAnswer:\n(b) Grass, goat and human.\nQuestion 3\nWhich of the following are environment friendly practices ?\n(a) Carrying cloth-bags to put purchases in while shopping\n(b) Switching off unnecessary lights and fans\n(c) Walking to school instead of getting your mother to drop on her scooter\n(d) All of the above\nAnswer:\n(d) All of the above.\nQuestion 4\nWhat will happen if we kill all the organisms in one trophic level ?\nAnswer:\nThe food chain would end and ecological balance would be affected.\nIf the herbivores are killed, then the carnivores would not be able to get food and would die.\nIf carnivores are killed, then the population of herbivores would increase to unsustainable level.\nIf producers are killed, then the nutrient cycle in that area would not be completed.\nQuestion 5", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nIf carnivores are killed, then the population of herbivores would increase to unsustainable level.\nIf producers are killed, then the nutrient cycle in that area would not be completed.\nQuestion 5\nWill the impact of removing all the organisms in a trophic level be different for different trophic levels ? Can the organisms of any trophic level be removed without causing any damage to the ecosystem ?\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nAnswer:\nYes, the impact of removing all the organisms in a trophic level will be different for different trophic levels. For example, on removing producers; herbivores would not be able to survive or they would migrate and ecosystem would collapse. If herbivores are removed, producers would grow unchecked and carnivores would not get food. If carnivores are removed, herbivores would increase to unsustainable levels and could destroy the producers. If decomposers are removed, the dead animals would pile up due to which the environment would become polluted. In addition to this, if dead animals will not decompose, the recycling of nutrients in the soil will be stopped and its fertility will be reduced. As a result the green cover of the earth will be lost. Thus to maintain the balance of the ecosystem the presence of organisms is necessary at each trophic level.\nQuestion 6", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nQuestion 6\nWhat is biological magnification ? Will the levels of this magnification be different at different levels of the ecosystem ?\nAnswer:\nBiological magnification : The increase in concentration of harmful chemical substances like pesticides in the body of living organisms at each trophic level of a food chain is called biological magnification.\nYes, levels of bio-magnification would increase as the trophic level increases and would be the highest for topmost trophic level. It would affect their biological process such as growth, reproduction, etc.\nQuestion 7\nWhat are the problems caused by the non-biodegradable wastes that we generate ?\nAnswer:\nThe problems caused by the non-biodegradable wastes are :\nIf the quantity of non-biodegradable matter increases in the nature then bio-magnification of poisonous chemicals in our body increases.\nIf the non-biodegradal waste keeps on increasing there will not be left any substance for new organisms.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nIf the non-biodegradal waste keeps on increasing there will not be left any substance for new organisms.\nThe increasing quantity of non-biodegradable waste will cause imbalance of ecosystem.\nQuestion 8\nIf all the waste we generate is biodegradable, will this have no impact on the environment ? [CBSE 2011, 2013]\nAnswer:\nIf all the waste we generate is biodegradable, it will also have impact on the environment. If it is disposed off properly, the problem of air, water and soil pollution can be lessened to an extent. There would be less health problems and humans would be disease-free.\nBut if it is not disposed off properly, it will affect the environment adversely.\nQuestion 9\nWhy is damage to the ozone layer a cause for concern ? What steps are being taken to limit this damage ?\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nQuestion 9\nWhy is damage to the ozone layer a cause for concern ? What steps are being taken to limit this damage ?\nAnswer:\nThe damage to the ozone layer is a cause for concern because if the ozone layer in the atmosphere disappears completely, then all the extremely harmful ultraviolet radiations coming from the sun would reach the earth. These ultraviolet radiations would cause skin cancer and other ailments in men and animals and also damage the plants.\nIn an attempt to protect the ozone layer, the United Nations Environment Programme (UNEP) unanimously forged an agreement among its member countries to freeze CFC production at 1986 levels.\nNCERT Solutions for Class 10 Science Chapter 15 Our Environment\nOur environment: Eco-system, Environmental problems, Ozone depletion, waste production and their solutions. Biodegradable and non-biodegradable substances.\nFormulae Handbook for Class 10 Maths and Science\nPage 257\nQuestion 1.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nFormulae Handbook for Class 10 Maths and Science\nPage 257\nQuestion 1.\nWhy are some substances biodegradable and some non-biodegradable ?\nAnswer:\nSubstances that are broken down by biological processes are said to be biodegradable. In our environment, many of the substances are broken easily by decomposers (bacteria and fungi) as they possess specific enzymes for such activity. However, there are other substances also which are not broken down in this manner and are known as non-biodegradable substances. Since these substances are not degraded by bacteria and fungi, so they persist for a long time. These non-biodegradable substances will be acted upon by physical processes like heat and pressure.\nQuestion 2.\nGive any two ways in which biodegradable substances would affect the environment.\nAnswer:\nThey may produce foul smell during decomposition process.\nThey may produce some harmful gases such as ammonia, methane, carbon dioxide, etc., which can further-cause global warming.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nAnswer:\nThey may produce foul smell during decomposition process.\nThey may produce some harmful gases such as ammonia, methane, carbon dioxide, etc., which can further-cause global warming.\nQuestion 3.\nGive any two ways in which non-biodegradable substances would affect the environment.\nAnswer:\nThese inert substances simply persist in the environment. This means that these substances require land area for dumping.\nExcess of fertilizers, pesticides and other chemicals changes soil chemistry and also affects aquatic life.\nMost of these chemicals and heavy metal are easily absorbed by the organisms. This causes biological magnification.\nDownload NCERT Solutions for Class 10 Science Chapter 15 Our Environment PDF\nPage : 261\nQuestion 1.\nWhat are trophic levels? Give an example of a food chain and state the different trophic levels in it.\nAnswer:\nEach step or level of the food chain forms a trophic level. Consider the following food chain:\n\nQuestion 2.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nAnswer:\nEach step or level of the food chain forms a trophic level. Consider the following food chain:\n\nQuestion 2.\nWhat is the role of decomposers in the ecosystem?\nAnswer:\nRole of decomposers in the ecosystem :\nThey help in breaking down the complex organic into simple inorganic that go into the soil and are used up by the plants.\nThey the nutrient pool of the putting In this way, ad as cleansing agents of nature.\nThey help in maintaining the fertility of by adding humus content to it.\nPage : 264\nQuestion 1.\nWhat is ozone and how does it affect any ecosystem?\nAnswer:\nOzone (O3) is a molecule formed by three atoms of oxygen. At the higher of the atmosphere, it shields the surface of the earth from ultraviolet (UV) radiation from the Sun. It may affect any ecosystem in the following ways :\nAt the surface of the earth, it is a deadly poison for all lower forms of life.\nIf this layer gets depleted, then it may cause cancer in human beings including other plants and animals.\nQuestion 2.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nAt the surface of the earth, it is a deadly poison for all lower forms of life.\nIf this layer gets depleted, then it may cause cancer in human beings including other plants and animals.\nQuestion 2.\nHow can you help in reducing the problem of waste disposal? Give any two methods?\nAnswer:\nBy changing our lifestyle and change in attitude will reduce disposable waste.\nReducing packaging.\nRecycling of waste.\nPreparing compost of biodegradable waste.\nExcercise:\nQuestion 1.\nWhich of the following groups contain only biodegradable items?\n(a) Grass, flowers and leather\n(b) Grass, wood and plastic\n(c) Fruit-peels, cake and lime-juice\n(d) Cake, wood and grass\nAnswer:\n(c) Fruit-peels, cake and lime-juice and (d) Cake, wood and grass\nQuestion 2.\nWhich of the following constitute a food-chain?\n(a) Grass, wheat and mango\n(b) Grass, goat and human\n(c) Goat, cow and elephant\n(d) Grass, fish and goat\nAnswer:\n(b) Grass, goat and human\nQuestion 3.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nWhich of the following constitute a food-chain?\n(a) Grass, wheat and mango\n(b) Grass, goat and human\n(c) Goat, cow and elephant\n(d) Grass, fish and goat\nAnswer:\n(b) Grass, goat and human\nQuestion 3.\nWhich of the following are environment-friendly practices?\n(a) Carrying cloth-bags to put purchases in while shopping\n(b) Switching off unnecessary lights and fans\n(c) Walking to school instead of getting your mother to drop you on her scooter\n(d) All of the above\nAnswer:\n(d) All of the above\nQuestion 4.\nWhat will happen if we kill all the organisms in one trophic level?\nAnswer:\nIf we kill all the organisms in one trophic level, then transfer of energy as well as matter to next higher level will stop. It will lead to over- population at one particular level causing amongst the individuals. This would seriously disturb the food chain and can cause the collapse of an ecosystem even.\nQuestion 5.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nQuestion 5.\nWill the impact of removing all the organisms in a trophic level be different for different trophic levels? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?\nAnswer:\nYes, the impact Of removing all the Organisms in a trophic level will be different for different trophic levels. It will not be possible to remove any organism in any trophic level without causing damage to the ecosystem.\nQuestion 6.\nWhat is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?\nAnswer:\nThe accumulation of harmful chemicals in the body of living organisms at different trophic levels in a food chain is called biological magnification. Yes, the concentration of these harmful chemicals will be different at different trophic levels. It will be maximum at the last trophic levels which is mostly of the top carnivores (quaternary consumers).\nQuestion 7.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nQuestion 7.\nWhat are the problems caused by the non-biodegradable wastes that we generate?\nAnswer:\n(i) Non-biodegradable wastes persist in the environment for a long time and cause greater harm to the various members of the ecosystem by causing biological magnification.\n(ii) Non-biodegradable waste such as fertilizers, pesticides, weedicides, etc., changes the soil chemistry. in turn affects the fertility of soil and subsequently reduces the crop yield.\nQuestion 8.\nIf all the waste we generate is biodegradable, will this have no impact on the environment?\nAnswer:\nBiodegradable waste will be recycled easily by the decomposers such as bacteria and fungi. It will have only this bad impact on our environment that, many Of the gases released during decomposition process may result in global warming.\nQuestion 9.\nWhy is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage?\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nQuestion 9.\nWhy is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage?\nAnswer:\nThe ozone shields the surface of the earth from ultraviolet (UV) radiation from the sun. These radiations are highly damaging as they can cause cancer in both plants and animals, damage to eyes and immune system. They can also lead to variations in global rainfall, ecological disturbances and dwindling of global food supplies. Due to these reasons, damage to the ozone layer is a major cause for concern.\nSteps which are taken to limit this damage :\nTo decrease the use of synthetic chemicals like chlorofluorocarbons (CFCs) which are used as refrigerants and in fire extinguishers.\nIn 1987, the United Nations Environment Programme (UNEP) succeeded in reaching an agreement to freeze CFC production at 1986\nMultiple Choice Questions (MCQs) [1 Mark each]\nQuestion 1.\nIdentify the original source of the energy which flows through a food chain?\n(a) Carbon dioxide\n(b) Glucose", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nMultiple Choice Questions (MCQs) [1 Mark each]\nQuestion 1.\nIdentify the original source of the energy which flows through a food chain?\n(a) Carbon dioxide\n(b) Glucose\n(c) Oxygen\n(d) Sunlight\nAnswer:\n(d) All the living organisms get energy directly or indirectly from the sunlight reaching the surface of Earth.\nQuestion 2.\nA teacher draws the pyramid of energy on board and writes A, B, C and D, in each trophic level as shown in the diagram given alongside. Which level represents the herbivores?\n\n(a) A\n(b) B\n(c) C\n(d) D\nAnswer:\n(c) The trophic levels represent the transfer of food or energy through various steps or levels in the food chain. Producers harness the maximum energy followed by primary consumers, i.e. a herbivore, which is represented by C in thg given diagram.\nQuestion 3.\nWhich of the following groups contains only biodegradable items?", + "Producers harness the maximum energy followed by primary consumers, i.e. a herbivore, which is represented by C in thg given diagram.\nQuestion 3.\nWhich of the following groups contains only biodegradable items? [NCERT]\n(a) Grass, flowers and leather\n(b) Grass, wood and plastic\n(c) Fruit-peels, cake and lime-juice\n(d) Cake, wood and grass\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\n(a) Grass, flowers and leather\n(b) Grass, wood and plastic\n(c) Fruit-peels, cake and lime-juice\n(d) Cake, wood and grass\nAnswer:\n(a), (c) and (d) substances that are broken down (decomposed) by the biological processes are said to be biodegradable e.g. fruit-peels, cake, lime-juice, wood, grass, leather, flowers, etc.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nQuestion 4.\nWhich of the following constitutes a food-chain? [NCERT]\n(a) Grass, wheat and mango\n(b) Grass, goat and human\n(c) Goat, cow and elephant\n(d) Grass, fish and goat\nAnswer:\n(b) Each step of food chain form a trophic level. Producers (grass) forms the first trophic level, herbivore (goat) the second and carnivore (human) the third trophic level.\nQuestion 5.\nWhich of the following are environment- friendly practices? [NCERT]\n(a) Carrying cloth-bags to put purchases \u2018 while shopping .\n(b) Switching off unnecessary light and fans\n(c) Walking to school instead of getting your mother to drop you on her scooter\n(d) All of the above\nAnswer:\n(d) Cloth-bags are biodegradable, switching off unnecessary light and fans conserves electricity and limited use of petrol/diesel causes less pollution. Hence, all these practices are considered as environment friendly.\nQuestion 6.\nWhich of the following statements about food chain are correct?", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nQuestion 6.\nWhich of the following statements about food chain are correct?\n(a) It includes repeated eating, i.e. each group eats the other and is subsequently eaten by some other group of organisms.\n(b) It shows a series of branching lines and unidirectional flow of energy.\n(c) It shows the unidirectional flow of energy and proceeds in a progressive straight line.\n(d) Both (a) and (c)\nAnswer:\n(d) A food chain is a series of organisms in an environment through which energy transfer occurs starting with a producer. It proceeds in straight line. Food chain does not consist of branching lines.\nQuestion 7.\nIn class, the teacher explained the concept of food chain and energy flow. She made a diagram as given below and asked the students to identify the producer organism in the chain. What do you think will be the student\u2019s answer?\n\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nAnswer:\n(b) Cabbage is the producer component of this chain. It produces food using sunlight and other components by photosynthesis process. Others are all consumers.\nQuestion 8.\nAn ecosystem includes\n(a) all living organisms\n(b) non-living objects ,\n(c) both living organisms and non-living objects\n(d) sometimes living organisms and sometimes non-living objects\nAnswer:\n(c) All the interacting organisms in an area taken together with the non-living constituents of the environment form an ecosystem. Thus, an ecosystem consists of biotic components including all living organisms and abiotic components constituting physical factors like temperature, rainfall, wind, soil and minerals.\nQuestion 9.\nIn the following groups of materials, which group (s) contains only non-biodegradable items?\n(i) Wood, paper, leather\n(ii) Polythene, detergent, PVC\n(iii) Plastic, detergent, grass\n(iv) Plastic, bakelitC DDT\n(a) (iii)\n(b) (iv)\n(c) (i) and (iii)\n(d) (ii) and (iv)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\n(i) Wood, paper, leather\n(ii) Polythene, detergent, PVC\n(iii) Plastic, detergent, grass\n(iv) Plastic, bakelitC DDT\n(a) (iii)\n(b) (iv)\n(c) (i) and (iii)\n(d) (ii) and (iv)\nAnswer:\n(d) Substances that cannot be broken down by biological processes in nature are non-biodegradable. e.g. polythene, detergent, PVC, plastics, bakelite, DDT, etc. On the other hand, substances that are broken down (decomposed) by biological processes are said to be biodegradable, e.g. wood, paper, leather, grass, animal bones, etc.\nQuestion 10.\nWhich of the following statement is incorrect?\n(a) All green plants and blue-green algae are producers.\n(b) Green plants get their food from organic compounds.\n(c) Producers prepare their own food from inorganic compounds.\n(d) Plants convert solar energy into chemical energy.\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\n(b) Green plants get their food from organic compounds.\n(c) Producers prepare their own food from inorganic compounds.\n(d) Plants convert solar energy into chemical energy.\nAnswer:\n(b) Green plants prepare their food from inorganic compounds using radiant energy of the sun in the presence of chlorophyll. All green plants and blue-green algae are called producers as they can prepare food from inorganic substances by photosynthesis. Producers capture the solar energy and convert it into chemical energy.\nQuestion 11.\nWhat will happen if deer is missing in the food chain given below?\nGrass \u2192 Deer \u2192 Tiger\n(a) The population of tiger increases.\n(b) The population of grass decreases.\n(c) Tiger will start eating grass.\n(d) The population of tiger decreases and the population of grass increases.\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\n(a) The population of tiger increases.\n(b) The population of grass decreases.\n(c) Tiger will start eating grass.\n(d) The population of tiger decreases and the population of grass increases.\nAnswer:\n(d) If deer is missing in the given food chain, there will not be sufficient food for the tigers. Some of the tigers will die because of starvation and hence, the population of tigers will decrease. Since, grass is eaten by deers, the population of grass will also increase whefl deer is missing.\nQuestion 12.\nIn a class activity, two students were asked to collect different items from their fellow mates and classify them as biodegradable and non- biodegradable. All the items have been identified except three. Find out which one is non-biodegradable among these?\n(a) Jute crafted bag\n(b) A sharpner\n(c) Empty fevistick\n(d) Both (b) and (c)\nAnswer:\n(d) Both (b) and (c), i.e. the sharpener and the empty fevistick container. These products are made from plastic and hence are non-biodegradable.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\n(c) Empty fevistick\n(d) Both (b) and (c)\nAnswer:\n(d) Both (b) and (c), i.e. the sharpener and the empty fevistick container. These products are made from plastic and hence are non-biodegradable.\nQuestion 13.\nThe diagram shows excretory losses from a rat to the environment.\n\nWhich of the following will not be returned to the ecosystem and recycled?\n(a) carbon dioxide\n(b) heat energy\n(c) salts\n(d) urea\nAnswer:\n(b) Heat energy cannot be recycled, it gets lost in the environment. The generated is returned through the carbon cycle. Salts are used by living organisms present in the ecosystem. Urea also returns to the nitrogen cycle.\nQuestion 14.\nWhich of the following limits the number of trophic levels in a food chain?\n(a) Decrease in energy at higher trophic levels\n(b) Deficient food supply\n(c) Polluted air\n(d) Water\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nQuestion 14.\nWhich of the following limits the number of trophic levels in a food chain?\n(a) Decrease in energy at higher trophic levels\n(b) Deficient food supply\n(c) Polluted air\n(d) Water\nAnswer:\n(a) Decrease in energy at higher trophic levels limits the number of trophic levels in a food chain. At each trophic level, a large portion of energy is utilised for the maintenance of organisms that occur at that trophic level. So, organisms at higher level gets less and less energy at successive levels. The. number of trophic levels are limited to 3-4 because after that, the energy available for the next level will be too small, i.e. it will be insufficient to sustain life of the organisms.\nQuestion 15.\nIf a grasshopper is eaten by a frog, then the energy transfer will be from\n(a) producer to decomposer\n(b) producer to primary consumer\n(c) primary consumer to secondary consumer\n(d) secondary consumer to primary consumer\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\n(a) producer to decomposer\n(b) producer to primary consumer\n(c) primary consumer to secondary consumer\n(d) secondary consumer to primary consumer\nAnswer:\n(c) In a food chain, if a grasshopper is eaten by a frog, then the energy transfer will be from primary consumer to secondary consumer. Grasshopper feeds on producers i.e. the grass/plants. So, it occupies the level of primary consumer. Frogs, eating grasshopper thus become the secondary consumer.\nQuestion 16.\nIn the given food chain, suppose the amount of energy at fourth trophic level is 5 kJ, what will be the energy available at the producer level?\nGrass \u2192 Grasshopper \u2192 Frog \u2192 Snake \u2192 Hawk\n(a) 5 kJ\n(b) 50 kJ\n(c) 500 kJ\n(d) 5000 kJ\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nGrass \u2192 Grasshopper \u2192 Frog \u2192 Snake \u2192 Hawk\n(a) 5 kJ\n(b) 50 kJ\n(c) 500 kJ\n(d) 5000 kJ\nAnswer:\n(d) According to 10% law, only 10% of the energy entering a particular trophic level of organisms is available for transfer to the next higher trophic level. In this food chain, at the 4th trophic level, only 5 kJ energy is available to the snake. So, the energy available at the producer level will be 5000 kJ.\nIt can be shown as\n\nClass 10 Science Our Environment Mind Map\nEco-system\nIt is defined as functional unit of nature, where living organisms interact among themselves and also with the surrounding physical environment.\nHence, there are two main components of the ecosystem:\nBiotic: Living organisms such as plants, animals, microorganisms and humans.\nAbiotic: It includes physical factors such as temperature, rainfall, wind, soil and minerals.\nExamples of natural ecosystem: forests, ponds, lakes, etc, and human made or artificial ecosystems are gardens, crop-fields, aquarium etc.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nExamples of natural ecosystem: forests, ponds, lakes, etc, and human made or artificial ecosystems are gardens, crop-fields, aquarium etc.\nEcosystem consists of various organisms which can be classified as producers and consumers.\nProducers are the organisms which make organic compounds like sugar, starch, etc from inorganic substances with the help of sunlight and chlorophyll.\nConsumers are the organisms which are dependent on producers for their nutrition. They can be grouped as herbivores, carnivores, omnivores, parasites, and decomposers.\nDecomposers are the microorganisms (bacteria & fungi) which break down complex organic substances (dead remains & waste material of living organisms) into simpler inorganic substances that go back into the soil and are used up again by the plants.\nThus, they help in proper cycling of the nutrients in an ecosystem.\nHave you ever wondered why we need to clean aquarium but not lakes or ponds?", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nThus, they help in proper cycling of the nutrients in an ecosystem.\nHave you ever wondered why we need to clean aquarium but not lakes or ponds?\nIt is because aquarium is an artificial and incomplete system which lacks producers, food chains, and decomposers.\nHence, it lacks natural nutrients recycling and self\u00accleaning abilities. In contrast lake or ponds are natural and complete ecosystem where there is perfect recycling of nutrients and thus does not need to be cleaned very often.\nWaste Material\nWaste material can be broadly classified into two categories depending on their degradation:\nBiodegradable Waste\nThese are the wastes that can be broken down into simpler compounds by the action of bacteria or other saprophytes.\nIn addition, physical processes such as high temperature and pressure also act on them however, under ambient conditions these substances persist in our environment for a very long time.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nIn addition, physical processes such as high temperature and pressure also act on them however, under ambient conditions these substances persist in our environment for a very long time.\nSome examples of such wastes are food materials, kitchen wastes, and other natural wastes.\nNon-biodegradable Waste\nThese are the substances that are not broken down into simpler compounds by the action of microorganisms.\nThese substances may be inert and simply persist in the environment for a long time or may harm the various members of the eco-system.\nThey are the main causes of air, water and soil pollution and diseases like cancer.\nSome examples of such waste are plastic, cans, metals, and chemicals for agricultural and industrial purposes.\nFood Chains And Webs", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nSome examples of such waste are plastic, cans, metals, and chemicals for agricultural and industrial purposes.\nFood Chains And Webs\nIn an ecosystem, there exists a series of organisms feeding on one another. This series or organisms taking part at various biotic levels form a food chain. Alternatively, food chain can be defined as a linear network of food or energy flow starting from producer and ending at apex predator.\nTrophic Level\nBased on the source of their nutrition or food, organisms occupy a specific place in food chain that is known as their trophic level. There are usually four trophic levels:\nI trophic level: It includes producers or autotrophs for e.g. phytoplankton, grass, trees etc.\nII trophic level: It includes primary consumer or herbivores for e.g. zooplanktons, grasshoppers, cow etc.\nIII trophic level: It includes secondary consumer or small carnivore for e.g. birds, fishes, wolf etc.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nIII trophic level: It includes secondary consumer or small carnivore for e.g. birds, fishes, wolf etc.\nIV trophic level: It includes tertiary consumers or larger carni vores for e.g. level lion, tiger, man etc.\nEnergy Flow\nThe flow of energy is unidirectional.\nThe green plants in a terrestrial ecosystem capture about 1 % of the energy of sunlight that falls on their leaves and convert it into food energy.\nOn an average only 10% organic matter is present at each step and reaches the next level of consumers. It is because a great deal of energy is lost as heat to the environment and rest goes into digestion, in doing work and in growth & reproduction.\nIn addition, the loss of energy at each step is so great that very little usable energy remains after four trophic levels and this is the reason that a food chain usually contain maximum of four trophic levels.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nGenerally, there are greater number of individuals at lower trophic levels of an ecosystem (the greatest number is of the producers)\nThe length and complexity of food chains vary greatly.\nEach organism is generally eaten by two or more other kinds of organisms which in turn are eaten by several other organisms.\nSo instead of a straight line food chain, the relationship can be shown as a series of branching lines called a food web.\n\nEnvironmental Problems\nDepletion of the ozone layer: Ozone (O3) is a molecule formed by three atoms of oxygen. It is located in upper part of the atmosphere called stratosphere and it acts as shield absorbing UV radiation from sun.\nAt higher levels of atmosphere, high energy UV radiations split apart some moleculer oxygen (O2) into free oxygen (O) atoms. These atoms then combine with the molecular oxygen to form ozone.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nAt higher levels of atmosphere, high energy UV radiations split apart some moleculer oxygen (O2) into free oxygen (O) atoms. These atoms then combine with the molecular oxygen to form ozone.\nOzone depletion permits entry of harmful UV radiations which lead to diseases like skin aging, irritation & cancer, snow- blindness, cataract, etc.\nOzone-depleting substances: CFCs, HCFCs, hydrobromofluorocarbons, etc.\nWaste disposal: Improvements in our life-style have resulted in greater amounts of waste material generation. For e.g. increased use of disposable items, plastic bags, packing materials etc have resulted in much of our waste becoming non-biodegradable.\nFollowing methods can be helpful in managing the garbage we produce: categorization of waste materials into biodegradable, recyclable & non-biodegradable, reduction in use of non- biodegradable substances such as plastics, thermocol etc, burning & proper dumping of waste.", + "CHAPTER 13-OUR ENVIRONMENT INTEXT QUESTIONS.txt\nIt is interesting note that how unknowingly some harmful chemicals enter our bodies through the food chain.\nIt happens by two ways:\nOveruse of several pesticides: These chemicals are either washed down into the soil or into the waterbodies.\nFrom the soil, these are absorbed by the plants along with water and minerals, and from the water bodies these are taken up by aquatic plants and animals\nBiological magnification: It is defined as an increase in the concentration of the toxicant at successive trophic levels.\nThese chemicals are not degradable and the organism can neither metabolize nor excrete them and thus they get accumulated progressively at each trophic level.\nIn addition, human beings occupy the top level in any food chain and thus a maximum concentration of these chemicals gets accumulated in our bodies.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nImportant Questions of Our Environment Class 10 Science Chapter 15\nHuman body is made up of five important components ofwhich water is the main component. Food as well as potable water are essential for every human being. The food is obtained from plants through agriculture. Pesticides are being used extensively for a high yield in the fields. These pesticides are absorbed by the plants from the soil along with water and minerals and from the water bodies these pesticides are taken up by the aquatic animals and plants. As these chemicals are not biodegradable, they get accumulated progressively at each trophic level The maximum concentration of these chemicals gets accumulated in our bodies and greatly affects the health of our mind and body.\nQuestion 1.\nWhy is the maximum concentration of pesticides found in human beings?\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 1.\nWhy is the maximum concentration of pesticides found in human beings?\nAnswer:\nThe pesticides are not biodegradable, they get accumulated progressively at each trophic level. As human beings occupy the topmost level in food chain, their concentration becomes maximum in our bodies.\nQuestion 2.\nGive one method which could be applied to reduce our intake of pesticides through food to some extent.\nAnswer:\nBy using biological methods for controlling insects in fields and by washing fruits and vegetables before eating could help to reduce our intake of pesticides through food to some extent.\nQuestion 3.\nVarious steps in a food chain represent:\n(a) food web\n(b) trophic level\n(c) ecosystem\n(d) biomagnification.\nAnswer:\n(b) trophic level\nQuestion 4.\nWith regard to various food chains operating in an ecosystem, man is a:\n(a) Consumer\n(b) Producer\n(c) Producer and consumer\n(d) Producer and decomposer. (2020)\nAnswer:\n(a) consumer\nQuestion 5.\nFood web is constituted by", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\n(a) Consumer\n(b) Producer\n(c) Producer and consumer\n(d) Producer and decomposer. (2020)\nAnswer:\n(a) consumer\nQuestion 5.\nFood web is constituted by\n(a) relationship between the organisms and the environment\n(b) relationship between plants and animals\n(c) various interlinked food chains in an ecosystem\n(d) relationship between animals and environment. (2020)\nAnswer:\n(c) various interlinked food chains in an ecosystem\nQuestion 6.\nWhat is an ecosystem? (Delhi 2017)\nAnswer:\nAn ecosystem is defined as a structural and functional unit of the biosphere. It comprises of living organisms and their non-living environment that interact by means of food chains and biogeo-chemical cycles resulting in energy-flow, biotic diversity and material cycling to form stable self-supporting system.\nQuestion 7.\nWhy is a lake considered to be a natural ecosystem? (Delhi 2017)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 7.\nWhy is a lake considered to be a natural ecosystem? (Delhi 2017)\nAnswer:\nLake is an ecosystem where living organisms grow, reproduce and interact among each other as well as with abiotic components and carry out other activities in nature by themselves without any human interference, therefore it is referred to as a natural ecosystem.\nQuestion 8.\nIn the following food chain, plants provide 500 J of energy to rats. How much energy will be available to hawks from snakes?\nPlants \u2192 Rats \u2192 Snakes \u2192 Hawks (AI 2017)\nAnswer:\nIn an ecosystem, only 10% of energy is transferred from one trophic level to next, i.e. 10 percent law and rest is dissipated into the environment. Therefore, if plants (being producers-1st trophic level)-transfer 500 J of energy to rats (2nd trophic level) then rats would transfer 50 J of energy to snakes (3rd trophic level) which in turn will transfer only 5 J of energy to hawks (4th or last trophic level) in a food chain.\n\nQuestion 9.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 9.\nIn the following food chain, 100 J of energy is available to the lion. How much energy was available to the producers?\nPlants \u2192 Deer \u2192 Lion (AI 2017)\nAnswer:\nAs per 10% law of flow of energy in an ecosystem only 10% of energy is received by the next trophic level. Hence, in the given food chain : If 100 .J of energy is available to lion, the plants or producers have 10,000 J of energy available to them.\n\nQuestion 10.\nList two biotic components of a biosphere. (Delhi 2016)\nAnswer:\nTwo biotic components of a biosphere are:\n(i) Producers \u2013 Include organisms which can produce their food using simple inorganic compounds, e.g., all green plants, blue green algae (cyanobacteria).\n(ii) Consumers \u2013 Include organisms which are unable to synthesise their food, therefore, utilise materials and energy stored by the producers or eat other organisms, e.g., all the animals.\nQuestion 11.\nWhy are green plants called producers? (Delhi 2016)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 11.\nWhy are green plants called producers? (Delhi 2016)\nAnswer:\nGreen plants are called producers because they manufacture their own food with the help of CO2 and H2O in the presence of sunlight and chlorophyll.\nQuestion 12.\nIn a food chain of frog, grass, insect and snake, assign trophic level to frog. (AI 2016)\nAnswer:\nIn the given food chain, frog belongs to the third trophic level as shown here :\n\nQuestion 13.\nWhy do producers always occupy the first trophic level in every food chain? (Foreign 2016)\nAnswer:\nProducers are the green plants that can manufacture food using CO2 and H2O in the presence of sunlight, i.e., they are autotrophs. They serve as a source of food for all non-producers or consumers directly or indirectly. Hence, producers occupy the first trophic level in a food chain.\nQuestion 14.\nWe often use the word environment. What does it mean? (Foreign 2016)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 14.\nWe often use the word environment. What does it mean? (Foreign 2016)\nAnswer:\nEnvironment can be defined as the physical or biological world where an organism lives. Literally speaking, an organisms immediate surrounding constitutes its environment which includes both biotic and abiotic components around him.\nQuestion 15.\nWhich of the following are always at the second trophic level of food chains? Carnivores, Autotrophs, Herbivores (AI 2015)\nAnswer:\nHerbivores always occupy the second trophic level of food chains.\nQuestion 16.\nThe following organisms form a food chain. Which of these will have the highest concentration of non-biodegradable chemicals? Name the phenomenon associated with it. Insects, Hawk, Grass, Snake, Frog (Foreign 2015)\nAnswer:\nAmong the following organisms of the food chain, hawk being top consumer is present at topmost trophic level, hence will have the highest concentration of non-biodegradable chemicals due to a phenomenon known as biomagnification.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 17.\nList two examples of natural ecosystem. (Foreign 2015)\nAnswer:\nThe two examples of natural ecosystem are :\nForest ecosystem\nRiver ecosystem\nQuestion 18.\nWhat is meant by the term \u2018Biomass? (Board Term I, 2014)\nAnswer:\nBiomass is the total amount of living or organic matter in an ecosystem at any time.\nQuestion 19.\nBacteria and fungi are called decomposers. Why? (Delhi 2012, Foreign 2011)\nAnswer:\nBacteria and fungi are called decomposers because these microorganisms break down the complex organic matter present in dead plants and animals into simpler substances.\nQuestion 20.\nIn a food chain, if 10,000 joules of energy is available to the producer, how much energy will be available to the secondary consumer to transfer it to the tertiary consumer? (AI 2012)\nAnswer:\nAccording to ten percent law, 10% of the energy of producer will be available to primary consumer, and 10% of this energy will be available to secondary consumer and so on.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nAnswer:\nAccording to ten percent law, 10% of the energy of producer will be available to primary consumer, and 10% of this energy will be available to secondary consumer and so on.\n\nHence, 100 J of energy will be available to the secondary consumer to transfer it to tertiary consumer.\nQuestion 21.\nConsider the following food chain which occurs in a forest:\nGrass \u2192 Deer \u2192 Lion\nIf 10000 J of solar energy is available to the grass, how much energy would be available to the deer to transfer it to the lion? (Foreign 2012)\nAnswer:\nThe energy available to the deer is 1000 J to transfer it to the lion. This can be depicted as follows:\n\nQuestion 22.\nWhat is meant by biological magnification? (AI 2011)\nAnswer:\nBiological magnification or biomagnification refers to the process of accumulation of non- biodegradable chemicals (pesticides, etc.) into the body of organisms through the food chain which go on increasing in its concentration at each trophic level.\nQuestion 23.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 23.\nGive an example to illustrate that indiscriminate use of pesticides may result in the degradation of the environment. (AI 2011)\nAnswer:\nPesticides are the chemicals used to kill plant and animal pests. They are non-biodegradable and toxicants. For example, excessive use of DDT resulted in reduced population of fish eating birds. DDT accumulated in such birds through the food chain. It interfered with the egg shell formation. The shell being thin broke due to weight of the bird during incubation. Hence, their population declined.\nQuestion 24.\nWhat are decomposers? List two important roles they play in the environment. (AI 2014)\nAnswer:\nDecomposers are microorganisms including bacteria and fungi which decompose or break-down the complex organic compound present in dead plants and animals into simpler substances. Role of decomposers in environment are-\nThey help in decomposing dead bodies of plants and animals and hence act as cleansing agents of environmcnl.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nThey help in decomposing dead bodies of plants and animals and hence act as cleansing agents of environmcnl.\n\u2018They help in recycling of materials in the ecosystem lo maintain its stability.\nQuestion 25.\nList two reasons to show that the existence of decomposers is essential in an ecosystem. (AI 2014)\nAnswer:\nThe existence of decomposers in an ecosystem is essential because:\n(i) If there were no decomposers, then the dead bodies of plants and animals would keep lying as such and the elements constituting plant and animal bodies would never be returned to their original pools like soil, air and water. In such case the organic waste go on accumulating and the cyclic process of life and death would be disrupted.\n(ii) Decomposers make the soil fertile by providing/replenishing nutrients to it, thus forming the integral part of ecosystem.\nQuestion 26.\nState with reason any two possible consequences of elimination of decomposers from the earth. (AI 2014)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 26.\nState with reason any two possible consequences of elimination of decomposers from the earth. (AI 2014)\nAnswer:\nConsequences of elimination of decomposers are:\n(i) There would be no recycling of nutrients and therefore, raw materials to produce food will not be available to producers. Hence, the food chains will get affected.\n(ii) The dead bodies of plants and animals will go on accumulating in the absence of decomposition thereby polluting the environment.\nQuestion 27.\nWhat does a trophic level represent in a food chain? State the position of autotrophs and herbivores in a food chain. (Delhi 2013C)\nAnswer:\nTrophic level represents each of several hierarchical levels of a food chain operating in an ecosystem, consisting of organism sharing the same function in the food chain and the same nutritional relationship to the primary sources of energy.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nThe position of producers (or autotrophs) in a food chain constitute the first trophic level. They fix up sun\u2019s energy and make it available for consumers. The herbivores or primary consumers (which feed upon plants) constitute the second trophic level in a food chain.\nQuestion 28.\n(a) From the following group of organisms create a food chain which is most advantageous for human beings in terms of energy.\nHawk, Rat, Cereal plant, Goat, Snake, Human being\n(b) State the possible disadvantage if the cereal plant is growing in soil rich in pesticides.\n(c) Construct a food web using the organisms mentioned above. (2020)\nAnswer:\n(a) A food chain which is most advantageous for human beings in terms of energy is:\nCereal plant \u2192 Human being", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\n(c) Construct a food web using the organisms mentioned above. (2020)\nAnswer:\n(a) A food chain which is most advantageous for human beings in terms of energy is:\nCereal plant \u2192 Human being\n(b) If the cereal plant is growing in soil rich in pesticides, these pesticides are absorbed by growing plants along with water and minerals, when animals eat these cereal plants, these poisonous chemical pesticides go into their bodies through food. This increase in concentration of harmful pesticides in the body of living organisms at each trophic level of a food chain is called biological magnification. Pesticides are lethal to non-target species also. The extensive use of pesticides in agriculture can change the community of microorganisms living in soil.\n\nQuestion 29.\n(a) Create a food chain of the following organisms.\nInsect, Hawk, Grass, Snake, Frog\n(b) Name the organism at the third trophic level of the created food chain.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 29.\n(a) Create a food chain of the following organisms.\nInsect, Hawk, Grass, Snake, Frog\n(b) Name the organism at the third trophic level of the created food chain.\n(c) Which organism of this food chain will have the highest concentration of non- biodegradable chemicals?\n(d) Name the phenomenon associated with it.\n(e) If 10,000 Joules of energy is available to frogs, how much energy will be available to snakes in this food chain? (2020)\nAnswer:\n(a) Grass \u2192 Insect \u2192 Frog \u2192 Snake \u2192 Hawk\n(b) Frog is present in the above created food chain.\n(c) Hawk is the top consumer of the food chain, so, it will have high concentration of non- biodegradable chemicals.\n(d) Biological magnification\n(e) As per 10% law of flow of energy in an ecosystem, only 10% of energy is received by the next trophic level. Hence, in the given food chain, if 10,000 Joules of energy is available to frog, then the energy available to snakes will be 1000 Joule.\n\nQuestion 30.", + "Hence, in the given food chain, if 10,000 Joules of energy is available to frog, then the energy available to snakes will be 1000 Joule.\n\nQuestion 30.\n(a) What is an ecosystem?", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 30.\n(a) What is an ecosystem?\n(b) List any two natural ecosystems.\n(c) We do not clean ponds or lakes but an aquarium needs to be cleaned regularly. Why? (2020)\nAnswer:\n(a) An ecosystem is defined as a structural and functional unit of the biosphere comprising of living organisms and their non-living environment.\n(b) Two examples of natural ecosystem are: pond ecosystem and grassland ecosystem.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\n(b) Two examples of natural ecosystem are: pond ecosystem and grassland ecosystem.\n(c) Ponds or lakes being natural ecosystems are self sufficient and do not need to be maintained regularly. On the other hand, aquarium being an artificial ecosystem needs to be cleaned and maintained regularly as it contains mainly fishes as living organisms, as compared to natural ecosystems where presence of other organisms maintains a balance. Also, because the producers and decomposers are absent in aquarium the fish waste or excretory products, i.e., ammonia may turn into toxic compounds and accumulate to dangerous levels causing the fishes to die. Hence, the artificial ecosystems like aquarium needs to be cleaned regularly.\nQuestion 31.\nWhat is meant by trophic level in a food chain? Construct a terrestrial food chain with trophic levels. The energy flow in a food chain is always unidirectional. Why? (2020)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 31.\nWhat is meant by trophic level in a food chain? Construct a terrestrial food chain with trophic levels. The energy flow in a food chain is always unidirectional. Why? (2020)\nAnswer:\nThe various steps representing organisms in a food chain at which the transfer of food and energy takes place are called trophic levels.\nFour trophic levels in a terrestrial food chain:\nGrass \u2192 Rabbit \u2192 Wild cat \u2192 Tiger\nThere is a unidirectional llow of energy from sun to producers and subsequenllv to series of different types of consumers, i.e.,\nSolar radiations \u2192 Producers \u2192 Herbivores \u2192 Carnivores\nIt cannot pass in reverse direction, lhcre is always a decrease in the flow of energy and content with rise in trophic level. Large quantity of energy is lost at each step in the form of heat and is also used up in various metabolic activities.\nQuestion 32.\nComplete the following flow chart based on ecosystem and its components.\n\nAnswer:\nAquatic\nAbiotic\nInorganic substances\nProducers", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 32.\nComplete the following flow chart based on ecosystem and its components.\n\nAnswer:\nAquatic\nAbiotic\nInorganic substances\nProducers\nStructural and functional unit of biosphere\nQuestion 33.\n(a) Construct a terrestrial food chain comprising four trophic levels.\n(b) What will happen if we kill all organisms in one trophic level?\n(c) Calculate the amount of energy available to the organisms at the fourth trophic level. If the energy available to the organisms at the second trophic level is 2000 J. (2020)\nAnswer:\n(a) A terrestrial food chain with four trophic levels is :\nGrass \u2192 Insect \u2192 Frog \u2192 Eagle\n(b) Removal of the organisms of any trophic level will always adversely affect the ecosystem, e.g., the removal of lions and tigers (top carnivores) will cause rapid increase in deer population, which leads to rapid consumption of vegetation resulting in scarcity of vegetation and population crash of deer.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\n(c) According to ten percent law, only 10% of the energy is received by the next trophic level.\n\nIf the energy available at second trophic level (T2) is 2000 J, so the 20 J of energy will be at fourth trophic level (T4).\nQuestion 34.\nDefine an ecosystem. Draw a block diagram to show the flow of energy in an ecosystem. (Delhi 2019)\nAnswer:\nAn ecosyslcm is defined as a structural and functional unil of ihe biosphere. It comprises of living organisms and their non-living environment that interact by means of food chains and biogeo-chemical cycles resulting in energy-flow, biotic diversity and material cycling to form stable self-supporting system.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nGreen plants capture about 1% of the solar energy incident on the earth to carry out the process of photosynthesis. A part of this trapped energy is used by plants in performing their metabolic activities and some energy is released as heat into the atmosphere. The remaining energy is chemical energy stored in the plants as photosynthetic products. When these green plants are eaten up by herbivores, the chemical energy stored in the plants is transferred to these animals. These animals (herbivores) utilise some of this energy for metabolic activities and some energy is released as heat while the remaining energy is stored in their body. This process of energy transfer is repeated till top carnivores. In an ecosystem, transfer of energy follows 10 percent law, i.e,, only 10 percent of the energy is transferred to each trophic level from the lower trophic level. Nearly 90 percent of energy is lost when it moves from one trophic level to the next.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nThe given block diagram shows unidirectional flow of energy at different trophic levels in a freshwater ecosystem:\n\nQuestion 35.\nWhat is a food chain? Why is the flow of energy in an ecosystem unidirectional? Explain briefly. (AI 2019)\nAnswer:\nThe sequential interlinking of organisms involving transfer of food energy from the producers, through a series of organisms with repeated eating and being eaten is called the food chain. A food chain involves a nutritive interaction between the living organisms of an ecosystem. There is a unidirectional flow of energy from sun to producers and subsequently to series of different types of consumers, i.e.,\nSolar radiations \u2192 Producers \u2192 Herbivores \u2192 Carnivores\nIt cannot pass in reverse direction. There is always a decrease in the flow of energy and content with rise in trophic level. Large quantity of energy is lost at each step in the form of heat and is also used up in various metabolic activities.\nQuestion 36.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 36.\n\u201cEnergy flow in food chains is always unidirectional.\u201d Justify this statement.\nExplain how the pesticides enter a food chain and subsequently get into our body. (Foreign 2015, AI 2014)\nAnswer:\nRefer to answer 35.\nSome harmful non-biodegradable chemicals (pesticides, e.g., D.D.T.) enter the bodies\nof organisms through the food chains and get concentrated at each trophic level. This phenomenon is called biomagnification or biological magnification. For example, in a food chain operating in a pond, river or lake, the water contains a small amount \u2014 0.02 ppb (parts per billion) of harmful pesticides, i.e., D.D.T. When this water is consumed by phytoplanktons and zooplanktons, the concentration of these chemicals increases to 5 ppm. Fishes feeding on these, accumulate 240 ppm. Birds and humans feeding on these fishes were found to contain 1600 ppm of these chemicals. Thus, there is an increase in the concentration of the chemicals at each trophic level.\nQuestion 37.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 37.\n\u201cOur food grains such as wheat and rice, the vegetables and fruits and even meat are found to contain varying amounts of pesticide residues.\u201d State the reason to explain how and why it happens. (Delhi 2014)\nAnswer:\nPesticides are poisonous chemical substances which are sprayed over crop plants to protect them from pests and diseases. These chemical pesticides mix up with soil and water. From soil and water, these pesticides are absorbed by the growing plants along with water and other minerals. When herbivorous animals feed on these plants the poisonous pesticides enter their bodies through the food chain. Similarly, when the carnivorous animals eat these herbivores, the pesticides get transferred to their bodies. Therefore, the plant products such as food grains, vegetables and fruits as well as meat of animals contain varying amounts of pesticide residues in them depending upon the trophic level they occupy in a food chain.\nQuestion 38.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 38.\nWhat is meant by food chain? \u201cThe number of trophic levels in a food chain is limited.\u201d Give reason to justify this statement. (Foreign 2014)\nAnswer:\nThe sequence of living organisms in a community in which one organism consumes or feeds upon another organism to transfer food energy, is called a food chain. The various steps in a food chain at which the transfer of food (or energy) takes place are called trophic levels. In fact, in a food chain, each step representing an organism forms a trophic level.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nThe number of trophic levels in a food chain are limited because at each trophic level only 10% of energy is utilised for the maintenance of organism which occur at that trophic level and the remaining large portion is lost as heat. As a result organisms at each trophic level pass on lesser energy to the next trophic level, than they receive. The longer the food chain, the lesser is the energy available to the final member of food chain. Food chains generally consist of three or four trophic levels because beyond that the energy available to the next organism will be too small and insufficient to sustain the life of that organism.\nQuestion 39.\n(a) What is an ecosystem? List its two main components.\n(b) We do not clean ponds or lakes, but an aquarium needs to be cleaned regularly. Explain. (Delhi 2013)\nAnswer:\n(a) Refer to answer 34.\nThe two main components of ecosystem are :\nAbiotic components \u2013 Non-living components of ecosystem, e.g., soil, water, air, light, etc.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Refer to answer 34.\nThe two main components of ecosystem are :\nAbiotic components \u2013 Non-living components of ecosystem, e.g., soil, water, air, light, etc.\nBiotic components \u2013 Living components of ecosystem, e.g., plants, animals and microbes.\n(b) Refer to answer 30 (c).\nQuestion 40.\nChoose the incorrect statement from the following:\n(a) Ozone is a molecule formed by three atoms of oxygen.\n(b) Ozone shields the surface of the earth from ultraviolet radiations.\n(c) Ozone is deadly poisonous.\n(d) Ozone gets decomposed by UV radiations. (2020)\nAnswer:\n(d) Ozone gets decomposed by UV radiations. (2020)\nAnswer questions numbers 41 to 44 on the basis of your understanding of the following paragraph and the related studied concepts:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nAnswer questions numbers 41 to 44 on the basis of your understanding of the following paragraph and the related studied concepts:\nIndia today is facing the problem of overuse of resources, contamination of water and soil and lack of methods of processing the waste. The time has come for the world to say goodbye to \u201csingle use plastics\u201d. Steps must be undertaken to develop environment-friendly substitutes, effective plastics waste collection and methods of its disposal. Indore treated 15 lakhs metric tonnes of waste in just 3 years, through biomining and bioremediation techniques. Bioremediation involves introducing microbes into a landfill to naturally \u2018break\u2019 it down and biomining involves using trommel machines to sift through the waste to separate to \u2018soil\u2019 and the waste component. The city managed to chip away 15 lakh metric tonnes of waste at a cost of around t 10 crore. A similar experiment was successfully carried out in Ahmedabad also.\nQuestion 41.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 41.\nState two methods of effective plastic waste collection in your school.\nAnswer:\nTwo methods of effective plastic waste collection : (i) Use of separate dustbins for plastic collection (ii) Use of reusable for the canteen and school events (iii) Encourage less plastic in packed lunches.\nQuestion 42.\nName any two uses of \u2018single use plastic\u2019 in daily life.\nAnswer:\nSingle use plastic, often also referred to as disposable plastics are commonly used for plastic packaging and include items intended to be use only once before they are thrown away or recycled. These include grocery bags, food packaging, bottles, straws, containers, cups and cutlery.\nQuestion 43.\nIf we discontinue the use of plastic, how can an environment-friendly substitute be provided?\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 43.\nIf we discontinue the use of plastic, how can an environment-friendly substitute be provided?\nAnswer:\nBest alternatives\u2019can be use of stainless steel, glass and platinum. Silicone storage containers. Cloth bags can be used in place of plastic bags. Use of wooden cleaning brushes, kitchen utensils and cutting board, pottery and other ceramics products, etc.\nQuestion 44.\nDo you think microbes will work similarly in landfill sites as they work in the laboratory? Justify your answer. (2020)\nAnswer:\nMicrobes may not work exactly the same way in landfill sites as they work in laboratories because it is difficult to replicate the exact ambient conditions required for the microbes to thrive in these two conditions. But due to same downstream processing and other mechanism, they will work similarly in landfill sites.\nQuestion 45.\nThe depletion of ozone layer is a cause of concern. Why? (AI 2016)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 45.\nThe depletion of ozone layer is a cause of concern. Why? (AI 2016)\nAnswer:\nOzone layer is the ozone rich area in the stratospheric layer of atmosphere which acts as a protective shield by preventing harmful UV radiations from entering the Earth surface. Hence, the depletion of ozone layer is a cause of concern.\nQuestion 46.\nWrite one negative effect on the environment, of affluent life style of few persons of a society. (Al 2016, 2014)\nAnswer:\nAffluent lifestyle of few persons leads to exploitation and over consumption of resources leading to their scarcity and generation of greater amount of waste materials which causes imbalance in environment.\nQuestion 47.\nWhy is excessive use of CFCs a cause of concern? (Foreign 2016)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 47.\nWhy is excessive use of CFCs a cause of concern? (Foreign 2016)\nAnswer:\nCFCs or Chlorofluorocarbons are potent compounds that release active chlorine in the atmosphere which reacts with ozone molecules present there to convert them to oxygen. This results in thinning of ozone layer. Hence, excessive use of CFCs is a cause of concern.\nQuestion 48.\nWhat is the function of ozone in the upper atmosphere? (Delhi 2015, Foreign 2012)\nAnswer:\nOzone (O3) gas forms a protective shield in the upper atmosphere that absorbs most of the harmful ultraviolet radiations coming from Sun that can harm human beings, animals and plants. It protects us from various health hazards.\nQuestion 49.\nWhy should biodegradable and non- biodegradable wastes be discarded in two separate dustbins? (Delhi 2015, 2013)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 49.\nWhy should biodegradable and non- biodegradable wastes be discarded in two separate dustbins? (Delhi 2015, 2013)\nAnswer:\nBiodegradable wastes are decomposed naturally by the action of microbes which degrade them to their simple constituents enabling their nutrients to recycle among the biotic and abiotic components of ecosystem. However, non-biodegradable wastes cannot be disposed off naturally since they cannot be decomposed by microbes. Such wastes are either recycled, incinerated or put in landfills, etc. As the disposal methods of the two types of waste is different, it is advisable to discard the two types of waste in two separate dustbins.\nQuestion 50.\nWrite the full name of the group of compounds mainly responsible for the depletion of ozone layer. (Foreign 2015, Delhi 2013C)\nAnswer:\nChlorofluorocarbons (CFCs) are the group of compounds mainly responsible for ozone depletion.\nQuestion 51.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nAnswer:\nChlorofluorocarbons (CFCs) are the group of compounds mainly responsible for ozone depletion.\nQuestion 51.\nWhich class of chemical is linked to the decrease in the amount of ozone in the upper atmosphere of the Earth? (Delhi 2012)\nAnswer:\nThe decrease in the amount of ozone in the atmosphere has been linked to synthetic chemicals like chlorofluorocarbons (CFCs) which are used as coolant and in fire extinguishers, etc.\nQuestion 52.\nWhat happens when higher energy ultraviolet radiations act on the oxygen at the higher level of the atmosphere? (Al 2012)\nAnswer:\nWhen high energy ultraviolet radiations react with oxygen present in stratosphere (the higher level of atmosphere) it splits into its constituent atoms. Since these atoms produced are very reactive they react with molecular oxygen (O2) to form ozone (O3).\nQuestion 53.\nWhy did United Nations act to control the production of chlorofluorocarbons (CFCs) used in refrigerators? (Delhi 2011)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 53.\nWhy did United Nations act to control the production of chlorofluorocarbons (CFCs) used in refrigerators? (Delhi 2011)\nAnswer:\nUnited Nations act to control the production of CFCs as they are the main chemicals responsible for ozone depletion.\nQuestion 54.\nWhy is decrease of ozone in our ozone layer a cause for our worry? (Foreign 2011)\nAnswer:\nRefer to answer 45.\nQuestion 55.\n\u201cIndustrialisation has adversely deteriorated the environment.\u201d Give four reasons in support of this statement. (Foreign 2016)\nAnswer:\nIndustrialisation has deteriorated our environment in the following ways :\nRapid industrialisation has increased the demand of more land area for setting up of new factories. This demand is being fulfilled by clearing up of forest area. Deforestation is one of the major causes of ecological imbalance, biodiversity loss and ecosystem unstability.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nIndustries release various harmful gases in the environment which pollute the air. These gases when inhaled by people around, cause various respiratory diseases in them.\nA lot ol effluent and liquid waste is discharged from various industries which is mostly dumped into nearby water bodies. This causes water pollution. Polluted water causes death of various aquatic organisms and consumption of this polluted water causes various diseases in humans.\nThe solid waste released from factories is dumped on open land and not treated properly to ensure their proper decomposition. Ibis leads to land pollution that degrades quality of soil and also causes various kinds of diseases in humans and animals.\nQuestion 56.\nWhy is Government of India imposing a ban on the use of polythene bags? Suggest two alternatives to these bags and explain how this ban is likely to improve the environment. (Delhi 2014)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nWhy is Government of India imposing a ban on the use of polythene bags? Suggest two alternatives to these bags and explain how this ban is likely to improve the environment. (Delhi 2014)\nAnswer:\nGovernment of India is imposing a ban on the use of polythene hags because these are non-bindegradahle substances which are not acted upon by the microbes. So, they cannot be decomposed and therefore persist in the environment for a long time thereby causing harm to the ecosystem. I\u2019oh bags choke drains which results in water logging, that allows breeding of mosquitoes and hence leads to various diseases like malaria, dengue, etc. Jute bags and cloth bags are the alternatives to the polythene bags.\nQuestion 57.\nIn some states of our country there is a ban on the use of polythene bags for shopping. Why? List three advantages of using jute or cloth bags over polythene bags. (Delhi 2014)\nAnswer:\nRefer to answer 56.\nThe three advantages of using jute or cloth bags over polythene bags are :", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 56.\nThe three advantages of using jute or cloth bags over polythene bags are :\nThey are made of biodegradable material.\nThey can be reused.\nThey do not pollute our environment.\nQuestion 58.\n\u201cAffluent life style has a negative effect on the environment.\u201d Justify this statement with the help of an example. (Delhi 2014)\nAnswer:\nWith the advancement in technology over time, there has been improvement in lifestyle of people. Such changes in peoples lives have also changed their attitudes. When people have more resources at their end they tend to overuse and misuse it thereby generating huge amounts of wa>le material. for example, the atllueiit life style has forced people to start using more of disposable articles, e.g., plastic cups, bags, etc., which keep on accumulating in the environment and lie undecomposed, therein negatively a Heeling the environment.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nSimilarly, excessive use of refrigerators and air conditioners, plastic foams, etc., also releases high quantities of CFCs which are responsible for ozone depletion.\nQuestion 59.\nGive one example each from your daily life where the domestic waste can be effectively reused and recycled. (AI 2014)\nAnswer:\nWe can reuse plastic and glass jars of jams and pickles, etc, for the purpose of storage of things like salt, sugar, tea, etc., whereas we can recycle newspapers, plastic of some types, broken glass and metal wares for making fresh paper, plastic, glass and metal objects.\nQuestion 60.\n\u201cTo discard the household waste we should have two separate dust-bins, one for the biodegradable waste and the other for the non-biodegradable waste.\u201d Justify this statement suggesting the proper way of disposal of these wastes. (Foreign 2014)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nAnswer:\n\u201cTo discard the household waste we should have two separate dust-bins, one for the biodegradable waste and the other for the non-biodegradable waste.\u201d Segregation of biodegradable and non- biodegradable waste is utmost important for their proper disposal, as different methods are adopted for their disposal, biodegradable wastes can be composted whereas non-biodegradable wastes can be recycled, incinerated or landfilled.\nQuestion 61.\nWe often observe domestic waste decomposing in the bylanes of residential colonies. Suggest ways to make the residents realise that the improper disposal of their waste is harmful to the environment. (Foreign 2014, Delhi 2013)\nAnswer:\nSome of the ways to make people realise that the improper disposal of waste is harmful to the environment includes making people aware of negative impacts of waste disposal. They can be made aware by:\n(i) Conducting seminars about the negative effects of the wastes on environment.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\n(i) Conducting seminars about the negative effects of the wastes on environment.\n(ii) Usage of pamphlets and posters for providing information.\n(iii) Forming an eco-club in the society for spreading awareness about the ill effects of waste on the surroundings such as :\nImproper disposal of waste will release harmful gases in the environment that make it unclean and unhygienic for the living organisms.\nThe waste will flow to water bodies along with rain water and become a threat to aquatic life and pollute the water bodies.\nIt provides space for breeding of the mosquitoes and which results in spread of malaria, filariasis, dengue, etc.\nHazardous chemicals from wastes get into the soil and can harm the plants when they take up the contamination through their roots. This will affect the health of other animals and humans and will have negative impact on environment.\nQuestion 62.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nQuestion 62.\nSuppose you find a heap of domestic waste, in a nearby park, which is decomposing. What would you do to make the people of the surrounding area realise that such type of disposal of domestic waste is harmful to the environment? (Foreign 2014)\nAnswer:\nRefer to answer 61.\nQuestion 63.\n(a) Write two harmful effects of using plastic bags on the environment. Suggest alternatives to the usage of plastic bags.\n(b) List any two practices that can be followed to dispose off the waste produced in our homes. (2020)\nAnswer:\n(a) Two harmful effects of using plastic bags on the environment:\n(i) Plastic bags are non-biodegradable substances which are not acted upon by microbes. So, they cannot be decomposed and therefore persist in the environment for a long time causing harm to the soil fertility and quality.\n(ii) Plastic bags choke drains which result in waterlogging, that allows breeding of mosquitoes and hence leads to various diseases.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\n(ii) Plastic bags choke drains which result in waterlogging, that allows breeding of mosquitoes and hence leads to various diseases.\nJute bags and cloth bags are the alternatives to the polyethene bags.\n(b) Practices that can be followed to dispose off the waste produced in our homes:\nSeparation of biodegradable and non- biodegradable wastes.\nThe biodegradable waste can be converted to manure.\nNon-biodegradable waste should be disposed off at suitable places from where municipal authorities can pick them up and dispose properly and scientifically.\nUse discarded bottles and jars to store food, items.\nQuestion 64.\n(a) Complete the following table :\n(b) How is ozone formed at the higher level of atmosphere? (2020)\nAnswer:\n(a) (i) O2\n(ii) O3\n(iii) Respiration\n(iv) Absorbs harmful ultraviolet radiations coming from the sun.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\n(b) How is ozone formed at the higher level of atmosphere? (2020)\nAnswer:\n(a) (i) O2\n(ii) O3\n(iii) Respiration\n(iv) Absorbs harmful ultraviolet radiations coming from the sun.\n(b) When high energy ultraviolet radiations react with oxygen present in stratosphere (the higher level of atmosphere) it splits into its constituent atoms. Since these atoms produced are very reactive they read with molecular oxygen (O2) to form ozone (O3).\n\nQuestion 65.\nYou have been selected to talk on \u201cOzone layer and its protection\u201d in the school assembly on \u2018Environment Day\u2019.\n(a) Why should ozone layer be protected to save the environment?\n(b) List any two ways that you would stress in your talk to bring in awareness amongst your fellow friends that would also help in protection of ozone layer as well as the environment. (Delhi 2017)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nAnswer:\n(a) The ozone layer is very important for the existence of life on earth because it forms a protective shield around earth by absorbing most of the harmful ultraviolet (UV) radiations coming from the Sun and prevents them from reaching the Earth.\nThe UV radiations have extremely harmful effects on human beings, animals and plants as well, i.e., cause mutations, skin cancer, cataract, damage immune system, etc. So, ozone layer must be protected to save the environment.\n(b) The two ways which can help in protection of ozone layer and environment are :\n(i) The use of chemicals like chlorofluorocarbons (CFCs) which are widely used in refrigerators and air conditioners (as a coolant), in fire extinguishers and in aerosol sprayers destroy the ozone layer gradually. We can protect our ozone layer by avoiding the use of such objects which are releasing UFCs.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\n(ii) Nitrous oxide is the largest ozone depleting substance as well as greenhouse gas released by human activities, such as from motor vehicles, fertilisers. People should be encouraged to use more public transport, car pooling, using hybrid or electric cars and use of fertiliser formulations to reduce emission of nitrous oxide.\nQuestion 66.\nAfter the examinations Rakesh with his friends went on a picnic to a nearby park. All friends carried cooked food packed in plastic bags or plastic cans. After eating the food some friends collected the leftover food and plastic bags etc., and planned to dispose them off by burning. Rakesh immediately checked them and suggested to segregate the leftover food and peels of fruits from the plastic materials and respectively dispose them off separately in the green and red dustbins placed in the corner of the park.\n(a) In your opinion, is burning plastic an eco-friendly method of waste disposal? Why? State the advantage of method suggested by Rakesh.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\n(a) In your opinion, is burning plastic an eco-friendly method of waste disposal? Why? State the advantage of method suggested by Rakesh.\n(b) How can we contribute in maintaining the parks and roads neat and clean? (Delhi 2015)\nAnswer:\n(a) No, burning plastic is not an eco-friendly method of waste disposal, burning plastics can produce toxic fumes and cause air pollution. Plastic, being non-biodegradable cannot be dumped (land filled), so the best way to dispose plastic items is to recycle them. Recycling is a less polluting and more sustainable option.\nRakesh segregated left over food items, fruit peels, i.e., biodegradable waste and plastic material, i.e., non-biodegradable waste. The biodegradable waste can be converted to manure whereas non-biodegradable waste can be recycled. This contributes in decreasing the level of pollution and easy disposal and treatment of waste.\n(b) We can contribute to keep our roads and parks clean by adopting following habits:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\n(b) We can contribute to keep our roads and parks clean by adopting following habits:\nWe should recycle non-biodegradable waste products instead of dumping it in garbage.\nWe should use carry bags made of natural fibre as jute, cloth, instead of plastic bags.\nWe should make compost of biomass collected from park such as food waste, leaf litters, etc,.\nWe should stop littering, throwing garbage and spitting on road sides, parks, etc.\nQuestion 67.\nDifferentiate between biodegradable and non-biodegradable substances with the help of one example each. List two changes in habit that people must adopt to dispose non-biodegradable waste, for saving the environment. (A I2015, Delhi 2013C)\nAnswer:\nDifferences between biodegradable and non- biodegradable wastes are as follows:\nPeople should adopt the following changes in their habit to dispose off non-biodegradable waste, so as to save the environment.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nPeople should adopt the following changes in their habit to dispose off non-biodegradable waste, so as to save the environment.\nNon-biodegradable household waste should be disposed in separate dustbins and should not be mixed with biodegradable waste.\nRecyclable \u201e non-biodegradable wastes like glass, paper, metal, various types of plastics etc., can be sent to local recycling units.\nPeople should try to reuse items as much as possible instead of disposing them as this reduces need of new materials and keeps ecological impact down.\nUse of cloth bags/gunny bags/paper bags instead of polythene /plastic bags.\nUse of compost, vermicompost instead of fertilisers.\nQuestion 68.\nWhat is ozone ? How and where is it formed in the atmosphere ? Explain how does it affect an ecosystem. (Foreign 2015)\nAnswer:", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nUse of compost, vermicompost instead of fertilisers.\nQuestion 68.\nWhat is ozone ? How and where is it formed in the atmosphere ? Explain how does it affect an ecosystem. (Foreign 2015)\nAnswer:\nOzone is a form of oxygen. It is a made up of three atoms of oxygen. It is highly poisonous. However, good amount of ozone is present in upper part of the atmosphere called stratosphere. In the stratosphere ozone is being photo- dissociated and generated simultaneously by absorption of harmful ultraviolet (UV) radiations coming from sun.\n\nThe two reactions are in equilibrium thereby maintaining a steady concentration of ozone in the stratosphere. Ozone layer is commonly called ozone blanket. It acts as a protective shield to protect all types of life from the harmful effect of UV radiation. Therefore, any thinning or depletion of ozone layer allows entry of high energy UV radiations into the earth\u2019s surface, thereby causing harmful effects on plants, animals and human beings.", + "CHAPTER 13-OUR ENVIRONMENT IMPORTANT QUESTIONS.txt\nThe harmful effects of ozone depletion on man, animals and plants includes :\nIncidences of skin cancer and herps.\nDamage to eye sight, photoburning as well as increased incidences of cataract in eyes.\nDamage to immune system and hence lowering the body\u2019s resistance to disease.\nIncreased embryonic mortality.\n10-25% decline of photosynthesis in plants.\nGlobal warming.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nCBSE Class 10 Science Notes Chapter 15 Our Environment\nBiodegradable and Non-biodegradable Wastes, Ecosystem, Components of Ecosystem. The environment includes our physical surroundings like air (or atmosphere), water bodies, soil (land and all the organisms such as plants, animals, human beings and micro-organisms like bacteria and fungi (called decomposers). The waste materials produced by the various activities of man \u00a3nd animals are poisonous to some extent and can be divided into two main groups\n1. Biodegradable Wastes: Substances that are broken down by the biological processes are said to be biodegradable. These substances are decomposed through the actions of fungi, bacteria, and other living organisms. Temperature and sunlight also play an important role in the decomposition of biodegradable substances.\nFor Examples: Food waste, trees leaves, urine and fecal matter, sewage agricultural residue, paper, wood, cloth, cow-dung etc.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nFor Examples: Food waste, trees leaves, urine and fecal matter, sewage agricultural residue, paper, wood, cloth, cow-dung etc.\n2. Non-Biodegradable Wastes: Substances that are not broken down by biological\nprocesses. These substances may be in solid, liquid or gaseous form. These substances are inert and simply persist in the environment for a long time or may harm the various members of the ecosystem.\n\nFor Examples: These includes DDT (Di-chloro-di phenyl trichloro ethane-in-pheneyle the cheoro ethane), insecticides, pesticides, mercury, lead, arsenic aluminum, plastics, polythene bags, glass, radioactive wastes. These non-biodegradable wastes are major pollutants of the environment.\nHarmful effects of biodegradable and Non-Biodegradable Substance\nThe waste destroys the natural beauty and our surroundings become dirty.\nDecomposition of these wastes results in the production of foul smell, which spreads to the surrounding areas.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nThe waste destroys the natural beauty and our surroundings become dirty.\nDecomposition of these wastes results in the production of foul smell, which spreads to the surrounding areas.\nThese wastes may also block the drains creating pools of waste, which becomes the breeding sites of mosquitoes. The latter is carriers of diseases like malaria and dengue.\nDifference between Biodegradable and Non-Biodegradable wastes\nEcosystem: An ecosystem is a self-contained unit of living things (plants, animals and decomposers), and their non-living environment (soil, air and water). For example; a forest, a pond, a lake, a green land etc.\nIn an ecosystem, energy and matter are continuously exchanged between living and non\u00acliving components.\nAn ecosystem can be both natural or man-made. Some examples of natural ecosystems are grass land, forest, sea, river, desert, mountain, pond, lake etc.\nThe desert, grass land and mountains represent the terrestrial ecosystem (land-based ecosystem).", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nThe desert, grass land and mountains represent the terrestrial ecosystem (land-based ecosystem).\nThe ponds, rivers, lakes and sea represent the aquatic ecosystem (water-based ecosystem). Man-made artificial ecosystems are garden, crop fields, park, aquarium, etc.\n\nComponents of Ecosystem: There are two components of an ecosystem : (i) biotic component and (ii) abiotic component.\n1. Biotic component: It includes three types of organisms :\n(a) Producers: All green plants, blue green algae can produce their food (Sugar and starch) from inorganic substance using light energy (Photosynthesis). Therefore, all green plants are called producers. They are also called autotrophs.\nPlanktons are very minute or microscopic organisms freely floating on the surface of water in a pond, lake, river or ocean. Planktons are of two types : Phytoplanktons and Zooplanktons.\nThe microscopic aquatic plants freely floating on the surface of water are called phytoplanktons.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nThe microscopic aquatic plants freely floating on the surface of water are called phytoplanktons.\nThe microscopic aquatic animals freely floating on water are called zooplanktons. The freely floating protozoa are an example of zooplankton.\n(b) Consumers: They are organisms which consume other organisms or their products as their food. All animals belong to this category. The consumers depend upon producers for their food directly or indirectly. They get their food by eating other organisms or their products. For example, man, goat, deer, fish, lion, cow, buffalo, etc., are common consumers.\nThe consumers can be classified into the following types :\nHerbivores.\nCarnivores.\nParasite.\nOmnivores.\n\n(i) Herbivores: These are organisms (animals) which get their food by eating the producers (or plant) directly. Herbivores are also called first order consumers. Some common examples of herbivores are : deer, rabbit, rat, squirrel, goat, cattle, etc.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\n(ii) Carnivores: These are organisms (animals) which consume other animals. Therefore, carnivores feed on the flesh of herbivores. These are also called primary carnivores or second order consumers. Some common examples are snake, wild cat, jackal, frog, some birds, fishes, etc.\nThere are animals which prey upon primary carnivores. They are called second order consumers or third order consumers. For example, owl, peacock, tiger, lion, etc., are some second order carnivores and may be eaten by third order carnivores. The carnivores which are not preyed upon further are called top carnivores. For example, lion is a top carnivore.\n(iii) Omnivores: The organisms which feed on both plants and animals are called omnivores. Human beings are common example of omnivores because they eat both plants (For example; pulses, grams, oilseeds, fruit, etc.) and animal products (milk, meat, egg, etc.).", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\n(c) Decomposers: Fungi and bacteria which break down (decompose) the dead plants, animals complex compounds into the simpler one. The decomposers help in the replenishment of natural resources. These are also known as microorganism or saprotrophs. These are also called reducers.\nImportance of Decomposers\nDecomposers help in disposing of the wastes and dead bodies of plants and animals. Therefore, they clean the environment and create space for a living of newer generations of organisms.\nThe decomposers release minerals and other raw materials trapped in organic matter. These are picked up by plants. This also helps to maintain the fertility of soil.\nThe decomposers produce some acids which are useful in solubilization of some minerals.\nDecomposers help in recycling the materials in the biosphere so that, the process of life may go on and on like an unending chain.\n2. Abiotic Components: These are non-living components of an ecosystem. These include the physical environment.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\n2. Abiotic Components: These are non-living components of an ecosystem. These include the physical environment.\nEdaphic factors like soil texture, topography, water, and air.\nInorganic substances like carbon dioxide, nitrogen, oxygen, water, phosphorus, sodium, potassium, and calcium. These are involved in the cyclic of materials in the ecosystem.\nOrganic compounds like proteins, carbohydrates, and lipids. These largely form the living body and link the abiotic and biotic components.\nClimatic factors: These are sunlight temperature, pressure humidity, moisture, rainfall, etc. these factors affect the distribution of the organisms.\nFunctions of an Ecosystem\nEcosystem indicates available solar energy and the efficiency of an ecosystem to trap the same.\nIt gives information about the available essential minerals and their recycling periods.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nEcosystem indicates available solar energy and the efficiency of an ecosystem to trap the same.\nIt gives information about the available essential minerals and their recycling periods.\nIt provides knowledge about the web of interactions and inter-relationship among the various population as well as between the population and the abiotic environment.\nIt helps human beings to know about conservation of resources, protection from pollution and inputs required for maximizing productivity.\nIn the ecosystem, two processes of energy flow and biogeochemical cycles (nutrients movement) proceed side by side. The energy flow is unidirectional while the movement of nutrients is cyclic.\nFood chain, Food web, Trophic levels. Flow of energy ten percent law, Depletion of the ozone layer, Biological magnification. Mode of waste disposal.\nFood Chain: The sequence of living organisms in a community in which one organism consumes another organism to transfer food energy, is called a food chain.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nFood Chain: The sequence of living organisms in a community in which one organism consumes another organism to transfer food energy, is called a food chain.\nA food chain is unidirection where transfer of energy takes place in only one direction.\nOR\nFood chain is sequential process which represents \u201cwho eats whom\u201d.\nOR\nFood chain refers to an arrangement of different biotic groups in a sequence of energy transfer. These biotic groups are producer herbivores, carnivores.\nFor example, T1(Grass) \u2192 T2(Deer) \u2192 T3(Lion)\nExamples of Food Chains: Simple food chain operating in a grass land or forest\nGrass(Producer) \u2192 Deer(Herbivore) \u2192 Lion(Carnivore)\nIn this food chain, grass represent the producers (first tropic level). Grass synthesize their own food by the process of photosynthesis. Grass is eaten up by deer, which represents the herbivores or the primary consumers. Deer in turn is consumed by lion, the carnivores or the secondary consumers.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nA food chain in grassland which has four steps is :\nGrass(Producers) \u2192 Insect(Herbivores) \u2192 Frog(Carnivores) \u2192 Eagle(Secondary Carnivore)\nSignificance of Food Chains\nThe study of food chains helps in understanding food relationships and interactions among the various organisms in an ecosystem. The food chains, transfer energy and materials between various living components of an ecosystem.\nThe food chains transfer energy and materials between various living components in an ecosystem or biosphere.\nThe food chains give dynamicity to an ecosystem or biosphere.\nThe movement of toxic substances like pesticides, weedicides, etc., through food chains, can prove very harmful.\nFood Web: The inter-connected food chains operating in an ecosystem which establish a network of relationship between various species, are called a food web.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nFood Web: The inter-connected food chains operating in an ecosystem which establish a network of relationship between various species, are called a food web.\nIn a food web, one organism may occupy a position in more than one food chain. An organism can obtain its food from different sources and in turn, may be eaten up by different types of organisms.\n\nTrophic Levels: The various steps in the food chain at which the transfer of food (or energy) takes place is called trophic levels.\nThere is a gradual decrease in the amount of energy transfer from one trophic level to the next trophic level in a food chain.\n\nSo only 10% of energy is transferred to next trophic level while 90% of energy is used by present trophic level in its life processes.\nThe various trophic levels are given below :\nThe plant or the producers constitute the first trophic level.\nThe herbivores or primary consumers form the second trophic level.\nCarnivores or secondary consumers make up the third trophic level.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nThe plant or the producers constitute the first trophic level.\nThe herbivores or primary consumers form the second trophic level.\nCarnivores or secondary consumers make up the third trophic level.\nLarge carnivores or the tertiary consumers which feed upon the small carnivores constitute the fourth trophic level.\nFlow Open Energy", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nLarge carnivores or the tertiary consumers which feed upon the small carnivores constitute the fourth trophic level.\nFlow Open Energy\nEnergy is used and conveyed from one trophic level to another in a food chain. This is called flow of energy. Green plants capture about 1% of the solar energy incident on the Earth through the biochemical process of photosynthesis. A part of this trapped energy is used by plants in performing their metabolic activities and some energy is released as heat into the atmosphere. The remaining energy is chemical energy stored in the plants as \u2018carbohydrates\u2019. When plants are eaten up by herbivores, the chemical energy stored in the plants is transferred to these animals. These animals (herbivores) utilize some of this energy for metabolic activities, some energy is \u201creleased as heat and the remaining energy is stored. The process of energy transferred is similarly repeated with carnivores and so on.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nTen percent law: Ten percent law states that only 10 percent of the energy entering a particular trophic level of organisms is available for transfer to the next higher trophic level.\n\nFor example, Suppose 1000 J of solar energy is received by green plants, then only 1% of solar energy available on earth is utilized by plants. So only 10 J (1% of 1000 J) is trapped by plants and the rest 990 J of energy is lost to the environment. So, plants utilizes only 10 J of energy. Next, only 10% of the 10 J energy of plant, that is, 1 J, is available to the herbivore animal while 9 J is lost to the environment. Again, just 10% of the 1 J of energy of herbivore animals is utilized by carnivore animals. Thus, carnivore animals have only 0.1 J of energy while 0.9 J is lost to the environment.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nEnvironmental Problems: Changes in the environment affect us and our activities change the environment around us. This led to the slow degradation of the environment that arose many environmental problems. For Example; depletion of the Ozone Layer and waste disposal.\nDepletion of Ozone Layer: Ozone (O3) layer is largely found in the stratosphere which is a part of our atmosphere from 12 km -50 km above sea level. This region is called ozonosphere. Ozone is deadly poisonous at the ground level.\nOzone is formed as a result of the following photochemical reaction.\n\nOzone layer is a protective blanket around earth which absorbs most of the harmful U.V. (Ultraviolet) radiation of the Sun, thus, protecting the living beings of the Earth from health hazards like skin cancer, cataract in eyes, weaken immune system, destruction of plants etc. The decline of Ozone layer thickness in Antarctica was first discovered in 1985 and was termed as OZONE HOLE.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nSteps taken to limit damage of ozone layer: Excessive use of CFCs (Chloro Flouro Carbon) a synthetic, inert chemical. For example; Freon which are used as refrigerants and also in fire extinguishers caused Ozone depletion in the upper atmosphere. A single chlorine atom can destroys 1,00,000 Ozone molecules. U.N.E.P. (United Nation Environment Programme) did an excellent job in forging an agreement to freeze CFC production at 1986 levels (KYOTO Protocol) by all countries.\nBiological Magnification: The increase in concentration of harmful chemical substances like pesticides in the body of living organisms at each trophic level of a food chain is called biological magnification.\nExample:\n\nMaximum concentration of such chemicals gets accumulated in human bodies.\nGarbage Disposal: Industrialization and rise in demand of consumer goods have created a major problem in the form of wastes/garbage accumulation and its disposal especially in urban areas.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nGarbage Disposal: Industrialization and rise in demand of consumer goods have created a major problem in the form of wastes/garbage accumulation and its disposal especially in urban areas.\nThe disposal of waste should be done in a scientific way. There are different methods of waste disposal. The method to be used depends on the nature of the waste. Some of the important modes of waste disposal are :\nIncineration: Burning of waste on high temperature to form ash is called incineration. This process is carried out in an incinerator. Incineration is used to destroy household, chemical and biological wastes.\nOpen dumping: A conventional method in which solid waste are dumped in selected areas of a town. It actually cause pollution\nLand fillings: Wastes are dumped in low living areas and are compacted by rolling with bulldozers", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nLand fillings: Wastes are dumped in low living areas and are compacted by rolling with bulldozers\nComposting: Organic wastes are filled into a compost pit (2m \u00d7 1m \u00d7 1m). It is then covered with a thin layer of soil. After about three months the same garbage filled inside the pit changes into organic manure.\nRecycling: The solid wastes is broken down into its constituent simpler materials. These materials are then used to make new items. Even non-bio degradable solid wastes like plastic, metal can be recycled.\nReuse: A very simple conventional technique of using an item again and again. For example; paper can be reused for making envelops, etc\u2026\nEnvironment: The combination of all the physical and biological conditions affecting the responses of living organisms is called environment.\nBiodegradable wastes: The wastes which are broken down by the activity of micro\u00acorganisms and enter into the biogeochemical cycle are known as biodegradable wastes.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nBiodegradable wastes: The wastes which are broken down by the activity of micro\u00acorganisms and enter into the biogeochemical cycle are known as biodegradable wastes.\nNon-biodegradable wastes: The wastes which cannot be broken down by the enzymes produced by microorganisms into simpler and harmless products in nature are called non- biodegradable wastes.\nGarbage: Domestic wastes including the kitchen waste are termed as garbage.\nIncineration: Destruction of waste materials by burning at high temperature is called incineration.\nBiotic Community: A group of various populations of organisms living in a region is called the biotic community.\nEcosystems: The self-contained and distinct functional unit capable of independent existence made by the interaction of living and non-living components is called an ecosystem.\nEcosystem component consists of two components- Abiotic and biotic\nAbiotic: Components consist of inorganic and organic substances and climatic factors.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nEcosystem component consists of two components- Abiotic and biotic\nAbiotic: Components consist of inorganic and organic substances and climatic factors.\nBiotic: Components consist of a living organism.\nAutotrophs: Those organisms which can produce their own food are called autotrophs or producers. All green plants are producers.\nConsumers: Those organisms which are unable to synthesise their food themselves and consume the food produced by producers or eat other organisms as food, are known as consumers.\nDecomposers: Bacteria and fungi which break down the complex organic compounds present in the dead plants and animals and their products into simpler substances are known as decomposers.\nFood Chain: A series of organisms in a community in which one organism consumes another organism to transfer food energy is called a food chain.\nCharacteristics:\nA food chain helps in understanding the food relationship and interactions among various organisms in an ecosystem.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nCharacteristics:\nA food chain helps in understanding the food relationship and interactions among various organisms in an ecosystem.\nThere is a progressive decline in the amount of energy available as we move from one trophic level to another in a food chain.\nTrophic levels: Each step of the food chain is known as a trophic level. 10% of food taken by one trophic level is available for the next trophic level.\nFood Web: The web formed by the interconnection of food-chains of the various trophic levels is called a food web.\nBiomagnification: The increase in the concentration of the harmful chemicals in the body of an\norganism per unit its mass at each successive trophic level in a food chain is known as biomagnification.\nOzone Depletion: The thinning of the ozone layer is called ozone depletion.\nOzone-depleting substances: Chlorofluorocarbons (CFCs), oxides of nitrogen, methane, carbon tetrachloride and chlorine are the ozone-depleting substances.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nOzone-depleting substances: Chlorofluorocarbons (CFCs), oxides of nitrogen, methane, carbon tetrachloride and chlorine are the ozone-depleting substances.\nFour practices which can help in the protection of our environment:\nDisposal of wastes after separating them into biodegradable and non-biodegradable waste material.\nUse of unleaded petrol and alternate sources of energy, and keeping the engine properly tuned and serviced and the tyres inflated to the right pressure so that the vehicle runs efficiently.\nThe use of gunny bags and paper bags in place of polythene/plastic bags.\nActivities such as gardening, rain-water harvesting and use of compost in place of fertilizers will help protect our environment from further damage.\nHarmful effects of agricultural practices on the environment.\nExcessive use of fertilisers changes the chemistry of soil and kills useful microbes.\nExcessive use of non-biodegradable chemical pesticides leads to biological magnification.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nExcessive use of fertilisers changes the chemistry of soil and kills useful microbes.\nExcessive use of non-biodegradable chemical pesticides leads to biological magnification.\nExtensive cropping causes loss of soil fertility.\nExcess use of groundwater for agriculture lowers the water table.\nAgricultural practices lead to some amount of damage to the natural ecosystem/habitat.\n1. Environment: The physical, chemical and \u2018 biological conditions of the region in which an organism lives is called its Environment. It includes air, light, soil, temperature, water and the presence or absence of other organisms, i.e., the conditions for development or growth.\nThe environment has three main components, viz :\nPhysical surroundings [soil, air and water bodies]\nLiving organisms [plants, animals, decomposers (bacteria and fungi)]\nMeteorological factors (or climatic factors) . [sunlight, temperature, rainfall, humidity, pressure and wind speed].", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nLiving organisms [plants, animals, decomposers (bacteria and fungi)]\nMeteorological factors (or climatic factors) . [sunlight, temperature, rainfall, humidity, pressure and wind speed].\n2. Physical environment: It is also called a abiotic or non-living environment. It includes :\nSoil, water bodies and air on the surface of the earth.\nMeteorological factors.\nThe physical environment is essential for :\nSupply of nutrient elements to the living beings.\nProviding space to the organisms for living.\nControlling weather of a place.\n3. Biotic (or biological) environment: It includes :\nPlants.\nAnimals (including human beings).\nDecomposers (bacteria and.fungi).\nOther important constituents of the biotic environment includes Kites and vultures as they feed on dead organisms and act as scavengers (cleansing agents) of the environment.\n4. Ecosystem: A community of organisms, interacting with each other, plus the environment in which they", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\n4. Ecosystem: A community of organisms, interacting with each other, plus the environment in which they\nlive and with which they also interact. The examples of the ecosystem are a pond; a desert; a forest; a lake; a river; a mountain; the sea.\nAll the above ecosystems are made up of two main components.\n\n5. Autotrophs (Producers) and Heterotrophs (consumers):\n\n6. Food chain: The sequence of living organisms in a . community in which one organism.e^ts other and is\nitself eaten by another organism to transfer energy is called a food chain. It is also defined as, \u201cchain of organisms, existing in any natural community, through which energy is transferred\u201d.\n7. Ozone layer : Ozone (O3) is a molecule formed by three atoms of oxygen unlike oxygen which is required for respiration by aerobic forms, ozone is a deadly poison. However, at the higher levels of the atmosphere,", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nozone performs an essential function. It shields the surface of the earth from ultraviolet (UV) radiation from the Sun. This radiation is highly damaging to organisms, for example, it is known to cause skin cancer in human beings.\nOzone at the higher levels of the atmosphere is a product of UV radiation, acting on oxygen (O2) molecule. The higher energy UV radiations split apart some molecular\noxygen (O2) into free oxygen (O) atoms. These atoms then combine with the molecular oxygen to form ozone as shown:\n\nDepletion of ozone layer: Ozone layer gets depleted \u2013 due to the use of chemicals called aerosol, spray propellants like chlorofluorocarbons. Depletion of \u2018 ozone layer would cause skin cancer in men and animals and severe damage to the plants.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\n8. Biological magnification: It means accumulation of non-biodegradable chemicals (like pesticides) in the living organisms (like plants, animals, including man) in a food chain. \u201cThe increase in the concentration of harmful chemicals in the body of living organisms at each trophic level of a food chain is called biological magnification\u201d.\n9. Biodegradable wastes and Non-biodegradable wastes:\n\n10. A generalised food chain:\n\n11. Food web: A food web is a network of food chains which establish a network of relationships between various species. Food web showing 8 interlinked food chains.\n\n\n13. The flow of energy between various components of the environment:\nGreen plants capture about 1 % of the energy and convert it into food energy.\nAbout f % of the food eaten is turned into the body of an organism and made available for the next level of consumers.\nAbout 10% of organic matter is present at each step and reaches the next level of consumers.", + "CHAPTER 13-OUR ENVIRONMENT NOTES.txt\nAbout 10% of organic matter is present at each step and reaches the next level of consumers.\nSince so little energy is available for the next level of consumers, food chains generally consist of only three or four steps. The loss of energy at each step is so great that very little usable energy remains after four trophic levels.\nThere are generally a greater number of individuals at the lower trophic levels of an ecosystem, the greatest number is of the producers.", + "CHAPTER 13-OUR ENVIRONMENT EXPLEMPLAR SOLUTIONS.txt\nNCERT Exemplar Class 10 Science Chapter 15 Our Environment\nShort Answer Questions\nQuestion 1. Why is improper disposal of waste a curse to environment?\nAnswer. Improper disposal of waste is a curse to environment because it pollutes the environment, air, water, soil and cause harmful effects on living organisms.\n\nFor example, passage of sewage into water body cause eutrophication, killing of animals and source of water-borne pathogens.\nQuestion 2. Write the common food chain of a pond ecosystem.\nAnswer. Phytoplanktons and aquatic plants \u2014\u2014\u2014\u2013> Zooplanktons and small aquatic animals and larvae, insects, etc. \u2014\u2014\u2014\u2013> Fish \u2014\u2014\u2014\u2013>Bird.\nQuestion 3. What are the advantages of cloth bags over plastic bags during shopping?\nAnswer. Advantages of cloth bags:\nThey are made of biodegradable material.\nThey can be recycled and reused.\nThey do not pollute the environment.\nThey are strong and more durable than plastic bags.\nThey are capable of carrying more things.\nThey are washable.", + "CHAPTER 13-OUR ENVIRONMENT EXPLEMPLAR SOLUTIONS.txt\nThey can be recycled and reused.\nThey do not pollute the environment.\nThey are strong and more durable than plastic bags.\nThey are capable of carrying more things.\nThey are washable.\nQuestion 4. Why are crop fields known as artificial ecosystems?\nAnswer. Crop fields are known as artificial ecosystems because they are man-made and some biotic and abiotic components are maintained, nourished and reaped by human beings.\nQuestion 5. We do not clean ponds or lakes, but an aquarium needs to be cleaned. Why?\nAnswer. An aquarium is an artificial and incomplete ecosystem compared to pond or lakes which are natural, self-sustaining and complete ecosystem where there is a perfect recycling of materials. Therefore, it needs to be cleaned.\nLong Answer Questions\nQuestion 6. Suggest any four activities in daily life which are eco-friendly.\nAnswer.\nUse of cloth bags instead of plastic bags.\nGardening.\nHarvesting of rainwater and preventing wastage of resources.\n Use of compost.", + "CHAPTER 13-OUR ENVIRONMENT EXPLEMPLAR SOLUTIONS.txt\nAnswer.\nUse of cloth bags instead of plastic bags.\nGardening.\nHarvesting of rainwater and preventing wastage of resources.\n Use of compost.\nQuestion 7. Indicate the flow of energy in an ecosystem. Why is it unidirectional ? Justify.\nAnswer. The flow of energy generally is as follows:\nSun \u2014\u2014\u2014>Producers \u2014\u2014\u2014> Herbivores \u2014\u2014\u2014> Carnivores\nSince, flow is progressively from one trophic level to another and does not revert back, it is said to be unidirectional, i.e. from sun to plants, plants to animals,\nanimals to other animals and organic remains to decomposers. Thus, the available energy decreases in the higher trophic levels making impossible for energy flow in reverse direction.\nQuestion 8. Name the wastes which are generated in your house daily. What measures would you take for their disposal?", + "CHAPTER 13-OUR ENVIRONMENT EXPLEMPLAR SOLUTIONS.txt\nQuestion 8. Name the wastes which are generated in your house daily. What measures would you take for their disposal?\nAnswer. The wastes generated daily are kitchen wastes, paper wastes like newspapers, bags, envelopes, plastic bags, vegetable and fruit peels, dust and empty cartons, etc. Measures for disposal are:\n Separation of biodegradable and non-biodegradable recyclable and non-recyclable wastes.\nSafe disposal of plastic bags.\nPreparation of compost from kitchen wastes for home garden or given to waste collector for disposal.\nPaper waste can be given for recycling.\nQuestion 9. Explain some harmful effects of agricultural practices on the environ-ment.\nAnswer. Some harmful effects of agricultural practices on the environment are as follows:\nUse of fertilisers change the chemistry of soil and kills useful microbes.\nUse of non-biodegradable pesticides leads to bio-magnification.\nExtensive cropping causes loss of soil fertility.", + "CHAPTER 13-OUR ENVIRONMENT EXPLEMPLAR SOLUTIONS.txt\nUse of fertilisers change the chemistry of soil and kills useful microbes.\nUse of non-biodegradable pesticides leads to bio-magnification.\nExtensive cropping causes loss of soil fertility.\nUse of ground water for agriculture \u2018 has resulted in lowering water table,\nNatural ecosystems and habitat have been damaged during clearing of land for agriculture.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 3 Intext Questions\nClass 10 Metals and Non Metals NCERT Book Page Number: 40\nQuestion 1\nGive an example of a metal which :\n(i) is a liquid at room temperature.\n(ii) can be easily cut with a knife.\n(iii) is the best conductor of heat.\n(iv) is a poor conductor of heat.\nAnswer:\n(i) Mercury\n(ii) Sodium\n(iii) Silver\n(iv) Lead\nQuestion 2\nExplain the meanings of malleable and ductile.\nAnswer:\nMalleable : A metal that can be beaten into thin sheets on hammering is called malleable.\nDuctile : A metal which can be drawn into thin wires is called ductile.", + "Answer:\nMalleable : A metal that can be beaten into thin sheets on hammering is called malleable.\nDuctile : A metal which can be drawn into thin wires is called ductile.\nMalleable Meaning in Hindi\n\u0915\u0941\u091b \u0927\u093e\u0924\u0941\u0913 \u0915\u094b \u092a\u0940\u091f\u0915\u0930 \u092a\u0924\u0932\u0940 \u091a\u093e\u0926\u0930 \u092c\u0928\u093e\u092f\u093e \u091c\u093e \u0938\u0915\u0924\u093e \u0939\u0948 | \u0907\u0938 \u0917\u0941\u0923\u0927\u0930\u094d\u092e \u0915\u094b \u0906\u0918\u093e\u0924\u0935\u0930\u094d\u0927\u094d\u092f \u0915\u0939\u0924\u0947 \u0939\u0948 | \u0915\u0941\u091b \u0927\u093e\u0924\u0941\u0913 \u0915\u0947 \u092a\u0924\u0932\u0947 \u0924\u093e\u0930 \u0915\u0947 \u0930\u0942\u092a \u092e\u0947\u0902 \u0916\u0940\u091a\u0928\u0947 \u0915\u093f \u0915\u094d\u0937\u092e\u0924\u093e \u0915\u094b \u0924\u0928\u094d\u092f\u0924\u093e \u0915\u0939\u0924\u0947 \u0939\u0948 |\nClass 10 Metals and Non Metals NCERT Book Page Number: 46\nQuestion 1\nWhy is sodium kept immersed in kerosene oil ?\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nClass 10 Metals and Non Metals NCERT Book Page Number: 46\nQuestion 1\nWhy is sodium kept immersed in kerosene oil ?\nAnswer:\nSodium is highly reactive. So it is kept immersed in kerosene oil to prevent its reaction with oxygen, moisture and carbon dioxide of air to prevent accidental fires.\nQuestion 2\nWrite equations for the reactions of\n(i) iron with steam.\n(ii) calcium and potassium with water.\nAnswer:\n\nQuestion 3\nSamples of four metals A, B, C and D were taken and added to the following solution one by one.\nThe results obtained have been tabulated as follows :\nUse the Table above to answer the following questions about metals A, B, C and D.\n(i) Which is the most reactive metal ?\n(ii) What would you observe if B is added to a solution of copper (II) sulphate?\n(iii) Arrange the metals A, B, C and D in the order of decreasing reactivity.\nAnswer:\n(i) B is the most reactive metal because it gives displacement reaction with iron (II) sulphate.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\n(iii) Arrange the metals A, B, C and D in the order of decreasing reactivity.\nAnswer:\n(i) B is the most reactive metal because it gives displacement reaction with iron (II) sulphate.\n(ii) When metal B is added to copper (II) sulphate solution, a displacement reaction will take place due to which the blue colour of copper (II) sulphate solution will fade and a red-brown deposit of copper will be formed on metal B.\n(iii) Metal B is the most reactive because it displaces iron from its salt solution. Metal A is less reactive because it displaces copper from its salt solution. Metal C is still less reactive because it can displace only silver from its salt solution and metal D is the least reactive because it cannot displace any metal from its salt solution. Hence, the decreasing order of reactivity of the metals is B > A > C > D.\nQuestion 4\nWhich gas is produced when dilute hydrochloric acid is added to a reactive metal ? Write the chemical reaction when iron reacts with dilute H2SO4.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nQuestion 4\nWhich gas is produced when dilute hydrochloric acid is added to a reactive metal ? Write the chemical reaction when iron reacts with dilute H2SO4.\nAnswer:\nHydrogen gas is produced when dilute hydrochloric acid is added to a reactive metal.\nChemical reaction when iron reacts with dilute H2SO4 :\nFe(s) + H2SO4(aq) \u2192 FeSO4(aq) + H2(g)\nQuestion 5\nWhat would you observe when zinc is added to a solution of iron (II) sulphate ? Write the chemical reaction that takes place.\nAnswer:\nZinc is more reactive than iron. Therefore, when zinc is added to a solution of iron (II) sulphate, then the greenish colour of iron (II) sulphate solution fades gradually due to the formation of colourless zinc sulphate solution and iron metal is deposited on zinc.\n\nClass 10 Metals and Non Metals NCERT Book Page Number: 49\nQuestion 1\n(i) Write the electron dot structures for sodium, oxygen and magnesium.", + "Class 10 Metals and Non Metals NCERT Book Page Number: 49\nQuestion 1\n(i) Write the electron dot structures for sodium, oxygen and magnesium.\n(ii) Show the formation of Na2O and MgO by the transfer of electrons.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nQuestion 1\n(i) Write the electron dot structures for sodium, oxygen and magnesium.\n(ii) Show the formation of Na2O and MgO by the transfer of electrons.\n(iii) What are ions present in these compounds?\nAnswer:\n\n(ii) Formation of Na2O and MgO\n\n(iii) In Na2O, ions present are Na+ and O2-.\nIn MgO, ions present are Mg2+ and O2-.\nQuestion 2\nWhy do ionic compounds have high melting points ?\n(iii) What are ions present in these compounds?\nAnswer:\nThe ionic compounds are made up of positive and negative ions. There is a strong force of attraction between the oppositely charged ions, so a lot of heat energy is required to break this force of attraction and melt the ionic compound. Due to this, ionic compounds have high melting points.\nClass 10 Metals and Non Metals NCERT Book Page Number: 53\nQuestion 1\nDefine the following terms : (i) Mineral, (ii) Ore and (iii) Gangue.", + "Due to this, ionic compounds have high melting points.\nClass 10 Metals and Non Metals NCERT Book Page Number: 53\nQuestion 1\nDefine the following terms : (i) Mineral, (ii) Ore and (iii) Gangue.\nAnswer:\n(i) Mineral : The natural materials in which the metals or their compounds are found in earth are called minerals.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nDefine the following terms : (i) Mineral, (ii) Ore and (iii) Gangue.\nAnswer:\n(i) Mineral : The natural materials in which the metals or their compounds are found in earth are called minerals.\n(ii) Ore : Those minerals from which the metals can be extracted conveniently and profitably are called ores.\n(iii) Gangue : The unwanted impurities like sand, rocky material, earth particles, lime stone, mica, etc in an ore are called gangue.\nQuestion 2\nName two metals which are found in nature in the free state.\nAnswer:\nGold and platinum\nQuestion 3\nWhat chemical process is used for obtaining a metal from its oxide.\nAnswer:\nReduction process is used for obtaining a metal from its oxide.\nFor example, zinc oxide is reduced to metallic zinc by heating with carbon.\nZnO(s) + C(s) \u2192 Zn(s) + CO(g)\nBesides carbon, highly reactive metals like sodium, calcium, aluminium etc. are used as reducing agents. These displace metals of low reactivity from their oxides.\nFor example,", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nBesides carbon, highly reactive metals like sodium, calcium, aluminium etc. are used as reducing agents. These displace metals of low reactivity from their oxides.\nFor example,\nFe2O3(s) + 2Al(s) \u2192 2Fe(l) + Al2O3(s) + Heat\nGold is Metal or Nonmetal ?\nGold is a metal found in nature in the free state\nClass 10 Metals and Non Metals NCERT Book Page Number: 55\nQuestion 1\nMetallic oxides of zinc, magnesium and copper were heated with the following metals :\nIn which cases will you find displacement reactions taking place ?\nAnswer:\nA more reactive metal can displace a less reactive metal from its oxide. But out of zinc, magnesium, and copper metals, magnesium is the most reactive, zinc is less reactive whereas copper is the least reactive metal.\nThe displacement will take place in the following cases :\nQuestion 2\nWhich metals do not corrode easily ?\nAnswer:\nGold and Platinum.\nQuestion 3\nWhat are alloys ?\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nThe displacement will take place in the following cases :\nQuestion 2\nWhich metals do not corrode easily ?\nAnswer:\nGold and Platinum.\nQuestion 3\nWhat are alloys ?\nAnswer:\nAn alloy is a homogeneous mixture of two or more metals, or a metal and a non-metal. For example, bronze is an alloy of copper and tin.\nNCERT Solutions for Class 10 Science Chapter 3 Textbook Chapter End Questions\nMetals and Nonmetals Class 10 Question 1.\nWhich of the following pairs will give displacement reactions ?\n(a) NaCl solution and copper metal.\n(b) MgCl2 solution and aluminium metal.\n(c) FeSO4 solution and silver metal.\n(d) AgNO3 solution and copper metal.\nAnswer:\n(d) AgNO3 solution and copper metal.\nQuestion 2.\nWhich of the following methods is suitable for preventing an iron frying pan from rusting ?\n(a) Applying grease\n(b) Applying paint.\n(c) Applying a coating of zinc\n(d) All the above.\nAnswer:\n(c) Applying a coating of zinc.\nQuestion 3.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\n(a) Applying grease\n(b) Applying paint.\n(c) Applying a coating of zinc\n(d) All the above.\nAnswer:\n(c) Applying a coating of zinc.\nQuestion 3.\nAn element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be\n(a) calcium\n(b) carbon\n(c) silicon\n(d) iron\nAnswer:\n(a) Calcium.\nQuestion 4.\nFood cans are coated with tin and not with zinc because\n(a) zinc is costlier than tin\n(b) zinc has a higher melting point than tin\n(c) zinc is less reactive than tin\n(d) zinc is more reactive than tin.\nAnswer:\n(d) Zinc is more reactive than tin.\nMetals and Non metals Class 10 Question 5.\nYou are given a hammer, a battery, a bulb, wires and a switch.\n(a) How could you use them to distinguish between samples of metals and non-metals?\n(b) Assess the usefulness of these tests in distinguishing between metals and non-metals.\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\n(a) How could you use them to distinguish between samples of metals and non-metals?\n(b) Assess the usefulness of these tests in distinguishing between metals and non-metals.\nAnswer:\n(a) Metals can be beaten into thin sheets with a hammer without breaking. Non-metals cannot be beaten with a hammer to form thin sheets. Non-metals break into pieces when hammered. Metals are malleable, while non-metals are non-melleable. When metals are connected into circuit using a battery, bulb, wires and switch, current passes through the circuit and the bulb glows. When non-metals (like sulphur) are connected, the bulb does not light up at all. Metals are good conductors of electricity.\n(b) Because of malleability, metals can be casted into sheets. Metals are good conductors of electricity so these can be used for electrical cables.\nQuestion 6.\nWhat are amphoteric oxides ? Give two examples of amphoteric oxides ?\nOR", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nQuestion 6.\nWhat are amphoteric oxides ? Give two examples of amphoteric oxides ?\nOR\nWrite chemical equations that show aluminium oxide reacts with acid as well as base. [CBSE2011]\nAnswer:\nThose metal oxides which show basic as well as acidic behaviour are known as amphoteric oxides. In other words, metal oxides that react wtih both acids and bases to form salt and water are called amphoteric oxides. Aluminium oxide and zinc oxide are amphoteric in nature.\n\nQuestion 7.\nName two metals which will displace hydrogen from dilute acids and two metals which will not.\nAnswer:\n(i) Metals above hydrogen in the activity series like sodium and magnesium displace hydrogen from dilute acids.\n(ii) Metals below hydrogen in the activity series like copper, silver do not displace hydrogen from dilute acids.\nQuestion 8.\nIn the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte ?\nAnswer:\nCathode \u2013 Pure metal", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nQuestion 8.\nIn the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte ?\nAnswer:\nCathode \u2013 Pure metal\nAnode \u2013 Impure metal\nElectrolyte \u2013 Metal salt solution\nQuestion 9.\nPratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test tube over it, as shown in the figure.\n(a) What will be the action of gas on\n(i) dry litmus paper ?\n(ii) moist litmus paper ?\n(b) Write a balanced chemical equation for the reaction taking place.\n\nAnswer:\n(i) Dry litmus paper \u2013 no action.\n(ii) Moist litmus paper \u2013 becomes red.\n\nQuestion 10.\nState two ways to prevent the rusting of iron.\nAnswer:\nWays to prevent rusting of iron are :\n(a) By painting\n(b) By galvanizing\nQuestion 11.\nWhat type of oxides are formed when non-metals combine with oxygen ?", + "Question 10.\nState two ways to prevent the rusting of iron.\nAnswer:\nWays to prevent rusting of iron are :\n(a) By painting\n(b) By galvanizing\nQuestion 11.\nWhat type of oxides are formed when non-metals combine with oxygen ?\nAnswer:\nNon-metals combine with oxygen to form acidic oxides or neutral oxides.\n\nQuestion 12.\nGive reasons :", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nQuestion 11.\nWhat type of oxides are formed when non-metals combine with oxygen ?\nAnswer:\nNon-metals combine with oxygen to form acidic oxides or neutral oxides.\n\nQuestion 12.\nGive reasons :\n(a) Platinum, gold and silver are used to make jewellery.\n(b) Sodium, potassium and lithium are stored under oil.\n(c) Aluminium is a highly reactive metal, yet it is used to make utensils for cooking.\n(d) Carbonate and sulphide ores are usually converted into oxides during the process of extraction.\nAnswer:\n(a) Platinum, gold and silver are used to make jewellery because these are malleable and ductile. These are highly resistant to corrosion.\n(b) Sodium, potassium and lithium are very reactive and catch fire when exposed to air. This is due to their low ignition temperature and high reactivity.\n(c) Aluminium forms a non-reactive layer of aluminium oxide on its surface. This layer prevents aluminium to react with other substances. That\u2019s why aluminium is used to make cooking utensils.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\n(c) Aluminium forms a non-reactive layer of aluminium oxide on its surface. This layer prevents aluminium to react with other substances. That\u2019s why aluminium is used to make cooking utensils.\n(d) It is easier to reduce a metal oxide into free metal. Since it is easier to obtain metals from their oxides than from their carbonates or sulphides directly, therefore, the carbonate and sulphide ores are first converted to oxides for extracting the metals.\nQuestion 13.\nYou must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.\nAnswer:\nThe sour substances such as lemon or tamarind juice contain acids. These acids dissolve the coating of copper oxide or basic copper carbonate present on the surface of tarnished copper vessels and makes them shining red-brown again.\nQuestion 14.\nDifferentiate between metal and non-metal on the basis of their chemical properties. [CBSE 2017 (Delhi)]\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nQuestion 14.\nDifferentiate between metal and non-metal on the basis of their chemical properties. [CBSE 2017 (Delhi)]\nAnswer:\nDifference between metals and non-metals\nQuestion 15.\nA man went door-to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty repeat. Can you play the detective to find out the nature of the solution he has used ?\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nAnswer:\nThe dishonest goldsmith dipped the gold bangles in aqua-regia (which contains 1 part of concentrated nitric acid and 3 parts of concentrated hydrochloric acid, by volume). Aqua-regia dissolved a considerable amount of gold from gold bangles and hence reduced their weight drastically. The dishonest goldsmith can recover the dissolved gold from aqua-regia by a suitable treatment.\nQuestion 16.\nGive reasons why copper is used to make hot water tanks and not steel (analloy of iron).\nAnswer:\n(i) Copper is a better conductor of heat than steel.\n(ii) Copper does not corrode easily. But steel corrodes easily.\n(iii) Copper does not react with water at any temperature, whereas iron reacts with water on heating.\nNCERT Solutions for Class 10 Science Chapter 3 Metals and Non-metals\nMetals and non metals: Properties of metals and non-metals, reactivity series, Formation and properties of ionic compounds, Basic metallurgical processes, corrosion and its prevention.\nQuestion 1", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nMetals and non metals: Properties of metals and non-metals, reactivity series, Formation and properties of ionic compounds, Basic metallurgical processes, corrosion and its prevention.\nQuestion 1\nWhat are amphoteric oxides? Give two examples of amphoteric oxides.\nSolution:\nAmphoteric oxides are the oxides, which react with both acids and bases to form salt and water. E.g. ZnO and Al2O3.\nQuestion 2\nName two metals, which will displace hydrogen from dilute acids, and two metals which will not.\nSolution:\nVery reactive metals like Zn and Mg displace hydrogen from dilute acids. On the other hand less reactive metals like Cu, Ag, etc. do not displace hydrogen from dilute acids.\nQuestion 3\nIn the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?", + "On the other hand less reactive metals like Cu, Ag, etc. do not displace hydrogen from dilute acids.\nQuestion 3\nIn the electrolytic refining of a metal M, what would you take as the anode, the cathode and the electrolyte?\nSolution:\nAnode is impure, thick block of metal M.\nCathode is a thin strip/wire of pure metal M.\nElectrolyte is a suitable salt solution of metal M.\nMetals and nonmetals Class 10 PDF\nQuestion 4", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nSolution:\nAnode is impure, thick block of metal M.\nCathode is a thin strip/wire of pure metal M.\nElectrolyte is a suitable salt solution of metal M.\nMetals and nonmetals Class 10 PDF\nQuestion 4\nState two ways to prevent the rusting of iron.\nSolution:\nBy coating the surface of iron by rust proof paints.\nBy applying oil or grease to the surface of iron objects so that supply of air consisting of moisture is cut off form the surface.\nQuestion 5\nWhat types of oxides are formed when non-metals combine with oxygen?\nSolution:\nWhen non-metals combine with oxygen it forms either neutral or acidic oxides. CO is a neutral oxide; N2O5 or N2O3 is an acidic oxide.\nextraction of metals from ores class 10 Question 6\nGive reason\ni. Metals replace hydrogen from dilute acids, where as non-metals do not.\nii. Carbonate and sulphide ores are usually converted into oxides during the process of extraction.\nSolution:", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nGive reason\ni. Metals replace hydrogen from dilute acids, where as non-metals do not.\nii. Carbonate and sulphide ores are usually converted into oxides during the process of extraction.\nSolution:\ni. Metals are electropositive in nature. They readily lose electrons. These electrons reduce the protons liberated from the acid to liberate hydrogen gas, where as non-metals possess a tendency to gain electrons and hence they do not furnish electrons to protons liberated from acids. Hence H2 gas is not liberated.\nii. As it is easier to reduce metal oxides to metal, prior to reduction, metal sulphides and carbonates must be converted to oxides.\nQuestion 7\nDifferentiate between metals and non-metals on the basis of their chemical properties.\nSolution:\n\nQuestion 8\nExplain why the surface of some metals acquires a dull appearance when exposed to air for a long time.\nSolution:", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nSolution:\n\nQuestion 8\nExplain why the surface of some metals acquires a dull appearance when exposed to air for a long time.\nSolution:\nThis is due to the surface oxidation of metals when exposed to moist air. For e.g. copper turns green on its surface due to the formation of basic copper carbonate Cu(OH) 2. CuCO3. Similarly silver becomes black due to the formation of black Ag2S and Aluminium forms a white coating of Al2O3 on its surface.\nQuestion 9\nState which of the following metals would give hydrogen when added to dilute hydrochloric acid. i. Iron, ii. Copper iii. Magnesium\nCopper does not react with dilute hydrochloric acid at all. This shows that copper is even less reactive than iron.\n\nQuestion 10\nName a non-metallic element, which conducts electricity.\nSolution:\nCarbon in the form of graphite conducts electricity, as there is a free electron in each carbon atom, which moves freely in between the hexagonal layers.\nQuestion 11\nWhich metals do not corrode easily?\nSolution:", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nQuestion 11\nWhich metals do not corrode easily?\nSolution:\nGold and platinum and other noble metals do not corrode in air.\nQuestion 12\nWhat are alloys?\nSolution:\nAlloys are homogeneous mixtures of two or more metals, or a metal and a non-metal.E.g. steel, brass, bronze, etc.\nQuestion 13\nDefine the following terms.\n(i) Minerals\n(ii) Ores\n(iii) Gangue\nSolution:\n(i) Minerals\nAll compounds or elements, which occur naturally in the earth\u2019s crust, are called minerals. Example: Alums, K2SO4.Al2(SO4)3 . 24 H2O, Bauxite Al2O3.2H2O\n(ii) Ores\nThose minerals from which a metal can be profitably extracted are called ores. Bauxite (Al2O3.2H2O) is the ore of Al, copper pyrite CuFeS2. All minerals are not ores but all ores are minerals.", + "Bauxite (Al2O3.2H2O) is the ore of Al, copper pyrite CuFeS2. All minerals are not ores but all ores are minerals.\n(iii) Gangue\nWhen an ore is mined from the earth, it is always found to be contaminated with sand rocky materials. The impurity of sand and rock materials present in the ore is known as gangue.\nQuestion 14\nName two metals that are found in nature in the free state.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nQuestion 14\nName two metals that are found in nature in the free state.\nSolution:\nGold and platinum are found in the free state in nature.\nQuestion 15\nWhat is chemical process used for obtaining a metal from its oxide?\n\nQuestion 16\nName two metals, which can form hydrides with metals.\nSolution:\nSodium and calcium form stable hydrides on reacting with hydrogen.\nQuestion 17\nDoes every mineral have a definite and a fixed composition? Explain.\nSolution:\nYes, every mineral has a definite and a fixed composition. Minerals are widely distributed in the earth\u2019s crust in the form of oxides, carbonates, sulphides, sulphates, nitrates, etc. These minerals are formed as a result of chemical changes taking place during the formation of earth.\nClass 10 metals and nonmetals Question 18\nExplain the meaning of malleable and ductile.\nSolution:\nMalleable is being able to be beaten/hammered into thin sheets.\nDuctile is being able to be drawn into thin wires.\nQuestion 19", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nExplain the meaning of malleable and ductile.\nSolution:\nMalleable is being able to be beaten/hammered into thin sheets.\nDuctile is being able to be drawn into thin wires.\nQuestion 19\ni. Write the electron dot structures for sodium, oxygen and magnesium.\nii. Show the formation of MgO and Na2O by the transfer of electrons.\niii. What are the ions present in these compounds?\nSolution:\n\nii. Formation of Magnesium oxide\nWhen magnesium reacts with oxygen, the magnesium atom transfers its two outermost electrons to an oxygen atom. By losing 2 elections, the magnesium atoms form a magnesium ion (Mg2+) and by gaining 2 electrons, the oxygen atom forms an oxide ion (O2-).\n\nFormation of Sodium oxide\nTwo sodium atoms transfer their 2 outermost electrons to an oxygen atom. By losing two electrons, the two sodium atoms form two sodiumions (2Na+). And by gaining two electrons, the oxygen atom forms an oxide ion (O2-.)", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\niii. The ions present in sodium oxide compound (Na20) aie sodium ions (2Na+ and oxide ions (O2-).\nThe ions present in Magnesium oxide compound (MgO) are magnesiumions Mg2+ and oxide ions (O2-).\nQuestion 20\nYou must have seen tarnished copper vessels being cleaned with lemon or tamarind juice. Explain why these sour substances are effective in cleaning the vessels.\nSolution:\nThe sour substances such as lemon (or tamarind juice) contain acids. These acids dissolve the coating of copper oxide or basic copper carbonate present on the surface of tarnished copper vessels and make them shining red-brown again.\nQuestion 21\nGive an example of a metal which\ni. is a liquid at room temperature.\nii. can be easily cut with a knife.\niii. is the best conductor of heat.\niv. is a poor conductor of heat.\nSolution:\ni. Mercury is in liquid state at room temperature.\nii. Sodium and potassium are soft metals which can be easily cut with a knife.\niii. Silver is the best conductor of electricity.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nSolution:\ni. Mercury is in liquid state at room temperature.\nii. Sodium and potassium are soft metals which can be easily cut with a knife.\niii. Silver is the best conductor of electricity.\niv. Mercury is a poor conductor of heat.\nQuestion 22\nWhy is sodium kept immersed in kerosene?\nSolution:\nSodium metal is kept immersed in kerosene to prevent their reaction with oxygen, moisture and carbon dioxide of air.\nQuestion 23\nWhy do ionic compounds have high melting points?\nSolution:\nThese compounds are made up of positive and negative ions. There is a strong force of attraction between the oppositively charged ions, so a lot of heat energy is required to break this force of attraction and melt the ionic compounds. This is why ionic compounds have high melting points.\nQuestion 24", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nQuestion 24\nA man went door to door posing as a goldsmith. He promised to bring back the glitter of old and dull gold ornaments. An unsuspecting lady gave a set of gold bangles to him which he dipped in a particular solution. The bangles sparkled like new but their weight was reduced drastically. The lady was upset but after a futile argument the man beat a hasty retreat. Can you play the detective to find out the nature of the solution he had used?\nSolution:\nAqua regia (By volume, this contains 3 parts of concentrated hydrochloric acid and 1 part of concentrated nitric acid) is the solution, which is used to sparkle the bangles like new, but their weight will be reduced drastically.\nQuestion 25\nWrite equations for the reactions of\n(i) iron with water\n(ii) calcium and potassium with water\nSolution:\n\nQuestion 26\nWhat would you observe when zinc is added to a sodium of iron(II) sulphate? Write the chemical reaction that takes place?\nSolution:", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\n(ii) calcium and potassium with water\nSolution:\n\nQuestion 26\nWhat would you observe when zinc is added to a sodium of iron(II) sulphate? Write the chemical reaction that takes place?\nSolution:\nZinc is more reactive (more electro positive) than iron. Therefore it displaces iron from its salt solution. The colour of ferrous sulphate is pale green which becomes colourless.\n\nMetals and nonmetals class 10 Question 27\nPratyush took sulphur powder on a spatula and heated it. He collected the gas evolved by inverting a test-tube over the burning sulphur.\nWhat will be the action of this gas on:\nDry litmus paper?\nMoist litmus paper?\nWrite a balanced chemical equation for the reaction taking place.\nSolution:\na) When sulphur is brunt in air then sulphur dioxide gas is formed.\n(i) Sulphur dioxide gas has no action on dry litmus paper.\n(ii) Sulphur dioxide gas turns moist blue litmus paper to red.", + "Write a balanced chemical equation for the reaction taking place.\nSolution:\na) When sulphur is brunt in air then sulphur dioxide gas is formed.\n(i) Sulphur dioxide gas has no action on dry litmus paper.\n(ii) Sulphur dioxide gas turns moist blue litmus paper to red.\n(b) S(s) + O2(g) \u2014> SO2(g)\nMultiple Choice Questions (MCQs) [1 Mark each]", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\n(i) Sulphur dioxide gas has no action on dry litmus paper.\n(ii) Sulphur dioxide gas turns moist blue litmus paper to red.\n(b) S(s) + O2(g) \u2014> SO2(g)\nMultiple Choice Questions (MCQs) [1 Mark each]\nMetals and nonmetals class 10 Question 1.\nWhat is the colour of aqueous solution of CuSO4 and FeSO4 as observed in the laboratory?\n(a) CuSO4 \u2013 blue; FeSO4 \u2013 light green\n(b) CuSO4 \u2013 blue; FeSO4 \u2013 dark green\n(c) CuSO4 \u2013 green; FeSO4 \u2013 blue\n(d) CuSO4 \u2013 green; FeSO4 \u2013 colourless\nAnswer:\n(a) Colour of CuSO4 solution is blue and FeSO4 solution is light green.\nMetals and nonmetals class 10 Question 2.\nA student took four test tubes I, II, III and IV containing aluminium sulphate, copper sulphate? ferrous sulphate and zinc sulphate solutions respectively. He placed an iron strip in each of them.", + "Metals and nonmetals class 10 Question 2.\nA student took four test tubes I, II, III and IV containing aluminium sulphate, copper sulphate? ferrous sulphate and zinc sulphate solutions respectively. He placed an iron strip in each of them.\n\nIn which test tube, he found a brown deposit?\n(a) I\n(b) II\n(c) III\n(d) IV\nAnswer:\n(b) In test tube II, because Fe is more reactive than copper but less reactive than Al arid Zn.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nIn which test tube, he found a brown deposit?\n(a) I\n(b) II\n(c) III\n(d) IV\nAnswer:\n(b) In test tube II, because Fe is more reactive than copper but less reactive than Al arid Zn.\nMetals and nonmetals class 10 Question 3.\nAluminium sulphate and copper sulphate solutions were taken in two test tubes I and II respectively. A few pieces of iron filings were then added to both the solutions. The four students A, B, C and D recorded their observations in the form of a table as given below:\nWhich student has recorded the correct observation?\n(a) D\n(b) C\n(c) B\n(d) A\nAnswer:\n(c) Student B\nIron does not react with Al2(SO4)3 solution because iron is less reactive than aluminium. But Fe being more reactive than Cu displaces Cu from CuSO4 solution.\n\nMetals and nonmetals class 10 Question 4.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nMetals and nonmetals class 10 Question 4.\nAqueous solutions of zinc sulphate and iron sulphate were taken in test tubes I and II by four students A, B, C and D. Metal pieces of iron and zinc were dropped in the two solutions and observations made after several hours were recorded in the form of table as given below:\nWhich student has given the correct report?\n(a) B\n(b) D\n(c) A\n(d) C\nAnswer:\n(d) Student C\n(i) Fe is less reactive than zinc. So,\n\n(ii) Zn is more reactive than Fe, so it displaces iron as follows:\n\nMetals and nonmetals class 10 Question 5.\n2 mL each of cone. HCl, cone. HNO3 and a mixture of cone. HCl and cone. HNO3 in the ratio of 3 : 1 were taken in test tubes labelled as A, B and C. A small piece of metal was put in each test tube.", + "HCl, cone. HNO3 and a mixture of cone. HCl and cone. HNO3 in the ratio of 3 : 1 were taken in test tubes labelled as A, B and C. A small piece of metal was put in each test tube. No change occurred in test tubes A\u2019and Bbut the metal got dissolved in test tube C. The metal could be [NCERT Exemplar]\n(a) Al\n(b) Au\n(c) Cu\n(d) Pt\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\n(a) Al\n(b) Au\n(c) Cu\n(d) Pt\nAnswer:\n(b, d) A mixture of cone. HCl and cone. HNO3 in the ratio of 3 : 1 is known as aqua-regia. Gold (Au) and platinum (Pt) dissolve only in aqua-regia as these metals are very less reactive.\nMetals and nonmetals class 10 Question 6.\nWhen an aluminium strip is kept (a) Green solution of FeSO4 slowly turns brown\n(b) Green solution of FeSO4 rapidly turns brown\n(c) No change in colour of FeSO4\n(d) Green solution of FeSO4 slowly turns colourless\nAnswer:\n(a) The green solution of ferrous sulphate slowly turns brown. As aluminium is more reactive than iron, it displaces iron from ferrous sulphate solution.\n\nMetals and nonmetals class 10 Question 7.\nAluminium is used for making cooking utensils. Which of the following properties of aluminium are responsible for the same?", + "As aluminium is more reactive than iron, it displaces iron from ferrous sulphate solution.\n\nMetals and nonmetals class 10 Question 7.\nAluminium is used for making cooking utensils. Which of the following properties of aluminium are responsible for the same?\n(i) Good thermal conductivity\n(ii) Good electrical conductivity\n(iii) Ductility\n(iv) Fligh melting point [NCERT Exemplar]\n(a) (i) and (ii)\n(b) (i) and (iii)\n(c) (ii) and (iii)\n(d) (i) and (iv)", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\n(i) Good thermal conductivity\n(ii) Good electrical conductivity\n(iii) Ductility\n(iv) Fligh melting point [NCERT Exemplar]\n(a) (i) and (ii)\n(b) (i) and (iii)\n(c) (ii) and (iii)\n(d) (i) and (iv)\nAnswer:\n(d) Good thermal conductivity, malleability, light weight and high melting point are the properties/of aluminium due to which it-is used for making cooking utensils.\nMetals and nonmetals class 10 Question 8.\nIf copper is kept open in air, it slowly loses its shining brown surface and gains a green coating.", + "Metals and nonmetals class 10 Question 8.\nIf copper is kept open in air, it slowly loses its shining brown surface and gains a green coating. It is due to the formation of [NCERT Exemplar]\n(a) CuSO4\n(b) CuCO3\n(c) CU(NO3)2\n(d) CuO\nAnswer:\n(b) Copper reacts with CO2 present in air and forms a green coating on its surface due to the formation of basic copper carbonate [CuCO3.Cu(OH)2] as:\n\nMETALs and Non-metals Mind Maps\nMetals and Non-Metals\nThere 92 well known naturally occurring minerals of which 70 are metals and rest 20 are the non-metals.\nPhysical Properties of Metals & Non-Metals\nReaction between Metals and Non Metals", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nThere 92 well known naturally occurring minerals of which 70 are metals and rest 20 are the non-metals.\nPhysical Properties of Metals & Non-Metals\nReaction between Metals and Non Metals\nReactivity of an element can be explained as tendency to attain a completely filled outermost shell.\nMetals have 1, 2 or 3 e- in outermost shell and thus it is easier for them to loss e- rather than to gain. They loss e- & gains positive charge & are tenned as cation.\nIn contrast, non-metals have 4-8 e- in outermost shell & thus they gain e- to achieve their octet. They gain e- as well as negative charge & tenned as anion.\nCations & anions attract each other & are held by strong electrostatic force of attraction.\nThe compounds fonned by the transfer of electrons from metal to non-non-metal are known as ionic compounds or electrovalent compound.\n\nOccurrence of Metals\nThe elements or compounds, which occur naturally in the earth\u2019s crust, are known as minerals.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nOccurrence of Metals\nThe elements or compounds, which occur naturally in the earth\u2019s crust, are known as minerals.\nAt some places, minerals contain a very high percentage of a particular metal and the metal can be profitably extracted from it. These minerals arc called ores.\nCorrosion\nCorrosion is the deterioration of materials by chemical interaction with their environment for e.g. darkening of silver articles when exposed to air, gaining of green coat on copper, rusting of iron.\nPrevention: The rusting of iron can be prevented by painting, oiling, greasing, galvanising, chrome plating, anodising or making alloys.\n\u2022 Galvanisation is a method of protecting steel and iron from rusting by coating them with a thin layer of zinc.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\n\u2022 Galvanisation is a method of protecting steel and iron from rusting by coating them with a thin layer of zinc.\n\u2022 Alloy is a homogeneous mixture of two or more metals, or a metal & nonmetal. For e.g. stainless steel (alloy of Fe. Ni, & Cr), amalgam (alloy of Hg), brass (alloy of Cu & Zn) etc. The electrical conductivity & melting point of an alloy is less than that of pure metals.\nEnrichment of Ores\nOres mined from the earth are usually contaminated with large amounts of impurities such as soil, sand, etc., called gangue. The impurities must be removed from the ore prior to the extraction of the metal.\nThe processes used for removing the gangue from the ore are based on the differences between physical or chemical properties of the gangue and the ore.\nExtraction of Metals\n\nExtracting Metals towards the Top of the Activity Series\nThese metals are highly reactive & are obtained by electrolytic reduction. For e.g. Na, Mg, & Ca are obtained by the electrolysis of their molten chlorides.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nThese metals are highly reactive & are obtained by electrolytic reduction. For e.g. Na, Mg, & Ca are obtained by the electrolysis of their molten chlorides.\nThe metals are deposited at the cathode whereas chlorine is liberated at anode.\nAt cathode Na+ + e\u2013 \u2192Na\nAt anode 2Cl\u2013 \u2192 Cl2 + 2e\u2013\nSimilarly, aluminium is obtained by the electrolytic reduction of aluminium oxide.\n\nExtracting Metals Low in the Activity Series\nThese metals are the least reactive & are often found in a free state for e.g. An, Ag. Pt & Cu are found in the free state.\nHowever, Cu & Ag are also found in the combined state as their sulphide or oxide ores.\nThe oxides of these metals can be reduced to metals by heating alone. For e.g. cinnabar (HgS), ore of mercury it is heated in air to converted it in mercuric oxide (HgO) which is then reduced to mercury by further heating.", + "The oxides of these metals can be reduced to metals by heating alone. For e.g. cinnabar (HgS), ore of mercury it is heated in air to converted it in mercuric oxide (HgO) which is then reduced to mercury by further heating.\n2HgS(s) + 3O2(g) Heat 2HgO(s) + 2SO2(g)\n2HgO(s)Heat 2Hg(l) + O(g)\nAnother instance is reduction of Cu2S (ore of copper) to copper by heating.", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\n2HgS(s) + 3O2(g) Heat 2HgO(s) + 2SO2(g)\n2HgO(s)Heat 2Hg(l) + O(g)\nAnother instance is reduction of Cu2S (ore of copper) to copper by heating.\n2Cu2S + 3O2(g)Heat 2Cu2O(s) + 2SO2(g)\n2CU2O + Cu2S Heat 6Cu(s) + SO2\nExtracting Metals Middle in the Activity Series\nThese metals such as Fe, Zn, Pb, Cu, etc are moderately reactive & are usually present as sulphides or carbonates in nature.\nThe sulphide ores are converted into oxides by heating strongly in the presence of excess air which is known as roasting.\nThe carbonate ores are changed into oxides by heating strongly in limited air which is known as calcination.\nThe metal oxides are then reduced to the corresponding metals by using suitable reducing agents such as carbon.\nFor e.g.", + "The carbonate ores are changed into oxides by heating strongly in limited air which is known as calcination.\nThe metal oxides are then reduced to the corresponding metals by using suitable reducing agents such as carbon.\nFor e.g. extraction of Zn\nRoasting: 2ZnS(s) + 3O2(g) Heat . 2ZnO(s) + 2SO2(g)\nCalcination: ZnCO3(s) Heat ZnO(s) + CO2(g)\nReduction: ZnO(s) + C(s) \u2192 Zn(s) + CO(g)", + "CHAPTER 3- METALS AND NON METALS- INTEXT QUESTIONS.txt\nFor e.g. extraction of Zn\nRoasting: 2ZnS(s) + 3O2(g) Heat . 2ZnO(s) + 2SO2(g)\nCalcination: ZnCO3(s) Heat ZnO(s) + CO2(g)\nReduction: ZnO(s) + C(s) \u2192 Zn(s) + CO(g)\nSometimes displacement reactions can also be used in place of reduction & highly reactive metals such as Na, Ca, Al, etc., are used as reducing agents.\nFor e.g. 3MnO2(s) + 4Al(s) \u2192 3Mn(l) + 2Al2O3(s) + Heat\nFe2O3(s) + 2Al(s) \u2192 2Fe(l)+Al2O3(s) + Heat\nThis reaction is used to join railway tracks or cracked machine parts and is known as the thermit reaction.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nImportant Questions of Metals and Non-Metals Class 10 Science Chapter 3\nQuestion 1.\nReverse of the following chemical reaction is not possible:\nZn(s) + CuSO4(aq) \u2192 ZnSO4(aq) + Cu(s)\nJustify this statement with reason. (Board Term I, 2016)\nAnswer:\nIf a strip of zinc metal is put in copper sulphate solution, then the blue colour of copper sulphate fades gradually due to the formation of colourless zinc sulphate solution and reddish-brown copper metal is deposited on zinc strip.\n\nIn this reaction, zinc metal being more reactive than copper displaces copper from copper sulphate solution. If however, a strip of copper metal is placed in zinc sulphate solution, then no reaction occurs. This is because copper metal is less reactive than zinc metal and hence, cannot displace zinc from its salt solution.\n\nQuestion 2.\nName a metal which:\n(a) is the best conductor of heat.\n(b) has a very low melting point.\n(c) does not react with oxygen even at high temperature.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nQuestion 2.\nName a metal which:\n(a) is the best conductor of heat.\n(b) has a very low melting point.\n(c) does not react with oxygen even at high temperature.\n(d) is most ductile. (Board Term I, 2015)\nAnswer:\n(a) Metal which is the best conductor of heat is silver.\n(b) Gallium has a very low melting point.\n(c) Silver and gold do not react with oxygen even at high temperature.\n(d) Gold is the most ductile metal.\nQuestion 3.\nWhat is meant by amphoteric oxides? Choose the amphoteric oxides from the following :\nNa2O, ZnO, CO2, Al2O3, H2O (Board Term I, 2014)\nAnswer:\nAmphoteric oxides are those which show acidic as well as basic character, i.e., they react with bases as well as acids. ZnO and Al2O3 are amphoteric oxides.\nQuestion 4.", + "ZnO and Al2O3 are amphoteric oxides.\nQuestion 4.\nComplete the following equation with balancing: (Board Term I, 2013)\n(i) Al + HCl \u2192\n(ii) Mg + HNO3 \u2192\nAnswer:\n\nQuestion 5.\nCompare in tabular form the reactivities of the following metals with cold and hot water: (2020)\n(a) Sodium\n(b) Calcium\n(c) Magnesium", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(i) Al + HCl \u2192\n(ii) Mg + HNO3 \u2192\nAnswer:\n\nQuestion 5.\nCompare in tabular form the reactivities of the following metals with cold and hot water: (2020)\n(a) Sodium\n(b) Calcium\n(c) Magnesium\nAnswer:\n\nQuestion 6.\nGive reason for the following:\n(i) Hydrogen gas is not evolved when most of the metals react with nitric acid.\n(ii) Zinc oxide is considered as an amphoteric oxide.\n(iii) Metals conduct electricity. (Board Term I, 2016)\nAnswer:\n(i) Hydrogen gas is not evolved when most metals react with nitric acid. It is because HNO3 is a strong oxidising agent. It oxidises the H2 produced to water and itself gets reduced to any of the nitrogen oxides (N2O, NO, NO2).\n(ii) ZnO reacts both with acids as well as bases to form salt and water. Thus, ZnO is an amphoteric oxide.\n\n(iii) Metals conduct electricity due to the flow of free electrons present in them.\nQuestion 7.", + "(ii) ZnO reacts both with acids as well as bases to form salt and water. Thus, ZnO is an amphoteric oxide.\n\n(iii) Metals conduct electricity due to the flow of free electrons present in them.\nQuestion 7.\n(a) Why does calcium start floating when it reacts with water? Write the balanced chemical equation of the reaction.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nQuestion 7.\n(a) Why does calcium start floating when it reacts with water? Write the balanced chemical equation of the reaction.\n(b) Name two metals which do not react with water. (Board Term I, 2015)\nAnswer:\n(a) Calcium reacts with cold water to form calcium hydroxide and hydrogen gas.\n\nThe bubbles of hydrogen gas produced stick to the surface of calcium and hence, it starts floating on the surface of water.\n(b) Gold and silver do not react with water.\nQuestion 8.\nState what would happen if:\n(i) some zinc pieces are placed in blue copper sulphate solution.\n(ii) some\u2019copper pieces are placed in green ferrous sulphate solution.\n(iii) an iron nail is dipped in a solution of copper sulphate for some time. (Board Term I, 2014)\nAnswer:\n(i) Refer to answer 1.\n(ii) Cu + FeSO4 No Reaction\nCu is less reactive than Fe, thus, it cannot displace Fe from FeSO4 solution.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nAnswer:\n(i) Refer to answer 1.\n(ii) Cu + FeSO4 No Reaction\nCu is less reactive than Fe, thus, it cannot displace Fe from FeSO4 solution.\n(iii) When an iron nail is dipped in copper sulphate solution, then the blue colour of copper sulphate fades gradually and a reddish brown coating is formed on the iron nail.\n\nAs iron is more reactive than copper, it displaces copper from copper sulphate solution.\nQuestion 9.\nGive reason:\n(a) Aluminium is a reactive metal but is still used for packing food articles.\n(b) Calcium starts floating when water is added to it. (Board Term I, 2014)\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nQuestion 9.\nGive reason:\n(a) Aluminium is a reactive metal but is still used for packing food articles.\n(b) Calcium starts floating when water is added to it. (Board Term I, 2014)\nAnswer:\n(a) Aluminium is a strong and cheap metal. It is also a good conductor of heat. But it is highly reactive. When it is exposed to moist air, its surface is covered with a thin impervious layer of aluminium oxide (Al2O3). This layer does not allow moist air to come in contact with the fresh metal and hence, protects the metal underneath from further damage or corrosion. Thus, after the formation of this protective layer of Al2O3, aluminium becomes resistant to corrosion. It is because of this reason that although aluminium is a highly reactive metal, it is still used in food packaging.\n(b) Refer to answer 7(a).\nQuestion 10.\n(a) Complete and balance the following chemical equations:\n(i) Al2O3 + HCl \u2192\n(ii) K2O + H2O \u2192\n(iii) Fe + H2O \u2192", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(b) Refer to answer 7(a).\nQuestion 10.\n(a) Complete and balance the following chemical equations:\n(i) Al2O3 + HCl \u2192\n(ii) K2O + H2O \u2192\n(iii) Fe + H2O \u2192\n(b) An element \u2018X\u2019 displaces iron from the aqueous solution of iron sulphate. List your observations if the element \u2018X\u2019 is treated with the aqueous solutions of copper sulphate, zinc sulphate and silver nitrate. Based on the observations arrange X, Zn, Cu and Ag in increasing order of their reactivities. (2020)\nAnswer:\n(a) (i) Al2O3 + 6HCl \u2192 2AlCl3 + 3H2O\n(ii) K2O + H2O \u2192 2KOH\n(iii) 3Fe + 4H2O \u2192 Fe3O4 + 4H2\n(b) As X displaces iron from its salt solution hence X is more reactive than iron.", + "It will also displace copper from copper sulphate and silver from silver nitrate as both are less reactive than iron. As zinc is more reactive than iron hence, X can be more or less reactive than zinc. Then the order of their reactivities can be\nAg < Cu < Fe < Zn < X or Ag < Cu < Fe < X < Zn.\nQuestion 11.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nAg < Cu < Fe < Zn < X or Ag < Cu < Fe < X < Zn.\nQuestion 11.\nA metal \u2018X\u2019 combines with a non-metal \u2018Y\u2019 by the transfer of electrons to form a compound Z.\n(i) State the type of bond in compound Z.\n(ii) What can you say about the melting point and boiling point of compound Z?\n(iii) Will this compound dissolve in kerosene or petrol?\n(iv) Will this compound be a good conductor of electricity? (Board Term I, 2017)\nAnswer:\nX being a metal loses electrons and Y being a non-metal gains electrons to form Z.\n(i) The chemical bond formed by the transfer of electrons from one atom to another is known as an ionic bond. Hence, Z is an ionic compound.\n(ii) Compound Z is an ionic compound thus, it has high melting and boiling points.\n(iii) Ionic compounds are insoluble in non-polar solvents such as kerosene or petrol.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(ii) Compound Z is an ionic compound thus, it has high melting and boiling points.\n(iii) Ionic compounds are insoluble in non-polar solvents such as kerosene or petrol.\n(iv) As Z is an ionic compound, it does not conduct electricity in the solid state because movement of ions in the solid is not possible due to their rigid structure. But it conducts electricity in the molten state or in aqueous solution due to the movement of ions freely.\nQuestion 12.\n(i) By the transfer of electrons, illustrate the formation of bond in magnesium chloride and identify the ions present in this compound.\n(ii) Ionic compounds are solids. Give reasons.\n(iii) With the help of a labelled diagram show the experimental set up of action of steam on a metal. (2020)\nAnswer:\n\n(ii) Ionic compounds are solids because the particles which make up ionic compounds are held together by strong electrostatic bonds.\n\nQuestion 13.\n(a) (i) Write two properties of gold which make it the most suitable metal for ornaments.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nQuestion 13.\n(a) (i) Write two properties of gold which make it the most suitable metal for ornaments.\n(ii) Name two metals which are the best conductors of heat.\n(iii) Name two metals which melt when you keep them on your palm.\n(b) Explain the formation of ionic compound CaO with electron-dot structure. Atomic numbers of calcium and oxygen are 20 and 8 respectively. (2020)\nAnswer:\n(a) (i) The malleability and ductility properties of gold make it suitable for ornaments.\n(ii) Silver and gold.\n(iii) Gallium and caesium have so low melting points that they melt even on keeping them on palm.\n\nQuestion 14.\n(i) Write down the electronic configuration\nof magnesium and oxygen.\n(ii) Give two general properties of the compound formed by combination of magnesium and oxygen.\n(iii) Show the formation of this compound by the transfer of electrons. (Board Term 1,2014)\nAnswer:\n(i) Atomic number of magnesium (Mg) = 12\n\u2234 Its electronic configuration = 2, 8, 2\nAtomic number of oxygen = 8", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nAnswer:\n(i) Atomic number of magnesium (Mg) = 12\n\u2234 Its electronic configuration = 2, 8, 2\nAtomic number of oxygen = 8\nElectronic configuration of oxygen = 2, 6\n(ii) Magnesium (Mg) reacts with oxygen (O2) to form magnesium oxide (MgO).\n2Mg + O2 \u2192 2MgO\nProperties of MgO are :\n(a) It involves ionic bonding.\n(b) It has high melting point due to strong electrostatic forces of attraction between Mg2+ and O2- ions.\n(iii) In the formation of magnesium oxide, two electrons are transferred from magnesium atom to oxygen atom as represented :\n\nQuestion 15.\nAn ore on treatment with dilute hydrochloric acid produces brisk effervescence. Name the type of ore with one example. What steps will be required to obtain metal from the enriched ore? Also write the chemical equations for the reactions involved in the process. (AI 2019)\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nAnswer:\nThe ore on treatment with dilute hydrochloric acid produces brisk effervescence hence, it must be a carbonate ore. Calamine (ZnCO3) is an important carbonate ore of zinc.\nSteps required to obtain metal from the enriched carbonate Are:\n(a) Conversion of the carbonate ore into metal oxide : This is done by calcination (for carbonate ores).\nCalcination is the process of heating the ore strongly in the absence or limited supply of air. The zinc carbonate on heating decomposes to form zinc oxide as shown :\n\n(b) Reduction of the metal oxide to metal : As zinc is moderately reactive, zinc oxide cannot be reduced by heating alone. Hence, it is reduced to zinc by using a reducing agent such as carbon.\n\nThe reduction of metal oxides by heating with coke is called smelting.\nQuestion 16.\n(i) Carbonate of metal \u20182T is abundant in earth crust and its hydroxide is used in \u2018white washing\u2019. Identify metal \u2018XI", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nQuestion 16.\n(i) Carbonate of metal \u20182T is abundant in earth crust and its hydroxide is used in \u2018white washing\u2019. Identify metal \u2018XI\n(ii) How will you convert this carbonate into its oxide? Name the process and write its equation. (Board Term I, 2014)\nAnswer:\n(i) Calcium carbonate is abundant in earth\u2019s crust and calcium hydroxide is used in white washing. Hence, metal X is calcium (Ca).\n(ii) CaCO3 is strongly heated in the absence of air to get the metal oxide. This process is called calcination.\n\nQuestion 17.\nZinc is a metal found in the middle of the activity series of metals. In nature, it is found as a carbonate ore, ZnCO3. Mention the steps carried out for its extraction from the ore. Support with equations. (Board Term 1,2013)\nAnswer:\nRefer to answer 15.\nQuestion 18.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 15.\nQuestion 18.\nCarbon cannot reduce the oxides of sodium, magnesium and aluminium to their respective metals. Why? Where are these metals placed in the reactivity series? How are these metals obtained form their ores? Take an example to explain the process of extraction along with chemical equations. (2020)\nAnswer:\nSodium, magnesium and aluminium have higher affinity towards oxygen than that of carbon because these are highly reactive metals. Hence, carbon cannot reduce the oxides of sodium, magnesium and aluminium to their respective metals. These metals are placed at the top of the reactivity series. The highly reactive metals like Na, Mg, Al, etc. are extracted by electrolytic reduction of their molten chlorides or oxides. Electrolytic reduction is brought about by passing electric current through the molten state. Metal gets deposited at the cathode.\nNaCl \u21cc Na+ + Cl\u2013\nAt cathode : Na+ + e\u2013 \u2192 Na\nAt anode : 2Cl\u2013 \u2192 Cl2 + 2e\u2013\nQuestion 19.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nNaCl \u21cc Na+ + Cl\u2013\nAt cathode : Na+ + e\u2013 \u2192 Na\nAt anode : 2Cl\u2013 \u2192 Cl2 + 2e\u2013\nQuestion 19.\nWrite balanced chemical equations to explain what happens, when\n(i) Mercuric oxide is heated.\n(ii) Mixture of cuprous oxide and cuprous sulphide is heated.\n(iii) Aluminium is reacted with manganese dioxide.\n(iv) Ferric oxide is reduced with aluminium.\n(v) Zinc carbonate undergoes calcination. (2020)\nAnswer:\n(i) On heating, mercuric oxide decomposes to give mercury .and oxygen.\n\n(ii) On heating mixture of cuprous oxide and cuprous sulphide, copper and sulphur dioxide are produced.\n\n(iii) When aluminium is heated with manganese dioxide, manganese and aluminium oxide are formed.\n\n(iv) Ferric oxide reacts with aluminium to produce aluminium oxide and iron.\n\n(v) On calcination, zinc carbonate produces zinc oxide and carbon dioxide.\n\nQuestion 20.\n(a) List in tabular form three chemical properties on the basis of which we can differentiate between a metal and a non\u00acmetal.", + "(iv) Ferric oxide reacts with aluminium to produce aluminium oxide and iron.\n\n(v) On calcination, zinc carbonate produces zinc oxide and carbon dioxide.\n\nQuestion 20.\n(a) List in tabular form three chemical properties on the basis of which we can differentiate between a metal and a non\u00acmetal.\n(b) Give reasons for the following :", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nQuestion 20.\n(a) List in tabular form three chemical properties on the basis of which we can differentiate between a metal and a non\u00acmetal.\n(b) Give reasons for the following :\n(i) Most metals conduct electricity well.\n(ii) The reaction of iron (III) oxide [Fe2O3] with heated aluminium is used to join cracked machine parts. (Delhi 2019)\nAnswer:\n(b) (i) Refer to answer 6 (iii).\n(ii) The reaction of iron (III) oxide, Fe2O3 with aluminium is highly exothermic and the iron produced melts. This molten iron is used to join cracked iron parts of machines and railway tracks.\n\nQuestion 21.\n(a) Write the steps involved in the extraction of pure metals in the middle of the activity series from their carbonate ores.\n(b) How is copper extracted from its sulphide ore? Explain the various steps supported by chemical equations. Draw labelled diagram for the electrolytic refining of copper. (2018)\nAnswer:\n(a) Extraction of metals of medium reactivity:", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Extraction of metals of medium reactivity:\nThe metals in the middle of the reactivity series are zinc, iron, lead, etc. The carbonate ores first need to get converted to oxides as it is easier to get metal from their oxides.\nRefer to answer 15.\n(b) Copper glance (Cu2S) when heated in air gets partially oxidised to copper oxide which further reacts with the remaining copper glance to give copper metal.\n\nQuestion 22.\nDraw a schematic diagram of the various steps involved in the extraction of metals from ores for metals of medium reactivity and for metals of low reactivity. (Board Term I, 2018)\nAnswer:\nVarious steps involved in the extraction of a metal from its ore followed by refining of the metal is called \u2018metallurgy\u2019. The steps involved are summarised as follows :\n\nQuestion 23.\n(a) Describe an activity to show that metals are good conductors of electricity.\n(b) Account for the following :\n(i) Hydrogen gas is not evolved when a metal reacts with nitric acid.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nQuestion 23.\n(a) Describe an activity to show that metals are good conductors of electricity.\n(b) Account for the following :\n(i) Hydrogen gas is not evolved when a metal reacts with nitric acid.\n(ii) For storing sodium metal, it is kept immersed in kerosene.\n(iii) The reaction of iron (III) oxide with aluminium is used to join cracked iron parts of machines. (Board Term I, 2016)\nAnswer:\n(a) Activity : (i) Set up an electric circuit as shown in the figure.\n(ii) Place the metal to be tested (Cu, Al, Ag, Fe, etc.) in the circuit, between the terminal A and B.\n(iii) Switch on the battery.\n\nObservations : Bulb begins to glow.\nConclusion : This indicates that the current is flowing through the metal wire. Hence, metals are good conductors of electricity.\n(b) (i) Refer to answer 6(i).\n(ii) Sodium reacts vigorously with air and catches fire. Also, sodium reacts with water and the hydrogen gas is evolved which catches fire. Therefore, sodium is kept under kerosene.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(ii) Sodium reacts vigorously with air and catches fire. Also, sodium reacts with water and the hydrogen gas is evolved which catches fire. Therefore, sodium is kept under kerosene.\n(iii) Refer to answer 20(b) (ii).\nQuestion 24.\nHow is copper obtained from its ore (Cu2S)?\nWrite only the chemical equations. How is copper thus obtained refined? Name and explain the process along with a labelled diagram. (Board Term I, 2015)\nAnswer:\nRefer to answer 21(b).\nCopper obtained is refined by electrolytic refining.\nElectrolytic refining of crude copper :\nThick block of impure metal acts as anode and a thin strip of pure copper metal acts as cathode. The electrolyse used is aqueous solution of copper sulphate containing a small amount of sulphuric acid. On passing electric current through the electrolyte, the metal from the anode dissolves into the electrolyte. An equivalent amount of copper metal from copper sulphate solution gets deposited on cathode.\n\nQuestion 25.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nQuestion 25.\n(a) Copper produced by heating the ore in air is not very pure. Describe the method used for refining impure copper. Draw labelled diagram of the process.\n(b) Write chemical equations for the reactions taking place when :\n(i) zinc sulphide is heated in air.\n(ii) zinc carbonate is calcined. (Board Term I, 2014)\nAnswer:\nRefer to answer 24.\n\nQuestion 26.\nAssertion (A) : The metals and alloys are good conductors of electricity.\nReason (R) : Bronze is an alloy of copper and tin and it is not a good conductor of electricity.\n(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true.\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true.\nAnswer:\n(c) : Metals and alloys have free electrons in them which can move freely inside them, so they conduct electricity very easily. Bronze is an alloy of copper and tin and it is a very good conductor of electricity.\nQuestion 27.\nName first discovered alloy. Give its composition also. (Board Term I, 2014)\nAnswer:\nBronze is the first discovered alloy. Its composition is copper (90%) and tin (10%).\nQuestion 28.\nList three differentiating features between the processes of galvanisation and alloying.\nAnswer:\nQuestion 29.\nDescribe an activity to find out the conditions under which iron rusts. (Board Term I, 2017)\nAnswer:\nActivity:\n(i) Take three test tubes and put clean nails in each of the three tubes. Label them as A, B and C.", + "Answer:\nQuestion 29.\nDescribe an activity to find out the conditions under which iron rusts. (Board Term I, 2017)\nAnswer:\nActivity:\n(i) Take three test tubes and put clean nails in each of the three tubes. Label them as A, B and C.\n(ii) Pour some water in test tube A and cork it.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nAnswer:\nActivity:\n(i) Take three test tubes and put clean nails in each of the three tubes. Label them as A, B and C.\n(ii) Pour some water in test tube A and cork it.\n(iii) In tube B, pour some boiled distilled water along with some turpentine oil and cork it.\n(iv) In test tube C, add some anhydrous calcium chloride and cork it.\n(v) Look these test tubes properly and keep them undisturbed for a few days.\nObservation : Only in test tube A, iron nails get rusted since the nails in this test tube are exposed to both air and water.\nConclusion : Both air and water are required for rusting of iron.\n\nQuestion 30.\nWhy some metal surfaces acquire a dull appearance when they are exposed to moist air? Write colour acquired by the surfaces of copper and silver in such situation and also write the chemical names of the substances due to which it happens. (Board Term I, 2016)\nAnswer:", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nAnswer:\nWhen a metal has been kept exposed to air for a long time, then it gets a dull appearance. The metals lose their shine or brightness due to the formation of a thin layer of oxide, carbonate or sulphide on their surface and thus, the metal surface gets corroded. The surface of copper gets coated with a green layer in moist air due to the formation of basic copper carbonate, silver articles acquire a blackish tinge due to the formation of silver sulphide.\nQuestion 31.\n(a) Name the following :\n(i) Metal that can be cut by knife\n(ii) Lustrous non-metal\n(iii) Metal that exists in liquid state at room temperature\n(iv) Most malleable and ductile metal\n(v) Metal that is best conductor of electricity\n(vi) Non-metal that can exist in different forms\n(b) How are alloys better than metals? Give composition of solder and amalgam.\nAnswer:\n(a) (i) Sodium\n(ii) Iodine\n(iii) Mercury\n(iv) Gold\n(v) Silver\n(vi) Carbon", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(b) How are alloys better than metals? Give composition of solder and amalgam.\nAnswer:\n(a) (i) Sodium\n(ii) Iodine\n(iii) Mercury\n(iv) Gold\n(v) Silver\n(vi) Carbon\n(b) Alloys are stronger than the metals from which they are made, more resistant to corrosion, have lower melting point, have lower electrical conductivity. Solder is an alloy of lead and tin. An amalgam is an alloy of mercury with another metal.\nQuestion 32.\n(a) Define corrosion.\n(b) What is corrosion of iron called?\n(c) How will you recognise the corrosion of silver?\n(d) Why corrosion of iron is a serious problem?\n(e) How can we prevent corrosion of iron? (Board Term I, 2017)\nAnswer:\n(a) The process of slowly eating up of metals due to their conversion into oxides, carbonates, sulphides, etc., by the action of atmospheric gases and moisture is called corrosion.\n(b) The corrosion of iron is called rusting.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(b) The corrosion of iron is called rusting.\n(c) Silver articles become black after sometime when exposed to air. This is due to formation of a coating of black silver sulphide (Ag2S) on its surface by the action of H2S gas present in the air.\n(d) Corrosion of iron is a serious problem. Every year large amount of money is spent to replace damaged iron articles. Corrosion causes damage to car bodies, bridges and iron railings, ships and to all objects made of metals specially those of iron.\n(e) Corrosion of iron is prevented by coating it with a layer of oil. The reason being that the layer of oil does not allow air and water to react the surface of iron. Corrosion of iron can also be prevented by painting, greasing, galvanising, anodising, electroplating or making alloys.\nQuestion 33.\nGive reason for the following :\n(a) Ionic compounds have higher melting point and higher boiling point.\n(b) Sodium is kept immersed in kerosene.\n(c) Reaction of calcium with water is less violent.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nGive reason for the following :\n(a) Ionic compounds have higher melting point and higher boiling point.\n(b) Sodium is kept immersed in kerosene.\n(c) Reaction of calcium with water is less violent.\n(d) Silver articles become black after some time when exposed to air.\n(e) Prior to reduction the metal sulphides and carbonates must be converted into metal oxides for extracting metals. (Board Term I, 2015)\nAnswer:\n(a) Due to strong forces of attraction, the ions are bound to each other very firmly. As a result, the electrovalent or ionic solids have high melting and boiling points.\n(b) Refer to answer 23(b) (ii).\n(c) Calcium reacts with cold water but the reaction is less violent. The heat evolved is not sufficient for the hydrogen to catch fire.\n\n(d) Refer to answer 32(c).\n(e) The reduction of metal oxides to metal is easier than the reduction of metal sulphides and metal carbonates. Hence, these are first reduced to their corresponding metal oxides.\nQuestion 34.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(e) The reduction of metal oxides to metal is easier than the reduction of metal sulphides and metal carbonates. Hence, these are first reduced to their corresponding metal oxides.\nQuestion 34.\n(a) Metals like iron, silver and copper get corroded on exposure to air. Write the chemical name of the substance deposited on their surface respectively with it\u2019s colour, in each case.\n(b) List four ways by which rusting can be prevented. (Board Term I, 2013)\nAnswer:\n(a) Iron gets corroded and forms ferric oxide which is rust, it is reddish brown in colour. For silver and copper, refer to answer 30.\n(b) The various methods used for preventing the rusting of iron are given below:\n(i) By applying paint : Materials like railings, iron gates, iron bridges, bodies of cars, buses and trucks, etc. are all painted to protect them from rusting. Painting the metal surface does not allow them to come in contact with the moist air and thus, prevents rusting.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(ii) Greasing and oiling : When some grease or oil is applied on the surface of an iron object, then moisture and air cannot come in contact with it and hence, rusting is prevented.\n(iii) Galvanization : It is a method of protecting iron from rusting by coating them with a thin layer of zinc. The iron coated with zinc is called galvanized iron.\n(iv) Electroplating : It is another technique used to prevent articles from rusting. In this process, metals like tin, nickel and chromium which do not corrode are electroplated on iron.\nShort Answer Type Questions[l] [2 Marks]-Year 2015\n35.Write one example of each of\n(i) a metal which is so soft that, it can be cut with knife and a non-metal which is the hardest sustance.\n(ii) a metal and a non-metal which exist as liquid at room temperature.\nAnswer.\n(i) Sodium, carbon (diamond).\n(ii)Mercury is liquid metal, bromine is liquid non-metal.\n36.Mention the names of the metals for the following:", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nAnswer.\n(i) Sodium, carbon (diamond).\n(ii)Mercury is liquid metal, bromine is liquid non-metal.\n36.Mention the names of the metals for the following:\n(i) Two metals which are alloyed with iron to make stainless steel.\n(ii) Two metals which are used to make jewellary.\nAnswer.\n(i) Nickel and chromium.\n(ii) Gold and platinum.\nCBSE Class 10 Science \u2013 More Resources\nShort Answer Type Question[l] [2 Marks] -Year 2013\n37.Give reason for the following:\n(a) School bells are made up of metals.\n(b) Electric wires are made up of copper.\nAnswer.\n(a) It is because metals are sonorous, i.e. they produce sound when struk with a hard substance.\n(b) It-is because copper is good conductor of electricity.\nShort Answer Type Question[ll] [3 Marks] -Year 2013\n38. Suggest a method of reduction for the following metals during their metallurgical processes:\n(i) metal \u2018A\u2019 which is one of the last, second or third position in the reactivity.", + "Short Answer Type Question[ll] [3 Marks] -Year 2013\n38. Suggest a method of reduction for the following metals during their metallurgical processes:\n(i) metal \u2018A\u2019 which is one of the last, second or third position in the reactivity.\n(ii) metal \u2018B\u2019 which gives vigorous reaction even with water and air.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(i) metal \u2018A\u2019 which is one of the last, second or third position in the reactivity.\n(ii) metal \u2018B\u2019 which gives vigorous reaction even with water and air.\n(iii) metal \u2018C\u2019 which is kept in the middle df activity series.\nAnswer.\n(i) \u2018A\u2019 can be obtained by chemical reduction using carbon or carbon monoxide as reducing agent.\n(ii) \u2018B\u2019 can be obtained by electrolytic reduction.\n(iii) \u2018C\u2019 can be reduced by reducing agent like \u2018Al\u2019.\nVery Short Answer Type Question [1 Mark] -Year 2012\n39.A green layer is gradually formed on a copper plate left exposed to air for a week in a bathroom. What could this green substance be?\nAnswer. It is due to the formation of basic copper carbonate [CuC03.Cu(0H)2].\nShort Answer Type Questions[l] [2 Marks] -Year 2012\n40.Name the following:\n(a) A metal, which is preserved in kerosene.\n(b) A lustrous coloured non-metal.\n(c) A metal, which can melt while kept on palm.", + "Short Answer Type Questions[l] [2 Marks] -Year 2012\n40.Name the following:\n(a) A metal, which is preserved in kerosene.\n(b) A lustrous coloured non-metal.\n(c) A metal, which can melt while kept on palm.\n(d) A metal, which is a poor conductor of heat.\nAnswer.\n(a) Sodium is preserved in kerosene.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(b) A lustrous coloured non-metal.\n(c) A metal, which can melt while kept on palm.\n(d) A metal, which is a poor conductor of heat.\nAnswer.\n(a) Sodium is preserved in kerosene.\n(b) Iodine is lustrous coloured non-metal.\n(c) Gallium. \u2018\n(d) Lead.\n41. Give reason for the following:\n(a) Aluminium oxide is considered as an amphoteric oxide.\n(b) Ionic compounds conduct electricity in molten state.\nAnswer.\n(a) It is because it reacts with acids as well as bases to produce salts and water.\u2019Al\u2019 is less electropositive metal. So, it forms amphoteric oxide which can react with acid as well as base.\n(b) Ionic compounds can conduct electricity in molten state because ions \u2019 become free to move in molten state.\nShort Answer Type Questions[ll] [3 Marks] -Year 2012\n42. A metal \u2018X\u2019 acquires a green colour coating on its surface on exposure to air.\n(i) Identify the metal \u2018X\u2019 and name the process responsible for this change.", + "Short Answer Type Questions[ll] [3 Marks] -Year 2012\n42. A metal \u2018X\u2019 acquires a green colour coating on its surface on exposure to air.\n(i) Identify the metal \u2018X\u2019 and name the process responsible for this change.\n(ii)Name and write chemical formula of the green coating formed on the metal.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(i) Identify the metal \u2018X\u2019 and name the process responsible for this change.\n(ii)Name and write chemical formula of the green coating formed on the metal.\n(iii) List two important methods to prevent the process.\nAnswer.\n(i) Metal is copper. The process is corrosion.\n(ii)Basic copper carbonate [CuCO3.Cu(0H)2].\n(iii)\n\u2022 It should be coated with tin\n\u2022 It should be mixed with other metals to form alloys.\n43.What are amphoteric oxides? Choose the amphoteric oxides from amongst the following oxides:\nNa2O, ZnO, Al2O3, CO2, H2O\nAnswer.Those oxides which reacts with acids as well as bases to produce salts and water are called amphoteric oxides, e.g. Na2O, ZnO, are amphoteric oxides among given oxides.\n44.Define the terms:\n(i) mineral\n(ii) ore, and\n(iii) gangue.\nAnswer.", + "Na2O, ZnO, are amphoteric oxides among given oxides.\n44.Define the terms:\n(i) mineral\n(ii) ore, and\n(iii) gangue.\nAnswer.\n(i) Mineral: It is a naturally occurring substance from which metal may or may not be extracted profitably or economically, e.g. A1 cannot be extracted profitably from mica.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(iii) gangue.\nAnswer.\n(i) Mineral: It is a naturally occurring substance from which metal may or may not be extracted profitably or economically, e.g. A1 cannot be extracted profitably from mica.\n(ii)Ore: It is a rocky material which contains sufficient quantity of mineral so that metal can be extracted profitably, e.g. zinc blende is an ore of zinc from which zinc can be extracted profitably.\n(iii) Gangue: It is a rocky material which is present along with the mineral in the ore, e.g. FeO is gangue in extraction of copper.\nLong Answer Type Questions [5 Marks] -Year 2012\n45. (a) Write the chemical name of the coating that forms on silver and copper articles when these are left exposed to moist air.\n(b) Explain what is galvanisation. What purpose is served by it?\n(c) Define an alloy. How are alloys prepared? How do the properties of iron change when:\n(i) small quantity of carbon,\n(ii) nickel and chromium are mixed with it.\nAnswer.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(c) Define an alloy. How are alloys prepared? How do the properties of iron change when:\n(i) small quantity of carbon,\n(ii) nickel and chromium are mixed with it.\nAnswer.\n(a) Ag2S (silver sulphide) is formed on silver, basic copper carbonate CuCO3. CU(OH)2 is formed on copper.\n(b) The process of coating zinc over iron is called galvanisation. It is used to prevent rusting of iron.\n(c) Alloy is a homogeneous mixture of two or more metals. One of them can be non-metal. Alloys are prepared by melting two or more metals together.\n(?) Iron does not rust on adding small,quantity of carbon.\n(ii) When we form alloy of iron with nickel and chromium, we get stainless steel which is malleable and does not get rusted.\n46.(a) Carbon cannot be used as reducing agent to obtain Mg from MgO. Why?\n(b) How is sodium obtained from molten sodium chloride? Give equation of the reactions.\n(c) How is copper obtained from its sulphide ore? Give equations of the reactions.\nAnswer.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(b) How is sodium obtained from molten sodium chloride? Give equation of the reactions.\n(c) How is copper obtained from its sulphide ore? Give equations of the reactions.\nAnswer.\n(a) It is because \u2018Mg\u2019 is stronger reducing agent than carbon.\n(b) Sodium is obtained from molten NaCl by electrolysis.\n\nBlister Copper is purified by electrolytic refining.\n47.Give reasons for the following:\n(i) Silver and copper lose their shine when they are exposed to air. Name the substance formed on their surface in each case.\n(ii) Tarnished copper vessels are cleaned with tamarind juice.\n(iii) Aluminium is more reactive than iron yet there is less corrosion of aluminium as compared to iron when both are exposed to air.\nAnswer.\n(i) These metals get corroded. Silver forms black Ag2S (silver sulphide) and copper form greenish layer of basic copper carbonate CuCO3. CU(OH)2 .\n(ii) Tamarind contains acid which reacts with basic copper carbonate and product gets dissolved and removed from copper vessel.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(ii) Tamarind contains acid which reacts with basic copper carbonate and product gets dissolved and removed from copper vessel.\n(in) Aluminium forms oxide layer on its surface which does not further react with air.\n48. What are alloys? How are they made? Name the constituents and uses of brass, bronze and solder.\nAnswer. Alloys are homogeneous mixtures of two or more metals. One of them can be a non-metal also. They are made by melting a metal which is in large amount first and then adding the other metal. ,\nBrass contains copper and zinc. It is used for making decorative articles. Bronze contains copper and tin. It is used for making statues and medals. Solder contains lead and tin. It is used for soldering purposes.\nVery Short Answer Type Questions [1 Mark] -Year 2011\n49. A non-metal X exists in two different forms Y and Z. Y is the hardest natural substance, whereas Z is a good conductor of electricity. Identify X, Y and Z.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n49. A non-metal X exists in two different forms Y and Z. Y is the hardest natural substance, whereas Z is a good conductor of electricity. Identify X, Y and Z.\nAnswer. \u2018X\u2019 is carbon, \u2018Y\u2019 is diamond as it is the hardest natural substance and \u2018Z\u2019 is graphite as it is good conductor of electricity.\n50. Why does calcium float in water?\nAnswer. It is because hydrogen gas is formed which sticks to surface of calcium, therefore it floats. \u2013\n51.Name a non-metal which is lustrous and a metal which is non-lustrous. Iodine is a non-metal which is lustrous,\nAnswer.lead is a non-lustrous metal.\n52.Which gas is liberated when a metal reacts with an acid? How will you test the presence of this gas?\nAnswer. Hydrogen gas is formed. Bring a burning matchstick near to it, H2 will burn explosively with \u2018pop\u2019 sound.\n53.Name the metal which reacts with a very dilute HNOs to evolve hydrogen gas.\nAnswer.", + "How will you test the presence of this gas?\nAnswer. Hydrogen gas is formed. Bring a burning matchstick near to it, H2 will burn explosively with \u2018pop\u2019 sound.\n53.Name the metal which reacts with a very dilute HNOs to evolve hydrogen gas.\nAnswer. Magnesium\n54.Name two metals which are found in nature in the free state.\nAnswer.(i) Gold (ii) Silver", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n53.Name the metal which reacts with a very dilute HNOs to evolve hydrogen gas.\nAnswer. Magnesium\n54.Name two metals which are found in nature in the free state.\nAnswer.(i) Gold (ii) Silver\nAnswer Type Questions[l] [2 Marks] -Year 2011\n55.The way, metals like sodium, magnesium and iron react with air and water is an indication of their relative positions in the \u2018reactivity series\u2019. Is this statement true ? Justify your answer with examples.\nAnswer.Yes, sodium reacts explosively even with cold water, it is most reactive. Magnesium reacts with hot water, it is less reactive than Na. Iron reacts only with steam which shows it is least reactive among the three.\n56. X + YSO4\u2014\u2014-> XSO4+ Y\nY + XSO4 \u2014\u2014\u2013> No reaction\nOut of the two elements, \u2018X\u2019 and \u2018Y\u2019, which is more reactive and why?\nAnswer. \u2018X\u2019 is more reactive than \u2018Y\u2019 because it displaces \u2018Y\u2019 from its salt solution.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nY + XSO4 \u2014\u2014\u2013> No reaction\nOut of the two elements, \u2018X\u2019 and \u2018Y\u2019, which is more reactive and why?\nAnswer. \u2018X\u2019 is more reactive than \u2018Y\u2019 because it displaces \u2018Y\u2019 from its salt solution.\n57.What is an alloy? State the constituents of solder. Which property of solder makes it suitable for welding electrical wires?\nAnswer. Alloy is a homogeneous mixture of two or more metals. One of them can be a non-metal also. Solder consists of lead and tin. It has low melting point which makes it suitable for welding electrical wires.\n58. Write chemical equations that shows aluminium oxide reacts with acid as well as base.\nAnswer.\n\nLong Answer Type Questions [5 Marks] -Year 2011\n59.(a) How can the metals at the top of the reactivity series be extracted from their ores? Explain with an example.\n(b) Name any one alloy made from\n(i) a metal and a non-metal, and\n(ii) two metals.\nAnswer.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(b) Name any one alloy made from\n(i) a metal and a non-metal, and\n(ii) two metals.\nAnswer.\n(a) These metals are extracted by electrolytic reduction, e.g. aluminium is obtained from bauxite by electrolytic reduction.\n(b) (i) Steel is made up of iron and carbon.\n(ii) Brass is made up of copper and zinc.\nVery Short Answer Type Questions [1 Mark] -Year 2010\n60.What is the valency of silicon with atomic number 14?\nAnswer. Its valency is equal to 4.\n61.What is the valency of phosphorus with atomic number 15?\nAnswer. Phosphorus has valency 3.\n62.What is the valency of an element with atomic number 35?\nAnswer. Its valency is 1.\nShort Answer Type Question[I] [2 Marks] -Year 2010\n63.Elements magnesium and oxygen respectively belong to group 2 and group 16 of the Modern Periodic Table.", + "62.What is the valency of an element with atomic number 35?\nAnswer. Its valency is 1.\nShort Answer Type Question[I] [2 Marks] -Year 2010\n63.Elements magnesium and oxygen respectively belong to group 2 and group 16 of the Modern Periodic Table. If the atomic numbers of magnesium and oxygen are 12 and 8 respectively, draw their electronic configurations and show the process of formation of their compound by transfer of electrons.\nAnswer.(Mg)=2,8,2", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nAnswer.(Mg)=2,8,2\n\nVery Short Answer Type Question [1 Mark] -Year 2009\n64. Arrange the following metals in the decreasing order of reactivity: Na, K, Cu, Ag.\nAnswer. K > Na > Cu > Ag\nShort Answer Type Questions[ll] [3 Marks] -Year 2009\n65. Give reasons for the following observations:\n(i) Ionic compounds in general have high melting and boiling points.\n(ii) Highly reactive metals cannot be obtained from their oxides by heating\nthem with carbon.\n(iii) Copper vessels get a green coat when left exposed to air in the rainy season.\nAnswer. (i) Ionic compounds have high melting and boiling points due to strong force of attraction between oppositely charged ions.\n(ii) It is because these metals themselves are strong reducing agents. Therefore, cannot be reduced by reducing agent like carbon.\n(iii) Copper vessels react with CO2, O2 and moisture to form green-coloured basic copper carbonate\n[CuCO3.Cu(OH)2].\n66. State reasons for the following observations:", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n(iii) Copper vessels react with CO2, O2 and moisture to form green-coloured basic copper carbonate\n[CuCO3.Cu(OH)2].\n66. State reasons for the following observations:\n(i) The shining surface of some metals becomes dull when exposed to air for a long time.\n(ii) Zinc fails to evolve hydrogen gas on reacting with dilute nitric acid.\n(iii) Metal sulphides occur mainly in rocks but metal halides occur mostly in sea and lake waters.\nAnswer.\n(i) It is because metal reacts with substances present in atmosphere to form surface compounds which make it dull.\n(ii) It is because dil. HNOs is an oxidising agent therefore zinc gives NO and notH2 with dil. HNOs.\n(iii) It is because sea water contains sodium chloride due to which metal halides are formed, whereas sulphur is found below rocks. Therefore, metal \u2013 sulphides are formed in rocks.\n67. State reasons for the following:\n(i) Electric wires are covered with rubber like material.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n67. State reasons for the following:\n(i) Electric wires are covered with rubber like material.\n(ii)From dilute hydrochloric acid, zinc can liberate hydrogen gas but copper cannot.\n(iii) Sulphide ore of a metal is first converted to its oxide to extract the metal from it.\nAnswer.\n(i) It is because rubber is an insulator and does not allow current to flow through it.\n(ii) Zinc is more reactive than hydrogen. Therefore, it can displace hydrogen from dilute HCl whereas copper cannot, because it is less reactive than hydrogen. ,\n(iii) It is because it is easier to reduce oxide ore as compared to sulphide ore.\nLong Answer Type Questions [5 Marks] -Year 2009\n68.(a) What is meant by corrosion? Name any two methods used for the prevention\nof corrosion.\n(b) Suppose you have to extract metal M from its enriched sulphide ore. If M is in the middle of the reactivity series, write various steps used in extracting this metal.", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\nof corrosion.\n(b) Suppose you have to extract metal M from its enriched sulphide ore. If M is in the middle of the reactivity series, write various steps used in extracting this metal.\nAnswer. (a) Corrosion is a process in which metal reacts with substances present in the environment to form surface compounds.\nPrevention:\n(i) Galvanisation is a process to prevent corrosion of iron.\n(ii)Electroplating is also used to prevent corrosion.\n(b)(i) Concentration of ores: Sulphide ore will be concentrated by froth- floatation process. Sulphide ore will be collected in froth whereas gangue will be left behind.\n(ii) Roasslng: Sulphide ore is heated strongly in the presence of O2 to form metal oxide and sulphur dioxide.\n2MS + 3O2 \u2014\u2014\u2014\u2014- \u25ba 2MO + 2SO2\n(iii) Reduction: MO reacts with carbon (acts as reducing agent) to form metal and CO.\nMO + C \u2014-> M + CO", + "CHAPTER 3- METALS AND NON METALS- IMPORTANT QUESTIONS.txt\n2MS + 3O2 \u2014\u2014\u2014\u2014- \u25ba 2MO + 2SO2\n(iii) Reduction: MO reacts with carbon (acts as reducing agent) to form metal and CO.\nMO + C \u2014-> M + CO\n(iv) Electrolytic refining: Impure metal \u2018M\u2019 is purified by electrolytic refining. Impure metal is taken as anode, pure metal is taken as cathode, soluble salt of metal is taken as electrolyte. Impure metal forms metal ions which gain electrons and form pure metal at cathode.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nMetals and Non-metals Class 10 Science Notes\nMetals: Physical properties of metals, chemical properties of metals and non-metal oxide.\nMetals are the elements that conduct heat and electricity and are malleable and ductile. Examples are Iron (Fe), Aluminium (Al), Silver (Ag), Copper (Cu), Gold (Au), Platinum (Pt), Lead (Pb), Potassium (K), Sodium (Na), Calcium (Ca) and Magnesium (Mg) etc.\nMetals are the elements which form positive ions by losing electrons. Thus, metals are known as Electropositive Elements.\nPhysical Properties of Metals\nHardness: Most of the metals are hard, except alkali metals, such as sodium, potassium, lithium, etc. are very soft metals. These can be cut by using a knife.\nStrength: Most of the metals are strong and have high tensile strength. Because of this, big structures are made using metals, such as copper (Cu) and iron (Fe). (Except Sodium (Na) and potassium (K) which are soft metals).\nState: Metals are solid at room temperature except for mercury (Hg).", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nState: Metals are solid at room temperature except for mercury (Hg).\nSound: Metals produce ringing sound, so, metals are called Sonorous. Sound of metals is also known as Metallic sound. This is the cause that metal wires are used in making musical instruments.\nConduction: Metals are a good conductor of heat and electricity. This is the cause that electric wires are made of metals like copper and aluminium.\nMalleability: Metals are malleable. This means metals can be beaten into a thin sheet. Because of this property, iron is used in making big ships.\nDuctility: Metals are ductile. This means metals can be drawn into thin wire. Because of this property, a wire is made of metals.\nMelting and Boiling Point: Metals have generally high melting and boiling points. (Except sodium and potassium metals which have low melting and boiling point.)\nDensity: Most of the metals have a high density.\nColour: Most of the metals are grey in colour. But gold and copper are exceptions.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nDensity: Most of the metals have a high density.\nColour: Most of the metals are grey in colour. But gold and copper are exceptions.\nChemical Properties of Metals\n1. Reaction with oxygen: Most of the metals form respective metal oxides when reacting with oxygen.\nMetal + Oxygen \u2192 Metal Oxide\nExamples:\nReaction of Potassium with Oxygen: Potassium metal forms potassium oxide when reacts with oxygen.\n\nReaction of Sodium with Oxygen: Sodium metal forms sodium oxide when reacts with oxygen.\n\nLithium, potassium, sodium, etc. are known as Alkali-metals. Alkali metals react vigorously with oxygen.\nReaction of Copper metal with Oxygen: Copper does not react with oxygen at room temperature but when burnt in air, it gives oxide.\n\nSilver, gold and platinum do not combine with the oxygen of air even at high temperature. They are the least reactive.\n2. Reaction of metals with water: Metals form respective hydroxide and hydrogen gas when reacting with water.\nMetal + Water \u2192 Metal hydroxide + Hydrogen", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\n2. Reaction of metals with water: Metals form respective hydroxide and hydrogen gas when reacting with water.\nMetal + Water \u2192 Metal hydroxide + Hydrogen\nMost of the metals do not react with water. However, alkali metals react vigorously with water.\nReaction of Sodium metal with Water: Sodium metal forms sodium hydroxide and liberates hydrogen gas along with lot of heat when reacting with water.\n\nReaction of Calcium metal with Water: Calcium forms calcium hydroxide along with hydrogen gas and heat when react with water.\n\nReaction of Magnesium metal with Water: Magnesium metal reacts with water slowly and forms magnesium hydroxide and hydrogen gas.\n\nWhen steam is passed over magnesium metal, magnesium oxide and hydrogen gas are formed.\n\nReaction of Aluminium metal with Water: Reaction of aluminium metal with cold water is too slow to come into notice. But when steam is passed over aluminium metal, aluminium oxide and hydrogen gas are produced.\n2Al + 3H2O \u2192 Al2O3 + 2H2", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\n2Al + 3H2O \u2192 Al2O3 + 2H2\nReaction of Zinc metal with Water: Zinc metal produces zinc oxide and hydrogen gas when steam is passed over it. Zinc does not react with cold water.\n\nReaction of Iron with Water: Reaction of iron with cold water is very slow and comes into notice after a long time. Iron forms rust (iron oxide) when reacts with moisture present in the atmosphere. Iron oxide and hydrogen gas are formed by passing of steam over iron metal.\n\nBoth calcium (Ca) and magnesium (Mg) are heavier than water but still float over it: Both calcium and magnesium float over water surface because hydrogen gas is evolved when these metals react with water. It is in the form of bubbles which stick on the metal surface. Therefore, they float over it.\n\nOther metals usually do not react with water or react very slowly. Lead, copper, silver and gold do not react with steam. Thus, the order of reactivity of different metals towards water may be written as :", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nK > Na > Ca > Mg > Ae > Zn > Fe > Pb > Cu > Ag > Au\n3. Reaction of metals with dilute acid: Metals form respective salts when reacting with dilute acid.\nMetal + dil. acid \u2192 Metal salt + Hydrogen\nReaction of Sodium metal with dilute hydrochloric acid: Sodium metal gives sodium chloride and hydrogen gas when react with dilute hydrochloric acid.\n\nReaction of Magnesium metal with dilute hydrochloric acid: Magnesium chloride and hydrogen gas are formed when magnesium reacts with dilute hydrochloric acid.\n\nReaction of Zinc with dilute sulphuric acid: Zinc sulphate and hydrogen gas are formed when zinc reacts with dilute sulphuric acid. This method is used in the laboratory to produce hydrogen gas.\n\nHydrogen (H2) gas is not evolved when metal is treated with nitric acid (HNO3):", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nHydrogen (H2) gas is not evolved when metal is treated with nitric acid (HNO3):\nNitric acid is strong oxidising agent and it oxidises the hydrogen gas (H2) liberated into water (H2O) and itself get reduced to some oxide of nitrogen like nitrous oxide (N2O)3 nitric oxide (NO) and nitrogen dioxide (NO2).\nCopper, gold, silver are known as noble metals. These do not react with water or dilute acids.\nThe order of reactivity of metal towards dilute hydrochloric acid or sulphuric acid is in the order;\nK > Na > Ca > Mg > Al > Zn > Fe > Cu > Hg > Ag\nMetal Oxides\nChemical Properties: Metal oxides are basic in nature. The aqueous solution of metal oxides turns red litmus blue.\nReaction of Metal oxides with Water: Most of the metal oxides are insoluble in water. Alkali metal oxides are soluble in water. Alkali metal oxides give strong base when dissolved in water.", + "The aqueous solution of metal oxides turns red litmus blue.\nReaction of Metal oxides with Water: Most of the metal oxides are insoluble in water. Alkali metal oxides are soluble in water. Alkali metal oxides give strong base when dissolved in water.\nReaction of Sodium oxide with Water: Sodium oxide gives sodium hydroxide when reacts with water.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nReaction of Sodium oxide with Water: Sodium oxide gives sodium hydroxide when reacts with water.\n\nReaction of Potassium oxide with Water: Potassium oxide gives potassium hydroxide when reacts with water.\n\nReaction of Zinc oxide and Aluminium oxide: Aluminium oxide and zinc oxide are insoluble in water. Aluminium oxide and zinc oxide are amphoteric in nature. An amphoteric substance shows both acidic and basic characters. It reacts with base like acid and reacts with an acid like a base.\nWhen zinc oxide reacts with sodium hydroxide, it behaves like an acid. In this reaction, sodium zincate and water are formed.\n\nZinc oxide behaves like a base when reacts with acid. Zinc oxide gives zinc chloride and water on reaction with hydrochloric acid.\n\nIn a similar way, aluminium oxide behaves like a base when reacts with acid and behaves like acid when reacts with a base.\nAluminium oxide gives sodium aluminate along with water when reacts with sodium hydroxide.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nAluminium oxide gives sodium aluminate along with water when reacts with sodium hydroxide.\n\nAluminium oxide gives aluminium chloride along with water when it reacts with hydrochloric acid.\n\nReactivity Series of Metals: The order of intensity or reactivity of metal is known as Reactivity Series. Reactivity of elements decreases on moving from top to bottom in the given reactivity series.\nIn the reactivity series, copper, gold, and silver are at the bottom and hence, least reactive. These metals are known as Noble metals. Potassium is at the top of the series and hence, most reactive.\nReactivity of some metals are given in descending order :\nK > Na > Ca > Mg > Al > Zn > Fe > Pb > Cu\n\n4. Reaction of metals with solution of other metal salts: Reaction of metals with the solution of other metal salt is displacement reaction. In this reaction, more reactive metal displaces the less reactive metal from its salt.\nMetal A + Salt of metal B \u2192 Salt of metal A + Metal B\nExamples :", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nMetal A + Salt of metal B \u2192 Salt of metal A + Metal B\nExamples :\nIron displaces copper from copper sulphate solution.\n\nSimilarly, aluminium and zinc displace copper from the solution of copper sulphate.\n\nIn all the above examples, iron, aluminium and zinc are more reactive than copper. This is why they displace copper from its salt solution.\nWhen copper is dipped in the solution of silver nitrate, it displaces silver and forms copper nitrate.\n\nIn the reaction, copper is more reactive than silver and hence, displaces silver from silver nitrate solution.\nSilver metal does not react with copper sulphate solution because silver is less reactive than copper and not able to displace copper from its salt solution.\n\nSimilarly, when gold is dipped in the solution of copper nitrate, no reaction takes place because copper is more reactive than gold.\n\nIn similar way, no reaction takes place when copper is dipped in the solution of aluminium nitrate because copper is less reactive than aluminium.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nIn similar way, no reaction takes place when copper is dipped in the solution of aluminium nitrate because copper is less reactive than aluminium.\n\nNon-Metals: Physical Properties of non-metals, chemical properties of non-metals, non\u00acmetal oxides, Reaction of metal and Non-metal, Ionic bonds and formation of an ionic bond. Non-metals are the elements that do not conduct electricity and are neither malleable nor ductile.\nExamples: Carbon (C), Sulphur (S), Phosphorous (P), Silicon (Si), Hydrogen (H), Oxygen (O), Nitrogen (N), Chlorine (Cl), Bromine (Br), Neon (Ne) and Argon (Ar) etc.\nNon-metals are the elements which form negative ions by gaining an electron. Thus, non\u00acmetals are also known as Electronegative Elements.\nPhysical properties of non-metals\nHardness: Non-metals are not hard rather they are generally soft. But the diamond is an exception; it is the hardest naturally occurring substance.\nState: Non-metals may be solid, liquid or gas.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nHardness: Non-metals are not hard rather they are generally soft. But the diamond is an exception; it is the hardest naturally occurring substance.\nState: Non-metals may be solid, liquid or gas.\nLustre: Non-metals have a dull appearance. Diamond and iodine are exceptions.\nSonority: Non-metals are not sonorous, i.e., they do not produce a typical sound on being hit.\nConduction: Non-metals are a bad conductor of heat and electricity. Graphite which is allotrope of carbon is a good conductor of electricity and is an exception.\nMalleability and ductility: Non-metals are brittle.\nMelting and boiling point: Non-metals have generally low melting and boiling points.\nDensity: Most of the non-metals have low density.\nColour: Non-metals are in many colours.\nCarbon in the form of graphite is non-metal which conduct electricity.\nIodine is non-metal which is lustrous having a shining surface.\nCarbon in the form of diamond is a non-metal which is extremely hard.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nIodine is non-metal which is lustrous having a shining surface.\nCarbon in the form of diamond is a non-metal which is extremely hard.\nDiamond is a non-metal which has a very high melting point and boiling point.\nChemical properties of Non-metals\n1. Reaction of Non-metals with Oxygen: Non-metals form respective oxide when reacting with oxygen.\nNon-metal + Oxygen \u2192 Non-metallic oxide\nWhen carbon reacts with oxygen, carbon dioxide is formed along with the production of heat.\n\nWhen carbon is burnt in an insufficient supply of air, it forms carbon monoxide. Carbon monoxide is a toxic substance. Inhaling of carbon monoxide may prove fatal.\n\nSulphur gives sulphur dioxide when reacting with oxygen. Sulphur catches fire when exposed to air.\n\nWhen hydrogen reacts with oxygen it gives water.\n\nNon-metallic Oxide: Non-metallic oxides are acidic in nature. The solution of non-metal oxides turns blue litmus red.\nCarbon dioxide gives carbonic acid when dissolved in water.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nNon-metallic Oxide: Non-metallic oxides are acidic in nature. The solution of non-metal oxides turns blue litmus red.\nCarbon dioxide gives carbonic acid when dissolved in water.\n\nSulphur dioxide gives sulphurous acid when dissolved in water.\n\nSulphur dioxide gives sulphuric acid when reacts with oxygen.\n\n2. Reaction of Non-metal with Chlorine: Non-metal gives respective chloride when they react with chlorine gas.\nNon-metal + Chlorine \u2192 Non-metal chloride\nHydrogen gives hydrogen chloride and phosphorous gives phosphorous trichloride when reacting with chlorine.\n\n3. Reaction of Non-metals with Hydrogen: Non-metals reactive with hydrogen to form covalent hydrides.\nNon-metal + Hydrogen \u2192 Covalent Hydride\nSulphur combines with hydrogen to form a covalent hydride is called Hydrogen sulphide.\n\nNitrogen combines with hydrogen in presence of an iron catalyst to form covalent hydride ammonia.\n\nNon-metals do not react with water (or steam) to evolve Hydrogen gas.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nNitrogen combines with hydrogen in presence of an iron catalyst to form covalent hydride ammonia.\n\nNon-metals do not react with water (or steam) to evolve Hydrogen gas.\nNon-metals do not react with dilute acids.\n4. Reaction of Metal and Non-metal: Many metals form ionic bonds when they react with non-metals. Compounds so formed are known as Ionic Compounds.\nIons: Positive or negative charged atoms are known as ions. Ions are formed because of loss or gain of electrons. Atoms form ions obtain by the electronic configuration of the nearest noble gas.\nPositive ion: A positive ion is formed because of the loss of electrons by an atom.\nFollowing are some examples of positive ions:\nSodium forms sodium ion because of the loss of one electron. Because of the loss of one electron, one positive charge comes over sodium.\n\nMagnesium forms positive ion because of the loss of two electrons. Two positive charges come over magnesium because of loss of two electrons.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nMagnesium forms positive ion because of the loss of two electrons. Two positive charges come over magnesium because of loss of two electrons.\n\nNegative ion: A negative ion is formed because of the gain of an electron.\nSome examples are given below :\nChlorine gains one electron in order to achieve a stable configuration. After the loss of one electron, chlorine gets one negative charge over it forming chlorine ion.\n\nIonic Bonds: Ionic bonds are formed because of transfer of electrons from metal to non\u00acmetal. In this course, metals get positive charge because of transfer of electrons and non-metal gets negative charge because of acceptance of electrons. In other words, bond formed between positive and negative ion is called Ionic Bond.\nSince, a compound is electrically neutral, so to form an ionic compound, negative and positive both ions must be combined.\nSome examples are given below:", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nSince, a compound is electrically neutral, so to form an ionic compound, negative and positive both ions must be combined.\nSome examples are given below:\nFormation of Sodium Chloride (NaCl): In sodium chloride, sodium is a metal (alkali metal) and chlorine is a non-metal.\nAtomic number of sodium = 11\nElectronic configuration of sodium : 2, 8, 1\nNumber of electrons in outermost orbit = 1\nValence electrons = Electrons in outermost orbit = 1\nAtomic number of chlorine = 17\nElectronic configuration of chlorine : 2, 8, 7\nElectrons in outermost orbit = 7\nTherefore, valence electrons = ?", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nValence electrons = Electrons in outermost orbit = 1\nAtomic number of chlorine = 17\nElectronic configuration of chlorine : 2, 8, 7\nElectrons in outermost orbit = 7\nTherefore, valence electrons = ?\n\nSodium has one valence electron and chlorine has seven valence electrons. Sodium requires losing one electron to obtain stable configuration and chlorine requires gaining one electron in order to obtain stable electronic configuration. Thus, in order to obtain stable configuration, sodium transfers one electron to chlorine. After loss of one electron, sodium gets one positive charge (+) and chlorine gets one negative charge after gain of one electron. Sodium chloride is formed because of transfer of electrons. Thus, ionic bond is formed between sodium and chlorine. Since, sodium chloride is formed because of ionic bond, thus, it is called Ionic compound. In similar way, potassium chloride (KCl) is formed.\nProperties of Ionic compound", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nProperties of Ionic compound\nIonic compounds are solid. Ionic bond has a greater force of attraction because of which ions attract each other strongly. This makes ionic compounds solid.\nIonic compounds are brittle.\nIonic compounds have high melting and boiling points because force of attraction between ions of ionic compounds is very strong.\nIonic compounds generally dissolve in water.\nIonic compounds are generally insoluble in organic solvents; like kerosene, petrol, etc.\nIonic compounds do not conduct electricity in the solid state.\nThe solution of ionic compounds in water conduct electricity. This happens because ions present in the solution of ionic compound facilitate the passage of electricity by moving towards opposite electrodes.\nIonic compounds conduct electricity in the molten state.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nIonic compounds conduct electricity in the molten state.\nOccurrence and Extraction of Metals: Minerals, ores, extraction of metals of least reactivity, extraction of metals of middle reactivity, extraction of metals of high reactivity, refining or purification of metals and corrosion.\nOccurrence and Extraction of Metals:\nSource of metal: Metals occur in Earth\u2019s crust and in seawater; in the form of ores. Earth\u2019s crust is the major source of metal. Seawater contains many salts such as sodium chloride, magnesium chloride, etc.\nMineral: Minerals are naturally occurring substances which have a uniform composition.\nOres: The minerals from which a metal can be profitably extracted are called Ores.\nMetals found at the bottom of reactivity series are least reactive and they are often found in nature in free-state; such as gold, silver, copper, etc. Copper and silver are also found in the form of sulphide and oxide ores.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nMetals found in the middle of reactivity series, such as Zn, Fe, Pb, etc. are usually found in the form of oxides, sulphides or carbonates.\nMetals found at the top of the reactivity series are never found in free-state as they are very reactive, example; K, Na, Ca, Mg and Al, etc.\nMany metals are found in the form of oxides because oxygen is abundant in nature and is very reactive.\nExtraction of Metals: Metals can be categorised into three parts on the basis of their reactivity: Most reactive, medium reactive and least reactive.\nThe three major steps involved in the extraction of a metal from its ore are\nConcentration or enrichment of ores.\nConversion of concentrated ore into crude metal and,\nRefining of impure or crude metal.\n1. Concentration of Ores: Removal of impurities, such as soil, sand, stone, silicates, etc. from mines ore is known as Concentration of Ores.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nRefining of impure or crude metal.\n1. Concentration of Ores: Removal of impurities, such as soil, sand, stone, silicates, etc. from mines ore is known as Concentration of Ores.\nOres which are mined often contain many impurities. These impurities are called gangue. First of all, concentration is done to remove impurities from ores. The concentration of ores is also known as enrichment of ores. Process of concentration depends upon physical and chemical properties of ores. Gravity separation, electromagnetic separation, froth flotation process, etc. are some examples of the processes which are applied for concentration of ores.\n2. Conversion of Concentrated Ore into Crude Metal\nConversion of metals ores into oxides: It is easy to obtain metals from their oxides. So, ores found in the form of sulphide and carbonates are first converted to their oxides by the process of roasting and calcination. Oxides of metals so obtained are converted into metals by the process of reduction.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nRoasting: Heating of sulphide ores in the presence of excess air to convert them into oxides is known as Roasting.\n\nCalcination: Heating of carbonate ores in the limited supply of air to convert them into oxides is known as Calcination.\n\n3. Reduction: Heating of oxides of metals to turn them into metal is known as Reduction.\n(i) Extraction of Metals of Least Reactivity: Mercury and copper, which belong to the least reactivity series, are often found in the form of their sulphide ores. Cinnabar (HgS) is the ore of mercury. Copper glance (Cu2S) is the ore of copper.\nExtraction of Mercury Metal: Cinnabar (HgS) is first heated in air. This turns HgS (mercury sulphide or cinnabar) into HgO (mercury oxide) by liberation of sulphur dioxide. Mercury oxide so obtained is again heated strongly. This reduces mercury oxide to mercury metal.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nExtraction of Copper Metal: Copper glance (Cu2S) is roasted in the presence of air. Roasting turns copper glance (ore of copper) into copper (l) oxide. Copper oxide is then heated in the absence of air. This reduces copper (l) oxide into copper metal.\n\n(ii) Extraction of Metals of Middle Reactivity: Iron, zinc, lead, etc. are found in the form of carbonate or sulphide ores. Carbonate or sulphide ores of metals are first converted into respective oxides and then oxides are reduced to respective metals.\nExtraction of Zinc: Zinc blende (ZnS: zinc sulphide) and smithsonite or zinc spar or calamine (ZnCO3: zinc carbonate) are ores of zinc. Zinc blende is roasted to be converted into zinc oxide. Zinc spar is put under calcination to be converted into zinc oxide.\n\nZinc oxide so obtained is reduced to zinc metal by heating with carbon (a reducing agent).\n\nExtraction of Iron from Haematite (Fe2O3): Haematite ore is heated with carbon to be reduced to iron metal.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nExtraction of Iron from Haematite (Fe2O3): Haematite ore is heated with carbon to be reduced to iron metal.\n\nExtraction of Lead from Lead oxide: Lead oxide is heated with carbon to be reduced to lead metal.\n\nReduction of Metal oxide by Heating with Aluminium: Metal oxides are heated with aluminium (a reducing agent) to be reduced to metal. Following is an example: Manganese dioxide and copper oxide are reduced to respective metals when heated with aluminium.\n\nThermite Reaction: Ferric oxide; when heated with aluminium; is reduced to iron metal. In this reaction, a lot of heat is produced. The thermite reaction is used in the welding of electric conductors, iron joints, etc. such as joints in railway tracks. This is also known as Thermite Welding (TW).", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\n(iii) Extraction of Metals of High Reactivity: Metals of high reactivity; such as sodium, calcium, magnesium, aluminium, etc. are extracted from their ores by electrolytic reduction. These metals cannot be reduced using carbon because carbon is less reactive than them.\nElectrolytic Reduction: Electric current is passed through the molten state of metal ores. Metal being positively charged is deposited over the cathode.\nExample: When an electric current is passed through molten state or solution of sodium chloride, sodium metal gets deposited over the cathode.\n\nMetals obtained from the process of electrolytic reduction are pure in form.\n4. Refining or purification of metals: Metals extracted from various methods contains some impurities, thus, they are required to be refined. Most of the metals are refined using electrolytic refining.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nElectrolytic Refining: In the process of electrolytic refining, a lump of impure metal and a thin strip of pure metal are dipped in the salt solution of metal to be refined. When an electric current is passed through the solution, pure metal is deposited over a thin strip of pure metal\nfrom a lump of impure metal. In this, impure metal is used as anode and pure metal is used as a cathode.\nElectrolytic Refining of Copper: A lump of impure copper metal and a thin strip of pure copper are dipped in the solution of copper sulphate. Impure lump of metal is connected with the positive pole and thin strip of pure metal is connected with negative pole. When electric current is passed through the solution, pure metal from anode moves towards cathode and is deposited over it. Impurities present in metal are settled near the bottom of anode in the solution. Settled impurities in the solution are called Anode Mud.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\n5. Corrosion: Most of the metals keep on reacting with the atmospheric air. This leads to the formation of a layer over the metal. In the long run, the underlying layer of metal keeps on getting lost due to conversion into oxides or sulphides or carbonate, etc. As a result, the metal gets eaten up. The process is called Corrosion.\nRusting of Iron: Rusting of iron is the most common form of corrosion. When iron articles like the gate, grill, fencing, etc. come in contact with moisture present in the air, the upper layer of iron turns into iron oxide. Iron oxide is brown-red in colour and is known as Rust. The phenomenon is called Rusting of Iron.\nIf rusting is not prevented in time, the whole iron article would turn into iron oxide. This is also known as Corrosion of Iron. Rusting of iron gives a huge loss every year.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nIf rusting is not prevented in time, the whole iron article would turn into iron oxide. This is also known as Corrosion of Iron. Rusting of iron gives a huge loss every year.\nPrevention of Rusting: For rusting, iron must come in contact with oxygen and water. Rusting is prevented by preventing the reaction between atmospheric moisture and the iron article. This can be done by:\nPainting\nGreasing\nGalvanization\nElectroplating\nAlloying\n6. Alloys: The homogeneous mixture of two or more metals, or a metal and a non-metal is called Alloy.\nTypes of alloys :\nFerrous alloys: An alloy in which iron (Fe) is present. For example : manganese steel (Fe = 86% ; Mn = 13% ; C = 1%) and Nickle steel (Fe = 98% ; Ni = 2%).\nNon-ferrous alloys: An alloy does not contain iron. For example : Brass (Cu = 80% ; Zn = 20%), and Bronze (Cu = 90% ; Sn = 10%).", + "Non-ferrous alloys: An alloy does not contain iron. For example : Brass (Cu = 80% ; Zn = 20%), and Bronze (Cu = 90% ; Sn = 10%).\nAmalgams: An alloy in which mercury (Hg) is present. For example Sodium amalgams [Na(Hg)] and Zinc amalgams [Zn(Hg)].\nProperties of an Alloy", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nAmalgams: An alloy in which mercury (Hg) is present. For example Sodium amalgams [Na(Hg)] and Zinc amalgams [Zn(Hg)].\nProperties of an Alloy\nAlloys are stronger than the metal from which they are obtained.\nIt is harder than the constituent metals.\nMore resistance to corrosion.\nThe melting point of alloys is lower than the constituent metals.\nExample: Solder [Sn(80%) + Pb(50%)] has lower m. p. than Pb and Sn.\nThe electrical conductivity of alloys is lower than the constituent metals.", + "It is harder than the constituent metals.\nMore resistance to corrosion.\nThe melting point of alloys is lower than the constituent metals.\nExample: Solder [Sn(80%) + Pb(50%)] has lower m. p. than Pb and Sn.\nThe electrical conductivity of alloys is lower than the constituent metals.\nSome examples of Alloys:\nBrass: [80% Cu + 20% Zn ]\nBronze: [90% Cu + 20% Sn]\nSolder: [50% Pb + 50% Sn]\nDuralumin: [95% Al + 4% Cu + 0.5% Mg + 0.5 Mn]\nSteel: [99.95% Fe + 0.05% C]\nStainless steel: [74% Fe + 18% Cr + 8% Ni]\nMagnesium: [95% Al + 5% Mg]\nGerman Silver: [60% Cu + 20% Zn + 20% Ni]\nAlloys of Gold: Pure gold is said to be of 24 carats. Gold is alloyed with a small amount of silver or copper to make it hard.", + "Gold is alloyed with a small amount of silver or copper to make it hard.\nMetals and Non-metals:\nChemical Properties of Metals and Non-metals.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nAlloys of Gold: Pure gold is said to be of 24 carats. Gold is alloyed with a small amount of silver or copper to make it hard.\nMetals and Non-metals:\nChemical Properties of Metals and Non-metals.\nThe reaction of metals with oxygen. Metals form their oxides when reacting with oxygen.\nMetal + Oxygen \u2192 Metal oxide\nMetal oxides are basic in nature. Example, Reaction of Iron metal with oxygen When iron reacts with moist air, it forms rust.\nRust is iron oxide. Articles made of iron, such as grills, fencing, etc. are getting rusted because of reaction with moist air.\nIron (Fe) + Water (H2O) + Oxygen (O2) \u2192 Fe3O4n.H2O (Iron II, III) Oxide (Rust)\nRust is reddish brown in colour and is iron oxide. Iron oxide is basic in nature. It turns red litmus blue.\nRusting of iron can be prevented:\nby galvanizing the iron articles with zinc coating.\nby painting and applying grease on the articles.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nRusting of iron can be prevented:\nby galvanizing the iron articles with zinc coating.\nby painting and applying grease on the articles.\nThe reaction of Magnesium metal with oxygen: When magnesium is burnt in air, it forms magnesium oxide. Burning in the air means reaction with oxygen.\nMagnesium + Oxygen (O2) \u2192 MgO (Magnesium oxide)\nMagnesium oxide forms magnesium hydroxide with water. The solution of Magnesium oxide turns red litmus paper blue. This means magnesium oxide is basic in nature.\nMgO + H2O \u2192 Mg(OH)2 (Magnesium Hydroxide)\nThe reaction of Non-metals with oxygen: Non-metals form their oxides when they react with oxygen.\nNon-metal + Oxygen \u2192 Non-metal oxide\nNon-metal oxides are acidic in nature.\nExample., Reaction of sulphur with oxygen.\nWhen sulphur is burnt in air, it forms sulphur dioxide.\nSulphur + Oxygen (O2) \u2192 SO2 (Sulphur dioxide)", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nNon-metal oxides are acidic in nature.\nExample., Reaction of sulphur with oxygen.\nWhen sulphur is burnt in air, it forms sulphur dioxide.\nSulphur + Oxygen (O2) \u2192 SO2 (Sulphur dioxide)\nThe solution of sulphur dioxide turns blue litmus paper red. Sulphur dioxide forms sulphurous acid when dissolved in water. Thus, sulphur dioxide is acidic in nature.\nSO2 + H2O \u2192 Sulphurous acid (H2SO3)\nThe reaction of carbon with oxygen\u2014When carbon is burnt in air, it forms carbon dioxide.\nCarbon + Oxygen (O2) \u2192 CO2 (Carbon dioxide)\nYou can observe that when coal (carbon) is burnt it forms smoke, which contains carbon dioxide. Carbon dioxide is acidic in nature. The solution of carbon dioxide in water turns blue litmus paper red.\nCO2 + H2O \u2192 Carbonic acid (H2CO3)\nThe reaction of Metals and Non-metals with water: Generally, metals form respective hydroxides when they react with water.\nMetal + Water \u2192 Metal hydroxide", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nCO2 + H2O \u2192 Carbonic acid (H2CO3)\nThe reaction of Metals and Non-metals with water: Generally, metals form respective hydroxides when they react with water.\nMetal + Water \u2192 Metal hydroxide\nThe reaction of sodium metal with water: Sodium metal vigorously reacts with water and forms sodium hydroxide along with a lot of heat.\nNa + H2O \u2192 NaOH (Sodium hydroxide) + H2 (Hydrogen) + Heat\nNon-metals generally do not react with water. Rather some non-metals which react with air vigorously are stored in water. The reaction of metals and non-metals with dilute acid. Metals give hydrogen gas when they react with dilute acid.\nMetal + Acid \u2192 Hydrogen gas + Salt\nThe reaction of zinc with dilute acid. Zinc gives hydrogen gas along with zinc chloride when it reacts with hydrochloric acid. Similarly, zinc gives hydrogen gas along with zinc sulphate when it reacts with sulphuric acid. This method is used to produce hydrogen gas in the laboratory.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nZn + H2SO4 (Sulphuric acid) \u2192 ZnSO4 (Zinc sulphate) + H2 (Hydrogen)\nThe reaction of Aluminium with dilute acid. Aluminium gives hydrogen gas along with aluminium chloride when it reacts with dilute hydrochloric acid.\n2Al + 6HCl (Hydrochloric acid) \u2192 2AlCl3 (Aluminium Chloride) + 3H2 (Hydrogen)\nCopper does not react with dilute sulphuric acid even on heating, but it reacts with concentrated sulphuric acid. Copper, silver and gold are considered as noble metals as do not react with dilute acid.\nGenerally, non-metals do not react with dilute acid.\nThe reaction of metals and non-metals with the base. Metals give hydrogen gas when they react with a base.\nMetal + Base \u2192 Hydrogen gas + Salt\nThe reaction of aluminium metal with sodium hydroxide.\nAl + NaOH (Sodium hydroxide) \u2192 NaAlO2 (Sodium aluminate) + H2 (Hydrogen)", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nMetal + Base \u2192 Hydrogen gas + Salt\nThe reaction of aluminium metal with sodium hydroxide.\nAl + NaOH (Sodium hydroxide) \u2192 NaAlO2 (Sodium aluminate) + H2 (Hydrogen)\nAluminium metal forms hydrogen gas and sodium aluminate when it reacts with sodium hydroxide. Similarly, zinc gives sodium zincate and hydrogen gas when it reacts with sodium hydroxide.\nDisplacement Reaction: When a more reactive metal reacts with the salt solution of less reactive metal, more reactive metal displaces the less reactive metal from its solution.\nMetal A + Salt Solution of metal B \u2192 Salt Solution of metal A + metal B\nIn the above equation, metal A is more reactive than metal B.\nExample., When aluminium metal is dipped in the solution of copper sulphate, it forms aluminium sulphate and copper.\nAl + CuSO4 (Copper sulphate) \u2192 Al2(SO4)3 (Aluminium Sulphate) + Cu (Copper)\nIn the above reaction, aluminium is more reactive than copper, that is why it replaces copper from the solution of copper sulphate.", + "CHAPTER 3- METALS AND NON METALS- NOTES.txt\nIn the above reaction, aluminium is more reactive than copper, that is why it replaces copper from the solution of copper sulphate.\nWhen copper metal is dipped in the solution of aluminium nitrate, no reaction takes place. Because copper is less reactive than aluminium.\n\nRoasting and Calcination:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- EXTRA QUESTIONS.txt\nExtra Questions Carbon and its Compounds \u2013 CBSE Class 10 Science\nAccording to new CBSE Exam Pattern, MCQ Questions for Class 10 Science pdf Carries 20 Marks.\nTopics and Subtopics in NCERT Solutions for Class 10 Science Chapter 4 Carbon and its Compounds:\nQuestion-1\nA piece of black electrode used in dry cell on strong heating in air gave a colourless gas which turned lime -water milky. What was the material of the electrode?\nSolution:\nWe know that graphite is used for making the electrodes. So, the piece of black electrode used in the dry cell is made of graphite (which is an allotrope of carbon element). This is confirmed by the fact that the piece of electrode, on strong heating in air, gave a colourless gas carbon dioxide which turned lime- water milky. Thus, the material of the electrode is graphite.\nQuestion-2\nWhy does graphite conducts electricity, but not diamond?\nSolution:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- EXTRA QUESTIONS.txt\nQuestion-2\nWhy does graphite conducts electricity, but not diamond?\nSolution:\nIn case of diamond, each carbon atom of a single crystal is surrounded by four other carbon atoms by covalent bonds such that they form four corners of a regular tetrahedron. Because of four covalent bonds with each carbon atoms there are no free electrons available. Due to the non-availability of free electrons within crystalline structure, diamond acts as a bad conductor of electricity.\nIn case of graphite, every carbon atom in a single crystal is covalently bonded to three carbon atoms. As each carbon atoms has four valence electrons one valence electron is left free for each carbon atom. These free electrons can be easily made to flow within the crystalline structure of graphite by applying electric potential. Thus, graphite is a good conductor of electricity.\nQuestion-3\nWrite three important uses of ethanol.\nSolution:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- EXTRA QUESTIONS.txt\nQuestion-3\nWrite three important uses of ethanol.\nSolution:\nEthanol is used as a solvent to dissolve varnishes, medicines and other organic compounds\nIt is used as beverage (for drinking as an intoxicant) in different forms, viz; Brandy, Whisky etc.,\n\nIt is used for industrial uses in the name of denatured spirit.\nQuestion-4\nState what you will observe when sugar crystals is heated strongly.\nState what you will observe when sugar crystals is treated with conc. Sulphuric acid.\nSolution:\nThe sugar crystal will initially melt. Gradually, they turn brown and start swelling up. They give off large amount of steam. Finally black porous residue of carbon is left behind.\nThe sugar crystals will initially turn brown. Lot of frothing takes place with the evolution of large amount of heat and steam is given off. Finally a black porous residue of carbon is left behind.\nQuestion-5\nHow are the molecules of aldehyde and Ketone structurally different?\nSolution:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- EXTRA QUESTIONS.txt\nQuestion-5\nHow are the molecules of aldehyde and Ketone structurally different?\nSolution:\nIn aldehyde, the carbon atom of the carbonyl group is attached to one alkyl group (R) and one hydrogen atom but in ketone, the carbonyl group is attached to two alkyl groups.\nQuestion-6\nWhat change has been made in the composition of detergents to make them biodegradable?\nSolution:\nDetergents made from long chain hydrocarbons having the minimum branching in their molecules are degraded more easily.\nQuestion-7\nA hydrocarbon molecule contains 4 hydrogen atoms. Give its molecular formula, if it is an: (i) alkane, (ii) alkene (iii) alkyne.\nSolution:\n(i) An alkane containing 4 hydrogen atoms in its molecule is methane, CH4.\n(ii) An alkene containing 4 hydrogen atoms in its molecule is ethane, C2H4\n\n(iii) An alkyne containing 4 hydrogen atoms in its molecule is propyne, C3H4.\nQuestion-8\nWhy common salt is added in soap making?\nSolution:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- EXTRA QUESTIONS.txt\n(iii) An alkyne containing 4 hydrogen atoms in its molecule is propyne, C3H4.\nQuestion-8\nWhy common salt is added in soap making?\nSolution:\nCommon salt is added to the mixture to make the soap come out of solution. Though most of the soap separates out on its own but some of it remains in solution. Common salt is added to precipitate out all the soap from the aqueous solution. Actually, when we add common salt to the solution, then the solubility of soap present in it decreases, due to which all the soap separates out from the solution in the form of a solid.\nQuestion-9\nWhat is meant by denatured alcohol? What is the need to denature alcohol?\nSolution:\nThe alcohol which is rendered unfit by mixing it with some poisonous substances, such as methanol, pyridine, copper sulphate, etc is known as denatured alcohol. Ethanol is an important industrial chemical. Therefore, it subjected to very small excise duty. To prevent its misuse for drinking purpose, there is a need to denature alcohol.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- EXTRA QUESTIONS.txt\nQuestion-10\nWhat is meant by the term \u201cfunctional group\u201d?\nSolution:\nA functional group in an organic compound is an atom or a group of atoms binded together in a unique fashion, which is usually the site of chemical reactivity in an organic molecule.\nFiled Under: CBSE", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- EXPLEMPLAR QUESTIONS.txt\nNCERT Exemplar Class 10 Science Chapter 4 Carbon and its Compounds\nShort Answer Questions\nQuestion.1 Draw the electron dot structure of ethyne and also draw its structural formula.\nAnswer.\n\nQuestion.2 Write the names of the following compounds.\n\nAnswer.\n(a) Pentanoic acid (b) 1-Butyne\n(c) Heptanal (d) 1-Pentanol\nQuestion.3 Identify and name the functional groups present in the following compounds.\n\nAnswer.\n\nQuestion.4 A compound X is formed by the reaction of a carboxylic acid C2H4O2 and an alcohol in presence of a few drops of H2SO4. The alcohol on oxidation with alkaline KMnO4 followed by acidification gives the same carboxylic acid as used in this reaction. Give the names and structures of\n(a) carboxylic acid, (6) alcohol and (c) the compound X. Also write the\nreaction.\nAnswer.\n\nQuestion.5 Why detergents are better cleansing agents than soaps? Explain.\nAnswer.", + "Give the names and structures of\n(a) carboxylic acid, (6) alcohol and (c) the compound X. Also write the\nreaction.\nAnswer.\n\nQuestion.5 Why detergents are better cleansing agents than soaps? Explain.\nAnswer. It is because detergents form lot of lather even with hard water.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- EXPLEMPLAR QUESTIONS.txt\nreaction.\nAnswer.\n\nQuestion.5 Why detergents are better cleansing agents than soaps? Explain.\nAnswer. It is because detergents form lot of lather even with hard water.\nHard water contains Ca2+ and Mg2+ ions which react with soap to form insoluble salts of calcium and magnesium called scum and soap goes waste. Detergents do not form insoluble compounds with Ca2+ and Mg2+ ions, therefore, these are more effective.\nQuestion.6 Intake of small quantity of methanol can be lethal. Comment.\nAnswer. Methanol is oxidised to methanal in liver.\u2019 Methanal is highly reactive and good reducing agent. It causes protoplasm to coagulate. It also affects optic nerve and leads to blindness.\nQuestion.7 A gas is evolved when ethanol reacts with sodium. Name the gas evolved and also write the balanced chemical equation of the reaction involved.\nAnswer. Hydrogen gas is evolved.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- EXPLEMPLAR QUESTIONS.txt\nQuestion.7 A gas is evolved when ethanol reacts with sodium. Name the gas evolved and also write the balanced chemical equation of the reaction involved.\nAnswer. Hydrogen gas is evolved.\n\nQuestion.8 Catenation is the ability of an atom to form bonds with other atoms of the same element. It is exhibited by both carbon and silicon. Compare the ability of catenation of the two elements. Give reasons.\nAnswer. Carbon shows catenation to large extent as compared to silicon as well as any other element due to smaller size of carbon. C\u2014C bond is stronger than Si-Si bond because Si is larger in size, forms wea\u2019ker bond.\nQuestion.9 Match the reactions given in Column (A) with the names given in Column (B).\n\nAnswer.\n(a) \u2013 (iv) is esterification reaction because ester is being formed from carboxylic acid and alcohol.\n(b) \u2013 (i) is addition reaction as hydrogen is being added.\n\n(c) \u2013 (ii) is substitution reaction because hydrogen of methane is being substituted by\nchlorine atom.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- EXPLEMPLAR QUESTIONS.txt\n(b) \u2013 (i) is addition reaction as hydrogen is being added.\n\n(c) \u2013 (ii) is substitution reaction because hydrogen of methane is being substituted by\nchlorine atom.\n(d) \u2013 (iii) is neutralisation reaction because acetic acid reacts with sodium hydroxide to form salt and water.\nQuestion.10 What is the role of metal or reagents written on arrows in the given chemical reactions?\n\nAnswer.\n(a) Ni is used as a catalyst.\n(b) Cone. H2SO4 acts as a dehydrating agent.\n(c) Alkaline KMnO4 is an oxidising agent.\nLong Answer Questions\nQuestion.11 (a) Write the formula and draw electron dot structure of carbon tetrachloride.\n(b) What is saponification? Write the reaction involved in this process.\nAnswer.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- EXPLEMPLAR QUESTIONS.txt\nLong Answer Questions\nQuestion.11 (a) Write the formula and draw electron dot structure of carbon tetrachloride.\n(b) What is saponification? Write the reaction involved in this process.\nAnswer.\n\nQuestion.12 A compound \u2018C\u2019 (molecular formula, C2H4O2) reacts with Na-metal to form a compound \u2018R\u2019 and evolves a gas which burns with a pop sound. Compound \u2018C\u2019 on treatment with an alcohol \u2018A\u2019 in presence of an acid forms a sweet smelling compound \u2018S\u2019 (molecular formulaC3H6O2). On addition of NaOH to \u2018C\u2019, it also gives \u2018R\u2019 and water. \u2018S\u2019 on treatment with NaOH solution gives back \u2018R\u2019 and \u2018A\u2019.\nIdentify \u2018C\u2019, \u2018R\u2019, \u2018A\u2019, \u2018S\u2019 and write down the reactions involved.\nAnswer.\n\nQuestion.13 How would you bring about the following conversions? Name the process and write the reaction involved.\n(a) Ethanol to ethene\n(b) Propanol to propanoic acid\nAnswer.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nCarbon and its Compounds Notes CBSE Class 10 Science Chapter 4\nBonding in Carbon: The Covalent bond, Electron dot structure, Physical properties of organic compounds, Allotropes of Carbon.\nCovalent Bond: The atomic number of carbon is 6. Its electronic configuration is 2, 4. It requires, 4 electrons to achieve the inert gas electronic configuration. But carbon cannot form an ionic bond\nIt could gain four electrons forming C4- cation. But it would be difficult for the nucleus with six protons to hold on to ten electrons.\nIt could lose four electrons forming C4+ cations. But it requires a large amount of energy to remove four electrons.\nThus, carbon overcomes this problem by sharing of its valence electrons with other carbon atoms or with atoms of other elements.\nThe bond formed by mutual sharing of electron pairs between two atoms in a molecule is known as Covalent Bond.\nTypes of Covalent Bond:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nThe bond formed by mutual sharing of electron pairs between two atoms in a molecule is known as Covalent Bond.\nTypes of Covalent Bond:\nSingle Covalent Bond: When a single pair of electrons are shared between two atoms in a molecule. For example; F2, Cl2, H2 etc.\nDouble Covalent Bond: When two pairs of electrons are shared between two atoms in a molecule. For example; O2, CO2 etc.\nTriple Covalent Bond: When three pairs of electrons are shared between two atoms in a molecule. For example; N2 etc.\nElectron Dot Structure: The electron dot structures provides a picture of bonding in molecules in terms of the shared pairs of electrons and octet rule.", + "Triple Covalent Bond: When three pairs of electrons are shared between two atoms in a molecule. For example; N2 etc.\nElectron Dot Structure: The electron dot structures provides a picture of bonding in molecules in terms of the shared pairs of electrons and octet rule.\nFormation of Hydrogen Molecule\nAtomic number of Hydrogen = 1\nNumber of valence electrons = 1\n\nFormation of CH4 Molecule\nAtomic number of Carbon = 6 [2, 4]\nNumber of valence electrons = 4\nAtomic number of Hydrogen = 1\nNumber of valence electrons = 1\n\nFormation of CO2 Molecule\nAtomic number of Carbon = 6 [2, 4]\nNumber of valence electrons = 4", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nNumber of valence electrons = 4\nAtomic number of Hydrogen = 1\nNumber of valence electrons = 1\n\nFormation of CO2 Molecule\nAtomic number of Carbon = 6 [2, 4]\nNumber of valence electrons = 4\nAtomic number of Oxygen = 8 [2, 6]\nNumber of valence electrons = 6\n\nFormation of H2S Molecule\nAtomic number of Sulphur = 16 [2, 8, 6]\nNumber of valence electrons = 6\n\nPhysical Properties of Organic Compounds\nMost of the organic compounds have low boiling and melting point, due to the weak force of attraction (i.e., the inter-molecular force of attraction) between these molecules.\nMost carbon compounds are poor conductors of electricity, due to the absence of free electrons and free ions.\nAllotropes of Carbon\nAllotropy: The phenomenon in which the element exists in two or more different physical states with similar chemical properties are called Allotropy.\nCarbon has Three Main Allotropes", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nAllotropy: The phenomenon in which the element exists in two or more different physical states with similar chemical properties are called Allotropy.\nCarbon has Three Main Allotropes\nDiamond: In this, carbon, an atom is bonded to four other atoms of carbon forming three-dimensional structures. It is the hardest substance and an insulator. It is used for drilling rocks and cutting. It is also used for making jewellery.\nGraphite: In this, each carbon atom is bonded to three other carbon atoms. It is a good conductor of electricity and used as a lubricant.\nBuckminster Fullerene: It is an allotrope of the carbon-containing cluster of 60 carbon atoms joined together to form spherical molecules. It is dark solid at room temperature.\nVersatile nature of Carbon, Hydrocarbons, Isomerism, Homologous series, Functional groups, Nomenclature of functional groups.\nVersatile Nature of Carbon: The existence of such a large number of organic compounds is due to the following nature of carbon,", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nVersatile Nature of Carbon: The existence of such a large number of organic compounds is due to the following nature of carbon,\nCatenation\nTetravalent nature.\n(i) Catenation: The self linking property of an element mainly carbon atom through covalent bonds to form long straight, branched and rings of different sizes are called Catenation.\nThis property is due to\nThe small size of the carbon atom.\nThe great strength of the carbon-carbon bond.\nCarbon can also form stable multiple bonds (double or triple) with itself and with the atoms of other elements.\nStraight Chain\n\nBranched Chain\n\nRings\n\n(ii) Tetravalent Nature: Carbon has valency of four. It is capable of bonding with four other atoms of carbon or some other heteroatoms with single covalent bond as well as double or triple bond.\n\nHydrocarbons: Compounds of carbon and hydrogen are known as hydrocarbons.\nFor example; Methane (CH4), Ethane (C2H6), Ethene (C2H4), Ethyne (C2H2) etc.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nHydrocarbons: Compounds of carbon and hydrogen are known as hydrocarbons.\nFor example; Methane (CH4), Ethane (C2H6), Ethene (C2H4), Ethyne (C2H2) etc.\n\nSaturated Hydrocarbon (Alkanes): General formula is CnH2n+2.\nn = number of carbon atoms.\nIn this, the carbon atoms are connected by only a single bond.\nFor example; Methane (CH4), Ethane (C2H6) etc.\n\nUnsaturated Hydrocarbons\nAlkenes: General formula is CnH2n, where n = number of carbon atoms.\nIn this, the two carbon atoms are connected by double bond.\n\nAlkynes: General formula is CnH2n-2, where n = number of carbon atoms. In this, the two carbon atoms are connected by triple bond.\n\nElectron Dot Structure of Hydrocarbons\n\nIsomerism: Compounds having the same molecular formula but different structural formula and properties are known as Isomers and this phenomenon is known as Isomerism.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nIsomerism: Compounds having the same molecular formula but different structural formula and properties are known as Isomers and this phenomenon is known as Isomerism.\nStructural Isomerism: Compounds having the same molecular formula but different structures are called Structural isomers. Example: Isomers of butane (C4H10)\n\n\nHomologous Series: Series of organic compounds having the same functional group and chemical properties and successive members differ by a CH2 unit or 14 mass units are known as Homologous series.\nHomologous series of Alkanes, Alkenes and Alkynes\n\nCharacteristic of Homologous Series\nThe successive members in homologous series differ by CH2 unit or 14 mass unit.\nMembers of given homologous series have the same functional group.\nAll the members of homologous series shows similar chemical properties.\nFunctional Group: An atom or group of atoms present in a molecule which largely determines its chemical properties are called Functional Group.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nFunctional Group: An atom or group of atoms present in a molecule which largely determines its chemical properties are called Functional Group.\n\n\nNomenclature of Organic Compounds: It is difficult to remember millions of compounds by their individual common name. Thus, to systematize the nomenclature of organic compounds IUPAC (International Union of Pure and Applied Chemistry) has given certain rule which is as follows:\n1. Identify the Number of Carbon Atoms in the Compound\n2. Identify the functional group\n3. Name the Compounds By Following Order\nPrefix + Word Root + Suffix\n\n\nChemical Properties of Carbon Compounds: The important chemical properties are as follows:\n1. Combustion: The complete combustion of carbon compounds in the air gives carbon dioxide water, heat and light.\nCH3CH2OH(l) + O2(g) \u2192 CO2(g) + H2O(l) + Heat and light\nCarbon burns in air or oxygen to give carbon dioxide and heat and light.\nC(s) + O2(g) \u2192 CO2(g) + Heat and light", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nCH3CH2OH(l) + O2(g) \u2192 CO2(g) + H2O(l) + Heat and light\nCarbon burns in air or oxygen to give carbon dioxide and heat and light.\nC(s) + O2(g) \u2192 CO2(g) + Heat and light\nSaturated hydrocarbons burn with a blue flame in the presence of a sufficient supply of air or oxygen.\nCH4(g) + 2O2(g) \u2192 CO2(g) + 2H2O(l) + Heat and light\nIn presence of limited supply of air, saturated hydrocarbon forms a sooty flame.\nUnsaturated hydrocarbons burn with a yellow smoky flame.\nThe gas and kerosene stove used at home has inlet for air so that, burnt to given clean blue flame.\nDue to presence of small amount of nitrogen and sulphur, coal and petroleum produces carbon dioxide with oxides of nitrogen and sulphur which are major pollutant.\n2. Oxidation: Oxidation of ethanol in presence of oxidizing agents gives ethanoic acid.", + "Due to presence of small amount of nitrogen and sulphur, coal and petroleum produces carbon dioxide with oxides of nitrogen and sulphur which are major pollutant.\n2. Oxidation: Oxidation of ethanol in presence of oxidizing agents gives ethanoic acid.\n\nOxidizing Agent: Some substances are capable of adding oxygen to others, are known as Oxidising Agent.\nExample: Alkaline KMnO4 (or KMnO4\u2014KOH)\nAcidified K2Cr2O7 (or K2Cr2O7\u2014H2SO4)", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nOxidizing Agent: Some substances are capable of adding oxygen to others, are known as Oxidising Agent.\nExample: Alkaline KMnO4 (or KMnO4\u2014KOH)\nAcidified K2Cr2O7 (or K2Cr2O7\u2014H2SO4)\nKMnO4 \u2013 Potassium permanganate\nK2Cr2O7 \u2013 Potassium dichromate\n3. Addition Reaction: Addition of dihydrogen with unsaturated hydrocarbon in the presence of catalysts such as nickel or platinum or palladium are known as Hydrogenation (addition) reaction.\n\nCatalyst: Substances that cause a reaction to occur or proceeds to different rate without consuming in it are called a catalyst. For example; Ni, Pt, Pd, etc.\nProcess of converting vegetable oil into solid fat (vegetable ghee) is called Hydrogenation of Oil.\nVegetable oil + H2 \nUndefined control sequence \\xrightarrow\n Vegetable ghee\nVegetable fats are saturated fats which are harmful for health.\nVegetable oil containing unsaturated fatty acids are good for health.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nUndefined control sequence \\xrightarrow\n Vegetable ghee\nVegetable fats are saturated fats which are harmful for health.\nVegetable oil containing unsaturated fatty acids are good for health.\n4. Substitution Reaction: Replacement of one or more hydrogen atom of an organic molecule by another atom or group of the atom is known as Substitution Reaction.\n\nSome Important Carbon Compounds :\nEthanol (CH3CH2\u2014OH): Commonly known as Ethyl Alcohol.\nPhysical Properties\nIt is colourless, inflammable liquid.\nIt is miscible with water in all proportions.\nIt has no effect on the litmus paper.\nChemical Properties\nReaction with sodium\n\nReaction with concentrated H2SO4 (Dehydration Reaction)\n\nDehydrating agent: Substances which removes water from ethanol (alcohols) is known as Dehydrating agent. For example; Cone. H2SO4.\nUses: As solvent, as antiseptic (tincture iodine), as anti-freeze in automobiles.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nUses: As solvent, as antiseptic (tincture iodine), as anti-freeze in automobiles.\nEthanoic Acid (CH3COOH): Commonly known as Acetic acid. 5-8% of ethanoic acid in water is called Vinegar. The melting point of pure ethanoic acid is 290 K and hence, it often freezes in cold climate so named as glacial acetic acid.\nPhysical Properties\nIt is a colourless, pungent-smelling liquid.\nMiscible with water in all proportions.\nTurns blue litmus to red.\nChemical Properties\n(i) Esterification Reaction: Reaction of ethanoic acid with an alcohol in the presence of a few drops of conc. H2SO4 as catalyst gives a sweet-smelling substance known as Esters, called Esterification reaction.\n\nEsters are used in making perfumes and flavouring agents.\nSaponification Reaction: Reaction of esters with sodium hydroxide, gives alcohol and sodium salt of carboxylic acid (soap). This reaction is known as Saponification Reaction.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nSaponification Reaction: Reaction of esters with sodium hydroxide, gives alcohol and sodium salt of carboxylic acid (soap). This reaction is known as Saponification Reaction.\n\n(ii) Reaction with Carbonates and Hydrogen Carbonates: Ethanoic acid reacts with sodium carbonates and sodium hydrogen carbonates to give rise to a salt, carbon dioxide and water.\n\nUsed as vinegar.\nUsed as raw material for the preparation of acetyl chloride and esters.\nSoap: Sodium or potassium salts of long chain fatty acids is called Soap.\nGeneral formula: RCOO\u2013Na+\nDetergent: Ammonium and sulphonate salts of long chain fatty acids are called Detergent.\nExample: CH3\u2014(CH2)11\u2014C6H4\u2014SO3Na.\nHard and Soft Water: Water that does not produce lather with soap readily is called Hard water and which produces lather with soap is called Soft Water.\nHardness of water is due to the presence of bicarbonates, chlorides and sulphate salt of calcium and magnesium.\nDifference between soaps and detergents", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nHardness of water is due to the presence of bicarbonates, chlorides and sulphate salt of calcium and magnesium.\nDifference between soaps and detergents\nAdvantage of Detergents: The main advantage of detergent over soaps is that soaps cannot be used in hard water for washing because hard water reacts with soap to form curdy white precipitate called Scum.\n\nThus, in hard water, soap does not give lather while detergent does.\nCleansing Action of Soaps and Detergents: Both soaps and detergents cantains two parts. A long hydrocarbon part which is hydrophobic (water repelling) in nature and a short ionic part which is hydrophillic (water attracting) in nature.\nThe hydrocarbon part of the soap molecule links itself to the oily (dirt) drop and ionic end orients itself towards water and forms a spherical structure called micelles. The soap micelles helps in dissolving the dirt in water and wash our clothes.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nCovalent Bond: A chemical bond formed between two atoms by sharing of valence electrons between two atoms so that each atom acquires the stable electronic configuration of the nearest noble gas.\nCovalency: The number of electrons contributed by each atom for sharing.\nCarbon always forms a covalent bond: Atomic no of carbon is 6. So, its configuration is K-2, L-4. Therefore, it should either lose or gain 4 electrons to achieve the noble gas configuration and become stable.\nHowever, it is difficult for carbon to gain or lose four electrons because of the following reasons:\nIt cannot gain 4 electrons to form C4- ion having Neon gas (2, 8) configuration because this anion would be highly unstable due to a large amount of energy required to overcome the forces of repulsion between the four electrons being added and the six electrons already present in the carbon atom.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nIt cannot lose 4 electrons to form C4+ ion having Helium gas (2) configuration because this cation would be highly unstable due to a large amount of energy required to remove four electrons from the carbon atom.\nTetravalency of Carbon: A carbon atom has four electrons in the valence shell. Therefore, carbon forms four covalent bonds, i.e., carbon is tetravalent.\nAllotropic forms of Carbon: The phenomenon of existence of an element in two or more forms which have different physical properties but identical chemical properties is called allotropy.\nThree allotropic forms of carbon:\nDiamond\nGraphite\nFullerenes\nHydrocarbon: Organic compounds of carbon and hydrogen are called hydrocarbons.\nSaturated Compound: Compounds of carbon which have only single bonds between the carbon atoms are called saturated compounds e.g., Ethane, Propane, Butane etc.\nUnsaturated Compound: Compounds of carbon which contain one or more double or triple bonds between carbon atoms are called unsaturated compounds", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nUnsaturated Compound: Compounds of carbon which contain one or more double or triple bonds between carbon atoms are called unsaturated compounds\ne.g., Ethene, Propene, Butyne, etc.\nAlkanes\nGeneral formula \u2013 CnH2n+2\nSaturated hydrocarbons\nMethane \u2013 CH4\nEthane \u2013 C2H6\nAlkenes\nGeneral formula \u2013 CnH2n\nUnsaturated hydrocarbon.\nEthene \u2013 C2H4\nPropene \u2013 C3H6\nHomologous series: A family of organic compounds having the same functional group, similar chemical properties and the successive (adjacent) members differ by a CH2 unit or 14 mass unit.\nCharacteristics of a homologous series:\nAll the members of a homologous series can be represented by a general formula.", + "Characteristics of a homologous series:\nAll the members of a homologous series can be represented by a general formula.\nAlkane \u2013 CnH2n+2\nAlkyne \u2013 CnH2n-2\nAlcohol \u2013 CnH2n+1OH\nKetone \u2013 CnH2n+1COCnH2n+1\nAlkene \u2013 CnH2n\nHaloalkane \u2013 CnH2n+1X\nAldehyde \u2013 CnH2n+1CHO\nCarboxylic acid \u2013 CnH2n+1COOH\nThe molecular formula of two successive (adjacent) members of a homologous series differs by a CH2 unit.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nAlkene \u2013 CnH2n\nHaloalkane \u2013 CnH2n+1X\nAldehyde \u2013 CnH2n+1CHO\nCarboxylic acid \u2013 CnH2n+1COOH\nThe molecular formula of two successive (adjacent) members of a homologous series differs by a CH2 unit.\nThe molecular masses of any two successive members of a homologous series differ by 14 u.\nAll the members of a given homologous series have the same functional group.\nAll the members of a series show similar chemical properties.\nThe members of a homologous series show a gradation in physical properties.\nNomenclature of carbon compound: International Union of Pure and Applied Chemistry (IUPAC) decided some rules to name the carbon compounds. This was done to maintain uniformity throughout the world. Names which are given on this basis are popularly known as IUPAC name. The rules for nomenclature are as follows:\n(i) Identify the number of carbon atoms in the carbon compound. Name the carbon compounds according to the number of carbon atoms.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\n(i) Identify the number of carbon atoms in the carbon compound. Name the carbon compounds according to the number of carbon atoms.\nExample, Saturated hydrocarbon having one carbon atom is named as Methane. Saturated hydrocarbon having two carbon atoms is named as Ethane.\nAn unsaturated hydrocarbon with a double bond having two carbon atoms is named as Ethene.\nAn unsaturated hydrocarbon with a triple bond between carbon atoms is named as Ethyne.\n(ii) If the structure has a branched chain, identify the longest chain and then identify the number of carbon atoms.\n(iii) In the case of a functional group present, write the prefix or suffix of the functional group as given below. Then write the name of the parent compound:\n\nChemical properties of Ethanol\nEthanol (C2H5OH) compound is a colourless liquid at room temperature. It is the second member of the homologous series of alcohols. Its common name is ethyl alcohol. Its functional group is \u2013 OH.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nEthanol (C2H5OH) compound is a colourless liquid at room temperature. It is the second member of the homologous series of alcohols. Its common name is ethyl alcohol. Its functional group is \u2013 OH.\nIt has a very low melting point (156 K) and low boiling point (351 K or 78\u00b0C).\nEthanol is highly soluble in water.\nEthanol is one of the main components of alcoholic drinks.\nIt is a good organic solvent.\nIt is a neutral substance, so it does not have any effect on either blue litmus solution or red litmus solution.\nIt bums with a blue flame in the presence of O2 of air. This combustion is an oxidation process.\n\nIn the presence of alkaline KMnO4, it is oxidised to ethanoic acid.\n\nEthanol alcohol reacts with sodium(Na) metal vigorously to form sodium ethoxide and evolves H2 gas.\n2C2H5OH + 2Na \u2192 2C2H5ONa (Sodium ethoxide) + H2 (g)", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nEthanol alcohol reacts with sodium(Na) metal vigorously to form sodium ethoxide and evolves H2 gas.\n2C2H5OH + 2Na \u2192 2C2H5ONa (Sodium ethoxide) + H2 (g)\nEthanol on dehydration in the presence of cone. H2SO4 acid at 443 K forms ethene gas. H2SO4 acid absorbs water molecules from the alcohol molecules and acts as a strong dehydrating agent.\n\nChemical properties of Ethanoic acid\nEthanoic acid commonly called acetic acid (CH3COOH) is a colourless liquid. The functional group present in it is carboxylic acid \u2013 COOH.\nIt\u2019s melting point is 290 K and the boiling point is 391 K.\nBeing an acid, it turns blue litmus red.\nIt is sour in taste.\nEthanoic acid reacts with alcohols in the presence of cone. H2SO4 acid to form sweet smelling compounds called esters.\n\nEthanoic acid reacts with bases to form its salt and water.", + "It is sour in taste.\nEthanoic acid reacts with alcohols in the presence of cone. H2SO4 acid to form sweet smelling compounds called esters.\n\nEthanoic acid reacts with bases to form its salt and water.\n\nIt reacts with carbonate and hydrogen-carbonate compounds of metals to form its salt (sodium ethanoate commonly called sodium acetate) and release CO2 gas.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nIt reacts with carbonate and hydrogen-carbonate compounds of metals to form its salt (sodium ethanoate commonly called sodium acetate) and release CO2 gas.\n\nCleansing action of soap: The dirt is generally held to the surface of a dirty cloth by a thin film of oil or grease.\nWhen a dirty cloth is treated with soap or detergent solution, the non- polar tail of the soap or the detergent dissolve in oil or grease while the polar heads are held by the surrounding water. Soap or detergent micelle is formed with the oily or greasy dirt lying at their Centre (Soap or detergent is attracted both by the greasy dirt and water.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nWhen the surface of the cloth is mechanically scrubbed or beaten on a stone or with a wooden paddle or agitated in a washing machine, the loosened oily particle is removed from the dirty surface and the cloth is cleaned. Detergents lower the surface tension of water to a greater extent than soap, therefore the cleansing action of detergent is much higher than those of soaps.\n1. The earth\u2019s crust has only 0.02% carbon in the form of minerals (like carbonates^bicarbonates, coal, and petroleum).\n2. The atmosphere has 0.03% of carbon dioxide.\n3. In spite of its small amount available in nature, carbon is a versatile element as it forms the basis for all living organisms and many things which we use.\n4. Bonding in carbon :\nThe atomic number of carbon = 6\nAn electronic configuration has 2 electrons in K shell and 4 electrons in L shell.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\n4. Bonding in carbon :\nThe atomic number of carbon = 6\nAn electronic configuration has 2 electrons in K shell and 4 electrons in L shell.\nIn order to attain the noble gas configuration, carbon should either gain 4 electrons or lose 4 electrons or can share it\u2019s 4 electrons with some other element.\nThe gain of 4 electrons (to form an octet, i.e., 8 electrons in C4- anion) is difficult because then a nucleus with 6 protons will have to hold extra four electrons.\nLoss of 4 electrons (to attain duplet, i.e., 2 electrons like He atom in C4+ cation) is difficult as it requires a large amount of energy to remove four electrons.\nCarbon, hence, overcomes this difficulty by sharing it\u2019s four valence electrons with other atoms of carbon or with atoms of other elements. These electrons contributed by the atoms for mutual sharing in order to acquire the stable noble gas configuration is called covalency of that atom. Hence, carbon shows TETRACOVALENCY.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nThe simplest molecule formed by sharing of electrons (i.e., covalent bonds), can be represented by electron dot structure.\n\n5. Allotropes of carbon: The phenomenon by means of which an element can exist in two or more forms, with similar chemical properties but different physical properties are called allotropy and the different forms are called allotropes. Carbon shows three allotropic forms :\n\nNote:\nDiamond is the hardest substance whereas graphite is very soft.\nDiamond is used for grinding and polishing of \u2018 hard materials and graphite is used as a lubricant.\nDiamond has a three-dimensional rigid structure but graphite has a hexagonal sheet layer structure.\nDiamond is a bad conductor of electricity but graphite is a very good conductor of electricity.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nDiamond has a three-dimensional rigid structure but graphite has a hexagonal sheet layer structure.\nDiamond is a bad conductor of electricity but graphite is a very good conductor of electricity.\n6. Fullerenes: A new category of carbon allotrope, fullerenes are spherical in shape or a soccer ball like. The first fullerene identified was C-60 with 60 carbon atoms arranged like the geodesic dome designed by US architect, Buckminster Fuller, hence these are also known as Buckminster Fullerenes or Bucky Ball structures.\n7. Cause of versatile nature of carbon: Four main reasons for the versatile nature of carbon are:\n(a) Catenation: It is the unique property of self-linkage of carbon atoms by means of covalent bonds to form straight chains, or branched chains, or the rings of different sizes (as shown below):", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\n(a) Catenation: It is the unique property of self-linkage of carbon atoms by means of covalent bonds to form straight chains, or branched chains, or the rings of different sizes (as shown below):\n\n\n(b) Tetracovalency: Due to small size, and presence of four valence electrons, carbon can form strong bonds with other carbon atoms, hydrogen, oxygen, nitrogen, or sulphur, etc. For example, compounds of carbon with hydrogen are called hydrocarbons.\n(c) Multiple Bond Formation : Small size of carbon also enables it to form multiple bonds, (i.e., double bonds or triple bonds) with other elements as well as with its own atoms. This increases the number of carbon compounds.\nNote:\nCompounds of carbon with double bonds and triple bonds are called as unsaturated compounds while those with carbon-carbon single bonds are called saturated compounds.\nAlkenes (with \u2014C = C \u2014) and Alkynes (with \u2014C = C\u2014) are hence unsaturated, whereas Alkanes (with \u2014 C \u2014 C\u2014) are saturated compounds.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nAlkenes (with \u2014C = C \u2014) and Alkynes (with \u2014C = C\u2014) are hence unsaturated, whereas Alkanes (with \u2014 C \u2014 C\u2014) are saturated compounds.\n(d) Isomerism: The phenomenon by means of which the carbon compounds with same molecular formula show different structures, and properties, e.g., A chain of 4 carbon atoms can be written in two ways :\n\nHence, the number of carbon compounds increases to a huge number.\n8. Hydrocarbons: a Large number of hydrocarbons can be classified as:\n\nNote: In an open chain, the name of parent chain is derived from the root word and suffix ane, ene or yne is added depending on the type of bond present in a chain :\n\nImportant: No alkene or alkyne is possible with single carbon atom because double or triple bond is not possible between carbon and hydrogen atom. It is only between two carbon atoms.\n9. Functional Group:\nAn atom or a group of atoms which when present in a compound gives specific properties to it, is called a functional group.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\n9. Functional Group:\nAn atom or a group of atoms which when present in a compound gives specific properties to it, is called a functional group.\nA single line shown along with a functional group is called as its free valency by which it gets attached to a compound by replacing one hydrogen atom or atoms, e.g., -Cl.\nFunctional group, replacing the hydrogen is also called as heteroatom because it is different from carbon, and can be nitrogen, sulphur, or halogen, etc.\nImportant: Replacement of hydrogen atom by a functional group is always in such a manner that valency of carbon remains satisfied.\n\nNote : Cl is named as prefix Chloro; Br as Bromo; NH2 as Amino and N02 as Nitro.\nImportant Note: Symbol \u2018R\u2019 in a formula represents an Alkyl Group which is formed by the removal of one hydrogen atom from an alkane.\n\n10. Homologous series: A series of organic compounds in which every succeeding member differs from the previous one by -CH2 group or 14 a.m.u.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\n10. Homologous series: A series of organic compounds in which every succeeding member differs from the previous one by -CH2 group or 14 a.m.u.\nNote : As the molecular mass increases in a series, : so physical properties of the compounds\nshow a variation, but chemical properties which are determined solely by a functional group, remains same within a series.\n11. Nomenclature of Organic Compounds\nTrivial or common names: These names were given after the source from which the organic compounds were first isolated, e.g., If a compound has one carbon atom, then its common name will have root word form and so on (see table).\n\nIUPAC name: International Union of Pure and Applied Chemistry gave following rules for naming various compounds :\nIdentify the number of carbon atoms and write the word root corresponding to it. e.g., If a number of carbon atoms is three, then the word root is a prop.\nPresence of a functional group is indicated by prefix or suffix as given in table 2, and table 3.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nPresence of a functional group is indicated by prefix or suffix as given in table 2, and table 3.\nIf the name of the functional group is to be given as a suffix, the last letter \u2018e\u2019 in the name of the compound is deleted and the suffix is added. e.g., a ketone with three carbon atoms is named as :\nPropane \u2013 e = Propan + \u2018one\u2019 = Propanone. Alcohol with three carbons is propanol. Carboxylic acid with three carbons is propanoic acid.\nHalogens, in IUPAC, are written as Prefixes, e.g., Compound With two carbons and one chloro group is named as chloroethane (CH3CH2CI).\n12. Chemical properties of carbon compounds :\nMain properties of carbon compounds are :\n(a) Combustion Reaction\n(b) Oxidation Reaction\n(c) Addition Reaction.\n(d) Substitution Reaction\n(a) Combustion Reaction: A chemical reaction in which a substance burns in the presence of air or oxygen is called combustion reaction.", + "(d) Substitution Reaction\n(a) Combustion Reaction: A chemical reaction in which a substance burns in the presence of air or oxygen is called combustion reaction.\nNote: Combustion is always an EXOTHERMIC reaction, e.g.,\n\nRemember:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nNote: Combustion is always an EXOTHERMIC reaction, e.g.,\n\nRemember:\nSaturated hydrocarbons generally give clean flame whereas unsaturated hydrocarbons give sooty flame (because carbon content is more than hydrogen content in these, and hence carbon shows incomplete combustion and appears as soot).\nSaturated hydrocarbons can give sooty flame in a limited supply of oxygen.\n(b) Oxidation Reaction: The addition of oxygen in a compound upon combustion is called oxidation.\nIn addition to combustion, oxidation can also be : brought about by some substances which are\ncapable of giving oxygen to others, i.e., Oxidising agents, e.g., Acidified K2Cr207 (Potassium dichromate) and alkaline KMn04 (Potassium permanganate).\nNote:\n\n(c) Addition Reaction: Addition of a molecule in unsaturated compounds in the presence of a catalyst, to give saturated compound is called an addition reaction, e.g.,\nHydrogenation of vegetable oils as shown in the reaction below :", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nHydrogenation of vegetable oils as shown in the reaction below :\n\n(d) Substitution Reaction: The reactions which involve the replacement of an atom or group of atoms from a molecule by another atom without any change in structure in the remaining part of the molecule.\n\n13. Ethanol: (or alcohol)\nColourless liquid, soluble in water, and has a distinct smell and burning taste. Its consumption in small quantities causes drunkenness and can be lethal.\n\n14. Ethanoic Acid: CH3COOH\nCommon Name: Acetic Acid.\n5-8% solution of acetic acid in water is called Vinegar. And 100% pure acetic acid is called Glacial acetic acid because it has m.pt. 290 K and freezes forming glacier like crystals.\nReactions of ethanoic acid :\n\nSaponification : Esters in the presence of acid or base react to give back alcohol and carboxylic acid is called saponification.\n\n15. Soaps and Detergents :\nSoaps and Synthetic Detergents: Soaps and detergents are substances used for cleaning.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\n15. Soaps and Detergents :\nSoaps and Synthetic Detergents: Soaps and detergents are substances used for cleaning.\nSoap: Soaps are sodium or potassium salts of higher fatty acids, such as Oleic acid (C17H33COOH), Stearic acid (C17H35COOH), Palmitic acid (C15H31COOH), etc. These acids are present in the form of their esters along with glycerol (alcohol containing three hydroxyl groups). These esters, called \u2018glycerides\u2019 are present in fats and oils of animal and vegetable origin.\nPreparation of Soap: When an oil or fat (glyceride) is treated with sodium hydroxide solution, it gets converted to sodium salt of the acid (soap) and glycerol. The reaction is known as saponification.", + "Preparation of Soap: When an oil or fat (glyceride) is treated with sodium hydroxide solution, it gets converted to sodium salt of the acid (soap) and glycerol. The reaction is known as saponification.\n\nDetergents: Chemically, detergents are sodium salts of sulphonic acids, i.e., detergents contain a sulphonic acid group (\u2014S03H), instead^of a carboxylic acid group (\u2014COOH), on one end of the hydrocarbon.\n\nThe cleansing action of detergent is considered to be more effective than a soap.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\nThe cleansing action of detergent is considered to be more effective than a soap.\nCleansing Action of Soaps and Detergents: The cleansing action of soaps and detergents follows the same principle.\n\nWhen soap or detergent is dissolved in water, the molecules gather together as clusters, called micelles. The tails stick inwards and the heads outwards.\nIn cleansing, the hydrocarbon tail attaches.itself to oily dirt. When water is agitated, the oily dirt tends to lift off from the dirty surface and dissociates into fragments. This gives an opportunity to other tails to stick to oil. The solution now contains small globules of oil surrounded by detergent molecules. The negatively charged heads present in water prevent the small globules from coming together and form aggregates. Thus, the oily dirt is removed from the object.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- NOTES.txt\n16. Scum: The insoluble precipitates formed by soap molecule when they react with calcium and magnesium ions present in hard water. Due to this, a lot of soap gets wasted and cleansing action gets reduced to a larger extent.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nImportant Questions of Carbon and its Compounds Class 10 Science Chapter 4\nQuestion 1.\nCovalent compounds have low melting and boiling point. Why? (2020)\nAnswer:\nCovalent compounds have low melting and boiling points because the forces of attraction between molecules of covalent compounds are very weak. On applying a small amount of heat these molecular forces break.\nQuestion 2.\nWhat are covalent compounds? Why are they different from ionic compounds? List their three characteristic properties. (Delhi 2016)\nAnswer:\nCovalent compounds are those compounds which are formed by sharing of valence electrons between the atoms e.g., hydrogen molecule is formed by mutual sharing of electrons between two hydrogen atoms.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nCovalent compounds are those compounds which are formed by sharing of valence electrons between the atoms e.g., hydrogen molecule is formed by mutual sharing of electrons between two hydrogen atoms.\nThey are different from ionic compounds as ionic compounds are formed by the complete transfer of electrons from one atom to another e.g., NaCl is formed when one valence electron of sodium gets completely transferred to outer shell of chlorine atom. The characteristic properties of covalent compounds are:\n(i) They are generally insoluble or less soluble in water but soluble in organic solvents.\n(ii) They have low melting and boiling points.\n(iii) They do not conduct electricity as they do not contain ions.\nQuestion 3.\nWhat are covalent bonds? Show their formation with the help of electron dot structure of methane. Why are covalent compounds generally poor conductors of electricity? (Delhi 2013C)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nWhat are covalent bonds? Show their formation with the help of electron dot structure of methane. Why are covalent compounds generally poor conductors of electricity? (Delhi 2013C)\nAnswer:\nCovalent bonds are those bonds which are formed by sharing of the valence electrons between two atoms. Electron dot structure of methane is shown in the figure.\n\nCovalent compounds are generally poor conductors ol electricity because they do not have tree electrons or ions.\nQuestion 4.\nGive reasons for the following:\n(i) Element carbon forms compounds mainly by covalent bonding.\n(ii) Diamond has high melting point.\n(iii) Graphite is a good conductor of electricity. (3/5, Foreign 2011)\nAnswer:\n(i) As carbon has four valence electrons and it can neither loose nor gain lour electrons thus, it attains noble gas configuration only by sharing of electrons. I bus, it forms covalent compounds.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(i) As carbon has four valence electrons and it can neither loose nor gain lour electrons thus, it attains noble gas configuration only by sharing of electrons. I bus, it forms covalent compounds.\n(ii) In diamond, each carbon atom is bonded to four other carbon atoms forming a rigid three-dimensional structure. This makes diamond the hardest known substance. Thus, it has high melting point.\n(iii) In graphite, each carbon atom is bonded to three other carbon atoms by covalent bonds in the same plane giving a hexagonal array. Thus, only three valence electrons are used for bond formation and hence, the fourth valence electron is free to move. As a result, graphite is a good conductor of electricity.\nQuestion 5.\nWhat is methane? Draw its electron dot structure. Name the type of bonds formed in this compound. Why are such compounds\n(i) poor conductors of electricity and\n(ii) have low melting and boiling points?\nWhat happens when this compound burns in oxygen? (Delhi 2019)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(i) poor conductors of electricity and\n(ii) have low melting and boiling points?\nWhat happens when this compound burns in oxygen? (Delhi 2019)\nAnswer:\nMethane is the first member of alkane series having formula CH4.\nRefer to answer 3.\n(ii) Refer to answer 1.\nWhen methane is burnt in presence of oxygen then carbon dioxide will be produced.\nCH4 + O2 \u2192 CO2 + H2O + heat + light\nQuestion 6.\nElements forming ionic compounds attain noble gas electronic configuration by either gaining or losing electrons from their valence shells. Explain giving reason why carbon cannot attain such a configuration in this manner to form its compounds. Name the type of bonds formed in ionic compounds and in the compounds formed by carbon. Also explain with reason why carbon compounds are generally poor conductors of electricity. (Foreign 2015, AI 2014)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nIonic compounds are formed either by gaining or losing electrons from the outermost shells, but carbon which has four electrons in its outermost shell cannot form ionic bonds because\n1. If carbon forms ionic bonds by gaining four electrons to attain a noble gas configuration then it would be difficult for six protons in the nucleus to hold ten electrons.\n2. If carbon forms ionic bonds by loss of four electrons then it would require a lot of energy to remove these electrons from outermost shell.\nDue to these reasons carbon forms covalent bonds by sharing the valence electrons.\nType of bonds formed in ionic compounds are called electrovalent bonds and the type of bonds formed in carbon compounds are called covalent bonds.\nRefer to answer 3.\nQuestion 7.\nState the reason why carbon can neither form C4+ cations nor C4- anions, but forms covalent compounds. Also state reasons to explain why covalent compounds :\n(i) are bad conductors of electricity?", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nState the reason why carbon can neither form C4+ cations nor C4- anions, but forms covalent compounds. Also state reasons to explain why covalent compounds :\n(i) are bad conductors of electricity?\n(ii) have low melting and boiling points? (Delhi 2014)\nAnswer:\nRefer to answer 6.\n(i) Refer to answer 3.\n(ii) Refer to answer 1.\nQuestion 8.\nName a cyclic unsaturated carbon compound. (2020)\nAnswer:\n\nQuestion 9.\nAssertion (A) : Following are the members of a homologous series :\nCH3OH, CH3CH2OH, CH3CH2CH2OH\nReason (R) : A series of compounds with same functional group but differing by -CH2 unit is called homologous series.\n(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).", + "(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true. (2020)\nAnswer:\n(a): The given compounds are members of homologous series of alcohol.\nQuestion 10.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true. (2020)\nAnswer:\n(a): The given compounds are members of homologous series of alcohol.\nQuestion 10.\nWrite the molecular formula of first two members of homologous series having functional group -Cl. (Delhi 2017)\nAnswer:\nThe molecular formula of first two members of homologous series having -Cl functional group are CH3Cl and CH3CH2Cl.\nQuestion 11.\nWrite the molecular formula of first two members of homologous series having functional group -OH. (Delhi 2017)\nAnswer:\nThe molecular formula of first two members of homologous series having -OH functional group are CH3OH and CH3CH2OH.\nQuestion 12.\nWrite the molecular formula of the 2nd and 3rd member of the homologous series whose first member is ethene. (AI 2017)\nAnswer:\nHomologous series of alkenes have general formula, CnH2n whose first member is ethene.", + "Question 12.\nWrite the molecular formula of the 2nd and 3rd member of the homologous series whose first member is ethene. (AI 2017)\nAnswer:\nHomologous series of alkenes have general formula, CnH2n whose first member is ethene.\n2nd member of homologous series of alkenes is C3H6 i.e., propene.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nHomologous series of alkenes have general formula, CnH2n whose first member is ethene.\n2nd member of homologous series of alkenes is C3H6 i.e., propene.\n3rd member of homologous series of alkenes is C4H8 i.e., butene.\nQuestion 13.\nWrite the molecular formula of the 2nd and 3rd member of the homologous series whose first member is methane. (AI 2017)\nAnswer:\nMethane, CH4 is an alkane. Alkanes have general formula, CnH2n+2.\n2nd member of homologous series of alkanes is C2H6 i.e., ethane.\n3rd member of homologous series of alkanes is C3H8 i.e., propane.\nQuestion 14.", + "Alkanes have general formula, CnH2n+2.\n2nd member of homologous series of alkanes is C2H6 i.e., ethane.\n3rd member of homologous series of alkanes is C3H8 i.e., propane.\nQuestion 14.\nWrite the next homologue of each of the following:\n(i) C2H4\n(ii) C4H6 (Delhi 2016)\nAnswer:\n(i) C2H4 belongs to alkene series having general formula, CnH2n.\nThus, next homologue will be C3H2\u00d73 = C3H6\n(ii) C4H6 belongs to alkyne series having general formula, CnH2n-2.\nThus, next homologue will be C5H2\u00d75-2 = C5H8\nQuestion 15.\nName the following compounds :\n(a) CH3 \u2013 CH2 \u2013 OH\n\nAnswer:\n(a) CH3 \u2013 CH2 \u2013 OH : Ethanol", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nThus, next homologue will be C5H2\u00d75-2 = C5H8\nQuestion 15.\nName the following compounds :\n(a) CH3 \u2013 CH2 \u2013 OH\n\nAnswer:\n(a) CH3 \u2013 CH2 \u2013 OH : Ethanol\n\nQuestion 16.\nSelect saturated hydrocarbons from the following : C3H6; C5H10; C4H10; C6H14; C2H4\nAnswer:\nSaturated hydrocarbons have general formula, CnH2n+2.\nAmong the given compounds only C4H10 and C6H14 satisfy the above formula. Thus, these are saturated hydrocarbons.\nQuestion 17.\nWrite the name and structure of an alcohol with three carbon atoms in its molecule. (AI 2016)\nAnswer:\nAn alcohol with three carbon atoms in its molecule is propanol. The structure of propanol is\n\nQuestion 18.\nWrite the name and structure of an alcohol with four carbon atoms in its molecule.", + "(AI 2016)\nAnswer:\nAn alcohol with three carbon atoms in its molecule is propanol. The structure of propanol is\n\nQuestion 18.\nWrite the name and structure of an alcohol with four carbon atoms in its molecule. (AI 2016)\nAnswer:\nAn alcohol with four carbon atoms is butanol and its structure is :\n\nQuestion 19.\nWrite the name and structure of an aldehyde with four carbon atoms in its molecule. (AI 2016)\nAnswer:\nAn aldehyde with four carbon atoms is butanal and its structure is.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 19.\nWrite the name and structure of an aldehyde with four carbon atoms in its molecule. (AI 2016)\nAnswer:\nAn aldehyde with four carbon atoms is butanal and its structure is.\n\nQuestion 20.\nWhich element exhibits the property of catenation to maximum extent and why? (Foreign 2016)\nAnswer:\nCarbon has the unique ability to form bonds with other atoms of carbon, giving rise to large molecules. This property is called catenation. Carbon shows catenation due to its small size and Stronger carbon-carbon bond strength.\nQuestion 21.\nWrite the name and molecular formula of the fourth member of alkane series. (Foreign 2016)\nAnswer:\nThe general formula of the alkane series is CnH2n+2. For fourth member of alkane series, n = 4\n\u2234 C4H2\u00d74+2 = C4H10 i.e., butane.\nQuestion 22.\nWhat is homologous series of carbon compounds? (Foreign 2016)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n\u2234 C4H2\u00d74+2 = C4H10 i.e., butane.\nQuestion 22.\nWhat is homologous series of carbon compounds? (Foreign 2016)\nAnswer:\nA homologous series is the family of organic compounds having the same functional group, similar chemical properties but the successive (adjacent) members of the series differ by a -CH2 unit or 14 mass units.\nQuestion 23.\nWrite the name and formula of the 2nd member of homologous series having general formula CnH2n. (Delhi 2015)\nAnswer:\nRefer to answer 12.\nQuestion 24.\nWrite the name and formula of the 2nd member of homologous series having general formula CnH2n+2. (Delhi 2015)\nAnswer:\nRefer to answer 13.\nQuestion 25.\nWrite the name and formula of the 2nd member of homologous series having general formula CnH2n-2.", + "(Delhi 2015)\nAnswer:\nRefer to answer 13.\nQuestion 25.\nWrite the name and formula of the 2nd member of homologous series having general formula CnH2n-2. (Delhi 2015)\nAnswer:\nGeneral formula, CnH2n-2 belongs to alkyne series. The second member of this series is propyne i.e., (C3H4) or CH3 \u2013 C \u2261 CH.\nQuestion 26.\nWrite the number of covalent bonds in the molecule of ethane. (AI2015, Delhi 2014)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 26.\nWrite the number of covalent bonds in the molecule of ethane. (AI2015, Delhi 2014)\nAnswer:\nThe structural formula of ethane (C2H6) is\n\nThere are total 7 covalent bonds. Six C \u2013 H covalent bonds and one C \u2013 C covalent bond.\nQuestion 27.\nWrite the number of covalent bonds in the molecule of butane, C4H10. (AI 2015)\nAnswer:\nButane (C4H10) has the following structural formula as:\n\nTotal number of covalent bonds is 13 in which there are 10 C \u2013 H and 3 C \u2013 C covalent bonds.\nQuestion 28.\nWrite the name of each of the following functional groups: (Foreign 2015, Delhi 2013)\n(a) -OH\n\nAnswer:\n(a) -OH : Alcohol\n\nQuestion 29.\nWrite the name and molecular formula of the first member of the homologous series of alkynes.", + "Write the name of each of the following functional groups: (Foreign 2015, Delhi 2013)\n(a) -OH\n\nAnswer:\n(a) -OH : Alcohol\n\nQuestion 29.\nWrite the name and molecular formula of the first member of the homologous series of alkynes. (Foreign 2015)\nAnswer:\nGeneral formula for alkyne is CnH2n-2\nFirst member of homologous series of alkyne has the formula, C2H2\u00d72-2 = C2H2 i.e., ethyne.\nQuestion 30.\nDefine the term functional group. Identify the functional group present in\n\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nFirst member of homologous series of alkyne has the formula, C2H2\u00d72-2 = C2H2 i.e., ethyne.\nQuestion 30.\nDefine the term functional group. Identify the functional group present in\n\nAnswer:\nAn atom or a group of atoms present in a molecule which largely determines its chemical properties, is called functional group.\n\nQuestion 31.\nName the functional group present in each of the following organic compounds:\n(i) C2H5Cl\n(ii) C2H5OH (Delhi 2012)\nAnswer:\n(i) C2H5Cl contains -Cl (chloro) group which belongs to halo functional group.\n(ii) C5H5OH contains -OH group which belongs to alcoholic functional group.\nQuestion 32.\nWrite the name and formula of the second member of the carbon compounds having functional group -OH. (AI 2012)\nAnswer:\nThose having -OH as functional group belong to alcohol family. Second member of this family is ethanol, C2H5OH.\nQuestion 33.", + "Question 32.\nWrite the name and formula of the second member of the carbon compounds having functional group -OH. (AI 2012)\nAnswer:\nThose having -OH as functional group belong to alcohol family. Second member of this family is ethanol, C2H5OH.\nQuestion 33.\nWrite the name and formula of the first member of the series of carbon compounds having functional group (Foreign 2012)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 33.\nWrite the name and formula of the first member of the series of carbon compounds having functional group (Foreign 2012)\nAnswer:\nCarbon compound containing\n\ngroup is called carboxylic acid. The first member of this family is methanoic acid (HCOOH).\nQuestion 34.\nButanone is a four-carbon per molecule compound. Name the functional group present in it. (Foreign 2011)\nAnswer:\n\nQuestion 35.\nState two properties of carbon which lead to a very large number of carbon compounds. (2/5, AI 2011)\nAnswer:\nCarbon forms a large number of carbon compounds like long chains which may be straight or branched chains or ring of different sizes due to its tetravalency ahd unique property of catenation. Carbon due to its small size forms exceptionally stable compounds by forming strong bonds.\nQuestion 36.\nCarbon, a member of group 14, forms a large number of carbon compounds estimated to be about three million. Why is this property not exhibited by other elements of this group? Explain. (2020)", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nCarbon, a member of group 14, forms a large number of carbon compounds estimated to be about three million. Why is this property not exhibited by other elements of this group? Explain. (2020)\nAnswer:\nRefer to answer 20.\nAs we move down the group, the element-element bond energies decrease rapidly. For this reason other elements of this group show little or no catenation property.\nQuestion 37.\n(a) Why are most carbon compounds poor conductors of electricity?\n(b) Write the name and structure of a saturated compound in which the carbon atoms are arranged in a ring. Give the number of single bonds present in this compound. (2018)\nAnswer:\n(a) Due lo catenation, carbon forms covalent bonds with the constituent elements in the carbon compounds, hence it does not have mobile electrons and carbon compounds do not dissociate themselves into ions and hence, they are poor conductor of electricity.\n\nName: Cyclopentane\nNumber of single bonds : 15\nQuestion 38.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nName: Cyclopentane\nNumber of single bonds : 15\nQuestion 38.\nAn aldehyde as well as a ketone can be represented by the same molecular formula, say C3H6O. Write their structures and name them. State the relation between the two in the language of science. (AI 2016)\nAnswer:\nThe aldehyde and ketone represented by the\nmolecular formula, C3H6O.\n\nIn the language of science, they are called as isomers because both have same molecular formula but different structural formulae (having different functional groups.)\nQuestion 39.\nWhat is meant by isomers? Draw the structures of two isomers of butane, C4H10. Explain why we cannot have isomers of first three members of alkane series. (Delhi 2015, Foreign 2014)\nAnswer:\nIsomers are those molecules which have the same molecular formula but different structural formula i.e., show different properties.", + "Explain why we cannot have isomers of first three members of alkane series. (Delhi 2015, Foreign 2014)\nAnswer:\nIsomers are those molecules which have the same molecular formula but different structural formula i.e., show different properties.\nThe structures of possible isomers of butane (C4H10) are:\n\nThe first three members of alkane series are :\n(i) CH4 (methane)\n(ii) C2H6 (ethane)", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nThe structures of possible isomers of butane (C4H10) are:\n\nThe first three members of alkane series are :\n(i) CH4 (methane)\n(ii) C2H6 (ethane)\n(iii) C3H8 (propane)\nIn the above members of alkane series, it is not possible to have different arrangements of carbon atoms. Thus, we cannot have isomers of first three members of alkane series.\nQuestion 40.\nWrite the molecular formula of the following compounds and draw their electron-dot structures:\n(i) Ethane\n(ii) Ethene\n(iii) Ethyne (Foreign 2015)\nAnswer:\n(i) Molecular formula of ethane is C2H6.\nIts electron dot structure is :\n\n(ii) Molecular formula of ethene is C2H4. Its electron dot structure is :\n\n(iii) Molecular formula of ethyne is C2H2. Its electron dot structure is :\n\nQuestion 41.\nWhat is meant by functional group in carbon compounds?", + "Its electron dot structure is :\n\n(ii) Molecular formula of ethene is C2H4. Its electron dot structure is :\n\n(iii) Molecular formula of ethyne is C2H2. Its electron dot structure is :\n\nQuestion 41.\nWhat is meant by functional group in carbon compounds? Write in tabular form the structural formula and the functional group present in the following compounds :\n(i) Ethanol\n(ii) Ethanoic acid (Foreign 2015)\nAnswer:\nRefer to answer 30.\n\nQuestion 42.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(i) Ethanol\n(ii) Ethanoic acid (Foreign 2015)\nAnswer:\nRefer to answer 30.\n\nQuestion 42.\nWhy is homologous series of carbon compounds so called? Write the chemical formula of two consecutive members of any homologous series and state the part of these compounds that determines their (i) physical and (ii) chemical properties. (Foreign 2015, AI2014, Delhi 2013)\nAnswer:\nRefer to answer 22.\nConsecutive members of the homologous series of alcohols are:\n\nThe physical properties are determined by alkyl group/hydrocarbon part/part other than the functional group.\nThe chemical properties are determined by functional group such as -OH group.\nQuestion 43.\nState the meaning of functional group in a carbon compound. Write the functional group present in (i) ethanol and (ii) ethanoic acid and also draw their structures. (Delhi 2014)\nAnswer:\nRefer to answer 30 and 41.\nQuestion 44.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 30 and 41.\nQuestion 44.\nState the meaning of the functional group in an organic compound. Write the formula of the functional group present in alcohols, aldehydes, ketones and carboxylic acids. (Delhi 2014)\nAnswer:\nRefer to answer 30.\nThe formulae for different functional groups are : alcohols : -OH group\n\nQuestion 45.\nWhat is meant by homologous series of carbon compounds? Write the general formula of (i) alkenes, and (ii) alkynes. Draw the structures of the first member of each series to show the bonding between the two carbon atoms. (AI 2014)\nAnswer:\nRefer to answer 22.\nThe general formula for alkenes is CnH2n and for alkynes is CnH2n-2\nFirst member of alkene is ethene, C2H4 and its structure is\n\nFirst member of alkyne is ethyne, C2H2 and its structure is H \u2013 C \u2261 C \u2013 H\nQuestion 46.", + "Define the term structural isomerism\u2019. Explain why propane cannot exhibit this property. Draw the structures of possible isomers of butane, C4H10. (AI 2014)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 46.\nDefine the term structural isomerism\u2019. Explain why propane cannot exhibit this property. Draw the structures of possible isomers of butane, C4H10. (AI 2014)\nAnswer:\nTwo or more organic compounds having the same molecular formula but different structures, are called structural isomers and the phenomenon is known as structural isomerism.\nThere is no possible isomers for propane as it contains three carbon atoms and it is not possible to have different arrangements of these carbon atoms.\nRefer to answer 39.\nQuestion 47.\n(a) What is a homologous series of compounds? List any two of its characteristics. (Foreign 2011)\n(b) What is the next higher homologue of C3H7OH? What is its formula and what is it called? (Foreign 2011)\nAnswer:\n(a) Refer to answer 22.\nTwo characteristics of homologous series are :\n(i) The successive compounds of the homologous series differ by -CH2 unit i.e. 14 mass units.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Refer to answer 22.\nTwo characteristics of homologous series are :\n(i) The successive compounds of the homologous series differ by -CH2 unit i.e. 14 mass units.\n(ii) Each homologous series belongs to similar class of compounds which shows the same chemical properties.\n(b) Next higher homologue of C3H7OH is C4H9OH i.e., butanol.\nQuestion 48.\n(a) State the reason why carbon can neither form C4+ cations nor C4- anions, but forms covalent bonds. Also state reasons to explain why covalent compounds\n(i) are bad conductors of electricity\n(ii) have (ow melting and boiling points.\n(b) Write the structural formula of benzene, C6H6. (AI2019)\nAnswer:\n(a) Refer to answer 6.\n(i) Refer to answer 3.\n(ii) Refer to answer 1.\n\nQuestion 49.\nExplain why carbon forms compounds mainly by covalent bond. Explain in brief two main reasons for carbon forming a large number of compounds.", + "(i) Refer to answer 3.\n(ii) Refer to answer 1.\n\nQuestion 49.\nExplain why carbon forms compounds mainly by covalent bond. Explain in brief two main reasons for carbon forming a large number of compounds. Why does carbon form strong bond with most other elements? (Delhi 2015)\nAnswer:\nRefer to answers 6 and 35.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answers 6 and 35.\nDue to the small size of carbon atom, its nucleus holds the shared pair of electrons between atoms strongly. Thus, carbon forms strong covalent bonds with elements such as hydrogen, oxygen, nitrogen, sulphur, chlorine and other elements.\nQuestion 50.\nWhat are hydrocarbons? Distinguish alkanes from alkenes and each of them from alkynes, giving one example of each. Draw the structure of each compound cited as example to justify your answer. (Foreign 2014)\nAnswer:\nHydrocarbons are the compounds of carbon and hydrogen atoms. Those hydrocarbons which contain only single carbon-carbon bonds are called alkanes (saturated hydrocarbons) while those having double and triple bonds are called alkenes and alkynes respectively (unsaturated hydrocarbon).\nStructures of the above examples are:\n\nC2H2 (ethyne): H \u2013 C \u2261 C \u2013 H\nQuestion 51.\n(a) Define the term\u2018isomers\u2019.\n(b) Draw two possible isomers of the compound with molecular formula C3H6O and write their names.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nC2H2 (ethyne): H \u2013 C \u2261 C \u2013 H\nQuestion 51.\n(a) Define the term\u2018isomers\u2019.\n(b) Draw two possible isomers of the compound with molecular formula C3H6O and write their names.\n(c) Give the electron dot structures of the above two compounds. (Delhi 2013)\nAnswer:\n(a) Refer to answer 39.\n(b) Two possible isomers of the compound, C3H6O are:\n\n(c) The electron dot structures of propanone and Propanal are\n\nQuestion 52.\nExplain isomerism. State any four characteristics of isomers. Draw the structures of possible isomers of butane, C4H10. (AI 2011)\nAnswer:\nIsomers are those compounds which have same molecular formula but different structures. The phenomenon of existing these isomers are called isomerism.\nFour characteristics of isomers are :\n(i) They have same molecular formula but different structures.\n(ii) For hydrocarbons, isomers is possible only with hydrocarbons having four or more carbon atoms.", + "The phenomenon of existing these isomers are called isomerism.\nFour characteristics of isomers are :\n(i) They have same molecular formula but different structures.\n(ii) For hydrocarbons, isomers is possible only with hydrocarbons having four or more carbon atoms.\n(iii) Due to isomerism, a given molecular formula can represent two or more different compounds.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(ii) For hydrocarbons, isomers is possible only with hydrocarbons having four or more carbon atoms.\n(iii) Due to isomerism, a given molecular formula can represent two or more different compounds.\n(iv) Due to isomerism, the different compounds have different properties.\nRefer to answer 39.\nQuestion 53.\nName the process by which unsaturated fats are changed to saturated fats. (Foreign 2015)\nAnswer:\nHydrogenation is the process in which unsaturated fats are changed to saturated fats.\nQuestion 54.\nWrite the chemical equation to show what happen when methane is treated with chlorine in the presence of sunlight ? (1/3, Foreign 2014)\nAnswer:\nWhen methane is treated with chlorine in the presence of sunlight then substitution reaction takes place. In this, chlorine replaces the hydrogen atom of methane.\n\nQuestion 55.\nWrite the respective chemical reaction to show what happens when methane is burnt in presence of oxygen? (1/3, Foreign 2014)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 55.\nWrite the respective chemical reaction to show what happens when methane is burnt in presence of oxygen? (1/3, Foreign 2014)\nAnswer:\nWhen methane is burnt in presence of oxygen then carbon dioxide will be produced.\nCH4 + O2 \u2192 CO2 + H2O + heat + light\nQuestion 56.\nWrite one chemical equation to represent the following type of reaction of organic substances: substitution. (1/3, Foreign 2014)\nAnswer:\nSubstitution : In this type of reaction one or more hydrogen atoms of a hydrocarbon is replaced by some other atoms.\n\nQuestion 57.\nGive reason for the following : Acetylene burns with a sooty flame. (1/5, Foreign 2011)\nAnswer:\nThe formula of acetylene is HC \u2261 CH. It is an unsaturated hydrocarbon where carbon content is more than the hydrogen content. Hence, carbon is not completely burnt and the unburnt carbon deposits as a soot.\nQuestion 58.\nGive reason for the following : Kerosene does not decolourise bromine water while cooking oils do.", + "Hence, carbon is not completely burnt and the unburnt carbon deposits as a soot.\nQuestion 58.\nGive reason for the following : Kerosene does not decolourise bromine water while cooking oils do. (1/5, Foreign 2011)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 58.\nGive reason for the following : Kerosene does not decolourise bromine water while cooking oils do. (1/5, Foreign 2011)\nAnswer:\nCooking oils (unsaturated compounds) decolourise bromine water due to formation of addition products whereas kerosene (saturated compound) does not decolourise bromine water.\nQuestion 59.\nWhat happens when 5% alkaline KMnO4 solution is added drop by drop to warm ethanol taken in a test tube? State the role of alkaline KMnO4 solution in this reaction. (2/3, Foreign 2016)\nAnswer:\nWhen 5% alkaline KMnO4 solution is added drop by drop to warm ethanol then it gets oxidised to ethanoic acid.\n\nHere, alkaline KMnO4 acts as an oxidising agent i.e., the substance which is capable of adding oxygen to others. Thus, alkaline KMnO4 provides oxygen to ethanol to form ethanoic acid.\nQuestion 60.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nHere, alkaline KMnO4 acts as an oxidising agent i.e., the substance which is capable of adding oxygen to others. Thus, alkaline KMnO4 provides oxygen to ethanol to form ethanoic acid.\nQuestion 60.\n3 mL of ethanol is taken in a test tube and warmed gently in a water bath. A 5% solution of alkaline potassium permanganate is added first drop by drop to this solution, then in excess.\n(i) How is 5% solution of KMnO4 prepared?\n(ii) State the role of alkaline potassium permanganate in this reaction. What happens on adding it in excess?\n(iii) Write chemical equation of this reaction. (2020)\nAnswer:\n(i) 5% solution of KMnO4 is prepared by adding 5 g of KMnO4 in 95 g of water.\n(ii) Here alkaline KMnO4 acts as an oxidising agent. It oxidises ethanol to ethanoic acid by donating nascent oxygen.", + "(ii) Here alkaline KMnO4 acts as an oxidising agent. It oxidises ethanol to ethanoic acid by donating nascent oxygen. If excess of KMnO4 is added the purple colour will persist indicating no more alcohol is left and there is no reaction.\n\nQuestion 61.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 61.\nTwo carbon compounds X and Y have the molecular formula C4H8 and C5H12 respectively. Which one of these is most likely to show addition reaction? Justify your answer. Also give the chemical equation to explain the process of addition reaction in this case. (Delhi 2017)\nAnswer:\nAll unsaturated hydrocarbons (containing double or triple bonds) have tendency to get converted to saturated hydrocarbons (single bonds) by adding small molecules such as hydrogen (H2), halogens (X2), etc. Such reactions are called addition reactions.\nCompound X i.e., C4H8 belongs to alkene series (CnH2n) while compound Y i.e., C5H12 belongs to alkane series (CnH2n+2). Thus, compound X will undergo addition reaction.\n\nQuestion 62.\nThe molecular formula of two carbon compounds are C4H8 and C3H8. Which one of the two is most likely to show addition reaction? Justify your answer.", + "Thus, compound X will undergo addition reaction.\n\nQuestion 62.\nThe molecular formula of two carbon compounds are C4H8 and C3H8. Which one of the two is most likely to show addition reaction? Justify your answer. Also give the chemical equation to explain the process of addition reaction in this case. (Delhi 2017)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nC3H8 belongs to alkane series (CnH2n+2) Refer to answer 61.\nQuestion 63.\nWhat is an oxidising agent? What happens when an oxidising agent is added to propanol? Explain with the help of a chemical equation. (Delhi 2016)\nAnswer:\nThe substance that supply oxygen in a reaction for oxidation is called oxidising agent e.g., potassium permanganate, potassium dichromate, etc.\nWhen propanol is heated with alkaline KMnO4, it gets oxidised to propanoic acid.\n\nQuestion 64.\nDraw the electron-dot structure for ethyne. A mixture of ethyne and oxygen is burnt for welding. In your opinion, why cannot we use a mixture of ethyne and air for this purpose? (AI 2015)\nAnswer:\nThe formula for ethyne is C2H2 and its electron dot structure is :", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nThe formula for ethyne is C2H2 and its electron dot structure is :\n\nA mixture of ethyne and oxygen is burnt for welding so that complete oxidation of ethyne takes place. If in place of oxygen, air is taken which contains less amount of oxygen then incomplete combustion of oxygen takes place and temperature required for welding will not be attained.\nQuestion 65.\nWrite the name and general formula of a chain of hydrocarbons in which an addition reaction with hydrogen is possible. State the essential condition for an addition reaction. Stating this condition, write a chemical equation giving the name of the reactant and the product of the reaction. (AI 2015, Delhi 2014)\nAnswer:\nAlkenes, having general formula as CnH2n and alkynes, having general formula as CnH2n-2 are the class of hydrocarbons in which addition reaction is possible.\nThe essential conditions for addition reaction are :\n(i) Presence of unsaturated hydrocarbon.\n(ii) Presence of catalyst such as Ni/Pt/Pd.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nThe essential conditions for addition reaction are :\n(i) Presence of unsaturated hydrocarbon.\n(ii) Presence of catalyst such as Ni/Pt/Pd.\nLet us take an example of ethene. It undergoes addition reaction with hydrogen when it is heated in the presence of nickel catalyst to form ethane. The reaction is known as hydrogenation.\n\nQuestion 66.\nWhy are certain compounds called hydrocarbons? Write the general formula for homologous series of alkanes, alkenes and alkynes and also draw the structure of the first member of each series. Write the name of the reaction that converts alkenes into alkanes and also write a chemical equation to show the necessary conditions for the reaction to occur. (AI 2017)\nAnswer:\nRefer to answers 50 and 65.\nQuestion 67.\nWhat are hydrocarbons? Write the name and general formula of\n(i) saturated hydrocarbons\n(ii) unsaturated hydrocarbons, and draw the structure of one hydrocarbon of each type. How can an unsaturated hydrocarbon be made saturated? (AI 2012)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(i) saturated hydrocarbons\n(ii) unsaturated hydrocarbons, and draw the structure of one hydrocarbon of each type. How can an unsaturated hydrocarbon be made saturated? (AI 2012)\nAnswer:\nRefer to answer 50.\nUnsaturated hydrocarbons can be made to saturated hydrocarbons by hydrogenation reaction using nickel (Ni) as a catalyst.\n\nQuestion 68.\n(a) With the help of a suitable example, explain the process of hydrogenation mentioning the conditions of the reaction and any one change in physical property with the formation of the product. (Delhi 2015, 2013, Foreign 2012)\n(b) How does a saturated hydrocarbon react with chlorine? Write chemical equation for it. What type of reaction is it called and why? (Foreign 2012)\nAnswer:\n\ngroups that can be hydrogenated.\nHydrogenation is the addition of hydrogen to an unsaturated hydrocarbon to obtain a saturated hydrocarbon.\n\nHere R can be any alkyl group.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\n\ngroups that can be hydrogenated.\nHydrogenation is the addition of hydrogen to an unsaturated hydrocarbon to obtain a saturated hydrocarbon.\n\nHere R can be any alkyl group.\nThere is the change of unsaturated compound from the liquid state to saturated compound in the solid state thus, melting point increases.\n(b) Saturated hydrocarbon reacts with chlorine to form a substituted product, e.g.,\n\nThis reaction is called substitution reaction as here one hydrogen of methane is substituted by one chlorine atom.\nQuestion 69.\nAssertion (A) : Esterification is a process in which a sweet smelling substance is produced.\nReason (R): When esters react with sodium hydroxide, an alcohol and sodium salt of carboxylic acid are obtained.\n(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.", + "(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true. (2020)", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true. (2020)\nAnswer:\n(b): When an ester reacts with the base saponification reaction occurs.\nQuestion 70.\nAssertion (A) : Ethanoic acid is also known as glacial acetic acid.\nReason (R) : The melting point of pure ethanoic acid is 290 K and hence it often freezes during winters in cold climates.\n(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true.", + "(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true. (2020)\nAnswer:\n(a): Pure ethanoic acid or acetic acid freezes below room temperature into white crystals that resemble glaciers.\nQuestion 71.\nDraw the structure for ethanoic acid molecule, CH3COOH. (AI 2011)\nAnswer:\nStructure of ethanoic acid is\n\nQuestion 72.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 71.\nDraw the structure for ethanoic acid molecule, CH3COOH. (AI 2011)\nAnswer:\nStructure of ethanoic acid is\n\nQuestion 72.\nA compound \u2018X\u2019 on heating with excess cone, sulphuric acid at 443 K gives an unsaturated compound \u2018Y\u2019. \u2018X\u2019 also reacts with sodium metal to evolve a colourless gas \u2018Z\u2019. Identify \u2018X\u2019, \u2018Y\u2019 and \u2018Z\u2019. Write the equation of the chemical reaction of formation of \u2018Y\u2019 and also write the role of sulphuric acid in the reaction. (2018)\nAnswer:\nAs X reads with cone. H2SO4 to give an alkene so it should be an alcohol as cone. H2SO4 acts as a dehydrating agent. The reaction of X with Na also confirms that it is an alcohol because alcohols react with Na metal to evolve colourless hydrogen gas.\n\nHere, conc. H2SO4 acts as a dehydrating agent i.e., helps in the removal of water.\n\nQuestion 73.", + "The reaction of X with Na also confirms that it is an alcohol because alcohols react with Na metal to evolve colourless hydrogen gas.\n\nHere, conc. H2SO4 acts as a dehydrating agent i.e., helps in the removal of water.\n\nQuestion 73.\nWrite the chemical equations to show what happens when\n(i) an ester reacts with a base?\n(ii) ethanol reacts with ethanoic acid in the presence of sulphuric acid? (2/3, Foreign 2014)", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 73.\nWrite the chemical equations to show what happens when\n(i) an ester reacts with a base?\n(ii) ethanol reacts with ethanoic acid in the presence of sulphuric acid? (2/3, Foreign 2014)\nAnswer:\n(i) When an ester reacts with the base then it gives sodium salt of carboxylic acid and an alcohol. It is known as saponification reaction.\n\n(ii) Carboxylic acids react with alcohols in the presence of a little concentrated sulphuric acid to form pleasant smelling esters. This reaction is called esterification reaction.\n\nQuestion 74.\nWrite the respective chemical equations to show what happens when\n(i) ethanol is heated with concentrated sulphuric acid at 443 K ?\n(ii) ethanol reacts with ethanoic acid in the presence of an acid acting as a catalyst? (2/3, Foreign 2014)\nAnswer:\n\n(ii) Refer to answer 73(ii).\nQuestion 75.", + "(ii) ethanol reacts with ethanoic acid in the presence of an acid acting as a catalyst? (2/3, Foreign 2014)\nAnswer:\n\n(ii) Refer to answer 73(ii).\nQuestion 75.\nWrite one chemical equation to represent each of the following types of reactions of organic substances:\n(i) Esterification\n(ii) Saponification (2/3, Delhi 2011)", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 75.\nWrite one chemical equation to represent each of the following types of reactions of organic substances:\n(i) Esterification\n(ii) Saponification (2/3, Delhi 2011)\nAnswer:\n(i) Refer to answer 73(ii).\n(ii) Refer to answer 73(i).\nQuestion 76.\nComplete the following chemical equations : (Delhi 2017)\n(i) CH3COOC2H5 + NaOH \u2192\n(ii) CH3COOH + NaOH \u2192\n\nAnswer:\n\nQuestion 77.\nComplete the following chemical equations: (Delhi 2017)\n(i) C2H5OH + O2 \u2192\n\n(iii) CH3COOH + NaHCO3 \u2192\nAnswer:\n\n(ii) Refer to answer 74(i).\n\nQuestion 78.\nWrite the structural formula of ethanol. What happens when it is heated with excess of cone. H2SO4 at 443 K? Write the chemical equation for the reaction stating the role of cone. H2SO4 in this reaction.", + "Question 78.\nWrite the structural formula of ethanol. What happens when it is heated with excess of cone. H2SO4 at 443 K? Write the chemical equation for the reaction stating the role of cone. H2SO4 in this reaction. (AI 2017, Delhi 2015, 2013)\nAnswer:\nThe structural formula of ethanol (C2H5OH) is\n\nWhen ethanol is heated with conc. H2SO4 at 443 K then it looses a water molecule to form unsaturated alkene (ethene) as a product.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nThe structural formula of ethanol (C2H5OH) is\n\nWhen ethanol is heated with conc. H2SO4 at 443 K then it looses a water molecule to form unsaturated alkene (ethene) as a product.\n\nHere conc. H2SO4 acts as a dehydrating agent i.e., helps in the removal of water.\nQuestion 79.\nWhat happens when (write chemical equation in each case)\n(a) ethanol is burnt in air?\n(b) ethanol is heated with excess cone. H2SO4 at 443 K?\n(c) a piece of sodium is dropped into ethanol? (AI 2017)\nAnswer:\n(a) Refer to answer 77(i).\n(b) Refer to answer 74(i).\n(c) When a small piece of sodium is dropped into ethanol then hydrogen gas is liberated which burns with a pop sound.\n2C2H5OH + 2Na \u2192 2C2H5O\u2013Na+ + H2 \u2191\nQuestion 80.\nDistinguish between esterification and saponification reaction with the help of the chemical equations for each.", + "2C2H5OH + 2Na \u2192 2C2H5O\u2013Na+ + H2 \u2191\nQuestion 80.\nDistinguish between esterification and saponification reaction with the help of the chemical equations for each. State one use of each (i) esters, and (ii) saponification process. (AI 2017, Foreign 2012)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\n\nUse of esters: They are used for making perfumes or used as artificial flavouring substances.\nUse of saponification process : This process is used in making soaps.\nQuestion 81.\nExplain esterification reaction with the help of a chemical equation. Describe an activity to show esterification. (AI 2017)\nAnswer:\nRefer to answer 80.\nAim : To demonstrate esterification process using ethanol and acetic acid.\nMaterials required : Beaker, water, test tube, ethanol, acetic acid, cone. H2SO4, tripod stand, burner, wire gauze, etc.\n\nProcedure:\n\u2013 Take 2 mL of ethanol in a test tube.\n\u2013 Take 2 mL of ethanoic acid (acetic acid) into it.\n\u2013 Add few drops of cone. H2SO4\n\u2013 Warm it in a beaker containing water.\n\u2013 Observe the smell of the products formed. Observations: Pleasant fruity smelling compound (called ester) is formed.\nChemical reaction:\n\nConclusion : Carboxylic acid reacts with alcohol in presence of cone.", + "H2SO4\n\u2013 Warm it in a beaker containing water.\n\u2013 Observe the smell of the products formed. Observations: Pleasant fruity smelling compound (called ester) is formed.\nChemical reaction:\n\nConclusion : Carboxylic acid reacts with alcohol in presence of cone. H2SO4 which acts as a dehydrating agent to form esters.\nQuestion 82.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nChemical reaction:\n\nConclusion : Carboxylic acid reacts with alcohol in presence of cone. H2SO4 which acts as a dehydrating agent to form esters.\nQuestion 82.\nWhen ethanol reacts with ethanoic acid in the presence of cone. H2SO4, a substance with fruity smell is produced. Answer the following:\n(i) State the class of compounds to which the fruity smelling compounds belong. Write the chemical equation for the reaction and write the chemical name of the product formed.\n(ii) State the role of cone. H2SO4 in this reaction. (Delhi 2016)\nAnswer:\n(i) When ethanol reacts with ethanoic acid in presence of cone. H2SO4, ethyl ethanoate is formed which belongs to the class of ester compounds, having fruity smell.\n\n(ii) The above reaction is called esterification which occurs in presence of cone. H2SO4 which acts as a dehydrating agent and helps in the removal of water. Cone. H2SO4 also acts as a catalyst to speed up the reaction.\nQuestion 83.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 83.\nName the compound formed when ethanol is heated in excess of cone, sulphuric acid at 443 K. Also write the chemical equation of the reaction stating the role of cone, sulphuric acid in it. What would happen if hydrogen is added to the product of this reaction in the presence of catalyst such as palladium or nickel? (Delhi 2016, Foreign 2015)\nAnswer:\nRefer to answer 78.\nIf hydrogen is added to ethene in presence of palladium or nickel catalyst then one atom of hydrogen adds to each carbon atom of ethene to form ethane.\n\nQuestion 84.\nWrite chemical equation of the reaction of ethanoic acid with the following :\n(a) Sodium;\n(b) Sodium hydroxide;\n(c) Ethanol\nWrite the name of one main product of each reaction. (AI 2016)\nAnswer:\nEthanoic acid reacts with sodium as well as sodium hydroxide to form sodium ethanoate.\n\n(b) Refer to answer 76(ii).\n(c) Refer to answer 76(iii).\nQuestion 85.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nEthanoic acid reacts with sodium as well as sodium hydroxide to form sodium ethanoate.\n\n(b) Refer to answer 76(ii).\n(c) Refer to answer 76(iii).\nQuestion 85.\nOn dropping a small piece of sodium in a test tube containing carbon compound \u2018X\u2019 with molecular formula C2H6O, a brisk effervescence is observed and a gas \u2018Y\u2019 is produced. On bringing a burning splinter at the mouth of the test tube the gas evolved burns with a pop sound. Identify \u2018X\u2019 and \u2018Y\u2019. Also write the chemical equation for the reaction. Write the name and structure of the product formed, when you heat \u2018X\u2019 with excess cone, sulphuric acid. (AI 2016)\nAnswer:\nEthanol reacts with sodium to form sodium ethoxide and hydrogen gas is liberated which burns with a pop sound.\n\nThus, compound X is ethanol and gas Y is hydrogen gas.\nWhen ethanol is heated with excess of concentrated sulphuric acid then it gets dehydrated to form ethene.\n\nQuestion 86.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nThus, compound X is ethanol and gas Y is hydrogen gas.\nWhen ethanol is heated with excess of concentrated sulphuric acid then it gets dehydrated to form ethene.\n\nQuestion 86.\nWrite three different chemical reactions showing the conversion of ethanoic acid to sodium ethanoate. Write balanced chemical equation in each case. Write the name of the reactants and the products other than ethanoic acid and sodium ethanoate in each case. (AI 2016)\nAnswer:\nEthanoic acid reacts with Na2CO3 to form sodium ethanoate and CO2 gas is liberated.\n\nWith sodium hydrogen carbonate it forms sodium ethanoate.\n\nWith NaOH it forms sodium ethanoate.\n\nQuestion 87.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nEthanoic acid reacts with Na2CO3 to form sodium ethanoate and CO2 gas is liberated.\n\nWith sodium hydrogen carbonate it forms sodium ethanoate.\n\nWith NaOH it forms sodium ethanoate.\n\nQuestion 87.\nWrite the nam e and molecular formula of an organic compound having its name suffixed with \u2018ol\u2019 and having two carbon atoms in its molecule. Write balanced chemical equation to indicate what happens when this compound is heated with excess cone. H2SO4 and the narpe of main product formed. Also state the role of cone. H2SO4 in the reaction. (Foreign 2016)\nAnswer:\nThose organic compounds having suffix \u2018oF are alcohols. As the alcohol is having two carbon atoms in its molecule so, it is ethanol.\n\nRefer to answer 78.\nQuestion 88.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nThose organic compounds having suffix \u2018oF are alcohols. As the alcohol is having two carbon atoms in its molecule so, it is ethanol.\n\nRefer to answer 78.\nQuestion 88.\nAn organic compound \u2018P\u2019 is a constituent of wine. \u2018P\u2019 on reacting with acidified K2Cr2O7 forms another compound \u2018Q\u2019. When a piece of sodium is added to \u2018Q\u2019, a gas \u2018R\u2019 evolves which burns with a pop sound. Identify P, Q and R and write the chemical equations of the reactions involved. (Foreign 2016)\nAnswer:\n\u2018P\u2019 is ethanol which is a constituent of wine. Ethanol on reacting with acidified potassium dichromate (K2Cr2O7) solution gives ethanoic acid \u2018Q\u2019.\n\nWhen a piece of sodium is added to ethanoic acid then sodium salt of ethanoic acid is formed with the liberation of hydrogen gas which burns with a pop sound.\n\nQuestion 89.\nList two tests for experimentally distinguishing between an alcohol and a carboxylic acid and describe how these tests are performed. (AI 2015)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 89.\nList two tests for experimentally distinguishing between an alcohol and a carboxylic acid and describe how these tests are performed. (AI 2015)\nAnswer:\nTests for distinguishing between an alcohol and a carboxylic acid are :\n(i) Litmus test : When we place a drop of carboxylic acid on blue litmus paper it turns red while alcohol will not change the colour of blue litmus paper.\n(ii) Sodium hydrogen carbonate test/sodium carbonate test: If a pinch of NaHCO3 or Na2CO3 is added to two test tubes containing alcohol and carboxylic acid respectively, then test tube containing carboxylic acid will show the evolution of colourless gas with brisk effervescence while test tube containing alcohol does not show any reaction.\nQuestion 90.\nWhat are esters? How are they prepared? List two uses of esters. (Delhi 2014)\nAnswer:\nEsters are generally volatile liquids which have pleasant fruity smell.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 90.\nWhat are esters? How are they prepared? List two uses of esters. (Delhi 2014)\nAnswer:\nEsters are generally volatile liquids which have pleasant fruity smell.\nEsters are prepared when a carboxylic acid reacts with an alcohol in the presence of small amount of concentrated H2SO4. For example, when ethanoic acid reacts with ethanol it forms an ester (i.e. ethyl ethanoate).\n\nUses of ester:\n1. It is used in making perfumes.\n2. It is used in making artificial flavours and essences used in ice-creams, sweets and cold drinks.\nQuestion 91.\nA carboxylic acid (molecular formula, C2H4O2) reacts with an alcohol in the presence of an acid catalyst to form a compound \u2018X\u2019. The alcohol on oxidation with alkaline KMnO4 followed by acidification gives the same carboxylic acid C2H4O2.", + "The alcohol on oxidation with alkaline KMnO4 followed by acidification gives the same carboxylic acid C2H4O2. Write the name and structure of (i) carboxylic acid, (ii) alcohol and (iii) the compound \u2018X\u2019 (AI 2014)\nAnswer:\nThe molecular formula of carboxylic acid is C2H4O2. Thus, it should be acetic acid (ethanoic acid).", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nThe molecular formula of carboxylic acid is C2H4O2. Thus, it should be acetic acid (ethanoic acid).\n\nIt reacts with alcohol in presence of acid catalyst to give compound \u2018X\u2019.\nAs alcohol on oxidation with alkaline KMnO4 gives the same acid i.e. ethanoic acid, hence alcohol must contain two carbon atoms. Thus, formula for alcohol is CH3CH2OH i.e. ethanol.\nReactions involved are:\n\n(i) Structure of ethanoic acid :\n\n(ii) Structure of ethanol:\n\n(iii) Structure of ethyl ethanoate (X):\n\nQuestion 92.\nWrite the chemical equation to explain what happens when ethanol is heated with alkaline solution, of potassium permanganate. Mention two physical properties and two uses of ethanol. (Foreign 2014)\nAnswer:\nWhen ethanol is heated with alkaline solution of potassium permanganate then oxidation of ethanol takes place to form ethanoic acid.\n\nTwo physical properties of ethanol are:\n1. It is liquid at room temperature.\n2. It is soluble in water in all proportions.\nTwo uses of ethanol are :", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nTwo physical properties of ethanol are:\n1. It is liquid at room temperature.\n2. It is soluble in water in all proportions.\nTwo uses of ethanol are :\n1. It is used as a liquor for drinking purpose.\n2. It is a good solvent and hence, it is used in medicines such as tincture of iodine, cough syrup and many tonics.\nQuestion 93.\nWrite chemical equations to describe two examples of different oxidations of ethanol. List two uses of ethanol. (Foreign 2014)\nAnswer:\nAddition of oxygen to any substance is called oxidation.\nEthanol gets oxidised to ethanoic acid as :\n\nRefer to answer 92.\nQuestion 94.\nWrite the chemical equations to show what happens when\n(i) sodium hydroxide is added to ethanoic acid?\n(ii) solid sodium hydrogen carbonate is added to ethanoic acid?\n(iii) ethanol reacts with sodium? (Foreign 2014)\nAnswer:\n(i) Refer to answer 76(ii).\n(ii) Refer to answer 77(iii).\n(iii) Refer to answer 79(c).\nQuestion 95.\nWrite chemical equations for what happens when", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\n(i) Refer to answer 76(ii).\n(ii) Refer to answer 77(iii).\n(iii) Refer to answer 79(c).\nQuestion 95.\nWrite chemical equations for what happens when\n(i) sodium metal is added to ethanoic acid?\n(ii) solid sodium carbonate is added to ethanoic acid?\n(iii) ethanoic acid reacts with a dilute solution of sodium hydroxide? (AI 2011)\nAnswer:\n(i) Refer to answer 84(a).\n(ii) Refer to answer 86.\n(iii) Refer to answer 76(ii).\nQuestion 96.\n(a) What is a homologous series? Explain with an example.\n(b) Define the following terms giving one example of each.\n(i) Esterification (ii) Addition reaction (2020)\nAnswer:\n(a) Refer to answer 22.\nFor example, alkane series has general formula CnH2n + 2.\nFirst member of homologous series of alkane is .methane, i.e., CH4.", + "(i) Esterification (ii) Addition reaction (2020)\nAnswer:\n(a) Refer to answer 22.\nFor example, alkane series has general formula CnH2n + 2.\nFirst member of homologous series of alkane is .methane, i.e., CH4.\nSecond member of homologous series of alkane is ethane, i.e., C2H6.\nThird member of homologous series of alkane is propane i.e., C3H8.\n(b) (i) Refer to answer 73(ii).", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nSecond member of homologous series of alkane is ethane, i.e., C2H6.\nThird member of homologous series of alkane is propane i.e., C3H8.\n(b) (i) Refer to answer 73(ii).\n(ii) Addition reactions : Those reactions in which atoms or group of atoms are simply added to a double or triple bond without the elimination of any atom or molecule, are known as addition reactions.\n\nQuestion 97.\n(a) Carry out following conversions :\n(i) Ethanol to ethene\n(ii) Ethanol to ethanoic acid\n(b) Differentiate between addition reaction and substitution reaction. Give one example of each. (2020)\nAnswer:\n(a) (i) When ethanol is heated with cone. H2SO4 at 443 K, ethene is obtained due to dehydration of ethanol.\n\n(ii) When 5 % alkaline KMnO4 solution is added drop by drop to warm ethanol then it gets oxidised to ethanoic acid.\n\n(b) Refer to answer 96(ii).", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(ii) When 5 % alkaline KMnO4 solution is added drop by drop to warm ethanol then it gets oxidised to ethanoic acid.\n\n(b) Refer to answer 96(ii).\nSubstitution reactions : The reactions which involve the displacement or substitution of an atom or a group of atoms in an organic compound by another atom or group of atoms, are known as substitution reactions.\nSaturated hydrocarbons are fairly unreactive and inert in the presence of most of the reagents. However, in presence of sunlight, hydrocarbons undergo rapid substitution reactions, e.g.,\n\nQuestion 98.\nWrite the chemical formula and name of the compound which is the active ingredient of all alcoholic drinks. List its two uses. Write chemical equation and name of the product formed when this compound reacts with\n(i) sodium metal\n(ii) hot concentrated sulphuric acid. (Delhi 2019)\nAnswer:\nEthanol having chemical formula C2H5OH is the active ingredient of all alcoholic drinks.\nUses of ethanol:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(i) sodium metal\n(ii) hot concentrated sulphuric acid. (Delhi 2019)\nAnswer:\nEthanol having chemical formula C2H5OH is the active ingredient of all alcoholic drinks.\nUses of ethanol:\n1. Ethanol is widely used in industry as a solvent.\n2. Ethanol is used as an antiseptic for wounds in the form of rectified spirit.\nChemical equations:\n(i) Refer to answer 79(c).\n(ii) Refer to answer 74(i).\nQuestion 99.\n(a) Define the term isomer.\n(b) Two compounds have same molecular formula C3H6O. Write the name of these compounds and their structural formula.\n(c) How would you bring the following conversions:\n(i) Ethanol to ethene\n(ii) Propanol to propanoic acid? (AI 2019)\nAnswer:\n(a) Refer to answer 39.\n(b) Refer to answer 51 (b).\n(c) (i) Refer to answer 97(a)(i).\n\nQuestion 100.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(i) Ethanol to ethene\n(ii) Propanol to propanoic acid? (AI 2019)\nAnswer:\n(a) Refer to answer 39.\n(b) Refer to answer 51 (b).\n(c) (i) Refer to answer 97(a)(i).\n\nQuestion 100.\nA carbon compound \u2018P\u2019 on heating with excess cone. H2SO4 forms another carbon compound \u2018Q\u2019 which on addition of hydrogen in the presence of nickel catalyst forms a saturated carbon compound \u2018R\u2019 One molecule of \u2018R\u2019 on combustion forms two molecules of carbon dioxide and three molecules of water. Identify P, Q and R and write chemical equations for the reactions involved. (AI 2016)\nAnswer:\nWhen ethanol is heated with excess of concentrated H2SO4 it gets dehydrated to form ethene.\n\nWhen ethene is heated with hydrogen in presence of nickel catalyst it forms ethane.\n\nEthane on oxidation gives two moles of carbon dioxide and three moles of water.\nCH3CH3 + \n7\n2\n O2 \u2192 2CO2 + 3H2O + heat + light\nQuestion 101.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nEthane on oxidation gives two moles of carbon dioxide and three moles of water.\nCH3CH3 + \n7\n2\n O2 \u2192 2CO2 + 3H2O + heat + light\nQuestion 101.\nList in tabular form three physical and two chemical properties on the basis of which ethanol and ethanoic acid can be differentiated. (Delhi 2012)\nAnswer:\n(ii)\nQuestion 102.\n(a) In a tabular form, differentiate between ethanol and ethanoic acid under the following heads:\n(i) Physical state\n(ii) Taste\n(iii) NaHCO3 test\n(iv) Ester test\n(b) Write a chemical reaction to show the dehydration of ethanol. (Delhi 2011)\nAnswer:\n(a)\n\n(b) Refer to answer 97(a)(i).\nQuestion 103.\nSeveral factories were pouring their wastes in rivers A and B. Water samples were collected from these two rivers. It was observed that sample collected from river A was acidic while that of river B was basic.", + "Question 103.\nSeveral factories were pouring their wastes in rivers A and B. Water samples were collected from these two rivers. It was observed that sample collected from river A was acidic while that of river B was basic. The factories located near A and B are\n(a) Soaps and detergents factories near A and alcohol distillery near B.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(a) Soaps and detergents factories near A and alcohol distillery near B.\n(b) Soaps and detergents factories near B and alcohol distillery near A.\n(c) Lead storage battery manufacturing factories near A and soaps and detergents factories near B.\n(d) Lead storage battery manufacturing factories near B and soaps and detergents factories near A. (2020)\nAnswer:\n(c) Lead storage battery manufacturing factories near A and soaps and detergents factories near B.\nQuestion 104.\nWhy does micelle formation take place when soap is added to water? Why are micelles not formed when soap is added to ethanol? (3/5, AI 2011)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nQuestion 104.\nWhy does micelle formation take place when soap is added to water? Why are micelles not formed when soap is added to ethanol? (3/5, AI 2011)\nAnswer:\nA soap molecule has two ends with different properties, one end is polar i.e., water soluble or hydrophilic while other end is non-polar i.e., water insoluble or hydrophobic. When soap is added to water, the polar ends get dissolve in water and non-polar ends get dissolved in each other and directed towards the centre. As a result, a spherical ionic molecule known as micelles, formation takes place. Since, soaps are soluble in ethanol, therefore, micelles formation does not occur.\nQuestion 105.\nSoaps and detergents are both, types of salts. State the difference between the two. Write the mechanism of the cleansing action of soaps. Why do soaps not form lather (foam) with hard water? Mention any two problems that arise due to the use of detergents instead of soaps. (Delhi 2017, AI 2015)\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nSoaps are the sodium or potassium salts of higher fatty acids. The ionic group in soaps is -COO\u2013Na+.\nOn the other hand, synthetic detergents are the sodium salts of a long chain alkylbenzenesulphonic acids or long chain alkyl hydrogen sulphates. The ionic group in synthetic detergents is\n-SO3\u2013 Na+ or -OSO3 \u2013Na+\nCleansing action of soap :\nA soap molecule contains a polar part (COO\u2013Na+) called polar end and a non-polar part consisting of a long chain carbon atoms. This part is called hydrocarbon end.\nThe polar end is water soluble whereas hydrocarbon part is water-repellent and oil soluble.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nThe polar end is water soluble whereas hydrocarbon part is water-repellent and oil soluble.\n\nWhen an oily (dirty) piece of cloth is put into soap solution, the hydrocarbon part of the molecule attaches itself to the oily drop and the -COO\u2013 end orients itself towards water. Na+ ions in solution arrange themselves around the -COO\u2013 ions. The negatively charged micelle so formed entraps the oily dirt. The negatively charged micelle repel each other due to the electrostatic repulsion. As a result, the tiny oily dirt particles do not come together and get washed away in water during rinsing.\n\nIn hard water, soap does not form lather as hard water contains Ca2+and Mg2+ ions. Soap reacts with these ions to form insoluble calcium and magnesium salts of fatty acids.\n\nTwo problems which arise due to the use of detergents instead of soaps are :\n(i) Synthetic detergents are non-biodegradable and hence, cause water pollution.\n(ii) Synthetic detergents also cause skin related problems.\nQuestion 106.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(i) Synthetic detergents are non-biodegradable and hence, cause water pollution.\n(ii) Synthetic detergents also cause skin related problems.\nQuestion 106.\nWhat are micelles? Why does it form when soap is added to water? Will a micelle be formed in other solvents such as ethanol also? State briefly how the formation of micelles help to clean the clothes having oily spots. (Foreign 2016)\nAnswer:\nRefer to answer 104.\nRefer to answer 105.\nQuestion 107.\n(a) You have three unlabelled test tubes containing ethanol, ethanoic acid and soap solution. Explain the method you would use to identify the compounds in different test tubes by chemical tests using litmus paper and sodium metal.\n(b) Give the reason of formation of scum when soaps are used with hard water. (Foreign 2016)\nAnswer:\n(a) The tests may be tabulated as below:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(b) Give the reason of formation of scum when soaps are used with hard water. (Foreign 2016)\nAnswer:\n(a) The tests may be tabulated as below:\n(b) Hard water contains hydrogen carbonates, chlorides and sulphates of calcium and magnesium. When soap is added to hard water it reacts with these salts to form scum which is insoluble in water and floats on the top of the water surface. The scum is formed due to the formation of insoluble calcium or magnesium salts of fatty acids.\n\nQuestion 108.\nWhat is the difference between the molecules of soaps and detergents, chemically? Explain the cleansing action of soaps. (Delhi 2015)\nAnswer:\nRefer to answer 105.\nQuestion 109.\nWhat is the difference between the chemical composition of soaps and detergents? State in brief the action of soaps in removing an oily spot from a shirt. Why are soaps not considered suitable for washing where water is hard? (Delhi 2012)\nAnswer:\nRefer to answers 105 and 107(b).\nQuestion 110.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answers 105 and 107(b).\nQuestion 110.\nWhat are detergents chemically? List two merits and two demerits of using detergents for cleansing. State the reason for the suitability of detergents for washing, even in the case of water having calcium and magnesium ions. (AI 2012)\nAnswer:\nDetergents are generally ammonium or sulphonate or sulphate salts of long chain carboxylic acids. The more common detergents are sodium salts of long chain alkyl benzene sulphonic acids.\nMerits of using detergents :\n(i) Detergents are very strong cleansing agents.\n(ii) They can form lather well even in hard water as they do not form insoluble calcium or magnesium salts.\nDemerits of using detergents :\n(i) As detergents are sodium salts of long chain alkyl benzene sulphonic acids which are very bulky molecules, are not easily degraded by bacteria and hence, they are non-biodegradable.\n(ii) They are highly basic in nature and cause damage to skin.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(ii) They are highly basic in nature and cause damage to skin.\nSynthetic detergents can be used even in hard water because they do not react with Ca2+ and Mg2+ ions present in hard water. They do not form curdy white precipitates (scum) of calcium and magnesium salts of fatty acids.\nQuestion 111.\nWhat are soaps and detergents chemically? Explain the action of cleaning by soaps. State the reason why we can wash our clothes even in hard water using detergents. (Foreign 2012)\nAnswer:\nRefer to answers 105 and 110.\nQuestion 112.\n(a) What is a soap? Why are soaps not suitable for washing clothes when the water is hard?\n(b) Explain the action of soap in removing an oily spot from a piece of cloth. (Delhi 2011)\nAnswer:\n(a) Refer to answers 105 and 107(b).\n(b) Refer to answer 105.\nQuestion 113.\n(a) What is a detergent? Name one detergent.", + "(b) Explain the action of soap in removing an oily spot from a piece of cloth. (Delhi 2011)\nAnswer:\n(a) Refer to answers 105 and 107(b).\n(b) Refer to answer 105.\nQuestion 113.\n(a) What is a detergent? Name one detergent.\n(b) Write two advantages and two dis-advantages of using detergents over soaps.\n(c) Why, by using a detergent, can we wash clothes even in hard water?", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- IMPORTANT QUESTIONS.txt\n(a) What is a detergent? Name one detergent.\n(b) Write two advantages and two dis-advantages of using detergents over soaps.\n(c) Why, by using a detergent, can we wash clothes even in hard water?\nAnswer:\n(a) Detergents are ammonium or sulphonate or sulphate salts of long chain hydrocarbons containing 12-18 carbon atoms e.g., dodecyl benzene sulphonate.\n(b) Refer to answer 110.\n(c) Refer to answer 110.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 4 Intext Questions\nPage Number: 61\nQuestion 1\nWhat would be the electron dot structure of carbon dioxide which has the formula CO2 ?\nAnswer:\n\nQuestion 2\nWhat would be electron dot structure of sulphur which is made up of eight atoms of sulphur.\nAnswer:\n\nPage Number: 68 \u2013 69\nQuestion 1\nHow many structural isomers can you draw for pentane ?\nAnswer:\nThree, these are n-pentane, iso-pentane and neo-pentane.\n\nQuestion 2\nWhat are the two properties of carbon which lead to the huge number of carbon compounds we see around us ?\nAnswer:\n(i) Tetravalency\n(ii) Catenation.\nQuestion 3\nWhat will be the formula and electron dot structure of cyclopentane ?\nAnswer:\nThe molecular formula of cyclopentane is C5 H10 .\nThe electron dot structure of cyclopentane is given on the next page.", + "Question 3\nWhat will be the formula and electron dot structure of cyclopentane ?\nAnswer:\nThe molecular formula of cyclopentane is C5 H10 .\nThe electron dot structure of cyclopentane is given on the next page.\n\nQuestion 4\nDraw the structures for the following compounds :\n(i) Ethanoic acid\n(ii) Bromopentane\n(iii) Butanone\n(iv) Hexanal\nAnswer:\n(i) Ethanoic acid (CH3COOH)", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nQuestion 4\nDraw the structures for the following compounds :\n(i) Ethanoic acid\n(ii) Bromopentane\n(iii) Butanone\n(iv) Hexanal\nAnswer:\n(i) Ethanoic acid (CH3COOH)\n\n(ii) Bromopentane (C5H11Br)\n\n(iii) Butanone (CH3 \u2014 CH2 \u2014 COCH3)\n\n(iv) Hexanal (C5H11CHO)\n\nStructural isomers for bromopentane: There are three structural isomers for bromopentane depending on the position of Br at carbon 1, 2, 3.\n\nPositions 4 and 5 are same as 1, 2.\nQuestion 5\nHow would you name the following compounds ?\n\nAnswer:\n(i) Bromoethane\n(ii) Methanal\n(iii) 1 \u2013 Hexyne\nPage Number: 71\nQuestion 1\nWhy is the conversion of ethanol to ethanoic acid an oxidation reaction ?\nAnswer:\nConversion of ethanol into ethanoic acid is an oxidation reaction because addition of oxygen to a substance is called oxidation.", + "Answer:\nConversion of ethanol into ethanoic acid is an oxidation reaction because addition of oxygen to a substance is called oxidation. Here, oxygen is added to ethanol by oxidising agent like alkaline potassium permanganate or acidified potassium dichromate and it is converted into acid.\n\nQuestion 2", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nQuestion 2\nA mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used ?\nAnswer:\nA mixture of ethyne and air is not used for welding because burning of ethyne in air produces a sooty flame due to incomplete combustion, which is not enough to melt metals for welding.\nPage Number: 74\nQuestion 1\nHow would you distinguish experimentally between an alcohol and a carboxylic acid ?\nAnswer:\nDifferences between alcohol and carboxylic acid\nQuestion 2\nWhat are oxidising agents ?\nAnswer:\nOxidising agents are the substances which give oxygen to another substances or which remove hydrogen from a substance.\nFor example, acidic K2Cr2O7 is an oxidising agent, that converts (oxidises) ethanol into ethanoic acid.\nPage Number: 76\nQuestion 1\nWould you be able to check if water is hard by using a detergent ?\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nPage Number: 76\nQuestion 1\nWould you be able to check if water is hard by using a detergent ?\nAnswer:\nNo, because detergents can lather well even in hard water. They do not form insoluble calcium or magnesium salts (scum). On reacting with the calcium ions and magnesium ions present in the hard water.\nQuestion 2\nPeople use a variety of methods to wash clothes. Usually after adding the soap, they \u2018beat\u2019 the clothes on a stone, or beat it with a paddle, scrub with a brush or the mixture is agitated in a washing machine. Why is agitation necessary to get clean clothes ?\nAnswer:\nIt is necessary to agitate to get clean clothes because the soap micelles which entrap oily or greasy particles on the surface of dirty cloth have to be removed from its surface. When the cloth wetted in soap solution is agitated or beaten, the micelles containing oily or greasy dirt get removed from the surface of dirty cloth and go into water and the dirty cloth gets cleaned.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 4 Textbook Chapter End Questions\nQuestion 1\nEthane, with the molecular formula C2H6 has\n(a) 6 covalent bonds\n(b) 7 covalent bonds\n(c) 8 covalent bonds\n(d) 9 covalent bonds\nAnswer:\n(b) 7 covalent bonds.\nQuestion 2\nButanone is a four-carbon compound with the functional group\n(a) carboxylic acid\n(b) aldehyde\n(c) ketone\n(d) alcohol\nAnswer:\n(c) Ketone.\nQuestion 3\nWhile cooking, if the bottom of the vessel is getting blackened on the outside, it means that\n(a) the food is not cooked completely.\n(b) the fuel is not burning completely.\n(c) the fuel is wet.\n(d) the fuel is burning completely.\nAnswer:\n(b) The fuel is not burning completely.\nQuestion 4\nExplain the nature of the covalent bond using the bond formation in CH3Cl.", + "(b) the fuel is not burning completely.\n(c) the fuel is wet.\n(d) the fuel is burning completely.\nAnswer:\n(b) The fuel is not burning completely.\nQuestion 4\nExplain the nature of the covalent bond using the bond formation in CH3Cl.\nAnswer:\nCovalent bond is formed by sharing of electrons so that the combining atoms complete their outermost shell.\nIn CH3Cl : C = 6, H = 1 and Cl = 17 And their electronic configuration is C \u2013 2,4, H \u2013 1 and Cl \u2013 2, 8, 7", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nIn CH3Cl : C = 6, H = 1 and Cl = 17 And their electronic configuration is C \u2013 2,4, H \u2013 1 and Cl \u2013 2, 8, 7\n\nThree hydrogen atoms complete their shells by sharing three electrons (one electron each) of carbon atom.\nChlorine completes its outer shell by sharing its one out of seven electrons with one electron of carbon atom.\nThus carbon atom shares all its four electrons with three hydrogen atoms and one of chlorine atom and completes its outermost shell and single covalent bonds are formed in CH3Cl.\nQuestion 5\nDraw the electron dot structures for\n(a) ethanoic acid\n(b) propanone\n(c) H2S\n(d) F2.\nAnswer:\n\nQuestion 6\nWhat is a homologous series ? Explain with an example.\nAnswer:\nHomologous series : A homologous series is a group of organic compounds having\nsimilar structures and similar chemical properties in which the successive compounds differ by -CH2 group.\nCharacteristics of homologous series :", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nsimilar structures and similar chemical properties in which the successive compounds differ by -CH2 group.\nCharacteristics of homologous series :\n(i) All members of a homologous series can be represented by the same general formula. For example, the general formula of the homologous series of alkanes is CnH2n+2, in which \u2018n\u2019 denotes number of carbon and hydrogen atoms in one molecule of alkane.\n(ii) Any two adjacent homologues differ by one carbon atom and two hydrogen atoms in their molecular formulae.\n(iii) The difference in the molecular masses of any two adjacent homologues is 14u.\n(iv) All the compounds of a homologous series show similar chemical properties.\n(v) The members of a homologous series show a gradual change in their physical properties with increase in molecular mass.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\n(v) The members of a homologous series show a gradual change in their physical properties with increase in molecular mass.\nFor example, general formula of the homologous series of alkanes is CnH2n+2, in which \u2018n\u2019 denotes number of carbon atoms in one molecule of alkane. Following are the first five members of the homologous series of alkanes (general formula CnH2n+2).\nQuestion 7\nHow can ethanol and ethanoic acid he differentiated on the basis of their physical and chemical properties ?\nAnswer:\nDifference on the basis of physical properties\nDifference on the basis of chemical properties\nQuestion 8\nWhy does micelle formation take place when soap is added to water ? Will a micell be formed in other solvents such as ethanol also ?\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nDifference on the basis of chemical properties\nQuestion 8\nWhy does micelle formation take place when soap is added to water ? Will a micell be formed in other solvents such as ethanol also ?\nAnswer:\nMicelle formation takes place when soap is added to water because the hydrocarbon chains of soap molecules are hydrophobic (water repelling) which are insoluble in water, but the ionic ends of soap molecules are hydrophilic (water attracting) and hence soluble in water.\nSuch micelle formation will not be possible in other solvents like ethanol in which sodium salt of fatty acids do not dissolve.\nQuestion 9\nWhy are carbon and its compounds used as fuels for most applications ?\nAnswer:\nCarbon and its compounds give a large amount of heat per unit weight and are therefore, used as fuels for most applications.\nQuestion 10\nExplain the formation of scum when hard water is treated with soap.\nAnswer:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nQuestion 10\nExplain the formation of scum when hard water is treated with soap.\nAnswer:\nHard water contains salts of calcium and magnesium. Calcium and magnesium on reacting with soap form insoluble precipitate called scum. The scum formation lessens the cleansing property of soaps in hard water.\nQuestion 11\nWhat change will you observe if you test soap with litmus paper (red and blue)?\nAnswer:\nRed litmus will turn blue because soap is alkaline in nature. Blue litmus remains blue in soap solution.\nQuestion 12\nWhat is hydrogenation ? What is its industrial application ?\nAnswer:\nThe addition of hydrogen to an unsaturated hydrocarbon to obtain a saturated hydro-carbon is called hydrogenation. The process of hydrogenation takes place in the presence of nickel (Ni) or palladium (Pd) metals as catalyst.\n\nApplication : The process of hydrogenation has an important industrial application. It is used to prepare vegetable ghee (or vanaspati ghee) from vegetable oils.\nQuestion 13", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nApplication : The process of hydrogenation has an important industrial application. It is used to prepare vegetable ghee (or vanaspati ghee) from vegetable oils.\nQuestion 13\nWhich of the following hydrocarbons undergo addition reactions :\nC2H6, C3H8, C3H6, C2H2 and CH4\nAnswer:\nAddition reactions take place only in unsaturated hydrocarbons. So addition reaction take place only in C3H6 and C2H2.\nQuestion 14\nGive a test that can be used to differentiate chemically between butter and cooking oil.\nAnswer:\nButter is a saturated carbon compound while cooking oil is an unsaturated carbon compound. An unsaturated compound decolourises bromine water, while a saturated compound cannot decolourise it. So we can distinguish chemically between a cooking oil and butter by the bromine water. Add bromine water to a little of cooking oil and butter taken in separate test-tubes.\nCooking oil decolourises bromine water showing that it is an unsaturated compound.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nCooking oil decolourises bromine water showing that it is an unsaturated compound.\nButter does not decolourise bromine water showing that it is a saturated compound.\nQuestion 15\nExplain the mechanism of the cleaning action of soaps.\nOR\nExplain the cleansing action of soaps. [CBSE 2015 (Delhi)]\nAnswer:\nWhen a dirty cloth is put in water containing dissolved soap, then the hydrocarbon end of the soap molecules in micelle attach to the oil or grease particles present on the surface of dirty cloth. In this way the soap micelle entraps the oily or greasy particles by using its hydrocarbon ends. The ionic ends of the soap molecules in the micelles, however, remain attached to water. When the dirty cloth is agitated in soap solution, the oily and greasy particles present on its surface and entrapped by soap micelles get dispersed in water due to which the soap water becomes dirty but the cloth gets cleaned. The cloth is cleaned thoroughly by rinsing in clean water a number of times.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 4 Carbon and its Compounds\nCarbon compounds: Covalent bonding in carbon compounds, Versatile nature of carbon, Homologous series, Nomenclature of carbon compounds containing functional groups, (halogens, alcohol, ketones, aldehydes, alkanes, and alkynes), difference between saturated hydrocarbons and unsaturated hydrocarbons. Chemical properties of carbon compounds (combustion, oxidation, addition and substitution reaction). Ethanol (only properties and uses), Ethanoic acid (only properties and uses), soaps and detergents.\nFormulae Handbook for Class 10 Maths and Science\nQuestion 1\nWhat would be the electron dot structure of carbon dioxide which has the formula CO2?\nSolution:\n\nQuestion 2\nWhat would be the electron dot structure of a molecule of sulphur, which is made up of eight atoms of sulphur? (Hint \u2013 The eight atoms of sulphur are joined together in the form of a ring.)\nSolution:\n\nMore Resources\nNCERT Solutions", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nSolution:\n\nMore Resources\nNCERT Solutions\nNCERT Solutions for Class 10 Science\nNCERT Solutions for Class 10 Maths\nNCERT Solutions for Class 10 Social\nNCERT Solutions for Class 10 English\nNCERT Solutions for Class 10 Hindi\nNCERT Solutions for Class 10 Sanskrit\nNCERT Solutions for Class 10 Foundation of IT\nRD Sharma Class 10 Solutions\nQuestion 3\nHow many structural isomers can you draw for pentane?\nSolution:\nWe can draw 3 structural isomers for pentane.\nQuestion 4\nWhat are the two properties of carbon that lead to the huge number of carbon compounds we see around us?\nSolution:\nDue to its large valency, carbon atoms can form covalent bonds with a number of carbon atoms as well as with a large number of other atoms such as hydrogen, oxygen, nitrogen, sulphur, chlorine and many more atoms. This leads to the formation of a large number of organic compounds.\nDownload NCERT Solutions for Class 10 Science Chapter 4 Carbon and its Compounds PDF\nQuestion 5", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nDownload NCERT Solutions for Class 10 Science Chapter 4 Carbon and its Compounds PDF\nQuestion 5\nWhat will be the formula and electron dot structure of Cyclopentane?\nSolution:\n\nQuestion 6\nDraw the structures for the following compounds.\ni. Ethanoic acid\nii. Bromopentane\niii. Butanone\niv. Hexanal\nSolution:\n\nQuestion 7\nHow would you name the following compounds?\nSolution:\ni. Ethyl bromide\nii. Formaldehyde\niii. Hexyne\nQuestion 8\nWhy is the conversion of ethanol to Ethanoic acid an oxidation reaction?\nSolution:\nThe conversion of ethanol into ethanoic acid is called an oxidation reaction because oxygen is added to it during this conversion.\n\nQuestion 9\nA mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?\nSolution:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nQuestion 9\nA mixture of oxygen and ethyne is burnt for welding. Can you tell why a mixture of ethyne and air is not used?\nSolution:\nWhen a mixture of oxygen and ethyne is burnt, it burns completely producing a blue flame. This blue flame is extremely hot which produced a very high temperature which is used for welding metals. But the mixture of ethyne and air is not used for welding purposes because burning of ethyne in air produces a sooty flame, which is not enough to melt metals for welding.\nQuestion 10\nWhat are oxidizing agents?\nSolution:\nOxidizing agents are the substances that gain electrons in an redox reaction and whose oxidation number is reduced.\nQuestion 11\nExplain the nature of the covalent bond using the bond formation of CH3Cl.\nSolution:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nQuestion 11\nExplain the nature of the covalent bond using the bond formation of CH3Cl.\nSolution:\nCH3Cl(methyl chloride) is made up of one carbon atom, three hydrogen atoms and one chlorine atom. Carbon atom has 4 valence electrons, each hydrogen atom has one valence electron, and a chlorine atom has 7 valence electrons. Carbon atom shares its four valence electrons with three hydrogen atoms and 1 chlorine atom to form methyl chloride as follows:\n\nFrom the above reaction, in the dot structure of methyl chloride (CH3Cl) there are four pairs of shared electrons between carbon and other atoms. Each pair of shared electrons constitutes one single covalent bond. So, methyl chloride has four single covalent bonds.\nQuestion 12\nDraw the electron dot structures for-\nSolution:\n\nQuestion 13\nWhat is a homologous series? Explain with an example.\nSolution:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nQuestion 12\nDraw the electron dot structures for-\nSolution:\n\nQuestion 13\nWhat is a homologous series? Explain with an example.\nSolution:\nHomologous series is a series of compounds with a similar general formula, possessing similar chemical properties due to the presence of the same functional group, and shows a gradation in physical properties as a result of increase in molecular size and mass. For example, methane has a lower boiling point than ethane since it has more intermolecular forces with neighbouring molecules. This is because of the increase in the number of atoms making up the molecule.\nQuestion 14\nHow can ethanol and Ethanoic acid be differentiated on the basis of their physical and chemical properties?\nSolution:\n(i) Ethanol has a pleasant smell whereas ethanoic acid has the smell of vinegar.\n(ii) Ethanol has a burning taste whereas ethanoic acid has a sour taste.\n(iii) Ethanol has no action on litmus paper whereas ethanoic acid turns blue litmus paper red.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\n(ii) Ethanol has a burning taste whereas ethanoic acid has a sour taste.\n(iii) Ethanol has no action on litmus paper whereas ethanoic acid turns blue litmus paper red.\n(iv) Ethanol has no reaction with sodium hydrogencarbonate but Ethanoic acid gives brisk effervescence with sodium hydrogencarbonate.\nQuestion 15\nWhy does micelle formation take place when soap is added to water? Will a micelle be formed in other solvents such as ethanol also?\nSolution:\nMicelle formation takes place when soap is added to water. This is because when soap is added to water in which dirty clothes are soaked, the two parts of the soap molecule dissolves in two different mediums. The organic tail dissolves in the dirt, grime or grease and the ionic head dissolves in water. When the clothes are rinsed or agitated, the dirt gets pulled out of the clothes in the water by the soap molecule. In this way the soap does its cleaning work on dirty and grimy clothes or hands.", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nThe soap molecules actually form a closed structure because of mutual repulsion of the positively charged heads. This structure is called a micelle.\nQuestion 16\nWhy are carbon and its compounds used as fuels for most applications?\nSolution:\nCarbon and its compounds are used as fuels for most of the applications because they burn in air releasing a lot of heat energy.\nQuestion 17\nExplain the formation of scum when hard water is treated with soap.\nSolution:\nThe precipitate form of scum is formed when soap is used for washing clothes. With hard water, a large amount of soap is wasted in reacting with the calcium and magnesium ions of hard water to form an insoluble precipitate. The precipitate form formed by the action of hard water on soap, sticks to the clothes being washed and interferes with the cleaning ability of the additional soap. This makes the cleaning of clothes difficult.\nQuestion 18\nWhat change will you observe if you test soap with litmus paper (red and blue)?\nSolution:", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nQuestion 18\nWhat change will you observe if you test soap with litmus paper (red and blue)?\nSolution:\nSoap is the salt of a strong base (NaOH) and a weak acid (carboxylic acid), so a solution of soap in water is basic in nature. Being basic, a soap solution turns red litmus paper blue.\nQuestion 19\nWhat is hydrogenation? What is its industrial application?\nSolution:\nIt is a class of chemical reactions in which the net result is addition of hydrogen (H2) to unsaturated organic compounds such as alkenes, alkynes, etc. Hydrogenation is widely applied to the processing of vegetable oils and fats. Complete hydrogenation converts unsaturated fatty acids to saturated ones.\nQuestion 20\nC2H5, C3H8, C3H6, C2H2 and CH4\nSolution:\nAlkenes and alkynes (unsaturated hydrocarbons) undergo addition reactions. From the above hydrocarbons C2H2 is an alkyne, whereas C3H6 is an alkene.", + "From the above hydrocarbons C2H2 is an alkyne, whereas C3H6 is an alkene. So, C3H6 and C2H2 will undergo addition reactions.\nQuestion 21", + "CHAPTER 4-CARBON AND ITS COMPOUNDS- INTEXT QUESTIONS.txt\nQuestion 21\nGive a test that can be used to differentiate chemically between butter and cooking oil.\nSolution:\nBromine water test can be used to differentiate chemically between butter and cooking oil. Add bromine water to a little of cooking oil and butter taken in separate test tubes. motor/ output nerve)\nCoordination In Plants\nPlants have neither nervous system nor muscles. Their movements or responses are either growth dependent or growth independent.\nImmediate Response To Stimulus Or Growth Independent", + "CHAPTER 6-CONTROL AND COORDINATION -INTEXT QUESTIONS.txt\nPlants have neither nervous system nor muscles. Their movements or responses are either growth dependent or growth independent.\nImmediate Response To Stimulus Or Growth Independent\nThe plants also use electrical-chemical means to convey information from cell to cell, but unlike in animals, there is no specialised tissue in plants for the conduction of information.\nSecondly, plant cells change shape by changing the amount of water in them, resulting in swelling or shrinking. In contrast, animal muscle cells have special proteins that change both their shape and their arrangement in the cell in response to nervous electrical impulses.\nExample: folding up and drooping of leaves of chhui-mui (the \u2018sensitive\u2019 or \u2018touch-me-not\u2019 plant of the Mimosa family) in response to touch.\nMovement Due To Growth", + "CHAPTER 6-CONTROL AND COORDINATION -INTEXT QUESTIONS.txt\nExample: folding up and drooping of leaves of chhui-mui (the \u2018sensitive\u2019 or \u2018touch-me-not\u2019 plant of the Mimosa family) in response to touch.\nMovement Due To Growth\nThe movement of a plant in response to the stimulus is called as tropism. The movement in the direction of the response is called as positive tropism and movement away from the stimulus is termed as negative tropism.\nTypes:\nPhototropism: Bending of plant in response to the light e.g. movement of sunflowers in response to day or night.\nGeotropism: Movement of plant in response to gravity. Shoots show negative geotropism and roots show positive geotropism.\nChemotropism: Movement of plant in response to chemical as observed in case of growth of pollen tube.\nHydrotropism: e.g. roots beneath the Earth\u2019s surface bend in the direction of underground w^ater.", + "CHAPTER 6-CONTROL AND COORDINATION -INTEXT QUESTIONS.txt\nChemotropism: Movement of plant in response to chemical as observed in case of growth of pollen tube.\nHydrotropism: e.g. roots beneath the Earth\u2019s surface bend in the direction of underground w^ater.\nOther example may include the climbing of tendrils in response to touch. When they come in contact with any support, the part of the tendril in contact with the object does not grow as rapidly as the part of the tendril away from the object. This causes the tendril to circle around the object and thus cling to it.\nPlant Hormones\nThere are five main types of plant hormones or growth regulators:\nAuxins: Helps in cell elongation and thus phototropism, geotropism, and other plant responses.\nGibberellins: Stimulate growth of the stem and flowering.\nCytokinins: They cause cell division, enlargement, and organ formation. They are present in greater concentration in areas of rapid cell division, such as in fruits and seeds.\nEthylene: Promotes ripening of fruits.", + "CHAPTER 6-CONTROL AND COORDINATION -INTEXT QUESTIONS.txt\nEthylene: Promotes ripening of fruits.\nAbscisic acid: Inhibits growth and causes wilting of leaves and fruits.\nHormones In Animals\nHormones are non-nutrient chemicals which act as intercellular messengers & are produced in trace amounts. The timing and amount of hormone released are regulated by feedback mechanisms Examples:\nAdrenaline: Secreted from adrenal gland which prepare the body for fight or flight situation.\nThyroxin: Secreted from thyroid gland and regulates carbohydrate, protein and fat metabolism so as to provide the best balance for growth.\nDeficiency of iodine in diet may cause hypo-secretion of thyroxin which results in goiter.\nGrowth hormones which regulate growth and. development of the body are secreted from\npituitary gland.\nHyper-secretion may cause gigantism and hypo-secretion may cause dwarfism.\nTestosterone in males and oestrogen in females lead to changes take place during puberty\u2019.\nInsulin produced by pancreas regulates the blood sugar level in the body.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\n2\nExtra Questions \u2013 How do Organisms Reproduce \u2013 CBSE Class 10 Science\nQUESTION1 \n1.Name the various types of asexual reproduction.\nSolution:\nThe various types of asexual reproduction are\n\u2022 Fission \u2013 binary and multiple\n\u2022 Budding\n\u2022 Spore formation\n\u2022 Regeneration\n\u2022 Vegetative propagation \u2013 It is a process by which new organisms arise without production of seeds or spores. It can occur naturally or be induced by horticulturists.\nQuestion-2\nDifferentiate between binary fission and multiple fission.\nSolution:\nBinary Fission:\n1.In this type of fission, the parent cell gives rise to two daughter cells.\n2.It is expressed by organisms like amoeba, paramecium, etc.\nMultiple Fission:\n1.In this type of fission , the parent cell gives rise to more than two daughter cells.\n2.It is expressed by organisms plasmodium.\nQuestion-3\nList the advantages of vegetative propagation.\nSolution:\nThe advantages of vegetative propagation are as follows\n\u2022 It helps in the easy propagation of non\u2013flowering plants.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\nQuestion-3\nList the advantages of vegetative propagation.\nSolution:\nThe advantages of vegetative propagation are as follows\n\u2022 It helps in the easy propagation of non\u2013flowering plants.\n\n\u2022 It helps in producing hybrids of various plants, with improved qualities.\n\n\u2022 It helps in the propagation of a large number of populations in a very short duration.\n\u2022 It helps in the propagation of plants that do not produce seeds or produce them in very small quantities.\nQuestion-4\nName the male and the female gametes in human beings.\nSolution:\nThe male gamete is called sperm and the female gamete is called ovum in human beings.\nQuestion-5\nDefine fertilization.\nSolution:\nThe process of fusion of the male gamete with the female gamete of the same species, i.e., the sperm with the ovum is known as fertilization. After fertilization, the product obtained is the zygote, which eventually develops into a complete organism.\nQuestion-6\nDefine sexual reproduction.\nSolution:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\nQuestion-6\nDefine sexual reproduction.\nSolution:\nThe type of reproduction that involves both the sexes of the species is known as sexual reproduction. This kind of reproduction requires the fertilization of the gametes of both the sexes.\nQuestion-7\nMention any two functions of human ovary.\nSolution:\nThe two functions of human ovary are as follows\n(i) It is responsible for the production of the female gamete, ovum.\n(ii) It secretes hormones like estrogen and progesterone.\nQuestion-8\nWhat is syngamy?\nSolution:\nThe process of fusion of the two gametes is known as syngamy. It is the initial step in the process of fertilization.\nQuestion-9\nWhere does fertilization take place in human females?\nSolution:\nFertilization in the case of humans is internal. The process of fertilization takes place inside the fallopian tube of females.\nQuestion-10\nDefine menopause.\nSolution:\nAt the age of around 45-50 years, the ovaries of the females stop producing ova.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\nQuestion-10\nDefine menopause.\nSolution:\nAt the age of around 45-50 years, the ovaries of the females stop producing ova.\nThe stoppage of menstrual flow and other events like the changes in the hormonal composition is known as menopause.\nQuestion-11\nDiscuss briefly the types of reproduction.\nSolution:\nReproduction is a unique biological process that is essential for the continuity and propagation of species. It is a process by which new individuals of the same species are produced by parental organisms. It takes place by two methods: \u2013\nAsexual Reproduction\nThis method of reproduction generally involves only one parent and is observed only in lower organisms. Fission, budding, spore formation and vegetative propagation are some of the types of asexual propagation.\nSexual Reproduction\nThis method of reproduction involves two parents. It requires the fusion of gametes of both the sexes of the species. Almost all animals and higher forms of plants reproduce by this method.\nQuestion-12", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\nQuestion-12\nDescribe asexual reproduction in amoeba.\nSolution:\nAmoeba reproduces by the process of binary fission. During this process, the nucleus divides first, followed by the formation of a constriction in the cellular membrane of amoeba. This gradually increases inwards and divides the cytoplasm into two parts. Finally two daughter organisms are obtained.\nQuestion-13\nExplain various steps of budding in yeast.\nSolution:\nBudding is a form of asexual reproduction usually observed in yeast.\n\u2022 During this process, a small protrusion appears on the upper portion of the body of the organism. This bulge is called a bud.\n\u2022 The bud gradually grows in size and forms an individual cell.\n\u2022 From this newly budded cell, another bud appears at the tip.\n\u2022 This process continues and a chain of yeast cells is obtained.\nQuestion-14\nWhat are the various methods of vegetative propagation? Discuss any one method with example.\nSolution:\nThe various types of vegetative propagation are\n\u2022 Cutting", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\nQuestion-14\nWhat are the various methods of vegetative propagation? Discuss any one method with example.\nSolution:\nThe various types of vegetative propagation are\n\u2022 Cutting\n\u2022 Layering\n\u2022 Grafting\n\u2022 Parthenogenesis\n\u2022 Micro-propagation in vitro\nGrafting\n\u2022 It is a method in which two parts of different plants are joined together in such a way that they unite and grow as one plant.\n\u2022 The portion of the plant that is grafted onto another plant is known as the scion, and the plant on which grafting is performed is known as the stock.\n\u2022 The stock and the scion are tied in such a way that the cambium of the scion and the stock come in contact with each other.\n\u2022 The stock is so chosen that it possesses qualities like disease resistance, high water absorbing capacity, deep penetrating roots for a firm hold, etc. The scion is so chosen that it possesses qualities like high yield of pulp or seeds (as desired) from a single fruit, etc.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\n\u2022 For example citrus-root stock is used for a variety of grafts like sweet orange, lime, grape, etc.\nQuestion-15\nDefine the terms unisexual and bisexual giving examples of each.\nSolution:\nAn animal or an organism, which can be differentiated as male or female.\ni.e., The two sexes are present in separate individuals, then such an organism is said to be unisexual.For example, human beings, all mammals. An animal or an organism that possesses both the sexes in a single body is said to be a bisexual animal. For example, earthworm, starfish, etc.\nQuestion-16\nMention the reproductive parts of a flower.\nSolution:\nThe reproductive parts of a flower are as follows\nMale reproductive part \u2013 Stamens\nFemale reproductive part \u2013 Pistil.\nQuestion-17\nWhat is pollination? Describe cross-pollination.\nSolution:\nThe transfer of pollen grains from anther to stigma is called pollination. Pollen grains are shed from the anther and reach the stigma of either the same flower or a different flower.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\nSolution:\nThe transfer of pollen grains from anther to stigma is called pollination. Pollen grains are shed from the anther and reach the stigma of either the same flower or a different flower.\nCross-pollination\nThe transfer of pollen grains from the anther of one flower to the stigma of another flower of the same species is termed as cross-pollination.\nQuestion-18\nDescribe double fertilization in plants.\nSolution:\nIn the case of plants, the pollen grain releases two male gametes. One fuses with the egg and forms the zygote. The other male gamete fuses with the two polar nuclei. This fusion is said to be triple fusion. Thus, inside an embryo sac, two fusions, syngamy and triple fusion, take place. This mechanism of two fusions occurring in an embryo sac is called double fertilization.\nQuestion-19\nWhat are the male and the female gonads in the human beings? Mention their functions.\nSolution:\nTestes are the male gonads in human beings. Their functions are", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\nQuestion-19\nWhat are the male and the female gonads in the human beings? Mention their functions.\nSolution:\nTestes are the male gonads in human beings. Their functions are\n\u2022 To produce male hormones like testosterone.\n\u2022 To produce the male gamete \u2013 sperm.\nOvaries are the female gonads in human beings. Their functions are\n\u2022 To produce female hormones like estrogen and progesterone.\n\u2022 To produce the female gamete \u2013 ovum.\nQuestion-20\nMention the secondary sexual characters in human males and females.\nSolution:\nThe male secondary sexual characters are the growth of hair on the chest, near the genitals, on the face in the form of beard and moustache, development of deep hoarse voice, etc.\nThe female secondary sexual characters are the development of breasts and mammary glands, axial and pubic hair, etc.\nQuestion-21\nWrite names of one male and female sex hormones.\nSolution:\nThe names of one male and one female sex hormones are as follows\nMale sex hormone \u2013 testosterone", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\nQuestion-21\nWrite names of one male and female sex hormones.\nSolution:\nThe names of one male and one female sex hormones are as follows\nMale sex hormone \u2013 testosterone\nFemale sex hormone \u2013 progesterone.\nQuestion-22\nDescribe the menstrual cycle.\nSolution:\n\u2022 The commencement of menstruation at puberty is termed menarche and marks the beginning of the reproductive life of a female.\n\u2022 Initially, inside the ovary, an oogonium or an ovarian follicle is developed into a mature ovum and released into the fallopian tube. This is termed as ovulation.\n\u2022 After the release, the uterus and the fallopian tubes undergo changes like the thickening of the uterine wall, development of the endometrial lining, etc. to prepare the body for a probable pregnancy.\n\u2022 If fertilization does not occur, then the thickened inner wall of the uterus breaks down along with its blood vessels and moves out of the vagina in the form of bleeding, called menstrual flow (menstruation).\n\u2022 It lasts for about 4-7 days.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\n\u2022 It lasts for about 4-7 days.\n\u2022 This cycle of events taking place in the ovaries and the uterus every 28 days and marked by the menstrual flow is known as the menstrual cycle.\n\u2022 In a normal healthy woman, ovulation takes place around the 14th day and menstruation occurs every 28 to 30 days.\nQuestion-23\nDifferentiate between menarche and menopause.\nSolution:\n\u2022 Menarche marks the beginning of the reproductive cycle of a female whereas menopause marks the end of the reproductive cycle.\n\u2022 Menarche generally takes place at puberty, i.e., at the age of 12-16 years while menopause takes place around the age of 45-50 years.\n\u2022 After menarche, there is an increased amount of production of female sex hormones whereas after menopause, the secretion of female sex hormones declines and sometimes completely stops.\nQuestion-24\nWhat is ovulation?\nSolution:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\nQuestion-24\nWhat is ovulation?\nSolution:\nThe ovarian follicles present inside the ovaries develop into mature follicles. Usually, one mature ovarian follicle develops into a mature ovum. It pinches off from the surface of the ovary and enters the fallopian tube. This process is termed as ovulation. Generally ovulation takes place in only one ovary at a time.\nQuestion-25\nWrite whether the following statements are true or false.\n1. Fertilization is the fusion of the sperm and the ovum.\n2. Duration of menstrual cycle in human female is 20 days.\n3. Onset of menstruation is termed menopause.\n4. In human beings, male attains puberty little later than females.\nSolution:\n1. True\n2. False\n3. False\n4. True.\nQuestion-26\nFertilization is possible if copulation has taken place during the middle of the menstrual cycle.Give reasons.\nSolution:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\nSolution:\n1. True\n2. False\n3. False\n4. True.\nQuestion-26\nFertilization is possible if copulation has taken place during the middle of the menstrual cycle.Give reasons.\nSolution:\nDuring copulation, a large number of sperms enter the fallopian tube. If copulation takes place during the middle of the menstrual cycle, then fertilization is possible as at this time the ovum would have entered the fallopian tube from the ovary. If copulation takes place before ovulation then fertilization is not possible. Fertilization is marked by the absence of the regular menstrual flow.\nQuestion-27\nMention the methods used for regulation of childbirth.\nSolution:\nA number of techniques are employed for the regulation of childbirth. They are classified as follows\nBarrier methods \u2013 Use of physical devices to prevent the entry of sperms inside the female genitals. They include:\n1. Condoms\n2. Diaphragms\n3. Cervical caps\n4. Intrauterine contraceptive devices (IUCD) like Copper-T.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\n1. Condoms\n2. Diaphragms\n3. Cervical caps\n4. Intrauterine contraceptive devices (IUCD) like Copper-T.\nChemical methods \u2013 Use of specific drugs by females.\n1. Oral pills \u2013 Oral contraceptives of mainly hormonal preparations.\n2. Vaginal pills\nSurgical methods \u2013\n1. Vasectomy \u2013 Removal or ligation of a small portion of the vas deferens in males.\n2. Tubectomy \u2013 Removal or ligation of a small portion of fallopian tube in females.\nQuestion-28\nWrite the full forms of IUCD, AIDS, HIV, and OC.\nSolution:\nThe full forms for the above abbreviations are as follows\nIUCD \u2013 Intrauterine Contraceptive Devices\nAIDS \u2013 Acquired Immuno-Deficiency Syndrome\nHIV \u2013 Human Immuno Virus\nOC \u2013 Oral Contraceptives.\nQuestion-29\nDescribe the surgical method of birth control.\nSolution:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXTRA QUESTIONS.txt\nAIDS \u2013 Acquired Immuno-Deficiency Syndrome\nHIV \u2013 Human Immuno Virus\nOC \u2013 Oral Contraceptives.\nQuestion-29\nDescribe the surgical method of birth control.\nSolution:\nIn the surgical methods of birth control, a small portion of vas deferens in males and the fallopian tube in females is surgically removed or ligated (tied). This process is called Vasectomy in males and Tubectomy in females. As a result of this surgery, gametes cannot be released from the body (in males) and cannot enter the uterus (in females).\nQuestion-30\nDefine sexually transmitted disease and give two examples.\nSolution:\nThere are certain infectious diseases that spread by sexual contact. Such diseases are called STDs (Sexually Transmitted Diseases). In most cases, the symptoms are burning sensation on urination and urethral discharge. Gonorrhea and Syphilis are the two examples for sexually transmitted diseases. These diseases are curable.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXEMPLER QUESTIONS.txt\nNCERT Exemplar Class 10 Science Chapter 8 How do Organisms Reproduce?\nShort Answer Questions\nQuestion.1 In a bisexual flower, inspite of the young stamens being removed artificially, the flower produces fruit. Provide a suitable explanation for the above situation.\nAnswer. Removal of stamens of a bisexual flower will not affect pollination of its intact pistil and formation of fruit. Cross-pollination has occurred leading to fertilisation.\n Question. 2 Can you consider cell division as a type of reproduction in unicellular organism? Give one reason.\nAnswer. Yes, because cell division in a unicellular organism results in the formation of two daughter cells which means it produces more individuals of the organisms.\nQuestion.3 Explain how, offspring and parents of organisms reproducing sexually have the same number of chromosomes.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXEMPLER QUESTIONS.txt\nQuestion.3 Explain how, offspring and parents of organisms reproducing sexually have the same number of chromosomes.\nAnswer. The parents are diploid (2n) as each of them has two sets of chromosomes. They form haploid (1n) male and female gametes through the process of meiosis. The haploid gametes have one set of chromosomes. Since, these two gametes fuse during fertilisation, the original number of chromosomes is restored in the offspring.\nQuestion. 4 Colonies of yeast fail to multiply in water, but multiply in sugar solution. Give one reason for this.\nAnswer. Energy is essential for any activity in living organisms. Sugar provides this energy for sustaining all life activities in yeasts. In water, it fails to reproduce because of inadequate energy in its cells. So, colonies of yeast fail to multiply in water but multiply in sugar solution.\nQuestion. 5 Why does bread mould grow profusely on a moist slice of bread rather than on a dry slice of bread?", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXEMPLER QUESTIONS.txt\nQuestion. 5 Why does bread mould grow profusely on a moist slice of bread rather than on a dry slice of bread?\nAnswer.Bread mould require moisture and nutrients for its growth. A moist slice of bread contains both moisture and nutrients hence, it grows profusely as compared to a dry slice of bread which contains only nutrients but no moisture.\nQuestion. 6 Give two reasons for the appearance of variations among the progeny formed by sexual reproduction.\nAnswer.\n In sexual reproduction, two parents have different sets of characters.\nGene combinations are different in gametes.\nQuestion. 7 In tobacco plant, the male gametes have twenty four chromosomes. What is the number of chromosomes in the female gamete?\nWhat is the number of chromosomes in the zygote?\nAnswer. The number of chromosomes in the female gamete is 24. The number of chromosomes in the zygote is 48.\nQuestion. 8 Is the chromosome number of zygote,", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXEMPLER QUESTIONS.txt\nAnswer. The number of chromosomes in the female gamete is 24. The number of chromosomes in the zygote is 48.\nQuestion. 8 Is the chromosome number of zygote,\nembryonal cells and adult of a particular organism always constant? How is the constancy maintained in these three stages?\nAnswer. Yes, the chromosome number of zygote, embryonal cells and adult of a particular organism is always constant.\nThe constancy is maintained because the cells in all these three structures undergo only mitotic divisions.\nQuestion. 9 How are general growth and sexual v maturation different from each other?\nAnswer. General growth is the growth of different types of developmental process in the body like increase in height, weight gain, changes in shape and size of the body.\nSexual maturation is a set of changes in body of an individual at puberty like cracking of voice, new hair patterns, development of breast in female, etc.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXEMPLER QUESTIONS.txt\nSexual maturation is a set of changes in body of an individual at puberty like cracking of voice, new hair patterns, development of breast in female, etc.\nQuestion. 10 What would be the ratio of chromosome number between an egg and its zygote? How is the sperm genetically different from the egg?\nAnswer. The ratio of chromosome number between an egg and its zygote is 1: 2.\nSperm is genetically different from the egg in the way that it contains either X or Y chromosome whereas, an egg always contains an X chromosome.\nLong Answer Questions\nQuestion. 11 Distinguish between a gamete and zygote. Explain their roles in sexual reproduction.\nAnswer.\nRole of Gamete in sexual reproduction \u2014 Gamete is the sex or germ cell specialised to take part in sexual reproduction. Fusion of male gamete with a female gamete produces a zygote. Role of Zygote in sexual reproduction \u2014 Zygote develops into embryo that later forms the new individual.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXEMPLER QUESTIONS.txt\nQuestion. 12 What is placenta? Mention its role during pregnancy?\nAnswer. Placenta is a special tissue that develops between the uterine wall and the embryo (foetus).\n\nThe role of placenta during pregnancy is as follows:\n Placenta attaches the foetus to uterine wall.\n It possesses villi that increases the surface area for fixation and absorption.\n It facilitates the passage of nutrition and oxygen to embryo from mother through blood.\n Waste substances produced by embryo (foetus) are removed through placenta into mother\u2019s blood.\nQuestion. 13 How does fertilisation take place? Fertilisation occurs once in a month. Comment.\nAnswer. Fertilisation takes place in the following ways:\nThe sperm enters through the vaginal passage during sexual intercourse and moves upward.\nEgg released from the ovary reaches the oviduct.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-EXEMPLER QUESTIONS.txt\nAnswer. Fertilisation takes place in the following ways:\nThe sperm enters through the vaginal passage during sexual intercourse and moves upward.\nEgg released from the ovary reaches the oviduct.\nSperm encounters egg in the oviduct and fertilisation takes place. Fertilisation occurs once in a month because egg is released by female ovary once every month in the middle of menstrual cycle.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nImportant Questions of How Do Organisms Reproduce Class 10 Science Chapter 8\nQuestion 1.\nNewly formed DNA copies may not be identical at times. Give one reason. (AI2017)\nAnswer:\nWhen a cell reproduces, DNA replication occurs which results in formation of two similar copies of DNA. The process of copying the DNA leads to some variations each time. As a result, the DNA copies produced are similar to each other but sometimes may not identical.\nQuestion 2.\nWhen a cell reproduces, what happens to its DNA? (AI 2017)\nAnswer:\nWhen a cell reproduces, DNA replication occurs which forms two similar copies of DNA..\nQuestion 3.\nWhat is DNA? (Delhi 2016, Foreign 2015)\nAnswer:\nDNA (deoxyribonucleic acid) is a polymer made up of large number of nucleotide units. It carries genetic information from generation to generation.\nQuestion 4.\nName the life process of an organism that helps in the growth of its population. (AI 2015)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 4.\nName the life process of an organism that helps in the growth of its population. (AI 2015)\nAnswer:\nReproduction is a life process that helps in multiplication of an organism and growth of its population.\nQuestion 5.\nReproduction is one of the most important characteristic \u2018of living beings. Give three reasons in support of the statement. (AI 2017)\nAnswer:\nReproduction is one of the most important characteristics of living beings because :\nit is essential for existence and continuity of a species.\nit helps to pass genetic information to next generation.\nit brings variations in next generation which is the basis for evolution.\nQuestion 6.\nDefine reproduction. How does it helps in providing stability to the population of species? (AI 2016)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nit brings variations in next generation which is the basis for evolution.\nQuestion 6.\nDefine reproduction. How does it helps in providing stability to the population of species? (AI 2016)\nAnswer:\nThe production of new organisms by the existing organisms of the same species is known as reproduction. It is linked to the stability of population of a species. DNA replication during reproduction ensures transfer of specific characters or body design features that is essential for an individual of a population to live and use that particular niche. Some variations present in a few individuals of population caused due to reproduction which also helps in their survival at changing niches.\nQuestion 7.\nWhat is DNA copying? State its importance. (Delhi 2015)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 7.\nWhat is DNA copying? State its importance. (Delhi 2015)\nAnswer:\nDNA copying is the production of similar copies of DNA present in a cell using various chemical reactions. DNA copying is essential for reproduction through which the organisms pass on their body features to their offspring. Moreover, minor alternations during the process of DNA copying result in the production of variations. Such variations are useful for the survival of species over time.\nQuestion 8.\nWhat is the effect of DNA copying, which is not perfectly accurate, on the reproduction process? How does the amount of DNA remain constant through each new generation is a combination of DNA copies of two individuals? (AI 2014)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nIn the process of reproduction, if DNA copying is not perfectly accurate, variation occurs. These in turn may allow few individuals of a population to survive in an altered niche and becomes the basis of evolution and over time. Such variations are useful for the survival of species.\nThe combination of DNA copies of two individuals, (male and female) occurs during sexual reproduction. Reduction division (meiosis) during gamete formation halves the chromosome number in both male and female gametes. Since these two gametes fuse during fertilisation, the original number of chromosomes (as in the parent) is restored in the offspring. By this way the amount of DNA remains constant in each new generation.\nQuestion 9.\nName the method by which Spirogyra reproduces under favourable conditions. Is this method sexual or asexual? (Delhi 2017)\nAnswer:\nThe method by which Spirogyra reproduces under favorable conditions is fragmentation. This is an asexual mode of reproduction.\nQuestion 10.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nThe method by which Spirogyra reproduces under favorable conditions is fragmentation. This is an asexual mode of reproduction.\nQuestion 10.\nHow does Plasmodium reproduce. Is this method sexual or asexual? (Delhi 2017)\nAnswer:\nPlasmodium reproduces through multiple fission method. In this method, the parent organism splits to form many new organisms at the same time. This is an asexual method of reproduction.\nQuestion 11.\nName the part of Bryophyllum where the buds are produced for vegetative propagation. (Delhi 2016)\nAnswer:\nBryophyllum propagates vegetatively by the buds produced at the margins of leaves.\nQuestion 12.\nWhat happens when a Planaria gets cut into two pieces? (Delhi 2016)\nAnswer:\nWhen Planaria is cut into two pieces then each piece grows into a complete organism. This is known as regeneration.\nQuestion 13.\nWhat happens when a mature Spirogyra filament attains considerable length? (AI 2016)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 13.\nWhat happens when a mature Spirogyra filament attains considerable length? (AI 2016)\nAnswer:\nWhen a mature Spirogyra filament attains considerable length it simply breaks into two or more fragments and each fragment then grows into a new Spirogyra.\nQuestion 14.\nName the method by which Hydra reproduces. Is this method sexual or asexual ? (Foreign 2016)\nAnswer:\nHydra generally reproduces through budding. It is an asexual method of reproduction.\nQuestion 15.\nName two simple organisms having the ability of regeneration. (AI 2015)\nAnswer:\nHydra and Planaria are two organisms that have the ability to regenerate.\nQuestion 16.\nName the causative agent of the disease \u201ckala- azar\u201d and its mode of asexual reproduction. (Foreign 2015)\nAnswer:\nCausative agent of the disease Kala-azar is Leishmania. It reproduces asexually by binary fission.\nQuestion 17.\nWrite two differences between binary fission and multiple fission in a tabular form. (Delhi 2015)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 17.\nWrite two differences between binary fission and multiple fission in a tabular form. (Delhi 2015)\nAnswer:\nDifferences between binary fission and multiple fission are as follows:\nQuestion 18.\nList four modes of asexual reproduction other than fission in the living organisms. (Delhi 2014)\nAnswer:\nThe four modes of asexual reproduction other than fission in living organisms are :\nbudding\nspore formation\nregeneration and\nfragmentation.\nQuestion 19.\nList four advantages of vegetative propagation. (Delhi 2014)\nAnswer:\nThe following are the advantages of vegetative propagation:\nThe characters of the parent plants are preserved hence a good variety produced can be propagated by vegetative means.\nThe plants, which do not produce viable seeds or produce very few seeds, can be reproduced by this method. For example, banana, potato, grapes, sugarcane, rose, orange, etc.\nIt is an easier, quicker and cheaper method of propagation.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nIt is an easier, quicker and cheaper method of propagation.\nIt is easier to get rid of pathogen from any part of plant by vegetative propagation.\nQuestion 20.\nList four modes of asexual reproduction. (Delhi 2014)\nAnswer:\nThe four modes of asexual reproduction are :\nbinary fission\nbudding\nregeneration and\nvegetative propagation.\nQuestion 21.\nDraw labelled diagrams to illustrate budding in Hydra. (AI 2014)\nAnswer:\nThe given diagram illustrates budding in Hydra:\n\nQuestion 22.\nHow do Plasmodium and Leishmania reproduce? Write one difference in their mode of reproduction. (Foreign 2014)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nThe given diagram illustrates budding in Hydra:\n\nQuestion 22.\nHow do Plasmodium and Leishmania reproduce? Write one difference in their mode of reproduction. (Foreign 2014)\nAnswer:\nPlasmodium and Leishmania reproduce by the process of fission which is an asexual mode of reproduction. Plasmodium reproduces by multiple fission. About 1000 daughter cells are produced by the multiple fission of a Plasmodium. Leishmania reproduces by the process of binary fission. In Leishmania, the splitting of parent cell takes place in a definite plane (longitudinally) with respect to flagellum at its end to produce two daughter cells.\nQuestion 23.\nDefine multiple fission. Give its one example. (Foreign 2014)\nAnswer:\nMultiple fission is an asexual mode of reproduction in which the parent organism splits to form many new organisms at the same time. Multiple fission occurs in Plasmodium.\nQuestion 24.\nList two advantages ofvegetative reproduction practised in case of an orange plant. (Delhi 2012)", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 24.\nList two advantages ofvegetative reproduction practised in case of an orange plant. (Delhi 2012)\nAnswer:\nThe two advantages of vegetative propagation practised in case of an orange plant are :\nThe new plants produced by vegetative propagation will be exactly like the parent plant. Therefore, any desirable features of the parent plant will be replicated in the new plants.\nThe orange plants that have lost the capacity to produce seeds, can also be propagated.\nQuestion 25.\nName an organism which reproduces by spore formation. List three conditions favourable for spores to germinate and grow. (AI 2012)\nAnswer:\nRhizopus reproduce by the method of spore formation.\nThe three conditions favourable for spores to germinate and grow are moisture, suitable temperature and food (nutrition).\nQuestion 26.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nRhizopus reproduce by the method of spore formation.\nThe three conditions favourable for spores to germinate and grow are moisture, suitable temperature and food (nutrition).\nQuestion 26.\nList two advantages of practising vegetative propagation in plants. Select two plants raised by this method from the list given below : Banana, Gram, Pea, Rose, Tomato, Wheat (Foreign 2012)\nAnswer:\nTwo advantages of the vegetative propagation of plants are:\nAny desirable features of the parent plant can be replicated in the new plants.\nFlowers and fruits can be grown in a shorter time as compared to the plants grown from seeds. The two plants raised by this method are banana and rose.\nQuestion 27.\nWrite any two differences between binary fission and multiple fission in a tabular form as observed in cells of organisms. (Delhi 2011)\nAnswer:\nRefer to answer 17.\nQuestion 28.\nList any four reasons for vegetative propagation being practised in the growth of some type of plants. (AI2011)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 17.\nQuestion 28.\nList any four reasons for vegetative propagation being practised in the growth of some type of plants. (AI2011)\nAnswer:\nRefer to answer 19.\nQuestion 29.\nWhat is vegetative propagation? Write two of its advantages. (Foreign 2011)\nAnswer:\nVegetative propagation is a method of asexual reproduction in plants in which the parts other than seeds are used as propagules. Also refer to answer 26.\nQuestion 30.\nWrite one main difference between asexual and sexual mode of reproduction. Which species is likely to have comparatively better chances of survival \u2013 the one reproducing asexually or the one reproducing sexually? Give reason to justify your answer. (2018)\nAnswer:\nDifference between asexual and sexual mode of reproduction is as follows :\nAsexual reproduction:\nGametes are not formed hence fertilisation does not take place.\nSexual reproduction:\nGametes are always formed and fertilisation takes place to form a zygote.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAsexual reproduction:\nGametes are not formed hence fertilisation does not take place.\nSexual reproduction:\nGametes are always formed and fertilisation takes place to form a zygote.\nSpecies reproducing sexually has a better chance of survival as variations occur only during the sexual reproduction. Variations increase the chances of survival of an individual by making them more fit. Selection of variations by environmental factors forms the basis of evolution.\nQuestion 31.\nWhat happens when\n(a) accidently, Planaria gets cut into many pieces-\n(b) Bryophyllum leaf falls on the wet soil\n(c) on maturation sporangia of Rhizopus bursts? (Delhi 2017)\nAnswer:\n(a) When Planaria accidently gets cut into many pieces then its each piece grows into a complete organism. This is known as regeneration.\n(b) When the Bryophyllum leaf falls on the wet soil, the buds present in the notches along the leaf margin develop into new plants. This is known as vegetative propagation.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(b) When the Bryophyllum leaf falls on the wet soil, the buds present in the notches along the leaf margin develop into new plants. This is known as vegetative propagation.\n(c) The sporangia of Rhizopus contain cells or spores that can eventually develop into new Rhizopus individuals when it bursts on maturation.\nQuestion 32.\nDescribe reproduction by spores in Rhizopus. (AI 2017)\nAnswer:\nFungus Rhizopus reproduces by spore formation. During the growth of Rhizopus, small rounded, bulb-like structures develop at the top of the erect hyphae. Such structures are called sporangia. Inside each sporangium, nucleus divides several times. Each nucleus gets surrounded by a little amount of cytoplasm to become spore. Large number of spores are formed inside each sporangium. After sometime sporangium bursts and spores are released in the air. When these spores land on food or soil, under favourable conditions, they germinate into new individuals.\nQuestion 33.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 33.\nWhat is vegetative propagation? State two advantages and two disadvantages of this method. (AI 2017)\nAnswer:\nVegetative propagation is a type of asexual reproduction in which the plant parts other than seeds are used as a propagule.\nAdvantages of vegetative propagation :\nDesirable character of the plant can be preserved through generation.\nSeedless plants can be grown through this method.\nDisadvantages of vegetative propagation :\nPlants produced by this method posses less vigour and are more prone to diseases.\nPlants produced by this method show no genetic variation.\nQuestion 34.\nWhat is multiple fission? How does it occur in an organism? Explain briefly. Name one organism which exhibits this type of reproduction. (Delhi 2016)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 34.\nWhat is multiple fission? How does it occur in an organism? Explain briefly. Name one organism which exhibits this type of reproduction. (Delhi 2016)\nAnswer:\nMultiple fission refers to the process of asexual reproduction in which many individuals are formed from a single parent. This method of reproduction occurs in unfavourable conditions. The unicellular organism develops a protective covering called cyst, over the cell. The nucleus of the cell divides repeatedly producing many nuclei. Later on, each nucleus is surrounded by small amount of cytoplasm and many daughter cells are produced within the cyst.\nWhen conditions are favourable the cyst breaks and small offspring are liberated. This type of reproduction is seen in some protozoans, e.g., malarial parasite (Plasmodium).\nQuestion 35.\nExplain the term \u201cregeneration\u201d as used in relation to reproduction of organisms. Describe briefly how regeneration is carried out in multicellular organisms like Hydra. (AI 2016)", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 35.\nExplain the term \u201cregeneration\u201d as used in relation to reproduction of organisms. Describe briefly how regeneration is carried out in multicellular organisms like Hydra. (AI 2016)\nAnswer:\nThe process of formation of entire organism from the body parts of a fully differentiated organism is called regeneration. It occurs by process of growth and development.\nSimple animal like Hydra shows regeneration. When a small piece of Hydra breaks off it grows into complete new Hydra.\nDuring regeneration, the cells of cut body part of the organism divide rapidly to make a mass of cells. The cells here move to their proper places within the mass where they have to form different types of tissues. In this way complete organism is regenerated.\nQuestion 36.\nIn the context of reproduction of species state the main difference between fission and fragmentation. Also give one example of each. (AI 2016)\nAnswer:\nThe main differences between fission and fragmentation are as follows:\nQuestion 37.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nThe main differences between fission and fragmentation are as follows:\nQuestion 37.\nWhat happens when\n(a) Planaria gets cut into two pieces\n(b) a mature Spirogyra filament attains considerable length\n(c) on maturation sporangia burst? (Foreign 2016)\nAnswer:\n(a) When Planaria is cut into two pieces then each piece grows into a complete organism. This is known as regeneration.\n(b) When a mature Spirogyra filament attains a considerable length it breaks into small pieces called fragments. These fragments grow into new individuals and this mode of reproduction is called fragmentation.\n(c) When a sporangium burst, large number of spores are released in the air. When these spores land on food or soil, under favourable conditions they germinate into new individuals.\nQuestion 38.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(c) When a sporangium burst, large number of spores are released in the air. When these spores land on food or soil, under favourable conditions they germinate into new individuals.\nQuestion 38.\nWhat is vegetative propagation? List with brief explanation three advantages of practising this process for growing some types of plants. Select two plants from the following which are grown by this process : Banana, Wheat, Mustard, Jasmine, Gram (Foreign 2016)\nAnswer:\nVegetative propagation is an asexual method of reproduction in plants. In this method, new plants are obtained from the parts of old plants (like stems, roots and leaves), without the help of any reproductive organs.\nAdvantages of vegetative propagation are as follows:\nVegetative propagation is usually used for the propagation of those plants which produce either very few seeds or do not produce viable seeds.\nSeedless plants can be obtained by artificial vegetative propagation.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nSeedless plants can be obtained by artificial vegetative propagation.\nGrafting is a propagation method which is very useful for fruit trees and flowering bushes. It enables to combine the most desirable characteristics of two plants.\nPlants like rose, sugarcane, cactus, etc., can be rapidly propagated through stem cuttings as this method produces new plants from just one plant quickly without waiting for flowers and seeds. Banana and jasmine are generally grown through vegetative propagation method.\nQuestion 39.\nExplain budding in Hydra with the help of labelled diagrams only. (Delhi 2015)\nAnswer:\nRefer to answer 21.\nQuestion 40.\n(a) Name the following:\n(i) Thread like non-reproductive structures present in Rhizopus.\n(ii) \u2018Blobs\u2019 that develop at the tips of the non- reproductive threads in Rhizopus.\n(b) Explain how these structures protect themselves and what is the function of the structures released from the \u2018blobs\u2019 in Rhizopus. (Delhi 2015)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(b) Explain how these structures protect themselves and what is the function of the structures released from the \u2018blobs\u2019 in Rhizopus. (Delhi 2015)\nAnswer:\n(a) (i)Threadlike non-reproductivestructures present in Rhizopus are called hyphae.\n(ii) \u2018Blobs\u2019 developing at the tip of hyphae are called sporangia which contain spores.\n(b) \u2018Ihe structures called spores (released from \u2018blobs\u2019) are present in sporangia which can develop into new Rhizopus individuals. These spores are covered with thick walls that protect them until they come in contact with another moist surface and can begin to grow.\nQuestion 41.\nExplain any three advantages of vegetative propagation. (Foreign 2015)\nAnswer:\nRefer to answer 19.\nQuestion 42.\nExplain the process of regeneration in Planaria. How is this process different from reproduction? (Foreign 2015, AI 2014)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 19.\nQuestion 42.\nExplain the process of regeneration in Planaria. How is this process different from reproduction? (Foreign 2015, AI 2014)\nAnswer:\nPlanaria possesses great power of regeneration. If the body of Planaria somehow gets cut into a number of pieces, then each body piece can regenerate into a complete Planaria by growing all the missing parts. This is shown in following figure:\n\nDuring the process of reproduction, new organism is formed from the complete parent organism. However, during fragmentation, a fragment of original parent body grows into new individual.\nQuestion 43.\nOn cutting the body of an organism into many pieces, it was observed that many of these pieces developed as new individuals. Name the process and list two organisms in which this process may be observed. Draw a schematic diagram to illustrate the changes that are likely to be observed during the development of new individuals in any one of the organisms named. (Delhi 2014)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nOn cutting the body of an organism into many pieces, each of these pieces develop as new individuals. This process is known as regeneration.\nHydra and Planaria are the organism in which this process may be observed. Poliowing is the diagram showing development of new individuals by regeneration of body parts of a parent Hydra:\n\nQuestion 44.\nDraw diagrams to explain the regeneration that takes place in each of the body parts of Planaria when its body is cut into three pieces. Name any other organism in which a similar process can be observed. (Delhi 2014)\nAnswer:\nRefer to answer 42.\nHydra is the organism in which regeneration is observed.\nQuestion 45.\nList any two modes of asexual reproduction in animals. Under which mode of reproduction is vegetative propagation placed and why? List two advantages of vegetative propagation. (AI 2014)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nList any two modes of asexual reproduction in animals. Under which mode of reproduction is vegetative propagation placed and why? List two advantages of vegetative propagation. (AI 2014)\nAnswer:\nThe two modes of asexual reproduction in animals are : (i) fission and (ii) fragmentation. Vegetative propagation is placed under asexual mode of reproduction because in this mode new plants are obtained from the parts of old plants (like stems, roots and leaves), without the help of any reproductive organs. Also refer to answer 38.\nQuestion 46.\nWhat is vegetative propagation? List its two advantages. Select two plants raised by this method from the list given below:\nWheat, Tomato, Rose, Pea, Gram, Corn, Banana (Foreign 2014)\nAnswer:\nRefer to answers 38 and 26.\nQuestion 47.\n(a) Name the mode of reproduction of the following organisms and state the important feature of each mode :\n(i) Planaria\n(ii) Hydra\n(iii) Rhizopus\n(b) We can develop new plants from the leaves of Bryophyllum. Comment.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(i) Planaria\n(ii) Hydra\n(iii) Rhizopus\n(b) We can develop new plants from the leaves of Bryophyllum. Comment.\n(c) List two advantages of vegetative propagation over other modes of reproduction. (2020)\nAnswer:\n(a) (i) Planaria \u2013 Regeneration\nRegeneration of organism from its cut body parts occurs by the process of growth and development.\nRegeneration is an asexual mode of reproduction common in lower plants and animals.\n(ii) Hydra \u2013 Budding\nIn budding, a small part of the body of the parent organism grows out as a bud which on detaching forms a new organism.\nBudding occurs in yeast, some protozoans and certain lower animals.\n(iii) Rhizopus \u2013 Spores\nSpores are usually produced in sporangia.\nSpore formation is a common method of an asexual reproduction in bacteria and most of the fungi.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(iii) Rhizopus \u2013 Spores\nSpores are usually produced in sporangia.\nSpore formation is a common method of an asexual reproduction in bacteria and most of the fungi.\n(b) The leaves of a Bryophyllum have special type of buds in their margins. These buds may get detached from the leaves, fall to ground and then grow to produce new Bryophyllum plants. The buds can also drop to the ground together with the leaf and then grow to produce new plants.\n(c) Advantages of vegetative propagation are :\nIt is a quick method of propagation.\nThe new plants produced by artificial vegetative propagation are exactly like the parent plants.\nMany plants can be grown from one plant by vegetative propagation.\nQuestion 48.\n(a) What is fragmentation in organisms? Name a multicellular organism which reproduces by this method.\n(b) What is regeneration in organism? Describe regeneration in Planaria with the help of a suitable diagram. (Delhi 2011)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(b) What is regeneration in organism? Describe regeneration in Planaria with the help of a suitable diagram. (Delhi 2011)\nAnswer:\n(a) Fragmentation is the mode of reproduction in which parent body breaks into two or more fragments and each fragment develops into a new individual. It is a method of reproduction in many filamentous algae, mycelial fungi and thalloid bryophytes, e.g., Spirogyra.\nThe given figure shows the process of fragmentation in Spirogyra:\n\n(b) Regeneration may be defined as the ability of an organism to regenerate lost part of the body which have been removed as by injury or autotomy. Many fully differentiated organisms use this ability as a mode of reproduction and give rise to new individual organisms from their body parts. It is common in Hydra, Planaria, etc.\nThe process of regeneration in Planaria is described in the figure given below:\n\nQuestion 49.\nWith the help of suitable diagrams, explain the various steps of budding in Hydra. (AI 2011)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nThe process of regeneration in Planaria is described in the figure given below:\n\nQuestion 49.\nWith the help of suitable diagrams, explain the various steps of budding in Hydra. (AI 2011)\nAnswer:\nIn multicellular organisms, such as Hydra, a small protuberance arises from one side of the body. The protuberance grows and develops adult like structure. In Hydra, it develops a hypostome and tentacles at its free end. It develops a basal disc at the point of attachment with the parent organism and finally gets detached to lead an independent life. Also refer to answer 21.\nQuestion 50.\nWhat is binary fission in organisms? With the help of suitable diagrams, describe the mode of reproduction in Amoeba. (AI2011)\nAnswer:\nBinary fission is the division of adult parental body into two nearly equal daughter cells. It is the simplest and most common method of asexual reproduction found in protistan protozoans, i.e., Amoeba, Paramecium, etc.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAmoeba reproduces by binary fission by dividing its body into two parts. When the Amoeba cell has reached its maximum size of growth, then first the nucleus of Amoeba lengthens and divides into two parts. After that the cytoplasm of Amoeba divides to form two smaller Amoeba (called daughter amoebae).\nDiagrammatic representation of binary fission in Amoeba is as follows :\n\nQuestion 51.\n(a) What is spore formation?\n(b) Draw a diagram showing spore formation in Rhizopus.\n(c) List two advantages for organisms to reproduce themselves through spores. (Foreign 2011)\nAnswer:\n(a) Spore formation is the process of formation of microscopic reproductive structures called spores. These spores when detaches from the parent gives rise to a new individual. Reproduction by the formation of spores is a common method of asexual reproduction in some bacteria and most of the fungi.\n(b) Following figure shows the process of spore formation in Rhizopus:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(b) Following figure shows the process of spore formation in Rhizopus:\n\n(c) Two advantages to spore producing organism are as follows:\nSpores help organism to survive harsh environmental conditions as spores are covered by thick walls which protect them until they come in contact with moist surface and germinate.\nSpores are generally very small and light. Therefore, it ensures easy dispersal by wind, water and animal.\nQuestion 52.\nFertilisation is the process of\n(a) transfer of male gamete to female gamete\n(b) fusion of nuclei of male and female gamete\n(c) adhesion of male and female reproductive organs\n(d) the formation of gametes by a reproductive organ. (2020)", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(b) fusion of nuclei of male and female gamete\n(c) adhesion of male and female reproductive organs\n(d) the formation of gametes by a reproductive organ. (2020)\nThe growing size of the human population is a cause of concern for all people. The rate of birth and death in a\u2019given population will determine its size. Reproduction is the process by which organisms increase their population. The process of sexual maturation for reproduction is gradual and takes place while general body growth is still going on. Some degree of sexual maturation does not necessarily mean that the mind or body is ready for sexual acts or for having and bringing up children. Various contraceptive devices are being used by human beings to control the size of population.\nAnswer:\n(b) fusion of nuclei of male and female gamete\nQuestion 53.\nList two common signs of sexual maturation in boys and girls.\nAnswer:\n(a) Two common signs of sexual maturation in boys and girls are :", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(b) fusion of nuclei of male and female gamete\nQuestion 53.\nList two common signs of sexual maturation in boys and girls.\nAnswer:\n(a) Two common signs of sexual maturation in boys and girls are :\nGrowth of pubic hair and extra hair in the armpits.\nDevelopment of oily skin and pimples.\nQuestion 54.\nWhat is the result of reckless female feticide?\nAnswer:\nFemale feticide is reducing the number of girls drastically in our country, which is also declining male-female sex ratio.\nQuestion 55.\nWhich contraceptive method changes the hormonal balance of the body?\nAnswer:\nChemical contraceptive method changes the hormonal balance of the body.\nQuestion 56.\nWrite two factors that determine the size of a population. (2020)\nAnswer:\nThe rate of birth and death in a given population will determine the size of a population.\nQuestion 57.\nWhat are all organisms called which bear both the sex organs in the same individual? Give one example of such organism. (AI 2016)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 57.\nWhat are all organisms called which bear both the sex organs in the same individual? Give one example of such organism. (AI 2016)\nAnswer:\nOrganisms which bear both male and female sex organs in the same individual are called bisexual. For example, Hibiscus.\nQuestion 58.\nList two functions of ovary of human female reproductive system. (AI 2016)\nAnswer:\nTwo functions of ovary of human female are:\nproduction of female gametes, i.e., ova\nsecretion of female hormones, i.e., estrogen and progesterone.\nQuestion 59.\nList two unisexual flowers. (Foreign 2016)\nAnswer:\nFlowers of papaya and cucumber are unisexual.\nQuestion 60.\nWhy is fertilisation not possible without pollination? (Foreign 2016)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nList two unisexual flowers. (Foreign 2016)\nAnswer:\nFlowers of papaya and cucumber are unisexual.\nQuestion 60.\nWhy is fertilisation not possible without pollination? (Foreign 2016)\nAnswer:\nThe process of pollination (in plants) ensures that male gametes bearing structure called pollen comes in contact with the female reproductive structure of the plant. Once the male and female gametes are in close vicinity, they fuse and fertilisation is accomplished. Hence, fertilisation cannot take place without pollination.\nQuestion 61.\nName the parts of a bisexual flower that are not directly involved in reproduction. (Foreign 2015)\nAnswer:\nCalyx and corolla are parts of a flower that are not directly involved in reproduction.\nQuestion 62.\nNo two individuals are absolutely alike in a population. Why? (Delhi 2014)\nAnswer:\nNo two individuals are absolutely alike in a population because sexual reproduction promotes diversity of characters in the offspring by providing genetic variation.\nQuestion 63.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nNo two individuals are absolutely alike in a population because sexual reproduction promotes diversity of characters in the offspring by providing genetic variation.\nQuestion 63.\nIdentify and write the male reproductive parts from the list of different parts of a flower given below:\nStigma, Sepal, Anther, Petal, Ovule, Filament (Delhi 2013C)\nAnswer:\nThe male reproductive parts that are present in a flower frorti given list are:\n(i) anther and (ii) filament.\nQuestion 64.\nWhat is the main difference between sperms and eggs of humans? Write the importance of this difference. (AI 2014)\nAnswer:\nThe main difference between sperms and eggs of humans is that a sperm has X or Y chromosome whereas egg has X chromosome. This helps in determination of the sex of a person and maintaining the genetic continuity in the organisms.\nQuestion 65.\n\u201cThe chromosomal number of the sexually producing parents and their offspring is the same\u201d. Justify this statement. (AI 2014)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 65.\n\u201cThe chromosomal number of the sexually producing parents and their offspring is the same\u201d. Justify this statement. (AI 2014)\nAnswer:\nIn sexual reproduction, two gametes, male and female, combines together to form a new cell \u2018zygote\u2019. The reproductive cells or gametes contain only half the amount of DNA as compared to the non-reproductive cells of an organism. So, when a male gamete combines with a female gamete during sexual reproduction, then the new cell \u2018zygote\u2019 will have the normal amount of DNA. For example, the human sperm has 23 chromosomes and the human egg has also 23 chromosomes. So when a sperm and an egg fuse together during fertilisation, then the zygote formed will have 23 + 23 = 46 chromosomes, which is the normal number of chromosomes in humans.\nQuestion 66.\nList two preparations shown every month by the uterus in anticipation of pregnancy in humans. (Foreign 2014)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 66.\nList two preparations shown every month by the uterus in anticipation of pregnancy in humans. (Foreign 2014)\nAnswer:\nThe two preparations shown every month by the uterus in anticipation of pregnancy in human are:\n(i) the wall of uterus becomes thick to receive the fertilised egg and\n(ii) the uterine wall is richly supplied with blood to nourish the growing embryo.\nQuestion 67.\nState one genetically different feature between sperms and eggs of humans. What is its consequence? (Delhi 2012)\nAnswer:\nA sperm may have X or Y chromosomes whereas egg have X chromosomes. The consequence of this is that sperm decides the sex of the child because eggs contribute only X chromosome while sperms contribute either X or Y chromosomes to the offspring. Therefore, if a child inherits X chromosome from her father, will be a girl and the one that inherit Y chromosome will be a boy.\nQuestion 68.\nState the role of placenta in the development of embryo. (AI 2012)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 68.\nState the role of placenta in the development of embryo. (AI 2012)\nAnswer:\nPlacenta is a physiological connection between an embryo and uterine wall of the mother through which nutrients and other useful substances enter into fetus from mother\u2019s blood and waste products like urea and carbon dioxide are expelled into mother\u2019s blood from fetus.\nQuestion 69.\nList the parts of human male reproductive system which contribute fluid to the semen. State two advantages semen offers to the sperms. (Foreign 2012)\nAnswer:\nThe secretion of male accessory reproductive glands, i.e., seminal vesicles, prostate gland and Cowper\u2019s glands contribute fluid to the semen. The two advantages that semen offers to the sperms are:\n(i) it provides nutrition to the sperms and\n(ii) it also activates the sperms and make their transport easier into the vagina of female during sexual act.\nQuestion 70.\nExplain giving one example of each, the unisexual and the bisexual flowers. (Delhi 2011)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 70.\nExplain giving one example of each, the unisexual and the bisexual flowers. (Delhi 2011)\nAnswer:\n(i) Unisexual flowers : These flowers contain either stamens (male reproductive part) or carpel (female reproductive part). Example: Papaya, watermelon.\n(ii) Bisexual flowers : The flower is said to be bisexual when both male and females parts, i.e., stamens and carpels, are present on the same flower. Example: Hibiscus, mustard.\nQuestion 71.\nDescribe the role of fallopian tubes in the female reproductive system. (AI2011)\nAnswer:\nFallopian tubes are a pair of elongated, ciliated, muscular and tubular structures extending from close to ovaries to uterus. It is the site of fertilisation and helps in the conduction of ovum or zygote towards uterus by ciliary action and peristalsis.\nQuestion 72.\nExplain the terms:\n(i) Implantation of zygote\n(ii) Placenta (Foreign 2011)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 72.\nExplain the terms:\n(i) Implantation of zygote\n(ii) Placenta (Foreign 2011)\nAnswer:\n(i) Implantation of zygote refers to the process of attachment of the blastocyst on the inner wall of the uterus, It occurs on 7th day after fertilisation and is controlled by estrogen and progesterone hormones.\n(ii) Refer to answer 68.\nQuestion 73.\nDefine the term pollination. Differentiate between self-pollination and cross-pollination. What is the significance of pollination? (2020)\nAnswer:\nThe process of transfer of pollen grains from anther of a flower to the stigma of the same flower or another flower of the same species is known as pollination. The two inodes of pollination are self pollination and cross pollination.\nDifferences between self pollination and cross pollination are as follows:\nPollination is important because it brings pollen grains to the female reproductive part (carpel) of the plant that leads to fertilisation.\nQuestion 74.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nPollination is important because it brings pollen grains to the female reproductive part (carpel) of the plant that leads to fertilisation.\nQuestion 74.\n(a) What provides nutrition to human sperms? State the genetic constitution of a sperm.\n(b) Mention the chromosome pair present in a zygote which determines the sex of (i) a female child and (ii) a male child. (2020)\nAnswer:\n(a) The secretions of seminal vesicles and prostate gland provides nutrition to the human sperms and also make their further transport easier. The genetic constitution of a sperm can be 50% have X chromosome and 50% have Y chromosome.\n(b) (i) XX \u2013 Female child\n(ii) XY \u2013 Male child\nQuestion 75.\nState the basic requirement for sexual reproduction. Write the importance of such reproductions in nature. (Delhi 2017)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(b) (i) XX \u2013 Female child\n(ii) XY \u2013 Male child\nQuestion 75.\nState the basic requirement for sexual reproduction. Write the importance of such reproductions in nature. (Delhi 2017)\nAnswer:\nThe basic requirement for sexual reproduction is involvement of both sexes, i.e., male and female, to produce an offspring. It takes place by the combination of gametes which come from two different parents.\nThe importance of sexual reproduction in nature are :\n(i) Fusion of male and female gametes coming from two different and sexually distinct individuals, exhibit diversity of characters in offspring.\n(ii) Meiosis during gametogenesis provides opportunities for new combination of genes, which leads to variation required for evolution and plays a prominent role in the origin of new species. Variations lead to the appearance of such characters, which fit to the changing environment, resulting in the survival of the species.\nQuestion 76.\nState the changes that take place in the uterus when:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 76.\nState the changes that take place in the uterus when:\n(a) Implantation of embryo has occurred.\n(b) Female gamete/egg is not fertilised. (Delhi 2017)\nAnswer:\n(a) Implantation is the close attachment of the blastocyst (young multicellular embryo) to the uterine wall. It is fullowed by a number of developmental changes in the thickened wall of uterus. An intimate connection between the fetal membrane and the uterine wall called placenta is formed. This is a disc which is embedded in the uterine wall. The placenta serves as the nutritive, respiratory and excretory organ of the fetus.\n(b) When the female gamete/egg is not fertilised, this lining is not needed any longer. So, the lining slowly breaks and comes out through vagina as blood and mucus. This cycle takes place every month and is known as menstrual cycle.\nQuestion 77.\nList any two steps involved in sexual reproduction and write its two advantage. (Delhi 2017)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 77.\nList any two steps involved in sexual reproduction and write its two advantage. (Delhi 2017)\nAnswer:\nThe two main steps involved in sexual reproduction are:\nformation of male and female gametes.\nFusion of a male gamete with a female gamete to form a new cell called zygote by the process of fertilisation.\nThe two important advantages of sexual reproduction are:\nIt promotes diversity of characters in the offspring through genetic variations.\nIt plays an important role in continuous evolution of better organisms that may lead to the origin of new species.\nQuestion 78.\nList three techniques that have been developed to prevent pregnancy. Which one of these techniques is not meant for males? How does the use of these techniques have a direct impact on the health and prosperity of a family? (AI 2017)\nAnswer:\nMethods developed to prevent pregnancy are:\nbarrier method, i.e., use of condoms, diaphragm, etc.\nchemical method, i.e., use of oral pills or vaginal pills.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nMethods developed to prevent pregnancy are:\nbarrier method, i.e., use of condoms, diaphragm, etc.\nchemical method, i.e., use of oral pills or vaginal pills.\nsurgical method, i.e., vasectomy and tubectomy. Out of these methods, chemical method is not meant for males.\nUse of these techniques help to keep control over number of children in a family, which directly effects prosperity of a family. One of the most common reason for deterioralion of women\u2019s health is frequent conception and child bearing. Controlled childbirth will directly affect women health and this will indirectly affect the prosperity of family and nation.\nQuestion 79.\nHow do organisms, whether reproduced asexually or sexually maintain a constant chromosome number through several generations? Explain with the help of suitable example. (Delhi 2016)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nHow do organisms, whether reproduced asexually or sexually maintain a constant chromosome number through several generations? Explain with the help of suitable example. (Delhi 2016)\nAnswer:\nIn organisms reproducing asexually, only single parent is involved in reproduction. Therefore, amount of DNA remains same from parent to offspring. For example in Amoeba, whole organism divides into two daughter individuals by binary fission. Therefore, amount of DNA remain constant.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nIn organisms reproducing sexually, reproduction take place with the help of formation of haploid gametes. Gametes are special type of cells called reproductive cells which contain only half the amount of DNA as compared to the normal body cells of an organism. So, when a male gamete combines with a female gamete during sexual reproduction, then the new cell \u2018zygote\u2019 will have the normal amount of DNA. For example, the human sperm has 23 chromosomes and the human egg (or ovum) has also 23 chromosomes. So, when a sperm and an egg fuse together during fertilisation, then the zygote formed will have 23 + 23 = 46 chromosomes, which is the normal number of chromosomes.\nQuestion 80.\nName the parts A, B and C shown in the following diagram and state one function of each. (Delhi 2016)\n\nAnswer:\nIn the given figure, part A is anther, part B is style and part C is ovule.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 80.\nName the parts A, B and C shown in the following diagram and state one function of each. (Delhi 2016)\n\nAnswer:\nIn the given figure, part A is anther, part B is style and part C is ovule.\nAnther (A) is a part of male reproductive organ of flower called stamen. Large number of pollen grains are formed inside anther. Style (B) and ovule (C) are parts of female reproductive organ of flower called carpel / pistil.\nStyle is a long conducting tube which gives the passage to pollen tube carrying male gametes so that it reaches ovary which contains one or more ovules. Ovules contain female gamete or egg. On fertilisation ovary converts into fruit and ovules give rise to seeds.\nQuestion 81.\nSuggest three contraceptive methods to control the size of human population which is essential for the health and prosperity of a country. State the basic principle involved in each. (Delhi 2016)\nAnswer:\nThree contraceptive methods which can help to control human population are:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nThree contraceptive methods which can help to control human population are:\n(i) Condom : It is a mechanical barrier which does not allow sperms and ovum to meet, hence prevents fertilisation. Condoms are made of thin rubber/latex sheath used to cover the penis in the male and vagina/cervix in female just before coitus (intercourse) so that the ejaculated semen is not released in the female reproductive tract.\n(ii) Intrauterine devices (IUDs): These are devices inserted by doctors or expert nurses in the uterus through vagina. These are presently available as non-medicated IUDs, copper releasing IUDs (CuT, etc.) and hormone releasing IUDs. They increase phagocytosis of sperms within uterus and suppress sperm motility and its fertilising capacity. They also make uterus unsuitable for implantation and cervix hostile to sperms.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(iii) Oral pills : These pills contain progesterone alone or a combination of progestogen and estrogen. They inhibit ovulation and make uterus unsuitable for implantation, hence prevent fertilisation.\nQuestion 82.\nWhat are the functions of testes in the human male reproductive system? Why are these located outside the abdominal cavity? Who is responsible for bringing about changes in appearance seen in boys at the time of puberty? (Delhi 2016)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nTestes, in human males, are the primary reproductive organs. They are the site of sperm formation. The testes also produce male sex hormone testosterone. Testes are located outside the abdominal cavity because sperm formation requires a lower temperature than normal body temperature. The temperature of the testes in the scrotum is about 2-2.5\u00b0C lower than normal body temperature. This temperature is ideal for sperm formation and development. Hormone testosterone brings about the development of secondary sexual characters during puberty in boys like growth of facial hair, deepening of voice, growth of scrotum and penis, accumulation of muscle mass, etc., and also regulates formation of sperms.\nQuestion 83.\nWhat is meant by pollination? Name and differentiate between the two modes of pollination in flowering plants. (Delhi 2016)\nAnswer:\nRefer to answer 73.\nQuestion 84.\n(a) List two reasons for the appearance of variations among the progeny formed by sexual reproduction.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 73.\nQuestion 84.\n(a) List two reasons for the appearance of variations among the progeny formed by sexual reproduction.\n\n(i) Name the part marked A in the diagram.\n(ii) How does A reaches part B?\n(iii) State the importance of the part C.\n(iv) What happens to the part marked D after fertilisation is over? (AI 2016)\nAnswer:\n(a) Variations arise in sexually reproducing organisms on account of the following:\n(i) Genetic variations occur of because DNA copying mechanism is not absolutely accurate.\n(ii) Creation of new combinations of genetic variations because variations from two individuals combine during fusion of gametes.\n(b) (i) A is pollen grain.\n(ii) Part B is stigma. It is the part of pistil (female reproductive organ) that receives pollen grains. Pollen grains reach stigma through various agencies like wind, water, insect, etc.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(ii) Part B is stigma. It is the part of pistil (female reproductive organ) that receives pollen grains. Pollen grains reach stigma through various agencies like wind, water, insect, etc.\n(iii) Pollen tube (C) carries male gametes to the ovule present in ovary. Male gametes fuse with egg and secondary nucleus to give rise to zygote and endosperm.\n(iv) Female gamete (D) fuses with male gamete and converts to embryo after fertilisation.\nQuestion 85.\nWhat is pollination? List its two types and write a distinguishing feature between the two. (Foreign 2016)\nAnswer:\nRefer to answer 73.\nQuestion 86.\nWhat is sexual reproduction? List its four significances. (Foreign 2016)\nAnswer:\nSexual reproduction is the process of production of offspringby the fusion of male and female gametes. Here, haploid gametes fuse to form diploid zygote which develop into a mature organism.\nSignificance of sexual reproduction are as follows:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nSignificance of sexual reproduction are as follows:\nSexual reproduction gives rise to genetic variations because of genetic recombination that takes place during fusion of gametes.\nProgenies arising through sexual reproduction sometimes show better combination of traits and get better adapted to their surroundings.\nGenetic recombination, interaction, etc., during sexual reproduction provide vigour and vitality to the offspring.\nVariations in genes play an important role in evolution.\nQuestion 87.\nName the reproductive parts of an angiosperm. Where are these parts located? Explain in brief the structure of its female reproductive parts. (Foreign 2016)\nAnswer:\nThe reproductive parts of an angiosperm are stamen (male reproductive part) and carpel/pistil (female reproductive part). These are located in the flowers of an angiospermic plant.\nThe given diagram shows the structure of female reproductive part of a flower:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nThe given diagram shows the structure of female reproductive part of a flower:\n\nA Carpel is made of three parts: stigma, style and ovary. The top part of carpel is called stigma. Stigma is for receiving the pollen grains during pollination. Stigma is sticky so that pollen can stick to it. The middle part of carpel is called style. Style is a tube which connects stigma to the ovary. The swollen part at the bottom of a carpel is called ovary. The ovary contains ovules. Ovules contain the female gametes or female sex cells (egg) of the plant. There are usually many ovules in the ovary. Each ovule contains only one female gamete of the plant.\nQuestion 88.\n(a) Mention the role of the following organs of human male reproductive system.\n(i) Testes (ii) Scrotum (iii) Vas deferens\n(iv) Prostate gland\n(b) What are the two roles of testosterone? (Foreign 2016)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(i) Testes (ii) Scrotum (iii) Vas deferens\n(iv) Prostate gland\n(b) What are the two roles of testosterone? (Foreign 2016)\nAnswer:\n(a) (i) Testes : The two testes in male are the sites where male gametes, i.e., sperms are formed. Testes also produce the male sex hormone called testosterone.\n(ii) Scrotum : The scrotum is a pouch of skin that lie outside abdominal cavity. The two testes lie in respective scrotal sacs. The scrotum acts as a thermoregulator and provides an optimal temperature for the formation of sperms.\n(iii) Vas deferens : This is a straight tube, about 40 cm long, which carries the sperms to the seminal vesicles, where mucus and a watery alkaline fluid containing fructose, mix with the sperms.\n(iv) Prostate gland : It is a single large gland that surrounds the urethra and produces a milky, slightly acidic secretion.", + "(iv) Prostate gland : It is a single large gland that surrounds the urethra and produces a milky, slightly acidic secretion. Secretion of prostate gland nourishes and activates the sperm to swim.\n(b) Two roles of testosterone are:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(b) Two roles of testosterone are:\n(i) It plays a key role in development of male secondary sex organs such as prostate, etc.\n(ii) It promotes the secondary sexual character-istics in males such as increased muscle and bone mass, growth of body hair, etc.\nQuestion 89.\nList any four methods of contraception used by humans. How does their use have a direct effect on the health and prosperity of a family? (Delhi 2015, 2014)\nAnswer:\nThe four methods of birth control which deliberately prevent fertilisation in humans are:\nBarrier method- These are physical devices to prevent the entry of sperm in the female, e.g., condoms.\nChemical method \u2013 It involves the use of oral pills that check ovulation. These are mainly hormonal preparations and contain estrogen and progesterone.\nIntrauterine contraceptive device-These devices are implemented into uterus, e.g., copper \u2013 T, to prevent fertilisation.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nIntrauterine contraceptive device-These devices are implemented into uterus, e.g., copper \u2013 T, to prevent fertilisation.\nSurgical methods : These methods involves removal of a small jJortion of vas deferens in males or fallopian tube in females to prevent fertilisation. Contraception prevents frequent pregnancies and sexually transmitted diseases thus supports good health and prosperity of a family.\nQuestion 90.\nWhat are sexually transmitted diseases? List two examples each of diseases caused due to\n(i) bacterial infection and (ii) viral infection. Which device or devices may be used to prevent the spread of such diseases? (Delhi 2015)\nAnswer:\nThe diseases that are spread by sexual contact with an infected person are called sexually transmitted disease (STDs).\n(i) Bacterial infection causes gonorrhoea, syphilis.\n(ii) Viral infection causes AIDS, genital herpes. STDs can be prevented by using male and female condoms.\nQuestion 91.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(i) Bacterial infection causes gonorrhoea, syphilis.\n(ii) Viral infection causes AIDS, genital herpes. STDs can be prevented by using male and female condoms.\nQuestion 91.\nList six specific characteristics of sexual reproduction. (AI 2015)\nAnswer:\nSix specific characteristics of sexual reproduction are as follows:\nTwo different sexes, i.e., male and female are involved in this process.\nSexual reproduction involves formation of special sex cells called gametes.\nFusion of gametes or fertilisation takes place in the body of female (internal fertilisation) or outside (external fertilisation).\nOffspring inherit traits from both parents (heredity) and also show some new traits of their own (variation), hence they are not clones of the parents.\nVariations in sexually reproducing organisms arises on account of crossing over during meiotic division during gamete formation.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nVariations in sexually reproducing organisms arises on account of crossing over during meiotic division during gamete formation.\nIt plays a prominent role in origin of new species as it leads to variations which accumulate over a period of time and get carried to successive generations.\nQuestion 92.\nList four points of significance of reproductive health in a society. Name any two areas related to reproductive health which have improved over the past 50 years in our country. (AI 2015)\nAnswer:\nThe general awareness regarding reproductive health in a society is significant as :\nMaintenance of personal hygiene among youngsters and proper knowledge of their reproductive parts helps them adjust with the physical changes and cope with emotional disturbances.\nReproductively healthy society must be free from the curse of child marriage which begets many complications at the level of individual and society both.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nReproductively healthy society must be free from the curse of child marriage which begets many complications at the level of individual and society both.\nProper care of expecting mothers, monitoring their health after child birth and care of new born help in building a healthy society.\nMarried couples aware of contraceptive methods lead a better married life as they are capable of avoiding unwanted pregnancies and have negligible chances of contracting sexually transmitted diseases.\nIn past 50 years, various areas related to reproductive health have been launched which have improved the reproductive health of our society in following ways: Two of them are :\n(i) reduced mortality rate of mother and infant\n(ii) birth control due to easily available contraceptive and reduced STDs cases.\nQuestion 93.\nDraw longitudinal section of a bisexual flower and label the following parts on it.\n(i) Anther\n(ii) Ovary\n(iii) Stigma\n(iv) Style (Foreign 2015)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 93.\nDraw longitudinal section of a bisexual flower and label the following parts on it.\n(i) Anther\n(ii) Ovary\n(iii) Stigma\n(iv) Style (Foreign 2015)\nAnswer:\nLongitudinal section of a bisexual flower is as follows:\n\nQuestion 94.\nWhat is placenta? Explain its function in humans. (Foreign 2015, AI 2014)\nAnswer:\nPlacenta is an intimate connection between fetus and uterine wall of the mother to exchange the materials. It is a disc shaped structure embedded in the uterine wall. It contains villi on embryo\u2019s side and blood spaces towards mothers side. Blood spaces surround villi.\nPlacenta performs the following functions:\nAll nutritive elements from maternal blood pass into the fetus through it.\nPlacental helps in respiration, i.e., supply of oxygen and removal of CO2 from fetus to maternal blood.\nFetal excretory products diffuse out into maternal blood through placenta and are excreted by mother.\nPlacenta also secretes hormone.\nQuestion 95.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nFetal excretory products diffuse out into maternal blood through placenta and are excreted by mother.\nPlacenta also secretes hormone.\nQuestion 95.\nWrite one main difference between asexual and sexual mode of reproduction. Which species is likely to have comparatively better chances of survival \u2013 the one reproducing asexually or the one reproducing sexually? Justify your answer. (Foreign 2015, AI 2014)\nAnswer:\nDifference between asexual and sexual mode of reproduction is as follows :\nAsexual reproduction :\nGametes are not formed hence fertilisation does not take place.\nSexual reproduction :\nGametes are always formed and fertilisation takes place to form a zygote.\nSpecies reproducing sexually have a better chance of survival as variation occurs only during the sexual reproduction. Variations are necessary for evolution and to increase chances of survival in changed environmental conditions.\nQuestion 96.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 96.\nWhy is DNA copying an essential part of the process of reproduction? What are the advantages of sexual reproduction over asexual reproduction? (Foreign 2015)\nAnswer:\nDNA copying is an essential part of the process of reproduction as it results in passing of nearly same genetic information from parents to the offsprings. DNA replication also ensures that same number of chromosomes are passed from parents to offspring.\nAdvantages of sexual reproduction over asexual reproduction is that sexual reproduction provides variations which is a major factor for evolution that helps in survival of species in changing environment.\nQuestion 97.\nDraw a diagram of the longitudinal section of a flower exhibiting germination of pollen on stigma and label (i) ovary (ii) male germ cell,\n(iii) female germ cell and (iv) ovule on it. (Foreign 2015)\nAnswer:\nThe diagram of the longitudinal section of flower is as follows:\n\nQuestion 98.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(iii) female germ cell and (iv) ovule on it. (Foreign 2015)\nAnswer:\nThe diagram of the longitudinal section of flower is as follows:\n\nQuestion 98.\nWrite names of those parts of a flower which serve the same function as the following do in the animals\n(i) testes\n(ii) sperm\n(iii) ovary\n(iv) egg. (Delhi 2014)\nAnswer:\nThe parts of a flower which serve the same function as following do in the animals are\n(i) testes \u2013 anther of stamen\n(ii) sperm \u2013 pollen\n(iii) ovary \u2013 ovary of pistil\n(iv) egg \u2013 female germ cell present in ovule.\nQuestion 99.\nList four methods of contraception used by humans.\nJustify the following statement.\n\u201cThe use of contraceptive methods has a direct effect on the health and prosperity of a family.\u201d (AI 2014)\nAnswer:\nRefer to answer 89.\nQuestion 100.\nDescribe in brief the function of the various parts of the female reproductive part of a bisexual flower.", + "Justify the following statement.\n\u201cThe use of contraceptive methods has a direct effect on the health and prosperity of a family.\u201d (AI 2014)\nAnswer:\nRefer to answer 89.\nQuestion 100.\nDescribe in brief the function of the various parts of the female reproductive part of a bisexual flower. (Foreign 2014)\nAnswer:\nRefer to answer 87.\nQuestion 101.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nRefer to answer 89.\nQuestion 100.\nDescribe in brief the function of the various parts of the female reproductive part of a bisexual flower. (Foreign 2014)\nAnswer:\nRefer to answer 87.\nQuestion 101.\nOn the notice board of ultrasound clinics it is generally stated. \u201cHere prenatal sex determination and disclosure of sex (boy or girl before birth) of fetus is not done. It is prohibited and punishable under law.\u201d\n(a) List two advantages of imposing ban on prenatal sex determination.\n(b) What can students do to educate the society about the following?\n(i) The ill-effects of indiscriminate female feticide.\n(ii) Adopting small family norms. (Foreign 2014)\nAnswer:\n(a) The two advantages of imposing ban on prenatal sex determination are\n(i) check on female feticide\n(ii) improving sex ratio in the country.\n(b) Students should educate the society as that", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) The two advantages of imposing ban on prenatal sex determination are\n(i) check on female feticide\n(ii) improving sex ratio in the country.\n(b) Students should educate the society as that\n(i) female feticide is reducing the number of girls drastically in some societies. For a healthy society, the male-female sex ratio must be maintained at almost the same level. Due to reckless female feticide, the male-female child sex ratio is declining at an alarming rate in some sections of our society.\n(ii) Children in a small family can be provided with all the resources from education, good amenities like food, clothing and healthy life style. As the family grows larger, the resources should be shared with increased number of member. Having fewer children also keeps the mother in good health.\nQuestion 102.\nName the two reproductive parts of a bisexual flower which contain the germ cells. State the location and function of its female reproductive part. (Foreign 2014)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 102.\nName the two reproductive parts of a bisexual flower which contain the germ cells. State the location and function of its female reproductive part. (Foreign 2014)\nAnswer:\nThe two reproductive parts of a bisexual flower which contain the germ cells are carpel (female reproductive part) and stamen (male reproductive part). Carpel is situated in the centre of the flower as a flask-shaped structure. A carpel is made up of three parts-stigma, style and ovary. The distal part of a carpel is called stigma. Stigma is responsible for receiving pollen during pollination. Style is an elongated tubular structure which connects stigma with ovary. The basal swollen part of carpel is ovary. Ovary bears several ovules. After fertilisation ovules form seeds and ovary forms the fruit.\nQuestion 103.\nWrite two examples each of sexually transmitted diseases causes by (i) virus, (ii) bacteria. Explain how the transmission of such diseases be prevented? (Delhi 2013)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 103.\nWrite two examples each of sexually transmitted diseases causes by (i) virus, (ii) bacteria. Explain how the transmission of such diseases be prevented? (Delhi 2013)\nAnswer:\nThe infectious (communicable) diseases, which are spread from an infected person to a healthy person by sexual contact, are called sexually transmitted diseases.\nSexually transmitted diseases caused by virus are : AIDS (Acquired Immunodeficiency Syndrome) and genital warts while caused by bacteria are gonorrhoea and syphilis.\nPreventive measures for these diseases are:\neducating people in high risk groups.\nmutually faithful monogamous relationship.\navoiding prostitution, multipartner sex and homosexuality.\nusing condoms, etc.\nQuestion 104.\nExplain the meaning of sexually transmitted diseases (STDs). Give two examples of STDs each, caused due to (i) bacterial infection and (ii) viral infection. State in brief how the spread of such diseases may be prevented. (Delhi 2012)\nAnswer:\nRefer to answer 103.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 103.\nQuestion 105.\nList and explain in brief three methods of contraception. (AI 2012)\nAnswer:\nRefer to answer 81.\nQuestion 106.\nWhat is AIDS? Which microbe is responsible for AIDS infection? State one mode of transmission of this disease. Explain in brief one measure for prevention of AIDS. (Foreign 2012)\nAnswer:\nAIDS (Acquired Immunodeficiency Syndrome), is an infectious viral disease which weakens the immune system of human body and generally leads to death. It is caused by a retrovirus called HIV (Human Immunodeficiency Virus). AIDS can be transmitted by having sexual contact with an infected person.\nUse of condoms which are physical barriers can reduce the risk of a sexual exposure to HIV.\nQuestion 107.\n(a) List two sexually transmitted diseases in each of the following cases:\n(i) Bacterial infections\n(ii) Viral infections\n(b) How may the spread of such diseases be prevented? . (Delhi 2011)\nAnswer:\nRefer to answer 103.\nQuestion 108.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(i) Bacterial infections\n(ii) Viral infections\n(b) How may the spread of such diseases be prevented? . (Delhi 2011)\nAnswer:\nRefer to answer 103.\nQuestion 108.\nWhat does HIV stands for? Is AIDS an infectious disease? List any four modes of spreading AIDS. (AI 2011)\nAnswer:\nHIV stands for Human Immunodeficiency Virus. Yes, AIDS is an infectious disease. It is transmitted sexually or through exposure to contaminated blood.\nFour modes of spreading AIDS are :\nunprotected sex with an infected partner\nuse of contaminated needle and syringes\nuse of contaminated razors for shaving\ntransfusion of infected blood or blood products.\nQuestion 109.\nExpand AIDS. List any four methods of prevention (control) of AIDS. (Foreign 2011)\nAnswer:\nAIDS is expanded as Acquired Immuno-deficiency Syndrome.\nFour methods of prevention or control of AIDS are :\nuse of sterilised needles and syringes for injecting drugs or vaccine\nto avoid sex with multiple partners\nuse of condoms during intercourse", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nFour methods of prevention or control of AIDS are :\nuse of sterilised needles and syringes for injecting drugs or vaccine\nto avoid sex with multiple partners\nuse of condoms during intercourse\navoid use of contaminated razor in barber shop.\nQuestion 110.\n(a) List three different categories of contraceptive methods.\n(b) Why has Government of India prohibited prenatal sex determination by law? State its benefits in the long run.\n(c) Unsafe sexual act can lead to various infections. Name two bacterial and two viral infections caused due to unsafe sex. (2020)\nAnswer:\n(a) Three different categories of contraceptive methods are :\n(i) Barrier methods, i.e., use of condoms, etc.\n(ii) Chemical methods, i.e., use of oral pills or vaginal pills.\n(iii) Surgical methods, i.e., vasectomy and tubectomy.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(i) Barrier methods, i.e., use of condoms, etc.\n(ii) Chemical methods, i.e., use of oral pills or vaginal pills.\n(iii) Surgical methods, i.e., vasectomy and tubectomy.\n(b) Prenatal sex determination was banned in India in 1994. This was done to prevent sex selective abortion. It is being used to kill the normal female fetus. This killing of the unborn girl child is called female feticide which is reducing the number of girls drastically in some societies of our country. Due to reckless female feticide, male-female sex ratio is declining at an alarming rate. Its benefit in the long run is that the female-male ratio could be maintained for a healthy society.\n(c) Bacterial diseases due to unsafe sex are gonorrhoea, syphilis.\nViral diseases due to unsafe sex are AIDS, genital herpes.\nQuestion 111.\n(a) In the female reproductive system of human beings, state the functions of:\n(i) Ovary\n(ii) Oviduct.", + "(c) Bacterial diseases due to unsafe sex are gonorrhoea, syphilis.\nViral diseases due to unsafe sex are AIDS, genital herpes.\nQuestion 111.\n(a) In the female reproductive system of human beings, state the functions of:\n(i) Ovary\n(ii) Oviduct.\n(b) Mention the changes which the uterus undergoes, when\n(i) it has to receive a zygote.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(a) In the female reproductive system of human beings, state the functions of:\n(i) Ovary\n(ii) Oviduct.\n(b) Mention the changes which the uterus undergoes, when\n(i) it has to receive a zygote.\n(ii) no fertilisation takes place.\n(c) State the functions of placenta. (2020)\nAnswer:\n(a) (i) Tire ovaries in female are primary sex organs (or female gonads) which perform the dual function \u2013 production of female gametes (eggs or ova) and secretion of female sex hormones (estrogen and progesterone).\n(ii) Oviducts or fallopian tube are paired tubes originating near to the ovaries of their respective sides and extend upto uterus. The terminal part of fallopian tube is funnel-shaped with finger-like projections called fimbriae lying near ovary. Fimbriae pick up the ovum released from ovary and push it into fallopian tube. Fertilisation also takes place in the oviduct.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(b) (i) As the ovary releases one egg every month, the uterus also prepares itself, every month to receive fertilised egg by making its lining thick and spongy to nourish the zygote if fertilisation takes place.\n(ii) When the female gamete/egg is not fertilised, this lining is not needed any longer. So, the lining slowly breaks and comes out through vagina as blood and mucus through menstrual cycle that takes place every month.\n(c) Placenta performs the following functions :\nAll nutritive elements from maternal blood pass into the fetus through it.\nPlacenta helps in respiration i.e., supply of oxygen and removal of CO2 from fetus to maternal blood.\nFetal excretory products diffuse out into maternal blood through placenta and are excreted by mother.\nPlacenta also secretes hormone.\nQuestion 112.\n(a) Draw a diagram showing germination of pollen on stigma of a flower and mark on it the following organs/parts:\n(i) Pollen grain\n(ii) Pollen tube\n(iii) Stigma\n(iv) Female germ cell", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 112.\n(a) Draw a diagram showing germination of pollen on stigma of a flower and mark on it the following organs/parts:\n(i) Pollen grain\n(ii) Pollen tube\n(iii) Stigma\n(iv) Female germ cell\n(b) State the significance of pollen tube.\n(c) Name the parts of flower that develop after fertilisation into:\n(i) Seed\n(ii) Fruit. (2020)\nAnswer:\n(a) Diagram showing germination of pollen on stigma of a flower is:\n\n(b) The pollen tube acts as a conduit to transport the male gamete cells from the pollen grain at stigma to the ovules at the base of the carpel for the process of fertilisation.\n(c) After fertilisation,\n(i) ovule develops into seed and\n(ii) ovary develops into fruit.\nQuestion 113.\nDraw a neat diagram showing fertilisation in a flower and label (a) pollen tube (b) Male germ cell and (c) Female germ cell on it. Explain the process of fertilisation in a flower.", + "Question 113.\nDraw a neat diagram showing fertilisation in a flower and label (a) pollen tube (b) Male germ cell and (c) Female germ cell on it. Explain the process of fertilisation in a flower. What happens to the (i) ovary and (ii) ovule after fertilisation? (2020)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nDiagram showing fertilisation in a flower:\n\nFertilisation, in plants, occurs when the male gamete present in pollen grain fuses with the female gamete (or egg) present in ovule. When a pollen grain falls on the stigma of the carpel, it bursts open and grows a pollen tube downwards through the style towards the female gamete in the ovary. Male gametes move down the pollen tube. The pollen tube enters the ovule in the ovary. The tip of pollen tube bursts and male gametes comes out of pollen tube. In ovary, the male gamete of pollen combines with the female gamete or egg present in ovule to form a fertilised egg.\nAfter fertilisation,\n(i) ovule develops into seed\n(ii) ovary develops into fruit.\nQuestion 114.\n(a) What is puberty?\n(b) Describe in brief the functions of the following parts in the human male reproductive system.\n(i) Testes\n(ii) Seminal vesicle\n(iii) Vas deferens\n(iv) Urethra\n(c) Why are testes located outside the abdominal cavity?", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(i) Testes\n(ii) Seminal vesicle\n(iii) Vas deferens\n(iv) Urethra\n(c) Why are testes located outside the abdominal cavity?\n(d) State how sperms move towards the female germ cell. (2020)\nAnswer:\n(a) The age at which the sex hormones begin to be produced and the boy and girl becomes sexually mature, i.e., able to reproduce is called puberty.\n(b) (i) Testes : The two testes in male are the sites where male gametes, i.e., sperms are formed. Testes also produce the male sex hormone called testosterone.\n(ii) Seminal vesicles are one pair of sac-like structures near the base of bladder. Seminal fluid is a watery alkaline fluid that contains nutrients (fructose) which serve as a source of energy for the sperm. Each seminal vesicle releases its contents into the ejaculatory duct during ejaculation.", + "(ii) Seminal vesicles are one pair of sac-like structures near the base of bladder. Seminal fluid is a watery alkaline fluid that contains nutrients (fructose) which serve as a source of energy for the sperm. Each seminal vesicle releases its contents into the ejaculatory duct during ejaculation.\n(iii) Vas deferens : This is a straight tube, about 40 cm long, which carries the sperms to the seminal vesicles, where mucus and a watery alkaline fluid containing fructose, mix with the sperms.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(iii) Vas deferens : This is a straight tube, about 40 cm long, which carries the sperms to the seminal vesicles, where mucus and a watery alkaline fluid containing fructose, mix with the sperms.\n(iv) Urethra : It is a long tube that arises from urinary bladder. Urethra carries urine from the bladder as well as sperms from the vas deferens, through the penis.\n(c) Testes are located outside the abdominal cavity because sperm formation requires a lower temperature than normal body temperature. The temperature of the testes in the scrotum is about 2-2.5\u00b0C lower than normal body temperature. This temperature is ideal for sperm formation and development.\n(d) The sperms present in the testes of man are introduced into the vagina of the woman through penis during copulation. Millions of sperms are released into the vagina at one time. The sperms are highly active and mobile. They travel from here upward through the uterus at the top of fallopian tube within five minutes.\nQuestion 115.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 115.\nBased on the given diagram answer the questions given below:\n\n(a) Label the parts A, B, C and D.\n(b) Name the hormone secreted by testis and mention its role.\n(c) State the functions of B and C in the process of reproduction. (2020)\nAnswer:\n(a) A \u2013 Ureter\nB \u2013 Seminal vesicle\nC \u2013 Urethra\nD \u2013 Vas deferens\n(b) Testes produce male sex hormone testo-sterone. Hormone Testosterone brings about the development of secondary sexual characters during puberty in boys like growth of facial hair, deepening of voice, build up of muscle mass and also regulates formation of sperms.\n(c) Seminal vesicles (B) release its contents into the ejaculatory duct during ejaculation. Urethra (C) carries sperms from the vas deferens through the penis.\nQuestion 116.\n(a) List two reasons of using contraceptive methods by married couples.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 116.\n(a) List two reasons of using contraceptive methods by married couples.\n(b) Write in proper sequence the processes going on in the different organs of the reproductive system of a human female starting from the time of egg production to childbirth. (2020)\nAnswer:\n(a) Two reasons of using contraceptive methods by married couples are :\n(i) They prevent frequent or unwanted pregnancies.\n(ii) They prevent the transfer of sexually transmitted diseases (STDs).\n(b) The female germ cells or eggs are made in ovaries. They are also responsible for the production of some hormones. When a girl is born, the ovaries already contain thousands of immature eggs on reaching puberty.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nSome of these start maturing, one egg is produced every month by one of the ovaries. The egg is carried from the ovary to the womb through a thin oviduct or fallopian tube. The oviducts unite into an elastic bag-like structure known as uterus. The uterus opens into the vagina through the cervix. The sperms enter through vagina passage during sexual intercourse. They travel upwards and reach the oviduct where they may meet the egg. The fertilised egg, the zygote gets implanted in the lining of uterus and starts dividing. The mothers body is designed to undertake the development of the child. So, the uterus prepares itself every month to receive and nurture the growing embryo. The lining thickens and is richly supplied with blood to nourish the growing embryo.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAfter implantation, a disc-like special tissue develops between the uterus wall (called uterine wall) and the embryo (or fetus), which is called placenta. The fetus is connected to placenta in mothers body through umbilical cord. It is through the placenta that all the requirements of the developing fetus like nutrition, respiration and excretion, etc., are met from the mothers body. The development of the child inside the mothers body takes approximately nine months. The child is born as a result of rhythmic contractions of the muscles in the uterus.\nQuestion 117.\n(a) Identify the modes of asexual reproduction in each of the following organisms:\n(i) Hydra\n(ii) Planaria\n(iii) Amoeba\n(iv) Spirogyra\n(v) Rhizopus\n(b) List three advantages of vegetative propagation.\n(c) Why cannot fertilisation take place in flowers if pollination does not occur?", + "(c) Why cannot fertilisation take place in flowers if pollination does not occur? (2020)\nAnswer:\n(a) (i) Hydra \u2013 Budding\n(ii) Planaria \u2013 Regeneration\n(iii) Amoeba \u2013 Fission\n(iv) Spirogyra \u2013 Fragmentation\n(v) Rhizopus \u2013 Spores", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) (i) Hydra \u2013 Budding\n(ii) Planaria \u2013 Regeneration\n(iii) Amoeba \u2013 Fission\n(iv) Spirogyra \u2013 Fragmentation\n(v) Rhizopus \u2013 Spores\n(b) Refer to answer 47(c).\n(c) Pollination brings male gametes in close proximity to the female reproductive part of flower. In the absence of pollination, there will be no male gametes available for fertilisation with female gametes.\nQuestion 118.\nDefine pollination. Explain the different types of pollination. List two agents of pollination. How does suitable pollination lead to fertilisation? (Delhi 2019)\nAnswer:\nThe process of transfer of pollen grains from anther of a flower to the stigma of the same flower or another flower of the same species is known as pollination. Pollination may be of two major types- (i) self pollination and (ii) cross pollination.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(i) Self pollination is the transfer of pollen grains from the anther to the stigma of the same flower, or to the stigma of another flower of the same plant. This pollination generally takes place in bisexual flowers because they have both male and female gametes in them.\n(ii) Cross pollination is the transfer of pollen grains from the anther of a flower of one plant to the stigma of a flower of another plant of the same species. This occurs in unisexual as well as bisexual flowers.\nTwo agents of pollination are wind and water.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nTwo agents of pollination are wind and water.\nPollination results in the deposition of related pollen grains over the receptive stigma of the carpel. Pollen grains after landing on stigma, absorb water, swell and then germinate to produce pollen tubes. Many pollen tubes grow into the stigma, but only one passes through the style and then moves towards the ovary. Two non-motile male gametes are formed inside the tube during its growth through the style. After reaching the ovary, pollen tube enters the ovule through the micropyle. The tip of the tube finally pierces the micropylar end of the embryo sac. After penetration, the tip of pollen tube ruptures releasing two male gametes into the embryo sac. The mature embryo sac consists of an egg apparatus (one haploid egg and two synergids), two polar nuclei and three antipodal cells. During the act of fertilisation, one male gamete fuses with the egg to form the diploid zygote.\nQuestion 119.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 119.\n(a) Identify the given diagram. Name the parts 1 to 5.\n\n(b) What is contraception? List three advantages of adopting contraceptive measures. (Delhi 2019)\nAnswer:\n(a) The given diagram is the sectional view of human female reproductive system.\nThe labelled parts are:\n1. Funnel of fallopian tube or oviduct\n2. Ovary\n3. Uterus or womb\n4. Cervix\n5. Vagina\n(b) Contraception is the avoidance of pregnancy. There are several methods of contraception such as:\nBarrier methods (condoms, diaphragm, etc.)\nChemical methods (spermicide creams and jellies)\nIntrauterine Contraceptive Devices (IUCDs) (Lippes loop, CuT, etc.)\nNatural methods (rhythm method, coitus interruptus)\nSurgical methods (vasectomy, tubectomy) Three advantages of adopting contraceptive methods are:\nThey prevent frequent or unwanted pregnancies.\nThey prevent the transfer of sexually transmitted diseases (STDs).\nThey help to regulate the population growth.\nQuestion 120.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nThey prevent frequent or unwanted pregnancies.\nThey prevent the transfer of sexually transmitted diseases (STDs).\nThey help to regulate the population growth.\nQuestion 120.\n(a) Distinguish between cross-pollination and self-pollination. Mention the site and product of fertilisation in a flower.\n(b) Draw labelled diagram of a pistil showing the following parts:\nStigma, Style, Ovary, Female germ cell (AI 2019)\nAnswer:\n(a) Refer to answer 73.\nFertilisation takes place in the ovary of a female flower. Inside the ovary, the ovule is fertilised by pollen. After the process of fertilisation, the ovary in the flower thickens and enlarges to form the fruit, whereas the ovule becomes the seed containing the embryo.\n(b) The labelled diagram of a pistil is as follows :\n\nQuestion 121.\n(a) Draw a diagram of human female reproductive system and label the parts:\n(i) which produce an egg\n(ii) where fertilisation takes place\n(b) List two bacterial diseases which are transmitted sexually.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(i) which produce an egg\n(ii) where fertilisation takes place\n(b) List two bacterial diseases which are transmitted sexually.\n(c) What are contraceptive devices? Give two reasons for adopting contraceptive devices in humans. (AI 2019)\nAnswer:\n(a) The sectional view of human female reproductive system is as follows:\n\n(i) Ovary is the part where eggs develop.\n(ii) Fallopian tube is the part where fertilisation takes place.\n(b) Gonorrhoea and syphilis are two bacterial diseases which are transmitted sexually by bacterium Neisseria gonorrhoeae and Treponema pallidum respectively.\n(c) Contraceptive devices are those devices which are used to prevent pregnancy. It includes diaphragm, condom and intrauterine devices. Contraceptive methods are adopted:\n(i) to avoid unwanted birth.\n(ii) to keep the population of a country under control.\nQuestion 122.\n(a) Write the function of following parts in human female reproductive system:\n(i) Ovary (ii) Oviduct (iii) Uterus.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(ii) to keep the population of a country under control.\nQuestion 122.\n(a) Write the function of following parts in human female reproductive system:\n(i) Ovary (ii) Oviduct (iii) Uterus.\n(b) Describe in brief the structure and function of placenta. (2018)\nAnswer:\n(a) Function of human female reproductive part are as follows :\n(i) Ovary: The ovaries are primary sex organs (or female gonads), which perform the dual function :\nProduction of female gametes (eggs or ova)\nSecretion of female sex hormones (estrogen and progesterone).\n(ii) Oviduct (Fallopian tube): It is a site where egg and sperms meet and fertilisation takes place.\nIt also conducts the ovum or zygote towards uterus by ciliary action and peristalsis.\n(iii) Uterus : It is the seat of implantation, placentation and foetal development. It\u2019s muscular wall helps in expelling of baby during childbirth.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(iii) Uterus : It is the seat of implantation, placentation and foetal development. It\u2019s muscular wall helps in expelling of baby during childbirth.\n(b) Placenta is an intimate mechanical and physiological connection between fetus and uterine wall of the mother. It is a disc shaped structure embedded in the uterine wall. It contains chorion having villi on the embryo\u2019s side and blood spaces towards mother\u2019s side. Blood spaces surround the villi. Placenta is connected to the fetus by umbilical cord.\nPlacenta performs the following functions :\nAll nutritive elements from maternal blood passes into the fetus through it.\nPlacenta helps in respiration, i.e., supply of O2 to the fetus and removal of CO2 from fetus to maternal blood.\nFetal excretory products diffuse out into maternal blood through placenta and are excreted by mother.\nPlacenta also secretes hormones.\nPlacenta acts as a barrier between mother and fetus blood and only allows necessary materials to pass through it.\nQuestion 123.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nPlacenta also secretes hormones.\nPlacenta acts as a barrier between mother and fetus blood and only allows necessary materials to pass through it.\nQuestion 123.\n(a) Name the organ that produces sperms as well as secretes a hormone in human males. Name the hormone it secretes and write its functions.\n(b) Name the part of the human female reproductive system where fertilisation occurs.\n(c) Explain how the developing embryo gets nourishment inside the mother\u2019s body. (Delhi 2017)\nAnswer:\n(a) Refer to answers 88 (a)(i) and (b).\n(b) In human female the fertilisation occurs in the oviducts or Fallopian tube.\n(c) The developing embryo gets nourishment from the mother\u2019s blood with the help of a special tissue called placenta. This is a disc like structure embedded in uterine wall. It contains villi that provides a large surface area to pass glucose and oxygen from mother to embryo. Placenta links the embryo to the mother through umbilical cord.\nQuestion 124.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 124.\n(a) Write the functions of each of the following parts in a human female reproductive system.\n(i) Ovary\n(ii) Uterus\n(iii) Fallopian tube\n(b) Write the structure and functions of placenta in a human female. (AI 2017)\nAnswer:\n(a) (i) Refer to answer 11 l(a)(i).\n(ii) Uterus is a single, pear-shaped, highly muscular, hollow structure present in the pelvic cavity, lying between urinary bladder and rectum. If fertilisation takes place, the embryo gets implanted to the wall of uterus and grows there until birth. Development of fetus occurs inside uterus, hence it is also called womb.\n(iii) Refer to answer 111 (a)(ii).\n(b) Refer to answer 122(b).\nQuestion 125.\n(a) Write the functions of the following parts in human female reproductive system :\n(i) Ovary\n(ii) Oviduct\n(iii) Uterus\n(b) Describe the structure and function of placenta.", + "(b) Refer to answer 122(b).\nQuestion 125.\n(a) Write the functions of the following parts in human female reproductive system :\n(i) Ovary\n(ii) Oviduct\n(iii) Uterus\n(b) Describe the structure and function of placenta. (Delhi 2016)\nAnswer:\n(a) Refer to answers 111(a)(i), (ii) and 124(a)(ii). (b) Refer to answer 122(b).\nQuestion 126.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(ii) Oviduct\n(iii) Uterus\n(b) Describe the structure and function of placenta. (Delhi 2016)\nAnswer:\n(a) Refer to answers 111(a)(i), (ii) and 124(a)(ii). (b) Refer to answer 122(b).\nQuestion 126.\nWhat is placenta? Describe its structure. State its functions in case of a pregnant human female. (AI 2016)\nAnswer:\nRefer to answer 122(b).\nQuestion 127.\n(a) State in brief the functions of the following organs in the human female reproductive system.\nOvary, Fallopian tube, Uterus\n(b) What is menstruation? Why does it occur? (Foreign 2016)\nAnswer:\n(a) Refer to answers 111(a)(i), (ii) and 124(a) (ii).", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nOvary, Fallopian tube, Uterus\n(b) What is menstruation? Why does it occur? (Foreign 2016)\nAnswer:\n(a) Refer to answers 111(a)(i), (ii) and 124(a) (ii).\n(b) Menstruation is the cyclic discharge of blood along with endometrial lining of the uterus and unfertilised egg in women. It last for 3-5 days. After the release of egg in the females, the uterine lining becomes thickened for the implantation of fertilised egg or zygote. In the absence of fertilisation, the egg along with endometrial lining is expelled out of the body in the form of menstruation.\nQuestion 128.\nWrite the functions of the following in human female reproductive system.\nOvary, oviduct, uterus\nHow does the embryo get nourishment inside the mothers body? Explain in brief. (Delhi 2015)\nAnswer:\nRefer to answers 111(a) (i), (ii), 124(a)(ii) and 123 (c).", + "Ovary, oviduct, uterus\nHow does the embryo get nourishment inside the mothers body? Explain in brief. (Delhi 2015)\nAnswer:\nRefer to answers 111(a) (i), (ii), 124(a)(ii) and 123 (c).\nQuestion 129.\n(a) Name the human male reproductive organ that produce sperms and also secretes a hormone. Write the functions of the secreted hormone.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 129.\n(a) Name the human male reproductive organ that produce sperms and also secretes a hormone. Write the functions of the secreted hormone.\n(b) Name the parts of the human female reproductive system where, (i) fertilisation takes place (ii) implantation of the fertilised egg occurs.\nExplain how the embryo gets nourishment inside the mother\u2019s body. (AI 2015)\nAnswer:\n(a) Refer to answers 88(a)(i) and (b).\n(b) Refer to answers 122(a) and 123(c).\nQuestion 130.\n(a) Identify A, B, C and D in the given diagram and write their names.\n\n(b) What is pollination? Explain its significance.\n(c) Explain the process of fertilisation in flowers. Name the parts of the flower that develop after fertilisation into\n(i) seed\n(ii) fruit. (Foreign 2015)\nAnswer:\n(a) In the given diagram A is stigma, B is pollen tube, C is ovary and D is female germ cell.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(i) seed\n(ii) fruit. (Foreign 2015)\nAnswer:\n(a) In the given diagram A is stigma, B is pollen tube, C is ovary and D is female germ cell.\n(b) The process of transfer of pollen grains from the anther of a flower to the stigma of the same flower or another flower is known as pollination. Pollination is important because it brings pollen grains to the female reproductive part (carpel) of the plant that leads to fertilisation.\n(c) Fertilisation, in plants, occurs when the male gamete present in pollen grain fuses with the female gamete (or egg) present in ovule. When a pollen grain falls on the stigma of the carpel, it bursts open and grows a pollen tube downwards through the style towards the female gamete in the ovary.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nMale gametes move down the pollen tube. The pollen tube enters the ovule in the ovary. The tip of pollen tube bursts and male gametes comes out of pollen tube. In ovary, the male gamete of pollen combines with the female gamete or egg present in ovule to form a fertilised egg.\nAfter fertilisation,\n(i) ovule develops into seed\n(ii) ovary develops into fruit.\nQuestion 131.\n(a) Give one example each of unisexual and bisexual flower.\n(b) Mention the changes a flower undergoes after fertilisation.\n(c) How does the amount of DNA remain constant though each new generation is a combination of DNA copies of two individuals. (Delhi 2014)\nAnswer:\n(a) Unisexual flowers bear organs of only one sex, i.e., either stamen or pistil, e.g., papaya. Bisexual flowers contain both stamen and pistil, e.g., Hibiscus.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Unisexual flowers bear organs of only one sex, i.e., either stamen or pistil, e.g., papaya. Bisexual flowers contain both stamen and pistil, e.g., Hibiscus.\n(b) After fertilisation, the fertilised egg (or zygote) divides several times to form an embryo within the ovule. The ovule develops a tough coat around it and is gradually converted into a seed. The ovary of flower develops and becomes a fruit (with seeds inside it). The other parts of flower like sepals, petals, stamens, stigma and style dry up and fall off. Only the ovary is left behind. So, at the place on plant vyhere we had a flower originally, we now have a fruit (which is the ovary of the flower containing seeds). A fruit protects the seeds.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(c) The amount of DNA remain constant in each new generation because of formation of haploid gametes. Gametes are special type of cells called reproductive cells which contain only half the amount of DNA as compared to the normal body cells of an organism. So, when a male gamete combines with a female gamete during sexual reproduction, then the new cell \u2018zygote\u2019 will have the normal amount of DNA. For example, the human sperm has 23 chromosomes and the human egg (or ovum) has also 23 chromosomes. So, when a sperm and an egg fuse together during fertilisation, then the zygote formed will have 23 + 23 = 46 chromosomes, which is the normal number of chromosomes.\nQuestion 132.\n(a) Write the name of the human male reproductive organ that produces sperms and secretes a hormone. Name the hormone secreted and state its function.\n(b) Write the site of fertilisation and the part where the zygote gets implanted in the human female.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(b) Write the site of fertilisation and the part where the zygote gets implanted in the human female.\n(c) State, in brief, how an embryo gets its nourishment inside the mothers body. (Delhi 2014)\nAnswer:\n(a) Refer to answers 88 (a)(i) and (b).\n(b) Refer to answer 122 (a) (ii), (iii).\n(c) Refer to answer 123 (c).\nQuestion 133.\n(a) Name the respective part of human female reproductive system :\n(i) that produces egg\n(ii) where fusion of egg and sperm takes place, and\n(iii) where zygote gets implanted.\n(b) Describe in brief what happens to the zygote after it gets implanted. (Delhi 2014)\nAnswer:\n(a) Refer to answer 122 (a).", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(iii) where zygote gets implanted.\n(b) Describe in brief what happens to the zygote after it gets implanted. (Delhi 2014)\nAnswer:\n(a) Refer to answer 122 (a).\n(b) After implantation of zygote or embryo in the thick lining of the uterus, a disc-like special tissue develops between the uterus wall and the embryo, which is called as placenta. Placenta meets all the requirements for developing the fetus like nutrition, respiration, excretion, etc. When fetus (embryo) develops completely, the rhythmic contraction of uterus muscles gradually pushes the baby out of the mother s baby through vagina a baby is born.\nQuestion 134.\n(a) Draw a sectional view of human female reproductive system and label that part where\n(i) eggs develop\n(ii) fertilisation take place\n(iii) fertilised egg gets implanted\n(b) Describe, in brief, the changes that uterus undergoes\n(i) to receive the zygote\n(ii) if zygote is not formed. (AI 2014)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(ii) fertilisation take place\n(iii) fertilised egg gets implanted\n(b) Describe, in brief, the changes that uterus undergoes\n(i) to receive the zygote\n(ii) if zygote is not formed. (AI 2014)\nAnswer:\n(a) The sectional view of human female reproductive system is as follows:\n\n(i) Ovary is the part where eggs develop.\n(ii) Fallopian tube is the part where fertilisation takes place.\n(iii) Uterus is the part where fertilised egg gets implanted.\n(b) (i) When the ovum (or egg) is fertilised in the oviduct, then a zygote is formed. The uterus prepares itself every month to receive a zygote. The inner lining of uterus becomes thick and spongy with lot of blood capillaries in it. This would be required for nourishment and further development of embryo.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(ii) If a sperm is not available at the time of ovulation, then fertilisation of ovum does not take place. Since the ovum (or egg) is not fertilised, so the thick and soft uterus lining having lot of blood capillaries in it is not required. The unfertilised ovum dies within a day and the uterus lining also breaks down. The breakdown and removal of the inner, thick and soft lining of the uterus alongwith its blood vessels in the form of vaginal bleeding is called menstrual flow or menstruation.\nQuestion 135.\n(a) Name the parts labelled as A, B, C and D in the diagram given below:\n\n(b) What is pollination? State its significance.\n(c) How does fertilisation occur in flowers? Name the parts of the flower that develop into (i) seed, and (ii) fruit after fertilisation. (AI 2014)\nAnswer:\n(a) A represents pollen grains, B represents pollen tubes, C represents ovary and D represents female germ cell.\n(b) Refer to answer 130(b).\n(c) Refer to answer 130 (c).\nQuestion 136.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) A represents pollen grains, B represents pollen tubes, C represents ovary and D represents female germ cell.\n(b) Refer to answer 130(b).\n(c) Refer to answer 130 (c).\nQuestion 136.\nList in tabular form the two differences between asexual and sexual mode of reproduction. Name and explain with the help of labelled diagram the process by which Hydra reproduces asexually. (Foreign 2014, Delhi 2013C)\nAnswer:\nDifferences between asexual and sexual forms of reproduction are follows :", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\nDifferences between asexual and sexual forms of reproduction are follows :\nHydra is simple multicellular animal. It reproduces asexually by the process of budding. In Hydra first a small outgrowth called \u2018bud\u2019 is formed on the side of its body by the repeated mitotic divisions of its cells. This bud then grows gradually to form a small Hydra by developing a mouth. Finally, the tiny new Hydra detaches itself from the body of parent and lives as a separate organism. In this way, the parent Hydra produce a new Hydra. The following figure shows Hydra reproducing by the method of budding.\n\nQuestion 137.\n(a) Name the human male reproductive organ that produces sperms and also secretes hormones. Write the functions of the hormone secreted.\n(b) Name the parts of the human female reproductive system where (i) fertilisation and (ii) implantation occur respectively. Explain how the embryo gets nutrition inside the mother\u2019s body. (Foreign 2014)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Refer to answers 88 (a) (i) and (b).\n(b) Refer to answers 122 (a) and 123 (c).\nQuestion 138.\n(a) List three distinguishing features between sexual and asexual types of reproduction.\n(b) Explain why variations are observed in the offsprings of sexually reproducing organisms? (Delhi 2013)\nAnswer:\n(a) Refer to answer 136.\n(b) Sexual reproduction involves fusion of male and female gametes coming from male and female parents. Variations occur due to (i) fusion of gametes which come from two different and sexually distinct individuals and (ii) meiosis which occurs during gametogenesis and create a new combination of genes. It plays a prominent role in the origin of new species and leads to variations required for evolution. Therefore, offsprings of sexually reproducing organisms show variation.\nQuestion 139.\n(a) Identify A, B and C in the given diagram , and write their functions.\n\n(b) Mention the role of gamete and zygote in sexually reproducing organisms. (Delhi 2013)\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 139.\n(a) Identify A, B and C in the given diagram , and write their functions.\n\n(b) Mention the role of gamete and zygote in sexually reproducing organisms. (Delhi 2013)\nAnswer:\n(a) A represents stigma. It receives the pollen grains from the anther of stamen. Stigma is sticky so that pollen can stick to it so that fertilisation can occur. B represents pollen tube. Pollen tube acts as a conduit to transport the male gamete cells from the pollen grain at stigma to the ovules at the base of the carpel for the process of fertilisation. C represent female germ cell. Female germ cell fertilise with male germ cells to forms zygote which develops into an embryo within the ovule. Ovule converts into a seed that gives rise a new individual.\n(b) Role of Gamete : Gametes are the reproductive cells involved in sexual reproduction having half of the chromosome. Gametes carry variations generated during its formation (meiosis). A male gamete and a female gamete fuses to form zygote.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nRole of Zygote : The fusion of male gamete with female gamete forms a zygote during sexual reproduction. Zygote has normal number of chromosomes and new combinations of variation that express in new generation. The zygote undergoes repeated mitotic divisions to form the embryo which has the potential to form a complete individual.\nQuestion 140.\n(a) Give an example of a bisexual flower.\n(b) Draw a longitudinal section of a pistil showing the germination of pollen grains. Label the following parts:\n(i) Stigma\n(ii) Male germ cell\n(iii) Female germ cell\n(iv) Ovary\n(v) Style\n(vi) Pollen tube\n(c) Mention the site and the product of fertilisation in a flower. (Delhi 2013 C)\nAnswer:\n(a) Hibiscus is an example of a bisexual flower.\n(b) Refer to answer 120 (b).\n(c) Refer to answer 130 (c).\nQuestion 141.\nDefine the terms pollination and fertilisation.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Hibiscus is an example of a bisexual flower.\n(b) Refer to answer 120 (b).\n(c) Refer to answer 130 (c).\nQuestion 141.\nDefine the terms pollination and fertilisation.\nDraw a diagram of a pistil showing pollen tube growth into the ovule and label the following: pollen grain, male gamete, female gamete, ovary. (Delhi 2012)\nAnswer:\nThe process of transfer of pollen grains from the anther of a flower to the stigma of the same flower or another flower is known as pollination. Fertilisation is the fusion of male gamete with the female gamete (or egg).\nRefer to answer 120(b) for figure.\nQuestion 142.\nDescribe in brief the role of (i) testes (ii)\nseminal vesicle, (iii) vas deferens, (iv) ureter and (v) \u2018prostate gland in human male reproductive system. (Delhi 2012)\nAnswer:\n(i) Refer to answer 88 (a) (i).", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nseminal vesicle, (iii) vas deferens, (iv) ureter and (v) \u2018prostate gland in human male reproductive system. (Delhi 2012)\nAnswer:\n(i) Refer to answer 88 (a) (i).\n(ii) Seminal vesicles are one pair of sac like structures near the base of bladder. Seminal fluid is a watery alkaline fluid that contains nutrients (fructose) which serve as a source of energy for the sperm. Each seminal vesicle releases its contents into the ejaculatory duct during ejaculation.\n(iii) Refer to answer 88(a) (iii).\n(iv) Ureter : From hilum of each kidney emerges out a slender, whitish tube called ureter. Ureter of each kidney leaves from the renal pelvis. Each ureter is about 30 cm long, 3 \u2013 4 mm in diameter and opens into urinary bladder by slit-like aperture. The ureters carry urine from kidneys to urinary bladder.\n(v) Refer to answer 88(a) (iv).\nQuestion 143.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(v) Refer to answer 88(a) (iv).\nQuestion 143.\nDistinguish between unisexual and bisexual flowers giving one example of each. Draw a diagram showing process of germination of pollen grains on stigma and label the following parts :\n(i) Female germ cell\n(ii) Male germ cell\n(iii) Ovary (AI 2012)\nAnswer:\nThe flowers which contain only one sex organ, either stamens or carpels are called unisexual flowers. E.g., flowers of papaya and watermelon whereas the flowers which contain both the sex organs, stamens as well as carpel, are called bisexual flowers, e.g., flowers of Hibiscus and mustard.\nRefer to answer 120(b) for figure.\nQuestion 144.\nDraw a diagram of human female reproductive system and label the part\n(i) that produces eggs\n(ii) where fusion of egg and sperm takes place\n(iii) where zygote is implanted.\nWhat happens to human egg when it is not fertilised? (AI 2012)\nAnswer:\nRefer to answer 134 (a) and b (ii).\nQuestion 145.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\n(iii) where zygote is implanted.\nWhat happens to human egg when it is not fertilised? (AI 2012)\nAnswer:\nRefer to answer 134 (a) and b (ii).\nQuestion 145.\n(a) Explain with the help of a diagram how pollen after landing on the stigma of a flower helps male germ cell to reach the female germ cell. Label the following : ovary, female germ cell, male germ cell and pollen grain.\n(b) State the meaning of pollination and mention a distinguishing feature between self-pollination and cross-pollination.\n(Foreign 2012)\nAnswer:\n(a) After landing of stigma, the pollen grain does not pass down the stigma. Only its pollen tube does so. The pollen tube eats its way through the solid part of the stigma and style by secreting pectinases and hydrolytic enzymes. It travels intercellularly and chemotropically along the concentration gradient of calcium \u2013 boron \u2013 inositol sugar complex.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nThe contents of the pollen grains shift into pollen tube with the tube or vegetative nucleus moving to its tip followed by the two gametes. Further growth of the pollen tube occurs only towards its tip.\nIn this way, pollen tube helps the male germ cell to reach the female germ cell.\nRefer to answer 120 (b) for figure.\n(b) Refer to answer 73.\nQuestion 146.\nState in brief the changes that take place in a fertilised egg(zygote) till birth of the child in the human female reproductive system. What happens to the egg when it is not fertilised? (Foreign 2012)\nAnswer:\nWhen the ovum (or egg) is fertilised in the oviduct, then a zygote is formed. The zygote divides rapidly by mitosis as it moves down slowly in the oviduct and forms a ball of cells. This hollow ball of cells, called an embryo sinks into the soft and thick lining of the uterus and gets embedded in it. The embedding of embryo in the thick lining of the uterus is called implantation.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nAfter implantation, a disc-like special tissue develops between the uterus wall (called uterine wall) and the embryo (or fetus), which is called placenta. The fetus is connected to placenta in mothers body through umbilical cord. It is through the placenta that all the requirements of the developing fetus like nutrition, respiration and excretion, etc., are met from the mothers body.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nThe time period from the fertilisation upto the birth of the baby is called gestation. The average gestation period in humans (or the average during of human pregnancy) is about nine months. During the gestation period, the fetus grows to become a baby. Birth begins when the strong muscles in the walls of the uterus start to contract rhythmically. The rhythmic contraction of uterus muscles gradually pushes the baby out of the mother\u2019s body through vagina and a baby is born. If, a sperm is not available at the time of ovulation, then fertilisation of ovum (or egg) does not take place. Since the ovum (or egg) is not fertilised, so the thick and soft uterus lining having lot of blood capillaries in it is not required. The unfertilised ovum dies within a day and the uterus lining also breaks down. The breakdown and removal of the inner, thick and soft lining of the uterus alongwith its blood vessels is called menstrual flow or menstruation.\nQuestion 147.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-IMPORTANT QUESTIONS.txt\nQuestion 147.\n(a) Draw a diagram of the longitudinal section of a flower and label on it sepal, petal, ovary and stigma.\n(b) Write the names of male and female reproductive parts of a flower. (Delhi 2011)\nAnswer:\n(a) The longitudinal section of a flower is as follows :\n\n(b) Refer to answer 87.\nQuestion 148.\n(a) Draw a diagram illustrating fertilisation in a flowering plant and label on it: male germ cell, ovary, female germ cell and pollen grain.\n(b) Distinguish between self pollination and cross pollination. (Foreign 2011)\nAnswer:\n(a) Refer to answer 120 (b). (b) Refer to answer 73.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 8 Intext Questions\nPage Number: 128\nQuestion 1\nWhat is the importance of DNA copying in reproduction?\nAnswer:\nDNA copying has following importance in reproduction:\nIt maintains the characteristics of species.\nIt maintains the continuity of life.\nFrom this, the characteristics and features of organisms are transformed to their progeny.\nIt produces variations in organisms which is the basis of evolution of new species.\nQuestion 2\nWhy is variation beneficial to the species but not necessarily for the individual?\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nIt produces variations in organisms which is the basis of evolution of new species.\nQuestion 2\nWhy is variation beneficial to the species but not necessarily for the individual?\nAnswer:\nThe various populations of organisms interact with many types of ecological niches. This is important for them to survive in given conditions. In case of any damage caused to the ecological conditions of the population, the population gets adversely affected. The organisms which are able to survive, may reproduce to develop population which is adapted or suited to the varied conditions. Hence variation is beneficial to species, but not to the individuals.\nPage Number: 133\nQuestion 1\nHow is the process of pollination different from fertilisation ?\nAnswer:\nQuestion 2\nHow will an organism be benefited if it reproduces through spores ?\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nPage Number: 133\nQuestion 1\nHow is the process of pollination different from fertilisation ?\nAnswer:\nQuestion 2\nHow will an organism be benefited if it reproduces through spores ?\nAnswer:\nAn organism is benefited by reproducing through the spores because spores are surrounded by a thick layer which protects them in adverse conditions. When the favourable conditions occur, these spores start to grow again. In this way they are successfully live in unfavourable conditions.\nQuestion 3\nCan you think of reasons why more complex organisms cannot give rise to new individuals through regeneration ?\nAnswer:\nIn complex multicellular organisms, specialised cells make up tissues, tissue make up organs, organs make up organ systems and finally organ systems make up organisms. Since complex multicellular organisms have a very high degree of organisation in their body, they cannot be reproduced from their cut body parts by the process of regeneration.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nFor example, a dog is a complex multicellular organism which cannot be regenerated from its cut body part say, a cut tail. This is because the cells present in the cut tail of a dog cannot produce dog\u2019s organs like heart brain, lungs, stomach, intestines and limbs, etc, needed for the making of a complete dog.\nQuestion 4\nWhy is vegetative propagation practised for growing some types of plants ?\nAnswer:\nVegetative propagation is practiced for growing such plants which usually do not produce seeds or produce non-viable seeds.\nQuestion 5\nWhy is DNA copying an essential part of the process of reproduction ?\nAnswer:\nDNA copying is essential part of the process of reproduction so that the characteristics of the parent organisms are transmitted to its offspring and at the same time some occasional variations are also produced in the offspring. The changes in the copy of DNA provide an organism the capability to survive in changing conditions.\nPage Number: 140\nQuestion 1", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nPage Number: 140\nQuestion 1\nHow is the process of pollination different from fertilisation ?\nAnswer:\nQuestion 2\nWhat is the role of the seminal vesicles and the prostate gland ?\nAnswer:\n(i) Both seminal vesicle and prostate gland secretes fluids which forms a part of the semen. The fluid secreted from seminal vesicle forms 60% of semen while the fluid secreted from the prostate gland forms 30% of the semen. It makes the path smooth through which the sperms travel.\n(ii) This fluid protects the sperms from the acids present in the urethra.\n(iii) This fluid provides nutrition to sperms in the form of fructose, calcium and some enzymes.\nQuestion 3\nWhat are the changes seen in girls at the time of puberty ?\nAnswer:\nThe various changes occur in girls at puberty are :\nHair grow under armpits and pubic region.\nMammary glands (or breasts) develop and enlarge.\nThe hips broaden.\nExtra fat is deposited in various parts of the body like hips and thighs.\nFallopian tube, uterus and vagina enlarge.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nMammary glands (or breasts) develop and enlarge.\nThe hips broaden.\nExtra fat is deposited in various parts of the body like hips and thighs.\nFallopian tube, uterus and vagina enlarge.\nOvaries start to release eggs.\nMenstruation (monthly periods) starts.\nFeelings and sexual drives associated with adulthood begin to develop.\nQuestion 4\nHow does the embryo get nourishment inside the mother\u2019s body ?\nAnswer:\nIn mother\u2019s body, the embryo gets nutrition from the mother\u2019s blood. For this, there is a special structure, called placenta. Placenta contains villi. There are empty spaces in mother\u2019s tissues that cover the villi. It provides a large surface area for the transfer of glucose, oxygen and other substances from the mother to the embryo.\nQuestion 5\nA woman is using a copper-T. Will it help in protecting her from sexually transmitted diseases ?\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nQuestion 5\nA woman is using a copper-T. Will it help in protecting her from sexually transmitted diseases ?\nAnswer:\nCopper-T is a contraceptive method which prevents implantation of the zygote inside the uterus. It cannot prevent a women from sexually transmitted diseases. These diseases are transmitted by contact which cannot be prevented by copper-T.\nNCERT Solutions for Class 10 Science Chapter 8 Textbook Chapter End Questions\nQuestion 1\nAsexual reproduction takes place through budding in\n(a) amoeba\n(b) yeast\n(c) plasmodium\n(d) leishmania\nAnswer:\n(b) Yeast\nQuestion 2\nWhich of the following is not a part of the female reproductive system in human beings ?\n(a) Ovary\n(b) Uterus\n(c) Vas deferens\n(d) Fallopian tube\nAnswer:\n(c) Vas deferens\nQuestion 3\nThe anther contains\n(a) sepals\n(b) ovules\n(c) carpel\n(d) pollen grains\nAnswer:\n(d) Pollen grains\nQuestion 4\nWhat are the advantages of sexual reproduction over asexual reproduction ?\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nThe anther contains\n(a) sepals\n(b) ovules\n(c) carpel\n(d) pollen grains\nAnswer:\n(d) Pollen grains\nQuestion 4\nWhat are the advantages of sexual reproduction over asexual reproduction ?\nAnswer:\n(i) In asexual reproduction, the offspring are almost identical to their parent because they have the same genes as their parent. So, much genetic variation is not possible in asexual reproduction. This is a disadvantage because it inhibits the further evolution of the organism.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\n(ii) In sexual reproduction the offspring, although similar to their parents, are not identical to them or to one another. This is because the offspring receive some genes from the mother and some from the father. Because of the mixing of genes of mother and father in various different combinations, all of the offspring have genetic variations. In this way, sexual reproduction leads to a greater variety in population. This means that a species (animal or plant) can adapt quickly to changes in its surroundings. This is because there are always likely to be some individuals which are more suited to the changes than others, and these individuals will survive and reproduce themselves.\nQuestion 5\nWhat are the functions performed by the testis in human beings ?\nAnswer:\nThe functions of testes in humans are following :\n(i) After the stage of adolescent, testes produce male gametes in the human males which are called sperms.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nAnswer:\nThe functions of testes in humans are following :\n(i) After the stage of adolescent, testes produce male gametes in the human males which are called sperms.\n(ii) A hormone called testosterone is produced in testes. Testosterone controls the development of reproductive organs and secondary sexual characters.\nQuestion 6\nWhy does menstruation occur ?\nAnswer:\nIf the ovum (or egg) does not get fertilised (due to non-availability of sperm in the female body) then the thick and soft inner lining of uterus is no longer needed and hence it breaks. So, the thick and soft inner lining of uterus alongwith the blood vessels and the dead ovum (or egg) comes out of the vagina in the form of blood called menstruation. Menstruation occurs after the interval of every 28 days and the time period between ovulation and menstruation is about 14 days.\nQuestion 7\nDraw a labelled diagram of the longitudinal section of a flower.\nAnswer:\n\nQuestion 8\nWhat are the different methods of contraception ?", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nQuestion 7\nDraw a labelled diagram of the longitudinal section of a flower.\nAnswer:\n\nQuestion 8\nWhat are the different methods of contraception ?\nAnswer:\nThe different methods of contraception are as follow :\n(i) Barrier method : In this method, condom, diaphragm and cervical caps are used. These prevent the entry of sperms in the female genital tract during sexual intercourse.\n(ii) Chemical method : In this method a woman uses two kinds of pills (oral and vaginal pills). The oral pills are hormonal preparations which suppress the release of ovum in fallopian tube. These are called oral contraceptives. The vaginal pills/ creams are spermicidal. The chemicals in these spermicidals kill the sperms during their journey in the vaginal tract.\n(iii) Intrauterine contraceptive devices : Intrauterine contraceptive devices such as copper-T are placed safely in the uterus by a skilled doctor. It prevents the sperms to reach the uterus.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\n(iii) Intrauterine contraceptive devices : Intrauterine contraceptive devices such as copper-T are placed safely in the uterus by a skilled doctor. It prevents the sperms to reach the uterus.\n(iv) Surgical method : In this method, a small part of vas deferens of male and fallopian tube of female is cut or tied by surgery. It is called vasectomy in males and tubectomy in females.\nQuestion 9\nHow are the modes for reproduction different in unicellular and multicellular organisms ?\nAnswer:\nQuestion 10\nHow does reproduction help in providing stability to populations of species ?\nAnswer:\nThe introduction of variations during reproduction provides stability to the populations of various species by preventing them from getting wiped out during adverse conditions. Reproduction also helps to generate copies of individuals which are suited to a particular environment.\nQuestion 11\nWhat could be the reasons for adopting contraceptive methods ?\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nQuestion 11\nWhat could be the reasons for adopting contraceptive methods ?\nAnswer:\nThe reasons for adopting contraceptive devices are as follow:\nTo control the birth rate and prevent the increase in population.\nTo reduce the adverse effects on mother\u2019s body due to frequent pregnancy.\nTo provide safety from sexually transmitted diseases.\nNCERT Solutions for Class 10 Science Chapter 8 How Do Organisms Reproduce?\nReproduction: Reproduction in animals and plants (asexual) and (sexual) reproductive health \u2013 need and methods of family planning. Safe sex vs HIV/ AIDS. Child bearing and women\u2019s health.\nFormulae Handbook for Class 10 Maths and Science\nPage 128\nQuestion 1.\nWhat is the importance of DNA copying in reproduction?\nAnswer:\nDNA copying in reproduction is important for maintenance of body designs and features. Moreover, DNA copying leads to variations. Variation is useful for the survival of species.\nQuestion 2.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nAnswer:\nDNA copying in reproduction is important for maintenance of body designs and features. Moreover, DNA copying leads to variations. Variation is useful for the survival of species.\nQuestion 2.\nWhy is variation beneficial to the species but not necessarily for the individual?\nAnswer:\nPopulation of organisms reside in well-defined places or niches in the ecosystem. However, niches can change because of reasons beyond the control of the organisms, e.g., temperature changes, water level changes, etc. If population of reproducing organisms suited to particular niche and if the niche is drastically altered, the population can be wiped out. However, if some variations are present in a few inAividuals in these populations, there will be chances for their survival. The surviving individual can further reproduce and develop a population according to the changed niche, Thus, variation is beneficial to the species but not necessarily for the individual.\nPage 133:\nQuestion 1.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nPage 133:\nQuestion 1.\nHow does binary fission differ from multiple fission ?\nAnswer:\nWhen two new daughter cells are formed as a result of fission. It is called binary fission, e.g., Amoeba.When many daughter cells are formed as a result of fission, this is called multiple fission, e.g., Malarial parasite.\nQuestion 2.\nHow will an organism be benefited if it reproduces through spores?\nAnswer:\nSpore formation is an asexual mode of reproduction. Spores formed are covered by\nthick walls that protect them from adverse conditions. During favourable condition thick resistant wall breaks down and new organism grows from it.\nSpores are very light weight and they easily get dispersed through winds which give them more variations and thus better chances of survival.\nDownload NCERT Solutions for Class 10 Science Chapter 8 How Do Organisms Reproduce PDF\nQuestion 3.\nCan you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nQuestion 3.\nCan you think of reasons why more complex organisms cannot give rise to new individuals through regeneration?\nAnswer:\nComplex organisms are not simply a random collection of cells where sPecialized cells are organised as tissues, and tissues ale organised into organs which then have to be placed at definite positions in the body. In such a carefully organised situation, it is not easy to develoP organism through regeneration\nQuestion 4.\nWhy is vegetative propagation practised for growing some type of plants?\nAnswer:\nVegetative propagation makes possible for the propagation of plants such as banana, orange, rose and jasmine that have lo6t the capacity to Produce seeds. Moreover, all plants produced through vegetative propagation are genetically similar to the parent plant.\nQuestion 5.\nWhy is DNA copying an essential part of the process of reproduction?\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nQuestion 5.\nWhy is DNA copying an essential part of the process of reproduction?\nAnswer:\nThe process of reproduction results in the production of off springs which resemble to their parents. This means during the reproduction there must be a transfer of the blueprint of the body design from parent to the off springs. As we know DNA contains all the information that passes from parents to the next generation, so before reproduction, DNA is copied in the parent cell. Out of these two copies, one copy is passed to the newly formed individual.\nPage 140\nQuestion 1.\nHow is the process Of pollination different from fertilization ?\nAnswer:\nPollination is the transfer of pollen grains from the anther to the stigma Of a flower whereas fertilization is the fusion Of male gamete with female gamete (egg).\nQuestion 2.\nWhat is the role of the seminal vesicles and the prostate gland ?\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nQuestion 2.\nWhat is the role of the seminal vesicles and the prostate gland ?\nAnswer:\nSeminal vesicles and the prostate gland add their secretions so that the sperms are in a fluid (semen) which makes their transport easier and this fluid also provides nutrition.\nQuestion 3.\nWhat are the changes seen in girls at the time Of puberty ?\nAnswer:\nChanges seen in girls at the time of puberty are :\n1. Breast size begins to increase.\n2. Girls begin to menstruate.\n3.Growth Of pubic hair.\n4.Skin becomes oily.\nQuestion 4.\nHow does the embryo get nourishment inside the Mother\u2019s body?\nAnswer:\nembryo gets nutrition from mother\u2019s blood with the help Of a special tissue called placenta. Through placenta, glucose and oxygen pass from mother to the embryo. Moreover, waste substance of embryo are removed through placenta into the\nmother\u2019s blood.\nQuestion 5.\nIf a woman is using a copper-T will it help in protecting her from sexually transmitted diseases?\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nmother\u2019s blood.\nQuestion 5.\nIf a woman is using a copper-T will it help in protecting her from sexually transmitted diseases?\nAnswer:\nNo, copper-Twill not protect her from Only Barrier methods protect from sexually transmitted diseases.\nPage 141\nQuestion 1.\nAsexual reproduction takes place through budding in :\n(a) Amoeba\n(b) Yeast\n(c) Plasmodium\n(d) Leishmania.\nAnswer:\n(b) Yeast.\nQuestion 2.\nWhich of the following is not a pan Of the female reproductive system in human beings ?\n(a) Ovary\n(b) Uterus\n(c) Vas deferens\n(d) Fallopian tube\nAnswer:\n(c) Vas deferens.\nQuestion 3.\nThe anther contains :\n(a) Sepals\n(b) Ovules\n(c) Carpel\n(d) Pollen grains.\nAnswer:\n(d) Pollen grains\nQuestion 4.\nWhat are the advantages of sexual reproduction over asexual reproduction?\nAnswer:\nSexual reproduction leads to new combination Of genes as it involves two parents and meiosis. This produces variation in offspring.", + "Answer:\n(d) Pollen grains\nQuestion 4.\nWhat are the advantages of sexual reproduction over asexual reproduction?\nAnswer:\nSexual reproduction leads to new combination Of genes as it involves two parents and meiosis. This produces variation in offspring. Variations are the basis for evolution.\nQuestion 5.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nAnswer:\nSexual reproduction leads to new combination Of genes as it involves two parents and meiosis. This produces variation in offspring. Variations are the basis for evolution.\nQuestion 5.\nWhat are the functions performed by the testes in human beings?\nAnswer:\nFunctions Of testes are .\n(i) Testes produce sperms.\n(ii) Testosterone (male sex hormone) is also produced by testes.\nQuestion 6.\nWhy does menstruation occur?\nAnswer:\nIf the egg is not fertilized and uterus does not get zygote, the developed lining slowly breaks and menstruation occur.\nQuestion 7.\nWhat are the different methods of contraception?\nAnswer:\nThere are three main methods of contraception :\nBarrier methods,\nChemical methods, and\nSurgical methods.\n1. Barrier methods: In barrier methods, physical devices such as condom, diaphragm and cervical caps are used. They prevent the entry Of sperms in the female genital tract during copulation.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\n1. Barrier methods: In barrier methods, physical devices such as condom, diaphragm and cervical caps are used. They prevent the entry Of sperms in the female genital tract during copulation.\n2. Chemical methods: The chemical methods make use of specific drugs by females. There are two types of such drugs, Oral pills and vaginal pills. Oral pills are mainly hormonal preparation, and are called oral contraceptives (OCS).\n3. Surgical methods: In the surgical methods, a small portion of vas deferens in male, and the fallopian tube in female, in surgically removed or ligated (tied). It is called vasectomy in males and tubectomy in females.\nApart from these three methods the intrauterine contraceptive devices are used to prevent pregnancies. The use of Intra Uterine Contraceptive Devices (IUCDs) is also very effective and popular. A copper-T is placed safely inside the uterus by a practising doctor or a skilled nurse. IUCDs prevent implantation in the uterus.\nQuestion 8.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nQuestion 8.\nHow are the modes for reproduction different in unicellular and multicellular organisms ?\nAnswer:\nUnicellular organisms reproduce asexually whereas multicellular organisms reproduce manly by n 1 reproduction.\nQuestion 9.\nHow does reproduction help in providing stability to populations Of species?\nAnswer:\nThe rate Of birth and death in a given population determine its stability. The rate of birth should be approximately equal to the rate of death. So, by checking birth rate, which is increasing at an alarming rate, stability to population of species can be provided\nQuestion 10.\nWhat could be the reasons for adopting contraceptive methods?\nAnswer:\nFrequent pregnancies have an adverse effect on the health of a woman. Frequent and unwanted pregnancies can be avoided by adopting contraceptive methods. Also, these methods check population growth by controlling child birth rate.\nMultiple Choice Questions (MCQs) [1 Mark each]\nQuestion 1.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nMultiple Choice Questions (MCQs) [1 Mark each]\nQuestion 1.\nAsexual reproduction takes place through budding in [NCERT]\n(a) Amoeba\n(b) Yeast\n(c) Plasmodium\n(d) Leishmania\nAnswer:\n(b) Asexual reproduction in Hydra and yeast takes place by budding.\nQuestion 2.\nThe ability of a cell to divide into several cells during reproduction in Plasmodium is called [NCERT Exemplar]\n(a) budding\n(b) reduction division\n(c) binary fission\n(d) multiple fission\nAnswer:\n(d) Multiple fission Organisms divide into many daughter cells simultaneously, e.g. Plasmodium.\nQuestion 3.\nThe anther contains [NCERT]\n(a) sepals\n(b) ovules\n(c) carpels\n(d) pollen grains\nAnswer:\n(d) Anther is the male reproductive part in plants. It contains pollen grains, having male germ cells.\nQuestion 4.", + "Question 3.\nThe anther contains [NCERT]\n(a) sepals\n(b) ovules\n(c) carpels\n(d) pollen grains\nAnswer:\n(d) Anther is the male reproductive part in plants. It contains pollen grains, having male germ cells.\nQuestion 4.\nCharacters that are transmitted from parents to offspring during reproduction show [NCERT Exemplar]\n(a) only similarities with parents\n(b) only variations with parents\n(c) both similarities and variations with parents", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\n(a) only similarities with parents\n(b) only variations with parents\n(c) both similarities and variations with parents\n(d) neither similarities nor variations\nAnswer:\n(c) In sexual reproduction, the offsprings are not exactly identical to the parents or to one another. This is because the offsprings receive some genes from mother and some from father. Because of mixing of genes on re-establishment of the exact number of chromosomes as in the parents, the offsprings show both similarities and variations with their parents.\nQuestion 5.\nWhich among the following diseases is not sexually transmitted? [NCERT Exemplar]\n(a) Syphilis\nb) Hepatitis\n(c) HIV-AIDS\n(d) Gonorrhoea\nAnswer:\n(b) The diseases, which are spread by sexual contact with an infected person are called Sexually Transmitted Diseases or STDs, e.g. gonorrhoea, syphilis and AIDS. Hepatitis is a water borne viral disease which affects liver.\nQuestion 6.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nQuestion 6.\nWhich of the following is not a part of the female reproductive system in human beings? [NCERT]\n(a) Ovary\n(b) Uterus\n(c) Vas deferens\n(d) Fallopian tube\nAnswer:\n(c) Vas deferens is a part of male reproductive system in humans.\nQuestion 7.\nA feature of reproduction that is common to Amoeba, Spirogyra and yeast is that [NCERT Exemplar]\n(a) they reproduce asexualiy\n(b) they are all unicellular\n(c) they reproduce only sexually\n(d) they are all multicellular\nAnswer:\n(a) Amoeba and yeast are unicellular while Spirogyra is multicellular. But, all the three reproduce asexualiy.\nQuestion 8.\nWhich among the following statements are true for unisexual flowers? [NCERT Exemplar]\nI. They possess both stamen and pistil.\nII. They possess either stamen or pistil.\nIII. They exhibit cross-pollination.\nIV. Unisexual flowers possessing only stamens cannot produce fruits.", + "[NCERT Exemplar]\nI. They possess both stamen and pistil.\nII. They possess either stamen or pistil.\nIII. They exhibit cross-pollination.\nIV. Unisexual flowers possessing only stamens cannot produce fruits.\n(a) I and IV\n(b) II, III and IV\n(c) III and IV\n(d) I, III and IV\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nIII. They exhibit cross-pollination.\nIV. Unisexual flowers possessing only stamens cannot produce fruits.\n(a) I and IV\n(b) II, III and IV\n(c) III and IV\n(d) I, III and IV\nAnswer:\n(b) The flowers which are unisexual (papaya, watermelon) contain either stamens or carpels. Since, only one reproductive organ is present in them, they depend on cross-pollination to form zygote after fertilisation. Both stamens and carpels are required for fertilisation, so only one of them cannot produce fruits.\nQuestion 9.\nLength of pollen tube depends on the distance between [NCERT Exemplar]\n(a) pollen grain and upper surface of stigma.\n(b) pollen grain on upper surface of stigma and ovule.\n(c) pollen grain in anther and upper surface of stigma.\n(d) upper surface of stigma and lower part of style.\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\n(b) pollen grain on upper surface of stigma and ovule.\n(c) pollen grain in anther and upper surface of stigma.\n(d) upper surface of stigma and lower part of style.\nAnswer:\n(b) Length of pollen tube depends on the distance between pollen grain on upper surface of stigma and ovule. A pollen grain falls on the stigma of the carpel, bursts open and develops a pollen tube downwards through the style towards the ovule in the ovary.\nQuestion 10.\nWhich among the following statements arer true for sexual reproduction in flowering plants? [NCERT Exemplar]\nI. It requires two types of gametes.\nII. Fertilisation is a compulsory event.\nIII. It always results in formation of zygote.\nIV. Offsprings formed are clones.\n(a) I and IV\n(b) I and II\n(c) I, II and III\n(d) I, II and IV\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nII. Fertilisation is a compulsory event.\nIII. It always results in formation of zygote.\nIV. Offsprings formed are clones.\n(a) I and IV\n(b) I and II\n(c) I, II and III\n(d) I, II and IV\nAnswer:\n(c) Sexual reproduction creates variation in organisms, so, clones cannot be produced through it. Clones are identical copy of parent organism. Sexual reproduction needs two type of gametes, i.e. male and female to form zygote after fertilisation.\nQuestion 11.\nFactors responsible for the rapid spread of bread mould on slices of bread are [NCERT Exemplar]\nI. large number of spores.\nII. availability of moisture and nutrients in bread.\nIII. presence of tubular branched hyphae.", + "Question 11.\nFactors responsible for the rapid spread of bread mould on slices of bread are [NCERT Exemplar]\nI. large number of spores.\nII. availability of moisture and nutrients in bread.\nIII. presence of tubular branched hyphae.\nIV formation of round-shaped sporangia\n(a) I and III\n(b) II and IV\n(c) I and II\n(d) III and IV\nAnswer:\n(c) Under favourable Conditions (like damp and warm conditions, availability of nutrients), the fungal spores present in the air, lands on food, germinate and produce new plaints.\nQuestion 12.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\n(c) Under favourable Conditions (like damp and warm conditions, availability of nutrients), the fungal spores present in the air, lands on food, germinate and produce new plaints.\nQuestion 12.\nDuring adolescence various changes occur in the body of humans. Mark one change associated with sexual maturation in males. [NCERT Exemplar]\n(a) Loss of milk teeth\n(b) Increase in body height\n(c) Cracking of voice\n(d) Weight gain\nAnswer:\n(c) Hypertrophy of larynx results in low pitched. cracking voice in human males during adolescence.\nQuestion 13.\nObserve the diagram given along side.\n\nWhat happens after the above stage?\n(a) The ovary splits open\n(b) Ovary develops into a fruit and ovules into seeds\n(c) The pvules are dispersed\n(d) Germination of seeds takes place\nAnswer:\n(b) Ovary develops into a fruit and ovulesdnto seeds as in the above given diagram fertilisation has already taken place.\nQuestion 14.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\n(d) Germination of seeds takes place\nAnswer:\n(b) Ovary develops into a fruit and ovulesdnto seeds as in the above given diagram fertilisation has already taken place.\nQuestion 14.\nWhat in your opinion could be the best reason to explain why menstruation is not taking place in a healthy woman?\n(a) Early release of ovum\n(b) Psychological reason\n(c) Fertilisation of ovum\n(d) Build up of female sex hormones in the blood stream\nAnswer:\n(c) If a woman is not having her menstruation on time the probable reason from the given option is that fertilisation of ovum has taken place. Because, during gestation period of pregnancy, menstruation does not take place.\nQuestion 15.\nThe correct sequence of reproductive stages seen in flowering plants is [NCERT Exemplar]\n(a) gametes, zygote, embryo, seedling\n(b) zygote, gametes, embryo, seedling\n(c) seedling, embryo, zygote, gametes\n(d) gametes, embryo, zygote, seedling\nAnswer:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\n(a) gametes, zygote, embryo, seedling\n(b) zygote, gametes, embryo, seedling\n(c) seedling, embryo, zygote, gametes\n(d) gametes, embryo, zygote, seedling\nAnswer:\n(a) Correct sequence of reproductive stages in flowering plants is \u2192 formation of gametes \u2192 fusion of gametes to form zygote \u2192 zygote develops into embryo in the ovary \u2192 ovule develops a tough coat and converts into a seed.\nQuestion 16.\nOffsprings formed by asexual method of reproduction have greater similarity among themselves because [NCERT Exemplar]\nI. asexual reproduction involves only one parent.\nII. asexual reproduction does not involve gametes.\nIII. asexual reproduction occurs before sexual reproduction.\nIV. asexual reproduction occurs after sexual reproduction.\n(a) I and II\n(b) I and III\n(c) II and IV\n(d) III and IV\nAnswer:\n(a) Offsprings have greater similarity as only one parent is involved in asexual reproduction thus, no gametes are formed.\nQuestion 17.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\n(a) I and II\n(b) I and III\n(c) II and IV\n(d) III and IV\nAnswer:\n(a) Offsprings have greater similarity as only one parent is involved in asexual reproduction thus, no gametes are formed.\nQuestion 17.\nTwo flowers are identified by a botanist with the following features that flower A is having only stamen and flower B is having both stamen and pistil. Which of the following statements is correct?\n(a) Flower A will bear seeds and flower B cannot bear seeds after fertilisation.\n(b) Flower A will produce pollen grains and flower B cannot produce pollen grains.\n(c) Flower A cannot be fertilised and flower B can show fertilisation.\n(d) Neither flower A and nor flower B can show self-pollination.\nAnswer:\nOption (c) is correct. Since, flower A bears only stamen, i.e. male reproductive part so, it cannot get fertilised. And flower B bears both male and female reproductive parts, therefore it can get fertilised by pollination and can change into fruit.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\n10 Science How Do Organisms Reproduce Mind Map\nReproduction is a biological process in which an organism gives rise to young ones similar to themselves.\nBasic event in reproduction is the creation of a DN A copy.\nCells use chemical reactions to build two copies of the DNA in a reproducing cell.\nIn addition, DNA copying is accompanied by the creation of an additional cellular apparatus.\nThen each DNA copy is separated with its own cellular apparatus.\nEffectively, a cell divides to give rise to two cells.\nAsexual Reproduction\nWhen offspring is produced by single parent with or without the involvement of gamete formation.\nFragmentation\nParent organism breaks into smaller fragments upon maturation, each fragment grows into a new individual.\nIt is shown by multi-cellular organisms with simple body organization for e.g. Spirogyra\nFission", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nIt is shown by multi-cellular organisms with simple body organization for e.g. Spirogyra\nFission\nOrganisms divide mitotically into two halves, each behaves like independent individual. It is termed as binary fission and is mostly shown by bacteria and protozoa.\nBinary fission can take place in any plane as observed in Amoeba, or it can occur in a definite orientation for e.g. Leishmania (causes kala-azar), Euglena (longitudinal), Paramecium (transverse) etc.\nIn few organisms parent cell divides into many daughter cells simultaneously which is termed as multiple fission. It is observed in Plasmodium.\nBudding\nFormation of daughter organism takes place from a small projection called as bud. For e.g. Hydra, Yeast etc\nOrganisms such as Hydra use regenerative cells for reproduction in the process of budding.\nRepeated cell division at one specific site leads to the formation of an outgrowth called as bud.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nOrganisms such as Hydra use regenerative cells for reproduction in the process of budding.\nRepeated cell division at one specific site leads to the formation of an outgrowth called as bud.\nThese buds develop into tiny individuals and detach form parent body once they become fully mature.\nDetached organism acts as an independent organism.\nRegeneration\nIt is an ability of simple organisms to re-grow their lost body parts.\nIn asexual reproduction, this ability is used by many organisms to give rise to new individual from their body parts. That is, if the individual is somehow\u2019 cut or broken up into many pieces, many of these pieces grow into separate individuals.\nFor e.g. Hydra and Planaria\nIt is carried out by specialized cells which proliferate & differentiate to make various cell types tissues.\nThese changes take place in an organized sequence referred to as development.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nIt is carried out by specialized cells which proliferate & differentiate to make various cell types tissues.\nThese changes take place in an organized sequence referred to as development.\nHowever, regeneration is not the same as reproduction, since most organisms would not normally depend on being cut up to be able to reproduce.\nSpore Formation\nAn individual divides into no. of small spores, each spore giving rise to new individual.\nSpores are covered by thick walls that protect them until they come into contact with moist surface or suitable environment and can begin to grow. E.g. Spore formation in Rhizopus\nVegetative Propagation\nIt refers to the formation of new plants from parts of parent plants such as root, stem, leave etc. These parts are termed as vegetative units or vegetative propagules. For e.g. buds produced in the notches along the leaf margin of Bryophyllum fall on the soil and develop into new plants\nAdvantages of vegetative propagation:", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nAdvantages of vegetative propagation:\nVegetative propagation is used in methods such as layering, cutting, grafting to grow many plants like sugarcane, roses, or grapes for agricultural purposes.\nPlants raised by vegetative propagation can bear flowers and fruits earlier than those produced from seeds.\nIt makes it possible to propagate plants that have lost the capacity to produce seeds such as banana, orange, rose and jasmine.\nPlants produced are genetically similar enough to the parent plant to have all its characteristics.\nSexual Reproduction\nIt involves the formation and fusion of the gametes. It leads to formation of variations in individuals. Variations form the basis of evolution of the species and ensure the survival of the species.\nReproduction in Human Beings\nPuberty: The period during which adolescents reach sexual maturity and become capable of reproduction.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nReproduction in Human Beings\nPuberty: The period during which adolescents reach sexual maturity and become capable of reproduction.\nChanges in girls during puberty: breast size begins to increase, darkening of skin of nipples, & girls begin to menstruate.\nChanges in boy during puberty: hair growth on face, voices begin to crack & occasional erection & enlargement of penis\nChanges common to both boys & girls: hair growth in various parts such as armpits, genital area, thin hairs on arms & legs and skin may become oily.\nFemale Reproductive System\nIt consists of a pair of ovaries, pair of oviduct (fallopian tube), uterus, cervix, & vagina.\nOne egg is produced every month by one of the ovaries after reaching the age of puberty. The egg is carried from the ovary to the womb through a thin oviduct or fallopian tube.\nUterus serves as womb and is richly supplied with blood vessels to nurture the developing embryo.\nVagina serves as the site of entry of sperm during sexual intercourse.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nUterus serves as womb and is richly supplied with blood vessels to nurture the developing embryo.\nVagina serves as the site of entry of sperm during sexual intercourse.\nMale Reproductive System\nIt consists of a pair of testes located outside the body in a pouch called scrotum. It helps in maintaining lower temperature which favors the perm formation. Testes are responsible for synthesizing sperms and testosterone.\nSperms are tiny bodies consists of mainly genetic material and a tail that helps them to move towards the female germ-cell.\nSperms are then delivered through vas deferens which unites with urethra to form a common passage for sperm and urine.\nAccessory glands like seminal vesicles, prostate, and bulbourethral add their secretion to sperm. It makes the sperm fluid in nature that not only eases the transportation of sperm but also provide the nutrition to it.\nReproductive Health", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nReproductive Health\nSexually transmitted diseases include bacterial infections (gonorrhea, syphilis etc) & viral infections (warts, AIDS etc).\nContraception refers to the act of preventing the unwanted pregnancies. Contraceptive methods may fall in following categories:\nMechanical barrier: e.g. condom, diaphragms. They also prevent STDs.\nOral contraceptives: they change hormonal balance, inhibits ovulation & thus fertilization e.g. saheli, iPill, etc.\nIntra uterine device: plastic or metal devices placed in the uterus for e.g. loop, copper-T etc.\nSurgical methods such as vasectomy & tubectomey.\nEvents of Reproduction\nThe sperms after entering the vaginal passage travel upwards and reach the oviduct where they may fertilize the egg.\nPost-fertilization, the zygote gets implanted in the lining of the uterus, and starts dividing.\nSpecial tissue called placenta is developed to provide nutrition to the developing embryo as well as for removing waste from it.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\nSpecial tissue called placenta is developed to provide nutrition to the developing embryo as well as for removing waste from it.\nThe development of a child takes up approx, nine months.\nThe child is born as a result of rhythmic contractions of the muscles in the uterus.\nMenstruation\nMenstrual cycle is a cyclic event that places roughly every month in females after puberty. Unfertilized egg lives for 1 day after which it degenerates. Consequently, uterus lining slowly breaks & comes out through the vagina as blood & mucus. This discharge is known as menstruation which lasts for about 2-8 days.\nSexual Reproduction in Flowering Plants\n\u2022 Stamens and carpels are the reproductive parts of a flower which contain the germ-cells. Stamen is the male reproductive part and it produces pollen grains. Carpel is the female reproductive part made up of three parts: ovary, style and stigma.\n\u2022 The ovary contains ovules and each ovule has an egg cell.", + "CHAPTER 7-HOW DO ORGANISMS REPRODUCE-INTEXT QUESTIONS.txt\n\u2022 The ovary contains ovules and each ovule has an egg cell.\n\u2022 The flower may be unisexual i.e. contains either stamens or carpels e.g. papaya, watermelon or bisexual i.e. contains both stamens and carpels e.g. Hibiscus, mustard.\n\u2022 Transfer of pollen grains (shed from the anther) to the stigma of a pistil is termed pollination. Two types: self-pollination and cross pollination. Pollinating agents are air, water, insects, & animals.\n\u2022 Compatible pollen grain germinates on stigma to produce pollen tube. Pollen tube grows through tissues of stigma, style & reaches ovary.\n\u2022 Fertilization results in the formation of zygote which develops into an embryo.\n\u2022 The ovule develops a tough coat and is gradually converted into a seed. The ovary grows rapidly and ripens to form a fruit.\n\u2022 The petals, sepals, stamens, style and stigma may shrivel and fall off.\n\u2022 The seed develops into a seedling under appropriate conditions which is known as germination.", + "CHAPTER 8-HEREDITY -EXTRA QUESTIONS.txt\nQuestion-1\nWhat is heredity?\nSolution:\nThe continuity of features from one generation to another is known as heredity. It is also defined as the transmission of traits from parents to the offsprings.\nQuestion-2\nName the plant on which Mendel performed his experiments.\nSolution:\nMendel performed his experiments on the plant, Pisum sativum \u2013 the garden pea plant.\nQuestion-3\nDefine variation.\nSolution:\nChildren do not resemble their parents completely. They possess characters obtained from both the parents. These changes in the phenotypic and genotypic characters are known as variations. Thus a given population of a species has indefinite variants.\nQuestion-4\nDefine a gene.\nSolution:\nA gene is a small portion of the DNA, with codes for a particular polypeptide or a protein. In other words, it is the functional unit of the DNA. It is also responsible for the transmission of hereditary characters from the parents to the offspring.\nQuestion-5\nWrite the expanded form of DNA.\nSolution:", + "CHAPTER 8-HEREDITY -EXTRA QUESTIONS.txt\nQuestion-5\nWrite the expanded form of DNA.\nSolution:\nThe expansion for DNA is Deoxyribonucleic Acid.\nQuestion-6\nWhat are the components of the chromosome?\nSolution:\nA chromosome consists of two Chromatids joined together at the centromere. The chromaid consists of the DNA material wound over small protein molecules called histones.\nQuestion-7\nWhat is a retrovirus?\nSolution:\nA retrovirus is a virus that has RNA as its genetic material instead of DNA. HIV virus is a retrovirus.\nQuestion-8\nWhat is a sex chromosome?\nSolution:\nChromosomes that are responsible for the determination of sex in an individual are known as sex chromosomes. There are two types sex chromosomes \u2013 the X chromosome and the Y chromosome. Males possess one X and one Y chromosome and females possess two X chromosomes.\nQuestion-9\nHow is sex determined in human beings?\nSolution:", + "CHAPTER 8-HEREDITY -EXTRA QUESTIONS.txt\nQuestion-9\nHow is sex determined in human beings?\nSolution:\nThe male sex gametes have X and Y chromosomes, whereas the female sex gametes have two X chromosomes. When a sperm containing a Y chromosome fuses with the ovum containing X chromosome, the zygote develops into a male. When a sperm containing the X chromosome fuses with an ovum containing X chromosome, the zygote develops into a female. Thus the sex of an individual is determined by the sex chromosomes X and Y, which is present in the male chromosomes.\nQuestion-10\nDefine homologous organs.\nSolution:\nHomologous organs are those organs, which have similar origin and basic plan of development, but may or may not differ in their functions. The forelimbs of a human, a bird and a horse are homologous organs.\nQuestion-11\nExplain Darwin\u2019s theory of evolution.\nSolution:", + "CHAPTER 8-HEREDITY -EXTRA QUESTIONS.txt\nQuestion-11\nExplain Darwin\u2019s theory of evolution.\nSolution:\nDarwin\u2019s theory of evolution is also known as the Theory of Natural Selection or Darwinism. Darwin explained that despite having enormous potential of fertility, the population size of any kind of organism remains within a limit. It is due to the struggle between members of the same species and different species for food, space and mate. This struggle eliminates the unfit individuals. In other words, the fit organisms possess some variations, which are favorable and can leave the progeny to continue the variations. This is called Natural Selection.\nQuestion-12\nDefine Genetics. What is the contribution of Mendel in the field of genetics?\nSolution:", + "CHAPTER 8-HEREDITY -EXTRA QUESTIONS.txt\nQuestion-12\nDefine Genetics. What is the contribution of Mendel in the field of genetics?\nSolution:\nThe branch of biology that deals with the study of heredity and variations is known as Genetics. Gregor Johann Mendel was the first person to carry out experiments regarding the heredity of certain characters from one generation to another in a scientific manner. He worked mainly on the garden pea plant. His observations regarding the occurrence of contrasting characters in various generations of garden pea led him to interpret that these are controlled by units which he called, factors. These factors are today known as genes. He is also known as the Father of Genetics.\nQuestion-13\nWhere are the genes located? What is the chemical nature of genes?\nSolution:\nGenes are segments of the DNA, which is wound compactly into chromosomes. A gene is composed of a specific sequence of nucleotides. A nucleotide is made up of a nitrogenous base, a sugar molecule and a phosphate group.\nQuestion-14", + "CHAPTER 8-HEREDITY -EXTRA QUESTIONS.txt\nQuestion-14\nDuring which stage can the chromosomes be seen clearly? Write the features of the eukaryotic and prokaryotic chromosomes.\nSolution:\nChromosomes are distinctly visible during the Metaphase stage of mitosis of a cell.\n\nThe features of eukaryotic and prokaryotic chromosomes are as follows\nEukaryotic chromosomes\n1. They are present in large numbers.\n2. In higher organisms it occurs in paired condition.\n3. They are visible distinctly only during the metaphase stage of mitosis.\n\n4. Each chromosome consists of two chromatics attached together by a centromere.\nProkaryotic chromosomes\n1. They are simple in composition when compared to the eukaryotic chromosomes.\n2. They are generally circular in shape.\n3. There is only one chromosome in a cell.\nQuestion-15\nWho provided the evidence of DNA as the genetic material? Write the names of the components of the DNA molecule.\nSolution:", + "CHAPTER 8-HEREDITY -EXTRA QUESTIONS.txt\n3. There is only one chromosome in a cell.\nQuestion-15\nWho provided the evidence of DNA as the genetic material? Write the names of the components of the DNA molecule.\nSolution:\nFrederick Griffith, Avery, Mc Leod and Mc Carty established that DNA was the carrier of the genetic information.\nThe DNA molecule is a polynucleotide. A nucleotide is made up of a nitrogenous base, a sugar molecule and a phosphate group.\nQuestion-16\nWhat do you understand by the double helical structure of DNA? Who proposed this structure?\nSolution:\nJames Watson and Francis Crick proposed the double helical structure of the DNA. According to this structure,\n\u2022 DNA molecule consists of two polynucleotide strands forming a double helix. Each helical turn has a length of 3.4nm in which ten nucleotides are present.\n\n\u2022 Each polynucleotide strand has a backbone of sugar and phosphate. The nitrogenous base is attached to the sugar.", + "CHAPTER 8-HEREDITY -EXTRA QUESTIONS.txt\n\u2022 Each polynucleotide strand has a backbone of sugar and phosphate. The nitrogenous base is attached to the sugar.\n\n\u2022 The nitrogenous bases of the two strands of a double helix form a pair with the help of hydrogen bonds. Adenine pairs with thymine by two hydrogen bonds, whereas guanine pairs with cytosine by three hydrogen bonds.\n\u2022 The hydrogen bonds hold the two strands of the helix together.\nQuestion-17\nDescribe the different types of chromosomes.\nSolution:\nThe different types of chromosomes are,\nMetacentric\nIt is a chromosome with the centromere near the middle and the two chromatics are of equal lengths.\nSub-metacentric\nHere the centromere is situated slightly closer to one end than the other. Thus one chromatic is slightly longer than the other.\nAcrocentric\nHere the chromosome is situated near one end of the chromosome. Thus one chromatic is very long while the other is very small.\nTelocentric\nHere the centromere is situated at the tip of the chromosome.\nQuestion-18", + "CHAPTER 8-HEREDITY -EXTRA QUESTIONS.txt\nTelocentric\nHere the centromere is situated at the tip of the chromosome.\nQuestion-18\nHow do embryological studies provide evidence for evolution?\nSolution:\nThe embryology of different vertebrates provide very strong evidence favoring organic evolution. The early embryos of different vertebrates show striking similarities. This indicates common origin and ancestry of different vertebrates. Thus embryological studies provide direct evidence for evolution.\nQuestion-19\nDefine evolution. Describe the contribution of Lamarck.\nSolution:\nEvolution is referred to as the changes acquired by a species or a certain population of a species gradually over a long period of time. These changes should be heritable.\nContribution of Lanark", + "CHAPTER 8-HEREDITY -EXTRA QUESTIONS.txt\nEvolution is referred to as the changes acquired by a species or a certain population of a species gradually over a long period of time. These changes should be heritable.\nContribution of Lanark\nAccording to the Theory of inheritance of acquired characters or Lamarckism, put forward by Lanark, the use and disuse of an organ leads to acquiring change in the features of that organ. These changes are also inherited by the offspring\u2019s. The favorable variations caused due to use and disuse after a considerably long period of time, results in evolution of a new species.\nQuestion-20\nHow do homologous organs provide evidence in support of evolution?\nSolution:\nThe presence of homologous organs indicates that all vertebrates have a common ancestry. Similarly, all organs and systems of the vertebrates show fundamental similarities, which point towards common ancestry.", + "CHAPTER 8-HEREDITY -EXEMPLER QUESTIONS.txt\nNCERT Exemplar Class 10 Science Chapter 9 Heredity and Evolution\nShort Answer Questions\nQuestion 1. Do genetic combination of mothers play a significant role in determining the sex of new born?\nAnswer. No, mothers have no role in determining the sex of the new born. Mothers have a pair of X chromosome. And all children will inherit an \u2018X\u2019 chromosome from their mother regardless of whether they are boys or girls.\nQuestion 2. Mention three important features of fossils which help in the study of evolution.\nAnswer. The three important features of fossils that help in the study of evolution are as follows:\n It helps in establishing the time period in which the organisms lived.\n It helps in establishing evolutionary traits among organisms and their ancestors.\n It is the mode of preservation of the remains of ancient species.\nQuestion 3. In human beings, the statistical probability of getting either a male or female child is 50 : 50. Give a suitable explanation.", + "CHAPTER 8-HEREDITY -EXEMPLER QUESTIONS.txt\nQuestion 3. In human beings, the statistical probability of getting either a male or female child is 50 : 50. Give a suitable explanation.\nAnswer. The sex of a child is determined by the type of sex chromosome contributed by male gamete. The statistical probability of getting either a male or female child is 50: 50, because the ratio of male gametes containing X chromosome and those containing Y chromosome is 50 : 50.\nQuestion 4. What are homologous structures? Give an example. Is it necessary that homologous structures always have a common ancestor?\nAnswer. Homologous organs are those organs which have the same basic structural design and developmental origin but have different functions and appearance. Example: The forelimb of a frog, a lizard, a bird and a man seem to be built from the same basic design of bones, but they perform different functions.", + "CHAPTER 8-HEREDITY -EXEMPLER QUESTIONS.txt\nYes, it is necessary that homologous structures always have a common ancestor in order to carry out the different activities. Otherwise there cannot be any similarity in basic plan, internal structure, development or origin.\nQuestion 5. Does the occurrence of diversity of animals on earth suggest their diverse ancestry also? Discuss this point in the light of evolution.\nAnswer. Diversity of animals does not mean that they have diverse ancestry because common ancestry greatly limit the extent of diversity. If the animals are inhabiting in the same habitat their evolution by speciation and geographical isolation is also not likely. Animals having a common ancestor have developed new traits forming various groups of animals.\nQuestion 6. A woman has only daughters. Analyse the situation genetically and provide a suitable explanation.\nAnswer. The women produces ova with \u2018X\u2019 chromosome and man produces sperms with X and Y chromosome.", + "CHAPTER 8-HEREDITY -EXEMPLER QUESTIONS.txt\nAnswer. The women produces ova with \u2018X\u2019 chromosome and man produces sperms with X and Y chromosome.\nIf the husband of the woman transfer X chromosome, then child will be a girl. On the other hand, if the husband transfer Y chromosome, the child will be a boy. In the case, the husband is always transferring X chromosome and hence, all the children are girl.\nLong Answer Questions\nQuestion 7. Give reasons why acquired characters are not inherited.\nAnswer. Acquired characters are not inherited because they do not produce change in the DNA of germ cells, only those characters which have a gene for them can be inherited.\nQuestion 8. Does geographical isolation of individual of a species lead to formation of a new species? Provide a suitable explanation.\nAnswer. Yes, geographical isolation of individual of a species lead to the formation of a new species.", + "CHAPTER 8-HEREDITY -EXEMPLER QUESTIONS.txt\nAnswer. Yes, geographical isolation of individual of a species lead to the formation of a new species.\n\nGeographical isolation of a population leads to genetic drift and there will be no gene flow between it and the parent species. Inbreeding in small population will reproduce among themselves and generate new variations. Accumulation of those variations over several generations will lead to formation of new species.\nQuestion 9. Bacteria have a simpler body plan when compared with human beings. Does it mean that human beings are more evolved than bacteria? Provide a suitable explanation.\nAnswer. Both bacteria and human beings perform all activities of life to live in their environment. Human beings have more complex organisation and differentiation which are absent in bacteria. Since, complexity and differentiation develop only through evolution, humans are more evolved than bacteria.", + "CHAPTER 8-HEREDITY -EXEMPLER QUESTIONS.txt\nQuestion 10. Evolution has exhibited a greater stability of molecular structure when compared with morphological structures. Comment on the statement and justify your opinion.\nAnswer. There is immense diversity in size, form, structure and morphological features in the living world. At the molecular level, these diverse types of organisms exhibit similarity of the basic bio molecules like DNA, RNA, carbohydrates, proteins, etc.\nQuestion 11. Study the following cross and showing self-pollination in Flf fill in the blank and answer the question that follows (Q12, Q13 and Q14):\n\nAnswer. Rr Yy Round, yellow.\nQuestion 12. In previous question, what are the combinations of character in the F2 progeny? What are their ratios?\nAnswer. Round yellow \u2013 9, Round green \u2013 3, Wrinkled yellow \u2013 3 Wrinkled green \u2013 1, Le. 9 : 3 : 3 : 1.\nQuestion 13. Give the basic features of the mechanism of inheritance.\nAnswer. Basic features for the mechanism of Inheritance are as follows:", + "CHAPTER 8-HEREDITY -EXEMPLER QUESTIONS.txt\nQuestion 13. Give the basic features of the mechanism of inheritance.\nAnswer. Basic features for the mechanism of Inheritance are as follows:\nCharacters are controlled by genes,\nEach gene controls one character.\nGenes are located oh chromosomes.\nThere may be two or more forms of gene.\nAn individual possess two forms of genes whether similar or dissimilar.\n One form may be dominant over the other.\nThe two alleles separate at the time of gamete formation.\nThe two forms are brought together in the zygote.\nAlleles of different genes located on separate chromosomes behave independent of one another.\nQuestion 14. Give reasons for the appearance of new combinations of characters in the F2 progeny.\nAnswer. An organism can inherit each character independently. So, in the F2 progeny new combination of character appears. Tall/ Short and Round/Wrinkled seed trait are independently inherited.", + "CHAPTER 8-HEREDITY -NOTES.txt\nCBSE Class 10 Science Notes Chapter 9 Heredity and Evolution\nHeredity and Inherited Traits: Mendel\u2019s Experiment; Sex determination.\nHeredity refers to the transmission of characters from parents to offsprings. An inherited trait is a particular genetically determined feature that distinguishes a person from the others for example; attached or free ear lobes in human beings.\nRules for the inheritance of traits:\nMendel\u2019s contribution: The rules for inheritance of traits in human beings are related to the fact that both mother and father contribute an equal amount of genetic material i.e. DNA to their offspring. So an offspring will get two versions of that trait from the two parents. Mendel worked out rules for inheritance of these traits. Gregor Johann Mendel regarded as the \u2018Father of Genetics\u2019 performed his experiments with garden peas (Pisum sativum) in the garden behind his monastery. He observed a number of contrasting characters in garden peas and observed their inheritance.", + "CHAPTER 8-HEREDITY -NOTES.txt\nSome important terms\n1. Chromosomes are long thread-like structures present in the nucleus of a cell which contain hereditary information of the cell in the form of genes.\n2. DNA is a chemical in the chromosome which carries the traits in a coded form.\n3. Gene is the part of a chromosome which controls a specific biological function.\n4. Contrasting characters: A pair of visible charactes such as tall and dwarf, white and violet flowers, round and wrinkled seeds, green and yellow seeds etc.\n5. Dominant trait: The character which expresses itself in a (Ft) generation is dominant trait. Example : Tallness is a dominant character in pea plant.\n6. Recessive trait: The character which does not express itself but is present in a generation is recessive trait. Ex. dwarfism in the pea plant.\n7. Homozygous: A condition in which both the genes of same type are present for example; an organism has both the genes for tallness it is expressed as TT and genes for dwarfness are written as tt.", + "CHAPTER 8-HEREDITY -NOTES.txt\n7. Homozygous: A condition in which both the genes of same type are present for example; an organism has both the genes for tallness it is expressed as TT and genes for dwarfness are written as tt.\n8. Heterozygous: A condition in which both the genes are of different types for example; an organism has genes Tt it means it has a gene for tallness and the other for dwarfness only tall character is expressed.\n9. Genotype: It is genetic make up of an individual for example; A pure tall plant is expressed as TT and hybrid tall as Tt.\n10. Phenotype: It is external appearance of the organism for example; a plant having Tt composition will appear tall although it has gene for dwarfness.\n11. Homologous pair of characters are those in which one member is contributed by the father and the other member by the mother and both have genes for the same character at the same position.", + "CHAPTER 8-HEREDITY -NOTES.txt\n11. Homologous pair of characters are those in which one member is contributed by the father and the other member by the mother and both have genes for the same character at the same position.\nMendel\u2019s Experiment: Mendel started his experiment on the pea plants. He conducted first monohybrid and then dihybrid crosses.\nMonohybrid Cross: The cross in which Mendel showed inheritance of dominant and recessive characters is monohybrid cross. To observe inheritance of single pair of contrasting characters", + "CHAPTER 8-HEREDITY -NOTES.txt\nhe took pure tall (genotype TT) and pure dwarf (genotype tt) pea plants and cross pollinated them to obtain first generation or first filial generation. In this figuration (F1 generation) he obtained only tall plants. This meant that only one of the parental traits was seen, not the mixture of the two. The plants of F generation or progeny are then self pollinated to obtain F2 generation or progeny. Now all plants were not tall. He obtained 75% tall plants and 25% dwarf plants i.e. the phenotypic ratio was 3:1. This indicates that in the F, generation both tall and dwarf traits were inherited but tallness expressed it self. Tallness is a dominant trait and dwarfness is a recessive trait. F2 generation has a genotypic ratio of 1 : 2 : 1 of three types of plants represented by TT, Tt and tt as shown in the cross.\nConclusion: Phenotypic ratio\u2014Tall : Dwarf 3 : 1\nGenotype ratio\u2014Pure Tall : Hybrid Tall : Pure Dwarf 1 : 2 : 1", + "CHAPTER 8-HEREDITY -NOTES.txt\nConclusion: Phenotypic ratio\u2014Tall : Dwarf 3 : 1\nGenotype ratio\u2014Pure Tall : Hybrid Tall : Pure Dwarf 1 : 2 : 1\nLaw of Dominance: When parents having pure contrasting characters are crossed then only one character expresses itself in the Ft generation. This character is the dominant character and the character/factor which cannot express itself is called the recessive character.", + "CHAPTER 8-HEREDITY -NOTES.txt\nDihybrid Cross: Mendel also carried out experiments to observe inheritance of two pairs of contrasting characters, which is called dihybrid cross. He cross breed pea plants bearing round green seed with plants bearing wrinkled and yellow seeds. In the Fx generation he obtained all round and yellow seeds it means round and yellow traits of seeds are dominant features while wrinkled and green are recessive. He self-pollinated the plants of F: generation to obtain F2 generation, he obtained four different types of seeds round yellow, round green, wrinkled yellow and wrinkled green in the ratio of 9 : 3 : 3 : 1. He concluded that traits are independently inherited\nConclusion\nRound and yellow seeds-9.\nRound and green seeds-3.\nWrinkled and yellow seeds-3.\nWrinkled and green seeds-1.\nHow do traits get expressed?\nCellular DNA is the information source for making proteins in the cell.", + "CHAPTER 8-HEREDITY -NOTES.txt\nRound and green seeds-3.\nWrinkled and yellow seeds-3.\nWrinkled and green seeds-1.\nHow do traits get expressed?\nCellular DNA is the information source for making proteins in the cell.\nA part of DNA that provides information for one particular protein is called a gene for that protein for example; the height of a plant depends upon the growth hormone which is in turn controlled by the gene. If the gene is efficient and more growth hormone is secreted the plant will grow tall. If the gene for that particular protein gets altered and less of it is secreted when the plant will remain short. Both the parents contribute equally to the DNA of next generation during sexual reproduction. They actually contribute a copy of the same gene for example; when tall plant is crossed with short plant the gametes will have single gene either for tallness or for shortness. F1 generation will get one gene for tallness and other for shortness also.", + "CHAPTER 8-HEREDITY -NOTES.txt\nHow do germ cells i.e. gametes get single set of genes from parents who have two copies in them ?\nEach gene set is present, not as a single long thread of DNA, but as separate independent pieces each called a chromosome. Each cell gets two copies of the chromosome, one from each parent. Each germ cell or gamete has one copy of it because there is reductional division in the sex organs at the time of formation of gametes. When fertilization takes place normal number of chromosomes is restored in the progeny ensuring the stability of DNA of the species.\nHow is the sex of a newborn individual determined?\nIt is the process by which sex of a newborn can be determined.\nDifferent species use different strategies for this :\nIn some animals the temperature at which fertilized eggs are kept determines whether the developing animals will be males or females.\nSome animals like snails can change sex indicating that sex is not genetically determined.", + "CHAPTER 8-HEREDITY -NOTES.txt\nSome animals like snails can change sex indicating that sex is not genetically determined.\nIn human beings sex of the individual is determined genetically; means genes inherited from the parents decide the sex of the offspring.\nSex determination in human beings: In human beings, all chromosomes are not paired. 22 chromosomes are paired but one pair called sex chromosome is odd in not having a perfect pair in males. Females have a perfect pair both represented by XX. On the other hand males have a normal sized X but the other is short called Y so it is shown as XY. All gametes or ova formed by the homogenetic female are similar i.e. have X chromosome. Males heterogenetic form two types of sperms i.e. half with X chromosome and the other half with Y chromosome. Sex of the baby will depend on fertilization. There are two possibilities :\n\nAutosomes: Those chromosomes which do not play any role in sex determination.", + "CHAPTER 8-HEREDITY -NOTES.txt\nAutosomes: Those chromosomes which do not play any role in sex determination.\nSex chromosomes: Those chromosomes which play a role in determining sex of the newborn.\nIf the sperm having X chromosome fertilizes with ovum with X chromosome then the baby will have XX chromosome and it will be female.\nIf the sperm having Y chromosome fertilizes with ovum with X chromosome then the baby will have XY chromosomes and it will be male.\nEvolution: Acquired and inherited traits, Speciation, Evolution and classification, Evolution by stages, Human evolution.\nEvolution: It is the sequence, of gradual, irreversible changes which took place in the primitive organisms over millions of years to form new present-day species. Variations that resulted in formation of new species occurred basically due to errors in DNA copying as well as due to sexual reproduction.", + "CHAPTER 8-HEREDITY -NOTES.txt\nAn Illustration to show variations in a population: A group of twelve red beetles live in green bushes and reproduce sexually so are likely to develop variations. There are the following possibilities\nFirst situation: Crows eat these beetles as they can easily pick up red ones in the green bushes There is a colour variation during sexual reproduction and green beetles appears, it reproduces and its population increases. Crows are not able to see green beetles so their population continues to increase but that of red beetles decreases. This type of variation gives a survival advantage.", + "CHAPTER 8-HEREDITY -NOTES.txt\nSecond situation: Due to a colour variation few blue beetle appear forming blue population. Crows can see both red and blue and eat them. Initially there are more of red beetles and less of blue. There is sudden calamity, an elephant kills red beetles by stamping on bush, blue beetles survive reproduce and increase in number. In this case there is no survival advantage but provides diversity without any adaptation.\nThird situation: As the population of beetles increases, the bushes suffer from a disease and the availability of food for beetles decreases. The size of beetles decrease but after a few years as the plant disease is eliminated and enough food is available for the beetles they come back to their normal size. This type of change is not inherited.", + "CHAPTER 8-HEREDITY -NOTES.txt\nAcquired Traits: Acquired traits are those which are not inherited over generations as they are caused due to change in the non-reproductive tissue and are not passed on the DNA of the germ cells for example; the size of the beetles in the population decreased due to scarcity of food.\nInherited Trait: Inherited traits are caused due to changes in the DNA of germ cells which are inherited from generation to generation, for example; formation of green beetles in the population of red beetles.\nAcquired Traits and Inherited Traits\nCharles Darwin\u2019s Idea of Evolution: His concept of evolution was based on the idea that new species were formed due to variations that occurred in the organisms Nature played an important role in selecting the organisms having suitable variations.\nSpeciation: It means the development of one or more species from an existing species The factors that could lead to rise of a new species are :", + "CHAPTER 8-HEREDITY -NOTES.txt\nSpeciation: It means the development of one or more species from an existing species The factors that could lead to rise of a new species are :\nGene flow: It means the exchange of genetic material by interbreeding between populations of the same species or between individuals within a population. It increases the variation in the genetic composition of a population.\nGenetic drift: It is random change in the frequency of alleles in a populate over successive generation due to errors in the gametes. The process is rapid in smaller population. Genetic drift can lead to accumulation of changes in the generations.", + "CHAPTER 8-HEREDITY -NOTES.txt\nNatural selection: According to Darwin, natural selection also plays an important role in bringing about evolution of new species of plants and animals. According to him variations existed between the individuals of a population and some natural phenomena eliminated those individuals which were less adapted. The surviving population would pass the hereditary advantageous features to their offsprings. With time this process would give rise to organisms different from the original population and new species are formed.\nIsolation: When a population of a species splits into two, it cannot reproduce with each other and forms a new species, for example; when a population of beetles feed on bushes on a mountain range, some may start feeding on nearby bushes finding entry into a new subpopulation. They reproduce with them so genes enter in a new population. Ultimately the two groups will be incapable of reproducing with each other and new species will be formed.", + "CHAPTER 8-HEREDITY -NOTES.txt\nEvolution and Classification: The organisms show certain features, like appearance and behaviour which are called characteristics for example; Plants can perform photosynthesis. The basic characteristics are shared by a large number of organisms. More characteristics which two species have in common more closely are related, if they are more closely related then they have common ancestors (explain the example of brother sister and cousins). Evolutionary relationships can be traced with the help of the following :\nHomologous organs: Those organs which have the same basic structural design and developmental origin but perform different functions and appearance, for example; Forelimbs of frog, lizard, bird, bat and human beings. They have same design of bones but they perform different functions.", + "CHAPTER 8-HEREDITY -NOTES.txt\nAnalogous organs: Those organs which have different basic design and developmental origin but have similar appearance and perform a similar function, for example; wings of bat and bird. Wings of bat are folds of skin attached between fingers. But wing of birds are modified forelimbs.\n\nStudy of Fossils: Fossils are preserved remains of living organisms that lived in the past. When living organisms die their bodies decompose but some parts of their body may be in such an environment that they do not decompose for example; if a dead insect gets caught in hot mud it will not decompose quickly but the mud will harden and retain impressions of the body parts of the insects. These impressions are also called fossils: The age of fossil can be estimated in two ways :\nThe fossils that occur closer to earth surface are more recent to those found in deeper layers.\nThe second method is isotope dating i.e. detecting the ratio of different isotopes of the same element in the fossil material.", + "CHAPTER 8-HEREDITY -NOTES.txt\nThe second method is isotope dating i.e. detecting the ratio of different isotopes of the same element in the fossil material.\n\nSignificance of fossils: Fossils are formed layer by layer in the earths crust. The animals and plants which existed earlier are buried in the deeper layer which ones found in the upper layer. It is found that, deeper fossils have simpler structure than found than upper layer. Complete fossil record of animals like horse, camel, man has helped us to study the stages of evolution.", + "CHAPTER 8-HEREDITY -NOTES.txt\nEvolution by stages: Evolution is a continuous and gradual process, complicated organs did not evolve by a single DNA change but were formed by bit by bit change over generations for example; complex organs like eyes were created by bit by bit changes, in between the rudimentary eye in some insects also provided a fitness advantage. The structure of eye in all organisms is different enough to have evolutionary origins. Some organs even developed for one particular function but later become useful for quite a different function, e.g Feathers developed to provide warmth to the animal but later helped in flight.\nSome dinosaurs had feathers although they could not fly, this shows that birds are closely related to reptiles, since dinosaurs were reptiles Some dissimilar looking structures also evolved from common ancestors. The current example of such a process is wild cabbage plant from which different vegetables are generated by artificial selection rather than natural selection", + "CHAPTER 8-HEREDITY -NOTES.txt\nSelection of short distance between the leaves has led to formation of cabbage that, we eat.\nSelection for arrested flower development had led to broccoli,\nSelection for sterile flowers had made cauliflower,\nSelection for swollen-stem had formed kohlrabi.\nSelection for large leaves had formed leafy vegetable kale,\nSelection for colored leaves formed red cabbage.\nTo sum up we can say that evolutionary relationships can be established by\nStudy of Homologous organs\nStudy of Analogous organs\nStudy of fossils\nChanges in DNA during reproduction", + "CHAPTER 8-HEREDITY -NOTES.txt\nTo sum up we can say that evolutionary relationships can be established by\nStudy of Homologous organs\nStudy of Analogous organs\nStudy of fossils\nChanges in DNA during reproduction\nEvolution versus Progress: Evolution can not be called progress from lower forms to higher forms. It is basically forming more complex designs while the simpler once also keep growing. Evolution is generation of diversity with the help of environmental selection. Bacteria which were formed first have the capacity to live in diverse conditions and are still flourishing; on the other hand human beings which are highly evolved species can not be called the pinnacle of evolution but yet another species in the evolving life forms.", + "CHAPTER 8-HEREDITY -NOTES.txt\nHuman Evolution: Human evolution has been studied with the help of excavation; time dating and fossil study All human beings belong to single species i.e. Homo sapiens. Human species have come from Africa. Some of our ancestors left Africa while others stayed on. These migrants slowly spread across the planet i.e. West Asia, Central Asia, Eurasia, South Asia and East Asia They traveled to Indonesia, the Philippines, Australia and America They traveled forward and backward sometimes separating and sometimes coming back to mix with each other. They had come into being as an accident of evolution.\n\nAlthough there is a great diversity of human forms all over the world get all humans are single species.\nThey didn\u2019t go in a single line.\nThey went forward and backward.\nMoved in and out of Africa.\nSometimes come back to mix with each other.\n\nGenetics: Branch of science that deals with heredity and variation.", + "CHAPTER 8-HEREDITY -NOTES.txt\nThey went forward and backward.\nMoved in and out of Africa.\nSometimes come back to mix with each other.\n\nGenetics: Branch of science that deals with heredity and variation.\nHeredity: It means the transmission of features/characters/traits from one generation to the next generation.\nVariation: The differences among the individuals of a species/population are called variations.\nMendel and his work on Inheritance.\nGregor Johann Mendel started his experiments on plant breeding and hybridisation. Mendel was known as Father of Genetics.\nThe plant selected by Mendel was Pisutn sativum (garden pea). Mendel used a number of contrasting characters for garden pea.\nSex Determination: Phenomenon of decision or determination of sex of an offspring.\nFactors Responsible for Sex Determination:\nEnvironmental: In some animals, the temperature at which the fertilised eggs are kept decides the gender. Example, in turtle.", + "CHAPTER 8-HEREDITY -NOTES.txt\nFactors Responsible for Sex Determination:\nEnvironmental: In some animals, the temperature at which the fertilised eggs are kept decides the gender. Example, in turtle.\nGenetic: In some animals like humans gender or individual is determined by a pair of chromosomes called sex chromosomes (XX \u2013 female; XY \u2013 male).\nSex Chromosomes: In human beings, there are 23 pairs of chromosomes. Out of these 22 chromosome pairs are called autosomes and the last pair of chromosomes that help in deciding the gender of that individual are called sex chromosome.\nXX \u2013 female; XY \u2013 male\nThe cross done shows that half the children will be boys and half will be girls. All children will inherit an X chromosome from their mother regardless of whether they are boys or girls. Thus sex of children will be determined by what they inherit from their father, and not from their mother.\nAcquired Traits:\nThese are the traits which are developed in an individual due to special conditions.", + "CHAPTER 8-HEREDITY -NOTES.txt\nAcquired Traits:\nThese are the traits which are developed in an individual due to special conditions.\nThey cannot be transferred to the progeny.\nThey cannot direct evolution, for example, the low weight of starving beetles.\nInherited Traits:\nThese are the traits which are passed from one generation to the next.\nThey get transferred to the progeny.\nThey are helpful in evolution, for example, the colour of eyes and hair.\nMicroevolution: It is the evolution which takes place on a small scale. Example, change in body colour of beetles.\nSpeciation: It is the process of formation of new species. A species is a group of similar individuals that belong to a population that can interbreed and produce fertile offspring. Speciation takes place when the variation is combined with geographical isolation.", + "CHAPTER 8-HEREDITY -NOTES.txt\nGene flow: It is the exchange of genetic material by interbreeding between populations of the same species or individuals. Gene flow occurs between populations that are partly but not completely separated.\nGenetic Drift: It is the random change in the frequency of alleles (gene pair) in a population over successive generations.\nGenetic drift takes place due to:\nsevere changes in the DNA.\nchange in the number of chromosomes.\nNatural Selection: The process by which nature selects and consolidates those organisms which are more suitably adapted and possesses favourable variations.\nEvolution and classification. Both evolution and classification are interlinked.\nClassification of species is a reflection of their evolutionary relationship.\nThe more characteristics two species have in common the more closely they are related.\nThe more closely they are related, the more recently they have a common ancestor.", + "CHAPTER 8-HEREDITY -NOTES.txt\nThe more characteristics two species have in common the more closely they are related.\nThe more closely they are related, the more recently they have a common ancestor.\nSimilarities among organisms allow us to group them together and to study their characteristics.\nTracing Evolutionary Relationships:\nHomologous Organs: Morphological and anatomical evidences. These are the organs that have same basic structural plan and origin but different functions.\nExample, forelimb of a horse (running), wings of bat (flying), paw of a cat (walk/ scratch/ attack) \u2014 same basic structure but different functions.\nAnalogous Organs: These are the organs that have different origin and structural plan but same functions.\nExample, wings of a bat (elongated fingers with skin folds), wings of bird (feathery covering along the arm) \u2014 different structures but same functions.", + "CHAPTER 8-HEREDITY -NOTES.txt\nExample, wings of a bat (elongated fingers with skin folds), wings of bird (feathery covering along the arm) \u2014 different structures but same functions.\nFossils: The remains and relics of dead organisms that lived in the remote past. Fossils provide evidence of evolution. Example, a fossil called Archaeopteryx has feathered wings like birds but teeth and tail like reptiles hence suggesting that birds and reptiles had a common ancestor.\nArtificial Selection: Humans have been a powerful agent in modifying wild species to suit their own requirement throughout ages by using artificial selection. Example, wheat (many varieties obtained due to artificial selection).\n1. Heredity : It refers to the transmission of characters or traits from the parents to their offspring. Heredity is the continuity of features from one generation to another which are present in fertilised egg or zygote. The zygote develops into an organism of a particular type only.", + "CHAPTER 8-HEREDITY -NOTES.txt\n2. Genetics : It is the branch of biology which deals with heredity and variation. Genetics is to help our understanding of heredity by knowing how offspring inherit characteristics from their parents.\n3. Variation : It means the differences in the characters or traits among the individuals of a species. Variations occur during reproduction both because of error in DNA copying and as a result of sexual reproduction. Variations contribute to evolution.\nCauses of variations:\nDifferent combinations of genetic material.\nSome positive gene mutations.\nInteraction of genes with environmental changes (adaptations).\nImportance of variations:\nIt forms, the. basis of heredity.\nIt causes adaptations due to which organism can easily adjust to its changing environment.\nAccumulation of variations forms the basis of evolution.\nRemember!", + "CHAPTER 8-HEREDITY -NOTES.txt\nIt forms, the. basis of heredity.\nIt causes adaptations due to which organism can easily adjust to its changing environment.\nAccumulation of variations forms the basis of evolution.\nRemember!\nVariations are produced both in sexual and asexual reproduction but amount of variations produced in asexual reproduction are subtle (so little) that they are hardly noticeable as compared to variations caused due to sexual reproduction.\n4. Genotype : The genetic constitution of an organism e.g., Genotype of human male is 44 + XY and\ngenotype of human female is 44 + XX\n5. Phenotype : The appearance of the organism, i.e., the way in which genotype is expressed. Phenotype is the result of interaction of genes with the environment.\ne.g., Red colour may be controlled by a pair of genes RR. Now if genotype is RR phenotype will be red only but if genotype is Rr then also phenotype will be red since R is a dominant gene.", + "CHAPTER 8-HEREDITY -NOTES.txt\ne.g., Red colour may be controlled by a pair of genes RR. Now if genotype is RR phenotype will be red only but if genotype is Rr then also phenotype will be red since R is a dominant gene.\n6. Gene : It is the basic unit of inheritance by which characters are transferred from parents to their offspring. Gene consists of a specific length of DNA on a chromosome. A specific Segment of DNA that provides the information for one protein is called gene for that protein.\nAccording to Mendel, both parents must contribute equally to the DNA of the progeny during sexual reproduction. As both parents determine the trait in the progeny, so both parents must be contributing a copy of the same gene.\n7. Chromosomes : These are the long threads present in the nucleus of every cell. Chromosomes are made- up of DNA and protein. Each chromosome contains very long molecule of DNA.\nRemember!", + "CHAPTER 8-HEREDITY -NOTES.txt\n7. Chromosomes : These are the long threads present in the nucleus of every cell. Chromosomes are made- up of DNA and protein. Each chromosome contains very long molecule of DNA.\nRemember!\nEach gene set is present as separate independent pieces each called a chromosome. Each cell have two copies of each chromosome, one each from male and female parents. Every germ cell will take one chromosome from each pair and these may be of either maternal or paternal origin. When two germ cells combine, they will restore the normal number of chromosomes in the progeny, ensuring the stability of the DNA of the species. Such mechanism of inheritance is used by all sexually and asexually reproducing organisms.\n8. Allele: It is an alternative form of a gene occupying the same position on a chromosome and affecting the same characters but in two alternative ways, e.g., the free and attached ear lobe are the alleles of ear lobe character.\nExpressing allele of a gene :", + "CHAPTER 8-HEREDITY -NOTES.txt\nExpressing allele of a gene :\nHomozygous dominant in capital letters, e.g., tallness(TT)\nHomozygous recessive in smalMetters, e.g., shortness or dwarfness (tt)\nHeterozygous (Tt)-lt will be called hybrid tall.\n9. Dominant allele: An allele that affects the phenotype of an organism both in heterozygous and homozygous condition. It is denoted by a capital letter, e.g., tallness in pea plant is denoted by \u2018T.\n10. Recessive allele: An allele that affects the phenotype of the organism in absence of a dominant allele, i.e., in homozygous recessive individuals. It is denoted by a small alphabet, e.g., dwarfness in pea plant is denoted by\u2019t\u2019.\n11. Homozygous: When both alleles of a particular gene are the same, e.g., TT\n12. Heterozygous : When both alleles of a particular gene are different, e.g., Tt\n13.", + "11. Homozygous: When both alleles of a particular gene are the same, e.g., TT\n12. Heterozygous : When both alleles of a particular gene are different, e.g., Tt\n13. Diploid : Cells or organism containing two sets of genes, e.g., human body cells. Diploid cells have genetic constitution of 2n.", + "CHAPTER 8-HEREDITY -NOTES.txt\n13. Diploid : Cells or organism containing two sets of genes, e.g., human body cells. Diploid cells have genetic constitution of 2n.\n14. Haploid : Cells or organism containing one set of genes, e.g., human reproductive cells (sperms and ova). Haploid cells have genetic constitution of n.\n15. Monohybrid cross : A cross between two parents taking the alternative traits of one single character, e.g., A cross between tall and dwarf pea plants.\nMonohybrid Ratio :\nIn F1 generation : 100% hybrid\nIn F2 generation : phenotypic ratio is 3 : 1 and genotypic ratio is 1 : 2 : 1\n16. Dihybrid cross: A cross between two parents taking into consideration alternative traits of two different characters, e.g., A cross between two pea plants one having round, green seeds and the other having wrinkled, yellow seeds.\nDihybrid Ratio :\nF1 ratio is 100% Hybrid type.\nF2 ratio : Phenotypic is 9 : 3 : 3 : 1 and Genotypic . ratio is very complex.", + "CHAPTER 8-HEREDITY -NOTES.txt\nDihybrid Ratio :\nF1 ratio is 100% Hybrid type.\nF2 ratio : Phenotypic is 9 : 3 : 3 : 1 and Genotypic . ratio is very complex.\n17. Human Blood Groups: There are four types of blood groups A, B, AB or O. These are controlled by a gene which is denoted by symbols IA, IB and IO (sometimes also denoted as i). The genes IA and IB show no dominance over each other (they are codominant, i.e., both expresses themselves independently). But these both genes are dominant over the gene IO. Therefore, blood group of a person depends on the type of genes present, e.g., (i) Blood group A has the following gene types :\n18. Determining sex of a newborn individual genetically:\nIn human beings the sex of the individual is \u201d determined genetically.\nThere are 23 pairs of chromosomes of which 22 are similar in male and female and are known as autosomes.\nThe remaining one is sex chromosome which is XY in males and XX in females.", + "CHAPTER 8-HEREDITY -NOTES.txt\nThere are 23 pairs of chromosomes of which 22 are similar in male and female and are known as autosomes.\nThe remaining one is sex chromosome which is XY in males and XX in females.\nMales produce two types of sperms X and Y, while female produces one type of egg X.\nIf a X type of sperm fertilizers the egg then the sex of baby will be female (XX).\nIf Y type of sperm fertilizers the egg then the sex of the baby will be male (XY).\n\n19. Mendel\u2019s experiment to show that traits may be dominant or recessive:\nMendel conducted breeding experiments in garden pea.\nselected pure plant of a tall/short plant.\nproduced first generation plants by crossing them.\nfound that all plants were tall.\nproduced the second generation by self-fertilization of hybrids.\nfound that three-quarter of the plants was tall and one quarter was short.\n20. Homologous chromosomes: A pair of corresponding chromosomes of the same shape and size, one from each parent.", + "CHAPTER 8-HEREDITY -NOTES.txt\nfound that three-quarter of the plants was tall and one quarter was short.\n20. Homologous chromosomes: A pair of corresponding chromosomes of the same shape and size, one from each parent.\n21. Autosomes and Sex chromosomes : The identical \u00bb chromosome pairs are called autosomes. The\nchromosome pair which is different are called sex chromosomes. Humans have 23 pairs of chromosomes. 1-22 pairs are autosomes while 23rd pair (XX in females and XY in males) which are designated as X and Y are sex chromosomes.\n22. Molecular Phylogeny: It is the study of evolutionary relationships by comparing DNA of different species.", + "CHAPTER 8-HEREDITY -NOTES.txt\n22. Molecular Phylogeny: It is the study of evolutionary relationships by comparing DNA of different species.\n23. Natural selection : Natural selection is one of the basic mechanisms of evolution, along with mutation, migration and genetic drift. Natural selection means the environmental conditions prevailing around an organism against which organism adapts itself, grows \u2013 and reproduces further. This leads to a change in the composition of genes within a population further causing evolution. Thus, it can be said that,\nNatural selection results in adaptation in population to fit their environment better. Thus, natural selection direct evolution in the population of a particular species.\n24. Fossils of the information which they provide regarding evolution: Fossils are the remains of ancient life forms, which got preserved somehow in the layers of earth, snow or oil.\nInformation given by fossils:", + "CHAPTER 8-HEREDITY -NOTES.txt\nInformation given by fossils:\nThey reveal that the life forms which existed earlier do not exist today which indicate that the living forms are ever changing (evolving).\nThey are used to guess the time when a particular organism existed on earth. It is done through carbon dating.\n25. Genetic drift: The change in the frequency of some genes in a population which provides diversity without any survival advantage is called genetic drift.\n26. The various ways in which individuals with a particular trait may increase in a population : Differences in population are responsible for the diversity such as, colour of eyes, hair, shape of ear lobes. This occurs due to : (i) Sexual reproduction (ii) Inaccuracies during DNA replication (iii) Due to environmental changes. This diversity will increase with time as these variations can be passed on only through DNA/genes during reproduction through reproductive tissue (germ cells or gametes).", + "CHAPTER 8-HEREDITY -NOTES.txt\nIf these variations give survival advantage, then such traits are selected in nature and such traits increase in a population.\nDue to genetic drift. This occurs due to geographical or reproductive isolation. It results in the change in gene frequency in a particular: population.\nMigration which leads to gene flow in and out of the population.\nThe mutation caused due to particular type of environment. ,\nAcquired traits due to particular type of environment.\n27. Evidence of evolution: Errors in DNA copying (mutation) and sexual reproduction lead to variations which form the basis of evolution. Characteristics that\nare common in different kinds of living organisms provide evidence in favour of evolution.", + "CHAPTER 8-HEREDITY -NOTES.txt\nare common in different kinds of living organisms provide evidence in favour of evolution.\n28.Evolution : Evolution can be defined as a naturally occurring slow, continuous and irreversible process of change. The gradual change of living organisms from pre-existing organisms since the beginning of life is called organic evolution. Whereas, gradual change in elements from one form to another with time is termed as inorganic evolution, i.\n29.Inherited traits : are those traits which are passed from one generation to another through specific genes. Any change in DNA of the germ cells will be passed.\n30. Acquired traits : are those traits which are acquired by the organism in its lifetime, e.g., removal of tail cannot change the genes of the germ cells of the mice thus cannot be passed to next generation.\n31. Speciation : It means the origin of new species from the existing ones. It happens when different populations of the same species evolve along different lines.", + "CHAPTER 8-HEREDITY -NOTES.txt\n31. Speciation : It means the origin of new species from the existing ones. It happens when different populations of the same species evolve along different lines.\nHow speciation occurs ?\nIt occurs when two populations are isolated (both geographically and reproductively) leading to almost no gene flow between the two populations.\nOver generations, genetic drift will accumulate different changes in each sub-population.\nNatural selection may also operate differently in these different locations.\nTogether natural selection and genetic drift will cause such changes (severe changes in the DNA) that these two groups will not be able to reproduce with each other even if they happen to meet.\nWhen DNA changes occur to larger extent, it may lead to change in the number of chromosomes or gene expression, eventually the germ cells of the two groups cannot fuse with each other. This leads to emergence of new species.\n32. Estimating Age of Fossil: There are 2 methods :", + "CHAPTER 8-HEREDITY -NOTES.txt\n32. Estimating Age of Fossil: There are 2 methods :\nRelative method : On digging, the fossils which are closer to the surface are more recent than the fossils found in deeper layers.\nDating fossils (carbon dating method): It is done by detecting the ratios of different isotopes of the same element (i.e., isotope of C-14 which is radioactive) in the fossil material.\n33. Evolution by stages :\nComplex organs like eye has evolved from rudimentary organs, (e.g., rudimentary eye in flatworm might be useful enough to give only a fitness advantage and the structure of eye in different organisms is different indicating them to have different evolutionary origins) not by a single DNA change but created bit-by-bit over generations.\nA change that is useful for one property to start with can become useful later for quite a different", + "CHAPTER 8-HEREDITY -NOTES.txt\nA change that is useful for one property to start with can become useful later for quite a different\nfunction (e.g., Feathers might start as providing insulation in cold weather. But later, they might be useful for flight. Some heavy birds and reptiles also have feathers but they do not fly.\nSome very dissimilar looking structures evolve from a common ancestral design, e.g., wild cabbage was cultivated as a food plant and many different vegetables were generated by selection over last two thousand years, (a) Selection of very small distances between the leaves gave rise to cabbage we eat. (b) Selection for arrested flower development gave rise to broccoli, (c) Selection for sterile flowers gave rise to cauliflower (d) Selection for swollen parts gave rise to kohlrabi.(e) Selection for larger leaves gave rise to leafy vegetable kale. It suggests that, if these selections were not done then there would have been only wild cabbage.", + "CHAPTER 8-HEREDITY -NOTES.txt\n34. Homologous organs are organs having same origin and basic structure but they appear different and perform different functions in various organisms,\ne.g.,\nForelimbs of horse and arms of man.\nWings of birds and flippers of whale.\nSimilarities in basic structure of (homologous) organs in different organisms, indifferent groups indicate common ancestry.\n35. Analogous organs are organs, which look similar because they perform same function, but they do not have same origin and basic structure.\ne.g.,\nWings of birds and wings of insects.\nFins of fish and flippers of the whale.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nImportant Questions of Heredity and Evolution Class 10 Science Chapter 9\nQuestion 1.\nAll the variations in a species do not have equal chances of survival. Why? (Foreign 2014)\nAnswer:\nAll the variations do not have equal chances of survival in the environment in which they live. Depending on the nature of variations, different individuals would have different kinds of advantages. The organisms which are most adapted to the environment will survive.\nQuestion 2.\n\u201cOnly variations that confer an advantage to an individual organism will survive in a population.\u201d Justify this statement. (Foreign 2011)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 2.\n\u201cOnly variations that confer an advantage to an individual organism will survive in a population.\u201d Justify this statement. (Foreign 2011)\nAnswer:\nVariations are the structural, functional or behavioural changes from the normal characters developed in the living organisms. Inheritable variations participate in evolution. According to Darwin, natural selection sorts out individuals with favourable variations. Such organism will survive, reproduce more and thus, will leave more progenies. Hence, useful variations get established in nature.\nQuestion 3.\nAssertion (A) : The sex of a child in human beings will be determined by the type of chromosome he/she inherits from the father.\nReason (R) : A child who inherits \u2018X\u2019 chromosome from his father would be a girl (XX), while a child who inherits a \u2018Y\u2019 chromosome from the father would be a boy (XY).\n(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).\n\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is, false.\n(d) (A) is false, but (R) is true. (2020)\nAnswer:\n(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).\nQuestion 4.\nA Mendelian experiment consisted of breeding pea plants bearing violet flowers with pea plants bearing white flowers. What will be the result in F1 progeny? (2018)\nAnswer:\nAccording to the Mendelian experiment, violet coloqr (VV) is a dominant trait while white colour (ww) is a recessive trait. Hence, the colour of the flower in F1 progeny will be violet (Vw).\nQuestion 5.\nName the information source for making proteins in the cells.", + "Hence, the colour of the flower in F1 progeny will be violet (Vw).\nQuestion 5.\nName the information source for making proteins in the cells. (Delhi 2014)\nAnswer:\nDeoxyribonucleic acid (DNA) present in the chromosomes of cell nucleus is the information source for making proteins.\nQuestion 6.\nWhat is a gene? (AI 2014)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAnswer:\nDeoxyribonucleic acid (DNA) present in the chromosomes of cell nucleus is the information source for making proteins.\nQuestion 6.\nWhat is a gene? (AI 2014)\nAnswer:\nA gene is a unit of DNA on a chromosome which governs the synthesis of particular protein that controls specific characteristics (or traits) of an organism.\nQuestion 7.\nWhat is heredity? (AI 2014)\nAnswer:\nThe inheritance of characters (or trails) from the parents to their offsprings is called heredity.\nQuestion 8.\nWhy is the progeny always tall when a tall pea plant is crossed with a short pea plant? (Foreign 2014)\nAnswer:\nWhen a tall pea plant is crossed with a short pea plant, the resultant progeny is always tall because tall is dominant trait and short is recessive trait. Therefore, dominant trait expresses itself in the progeny.\nQuestion 9.\nWrite a difference between inherited traits and acquired traits giving one example of each. (Delhi 2013C)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 9.\nWrite a difference between inherited traits and acquired traits giving one example of each. (Delhi 2013C)\nAnswer:\nA trait (or characteristic) of an organism which is \u2018not inherited\u2019 but develops in response to the environment is called an acquired trait. For example, if a group of mice are normally bred, all their progeny will have tails. Now, if the tails of these mice are cut by surgery in each generation, tail-less mice will not be produced. This is so because removal of tail is an acquired character.\nA trait (or characteristic) of an organism which is caused by a change in its genes (or DNA) is called an inherited trait. Inherited traits can be passed on to the progeny of the organism because they have produced changes in the genes (or DNA) of the organism. For example, skin colour in human beings.\nQuestion 10.\n(a) Why did Mendel carry out an experiment to study inheritance of two traits in garden pea?", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 10.\n(a) Why did Mendel carry out an experiment to study inheritance of two traits in garden pea?\n(b) What were his findings with respect to inheritance of traits in F1 and F2 generation?\n(c) State the ratio obtained in the F2 generation in the above mentioned experiment. (2020)\nAnswer:\n(a) Mendel carried out crosses with two traits to see the interaction and basis of inheritance between them. In a dihybrid cross given by Mendel, it was observed that when two pairs of characters were considered each trait expressed independent of the other.\n(b) For example, a cross between round yellow and wrinkled green parents.\n\nIn F1 generation, all plants are with round yellow seeds. But in F2 generation, we find all types of plants : Round yellow, Round green, Wrinkled yellow, Wrinkled green.\nF2 generation ratio : Round-yellow = 9 : Round- green = 3 : Colour of stem in F1 progeny Wrinkled- yellow = 3 : Wrinkled-green = 1\nQuestion 11.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nF2 generation ratio : Round-yellow = 9 : Round- green = 3 : Colour of stem in F1 progeny Wrinkled- yellow = 3 : Wrinkled-green = 1\nQuestion 11.\nA green stemmed rose plant denoted by GG and a brown stemmed rose plant denoted by gg are allowed to undergo a cross with each other.\n(a) List your observations regarding :\n(i) Colour of stem in their F1 progeny\n(ii) Percentage of brown stemmed plants in F2 progeny if plants are self pollinated.\n(iii) Ratio of GG and Gg in the F2 progeny.\n(b) Based on the findings of this cross, what conclusion can be drawn? (2020)\nAnswer:\n(a) (i) Colour of stem in F1 progeny:\n\nThe colour in the F1 progeny is green stemmed as green stem colour is dominant.\n(ii) F1 progeny on self pollination:\n\nF2 generation Green stemmed: Brown stemmed\n\n1\n4\n or 25% of F2 progeny are brown stemmed rose plant.", + "(ii) F1 progeny on self pollination:\n\nF2 generation Green stemmed: Brown stemmed\n\n1\n4\n or 25% of F2 progeny are brown stemmed rose plant.\n(iii) Ratio of GG and Gg in F2 progeny:\nGenotype of F2 progeny \u2013 GG : Gg\n1 : 2", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nF2 generation Green stemmed: Brown stemmed\n\n1\n4\n or 25% of F2 progeny are brown stemmed rose plant.\n(iii) Ratio of GG and Gg in F2 progeny:\nGenotype of F2 progeny \u2013 GG : Gg\n1 : 2\n(b) This is a monohybrid cross. This shows that out of two contrasting traits only one dominant trait appears in F1 generation and the trait which does not express is recessive. On selfing the F1 plants, both the traits appear in next generation but in a definite proportion.\nQuestion 12.\n(a) Why is the F1 progeny always of tall plants when a tall plant is crossed with a short pea plant?\n(b) How is F2 progeny obtained by self-pollination of F1 progeny different from F1 progeny? Give reason for this observation.\n(c) State a conclusion that can be drawn on the basis of this observation. (2020)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(c) State a conclusion that can be drawn on the basis of this observation. (2020)\nAnswer:\n(a) When a tall plant (TT) is crossed with a short tea plant (tt), only tall plants are obtained in F1 progeny. It is because out of two contrasting traits only one appears in the progeny of first generation. This means that the trait which appears in F1 generation is dominant and the trait which does not express is recessive. The character TT for tall plant is dominant, so all the plants are tall.\n\n(b) On selfing F1 progeny\n\nIn F2 generation we obtained both tall and dwarf plants. Appearance of suppressed recessive trait in individuals of F2 generation in Mendelian cross indicates that characters of recessive traits are not lost. When the F1 generation plants were allowed to self-fertilise both the parental trait were expressed in definite proportion in F2 generation.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(c) Above observation shows that only one dominant allele is expressed. This is called law of dominance. In a heterozygous individual, two dissimilar alleles remain together and do not get mixed up. At the time of gamete formation, they separate so that each gamete receives only one allele is always pure. This is called \u201claw of purity of gametes\u201d.\nQuestion 13.\nName the plant Mendel used for his experiment. What type of progeny was obtained by Mendel in F1 and F2 generations when he crossed the tall and short plants? Write the ratio he obtained in F2 generation plants. (Delhi 2019)\nAnswer:\nMendel selected garden pea (Pisum sativum) for his series of hybridisation experiments.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAnswer:\nMendel selected garden pea (Pisum sativum) for his series of hybridisation experiments.\nHe first selected two pureline plants (tall plant having gene TT and short plant having gene tt) and then crossed such plants having contrasting characters. In the F1 generation, he observed that only one of the two contrasting character appeared, he called this character as dominant and the one which does not get expressed in F1 was called as recessive. He later selfed the F1 plants and observed that both the traits appear in next generation but in a definite proportion. This can be explained by the following cross :\n\nSo, the plants of F1 generation will be all tall plants and after selfing the ratio of tall and dwarf plants that Mendel obtained in F2 generation plants is 3 : 1.\nQuestion 14.\nHow did Mendels experiments show that different traits are inherited independently? Explain. (Delhi 2017)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 14.\nHow did Mendels experiments show that different traits are inherited independently? Explain. (Delhi 2017)\nAnswer:\nIn a dihybrid cross given by Mendel, it was observed that when two pairs of traits or characters were considered, each trait expressed independent of the other. Thus, Mendel was able to propose the Law of Independent Assortment which says about independent inheritance of traits. This could be explained clearly from the given cross:\n\nF2 generation ratio : Round-yellow = 9 : Round- green = 3: Wrinkled-yellow = 3: Wrinkled-green = 1\nQuestion 15.\nHow did Mendel explain that it is possible that a trait is inherited but not expressed in an organism? (AI 2017)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 15.\nHow did Mendel explain that it is possible that a trait is inherited but not expressed in an organism? (AI 2017)\nAnswer:\nMendel first selected two pure line plants. He then crossed such plants having contrasting characters. In the F1 generation, he observed that only one of the two contrasting character appeared, he called it dominant and the one which does not get expressed in F1 was recessive. He later selfed the F1 plants and observed that both the traits appear but in a definite proportion. It can be explained by the following cross :\n\nThis is how Mendel explained that a trait may be inherited but not expressed in the plant.\nQuestion 16.\nIn one of his experiments with pea plants Mendel observed that when a pure tall pea plant is crossed with a pure dwarf pea plant, in the first generation, F1 only tall plants appear.\n(a) What happens to the traits of the dwarf plants in this case?", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(a) What happens to the traits of the dwarf plants in this case?\n(b) When the F, generation plants were self- fertilised, he observed that in the plants of second generation, F2 both tall plants and dwarf plants were present. Why it happened? Explain briefly. (Delhi 2016)\nAnswer:\n(a) Mendel\u2019s monohybrid cross indicated that out of two contrasting traits only one appears in the progeny of first generation. This implies that the trait which appears in F1 generation is dominant and the trait which does not express is recessive. We can also say that gene controlling the dominant trait is dominant gene or allele and gene controlling the recessive trait is recessive gene or allele.\nIn F1 progeny although the dominant trait is expressed but genes for both dominant and recessive traits are present in a heterozygous condition. The recessive trait has a chance to express in next generation only if recessive genes come in homozygous condition. This can be illustrated by the given cross:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(b) Appearance of suppressed recessive trait in individuals of F2 generation in Mendelian cross indicates that the characters of recessive traits are not lost even when they are not expressed. When the F1 generation plants were allowed to self- fertilise, both the parental traits were expressed in definite proportion in F2 generation. This could be explained by the given cross by selfing the gametes obtained in F1 generation.\n\nQuestion 17.\nHow did Mendel interpret his result to show that traits may be dominant or recessive? Describe briefly. (Delhi 2016)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 17.\nHow did Mendel interpret his result to show that traits may be dominant or recessive? Describe briefly. (Delhi 2016)\nAnswer:\nMendel crossed the pea plant for two contrasting characters under consideration. The trait that expressed itself in F1 generation was dominant and the one not expressed in F1 generation was recessive. He later selfed the plants of F1 generation and recovered, both parental traits in a definite proportion in F2 generation. Mendel interpreted his results as, the trait that expressed itself in F1 was dominant and the one that reappeared in F2 generation was recessive. It can be demonstrated by the following cross:\n\nQuestion 18.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 18.\nIn a monohybrid cross between tall pea plants (TT) and short pea plants (tt) a scientist obtained only tall pea plants (Tt) in the F1 generation. However, on selfing the F1 generation pea plants, he obtained both tall and short plants in F2 generation. On the basis of above observations with other angiosperms also, can the scientist arrive at a law? If yes, explain the law. If not, give justification for your answer. (Delhi 2016)\nAnswer:\nIn the situation discussed in the question the scientist can arrive at two different laws, i.e., law of dominance and law of segregation (or law of purity of gametes). This can be explained with the help of following crosses:\n\nIn F1 hybrid two dissimilar alleles are present for one character, i.e., height T is for tallness and t is for dwarfness, out of which only one allele called dominant allele expresses itself and the one which remains unexpressed is called recessive allele. This is called \u201claw of dominance\u201d.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAlso the two dissimilar alleles that remain together in a heterozygous individual do not get mixed up and keep their distinct identity. Hence, at the time of gamete formation they separate so that each gamete receives only one allele and is always pure which enables reappearance of recessive trait in F2 progenies when the two recessive alleles come together. This is called \u201claw of purity of gametes.\u201d\nQuestion 19.\nHow do Mendel\u2019s experiment show that traits are inherited independently? (AI 2016)\nAnswer:\nRefer to answer 14.\nQuestion 20.\nWith the help of an example justify the following statement: \u201cA trait may be inherited, but may not be expressed.\u201d (AI 2016)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 14.\nQuestion 20.\nWith the help of an example justify the following statement: \u201cA trait may be inherited, but may not be expressed.\u201d (AI 2016)\nAnswer:\nA trait may be inherited but may not be expressed, this could be explained by the given example. When a tall pea plant was crossed with a dwarf pea plant, then F1 generation plants were all tall. When F1 plants were selfed, then F2 generation plants were both tall and dwarf. This shows that the F1 plants had inherited both the parental traits but did not express dwarfness or recessive trait in the presence of the trait for tallness or dominant trait. This could be explained by the given cross :\n\nQuestion 21.\nList two differences in tabular form between dominant trait and recessive traits. What percentage/proportion of the plants in the F2 generation/progeny were round, in Mendel\u2019s cross between round and wrinkled pea plants? (Foreign 2016)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAnswer:\nDifferences between dominant traits and recessive trait are given below:\nOut of total 4 genotypes possible in F2 generation 31 genotypes result in phenotypic expression of round seeds. So, the percentage of plants with round seeds will be 75%. This can be illustrated as follows:\n\nQuestion 22.\nExplain Mendel\u2019s experiment with peas on inheritance of characters considering only one visible contrasting character. (Foreign 2016, 2014)\nAnswer:\nMendel crossed a pure tall pea plant with pure dwarf pea plant. All the plants obtained in F1 generation were tall. When Mendel selfed plants from F1 generation then he obtained both tall and dwarf plants in F2 generation in the ratio of 3 : 1. This can be illustrated as follows :", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nThis explains that for each pair of contrasting characters there are two alleles. The trait which is expressed in F1 is dominant trait and is controlled by dominant allele and the trait which remains unexpressed in F1 is the recessive trait and is controlled by recessive gene. When both the contrasting alleles are present together in F1 individuals, no mixing of alleles occurs and they again segregate at the time of gamete formation Therefore, when the recessive alleles come together they result in reappearance of recessive trait in F2 generation.\nQuestion 23.\n\u201cIt is a matter of chance whether a couple will have a male or a female child.\u201d Justify this statement by drawing a flow chart. (Foreign 2015)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 23.\n\u201cIt is a matter of chance whether a couple will have a male or a female child.\u201d Justify this statement by drawing a flow chart. (Foreign 2015)\nAnswer:\nSex is determined at the time of fertilisation when male and female gametes fuse. Male produces two types of gametes, i.e., having X or Y chromosome and female produces same type of gametes containing X chromosomes. The sex of the child is determined at the time of fertilisation when male and female gametes fuse to form zygote.\nIf a sperm (male gamete) carrying X chromosome fertilises an egg or ovum (female gamete) carrying X chromosome, then the offspring will be a girl (female). This is because the offspring will have XX combination of sex chromosomes.\nIf a sperm (male gamete) carrying Y chromosome fertilises an egg or ovum (female gamete) which has X chromosome, then the offspring will be a boy (male). This is because the offspring will have XY combination of sex chromosomes.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nTherefore, there are 50% chance of a male child and 50% chance of a female child.\n\nQuestion 24.\n\u201cIt is possible that a trait is inherited but may not be expressed.\u201d Give a suitable example to justify this statement. (Foreign 2015)\nAnswer:\nRefer to answer 20.\nQuestion 25.\nA cross was made between pure breeding pea plants, one with round and green seeds and the other with wrinkled and yellow seeds.\n(a) Write the phenotype of F1 progeny. Give reason for your answer.\n(b) Write the different types of F2 progeny obtained along with their ratio when F1 progeny was selfed. (Delhi 2014, Delhi 2013C)\nAnswer:\n(a) The given cross was made between pure breeding pea plants, one with round and green seeds and the other with wrinkled and yellow seeds.\nYellow seed colour and round seed shape is dominant over green seed colour and wrinkled seed shape. In F1 generation, dominant traits express itself, whereas recessive traits get suppressed.\n\nTherefore, the phenotype of F1 progeny is round and yellow.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nTherefore, the phenotype of F1 progeny is round and yellow.\n(b) The different types of F2 progeny obtained along with their ratio when F1 progeny was selfed could be illustrated by the given cross.\n\nPhenotypic ratio : 9 : 3 : 3 : 1\nRound yellow seeds \u2013 9 ; Round green seeds \u2013 3;\nWrinkled yellow seeds \u2013 3; Wrinkled green seeds \u2013 1\nQuestion 26.\n(a) Mendel crossed tall pea plants with dwarf pea plants in his experiment. Write his observations giving reasons on the F1 and F2 generations.\n(b) List any two contrasting characters other than height that Mendel used in his experiments in pea plants. (Delhi 2014)\nAnswer:\n(a) The possible cross of Mendel\u2019s experiment is:\n\nHence, tall (T) is dominant whereas dwarf (t) is recessive. In F1 generation, only dominant trait expresses itself, whereas recessive trait gets suppressed.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nHence, tall (T) is dominant whereas dwarf (t) is recessive. In F1 generation, only dominant trait expresses itself, whereas recessive trait gets suppressed.\nIn F2 generation, both traits, i.e., dominant and recessive express themselves. In this way, Mendels experiment showed that the traits (tall and dwarf) are inherited independently.\n(b) The two contrasting characters other than height that Mendel used in his experiment in pea plants are round/wrinkled seeds and violet/white flowers.\nQuestion 27.\n\u201cA trait may be inherited, but may not be expressed\u201d. Justify this statement with the help of a suitable example. (AI 2014)\nAnswer:\nRefer to answer 20.\nQuestion 28.\n\u201cThe sex of a newborn child is a matter of chance and none of the parents may be considered responsible for it\u201d. Justify this statement with the help of flow chart showing determination of sex of a new born. (Delhi 2013)\nAnswer:\nRefer to answer 23.\nQuestion 29.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 23.\nQuestion 29.\nA blue colour flower plant denoted by BB is cross-bred with that of white colour flower plant denoted by bb.\n(a) State the colour of flower you would expect in their F1 generation plants.\n(b) What must be the percentage of white flower plants in F2 generation if flowers of F1 plants are self-pollinated?\n(c) State the expected ratio of the genotypes BB and Bb in the F2 progeny. (Delhi 2012)\nAnswer:\n(a) The colour of the flower in F1 generation will be blue.\n(b) If the flowers of F1 generation are self pollinated, then the percentage of white flowers in F2 generation must be 25%.\n(c) The expected ratio of the genotypes BB and Bb in the F2 generation progeny is 1 : 2.\nThe above results could be depicted by the given cross:\n\nQuestion 30.\nIf we cross pure-breed tall (dominant) pea plant with pure-breed dwarf (recessive) pea plant we will get pea plants of F1 generation.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 30.\nIf we cross pure-breed tall (dominant) pea plant with pure-breed dwarf (recessive) pea plant we will get pea plants of F1 generation.\nIf we now self-cross the pea plant of F1 generation, then we obtain pea plants of F2 generation.\n(a) What do the plants of F1 generation look like?\n(b) State the ratio of tall plants to dwarf plants in F2 generation.\n(c) State the type of plants not found in F1 generation but appeared in F2 generation, mentioning the reason for the same. (AI 2012)\nAnswer:\n(a) The plants of F1 generation will be all tall plants.\n(b) The ratio of tall plants to dwarf plants in F2 generation is 3 : 1.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAnswer:\n(a) The plants of F1 generation will be all tall plants.\n(b) The ratio of tall plants to dwarf plants in F2 generation is 3 : 1.\n(c) Dwarf plants are not found in F1 generation but appeared in F2 general ion. This is so because in F1 generation only dominant trait (tall) expresses itself and recessive trait (dwarf) gets suppressed. The dwarf plants appeared in F2 generation, because the traits whether dominant or recessive are independently inherited. In others words, a single copy of (T) is enough to make the plant tall, while both copies have to be (t) for the plant to be dwarf.\n\nQuestion 31.\nWith the help of a flow chart explain in brief how the sex of a newborn is genetically determined in human beings. Which of the two parents, the mother or the father, is responsible for determination of sex of a child? (Foreign 2012)\nAnswer:\nRefer to answer 23.\nQuestion 32.\nHow is the sex of the child fixed during the fertilisation, step in human beings? Explain. (Foreign 2011)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 23.\nQuestion 32.\nHow is the sex of the child fixed during the fertilisation, step in human beings? Explain. (Foreign 2011)\nAnswer:\nHuman beings have 23 pairs chromosomes (22 pairs of autosomes +1 pair of sex chromosome). A male has one X chromosome and one Y chromosome whereas a female has two X chromosomes. Sex of a child depends on the two conditions which takes place during fertilisation. The two conditions are given below:\nIf a sperm carrying X chromosome fertilises an ovum which carries X chromosome, then the child born will be girl.\nIf a sperm carrying Y chromosome fertilises an ovum which carries X chromosome, then the child born will be a boy.\nQuestion 33.\nHow do Mendels experiments show that\n(a) traits may be dominant or recessive?\n(b) inheritance of two traits is independent of each other? (Delhi 2017)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 33.\nHow do Mendels experiments show that\n(a) traits may be dominant or recessive?\n(b) inheritance of two traits is independent of each other? (Delhi 2017)\nAnswer:\n(a) Mendel first crossed pure-bred tall pea plants with pure-bred dwarf pea plants and found that only tall pea plants were produced in the first generation (F1). He then self crossed the tall pea plants of the F1 generation and found that tall plants and dwarf plants were obtained in the second generation or (F2) in the ratio of 3 : 1. Mendel said that the trait of dwarfness of one of the parent pea plant had not been lost, it was merely concealed or suppressed in the first generation to re-emerge in the second generation. He called the suppressed trait of \u2018dwarfness as \u2018recessive trait\u2019 and the expressed trait of \u2018tallness\u2019 as the \u2018dominant trait\u2019. In this way, Mendel\u2019s experiments with tall and dwarf pea plants showed that the traits may be dominant or recessive.\nHence this could be explained by the given cross :", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nHence this could be explained by the given cross :\n\n(b) Mendel observed two combinations of characteristics in seeds, round-yellow and wrinkled-green, and two new combinations of characteristics had appeared in the F2 generation, round-green and wrinkled- yellow. According to Mendel\u2019s second law of inheritance more than one pair of traits are considered in a cross simultaneously, the factors responsible for each pair of trait are distributed independently to the gametes.\nThe cross given below showing dihybrid cross explains that the inheritance of two traits is independent of each other.\n\nQuestion 34.\n(a) Why did Mendel choose garden pea for his experiments ? Write two reasons.\n(b) List two contrasting visible characters of garden pea Mendel used for his experiment.\n(c) Explain in brief how Mendel interpreted his results to show that the traits may be dominant or recessive. (Foreign 2016)\nAnswer:\n(a) Mendel choose garden pea for his experiments because:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(c) Explain in brief how Mendel interpreted his results to show that the traits may be dominant or recessive. (Foreign 2016)\nAnswer:\n(a) Mendel choose garden pea for his experiments because:\n(i) It was easy to grow and it shows some clear contrasting traits like some pea plants are tall whereas others are dwarf.\n(ii) Pea plants are self pollinating and many generation of pea plants can be produced in comparatively less time.\n(b) The contrasting characters of garden pea plant studied by Mendel are:\n(c) Refer to answer 17.\nQuestion 35.\nGive the respective scientific terms used for studying:\n(a) the mechanism by which variations are created and inherited and\n(b) the development of new type of organisms from the existing ones. (Delhi 2014)\nAnswer:\n(a) Heredity is the mechanism by which variations are created and inherited.\n(b) Evolution is the process in which development of new type of organisms takes place from the existing ones.\nQuestion 36.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(b) Evolution is the process in which development of new type of organisms takes place from the existing ones.\nQuestion 36.\nWrite the contribution of Charles Darwin in the field of evolution. (Delhi 2014)\nAnswer:\n\u2018Lhe contribution of Charles Darwin in the field of evolution is that only the organism which can adapt; to the changing environmental conditions can survive and he also stated that the natural selection is a process which plays an important role in evolution of plants and animals by selecting the organisms with traits favourable to the environment.\nQuestion 37.\nWhy do mice whose tails were surgically removed just after birth for generations, continue to produce mice with tails? (Foreign 2014)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 37.\nWhy do mice whose tails were surgically removed just after birth for generations, continue to produce mice with tails? (Foreign 2014)\nAnswer:\nA group of mice are normally bred and all their progeny possess tail. Now, if the tails of these mice are surgically removed in each generation, tail\u00acless mice will not be produced. This is so because removal of tail during the life cycle of mice is an acquired character which is not transferred to future generations. The removal of the tail cannot change the genes of the germ cells of the mice.\nQuestion 38.\nList two differences between acquired traits and inherited traits by giving an example of each. (Delhi 2019)\nAnswer:\nDifferences between the inherited traits and acquired traits are as follows :\nQuestion 39.\nDefine genetics. Why is decrease in the number of surviving tigers a cause of concern from the point of view of genetics? Explain briefly. (AI 2019)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 39.\nDefine genetics. Why is decrease in the number of surviving tigers a cause of concern from the point of view of genetics? Explain briefly. (AI 2019)\nAnswer:\nGenetics is the branch of biology that deals with the study of heredity and variations. The term genetics\u2019 was coined by William Bateson in 1906. When a population is small, the number and scope of variations is limited and hence diversity and traits are reduced. Small numbers of surviving tigers are a cause of worry from the point of genetics because of the following reasons:\nTheir loss would cause a loss of gene pool, i.e., many genes will be eliminated from a gene pool.\nTigers are surviving in limited numbers, so if some natural calamity kills these small population of tigers, they will suddenly become extinct as per genetic drift phenomenon,", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nTigers are surviving in limited numbers, so if some natural calamity kills these small population of tigers, they will suddenly become extinct as per genetic drift phenomenon,\nA disease may wipe out the leftover population, if the entire population is susceptible to the disease. This can cause sudden extinction of the tiger species and loss of their genes forever, thus, adversely affecting the diversity of nature.\nQuestion 40.\nDistinguish between the acquired traits and the, inherited traits in tabular form, giving one example for each. (Delhi 2017)\nAnswer:\nRefer to answer 38.\nQuestion 41.\nWith the help of two suitable examples, explain why certain experiences and traits earned by people during their lifetime are not passed on to their next generations. When can such traits be passed on? (AI 2017)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAnswer:\nCertain experiences and traits earned by people during their lifetime are not passed on to their next generations because all these characters are acquired by the man during his lifetime. The man is not born with these traits and he cannot pass on these traits to his children. These could be better explained by the given examples:\n(i) Child of a very good swimmer may not know how to swim. This is so because the technique of swimming is not inherited from parents but it is learnt by the person himself or herself.\n(ii) A person may have a scar on the face from a cut he got in an accident. This is also an example of acquired trait which cannot be passed to the next generation.\nTraits can be passed on to future generations in which changes have occurred in the genes present in the reproductive cells of the parent organisms. These traits or characters are known as inherited traits.\nQuestion 42.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 42.\nList three distinguishing features, in tabular form, between acquired traits and the inherited traits. (Delhi 2016)\nAnswer:\nRefer to answer 38.\nQuestion 43.\n\u201cWe cannot pass on to our progeny the experience and qualifications earned during our lifetime.\u201d Justify the statement giving reason and examples. (Delhi 2015)\nAnswer:\nRefer to answer 41.\nQuestion 44.\nList in tabular form, two distinguishing features between the acquired traits and the inherited traits with one example of each. (Delhi 2015, AI 2012)\nAnswer:\nRefer to answer 38.\nQuestion 45.\nDistinguish between inherited and acquired traits by giving one example of each. Give reason why the traits acquired during the lifetime of an individual are not inherited. (Foreign 2014)\nAnswer:\nRefer to answers 38 and 41.\nQuestion 46.\nWith the help of suitable examples, explain why certain traits cannot be passed on to the next generation? What are such traits called? (AI 2014)\nAnswer:\nRefer to answer 41.\nQuestion 47.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 46.\nWith the help of suitable examples, explain why certain traits cannot be passed on to the next generation? What are such traits called? (AI 2014)\nAnswer:\nRefer to answer 41.\nQuestion 47.\nTabulate two distinguishing features between acquired traits and inherited traits with one example of each. (Delhi 2013)\nAnswer:\nRefer to answer 38.\nQuestion 48.\n\u201cAn individual cannot pass on to its progeny the experiences of its life-time\u201d. Justify the statement with the help of an example and also give reason for the same. (Foreign 2012)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n\u201cAn individual cannot pass on to its progeny the experiences of its life-time\u201d. Justify the statement with the help of an example and also give reason for the same. (Foreign 2012)\nAnswer:\nIf we breed a group of mice, all the progeny of mice will have tails just like their parents. Now, if we remove the tails surgically and again breed them, we still get new mice with tails. This is because cutting the tails of mice does not change the genes of their reproductive cells (or gametes). And since the acquired trait of\u2018cut tails\u2019 does not bring about a change in the genes of mice, this trait cannot be passed on to their next generations. From this we conclude that the experiences acquired by an individual during his lifetime (called acquired traits) cannot be passed on to its progeny, and hence cannot lead to evolution because they are not caused by the change in genes.\nQuestion 49.\nDescribe any three ways in which individuals with a particular trait may increase in population. (AI 2011)", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 49.\nDescribe any three ways in which individuals with a particular trait may increase in population. (AI 2011)\nAnswer:\nThree ways in which individuals with a particular trait may increase in population are:\n(i) Genetic drift : It is the random change in gene frequency occurring by chance fluctuations. In this phenomenon, an event may increase the frequency of a particular trait having little adaptive value and survival advantage.\n(ii) Natural selection : It is the phenomenon wherein nature selects traits favourable to the species in the environment. Thus, a particular trait selected by the nature increases in number.\n(iii) Geographical isolation : Interbreeding populations are geographically isolated by barriers such as mountain ranges, rivers and sea. This geographical isolation leads to reproductive isolation and thus there is no gene flow between separated groups of population and therefore, population with particular character increases.\nQuestion 50.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 50.\n(a) What is the law of dominance of traits? Explain with an example.\n(b) Why are the traits acquired during the life time of an individual not inherited? Explain. (2020)\nAnswer:\n(a) Mendel\u2019s law of dominance states that for a particular trait one gene is usually expressed over the other. The dominant allele expresses itself and the one which remains unexpressed is called recessive allele. This is called \u201claw of dominance\u201d. E.g., for height, there are two alleles, T for tallness and t for dwarfness.\n(b) Refer to answer 41.\nQuestion 51.\nWhat is speciation?\nAnswer:\nThe process by which new species develop from the existing species is known as speciation.\nQuestion 52.\nWhat is speciation? Explain in brief the role of natural selection and genetic drift in this process. (Foreign 2016)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 52.\nWhat is speciation? Explain in brief the role of natural selection and genetic drift in this process. (Foreign 2016)\nAnswer:\n\u2018Hie process by which new species develop from the existing species is known as speciation. Both natural selection and genetic drift play a very important role in speciation. By natural selection traits favourable to the species in the given environment are selected by the nature.\nExample, beetles develop colour variation during reproduction. Consequently, in the progeny, one beetle develops green body colour instead of normal red colour. This beetle can inherit this variation in colour on to its progeny so that all its offspring are green in colour. Crows cannot easily locate green-coloured beetles on the green leaves of bushes and continue to eat red beetles which are easily located on green leaves. As a result, in population of green beetles increases in each generation.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nGenetic drift is the elimination of the genes of certain traits from the small population when a section of population dies of natural calamity or migrates to other region. It alters the gene frequency of the remaining population.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nSuppose in sexually reproducing red beetle population, a colour variation arises wherein one beetle develops blue body colour instead of red. This beetle can also pass this colour variation to its progeny so that all its offspring are blue coloured beetles. As the population of beetles expand, initially there would be few blue coloured beetles among the majority of red coloured beetles. At this point, an elephant comes by and stamps on the bushes where the beetles live. Consequently, most of the beetles get killed. By chance, most of the survived beetles are of blue colour. This population again slowly multiply and will contain mostly blue coloured beetles over a period of time. Survival of more blue beetles in the population changed the coloured characteristic from normal red to blue over a period of time. In small population, accidents can change the frequency of some genes in a population, even if it does not give any survival advantage to the possessors.\nQuestion 53.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 53.\nWhat is speciation? List four factors responsible for speciation. (Delhi 2015)\nAnswer:\nThe process by which new species develop from the existing species is known as speciation. The four factors responsible for speciation are:\nGeographical isolation of a population leads to reproductive isolation and hence speciation.\nGenetic drift leads to the formation of new species by causing drastic changes in the frequencies of particular gene by chance alone.\nNatural selection wherein nature in the wild selects traits favourable to the species in the given environment.\nRandom mating resulting in genetic variations in genotype frequencies within the population.\nQuestion 54.\nExplain the following:\n(a) Speciation\n(b) Natural selection. (Al 2015, Delhi 2011)\nAnswer:\n(a) Speciation : Origin of new species from existing one due to reproductive isolation of a part of its population is called speciation.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(b) Natural selection. (Al 2015, Delhi 2011)\nAnswer:\n(a) Speciation : Origin of new species from existing one due to reproductive isolation of a part of its population is called speciation.\nSuppose a large population of beetle living in an area gets split into two sub populations due to geographical barriers like river, mountain, sea, etc. Then members of these sub-populations will not be able to interbreed. Hence, there will be no gene flow between the members of these populations. After few generations, genetic drift will accumulate different variations in each of the two geographically separated sub-populations. Natural selection may also operate simultaneously in a different way in these geographically isolated sub-populations. This makes geographically isolated sub-populations to become more and more different from each other and ultimately reproductive isolation occurs between individuals of these groups and they transform into new species.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(b) Natural selection is the phenomenon wherein nature, in the wild, selects traits favourable to the species in its environment.\nAlso refer to answer 52.\nQuestion 55.\nList three main factors responsible for the speciation and briefly describe each one of them. (AI 2014)\nAnswer:\nSpeciation is the process by which new species develop from the existing species.\nThe three important factors responsible for the speciation are:\n(i) Geographical isolation of a population caused by various types of barriers such as mountain ranges, rivers and sea. Geographical isolation is the major factor in the speciation of sexually reproducing animals because it causes reproductive isolation and interrupts the flow of genes between their isolated populations through the gametes.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(ii) Genetic drift is the elimination of the genes of certain traits from the small population when a section of population dies of natural calamity or migrates to other region. It alters the gene frequency of the remaining population.\n(iii) Variations introduced by random mating and hybridisation. Random fusion of gametes adds new alleles and it result in formation of new species.\nQuestion 56.\nList three main factors responsible for the rise of a new species giving a brief description about each. (Foreign 2014)\nAnswer:\nRefer to answer 55.\nQuestion 57.\nWhat is meant by the term speciation? List four factors which could lead to speciation. (Delhi 2012)\nAnswer:\nRefer to answer 53.\nQuestion 58.\n(a) Classify the following as homologous or analogous pairs:\n(i) Broccoli and cabbage\n(ii) Ginger and radish\n(iii) Forelimbs of birds and lizard\n(iv) Wings of a bat and wings of a bird\n(b) State the main feature that categorises a given pair of organs as homologous or analogous. (2020)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(iii) Forelimbs of birds and lizard\n(iv) Wings of a bat and wings of a bird\n(b) State the main feature that categorises a given pair of organs as homologous or analogous. (2020)\nAnswer:\n(a) (i) Broccoli and cabbage \u2013 Homologous organs\n(ii) Ginger and radish \u2013 Analogous organs\n(iii) Forelimbs of birds and lizard \u2013 Homologous organs\n(iv) Wings of bat and wings of a bird \u2013 Analogous organs\n(b) Homologous organs have the same fundamental structure but different in function, while analogous organs have different internal structure but similar functions.\nQuestion 59.\n\u201cDuring the course of evolution, organs or features may be adapted for new functions\u201d. Explain this fact by choosing an appropriate example. (2020)\nAnswer:\nEvolution is the sequence of gradual changes from simple life forms to complex life form. During the evolution process, many organs are adapted for new functions.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAnswer:\nEvolution is the sequence of gradual changes from simple life forms to complex life form. During the evolution process, many organs are adapted for new functions.\nThe organs which perform different functions in different species but have similar basic structure and similar embryonic origin are called homologous organs. Homologous structures are a result of divergent evolution. Homology indicates common ancestry.\nFor example, the forelimbs of a frog, a lizard, a bird and a human being show similarity in basic structure. However, these organs perform entirely different functions.\nThe given figures showing homologous organs :", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nThe given figures showing homologous organs :\n\nThe forelimbs of a frog help the animal to prop up the front end of the body at rest, and also act as shock absorbers when the animal lands on the ground after a leap; the forelimbs of a lizard are modified for creeping movements; the forelimbs of a bird are modified for flying purposes, while the forelimbs of a human being are used for grasping. The basio similarity in the forelimbs of these different vertebrates indicates that all these vertebrates had common ancestry, i.e., they have evolved from a common ancestor who had \u2018five digited\u2019 or \u2018pentadactyl\u2019 limbs (basic plan of limbs).\nThis means that these vertebrates have modified according to the special needs of the subsequent generations during the course of evolution.\nQuestion 60.\nExplain with the help of an example each, how the following provide evidences in favour of evolution:\n(a) Homologous organs\n(b) Analogous organs\n(c) Fossils (Delhi 2017, AI 2015, Delhi 2011)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nExplain with the help of an example each, how the following provide evidences in favour of evolution:\n(a) Homologous organs\n(b) Analogous organs\n(c) Fossils (Delhi 2017, AI 2015, Delhi 2011)\nAnswer:\n(a) Homologous organs perform different functions in different species but have similar basic structure and similar embryonic origin. For example, forelimbs of a frog, lizard, bird and human being show similarity in basic structure. However, these organs perform entirely different functions but the basic similarity in the forelimbs of these different vertebrates indicates common ancestry. This means that these vertebrates have modified according to the special needs of the subsequent generations during the course of evolution.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(b) Analogous organs are different in fundamental structure and embryonic origin but have similar appearance and perform similar functions. For example, the wings of an insect and a bird have different structures but they perform the same function of flying. The presence of these organs indicates that they are not derived from common ancestors but they can still evolve to perform similar functions to survive, flourish and keep on evolving in the prevailing environment. This provides a mechanism for evolution.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(c) Fossils are the remains or impressions of the dead animals and plants that lived in the remote past. The fossils provide evidence for evolution. For example, a fossil bird called Archaeopteryx have characters of both reptiles and birds. It had feathers, fused bones and beak which are exclusive characters of birds and had teeth in the jaws, claws on fingers, a long tail, etc like reptiles. Thus, Archaeopteryx is a connecting link between the reptiles and birds, and hence suggests that the birds have evolved from the reptiles through the process of continuous evolution.\nQuestion 61.\n\u201cEvolution and classification of organisms are interlinked\u201d. Give reasons to justify this statement. (AI 2017)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAnswer:\nEvolution is the process by which newer types of organisms are developed from the pre-existing ones through modifications. Classification is the arrangement of organisms into a series of groups based on physiological, biological, anatomical or other relationships. All systems of classification are hierarchial. The more closely two species are related, the more recently they have a common ancestor. Classification of organisms necessarily involves organising them in different groups, based on the similarities and differences of characteristics. It helps in the recognisation of the basic arrangement of a hierarchical structure among diverse species. It facilitates studies or research of wide variety associated with organisms effortlessly. Thus, classification of species is infact a reflection of their evolutionary relationship or we can say that evolution and classification are interlinked.\nQuestion 62.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 62.\n\u201cTwo areas of study namely evolution\u2019 and classification are interlinked\u201d. Justify this statement. (AI 2016)\nAnswer:\nRefer to answer 61.\nQuestion 63.\nList three factors that provide evidences in favour of evolution in organisms and state the role of each in brief. (Foreign 2016)\nAnswer:\nSome of the important sources which provide evidences for evolution are homologous organs, analogous organs and fossils.\nAlso refer to answer 60.\nQuestion 64.\n(a) Planaria, insects, octopus and vertebrates all have eyes. Can we group eyes of these animals together to establish a common evolutionary origin? Justify your answer.\n(b) \u201cBirds have evolved from reptiles\u201d. State evidence to prove the statement. (Delhi 2015)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(b) \u201cBirds have evolved from reptiles\u201d. State evidence to prove the statement. (Delhi 2015)\nAnswer:\n(a) Planaria, insects, octopus and vertebrates cannot be grouped together on the basis of eye. Eyes of insects, octopus, Planaria and vertebrates are analogous organs which have developed over generation as an adaptation for similar function. They represent convergent evolution where distantly related groups develop similar functional structure as an adaptation for same function.\n(b) Fossil bird Archaeopteryx had features like feathers, fused bones, beak which are the characteristic features of birds. It also had some features of reptiles, like, teeth in jaws, claws on free fingers, a long tail, etc. Thus, it represents a connecting link between reptiles and birds. This example provides a clue that birds have evolved from reptiles.\nQuestion 65.\n(a) Cite the evidence on the basis of which it is concluded that birds have evolved from reptiles.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 65.\n(a) Cite the evidence on the basis of which it is concluded that birds have evolved from reptiles.\n(b) Insects, octopus, Planaria and vertebrates also possess eyes. Can these animals be grouped together on the basis of the eyes they possess. Why or why not ? Give reason to justify your answer. (Foreign 2015)\nAnswer:\n(a) Refer to answer 64(b).\n(b) Refer to answer 64(a).\nQuestion 66.\n(a) Give the evidence that the birds have evolved from reptiles.\n(b) Insects, octopus, Planaria and vertebrates possess eyes. Can we group these animals together on the basis of eyes that they possess? Justify your answer giving reason. (Delhi 2014)\nAnswer:\n(a) Refer to answer 64(b).\n(b) Refer to answer 64(a).\nQuestion 67.\nWhat are fossils? How do they help in the study of evolution? (Delhi 2013C, AI 2011)\nAnswer:\nRefer to answer 60(c).\nQuestion 68.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(a) Refer to answer 64(b).\n(b) Refer to answer 64(a).\nQuestion 67.\nWhat are fossils? How do they help in the study of evolution? (Delhi 2013C, AI 2011)\nAnswer:\nRefer to answer 60(c).\nQuestion 68.\nDistinguish between homologous organs and analogous organs. In which category would you place wings of a bird and wings of a bat? Justify your answer giving a suitable reason. (Delhi 2012)\nAnswer:\nThose organs which have the same basic structure and similar embryonic origin but different functions are called as homologous organs.\nThese organs follow same basic plant of organisation during development, but in adults got modified to perform different function as an adaptation to different environments. Those organs which have different basic structure and embryonic origin but perform similar functions are called analogous organs.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nThe wings of bird and wings of bat are analogous organs because the basic structure or design of the wings of bird and bat are different but they perform the similar function of flying.\nQuestion 69.\nHow are fossils formed? Describe, in brief, two methods of determining the age of fossils. (AI 2012)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nWhen organisms like plants or animals die, their bodies get decomposed by the action of microorganisms in the presence of oxygen, moisture, etc. Sometimes the conditions in the environment are such (like absence of oxygen, moisture, etc.), which do not let the body of the organism to get decompose completely. It is the body (or body part) of an organism which we get as fossil on digging the earth. In many cases the soft parts of the organisms get decomposed and we get skeleton of hard parts (like teeth, bones, etc.) as fossil. Even the soft parts of the plants and animals (which usually decompose quickly) are sometimes preserved as fossils in the form of their impressions inside the rocks. For example, if a dead leaf gets caught in mud, it will not decompose quickly. The mud around the leaf will set around it as a mould,\u2019gradually harden to form a rock and retain the impression of the leaf. This forms a leaf fossil which can be dug out from the earth a after a long time period.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nThe age of fossils can be determined in two ways as given below:\n(i) By the relative method: When we dig into the earth, we find fossils at different depths. The fossils which we find in layers closer to the surface of the earth are more recent and those fossils which are found in deeper layers are older; whereas the fossils found in the deepest layers of earth are the oldest ones.\n(ii) Carbon dating method: When a living object dies and forms fossil, its carbon-14 radioactivity goes on decreasing gradually. By this method, the age of fossils is found by comparing the carbon-14 radioactivity left in fossils with the carbon-14 radioactivity present in living objects today.\nQuestion 70.\n(a) We see eyes in Planaria, insects, octopus and vertebrates. Can eyes be grouped together in case of the above-mentioned animals to establish a common evolutionary origin? Why?\n(b) State one evidence to prove that birds have evolved from reptiles. (Foreign 2012)\nAnswer:\n(a) Refer to answer 64 (a).", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n(b) State one evidence to prove that birds have evolved from reptiles. (Foreign 2012)\nAnswer:\n(a) Refer to answer 64 (a).\n(b) Refer to answer 64 (b).\nQuestion 71.\nExplain how evolutionary relationship can be traced by the study of homologous organs. (Foreign 2011)\nAnswer:\nHomologous organs provide morphological and anatomical evidences of evolution. There are number of organs in different groups of animals or plants which have similar basic design but are used for different purposes. These are termed as homologous organs. For example, the forelimbs of frog, lizard, bird and human beings show similarity in basic structure. The basic similarity of forelimbs of these different vertebrates indicates that all these have evolved from a common ancestor who had five digited or pentadactyle limbs which became modified according to the special needs of subsequent generations during the course of evolution. Hence, homologous organs depict divergent evolution or adaptive radiation.\nQuestion 72.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 72.\n(a) How do the following provide evidences in favour of evolution in organisms? Explain with an example for each.\n(i) Homologous organs\n(ii) Analogous organs\n(iii) Fossils\n(b) Explain two methods to determine the age of fossils. (AI 2019)\nAnswer:\n(a) Refer to answer 60.\n(b) Refer to answer 69.\nQuestion 73.\nDefine evolution. How does it occur?\nDescribe how fossils provide us evidence in support of evolution. (AI 2016)\nAnswer:", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\n\nEvolution is the sequence of gradual changes, from simple life forms to complex life form, i.e., from primitive organisms that lived over millions of years ago to new organisms that exist today. Evolution occurs by changes, improvement and modification of simple life forms. Fossils are the remains or impressions of organisms that lived in the remote past. Fossils provide the evidence that the present animal have originated from previously existing ones through the process of continuous evolution. Fossils can be used to reconstruct evolutionary history of an organism. The distribution pattern of fossils shows that the ancient fossils present in the bottom rocks are simple, while the most recent fossils found in the upper strata are more highly evolved. It means fossils form and become more and more complex as we proceed from earliest to recent rocks. It gives us an idea of time in history when different species were formed or became extinct. Thus, fossils provide us evidences in support of evolution.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nQuestion 74.\nWhat are fossils? How are they formed? List two methods of determining the age of fossils. Explain in brief the importance of fossils in deciding the evolutionary relationships. (Foreign 2016)\nAnswer:\nRefer to answers 60(c) and 69.\nQuestion 75.\nDefine the term evolution. \u201cEvolution cannot be equated with progress\u201d. Justify this statement. (2020)\nAnswer:\nEvolution is the sequence of gradual changes which take place in the primitive organisms over millions of years in which new species are produced.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nAnswer:\nEvolution is the sequence of gradual changes which take place in the primitive organisms over millions of years in which new species are produced.\nEvolution should not be equated with progress because there is no real progress in the concept of evolution. Evolution is just the production of diversity of life forms and shaping of this diversity by the environmental selection. The only progress in evolution appears to be that more and more complex body designs of organisms have emerged over the ages. This will become clear from the following examples. When a new species is formed, it is not necessary that the old species will disappear or get eliminated from earth.", + "CHAPTER 8-HEREDITY -IMPORTANT QUESTIONS.txt\nIt will all depend on the environment. Also it is not as if the newly formed species are in any way better than the older ones. It is simply that genetic drift and natural selection processes have combined to form a population having different body design which cannot interbreed with the older population. It is a common belief that chimpanzees are the ancestors of human beings. It is, however, not true that human beings have evolved from chimpanzees. Actually, both chimpanzees and human beings had a common ancestor long time ago. The two offsprings of that ancestor evolved in their own separated ways to form the modern day chimpanzees and human beings.\nQuestion 76.\nDefine the term \u201cevolution\u201d. Evolution should not be equated with progress.\u201d Give reason to justify this statement. (Foreign 2014)\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 9 Intext Questions\nPage Number: 143\nQuestion 1.\nIf a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% of the same population, which trait is likely to have arisen earlier ?\nAnswer:\nTrait B, because it is present in more members of the population. It is likely to have arisen earlier and has now spread to 60% of the population. Trait A is new and has spread to only 10% of the population.\nQuestion 2.\nHow does the creation of variations in a species promote survival ?\nAnswer:\nThe variations provide stability to the population of various species by preventing them from getting wiped out during adverse conditions.\nThe natural environment also changes, and variations in species which become suited to the environment help it to survive.\nPage Number: 147\nQuestion 1.\nHow do Mendel\u2019s experiments show that traits may be dominant or recessive ? [AICBSE 2015]\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nPage Number: 147\nQuestion 1.\nHow do Mendel\u2019s experiments show that traits may be dominant or recessive ? [AICBSE 2015]\nAnswer:\nMendel took pea plants with contrasting characteristics tall plant and dwarf (or short) plant. On cross pollination, he got all tall plants in first generation (F1). But by the self\u00acpollination of F1 tall plants, the plants of second generation consisted of tall and short pants in the ratio of 3 : 1. On the basis of these experiments, the characteristics appeared in first generation were called dominant (i.e. tall plants) and the characteristics that did not appear were called recessive (dwarf i.e. plants).\nQuestion 2.\nHow do Mendel\u2019s experiments show that traits are inherited independently ? [AICBSE 2015]\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 2.\nHow do Mendel\u2019s experiments show that traits are inherited independently ? [AICBSE 2015]\nAnswer:\nMendel took two pairs of alternate expression of two traits and carried out dihybrid crosses by crossing them. The traits appeared in first generation were termed as dominant. When he used these F1 progeny to generate F2 progeny by self-pollination plants of different types were produced. In some plants both the traits were dominant, while in some plants both were recessive and some plants exhibited mixed traits. This indicates that traits are inherited independently.\nQuestion 3.\nA man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits \u2013 blood group A or O \u2013 is dominant ? Why or why not ?\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nAnswer:\nThis information is not enough. This is because each individual is carrying two alleles. The recessive trait can occur only when who alleles are similar. It blood group A is dominant and O is recessive, then daughter can have blood group O only when both recessive alleles occur together in mother, and father has one allele of O and other of A.\nQuestion 4.\nHow is the sex of the child determined in human beings ?\nOR\n\u201cThe sex of a newborn child is a matter of chance and none of the parents may be considered responsible for it.\u201d Justify this statement with the help of a flow chart showing determination of sex of a newborn. [CBSE (Delhi) 2013]\nAnswer:\nHalf of the male gametes (sperms) carry X chromosome and other half carry Y chromosomes. All the female gametes carry only X chromosomes. When a sperm fertilizes an egg, the following situations become possible.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\n(i) When a sperm carrying X chromosome fertilises an egg that contains only X chromosome), the resulting zygote develops into a female (XX condition).\n(ii) When a sperm carrying Y chromosome fertilises an egg (that contains only X chromosome), the resulting zygote develops into a male (XY condition).\nThus there are 50 \u2013 50 chances of a male or female child and none of the parents may Sex determination in humans be considered responsible for it.\nThe sex-determination mechanism is shown alongside.\n\nPage Number: 150\nQuestion 1.\nWhat are the different ways in which individuals with a particular trait may increase in a population ?\nAnswer:\nDifferent ways in which individuals with a particular trait may increase in a population are as follow :\nIf it gives the benefit of survival through natural selection.\nDue to a sudden increase in a particular trait in a population, i.e., by genetic drift.\nQuestion 2.\nWhy are traits acquired during the life-time of an individual not inherited ?\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nDue to a sudden increase in a particular trait in a population, i.e., by genetic drift.\nQuestion 2.\nWhy are traits acquired during the life-time of an individual not inherited ?\nAnswer:\nThe traits acquired during the life-time are changes in the non-reproductive cells of the organisms and are not capable of being passed on to the next generation.\nQuestion 3.\nWhy are the small numbers of surviving tigers a cause of worry from the point of view of genetics ?\nAnswer:\nThe small numbers of surviving tigers are a cause of worry from the point of view of genetics because in tigers there are negligible genetic variations. Due to this they are not well adapted. The rapid environmental changes cannot be favouable for them. If these changes are not controlled, tigers would be wiped out.\nPage Number: 151\nQuestion 1.\nWhat factors would lead to the rise of a new species ?\nAnswer:\nThe factors that would lead to the rise of a new species are the following :", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nPage Number: 151\nQuestion 1.\nWhat factors would lead to the rise of a new species ?\nAnswer:\nThe factors that would lead to the rise of a new species are the following :\nGeographical isolation of a population caused by various types of barriers (such as mountain ranges, rivers and sea). The geographical isolation leads to reproductive isolation due to which there is no flow of genes between separated groups of pupulation.\nGenetic drift caused by drastic changes in the frequencies of particular genes by chance alone.\nVariations caused in individuals due to a natural selection.\nQuestion 2.\nWill geographical isolation be a major factor in the speciation of a self- pollinating plant species ? Why or why not ?\nAnswer:\nThe geographical isolation cannot be major factor in the speciation of a self-pollinating plant species because it does not have to look the plants for its process of reproduction to be carried out.\nQuestion 3.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 3.\nWill geographical isolation be a major factor in the speciation of an organism that reproduces asexually ? Why or why not ?\nAnswer:\nGeographical isolation cannot be a major factor in the speciation of an asexually reproducing organism because it does not require any other organism to carry out reproduction.\nPage Number: 156\nQuestion 1.\nGive an example of characteristics being used to determine how close two species are in evolutionary terms.\nAnswer:\nIf similar characteristics are shown in different organisms, then these are considered to be inherited from the common ancestry. It also shows the closeness of the species.\nFor example, bats and birds have some similarity in their wings, so they are closely related, while lizard and squirrel do not have wings so these are not closely related to the birds and bats.\nQuestion 2.\nCan the wing of a butterfly and the wing of a bat be considered homologous organs ? Why or why not ?\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 2.\nCan the wing of a butterfly and the wing of a bat be considered homologous organs ? Why or why not ?\nAnswer:\nThe wings of a butterfly and the wings of a bat cannot be considered to be homologous organs because they have different basic designs though they are used for the same purpose of flying. They are analogous organs.\nQuestion 3.\nWhat are fossils ? What do they tell us about the process of evolution ?\nAnswer:\nFossils : Fossils are the remains or traces of a dead organism. These are formed through the formation of sedimentary rocks. They provide following information on the process of evolution.\nThey tell about the changes that occured on the earth\u2019s surface and the corresponding organisms.\nThey tell about the gradual development of complex structured organisms from simple structured organisms.\nIt is known through them that birds are evolved from reptiles.\nThey state that angiosperms are developed from pteriodophytes and gymnosperms.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nIt is known through them that birds are evolved from reptiles.\nThey state that angiosperms are developed from pteriodophytes and gymnosperms.\nThey exhibit the process of humana evolution.\nPage Number: 158\nQuestion 1.\nWhy are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species ?\nAnswer:\nThis is because although genetic make up of humans may be slightly different in different races of people, there is no reproductive isolation. Reproductive isolation differentiates one species from the other. Human beings different in size, colour and looks can marry among themselves and produce fertile offspring.\nQuestion 2.\nIn evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a \u2018better\u2019 body design ? Why or why not ?\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 2.\nIn evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a \u2018better\u2019 body design ? Why or why not ?\nAnswer:\nBacteria is a primitive organism as they came into being very early in evolution. But these organisms are still surviving in the present conditions after millions of years. This is because they have adapted well to the changing environment over these years. Same is the case for all other organisms like spiders, fishes and chimpanzees which have adapted to their environment and have survived. Therefore, all the organisms which exist have a body design which is better as it is suited to their environment.\nNCERT Solutions for Class 10 Science Chapter 9 Textbook Chapter End Questions\nQuestion 1.\nA Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nThis suggests that the genetic make-up of the tall parent can be depicted as:\n(a) TTWW\n(b) TTww\n(c) TtWW\n(d) TtWw\nAnswer:\n(c) TtWW\nQuestion 2.\nAn example of homologous organs is :\n(a) our arm and a dog\u2019s fore-leg\n(b) our teeth and an elephant\u2019s tusks\n(c) potato and runners of grass\n(d) all of the above\nAnswer:\n(d) All of the above\nQuestion 3.\nIn evolutionary terms, we have more in common with :\n(a) a Chinese school-boy\n(b) a chimpanzee\n(c) a spider\n(d) a bacterium\nAnswer:\n(a) A Chinese school-boy\nQuestion 4.\nA study found that children with light coloured eyes are likely to have parents with light coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive ? Why or why not ?\nAnswer:\nThis information is not complete. On the basis of this, it cannot be decided light colour trait is dominant or recessive.", + "On this basis, can we say anything about whether the light eye colour trait is dominant or recessive ? Why or why not ?\nAnswer:\nThis information is not complete. On the basis of this, it cannot be decided light colour trait is dominant or recessive. So it cannot be said until one does not know the nature of this trait in the parents.\nQuestion 5.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 5.\nHow are the areas of study-evolution and classification interlinked ?\nOR\n\u2018Two areas of study namely \u2018evolution\u2019 and \u2018classification\u2019 are interlinked\u201d. Justify this statement. [AICBSE 2016]\nAnswer:\nClassification of organisms is based on relative similarities and differences among organisms. Resemblances in organisms are because they have arisen from a common ancestor and differences in them are due to adaptations to different types of environment. Since the organisms can be graded in order of increasing complexity it indicates at the concept of evolution.\nQuestion 6.\nExplain the terms analogous and homologous organs with examples. [CBSE 2011,2013, 2014]\nAnswer:\nAnalogous organs : Those organs which have different basic structure (or different basic design) but have similar appearance and perform similar functions are called analogous organs.\nFor example, The wings of an insect and a bird are analogous organs.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nFor example, The wings of an insect and a bird are analogous organs.\nHomologous organs : Those organs which have the same basic structure (or same basic design) but different functions are called homologous organs.\nFor example, The wing of a bat, flipper of a seal, front leg of a horse and arm of a man are homologous organs.\nQuestion 7.\nOutline a project which aims to find the dominant coat colour in dogs.\nAnswer:\nSuppose a black homozygous male is mated with a white homozygous female. If the progeny has all black dogs then the dominant coat colour is black.\nQuestion 8.\nExplain the importance of fossils in deciding evolutionary relationships.\nAnswer:\nFossils play important role in providing evolutionary evidences because by knowning the age of fossils we can know about the evolution process of an organism.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nAnswer:\nFossils play important role in providing evolutionary evidences because by knowning the age of fossils we can know about the evolution process of an organism.\nFor example, a fossil bird called archaeopteryx that looked like a bird had many other features of reptiles. It had feathered wings like those of birds, but teeth and tail like those of reptiles. Archaeopteryx is, therefore, a connecting link between the reptiles and birds, and hence suggests that the birds have evolved from the reptiles.\nQuestion 9.\nWhat evidence do we have for the origin of life from inanimate matter ? [CBSE 2011, 2014]\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nAnswer:\nA British scientist J.B.S. Haldane at first in 1929 suggested that life is originated from inanimate matter. According to him life must have developed from the simple inorganic molecules which were present at that time. Later, Miller and Urey in 1953 presented its evidences. They assembled an apparatus to create an early earth atmosphere which was supposed to consist of gases like methane, ammonia and hydrogen sulphide, etc. over water. This was maintained at a temperature just below 100\u00b0C and electric sparks were then passed through the mixture of gases to stimulate lightning for about one week. At the end of one week, it was found that about 15 per cent of carbon (from methane) had been converted into simple compounds and amino acids which make up protein molecules formed in living organisms. This experiment provides the evidence that the life originated from inanimate matter (or lifeless matter) like inorganic molecules.\n\nQuestion 10.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 10.\nExplain how sexual reproduction gives rise to more viable variations than asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually ? [CBSE 2011,2014]\nAnswer:\nDuring sexual reproduction there is \u2018crossing over\u2019 of chromosomes, that gives rise to variations. These variations are inherited and increase the chances of survival of an organism.\nIn sexual reproduction variations may occur due to errors in DNA copying.\nThere may be variations due to interchange of homologous chromosomes during crossing over of male and female.\nIn sexual reproduction, it is not predetermined that which gamete would fuse with another gamete. It depends only on chance. It is also a reason of variation.\nThese variations enable the organisms to adapt themselves to the changing conditions and also help to give rise to new species.\nQuestion 11.\nHow is the equal genetic contribution of male and female parents ensured in the progeny ? [CBSE 2011, 2013]\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 11.\nHow is the equal genetic contribution of male and female parents ensured in the progeny ? [CBSE 2011, 2013]\nAnswer:\nGenetic material in most organisms is present in pairs of chromosomes. Gametes in the sexually reproducing organisms are formed by the process of meiosis during which half of the genetic material goes into each gamete. When the gametes from male and female parents fuse with each other during sexual reproduction, the normal complement is restored. Half of the genetic material comes from the female and half from the male.\nQuestion 12.\nOnly variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement ? Why or why not?\nAnswer:\nYes, variations that confer an advantage to an individual organism are inherited. The organism can survive longer in an environment and maintain its existence in the population.\nNCERT Solutions for Class 10 Science Chapter 9 Heredity and Evolution", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 9 Heredity and Evolution\nHeredity and Evolution: Heredity; Mendel\u2019s contribution- Laws for inheritance of traits, Sex determination : brief introduction; Basic concepts of evolution.\nFormulae Handbook for Class 10 Maths and Science\nPage 143\nQuestion 1.\nIf a trait A exists in 10% of a population of an asexually reproducing species and a trait B exists in 60% Of the same population, which trait is likely to have arisen earlier ?\nAnswer: As species are asexually reproducing, there would be only very minor differences generated due to small inaccuracies in DNA copying, so trait B, which exists in 60% of the same population may get inherited earlier while trait A, which exists in 10% of the population may be originated late due to variations. Thus, trait B have arisen earlier since it is present in 60% of the same population.\nQuestion 2.\nHow does the creation of variations in a species promote survival ?", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 2.\nHow does the creation of variations in a species promote survival ?\nAnswer: Natural selection selects the individuals having useful variations which ensure their survival in the prevailing conditions of environment. Variant individuals that can withstand or cope with prevailing environment will survive better and will increase in number through differential reproduction.\n\nPage 147\nQuestion 1.\nHow do Mendel\u2019s experiments show that traits may be dominant or recessive ?\nAnswer:\nMendel took pea plants with contrasting characteristics \u2013 tall plant and dwarf (short) plant. On cross pollination, he got all tall plants in F1 generation. Then by self pollination of F1 tall plants, he produced second generation (F2) consisting of tall and short plants in the ratio of 3 : 1. Then he concluded that, \u2018T\u2019 (tall) trait is dominant while \u2018t\u2019 trait for shortness is recessive.\nDownload NCERT Solutions for Class 10 Science Chapter 9 Heredity and Evolution PDF\nQuestion 2.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nDownload NCERT Solutions for Class 10 Science Chapter 9 Heredity and Evolution PDF\nQuestion 2.\nHow do Mendel\u2019s experiments show that traits are inherited independently ?\nAnswer:\nIn a dihybrid cross made by Mendel, it was observed that when two pairs of traits or characters were considered; each trait expressed independent of the other. Thus, Mendel was able to propose the Law of Independent Assortment which says about independent inheritance of traits.\nQuestion 3.\nA man with blood group A marries a woman with blood O and their daughter has blood group O. Is this information enough to tell you which of the traits \u2013 blood group A or O is dominant ? Why or why not ?\nAnswer:\nNo. This information is not sufficient to determine which of the traits \u2212 blood group A or O \u2212 is dominant. This is because we do not know about the blood group of all the progeny.Blood group A can be genotypically AA or AO. Hence, the information is incomplete to draw any such conclusion.\nQuestion 4.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 4.\nHow is the sex of the child determined in human beings?\n\nAnswer:\nThe females carry two X-chromosomes. Females produce one type of gametes (eggs) with same type of chromosomes (22 + X). Males have one X and one Y- chromosome. Among the male gametes, half of the sperms carry X-chromosome (22 X) and half\ncarry Y-chromosome (22 + Y). Thus, female is homogametic and male is heterogametic. When a sperm carrying X- chromosome fertilises an egg, the zygote develops into female (XX condition). When sperm carrying Y-chromosome fertilises an egg, the zygote develops into a male (XY condition). Thus, sex is determined at the time of fertilisation.\nPage 150\nQuestion 1.\nWhat are the different ways in which individuals with a particular trait may increase in a population ?\nAnswer:\nDifferent ways are : variation, natural selection and genetic drift (isolation).\nQuestion 2.\nWhy are traits acquired during the lifetime of an individual not inherited ?\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nAnswer:\nDifferent ways are : variation, natural selection and genetic drift (isolation).\nQuestion 2.\nWhy are traits acquired during the lifetime of an individual not inherited ?\nAnswer:\nBecause acquired characters bring changes only in non-reproductive tissues and cannot change the genes of the germ cells. Thus, acquired traits cannot be passed to next generation.\nQuestion 3.\nWhy are the small numbers of surviving tigers a cause of worry from the point of view of genetics ?\nAnswer:\n(i) If any natural calamity occurs and kills these small number of surviving tigers, they can become extinct resulting in the loss of some genes forever.\n(ii) Small number will lead to little recombination and, therefore, lesser variations. These both are very important for giving better survival chances to the species.\n(iii) Less number of species means lesser extent of diversity and lesser number Of traits which reduces the chances of adaptability with respect to the change in the environment.\nPage 151", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\n(iii) Less number of species means lesser extent of diversity and lesser number Of traits which reduces the chances of adaptability with respect to the change in the environment.\nPage 151\nQuestion 1.\nWhat factors could lead to the rise of a new species ?\nAnswer:\nGenetic variations, natural selection and reproductive isolation could lead to the rise of a new species.\nQuestion 2.\nWill geographical isolation be a major factor in the speciation of a self-pollinating plant species ? Why or why not ?\nAnswer:\nNo, because pollination occurs on the same plant in self-pollinating plant species.\nQuestion 3.\nWill geographical isolation be a major factor in the speciation of an organism that reproduces asexually ? Why or why not ?\nAnswer:\nNo, because asexual reproduction involves single parent or organism.\nPage 156\nQuestion 1.\nGive an example of characteristics being used to determine how close two species am in evolutionary terms ?\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nPage 156\nQuestion 1.\nGive an example of characteristics being used to determine how close two species am in evolutionary terms ?\nAnswer:\nHomologous organs, analogous organs and vestigial organs help to identify evolutionary relationships amongst the species.\nQuestion 2.\nCan the wing of butterfly and the wing of a bat be considered homologous organs ? Why or why not ?\nAnswer:\nNo, wing of a bat and wing of a bird cannot be considered as homologous organs because they have different basic structure.\nQuestion 3.\nWhat are fossils ? What do they tell us about the process of evolution ?\nAnswer:\nFossils are the impression or remains of ancient life found preserved in the sedimentary rocks. Fossils are direct evidences of evolution. Fossils also help to identify evolutionary relationship between apparently different species. They also tell about the extent of evolution that has taken place.\nPage 158\nQuestion 1.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nPage 158\nQuestion 1.\nWhy are human beings who look so different from each other in terms of size, colour and looks said to belong to the same species ?\nAnswer:\nThey look different because of interaction of genes with environment which results in change in their appearance. But they belong to the same species as they have same number of chromosomes and can breed among themselves.\nQuestion 2.\nIn evolutionary terms, can we say which among bacteria, spiders, fish and chimpanzees have a \u2018better body design\u2019 why or why not ?\nAnswer:\nNo, because different designs are the product of evolution and different species have different body design to suit or adapt to their environment.\nPage 159\nQuestion 1.\nA Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing whfte flowers. The progeny all bore violet flowers, but almost half of them are short. This suggests that the genetic make-up of the tall parent can be depicted as\n(a) TTWW\n(b) TTww", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\n(a) TTWW\n(b) TTww\n(c) TtWW\n(d) TtWw\nAnswer:\n(c) Genetic make-up of tall plant can be depicted by TtWW.\nQuestion 2.\nAn example of homologous organs is\n(a) our arm and a dogs fore-leg.\n(b) our teeth and an elephants tusks.\n(c) potato and runners of grass.\n(d) All of the above.\nAnswer:\n(d) Both organs in all options have same basic structural design but have different functions and appearance.\nQuestion 3.\nIn evolutionary terms, we have more in common with\n(a) a Chinese school-boy.\n(b) a chimpanzee.\n(c) a spider.\n(d) a bacterium.\nAnswer:\n(a) A Chinese school-bpy is also a human being.\nQuestion 4.\nA study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?\nAnswer:\nWe can say that light eye colour trait is dominant because only dominant traits are expressed in the first generation.", + "On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?\nAnswer:\nWe can say that light eye colour trait is dominant because only dominant traits are expressed in the first generation.\nQuestion 5.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nAnswer:\nWe can say that light eye colour trait is dominant because only dominant traits are expressed in the first generation.\nQuestion 5.\nHow are the areas of study \u2013 evolution and classification\u2014 inteilinked?\nAnswer:\nEvolution and classification are interlinked with each other in many ways. Classification is the most important term to explain evolution. It is based on the similarities and differences between two species or among two organisms. More closer the characteristics, the moe doser is the evolution and chances to be in the same group of classification. Thus, the classification of species is a reflection of their evolutionary relationship.\nQuestion 6.\nExplain the terms analogous and homologous organs with examples.\nAnswer:\nAnalogous organs are those organs which have different basic structural designs and developmental origins but have similar appearance and perform similar functions.\nExamples:\nWings of an insect and wings of a bat.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nExamples:\nWings of an insect and wings of a bat.\nHomologous organs are those organs which have the same basic structural design and developmenta\u2019 origin but have different functions and appearance.\nExamples: Forelimbs of frog and forelimbs of human.\nQuestion 7.\nOutline a project which alms to find the dominant coat colour in dogs.\nAnswer:\nA homozygous black (RB) male dog and a homozygous white (bb) female dog is taken and given to mate and produce offspring in F1 generation. If black colour is dominant out of every 4 dogs, 3 will be black and if white colour is dominant 3 out of 4 dogs will be white.\n\nQuestion 8.\nExplain the importance of fossils in deciding evolutionary relationships.\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 8.\nExplain the importance of fossils in deciding evolutionary relationships.\nAnswer:\nFossils and their study is useful to know about the species which are no longer alive. They provide evidence and missing links between two classes. They are helpful in forming a sequence of organisms in the pathway of evolution. Thus, fossils have importance in deciding evolutionary relationships.\nQuestion 9.\nWhat evidence do we have for the origin of life from inanimate matter?\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 9.\nWhat evidence do we have for the origin of life from inanimate matter?\nAnswer:\nStanley L. Miller and Harold C. Urey provided evidence regarding origin of life from inanimate matter. They assembled an atmosphere similar to that existed on early earth. The atmosphere had molecules like ammonia, methane, hydrogen sulphide and water, but no oxygen. The mixture was maintained at a temperature just below 100\u25e6C and sparks were passed through the mixture of gases. At the end of a week, 15% carbon from methane had been converted to simple compounds of carbon like aminoacids which make up protein molecules. So, life arose afresh on earth.\nQuestion 10.\nExplain how sexual reproduction gives rise to more viable variations that asexual reproduction. How does this affect the evolution of those organisms that reproduce sexually ?\nAnswer:\nVariations occurring during sexual reproduction may be due to:\nSeparation of homologous chromosomes (by chance only) during gamete formation.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nAnswer:\nVariations occurring during sexual reproduction may be due to:\nSeparation of homologous chromosomes (by chance only) during gamete formation.\nCrossing over (recombination) of homologous chromosomes.\nFertilisation of gametes to form zygote.\nErrors during DNA copying or mutations.\nIn asexually reproducing organisms only errors during DNA copying or mutations cause variations.\nSince the extent of variations is much larger in sexually reproducing organisms, therefore, the chances of evolution is also much in sexually reproducing These variations enable the organisms to adapt themselves to the changing conditions and also help to face the struggle for Over the time, they and rise to new species.\nQuestion 11.\nHow is the equal genetic contribution of male and female parents ensured in the progeny?\nAnswer:\nGenetically organisms are of types", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 11.\nHow is the equal genetic contribution of male and female parents ensured in the progeny?\nAnswer:\nGenetically organisms are of types\n(i) Haploid : They have single set of chromosomes, where each chromosome is represented singly. As the chromosomes are the bearer of genes so haploids have single set of genes. A single gene determines the expression of character.\n(ii) Diploid : \u2018They have two sets Of homologous chromosomes, where the chromosome occur in pair, one maternal contributed by the mother through her ovum and the second Of the pair is contributed by the male parent through his sperm. The resultant cell zygote produces by the fusion of male and female gametes have two sets of chromosomes \u2013 each set contributed\u2019 by each parent. In diploids a character is controlled by two genes/factors. Both the father and mother contribute practically equal amount of genetic material to the child. It means that each trait can be influenced by both paternal and maternal DNA.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 12.\nOnly variations that confer an advantage to an individual organism will survive in a population. Do you agree with this statement ? Why or why not ?\nAnswer:\nNo, many of the times the variations are not advantageous to an individual organism but still survive in a population, e.g., take the case of free ear lobe and attached ear lobe. Most of the other variations not only give survival advantage to an individual but also contribute to genetic drift. Thus, we can say that most of the variations lead to better adaptation of an organism to the changing environment. In this way, it gives survival advantage to that organism and will also survive in the coming population.\nMultiple Choice Questions (MCQs) [1 Mark each]\nQuestion 1.\nAn example of homologous organs is [NCERT]\n(a) our arm and a dog\u2019s foreleg\n(b) our teeth and an elephant\u2019s tusks\n(c) potato and runners of grass\n(d) All of the above\nAnswer:\n(a) Our arm and a dog\u2019s foreleg is the example of homologous organs.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\n(a) our arm and a dog\u2019s foreleg\n(b) our teeth and an elephant\u2019s tusks\n(c) potato and runners of grass\n(d) All of the above\nAnswer:\n(a) Our arm and a dog\u2019s foreleg is the example of homologous organs.\nQuestion 2.\nThe science, which deals with study of heredity and variations is called\n(a) phylogeny\n(b) embryology\n(c) genetics\n(d) palaeontology\nAnswer:\n(c) The genetics is the study of heredity and variations and includes their occurrence, causes, benefits, disadvantages, significance, etc.\nQuestion 3.\nArchaeopteryx is a connecting link between\n(a) reptiles and aquatic animals\n(b) birds and insects\n(c) reptiles and birds\n(d) birds and dinosaurs\nAnswer:\n(c) Archaeopteryx is a connecting link- between the reptiles and birds. It appears like a bird, but has other features which are present in reptiles, e.g. it has wings like bird, but teeth and tail like the reptilians.\nQuestion 4.", + "It appears like a bird, but has other features which are present in reptiles, e.g. it has wings like bird, but teeth and tail like the reptilians.\nQuestion 4.\nFor palaeontological studies a scientist will gather the evidences from\n(a) study of homology", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 4.\nFor palaeontological studies a scientist will gather the evidences from\n(a) study of homology\n(b) study of analogy\n(c) fossils\n(d) All of these\nAnswer:\n(d) Study of homologous and analogous organs indicates the origin and modification in organisms and study of fossils indicates the age and features of an organism.\nQuestion 5.\nIn evolutionary terms, we have more in common with [NCERT]\n(a) a Chinese school boy\n(b) a chimpanzee\n(c) a spider\n(d) a bacterium\nAnswer:\n(a) Chinese school boy because both of us belong to the same species, i.e. Homo sapiens.\nQuestion 6.\nAditya was observing some organisms in lab and tried to compare them. The presence of which organs will confirm to him that they share evolutionary history?\n(a) Analogous organs\n(b) Paralogous organs\n(c) Homologous organs\n(d) None of these\nAnswer:\n(c) Homologous organs are present in organisms who share evolutionary history. However, these organs perform different functions in different organisms.\nQuestion 7.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\n(d) None of these\nAnswer:\n(c) Homologous organs are present in organisms who share evolutionary history. However, these organs perform different functions in different organisms.\nQuestion 7.\nNew species may be formed if\nI. DNA undergoes significant changes in germ cells. .\nII. chromosome number changes in the gamete.\nIII. there is no change in the genetic material.\nIV. mating does not take place.\n(a) I and II\n(b) I and III\n(c) II, III and IV\n(d) I, II and III\nAnswer:\n(a) New species may be formed if the DNA changes are severe enough, such as a change in the number of chromosome. This leads to new variations.\nQuestion 8.\nWhich of the following statements is not true with respect to variation?\n(a) All variations in a species have equal chance of survival.\n(b) Change in genetic composition results in variation.\n(c) Selection of variants by environmental factors forms the basis of evolutionary processes.\n(d) Variation is minimum in asexual reproduction.\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\n(c) Selection of variants by environmental factors forms the basis of evolutionary processes.\n(d) Variation is minimum in asexual reproduction.\nAnswer:\n(a) All variations in a species do not have equal chances of survival. Some of the variations may be so drastic that the new DNA copy cannot work with the cellular apparatus it inherits. Such, a newborn cell dies soon.\nQuestion 9.\nSelect the statement that describes characteristics of genes. .\n(a) Genes are specific sequence of bases in a DNA molecule.\n(b) A gene does not code for proteins.\n(c) In individuals of a given species, a specific gene is located on a particular chromosome.\n(d) Each chromosome has only one gene.\nAnswer:\n(b) Genes are stretches of DNA found on chromosomes of a cell. A gene contains information for making proteins in a cell. A specific gene is located on a particular chromosome in individuals of a given species.\nQuestion 10.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 10.\nIf a round, green seeded pea plant (RRyy) is crossed with wrinkled, yellow seeded pea plant (rrYY), the seeds produced in F1 \u2013 generation will be [NCERT Exemplar]\n(a) round and yellow\n(b) round and green\n(c) wrinkled and green\n(d) wrinkled and yellow\nAnswer:\n(a) The cross between RRyy and rrYY seeds will produce RrYy (round and yellow) seeds in F1-generation, because round and yellow are the dominant traits.\nQuestion 11.\nFrom the list given below, select the character, which can be acquired but not inherited. [NCERT Exemplar]\n(a) colour of eye\n(b) colour of skin\n(c) size of body\n(d) nature of hair\nAnswer:\n(c) Acquired traits develop in response to the environment. The size of the body is an acquired trait because it can vary based on the availability of less or more food. The other three colour of eye and skin and nature of hair are characters inherited from the parents.\nQuestion 12.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 12.\nAccording to the evolutionary theory, formation of a new species is generally due to [NCERT Exemplar]\n(a) sudden creation by nature.\n(b) accumulation of variations over several generations.\n(c) clones formed during asexual reproduction.\n(d) movement of individuals from one habitat to another\nAnswer:\n(b) Accumulation of variations over several generations forms new species. Genetic drift accumulates different changes in sub-populations of a species. Also, natural selection may also operate differendy in the different geographic locations. Eventually, different groups of new species will be formed.\nQuestion 13.\nSelect the incorrect statement. [NCERT Exemplar]\n(a) Frequency of certain genes in a population changes over several generations resulting in evolution.\n(b) Reduction in weight of the organism due to starvation is genetically controlled.\n(c) Low weight parents can have heavy weight progeny.\n(d) Traits which are not inherited over generations do not cause evolution.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\n(c) Low weight parents can have heavy weight progeny.\n(d) Traits which are not inherited over generations do not cause evolution.\nAnswer:\n(b) The weight reduction due to starvation will not change the DNA of the germ cells, because low weight is not a trait that is genetically controlled or inherited. Also, low weight parents may have heavy weight progeny.\nQuestion 14.\nIn human males all the chromosomes are paired perfectly except one. This/these unpaired chromosome is/are\nI. large chromosome\nII. small chromosome\nIII. Y-chromosome IV X-chromosome\n(a) I and II\n(b) Only III\n(c) III and IV\n(d) II and IV\nAnswer:\n(c) In human males, one pair called the sex chromosomes are unpaired. Here, one is a normal-sized X-chromosome while other is a short Y-chromosome. Women have a perfect pair of sex chromosomes, both called X.\nQuestion 15.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nQuestion 15.\nRajneesh was studying the fossils of two different types, fossil A was found in upper layer of Earth and B in deeper layers. What can be predicted regarding the age of these fossils?\n(a) A has recently become extinct\n(b) B has become extinct recently\n(c) The time of extinction cannot be determined\n(d) None of the above\nAnswer:\n(a) Since, fossil A was found in upper layer of earth, it suggests that the organism has become extinct recently. Fossil B found in deeper layer must have become extinct long time ago and deposition of other layers occurred over it during this period.\nQuestion 16.\nA Mendelian experiment consisted of breeding tall pea plants bearing violet flowers with short pea plants bearing white flowers. The progeny all bore violet flowers, but almost half of them were short. This suggests that the genetic makeup of the tall parents can be depicted as [NCERT]\n(a) TTWW\n(b) TTww\n(c) TtWW\n(d) TtWw\nAnswer:", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\n(a) TTWW\n(b) TTww\n(c) TtWW\n(d) TtWw\nAnswer:\n(c) Parent with genotype TtWW produce two types of gametes TW and tW, while the other with genotype ttww produce only one type of gamete W.\n\n Heredity and Evolution Mind Map\nAccumulation Of Variation During Reproduction\nCharacters or features or traits are inherited from one generation to the next during reproduction.\nThis inheritance provides both a common basic body design & subtle changes in it for next generation.\nWhen this generation reproduces, the offspring would have differences they inherit from previous generation as well as newly created differences.\nAccumulation of these differences generation after generation leads to the development of variations in a population.\nDifferent variations provide different advantages to the population and the variation which provide best survival advantages are inherited to the next generation.\nFor e.g. bacteria having variation to tolerate heat will survive and multiply better in heat wave.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nFor e.g. bacteria having variation to tolerate heat will survive and multiply better in heat wave.\nHeredity\nHeredity refers to the transmission of characteristics from parent to offspring by means of genes in the chromosomes.\nMendel\u2019s Contributions\nMendel was the first scientist whose studies lead to the formulation of laws of inheritance.\nHe conducted cross hybridization experiments of garden pea plant (Pisum sativum) and studied the transmission of characters that had two contrasting traits such as round/wrinkled seeds, tall/short plants, white/\u2019violet flowers etc.\nExperiment 1: He cross pollinate pure breeds of tall (TT) & dwarf (tt) pea plant and calculated the percentages of tall & dwarf progeny.\nObservation: F1 generation was tall (Tt) with no halfway characteristics.\nF2 generation produced by self pollination of F1 included tall and short plants in 3:1. (Genotypic ratio 1:2:1 for TT:Tt:tt)", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nObservation: F1 generation was tall (Tt) with no halfway characteristics.\nF2 generation produced by self pollination of F1 included tall and short plants in 3:1. (Genotypic ratio 1:2:1 for TT:Tt:tt)\nInference: This indicates that both the tallness & shortness traits were inherited in the F, plants, but only the tallness trait was expressed.\nThus, two copies of the trait are inherited in each sexually reproducing organism. These two may be identical (TT or tt) or may be different (Tt), depending on the parentage.\nConclusion: This study leaded to the formulation of two laws:\nLaw of dominance: states that only one character expresses itself in F, generation.\nLaw of segregation: states that the two alleles of a character in an individual get separated or segregated during gamete formation and distributed randomly in gametes.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nLaw of segregation: states that the two alleles of a character in an individual get separated or segregated during gamete formation and distributed randomly in gametes.\nExperiment 2: He crossed the plant with two different characteristics such as tall plant with round seed and short plant with wrinkled seed. Other example may include round & green seeds (RRyy) and wrinkled & yellow seeds (rrYY).\nObservation: F1 generation; all were tall & round i.e. tall & round are dominant.\nF2 generation; tall plants with round seeds, tall with wrinkled seeds, short with round seeds, and short plants with wrinkled seeds in 9:3:3:1.\nSimilarly, round & yellow, round & green, wrinkled & yellow, and wrinkled & green in 9:3:3:1.\nInference: The tall/short trait and the round seed/wrinkled seed trait are independently inherited.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nSimilarly, round & yellow, round & green, wrinkled & yellow, and wrinkled & green in 9:3:3:1.\nInference: The tall/short trait and the round seed/wrinkled seed trait are independently inherited.\nConclusion: It formulated the law of independent assortment which states that genes of different characters located in different pairs of chromosomes are independent of one another in their segregation during gamete formation.\nSex Determination\nDifferent species use different strategies for this:\nEnvironment: for e.g. the temperature at which fertilised eggs are kept determines the sex of developing animals in the eggs. It is observed in animals like crocodile, turtle etc.\nSnails can change sex. indicating that sex is not genetically determined.\nSex of an individual is genetically determined for e.g. humans.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nSnails can change sex. indicating that sex is not genetically determined.\nSex of an individual is genetically determined for e.g. humans.\nHumans have 22 autosomal & 1 sex chromosome pairs. Females have XX & males have XY. Hence, sex of a child is determined by what he/she has inherited (X or Y) from the father since, child will always inherit X from the mother. If X is inherited from father then child will be a girl & if Y is inherited then a child will be a boy.\n\nEvolution\nIt refers to gradual change in the characteristics of the population (plants & animals) over successive generations.\nErrors in DNA copying during reproduction, mutations, & natural selection account for the evolution.\nEvolution gives rise to such a biodiversity at each level of biological organization such as at species level, among individuals, molecules etc.\nEvolution And Classification", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nEvolution gives rise to such a biodiversity at each level of biological organization such as at species level, among individuals, molecules etc.\nEvolution And Classification\nClassification is the process by which organisms are grouped into convenient categories based on some easily observable characters.\nCharacters such as cell type (prokaryote or eukaryote), single cell or multi cellular, presence or absence of nucleus, autotrophic (such as photosynthesis) or heterotrophic, sexual or asexual reproduction etc. are used to classify different organisms in different groups.\nThe more characteristics two species will have in common, the more closely they are related. And the more closely they are related, the more recently they will have had a common ancestor. For e.g. brother & sister are more closely related than a girl & her first cousin. Therefore, classification of species gives a reflection of their evolutionary relationship.\nTracing Evolutionary Relationships", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nTracing Evolutionary Relationships\nFew evidences which help us to trace evolutionary\nrelationships among different organisms or species:\nComparative anatomy and morphology: study of similarities & differences among organisms to understand the common ancestry\u2019.\nHomologous: Similar structure different functions. It indicates common ancestry for e.g. bones of forelimbs in frog, lizard, bird & human.\nAnalogous: Similar functions but different structure. Different structures evolved for same function & hence having similarity. For e.g. wings of bats & birds, eye of octopus & mammals, etc.\nFossils: Remains of hard parts of life-forms found in rocks. They represent extinct organisms (e.g., Dinosaurs).\nThe age of the fossils can be estimated by two ways; relative depth of the fossils, dating fossils i.e. detection of ratios of different isotopes of the same element in the fossil material\nSpeciation", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nThe age of the fossils can be estimated by two ways; relative depth of the fossils, dating fossils i.e. detection of ratios of different isotopes of the same element in the fossil material\nSpeciation\nIt refers to a gradual evolutionary process by which populations evolve to become different species.\nReproductive and geographical isolation play an important role in the process of speciation. They result in change in the frequency of an existing gene variant in a population i.e. genetic drift.\nOver generations, genetic drift along with natural selection results in the formation of new\u2019 species,\nOther factors that may result in speciation are sudden severe DNA changes (mutation) such as change in chromosomal no., variation such as female green beetle will not mate with red males. Her behavior ensures the reproductive isolation between them and thus results in generation of new\u2019 species.\nEvolution Should Not Be Equated With \u2018progress\u2019", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nEvolution Should Not Be Equated With \u2018progress\u2019\nEvolution is simply the generation of diversity & shaping of diversity by environmental selection.\nThe only progressive trend in evolution seems to be the emergence of more and more complex body designs over time. However, that doesn\u2019t mean that the older designs are inefficient.\nFor e.g. simplest life forms; bacteria inhabits the most inhospitable habitats like hot springs, deep-sea thermal vents & ice in Antarctica.\nHuman Evolution\nTolls, like excavating, time-dating and studying fossils, determining DNA sequences etc, have been used for studying human evolution.\nAll humans are a single species regardless of skin color or human races.\nThe earliest members of the human species, Homo sapiens, can be traced back to Africa i.e. we all come from Africa.\nA couple of hundred thousand years ago, some of our ancestors left Africa while others stayed on.", + "CHAPTER 8-HEREDITY -INTEXT QUESTIONS.txt\nA couple of hundred thousand years ago, some of our ancestors left Africa while others stayed on.\nThe migrants slowly spread across the planet; from Africa to West Asia, then to Central Asia, Eurasia, South Asia, & East Asia.\nThey travelled down the islands of Indonesia and the Philippines to Australia, and they crossed the Bering land bridge to the Americas.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER SOLUTIONS.txt\nNCERT Exemplar Class 10 Science Chapter 10 Light Reflection and Refraction\nShort Answer Questions\nQuestion 1. Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.\nAnswer. Due to variation in speed of light in different medium, light bends as it moves from One medium to another. Since, there are two surfaces for refraction which are parallel, the light ray should bend in opposite sense in them equally, the emerging light ray is always parallel to the incident ray.\n\nQuestion 2. A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appears to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine. Support your answer with reason.\nAnswer. No, it will depend on the refractive index of the liquid in which the pencil is dipped. The tip at the bottom will get elevated as per the relation.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER SOLUTIONS.txt\nAnswer. No, it will depend on the refractive index of the liquid in which the pencil is dipped. The tip at the bottom will get elevated as per the relation.\n\nQuestion 3. Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5. Find out the absolute refractive index of diamond.\nAnswer.\n\nQuestion 4. A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?\nAnswer. Yes, for getting virtual image the object has to be placed between the optical centre and the focus. For real and enlarged image, the object should be placed between F and 2F.\nQuestion 5. Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence. Show the same with the help of diagram.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER SOLUTIONS.txt\nAnswer. When the two mirrors are placed at 90\u00b0 to each other, the incident and the reflected rays will remain parallel to each other.\n\nLong Answer Questions\nQuestion 6. Draw ray diagrams showing the image formation by a convex mirror when an object is placed\n(a) at infinity\n(b) at finite distance from the mirror\nAnswer.\n\nQuestion 7. The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?\nAnswer.\n\nQuestion 8. Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of -50 cm. What is the nature of the lens and its power used by each of them?", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER SOLUTIONS.txt\nAnswer.\n\nQuestion 8. Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of -50 cm. What is the nature of the lens and its power used by each of them?\nAnswer. Power of a lens: The ability of a lens to converge or diverge the light rays is called power (P) of the lens. It is defined as the reciprocal of the focal length, i.e. f = 1/f .The SI unit of power of a lens is dioptre (D). A lens of focal length 100 cm has a power of 1 dioptre, i.e. 1 dioptre = 1 m-1. The lens of focal length +50 cm has a power of +2 D and is convex while the other is having a power of -2 D and is concave.\nNCERT Exemplar Solutions\nWe hope the NCERT Exemplar Class 10 Science Chapter 10 Light Reflection and Refraction help you.", + "NCERT Exemplar Solutions\nWe hope the NCERT Exemplar Class 10 Science Chapter 10 Light Reflection and Refraction help you. If you have any query regarding NCERT Exemplar Class 10 Science Chapter 10 Light Reflection and Refraction, drop a comment below and we will get back to you at the earliest.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\nCBSE Class 10 Science Notes Chapter 10 Light Reflection and Refraction\nREFLECTION\nReflection of Light: The phenomenon of bouncing back of light into the same medium by the smooth surface is called reflection.\nIncident light: Light which falls on the surface is called incident light.\nReflected light: Light which goes back after reflection is called reflected light.\nThe angle of incidence: The angle between the incident ray and the normal.\nAn angle of reflection: The angle between the reflected ray and the normal.\nMirror: The surface which can reflect the light is a mirror.\nPlane Mirror: If the reflecting surface is a plane then the mirror is plane.\nSpherical Mirror: If the reflecting surface is part of the hollow sphere then the mirror is a spherical mirror.\nThe spherical mirror is of two types:\nConvex mirror: In this mirror reflecting surface is convex. It diverges the light so it is also called a diverging mirror.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\nThe spherical mirror is of two types:\nConvex mirror: In this mirror reflecting surface is convex. It diverges the light so it is also called a diverging mirror.\nConcave mirror: In this mirror reflecting surface is concave. It converges the light so it is also called converging mirror.\nParameters of Mirror:\nCenter of Curvature: The centre of hollow sphere of which mirror is a part.\nThe radius of curvature: The radius of hollow sphere of which mirror is a part.\nPole: The centre of mirror (middle point) is pole.\nPrincipal axis: The line joining the pole and center of curvature is called principal axis.\nAperture: Size of mirror is called aperture of mirror.\nPrincipal Focus: The point on the principal axis, where all the incident rays parallel to principal axis converge or diverge after reflection through mirror.\nFocal Length: The distance between pole and focus point is focal length.\nSpecial Rays for Formation of Image:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\nFocal Length: The distance between pole and focus point is focal length.\nSpecial Rays for Formation of Image:\nA ray of light which is parallel to the principal axis of a spherical mirror, after reflection converges or diverges from focus.\nA ray of light passing through or appearing from the center of curvature of spherical mirror is reflected back along the same path.\nA ray of light passing through or appearing from the focus of spherical mirror becomes parallel to the principal axis.\nA ray of light which is incident at the pole of a spherical mirror is reflected back making same angle with principal axis.\nUse of Concave Mirror: It is used as a makeup mirror, the reflector in torches, in headlights of cars and searchlights, doctor\u2019s head-mirrors, solar furnace, etc.\nSign Conventions of Spherical Mirror\nAll the distances are measured from the pole of the mirror as the origin.\nDistances measured in the direction of incident rays are taken as positive.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\nSign Conventions of Spherical Mirror\nAll the distances are measured from the pole of the mirror as the origin.\nDistances measured in the direction of incident rays are taken as positive.\nDistances measured opposite to the direction of incident rays are taken as negative.\nDistances measured upward and perpendicular to the principal axis are taken as positive.\nDistances measured downward and perpendicular to the principal axis are taken as negative.\n\\frac { 1 }{ f } =\\frac { 1 }{ v } +\\frac { 1 }{ u } \u2026where f, v and u are focal length, image distance, object distance\nLinear Magnification: This is the ratio of the height of the image to the height of the object.\nm=\\frac { { h }^{ \u2018 } }{ h } \u2026where m = magnification, h = height of image, h\u2019 = height of object\nUse of Convex Mirror: Convex mirror used as rear view mirror in vehicles, as shop security mirrors, etc.\nREFRACTION\nRefraction of Light: The bending of light at the interface of two different mediums is called Refraction of light.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\nREFRACTION\nRefraction of Light: The bending of light at the interface of two different mediums is called Refraction of light.\nIf the velocity of light in medium is more, then medium is called optical rarer.\nExample, air or vacuum is more optical rarer.\nIf the velocity of light in medium is less, then medium is called optical denser.\nExample, glass is more denser than air.\nRefractive Index: It represents the amount or extent of bending of light when it passes from one medium to another.\nThere are two types of refractive index\nRelative refractive index and\nAbsolute refractive index.\nRefractive index of medium with respect to other medium is called Relative Refractive Index.\nRefractive index of medium 1 with respect to medium 2 = \\frac { Speed\\quad of\\quad light\\quad in\\quad medium\\quad 2(V2) }{ Speed\\quad of\\quad light\\quad in\\quad medium\\quad 1(V1) }\nRefractive index of medium with respect to air or vacuum is called Absolute Refractive Index.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\nRefractive index of medium with respect to air or vacuum is called Absolute Refractive Index.\nAbsolute refractive index of medium (m) = \\frac { Speed of light in air(c) }{ Speed of light in medium (Vm) }\nIncident ray: It is incoming ray on the refracting surface.\nRefracted ray: It is an outgoing ray from the refracting surface.\nAn angle of incidence (i): It is the angle between incident rays and perpendicular line (normal) at the point of incidence.\nAn angle of refraction (r): It is the angle between refracted rays and perpendicular line (normal) at the point of incidence.\nLaw of Refraction: According to this law\n\u201cThe incident ray, refracted ray and normal at the point of incidence all lie in the same plane.\u201d\n\u201cThe ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.\u201d\n\\frac { sin\\quad i }{ sin\\quad r } = constant (\u00b5)\nLens: The transparent refracting medium bounded by two surfaces in which at least one surface is curved is called lens.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\n\\frac { sin\\quad i }{ sin\\quad r } = constant (\u00b5)\nLens: The transparent refracting medium bounded by two surfaces in which at least one surface is curved is called lens.\nLenses are mainly two type\nConvex lens and\nConcave lens.\nCenter of Curvature: The centres of two spheres, of which lens is part is called the centre of curvature.\nRadii of Curvature: The radii of spheres, of which lens is part is called radius of curvature.\nPrincipal Axis: The line joining the centres of curvature of two surfaces of lens is called principal axis.\nOptical Center: It is a special point on the principal axis. Light incident on the optical centre passes through the lens without deviation.\nPrincipal Focus: The point on the principal axis at which all incident rays parallel to the principal axis converge or appear to diverge after refraction through the lens.\nSpecial Rays for Image Formation by Lens:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\nSpecial Rays for Image Formation by Lens:\nAn incident ray, parallel to the principal axis, after refraction passes through (or appears to come from), second focus of the lens.\nAn incident ray, passing through the optical center of the lens, goes undeviated from the lens.\nAn incident ray, passing through the (first) principal focus of the lens, or directed toward it, becomes parallel to the principal axis after refraction through lens.\nUse of Lens: In photographic cameras, magnifying glass, microscope, telescope, the human eye.\n1. Light travels in a straight line.\n2. Light gets reflected when it falls on polished surfaces; like mirrors.\n3. Light suffers refraction when it travels from one medium to another.\n4. There is a change in the wavelengths!light when it moves from one medium into another.\n5. The bouncing back of light when it strikes a smooth or polished surface is called reflection of light. Reflection is of two types; Specular or regular and Diffuse or irregular reflection.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\n5. The bouncing back of light when it strikes a smooth or polished surface is called reflection of light. Reflection is of two types; Specular or regular and Diffuse or irregular reflection.\n6. The angle of incidence is equal to the angle of reflection. Mathematically, we have \u2220i = \u2220r.\n7. The image is as far behind the mirror as the object is in front.\n8. The image is unmagnified, virtual and erect.\n9. The image has right-left reversal.\n10. Focal length of a plane mirror is infinity.\n11. Power of a plane mirror is zero.\n12. If a plane mirror is turned by an angle, the reflected ray turns by 2\u03b8.\n13. The least size of a plane mirror to view an object is equal to half the size of the object.\n14. Pole (Vertex): The central point of a mirror is called its pole.\n15. Centre of curvature : The centre of the sphere of which the mirror is a part is called the centre of curvature. It is denoted by C.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\n15. Centre of curvature : The centre of the sphere of which the mirror is a part is called the centre of curvature. It is denoted by C.\n16. Radius of curvature : The radius of the sphere of which the mirror is a part is called the radius of curvature. It is denoted by R.\n17. Principal axis : The straight line passing through the pole and the centre of curvature of the mirror is called the principal axis.\n18. Principal focus : It is a point on the principal axis at which the rays parallel to the principal axis meet after reflection or seem to come from. For a concave mirror, the focus lies in front of the mirror and for a convex mirror, it lies behind the mirror. In short, a concave mirror has a real focus while aconvex mirror has a virtual focus.\n19. Focal plane : A plane, drawn perpendicular to the principal axis and passing through the principal focus.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\n19. Focal plane : A plane, drawn perpendicular to the principal axis and passing through the principal focus.\n20. Focal length : The distance between the pole and the focus is called the focal length. It is represented by f. The focal length is half the radius of curvature.\n21. Aperture: The size of the mirror is called its aperture. It is also defined as the effective diameter of the light reflecting area of the mirror.\n22. Real image : When the rays of light, after reflection from a mirror, actually meet at a point, then the image formed by these rays is said to be real. Real images can be obtained on a screen.\n23. Virtual image: When the rays of light, after reflection from a mirror, appear to meet at a point, then the image formed by these rays is said to be virtual. Virtual images can\u2019t be obtained on a screen.\n24. The following rays are used while drawing ray diagrams to find the position of an image :", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\n24. The following rays are used while drawing ray diagrams to find the position of an image :\nA ray of light parallel to the principal axis after reflection passes through the focus. (1)\nA ray of light passing through the focus after reflection becomes parallel to the principal axis. (2)\nA ray of light incident on the centre of curvature retraces its path after reflection form the mirror.\n\n25. For mirrors, the following results hold :\nu is \u2013 ve, if the object is in front of the mirror.\n(Real object)\nu is + ve, if the object is behind the mirror.\n(Virtual object)\nv is \u2013 ve, if the image is in front of the mirror.\n(Real image)\nvis +ve, if the image is behind the mirror.\n(Virtual image)\nFocal length of a concave mirror is taken as \u2013 ve. Focal length of a convex mirror is taken as +ve.\n26. When the image formed by a spherical mirror is real, it is also inverted and is on the same side of the mirror as the object. Since both v and u are negative, the magnification is negative.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\n26. When the image formed by a spherical mirror is real, it is also inverted and is on the same side of the mirror as the object. Since both v and u are negative, the magnification is negative.\n27. When the image formed by a spherical mirror is virtual, it is also erect and is on the other side of the mirror as the object. In this case, u is \u2013 ve and v is + ve , therefore, m is positive.\n28. The expression for the mirror formula is 1/u+1/v = 1/f\n29. Linear magnification is given by the expression\n\n30. If m is positive, the image is erect w.r.t the object and if m is negative, the image is inverted w.r.t. the object.\n31.The position of the image for various positions of the object for a concave mirror is as shown in the table below. The table also shows the use of the mirror for different positions of the object.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\nThe position of the image for various positions of the object for a convex mirror is as shown in the table below. The table also shows the use of the mirror for different positions of the object.\n\n32. The bending of light when it travels from one medium into another is called refraction of light\n33.\n\n34. As light travels from ,one medium to another, the frequency of light does not change.\n35. Light refracts because it has different speeds in different media.\n36. The refraction of light obeys the following two laws :\nThe incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.\nThe ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant.This\nconstant is called the index of refraction or refractive index.\n\n37. If wng is the refractive index of glass w.r.t. water, ang be the refractive index of glass w.r.t. air and anw be the refractive index of water w.r.t. air ,then", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\n37. If wng is the refractive index of glass w.r.t. water, ang be the refractive index of glass w.r.t. air and anw be the refractive index of water w.r.t. air ,then\n\n38. The most familiar and widely used optical device is the lens. A lens is an optical system with two refracting surfaces. The simplest lens has two spherical surfaces close enough together that we can neglect the distance between them. Such a lens is called a thin lens. The two common types of lenses are Converging lens or Convex lens, Diverging lens or Concave lens.\n39. It should be noted that, if the above lenses are surrounded by .a material with a refractive index greater than that of the lens, the convex lens gets converted into a concave lens and vice-versa.\n40. Any lens that is thicker at its centre than at its edges is a converging lens with positive f, and any lens that is thicker at its edges than at the centre is a diverging lens with negative f.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\n40. Any lens that is thicker at its centre than at its edges is a converging lens with positive f, and any lens that is thicker at its edges than at the centre is a diverging lens with negative f.\n41. Optical centre : The central point C in the lens is called the optical centre. If a ray is incident towards the optical centre, it passes undeviated .through the lens.\n\n42.Principal axis: Since the lens contains two spherical surfaces, therefore, it has two centres of curvatures.\nThe line joining these centres and passing through the optical centre is called principal axis.\n43. Aperture: The effective width of a lens through which refraction takes place is called the aperture.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\n43. Aperture: The effective width of a lens through which refraction takes place is called the aperture.\n44. Focus and Focal Length : If a beam of light moving parallel to the principal axis of a convex lens is incident on it, the rays converge or meet at a point on the principal axis. This point F is called the focus. The distance CF is called the focal length. If a beam of light moving parallel to the principal axis is incident on a concave lens, the beam of light diverges. If these diverged rays are produced backward, they meet at a point F on the principal . axis. The transmitted rays appear to come from this point. This point F is called the focus and distance CF is called the focal length.\n45. For drawing the ray diagrams, we note the following :\n\n\nAll rays parallel to the principal axis after refraction pass through the principal focus or seem to come from it.\nA ray of light passing through the focus after refraction becomes parallel to the principal axis.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\nA ray of light passing through the focus after refraction becomes parallel to the principal axis.\nA ray of light passing through the optical centre of the lens after refraction passes undeviated.\n46. A convex and a concave lens can be supposed to be made-up of prisms.\n47. Image formation by a concave lens.\n\n48. Image formation by a convex lens.\n\n49. New Cartesian sign conventions :\nAll distances, object distance (u), image distance (v) and focal length f are measured from the optical centre.\nThe distances measured in the direction of incident ray are taken as positive and distances measured against the direction of incident ray are taken as negative.\nAll distances (heights) of objects and images above principal axis are taken as positive and those below the principal axis are taken as negative.\n50. For the two lenses, the sign conventions take the form\nu is- ve, if the object is in front of the lens. (Real object)\nu is +ve, if the object is virtual.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-EXEMPLER NOTES docx.txt\n50. For the two lenses, the sign conventions take the form\nu is- ve, if the object is in front of the lens. (Real object)\nu is +ve, if the object is virtual.\nv is \u2013 ve, if the image is on the same side as that of the object. (Virtual image )\nv is +ve, if the image is real.\nFocal length of a concave lens is taken as \u2013 ve.\nFocal length of a convex lens is taken as +ve.\n51. Lens formula for convex lens 1/v-1/u = 1/f\n52. The linear magnification produced by a lens is defined as the ratio of the size of the image (h\u2019) to the size of the object (h). It is represented by m i.e.,\n\n53. If the magnification of a lens is negative, then the image formed is inverted and real.\n54. If the magnification of a lens is positive, then the image formed is erect and virtual.\n55. Power is defined as the reciprocal of the focal length. Power is measured in dioptre.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nImportant Questions of Light Reflection and Refraction Class 10 Science Chapter 10\nQuestion 1.\nThe laws of reflection hold true for\n(a) plane mirrors only\n(b) concave mirrors only\n(c) convex mirrors only\n(d) all reflecting surface\nAnswer:\n(d) The laws of reflection holds true for all reflecting surface.\nQuestion 2.\nList four characteristics of the images formed by plane mirrors. (Delhi 2015, AI2011)\nAnswer:\nCharacteristics of the image formed by a plane mirror are\n(i) imagedistanceissameasthatofobjectdistance\n(ii) image formed is virtual and erect\n(iii) image formed is of the same size as that of the object\n(iv) image formed is laterally inverted (left appears right and right appears left).\nQuestion 3.\nState the two laws of reflection of light. (Delhi 2011)\nAnswer:\nLaws of reflection of light states that\n(i) The angle of incidence is equal to the angle of reflection.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nQuestion 3.\nState the two laws of reflection of light. (Delhi 2011)\nAnswer:\nLaws of reflection of light states that\n(i) The angle of incidence is equal to the angle of reflection.\n(ii) The incident ray, the reflected ray and the normal to the mirror at the point of incidence all lie in the same plane.\nQuestion 4.\nWhen an object is kept within the focus of a concave mirror, an enlarged image is formed behind the mirror. This image is\n(a) real\n(b) inverted\n(c) virtual and inverted\n(d) virtual and erect (2020)\nAnswer:\n(d) When an object is placed between the principal focus and pole of a concave mirror, an enlarged virtual and erect image is formed behind the mirror.\nQuestion 5.\nWhat is the magnification of the images formed by plane mirrors and why? (Delhi 2015)\nAnswer:\nMagnification of images formed by plane mirrors is unity because for plane mirrors, the size of the image formed is equal to that of the object.\nQuestion 6.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nMagnification of images formed by plane mirrors is unity because for plane mirrors, the size of the image formed is equal to that of the object.\nQuestion 6.\nDraw a labelled ray diagram to show the path of the reflected ray corresponding to an incident ray of light parallel to the principal axis of a convex mirror. Mark the angle of incidence and angle of reflection on it. (AI 2019)\nAnswer:\n\nQuestion 7.\nIf the image formed by a spherical mirror for all positions of the object placed in front of it is always erect and diminished , what type of mirror is it? Draw a labelled ray diagram to support your answer. (2018)\nAnswer:\nIf the image formed by a spherical mirror is always erect and diminished then it is convex mirror.\n\nQuestion 8.\nAn object is placed at a distance of 30 cm in front of a convex mirror of focal length 15 cm. Write four characteristics of the image formed by the mirror. (Delhi 2017)\nAnswer:\nFour characteristics of the image formed by the given convex mirror are :", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nFour characteristics of the image formed by the given convex mirror are :\n(i) Virtual\n(ii) Erect\n(iii) Diminished\n(iv) Image is always formed behind the mirror between pole and focus.\nQuestion 9.\nAn object is placed at a distance of 12 cm in front of a concave mirror of radius of curvature 30 cm. List four characteristics of the image formed by the mirror. (Delhi 2017)\nAnswer:\nRadius of curvature (R) = 30 cm, object distance is 12 cm in front of the mirror. Thus we can say that object is placed between focus and pole. Four characteristics of the image formed by die given concave mirror when object is placed between pole and focus are:\n(i) Virtual\n(ii) Erect\n(iii) Enlarged\n(iv) Image is formed behind the mirror\nQuestion 10.\nA ray of light is incident on a convex mirror as shown. Redraw the diagram and complete the path of this ray after reflection from the mirror. Mark angle of incidence and angle of reflection on it. (Delhi 2016)\n\nAnswer:\n\nQuestion 11.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n\nQuestion 11.\nName the type of mirrors used in the design of solar furnaces. Explain how high temperature is achieved by this device. (AI 2016)\nAnswer:\nConcave mirrors are used in the designing of solar furnaces.\nWhen a solar furnace is placed at the focus of a large concave mirror, it focuses a parallel beam of light on the furnace. Therefore, a high temperature is attained at the point after some time.\nQuestion 12.\n\u201cThe magnification produced by a spherical mirror is -3\u201d. List four informations you obtain from this statement about the mirror/ image. (AI 2016)\nAnswer:\nNegative sign of magnification indicates that the image is real and inverted. Since the image is real and inverted, the mirror is concave and magnification of -3 indicates that the image is magnified.\nQuestion 13.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nQuestion 13.\nAB and CD, two spherical mirrors, from parts of a hollow spherical ball with its centre at O as shown in the diagram. If arc AB = \\frac { 1 }{ 2 } arc CD, what is the ratio of their focal lengths? State which of the two mirrors will always form virtual image of an object placed in front of it and why? (Foreign 2016)\n\nAnswer:\nFocal length of a mirror is given by\nFocal length = \\frac { Radius of curvature }{ 2 }\nSince both the mirrors have same radius of curvature, therefore focal length of the two mirrors will be same, i.e.,\n\\frac { f_1 }{ f_2 } = \\frac { 1 }{ 1 }\nSince virtual image is always formed by convex mirror. The mirror AB will always form virtual image.\nQuestion 14.\nList two properties of the images formed by convex mirrors. Draw ray diagram in support of your answer. (Foreign 2016)\nAnswer:\nConvex mirrors always form diminished, virtual and erect images.\n\nQuestion 15.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nConvex mirrors always form diminished, virtual and erect images.\n\nQuestion 15.\nThe linear magnification produced by a spherical mirror is +3. Analyse this value and state the (i) type of mirror and (ii) position of the object with respect to the pole of the mirror. Draw a ray diagram to show the formation of image in this case. (Foreign 2016)\nAnswer:\nPositive value of the magnification indicates that image is virtual and erect.\n(i) Since the image is magnified, the mirror is concave.\n(ii) The object is between pole and focus of the mirror as shown\n\nThe image produced in second case will be real and inverted.\nQuestion 16.\nList four specific characteristics of the images of the objects formed by convex mirrors. (Delhi 2015)\nAnswer:\nRefer to answer 8\nQuestion 17.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nQuestion 16.\nList four specific characteristics of the images of the objects formed by convex mirrors. (Delhi 2015)\nAnswer:\nRefer to answer 8\nQuestion 17.\nDraw a ray diagram to show the path of the reflected ray corresponding to an incident ray which is directed towards the principal focus of a convex mirror. Mark on it the angle of incidence and the angle of reflection. (Delhi 2014)\nAnswer:\n\nQuestion 18.\nDraw a ray diagram to show the path of the reflected ray corresponding to an incident ray which is directed parallel to the principal axis of a convex mirror, Mark on it the angle of incidence and the angle of reflection. (Delhi 2014)\nAnswer:\nRefer to answer 6\nQuestion 19.\nDraw a ray diagram to show the path of the reflected ray corresponding to an incident ray of light parallel to the principal axis of a concave mirror. Mark the angle of incidence and angle of reflection on it. (Delhi 2014)\nAnswer:\n\nQuestion 20.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n\nQuestion 20.\nList two possible ways in which a concave mirror can produce a magnified image of an object placed in front of it. State the difference if any between these two images. (AI2014)\nAnswer:\nA concave mirror can produce a magnified image of an object when object is placed:\n(1) In between its pole and its focus\n(2) In between its focus and its centre of curvature.\nDifference,between these two images:\nThe image produced in first case will be virtual and erect.\nThe image produced in second case will be real and inverted.\nQuestion 21.\nThe image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should the position of the object be relative to the mirror? Draw ray diagram to justify your answer. (AI 2014)\nAnswer:\nThe position of the object should be between P and F\n\nQuestion 22.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nThe position of the object should be between P and F\n\nQuestion 22.\nThe linear magnification produced by a spherical mirror is +1/3. Analysing this value state the (i) type of mirror and (ii) position of the object with respect to the pole of the mirror. Draw any diagram to justify your answer. (AI 2014, Foreign 2014)\nAnswer:\n(i) Convex mirror\n(ii) Between infinity and the pole of the mirror.\n\nQuestion 23.\nThe linear magnification produced by a spherical mirror is -1. Analysing this value state the (i) type of mirror and (ii) position of the object with respect to the pole of the mirror. Draw any diagram to justify your answer. (Foreign 2014)\nAnswer:\n(i) Concave mirror because the image is real, inverted.\n(ii) Object is placed at C.\n\nQuestion 24.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(i) Concave mirror because the image is real, inverted.\n(ii) Object is placed at C.\n\nQuestion 24.\nThe linear magnification produced by a spherical mirror is -1/5. Analysing this value state the (i) type of spherical mirror and (ii) the position of the object with respect to the pole of the mirror. Draw ray diagram to justify your answer. (Foreign 2014)\nAnswer:\n(i) Concave mirror\n(ii) Object is placed beyond C.\n\nQuestion 25.\nDraw ray diagrams for the following cases when a ray of light:\n(i) passing through centre of curvature of a concave mirror is incident on it.\n(ii) parallel to principal axis is incident on convex mirror.\n(iii) is passing through focus of a concave mirror incident on it. (2020)\nAnswer:\n(i) Ray of light passing through centre of curvature of concave mirror, after reflection\n\n(ii) Ray of light parallel to the principal axis is incident on a convex mirror after reflection appear to diverge from the principal focus of a convex mirror.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(ii) Ray of light parallel to the principal axis is incident on a convex mirror after reflection appear to diverge from the principal focus of a convex mirror.\n\n(iii) Ray of light passing through focus of a concave mirror after reflection will emerge parallel\n\nQuestion 26.\nA concave mirror is used for image formation for different positions of an object. What inferences can be drawn about the following when an object is placed at a distance of 10 cm from the pole of a concave mirror of focal length 15 cm?\n(a) Position of the image\n(b) Size of the image\n(c) Nature of the image\nDraw a labelled ray diagram to justify your inferences. (2020)\nAnswer:\nGiven, f = -15 cm, u = -10 cm.\nThus the object is placed between the principal focus and pole of the mirror.\n(a) The position of the image will be behind the mirror.\n(b) The size of the image will be highly enlarged.\n(c) The nature of the image will be virtual and erect.\n\nQuestion 27.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(a) The position of the image will be behind the mirror.\n(b) The size of the image will be highly enlarged.\n(c) The nature of the image will be virtual and erect.\n\nQuestion 27.\nA concave mirror has a focal length of 20 cm. At what distance from the mirror should a 4 cm tall object be placed so that it forms an image at a distance of 30 cm from the mirror? Also calculate the size of the image formed. (AI 2019)\nAnswer:\nGiven f = -20 cm v = -30 cm u = ?\nUsing \\frac { 1 }{ v } + \\frac { 1 }{ u } = \\frac { 1 }{ f }\n\\frac { 1 }{ u } = \\frac { 1 }{ f } \u2013 \\frac { 1 }{ v } = \\frac { 1 }{ -20 } \u2013 \\frac { 1 }{ -30 } = \\frac { -3+2 }{ 60 }\n\u21d2 u = -60 cm\n\u2234 Object placed at 60 cm from the mirror.", + "Also magnification, m = \\frac { h\u2019 }{ h } = \\frac { -v }{ u }\n\u21d2 h\u2019 = \\frac { -(-30) }{ -60 } \u00d7 4 = -2 cm\n\u2234 The size of the image is 2 cm.\nQuestion 28.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n\u2234 Object placed at 60 cm from the mirror.\nAlso magnification, m = \\frac { h\u2019 }{ h } = \\frac { -v }{ u }\n\u21d2 h\u2019 = \\frac { -(-30) }{ -60 } \u00d7 4 = -2 cm\n\u2234 The size of the image is 2 cm.\nQuestion 28.\nThe image of a candle flame placed at a distance of 30 cm from a mirror is formed on a screen placed in front of the mirror at a distance of 60 cm from its pole. What is the nature of the mirror? Find its focal length. If the height of the flame is 2.4 cm, find the height of its image. State whether the image formed is erect or inverted. (Delhi 2017)\nGiven:\nObject distance, u = \u2013 30 cm, image size, h\u2019 = ?\nImage distance, v = \u2013 60 cm,\nObject size ,h = 2.4 cm,\nFocal length, f = ?", + "(Delhi 2017)\nGiven:\nObject distance, u = \u2013 30 cm, image size, h\u2019 = ?\nImage distance, v = \u2013 60 cm,\nObject size ,h = 2.4 cm,\nFocal length, f = ?\nUsing mirror formula,\n\\frac{1}{f}=\\frac{1}{v}+\\frac{1}{u} \\quad \\text { or } \\quad \\frac{1}{f}=\\frac{-1-2}{60}=\\frac{-3}{60}=-\\frac{1}{20}\nor f = \u2013 20 cm\nHence, focal length is 20 cm\nAlso, magnification, m = \\frac { h\u2019 }{ h } = \\frac { -v }{ u }\nor, m = \\frac { (-60)}{ (-30) } = -2 or \\frac { h\u2019 }{ h } = -2", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nor f = \u2013 20 cm\nHence, focal length is 20 cm\nAlso, magnification, m = \\frac { h\u2019 }{ h } = \\frac { -v }{ u }\nor, m = \\frac { (-60)}{ (-30) } = -2 or \\frac { h\u2019 }{ h } = -2\nh\u2019 = \u2013 2 \u00d7 2.4 = \u2013 4.8 cm\nAs the image formed is real, therefore the mirror is concave.\nThe height of the image is 4.8 cm.\nThe image formed is enlarged and inverted.\nQuestion 29.\nAn object 4 cm in height, is placed at 15 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror should a screen be placed to obtain a sharp image of the object. Calculate the height of the image. (Delhi 2017)\nAnswer:\nGiven : object distance, u = -15 cm,\nobject height, h = 4 cm, focal length f = -10 cm;\nImage distance, v = ?", + "Calculate the height of the image. (Delhi 2017)\nAnswer:\nGiven : object distance, u = -15 cm,\nobject height, h = 4 cm, focal length f = -10 cm;\nImage distance, v = ?\nUsing mirror formula,\n\\frac{1}{v}+\\frac{1}{u}=\\frac{1}{f} \\Rightarrow \\frac{1}{v}+\\frac{1}{(-15)}=\\frac{1}{-10} \\Rightarrow \\frac{1}{v}=\\frac{1}{15}-\\frac{1}{10}\nor \\frac{1}{v}=\\frac{10-15}{150}=\\frac{-5}{150}=\\frac{-1}{30} or v = -30", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nor \\frac{1}{v}=\\frac{10-15}{150}=\\frac{-5}{150}=\\frac{-1}{30} or v = -30\nIn order to obtain a sharp image of the object on the screen, screen should be placed at a distance of 30 cm in front of the mirror.\nAlso, magnification, m = \\frac{h^{\\prime}}{h}=\\frac{-v}{u}\nor \\frac{h^{\\prime}}{4}=-\\frac{(-30)}{(-15)} or h\u2019 = \\frac { -(30)\u00d74 }{ (15) } = -2 \u00d7 4\nor h\u2019 = -8 cm\nThus, the height of the image is 8 cm.\nQuestion 30.\nDraw the following diagram in which a ray of light is incident on a concave/convex mirror, on your answer sheet. Show the path of this ray, after reflection, in each case.\n\nAnswer:\nThe path of the rays are shown in figure.\n\nQuestion 31.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nThe path of the rays are shown in figure.\n\nQuestion 31.\nThe image of an object formed by a mirror is real, inverted and is of magnification -1. If the image is at a distance of 40 cm from the mirror, where is the object placed? Where would the image be if the object is moved 20 cm towards the mirror? State reason and also draw ray diagram for the new position of the object to justify your answer. (AI 2016)\nAnswer:\nSince the image formed by the mirror is real and inverted, therefore the mirror is concave and magnification of the mirror will be\nm = \u2013\\frac { v }{ u } \u21d2 -1 = \u2013\\frac { v }{ u } \u21d2 v = u\ni.e., object and image both are formed at the centre of curvature, i.e., 40 cm from the mirror.\nNow, if the object is moved 20 cm towards the mirror, the object will be at the focus of the mirror and therefore the image will be formed at infinity.\n\nQuestion 32.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nNow, if the object is moved 20 cm towards the mirror, the object will be at the focus of the mirror and therefore the image will be formed at infinity.\n\nQuestion 32.\nThe image formed by a spherical mirror is real, inverted and its magnification is -2. If the image is at a distance of 30 cm from the mirror, where is the object placed? Find the focal length of the mirror. List two characteristics of the image formed if the object is moved 10 cm towards the mirror. (AI 2016)\nAnswer:\nSince the image formed is real and inverted, the mirror is concave.", + "Find the focal length of the mirror. List two characteristics of the image formed if the object is moved 10 cm towards the mirror. (AI 2016)\nAnswer:\nSince the image formed is real and inverted, the mirror is concave.\nMagnification, m = \\frac { -v }{ u } \u21d2 -2 = \\frac { -v }{ u } \u21d2 v = 2u\nNow, if v = \u2013 30 cm then u = \u2013 15 cm\nAs focal length of the mirror is\nf = \\frac{u v}{u+v}=\\frac{-15 \\times-30}{-15-30}=f=\\frac{450}{-45} = -10 cm\nIf the object is shifted 10 cm towards the mirror, then the object is between principal focus and the optical centre and the image formed will be virtual and erect.\nQuestion 33.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nIf the object is shifted 10 cm towards the mirror, then the object is between principal focus and the optical centre and the image formed will be virtual and erect.\nQuestion 33.\nIf the image formed by mirror for all positions of the object placed in front of it is always virtual and diminished, state the type of the mirror. Draw a ray diagram in support of your answer. Where are such mirrors commonly used and why? (Foreign 2016, AI 2015)\nAnswer:\nRefer to answer 7.\nConvex mirrors are widely used as rear view mirrors in cars, motorcycles etc. It produces an erect image that is smaller in size than the object hence giving a wide view.\nQuestion 34.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nConvex mirrors are widely used as rear view mirrors in cars, motorcycles etc. It produces an erect image that is smaller in size than the object hence giving a wide view.\nQuestion 34.\nTo construct a ray diagram we use two rays of light which are so chosen that it is easy to determine their directions after reflection from the mirror. Choose these two rays and state the path of these rays after reflection from a concave mirror. Use these two rays to find the nature and position of the image of an object placed at a distance of 15 cm from a concave mirror of focal length 10 cm. (Delhi 2015, AI 2012)\nAnswer:\nWe use two rays of light, one passing through the centre of curvature of a concave mirror, and another is parallel to the principal axis. After reflection, the ray passing through the centre of a concave mirror is reflected back along the same path and the ray parallel to the principal axis will pass through the principal focus.\nu = -15 cm, f= -10 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nu = -15 cm, f= -10 cm\n\nFrom ray diagram, v = -30 cm, i.e., beyond C Nature of image is real, inverted and magnified.\nQuestion 35.\nDraw a ray diagram to show the path of the reflected ray in each of the following cases. A ray of light incident on a convex mirror :\n(a) strikes at its pole making an angle 0 from the principal axis.\n(b) is directed towards its principle focus.\n(c) is pardllel to its principal axis. (Foreign 2015)\nAnswer:\n\n(b) Refer to answer 17.\n(c) Refer to answer 6.\nQuestion 36.\nA spherical mirror produces an image of magnification -1 on a screen placed at a distance of 50 cm from the mirror.\n(a) Write the type of mirror.\n(b) Find the distance of the image from the object.\n(c) What is the focal length of the mirror?\n(d) Draw the ray diagram to show the image formation in this case.", + "(a) Write the type of mirror.\n(b) Find the distance of the image from the object.\n(c) What is the focal length of the mirror?\n(d) Draw the ray diagram to show the image formation in this case. (Delhi 2014, AI 2014)\nAnswer:\n(a) Concave mirror\n(b) Magnification, m = \u2013\\frac { v }{ u } or v = u\n\u2234 Distance of the image from the object is, v \u2013 u = 0", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Concave mirror\n(b) Magnification, m = \u2013\\frac { v }{ u } or v = u\n\u2234 Distance of the image from the object is, v \u2013 u = 0\n(c) As the image is formed at centre of curvature i.e., v = R.\n\u2234 focal length of the mirror, f = \\frac { -50 }{ 2 } = -25 cm\n(d) Refer to answer 23(ii).\nQuestion 37.\nA spherical mirror produces an image of magnification -1 on a screen placed at a distance of 40 cm from the mirror.\n(i) Write type of mirror.\n(ii) What is the nature of the image formed?\n(iii) How far is the object located from the mirror?\n(iv) Draw the ray diagram to show the image formation in this case. (Delhi 2014)\nAnswer:\n(i) This is a concave mirror.\n(ii) The image is real and inverted and of same size.", + "(iii) How far is the object located from the mirror?\n(iv) Draw the ray diagram to show the image formation in this case. (Delhi 2014)\nAnswer:\n(i) This is a concave mirror.\n(ii) The image is real and inverted and of same size.\n(iii) As m = \u2013 1\n\u2234 m = \u2013\\frac { v }{ u } \u21d2 -1 = \\frac { -v }{ u } \u21d2 u = v\nHence, object is located at centre of curvature i.e., at distance of 40 cm from the pole of the mirror,\n(iv) Refer to answer 23(ii).\nQuestion 38.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nHence, object is located at centre of curvature i.e., at distance of 40 cm from the pole of the mirror,\n(iv) Refer to answer 23(ii).\nQuestion 38.\nA spherical mirror produces an image of magnification -1.0 on a screen placed at a distance of 30 cm from the pole of the mirror.\n(i) Write the type of mirror in this case.\n(ii) What is the focal length of the mirror?\n(iii) What is the nature of the images formed?\n(iv) Draw the ray diagram to show the image formation in this case. (Delhi 2014)\nAnswer:\n(i) The mirror is concave mirror.\n(ii) Distance the image from the mirror = \u2013 30 cm\nMagnification, m = \\frac { -v }{ u }\nHere m = \u2013 1 and v = \u2013 30 cm\n-1 = \u2013\\frac { (-30) }{ u }\n\u2234 u = \u2013 30 cm\nAs v = u, object is placed at centre of curvature.", + "Therefore, focal length of the mirror,\nf = \\frac { -30 }{ 2 } = \u2013 15 cm\n(iii) Image formed is real and inverted and of the same size of the object.\n(iv) Refer to answer 23(ii).\nQuestion 39.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nf = \\frac { -30 }{ 2 } = \u2013 15 cm\n(iii) Image formed is real and inverted and of the same size of the object.\n(iv) Refer to answer 23(ii).\nQuestion 39.\nA student wants to project the image of a candle flame on a screen 48 cm in front of a mirror by keeping the flame at a distance of 12 cm from its pole.\n(a) Suggest the type of mirror he should use.\n(b) Find the linear magnification of the image produced.\n(c) How far is the image from its object?\n(d) Draw ray diagram to show the image formation in this case. (AI 2014)\nAnswer:\n(a) Concave mirror\n(b) Linear magnification,\nm = \\frac { -v }{ u } = \\frac { -(-48) }{ -12 } = -4\n(c) The distance between the image and the object\n= 48 \u2013 12 = 36 cm\n\nQuestion 40.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Concave mirror\n(b) Linear magnification,\nm = \\frac { -v }{ u } = \\frac { -(-48) }{ -12 } = -4\n(c) The distance between the image and the object\n= 48 \u2013 12 = 36 cm\n\nQuestion 40.\nA student wants to obtain an erect image of an object using a concave mirror of 12 cm focal length. What should be the range of distance of the candle flame from the mirror? State the nature and size of the image he is likely to observe. Draw a ray diagram to show the image formation in this case. (Foreign 2014)\nAnswer:\nTo obtain an erect image, the object is placed in between pole and the focus of the concave mirror. So range of distance of the candle llame from the mirror is in between 12 cm.\nNature of the image = Virtual and erect.\nSize of the image = Enlarged\n\nQuestion 41.\nA student wants to obtain an erect image of a candle flame using a concave mirror of focal length 15 cm.", + "Nature of the image = Virtual and erect.\nSize of the image = Enlarged\n\nQuestion 41.\nA student wants to obtain an erect image of a candle flame using a concave mirror of focal length 15 cm. What should be the range of distance of the candle flame from the mirror?", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nQuestion 41.\nA student wants to obtain an erect image of a candle flame using a concave mirror of focal length 15 cm. What should be the range of distance of the candle flame from the mirror?\nState the nature and size of the image he is likely to observe. Draw a ray diagram to show the image formation in this case. (Foreign 2014)\nAnswer:\nTo obtain an erect image of an object, the object should be placed in between pole and focus. Range of distance of the candle flame from the mirror is in between 15 cm.\nNature of the image = Virtual and erect\nSize of the image = Enlarged\nFor ray diagram, refer to answer 40.\nQuestion 42.\nA student has a concave mirror of 20 cm focal length and he wants to see an erect image of his face in the mirror. What should be the range of distance of the mirror from his face? State the nature and size of the image he is likely to observe. Draw a ray diagram to justify your answer. (Foreign 2014)\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nFocal length of a concave mirror = 20 cm Range will be in between 20 cm.\nNature of the image = Virtual and erect\nSize of the image = Enlarged\nFor ray diagram, refer to answer 40.\nQuestion 43.\nMention the types of mirrors used as (i) rear view mirrors, (ii) shaving mirrors. List two reasons to justify your answer in each case. (Delhi 2013, Delhi 2012)\nAnswer:\n(i) Convex mirror is used as rear view mirror because\n(a) it gives erect image.\n(b) it gives diminished image thus provides wider view of traffic behind the vehicle.\n(ii) Concave mirror is used as shaving mirror because\n(a) it gives erect image when mirror is close to the face.\n(b) it gives enlarged image of the face so that a person can shave safely.\nQuestion 44.\nCalculate the magnification of the image of an object placed perpendicular to the principal axis of a concave mirror of focal length 15 cm. The object is at a distance of 20 cm from the mirror.", + "Question 44.\nCalculate the magnification of the image of an object placed perpendicular to the principal axis of a concave mirror of focal length 15 cm. The object is at a distance of 20 cm from the mirror. (Delhi 2013)\nAnswer:\nGiven, focal length of concave mirror,", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nGiven, focal length of concave mirror,\nf = -15 cm\nObject distance, u = -20 cm\nImage distance, v = ?\nUsing mirror formula,\n\nUsing magnification formula,\nm = \u2013\\frac { v }{ u } = -(\\frac { -60 }{ -20 }) or m = -3\nSo, the magnification, m = -3.\nQuestion 45.\nTo construct ray diagram we use two light rays which are so chosen that it is easy to know their directions after reflection from the mirror. List these two rays and state the path of these rays after reflection. Use these rays to locate the image of an object placed between centre of curvature and focus of a concave mirror. (AI2012)\nAnswer:\nA ray parallel to the principal axis, after reflection, will pass through the principal focus in case of a concave mirror or appear to diverge from the principal focus in case of a convex mirror.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nA ray parallel to the principal axis, after reflection, will pass through the principal focus in case of a concave mirror or appear to diverge from the principal focus in case of a convex mirror.\nA ray passing through the centre of a curvature of a concave mirror or directed in the direction of the centre of curvature of a convex mirror, after reflection, is reflected back along the same path. The light rays come back along the same path because the incident rays fall on the mirror along the normal to the reflecting surface.\n\nQuestion 46.\nState the types of mirrors used for (i) headlights and (ii) rear view mirrors, in motorcycles. Give reason to justify your answer in each case. (AI 2012)\nAnswer:\n(i) Concave mirrors are used in headlights of cars to get powerful beams of light.\n(ii) Convex mirrors are used as rear-view mirrors of vehicle to get a wider field of view and and erect image of traffic behind.\nQuestion 47.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(ii) Convex mirrors are used as rear-view mirrors of vehicle to get a wider field of view and and erect image of traffic behind.\nQuestion 47.\nAn object is placed between infinity and the pole of a convex mirror. Draw a ray diagram and also state the position, the relative size and the nature of the image formed. (AI 2011)\nAnswer:\n\nPosition: Image is formed between pole and principal focus of the mirror.\nRelative size : Image formed is diminished.\nNature : Image formed is virtual and erect.\nQuestion 48.\nWith the help of a ray diagram explain why a convex mirror is preferred for rear view mirrors in the motor cars. (Foreign 2011)\nAnswer:\nConvex mirror is preferred for rear view mirrors in motor cars because no matter where the object is located in front of convex mirror, it always gives erect and diminished image of the object, so that driver is able to see the large traffic view in small area and the image is erect. This can be interpreted from the following diagram.\n\nQuestion 49.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nQuestion 49.\nAn object 4.0 cm in size, is placed 25.0 cm in front of a concave mirror of focal length 15.0 cm.\n(i) At what distance from the mirror should a screen be placed in order to obtain a sharp image?\n(ii) Find the size of the image.\n(iii) Draw a ray diagram to show the formation of image in this case. (2020)\nAnswer:\n(i) Given, h = 4 cm,\nu = -25 cm (concave mirror), f = -15 cm\nUsing mirror formula,\n\nThus, the image is real and inverted.\n\nQuestion 50.\n(a) A concave mirror of focal length 10 cm can produce a magnified real as well as virtual image of an object placed in front of it. Draw ray diagrams to justify this statement,\n(b) An object is placed perpendicular to the principal axis of a convex mirror of focal length 10 cm. The distance of the object from the pole of the mirror is 10 cm. Find the position of the image formed.", + "Draw ray diagrams to justify this statement,\n(b) An object is placed perpendicular to the principal axis of a convex mirror of focal length 10 cm. The distance of the object from the pole of the mirror is 10 cm. Find the position of the image formed. (2020)\nAnswer:\n(a) A magnified real image is produced in a concave mirror when the object is placed between principal focus and centre of curvature.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) A magnified real image is produced in a concave mirror when the object is placed between principal focus and centre of curvature.\n\nA magnified virtual image is produced in a concave mirror when the object is placed between the pole and the principle focus of the mirror.\n\n(b) Given, f = +10 cm (convex mirror) and u = -10 cm\nFrom mirror formula,\n\nQuestion 51.\n(a) A security mirror used in a big showroom has radius of curvature 5 m. If a customer is standing at a distance of 20 m from the cash counter, find the position, nature and size of the image formed in the security mirror.\n(b) Neha visited a dentist in his clinic. She observed that the dentist was holding an instrument fitted with a mirror. State the nature of this mirror and reason for its use in the instrument used by dentist. (2020)\nAnswer:\n(a) Given radius of curvature of the mirror,\nR = 5 m\n\u2234 Focal length, f = R/2 = 2.5 m (convex mirror) and u = -20 m\nFrom mirror formula,", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Given radius of curvature of the mirror,\nR = 5 m\n\u2234 Focal length, f = R/2 = 2.5 m (convex mirror) and u = -20 m\nFrom mirror formula,\n\\frac{1}{f}=\\frac{1}{v}+\\frac{1}{u} \\text { or } \\frac{1}{v}=\\frac{1}{f}-\\frac{1}{u}\n= \\frac{1}{2.5}-\\frac{1}{-20}=\\frac{-20-2.5}{-20 \\times 2.5}\n\u2234 v = 2.22 m\nThus, the image is formed 2.22 m behind the mirror. The image is diminished, virtual and erect.\n(b) Concave mirrors are used by dentist. Dentist use it as it is a converging mirror and when used at close range forms a highly enlarged, virtual and erect image of the object.\nQuestion 52.", + "The image is diminished, virtual and erect.\n(b) Concave mirrors are used by dentist. Dentist use it as it is a converging mirror and when used at close range forms a highly enlarged, virtual and erect image of the object.\nQuestion 52.\n(a) To construct a ray diagram we use two rays which are so chosen that it is easy to know their directions after reflection from the mirror. Use these two rays and draw ray diagram to locate the image of an object placed between pole and focus of a concave mirror.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(b) A concave mirror produces three times magnified image on a screen. If the objects placed 20 cm in front of the mirror, how far is the screen from the object? (Delhi 2017)\nAnswer:\n(a) Two lights rays whose path of reflection are priorly known are :\n(i) When the incident ray passes through the centre of curvature of a concave mirror, it gets reflected in the same path.\n(ii) When the ray is incident obliquely to the principal axis, towards the pole of mirror, it gets reflected back by making equal angles with the principal axis (laws of reflections).\nSuppose an object is placed between focus and pole of the concave mirror. Then by using the above two rays, the image of the object can be located as\n\nImage formed is virtual, erect, magnified and it is formed behind the mirror.\n(b) Given : Magnification, m = \u2013 3\nObject distance, u = \u2013 20 cm\nMagnification, m = \\frac { -v }{ u } or -3 = \\frac { -v }{ -20 }\nor v = -60 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(b) Given : Magnification, m = \u2013 3\nObject distance, u = \u2013 20 cm\nMagnification, m = \\frac { -v }{ u } or -3 = \\frac { -v }{ -20 }\nor v = -60 cm\nThe screen is placed in front of the mirror at a distance of 60 cm from the pole.\nThus, the screen is placed 40 cm (= 60 cm \u2013 20 cm) away from the object.\nQuestion 53.\n(a) If the image formed by a mirror for all positions of the object placed in front of it is always diminished, erect and virtual, state the type of the mirror and also draw a ray diagram to justify your answer. Write one use such mirrors are put to and why?\n(b) Define the radius of curvature of spherical mirrors. Find the nature and focal length of a spherical mirror whose radius of curvature is +24 cm. (AI2017)\nAnswer:\n(a) If the image formed by a mirror for all positions of the object placed in front of it is always diminished, erect and virtual then the mirror is convex mirror.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) If the image formed by a mirror for all positions of the object placed in front of it is always diminished, erect and virtual then the mirror is convex mirror.\nThe ray diagrams for the formation of image by a convex mirror for the first position when the object is at infinity and the second position when the object is at a finite distance from the mirror are shown.\n\nUse of Convex Mirrors\nConvex mirrors are commonly used as rear-view (wing) mirrors in vehicles because they always give an erect, though diminished image. Also, they have a wider field of view as they are curved outwards. Thus, convex mirrors enable the driver to view a large area.\n(b) Radius of Curvature: The radius of the sphere of which the reflecting surface of a spherical mirror forms a part, is called the radius of curvature of the mirror. It is represented by the letter R.\n\u2235 The radius of curvature is equal to twice the focal length.\n\u2234 R = 2f", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n\u2235 The radius of curvature is equal to twice the focal length.\n\u2234 R = 2f\nIf R = +24 cm \u2234 f = \\frac { R }{ 2 } = \\frac { 24 }{ 2 } = 12 cm\nSince the radius of curvature is positive, the mirror is convex mirror. Hence the nature of the image is virtual and erect.\nQuestion 54.\n(a) Define the following terms in the context of spherical mirrors:\n(i) Pole\n(ii) Centre of curvature\n(iii) Principal axis\n(iv) Principal focus\n(b) Draw ray diagrams to show the principal focus of a\n(i) Concave mirror (ii) Convex mirror\n(c) Consider the following diagram in which M is a mirror and P is an object and Q is its magnified image formed by the mirror.\n\nState the type of the mirror M and one characteristic property of the image Q. (Delhi 2016)\nAnswer:\n(a) (i) Pole : The centre of the reflecting surface of a spherical mirror is a point called the pole. It lies in the surface of the mirror and its represented by the letter P.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) (i) Pole : The centre of the reflecting surface of a spherical mirror is a point called the pole. It lies in the surface of the mirror and its represented by the letter P.\n(ii) Centre of curvature: The reflecting surface of a spherical mirror is a part of a sphere which has a centre. This point is called the centre of curvature of spherical mirror and is represented by the letter C.\n(iii) Principal axis : An imaginary line passing through the pole and the centre of curvature of a spherical mirror and normal to the mirror at its pole is called principal axis.\n(iv) Principal focus : Incident rays parallel to principal axis, after reflection either converge to as appear to diverge from a fixed point on the principal axis known as principal focus of the spherical mirror.\n\n(c) The object is placed between focus and pole of the mirror and a magnified image is formed behind the mirror.\n\u2234 The mirror is concave and image formed is virtual and erect.\nQuestion 55.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(c) The object is placed between focus and pole of the mirror and a magnified image is formed behind the mirror.\n\u2234 The mirror is concave and image formed is virtual and erect.\nQuestion 55.\nIt is desired to obtain an erect image of an object, using concave mirror of focal length of 12 cm.\n(i) What should be the range of distance of a , object placed in front of the mirror?\n(ii) Will the image be smaller or larger than the object? Draw ray diagram to show the formation of image in this case.\n(iii) Where will the image of this object be, if it is placed 24 cm in front of the mirror? Draw ray diagram for this situation also to justify your answer.\nShow the positions of pole, principal focus and the centre of curvature in the above ray diagrams. (AI2016)\nAnswer:\nGiven : focal length of the concave mirror f = 12 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nShow the positions of pole, principal focus and the centre of curvature in the above ray diagrams. (AI2016)\nAnswer:\nGiven : focal length of the concave mirror f = 12 cm\n(i) If the object is placed between the pole and focus of the concave mirror, then the image formed is virtual and erect. Iheretore, the range of distance of the object should be 0 < u <. 12 cm.\n(ii) The image formed will be enlarged as shown below.\n\n(iii) If the object is placed 24 cm in front of the mirror i.e., at the centre of the curvature then the image will also be formed at the centre of the curvature.\n\nQuestion 56.\nSuppose you have three concave mirrors A, B and C of focal lengths 10 cm, 15 cm and 20 cm. For each concave mirror you perform the experiment of image formation for three values of object distances of 10 cm, 20 cm and 30 cm. By giving reason, answer the following:\n(a) For the three object distances, identify the mirror/mirrors which will form an image of magnification -1.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(a) For the three object distances, identify the mirror/mirrors which will form an image of magnification -1.\n(b) Out of the three mirrors, identify the mirror which would be preferred to be used for shaving purposes/make up.\n(c) For the mirror B draw ray diagram for image formation for object distances 10 cm and 20 cm. (Foreign 2016)\nAnswer:\nGiven fa = 10 cm, fb = 15 cm, fc = 20 cm\nua = 10 cm, ub = 20 cm, uc = 30 cm\n(a) Magnification of -1 implies that size of image is same as that of object or image is formed at the same distance as of the object. This is the case when the object distance, u = 2f, i.e., when the object is at the centre of the curvature.\nFor fa, ub and for fb, uc, we get magnification \u2013 1.", + "This is the case when the object distance, u = 2f, i.e., when the object is at the centre of the curvature.\nFor fa, ub and for fb, uc, we get magnification \u2013 1.\n(b) Concave mirror forms virtual, erect and magnified image when the object is between focus and pole of the mirror, i.e., direct distance should be less than the focal length of the mirror.\nFor object distance 10 cm, mirrors of focal length fb = 15 cm and fc = 20 cm can be used.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nFor object distance 10 cm, mirrors of focal length fb = 15 cm and fc = 20 cm can be used.\n\nQuestion 57.\nA student has focused the image of a candle flame on a white screen using a concave mirror. The situation is a given below:\nLength of fhe flame = 1.5 cm\nFocal length of the mirror = 12 cm\nDistance of flame from the mirror = 18 cm\nIf the flame is perpendicular to the principal axis of the mirror, then calculate the following:\n(a) Distance of the image from the mirror\n(b) Length of the image\nIf the distance between the mirror and the flame is reduced to 10 cm, then what would be observed on the screen? Draw ray diagram to justify your answer from this situation. (Foreign 2015)\nAnswer:\nGiven: focal length of the concave mirror, f = \u2013 12 cm\nLength of the flame, h = 1.5 cm\nDistance of flame from the mirror, u = -18 cm\n\n(b) Let h\u2019 be the length of the image.", + "(Foreign 2015)\nAnswer:\nGiven: focal length of the concave mirror, f = \u2013 12 cm\nLength of the flame, h = 1.5 cm\nDistance of flame from the mirror, u = -18 cm\n\n(b) Let h\u2019 be the length of the image.\n\u2235 Magnification, m = \\frac { h\u2019 }{ h } = \\frac { -v }{ u }\n\u2234 h\u2019 = \\frac{-v \\times h}{u}=\\frac{-(-36) \\times 1.5}{-18} = -3 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(b) Let h\u2019 be the length of the image.\n\u2235 Magnification, m = \\frac { h\u2019 }{ h } = \\frac { -v }{ u }\n\u2234 h\u2019 = \\frac{-v \\times h}{u}=\\frac{-(-36) \\times 1.5}{-18} = -3 cm\nIf the distance between the mirror and the flame is reduced to 10 cm, then\n\\frac{1}{v}=\\frac{1}{f}-\\frac{1}{u}=\\frac{1}{-12}-\\frac{1}{-10}=\\frac{1}{60}\n\u2234 v = 60 cm\nHence, image is formed behind the mirror.\n\nQuestion 58.\nA student wants to project the image of a candle flame on the walls of school laboratory by using a mirror.\n(a) Which type of mirror should he use and why?\n(b) At what distance in terms of focal length \u2018f\u2019 of the mirror should he place the candle flame so as to get the magnified image on the wall?", + "(a) Which type of mirror should he use and why?\n(b) At what distance in terms of focal length \u2018f\u2019 of the mirror should he place the candle flame so as to get the magnified image on the wall?\n(c) Draw a ray diagram to show the formation of image in this case.\n(d) Can he use this mirror to project a diminished image of the candle flame on the same wall? State \u2018how\u2019 if your answer is \u2018yes\u2019 and \u2018why not\u2019 if your answer is \u2018no\u2019 (Delhi 2014)\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(d) Can he use this mirror to project a diminished image of the candle flame on the same wall? State \u2018how\u2019 if your answer is \u2018yes\u2019 and \u2018why not\u2019 if your answer is \u2018no\u2019 (Delhi 2014)\nAnswer:\n(a) He should use concave mirror to get image of candle flame on the walls of school laboratory. Because concave mirror is a converging mirror and produce real image.\n(b) He should place the candle flame in between centre of curvature C and principal focus F of the mirror to get the magnified image on the wall.\n\n(d) Yes, he can use concave mirror to project a diminished image of the candle flame on the same wall. He has to place the candle flame beyond centre of curvature to get diminished image.\nQuestion 59.\nDefine the following terms in case of a concave mirror:\n(a) Pole\n(b) Radius of curvature\n(c) Principal axis\n(d) Principal focus", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nQuestion 59.\nDefine the following terms in case of a concave mirror:\n(a) Pole\n(b) Radius of curvature\n(c) Principal axis\n(d) Principal focus\nSuppose you want to observe an erect image of a candle flame using a concave mirror of focal length 20 cm. State the range of distance of the candle flame from the mirror. List two other characteristics of the observed image. Draw a ray diagram to show the formation of image in this case. (Delhi 2013)\nAnswer:\n(a) Refer to answer 54(a) (i).\n(b) Refer to answer 53(b).\n(c) Refer to answer 54(a) (iii).\n(d) Refer to answer 54(a) (iv).\nIf we want to get an erect image of a candle flame using concave mirror of focal length 20 cm, then we must place the object between a pole and focus of the mirror.", + "(c) Refer to answer 54(a) (iii).\n(d) Refer to answer 54(a) (iv).\nIf we want to get an erect image of a candle flame using concave mirror of focal length 20 cm, then we must place the object between a pole and focus of the mirror. If u is the image distance, then 0 < u < 20 cm\nOther two characteristics of the observed image are\n(i) Image is virtual and erect\n(ii) Image is enlarged\nThe ray diagram for this situation can be drawn as follows:\n\nQuestion 60.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nOther two characteristics of the observed image are\n(i) Image is virtual and erect\n(ii) Image is enlarged\nThe ray diagram for this situation can be drawn as follows:\n\nQuestion 60.\nList the sign conventions for reflection of light by spherical mirrors. Draw a diagram and apply these conventions in the determination of focal length of a spherical mirror which forms a three times magnified real image of an object placed 16 cm infront of it. (Delhi 2012)\nAnswer:\nSign Convention for Reflection by Spherical Mirrors : While dealing with the reflection of light by spherical mirrors, we shall follow a set of sign conventions called the New Cartesian Sign Convention, the conventions are as follows:\n(i) The object is always placed to the left of the mirror. This implies that the light from the object falls on the mirror from the left-hand side.\n(ii) All distances parallel to the principal axis are measured from the pole of the mirror.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(ii) All distances parallel to the principal axis are measured from the pole of the mirror.\n(iii) All the distances measured to the right of the origin (along + x-axis) are taken as positive while those measured to the left of the origin (along \u2013 x-axis) are taken as negative.\n(iv) Distances measured perpendicular to and above the principal axis (along +y-axis) are taken as positive.\n(v) Distances measured perpendicular to and below the principal axis (along-y-axis) are taken as negative.\n\nGiven that m = -3 (real image), u = -16 cm\nMagnification, m = \\frac { -v }{ u }\n\nQuestion 61.\nList the new Cartesian sign convention for reflection of light by spherical mirrors. Draw a diagram and apply these conventions for calculating the focal length and nature of a spherical mirror which forms a 1/3 times magnified virtual image of an object placed 18 cm in front of it. (AI 2012)\nAnswer:\nRefer to answer 60.\nGiven that m = +\\frac { 1 }{ 3 } (virtual image), u = -18 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 60.\nGiven that m = +\\frac { 1 }{ 3 } (virtual image), u = -18 cm\nMagnification, m = \u2013\\frac { v }{ u }\n\\frac { 1 }{ 3 } = \\frac { -v }{ -18 } \u21d2 v = 6 cm\nmirror formula \\frac { 1 }{ v } + \\frac { 1 }{ u } = \\frac { 1 }{ f }\n\\frac { 1 }{ 6 } \u2013 \\frac { 1 }{ 18 } = \\frac { 1 }{ f } \u21d2 \\frac { 3-1 }{ 18 } = \\frac { 1 }{ f } \u21d2 f = 9 cm\nAs the value of focal length is positive, the mirror used is convex mirror.\nQuestion 62.\nName the type of mirror used in the following situations\n(i) Headlights of a car\n(ii) Rear-view mirror of vehicle\n(iii) Solar furnace\nSupport your answer with reason.", + "Question 62.\nName the type of mirror used in the following situations\n(i) Headlights of a car\n(ii) Rear-view mirror of vehicle\n(iii) Solar furnace\nSupport your answer with reason. (Foreign 2012)\nAnswer:\n(i) Refer to answer 46(i).\n(ii) Refer to answer 46(ii).\n(iii) Concave mirrors are used in solar furnaces to concentrate sunlight to produce heat.\nQuestion 63.\nWhat is meant by power of a lens? (Delhi 2015)\nAnswer:\nPower is the degree of convergence or divergence of light rays achieved by a lens.\nIt is defined as the reciprocal of its focal length.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nWhat is meant by power of a lens? (Delhi 2015)\nAnswer:\nPower is the degree of convergence or divergence of light rays achieved by a lens.\nIt is defined as the reciprocal of its focal length.\ni.e., P = \\frac { 1 }{ f }\nQuestion 64.\nAn object is placed at a distance of 15 cm from a convex lens of focal length 20 cm. List four characteristics (nature, position, etc.) of the image formed by the lens. (AI2017)\nAnswer:\nGiven : Object distance, u = \u2013 15 cm\nFocal length, f = + 20 cm\nUsing lens formula, As |u| < |f|\nThe object is placed between F and optical centre of lens.\nThus, the four characteristics of the image formed by the convex lens are:\n(i) Erect\n(ii) Virtual\n(iii) Enlarged image,\n(iv) Image is formed on the same side of the lens as the object.\nQuestion 65.\nWhat is meant by power of a lens? What does its sign (+ve or -ve) indicate?", + "Question 65.\nWhat is meant by power of a lens? What does its sign (+ve or -ve) indicate? State its S.I. unit related to focal length of a lens. (Delhi 2016)\nAnswer:\nRefer to answer 63.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nQuestion 65.\nWhat is meant by power of a lens? What does its sign (+ve or -ve) indicate? State its S.I. unit related to focal length of a lens. (Delhi 2016)\nAnswer:\nRefer to answer 63.\nPositive sign (+) of power indicates that lens is convex and negative sign (-) of power indicates that lens is concave.\nIf focal length (f) is expressed in metres, then, power is expressed in dioptres. The SI unit of power is dioptre. Thus, 1 dioptre is the power of lens whose focal length is 1 metre. 1 D = 1 m-1\nQuestion 66.\nThe refractive indices of glass and water with respect to air are 3/2 and 4/3 respectively. If speed of light in glass is 2 \u00d7 108 m/s, find the speed of light in water. (AI 2016)\nAnswer:\n\nQuestion 67.\nThe absolute refractive indices of glass and water are 4/3 and 3/2 respectively.", + "If speed of light in glass is 2 \u00d7 108 m/s, find the speed of light in water. (AI 2016)\nAnswer:\n\nQuestion 67.\nThe absolute refractive indices of glass and water are 4/3 and 3/2 respectively. If the speed of light in glass is 2 \u00d7 108 m/s, calculate the speed of light in (i) vacuum, (ii) water. (AI 2015)\nAnswer:\nGiven that: ng = \\frac { 4 }{ 3 }, nw = \\frac { 3 }{ 2 }, vg = 2 \u00d7 108 m/s", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nGiven that: ng = \\frac { 4 }{ 3 }, nw = \\frac { 3 }{ 2 }, vg = 2 \u00d7 108 m/s\nAbsolute refractive index of a medium, nm = \\frac { c }{ v }\nwhere, c is the speed of light in vacuum and v is the speed of light in medium.\n\nNote: The values given in question are not correct as the speed of light in vacuum is 3 \u00d7 108 m/s\nQuestion 68.\n\u201cA ray of light incident on a rectangular glass slab immersed in any medium emerges parallel to itself.\u201d Draw labelled ray diagram to justify the statement\u201d. (Delhi 2013)\nAnswer:\n\nQuestion 69.\nThe absolute refractive indices of glass and water are 1.5 and 1.33 respectively. In which medium does light travel faster? Calculate the ratio of speeds of light in the two media.", + "(Delhi 2013)\nAnswer:\n\nQuestion 69.\nThe absolute refractive indices of glass and water are 1.5 and 1.33 respectively. In which medium does light travel faster? Calculate the ratio of speeds of light in the two media. (Delhi 2013 C)\nAnswer:\nGiven : refractive index of glass, ng = 1.5\nRefractive index of water, nw = 1.33\nSince, refractive index of medium,\n\nFor glass ng = \\frac { c }{ v_g } \u2026\u2026\u2026 (i)\nFor water nw = \\frac { c }{ v_w } \u2026\u2026\u2026 (ii)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nRefractive index of water, nw = 1.33\nSince, refractive index of medium,\n\nFor glass ng = \\frac { c }{ v_g } \u2026\u2026\u2026 (i)\nFor water nw = \\frac { c }{ v_w } \u2026\u2026\u2026 (ii)\nSince velocity of light in medium is inversely proportional to its refractive index, the light will travel faster in optically rarer medium i.e., water.\nDividing (i) by (ii),\n\nSo, the ratio of vg and vw is 1.33 : 1.5.\nQuestion 70.\nTo construct a ray diagram we use two light rays which are so chosen that it is easy to know their directions after refraction from the lens. List these two rays and state the path of these rays after refraction. Use these two rays to locate the image of an object placed between \u2018f\u2019 and \u20182f\u2019 of a convex lens. (Foreign 2012)\nAnswer:\nThe two rays are :", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nThe two rays are :\n(i) A ray of light from the object, parallel to the principal axis, after refraction from a convex lens, passes through the principal focus on the other side of the leps and in case of concave lens, a ray appears to diverge from the principal focus located on the same side of the lens.\n\n(ii) A ray of light passing through the optical centre of a lens will emerge without any deviation.\n\nWhen object is placed between F and 2F.\n\nQuestion 71.\n(a) Water has refractive index 1.33 and alcohol has refractive index 1.36. Which of the two medium is optically denser? Give reason for your answer.\n(b) Draw a ray diagram to show the path of a ray of light passing obliquely from water to alcohol.\n(c) State the relationship between angle of incidence and angle of refraction in the above case. (2020)\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(c) State the relationship between angle of incidence and angle of refraction in the above case. (2020)\nAnswer:\n(a) Here, alcohol is optically denser medium as its refractive index is higher than that of water. When we compare the two media, the one with larger refractive index is called the optically denser medium than the other as the speed of light is lower in this medium.\n(b) Since light is travelling from water (rarer medium) to alcohol (denser medium), it slows down and bends towards the normal.\n\nwhere i = angle of incidence and r = angle of refraction.\n(c) According to Snell\u2019s law,\n\\frac{\\sin i}{\\sin r}=\\frac{\\mu_{\\text {alcohol }}}{\\mu_{\\text {water }}}=\\frac{1.36}{1.33} = 1.0225\n\u2234 sin i = 1.0225 \u00d7 sin r\nQuestion 72.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(c) According to Snell\u2019s law,\n\\frac{\\sin i}{\\sin r}=\\frac{\\mu_{\\text {alcohol }}}{\\mu_{\\text {water }}}=\\frac{1.36}{1.33} = 1.0225\n\u2234 sin i = 1.0225 \u00d7 sin r\nQuestion 72.\nThe refractive index of a medium V with respect to a medium \u2018y\u2019 is 2/3 and the refractive index of medium \u2018y\u2019 with respect to medium \u2018z\u2019 is 4/3. Find the refractive index of medium \u2018z with respect to medium V. If the speed of light in medium \u2018x\u2019 is 3 \u00d7 108 m s-1, calculate the speed of light in medium \u2018y\u2019.", + "Find the refractive index of medium \u2018z with respect to medium V. If the speed of light in medium \u2018x\u2019 is 3 \u00d7 108 m s-1, calculate the speed of light in medium \u2018y\u2019. (2020)\nAnswer:\nGiven, refractive index of medium x with respect to y,\ny\u00b5x = \\frac { 2 }{ 3 }\nRefractive index of medium y with respect to z,\nz\u00b5y = \\frac { 4 }{ 3 }\n\u2234 Refractive index of medium x with respect to z,\nz\u00b5x = y\u00b5x . z\u00b5y = \\frac { 2 }{ 3 } \u00d7 \\frac { 4 }{ 3 } = \\frac { 8 }{ 9 }\n\u2234 Refractive index of medium z with respect to x,\nx\u00b5y = \\frac { 1 }{ ^z\u00b5_x } = \\frac { 9 }{ 8 }\nNow speed of light in x = 3 \u00d7 108 m/s\nSpeed of light in y, vy = ?", + "\u21d2 vy = \\frac { 2 }{ 3 } \u00d7 3 \u00d7 108 = 2 \u00d7 108 m/s\nQuestion 73.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nx\u00b5y = \\frac { 1 }{ ^z\u00b5_x } = \\frac { 9 }{ 8 }\nNow speed of light in x = 3 \u00d7 108 m/s\nSpeed of light in y, vy = ?\n\n\u21d2 vy = \\frac { 2 }{ 3 } \u00d7 3 \u00d7 108 = 2 \u00d7 108 m/s\nQuestion 73.\nA real image 2/3rd of the size of an object is formed by a convex lens when the object is at a distance of 12 cm from it. Find the focal length of the lens. (AI 2019)\nAnswer:\nGiven, h\u2019 = \\frac { 2 }{ 3 }h, u = -12 cm\n\n\u2234 Focal length of the convex lens = 4.8 cm\nQuestion 74.\nState the laws of refraction of light. Explain the term absolute refractive index of a medium\u2019 and write an expression to relate it with the speed of light in vaccum.", + "State the laws of refraction of light. Explain the term absolute refractive index of a medium\u2019 and write an expression to relate it with the speed of light in vaccum. (2018)\nAnswer:\n(a) Laws of refraction of light:\n(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.\n(ii) The ratio of sine of angle of incidence to the sine of the angle of refraction is constant, for the light of a given colour and for the given pair of media.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(ii) The ratio of sine of angle of incidence to the sine of the angle of refraction is constant, for the light of a given colour and for the given pair of media.\nThis law is also known as Snell\u2019s law of refraction.\n\\frac { sin i }{ sin r } = constant,\nwhere i is the angle of incidence and r is the angle of refraction.\nThis constant value is called refractive index of the second medium with respect to the first when the light travels from first medium to second medium.\n\u21d2 constant = n21 = \\frac { v_1 }{ v_2 } \u2234\\frac { sin i }{ sin r } = \\frac { v_1 }{ v_2 }\nIf n is the absolute refractive index of the medium, c is the velocity of light in vacuum and v is the speed of light in a given medium, then n = cl v.\nQuestion 75.\nWhat is meant by power of a lens? Write its SI unit. A student uses a lens of focal length 40 cm and another of -20 cm.", + "What is meant by power of a lens? Write its SI unit. A student uses a lens of focal length 40 cm and another of -20 cm. Write the nature and power of each lens. (2018)\nAnswer:\nRefer to answer 65.\nGiven that: Focal length of lens A, fA = +40 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 65.\nGiven that: Focal length of lens A, fA = +40 cm\nFocal length of lens B, fB = -20 cm\nLens A is converging. Lens B is diverging.\n\nQuestion 76.\nDraw ray diagrams to show the formation of three times magnified (a) real, and (b) virtual image of an object by a converging lens. Mark the positions of O, F and 2F in each diagram. (AI 2017)\nAnswer:\n\n(b) Ray diagrams of an object placed between F1 and optical centre O of lens can be drawn as follows:\n\n(i) The image formed is virtual and erect.\n(ii) Image is formed in front of the lens.\n(iii) Image formed is enlarged.\nQuestion 77.\n(a) Draw a diagram to show the refraction of light through a glass slab and mark angle of refraction and the lateral shift suffered by a ray of light while passing through the slab.", + "(ii) Image is formed in front of the lens.\n(iii) Image formed is enlarged.\nQuestion 77.\n(a) Draw a diagram to show the refraction of light through a glass slab and mark angle of refraction and the lateral shift suffered by a ray of light while passing through the slab.\n(b) If the refractive index of glass for light going from air to glass is 3/2, find the refractive index of air for light going from glass to air. (Delhi 2016)\nAnswer:\n(a) Refer to answer 68.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(b) If the refractive index of glass for light going from air to glass is 3/2, find the refractive index of air for light going from glass to air. (Delhi 2016)\nAnswer:\n(a) Refer to answer 68.\n(b) Refractive index of glass w.r.t air is 3\ngna = \\frac { 3 }{ 2 }\nNow, refractive index of air w.r.t glass will be\nang = \\frac { 1 }{ _gn_a } = \\frac { 1 }{ (3/2) } = \\frac { 2 }{ 3 }\nQuestion 78.\nThe image of an object formed by a lens is of magnification -1. If the distance between the object and its image is 60 cm, what is the focal length of the lens? If the object is moved 20 cm towards the lens, where would the image be formed? State reason and also draw a ray diagram in support of your answer.", + "If the distance between the object and its image is 60 cm, what is the focal length of the lens? If the object is moved 20 cm towards the lens, where would the image be formed? State reason and also draw a ray diagram in support of your answer. (AI2016)\nAnswer:\nMagnification of -1 indicates that the image is real and inverted and is of the same size as of the object. The object must be at 2f and image also at 2f on the other side.\nTotal distance between image and object\nAlso 4f = 60 cm \u21d2 f = 15 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nTotal distance between image and object\nAlso 4f = 60 cm \u21d2 f = 15 cm\nIf object is moved 20 cm towards the lens, then the object will be between focus and optical centre of the lens and image formed will be virtual and erect and on the same side of the lens.\n\nQuestion 79.\n(a) Define focal length of a spherical lens.\n(b) A divergent lens has a focal length of 30 cm. At what distance should an object of height 5 cm from the optical centre of the lens be placed so that its image is formed 15 cm away from the lens? Find the size of the image also.\n(c) Draw a ray diagram to show the formation of image in the above situation. (AI 2016)\nAnswer:\n(a) Distance between the optical centre and the focus of the lens is known as the focal length of the lens.\n(b) Given f = -30 cm, v = -15 cm, h = 5 cm From the lens formula,\n\nObject should be placed 30 cm from the optical centre.", + "(b) Given f = -30 cm, v = -15 cm, h = 5 cm From the lens formula,\n\nObject should be placed 30 cm from the optical centre.\nAlso m = \\frac { h\u2019 }{ h } = \\frac { v }{ u } \u21d2 h\u2019 = h(\\frac { v }{ u })\nor h\u2019 = 5 \u00d7 \\frac { -15 }{ -30 } = 2.5 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nObject should be placed 30 cm from the optical centre.\nAlso m = \\frac { h\u2019 }{ h } = \\frac { v }{ u } \u21d2 h\u2019 = h(\\frac { v }{ u })\nor h\u2019 = 5 \u00d7 \\frac { -15 }{ -30 } = 2.5 cm\nSize of image formed is 2.5 cm\n\nQuestion 80.\nIf the image formed by a lens for all positions of the object placed in front of it is always virtual, erect and diminished, state the type of the lens. Draw a ray diagram in support of your answer. If the numerical value of focal length of such a lens is 20 cm, find its power in new cartesian sign conventions. (Foreign 2016)\nAnswer:\nConcave lens always forms virtual, erect and diminished image for all positions of the object.", + "If the numerical value of focal length of such a lens is 20 cm, find its power in new cartesian sign conventions. (Foreign 2016)\nAnswer:\nConcave lens always forms virtual, erect and diminished image for all positions of the object.\n\nFocal length of the concave lens\nf = -20 cm = \\frac { -20 }{ 100 } m\nPower of the lens, P = \\frac { 1 }{ f(in m) } = \\frac { -100 }{ 20 m } = -5D\nQuestion 81.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nFocal length of the concave lens\nf = -20 cm = \\frac { -20 }{ 100 } m\nPower of the lens, P = \\frac { 1 }{ f(in m) } = \\frac { -100 }{ 20 m } = -5D\nQuestion 81.\nState the laws of refraction of light. If the speed of light in vacuum is 3 \u00d7 108 m/s, find the absolute refractive index of a medium in which light travels with a speed of 1.4 \u00d7 108 m/s. (Foreign 2015)\nAnswer:\nLaws of refraction: Refer to answer 74.\nThe speed of light in vacuum = 3 \u00d7 108 m/s\nThe speed of light in a medium = 1.4 \u00d7 108 m/s\n\u2234 Absolute refractive index\n\nQuestion 82.\nState the laws of refraction of light. If the speed of light in vacuum is 3 \u00d7 108 m s-1, find the speed of light in a medium of absolute refractive index 1.5.", + "State the laws of refraction of light. If the speed of light in vacuum is 3 \u00d7 108 m s-1, find the speed of light in a medium of absolute refractive index 1.5. (Delhi 2014, AI 2014)\nAnswer:\nRefer to answer 74.\nThe speed of light in vacuum = 3 \u00d7 108 m/s\nAbsolute refractive index =1.5\n\u2234 The speed of light in a medium\n\nQuestion 83.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 74.\nThe speed of light in vacuum = 3 \u00d7 108 m/s\nAbsolute refractive index =1.5\n\u2234 The speed of light in a medium\n\nQuestion 83.\nThe image of a candle flame placed at a distance of 40 cm from a spherical lens is formed on a screen placed on the other side of the lens at a distance of 40 cm from the lens. Identify the type of lens and write its focal length. What will be the nature of the image formed if the candle flame is shifted 25 cm towards the lens? Draw a ray diagram to justify your answer. (Foreign 2014)\nAnswer:\nGiven : u = -40 cm, v = 40 cm\n\\frac{1}{f}=\\frac{1}{40}+\\frac{1}{40}=\\frac{2}{40} \u21d2 f = 20 cm\nType of lens : Convex lens\nFocal length = 20 cm\nNature of the image will be virtual and erect if the candle flame is shifted 25 cm towards the lens.\n\nQuestion 84.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nType of lens : Convex lens\nFocal length = 20 cm\nNature of the image will be virtual and erect if the candle flame is shifted 25 cm towards the lens.\n\nQuestion 84.\nAn object of height 6 cm is placed perpendicular to the principal axis of a concave lens of focal length 5 cm. Use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 10 cm. (Delhi 2013)\nAnswer:\nFocal length of given concave lens, f= \u2013 5 cm\nDistance, u = -10 cm, object size, h = 6 cm\nImage distance, v = ?", + "(Delhi 2013)\nAnswer:\nFocal length of given concave lens, f= \u2013 5 cm\nDistance, u = -10 cm, object size, h = 6 cm\nImage distance, v = ?\nUsing lens formula, \\frac { 1 }{ f } = \\frac { 1 }{ v } \u2013 \\frac { 1 }{ u }\n\\frac{1}{v}=\\frac{1}{f}+\\frac{1}{u}=\\frac{1}{-5}+\\frac{1}{-10}=\\frac{-3}{10}\nv = \u2013\\frac { 10 }{ u } = -3.33 cm\nSo, the image is located 3.33 cm from the lens. Magnification (m) of lens is given by\nm = \\frac { v }{ u } = \\frac{-\\frac{10}{3}}{-10}=\\frac{1}{3} = 0.33", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nv = \u2013\\frac { 10 }{ u } = -3.33 cm\nSo, the image is located 3.33 cm from the lens. Magnification (m) of lens is given by\nm = \\frac { v }{ u } = \\frac{-\\frac{10}{3}}{-10}=\\frac{1}{3} = 0.33\nm is positive implies that image is virtual and erect. Also, magnitude of m is less than one implies that image is diminished.\nSince m = \\frac { v }{ u } = \\frac { h\u2019 }{ h } \u21d2 \\frac { 1 }{ 3 } = \\frac { h\u2019 }{ 6 } or h\u2019 = 2 cm\nQuestion 85.\nDraw ray diagram to show the path of the refracted ray in each of the following cases. A ray of light incident on a concave lens\n(i) is parallel to its principal axis, (ii) is passing through its optical centre and (iii) is directed towards its principal focus.", + "Draw ray diagram to show the path of the refracted ray in each of the following cases. A ray of light incident on a concave lens\n(i) is parallel to its principal axis, (ii) is passing through its optical centre and (iii) is directed towards its principal focus. (Delhi 2013 C)\nAnswer:\n(i) A ray of light incident on a concave lens is parallel to its principal axis, the diagram can be drawn as follows:\n\nThe refracted ray appears to pass through focus on the same side of the lens.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(i) A ray of light incident on a concave lens is parallel to its principal axis, the diagram can be drawn as follows:\n\nThe refracted ray appears to pass through focus on the same side of the lens.\n(ii) If a ray of light incident on a concave lens is passing through its optical centre then the refracted ray will go without deviation.\n\n(iii) If a ray of light incident on a concave lens is directed towards its principal axis then it will go parallel to principal axis.\n\nQuestion 86.\nWhat is the principle of reversibility of light? Show that the incident of light is parallel to the emergent ray of light when light falls obliquely on a side of a rectangular glass slab. (AI 2011)\nAnswer:\nPrinciple of reversibility of light states that the light will follow exactly the same path if the direction is reversed.\nUsing Snell\u2019s law of refraction, \\frac{\\sin i}{\\sin r_{1}}=\\frac{\\sin e}{\\sin r_{2}}\nSince r1 = r2, so i = e\n\nso PQ is parallel to RS.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nUsing Snell\u2019s law of refraction, \\frac{\\sin i}{\\sin r_{1}}=\\frac{\\sin e}{\\sin r_{2}}\nSince r1 = r2, so i = e\n\nso PQ is parallel to RS.\nSo, we conclude that incident ray is parallel to the emergent ray.\nQuestion 87.\nWhat is understood by lateral displacement of light? Illustrate it with the help of a diagram. List any two factors on which the lateral displacement in a particular substance depends. (Foreign 2011)\nAnswer:\nWhen a ray of light is incident obliquely on a parallel sided glass slab, the emergent ray shifts laterally. The perpendicular distance between the direction of the incident ray and emergent ray is called lateral shift.\nDiagram : Refer to answer 68.\nTire factors on which the lateral displacement depends are:\n(i) thickness of the refracting material.\n(ii) the refractive index of the material.\nQuestion 88.\nDraw a ray diagram in each of the following cases to show the formation of image, when the object is placed:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(ii) the refractive index of the material.\nQuestion 88.\nDraw a ray diagram in each of the following cases to show the formation of image, when the object is placed:\n(i) between optical centre and principal focus of a convex lens.\n(ii) anywhere in front of a concave lens.\n(iii) at 2F of a convex lens.\nState the signs and values of magnifications in the above mentioned cases (i) and (ii). (2020)\nAnswer:\n(i) When an object is placed between Fj and optical centre, O of a convex lens, it forms a virtual and erect image. The ray diagram for this situation can be drawn as follows:\n\nIn case (i), the magnification, m is given by,\nm = \\frac { v }{ u } = \\frac { -v }{ -u } = positive\ni.e., the image formed virtual and erect.\nIn case (ii), the magnification,\nm = \\frac { v }{ u } = \\frac { -v }{ -u } = positive\ni.e., the image formed is virtual and erect.\nQuestion 89.", + "In case (ii), the magnification,\nm = \\frac { v }{ u } = \\frac { -v }{ -u } = positive\ni.e., the image formed is virtual and erect.\nQuestion 89.\n(a) Define the following terms :\n(i) Power of lens\n(ii) Principal focus of a concave mirror", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\ni.e., the image formed is virtual and erect.\nQuestion 89.\n(a) Define the following terms :\n(i) Power of lens\n(ii) Principal focus of a concave mirror\n(b) Write the relationship among the object distance (u), image distance (v) and the focal length (f) of a\n(i) Spherical lens\n(ii) Spherical mirror\n(c) An object is placed at a distance of 10 cm from optical centre of a convex lens of focal length 15 cm. Draw a labelled ray diagram to show the formation of image in this case. (2020)\nAnswer:\n(a) (i) Refer to answer 63.\n(ii) Refer to answer 54(a)(iv).\n(b) (i) for a spherical lens, according to lens formula,\n\\frac { 1 }{ f } = \\frac { 1 }{ v } + \\frac { 1 }{ u }\nwhere f is the focal length of the lens, v is the image distance and u is the object distance.", + "(ii) For a spherical mirror, according to mirror formula,\n\\frac { 1 }{ f } = \\frac { 1 }{ v } + \\frac { 1 }{ u }\nwhere f is the focal length of the mirror, v is the image distance and u is the object distance.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n\\frac { 1 }{ f } = \\frac { 1 }{ v } + \\frac { 1 }{ u }\nwhere f is the focal length of the mirror, v is the image distance and u is the object distance.\n(c) Given focal length, f = +15 cm (convex lens) and u = -10 cm\n\nQuestion 90.\nRishi went to a palmist to show his palm. The palmist used a special lens for this purpose.\n(i) State the nature of the lens and reason for its use.\n(ii) Where should the palmist place/hold the lens so as to have a real and magnified image of an object?\n(iii) If the focal length of this lens is 10 cm, the lens is held at a distance of 5 cm from the palm, use lens formula to find the position and size of the image.", + "(iii) If the focal length of this lens is 10 cm, the lens is held at a distance of 5 cm from the palm, use lens formula to find the position and size of the image. (2020)\nAnswer:\n(i) The lens used here is a convex lens and it is used as a magnifying glass because at close range, i.e., when the object is placed between optic centre and principal focus it forms an enlarged, virtual and erect image of the object.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(ii) When this lens is placed such that the object is between the centre of curvature and the principal focus, the palmist obtain a real and magnified image.\n(iii) Given focal length, f = 10 cm and u = -5 cm According to lens formula,\n\nThus, the image will be formed at 10 cm on the same side of the palm and the size of the image will be enlarged.\nQuestion 91.\nAn object is placed at a distance of 60 cm from a concave lens of focal length 30 cm.\n(i) Use lens formula to find the distance of the image from the lens\n(ii) List four characteristics of the image (nature, position, size, erect/inverted) formed by the lens in this case\n(iii) Draw ray diagram to justify your answer of pair (ii) . (Delhi 2019)\nAnswer:\n(i) Given, f = -30 cm, u = -60 cm, v = ?\nUsing lens formula,\n\n(ii) (a) The image is virtual in nature.\n(b) The image is formed at a distance of 20 cm on the left from the concave lens.", + "Using lens formula,\n\n(ii) (a) The image is virtual in nature.\n(b) The image is formed at a distance of 20 cm on the left from the concave lens.\n(c) Magnification, m = \\frac { v }{ u }\nHere, v = \u2013 20 cm; u = \u2013 60 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(ii) (a) The image is virtual in nature.\n(b) The image is formed at a distance of 20 cm on the left from the concave lens.\n(c) Magnification, m = \\frac { v }{ u }\nHere, v = \u2013 20 cm; u = \u2013 60 cm\nSo, m = \\frac { 20 }{ 60 } = \\frac { 1 }{ 3 } = + 0.3\nSince value of magnification is less than 1, therefore, the image is diminished.\n(d) The plus sign for the magnification shows that the image is erect.\n\nQuestion 92.\n(a) A 5 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find the position, nature and size of the image formed.\n(b) Draw a labelled ray diagram showing object distance, image distance and focal length in the above case.", + "The distance of the object from the lens is 30 cm. Find the position, nature and size of the image formed.\n(b) Draw a labelled ray diagram showing object distance, image distance and focal length in the above case. (AI 2019)\nAnswer:\n(a) Given, h = 5 cm, f = 20 cm, u = -30 cm\nUsing lens formula, \\frac { 1 }{ v } \u2013 \\frac { 1 }{ u } = \\frac { 1 }{ f }\n\\frac{1}{v}=\\frac{1}{u}+\\frac{1}{f}=\\frac{1}{-30}+\\frac{1}{20}=\\frac{-2+3}{60}=\\frac{1}{60}\n\u21d2 v = 60 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nUsing lens formula, \\frac { 1 }{ v } \u2013 \\frac { 1 }{ u } = \\frac { 1 }{ f }\n\\frac{1}{v}=\\frac{1}{u}+\\frac{1}{f}=\\frac{1}{-30}+\\frac{1}{20}=\\frac{-2+3}{60}=\\frac{1}{60}\n\u21d2 v = 60 cm\nNow, magnification, m = \\frac { h\u2019 }{ h } = \\frac { v }{ u }\n\u21d2 h\u2019 = \\frac { v }{ u } \u00d7 h = \\frac {60 }{ -30 } \u00d7 5 = -10 cm\nHence, the image formed at 60 cm, which is real and magnified.\n\nQuestion 93.\nAnalyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follows, without doing any calculations :\n(a) What is the focal length of the convex lens? Give reason in support of your answer.", + "Question 93.\nAnalyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follows, without doing any calculations :\n(a) What is the focal length of the convex lens? Give reason in support of your answer.\n(b) Write the serial number of that observation which is not correct. How did you arrive at this conclusion?\n(c) Take an appropriate scale to draw ray diagram for the observation at S. No. 4 and the approximate value of magnification. (Delhi 2017)\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(c) Take an appropriate scale to draw ray diagram for the observation at S. No. 4 and the approximate value of magnification. (Delhi 2017)\nAnswer:\n(a) When an object placed at 2F from a convex lens, then its image is formed on the other side of the lens at the same distance from the lens. Thus from S. No.(3) we can say that.\nf = v/2 \u21d2 f = \\frac {30}{ 2 } = + 15 cm\nThus, the focal length is + 15 cm.\n(b) In this case S.No. (6) is incorrect as the object distance is between focus and pole, for such case, the image formed is virtual and on the same side as the object, hence image distance is negative.\n(c) The approximate value of magnification for object distance -20 cm and image distance +60 cm is -3\n\nQuestion 94.\nAnalyse the following observation table showing variation of image-distance (v) with object-distance (u) in case of a convex lens and answer the questions that follow without doing any calculations.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnalyse the following observation table showing variation of image-distance (v) with object-distance (u) in case of a convex lens and answer the questions that follow without doing any calculations.\n(a) What is the focal length of the convex lens? Give reason to justify your answer.\n(b) Write the serial number of the observation which is not correct. On what basis have you arrived at this conclusion?\n(c) Select an appropriate scale and draw a ray diagram for the observation at S. No. 2. Also find the approximate value of magnification. (AI 2017)\nAnswer:\n(a) When an object is placed at 2F from the convex lens, then its image is formed on the other side of the lens at the same distance from the lens. Thus from S.No. (3), we can say that\n\u2234 f = \\frac {v}{ 2 } = \\frac {40}{ 2 } = 20 cm\n(b) In this case, S. No. (6) is incorrect as the object distance is between focus and optical centre for such cases, the image formed is virtual and image distance is negative.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(b) In this case, S. No. (6) is incorrect as the object distance is between focus and optical centre for such cases, the image formed is virtual and image distance is negative.\n\nThe approximate value of magnification for object distance -60 cm and image distance +30 cm is -1/2.\nQuestion 95.\n(a) Draw a ray diagram to show the formation of image by a convex lens when an object is placed in front of the lens between its optical centre and principal focus.\n(b) In the above ray diagram mark the object distance (u) and the image distance (v) with their proper sign (+ve or -ve as per the new Cartesian sign convention) and state how these distances are related to the focal length (f) of the convex lens in this case.\n(c) Find the power of a convex lens which forms a real and inverted image of magnification -1 of an object placed at a distance of 20 cm from its optical centre. (Delhi 2016)\nAnswer:\n(a) Refer to answer 88(i).\n(b) The lens formula is given as", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Refer to answer 88(i).\n(b) The lens formula is given as\n\\frac { 1 }{ v } \u2013 \\frac { 1 }{ u } = \\frac { 1 }{ f }\n(c) Magnification of the lens is given by\nm = \\frac { v }{ u } \u21d2 -1 = \\frac { -v }{ 20 } [u = -20 cm]\n\u2234 v = 20 cm\nAs v = u then\n\u2234 f = \\frac { 20 }{ 2 } cm = 10 cm = 0.1 m\nPower of the lens, P = \\frac { 1 }{ f(in m) } D = \\frac { 1}{ 0.1 } D = 10 D\nQuestion 96.\n(a) Draw a ray diagram to show the formation of image by a concave lens when an object is placed in front of it.", + "(a) Draw a ray diagram to show the formation of image by a concave lens when an object is placed in front of it.\n(b) In the above diagram mark the object distance (u) and the image distance (v) with their proper sign (+ve or -ve as per the new Cartesian sign convention) and state how these distances are related to the focal length (/) of the concave lens in this case.\n(c) Find the nature and power of a lens which forms a real and inverted image of magnification -1 at a distance of 40 cm from its optical centre. (Delhi 2016)\nAnswer:\n(a) Refer to answer 88(ii).\n(b) The lens formula is given by", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Refer to answer 88(ii).\n(b) The lens formula is given by\n\\frac { 1 }{ v } \u2013 \\frac { 1 }{ u } = \\frac { 1 }{ f }\n(c) Since, the nature of the image is real and inverted therefore the lens is convex.\nNow magnification of the lens is\nm = \\frac { v }{ u } \u21d2 -1 = \\frac { v }{ u} \u21d2 v = -u\nNow, from lens formula,\n\nQuestion 97.\n(a) Define optical centre of a spherical lens.\n(b) A divergent lens has a focal length of 20 cm. At what distance should an object of height 4 cm from the optical centre of the lens be placed so that its image is formed 10 cm away from the lens. Find the size of the image also.\n(c) Draw a ray diagram to show the formation of image in above situation. (AI 2016)\nAnswer:\n(a) Optical centre is the central part of the lens through which a ray of light passes without suffering any deviation.", + "Find the size of the image also.\n(c) Draw a ray diagram to show the formation of image in above situation. (AI 2016)\nAnswer:\n(a) Optical centre is the central part of the lens through which a ray of light passes without suffering any deviation. It is usually represented by the letter O.\n(b) Given f = \u2013 20 cm, h = 4 cm, v = \u2013 10 cm\nFrom lens formula,\n\\frac { 1 }{ v } \u2013 \\frac { 1 }{ u } = \\frac { 1 }{ f }", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(b) Given f = \u2013 20 cm, h = 4 cm, v = \u2013 10 cm\nFrom lens formula,\n\\frac { 1 }{ v } \u2013 \\frac { 1 }{ u } = \\frac { 1 }{ f }\n\u21d2 \\frac { 1 }{ u } = \\frac { -1 }{ 10 } + \\frac { 1 }{ 20 } \u21d2 u = -20 cm\nAlso, magnification of the lens,\nm = \\frac { h\u2019 }{ h } = \\frac { v }{ u } \u21d2 h\u2019 = \\frac { 10 }{ 20 } \u00d7 4 \u21d2 h\u2019 = 2 cm\n\nQuestion 98.\n(a) Define focal length of a divergent lens.\n(b) A divergent lens has a focal length of 30 cm forms the image of an object of size 6 cm on the same side as the object at a distance of 15 cm from its optical centre. Use lens formula to determine the distance of the object from the lens and the size of the image formed.", + "Use lens formula to determine the distance of the object from the lens and the size of the image formed.\n(c) Draw a ray diagram to show the formation of image in the above situation. (AI 2016)\nAnswer:\n(a) Distance between the principal focus and the optical centre is known as the focal length of the lens.\n(b) Given, f = -30 cm, v = -15 cm, h = 6 cm\nNow, from lens formula,\n\nQuestion 99.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Distance between the principal focus and the optical centre is known as the focal length of the lens.\n(b) Given, f = -30 cm, v = -15 cm, h = 6 cm\nNow, from lens formula,\n\nQuestion 99.\nAt what distance from a concave lens of focal length 20 cm a 6 cm tall object be placed so as to obtain its image at 15 cm from the lens? Also calculate the size of the image formed. Draw a ray diagram to justify your answer for the above situation and label it. (Foreign 2016)\nAnswer:\nFocal length of concave lens, f = -20\ncm Height of the object, h = 6 cm\nImage distance, v = -15 cm\nFrom lens formula,\n\nQuestion 100.\nAt what distance from a concave lens of focal length 25 cm a 10 cm tall object be placed so as to obtain its image at 20 cm from the lens. Also calculate the size of the image formed. Draw a ray diagram to justify your answer for the above situation and label it.", + "At what distance from a concave lens of focal length 25 cm a 10 cm tall object be placed so as to obtain its image at 20 cm from the lens. Also calculate the size of the image formed. Draw a ray diagram to justify your answer for the above situation and label it. (Foreign 2016)\nAnswer:\nFocal length of concave lens f = \u2013 25 cm\nImage distance, v = -20 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nFocal length of concave lens f = \u2013 25 cm\nImage distance, v = -20 cm\nHeight of the object, h = 10 cm\nNow, from lens formula,\n\nQuestion 101.\n\u201cA convex lens can form a magnified erect as well as magnified inverted image of an object placed in front of it\u201d. Draw ray diagram to justify this statement stating the position of the object with respect to the lens in each case. An object of height 4 cm is placed at a distance of 20 cm from a concave lens of focal length 10 cm. Use lens formula to determine the position of the image formed.", + "An object of height 4 cm is placed at a distance of 20 cm from a concave lens of focal length 10 cm. Use lens formula to determine the position of the image formed. (Delhi 2015)\nAnswer:\nMagnified erect image:\n\nMagnified inverted image:\n\nGiven that h = 4 cm, u = -20 cm, f = -10 cm\nLens formula:\n\\frac { 1 }{ v } \u2013 \\frac { 1 }{ u } = \\frac { 1 }{ f } \u2234 \\frac { 1 }{ v } \u2013 \\frac { 1 }{ (-20) } = \\frac { 1 }{ (-10) }\nor \\frac{1}{v}=\\frac{-1}{10}-\\frac{1}{20}=\\frac{-2-1}{20}=\\frac{-3}{20} or v = \\frac { -20 }{ 3 } cm\nQuestion 102.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nor \\frac{1}{v}=\\frac{-1}{10}-\\frac{1}{20}=\\frac{-2-1}{20}=\\frac{-3}{20} or v = \\frac { -20 }{ 3 } cm\nQuestion 102.\nThe image of a candle flame placed at a distance of 30 cm from a spherical lens is formed on a screen placed on the other side of the lens at a distance of 60 cm from the optical centre of the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 3 cm, find the height of its image.", + "Identify the type of lens and calculate its focal length. If the height of the flame is 3 cm, find the height of its image. (Delhi 2015)\nAnswer:\nGiven that u = -30 cm, v = 60 cm, h = 3 cm\nLens Formula:\n\\frac { 1 }{ v } \u2013 \\frac { 1 }{ u } = \\frac { 1 }{ f } \u2234 \\frac { 1 }{ 60 } \u2013 \\frac { 1 }{ (-30) } = \\frac { 1 }{ f }\n\u21d2 \\frac{1+2}{60}=\\frac{1}{f}=\\frac{3}{60}=\\frac{1}{20} or f = 20 cm\nAs focal length is positive, hence lens is convex lens.", + "Magnification, m = \\frac { v }{ u } = \\frac { h\u2019 }{ h }\n\u2234 \\frac { 60 }{ -30 } = \\frac { h\u2019 }{ 3 } or h\u2019 = \\frac { -60\u00d73 }{ 30 } = -6 cm\n\u21d2 The height of image is 6 cm and negative sign shows that the image is real and inverted.\nQuestion 103.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n\u2234 \\frac { 60 }{ -30 } = \\frac { h\u2019 }{ 3 } or h\u2019 = \\frac { -60\u00d73 }{ 30 } = -6 cm\n\u21d2 The height of image is 6 cm and negative sign shows that the image is real and inverted.\nQuestion 103.\n(a) State the laws of refraction of light. Explain the term absolute refractive index of a medium and write an expression to relate it with the speed of light in vacuum.\n(b) The absolute refractive indices of two media A and B are 2.0 and 1.5 respectively. If the speed of light in medium B is 2 \u00d7 108 m/s, calculate the speed of light in\n(i) vacuum\n(ii) medium A (Delhi 2015)\nAnswer:\n(a) Refer to answer 74.", + "If the speed of light in medium B is 2 \u00d7 108 m/s, calculate the speed of light in\n(i) vacuum\n(ii) medium A (Delhi 2015)\nAnswer:\n(a) Refer to answer 74.\n(b) Given that nA = 2.0, nA = 1.5, vA = 2 \u00d7 108 m/s\n(i) nB = \\frac { c }{ v_B }, where c is the speed of light in vacuum\n\nQuestion 104.\nWhat is meant by power of a lens? Define its S.I. unit.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(b) Given that nA = 2.0, nA = 1.5, vA = 2 \u00d7 108 m/s\n(i) nB = \\frac { c }{ v_B }, where c is the speed of light in vacuum\n\nQuestion 104.\nWhat is meant by power of a lens? Define its S.I. unit.\nYou have two lenses A and B of focal lengths + 10 and -10 cm respectively. State the nature and power of each lens. Which of the two lenses will form a virtual and magnified image of an object placed 8 cm from the lens? Draw a ray diagram to justify your answer. (AI 2015)\nAnswer:\nRefer to answer 65.\nGiven that:\nFocal length of lens A, fA = +10 cm\nFocal length of lens B, fB = -10 cm\nLens A is convex lens .", + "Draw a ray diagram to justify your answer. (AI 2015)\nAnswer:\nRefer to answer 65.\nGiven that:\nFocal length of lens A, fA = +10 cm\nFocal length of lens B, fB = -10 cm\nLens A is convex lens . Lens B is concave lens,\nPower of lens A = \\frac { 100 }{ f_A(in cm) } = \\frac { 100 }{ 10 } = +10 D\nPower of lens B = \\frac { 100 }{ f_B(in cm) } = \\frac { 100 }{ -10 } = -10 D\nLens A will form a virtual and magnified image.\n\nQuestion 105.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nPower of lens B = \\frac { 100 }{ f_B(in cm) } = \\frac { 100 }{ -10 } = -10 D\nLens A will form a virtual and magnified image.\n\nQuestion 105.\nOne half of a convex lens of focal length 10 cm is covered with a black paper. Can such a lens produce an image of a complete object placed at a distance of 30 cm from the lens? Draw a ray diagram to justify your answer.\nA 4 cm tall obj ect is placed perpendicular to the principal axis of a concave lens of focal length 20 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. (AI 2015)\nAnswer:\nYes, the lens will produce an image of complete object.", + "The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. (AI 2015)\nAnswer:\nYes, the lens will produce an image of complete object.\n\nGiven that h = 4 cm, f = 20 cm, u = -15 cm\nLens formula: = \\frac { 1 }{ v } \u2013 \\frac { 1 }{ u } = \\frac { 1 }{ f }\n\u2234 \\frac { 1 }{ v } \u2013 \\frac { 1 }{ (-15) } = \\frac { 1 }{ 20 } or \\frac { 1 }{ v } = \\frac { 1 }{ 20 } \u2013 \\frac { 1 }{ 15 } = \\frac { -1 }{ 60 }\n\u21d2 v = -60 cm\nMagnification, m = \\frac { v }{ u } = \\frac { -60 }{ -15 } = 4", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n\u21d2 v = -60 cm\nMagnification, m = \\frac { v }{ u } = \\frac { -60 }{ -15 } = 4\nNature of image: Virtual, erect and enlarged Position of image : In front of lens 60 cm from the lens\nSize of image : Four times of object, i.e.,\n(4 \u00d7 4) cm = 16 cm\nQuestion 106.\nWhat is meant by the power of a lens ? What is its S.I. unit ? Name the type of lens whose power is positive. The image of an object formed by a lens is real, inverted and of the same size as the object. If the image is at a distance of 40 cm from the lens, what is the nature and power of the lens? Draw ray diagram to justify your answer. (Foreign 2015)\nAnswer:\nRefer to answer 65.\nConvex lens has positive power. Since the image of an object formed by a lens is real, inverted and of the same size as the object.", + "Draw ray diagram to justify your answer. (Foreign 2015)\nAnswer:\nRefer to answer 65.\nConvex lens has positive power. Since the image of an object formed by a lens is real, inverted and of the same size as the object.\nGiven: Size of object = Size of image and h\u2019 = -h\n\u2234 Magnification, m = \\frac { h\u2019 }{ h} = \\frac { -h }{ h} = -1\n\u2234 -1 = \\frac { v }{ u} or v = -u\nFocal length of the lens,\n\nQuestion 107.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nGiven: Size of object = Size of image and h\u2019 = -h\n\u2234 Magnification, m = \\frac { h\u2019 }{ h} = \\frac { -h }{ h} = -1\n\u2234 -1 = \\frac { v }{ u} or v = -u\nFocal length of the lens,\n\nQuestion 107.\n(a) Explain the following terms related to spherical lenses:\n(i) optical centre\n(ii) centres of curvature\n(iii) principal axis (iv) aperture\n(v) principal focus (vi) focal length\n(b) A converging lens has focal length of 12 cm. Calculate at what distance should the object be placed from the lens so that it forms an image at 48 cm on the other side of the lens. (AI 2014)\nAnswer:\n(a) (i) Optical centre : The centre point of a lens is known as the optical centre. It always lies inside the lens. A light beam passing through the optical centre without any deviation.", + "(AI 2014)\nAnswer:\n(a) (i) Optical centre : The centre point of a lens is known as the optical centre. It always lies inside the lens. A light beam passing through the optical centre without any deviation.\n(ii) Centre of curvature : It is defined as the centre of the sphere of which the lens is originally a part of. Because the spherical lens consists of two spherical surfaces, the lens has two centre of curvature.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(iii) Principal axis : A straight line passing through the optical centre and principal focus of a spherical lens. This line is called the principal axis.\n(iv) Aperture : The diameter of the reflecting surface of spherical lens is called its aperture.\n(v) Principal focus : A number of rays parallel to the principal axis are falling on a lens.\nThese rays, after refraction from the lens, are appearing to converge to or diverge from a point on the principal axis. This point on the principal axis is called the principal focus of the lens.\n(vi) Focal length: The distance between the optical centre and the principal focus of a spherical lens is called the focal length. It is represented by the letter f.\n(b) Given f = 12 cm, v = 48 cm, u = ?\n\nQuestion 108.\n(i) Explain the following terms related to spherical lenses\n(a) Centres of curvature (b) Principal axis\n(c) Optical centre (d) Principal focus", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\n(b) Given f = 12 cm, v = 48 cm, u = ?\n\nQuestion 108.\n(i) Explain the following terms related to spherical lenses\n(a) Centres of curvature (b) Principal axis\n(c) Optical centre (d) Principal focus\n(ii) At what distance from a concave lens of focal length 20 cm, should a 6 cm tall object be placed so that it forms an image at 15 cm from the lens? Also determine the size of the image formed. (AI 2014)\nAnswer:\n(i) Refer to answer 107(a).\n(ii) Refer to answer 99.\nQuestion 109.\nWhat is meant by power of a lens? Name and define its S.I. unit.\nOne student uses a lens of focal length +50 cm and another of -50 cm. State the nature and find the power of each lens. Which of the two lenses will always give a virtual and diminished image irrespective of the position of the object? (Foreign 2014)\nAnswer:\nRefer to answer 65.\nA convex lens has the focal length +50 cm.", + "State the nature and find the power of each lens. Which of the two lenses will always give a virtual and diminished image irrespective of the position of the object? (Foreign 2014)\nAnswer:\nRefer to answer 65.\nA convex lens has the focal length +50 cm.\n\u2234 power = \\frac { 1 }{ f } = \\frac { +100 }{ 50 } = +2 D\nA concave lens has the focal length -50 cm.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 65.\nA convex lens has the focal length +50 cm.\n\u2234 power = \\frac { 1 }{ f } = \\frac { +100 }{ 50 } = +2 D\nA concave lens has the focal length -50 cm.\n\u2234 power = \\frac { 1 }{ f } = \\frac { -100 }{ 50 } = -2 D\nConcave lens always gives a virtual, erect and diminished image irrespective of the position of the object.\nQuestion 110.\n(a) State the laws of refraction of light. Give an expression to relate the absolute refractive index of a medium with speed of light in vacuum.\n(b) The refractive indices of water and glass with respect to air are 4/3 and 3/2 respectively. If the speed of light in glass is 2 \u00d7 108 m s-1, find the speed of light in (i) air, (ii) water. (Delhi 2013)\nAnswer:\n(a) Refer to answer 74.\n(b) Refer to answer 66.\nQuestion 111.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Refer to answer 74.\n(b) Refer to answer 66.\nQuestion 111.\nThe image of a candle flame placed at a distance of 45 cm from a spherical lens is formed on a screen placed at a distance of 90 cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2 cm, find the height of its image. (Delhi 2012)\nAnswer:\nGiven that u = -45 cm, v = +90 cm, h = 2 cm\n(as the image is formed on the screen, the image is real and hence image is formed by convex lens on the other side of the lens).", + "(Delhi 2012)\nAnswer:\nGiven that u = -45 cm, v = +90 cm, h = 2 cm\n(as the image is formed on the screen, the image is real and hence image is formed by convex lens on the other side of the lens).\nType of the lens used : Convex lens\nLens formula\n\\frac { 1 }{ v } \u2013 \\frac { 1 }{ u } = \\frac { 1 }{ f }\n\u2234 \\frac{1}{90}-\\frac{1}{(-45)}=\\frac{1}{f}=\\frac{3}{90} or f = 30 cm\nFocal length, f = 30 cm\nMagnification, m = \\frac { h\u2019 }{ h } = \\frac { v }{ u }\n\u2234 \\frac { h\u2019 }{ 2 } = \\frac { 90 }{ -45 } or h\u2019 = -4 cm\n\u21d2 height of image = 4 cm (inverted)\nQuestion 112.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nFocal length, f = 30 cm\nMagnification, m = \\frac { h\u2019 }{ h } = \\frac { v }{ u }\n\u2234 \\frac { h\u2019 }{ 2 } = \\frac { 90 }{ -45 } or h\u2019 = -4 cm\n\u21d2 height of image = 4 cm (inverted)\nQuestion 112.\nState the law of refraction of light that defines the refractive index of a medium with respect to the other. Express it mathematically. How is reffactive index of any medium \u2018A\u2019 with respect to a medium \u2018B\u2019 related to the speed of propagation of light in two media A and A? State the name of this constant when one medium is vacuum or air.\nThe refractive indices of glass and water with respect to vacuum are 3/2 and 4/3 respectively. If the speed of light in glass is 2 \u00d7 108 m/s, find the speed of light in (i) vacuum, (ii) water. (Delhi 2012)\nAnswer:\nRefer to answer 74 and 66.\nQuestion 113.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 74 and 66.\nQuestion 113.\nA 4 cm tall object is placed perpendicular to the principal axis of convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position, size and nature of the image formed, using the lens formula. (AI 2012, Foreign 2012)\nAnswer:\nGiven that :u = -16 cm, f = 24 cm, h = 4 cm\n\nNature of image : virtual and erect, enlarged\nQuestion 114.\nWith the help of a ray diagram state what is meant by refraction of light. State Snell\u2019s law for refraction of light and also express it mathematically.\nThe refractive index of air with respect to glass is 2/3 and the refractive index of water with respect to air is 4/3. If the speed of light in glass is 2 \u00d7 108 m/s, find the speed of light in (a) air, (b) water.", + "If the speed of light in glass is 2 \u00d7 108 m/s, find the speed of light in (a) air, (b) water. (AI 2012)\nAnswer:\nWhen travelling obliquely from one medium to another, the direction of propagation of light in the second medium changes. This phenomenon is known as refraction of light.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nWhen travelling obliquely from one medium to another, the direction of propagation of light in the second medium changes. This phenomenon is known as refraction of light.\nThe ratio of sine of angle of incidence to the sine of the angle of refraction is constant, for the light of a given colour and for the given pair of media. This law is also known as snell\u2019s law of refraction.\n\n\\frac { sin i }{ sin r } = constant,\nWhere i is the angle of incidence and r is the angle of refraction.\nGiven that: nag = 2/3, nwa = 4/3, vg = 2 \u00d7 108 m/s\n(a) Refer to answer 66.\n(b) Va = nga \u00d7 vg\n= \\frac { 1 }{ n_{ag} } \u00d7 Vg = \\frac { 3 }{ 2 } \u00d7 2 \u00d7 108 = 3 \u00d7 108 m/s\nQuestion 115.\nList the sign conventions that are followed in case of refraction of light through spherical lenses.", + "List the sign conventions that are followed in case of refraction of light through spherical lenses. Draw a diagram and apply these conventions in determining the nature and focal length of a spherical lens which forms three times magnified real image of an object placed 16 cm from the lens. (Foreign 2012)\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\nFor lenses, we follow sign convention, similar to one used for spherical mirrors. We apply the rules for signs of distances, except that all measurements are taken from the optical centre of the lens.\nFor sign convention of mirror : refer to answer 60.\n\nu = -16 cm, m = -3 (real)\n\nor f = 12 cm\nSpherical lens is convex lens or converging lens of focal length 12 cm.\nQuestion 116.\n(a) What is meant by \u2018power of a lens?\u2019\n(b) State and define the S.I unit of power of a lens.\n(c) A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with each other. Calculate the lens power of this combination. (AI 2011)\nAnswer:\n(a) Refer to answer 63.\n(b) Refer to answer 65.", + "Calculate the lens power of this combination. (AI 2011)\nAnswer:\n(a) Refer to answer 63.\n(b) Refer to answer 65.\n(c) Power of convex lens of focal length 25 cm is\np1 = \\frac { 100 }{ 25(in m) } = 4 D\nPower of concave lens of focal length 10 cm is 100\np2 = \\frac { 100 }{ -10(in m) } = -10 D\n\u2234 Power of the combination = P = P1 + P2\n\u2234 P = 4 \u2013 10 = -6D\nQuestion 117.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nPower of concave lens of focal length 10 cm is 100\np2 = \\frac { 100 }{ -10(in m) } = -10 D\n\u2234 Power of the combination = P = P1 + P2\n\u2234 P = 4 \u2013 10 = -6D\nQuestion 117.\n(a) Under what condition with a glass lens placed in a transparent liquid become invisible.\n(b) Describe and illustrate with a diagram, how we should arrange two converging lenses so that a parallel beam of light entering one lens emerges as a parallel beam after passing through the second lens.\n(c) An object is placed at a distance of 3 cm from a concave lens of focal length 12 cm. Find the (i) position and (ii) nature of the image formed. (Foreign 2011)\nAnswer:\n(a) If the refractive index of glass lens is equal to the refractive index of liquid then the glass lens placed in a transparent liquid will become invisible.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) If the refractive index of glass lens is equal to the refractive index of liquid then the glass lens placed in a transparent liquid will become invisible.\n(b) Suppose we have two converging lens of focal lengths f1 and f2. We will keep the two converging lens at a distance of f1 +f2 so that a parallel beam of light entering one lens emerges as a parallel beam after passing through the second lens.\n\nHere the focus of the two lenses should coincide,\n(c) (i) Focal length of concave lens, f = -12 cm\nObject distance, u = -3 cm Image distance, v = ?\nUsing lens formula,\n\nSo, the image is formed at 2.4 cm from the concave lens.\n(ii) Using magnification formula,\nm = \\frac { v }{ u } = \\frac { -2.4 }{ -3 } = + 0.8\nSince m is +ve and magnitude of m is less than 1, so the image formed is virtual and", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 10 Intext Questions\nPage Number: 168\nQuestion 1\nDefine the principal focus of a concave mirror.\nAnswer:\nThe principal focus of a concave mirror is a point on its principal axis to which all the light rays which are parallel and close to the axis, converge after reflection from the concave mirror.\nQuestion 2\nThe radius of curvature of a spherical mirror is 20 cm. What is its focal length?\nAnswer:\nFocal length = \\frac { 1 }{ 2 } x Radius of curvature = \\frac { 1 }{ 2 } x 20 cm = 10 cm\nQuestion 3\nName a mirror that can give an erect and enlarged image of an object.\nAnswer:\nConcave mirror.\nQuestion 4\nWhy do we prefer a convex mirror as a rear-view mirror in vehicles ?\nAnswer:\nWe prefer a convex mirror as a rear-view mirror in vehicles because of two reasons :\n A convex mirror always produces an erect image of the objects.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nAnswer:\nWe prefer a convex mirror as a rear-view mirror in vehicles because of two reasons :\n A convex mirror always produces an erect image of the objects.\n The image formed in a convex mirror is highly diminished or much smaller than the object, due to which a convex mirror gives a wide field of view of the traffic behind. A convex mirror enables the driver to view such larger area of the traffic behind him.\nPage Number: 171\nQuestion 1\nFind the focal length of a convex mirror whose radius of curvature is 32 cm.\nSolution:\nR = +32 cm and f=\\frac { R }{ 2 } =+\\frac { 32 }{ 2 } =+16cm\nQuestion 2\nA concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located ?\nSolution:\nBecause the image is real, so magnification m must be negative.\n\nThus the image is located at a distance of 30 cm from the mirror on the object side of the mirror.\nPage Number: 176\nQuestion 1", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nBecause the image is real, so magnification m must be negative.\n\nThus the image is located at a distance of 30 cm from the mirror on the object side of the mirror.\nPage Number: 176\nQuestion 1\nA ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal ? Why ?\nAnswer:\nThe light-ray bends towards the normal because the ray of light goes from a rarer medium to a denser medium.\nQuestion 2\nLight enters from air to glass having refractive index 1.50. What is the speed of light in the glass ? The speed of light in vacuum is 3 x 108 ms-1.\nSolution:\nRefractive index of glass, n8 = 1.50\n\nQuestion 3\nFind out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.\nAnswer:\nFrom table 10.3, diamond has highest refractive index (= 2.42), so it has highest optical density.", + "Also find the medium with lowest optical density.\nAnswer:\nFrom table 10.3, diamond has highest refractive index (= 2.42), so it has highest optical density.\nAir has lowest refractive index (= 1.0003),\nso it has lowest optical density.\nQuestion 4", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nAnswer:\nFrom table 10.3, diamond has highest refractive index (= 2.42), so it has highest optical density.\nAir has lowest refractive index (= 1.0003),\nso it has lowest optical density.\nQuestion 4\nYou are given kerosene, turpentine and water. In which of these does the light travel fastest ? Use the information given in Table 10.3.\nAnswer:\nFor kerosene, n = 1.44\nFor turpentine, n = 1.47\nFor water, n = 1.33\nBecause water has the lowest refractive index, therefore light travels fastest in this optically rarer medium than kerosene and turpentine oil.\nQuestion 5\nThe refractive index of diamond is 2.42. What is the meaning of this statement?\nAnswer:\nBy saying that the refractive index of diamond is 2.42, we mean that the speed of light in diamond is lower by a factor of 2.42 relative to that in vacuum.", + "What is the meaning of this statement?\nAnswer:\nBy saying that the refractive index of diamond is 2.42, we mean that the speed of light in diamond is lower by a factor of 2.42 relative to that in vacuum.\nPage Number: 184\nQuestion 1\nDefine 1 dioptre of power of a lens.\nAnswer:\nOne dioptre is the power of a lens whose focal length is 1 metre.\nQuestion 2", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nPage Number: 184\nQuestion 1\nDefine 1 dioptre of power of a lens.\nAnswer:\nOne dioptre is the power of a lens whose focal length is 1 metre.\nQuestion 2\nA convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object ? Also, find the power of the lens. , Sol. Here, u \u2014 +50 cm ..\nSolution:\nHere \u03bd = +50cm\nBecause the real image is of the same size as the object,\n\nQuestion 3\nFind the power of a concave lens of focal length 2 m.\nSolution:\nBecause the focal length of a concave lens is negative,\ntherefore f = -2 m\n\nNCERT Solutions for Class 10 Science Chapter 10 Textbook Chapter End Questions\nQuestion 1\nWhich one of the following materials cannot be used to make a lens ?\n(a) Water\n(b) Glass\n(c) Plastic\n(d) Clay\nAnswer:\n(d) Clay\nQuestion 2", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 1\nWhich one of the following materials cannot be used to make a lens ?\n(a) Water\n(b) Glass\n(c) Plastic\n(d) Clay\nAnswer:\n(d) Clay\nQuestion 2\nThe image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object ?\n(a) Between the principal focus and the centre of curvature\n(b) At the centre of curvature\n(c) Beyond the centre of curvature\n(d) Between the pole of the mirror and its principal focus.\nAnswer:\n(d) Between the pole of the mirror and its principal focus.\nQuestion 3\nWhere should an object be placed in front of a convex lens to get a real image of the size of the object ?\n(a) At the principal focus of the lens (b) At twice the focal length\n(c) At infinity\n(d) Between the optical centre of the lens and its principal focus.\nAnswer:\n(b) At twice the focal length.\nQuestion 4\nA spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be :", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nAnswer:\n(b) At twice the focal length.\nQuestion 4\nA spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be :\n(a) Both concave.\n(b) Both convex.\n(c) the mirror is concave and the lens is convex.\n(d) the mirror is convex, but the lens is concave.\nAnswer:\n(a) Both concave\nQuestion 5\nNo matter how far you stand from mirror, your image appears erect. The mirror is likely to be\n(a) plane\n(b) concave\n(c) convex\n(d) either plane or convex.\nAnswer:\n(d) Either plane or convex.\nQuestion 6\nWhich of the following lenses would you prefer to use while reading small letters found in a dictionary ?\n(a) A convex lens of focal length 50 cm.\n(b) A concave lens of focal length 50 cm.\n(c) A convex lens of focal length 5 cm.\n(d) A concave lens of focal length 5 cm.\nAnswer:\n(c) A convex lens of focal length 5 cm.\nQuestion 7", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\n(b) A concave lens of focal length 50 cm.\n(c) A convex lens of focal length 5 cm.\n(d) A concave lens of focal length 5 cm.\nAnswer:\n(c) A convex lens of focal length 5 cm.\nQuestion 7\nWe wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror ? What is the nature of the image ? Is the image larger or smaller than the object ? Draw a ray diagram to show the image formation in this case.\nAnswer:\nA concave mirror gives an erect image when the object is placed between the focus F and the pole P of the concave mirror, i.e., between 0 and 15 cm from the mirror. The image thus formed will be virtual, erect and larger than the object.\n\nQuestion 8\nName the type of mirror used in the following situations.\n(a) Headlights of a car.\n(b) Side/rear-view mirror of a vehicle.\n(c) Solar furnace.\nSupport your answer with reason.\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 8\nName the type of mirror used in the following situations.\n(a) Headlights of a car.\n(b) Side/rear-view mirror of a vehicle.\n(c) Solar furnace.\nSupport your answer with reason.\nAnswer:\n(a) Concave mirrors are used as reflectors in headlights of cars. When a bulb is located at the focus of the concave mirror, the light rays after reflection from the mirror travel over a large distance as a parallel beam of high intensity.\n(b) A convex mirror is used as a side/rear-view mirror of a vehicle because\nA convex mirror always forms an erect, virtual and diminished image of an object placed anywhere in front it.\nA convex mirror has a wider field of view than a plane mirror of the same size.\n(c) Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.\nQuestion 9\nOne-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object ? Verify your answer experimentally. Explain your observations.\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 9\nOne-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object ? Verify your answer experimentally. Explain your observations.\nAnswer:\nA convex lens forms complete image of an object, even if its one half is covered with black paper. It can be explained by considering following two cases.\nCase I : When the upper half of the lens is covered\nIn this case, a ray of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure.\n\nCase II: When the lower half of the lens Is covered\nIn this case, a ray of light coming from the object is refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the given figure.\n\nQuestion 10", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 10\nAn object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.\nAnswer:\nHere : Object distance, u= -25 cm,\nObject height, h = 5 cm,\nFocal length, f = +10 cm\nAccording to the lens formula, \\frac { 1 }{ f } =\\frac { 1 }{ \\nu } -\\frac { 1 }{ u } , we have\n\u21d2 \\frac { 1 }{ \\nu } =\\frac { 1 }{ f } -\\frac { 1 }{ u } =\\frac { 1 }{ 10 } -\\frac { 1 }{ 25 } =\\frac { 15 }{ 250 } \\quad or\\quad \\nu =\\frac { 250 }{ 15 } =16.66cm\nThe positive value of v shows that the image is formed at the other side of the lens.", + "The negative value of image height indicates that the image formed is inverted.\nThe position, size, and nature of image are shown alongside in the ray diagram.\n\nQuestion 11\nA concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens ? Draw the ray diagram.\nSolution:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 11\nA concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens ? Draw the ray diagram.\nSolution:\nFocal length, f = -15 cm, Image distance, \u03bd = -10 cm (as concave lens forms the image on the same side of the lens)\nFrom the lens formula \\frac { 1 }{ f } =\\frac { 1 }{ \\nu } -\\frac { 1 }{ u } , we have\n\nObject distance, u = -30 cm\nThe negative value of u indicates that the object is placed in front of the lens.\nQuestion 12\nAn object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.\nSolution:\nObject distance, u = -10 cm, Focal length, f = +15 cm, Image distance, \u03bd = ?\n\nThus, image distance, \u03bd = + 6 cm\nBecause \u03bd is +ve, so a virtual image is formed at a distance of 6 cm behind the mirror.", + "Solution:\nObject distance, u = -10 cm, Focal length, f = +15 cm, Image distance, \u03bd = ?\n\nThus, image distance, \u03bd = + 6 cm\nBecause \u03bd is +ve, so a virtual image is formed at a distance of 6 cm behind the mirror.\nMagnification, m=\\frac { -\\upsilon }{ u } =\\frac { -6 }{ -30 } =\\frac { 1 }{ 5 } (i.e. < 1)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nBecause \u03bd is +ve, so a virtual image is formed at a distance of 6 cm behind the mirror.\nMagnification, m=\\frac { -\\upsilon }{ u } =\\frac { -6 }{ -30 } =\\frac { 1 }{ 5 } (i.e. < 1)\nThe positive value of m shows that image erect and its value, which is less than 1, shows that image is smaller than the object. Thus, image is virtual, erect and diminished.\nQuestion 13\nThe magnification produced by a plane mirror is +1. What does this mean ?\nAnswer:\nSince magnification, m=\\frac { { h }^{ \u2018 } }{ h } =\\frac { -\\nu }{ u }. Given, m = +1, so h\u2019 = h and \u03bd = -u\n(i) m = 1 indicates the size of image is same as that of object.\n(ii) positive sign of m indicates that an erect image is formed.", + "Given, m = +1, so h\u2019 = h and \u03bd = -u\n(i) m = 1 indicates the size of image is same as that of object.\n(ii) positive sign of m indicates that an erect image is formed.\nThe opposite signs of \u03bd and u indicate that image is formed on the other side of the mirror from where the object is placed i.e., image is formed behind the mirror and thus image formed is virtual.\nQuestion 14", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 14\nAn object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.\nSolution:\nSince object size, h = +5 cm,\nobject distance, u = -20 cm\nand radius of curvature, R = +30 cm\n\nA virtual, erect image of height 2.2 cm is formed behind the mirror at a distance of 8.6 cm from the mirror.\nQuestion 15\nAn object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained ? Find the size and the nature of the image.\nAnswer:\nHere, object size, h = +7.0 cm,\nobject distance, u = -27 cm\nand focal length, f = -18 cm\nImage distance, \u03bd = ?\nand image size, h\u2019 = ?", + "Find the size and the nature of the image.\nAnswer:\nHere, object size, h = +7.0 cm,\nobject distance, u = -27 cm\nand focal length, f = -18 cm\nImage distance, \u03bd = ?\nand image size, h\u2019 = ?\nFrom the mirror formula, \\frac { 1 }{ f } =\\frac { 1 }{ \\nu } -\\frac { 1 }{ u }, we have\n\nThe screen should be placed at a distance of 54 cm on the object side of the mirror to obtain a sharp image.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nFrom the mirror formula, \\frac { 1 }{ f } =\\frac { 1 }{ \\nu } -\\frac { 1 }{ u }, we have\n\nThe screen should be placed at a distance of 54 cm on the object side of the mirror to obtain a sharp image.\n\nThe image is real, inverted and enlarged in size.\nQuestion 16\nFind the focal length of a lens of power -2.0 D. What type of lens is this ?\nAnswer:\nHere, P = -2.0 D\nThe type of lens is concave because the focal length is negative.\n\nQuestion 17\nA doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging ?\nAnswer:\nHere, P = +1.5 D\n\nBecause the focal length is positive, the prescribed lens is converging.", + "Question 17\nA doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging ?\nAnswer:\nHere, P = +1.5 D\n\nBecause the focal length is positive, the prescribed lens is converging.\nNCERT Solutions for Class 10 Science Chapter 10 Light Reflection and Refraction\nReflection of light by curved surfaces; Images formed by spherical mirrors, center of curvature, principal axis, principal focus, focal length, mirror formula (Derivation not required), magnification.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nRefraction; laws of refraction, refractive index.\nRefraction of light by spherical lens; Image formed by spherical lenses; Lens formula (Derivation not required); Magnification. Power of a lens;\nFormulae Handbook for Class 10 Maths and Science\nPage 168\nQuestion 1.\nDefine the principal focus of a concave mirror?\nAnswer:\nLight rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of the concave mirror.\nQuestion 2.\nThe radius of curvature of a spherical mirror is 20 cm. What is its focal length?\nAnswer:\nRadius of curvature, R = 20 cm\nRadius of curvature of a spherical mirror = 2 x Focal length (f)\nf = R/2 = 20/2 =10cm\nQuestion 1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.", + "Find the focal length of a convex mirror whose radius of curvature is 32 cm.\nAnswer: Radius of curvature, R = 32 cm\nRadius of curvature = 2 x Focal length (f)\nR = 2f\nf = R/2 = 32/2 = 16cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.\nAnswer: Radius of curvature, R = 32 cm\nRadius of curvature = 2 x Focal length (f)\nR = 2f\nf = R/2 = 32/2 = 16cm\nHence, the focal length of the given convex mirror is 16 cm.\nQuestion 2.\nA concave mirror produces three times magnified (enlarged) real image of object placed at 10 cm in front of it. Where is the image located?\nAnswer:\nGiven, u = \u2013 10 cm\nSince image is real inverted so, m = -3\nm = -v / u\n\u21d2 -3 = -v/ -10\nv= \u2013 30 cm\nNegative sign indicates the image will be real and image is formed at 30 cm in front of the mirror.\nPage: 176\nQuestion 1.\nA ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nPage: 176\nQuestion 1.\nA ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?\nAnswer:\nThe light ray bends towards the normal. When a ray of light travels from an optically rarer medium to an optically denser medium, it gets bent towards the normal. Since water is optically denser than air, a ray of light travelling from air into the water will bend towards the normal.\nQuestion 2.\nLight enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 \u00d7 108 m/s.\nAnswer:\nRefractive index of a medium nm is given by,\n\nQuestion 3.\nFind out, from Table, the medium having highest optical density. Also find the medium with lowest optical density.\nAnswer:\n\nHighest optical density = Diamond\nLowest optical density = Air", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 3.\nFind out, from Table, the medium having highest optical density. Also find the medium with lowest optical density.\nAnswer:\n\nHighest optical density = Diamond\nLowest optical density = Air\nOptical density of a medium is directly related with the refractive index of that medium. A medium which has the highest refractive index will have the highest optical density and vice-versa.\nIt can be observed from table 10.3 that diamond and air respectively have the highest and lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density\nQuestion 4.\nYou are given kerosene, turpentine and water. In which of these does the light travel fastest?\nAnswer:\n\nThe light can travel fast through water.\nQuestion 5.\nThe refractive index of diamond is 2.42. What is the meaning of this statement?\nAnswer:\nRefractive index of a medium nm is related to the speed of light in that medium v by the relation:\n\nWhere, c is the speed of light in vacuum/air", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nAnswer:\nRefractive index of a medium nm is related to the speed of light in that medium v by the relation:\n\nWhere, c is the speed of light in vacuum/air\nThe refractive index of diamond is 2.42. This suggests that the speed of light in diamond will reduce by a factor 2.42 compared to its speed in air.\nPage 184\nQuestion 1.\nDefine one dioptre of power of a lens?\nAnswer:\nOne dioptre is the power Of a lens Of focal length 1m.\nPower of lens is defined as the reciprocal of its focal length. If P is the power of a lens of focal length F in metres, then\nP = 1/ f (in meters)\nThe S.I. unit of power of a lens is Dioptre. It is denoted by D.\n1 dioptre is defined as the power of a lens of focal length 1 metre.\n1 D = 1 m\u22121\nQuestion 2.\nA convex lens forms a real and inverted image of a needle at a distance of 50 cm from it.", + "It is denoted by D.\n1 dioptre is defined as the power of a lens of focal length 1 metre.\n1 D = 1 m\u22121\nQuestion 2.\nA convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the lens if the image is equal to the size of the object? Also find the power of the lens.\nAnswer:\nv = + 50 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nAnswer:\nv = + 50 cm\nSince image is real and of same size. The position of image should be double the focal length.\nHence, the object should be at 2f.\nV = 2f = 50, f = 25 cm.\nPower = 1/f = 100/25 = 4D\nQuestion 3.\nFind the power of a concave lens of focal length 2 m.\nAnswer:\n\nQuestion 1.\nWhich one of the following materials cannot be used to make a lens?\n(a) Water (b) Glass\n(c) Plastic (d) Clay\nAnswer: (d) Clay\nQuestion 2.\nThe \u00a1mage formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?\n(a) Between the principal focus and the centre of Curvature\n(b) At the centre of curvature\n(c) Beyond the centre of curvature\n(d) Between the pole of the mirror and Its principal focus.\nAnswer: (d) Between the pole of the mirror and its principal focus.\nQuestion 3.", + "(a) Between the principal focus and the centre of Curvature\n(b) At the centre of curvature\n(c) Beyond the centre of curvature\n(d) Between the pole of the mirror and Its principal focus.\nAnswer: (d) Between the pole of the mirror and its principal focus.\nQuestion 3.\nWhere should an object b. placed In front of a convex lens to get a real\nimage of the size of the object?\n(a) At the principal focus of the lens", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 3.\nWhere should an object b. placed In front of a convex lens to get a real\nimage of the size of the object?\n(a) At the principal focus of the lens\n(b) At twice the focal length\n(c) At infinity\n(d) Between the optical centre of the lens and its principal focus\nAnswer:\n(b) At twice the focal length\nQuestion 4.\nA spherical mirror and a thin spherical lens have each a focal length of 15 cm. The mirror and the lens are likely to be:\n(a) both concave\n(b) both convex\n(c) the mirror is concave, but the lens is convex\n(d) the mirror is convex, but the lens is concave\nAnswer:\n(a) Both concave.\nQuestion 5.\nNo matter how far you stand from a mirror, your Image appears erect. The mirror is likely to be\n(a) plane\n(b) concave\n(c) convex\n(d) Either plane or convex\nAnswer:\n(d) Either plane or convex.\nQuestion 6.\nWhich of the following lenses would you prefer to use while reading small letters found \u00a1n a dictionary?", + "The mirror is likely to be\n(a) plane\n(b) concave\n(c) convex\n(d) Either plane or convex\nAnswer:\n(d) Either plane or convex.\nQuestion 6.\nWhich of the following lenses would you prefer to use while reading small letters found \u00a1n a dictionary?\n(a) A convex lens of focal length 50cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nAnswer:\n(d) Either plane or convex.\nQuestion 6.\nWhich of the following lenses would you prefer to use while reading small letters found \u00a1n a dictionary?\n(a) A convex lens of focal length 50cm\n(b) A concave lens of focal length 50cm\n(c) A convex lens of focal length 5 cm\n(d) A concave lens of focal length 5 cm.\nAnswer:\n(c) A convex lens of focal length 5 cm.\nQuestion 7.\nWe wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. what should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this\ncase.\nAnswer:\nWe are given the focal length cf the concave mirror as f = -15cm.\nFor getting an erect image using a concave mirror, the object should be placed at a distance less than the focal length.\ni.e. 15 cm from the pole.", + "Answer:\nWe are given the focal length cf the concave mirror as f = -15cm.\nFor getting an erect image using a concave mirror, the object should be placed at a distance less than the focal length.\ni.e. 15 cm from the pole. The image formed will be virtual, enlarged and erect.\n\nQuestion 8.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\ni.e. 15 cm from the pole. The image formed will be virtual, enlarged and erect.\n\nQuestion 8.\nName the type of mirror used in the following situations.\n(a) Headlights of a car\n(b) Side/rear-view mirror of a vehicle\n(c) Solar furnace\nSupport your answer with reason.\nAnswer:\n(a) Concave mirror, to get powerful and parallel beams of light.\n(b) Convex mirror because it always gives an erect image and enables the driver to view much larger area.\n(c) Concave or parabolic mirror because it can concentrate sunlight at the focus to produce heat in the solar furnace.\nQuestion 9.\nOne half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 9.\nOne half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.\nAnswer:\nYes, even when one half of the lens is covered with a black paper, complete image of the object will be formed. Take a convex lens and focus the light from a distant object onto a screen. As expected an image (sharp) is formed at a distance equal to the focal length Cover the lower or the upper half of the lens and focus the light from the same object onto the same screen. You will be able to get a sharp image again; however the brightness of the image will be less in the second case. The same effect w,ll be seen even if the lens is half covered with black strips.\nQuestion 10.\nAn object 5cm in length is held 25cm away from a converging lens of focal length 10 cm. Draw a ray diagram and find the position, size and the nature of the image formed.\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 10.\nAn object 5cm in length is held 25cm away from a converging lens of focal length 10 cm. Draw a ray diagram and find the position, size and the nature of the image formed.\nAnswer:\n\n\nTherefore, the mage \u00a1s formed between F2 and 2F2 on the other side of the lens. It is real and inverted, and smaller in size than the object.\nQuestion 11.\nA concave lens of focal length 15 cm forms an \u00a1mage 10 cm from the lens. How far is the object placed from the Pens? Draw the ray diagram.\nAnswer:\n\nQuestion 12.\nAn object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the \u00a1mage.\nAnswer:\nf = +15 cm. u = -1o cm\nFor mirror, we have\n\nThe image must be virtual and erect.\nQuestion 13.\nThe magnification produced by a plane mirror is +1. What does this mean?\nAnswer:\nThis means that size of the image is equal to the size of the object.\nQuestion 14.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 13.\nThe magnification produced by a plane mirror is +1. What does this mean?\nAnswer:\nThis means that size of the image is equal to the size of the object.\nQuestion 14.\nAn object 5.0 cm in length Is placedat a distanc, of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image nature and size.\nAnswer:\n\nQuestion 15.\nAn object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.\nAnswer:\n\nQuestion 16.\nFind the focal length of a lens of power -2.0 D. What type of lens is this?\nAnswer:\n\nQuestion 17.\nA doctor has prescribed a corrective lens of power +1.5 D. find the focal length of the lens. Is the prescribed lens diverging or converging?", + "Find the focal length of a lens of power -2.0 D. What type of lens is this?\nAnswer:\n\nQuestion 17.\nA doctor has prescribed a corrective lens of power +1.5 D. find the focal length of the lens. Is the prescribed lens diverging or converging?\nAnswer:\n\nMultiple Choice Questions (MCQs) [1 mark each]\nQuestion 1.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nAnswer:\n\nMultiple Choice Questions (MCQs) [1 mark each]\nQuestion 1.\nHold a highly polished steel spoon curved inwards close to your face and move it slowly away from your face. What will you observe?\n(a) Enlarged and erect image of your face\n(b) Smaller and inverted image of your face\n(c) Smaller and erect image of your face\n(d) Enlarged and inverted image of your face\nAnswer:\n(b) The inner curved surface of a highly polished steel spoon acts as a concave mirror. When the spoon is at a small distance from the face such that, the object lies between pole and focus of concave mirror, so an enlarged and erect image of your face will be observed but as the spoon is slowly moved away from the face, the image becomes smaller and appears inverted.\nQuestion 2.\nWhich one of the following materials cannot be used to make a lens? [NCERT]\n(a) Water\n(b) Glass\n(c) Plastic\n(d) Clay\nAnswer:\n(d) Clay can never be transparent, so it cannot be used to make lens.\nQuestion 3.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\n(a) Water\n(b) Glass\n(c) Plastic\n(d) Clay\nAnswer:\n(d) Clay can never be transparent, so it cannot be used to make lens.\nQuestion 3.\nNo matter how far you stand from a mirror, your image appears erect. The mirror is likely to be [NCERT]\n(a) plane\n(b) concave\n(c) convex\n(d) either plane or convex\nAnswer:\n(d) Plane mirrors and convex mirrors always form the erect images.\nQuestion 4.\nThe image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object? [NCERT]\n(a) Between principal focus and centre of curvature\n(b) At centre of curvature\n(c) Beyond centre of curvature\n(d) Between pole of the mirror and its principal focus\nAnswer:\n(d)\nQuestion 5.\nAn object AB is placed in front of a convex lens at its centre of curvature as shown in figure below.\n\nFour students traced the path of light ray after refraction through the lens. Which one of them is correct?", + "An object AB is placed in front of a convex lens at its centre of curvature as shown in figure below.\n\nFour students traced the path of light ray after refraction through the lens. Which one of them is correct?\n\n(a) Only I\n(b) Only II\n(c) Only III\n(d) Only IV\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nFour students traced the path of light ray after refraction through the lens. Which one of them is correct?\n\n(a) Only I\n(b) Only II\n(c) Only III\n(d) Only IV\nAnswer:\n(d) When the object is placed at centre of curvature (2Fx) of convex lens, the same sized image is formed at 2F2. The image formed is real and inverted.\nQuestion 6.\nA spherical mirror and a thin spherical lens have each of a focal length -15 cm. The mirror and lens are likely to be [NCERT]\n(a) both concave\n(b) both convex\n(c) mirror is concave and lens is convex\n(d) mirror is convex and lens is concave\nAnswer:\n(a) The focal length is taken as negative for both concave mirror and concave lens.\nQuestion 7.\nWhich of the following can make a parallel beam of light when light from a point source is incident on it?", + "Question 7.\nWhich of the following can make a parallel beam of light when light from a point source is incident on it? [NCERT Exemplar]\n(a) Concave mirror as well as convex lens\n(b) Convex mirror as well as concave lens\n(c) Two plane mirrors placed at 90\u00b0 to each other\n(d) Concave mirror as well as concave lens\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\n(a) Concave mirror as well as convex lens\n(b) Convex mirror as well as concave lens\n(c) Two plane mirrors placed at 90\u00b0 to each other\n(d) Concave mirror as well as concave lens\nAnswer:\n(a) A ray passing through the principal focus of a concave mirror or convex lens, after reflection/refraction, will emerge parallel to the principal axis.\n\nQuestion 8.\nUnder which of the following conditions, a concave mirror can form an image larger than the actual object? [NCERT Exemplar]\n(a) When an object is kept at a distance equal to its radius of curvature\n(b) When an object is kept at a distance less than its focal length\n(c) When an object is placed between the focus and centre of curvature\n(d) When an object is kept at a distance greater than its radius of curvature\nAnswer:\n(c) A concave mirror can form an image enlarged, real and inverted than the actual object, beyond centre of curvature (C) when object is placed between the focus (F) and centre of curvature.\nQuestion 9.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 9.\nA light ray enters from medium A to medium Bas shown in the figure. The refractive index of medium B relative to A will be [NCERT Exemplar]\n\n(a) greater than unity\n(b) less than unity\n(c) equal to unity\n(d) zero\nAnswer:\n(a) Since, light rays in the medium B goes towards normal. So, it has greater refractive index and lesser velocity of light w.r.t. medium A. So, refractive index of medium B w.r.t. medium A is greater than unity.\nQuestion 10.\nFigure shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is\n\nAnswer:\n(a) Given, angle of incidence, i = 60\u00b0, angle of refraction, r = 45\u00b0\nRefractive index of the medium B relative to medium A,\n\nQuestion 11.\nBeams of light are incident through the holes A and B and emerge out of box through the holes C \u2013 and D respectively, as Box shown in the figure.", + "Beams of light are incident through the holes A and B and emerge out of box through the holes C \u2013 and D respectively, as Box shown in the figure.\n\nWhich of the following could be inside the box? [NCERT Exemplar]\n(a) A rectangular glass slab\n(b) A convex lens", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nWhich of the following could be inside the box? [NCERT Exemplar]\n(a) A rectangular glass slab\n(b) A convex lens\n(c) A concave lens\n(d) A prism\nAnswer:\n(a) Here, the emergent rays are parallel to the direction of the incident ray. Therefore, a rectangular glass slab could be inside the box as the extent of bending of light ray at the opposite parallel faces AB (air-glass interface) and CD (glass-air interface) of the rectangular glass slab are equal and opposite. This is why the ray emerges parallel to the incident ray.\nQuestion 12.\nA beam of light is incident through the holes on side A and emerges out of the holes on the other face of the box as shown in the figure. Which of the following could be inside the box? [NCERT Exemplar]\n\n(a) Concave lens\n(b) Rectangular glass slab\n(c) Prism\n(d) Convex lens\nAnswer:\n(d) Since, in the figure all the parallel rays converge at a point. So, inside the box there must be a convex lens.\n\nQuestion 13.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\n(b) Rectangular glass slab\n(c) Prism\n(d) Convex lens\nAnswer:\n(d) Since, in the figure all the parallel rays converge at a point. So, inside the box there must be a convex lens.\n\nQuestion 13.\nWhich of the following statement is true? [NCERT Exemplar]\n(a) A convex lens has 4D power having a focal length 0.25 m\n(b) A convex lens has 4D power having a focal length -0.25 m\n(c) A concave lens has 4D power having a focal length 0.25 m\n(d) A concave lens has 4D power having a focal length -0.25 m\nAnswer:\n(a) The power P of a lens of focal length f is given by\nP = 1/f, where f is the focal length in metre and P is the power in dioptre.\nP= 1/f or f = 1/P = 1/4 = 0.25 m\nQuestion 14.", + "P= 1/f or f = 1/P = 1/4 = 0.25 m\nQuestion 14.\nMagnification produced by a rear view mirror fitted in vehicles [NCERT Exemplar]\n(a) is less than one\n(b) is more than one\n(c) is equal to one\n(d) can be more than or less than one depending upon the position of the object in front of it.\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\n(a) is less than one\n(b) is more than one\n(c) is equal to one\n(d) can be more than or less than one depending upon the position of the object in front of it.\nAnswer:\n(a) The convex mirror forms virtual, erect and diminished image of the object and rear view mirror also form same type of image. Therefore, magnification (m) produced by a rear view mirror fitted in vehicles is less than one, i.e. m < 1.\nQuestion 15.\nRays from the Sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed, so that size of its image is equal to the size of the object? [NCERT Exemplar]\n(a) 15 cm in front of the mirror\n(b) 30 cm in front of the mirror\n(c) between 15 cm and 30 cm in front of the mirror\n(d) more than 30 cm in front of the mirror\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\n(a) 15 cm in front of the mirror\n(b) 30 cm in front of the mirror\n(c) between 15 cm and 30 cm in front of the mirror\n(d) more than 30 cm in front of the mirror\nAnswer:\n(b) The rays from the Sun, i.e. from infinity, are parallel to principal axis after reflection converge at a point is known as focus. Therefore, focal length if) of concave mirror is 15 cm. And we know that, same size, real and inverted image is formed by concave mirror when object is placed at focus 2 A or centre of curvature, so to form same size of image, object will be placed at 15 x 2 =30 cm.\nQuestion 16.\nThe path of a ray of light coming from air passing through a rectangular glass slab traced by four students shown as I, II, III and IV in the figure. Which one of them is correct? [NCERT Exemplar]\n\n(a) Only I\n(b) Only II\n(c) Only III\n(d) Only IV\nAnswer:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\n(a) Only I\n(b) Only II\n(c) Only III\n(d) Only IV\nAnswer:\n(b) In a rectangular glass slab, the emergent rays are parallel to the direction of the incident ray, because the lateral deviation of bending of the ray of light at the opposite parallel faces (air-glass interface) and (glass-air interface) of the rectangular glass slab are equal and opposite. This is why the ray emerges are parallel to the incident ray.\nQuestion 17.\nYou are given water, mustard oil, glycerine and kerosene. In which of these media, a ray of light incident obliquely at same angle would bend the most? [NCERT Exemplar]\n(a) Kerosene\n(b) Water\n(c) Mustard oil\n(d) Glycerine\nAnswer:\n(d) The given material having their refractive index as kerosene is 1.44, water is 1.33, mustard oil is 1.46 and glycerine is 1.74. Thus, glycerine is most optically denser and hence have the largest refractive index.", + "Thus, glycerine is most optically denser and hence have the largest refractive index. Therefore, ray of light bend most in glycerine.\nQuestion 18.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nQuestion 18.\nA student placed a light bulb in midway between the two plane mirrors inclined at an angle of 60\u00b0. How many images will be observed by him?\n(a) 4\n(b) 6\n(c) 5\n(d) 8\nAnswer:\n(c) Number of images formed by two plane mirrors inclined at an angle 60\u00b0 when a light bulb is placed in midway between them is\nN = 360\u00b0/60\u00b0 \u2013 1 = 6 \u2013 1 = 5\nQuestion 19.\nWhere should an object be placed in front of a convex lens to get a real image of the size of the object? [NCERT]\n(a) At the principal focus of the lens\n(b) At twice the focal length\n(c) At infinity\n(d) Between the optical centre of the lens and its principal focus\nAnswer:\n(b) To set the real image of the size of the object, it should be placed at twice the focal length of a convex lens.\nQuestion 20.\nWhich of the following lenses would you prefer to use while reading small letters found in dictionary?", + "Question 20.\nWhich of the following lenses would you prefer to use while reading small letters found in dictionary? [NCERT]\n(a) A convex lens of focal length 50 cm\n(b) A concave lens of focal length 50 cm\n(c) A convex lens of focal length 5 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\n(a) A convex lens of focal length 50 cm\n(b) A concave lens of focal length 50 cm\n(c) A convex lens of focal length 5 cm\n(d) A concave lens of focal length 5 cm\nAnswer:\n(c) Convex lens is used as magnifying glass. For better performance its focal length should be small.\n 10 Science Light Reflection and Refraction Mind Map\nLIGHT REFLECTION & REFRACTION\nForm of energy produces the sensation of vision in eyes. Light (EM waves wave-length 400 nm to 750 nm).\nThe path of light (always travel in straight line) is ray of light\nCharacteristics of light\nRectilinear propagation of light\nLight travels with a speed of 3 \u00d7 108 m/s in air/vaccum.\nSpeed of light depends on the medium\nLight shows behaviour such as reflection, refraction, interference, diffraction, polarisation etc.\nLaw of Refraction\nRefraction of light: Bending of light ray while passing from one medium to another medium\nA ray of light bends towards the normal, while going from rarer to denser medium", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nLaw of Refraction\nRefraction of light: Bending of light ray while passing from one medium to another medium\nA ray of light bends towards the normal,", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nLaw of Refraction\nRefraction of light: Bending of light ray while passing from one medium to another medium\nA ray of light bends towards the normal, while going from rarer to denser medium\nAnd bends away from the normal while going from denser to rarer medium\nRefraction of light takes place because the speed of light is different in the two media\nTotal internal Reflection : Ray totally reflected back to denser medium\nPhenomena based on TIR\nMirage \u2013 optical illusion in deserts\nLooming \u2013 optical illusion in cold countries\nOptical fibre\nBrilliance of diamond\nNecessary conditions for TIR\n(i ) Ray of light must travel from denser to rarer medium\n(ii) \u2220i > \u2220c for two media\nCritical angle (c) Angle i in denser medium for which angle of refraction in rarer medium is 90\u00b0 \u03bc = \\frac{1}{\\sin C}\nSnell\u2019s law\n\u03bc = \\frac{\\sin i}{\\sin r}\n\nFor two media\n1\u03bc2 = \\frac{\\mu_{2}}{\\mu_{1}}=\\frac{\\sin i}{\\sin r}", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nSnell\u2019s law\n\u03bc = \\frac{\\sin i}{\\sin r}\n\nFor two media\n1\u03bc2 = \\frac{\\mu_{2}}{\\mu_{1}}=\\frac{\\sin i}{\\sin r}\nReflection of light: Turning back of light in the same medium after striking the reflecting surface or mirror\nAfter reflection, velocity, frequency and wavelength of light remains same but intensity decreases\nIf reflection takes place from denser medium then phase change \u2018\u03c0\u2019\nRegular Reflection\nReflection on smooth surface.\nDiffuse Reflection\nReflection on rough surface.\nLaws of Reflection\nThe incident ray the normal and the reflected ray all lie in the same plane The angle of incidence (i) is always equal to angle of reflection (r) i.e., \u2220i = \u2220r\nMirror formula\n\\frac{1}{f}=\\frac{1}{u}+\\frac{1}{v}\n\nWhen two plane mirrors are held at an angle 9 with their reflecting surfaces facing each other and an object is placed between them, images are formed by successive reflections. .\nfconcave = negative\nfconvex = positive\nand fplane = \u221e", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nfconcave = negative\nfconvex = positive\nand fplane = \u221e\nRelation between focal length (f) and radius of curvature, R\nf = \\frac{R}{2}\nMagnification\nm = \\frac{\\mathrm{v}}{\\mathrm{u}}=\\frac{\\text { height of image }}{\\text { height of object }}\n\nm = \\frac{f}{f-u}=\\frac{f-v}{f}\nThe incident ray, the normal and the refracted ray all lie in the same plane\nRefractive index,\n\u03bc = \\frac{c}{v}=\\frac{\\text { real depth }}{\\text { apparent depth }}\nPlane Mirror\nIs a looking glass, highly polished on one surface.\nForms virtual and erect image\nDistance of object from mirror = distance of image from mirror.\nThe size of the image is same as object.\nImage is laterally inverted.\nUsed in kaleidoscope periscope, etc.\nConcave Mirror\nSpherical glass polished on the outside. It is also known as a converging mirror.", + "The size of the image is same as object.\nImage is laterally inverted.\nUsed in kaleidoscope periscope, etc.\nConcave Mirror\nSpherical glass polished on the outside. It is also known as a converging mirror.\nImages produced are always real, inverted, can be enlarged based on the position except when object is placed between pole and focus.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nImages produced are always real, inverted, can be enlarged based on the position except when object is placed between pole and focus.\nUses: Make-up and shaving mirrors, dentist mirror, in floodlight etc.\nImage formation by a convex mirror for different positions of the object\nConvex Mirror\nSpherical glass polished inside. It is also known as diverging mirror.\nIt forms virtual, upright and small images.\nUses: for security\u2019 purposes, in vehicles as rear- view mirror and street lighting.", + "Image formation by a convex mirror for different positions of the object\nConvex Mirror\nSpherical glass polished inside. It is also known as diverging mirror.\nIt forms virtual, upright and small images.\nUses: for security\u2019 purposes, in vehicles as rear- view mirror and street lighting.\nImage formation by a concave mirror for different positions of the object\nAtmospheric Refraction\nEarth\u2019s atmosphere is thin at the top and dense at the bottom, thus leads to refraction of light,\n\u03bc = c/v\nTwinkling of stars\nRainbow\nAdvanced sunrise and delayed sunset\nRefraction Through a Glass Slab\nx = \\frac{t \\sin (i-r)}{\\cos r}\n\n\u2234 x \u221d \u03bc\nPower of a lens\nP = \\frac{1}{f(\\text { in metre })}\n\nUnit of power of lens is diopter (D)\nPconvex \u2192 Positive\nPconcave \u2192 Negative\nand Pplane \u2192 Zero\nLens", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nx = \\frac{t \\sin (i-r)}{\\cos r}\n\n\u2234 x \u221d \u03bc\nPower of a lens\nP = \\frac{1}{f(\\text { in metre })}\n\nUnit of power of lens is diopter (D)\nPconvex \u2192 Positive\nPconcave \u2192 Negative\nand Pplane \u2192 Zero\nLens\nPiece of transparent material with two refracting surfaces, at least one is curved and refractive index should different as that of the surrounding.\nLens formula\n\\frac{1}{f}=\\frac{1}{v}-\\frac{1}{u}\n\nfconvex \u2192 negative\nfconcave \u2192 positive\nand fplane \u2192 \u221e\nConcave Lens\nCental portion of lens is thinner than marginal. It as also known as diverging lens.\nConvex Lens\nCentral portion of lens is thicker than marginal. It is also known us converging lens.\nMagnification\nRatio of distance of image to the distance of object from the optical centre.", + "It as also known as diverging lens.\nConvex Lens\nCentral portion of lens is thicker than marginal. It is also known us converging lens.\nMagnification\nRatio of distance of image to the distance of object from the optical centre. Also equal to height of image to the height of object\nm = \\frac{\\mathrm{I}}{\\mathrm{o}}=\\frac{\\mathrm{v}}{\\mathrm{u}}=\\frac{\\mathrm{h}_{\\mathrm{I}}}{\\mathrm{h}_{\\mathrm{o}}}", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-INTEXT QUESTIONS.txt\nm = \\frac{\\mathrm{I}}{\\mathrm{o}}=\\frac{\\mathrm{v}}{\\mathrm{u}}=\\frac{\\mathrm{h}_{\\mathrm{I}}}{\\mathrm{h}_{\\mathrm{o}}}\nNature, position and relative size of the image formed by a concave lens for various position of the object\nNature, position and relative size of the image formed by a convex lens for various positions of the object\nNow that you are provided all the necessary information regarding NCERT Solutions For Class 10 Science Chapter 10 Light Reflection and Refraction and we hope this detailed article on light reflection and refraction class 10 NCERT solutions is helpful. If you have any questions related to this article, kindly ask your questions through the comment section below and we will get back to you as soon as possible.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nQuestion-1\nWhy is eye lens of telescope smaller than objective lens?\nSolution:\nThe eye lens of telescope is smaller than its objective lens to sharpen the image which is obtained using telescope.\nQuestion-2\nWhat is short sight? How can it be corrected?\nSolution:\nIf a person can see nearer objects clearly but not the far objects, then he is affected by short sight. It can be corrected using a suitable concave lens.\nQuestion-3\nGive the condition required to achieve a larger magnification of a small object by a compound microscope ?\nSolution:\nTo achieve a larger magnification of a small object, both the objective and the eyepiece of a compound microscope should have smaller focal lengths. Magnification of the order of 1000 is possible with very good design of the microscopes, such as in the oil-immersion type of microscopes.\nQuestion-4\nA person having a myopic eye used the concave lens of focal length 50cm. What is the power of the lens?\nSolution:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nQuestion-4\nA person having a myopic eye used the concave lens of focal length 50cm. What is the power of the lens?\nSolution:\nThe formula for the power of lens is Power (P) = 1/f.\nP = 1/50 = 0.02 dioptre.\nQuestion-5\nWhich part of human eye helps in the perception of colours?\nSolution:\nThe retina of the human eye has a large number of cone-shaped and rod-shaped cells. The rods respond to the intensity of light and the cones to the colours. These help us to perceive colours.\nQuestion-6\nDefine the term \u201caccommodation of the eye\u201d.\nSolution:\nThe ability of the eye to focus on near and distant objects, by adjusting its focal length, is called the accommodation of the eye.\nQuestion-7\nDefine \u201cleast distance of distinct vision\u201d.\nSolution:\nThe smallest distance, at which the eye can see objects clearly without strain, is called the near point or the least distance of distinct vision. For a young adult with normal vision, it is about 25cm.\nQuestion-8", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nQuestion-8\nDifferentiate between a microscope and a telescope based on its use.\nSolution:\nA microscope is an optical instrument which is used to see the finer details of tiny objects, and a telescope is used for viewing distant objects clearly.\nQuestion-9\nHow is the length of the telescope tube related to the distance between the objective and the eyepiece of the telescope? How can the magnification of the telescope be increased?\nSolution:\nThe length of the telescope tube is equal to the distance between the objective and eyepiece of the telescope. In order to increase the magnification of the telescope, the objective must have a large focal length and the eyepiece a small focal length.\nQuestion-10\nWhy does the objective lens of an astronomical telescope have a large light gathering power?\nSolution:\nThe objective lens of a telescope has a large aperture to have a large light gathering power.\nQuestion-11\nExplain clearly the working of a human eye?\nSolution:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nSolution:\nThe objective lens of a telescope has a large aperture to have a large light gathering power.\nQuestion-11\nExplain clearly the working of a human eye?\nSolution:\nThe light rays coming from the object kept in front of us enter the pupil of the eye and fall on the eye-lens. The eye-lens is a convex lens, so it converges the light rays and produces a real and inverted image of the object on the retina. The image formed on the retina is conveyed to the brain by the optic nerve and gives rise to the sensation of vision. The light-sensitive cells present in the retina gets activated and generates electrical signals. The retina sends these electrical signals to the brain through the optic nerve and gives rise to the sensation of vision. The mind interprets the inverted image formed on the retina as an erect object.\nQuestion-12\nHow does the eye adjust itself to deal with light of varying intensity?\nSolution:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nQuestion-12\nHow does the eye adjust itself to deal with light of varying intensity?\nSolution:\nThe amount of light entering the eye is controlled by the pupil. If the intensity of the outside light is low, then the pupil expands to allow more light to enter the eye. If the outside intensity is high, then the pupil contracts so that less light enters the eye.\nQuestion-13\nWhich phenomenon of vision is made use of in cinematography? Explain briefly how it is used.\nSolution:\nPersistence of vision is made use of in cinematography. It is the ability of an eye to continue to see the image of an object for a very short duration even after the removal of the object. We are able to see the movie pictures in a cinema due to the persistence of vision.\nThe still pictures taken by a movie camera are projected on the screen at a rate of about 24 pictures per second. Under these conditions, the image of one picture persists on the retina till the image of next picture falls on the screen and so on.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nDue to this, the slightly different images of the successive pictures present on the film merge smoothly into one another and give us the feeling of moving images. Thus the principle of persistence of vision is made use of, in the motion picture projection in cinematography.\nQuestion-14\nHow do we see colours?\nSolution:\nThe retina of our eyes has a large number of light sensitive cells. The cells on the retina are of two shapes:\n(i) Rod shaped and\n(ii) Cone shaped\nThe rod shaped cells of our retina respond to the intensity of light. The cone shaped cells of our retina respond to colours. These cells make us see colours and to distinguish between them. Thus cones make colour perception possible.\nQuestion-15\nHow does an eye focus the objects lying at various distances?\nSolution:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nQuestion-15\nHow does an eye focus the objects lying at various distances?\nSolution:\nAn eye can focus the images of the distant objects as well as that of the nearby objects on its retina by changing the focal length of its lens. The focal length of the eye-lens is changed by the action of the ciliary muscles. The ciliary muscles can change the thickness of the eye-lens and hence its focal length changes the converging power of the eye-lens.\nQuestion-16\nWhen is a person said to have developed cataract in his eye? How is the vision of a person having cataract restored?\nSolution:\nCataract develops when the eye-lens of a person becomes unclear due to the formation of a membrane over it. The vision of a person having cataract can be restored after getting cataract surgery done on the eye-lens having cataract. This defect cannot be corrected by any type of spectacle lenses.\nQuestion-17\nWhere should an object be placed in front of a convex lens so as to use it as a simple microscope?", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nQuestion-17\nWhere should an object be placed in front of a convex lens so as to use it as a simple microscope?\nSolution:\nThe object should be placed within the focus of the convex lens to use it as a simple microscope.\nQuestion-18\nIf we need magnification of 375 from a microscope of tube length 15 cm and objective of focal length 0.5 cm, what focal length of eye-piece should we use?\nSolution:\nMagnification m = 375\nLength of the tube L = 15 cm\nFocal length of the objective fo = 0.5 cm\nWe know that\n\nQuestion-19\nIf focal length of the objective and eye-piece of a telescope are 10 cm and 4 cm respectively, what will be its magnifying power in\n(i) normal adjustments,\n(ii) adjustment where image of distance object is formed at D? Given D = 25 cm.", + "Given D = 25 cm.\nSolution:\nFocal length of objective fo = 10 cm\nFocal length of eyepiece fe = 4 cm\n(i) In normal adjustment\n\nQuestion-20\nWhich of the two, objective or eye-piece, of a telescope has a large aperture? Give reason for your answer.\nSolution:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nFocal length of eyepiece fe = 4 cm\n(i) In normal adjustment\n\nQuestion-20\nWhich of the two, objective or eye-piece, of a telescope has a large aperture? Give reason for your answer.\nSolution:\nThe objective lens of a telescope has a large aperture, because it can gather more light from the distant object and form a bright image of the distant object.\nQuestion-21\nWhat are the common defects of vision that can be corrected by the use of suitable eyeglasses or spectacles?\nSolution:\nThere are mainly four common defects of vision that can be corrected by the use of suitable eye glasses or spectacles. These are\n(i) Myopia or near-sightedness,\n(ii) Hypermetropia or far-sightedness,\n(iii) Presbyopia, and\n(iv) Astigmatism.\nQuestion-22\nWhat property of the eye is the principle of motion, pictures?\nSolution:\nRefraction of light through the eye lens is the basic principle for the visualization of pictures and motion.\nQuestion-23\nWhat are coaxial lenses and where are they used?\nSolution:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nSolution:\nRefraction of light through the eye lens is the basic principle for the visualization of pictures and motion.\nQuestion-23\nWhat are coaxial lenses and where are they used?\nSolution:\nCoaxial lenses are lenses having common principal axes. A compound microscope, in its simple form, consists of coaxial lens (two lenses having common principal axes) mounted on the outer ends of two hollow metallic tubes. One of the two lenses acts as the objective and the other as the eyepiece.\nQuestion-24\nDraw a diagram of a simple microscope.\nSolution:\n\nQuestion-25\nHow does the eye adjust its focal length?\nSolution:\nThe eye adjust its focal length by means of its ciliary\u2019s muscles.\nQuestion-26\nExplain presbyopia and astigmatism.\nSolution:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nSolution:\n\nQuestion-25\nHow does the eye adjust its focal length?\nSolution:\nThe eye adjust its focal length by means of its ciliary\u2019s muscles.\nQuestion-26\nExplain presbyopia and astigmatism.\nSolution:\nPresbyopia is part of the natural aging process of the eye, and can be easily corrected. Technically, presbyopia is the loss of the eye\u2019s ability to change its focus to see objects that are near. It is not a disease. It\u2019s as natural as wrinkles, and it affects everybody at some point in life. Presbyopia generally starts to appear around age 40.\nAstimagtism: The cornea is the transparent layer over the colored part of the eye. It bends (refracts) light rays and helps focus the light onto the retina in the back of the eye so people can see. When the cornea is oblong shaped, it causes light rays to focus on two different points on the retina, instead of just one. As a result, people with significant astigmatism may have distorted or blurry vision.\nQuestion-27", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nQuestion-27\nA person uses convex lens spectacles. What vision defect does he have? Draw a diagram\n(i) to show the defective eye\n(ii) to show the correction with the lens.\nSolution:\nWhen a person uses convex lens to cure his long sight (hypermetropia).\nLong Sight\n\nQuestion-28\nHow is the terrestrial telescope different from the astronomical telescope?\nSolution:\nThe terrestrial telescopes have one more additional convex lens (erecting lens), placed between the objective and the eyepiece, to make the final image erect. This erecting lens is not required in the astronomical telescope as the celestial objects that are viewed using a astronomical telescope are spherical in shape, so it does not matter whether the final image is inverted or erect.\nQuestion-29\nHow will you distinguish between a compound microscope and a telescope.\nSolution:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nQuestion-29\nHow will you distinguish between a compound microscope and a telescope.\nSolution:\nA compound microscope uses a very short focal length objective lens to form a greatly enlarged image. This image is then viewed with a short focal length eyepiece used as a simple magnifier. The image should be formed at infinity to minimize eyestrain.\nA telescope is an instrument designed for the observation of remote objects. The term usually refers to optical telescopes, but there are telescopes for most of the spectrum of electromagnetic radiation and for other signal types.\nQuestion-30\nA lens used as a simple magnifier gives magnification of 6. What is its focal length?\nSolution:\n\nQuestion-31\nA compound microscope has an objective of focal length 1.0 cm and an eyepiece of focal length 4.0 cm. If the length of the tube is 20 cm, find the magnification of the compound microscope.\nSolution:\n\nQuestion-32", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nSolution:\n\nQuestion-32\nA telescope has an objective of focal length 140.0 cm and an eyepiece of focal length 5.0 cm. Find\n(i) the magnification of the telescope for viewing distant objects for normal adjustment\n(ii) separation between the objective lens and the eyepiece.\nSolution:\nObjective of focal length, f0 = 140.0 cm.\nEyepiece if focal length, fe = 5.0 cm.\n\nQuestion-33\nA 52-year old near-sighted person wears eye-glass with a power of \u20135.5 dioptres for distance viewing. His doctor prescribes a correction of +1.5 dioptres in the near-vision section of his bi-focals. This is measured relative to the main part of the lens (i) What is the focal length of his distance-viewing part of the lens? (ii) What is the focal length of the near-vision section of the lens?\nSolution:\n\nQuestion-34\nWhat is Astigmatism?\nSolution:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nSolution:\n\nQuestion-34\nWhat is Astigmatism?\nSolution:\nAstigmatism is the defect of an eye due to which a person cannot focus on both the horizontal and vertical lines, simultaneously. The reason for this defect is irregularities on the surface of the cornea.\nQuestion-35\nExplain the angle of prism.\nSolution:\nA triangular glass prism consists of three rectangular lateral surfaces and two triangular bases. Lateral faces are inclined to each other and the angle between two lateral surfaces is called the angle of prism.\nQuestion-36\nWhat is dispersion?\nSolution:\nThe splitting of white light into its seven colours on passing through a transparent medium (e.g. glass prism) is called dispersion of light.\nQuestion-37\nWhat is spectrum?\nSolution:\nWhen a white light is passed through a transparent medium, the white light splits into seven colours which are made to fall on a white screen. This coloured band is called spectrum.\nQuestion-38\nWhy is the colour of the clear sky blue?\nSolution:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nQuestion-38\nWhy is the colour of the clear sky blue?\nSolution:\nThe scattering of sunlight due to the air molecules in the atmosphere causes the sky to appear blue. Shorter wavelength (Blue colour) is scattered more compared to longer wavelength (Red colour). The scattered blue light enters our eyes.\nQuestion-39\nWhat is phorias?\nSolution:\nThis is a defect of eye. With this defect, the eye sees two images of one object. This defect is due to lack of balance among the muscles which hold the two eyes in their axes. This lack of balance produces cross-eyes. It is remedied by using Prismatic lenses.\nQuestion-40\nWrite the names of colours got from the word VIBGYOR.\nSolution:\nV \u2013 Violet\nI \u2013 Indigo\nB \u2013 Blue\nG \u2013 Green\nY \u2013 Yellow\nO \u2013 Orange\nR \u2013 Red.\nQuestion-41\nWhy is the colour of the clear sky blue?\nSolution:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXTRA QUESTIONS.txt\nWrite the names of colours got from the word VIBGYOR.\nSolution:\nV \u2013 Violet\nI \u2013 Indigo\nB \u2013 Blue\nG \u2013 Green\nY \u2013 Yellow\nO \u2013 Orange\nR \u2013 Red.\nQuestion-41\nWhy is the colour of the clear sky blue?\nSolution:\nThe scattering of sunlight due to the air molecules in the atmosphere causes the sky to appear blue. Shorter wavelength (Blue colour) is scattered more compared to longer wavelength (Red colour). The scattered blue light enters our eyes.\nQuestion-42\nWhat is optic axis?\nSolution:\nThe straight line passing through the centre of the cornea and the lens is called optic axis of the eye.\nQuestion-43\nWhat is visual axis?\nSolution:\nThe line passing through centre of the lens and fovea centralize is called visual axis of the eye. When an object is to be seen more minutely, it is brought on the visual axis to get its image on yellow spot which is the most sensitive part of the retina.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXEMPLER QUESTIONS.txt\nNCERT Exemplar Class 10 Science Chapter 10 Human Eye and Colourful World\nShort Answer Questions\nQuestion 1. A student sitting at the back of the classroom cannot read clearly the letters written on the blackboard. What advice will a doctor give to her? Draw ray diagram for the correction of this defect.\n\nAnswer. The student is suffering from short-sightedness. Doctor will advise to her to wear a concave lens having suitable power for correcting the vision.\nQuestion 2. How are we able to see nearby as well as the distant objects clearly?\nAnswer. Accommodation: The ability of the ciliary muscles to adjust the curvature and thereby the focal length to get clear view of objects is called accommodation. There is always a limit up to which ciliary muscles can increase or decrease the focal length of eye lens. This change enables us to see nearer and far-off objects clearly.\nQuestion 3. A person needs a lens of power \u2013 4.5 D for correction of her vision.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXEMPLER QUESTIONS.txt\nQuestion 3. A person needs a lens of power \u2013 4.5 D for correction of her vision.\n(a) What kind of defect in vision is she suffering from?\n\n(b) What is the focal length of the corrective lens?\n(c) What is the nature of the corrective lens?\nAnswer.\nQuestion 4. How will you use two identical prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw the diagram.\nAnswer. Consider a prism A. When white light falls on it, it splits into seven constituent colours. The violet colour deviates the most and red colour deviates the least, as shown.\n\nIf another prism B is placed such that they are as shown below, the light that emerges out of A will be made to merge together to come out as white light.\n>\nQuestion 5. Is the position of a star as seen by us its true position? Justify your answer.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXEMPLER QUESTIONS.txt\n>\nQuestion 5. Is the position of a star as seen by us its true position? Justify your answer.\nAnswer. Light from stars passing through the atmosphere, bends due to changing refractive index of different layers of the atmosphere and appears as if it comes from a higher level than they are actually. So, the stars appear slightly higher than the actual position.\nLong Answer Questions\nQuestion 6. Explain the structure and functioning of Human eye. How are we able to see nearby as well as distant objects?\nAnswer. Human Eye: The natural optical device through which one could see objects around him. It forms an inverted and real image on a light sensitive surface called the retina.\n\nParts of human eye are:\nIris: It is a dark muscular diaphragm that controls the size of the pupil.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXEMPLER QUESTIONS.txt\nParts of human eye are:\nIris: It is a dark muscular diaphragm that controls the size of the pupil.\nPupil: The black opening between the aqueous humour and the lens. Since, light does not get reflected from it, so its appearance is dark. The amount of light entering the eye is controlled by the size of the pupil. In dim light, it opens up completely through the iris, but in bright light, it becomes very small. \u2022\nCornea and Aqueous Humour: Acting as lens, they provide the refraction for light rays entering the eye. Cornea is a thin membrane covering the surface of eyeball, through which light enters. Aqueous humour is a transparent gelatinous fluid filled between cornea and eye-lens.\n Ciliary Muscles: These muscles hold the eye lens in vertical position and change the focal length of eye lens to form the sharp image of objects located at different distances on the retina.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXEMPLER QUESTIONS.txt\nCiliary Muscles: These muscles hold the eye lens in vertical position and change the focal length of eye lens to form the sharp image of objects located at different distances on the retina.\n Retina: The light sensitive surface of eye on which image is formed. It is equivalent of the photographic film in a camera. It contains rods and cones.\n Rods and Cones: The cells in retina, which are light and colour sensitive. Rods respond to the intensity of light. Cones respond to the colour. There are around 125 million rods and cones. The cells generate signals which are transmitted to the brain through optical nerves. The brain process the information via these electric signals and give the impression of erect image to us.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - EXEMPLER QUESTIONS.txt\nTo see nearby as well as distant objects, ciliary muscles modify the curvature of eye lens. This leads to variation in focal length. When the muscles are relaxed, the focal length of the lens has its maximum value, equal to the distance from the retina. So, parallel rays coming into the eye get focussed on the retina. When the eye looks at nearby objects, the ciliary muscles are strained and the focal length decreases. So, that sharp image again forms on the retina.\nQuestion 7. When do we consider a person to be myopic or hypermetropic? Explain using diagrams how the defects associated with myopic and hypermetropic eye can be corrected?\nAnswer. A person can be considered as myopic when he can see nearby objects clearly but cannot see distant objects distinctly. This defect of eye is called myopia.\n\nWhen the person cannot see nearby objects clearly but able to see distant objects clearly, the person can be treated as hypermetropic. The eye defect is called hypermetropia.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nCBSE Class 10 Science Notes Chapter 11 Human Eye and Colourful World\nHuman Eye: working of human eye, Persistence of vision, Power of accommodation of human eye, Defects of vision.\nThe Human Eye: It is a natural optical instrument which is used to see the objects by human beings. It is like a camera which has a lens and screen system.\nStructure of the Human Eye\n\nThe various parts of eye and their functions :\nRetina: It is a light sensitive screen inside the eye on which image is formed. It contains rods and cones.\nCornea: It is a thin membrane which covers the eye trail. It acts like a lens which refracts the light entering the eye.\nAqueous humour: It is fluid which fills the space between cornea and eye lens.\nEye lens: It is a convex lens made of transparent and flexible jelly like material. Its curvature can be adjusted with the help of ciliary muscles.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nEye lens: It is a convex lens made of transparent and flexible jelly like material. Its curvature can be adjusted with the help of ciliary muscles.\nPupil: It is a hole in the middle of iris through which light enters the eye. It appears black because light falling on it goes into the eye and does not come back.\nCiliary muscles: These are the muscles which are attached to eye lens and can modify the shape of eye lens which leads to the variation in focal lengths.\nIris: It controls the amount of light entering the eye by changing the size of the pupil.\nOptical nerve: These are the nerves which take the image to the brain in the form of electrical signals.\nThe human eye is roughly spherical in shape with a diameter of about 2.3 cm. It consists of a convex lens made up of living tissues. Hence, human lenses are living organs contrary to the simple optical lenses. The following table lists the main parts of the human eye and their respective functions.\nHow Pupil Works?", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nHow Pupil Works?\nFor Example, You would have observed that when you come out of the cinema hall after watching the movie in the bright sunlight, your eyes get closed. And when you entered the hall from the bright light, you won\u2019t be able to see and after some time you would be able to see. Here, the pupil of an eye provides a variable aperture, whose size is controlled by iris.\n(a) When the light is bright: Iris contracts the pupil, so that less light enters the eye.\n(b) When the light is dim: Iris expands the pupil, so that more light enters the eye.\nPupil opens completely when iris is relaxed.\nPersistence of Vision: It is the time for which the sensation of an object continue in the eye. It is about 1/16th of a second.\nPower of Accommodation: The ability of the eye lens to adjust its focal length accordingly as the distances is called power of accommodation.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nPower of Accommodation: The ability of the eye lens to adjust its focal length accordingly as the distances is called power of accommodation.\n\nColour Blindness: A person having defective cone cells is not able to distinguish between the different colours. This defect is known as Colour Blindness.\nDefects of Vision and their Correction\nMyopia (Short-sightedness): It is a kind of defect in the human eye due to which a person can see near objects clearly but he cannot see the distant objects clearly. Myopia is due to\n(i) excessive curvature of the cornea.\n(ii) elongation of eyeball.\n\nCorrection: Since a concave lens has an ability to diverge incoming rays, it is used to correct this defect of vision. The image is allowed to format the retina by using a concave lens of suitable power as shown in the given figure.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nHypermetropia (Long-sightedness): It is a kind of defect in the human eye due to which, a person can see distant objects properly but cannot see the nearby objects clearly. It happens due to\n(i) decrease in the power of eye lens i.e., increase in focal length of eye lens.\n(ii) shortening of eyeball.\n\nA hypermetropic eye has its least distance of distinct vision greater than 25 cm.\nCorrection: Since a convex lens has the ability to converge incoming rays, it can be used to correct this defect of vision, as you already have seen in the animation. The ray diagram for the corrective measure for a hypermetropic eye is shown in the given figure.\n\nPower of the correcting convex lens:\nThe Lens formula, \n1\nv\n\u2212\n1\nu\n=\n1\nf\n can be used to calculate the focal length and hence, the power of the myopia correcting lens.\nIn this case,\nObject distance, u = \u221e\nImage distance, v =\nperson\u2019s far point Focal length, f =?\nHence, lens formula becomes", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nIn this case,\nObject distance, u = \u221e\nImage distance, v =\nperson\u2019s far point Focal length, f =?\nHence, lens formula becomes\n\nIn case of a concave lens, the image is formed in front of the lens i.e., on the same side of the\nobject.\nFocal length = -Far point\nNow, power of the required lens (P) = \n1\nf(inm)\nPower of the correcting convex lens: Lens formula, \n1\nv\n\u2212\n1\nu\n=\n1\nf\n can be used to calculate focal length f and hence, power P of the correcting convex lens, where,\nObject distance, u = -25 cm, normal near point\nImage distance, v = defective near point\nHence, the lens formula is reduced to\n\n1\nv\n+\n1\n25\n=\n1\nf\nPresbyopia: It is a kind of defect in human eye which occurs due to ageing. It happens due to the following reasons\n(i) decrease in flexibility of eye lens.\n(ii) gradual weakening of ciliary muscles.", + "It happens due to the following reasons\n(i) decrease in flexibility of eye lens.\n(ii) gradual weakening of ciliary muscles.\nIn this, a person may suffer from both myopia and hypermetropia.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\n(i) decrease in flexibility of eye lens.\n(ii) gradual weakening of ciliary muscles.\nIn this, a person may suffer from both myopia and hypermetropia.\nCorrection: By using a bifocal lens with appropriate power. Bifocal lenses consist of both concave and convex lens, upper position consists of the concave lens and lower portion consists of a convex lens.\nAstigmatism: It is a kind of defect in human eye due to which a person cannot see (focus) simultaneously horizontal and vertical lines both.\nCorrection: By using a cylindrical lens.\nCataract: Due to the membrane growth over eye lens, the eye lens becomes hazy or even opaque. This leads to a decrease or loss of vision. This problem is called a cataract. It can be corrected only by surgery.\nRefraction of light through a prism, Dispersion of white light by a glass prism, Composition of white light, Recombination of spectrum colours, Rainbow.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nRefraction of light through a prism, Dispersion of white light by a glass prism, Composition of white light, Recombination of spectrum colours, Rainbow.\nRefraction of light through a prism: When a ray of light is incident on a rectangular glass slab, after refracting through the slab, it gets displaced laterally. As a result, the emergent ray comes out parallel to the incident ray.\nUnlike a rectangular slab, the side of a glass prism are inclined at an angle called the angle of prism.\nPrism: A prism has two triangular bases and three\nAngle of Prism: Angle between two lateral faces is\nAngle of Deviation: The angle between the incident deviation.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nPrism: A prism has two triangular bases and three\nAngle of Prism: Angle between two lateral faces is\nAngle of Deviation: The angle between the incident deviation.\nDispersion of white light by a glass prism: The phenomenon of splitting of white light into its seven constituent colours when it passes through a glass prism is called dispersion of white light. The various colours seen are Violet, Indigo, Blue, Green, Yellow, Orange and Red. The sequence of colours remembers as VIBGYOR. The band of seven colours is called the spectrum. The different component colour of light bends at a different angle with respect to the incident angle. The violet light bends the least while the red bends most.\n\nComposition of white light: White light consists of seven colours i.e., violet, indigo, blue, green, yellow, orange and red.\nMonochromatic light: Light consisting of single colour or wavelength is called monochromatic light, example; sodium light.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nMonochromatic light: Light consisting of single colour or wavelength is called monochromatic light, example; sodium light.\nPolychromatic light: Light consisting of more than two colours or wavelengths is called polychromatic light, example; white light.\nRecombination of white light: Newton found that when an inverted prism is placed in the path of dispersed light then after passing through the prism, they recombine to form white light.\nIssac Newton: He was the first, who obtained spectrum of sunlight by using glass prism. He tried to split the spectrum of white light more by using another similar prism, but he could not get any more colours.\nHe repeated the experiment using second prism in inverted position with respect to the first prism. It allowed all the colours of spectrum to pass through second prism. He found white light emerges on the other side of second prism.\n\nHe concluded that Sun is made up of seven visible colour VIBGYOR.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nHe concluded that Sun is made up of seven visible colour VIBGYOR.\nRainbow: It is the spectrum of sunlight in nature. It is formed due to the dispersion of sunlight by the tiny water droplet, present in the atmosphere.\nFormation of the rainbow: The water droplets act like small prism. They refract and disperse the incident sunlight, then reflect it internally, and finally refract it again when it comes out of the raindrop. Due to the dispersion of light and internal reflection, different colours reach the observer\u2019s eye.\nRed colour appears on top and violet at the bottom of rainbow.\nA rainbow is always formed in a direction opposite to that of Sun.\nAt \u2018A\u2019 \u2013 Refraction and dispersion take place.\nAt \u2018B\u2019 \u2013 Internal reflection takes place.\nAt \u2018C\u2019 \u2013 Refraction and dispersion take place.\n\nAtmospheric Refraction: The refraction of light caused by the Earth\u2019s atmosphere (having air layers of varying optical densities) is called Atmospheric Refraction.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nAtmospheric Refraction: The refraction of light caused by the Earth\u2019s atmosphere (having air layers of varying optical densities) is called Atmospheric Refraction.\nAppearance of Star Position: It is due to atmospheric refraction of star light.\nThe temperature and density of different layer of atmosphere keeps varying. Hence, we have different medium.\nDistant star act as point source of light. When the starlight enter the Earth\u2019s atmosphere, it undergoes refraction continuously, due to changing refractive index i.e. from Rarer to denser. It bends towards the normal.\nDue to this, the apparent position of the star is different from actual position. The star appear higher than its actual position.\n\nTwinkling of Star: It is also due to atmospheric refraction.\nDistant star act like a point source of light. As the beam of starlight keeps deviating from its path, the apparent position of star keeps on changing because physical condition of earth\u2019s atmosphere is not stationary.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nHence, the amount of light enters our eyes fluctuate sometimes bright and sometime dim. This is the \u201cTwinkling effect of star\u201d.\n\nWhy planets do not twinkle ?\nPlanets are closer to earth and are seen as extended source of light i.e. the collection of large number of point sized sources of light. Therefore the total amount of light entering our eyes from all individual point source will nullify the twinkling effect.\nWhy, the duration of day becomes approximately 4 minutes shorter if there is no atmosphere on earth: Actual sun rise happens when it is below the horizon in the morning. The rays of light from the sun below the horizon reach our eyes because of refraction of light. Similarly, the sun can be seen about few minutes after the actual sun set. Thus the duration of, day time will increase by 4 minutes.\nThis is due to atmospheric refraction. Because of this sun is visible about 2 minutes earlier than actual sunrise and about 2 minutes after the actual sun set.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nThis is due to atmospheric refraction. Because of this sun is visible about 2 minutes earlier than actual sunrise and about 2 minutes after the actual sun set.\nApparent flattering of the Sun\u2019s disc at sunset and sunrise is due to atmospheric refraction.\n\nScattering of light: According to Rayleigh\u2019 Law of Scattering, the amount of scattered light \u221d \n1\n\u03bb\n4\n (\u03bb = wavelength)\nScattering of light decreases with increase in wavelength.\nTyndall Effect: When a beam of light strikes, the minute particle of earth\u2019s atmosphere, suspended particles of dust and molecule of air the path of beam become visible. The phenomenon of scattering of light by the colloidal particle gives rise to Tyndall Effect.\nIt can be observed when sunlight passes through a canopy of a dense forest.\nThe colour of the scattered light depends on the size of the scattering particles.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nIt can be observed when sunlight passes through a canopy of a dense forest.\nThe colour of the scattered light depends on the size of the scattering particles.\n\nColour of Sunrise and Sunset: While sunset and sunrise, the colour of the sun and its surroundihg appear red. During sunset and sunrise, the sun is near to horizon, and therefore, the sunlight has to travel larger distance in atmosphere. Due to this, most of the blue light (shorter wavelength) is scattered away by the particles. The light of longer wavelength (red colour) reaches our eye. This is why sun appear red in colour.\nWhy the danger signal or sign is made of red colour?\nRed colour scatteres the most when strikes the small particle of fog and smoke because it has the maximum wavelength (visible spectrum). Hence, from large distance also, we can see the red colour clearly.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nAt noon sun appears white: At noon, the sun is overhead and sunlight would travel shorter distance relatively through the atmosphere. Hence, at noon, the sun appear white as only little of the blue and violet colours are scattered.\n\nHuman Eye: It is a wonderful gift of nature to the human body. Human eye is nearly spherical in shape of diameter about 2.5 cm.\nParts of Human Eye:\nCornea: It is the protective and front layer of the eye. It is made by a transparent membrane. Light enters the eye through the cornea.\nIris: Dark and a colourful muscular diaphragm is called iris. It is responsible for colour of the eye.\nPupil: Small circular hole in the centre of iris. It regulates the amount of light entering the eye by adjusting the size of the iris.\nCiliary Muscles: It holds the eye lens at its proper position. It changes the size of eye lens.\nEye lens: The eye lens is a convex lens made by the transparent jelly like material.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nCiliary Muscles: It holds the eye lens at its proper position. It changes the size of eye lens.\nEye lens: The eye lens is a convex lens made by the transparent jelly like material.\nRetina: It is the screen of the eye. A real and inverted image form on the retina.\nRods and Cones: These are colour sensitive rods and cones shaped cells. Rods are responsible for the vision in dim light while cones are responsible for colour.\nOptic Nerve: It converts information of the image into a corresponding electric signal and passes it to the brain.\nBlind Spot: The junction of the optic nerve and retina, where no rods and cones cells are present is called the blind spot. It is insensitive to light.\nNear Point: The nearest point from eye at which the eye can see clearly without strain is called near point. For normal eye it is 25 cm.\nFar Point: The farthest point, upto which the eye can see the object clearly is called far point. For normal eye it is infinity.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nFar Point: The farthest point, upto which the eye can see the object clearly is called far point. For normal eye it is infinity.\nThe range of Vision: Distance between near point and far point of eye is called range of vision.\nPower of Accommodation: The ability of the eye to see near as well as far objects clearly is called Power of Accommodation.\nMyopia (Nearsightedness): In this defect, the eye is unable to see far off objects clearly but is able to see near objects clearly.\nReason.\nDue to increase in size of eye ball\nDue to excessive curvature of cornea\nDue to increase in power (or decrease in focal length) of eye lens.\nCorrection: It is corrected by using concave lens of suitable focal length.\nHypermetropia (Far sightedness): In this defect eye is unable to see nearby objects clearly but is able to see far objects clearly.\nReason.\nDue to decrease in size of eye ball\nDue to decrease in power (or increase in focal length) of eye lens.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nReason.\nDue to decrease in size of eye ball\nDue to decrease in power (or increase in focal length) of eye lens.\nCorrection: It is corrected by using a convex lens of suitable focal length.\nDispersion of Light: Splitting of white light into seven colours is called dispersion. Example, the formation of Rainbow (VIBGYOR). Violet deviates the most but red deviates least.\n1. The human eye is one of the most valuable and sensitive sense organs. It enables us to see the wonderful world and the colours around us.\n2. The eyeball is approximately spherical in shape with a diameter of about 2.3 cm.\n3. Most of the refraction for the light rays entering the eye occurs at the outer surface of the cornea. The crystalline lens merely provides the finer adjustment of focal length required to focus.\n4. The human eye has the following parts :\nCornea : The transparent spherical membrane covering the front of the eye.\nIris: The coloured diaphragm between the cornea and lens.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\n4. The human eye has the following parts :\nCornea : The transparent spherical membrane covering the front of the eye.\nIris: The coloured diaphragm between the cornea and lens.\nPupil: The small hole in the iris.\nEye lens : It is a transparent lens made of jelly like material.\nCiliary muscles: These muscles hold the lens in position.\nRetina: The back surface of the eye.\nBlind spot: The point at which the optic nerve leaves the eye. An image formed at this point is not sent to the brain.\nAqueous humour: A clear liquid region between the cornea and the lens.\nVitreous humour: The space between eye lens and retina is filled with another liquid called vitreous humour.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nAqueous humour: A clear liquid region between the cornea and the lens.\nVitreous humour: The space between eye lens and retina is filled with another liquid called vitreous humour.\n5. In the eye, the image is formed on the retina by successive refractions at the cornea, the aqueous humour, the lens and the vitreous humour. Electrical signals then travel along the optic nerve to the brain to be interpreted. In good light, the yellow spot is most sensitive to detail and the image is automatically formed there.\n6. Accommodation: The ability of the eye to focus both near and distant objects, by adjusting its focal length, is called the accommodation of the eye or the ability of the ciliary muscles to change the focal length of the eye lens is called accommodation.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\n7. Defects of the Eye : Although the eye is one of the most remarkable organs in the body, it may have several abnormalities, which can often be corrected with eyeglasses, contact lenses, or surgery. The various defects from which an eye can suffer are (i) Hypermetropia or long sightedness, (ii) Myopia or shortsightedness and (iii) Astigmatism, (iv) Presbyopia.\n8. Hypermetropia, hyperopia, or long sightedness : A person suffering from this defect can see distant objects I clearly but cannot see nearby objects clearly. In this defect, the near point lies farther away than 25 cm. Hypermetropia (far sightedness \u2014 the image of nearby objects is focussed beyond the retina) is corrected by using a convex lens of suitable power. The eye loses\nits power of accommodation at old age.\n9. Hypermetropia is due to the following reasons :\nEither the hyperopic eyeball is too short or\nThe ciliary muscle is unable to change the shape", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nits power of accommodation at old age.\n9. Hypermetropia is due to the following reasons :\nEither the hyperopic eyeball is too short or\nThe ciliary muscle is unable to change the shape\nof the lens enough to properly focus the image i.e. the focal length of the eye lens increases.\n10. Myopia or short sightedness or near sightedness: A person suffering frommyopia or short sightedness can see nearby objects clearly but cannot see the far away objects clearly. Myopia (short sightedness \u2014 the image of distant objects is focussed before the retina) is corrected by using a concave lens of suitable power.\n11. This defect is due to the following reasons :\nEither the eyeball is longer than normal or\nThe maximum focal length (due to excessive curvature of the cornea) of the lens is insufficient to produce a clearly formed image on the retina.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\nEither the eyeball is longer than normal or\nThe maximum focal length (due to excessive curvature of the cornea) of the lens is insufficient to produce a clearly formed image on the retina.\n12. A person may also have an eye defect known as astigmatism, in which light from a point-source produces a line image on the retina. A person suffering from this defect cannot see in all directions equally well i.e., he cannot see the vertical and horizontal lines simultaneously. This condition arises either when the cornea or the crystalline lens or both are not perfectly spherical. Astigmatism can be corrected with lenses having different curvatures in two mutually perpendicular directions i.e., cylindrical lens.\n13. When a person suffers from both, the myopia as well as Hypermetropia, his spectacles for correction have bifocal lenses. The upper half is a concave lens for distant vision and lower half is a convex lens for reading.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\n14. Presbyopia is that defect of human eye, due to which an old person cannot read and write comfortably. That is why Presbyopia is also called old sight.\n15. To correct Presbyopia, an old person has to use spectacles with a convex lens of suitable focal length, or power as explained already.\n16. The cause of Hypermetropia is decrease in length of eyeball or increase In focal length of eye lens. But the cause of Presbyopia is only increase in focal length of eye lens. The eyeball, in Presbyopia, has normal length.\nthe vision of the eye decreases, leading sometimes to total loss of vision. The problem is overcome by cataract surgery i.e., removal of the eye lens, and its replacement by a lens of suitable focal length.\n18. We need two eyes because a human being has a horizontal field of view of about 150\u00b0 with one eye and of about 180\u00b0 with two eyes. Thus, two eyes provide us wider horizontal field of view.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\n18. We need two eyes because a human being has a horizontal field of view of about 150\u00b0 with one eye and of about 180\u00b0 with two eyes. Thus, two eyes provide us wider horizontal field of view.\nWith one eye, the world looks flat, i.e., two dimensional only. With two eyes, the view is three dimensional, i.e., dimension of depth is added to our view.\n19. As our two eyes are separated by a few centimetres, each eye observes a slightly different image. Our brain combines the two views into one and we get to know how close or far away the things seen are.\n20. By donating our eyes after we die, one pair of our eyes can give vision to two corneal blind people. Eye donors may belong to any sex or any age group. People suffering from diabetes, hypertension, asthma or any other non- communicable diseases can donate eyes. People who have been using spectacles or those operated for cataract can also donate eyes.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\n21. The smallest distance, at which the eye can see objects clearly without strain, is called the near point of the eye or the least distance of distinct vision. For a young adult with normal vision, it is about 25 cm.\n22. Persistence of vision of the eye: The image of an object persists on the retina for 1/16 second, even after the removal of the object. The sequence of still pictures taken by a movie camera is projected on a screen at a rate of about 24 images or more per second. The successive impressions of images on the screen appear to merge smoothly into one another to give us the feeling of moving images.\n23. The large numbers of light sensitive cells contained in the retina of the eye are of two types: rod shaped cells which respond to brightness or intensity of light and cone shaped cells, which respond to colour of light. Thus/cone shaped cells enable us to distinguish between different colours.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\n24. When a person cannot distinguish between different colours, he is said to be colour blind though his vision may otherwise be normal. Colour blindness is a genetic disorder which occurs by inheritance. So far, there is no cure for colour blindness.\n25. Farpoint: The farthest point upto which a short sighted eye can see clearly is called the far point of the eye. For a normal eye, the far point is infinity.\n26. Near point : The nearest point upto which a long sighted eye can see clearly is called the near point of the eye. For a normal human eye, of an adult, the near point is about 25 cm from the eye.\n27. Least cfistance of distinct vision: The minimum distance upto which an eye can see clearly is called the legist distance of distinct vision ; it is normally denoted by D. The least distance of distinct vision is equal to the distance between the eye and its near point. For a normal human eye, this distance is around 25 cm.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\n28. The distance between far point and near point of the eye is called range of vision of the eye.\n29. When white light passes through a prism, the violet light bends most and the red light bends the least. Dispersion of light is the phenomenon of splitting of white light into its constituent seven colours on passing through a glass prism. The band of seven colours so obtained is called visible spectrum.\n30. The seven colours of white light are violet, indigo, blue, green, yellow, orange and red. It is remembered by the acronym VIBGYOR.\n31. Isaac Newton was the first to use a prism to obtain a spectrum of sunlight.\n32. Spectrum is the band of distinct colours we obtain when white light is split by a prism.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\n31. Isaac Newton was the first to use a prism to obtain a spectrum of sunlight.\n32. Spectrum is the band of distinct colours we obtain when white light is split by a prism.\n33. Cause of dispersion : Every colour has its own characteristic wavelength/frequency. Different colours move with same speed in air/vacuum. But their speeds in refracting media like glass are different. Therefore, refractive index of the medium for different colours is different. As a result, different colours undergo different deviations on passing through the prism. Hence, different colours emerge from the prism along different directions.\n34. The speed of light in vacuum is same for all wavelengths, but the speed in a material substance is different for different wavelengths.\n35. In any medium other than air/vacuum red light travels the fastest and violet light travels the slowest.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\n35. In any medium other than air/vacuum red light travels the fastest and violet light travels the slowest.\n36. The most familiar form of electromagnetic radiation may be defined as that part of the spectrum that the human eye can detect. Light is produced by the rearrangement of electrons in atoms and molecules. The various wavelengths of visible light are classified with colours ranging from violet (\u03bb = 4 x 10-7 m) to red (\u03bb = 7 x 10-7 m). The eye\u2019s sensitivity is a function of wavelength, the sensitivity being a maximum at a wavelength of about \u03bb = 5.6 x 10-7 m (yellow-green).\n37. When we pass white light through two ideptical prisms held side by side with their refracting edges in opposite directions; the first prism disperses white light into seven colours and the second prism recombines the seven colours into white light. Thus, light emerging from 2nd prism is white.\n38. A rainbow is formed due to dispersion of light by tiny droplets of water which act as prisms.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - NOTES.txt\n38. A rainbow is formed due to dispersion of light by tiny droplets of water which act as prisms.\n39. Atmospheric refraction is the cause of twinkling of stars, advance sunrise and delayed sunset.\n40. Scattering of light causes the blue colour of sky and the reddening of the Sun at sunrise and sunset.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nImportant Questions of Light Reflection and Refraction Class 10 Science Chapter 11\nQuestion 1.\nState one function of iris in human eye. (AI 2012)\nAnswer:\nIrish adark muscular diaphragm that controls the size of the pupil.\nQuestion 2.\nState one function of the crystalline lens in the human eye. (Foreign 2012)\nAnswer:\nThe crystalline lens of human eye focuses the light that enters the eye and form the image on the retina.\nQuestion 3.\nDefine the term power of accommodation. Write the modification in the curvature of the eye lens which enables us to see the nearby objects clearly? (Delhi 2019)\nAnswer:\nThe ability of the eye lens to adjust its focal length is called power of accommodation. The ciliary muscles modifies the curvature to some extent. The change in the curvature of the eye lens can thus change its focal length. When the ciliary muscles contract, the lens becomes thick and its focal length decreases, thus enables us to see nearby objects clearly.\nQuestion 4.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 4.\nTrace the sequence of events which occur when a bright light is focused on your eyes. (Delhi 2019)\nAnswer:\nWhen a bright light enters the eye then most of the refraction for the light rays entering the eye occurs at the outer surface of the cornea. Then, the crystalline lens merely provides the finer adjustment of focal length required to focus object at different distances on the retina. The pupil regulates and controls the amount of light entering the eye. At retina, the light-sensitive cells get activated upon illumination and generate electric signals. These signals are sent to the brain via the optic nerves. The brain interprets these signals and finally, processes the information so that we perceive objects as they are.\nQuestion 5.\nWrite about power of accommodation of human eye. Explain why the image distance in the eye does not change when we change the distance of an object from the eye? (Delhi 2017)\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 5.\nWrite about power of accommodation of human eye. Explain why the image distance in the eye does not change when we change the distance of an object from the eye? (Delhi 2017)\nAnswer:\nThe ability of the eye lens to adjust its focal length is called power of accommodation.\nThe ciliary musqles modifies the curvature to some extent. The change in the curvature of the eye lens can thus change its focal length. Thus, the focal length of the human lens increases or decreases depending on the distance of the object value to this distance of the image does not change. For example, when the ciliary muscles are relaxed, the lens becomes thin and its focal length increases, thus enables us to see distant object clearly.\nQuestion 6.\nState the function of each of the following parts of human eye:\n(i) Cornea\n(ii) Iris\n(iii) Pupil (1.5/3, Delhi 2013 C)\nAnswer:\n(i) Cornea : It is a transparenl bulge on the front surface of eyeball which refracls most ol lhe light rays entering the eye.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(i) Cornea\n(ii) Iris\n(iii) Pupil (1.5/3, Delhi 2013 C)\nAnswer:\n(i) Cornea : It is a transparenl bulge on the front surface of eyeball which refracls most ol lhe light rays entering the eye.\n(ii) Iris : Refer to answer 1.\n(iii) Pupil: It controls the amount of light entering into the eye.\nQuestion 7.\nWrite the function of each of the following parts of human eye:\n(i) Cornea (ii) Iris (iii) Crystalline lens (iv) Ciliary muscles (2/5, 2018, Delhi 2016)\nAnswer:\n(i) Cornea : Refer to answer 6(i).\n(ii) Iris : Refer to answer 1.\n(iii) Crystalline lens : Refer to answer 2.\n(iv) Ciliary muscles: Ciliary muscles holds the eye lens and helps in the adjustment of its focal length.\nQuestion 8.", + "(ii) Iris : Refer to answer 1.\n(iii) Crystalline lens : Refer to answer 2.\n(iv) Ciliary muscles: Ciliary muscles holds the eye lens and helps in the adjustment of its focal length.\nQuestion 8.\nState the function of each of the following parts of the human eye :\n(i) Cornea (ii) Iris (iii) Pupil (iv) Retina (2/5, Foreign 2015)\nAnswer:\n(i) Cornea : Refer to answer 6(i).\n(ii) Iris : Refer to answer 1.\n(iii) Pupil: Refer to answer 6(iii).", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(i) Cornea (ii) Iris (iii) Pupil (iv) Retina (2/5, Foreign 2015)\nAnswer:\n(i) Cornea : Refer to answer 6(i).\n(ii) Iris : Refer to answer 1.\n(iii) Pupil: Refer to answer 6(iii).\n(iv) Retina: It capture light and convert it into electric signals that are translated into images by the brain.\nQuestion 9.\n(a) List the parts of the human eye that control the amount of light entering into it. Explain how they perform this function?\n(b) Write the function of retina in human eye. (3/5, AI2014)\nAnswer:\n(a) The part of the human eye that controls the amount of light entering into it is pupil.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(b) Write the function of retina in human eye. (3/5, AI2014)\nAnswer:\n(a) The part of the human eye that controls the amount of light entering into it is pupil.\nLight enters the eye through a thin membrane called the cornea. It forms the transparent bulge on the front surface of the eyeball most of the refraction for the light rays entering the eye occurs at the outer surface of the cornea, the crystalline lens merely provides the linear adjustment of focal length required to focus objects at different distances on the retina. Iris which is behind the cornea controls the size of the pupil. The pupil regulates and controls the amount of light entering the eye.\n(b) Refer to answer 8(iv).\nQuestion 10.\nPerson suffering from cataract has\n(a) elongated eyeball\n(b) excessive curvature of eye lens\n(c) weakened ciliary muscles\n(d) opaque eye lens\nAnswer:\n(d) A person suffering from cataract has cloudy opaque eye lens.\nQuestion 11.\n(a) List two causes of hypermetropia.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(c) weakened ciliary muscles\n(d) opaque eye lens\nAnswer:\n(d) A person suffering from cataract has cloudy opaque eye lens.\nQuestion 11.\n(a) List two causes of hypermetropia.\n(b) Draw ray diagrams showing (i) a hypermetropic eye and (ii) its correction using suitable optical device. (2020)\nAnswer:\n(a) Hypermetropia is caused due to following reasons:\n(i) Shortening of the eyeball\n(ii) Focal length of crystalline lens is too long.\n\nQuestion 12.\n(a) A person is suffering from both myopia and hypermetropia.\n(i) What kind of lenses can correct this defect?\n(ii) How are these lenses prepared?\n(b) A person needs a lens of power +3 D for correcting his near vision and -3 D for correcting his distant vision. Calculate the focal lengths of the lenses required to correct these defects. (2020)\nAnswer:\n(a) (i) The lens which can correct the vision of such a person suffering from both myopia and hypermetropia is a bifocal lens.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nAnswer:\n(a) (i) The lens which can correct the vision of such a person suffering from both myopia and hypermetropia is a bifocal lens.\n(ii) A common type of bifocal lens contains both concave and convex lens. It is prepared with the upper portion consisting of a concave lens facilitating distant vision and the lower portion consisting of convex lens facilitating near vision, (b) The power for correcting his near vision,\nPN = +3 D.\nAs P = \n1\nf(m)\n\n\u2234 Focal length of convex lens needed,\nfN = \n1\nP\nN\n = 0.33 m = +33.33 cm\nPower required to correct distant vision, PD = -3D\n\u2234 Focal length of concave lens,\nfD = \n1\nP\nD\n = -0.33 m = -33.33 cm.\nQuestion 13.\nA person may suffer from both myopia and hypermetropia defects.\n(a) What is this condition called?\n(b) When does it happen?", + "Question 13.\nA person may suffer from both myopia and hypermetropia defects.\n(a) What is this condition called?\n(b) When does it happen?\n(c) Name the type of lens often required by the persons suffering from this defect. Draw labelled diagram of such lenses. (2020)\nAnswer:\n(a) This condition is called presbyopia.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(c) Name the type of lens often required by the persons suffering from this defect. Draw labelled diagram of such lenses. (2020)\nAnswer:\n(a) This condition is called presbyopia.\n(b) It happens due to gradual weakening of ciliary muscles and diminishing flexibility of eye lens due to agening.\n(c) It can be corrected by using bifocal lenses.\n\nQuestion 14.\nWhat eye defect is myopia? Describe with a neat diagram how this defect of vision can be corrected by using a suitable lens. (AI 2011)\nAnswer:\nMyopia is also known as near-sightedness. A person with myopia can see nearby objects clearly but cannot see distant objects distinctly.\nMyopia can he corrected by using concave lens of appropriate local length.\n\nQuestion 15.\nName the three common defects of vision. What are their causes? Name the type of lens used to correct each of them. (Foreign 2011)\nAnswer:\nThree common defects of vision are\nMyopia\nHypermetropia\nPresbyopia\nMyopia can be caused due to following reasons.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nAnswer:\nThree common defects of vision are\nMyopia\nHypermetropia\nPresbyopia\nMyopia can be caused due to following reasons.\nElongation of eyeball.\nExcessive curvature of eye lens.\nHypermetropia can be caused due to following reasons.\nShortening of eyeball.\nFocal length of eye lens becomes too long.\nPresbyopia is caused due to gradual weakening of ciliary muscles and diminishing flexibility of eye lens due to ageing.\nCorrection of these defects:\nMyopia can be corrected by using concave lens of appropriate focal length.\nHypermetropia can he corrected by using convex lens of appropriate local length.\nPresbyopia can be corrected by using bifocal lens.\nQuestion 16.\nA student is unable to see clearly the words written on the black board placed at a distance of approximately 3 m from him. Name the defect of vision the boy is suffering from. State the possible causes of this defect and explain the method of correcting it. (3/5, 2018)\nAnswer:\nStudent is suffering from myopia.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nAnswer:\nStudent is suffering from myopia.\nThe two possible reasons due to which the defect of vision arises are : excessive curvature of the eye lens and elongation of the eye ball.\nA student with myopia has the far point nearer than infinity, thus, the image of a distant object is formed in front of the retina.\n\nCorrection of myopia: This defect can be corrected by using a concave lens of suitable power as it brings the image back on to the retina, thus the defect is corrected.\n\nQuestion 17.\nA student suffering from myopia is not able to see distinctly the objects placed beyond 5 m.\n(a) List two possible reasons due to which this defect of vision may have arisen. With the help of ray diagrams, explain\n(i) Why the student is unable to see distinctly the objects placed beyond 5 m from his eyes?\n(ii) The type of the corrective lens used to restore proper vision and how this defect is corrected by the use of this lens.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(ii) The type of the corrective lens used to restore proper vision and how this defect is corrected by the use of this lens.\n(b) If, in this case, the numerical value of the focal length of the corrective lens is 5 m, find the power of the lens as per the new Cartesian sign convention. (AI 2017)\nAnswer:\n(a) Refer to answer 16.\n(b) Focal length, f = -5 m\nP = \n1\nf(inmeters)\n or, p = \n1\n\u22125\n = -0.2D\nHence, the power is \u2013 0.2 D.\nQuestion 18.\nMillions of people of the developing countries of world are suffering from corneal blindness. These persons can be cured by replacing the defective cornea with the cornea of a donated eye. A charitable society of your city has organised a campaign in your neighbourhood in order to create awareness about this fact. If you are asked to participate in this mission how would you contribute in this noble cause?\n(a) State the objective of organising such campaigns.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(a) State the objective of organising such campaigns.\n(b) List two arguments which you would give to motivate the people to donate their eyes after death.\n(c) List two values which are developed in the persons who actively participate and contribute in such programmes. (VBQ, 3/5, Delhi 2016)\nAnswer:\nWe can encourage people to participate in the camp and also register ourselves as a donator.\n(a) The objective of organising such campaign is to make people aware and realize their duties towards society.\n(b) (i) By donating our eyes after we die, we can light the life of a blind person.\n(ii) One pair of eyes gives vision to two corneal blind people.\n(c) (i) It shows the concern for others.\n(ii) It also shows the responsible behavior towards the society.\nQuestion 19.\nA student is unable to see clearly the words written on the blackboard placed at a distance of approximately 4 m from him. Name the defect of vision the boy is suffering from.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 19.\nA student is unable to see clearly the words written on the blackboard placed at a distance of approximately 4 m from him. Name the defect of vision the boy is suffering from.\nExplain the method of correcting this defect. Draw ray diagram for the\n(i) defect of vision and also\n(ii) for its correction (Delhi 2015)\nAnswer:\nRefer to answer 16.\nQuestion 20.\nWrite the importance of ciliary muscles in the human eye. Name the defect of vision that arises due do gradual weakening of the ciliary muscles in old age. What type of lenses are required by the persons suffering from this defect to see the objects clearly?", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nAkshay, sitting in the last row in his class, could not see clearly the words written on the blackboard. When the teacher noticed it, he announced if any student sitting in the front row could volunteer to exchange his seat with Akshay. Salman immediately agreed to exchange his seat with Akshay. He could now see the words written on the blackboard clearly. The teacher thought it fit to send the message to Akshay s parents advising them to get his eyesight checked.\nIn the context of the above event, answer the following questions:\n(a) Which defect of vision is Akshay suffering from? Which type of lens is used to correct this defect?\n(b) State the values displayed by the teacher and Salman.\n(c) In your opinion, in what way can Akshay\nexpress his gratitude towards the teacher and Salman? (VBQ, AI 2015)\nAnswer:\nCiliary muscles modifies the curvature of eye lens and hence adjusts its focal length. I his enables us to see objects.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nexpress his gratitude towards the teacher and Salman? (VBQ, AI 2015)\nAnswer:\nCiliary muscles modifies the curvature of eye lens and hence adjusts its focal length. I his enables us to see objects.\nThe defect of vision arises due to weakening of ciliary muscles in old age is presbyopia, person suffering from this defect should wear bifocal lenses. These lenses consists of both concave and convex lenses.\n(a) Akshay is suffering from myopia or near-sightedness. He should use concave lens to correct this defect.\n(b) Teacher and Salman are concerned and caring.\n(c) Akshay can show his gratitude by saying thank you.\nQuestion 21.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(b) Teacher and Salman are concerned and caring.\n(c) Akshay can show his gratitude by saying thank you.\nQuestion 21.\nMillions of people of the developing countries are suffering from corneal blindness. This disease can be cured by replacing the defective cornea with the cornea of a donated eye. Your school has organised a campaign in the school and its neighbourhood in order to create awareness about this fact and motivate people to donate their eyes after death. How can you along with your classmates contribute in this noble cause? State the objectives of organising such campaigns in schools. (VBQ, 3/5, Foreign 2015)\nAnswer:\nRefer to answer 18.\nQuestion 22.\nDo you know that the corneal-impairment can be cured by replacing the defective cornea with the cornea of the donated eye?\nHow and why should we organise groups to motivate the community members to donate their eyes after death? (2/5, AI 2014)\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nHow and why should we organise groups to motivate the community members to donate their eyes after death? (2/5, AI 2014)\nAnswer:\nYes, we know that the corneal \u2013 impairment can be cured by replacing the objective cornea with the cornea of the donated eye. We can provide the importance of eye donation to the community members. Our eyes can live even after our death. By donating our eyes after die, we can light the life of a blind person. The human eye is one of the most valuable and sensitive sense organs. It enables us to see the wonderful world and colours around us. It is however, impossible to identify colours while closing the eyes. Thus of all the sense organs, the human eye is the most significant one as it enables as to see the beautiful colourful word around us. Hence, we should donate our eyes after death.\nQuestion 23.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 23.\nWhat is myopia? List two causes for the development of this defect? How can this defect be corrected using a lens? Draw ray diagrams to show the image formation in case (i) defective eye and (ii) corrected eye. (Foreign 2014)\nAnswer:\nMyopia is also known as near-sightedness defect in which a person can see nearby objects clearly but cannot see distant objects distinctly. This defect may arise due to\n(a) excessive curvature of the eye.\n(b) elongation of the eye ball.\nThis defect can be corrected by using a concave lens of suitable power.\nRefer to answer 16.\nQuestion 24.\n(a) A person cannot read newspaper placed nearer than 50 cm from his eyes. Name the defect of vision he is suffering from. Draw a ray diagram to illustrate this defect. List its two possible causes. Draw a ray diagram to show how this defect may be corrected using a lens of appropriate focal length.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(b) We see advertisements for eye donation on television or in newspapers. Write the importance of such advertisements. (Delhi 2013)\nAnswer:\n(a) The person is suffering from hyperme-tropia.\nHypermetropia : It is a defect in an eye in which a person is not able to see nearby object distinctly but can see far objects clearly.\nRefer to answer 11.\n(b) It is important to advertise for eye donation on television or in newspaper because\n(i) Few people are unaware about the fact that there can be an eye transplant through which blind people can see this colourful and beautiful world.\n(ii) To encourage them to donate their eye by spreading awareness about it through television or newspaper.\nQuestion 25.\nA student cannot see a chart hanging on a wall placed at a distance of 3 m from him. Name the defect of vision he is suffering from. How can it be corrected? Draw ray diagrams for the (i) defect of vision and also\n(ii) for its correction. (Delhi 2012)\nAnswer:\nRefer to answer 16.\nQuestion 26.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(ii) for its correction. (Delhi 2012)\nAnswer:\nRefer to answer 16.\nQuestion 26.\nAn old man cannot see objects closer than 1 m from the eye clearly. Name the defect of vision he is suffering from. How can it be corrected? Draw ray diagram for the (i) defect of vision and also (ii) for its correction. (AI 2012)\nAnswer:\nHe is suffering from hypermetropia. Refer to answer 24 (a).\nQuestion 27.\nDraw a diagram to show why distant objects cannot be seen distinctly by a myopic eye. List two reasons due to which this defect of vision may be caused.\nA person with a myopic eye cannot see objects clearly beyond a distance of 2 m. Name the type of the corrective lens that would be needed to correct the defect of vision and draw a ray diagram to show how the defect gets corrected. (Foreign 2012)\nAnswer:\nRefer to answer 16.\nQuestion 28.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 16.\nQuestion 28.\nDraw a ray diagram to show the refraction of light through a glass prism. Mark on it (a) the incident ray, (b) the emergent ray and (c) the angle of deviation. (AI 2011)\nAnswer:\n\ni = angle of incidence\n(a) PE = incident ray\n(b) FS = emergent ray\n(c) \u2220D = angle of deviation\nQuestion 29.\nDraw a ray diagram to explain the term angle of deviation. (1/5, Delhi 2017)\nAnswer:\nThe emergent ray bends at an angle to the direction of the\u2019incident, thus the angle between them is known as angle of deviation (D).\n\nQuestion 30.\nDraw a labelled diagram to explain the formation of a rainbow in the sky. (Foreign 2015)\nAnswer:\nA rainbow is a natural spectrum caused by dispersion of sunlight by tiny water droplets, present in the atmosphere.\n\nPoint A denotes dispersion and point B denotes internal reflection.\nQuestion 31.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nA rainbow is a natural spectrum caused by dispersion of sunlight by tiny water droplets, present in the atmosphere.\n\nPoint A denotes dispersion and point B denotes internal reflection.\nQuestion 31.\nHow will you use two identical glass prisms so that a narrow beam of white light incident on one prism emerges out of the second prism as white light? Draw and label the ray diagram. (2020)\nAnswer:\nNewton was the first to use a glass prism to obtain the spectrum of a white light. He then placed a second identical prism in an inverted position with respect to the first prism. This allowed all the colours of the white light to pass through the second prism combining to form a white light emerging from the other side of the second prism. This made him believe that white light was composed of different colours.\n\nQuestion 32.\nDifferentiate between a glass slab and a glass prism. What happens when a narrow beam of", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 32.\nDifferentiate between a glass slab and a glass prism. What happens when a narrow beam of\n(i) a monochromatic light and (ii) white light passes through (a) glass slab and (b) glass prism? (2020)\nAnswer:\nGlass slab:\nIt is a substance made of glass having three dimension and has cuboidal structure.\nIt does not deviate the path of light falling on it but produces a lateral displacement of the light ray after refraction. The incident and emergent ray are parallel to each other.\nGlass prism:\nA prism is a structure made of glass with two triangle bases and three rectangular lateral surfaces. These surfaces are inclined to each other.\nA prism deviates the path of light ray falling on it. Here the incident ray and emergent ray are not parallel to each other.\n(i) When a narrow beam of monochromatic light falls on a\n(a) glass slab, it gets refracted at its surface and the emergent ray is laterally displaced from the incident ray.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(i) When a narrow beam of monochromatic light falls on a\n(a) glass slab, it gets refracted at its surface and the emergent ray is laterally displaced from the incident ray.\n(b) prism, it gets refracted at the surface and the light gets deviated from its initial path. The angle between the incident ray and emergent ray is known as angle of deviation.\n(ii) When a white light passes through a\n(a) glass slab, the light does not undergo dispersion as its two refracting surfaces are parallel to each other. The white light is laterally displaced from its initial path.\n(b) prism, the white light undergoes dispersion and splits into its constituent colours along with deviation from its initial path.\nQuestion 33.\n(a) With the help of labelled ray diagram show the path followed by a narrow beam of monochromatic light when it passes through a glass prism.\n(b) What would happen if this beam is replaced by a narrow beam of white light? (2020)\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(b) What would happen if this beam is replaced by a narrow beam of white light? (2020)\nAnswer:\n\nHere, in the figure, \u2220D is the angle of deviation of the given monochromatic light by the glass prism.\n(b) If AO were a ray of white light, then on screen BC, a spectrum will be observed, consisting of seven colours arranged from bottom to top as follows. Violet, Indigo, Blue, Green, Yellow, Orange, Red (VIBGYOR)\nQuestion 34.\nWhat is rainbow? Draw a labelled diagram to show the formation of a rainbow. (Delhi 2019)\nAnswer:\nAfter a rain-shower, the sunlight gets dispersed by tiny droplets, present in the atmosphere. The water droplets acts like small glass prisms. They refract and disperse the incident sunlight, then reflect it internally, and finally refract it again when it comes out of the raindrop. Due to dispersion of light and internal reflection, different colours reaches the observers eye, which is called a rainbow.\nRefer to answer 30.\nQuestion 35.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nRefer to answer 30.\nQuestion 35.\nWhat is \u2018dispersion of white light\u2019? State its cause. Draw a ray diagram to show the dispersion of white light by a glass prism. (AI 2017)\nAnswer:\nSplitting of white light into its seven constituent colours due to refraction is known as dispersion of white light.\nCause of dispersion : When a beam of white light enters a prism, it gets refracted and splits into seven constituent colours. The splitting of the light ray occurs due to the different bending angle for each colour. Thus, each colour ray when passing through the prism bends at different angles with respect to the incident beam, thus giving rise to a spectrum.\n\nQuestion 36.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 36.\nState the cause of dispersion of white light passing through a glass prism. How did Newton showed that white light of sun contains seven colours using two identical glass prisms. Draw a ray diagram to show the path of light when two identical glass prisms are arranged together in inverted position with respect to each other and a narrow beam of white light is allowed to fall obliquely on one of the focus of the first prism. (Delhi 2016)\nAnswer:\nRefer to answer 35 and 31.\nQuestion 37.\nDescribe an activity to show that the colours of white light splitted by a glass prism can be recombined to get white light by another identical glass prism. Also draw ray diagrams to show the recombination of the spectrum of white light. (AI 2016)\nAnswer:\nRefer to answer 31.\nQuestion 38.\nA narrow PQ of white light is passing through a glass prism ABC as shown in the diagram. Trace it on your answer sheet and show the path of the emergent beam as observed on the screen DE.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nA narrow PQ of white light is passing through a glass prism ABC as shown in the diagram. Trace it on your answer sheet and show the path of the emergent beam as observed on the screen DE.\n\n(i) Write the name and cause of the phenomenon observed.\n(ii) Where else in nature is this phenomenon observed?\n(iii) Based on this observation, state the conclusion which can be drawn about the constituents of white light. (AI 2014)\nAnswer:\n\n(i) The phenomenon of the splitting up of the white light into its constituents colours is called dispersion of light. Dispersion of light is caused due to, different constituents colours of light after different refractive indices to the material of the prism.\n(ii) The formation of rainbow is caused by the dispersion of the white sunlight into its constituent colours.\n(iii) Based on the dispersion of white light into its constituents colours, we can conclude that\n(a) The white light consists of seven colours.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(iii) Based on the dispersion of white light into its constituents colours, we can conclude that\n(a) The white light consists of seven colours.\n(b) The violet light Millers maximum deviations and the red light suffers minimum deviation.\nQuestion 39.\nDefine the term dispersion of white light. Name the colour of light which bends (i) the most, (ii) the least while passing through a glass prism. Draw a ray diagram to justify your answer. (Foreign 2014)\nAnswer:\nRefer to answer 38.\nQuestion 40.\nWhat is a spectrum? How can we recombine the components of white light after a glass prism has separated them? Illustrate it by drawing a diagram. (Foreign 2014)\nAnswer:\nRefer to answer 36.\nQuestion 41.\nWhen we place a glass prism in the path of a narrow beam of white light, a spectrum is obtained. What happens when a second identical prism is placed in an inverted position with respect to the first prism? Draw a labelled diagram to illustrate it. (Delhi 2012)\nAnswer:\nRefer to answer 31.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 31.\nQuestion 42.\nDraw a labelled ray diagram to illustrate the dispersion of a narrow beam of white light when it passes through a glass prism. (AI 2012)\nAnswer:\nRefer to answer 35.\nQuestion 43.\nDraw a ray diagram to show the formation of a rainbow and mark the point where\n(i) dispersion, (ii) internal reflection occurs. (Foreign 2012)\nAnswer:\nRefer to answer 30.\nQuestion 44.\nWhat is meant by the dispersion of white light? Draw a diagram to show dispersion of white light by the glass prism. (Delhi 2011)\nAnswer:\nRefer to answer 35.\nQuestion 45.\nExplain the formation of rainbow in the sky with the help of a diagram. (Foreign 2011)\nAnswer:\nRefer to answer 30.\nQuestion 46.\nGive reasons:\n(i) The extent of deviation of a ray of light on passing through a prism depends on the colour.\n(ii) Lights of red colour are used for danger signals. (Foreign 2011)\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nGive reasons:\n(i) The extent of deviation of a ray of light on passing through a prism depends on the colour.\n(ii) Lights of red colour are used for danger signals. (Foreign 2011)\nAnswer:\n(i) The extent of deviation of a ray of light on passing through a prism depends on the colour because the refractive index of glass for different colour is different. It depends on wavelength of a particular light.\n(ii) Since the wavelength of light is maximum in the spectrum, its penetration power in the air is maximum and so we can see red colour from farther distances. Thus, danger signal uses red colour.\nQuestion 47.\n(a) Why do the component colours of incident white light split into a spectrum while passing through a glass prism, explain.\n(b) Draw a labelled ray diagram to show the formation of a rainbow. (4/5, Delhi 2017)\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(b) Draw a labelled ray diagram to show the formation of a rainbow. (4/5, Delhi 2017)\nAnswer:\n(a) When a beam of light incidents on a prism, it first gets refracted and splits into seven constituent colors. The splitting of the light ray occurs due to the different bending angle for each colour. Thus each colour ray when passing through the prism bends at different angles with respect to the incident beam. This gives rise to the formation of the spectrum.\n(b) Refer to answer 30.\nQuestion 48.\n(a) What is dispersion of white light? State its cause.\n(b) \u201cRainbow is an example of dispersion of sunlight.\u201d Justify this statement by explaining, with the help of a labelled diagram, the formation of a rainbow in the sky. List two essential conditions for observing a rainbow. (Foreign 2016)\nAnswer:\n(a) Refer to answer 35.\n(b) Refer to answer 34.\nTwo essential conditions for observing rainbow are\n(i) Sun should be at the back of the observer.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Refer to answer 35.\n(b) Refer to answer 34.\nTwo essential conditions for observing rainbow are\n(i) Sun should be at the back of the observer.\n(ii) Rainbow should be seen after rainfall or through a waterfall or water fountain.\nQuestion 49.\n(a) Trace on your answer sheet the path of a monochromatic ray AO incident on a glass prism and mark the angle of deviation.\n\n(b) If AO were a ray of white light,\n(i) describe what will you observe on the screen BC placed near the prism\n(ii) write the name of this phenomenon\n(iii) state the cause of this phenomenon\n(iv) what does it prove about the constituents of white light? (Delhi 2013 C)\nAnswer:\n(a) Refer to answer 33(a)\n(b) (i) Refer to answer 33(b)\n(ii) This phenomenon is known as dispersion of light.\n(iii) Refer to answer 35.\n(iv) It proves that a white light consists of seven colours and lower the wavelength higher will be the deviation of light.\nQuestion 50.\nWhy do stars appear to twinkle ?", + "(iii) Refer to answer 35.\n(iv) It proves that a white light consists of seven colours and lower the wavelength higher will be the deviation of light.\nQuestion 50.\nWhy do stars appear to twinkle ? Explain. (Foreign 2015)\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 50.\nWhy do stars appear to twinkle ? Explain. (Foreign 2015)\nAnswer:\n\nDue to atmospheric refraction, position of star visible from sun, is slightly different from its actual position. This apparent position of the star is not stationary, but keeps on changing with change in physical condition on earths atmosphere. Since the stars are very distant, they are approximately point-sized sources of light. As the path of rays of light coming from the star goes on varying slightly, the apparent position of the star fluctuates and the amount of starlight entering the eye flickers the star sometimes appears brighter, and at some other time, fainter, which is the twinkling effect.\nQuestion 51.\nExplain why the planets do not twinkle. (Foreign 2015)\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 51.\nExplain why the planets do not twinkle. (Foreign 2015)\nAnswer:\nPlanets do not emit light. However, they become visible due to reflection of light falling on them. The planets are much closer to the earth and thus can be considered as the extended source of light. The fluctuations in the light coming from various points of the planet due to atmospheric refraction get averaged out. As a result, no twinkling of planets is seen.\nQuestion 52.\nExplain in brief the reason for each of the following:\n(a) Advanced sun-rise\n(b) Delayed sun-set\n(c) Twinkling of stars (Foreign 2016)\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 52.\nExplain in brief the reason for each of the following:\n(a) Advanced sun-rise\n(b) Delayed sun-set\n(c) Twinkling of stars (Foreign 2016)\nAnswer:\n(a, b): The Sun is visible to us about 2 minutes before the actual sunrise, and about 2 minutes after the actual sunset because of atmospheric refraction. By actual sunrise, we mean the actual crossing of the horizon by the Sun. Figure shows the actual and apparent positions of the Sun with respect to the horizon. The time difference between actual sunset and the apparent sunset is about 2 minutes. The apparent flattening of the Sun\u2019s disc at sunrise and sunset is also due to the same phenomenon.\n\n(c) Refer to answer 50.\nQuestion 53.\nWhat is meant by advance sunrise and delayed sunset? Draw a labelled diagram to explain these phenomena. (Foreign 2015)\nAnswer:\nRefer to answer 52(a) and (b).\nQuestion 54.\nExplain with the help of a labelled diagram, the cause of twinkling of stars.", + "What is meant by advance sunrise and delayed sunset? Draw a labelled diagram to explain these phenomena. (Foreign 2015)\nAnswer:\nRefer to answer 52(a) and (b).\nQuestion 54.\nExplain with the help of a labelled diagram, the cause of twinkling of stars. (Delhi 2014)\nAnswer:\nRefer to answer 50.\nQuestion 55.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 52(a) and (b).\nQuestion 54.\nExplain with the help of a labelled diagram, the cause of twinkling of stars. (Delhi 2014)\nAnswer:\nRefer to answer 50.\nQuestion 55.\nA star sometimes appears brighter and some other times fainter. What is this effect called? State the reason for this effect. (Delhi 2012)\nAnswer:\nThis effect is called twinkling of stars.\nRefer to answer 50.\nQuestion 56.\nA star appears slightly higher (above) than its actual position in the sky. Illustrate it with the help of a labelled diagram. (AI2012)\nAnswer:\nA star appears slightly above than its actual position in the sky. Since the starlight, on entering the earth\u2019s atmosphere undergoes refraction continuously in a medium of gradually changing refractive index, before it reaches the earth. Since the atmosphere bends starlight towards the normal, the star appears slightly above than its actual position.\n\nQuestion 57.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 57.\n\u201cThe time difference between the actual sunset and the apparent sunset is about 2 minutes\u201d What is the reason for the same? Explain with the help of a diagram. (Foreign 2012)\nAnswer:\nRefer to answer 52(a) and (b).\nQuestion 58.\nExplain why the planets do not twinkle but the stars twinkle. (Delhi 2011)\nAnswer:\nRefer to answer 50 and 51.\nQuestion 59.\nWhy do stars twinkle ? Explain (2/3, 2018)\nAnswer:\nRefer to answer 50.\nQuestion 60.\nWhat is atmospheric refraction? Use this phenomenon to explain the following natural events.\n(a) Twinkling of stars\n(b) Advanced sun-rise and delayed sun-set. Draw diagrams to illustrate your answers. (AI 2016)\nAnswer:\nRefraction of the light by the different layers of the atmosphere having different refractive indices is known as atmospheric refraction.\n(a) Twinkling of stars : Refer to answer 50.", + "Draw diagrams to illustrate your answers. (AI 2016)\nAnswer:\nRefraction of the light by the different layers of the atmosphere having different refractive indices is known as atmospheric refraction.\n(a) Twinkling of stars : Refer to answer 50.\n(b) Advanced sun-rise and delayed sun-set : Refer to answer 52(a) and (b).\nQuestion 61.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(a) Twinkling of stars : Refer to answer 50.\n(b) Advanced sun-rise and delayed sun-set : Refer to answer 52(a) and (b).\nQuestion 61.\nThe sky appears dark to passengers flying at very high altitudes mainly because\n(a) Scattering of light is not enough at such heights.\n(b) There is no atmosphere at great heights.\n(c) The size of molecules is smaller than the wavelength of visible light.\n(d) The light gets scattered towards the earth. (2020)\nAnswer:\n(b) There is no atmosphere at great heights.\nQuestion 62.\nConsider the following reasons for the reddish appearance of the sun at the sunrise or the sunset:\n(A) Light from the sun near the horizon passes through thinner layers of air.\n(B) Light from the sun covers larger distance of the earths atmosphere before reaching our eyes.\n(C) Near the horizon, most of the blue light and shorter wavelengths are scattered away by the particles.\n(D) Light from the sun near the horizon passes through thicker layers of air.\nThe correct reasons are", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(D) Light from the sun near the horizon passes through thicker layers of air.\nThe correct reasons are\n(a) A and C only\n(b) B, C and D\n(c) A and B only\n(d) C and D only (2020)\nAnswer:\n(b) Near the horizon, the light rays from the sun has to travel a larger distance through the Earth\u2019s atmosphere as compared to when it is away from the horizon. Thus, when this light travels through the atmosphere, most of short wavelength lights are scattered away causing the reddish appearance of the sun.\nQuestion 63.\nWhat will be the colour of the sky when it is observed from a place in the absence of any atmosphere? (Delhi 2012)\nAnswer:\nIf the earth had no atmosphere, there would not have been any scattering. Then, the sky would look dark.\nQuestion 64.\nGive an example of a phenomenon where Tyndall effect can be observed. (AI 2011)\nAnswer:\nThe phenomenon of scattering of light by the colloidal particle give rise to Tyndall effect.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 64.\nGive an example of a phenomenon where Tyndall effect can be observed. (AI 2011)\nAnswer:\nThe phenomenon of scattering of light by the colloidal particle give rise to Tyndall effect.\nThis phenomenon is seen when a fine beam of sunlight enters a smoke-filled room through a small hole. This can also be observed when sunlight passes through a canopy of a dense forest.\nQuestion 65.\nWhy is the colour of clear sky blue? (Foreign 2011)\nAnswer:\nWhen sunlight passes the atmosphere, the fine particles in air scatter blue colour more strongly than red. This scattered blue light enters our eye and the colour of clear sky appears blue.\nQuestion 66.\nWhy is Tyndall effect shown by colloidal particles? State four instance of observing the Tyndall effect. (2020)\nAnswer:\nThe phenomenon of scattering of light by the colloidal particles give rise to Tyndall effect. When a beam of light strike colloidal particles, the path of the beam becomes visible. This is known as Tyndall effect.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nThis phenomenon can be observed when\nsunlight passes through a canopy of dense forest, when tiny water droplets in the mist scatter light.\ntorch light is switched on in a foggy environment, light rays are visible after being scattered by the fog particles in the surrounding air.\na fine beam of sunlight enters a smoke-filled room through a small hole.\nshining a flashlight beam into a glass of dilated milk produces Tyndall effect.\nQuestion 67.\nDraw a labelled diagram to show (i) reddish appearance of the sun at the sunrise or the sunset and (ii) white appearance of the sun at noon when it is overhead. (2020)\nAnswer:\n\nAt sun-rise and the sun-set, light from the sun passes through thicker layers of air and larger distance in the earth\u2019s atmosphere. As the red colour has longest wavelength hence, it is least scattered by the air and dust particles. So, the sun appears reddish.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(ii) At noon, when sun is overhead, the distance to be travelled is least. All wavelengths are scattered equally and hence sun appears white.\nQuestion 68.\n(a) State the relation between colour of scattered light and size of the scattering particle.\n(b) The apparent position of an object, when seen through the hot air, fluctuates or waves. State the basic cause of this observation.\n(c) Complete the path of white light when it passes through two identical prisms placed as shown\n\nAnswer:\n(a) The colour of scattered light depends on the size of the scattering particle. Very fine particles scatter short wavelengths such as blue and violet, lights. Large size particles scatter light of longer wavelengths.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\n(b) The basic cause of this observation is atmospheric refraction. As hot air is less denser then the colder air surrounding it, it has a slightly lower refractive index. Since the physical condition of the refracting medium, in air is not stationary, the apparent position of an object, when seen through hot air fluctuates.\n(c) Refer to answer 31.\nQuestion 69.\nWith the help of a labelled diagram, explain why the sun appears reddish at the sun-rise and the sun-set. (Delhi 2015)\nAnswer:\nRefer to answer 67(i).\nQuestion 70.\nWhat is meant by scattering of light? Use this phenomenon to explain why the clear sky appears blue or the sun appears reddish at sunrise. (AI 2015)\nAnswer:\nThe phenomenon by which a beam of light is redirected in many different directions when it interacts with a particle of matter is known as scattering of light.\nRefer to answer 65 and 67(i).\nQuestion 71.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nRefer to answer 65 and 67(i).\nQuestion 71.\nExplain giving reason why the sky appears blue to an observer from the surface of the Earth. What should the appearance of the sky be during the day for an astronaut staying in the international space station orbiting the Earth? State reason to justify your answer. (Foreign 2015)\nAnswer:\nRefer to answer 65.\nFor an astronaut staying in the international space station orbiting the Earth, the appearance of the sky will be black due to absence of air molecules to scatter the light coming from the Sun.\nQuestion 72.\nState the difference in colours of the sun observed during sunrise/sunset and noon. Give explanation for each. (Delhi 2013)\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - IMPORTANT QUESTIONS.txt\nQuestion 72.\nState the difference in colours of the sun observed during sunrise/sunset and noon. Give explanation for each. (Delhi 2013)\nAnswer:\nDuring sunrise or sunset, the suns rays pass through a maximum length of the atmosphere. Most of the blue and shorter wavelength get scattered. Only the red colour of light reaches the observer. That is why the Sun observed during sunset and sunrise appear red. At noon, the distance to be travelled is least. All wavelengths are scattered equally and hence sun appears white.\nQuestion 73.\nWhy does the sun appear reddish early in the morning? Will this phenomenon be observed by an astronaut on the Moon? Give reason to justify your answer. (3/5, 2018)\nAnswer:\nRefer to answer 67(i).\nThis phenomenon cannot be seen by an astronaut on the moon because there is no atmosphere on moon to scatter the light.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 11 Intext Questions\nPage Number: 190\nQuestion 1\nWhat is meant by power of accommodation of the eye ?\nAnswer:\nThe power of accommodation of the eye is the maximum variation of its power for focusing on near and far (distant) objects.\nQuestion 2\nA person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision ?\nAnswer:\nConcave lens.\nQuestion 3\nWhat is the far point and near point of the human eye with normal vision ?\nAnswer:\nFor a human eye with normal vision the far point is at infinity and near point is 25 cm from the eye.\nQuestion 4\nA student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from ? How can it be corrected?\nAnswer:\nThe child is suffering from myopia. The child should use concave lens of suitable focal length.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nAnswer:\nThe child is suffering from myopia. The child should use concave lens of suitable focal length.\nNCERT Solutions for Class 10 Science Chapter 11 Textbook Chapter End Questions\nQuestion 1\nThe human eye can focus objects at different distances by adjusting the focal length of the eye lens.", + "The child should use concave lens of suitable focal length.\nNCERT Solutions for Class 10 Science Chapter 11 Textbook Chapter End Questions\nQuestion 1\nThe human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to\n(a) presbyopia\n(b) accommodation\n(c) near \u2013 sightedness\n(d) far \u2013 sightedness\nAnswer:\n(b) Accommodation\nQuestion 2\nThe human eye forms the image of an object at its\n(a) cornea\n(b) iris\n(c) pupil\n(d) retina\nAnswer:\n(d) Retina\nQuestion 3\nThe least distance of distinct vision for a young adult with normal vision is about\n(a) 25 m\n(b) 2.5 cm\n(c) 25 cm\n(d) 2.5 m\nAnswer:\n(c) 25 cm\nQuestion 4\nThe change in focal length of an eye lens is caused by the action of the\n(a) pupil\n(b) retina\n(c) ciliary muscles\n(d) iris\nAnswer:\n(c) Ciliary muscles\nQuestion 5", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nAnswer:\n(c) 25 cm\nQuestion 4\nThe change in focal length of an eye lens is caused by the action of the\n(a) pupil\n(b) retina\n(c) ciliary muscles\n(d) iris\nAnswer:\n(c) Ciliary muscles\nQuestion 5\nA person needs a lens of power -5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision ?\nSolution:\n(i) \u2235 Power of distant viewing part of the lens, P1 = -5.5 D\n\u2234 Focal length of this part, f1 = \n1\np\n1\n = \n1\n\u22125.5\n m = -0.182 m = -18.2 cm\n(ii) For near vision,\n\nQuestion 6\nThe far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem ?", + "What is the nature and power of the lens required to correct the problem ?\nSolution:\nThe remedial lens should make the objects at infinity appear at the far point.\nTherefore, for object at infinity, u = \u221e\nFar point distance of the defected eye, \u03bd = \u2013 80 cm", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nSolution:\nThe remedial lens should make the objects at infinity appear at the far point.\nTherefore, for object at infinity, u = \u221e\nFar point distance of the defected eye, \u03bd = \u2013 80 cm\n\nNegative sign shows that the remedial lens is a concave lens.\nQuestion 7\nMake a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct the defect ? Assume that the near point of the normal eye is 25 cm.\nSolution:\n(i) The near point N of hypermetropic eye is farther away from the normal near point N.\n\n(ii) In a hypermetropic eye, the image of nearby object lying at normal near point N (at 25 cm) is formed behind the retina.\n\n(iii) Correction of hypermetropia : The convex lens forms a virtual image of the object (lying at normal near point N) at the near point N\u2019 of this eye.", + "(iii) Correction of hypermetropia : The convex lens forms a virtual image of the object (lying at normal near point N) at the near point N\u2019 of this eye.\n\nThe object placed at 25 cm from the correcting lens must produce a virtual image at 1 m or 100 cm.\nTherefore, u = \u2013 25 cm, \u03bd = 100 cm", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nThe object placed at 25 cm from the correcting lens must produce a virtual image at 1 m or 100 cm.\nTherefore, u = \u2013 25 cm, \u03bd = 100 cm\n\nThe positive sign shows that it is a convex lens.\nQuestion 8\nWhy is a normal eye not able to see clearly the objects placed closer than 25 cm ?\nAnswer:\nAt distance less than 25 cm, the ciliary muscles cannot bulge the eye lens any more, the object cannot be focused on the retina and it appears blurred to the eye, as shown in the given figure.\n\nQuestion 9\nWhat happens to the image distance in the eye when we increase the distance of an object from the eye ?\nAnswer:\nThe eye lens of a normal eye forms the images of objects at various distances on the same retina. Therefore, the image distance in the eye remains the same.\nQuestion 10\nWhy do stars twinkle ?\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nThe eye lens of a normal eye forms the images of objects at various distances on the same retina. Therefore, the image distance in the eye remains the same.\nQuestion 10\nWhy do stars twinkle ?\nAnswer:\nStars appear to twinkle due to atmospheric refraction. The light of star after the entry of light in earth\u2019s atmosphere undergoes refraction continuously till it reaches the surface of the earth. Stars are far away. So, they are the point source of light. As the path of light coming from stars keep changing, thus the apparent position of stars keep changing and amount of light from stars entering the eye keeps twinkling. Due to which a star sometimes appear bright and sometimes dim, which is the effect of twinkling.\nQuestion 11\nExplain why the planets do not twinkle ?\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nQuestion 11\nExplain why the planets do not twinkle ?\nAnswer:\nThe planets are much nearer to the earth than stars and because of this they can be considered as large source of light. If a planet is considered to be a collection of a very large number of point sources of light, then the average value of change in the amount of light entering the eye from all point size light sources is zero. Due to this the effect of twinkling is nullified.\nQuestion 12\nWhy does the sun appear reddish early in the morning ?\nAnswer:\nThe light coming from the sun passes through various denser layers of air in the earth\u2019s atmosphere before reaching our eyes near the horizon. Most of the part of blue light and light of small wavelength gets scattered by dust particles near the horizon. So, the light reaching our eyes is of large wavelength. Due to this the sun appears reddish at the time of sunrise and sunset.\nQuestion 13\nWhy does the sky appear dark instead of blue to an astronaut ?\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nQuestion 13\nWhy does the sky appear dark instead of blue to an astronaut ?\nAnswer:\nAs an astronaut moves away from the atmosphere of earth, the atmosphere becomes thin. Due to the absence of molecules (or dust particles) in air, the scattering of light does not take place. Thus, sky appears dark in the absence of scattering.\nNCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World\nFunctioning of a lens in human eye, defects of vision and their corrections, applications of spherical mirrors and lenses. Refraction of light through a prism, dispersion of light, scattering of light, applications in daily life.\nFormulae Handbook for Class 10 Maths and Science\nPage 190\nQuestion 1.\nWhat is meant by power of accommodation of the eye?\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nFormulae Handbook for Class 10 Maths and Science\nPage 190\nQuestion 1.\nWhat is meant by power of accommodation of the eye?\nAnswer:\nThe power Of accommodation of the eye is the ability of the eye to observe the distinct objects clearly which are situated at a large distance from the eye. The ciliary muscles are responsible to change the focal length Of the eye lens. The value of the power of accommodation Of the normal human eye is (d = 25 cm) = 100/f = 100/d = 100/25 = 4 dioptres. The value of power of accommodation Of human eye is about 4D\nQuestion 2.\nA person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the type of the corrective lens used to restore proper vision?\nAnswer:\nThe far point for myopic eye is 1.2m.\n\nQuestion 3.\nWhat is the far point and near point of the human eye with normal vision ?\nAnswer:\nFor human eye with normal vision, far point is at infinity and near point is at 25 cm from the eye.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nQuestion 3.\nWhat is the far point and near point of the human eye with normal vision ?\nAnswer:\nFor human eye with normal vision, far point is at infinity and near point is at 25 cm from the eye.\nDownload NCERT Solutions for Class 10 Science Chapter 11 Human Eye and Colourful World PDF\nQuestion 4.\nA student has difficulty reading the blackboard while sitting in the last row. What could be the defect the child is suffering from ? How can it be corrected ?\nAnswer:\nAs the child has difficulty in reading the blackboard, he is suffering from myopia or short sightedness. To correct this defect, he has to use spectacles with concave lens of suitable focal length.\nPage 197 \u2013 198\nQuestion 1.\nThe human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to\n(a) presbyopia\n(b) accommodation\n(c) near-sightedness\n(d) far-sightedness\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nThe human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to\n(a) presbyopia\n(b) accommodation\n(c) near-sightedness\n(d) far-sightedness\nAnswer:\n(b) Human eye can change the focal length of the eye lens to see the objects situated at various distances from the eye. This is possible due to the power of accommodation of the eye lens.\nQuestion 2.\nThe human eye forms the image of an object at its\n(a) cornea (b) iris (c) pupil (d) retina\nAnswer:\n(d) The human eye forms the image of an object at its retina.\nQuestion 3.\nThe least distance of distinct vision for a young adult with normal vision is about\n(a) 25 m\n(b) 2.5 cm\n(c) 25 cm\n(d) 2.5 m\nAnswer:\n(c) The least distance of distinct vision is the minimum distance of an object to see clear and distinct image. It is 25 cm for a young adult with normal visions.\nQuestion 4.", + "It is 25 cm for a young adult with normal visions.\nQuestion 4.\nThe change in focal length of an eye lens is caused by the action of the\n(a) pupil\n(b) retina\n(c) ciliary muscles", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nQuestion 4.\nThe change in focal length of an eye lens is caused by the action of the\n(a) pupil\n(b) retina\n(c) ciliary muscles\n(d) iris\nAnswer:\n(c) The relaxation or contraction of ciliary muscles changes the curvature of the eye lens. The change in curvature of the eye lens changes the focal length of the eyes. Hence, the change in focal length of an eye lens is caused by the action of ciliary muscles.\nQuestion 5.\nA person needs a lens of power \u2013 5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?\nAnswer:\n\nQuestion 6.\nThe far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?\nAnswer:\n\nQuestion 7.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nAnswer:\n\nQuestion 6.\nThe far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?\nAnswer:\n\nQuestion 7.\nMake a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye \u00a1s 1 m. What \u00a1s the power of a lens required to correct this defect? Assume that near point of the normal eye is 25 cm.\n\n\nQuestion 10.\nWhy do stars twinkle?\nAnswer: The stars twinkle at night, because the star light reaching Our eyes increases and decreases continuously due to atmospheric refraction. When star light reaching our eyes increases, the star looks bright and when the star light reaching our eyes decreases, it appears dim.\nQuestion 11.\nExplain why the planets do not twinkle ?\nAnswer:\nPlanets being close to earth appear larger in size. A planet can be Considered as a collection of large number of small sized objects. Twinkling effect Of these objects cancel each other. so, planets do not appear to twinkle.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nQuestion 12.\nWhy does the sun appear reddish early in the morning?\nAnswer: At sunrise, the sun looks almost reddish because only red colour which is least scattered is received by our eye and appears to come from sun. Hence the appearance Of sun at sunrise, near the horizon looks almost reddish.\nQuestion 13.\nWhy does the sky appear dark of blue to an astronaut?\nAnswer:\nAt such huge heights due to absence of atmosphere, no scattering of the light takes place. Therefore sky appears dark.\nMultiple Choice Questions (MCQs) [1 Mark each]\nQuestion 1.\nThe human eye forms the image of an object at its [NCERT]\n(a) cornea\n(b) iris\n(c) pupil\n(d) retina\nAnswer:\n(d) Retina is the light sensitive surface of eye on which the image is formed.\nQuestion 2.\nThe human eye can focus objects at different distances by adjusting the focal length of eye lens. This is due to [NCERT]\n(a) presbyopia\n(b) accommodation\n(c) nearsightedness\n(d) farsightednes\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nThe human eye can focus objects at different distances by adjusting the focal length of eye lens. This is due to [NCERT]\n(a) presbyopia\n(b) accommodation\n(c) nearsightedness\n(d) farsightednes\nAnswer:\n(b) Accommodation is the ability of eye lens to focus both near and distant objects by adjusting its focal length.\nQuestion 3.\nThe change in focal length of eye lens is caused by action of [NCERT]\n(a) pupil\n(b) retina\n(c) ciliary muscles\n(d) iris\nAnswer:\n(c) Ciliary muscles contract and extend in order to change the lens shape for focussing image ayretina.\nQuestion 4.\nThe least distance of distinct vision for a young adult with normal vision is about [NCERT]\n(a) 25 m\n(b) 2.5 cm\n(c) 25 cm\n(d) 2.5 m\nAnswer:\n(c) The minimum distance at which an object can be seen most distinctly without any strain is 25 cm.\nQuestion 5.", + "Question 5.\nAt noon the Sun appears white as [NCERT Exemplar]\n(a) light is least scattered\n(b) all the colours of the white light are scattered away\n(c) blue colour is scattered the most", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nQuestion 5.\nAt noon the Sun appears white as [NCERT Exemplar]\n(a) light is least scattered\n(b) all the colours of the white light are scattered away\n(c) blue colour is scattered the most\n(d) red colour is scattered the most\nAnswer:\n(a) At noon, the Sun appears white because the light from the Sun is directly over head and travel relatively shorter distance. The Sun appears white as only a little of the blue and violet colours are scattered.\nQuestion 6.\nA person cannot see distinctly objects kept beyond 2 m. This defect can be corrected by using a lens of power [NCERT Exemplar]\n(a) +0.5 D\n(b) -0.5 D\n(c) +0.2 D\n(d) -0.2 D\nIf a person cannot see distinctly, objects kept beyond 21 m, then he is suffering from myopia.\nAnswer:\n(b) As the person has the eye defect, myopia, therefore a concave lens has to be used whose focal length will be f = -2 m (using sign convention).", + "Answer:\n(b) As the person has the eye defect, myopia, therefore a concave lens has to be used whose focal length will be f = -2 m (using sign convention). Thus,\nPower, P = 1/f [where, f is focal length in metre.]\n= 1/-2 = -0.5D .\nQuestion 7.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nPower, P = 1/f [where, f is focal length in metre.]\n= 1/-2 = -0.5D .\nQuestion 7.\nWhich of the following phenomena of light are involved in the formation of a rainbow? [NCERT Exemplar]\n(a) Reflection, refraction and dispersion\n(b) Refraction, dispersion and total internal reflection\n(c) Refraction, dispersion and internal reflection\n(d) Dispersion, scattering and total internal reflection\nAnswer:\n(c) A rainbow is caused by dispersion, refraction and internal reflection of sunlight by tiny water droplets, present in the atmosphere and always formed in a direction opposite to that of the Sun. The water droplets act like small prisms. They refract and disperse the incident sunlight, then reflect it internally and finally refract it again when it comes out of the raindrop.\nQuestion 8.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nQuestion 8.\nA prism ABC (with BC as base) is placed in different orientations. A narrow beam of white light is incident on the prism as shown in figure. In which of the following cases, after dispersion, the third colour from the top corresponds to the colour of the sky? [NCERT Exemplar]\n\n(a) Only (i)\n(b) Only (ii)\n(c) Only (iii)\n(d) Only (iv)\nAnswer:\n(b) In (ii) case, after dispersion, the third colour from the top corresponds to colour of the sky, i.e. blue.\n\nQuestion 9.\nA student sitting on the last bench can read the letters written on the blackboard but is not able to read the letters written in his text book. Which of the following statements is correct?\n(a) The near point of his eyes has receded away\n(b) The near point of his eyes has come closer to him\n(c) The far point of his eyes has come closer to him\n(d) The far point of his eyes has receded away", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\n(b) The near point of his eyes has come closer to him\n(c) The far point of his eyes has come closer to him\n(d) The far point of his eyes has receded away\nHypermetropia may have blurred vision to a person when looking at an object close to them and clearer 1 vision while looking at an object at the distance.\nAnswer:\n(a) The student sitting on the last bench can read the letters written on the blackboard but is not able to read the letters written in his text book because he is suffering from hypermetropia or far sightedness. He can see distant objects clearly but cannot see nearby objects distinctly.\nQuestion 10.\nIn the following diagram, the path of a ray of light passing through a glass prism is shown below.\n\nIn this diagram, the angle of incidence, the angle of emergence and the angle of deviation respectively, are [CBSE2014]\n(a) X, R and T\n(b) Y, Q and T\n(c) X, Q and P\n(d) Y, Q and P\nAnswer:", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nIn this diagram, the angle of incidence, the angle of emergence and the angle of deviation respectively, are [CBSE2014]\n(a) X, R and T\n(b) Y, Q and T\n(c) X, Q and P\n(d) Y, Q and P\nAnswer:\n(d) Angle of incidence is the angle made by the incident ray with the normal to the first surface of prism, which is shown by angle Y. Angle of emergence is the angle made by the emergent ray with the normal to the surface when it comes out from the prism after refraction, which is shown by angle Q. Angle of deviation is the angle between the incident ray and the emergent ray, which is shown by angle P.\n 10 Science Human Eye and Colourful World Mind Map\nHuman Eye\n\nThe organ which gives us the sense of light or enables us to see. It interprets the shapes, colours and dimensions of the\nParts of the Human Eye\nRetina: It is a light sensitive screen on which image is formed.\nIt contains rods sensitive to intensity of light and cones sensitive to colour.", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nParts of the Human Eye\nRetina: It is a light sensitive screen on which image is formed.\nIt contains rods sensitive to intensity of light and cones sensitive to colour.\nCornea: Thin membrane acts like a lens which allow light to enter the eye.\nSclera: Outer part of the eye, protects interior of the eye.\nEye Lens: Convex lens made of transparent, crystalline and flexible jelly like material.\nRefractive index of eye lens is 1.437\nCiliary Muscles: Modify the shape of eye lens.\nPupil: Hole in the middle of iris through which light enters.\nIris: Controls the amount of light entering the eye by changing the size of pupil.\nOptical Nerve: Nerves take the image to the brain in the form of electrical signals.\nDefects of Human Eye\nMyopia or Short Sightedness: can see nearby objects but cannot see far off objects distinctly. Corrected by using a concave lens.\nHypermetropia or Long Sightedness: can see far off objects clearly but cannot see nearby objects clearly. Corrected by convex lens", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nHypermetropia or Long Sightedness: can see far off objects clearly but cannot see nearby objects clearly. Corrected by convex lens\nPresbyopia: It is due to lessening of the flexibility of the crystalline lens and weakening of ciliary muscles. Corrected by using bifocal lenses.\nAstigmatism: Refractive problem responsible for blurry vision. Corrected by using cylindrical lenses.\nCataract: It is a clouding of the lens in the eye. Corrected using cataract surgery\nPower of Accommodation\nThe ability of the eye lens to adjust its focal length so as to see the objects clearly located anywhere. Near point of the human eye is 25 cm and far point of the human eye is infinity.\nPersistence of Vision\nImage of any object seen persists on the retina for 1/16 second even after the removal of the object. This property is used in cinematography.\nReason for Hypermetropia\nIncrease in focal length of eye lens\nShortening of eye-ball\n\nReason for Myopia\nExcessive curvature of cornea\nElongation of eye-ball", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nReason for Hypermetropia\nIncrease in focal length of eye lens\nShortening of eye-ball\n\nReason for Myopia\nExcessive curvature of cornea\nElongation of eye-ball\n\nRefraction\nBending of light when it passes obliquely from one medium to another medium\nDispersion\nSplitting of white light into its component colours \u2013 VIBGYOR.\nRed colour deviates least and violet deviates most\n\nCause of Dispersion\nRefractive index of material for different wavelengths is different.\n\u00b5 \u221d\n1\n\u03bb\n \u00b5v > \u00b5r\nSpectrum\nBand of seven component colours VIBGYOR on a white screen\nRainbow\nSeven colours band of sunlight in the form of bow in the sky. It is formed due to reflection, refraction and dispersion of sunlight by tiny water droplets. To observe rainbow, observer should stand with its back towards sun.\nColour\nThe sensation received by the eye (rod cells of the eye) due to light coming from an object.", + "It is formed due to reflection, refraction and dispersion of sunlight by tiny water droplets. To observe rainbow, observer should stand with its back towards sun.\nColour\nThe sensation received by the eye (rod cells of the eye) due to light coming from an object.\nPrimary Rainbow\nTwo refraction and one total internal reflection\nSubtends an angle of 42\u00b0 at the eye of the observer", + "CHAPTER 10- HUMAN EYE AND THE COLOURFUL WORLD - INTEXT QUESTIONS.txt\nPrimary Rainbow\nTwo refraction and one total internal reflection\nSubtends an angle of 42\u00b0 at the eye of the observer\nInnermost arc is violet and outermost is red\nMore bright\nSecondary Rainbow\nTwo refraction and two total internal reflection\nSubtends an angle of 52.5\u00b0 at the eye\nInnermost arc is red and outermost is violet\nLess bright in comparison to primary rainbow\nAtmospheric Refraction\nPhenomena due to Refraction of light by atmosphere\nTwinkling of stars\nStars seen higher than they actually are\nAdvance sunrise and delayed sunset\nFlattering of the sun at morning and evening\nScatttering of Light\nRayleigh scattering Intensity of scattered light \u221d 1/\u03bb4\nTyndall effect The smoke particles become visible\nThe reddening of the sun at sunrise and sunset\nBlue colour of sky\nThe sky looks dark in absence of atmosphere\nDanger signals are of red coloured", + "CHAPTER 11-ELECTRCITY-EXEMPLER SOLUTONS.txt\nNCERT Exemplar Class 10 Science Chapter 12 Electricity\nShod Answer Type Questions\n1.Three 2 \u2126 resistors, A, B and C are connected as shown in figure. Each of them dissipates energy and can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors.\nAnswer.\n\nHere, P = 18 W\nSince A is in series with the parallel combination of B and C. So, it carries maximum current.\n\nLet IB and Ic be the current flowing through B and C respectively. As they are in parallel, potential difference across them will be same, so\n\n2.Should the resistance of an ammeter be low or high? Give reason.\nAnswer.The resistance of an ammeter should be low so that it will not disturb the magnitude of current flowing through the circuit when connected in series in a circuit.\n3.How does use of a fuse wire protect electrical appliances?", + "CHAPTER 11-ELECTRCITY-EXEMPLER SOLUTONS.txt\n3.How does use of a fuse wire protect electrical appliances?\nAnswer. The fuse wire is always connected in series with the live wire or electrical devices. If the flow of current exceeds the specified preset value due to some reason, the heat produced melts it and disconnects the circuit or the device from the mains. In this way, fuse wire protects the electrical appliances.\n4.What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?\nAnswer. The resistance offered by a metallic wire of unit length and unit cross-sectional area is called electrical resistivity.\nWe know that\n\nHence, when the length of wire is doubled, the resistance becomes double and current decreases to half.", + "CHAPTER 11-ELECTRCITY-EXEMPLER SOLUTONS.txt\nWe know that\n\nHence, when the length of wire is doubled, the resistance becomes double and current decreases to half.\n5.A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 D when connected to a 10 V battery. Calculate the resistance of the electric lamp.\nNow if a resistance of 10 \u2126 is connected in parallel with this series combination, what change (if any) in current flowing through 5 \u2126 conductor and potential difference across the lamp will take place? Give reason.\nAnswer. Given : In series circuit containing lamp and resistor,\n\nAccording to the given condition, circuit can be redrawn as shown.\n\nSo, effective resistance of parallel combination.\n\nAs in parallel combination, potential difference across them remains same. So\n\ni.e., current is divided in both the arms equally. So,\nl1 = l2 = lA.\nHence, there will be no change in the current through 5 n conductor.", + "CHAPTER 11-ELECTRCITY-EXEMPLER SOLUTONS.txt\ni.e., current is divided in both the arms equally. So,\nl1 = l2 = lA.\nHence, there will be no change in the current through 5 n conductor.\nAlso there will be no change in the potential difference across the lamp as in both cases, current through the lamp remains same i.e. l A.\n6.Why is parallel arrangement used in domestic wiring?\nAnswer.Parallel arrangement is used in domestic wiring because\nEach appliance gets the same voltage as that of the mains supply.\nIf one component is switched off, others can work properly.\n Fault in any branch of the circuit can be easily identified.\n7.B1 B2 and B3 are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A\nAnswer.\n\nWhat happens to the glow of the other two bulbs when the bulb B1 gets fused?\nWhat happens to the reading of A1, A2, A3 and A when the bulb gets fused?\nHow much power is dissipated in the circuit when all the three bulbs glow together?", + "CHAPTER 11-ELECTRCITY-EXEMPLER SOLUTONS.txt\nWhat happens to the reading of A1, A2, A3 and A when the bulb gets fused?\nHow much power is dissipated in the circuit when all the three bulbs glow together?\n(i) Since B1 ,B2 and B3 are in parallel, the potential difference across each of them will remain same. So when the bulb Bx gets fused,B1 ,B2 and B3 have the same potential and continues with the same energy dissipated per second, i.e. they will glow continuously as they were glowing before.\n(ii) Resistance of the parallel combination when all the three bulbs are glowing\n\nNow when bulb B2 gets fused, the equivalent resistance of parallel combination\nof B1 and B3 is\n\nSince resistance of each arm is same and p.d. is also same, current divides them equally. So lA current will pass through each bulb Bl and By\nTherefore, ammeter A1 and A3 reads l Acurrent while A2 will read zero and ammeter A read 2 A current.", + "is also same, current divides them equally. So lA current will pass through each bulb Bl and By\nTherefore, ammeter A1 and A3 reads l Acurrent while A2 will read zero and ammeter A read 2 A current.\n(iii) In parallel, total power consumed\nPeq =P1 + P2 + P3\nSo, when all the three bulbs glow together", + "CHAPTER 11-ELECTRCITY-EXEMPLER SOLUTONS.txt\n(iii) In parallel, total power consumed\nPeq =P1 + P2 + P3\nSo, when all the three bulbs glow together\n\nLong Answer Type Questions\n8. Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit, another set of three bulbs of the same wattage are connected in parallel to the same source.\n(\u0430) Will the bulb in the two circuits glow with the same brightness? Justify your answer.\n\n(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.\nAnswer. For three identical bulbs,\nIn series, Rs = 3R\nIn parallel, Rp = R/3\n(a) The bulbs in the two circuits will not glow equally bright as the current through them is not the same.", + "CHAPTER 11-ELECTRCITY-EXEMPLER SOLUTONS.txt\nAnswer. For three identical bulbs,\nIn series, Rs = 3R\nIn parallel, Rp = R/3\n(a) The bulbs in the two circuits will not glow equally bright as the current through them is not the same.\n\n(b) As one bulb fuses, the other bulbs in the series circuit will not glow because the circuit becomes an open circuit. While the rest of bulbs in parallel circuit will continue to glow without getting disturbed because in parallel combination, current gets additional paths to flow.\n9.Find out the following in the electric circuit given in figure:\n\n(a) Effective resistance of two 8 \u2126 resistors in the combination,\n(b) Current flowing through 4 \u2126 resistor,\n(c) Potential difference across 4 \u2126 resistance,\n(d) Power dissipated in 4 \u2126 resistor, and\n(e) Difference in ammeter readings, if any.\nAnswer.\n(a) Effective resistance, the two 8 \u2126 resistors in parallel,", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nImportant Questions of Electricity Class 10 Science Chapter 11\nQuestion 1.\nA current of 10 A flows through a conductor for two minutes.\n(i) Calculate the amount of charge passed through any area of cross section of the conductor.\n(ii) If the charge of an electron is 1.6 \u00d7 10-19 C, then calculate the total number of electrons flowing. (Board Term I, 2013)\nAnswer:\nGiven that: I = 10 A, t = 2 min = 2 \u00d7 60 s = 120 s\n(i) Amount of charge Q passed through any area of cross-section is given by I = \nQ\nt\n\nor Q = I \u00d7 t \u2234 Q = (10 \u00d7 120) A s = 1200 C\n(ii) Since, Q = ne\nwhere n is the total number of electrons flowing and e is the charge on one electron\n\u2234 1200 = n \u00d7 1.6 \u00d7 10-19\nor n = \n1200\n1.6\u00d7\n10\n\u221219\n = 7.5 \u00d7 1021\nQuestion 2.\nDefine electric current.", + "Define electric current. (1/5, Board Term 1,2017)\nAnswer:\nElectric current is the amount of charge flowing through a particular area in unit time.\nQuestion 3.\nDefine one ampere. (1/5, Board Term 1,2015)\nAnswer:", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nElectric current is the amount of charge flowing through a particular area in unit time.\nQuestion 3.\nDefine one ampere. (1/5, Board Term 1,2015)\nAnswer:\nOne ampere is constituted by the flow of one coulomb of charge per second.\n1 A = 1 C s-1\nQuestion 4.\nName a device that you can use to maintain a potential difference between the ends of a conductor. Explain the process by which this device does so. (Board Term I, 2013)\nAnswer:\nA cell or a battery can be used to maintain a potential difference between the ends of a conductor. The chemical reaction within a cell generates the potential difference across the terminals of the cell, even when no current is drawn from it. When it is connected to a conductor, it produces electric current and, maintain the potential difference across the ends of the conductor.\nQuestion 5.\nDraw the symbols of commonly used components in electric circuit diagrams for\n(i) An electric cell\n(ii) Open plug key\n(iii) Wires crossing without connection", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nQuestion 5.\nDraw the symbols of commonly used components in electric circuit diagrams for\n(i) An electric cell\n(ii) Open plug key\n(iii) Wires crossing without connection\n(iv) Variable resistor\n(v) Battery\n(vi) Electric bulb\n(vii) Resistance (4/5, Board Term 1,2017)\nAnswer:\n\nQuestion 6.\nA student plots V-I graphs for three samples of nichrome wire with resistances R1, R2 and R3. Choose from the following the statements that holds true for this graph. (2020)\n\n(a) R1 = R2 = R3\n(b) R1 > R2 > R3\n(c) R3 > R2 > R1\n(d) R2 > R1 > R3\nAnswer:\n(d) : The inverse of the slope of I-V graph gives the resistance of the material. Here the slope of -Rj is highest. Thus, R2 > R1 > R3\nQuestion 7.\nState Ohms law.", + "Here the slope of -Rj is highest. Thus, R2 > R1 > R3\nQuestion 7.\nState Ohms law. (AI 2019)\nAnswer:\nIt states that the potential difference V, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. Mathematically,\nV \u221d I\nV = RI\nwhere R is resistance of the conductor.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nV \u221d I\nV = RI\nwhere R is resistance of the conductor.\nQuestion 8.\nA V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph. (2020)\n\nAnswer:\nAs graph is a straight line, so it is clear from the graph that V \u221d I.\n\nThe shape of the graph obtained by plotting potential difference applied across conductor against the current flowing v. llmuigh il will be a straight line.\nAccording to ohms law,\n\nV = IR or R = \nV\nI\n\nSo, the slope of V\u2019-/ graph at any point represents the resistance of the given conductor.\nQuestion 9.\nStudy the V-I graph for a resistor as shown in the figure and prepare a table showing the values of I (in amperes) corresponding to four different values V (in volts). Find the value of current for V = 10 volts. How can we determine the resistance of the resistor from this graph? (Board Term I, 2016)\n\nAnswer:", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nSince, the graph is straight line so we can either extrapolate the data or simply mark the value from graph as shown in figure.\n\nHence, the value of current for V = 10 volts is 5 amperes (or 5 A).\nFrom Ohm\u2019s law, V = IR,\nWe can write, R = \nV\nI\n\nAt any point on the graph, resistance is the ratio of values of V and I. Since, the given graph is straight line (ohmic conductor) so, the slope of graph will also give the resistance of the resistor\nR = \n10V\n5A\n = 2\u03a9\nAlternately, R = \n(8\u22122)V\n(4\u22121)A\n = \n6V\n3A\n = 2 \u03a9\nQuestion 10.\nV-I graph for a conductor is as shown in the figure\n\n(i) What do you infer from this graph?\n(ii) State the law expressed here. (Board Term I, 2014)\nAnswer:\n(i) Refer to answer 8.\n(ii) Refer to answer 7.\nQuestion 11.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\n(i) What do you infer from this graph?\n(ii) State the law expressed here. (Board Term I, 2014)\nAnswer:\n(i) Refer to answer 8.\n(ii) Refer to answer 7.\nQuestion 11.\nState Ohm\u2019s law. Draw a labelled circuit diagram to verify this law in the laboratory. If you draw a graph between the potential difference and current flowing through a metallic conductor, what kind of curve will you get? Explain how would you use this graph to determine the resistance of the conductor. (Board Term I, 2016)\nAnswer:\nRefer to answer 7 and 8.\nQuestion 12.\nState and explain Ohm\u2019s law. Define resistance and give its SI unit. What is meant by 1 ohm resistance? Draw V-I graph for an ohmic conductor and list its two important features. (Board Term I, 2014)\nAnswer:\nOhm\u2019s law: Refer to answer 7.\nResistance : It is ihe properly of a conductor lo resist the How of charges through it.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nOhm\u2019s law: Refer to answer 7.\nResistance : It is ihe properly of a conductor lo resist the How of charges through it.\nIts SI unit is ohm (\u03a9). If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance R, of the conductor is 1 ohm (1 \u03a9). lvolt\n1 ohm = \n1volt\n1ampere\n\nV-I graph for an ohmic conductor can be drawn as given in figure.\n\nImportant feature of V-I graph are:\n(i) It is a straight line passing through origin.\n(ii) Slope of V-I graph gives the value of resistance of conductor slope = R = \nV\nI\nQuestion 13.\nAssertion (A) : The metals and alloys are good conductors of electricity.\nReason (R) : Bronze is an alloy of copper and tin and it is not a good conductor of electricity.\n(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).", + "Reason (R) : Bronze is an alloy of copper and tin and it is not a good conductor of electricity.\n(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true. (2020)\nAnswer:\n(c) : Metals and alloys are good conductors of electricity. Bronze is an alloy of copper and tin which are metals and thus is a good conductor of electricity.\nQuestion 14.\nA cylindrical conductor of length \u2018l\u2019 and uniform area of cross section \u2018A\u2019 has resistance \u2018R\u2019. The area of cross section of another conductor of same material and same resistance but of length \u20182l\u2019 is (2020)\n(a) \nA\n2\n\n(b) \n3A\n2\n\n(c) 2A\n(d) 3A\nAnswer:\n(c) : The resistance of a conductor of length!, and area of cross section, A is\nR = \u03c1\nl\nA\n\nwhere \u03c1 is the resistivity of the material.", + "Now for the conductor of length 21, area of cross-section A\u2019 and resistivity \u03c1.\nR\u2019 = \u03c1\nl\n\u2032\nA\n\u2032\n = \u03c1\n2l\nA\n\u2032\n\nBut given, R = R\u2019 \u21d2 \u03c1\nl\nA\n = \u03c1\n2l\nA\n or A\u2019 = 2A\nQuestion 15.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nNow for the conductor of length 21, area of cross-section A\u2019 and resistivity \u03c1.\nR\u2019 = \u03c1\nl\n\u2032\nA\n\u2032\n = \u03c1\n2l\nA\n\u2032\n\nBut given, R = R\u2019 \u21d2 \u03c1\nl\nA\n = \u03c1\n2l\nA\n or A\u2019 = 2A\nQuestion 15.\nAssertion (A) : Alloys are commonly used in electrical heating devices like electric iron and heater.\nReason (R): Resistivity of an alloy is generally higher than that of its constituent metals but the alloys have low melting points then their constituent metals.\n(a) Both (A) and (R) are true and (R) is the correct explanation of the assertion (A).\n(b) Both (A) and (R) are true, but (R) is not the correct explanation of the assertion (A).\n(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true. (2020)\nAnswer:\n(a)\nQuestion 16.", + "(c) (A) is true, but (R) is false.\n(d) (A) is false, but (R) is true. (2020)\nAnswer:\n(a)\nQuestion 16.\nHow is the resistivity of alloys compared with those of pure metals from which they may have been formed? (Board Term I, 2017)\nAnswer:\nThe resistivity of an alloy is generally higher than that of its constituent metals.\nQuestion 17.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nThe resistivity of an alloy is generally higher than that of its constituent metals.\nQuestion 17.\n(i) List three factors on which the resistance of a conductor depends.\n(ii) Write the SI unit of resistivity. (Board Term 1, 2015)\nAnswer:\n(i) Resistance of a conductor depends upon the following factors:\n(1) Length of the conductor : (Treater the length (I) of the conductor more will be the resistance (R).\nR \u221d I\n(2) Area ol cross section of the conductor: (Ireater the cross-sectional area of the conductor, less will be the resistance.\nR \u221d \n1\nA\n(3) Nature of conductor.\n(ii) SI unit of resistivity is \u03a9 m.\nQuestion 18.\nCalculate the resistance of a metal wire of length 2m and area of cross section 1.55 \u00d7 106 m\u00b2, if the resistivity of the metal be 2.8 \u00d7 10-8 \u03a9m.", + "Calculate the resistance of a metal wire of length 2m and area of cross section 1.55 \u00d7 106 m\u00b2, if the resistivity of the metal be 2.8 \u00d7 10-8 \u03a9m. (Board Term I, 2013)\nAnswer:\nFor the given metal wire,\nlength, l = 2 m\narea of cross-section, A = 1.55 \u00d7 10-6 m\u00b2\nresistivity of the metal, p = 2.8 \u00d7 10-8 \u03a9 m\nSince, resistance, R = \u03c1\nl\nA\n\nSo R = (\n2.8\u00d7\n10\n\u22128\n\u00d72\n1.55\u00d7\n10\n\u22126\n)\u03a9", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nFor the given metal wire,\nlength, l = 2 m\narea of cross-section, A = 1.55 \u00d7 10-6 m\u00b2\nresistivity of the metal, p = 2.8 \u00d7 10-8 \u03a9 m\nSince, resistance, R = \u03c1\nl\nA\n\nSo R = (\n2.8\u00d7\n10\n\u22128\n\u00d72\n1.55\u00d7\n10\n\u22126\n)\u03a9\n= \n5.6\n1.55\n \u00d7 10-2 \u03a9 = 3.6 \u00d7 10-2\u03a9 or R = 0.036\u03a9\nQuestion 19.\n(a) List the factors on which the resistance of a conductor in the shape of a wire depends.\n(b) Why are metals good conductors of electricity whereas glass is a bad conductor of electricity ? Give reason.\n(c) Why are alloys commonly used in electrical heating devices ? Give reason. (2018)\nAnswer:\n(a) Refer to answer 17 (i).\n(b) Metal have very low resistivity and hence they are good conductors of electricity.", + "Give reason.\n(c) Why are alloys commonly used in electrical heating devices ? Give reason. (2018)\nAnswer:\n(a) Refer to answer 17 (i).\n(b) Metal have very low resistivity and hence they are good conductors of electricity.\nWhereas glass has very high resistivity so glass is a bad conductor of electricity.\n(c) Alloys are commonly used in electrical heating devices due to the following reasons\n(i) Alloys have higher resistivity than metals\n(ii) Alloys do not get oxidised or burn readily.\nQuestion 20.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\n(c) Alloys are commonly used in electrical heating devices due to the following reasons\n(i) Alloys have higher resistivity than metals\n(ii) Alloys do not get oxidised or burn readily.\nQuestion 20.\nCalculate the resistivity of the material of a wire of length 1 m, radius 0.01 cm and resistance 20 ohms. (Board Term I, 2017)\nAnswer:\nWe are given, the length of wire, l = 1 m, radius of wire, r = 0.01 cm = 1 \u00d7 10-4 m and resistance, R = 20\u03a9 As we know,\nR = \u03c1\nl\nA\n, where \u03c1 is resistivity of the material of the wire.\n\u2234 20\u03a9.= \u03c1\nl\n\u03c0\nr\n2\n = \u03c1\n1 m\n3.14\u00d7\n(\n10\n\u22124\n)\n2\n m\n2\n\n\u2234 \u03c1 = 6.28 \u00d7 10-7 \u03a9 m\nQuestion 21.", + "A copper wire has diameter 0.5 mm and resistivity 1.6 \u00d7 10-8 \u03a9 m. Calculate the length of this wire to make it resistance 100 \u03a9. How much does the resistance change if the diameter is doubled without changing its length? (Board Term I, 2015)\nAnswer:\nGiven; resistivity of copper = 1.6 \u00d7 10-8 \u03a9 m, diameter of wire, d = 0.5 mm and resistance of wire, R = 100 \u03a9\nRadius of wire, r = \nd\n2\n = \n0.5\n2", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nGiven; resistivity of copper = 1.6 \u00d7 10-8 \u03a9 m, diameter of wire, d = 0.5 mm and resistance of wire, R = 100 \u03a9\nRadius of wire, r = \nd\n2\n = \n0.5\n2\n mm\n= 0.25 mm = 2.5 \u00d7 10-4 m\nArea of cross-section of wire, A = nr\u00b2\n\u2234 A = 3.14 \u00d7 (2.5 \u00d7 10-4)\u00b2\n= 1.9625 \u00d7 10-7 m\u00b2\n= 1.9 \u00d7 10-7 m\u00b2\nAs, R = \u03c1\nl\nA\n\n\u2234 100 \u03a9 = \n1.6\u00d7\n10\n\u22128\n\u03a9m\u00d7l\n1.9\u00d7\n10\n\u22127\n m\n2\n\nl = 1200 m\nIf diameter is doubled (d\u2019 = 2d),", + "6\u00d7\n10\n\u22128\n\u03a9m\u00d7l\n1.9\u00d7\n10\n\u22127\n m\n2\n\nl = 1200 m\nIf diameter is doubled (d\u2019 = 2d), then the area of cross-section of wire will become\nA\u2019 = \u03c0r\u00b2 = \u03c0(\nd\n\u2032\n2\n)\u00b2 = \u03c0(\n2d\n2\n)\u00b2 = 4A\nNow R \u221d \n1\nA\n, so the resistance will decrease by four times or new resistance will be\nR\u2019 = \nR\n4\n = \n100\n4\n = 25\u03a9\nQuestion 22.\nThe resistance of a wire of 0.01 cm radius is 10 \u03a9. If the resistivity of the material of the wire is 50 \u00d7 10-8 ohm meter, find the length of the wire.", + "The resistance of a wire of 0.01 cm radius is 10 \u03a9. If the resistivity of the material of the wire is 50 \u00d7 10-8 ohm meter, find the length of the wire. (Board Term I, 2014)\nAnswer:\nHere, r = 0.01 cm = 10-4 m, \u03c1 = 50 \u00d7 10-8 \u03a9 m and R = 10 \u03a9\nAs, R = \u03c1\nl\nA\n\nor l = \nRA\n\u03c1\n=\nR\n\u03c1\n(\u03c0\nr\n2\n)\n\nso l = \n10\n50\u00d7\n10\n\u22128\n3.14\u00d7\n(\n10\n\u22124\n)\n2\n\n= 0.628 m = 62.8 cm\nQuestion 23.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nHere, r = 0.01 cm = 10-4 m, \u03c1 = 50 \u00d7 10-8 \u03a9 m and R = 10 \u03a9\nAs, R = \u03c1\nl\nA\n\nor l = \nRA\n\u03c1\n=\nR\n\u03c1\n(\u03c0\nr\n2\n)\n\nso l = \n10\n50\u00d7\n10\n\u22128\n3.14\u00d7\n(\n10\n\u22124\n)\n2\n\n= 0.628 m = 62.8 cm\nQuestion 23.\nA wire has a resistance of 16 \u03a9. It is melted and drawn into a wire of half its original length. Calculate the resistance of the new wire. What is the percentage change in its resistance?", + "A wire has a resistance of 16 \u03a9. It is melted and drawn into a wire of half its original length. Calculate the resistance of the new wire. What is the percentage change in its resistance? (Board Term I, 2013)\nAnswer:\nWhen wire is melted, its volume remains same, so,\nV\u2019 = V or A\u2019l\u2019 = Al\nHere, l\u2019 = \nl\n2\n\nTherefore, A\u2019 = 2 A\nResistance, R = \u03c1\nl\nA\n = 16 \u03a9\nNow, R\u2019 = \n\u03c1\nl\n\u2032\nA\n\u2032\n=\u03c1\n(l/2)\n2A\n=\n1\n4\n\u03c1\nl\nA\n\nSo, R\u2019 = \nR\n4\n = \n16\n4\n = 4 \u03a9\nPercentage change in resistance,\n= \n(\nR\u2212\nR\n\u2032\nR\n)\u00d7100=(\n16\u22124\n16\n)\n \u00d7 100 = 75%\nQuestion 24.\nIf the radius of a current carrying conductor is halved, how does current through it change?", + "If the radius of a current carrying conductor is halved, how does current through it change? (2/5 Board Term I, 2014)\nAnswer:\nIf the radius of conductor is halved, the area of cross-section reduced to (\n1\n4\n) of its previous value.\nSince, R \u221d \n1\nA\n, resistance will become four times", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nIf the radius of conductor is halved, the area of cross-section reduced to (\n1\n4\n) of its previous value.\nSince, R \u221d \n1\nA\n, resistance will become four times\nFrom Ohm\u2019s law, V = IR\nFor given V, I \u221d \n1\nR\n\nSo, current will reduce to one-fourth of its previous value.\nQuestion 25.\nDefine resistance of a conductor. State the factors on which resistance of a conductor depends. Name the device which is often used to change the resistance without changing the voltage source in an electric circuit. Calculate the resistance of 50 cm length of wire of cross sectional area 0.01 square mm and of resistivity 5 \u00d7 10-8 \u03a9 m. (Board Term I, 2014)\nAnswer:\nResistance is the property of a conductor to resist the flow of charges through it.\nFactors affecting resistance of a conductor:\nRefer to answer 17(i)\nRheostat is the device which is often used to change the resistance without changing the voltage source in an electric circuit.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nFactors affecting resistance of a conductor:\nRefer to answer 17(i)\nRheostat is the device which is often used to change the resistance without changing the voltage source in an electric circuit.\nWe are given, length of wire, l = 50 cm = 50 \u00d7 10-2 m cross-sectional area, A = 0.01 mm\u00b2\n= 0.01 \u00d7 10-6 m\u00b2\nand resistivity, \u03c1 = 5 x 10-8 \u03a9 m.\nAs, resistance, R = \u03c1\nl\nA\n\n\u2234 R = \n(\n5\u00d7\n10\n\u22128\n\u00d750\u00d7\n10\n\u22122\n0.01\u00d7\n10\n\u22126\n)\n \u03a9\n= 2.5 \u03a9\nQuestion 26.", + "If a person has five resistors each of value \n1\n5\n \u03a9, then the maximum resistance he can obtain by connecting them is\n(a) 1 \u03a9\n(b) 5 \u03a9\n(c) 10 \u03a9\n(d) 25 \u03a9 (2020)\nAnswer:\n(a) The maximum resistance can be obtained from a group of resistors by connecting them in series. Thus,\nRs = \n1\n5\n+\n1\n5\n+\n1\n5\n+\n1\n5\n+\n1\n5\n 1 \u03a9\nQuestion 27.\nThe maximum resistance which can be made using four resistors each of 2 \u03a9 is\n(a) 2 \u03a9\n(b) 4 \u03a9\n(c) 8 \u03a9\n(d) 16 \u03a9 (2020)\nAnswer:\n(c) : A group of resistors can produce maximum resistance when they all are connected in series.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\n(a) 2 \u03a9\n(b) 4 \u03a9\n(c) 8 \u03a9\n(d) 16 \u03a9 (2020)\nAnswer:\n(c) : A group of resistors can produce maximum resistance when they all are connected in series.\n\u2234 Rs = 2 \u03a9 + 2 \u03a9 + 2 \u03a9 + 2 \u03a9 = 8 \u03a9\nQuestion 28.\nThe maximum resistance which can be made using four resistors each of resistance \n1\n2\n \u03a9 is\n(a) 2 \u03a9\n(b) 1 \u03a9\n(c) 2.5 \u03a9\n(d) 8 \u03a9 (2020)\nAnswer:\n(a) The maximum resistance can be produced from a group of resistors by connecting them in series.\nThus, Rs = \n1\n2\n \u03a9 + H \n1\n2\n \u03a9 + \n1\n2\n \u03a9 + \n1\n2\n \u03a9 = 2 \u03a9\nQuestion 29.\nThree resistors of 10 \u03a9, 15 \u03a9 and 5 \u03a9 are connected in parallel. Find their equivalent resistance.", + "Three resistors of 10 \u03a9, 15 \u03a9 and 5 \u03a9 are connected in parallel. Find their equivalent resistance. (Board Term I, 2014)\nAnswer:\nHere, R1 = 10 \u03a9, R2 =15 \u03a9, R3 = 5 \u03a9.\nIn parallel combination, equivalent resistance, (Req) is given by\n\nQuestion 30.\nList the advantages of connecting electrical devices in parallel with an electrical source instead of connecting them is series. (Board Term I, 2013)\nAnswer:", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nQuestion 30.\nList the advantages of connecting electrical devices in parallel with an electrical source instead of connecting them is series. (Board Term I, 2013)\nAnswer:\n(a) When a number of electrical devices are connected in parallel, each device gets the same potential difference as provided by the battery and it keeps on working even if other devices fail. This is not so in case the devices are connected in series because when one device fails, the circuit is broken and all devices stop working.\n(b) Parallel circuit is helpful when each device has different resistance and requires different current for its operation as in this case the current divides itself through different devices. This is not so in series circuit where same current flows through all the devices, irrespective of their resistances.\nQuestion 31.\nShow how would you join three resistors, each of resistance 9 \u03a9 so that the equivalent resistance of the combination is (i) 13.5 \u03a9, (ii) 6 \u03a9 (2018)\nAnswer:", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nQuestion 31.\nShow how would you join three resistors, each of resistance 9 \u03a9 so that the equivalent resistance of the combination is (i) 13.5 \u03a9, (ii) 6 \u03a9 (2018)\nAnswer:\n(i) The resistance of the series combination is higher than each of the resistances. A parallel combination of two 9 \u03a9 resistors is equivalent to 4.5 \u03a9. We can obtain 13.5 \u03a9 by coupling 4.5 \u03a9 and 9 \u03a9 in series. So, to obtain 13.5 \u03a9, the combination is as shown in figure (a).\n\n(ii) To obtain a equivalent resistance of 6 \u03a9, we have to connect two 9 \u03a9 resistors in series and then connect the third 9 \u03a9 resistor in parallel to the series combination as shown in the figure (b).\n\nQuestion 32.\nThree resistors of 3 \u03a9 each are connected to a battery of 3 V as shown. Calculate the current drawn from the battery.", + "Question 32.\nThree resistors of 3 \u03a9 each are connected to a battery of 3 V as shown. Calculate the current drawn from the battery. (Board Term I, 2017)\n\nAnswer:\nAs given in circuit diagram, two 3 \u03a9 resistors are connected in series to form R1; so R1 = 3 \u03a9 + 3 \u03a9 = 6 \u03a9", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nAs given in circuit diagram, two 3 \u03a9 resistors are connected in series to form R1; so R1 = 3 \u03a9 + 3 \u03a9 = 6 \u03a9\nAnd, R1 and R2 are in parallel combination, Hence, equivalent resistance of circuit (Req) given by\n\nReq = 2 \u03a9\nUsing Ohm\u2019s law, V = IR\nWe get,\n3 V = I \u00d7 2 \u03a9\nor I = \n3\n2\n A = 1.5 A\nCurrent drawn from the battery is 1.5 A.\nQuestion 33.\nTwo identical resistors are first connected in series and then in parallel. Find the ratio of equivalent resistance in two cases. (Board Term I, 2013)\nAnswer:\nLet resistance of each resistor be R.\nFor series combination,\nRs = R1 + R2\nSo, Rs = R + R = 2R\nFor parallel combination,\n\nQuestion 34.\n(a) A 6 \u03a9 resistance wire is doubled on itself. Calculate the new resistance of the wire.", + "(a) A 6 \u03a9 resistance wire is doubled on itself. Calculate the new resistance of the wire.\n(b) Three 2 \u03a9 resistors A, B and C are connected in such a way that the total resistance of the combination is 3 \u03a9. Show the arrangement of the three resistors and justify your answer. (2020)\nAnswer:\n(a) Given resistance of wire, R = 6 \u03a9", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Given resistance of wire, R = 6 \u03a9\nLet l be the length of the wire and A be its area of\ncross-section. Then\nR = \n\u03c1l\nA\n = 6 \u03a9\nNow when the length is doubled, l\u2019 = 2l and A\u2019 = \nA\n2\n\n\u2234 R\u2019 = \n\u03c1(2l)\nA/2\n=\n4\u03c1l\nA\n = 4 \u00d7 6 \u03a9 = 24 \u03a9\n(b) Given the total resistance of the combination = 3 \u03a9\nIn order to get a total resistance of 3 \u03a9, the three resistors has to be connected as shown.\n\nSuch that, \n1\nR\nP\n=\n1\n2\n+\n1\n2\n = 1\n\u21d2 Rp = 1 \u03a9\nand Rs = 2 \u03a9 + 1 \u03a9 = 3 \u03a9\nQuestion 35.", + "Such that, \n1\nR\nP\n=\n1\n2\n+\n1\n2\n = 1\n\u21d2 Rp = 1 \u03a9\nand Rs = 2 \u03a9 + 1 \u03a9 = 3 \u03a9\nQuestion 35.\nDraw a schematic diagram of a circuit consisting of a battery of 3 cells of 2 V each, a combination of three resistors of 10 \u03a9, 20 \u03a9 and 30 \u03a9 connected in parallel, a plug key and an ammeter, all connected in series. Use this circuit to find the value of the following :\n(a) Current through each resistor\n(b) Total current in the circuit\n(c) Total effective resistance of the circuit. (2020)\nAnswer:\nThe circuit diagram is as shown below.\n\n(a) Given, voltage of the battery = 2V + 2V + 2V = 6 V\nCurrent through 10 \u03a9 resistance,", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\n(c) Total effective resistance of the circuit. (2020)\nAnswer:\nThe circuit diagram is as shown below.\n\n(a) Given, voltage of the battery = 2V + 2V + 2V = 6 V\nCurrent through 10 \u03a9 resistance,\nI10 = \nV\nR\n=\n6\n10\n = 0.6 A\nCurrent through 20 \u03a9 resistance,\nI20 = \nV\nR\n=\n6\n20\n = 0.3 A\nCurrent through 30 \u03a9 resistance,\nI30 = \nV\nR\n=\n6\n30\n = 0.2 A\n(b) Total current in the circuit, 1= I10 + I20 + I30\n= 0.6 + 0.3 + 0.2 = 1.1 A\n(c) Total resistance of the circuit,\n\n1\nR\nP\n=\n1\n10\n+\n1\n20\n+\n1\n30\n=\n11\n60\nQuestion 36.", + "(a) With the help of a suitable circuit diagram prove that the reciprocal of the equivalent resistance of a group of resistances joined in parallel is equal to the sum of the reciprocals of the individual resistances.\n(b) In an electric circuit two resistors of 12 \u03a9 each are joined in parallel to a 6 V battery. Find the current drawn from the battery. (Delhi 2019)\nAnswer:\n(a) Resistors in parallel : When resistors are connected in parallel.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Resistors in parallel : When resistors are connected in parallel.\n\n(i) The potential difference across their ends is the same.\n(ii) The sum of current through them is the current drawn from the source of energy or cell.\nI = I1 + I1 + I3 or \nV\nR\nP\n=\nV\nR\n1\n+\nV\nR\n2\n+\nV\nR\n3\n\n(iii) The equivalent resistance is given by,\n\n1\nR\nP\n=\n1\nR\n1\n+\n1\nR\n2\n+\n1\nR\n3\n\nHence equivalent resistance in parallel combination is equal to the sum of reciprocals of the individual resistances.\n\nQuestion 37.\nFor the series combination of three resistors current in each resistor, establish the relation R = R1 + R2 + R3 where the symbols have their usual meanings. Calculate the equivalent resistance of the combination of three resistors of 6 \u03a9, 9 \u03a9 and 18 \u03a9 joined in parallel.", + "Calculate the equivalent resistance of the combination of three resistors of 6 \u03a9, 9 \u03a9 and 18 \u03a9 joined in parallel. (Board Term I, 2016)\nAnswer:\n\nGiven figure shows the series combination of three resistors R1, R2 and R3 connected across a voltage source of potential difference V.\nLet current I is flowing through the circuit.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\n\nGiven figure shows the series combination of three resistors R1, R2 and R3 connected across a voltage source of potential difference V.\nLet current I is flowing through the circuit.\nV1, V2 and V3 are the potential differences across resistors R1, R2 and R3 respectively.\nSince, the total potential difference across a combination of resistors in series is equal to the sum of potential difference across the individual resistors.\n\u2234 v = v1 + v2 + v3 \u2026(i)\nIn series current through each resistor is same. Applying the Ohms law,\nV1 = IR1, V2 = IR2 and V3 = IR1 \u2026\u2026..(ii)\nIf Rs is the equivalent resistance of the circuit, then\nV = IRs \u2026(iii)\nFrom eqns.", + "Applying the Ohms law,\nV1 = IR1, V2 = IR2 and V3 = IR1 \u2026\u2026..(ii)\nIf Rs is the equivalent resistance of the circuit, then\nV = IRs \u2026(iii)\nFrom eqns. (i), (ii) and (iii),\nwe can write IRs = IR1 + IR2 + IR3\nor Rs = R1 + R2 + R3\nWe can conclude that when several resistors are joined in series, the resistance of the combination Rs equals the sum of their individual resistances,\nR1, R2 and R3\nGiven : R1 = 6 \u03a9, R2 = 9 \u03a9,\nR3 = 18 \u03a9 are connected in parallel.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nR1, R2 and R3\nGiven : R1 = 6 \u03a9, R2 = 9 \u03a9,\nR3 = 18 \u03a9 are connected in parallel.\nEquivalent resistance, Req, is given by\n\nor Req = 3 \u03a9\nQuestion 38.\nState ohms law. Represent it graphically. In the given circuit diagram calculate\n(i) the total effective resistance of the circuit.\n(ii) the current through each resistor.\n\nAnswer:\nOhm\u2019s law: Refer to answer 7.\nGraphical representation of Ohm\u2019s law\n\nFor the given circuit\nR1 = 3 \u03a9, R2 = 4 \u03a9., R3 = 6 \u03a9 and V = 6V.", + "Answer:\nOhm\u2019s law: Refer to answer 7.\nGraphical representation of Ohm\u2019s law\n\nFor the given circuit\nR1 = 3 \u03a9, R2 = 4 \u03a9., R3 = 6 \u03a9 and V = 6V.\n(i) Total effective resistance of the circuit, Req is given by\n\n1\nR\neq\n=\n1\nR\n1\n+\n1\nR\n2\n+\n1\nR\n3\n=\n1\n3\n+\n1\n4\n+\n1\n6\n=\n9\n12\n\nor Req = \n12\n9\n \u03a9 = \n4\n3\n \u03a9 = 1.33 \u03a9\n(ii) Since, potential difference across each resistor connected in parallel is same.\nSo, V1 = V2 = V3 = 6 V\nApplying Ohm\u2019s law,\nV1 = I1R1 or I1 = \nV\n1\nR\n1\n or I3 = \n6\n3\n A = 2A\nSimilarly, I2 = \n6A\n4\n = 1.5 A and I3 = \n6\n6\n A = 1 A\nQuestion 39.", + "(a) Prove that the equivalent resistance of three resistors R1, R2 and R3 in series is R1 + R2 + R3", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nV\n1\nR\n1\n or I3 = \n6\n3\n A = 2A\nSimilarly, I2 = \n6A\n4\n = 1.5 A and I3 = \n6\n6\n A = 1 A\nQuestion 39.\n(a) Prove that the equivalent resistance of three resistors R1, R2 and R3 in series is R1 + R2 + R3\n(b) You have four resistors of 8 \u03a9 each. Show how would you connect these resistors to have effective resistance of 8 \u03a9? (4/5, Board Term I, 2015)\nAnswer:\n(a) Refer to answer 37.\n(b) If you have four 8 \u03a9 resistors and the effective resistance is also 8 \u03a9 then the two 8 \u03a9 resistors are connected in series. Now you have pair of two 16 \u03a9 resistors (8 \u03a9 + 8 \u03a9). If you connect these resistors in parallel, you will have net resistance 8 \u03a9.\n\nQuestion 40.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nQuestion 40.\nDraw a labelled circuit diagram showing three resistors R1, R2 and R3 connected in series with a battery (E), a rheostat (Rh), a plug key (K) and an ammeter (A) using standard circuit symbols. Use this circuit to show that the same current flows through every part of the circuit. List two precautions you would observe while performing the experiment. (Board Term I, 2014)\nAnswer:\n\nChange the positions of ammeter and note the reading of ammeter each time. You will find that all the reading obtained are same.\nSo, the value of the current in the ammeter is the same, independent of its position in the electric circuit. It means that in this circuit (series combination) the current is the same in every part of the circuit.\nPrecautions:\n(i) All the connections are neat and tight.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nPrecautions:\n(i) All the connections are neat and tight.\n(ii) Ammeter is connected with the proper polarity, i.e., positive terminal of the ammeter should go to positive terminal and negative terminal of ammeter to the negative terminal of the battery or cell used.\nQuestion 41.\nTwo wires A and B are of equal length and have equal resistances. If the resistivity of A is more than that of B, which wire is thicker and why ? For the electric circuit given below calculate:\n\n(i) current in each resistor\n(ii) total current drawn from the battery, and\n(iii) equivalent resistance of the circuit.", + "If the resistivity of A is more than that of B, which wire is thicker and why ? For the electric circuit given below calculate:\n\n(i) current in each resistor\n(ii) total current drawn from the battery, and\n(iii) equivalent resistance of the circuit. (Board Term I, 2014)\nAnswer:\nLet lA, aA and RA be the length, area of cross-section and resistance of wire A and lB, aB and RB are that of wire B.\nHere, lA = lB and RA = RB\nIf \u03c1A and \u03c1B are the resistivities of wire A and B respectively then\nRA = \u03c1A \nl\nA\na\nA\n and RB = \u03c1B \nl\nB\na\nB\n, As RA = RB\n\u2234 \u03c1A \nl\nA\na\nA\n, \u03c1B \nl\nB\na\nB\n\nor \n\u03c1\nA\n\u03c1\nB\n = \na\nA\na\nB", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nIf \u03c1A and \u03c1B are the resistivities of wire A and B respectively then\nRA = \u03c1A \nl\nA\na\nA\n and RB = \u03c1B \nl\nB\na\nB\n, As RA = RB\n\u2234 \u03c1A \nl\nA\na\nA\n, \u03c1B \nl\nB\na\nB\n\nor \n\u03c1\nA\n\u03c1\nB\n = \na\nA\na\nB\n\nSince \u03c1A > \u03c1B therefore aA > aB Hence, wire A is thicker than wire B.\nFor parallel combination,\nV1 = V2 = V3 = 6V\n(i) Using Ohm\u2019s law\nI1 = V1/R1 = 6/30 = 0.2 A\nI2 = V2/R2 = 6/10 = 0.6 A\nI3 = V3/R3 = 6/5 = 1.2 A\n(ii) Total current drawn from battery,\nI = I1 + I2 + I3 = 0.2 + 0.", + "6 A\nI3 = V3/R3 = 6/5 = 1.2 A\n(ii) Total current drawn from battery,\nI = I1 + I2 + I3 = 0.2 + 0.6 + 1.2 = 2 A\n(iii) Equivalent resistance of the circuit, Req can be obtained by Ohm\u2019s law\nV= I Req\nSo, 6 V = 2 A \u00d7 Req or, Req = \n6\n2\n = 3 \u03a9\nAliter, \n1\nR\neq\n=\n1\nR\n1\n+\n1\nR\n2\n+\n1\nR\n3\n\n\n1\n30\n+\n1\n10\n+\n1\n5\n=\n1+3+6\n30\n=\n10\n30\n=\n1\n3\n\nor Req = 3 \u03a9\nQuestion 42.\n(a) Derive an expression to find the equivalent resistance of three resistors connected in series. Also draw the schematic diagram of the circuit.\n(b) Find the equivalent resistance of the following circuit.\n\nAnswer:\n(a) Refer to answer 37.\n(b) For the given circuit,", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\n(b) Find the equivalent resistance of the following circuit.\n\nAnswer:\n(a) Refer to answer 37.\n(b) For the given circuit,\nR1 = 6 \u03a9, R2 = 10 \u03a9, R3 = 15 \u03a9.\nAs \n1\nR\neq\n=\n1\nR\n1\n+\n1\nR\n2\n+\n1\nR\n3\n\n\n1\nR\neq\n=\n1\n6\n+\n1\n10\n+\n1\n15\n\n= \n5+3+2\n30\n=\n10\n30\n=\n1\n3\n\nReq = 3 \u03a9\nQuestion 43.\nDraw a circuit diagram for a circuit consisting of a battery of five cells of 2 volts each, a 5 \u03a9 resistor, a 10 \u03a9 resistor and a 15 \u03a9 resistor, an ammeter and a plug key, all connected in series. Also connect a voltmeter to record the potential difference across the 15 \u03a9 resistor and calculate\n(i) the electric current passing through the above circuit and\n(ii) potential difference across 5 \u03a9 resistor when the key is closed.", + "Also connect a voltmeter to record the potential difference across the 15 \u03a9 resistor and calculate\n(i) the electric current passing through the above circuit and\n(ii) potential difference across 5 \u03a9 resistor when the key is closed. (Board Term 1, 2013)\nAnswer:\n\nPotential of the battery, V = (2 \u00d7 5) V = 10 V\nEquivalent resistance,\nReq = R1 + R2 + R3\n= (5 + 10 + 15)\u03a9 = 30 \u03a9\n(i) Current through circuit, I = \nV\nR\n=\n10\n30\n A=\n1\n3\n A\n\n(ii) Potential across 5 \u03a9 resistor, V1 = IR1\n= \n1\n3\n \u00d7 5 = \n5\n3\n V = 1.67 V\nQuestion 44.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nReq = R1 + R2 + R3\n= (5 + 10 + 15)\u03a9 = 30 \u03a9\n(i) Current through circuit, I = \nV\nR\n=\n10\n30\n A=\n1\n3\n A\n\n(ii) Potential across 5 \u03a9 resistor, V1 = IR1\n= \n1\n3\n \u00d7 5 = \n5\n3\n V = 1.67 V\nQuestion 44.\nThe resistance of a resistor is reduced to half of its initial value. In doing so, if other parameters of the circuit remain unchanged, the heating effects in the resistor will become\n(a) two times\n(b) half\n(c) one-fourth\n(d) four times (2020)\nAnswer:\n(a) : We know, H = I\u00b2Rt = \nV\n2\n4\n.", + "t\nNow when, R\u2019 = \nR\n24\n, V\u2019 = V and t\u2019 = t\nH\u2019 = \nV\n\u20322\nt\n\u2032\nR\n\u2032\n=\nV\n2\nt\nR/2\n=\n2\nV\n2\nt\nR\n = 2H\nQuestion 45.\n(a) Write the mathematical expression for Joules law of heating.\n(b) Compute the heat generated while transferring 96000 coulomb of charge in two hours through a potential difference of 40 V. (2020)\nAnswer:\n(a) The Joule\u2019s law of healing implies that heat produced in a resistor is\n(i) directly proportional to the square of current lor a given resistance,", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) The Joule\u2019s law of healing implies that heat produced in a resistor is\n(i) directly proportional to the square of current lor a given resistance,\n(ii) directly proportional to resistance for a given current, and\n(iii) directly proportional to the time for which the current flows through the resistor.\ni.e., H = I\u00b2 Rt\n(b) Given, charge q = 96000 C, time t = 2 h = 7200 s and potential difference V = 40 V\nWe know, H = I\u00b2Rt = \nQ\n2\nt\n2\n\u00d7\nV\nQ\n \u00d7 t \u00d7 t = VQ\n= 40 \u00d7 96000 = 3.84 \u00d7 106 J = 3.84 MJ\nQuestion 46.\nWrite Joules law of heating. (1/3, 2018)\nAnswer:\nRefer to answer 45(a).\nQuestion 47.\nExplain the use of an electric fuse. What type of material is used for fuse wire and why? (Board Term I, 2016)\nAnswer:", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 45(a).\nQuestion 47.\nExplain the use of an electric fuse. What type of material is used for fuse wire and why? (Board Term I, 2016)\nAnswer:\nElectric fuse protects circuits and appliances by stopping the flow of any unduly high electric current. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc. If a current larger than the specified value flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit.\nQuestion 48.\n(a) Why is tungsten used for making bulb filaments of incandescent lamps?\n(b) Name any two electric devices based on heating effect of electric current. (2/5, Board Term I, 2015)\nAnswer:\n(a) (i) Tungsten is a strong metal and has high melting point (3380\u00b0C).\n(ii) It emits light at high temperatures (about 2500\u00b0C).\n(b) Electric laundry iron and electric heater are based on heating effect of electric current.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\n(ii) It emits light at high temperatures (about 2500\u00b0C).\n(b) Electric laundry iron and electric heater are based on heating effect of electric current.\nQuestion 49.\nA fuse wire melts at 5 A. If it is desired that the fuse wire of same material melt at 10 A, then whether the new fuse wire should be of smaller or larger radius than the earlier one? Give reasons for your answer. (3/5, Board Term I, 2014)\nAnswer:\nLet the resistance of the wire be R, heat produced in the fuse at 5 A in Is is\nH=(5)\u00b2R ( H \u2013 I\u00b2Rt)\n50. fuse melts at (5)\u00b2R joules of heat.\nLet, the resistance of new wire is R\u2019\nSo, heat produced in 1 second = (10)\u00b2R\u2019\nTo prevent it from melting\n(5)\u00b2R = (10)\u00b2R\u2019 or R\u2019 = \nR\n4\n\nAs R \u221d \n1\nA\n\n\u2234 cross-sectional area of new fuse wire is four times the first fuse.", + "Now, A = \u03c0r\u00b2, so new radius is twice the previous one. So, at 10 A, the new fuse wire of same material and length has larger radius than the earlier one.\nQuestion 50.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nNow, A = \u03c0r\u00b2, so new radius is twice the previous one. So, at 10 A, the new fuse wire of same material and length has larger radius than the earlier one.\nQuestion 50.\nWhat is heating effect of current? List two electrical appliances which work on this effect. (2/5, Board Term I, 2013)\nAnswer:\nIf only resislors are connected to the battery, the source energy continually gets dissipated entirely in the form of heal. This is known as healing effect of current, \u2019file amount of heat (77) produced in time t is given by Joule\u2019s law of heating.\nH = I\u00b2Rt\nWhere, 7 is current flowing through resistor R.\nThe electric laundry iron, electric toaster, electric oven, electric kettle and electric heater are some common devices based on heating effect of current.\nQuestion 51.\nTwo bulbs of 100 W and 40 W are connected in series.", + "Question 51.\nTwo bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. The current through the 40 W bulb will be\n(a) 0.4 A\n(b) 0.6 A\n(c) 0.8 A\n(d) 1A (2020)\nAnswer:", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nTwo bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. The current through the 40 W bulb will be\n(a) 0.4 A\n(b) 0.6 A\n(c) 0.8 A\n(d) 1A (2020)\nAnswer:\n(d) : Given power of first bulb, P1 = 100 W and second bulb P2 = 40 W\nCurrent through 100 W bulb, I1 = 1 A\nCurrent through 40 W bulb, I2 = ?\nSince both the bulbs are connected in series, the electric current passing through both the bulbs are same i.e., I2 = 1 A.\nQuestion 52.\nWrite the relation between resistance (R) of filament of a bulb, its power (P) and a constant voltage V applied across it. (Board Term I, 2017)\nAnswer:\nP = \nV\n2\nR\nQuestion 53.\nPower of a lamp is 60 W. Find the energy in joules consumed by it in Is.", + "(Board Term I, 2017)\nAnswer:\nP = \nV\n2\nR\nQuestion 53.\nPower of a lamp is 60 W. Find the energy in joules consumed by it in Is. (Board Term I, 2016)\nAnswer:\nHere, power of lamp, P = 60 W time,\nt = 1 s\nSo, energy consumed = Power \u00d7 time = (60 \u00d7 1)J = 60 J\nQuestion 54.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nHere, power of lamp, P = 60 W time,\nt = 1 s\nSo, energy consumed = Power \u00d7 time = (60 \u00d7 1)J = 60 J\nQuestion 54.\nTwo lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220 V.", + "Two lamps, one rated 100 W; 220 V, and the other 60 W; 220 V, are connected in parallel to electric mains supply. Find the current drawn by two bulbs from the line, if the supply voltage is 220 V. (2/3, 2018, Board Term I, 2014)\nAnswer:\nSince both the bulbs are connected in parallel and to a 220 V supply, the voltage across each bulb is 220 V. Then\nCurrent drawn by 100 W bulb,\nI1 = \npowerrating\nvoltageapplied\n = \n100W\n220V\n = 0.454 A\nCurrent drawn by 60 W bulb,\nI2 = \n60W\n220V\n = 0.273 A\nTotal current drawn from the supply line,\nI = I1 + I2 = 0.454 A + 0.273 A = 0.727 A = 0.73 A\nQuestion 55.\nHow much current will an electric iron draw from a 220 V source if the resistance of its element when hot is 55 ohms? Calculate the wattage of the electric iron when it operates on 220 volts.", + "How much current will an electric iron draw from a 220 V source if the resistance of its element when hot is 55 ohms? Calculate the wattage of the electric iron when it operates on 220 volts. (Board Term I, 2016)\nAnswer:\nHere, V = 220 V, R = 55 \u03a9\nBy Ohm\u2019s law V = IR", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nHere, V = 220 V, R = 55 \u03a9\nBy Ohm\u2019s law V = IR\n\u2234 220 = 7 \u00d7 55 or I = 4A\nWattage of electric iron = Power\n= \nV\n2\nR\n=\n(220\n)\n2\n55\n = 880 W\nQuestion 56.\nAn electric iron has a rating of 750 W; 200 V. Calculate:\n(i) the current required.\n(ii) the resistance of its heating element.\n(iii) energy consumed by the iron in 2 hours.", + "An electric iron has a rating of 750 W; 200 V. Calculate:\n(i) the current required.\n(ii) the resistance of its heating element.\n(iii) energy consumed by the iron in 2 hours. [Board Term 1, 2015]\nAnswer:\nHere, P = 750 W, V = 200 V\n(i) As P = V7\nI = P/V= (750/200) A = 3.75A\n(ii) By Ohm\u2019s law V = IR or R = V/I\n\u2234 R = \n200\n3.75\n \u03a9 = 53.3 \u03a9\n(iii) Energy consumed by the iron in 2 hours\n= P \u00d7 t = 750 W \u00d7 2h = 1.5 kWh\nor E = (750 \u00d7 2 \u00d7 3600) J = 5.4 \u00d7 106 J\nQuestion 57.\nAn electric bulb is connected to a 220 V generator. The current is 2.5 A. Calculate the power of the bulb.", + "An electric bulb is connected to a 220 V generator. The current is 2.5 A. Calculate the power of the bulb. (1/3, Board Term I, 2015)\nAnswer:\nHere, V= 220 V,/= 2.5 A\nPower of the bulb P = VI = 220 \u00d7 2.5 W = 550 W\nQuestion 58.\n(a) Define power and state its SI unit.\n(b) A torch bulb is rated 5 V and 500 mA. Calculate its\n(i) power\n(ii) resistance", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nPower of the bulb P = VI = 220 \u00d7 2.5 W = 550 W\nQuestion 58.\n(a) Define power and state its SI unit.\n(b) A torch bulb is rated 5 V and 500 mA. Calculate its\n(i) power\n(ii) resistance\n(iii) energy consumed when it is lighted for 2 \n1\n2\n hours.\nAnswer:\n(a) Power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit.\nP = VI = I\u00b2R = V\u00b2/R\nThe SI unit of electric power is watt (W). It is the power consumed by a device that carries 1 A of current when operated at a potential difference of IV.", + "P = VI = I\u00b2R = V\u00b2/R\nThe SI unit of electric power is watt (W). It is the power consumed by a device that carries 1 A of current when operated at a potential difference of IV.\n1 W = 1 volt \u00d7 1 ampere = 1 V A\n(b) Given, V = 5 V and I = 500 mA = 0.5 A\n(i) Power, P = V \u00d7 7 = 5 \u00d7 0.5 = 2.5 W\n(ii) As, P = \nV\n2\nR\n\u21d2R=\nV\n2\nP\n=\n25\n2.5\n = 10 \u03a9\n(iii) Given, time t = 2.5 hrs = 9000 s\n\u2234 The energy consumed, E = P \u00d7 t\n= 2.5 \u00d7 9000 = 2.25 \u00d7 104 J\n= 6.25 Watt hour\nQuestion 59.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\n(ii) As, P = \nV\n2\nR\n\u21d2R=\nV\n2\nP\n=\n25\n2.5\n = 10 \u03a9\n(iii) Given, time t = 2.5 hrs = 9000 s\n\u2234 The energy consumed, E = P \u00d7 t\n= 2.5 \u00d7 9000 = 2.25 \u00d7 104 J\n= 6.25 Watt hour\nQuestion 59.\nTwo identical resistors, each of resistance 15 \u03a9, are connected in (i) series, and (ii) parallel, in turn to a battery of 6 V. Calculate the ratio of the power consumed in the combination of resistors in each case. (2020)\nAnswer:\nGiven, R1 = R1 = 15 \u03a9, V = 6 V\n(i) When connected in series,\nRs = R1 + R2 = 15 \u03a9 + 15 \u03a9 = 30 \u03a9\nPower, PS = \nV\n2\nR\nS\n=\n36\n30\n W\n(ii) When connected in parallel,\n\nQuestion 60.", + "An electric lamp of resistance 20 \u03a9 and a conductor of resistance 4 \u03a9. are connected to a 6 V battery as shown in the circuit. Calculate.\n\n(a) the total resistance of the circuit\n(b) the current through the circuit,\n(c) the potential difference across the (i) electric lamp and (ii) conductor, and\n(d) power of the lamp. (Delhi 2019)\nAnswer:\nResistance of the lamp = 20 \u03a9\nExternal resistance = 4 \u03a9", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\n(c) the potential difference across the (i) electric lamp and (ii) conductor, and\n(d) power of the lamp. (Delhi 2019)\nAnswer:\nResistance of the lamp = 20 \u03a9\nExternal resistance = 4 \u03a9\n(a) As both the lamp and external resistance are connected in series, therefore the total resistance,\nR = 20 + 4 = 24 \u03a9\n(b) Current, I = \nV\nR\n = \n6\n24\n = 0.25 A\n(c) (i) Potential difference across the electric lamp\n\nTotalvoltage\nTotalresistance\n \u00d7 resistance of lamp\n= \n6\n24\n \u00d7 20 = 5 V\n(ii) Potential difference across conductor\n\nTotalvoltage\nTotalresistance\n \u00d7 resistance of conductor\n= \n6\n24\n \u00d7 4 = 1 V\n(d) Power of the lamp\n= (current)\u00b2 \u00d7 resistance of lamp\n= (0.25)\u00b2 \u00d7 20 = 1.25 W\nQuestion 61.\nCompare the power used in 2 \u03a9. resistor in each of the following circuits.", + "Compare the power used in 2 \u03a9. resistor in each of the following circuits. (AI 2019)\n\nAnswer:\nIn circuit A,\nTotal resistance, R = l + 2 = 3\u03a9\nVoltage across 2 \u03a9 = \nV\nTotal\nR\nTotal\n \u00d7 2 \u03a9 = \n6\n3\n \u00d7 2 = 4 V\n\u2234 Power used in 2 \u03a9 resistor,\np = \nV\n2\nR\n = \n(4\n)\n2\n2\n = 8 w", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\nIn circuit A,\nTotal resistance, R = l + 2 = 3\u03a9\nVoltage across 2 \u03a9 = \nV\nTotal\nR\nTotal\n \u00d7 2 \u03a9 = \n6\n3\n \u00d7 2 = 4 V\n\u2234 Power used in 2 \u03a9 resistor,\np = \nV\n2\nR\n = \n(4\n)\n2\n2\n = 8 w\nIn circuit B, Voltage across both the resistance is same i.e. 4 V and both are connected in parallel combination.\n\u2234 Power used in 2 \u03a9 resistor = \nV\n2\nR\n = \n(4\n)\n2\n2\n = 8 w\n\u2234 Power used in 2 \u03a9 resistor in each case is same i.e. 8 W.\nQuestion 62.\nA bulb is rated 40 W; 220 V. Find the current drawn by it, when it is connected to a 220 V supply. Also find its resistance.", + "8 W.\nQuestion 62.\nA bulb is rated 40 W; 220 V. Find the current drawn by it, when it is connected to a 220 V supply. Also find its resistance. If the given bulb is replaced by a bulb of rating 25 W; 220 V, will there be any change in the value of current and resistance? Justify your answer and determine the change. (AI 2019)\nAnswer:\nIn first case, P = 40 W, V = 220 V\nCurrent drawn l = \nP\nV\n = \n40\n220\n = 0.18 A\nAlso, resistance of bulb,\nR = \nV\n2\nP\n=\n(220\n)\n2\n40\n = 1210 \u03a9\nIn second case, P = 25 W, V = 220 V\nCurrent drawn, I = \nP\nV\n = \n25\n220\n = 0.11 A\nAlso, resistance of the bulb,\nR = \nV", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\n= \n40\n220\n = 0.18 A\nAlso, resistance of bulb,\nR = \nV\n2\nP\n=\n(220\n)\n2\n40\n = 1210 \u03a9\nIn second case, P = 25 W, V = 220 V\nCurrent drawn, I = \nP\nV\n = \n25\n220\n = 0.11 A\nAlso, resistance of the bulb,\nR = \nV\n2\nP\n = \n(220\n)\n2\n25\n = 1936 \u03a9\nHence, by replacing 40 W bulb to 25 W bulb, having same source of voltage the amount of current flows decreases while resistance increases.\nQuestion 63.\n(a) How two resistors, with resistances R1 \u03a9 and R1 \u03a9 respectively are to be connected to a battery of emf V volts so that the electrical power consumed is minimum?", + "Question 63.\n(a) How two resistors, with resistances R1 \u03a9 and R1 \u03a9 respectively are to be connected to a battery of emf V volts so that the electrical power consumed is minimum?\n(b) In a house 3 bulbs of 100 watt each lighted for 5 hours daily, 2 fans of 50 watt each used for 10 hours daily and an electric heater of 1.00 kW is used for half an hour daily. Calculate the total energy consumed in a month of 31 days and its cost at the rate of Rs 3.60 per kWh. (Board Term I, 2017)\nAnswer:\n(a) Power consumed is minimum when current through the circuit is minimum, so the two resistors are connected in series.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\nAnswer:\n(a) Power consumed is minimum when current through the circuit is minimum, so the two resistors are connected in series.\n(b) Power of each bulb P1 = 100 watt\nTotal power of 3 bulbs, P1 = 3 \u00d7 100 = 300 watt\nEnergy consumed by bulbs in 1 day\nE1 = P1 \u00d7 t = 300 watt \u00d7 5 hours.", + "(b) Power of each bulb P1 = 100 watt\nTotal power of 3 bulbs, P1 = 3 \u00d7 100 = 300 watt\nEnergy consumed by bulbs in 1 day\nE1 = P1 \u00d7 t = 300 watt \u00d7 5 hours.\n= 1500 Wh = 1.5 kWh\nPower of each fan = 50 watt\nTotal power of 2 fans = 2 \u00d7 50 watt\nP2 = 100 watt\nEnergy consumed by fans in 1 day\nE2 = P2 \u00d7 t = 100 watt \u00d7 10 hours\n= 1000 watt hour = 1 kWh\nEnergy consumed by heater,\nE3 = 1 kW \u00d7 1/2 h = 0.5 kWh\nTotal energy consumed in one day\nE = E1 + E2 + E3 = (1.5 + 1 + 0.5) kWh = 3 kWh\nTotal energy consumed in a month of 31 days\n= E \u00d7 31 = (3 \u00d7 31) kWh = 93 kWh\nCost of energy consumed = Rs (93 \u00d7 3.60) = Rs 334.80\nQuestion 64.", + "(a) An electric bulb is connected to a 220 V generator. If the current drawn by the bulb is 0.50 A, find its power.\n(b) An electric refrigerator rated 400 W operates 8 hours a day. Calculate the energy per day in kWh.", + "CHAPTER 11-ELECTRCITY-IMPORTANT QUESTIONS.txt\n(b) An electric refrigerator rated 400 W operates 8 hours a day. Calculate the energy per day in kWh.\n(c) State the difference between kilowatt and kilowatt hour. (3/5, Board Term I, 2013)\nAnswer:\n(a) Here, V = 220 V, I = 0.50 A\nPower of the bulb, P = VI = (220 \u00d7 0.5)W = 110 W\n(b) Energy consumed by electric refrigerator in a day = Power x time\n= 400 W \u00d7 8 h = 3200 Wh = 3.2 kWh\n(c) Kilowatt is unit of power and kilowatt hour is a unit of energy.\nQuestion 65.\n(i) State one difference between kilowatt and kilowatt hour. Express 1 kWh in joules.\n(ii) A bulb is rated 5V; 500 mA. Calculate the rated power and resistance of the bulb when it glows. (Board Term I, 2013)\nAnswer:\n(i) Refer to answer 64(c).", + "Express 1 kWh in joules.\n(ii) A bulb is rated 5V; 500 mA. Calculate the rated power and resistance of the bulb when it glows. (Board Term I, 2013)\nAnswer:\n(i) Refer to answer 64(c).\n1 kWh = 1000 W \u00d7 1 h\n= 1000 W \u00d7 3600 s = 3600000 J = 3.6 \u00d7 106 J\n(ii) Here, V = 5 V, I = 500 mA = 0.5 A\nPower rating of bulb is\nP = VI = ( 5 \u00d7 0.5)W = 2.5W\nResistance of the bulb is R = V/I = (5/0.5) \u03a9 = 10 \u03a9", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 12 Intext Questions\nPage Number: 200\nQuestion 1\nWhat does an electric circuit mean ?\nAnswer:\nA continuous and closed path along which an electric current flows is called an electric circuit.\nQuestion 2\nDefine the unit of current.\nAnswer:\nUnit of current is ampere. If one coulomb of charge flows through any section of a conductor in one second then the current through it is said to be one ampere.\nI = \\frac { Q }{ t } or 1 A = I C s-1\nQuestion 3\nCalculate the number of electrons constituting one coulomb of charge.\nAnswer:\nCharge on one electron, e = 1.6 x 10-19 C\nTotal charge, Q = 1 C\nNumber of electrons, n = \\frac { Q }{ e } = \\frac { 1C }{ 1.6x{ 10 }^{ -19 } } = 6.25 x 1018\nPage Number: 202\nQuestion 1\nName a device that helps to maintain a potential difference across a conductor.", + "Answer:\nA battery.\nQuestion 2\nWhat is meant by saying that the potential difference between two points is IV?\nAnswer:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nName a device that helps to maintain a potential difference across a conductor.\nAnswer:\nA battery.\nQuestion 2\nWhat is meant by saying that the potential difference between two points is IV?\nAnswer:\nThe potential difference between two points is said to be 1 volt if 1 joule of work is done in moving 1 coulomb of electric charge from one point to the other.\nQuestion 3\nHow much energy is given to each coulomb of charge passing through a 6 V battery ?\nAnswer:\nEnergy given by battery = charge x potential difference\nor W = QV = 1C X 6V = 6J.\nPage Number: 209\nQuestion 1\nOn what factors does the resistance of a conductor depend ?\nOR\nList the factors on which the resistance of a conductor in the shape of a wire depends. [CBSE2018]\nAnswer:\nThe resistance of a conductor depends (i) on its length (ii) on its area of cross-section and (iii) on the nature of its material.\nQuestion 2", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\nThe resistance of a conductor depends (i) on its length (ii) on its area of cross-section and (iii) on the nature of its material.\nQuestion 2\nWill current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source ? Why ?\nAnswer:\nThe current will flow more easily through a thick wire than a thin wire of the same material. Larger the area of cross-section of a conductor, more is the ease with which the electrons can move through the conductor. Therefore, smaller is the resistance of the conductor.\nQuestion 3\nLet the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it ?\nAnswer:\nWhen potential difference is halved, the current through the component also decreases to half of its initial value. This is according to ohm\u2019s law i.e., V \u221d I.\nQuestion 4", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\nWhen potential difference is halved, the current through the component also decreases to half of its initial value. This is according to ohm\u2019s law i.e., V \u221d I.\nQuestion 4\nWhy are coils of electric toasters and electric irons are made of an-alloy rather than a pure metal ?\nOR\nWhy are alloys commonly used in electric heating devices? Given reason. [CBSE 2018]\nAnswer:\nThe coils of electric toasters, electric irons and other heating devices are made of an alloy rather than a pure metal because (i) the resistivity of an alloy is much higher than that of a pure metal, and (ii) an alloy does not undergo oxidation (or burn) easily even at high temperature, when it is red hot.\nQuestion 5\nUse the data in Table 12.2 (in NCERT Book on Page No. 207) to answer the following :\n(i) Which among iron and mercury is a better conductor ?\n(ii) Which material is the best conductor ?", + "Question 5\nUse the data in Table 12.2 (in NCERT Book on Page No. 207) to answer the following :\n(i) Which among iron and mercury is a better conductor ?\n(ii) Which material is the best conductor ?\nAnswer:\n(i) Resistivity of iron = 10.0 x 10-8 \u2126 m\nResistivity of mercury = 94.0 x 10-8 \u2126 m.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(i) Which among iron and mercury is a better conductor ?\n(ii) Which material is the best conductor ?\nAnswer:\n(i) Resistivity of iron = 10.0 x 10-8 \u2126 m\nResistivity of mercury = 94.0 x 10-8 \u2126 m.\nThus iron is a better conductor because it has lower resistivity than mercury.\n(ii) Because silver has the lowest resistivity (= 1.60 x 10-8 \u2126 m), therefore silver is the best conductor.\nPage Number: 213\nQuestion 1\nDraw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5\u2126 resistor, an 8 \u2126 resistor, and a 12 \u2126 resistor, and a plug key, all connected in series.\nAnswer:\nThe required circuit diagram is shown below :\n\nQuestion 2\nRedraw the circuit of Questions 1, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 \u2126 resistor. What would be the readings in the ammeter and the voltmeter ?", + "What would be the readings in the ammeter and the voltmeter ?\nSolution:\nThe required circuit diagram is shown on the right.\nTotal voltage, V = 3 x 2 = 6V", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nSolution:\nThe required circuit diagram is shown on the right.\nTotal voltage, V = 3 x 2 = 6V\nTotal resistance, R = 5\u2126 + 8\u2126 + 12\u2126 = 25\u2126", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nSolution:\nThe required circuit diagram is shown on the right.\nTotal voltage, V = 3 x 2 = 6V\nTotal resistance, R = 5\u2126 + 8\u2126 + 12\u2126 = 25\u2126\n\n\nPage Number: 216\nQuestion 1\nJudge the equivalent resistance when the following are connected in parallel :\n(i) 1 \u2126 and 106 \u2126,\n(if) 1 \u2126 and 103 \u2126 and 106 \u2126.\nAnswer:\nWhen the resistances are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance.\n(i) Equivalent resistance < 1 \u2126.\n(ii) Equivalent resistance < 1 \u2126.\nQuestion 2\nAn electric lamp of 100 \u2126, a toaster of resistance 50 \u2126, and a water filter of resistance 500 \u2126 are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it ?\nSolution:\nResistance of electric lamp, R1 = 100 \u2126\nResistance of toaster, R2 = 50 \u2126\nResistance of water filter, R3 = 500 \u2126", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nSolution:\nResistance of electric lamp, R1 = 100 \u2126\nResistance of toaster, R2 = 50 \u2126\nResistance of water filter, R3 = 500 \u2126\nEquivalent resistance Rp of the three appliances connected in parallel, is\n\nResistance of electric iron = Equivalent resistance of the three appliances connected in parallel = 31.25 \u2126\nApplied voltage, V = 220 V\nCurrent, I = \\frac { V }{ R } = \\frac { 220V }{ 31.25\\Omega }\nQuestion 3\nWhat are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ?\nAnswer:\nAdvantages of connecting electrical devices in parallel with the battery are :\nIn parallel circuits, if an electrical appliance stops working due to some defect, then all other appliances keep working normally.\nIn parallel circuits, each electrical appliance has its own switch due to which it can be turned on turned off independently, without affecting other appliances.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nIn parallel circuits, each electrical appliance has its own switch due to which it can be turned on turned off independently, without affecting other appliances.\nIn parallel circuits, each electrical appliance gets the same voltage (220 V) as that of the power supply line.\nIn the parallel connection of electrical appliances, the overall resistance of the household circuit is reduced due to which the current from the power supply is high.\nQuestion 4\nHow can three resistors of resistances 2\u2126, 3 \u2126, and 6\u2126 be connected to give a total resistance of (i) 4 \u2126, (ii) 1 \u2126 ?\nSolution:\n(i) We can get a total resistance of 4\u2126 by connecting the 2\u2126 resistance in series with the parallel combination of 3\u2126 and 6\u2126.\n\n(ii) We can obtain a total resistance of 1\u2126 by connecting resistors of 2 \u2126, 3 \u2126 and 6 \u2126 in parallel.", + "(ii) We can obtain a total resistance of 1\u2126 by connecting resistors of 2 \u2126, 3 \u2126 and 6 \u2126 in parallel.\n\nQuestion 5\nWhat is (i) the highest, (ii) the lowest total resistance that can be secured by combinations of four coils of resistance 4 \u2126, 8 \u2126, 12 \u2126, 24 \u2126?\nSolution:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nQuestion 5\nWhat is (i) the highest, (ii) the lowest total resistance that can be secured by combinations of four coils of resistance 4 \u2126, 8 \u2126, 12 \u2126, 24 \u2126?\nSolution:\n(i) Highest resistance can be obtained by connecting the four coils in series.\nThen, R = 4\u2126 + 8\u2126 + 12\u2126 + 24\u2126 = 48\u2126\n(ii) Lowest resistance can be obtained by connecting the four coils in parallel.\n\nPage Number: 218\nQuestion 1\nWhy does the cord of an electric heater not glow while the heating element does ?\nSolution:\nHeat generated in a circuit is given by I2R t. The heating element of an electric heater made of nichrome glows because it becomes red-hot due to the large amount of heat produced on passing current because of its high resistance, but the cord of the electric heater made of copper does not glow because negligible heat is produced in it by passing current because of its extremely low resistance.\nQuestion 2", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nQuestion 2\nCompute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.\nSolution:\nHere, Q = 96,000 C, t =1 hour = 1 x 60 x 60 sec = 3,600 s, V = 50 V\nHeat generated, H = VQ = 50Vx 96,000 C = 48,00,000 J = 4.8 x 106 J\nQuestion 3\nAn electric iron of resistance 20\u2126 takes a current of 5 A. Calculate the heat developed in 30 s.\nSolution:\nHere, R = 20 \u2126, i = 5 A, t = 3s\nHeat developed, H = I2 R t = 25 x 20 x 30 = 15,000 J = 1.5 x 104 J\nPage Number: 220\nQuestion 1\nWhat determines the rate at which energy is delivered by a current ?\nAnswer:\nResistance of the circuit determines the rate at which energy is delivered by a current.", + "Answer:\nResistance of the circuit determines the rate at which energy is delivered by a current.\nQuestion 2\nAn electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.\nAnswer:\nHere, I = 5 A, V = 220 V, t = 2h = 7,200 s\nPower, P = V I = 220 x 5 = 1100 W\nEnergy consumed = P x t = 100 W x 7200 s = 7,20,000 J = 7.2 x 105 J", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\nHere, I = 5 A, V = 220 V, t = 2h = 7,200 s\nPower, P = V I = 220 x 5 = 1100 W\nEnergy consumed = P x t = 100 W x 7200 s = 7,20,000 J = 7.2 x 105 J\nNCERT Solutions for Class 10 Science Chapter 12 Textbook Chapter End Questions\nQuestion 1\nA piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R\u2019, then the ratio R/R\u2019 is :\n(a) \\frac { 1 }{ 25 }\n(b) \\frac { 1 }{ 5 }\n(c) 5\n(d) 25\nAnswer:\n(d) 25\nQuestion 2\nWhich of the following terms does not represent electrical power in a circuit?", + "(a) I2R\n(b) IR2\n(c) VI\n(d) \\frac { { v }^{ 2 } }{ 2 }\nAnswer:\n(fa) IR2\nQuestion 3\nAn electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be :\n(a) 100 W\n(b) 75 W\n(c) 50 W\n(d) 25 W\nAnswer:\n(d) 25 W\nQuestion 4", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\n(fa) IR2\nQuestion 3\nAn electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be :\n(a) 100 W\n(b) 75 W\n(c) 50 W\n(d) 25 W\nAnswer:\n(d) 25 W\nQuestion 4\nTwo conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be :\n(a) 1 : 2\n(b) 2 : 1\n(c) 1 : 4\n(d) 4 : 1\nAnswer:\n(c) 1 : 4\nQuestion 5\nHow is a voltmeter connected in the circuit to measure the potential difference between two points ?\nAnswer:\nA voltmeter is connected in parallel to measure the potential difference between two points.", + "Answer:\nA voltmeter is connected in parallel to measure the potential difference between two points.\nQuestion 6\nA copper wire has diameter 0.5 mm and resistivity of 1.6 x 10-8 \u2126 m. What will be the length of this wire to make its resistance 10 \u2126 ? How much does the resistance change if the diameter is doubled ?\nAnswer:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\n\n\nIf a wire of diameter doubled to it is taken, then area of cross-section becomes four times.\nNew resistance = \\frac { 10 }{ 2 } = 2.5 \u2126, Thus the new resistance will be \\frac { 1 }{ 4 } times.\nDecrease in resistance = (10 \u2013 2.5) \u2126 = 7.5 \u2126\nQuestion 7\nThe values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below :\n\nPlot a graph between V and I and calculate the resistance of the resistor.\nSolution:\nThe graph between V and I for the above data is given below.\nThe slope of the graph will give the value of resistance.\nLet us consider two points P and Q on the graph.\nand from P along Y-axis, which meet at point R.\nNow, QR = 10.2V \u2013 34V = 6.8V\nAnd PR = 3 \u2013 1 = 2 ampere", + "Thus, resistance, R = 3.4 \u2126\nQuestion 8\nWhen a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.\nSolution:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nQuestion 8\nWhen a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.\nSolution:\nHere, V = 12 V and I = 2.5 mA = 2.5 x 10-3 A\n\u2234 Resistance, R = \\frac { V }{ I } = \\frac { 12V }{ 2.5\\times { 10 }^{ 3 }A } = 4,800 \u2126 = 4.8 x 10-3 \u2126\nQuestion 9\nA battery of 9V is connected in series with resistors of 0.2 O, 0.3 O, 0.4 Q, 0.5 Q and 12 \u00a31, respectively. How much current would flow through the 12 Q resistor?", + "How much current would flow through the 12 Q resistor?\nSolution:\nTotal resistance, R = 0.2 \u2126 + 0.3 \u2126 + 0.4 \u2126 + 0.5 \u2126 + 12 \u2126 \u2013 13.4 \u2126\nPotential difference, V = 9 V\nCurrent through the series circuit, I = \\frac { V }{ R } = \\frac { 12V }{ 13.4\\Omega } = 0.67 A\n\u2235 There is no division of current in series. Therefore current through 12 \u2126 resistor = 0.67 A.\nQuestion 10\nHow many 176 \u2126 resistors (in parallel) are required to carry 5 A on a 220 V line? [CBSE (Delhi) 2013]\nSolution:\nSuppose n resistors of 176 \u2126 are connected in parallel.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nQuestion 10\nHow many 176 \u2126 resistors (in parallel) are required to carry 5 A on a 220 V line? [CBSE (Delhi) 2013]\nSolution:\nSuppose n resistors of 176 \u2126 are connected in parallel.\n\nThus 4 resistors are needed to be connect.\nQuestion 11\nShow how you would connect three resistors, each of resistance 6 \u2126, so that the combination has a resistance of (i) 9 \u2126, (ii) 4\u2126\nSolution:\nHere, R1 = R2 = R3 = 6 \u2126.\n(i) When we connect R1 in series with the parallel combination of R2 and R3 as shown in Fig. (a).\nThe equivalent resistance is\n\n\n(ii) When we connect a series combination of R1 and R2 in parallel with R3, as shown in Fig. (b), the equivalent resistance is", + "(ii) When we connect a series combination of R1 and R2 in parallel with R3, as shown in Fig. (b), the equivalent resistance is\n\n\nQuestion 12\nSeveral electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A ?\nSolution:\nHere, current, I = 5 A, voltage, V = 220 V\n\u2234 Maxium power, P = I x V = 5 x 220 = 1100W", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nSolution:\nHere, current, I = 5 A, voltage, V = 220 V\n\u2234 Maxium power, P = I x V = 5 x 220 = 1100W\nRequired no. of lamps =\\frac { Max.Power }{ Power\\quad of\\quad 1\\quad lamp } \\quad =\\quad \\frac { 1100 }{ 10 } =110\n\u2234 110 lamps can be connected in parallel.\nQuestion 13\nA hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 \u2126 resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases ?\nSolution:\n(i) When the two coils A and B are used separately. R = 24 \u2126, V = 220 V\n\n(ii) When the two coils are connected in series,\n\n(iii) When the two coils are connected in parallel.", + "What are the currents in the three cases ?\nSolution:\n(i) When the two coils A and B are used separately. R = 24 \u2126, V = 220 V\n\n(ii) When the two coils are connected in series,\n\n(iii) When the two coils are connected in parallel.\n\nQuestion 14\nCompare the power used in the 2 \u2126 resistor in each of the following circuits\n(i) a 6 V battery in series with 1 \u2126 and 2 \u2126 resistors, and\n(ii) a 4 V battery in parallel with 12 \u2126 and 2 \u2126 resistors.\nSolution:\n(i) The circuit diagram is shown in figure.\nTotal resistance, R = 1\u2126 + 2\u2126 = 3\u2126\nPotential difference, V = 6 V", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(ii) a 4 V battery in parallel with 12 \u2126 and 2 \u2126 resistors.\nSolution:\n(i) The circuit diagram is shown in figure.\nTotal resistance, R = 1\u2126 + 2\u2126 = 3\u2126\nPotential difference, V = 6 V\n\nPower used in 2\u2126 resistor = I2R = (2)2 x 2 = 8 W\n(ii) The circuit diagram for this case is shown :\nPower used in 2 resistor = \\frac { { v }^{ 2 } }{ R } =\\frac { { 4 }^{ 2 } }{ 2 } = 8 W.\n\n[ \u2235 Current is different for different resistors in parallel combination.]\nQuestion 15\nTwo lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V ?", + "Question 15\nTwo lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V ? [CBSE 2018]\nSolution:\nPower of first lamp (P1) = 100 W\nPotential difference (V) = 220 V\n\nQuestion 16\nWhich uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes ?\nSolution:\nEnergy used by 250 W TV set in 1 hour = 250 W x 1 h = 250 Wh\nEnergy used by 1200 W toaster in 10 minutes = 1200 W x 10 min", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nSolution:\nEnergy used by 250 W TV set in 1 hour = 250 W x 1 h = 250 Wh\nEnergy used by 1200 W toaster in 10 minutes = 1200 W x 10 min\n= 1200 x \\frac { 10 }{ 60 } = 200 Wh 60\nThus, the TV set uses more energy than the toaster.\nQuestion 17\nAn electric heater of resistance 8 \u2126 draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.\nSolution:\nHere, R = 8 \u2126, 1 = 15 A, t = 2 h\nThe rate at which heat is developed in the heater is equal to the power.\nTherefore, P = I2 R = (15)2 x 8 = 1800 Js-1\nQuestion 18\nExplain the following:\n(i) Why is tungsten used almost exclusively for filament of electric lamps ?\n(ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal ?", + "(ii) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal ?\n(in) Why is the series arrangement not used for domestic circuits ?\n(iv) How does the resistance of a wire vary with its area of cross-section ?", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(in) Why is the series arrangement not used for domestic circuits ?\n(iv) How does the resistance of a wire vary with its area of cross-section ?\n(v) Why are copper and aluminium wires usually employed for electricity transmission?\nAnswer:\n(i) The tungsten is used almost exclusively for filament of electric lamps because it has a very high melting point (3300\u00b0C). On passing electricity through tungsten filament, its temperature reaches to 2700\u00b0C and it gives heat and light energy without being melted.\n(ii) The conductors of electric heating devices such as bread-toasters and electric irons, are made of an alloy rather than a pure metal because the resistivity of an alloy is much higher than that of pure metal and an alloy does not undergo oxidation (or burn) easily even at high temperature.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(iii) The series arrangement is not used for domestic circuits because in series circuit, if one electrical appliance stops working due to some defect, than all other appliances also stop working because the whole circuit is broken.\n(iv) The resistance of a wire is inversely proportional to its area of cross-section, i.e., Resistance R \u221d (1/\u03c0r2). If the area of cross section of a conductor of fixed length is increased, then resistance decreases because there are more free electrons for movement in conductor.\n(v) Copper and aluminium wires usually employed for electricity transmission because they have very low resistances. So, they do not become too hot on passing electric current.\nNCERT Solutions for Class 10 Science Chapter 12 Electricity", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 12 Electricity\nElectric current, potential difference and electric current, Ohms law, Resistance, Resistivity factors on which the resistance of a conductor depends; Series combination of resistors, parallel combination of resistors; and its application on daily life; Heating effect of Electric current, electric Power, Interrelation between P, V, and R.\nFormulae Handbook for Class 10 Maths and Science\nPage 200\nWhat does an electric circuit mean?\nElectric circuit is a continuous and closed path made of conducting wires, through which the electric current flows. It comprises a cell, ammeter, voltmeter, plug key, etc.\nDefine the unit of current.\nSI unit of electric current is ampere (A).\nAmpere is the flow of electric charges through an area at the rate of one coulomb per second, i.e. if 1 coulomb of electric charge flows through a cross-section of wire for 1 second, then it would be equal to 1 ampere.\n\nPage 202\nQuestion 1:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nPage 202\nQuestion 1:\nName a device that helps to maintain a potential difference across a conductor.\nAnswer:\nCell or battery eliminator.\nQuestion 2:\nWhat is meant by saying that the potential difference between two points is 1 V?\nAnswer:\nAs we know that V = W / q\nThus, the potential difference between two points is one volt when one joule of work is done to carry a charge of one coulomb between the two points in the electric field.\n\nQuestion 3:\nHow much energy is given to one coulomb of charge passing through a 6 V battery\nAnswer:\n\nPage 209\nQuestion 1:\nOn what factors does the resistance of a conductor depend\nAnswer:\nResistance of a conductor depends upon:\n(i) Resistivity of the material.\n(ii) Length of the conductor.\n(iii) Cross-sectional area of the conductor.\nQuestion 2:\nWill current flow more easily through a thick wire or thin wire of the same material when connected to the same source? Why\nAnswer:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(iii) Cross-sectional area of the conductor.\nQuestion 2:\nWill current flow more easily through a thick wire or thin wire of the same material when connected to the same source? Why\nAnswer:\nThe current flows more easily through a thick wire than through a thin wire because the resistance of thick wire is less than that of a thin wire as R \u221d 1/A.\nDownload NCERT Solutions for Class 10 Science Chapter 12 Electricity PDF\nQuestion 3:\nLet the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?\nAnswer:\n\nHence, the current through an electrical component also becomes half of its previous value.\nQuestion 4:\nWhy are the coils of electric toasters and electric irons made of an alloy rather than a pure metal\nAnswer:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nQuestion 4:\nWhy are the coils of electric toasters and electric irons made of an alloy rather than a pure metal\nAnswer:\nThe coils of electric toaster and electric iron are made of an alloy rather than a pure metal because of the following reasons;\n(i) The resistivity of an alloy is higher than that of a pure metal.\n(ii) It has high melting point and does not oxidise.\nQuestion 5:\nUse the data in Table 12.2 of NCERT book to answer the following:\n(a) Which among iron and mercury is a better conductor?\n(b) Which material is the best conductor? \u2018\nAnswer:\n(a) Iron because its resistivity is less than mercury.\n(b) Silver is the best conductor as it has least resistivity.\nPage 213\nQuestion 1:\nDraw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 \u2126 resistor, a 8 \u2126 resistor and a 12 \u2126 resistor and a plug key, all connected in series.\nAnswer:\n\nQuestion 2:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\n\nQuestion 2:\nRedraw the circuit of the above question, putting in an ammeter to measure the current through the resistors and a voltmeter to measure the voltage across the 12 resistor. What would be the reading in the ammeter and the voltmeter?\nAnswer:\n\nTotal resistance of the circuit = R\nSince all the three resistors are connected in series, so, the equivalent resistance R is equal to the sum of all resistance.\nR = 5 \u2126 + 8 \u2126 + 12 \u2126 = 25 \u2126\n\nPage 216\nQuestion 1:\nJudge the equivalent resistance when the following are connected in parallel.\n(a) 1 \u2126 and 106 \u2126\n(b) 1 \u2126 , 103 \u2126 and 106 \u2126\nAnswer:\nEquivalent resistance in parallel combination of resistors is always less than the least resistance of any resistor in the circuit.\nHence, in both the given cases, the equivalent resistance is less than 1 \u2126.\nQuestion 2:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nHence, in both the given cases, the equivalent resistance is less than 1 \u2126.\nQuestion 2:\nAn electric lamp of 100 \u2126, a toaster of resistance 50 \u2126 and a water filter of resistance 500 \u2126 are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances and what is the current flows through it?\nAnswer:\n\nQuestion 3:\nWhat are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series ?\nAnswer:\nAdvantages of connecting electrical devices in parallel:\nWhen the appliances are connected in parallel with the battery, each appliance gets the same potential difference as that-of battery which is not possible in series connection.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nWhen the appliances are connected in parallel with the battery, each appliance gets the same potential difference as that-of battery which is not possible in series connection.\nEach appliance has different resistances and requires different currents to operate properly. This is possible only in parallel connection, as in series connection, same current flows through all devices, irrespective of their resistances.\nIf one appliance fails to work, other will continue to work properly.\nQuestion 4:\nHow can three resistors of resistances 2 \u2126, 3 \u2126 and 6 \u2126 be connected to give a total resistance of (a) 4 \u2126 (b) 1 \u2126?\nAnswer:\n(a) In order to get 4 \u2126, resistance 2 \u2126 should be connected in series with the parallel combination of 3 \u2126 and 6 \u2126.\n\nQuestion 5:\nWhat is (a) the highest (b) the lowest total resistance that can be secured by combination of four coils of resistances 4 \u2126, 8 \u2126, 12 \u2126, 24 \u2126?\nAnswer:\n(a) The highest resistance is secured by combining all four coils of resistance in series.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\n(a) The highest resistance is secured by combining all four coils of resistance in series.\nRs = 4 \u2126+ 8 \u2126 + 12 \u2126 + 24 \u2126 = 48 \u2126\n(b) The lowest resistance is secured by combining all four coils of resistance in parallel.\n\nPage 218\nQuestion 1:\nWhy does the cord of an electric heater not glow while the heating element does?\nAnswer:\nThe cord of an electric heater is made up of metallic wire such as copper or aluminum which has low resistance while the heating element is made up of an alloy which has more resistance than its constituent metals. Also heat produced \u2018H\u2019 is\nH = I2Rt\nThus, for the same current H oc R, so for more resistance, more heat is produced by heating element and it glows.", + "Also heat produced \u2018H\u2019 is\nH = I2Rt\nThus, for the same current H oc R, so for more resistance, more heat is produced by heating element and it glows.\nQuestion 2:\nCompute the heat generated while transferring 96000 C of charge in one hour through a potential difference of 50 V.\nAnswer:\n\nQuestion 3:\nAn electric iron of resistance 20 Q takes a current of 5 A. Calculate the heat developed in 30 s.\nAnswer:\nGiven R = 20 \u2126, I = 5 A, t = 30 s", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\n\nQuestion 3:\nAn electric iron of resistance 20 Q takes a current of 5 A. Calculate the heat developed in 30 s.\nAnswer:\nGiven R = 20 \u2126, I = 5 A, t = 30 s\nH = I2Rt = (5)2 x 20 x 30 = 15000 J = 1.5 x 104 J\nPage 220\nQuestion 1:\nWhat determines the rate at which energy is delivered by a current?\nAnswer:\nElectric power determines the rate at which energy is delivered by a current.\nQuestion 2:\nAn electric motor takes 5 A from a 220 V line. Determine the power of the motor and the energy consumed in 2 h.\nAnswer:\nGiven I = 5 A, V = 220 V, t = 2 h Power,\np = VI = 220 x 5 = 1100 W\nEnergy consumed = Vlt = Pt\n= 1100 x 2 = 2200 Wh\nTextbook Questions\nQuestion 1:\nA piece of wire of resistance R is cut into five equal parts.", + "These parts are then connected in parallel. If the equivalent resistance of this combination is R\u2019, then the ratio R/R\u2019 is\n\nAnswer:\n\nQuestion 2:\nWhich of the following terms does not represent electrical power in a circuit?\n(a) I2R\n(b) IR2\n(c) VI\n(d) V2/R\nAnswer:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\n\nQuestion 2:\nWhich of the following terms does not represent electrical power in a circuit?\n(a) I2R\n(b) IR2\n(c) VI\n(d) V2/R\nAnswer:\n(b) P = V2/R = I2R = VI Option (b) does not represent electrical power.\nQuestion 3:\nAn electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be\n(a) 100 W\n(b) 75 W\n(c) 50 W\n(d) 25 W\nAnswer:\n\nQuestion 4:\nTwo conducting wires of same material and of equal lengths and diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be\n(a) 1:2\n(b) 2:1\n(c) 1:4\n(d) 4:1\nAnswer:\n\nQuestion 5:\nHow is a voltmeter connected in the circuit to measure the potential difference between two points?\nAnswer:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(a) 1:2\n(b) 2:1\n(c) 1:4\n(d) 4:1\nAnswer:\n\nQuestion 5:\nHow is a voltmeter connected in the circuit to measure the potential difference between two points?\nAnswer:\nA voltmeter is connected in parallel across any two points in a circuit to measure the potential difference between them with its +ve terminal to the point at higher potential and -ve terminal to the point at lower potential of the source.\nQuestion 6:\nA copper wire has a diameter 0.5 mm and resistivity of 1.6 X 10-8 \u2126m. What will be the length of this wire to make its resistance 10 \u2126? How much does the resistance change if the diameter is doubled?\nAnswer:\n\n\nQuestion 7:\nThe values of the current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below:\n\nPlot a graph between V and I and calculate the resistance of that resistor.\nAnswer:\n\nQuestion 8:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nPlot a graph between V and I and calculate the resistance of that resistor.\nAnswer:\n\nQuestion 8:\nWhen a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.\nAnswer:\n\nQuestion 9:\nA battery of 9 V is connected in series with resistors of 0.2 \u2126, 0.3 \u2126, 0.4 \u2126, 0.5 \u2126 and 12 \u2126, respectively. How much current would flow through the 12 \u2126 resistor?\nAnswer:\nSince all the resistors are in series, the same current, 0.67 A flows through the 12 \u2126 resistor.\n\nQuestion 10:\nHow many 176 \u2126 resistors (in parallel) are required to carry 5 A on a 220 V line?\nAnswer:\n\nQuestion 11:\nShow how you would connect three resistors, each of resistance 6 \u2126 , so that the combination has a resistance of (i) 9 \u2126 , (it) 4 \u2126 .", + "Answer:\n\nQuestion 11:\nShow how you would connect three resistors, each of resistance 6 \u2126 , so that the combination has a resistance of (i) 9 \u2126 , (it) 4 \u2126 .\nAnswer:\n(i) When two 6 \u2126 resistances are in parallel and the third is in combination to this, the equivalent resistance will be 9 \u2126.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\n(i) When two 6 \u2126 resistances are in parallel and the third is in combination to this, the equivalent resistance will be 9 \u2126.\n\n(ii) When two 6 \u2126 resistances are in series and the third is in parallel to them, then it will be 4 \u2126.\n\nQuestion 12:\nSeveral electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?\nAnswer:\nSince, N bulbs of power P each connected in parallel will make the total power of NP,\n\nQuestion 13:\nA hot plate of an electric oven connected to 220 V line has two resistance coils A and B, each of 24 Q resistance, which may be used separately, in series, or in parallel. What are the currents in the three cases?\nAnswer:\n\nQuestion 14:\nCompare the power used in the 2 \u2126 resistor in each of the following circuits.\n(i) a 6 V battery in series with 1 \u2126 and 2 \u2126 resistors, and", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\n\nQuestion 14:\nCompare the power used in the 2 \u2126 resistor in each of the following circuits.\n(i) a 6 V battery in series with 1 \u2126 and 2 \u2126 resistors, and\n(ii) a 4 V battery in parallel with 12 \u2126 and 2 \u2126 resistors.\nAnswer:\n\nQuestion 15:\nTwo lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?\nAnswer:\n\nQuestion 16:\nWhich uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?\nAnswer:\nEnergy consumed by 250 W TV set in 1 h = 250 x 1 = 250 Wh.\nEnergy consumed by 1200 W toaster in 10 min = 1200 X 1/6 = 200 Wh.\n\u2234 Energy consumed by TV set is more than the energy consumed by toaster in the given timings.", + "Energy consumed by 1200 W toaster in 10 min = 1200 X 1/6 = 200 Wh.\n\u2234 Energy consumed by TV set is more than the energy consumed by toaster in the given timings.\nQuestion 17:\nAn electric heater of resistance 8 f2 draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.\nAnswer:\n\nQuestion 18:\nExplain the following.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nQuestion 17:\nAn electric heater of resistance 8 f2 draws 15 A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.\nAnswer:\n\nQuestion 18:\nExplain the following.\n(a) Why is the tungsten used almost exclusively for filament of electric lamps?\n(b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?\n(c) Why is the series arrangement not used for domestic circuits?\n(d) How does the resistance of a wire vary with its area of cross-section?\n(e) Why are copper and aluminum wires usually employed for electricity transmission?\nAnswer:\n(a) It has high melting point and emits light at a high temperature.\n(b) It has more resistivity and less temperature coefficient of resistance.\n(c) (i) All appliances do not get same potential in series arrangement.\n(ii) All appliances cannot be individually operated.\n(d) R \u221d =1 / Area of cross \u2013 section.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(c) (i) All appliances do not get same potential in series arrangement.\n(ii) All appliances cannot be individually operated.\n(d) R \u221d =1 / Area of cross \u2013 section.\n(e) They are very good conductors of electricity.\nShort Answer Type Questions\nQuestion 1:\nThree 2 \u2126 resistors, A, B and C are connected as shown in figure. Each of them dissipates energy and can withstand a maximum power of 18 W without melting. Find the maximum current that can flow through the three resistors.\nAnswer:\n\nQuestion 2:\nShould the resistance of an ammeter be low or high? Give reason.\nAnswer:\nThe resistance of an ammeter should be low so that it will not disturb the magnitude of current flowing through the circuit when connected in series in a circuit.\nQuestion 3:\nHow does use of a fuse wire protect electrical appliances?\nAnswer:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nQuestion 3:\nHow does use of a fuse wire protect electrical appliances?\nAnswer:\nThe fuse wire is always connected in series with the live wire or electrical devices. If the flow of current exceeds the specified preset value due to some reason, the heat produced melts it and disconnects the circuit or the device from the mains. In this way, fuse wire protects the electrical appliances.\nQuestion 4:\nWhat is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?\nAnswer:\nThe resistance offered by a metallic wire of unit length and unit cross-sectional area is called electrical resistivity.\n\nHence, when the length of wire is doubled, the resistance becomes double and current decreases to half.\nQuestion 5:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nHence, when the length of wire is doubled, the resistance becomes double and current decreases to half.\nQuestion 5:\nA current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 \u2126 when connected to a 10 V battery. Calculate the resistance of the electric lamp.\nNow if a resistance of 10 \u2126 is connected in parallel with this series combination, what change (if any) in current flowing through 5 \u2126 conductor and potential difference across the lamp will take place? Give reason.\nAnswer:\n\n\nQuestion 6:\nWhy is parallel arrangement used in domestic wiring?\nAnswer:\nParallel arrangement is used in domestic wiring because\n(i) Each appliance gets the same voltage as that of the mains supply.\n(ii) If one component is switched off, others can work properly.\n(iii) Fault in any branch of the circuit can be easily identified.\nQuestion 7:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(ii) If one component is switched off, others can work properly.\n(iii) Fault in any branch of the circuit can be easily identified.\nQuestion 7:\nB1, B2 and B3 are three identical bulbs connected as shown in figure. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.\n\n(i) What happens to the glow of the other two bulbs when the bulb B j gets fused?\n(ii) What happens to the reading of A1 ,A2 , A3 and A when the bulb B2 gets fused?\n(iii) How much power is dissipated in the circuit when all the three bulbs glow together?\nAnswer:\n(i) Since B1 ,B2 and B3 are in parallel, the potential difference across each of them will remain same. So when the bulb B1 gets fused, B2 and B3 have the same potential and continues with the same energy dissipated per second, i.e. they will glow continuously as they were glowing before.\n\n\nLong Answer Type Questions\nQuestion 1:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nLong Answer Type Questions\nQuestion 1:\nThree incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit, another set of three bulbs of the same wattage are connected in parallel to the same source.\n(\u0430) Will the bulb in the two circuits glow with the same brightness? Justify your answer.\n(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason.\nAnswer:\n\n(a) The bulbs in the two circuits will not glow equally bright as the current through them is not the same.\n\n(b) As one bulb fuses, the other bulbs in the series circuit will not glow because the circuit becomes an open circuit. While the rest of bulbs in parallel circuit will continue to glow without getting disturbed because in parallel combination, current gets additional paths to flow.\nQuestion 2:\nFind out the following in the electric circuit given in figure:\n\nAnswer:", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nQuestion 2:\nFind out the following in the electric circuit given in figure:\n\nAnswer:\n\n(e) No difference, since the ammeters are connected in series and same current will pass through them, so reading of both ammeters will be same.\nMultiple Choice Questions (MCQs) [1 Mark each]\nQuestion 1.\nTo determine the equivalent resistance of two resistors when connected in series, a student arranged the circuit components as shown in the diagram. But he did not succeed to achieve the objective. [CCE 2010]\n\nWhich of the following mistakes has been committed by him in setting up the circuit?\n(a) Position of ammeter is incorrect\n(b) Position of voltmeter is incorrect\n(c) Terminals of ammeter are wrongly connected\n(d) Terminals of voltmeter are wrongly connected\nAnswer:\n(c) Because positive terminal of ammeter must be connected with positive terminal of cell and negative terminal of an ammeter must be connected to negative terminal of a cell.\nQuestion 2.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAnswer:\n(c) Because positive terminal of ammeter must be connected with positive terminal of cell and negative terminal of an ammeter must be connected to negative terminal of a cell.\nQuestion 2.\nFor the given circuit, name the components which are connected in parallel. [CCE 2011]\n\n(a) R1 and R2\n(b) R1, R2 and V\n(c) R2 and V\n(d) R1 and V\nAnswer:\n(b) The components R1, R2 and V are connected in parallel combination. Because terminals of the resistance and voltmeter are connected together.\nQuestion 3.\nA student arranges the following circuit to get equivalent resistance of a series combination of two resistors R1 and R2.\n\nWhich one of the following statements will be true for this circuit [CCE 2007]\n(a) It gives incorrect reading for current I as well as potential difference V\n(b) It gives correct reading for current I but incorrect reading for potential difference V\n(c) It gives correct reading for potential difference V but incorrect reading for current I", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(b) It gives correct reading for current I but incorrect reading for potential difference V\n(c) It gives correct reading for potential difference V but incorrect reading for current I\n(d) It gives correct reading for both I and V\nAnswer:\n(b) The voltmeter should be connected across the components of and R2 to give correct reading for potential difference.\nQuestion 4.\nAn ammeter has 20 divisions between 0 mark and 2A mark on its scale. The least count of ammeter is\n(a) 0.01A\n(b) 0.2A\n(c) 0.1A\n(d) 1A\nAnswer:\n(c) Number of divisions = 20\nMaximum reading of ammeter = 2 A\nLeast count of ammeter = 2/20 = 1/10 = 0.1 A\nQuestion 5.\nA student finds that there are 20 divisions between zero mark and 1V mark of a voltmeter.", + "A student finds that there are 20 divisions between zero mark and 1V mark of a voltmeter. The least count of voltmeter is\n(a) 0.1 V\n(b) 0.01 V\n(c) 0.05 V\n(d) 1.0 V\nAnswer:\n(c) Number of divisions = 20\nMaximum reading of the voltmeter = 1 V\nLeast count of voltmeter = 1/20 = 0.05 V\nQuestion 6.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(a) 0.1 V\n(b) 0.01 V\n(c) 0.05 V\n(d) 1.0 V\nAnswer:\n(c) Number of divisions = 20\nMaximum reading of the voltmeter = 1 V\nLeast count of voltmeter = 1/20 = 0.05 V\nQuestion 6.\nThe current flowing through a resistor connected in an electric circuit and the potential difference applied across its ends are shown in figure alongside.\n\nThe value of the resistance of the resistor is [CCE2013]\n(a) 1 \u2126\n(b) 5 \u2126\n(c) 8 \u2126\n(d) 10 \u2126\nAnswer:\n(d) Reading from ammeter (7) = 180 mA= 0.18 A,\nreading from voltmeter (V) = 1.8 V\nResistance of the resistor R = V/I = 1.8/0.18 = 180/18 = 10\u2126\nQuestion 7.", + "Which of the following is the correct method to connect the ammeter and voltmeter with resistance in the circuit to verify Ohm\u2019s law? [CCE 2012]\n(a) Ammeter and voltmeter in series\n(b) Ammeter in series and voltmeter in parallel\n(c) Ammeter in parallel and voltmeter in series\n(d) Ammeter and voltmeter in parallel\nAnswer:\n(b) In a circuit, ammeter should be connected in series, while voltmeter in parallel.", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(c) Ammeter in parallel and voltmeter in series\n(d) Ammeter and voltmeter in parallel\nAnswer:\n(b) In a circuit, ammeter should be connected in series, while voltmeter in parallel.\nQuestion 8.\nIn an experiment on studying the dependence of the current I flowing through a given resistor on the potential difference V applied across it, a student has to change the value of the current. For doing this, he should change the\n(a) number of cells used\n(b) resistor itself\n(c) ammeter used in the circuit\n(d) Voltmeter used in the circuit\nAnswer:\n(a) If we change the number of cells in electric circuit, the potential difference will change and as a result current flowing in the circuit changes.\nQuestion 9.\nA milliammeter had graduations marked 0, 100, 200, 300, 400 and 500. The space between 0 mark and 100 mark is divided into 20 divisions.", + "Question 9.\nA milliammeter had graduations marked 0, 100, 200, 300, 400 and 500. The space between 0 mark and 100 mark is divided into 20 divisions. If the pointer of the milliammeter is indicating the seventh graduation after 300 mark, the current flowing in the circuit is\n(a) 335 mA\n(b) 330 mA\n(c) 331mA", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(a) 335 mA\n(b) 330 mA\n(c) 331mA\n(d) 340 mA\nAnswer:\n(a) Number of divisions = 20\nLeast count of milliammeter = (100-0) / 20 = 5 mA\nMilliammeter reading = 300 + 7 x 5 = 335 mA\nQuestion 10.\nIf a student while studying the dependence of current on the potential difference keeps the circuit closed for a long time to measure the current and potential difference, then\n(a) ammeter\u2019s zero error will change\n(b) ammeter will give more reading\n(c) voltmeter will show constantly higher readings\n(d) resistor will get heated up and its value will change\nAnswer:\n(d) If the circuit is closed for a long time, then current flows in it for a long time which results that the resistor is heated.\nQuestion 11.", + "Question 11.\nTo determine the eguivalent resistance of two resistors connected in series, a student prepared two electric circuits, correct reading of ammeter in the circuits is [CCE 2015]\n\n(a) In circuit I, 1.0 A and in II, 0.1 A\n(b) In both circuits I and II, 1.0 A\n(c) In circuit I, 0.1 A and in II, 1.0 A", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(a) In circuit I, 1.0 A and in II, 0.1 A\n(b) In both circuits I and II, 1.0 A\n(c) In circuit I, 0.1 A and in II, 1.0 A\n(d) In both circuits I and II, 0.1 A\nAnswer:\n(b) Equivalent resistance of two resistors 3.5\u2126 and 1\u2126 in both the circuits I and II is R = 3.5 + 1 = 4.5 \u2126\nAs, I = V/R = 4.5/4.5 = 1A\nTherefore, current in both the circuits I and II is 1.0 A.\nQuestion 12.\nWhen parallel resistors are of three different values, the potential difference across its terminals is [CCE 2015]\n(a) greatest across smallest resistance\n(b) greatest across largest resistance\n(c) equal across each resistance\n(d) least across the smallest resistance\nAnswer:\n(c) Potential difference across each resistor is same in parallel combination of resistors.", + "Class 10 Science Electricity Mind Map\nElectricity\nStudy of Electric Charges at Rest and in Motion\nCharge", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n(c) Potential difference across each resistor is same in parallel combination of resistors.\nClass 10 Science Electricity Mind Map\nElectricity\nStudy of Electric Charges at Rest and in Motion\nCharge\nSomething associated with the matter due to which it produces and experiences electric and magnetic effects. Resides on the outer surface of the conductor.\nQ = ne S.I. unit coulomb (C)\nElectric Current (I)\nThe time rate of flow of charge (Q) through any cross-section\nI = \\frac{Q}{t} S.I. unit ampere (A)\nTypes of Current\nDirect Current\nCurrent whose magnitude and direction does not vary with time.\nAlternating Current\nCurrent whose magnitude and direction periodically changes with time.\nElectric Potential\nWork done per unit charge\nV = \\frac{W}{Q}\nS.I. unit volt\nOhm\u2019s law: If the physical conditions remain same, current I \u221d V => V = IR\nR-electric resistance Substances which obey ohm\u2019s law called ohmic and that do not obey called non-ohmic substances.\nDependence of Resistance", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nR-electric resistance Substances which obey ohm\u2019s law called ohmic and that do not obey called non-ohmic substances.\nDependence of Resistance\nOn length (l) and area of cross-section (A)\nR \u221d l\n\u221d \\frac{1}{A}\nR = \\rho \\frac{l}{A}\n\u03c1 = resistivity\nResistivity depends on the material of the conductor only.\nOn Temperature\nRt = R0( 1 + \u03b1t)\n\u03b1 = temperature coefficient of resistance\nResistance (R): Obstruction offered to flow of electrons.\nSI unit ohm\nResistance, R \u221d \\frac{\\ell^{2}}{m}\nl = length and\nm = mass of conducting wire\nAfter stretching, if length increases by n times then resistance will increase by n2 times i.e., R2 = n2 R1.", + "SI unit ohm\nResistance, R \u221d \\frac{\\ell^{2}}{m}\nl = length and\nm = mass of conducting wire\nAfter stretching, if length increases by n times then resistance will increase by n2 times i.e., R2 = n2 R1. Similarly\nif radius be reduced to \\frac{1}{n} times then area of cross-section decreases \\frac{1}{n^{2}} times so the resistance becomes n4 times i.e.. R2 = n4 R1\nAfter stretching, if length of a conductor increases by x%, then resistance will increase by 2x% (valid only if x< 10%).", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAfter stretching, if length of a conductor increases by x%, then resistance will increase by 2x% (valid only if x< 10%).\nUsing n conductors of equal resistance, the number of possible combinations is 2n-1 .\nIf the resistances of n conductors are totally different, then the number of possible combinations will be 2n .\nIf n identical resistances are first connected in series and then in parallel, the ratio of the equivalent resistance is given by\n\\frac{R_{s}}{R_{p}}=\\frac{n^{2}}{1}\nIf a wire of resistance R is cut in n equal parts and then these parts are collected to form a bundle, then equivalent resistance of combination will be \\frac{\\mathrm{R}}{\\mathrm{n}^{2}}\nIf equivalent resistance of R1 and R2 in series and parallel be Rs and Rp respectively, then\nR1 = \\frac{1}{2}\\left[\\mathrm{R}_{\\mathrm{s}}+\\sqrt{\\mathrm{R}_{\\mathrm{s}}^{2}-4 \\mathrm{R}_{\\mathrm{s}} \\mathrm{R}_{\\mathrm{p}}}\\right]", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nR1 = \\frac{1}{2}\\left[\\mathrm{R}_{\\mathrm{s}}+\\sqrt{\\mathrm{R}_{\\mathrm{s}}^{2}-4 \\mathrm{R}_{\\mathrm{s}} \\mathrm{R}_{\\mathrm{p}}}\\right]\nand R2 = =\\frac{1}{2}\\left[\\mathrm{R}_{\\mathrm{s}}-\\sqrt{\\mathrm{R}_{\\mathrm{s}}^{2}-4 \\mathrm{R}_{\\mathrm{s}} \\mathrm{R}_{\\mathrm{p}}}\\right]\nGrouping of Resistances\nSeries Grouping of Resistances\nEquivalent resistance, resistance,", + "resistance, Rs = R1 + R2 + \u2026 + Rn\nIn this case same current flows through each resistance but potential difference in the ratio of resistance\nParallel Grouping of Resistances\n\\frac{1}{R_{P}}=\\frac{1}{R_{1}}+\\frac{1}{R_{2}}+\\ldots+\\frac{1}{R_{n}}\n\nIn this case same potential across each resistance but current distributes in the reverse ratio of their resistances\nElectric Circuit\nThe arrangement of various electrical components along which electric current flow\n\nHeating Effect of Electric Current\nAs current flows through a conductor, the free electrons lose energy which is converted into heat.\nJoule\u2019s heating law\nH \u221d I2\nH \u221d R\nH \u221d t", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nHeating Effect of Electric Current\nAs current flows through a conductor, the free electrons lose energy which is converted into heat.\nJoule\u2019s heating law\nH \u221d I2\nH \u221d R\nH \u221d t\nH = I2Rt = VIt\nPractical Applications\nElectric heater, electric iron and water heater, etc. work on the principle of heating effect of current\nElectric bulb glows when electric current flows through the filament of the bulb\nElectric Power\nRate at which electric energy is dissipated or consumed in a circuit,\nP = VI ,\nor P = I2R = \\frac{\\mathrm{V}^{2}}{\\mathrm{R}}\nWatt is a smaller unit of power, its other bigger units are kilowatt (KW),\nMegawatt (MW) and Horsepower (HP)\n1 KW = 103W 1 MW = 106W\n1 hp =746 W\nThe commercial unit of electrical energy is 1 Kwh.\n1 Kwh = 3.6 \u00d7 106J\nElements of Circuit\nCell\nDirect current source of electromotive force.", + "1 Kwh = 3.6 \u00d7 106J\nElements of Circuit\nCell\nDirect current source of electromotive force. Combination of two or more cells is called battery.\n\nRheostat", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\n1 Kwh = 3.6 \u00d7 106J\nElements of Circuit\nCell\nDirect current source of electromotive force. Combination of two or more cells is called battery.\n\nRheostat\nWire of special type of alloy like manganin, Eureka, nichrome, etc. is wound on a hollow cylinder of china clay. It controls the current in the electric circuit.\n\nSwitch\nIt is used to close or open the electric circuit, controls the movement of electrons in a circuit.\n\nVoltmeter\nMeasures the potential difference between two points in the circuit. Its resistance is high and it is used in parallel with the resistance wire.\n\nFuse\nIt is a safety device having very thin wire which is made up of either tin or alloy of tin and lead.\nThis wire has low melting point so it melts and breaks the circuit easily if the current in the circuit exceeds.\n\nAmmeter\nMeasures the value of current flowing in the circuit.\nThe resistance of ammeter is small and it is used in series with the circuit.\n\nLED", + "CHAPTER 11-ELECTRCITY-INTEXT QUESTIONS.txt\nAmmeter\nMeasures the value of current flowing in the circuit.\nThe resistance of ammeter is small and it is used in series with the circuit.\n\nLED\nIt is a device which glows even if a weak electric current is allowed to flow through it", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-1\nDefine a solenoid. Compare the magnetic field produced by a solenoid with that of a bar magnet?\nSolution:\nA coil of many circular turns of wire wrapped in the shape of a cylinder, is called a solenoid. The magnetic field lines in a solenoid, through which current is passed, is very similar to that of a bar magnet. One end of the coil acts like a magnetic north pole, while the other acts like a south pole. The magnetic field produced by a long solenoid has all the properties of the field produced by a bar magnet.\nQuestion-2\nGive one important advantage of AC over DC\nSolution:\nA.C can be stepped up and stepped down which means that the voltage can be increased or decreased. Hence it can be transmitted to long distances without much loss of energy. So A.C is preferred over D.C.\n\nQuestion-3\nGive the circuit symbol for a fuse. Explain its importance in a circuit.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-3\nGive the circuit symbol for a fuse. Explain its importance in a circuit.\nSolution:\n\nA fuse is a very important device used for protecting electric circuits. It is a wire made out of a metal like tin or tin alloy having a very low melting point. When a high current flows through a circuit, the fuse wire gets heated or melts due to short circuiting or overloading. Hence the circuit is broken and the current stops flowing. This saves all the appliances of the circuit.\nFuse wires are of various capacities. A fuse with 5 ampere capacity will be thinner than a fuse with 15 ampere capacity. A fuse of 5 amps is used in circuits where lights and fans are connected whereas a fuse of 15 amps is used in power circuits where appliances like electric heater, geyser, electric iron and air conditioner are connected.\nQuestion-4\nGive a note on Magnetism in Human beings.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-4\nGive a note on Magnetism in Human beings.\nSolution:\nWhenever there is an electric current, there is a magnetic field. Even the extremely weak ion currents that travel along the nerve cells in our body produce magnetic fields. When we try to touch something, our nerves carry an electric impulse to the muscles we need to use. This impulse creates a temporary magnetic field. These fields are about one billionth as weak as the Earth\u2019s field. Two main organs in the human body where the magnetic field produced is significant are heart and brain.\nQuestion-5\nPQ is conductor \u00d7 represents magnetic is \u22a5 to the paper field and into the plane of the paper.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-5\nPQ is conductor \u00d7 represents magnetic is \u22a5 to the paper field and into the plane of the paper.\nSolution:\nFleming\u2019s left hand rule gives the direction of force experienced by a current carrying conductor kept in a magnetic field. According to it, when the thumb, first finger and second finger of the left hand are kept perpendicular to each other such that the first finger points towards the direction of magnetic field, the central finger is along the direction of current, then the thumb shows the direction of the force acting on the conductor.\n\nQuestion-6\nDefine Electromotive force.\nSolution:\nThe motion of a magnet, with respect to the coil, produces an induced potential difference. This induced potential difference is called electromotive force which sets up an induced electric current in the circuit. The motion of a magnet, with respect to the coil, produces an induced potential difference.\nQuestion-7\nWhat is meant by earthing? Why should electrical appliances be earthed?", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-7\nWhat is meant by earthing? Why should electrical appliances be earthed?\nSolution:\nThe metal body of appliances like fridge, cooler, mixer etc. are connected to a an earth wire so that any leakage of current to the body of the appliance goes to the earth and does not give electric shock. This is called earthing. It is used as a safety measure in order to prevent electric shocks to the users.\nQuestion-8\nWhat is a solenoid?\nSolution:\nA solenoid is a long cylindrical conductor coil, having a large number of turns of insulated copper wire.\nQuestion-9\nState Fleming\u2019s right Hand Rule. Give the principle, construction and working of the AC generator with a simple diagram. What modification will you suggest so that the output is DC\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nState Fleming\u2019s right Hand Rule. Give the principle, construction and working of the AC generator with a simple diagram. What modification will you suggest so that the output is DC\nSolution:\nAccording to Fleming\u2019s right hand rule, when the thumb and the central finger of right hand are kept perpendicular to each other, the thumb shows the direction of motion of the conductor, the first finger the direction of magnetic field when the current induced is in the direction of central finger.\n\nAC generator\nPrinciple: It works on the principle of electromagnetic induction. Induced current is produced, whenever current is produced.\nConstruction: A generator consists of mainly four parts, namely coil, magnets, slip rings and brushes just like an electric motor.\nCoil: A large number of insulated copper wires wound on a rectangular frame.\nMagnets: A large permanent magnet to provide a strong a magnetic field.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nCoil: A large number of insulated copper wires wound on a rectangular frame.\nMagnets: A large permanent magnet to provide a strong a magnetic field.\nSlip rings: Two solid rings connected the two ends of the coil used to convey the current produced to outside circuit.\nBrushes: Two carbon brushes remain in sliding contact with slip rings.\nWorking: The coil of the generator is rotated with the help of an axel. When coil rotates, it cuts through the magnetic filed of the magnet. So a current is induced in the coil by electromagnetic induction. The direction of this current is given by Fleming\u2019s right hand rule.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nAs the coil turns clockwise, arm AB moves up and arm CD goes down. The direction of the current is from A to B and C to D. When after half rotation CD starts going up and AB starts coming down, the direction of the current in the coil also reverses. Now it is from D to C and from B to A. This alternating current with the help of slip rings which are in sliding contact with brushes B1 and B2 is given out to the circuit. Hence the current produced by the generator is alternating and such a generator is called AC generator.\nTo get direct current in place of slip rings, split rings are used so that one brush is always in contact with the arm that goes downward. Then the current given out to the outer circuit is in the same direction. This type of generator is called DC generator.\nQuestion-10\nHow can you convert an A.C. into a D.C. generator?\nSolution:\nAn A.C. generator can be converted into a D.C. generator by replacing the solid ring arrangement with split ring arrangement.\nQuestion-11", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nHow can you convert an A.C. into a D.C. generator?\nSolution:\nAn A.C. generator can be converted into a D.C. generator by replacing the solid ring arrangement with split ring arrangement.\nQuestion-11\nWhat is a magnetic field?\nSolution:\nThe region around a magnet, in which the magnetic force of attraction and repulsion is felt, is called a magnetic field.\nQuestion-12\nDistinguish between a solenoid and a bar magnet. Draw the magnetic lines for both\nSolution:\n\nThe solenoid is a long coil containing a large number of close turns of insulated copper wire. The magnetic field produced by the current carrying solenoid is similar to the magnetic field produced by a bar magnet. A solenoid is used for making electromagnets.\nDifferences between a bar magnet and solenoid:\nBar magnet\nIt is a permanent magnet.\nThe strength of a bar magnet cannot be changed.\nThe polarity (North \u2013 South) of a bar magnet cannot be changed.\nSolenoid", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nBar magnet\nIt is a permanent magnet.\nThe strength of a bar magnet cannot be changed.\nThe polarity (North \u2013 South) of a bar magnet cannot be changed.\nSolenoid\nIt is a temporary magnet. It acts as a magnet only as long as the current passes through it.\nThe strength of a solenoid can be changed by changing the number of turns in its coil or by changing the current passing through it.\nThe polarity of a solenoid can be changed by changing the direction of current in its coil.\nQuestion-13\nWhat is electromagnetic induction? Explain how the movement of a magnet towards or away from a coil carrying a galvanometer produce current? Write the rule to find the direction of current in this above coil.\nSolution:\nWhenever the magnetic field through a conductor changes, and induced current and e. m. f. is set up in the conductor. This is known as electromagnetic induction.\nQuestion-14\nWhich effect of electric current is utilized in the working of an electric fuse?\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-14\nWhich effect of electric current is utilized in the working of an electric fuse?\nSolution:\nAn electric fuse works on the heating effect of current.\nQuestion-15\nWhat will you do if you see a person coming in contact with a live wire?\nSolution:\nSuch a person should be provided with an insulated support of wood, plastic or rubber.\nQuestion-16\nName an instrument in which the directive property of a magnet is used.\nSolution:\nCompass needle makes use of the directive property of a magnet.\nQuestion-17\nName the elements of Earth\u2019s magnetic field.\nSolution:\nThe elements of Earth\u2019s magnetic field are angle of dip, declination and horizontal component of earth\u2019s magnetic field.\nQuestion-18\nExplain why, two magnetic lines of force do not intersect.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-18\nExplain why, two magnetic lines of force do not intersect.\nSolution:\nThe magnetic lines of force do not intersect one another due to the fact that the resultant force on a north pole at any point can be only in one direction. But if the two magnetic lines of force intersect one another, then the resultant force on a north pole placed at the point of intersection will be along directions, which is not possible.\nQuestion-19\nState the right hand thumb rule.\nSolution:\n\nIf you hold the thumb, the forefinger and the centre finger of your right hand at right angles to one another. Adjust you hand in such a way that forefinger points in the direction of magnetic field, and the thumb points in the direction of motion of conductor, then the direction in which centre finger points, gives the direction of induced current in the conductor.\nQuestion-20\nWhat is the cause of earth\u2019s magnetism?\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-20\nWhat is the cause of earth\u2019s magnetism?\nSolution:\nEarth\u2019s magnetism is due to the magnetic effect of current which is flowing in the liquid core at the center of the earth.\nQuestion-21\nHow will you find out the direction of the magnetic field produced by current-carrying conductor?\nSolution:\nThe direction of lines of force of the magnetic field produced by a straight wire carrying current is obtained by Maxwell\u2019s right hand thumb rule. According to Maxwell\u2019s right-hand thumb rule, \u201cImagine that the current carrying wire is in the right hand so that the thumb points in the direction of current, then the direction in which the fingers encircle the wire gives the direction of magnetic lines of force around the wire.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nImagine a current carrying wire AB in which the current is flows vertically upwards. To find out the direction of magnetic lines of force produced by this current, we imagine the wire AB to be held in the right hand, so that the thumb points in the direction of current towards A. Now, the direction in which the fingers are folded gives the direction of the lines of force. In this case the fingers are folded in the anti-clockwise direction, so the magnetic lines of force are also in the anti-clockwise direction.\nQuestion-22\nWhat type of core should be put inside a current-carrying solenoid to make an electromagnet?\nSolution:\nA soft iron core is placed inside a solenoid to make an electromagnet. When a soft iron core is placed inside a solenoid, then the strength of the magnetic field becomes very large because the iron core gets magnetized by induction. This combination of a solenoid and a soft iron core is called an electromagnet.\nQuestion-23", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-23\nDistinguish between a bar magnet and an electromagnet.\nSolution:\nBar Magnets\nThe bar magnet is a permanent magnet.\nIt produces a comparatively weak force of attraction.\nThe strength of a bar magnet cannot be changed.\nThe polarity of a bar magnet is fixed and cannot be changed.\nElectromagnets\nAn electromagnet is a temporary magnet.\nIt produces a very strong magnetic force.\nThe strength of an electromagnet can be changed by changing the number of turns in its coil or by changing the current passing through it.\nThe polarity of an electromagnet can be changed by changing the direction of current in its coil.\nQuestion-24\nState the composition of the alloy called nipermag? Give an important use of this alloy.\nSolution:\nNipermag is an alloy of iron, nickel, aluminium and titanium. Permanent magnet of this alloy is more stronger than those made of ordinary steel. Hence it used in microphones and loudspeakers.\nQuestion-25", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-25\nDerive the formula for the force acting on a charged particle moving in a magnetic field.\nSolution:\nThe force acting on a current-carrying conductor placed in a magnetic field is,\nF = B \u00d7 I \u00d7 L\nThe current I is the rate of flow of charge.\nNow, if a charge Q flows in time t then the current I = Q/t. Hence substituting for I in\nthe above equation, we get,\nF = (B \u00d7 Q \u00d7 L)/t\nSuppose the particle carrying the charge Q travels a length L in time t, then the velocity\nv = L/t. So substituting this value, we get\nForce on moving charge F = B \u00d7 Q \u00d7 v.\nQuestion-26\nHow does alternating current differ from the direct current?\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nv = L/t. So substituting this value, we get\nForce on moving charge F = B \u00d7 Q \u00d7 v.\nQuestion-26\nHow does alternating current differ from the direct current?\nSolution:\nIf current always flows in the same direction, it is called a direct current. The current, which we get from the cell or a battery, is a direct current because it always flows in the same direction. The positive and negative polarity of a direct current is fixed. If the current changes direction after regular intervals of time, it is called alternating current. Most of the power stations in India generate alternating current. The alternating current produced in India changes its direction every 1/100 second. Thus, the positive and negative polarity of an alternating current is not fixed.\nQuestion-27\nGive two reasons why different electrical appliances in a domestic circuit are connected in parallel.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-27\nGive two reasons why different electrical appliances in a domestic circuit are connected in parallel.\nSolution:\n(i) If one of the appliances is switched off or gets fused, there is no effect on the other appliances and they keep on operating.\n(ii) The same voltage of the main line is available for all electrical appliances.\nQuestion-28\nWhy is a fuse wire made of a tin-lead alloy and not copper?\nSolution:\nA fuse wire is made of tin alloy because it has low melting point, so that it may melt easily, whereas a copper wire cannot be used as a fuse wire because it has a high melting point due to which it will not melt easily when a short circuit takes place.\nQuestion-29\nExplain the principle and working of an electric motor with the help of a diagram. What is the function of a split ring commutator?\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-29\nExplain the principle and working of an electric motor with the help of a diagram. What is the function of a split ring commutator?\nSolution:\nAn electric motor converts electrical energy into mechanical energy. It works on the principle that \u2013 a current carrying conductor placed in a magnetic field experiences a force.\nFollowing are the essential parts of an electric motor.\n(i) Coil: It is a rectangular coil of insulated copper wire having large number of turns.\n(ii) A large permanent magnet provides strong magnetic field between its pole pieces. The coil rotates between these pole pieces.\n(iii) Split rings: The two ends of the coil are connected to two split rings, which are two halves of a slip rings.\n(iv) Brushes: Two carbon brushes keep in sliding contact with split rings.\n\nWorking", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nWorking\nWhen a current is passed through the coil, the direction of current in AB and CD is in opposite direction but both are perpendicular to magnetic field. Therefore, by Fleming\u2019s left hand rule AB arm of the coil experiences an upward force and CD arm experiences a downward force. These two forces being equal and opposite to each other form a couple which rotates the coil. Arms BC and DA are parallel to the field and the force on them is zero. The forces, on AB and CD turn the coil in clockwise direction. After half revolution, the split rings change their position. Now S2 is in contact with brush B1 and S1 is in contact with B2. So the direction of current in the coil reverses. Therefore, AB now experiences downward force and CD upward force. The couple now acting on the coil again moves it in clockwise direction. Due to the function of split ring commutator and brushes, coil continues to turn in clockwise direction.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nSplit ring commutator changes direction after every half rotation, so that the direction of current going in the coil also reverses and the arm of the coil which goes up in the first half, goes down in second half. As a result, the coil continues to rotate in one direction. Anything connected to the axis of the coil also rotates. So, the electrical energy given to the coil changes into mechanical energy.\nQuestion-30\nWith the help of a labelled diagram, explain the working of an A.C. generator.\nSolution:\n\u201cA. C. generator\u201d means \u201cAlternating Current generator\u201d. That is, an A. C. generator produces alternating current, which alternates (changes) in polarity continuously. We will now describe the construction an working of the A. C. generator or A. C. dynamo.\nConstruction of an A. C. generator", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nConstruction of an A. C. generator\nA simple A. C. generator consists of a rectangular coil ABCD that can be rotated rapidly between the poles N and S of a strong horseshoe type magnet M. The coil is made of a large number of turns of insulated copper wire. The ends A and D of the rectangular coil are connected to two circular pieces of copper metal called slip rings R1 and R2. As the slip rings R1 and R2 rotate with the coil, the two pieces of carbon called brushes, B1 and B2, keep contact with them. So, the current produced in the rotating coil can be tapped out through slip rings into the carbon brushes. From the carbon brushes B1 and B2 we take the current into various electrical appliances like radio, T. V., electric iron, bulbs, etc. But in this figure, we have shown only a galvanometer G connected the two carbon brushes.\nWorking of an A. C. generator", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nWorking of an A. C. generator\nSuppose that the generator coil ABCD is initially in the horizontal position. Again suppose that he coil ABCD is being rotated in the anticlockwise direction between the poles N and S of a horseshoe type magnet.\n(i) As the coil rotates in the anticlockwise direction, the side AB of the coil moves down cutting the magnetic lines of force near the N-pole of the magnet, and side CD moves up, cutting the lines of force near the S-pole of the magnet. Due to this, induced current is produced in the sides AB and DC of the coil. On applying Fleming\u2019s right hand rule to the side AB and DC of the coil, we find that the currents are in the direction B to A and D to C respectively. Thus, the induced currents in the two sides of the coil are in the same direction, and we get an effective induced current in the direction BADC.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\n(ii) After half revolution, the sides AB and DC of the coil will interchange their positions. The side AB will come on the right hand side and DC will come on the left side. So, after half a revolution, side AB starts moving up and side DC starts coming down. As a result of this, the direction of induced current in each side of the coil is reversed after half a revolution. Since the direction of induced current in the coil is reversed after half revolution so the polarity (positive and negative) of the two ends of the coil also changes after half revolution. The end of coil which was positive in the first half of rotation becomes negative in the second in the second half. And the end which was negative in the first half revolution becomes positive in the second half of revolution. Thus, in 1 revolution of the coil, the current changes its direction 2 times.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nThe alternating current (A. C.) produced in India has a frequency of 50 Hz. That is, the coil is rotated at the rate of 50 revolutions per second. Since in 1 revolution of coil, the current changes its direction 2 times, so in 50 revolutions of coil, the current changes its direction 2 \u00d7 50 = 100 times. Thus, the A. C. supply in India changes its direction 100 times in 1 second. Another way of saying this is that the alternating current produced in India changes its direction every 1/100 second. That is, each terminal of the coil is positive (+) for 1/100 of a second and negative (-) for the next 1/100 of a second. This process is repeated again and again with the result that there is actually no positive and negative in an A. C. generator. We will now describe why the direction of induced current in the coil of an A. C. generator changes after every half revolution of the coil.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nAfter every half revolution, each side of the generator coil starts moving in the opposite direction in the magnetic field. The side of the coil which was initially moving downwards in a magnetic field, after half revolution, it starts moving in opposite direction \u2013 upwards. Similarly the side of coil which was initially moving upwards, after half revolution, it starts moving downwards. Due to the change in the direction of motion of the two sides of the coil in the magnetic field after every half revolution, the direction of current produced in them also changes after every half revolution.\nQuestion-31\nExplain the principle, construction and working of a DC Motor.\nSolution:\n\u201cD. C. generator\u201d means \u201cDirect Current generator\u201d. That is, a D. C. generator\nproduces direct current and not alternating current. We will now describe the construction and working of D. C. generator or D. C. Dynamo.\nConstruction of a D. C. generator", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nA simple D. C. generator consists of a rectangular coil ABCD which cab be rotated rapidly between the poles N and S of a strong horse-shoe type magnet M. The generator coil is made of a large number of turns of insulated copper wire. The two ends of the coil are connected to the two copper half rings (or split rings) R1 and R2 of a commutator. There are two carbon brushes B1 and B2 which press lightly against the two half rings. When the coil is rotated, the two half rings R1 and R2 touch the two carbon brushes B1 and B2 one by one. So the current produced in the rotating coil can be tapped out through the commutator half rings into the carbon brushes. From the carbon brushes B1 and B2, we can take the current into the various electrical appliances like radio, T. V., electric iron, bulbs, etc. But in this figure, we have shown only a galvanometer G connected between the two carbon brushes. The galvanometer is a current detecting and current measuring instrument.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nWorking of a D. C. generator\nSuppose that the generator coil ABCD is initially in the horizontal position. Again suppose that he coil ABCD is being rotated in the anticlockwise direction between the poles N and S of a horseshoe type magnet.\nAs the coil rotates in the anticlockwise direction, the side AB of the coil moves down cutting the magnetic lines of force near the N-pole of the magnet, and side DC moves up, cutting the lines of force near the S-pole of the magnet. Due to this, induced current is produced in the sides AB and DC of the coil. On applying Fleming\u2019s right hand rule to the side AB and DC of the coil we find that the currents in them are in the direction B to A and D to C respectively. Thus, the induced currents in the two sides of the coil are in the same direction, and we get an effective induced current in the direction BADC. Due to this the brush B1 becomes a positive (+) pole and brush B2 becomes negative (-) pole of the generator.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nAfter half revolution, the sides AB and DC of the coil will interchange their positions. The side AB will come on the right hand side and start moving up whereas side DC will come on then the two commutator half rings R1 and R2 automatically change their contacts from one carbon brush to the other. Due to this change, the current keeps flowing in the same direction in the other circuits. The brush B1 always remaining positive terminal and brush B2 always remaining negative terminal of the generator. Thus, a D. C. generator supplies a current in one direction by the use of a commutator consisting of two, half-rings of copper. In the above discussion we have used the word D. C. generator everywhere. Please note that we can also write D. C. dynamo in place of D. C. generator.\nQuestion-32\nWhat is a fuse wire? What is the advantage and disadvantage of using a thick fuse wire?\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-32\nWhat is a fuse wire? What is the advantage and disadvantage of using a thick fuse wire?\nSolution:\nA fuse is a very important device used for protecting electric circuits. It is a wire made out of a metal like tin or tin alloy having a very low melting point.\nWhen a high current flows through a circuit, the fuse wire gets heated or melts due to short-circuiting or overloading. Hence the circuit is broken and the current stops flowing. This saves all the appliances of the circuit.\nQuestion-33\nThe device used for producing current is called a,\n(i) Generator\n(ii) Voltmeter\n(iii) Ammeter\n(iv) Galvanometer.\nSolution:\n(i) Generator. The other devices are measuring instruments.\nQuestion-34\nWhat are magnetic field lines? How is the direction of a magnetic field at a point determined? Mention two important properties of the magnetic field lines.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-34\nWhat are magnetic field lines? How is the direction of a magnetic field at a point determined? Mention two important properties of the magnetic field lines.\nSolution:\nThe space surrounding a magnet in which magnetic force is exerted, is called a magnetic field. Magnetic field lines are the lines that are drawn at every point indicating the direction in which a north pole would move if placed at that point. They are determined by placing an imaginary hypothetical north pole at that point and finding the direction in which it would move due to the magnetic field at that point. A compass needle gets deflected when placed near a magnet due to the magnetic force exerted by the magnet on it.\n\nSome important properties of magnetic field lines are;\n(i) The tangent drawn at any point on the field line indicates the direction in which a north pole would move if placed at that point.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nSome important properties of magnetic field lines are;\n(i) The tangent drawn at any point on the field line indicates the direction in which a north pole would move if placed at that point.\n(ii) The relative strength of the field is proportional to the degree of closeness of the lines. The more clustered they are, the stronger the field in that region.\n(iii) The magnetic field lines never intersect. This is because a pole can move only in zone direction and if the lines intersect they would have to move in two direction simultaneously which is impossible.\nQuestion-35\nDraw a rough sketch of the pattern of field lines due to a\n(i) current flowing into a circular coil and\n(ii) solenoid carrying current.\nSolution:\n\nQuestion-36\nState the rule to determine the direction of a\n(i) magnetic field produced around a straight conductor-carrying current,\n(ii) force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\n(i) magnetic field produced around a straight conductor-carrying current,\n(ii) force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and\n(iii) current induced in a coil due to its rotation in a magnetic field.\n(iv) Current induced in a circuit by the changing magnetic flux due to the motion of a magnet.\nSolution:\n(i) The direction of a magnetic field produced around a current-carrying conductor can be obtained by using Maxwell\u2019s right-hand thumb rule. It states that \u201cif you hold the current carrying wire in your right hand with your thumb pointing in the direction of the magnetic field then the fingers will wrap around the conductor in the direction of the magnetic field lines due to the conductor\u201d.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\n(ii) The direction of the force experienced by a straight conductor carrying current placed in a magnetic field is determined using Fleming\u2019s left hand rule. It states that \u201cif you stretch the forefinger, the central finger and the thumb of your left hand mutually perpendicular to each other, the forefinger points in the direction of the magnetic field and the central finger points in the direction of current, and the thumb points in the direction of force acting on the conductor\u201d.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\n(iii) The direction of the current induced in a circuit by changing the magnetic flux due to motion of a conductor is given by Fleming\u2019s right hand rule. It states that \u201cif you hold the forefinger, the central finger and the thumb of your right hand mutually perpendicular to each other, the forefinger indicates the direction of the changing field / flux, the thumb indicates the direction of motion of the conductor and the middle finger gives the direction of the induced current\u201d. This phenomenon is called electromagnetic induction.\nQuestion-37\nOn what factors does the force experienced by a current\u2013carrying conductor placed in a uniform magnetic field depend?\nSolution:\nThe force on a current carrying conductor placed in a magnetic field is given by the Fleming\u2019s left hand rule.\nF = B I L\nWhere F is the force on the conductor\nB is the magnitude of the uniform magnetic field\nI is the current in the conductor\nL is the Length of the current carrying wire", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nF = B I L\nWhere F is the force on the conductor\nB is the magnitude of the uniform magnetic field\nI is the current in the conductor\nL is the Length of the current carrying wire\nTherefore from the above formula, force is directly proportional to the magnitude of the field, current in the wire and the length of the wire.\nQuestion-38\nExplain the principle and working of an electric motor with the help of a diagram.What is the function of a split\u2013ring commutator ?\nSolution:\n\nA motor is a device that converts the electrical energy into mechanical energy.\nPrinciple", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nSolution:\n\nA motor is a device that converts the electrical energy into mechanical energy.\nPrinciple\nAn electric motor is based on the fact that when a current carrying conductor is placed in a magnetic field the conductor experiences a force which is given by Fleming\u2019s Left Hand Rule. For example, when a rectangular coil is placed in the magnetic field and current is passed through it, a torque acts on the coil, which rotates it continuously. When the coil rotates, the shaft attached to it also rotates and therefore the electrical energy supplied to the motor is converted into the mechanical energy of rotation.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nAn electrical motor consists of a rectangular coil ABCD of insulated copper wire, wound on a soft iron core called armature. The coil is mounted between the poles of a magnet in such a way that it can rotate between the poles N and S. The two ends of the coil are soldered to the ends of a commutator whose main function is to reverse the direction of the current flowing through the coil every time the coil just passes the vertical position during its revolution.\nWorking\nSuppose the coil ABCD is initially at a horizontal position. When the switch is in ON position the current enters the coil through the carbon brushes and the half ring \u2018A\u2019 of the commutator.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nWorking\nSuppose the coil ABCD is initially at a horizontal position. When the switch is in ON position the current enters the coil through the carbon brushes and the half ring \u2018A\u2019 of the commutator.\nThe current flows in the direction DCBA and leaves via the half ring \u2018B\u2019. In the side PQ of the coil, the direction is from Q to P towards the south and the direction of the magnetic field is from the N to S pole towards the east. So, by applying Fleming\u2019s left hand rule, we find that it will experience a force in upward direction. Similarly, the side SR of the coil will experience a downward force. Thus we have two parallel wires experiencing forces in opposite directions. They form a couple tending to rotate the coil in the anticlockwise direction.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nWhen the coil goes beyond the vertical position, the two commutator half rings automatically changes contact from one brush to the other. This reverses the direction of current through the coil which, in turn, reverses the direction of forces acting on the two sides of the coil. The sides of the coil are interchanged, but rotate in the same anticlockwise direction. This process is repeated again and again and the coil continues to rotate as long as the current is passing.\nQuestion-39\nA coil of copper wire is connected to a galvanometer. What would happen if a bar magnet is\n(i) Pushed into the coil with its north pole entering first?\n(ii) Pulled out of the bar magnet?\n(iii) Held stationary inside the coil?\nSolution:\n(i) A deflection is observed in the galvanometer due to the induced current because of the changing magnetic flux (increasing) through the turns of the coil connected to the galvanometer.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nSolution:\n(i) A deflection is observed in the galvanometer due to the induced current because of the changing magnetic flux (increasing) through the turns of the coil connected to the galvanometer.\n(ii) A deflection is again observed in the galvanometer, as when it is pulled out, the flux linked with the coil due to the bar magnet decreases. Hence a current flows in the coil to reduce the change in flux. The deflection can be observed in the opposite direction as compared with the previous case.\n(iii) No deflection is observed in the galvanometer. The flux linked with the coil due to the magnetic field is at a constant. Hence no current is induced due to the bar magnet.\nQuestion-40\nDraw a labelled diagram to explain the principle underlying the working of an electric generator.\nSolution:\n\nQuestion-41\nWhat is the function of an earth wire? Why is it necessary to earth the metallic appliances?\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nSolution:\n\nQuestion-41\nWhat is the function of an earth wire? Why is it necessary to earth the metallic appliances?\nSolution:\nTo avoid electric shocks, the metal body of an electrical device is \u2018earthed\u2019. A wire called \u2018earth wire\u2019 is used to connect the metal body of the electrical device to the earth, which is at zero potential. In household circuits, we have three wires, the live wire, the neutral wire and the earth wire. One end of the earth wire is connected to the device and the other end of the wire is connected to the earth. We now say that the device is \u201cearthed\u201d or \u201cgrounded\u201d. Usually the three wires are connected to a three-pin plug. The neutral wire or the earth connection carries the high current to the earth from the device and prevents an electric shock.\nQuestion-42\nExplain what is short-circuiting and overloading in an electric supply.\nSolution:\nShort circuiting", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-42\nExplain what is short-circuiting and overloading in an electric supply.\nSolution:\nShort circuiting\nIf the plastic insulation of the live wire and neutral wire gets torn, then the two wires touch each other. This touching of the live wire and neutral wire directly is known as short-circuiting. The current passing through the circuit formed by these wires is very large and consequently a high heating effect is created which may lead to fire.\nOverloading\nThe current flowing in domestic wiring at a particular time depends on the power ratings of the appliances being used. If too many electrical appliances of high power rating are switched on at the same time, they draw an extremely large current from the circuit. This is known as overloading. Due to this large current flowing through them, the copper wires of household wiring get heated to a very high temperature and may lead to fire.\nQuestion-43\nDescribe an experiment to illustrate the action of an electric fuse.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXTRA QUESTIONS.txt\nQuestion-43\nDescribe an experiment to illustrate the action of an electric fuse.\nSolution:\nTake a thin fuse wire made of tin or tin-alloy having low melting point. Place this fuse wire on the porcelain fuse grip and insert the grip into the fuse holder. Now switch on all the electrical appliances of high power rating like electric iron, water heater, air conditioner, etc.,. Since the melting point of the fuse wire is much lower, it melts and breaks the circuit.\nQuestion-44\nImagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of the magnetic field?\nSolution:\nvertically downwards.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXEMPLER QUESTIONS.txt\nNCERT Exemplar Class 10 Science Chapter 13 Magnetic Effects of Electric Current\nShort Answer Type Questions\n1.A magnetic compass needle is placed in the plane of paper near point A as shown in figure. In which plane should a straight current carrying conductor be placed so that it passes through A and there is no change in the deflection of the compass? Under what condition is the deflection maximum and why?\n\nAnswer. In the given situation, a straight current-carrying conductor should be placed in the plane of paper. The held produced by it is perpendicular to the plane of paper and parallel to the vertical axis of compass needle. As a result there will be no change in the deflection of compass needle.\n\nThe deflection is maximum when the straight current carrying conductor through the point A is perpendicular to the plane of paper so that field due to it is parallel to the plane of paper.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXEMPLER QUESTIONS.txt\nThe deflection is maximum when the straight current carrying conductor through the point A is perpendicular to the plane of paper so that field due to it is parallel to the plane of paper.\n2. Under what conditions permanent electromagnet is obtained if a current carrying solenoid is used? Support your answer with the help of a labelled circuit diagram.\nAnswer. Required conditions are:\n(i) The magnitude of direct current through the solenoid should be large.\n(ii)The number of turns in the solenoid should be large and closely packed, so that a strong uniform magnetic field inside it is produced.\n\n(iii) The rod kept inside is made of a magnetic material such as steel.\n\n3.AB is a current carrying conductor in the plane of the paper as shown in figure. What are the directions of magnetic fields produced by it at points P and Q? Given r1 > r2, where will the strength of the magnetic field be larger?", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXEMPLER QUESTIONS.txt\nAnswer. Apply right hand thumb rule, the direction of magnetic field produced by a current carrying conductor at Point P \u2013 into the plane of paper\nPoint Q \u2013 out of the plane of paper.\nAs magnetic field is inversely proportional to distance, the field at Q will be larger than the field at P ( r1 > r2).\n4.It is established that an electric current through a metallic conductor produces a magnetic field around it. Is there a similar magnetic field produced around a thin beam of moving (i) alpha particles, (ii) neutrons? Justify your answer.\nAnswer.\n(i) Yes, similar magnetic field is produced around a thin beam of moving a-particle. a-particle being positively charged constitutes a current in the direction of beam motion.\n(ii) No, neutron being electrically neutral, does not constitute a current.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXEMPLER QUESTIONS.txt\n(ii) No, neutron being electrically neutral, does not constitute a current.\n5.Name four appliances wherein an electric motor, a rotating device that converts electrical energy to mechanical energy, is used as an important component. In what respect motors are different from generators?\nAnswer.The four appliances wherein electric motor, a rotating device that converts electrical energy into mechanical energy is used are electric fan, water pump, grinder and washing machine.\nDifference between electric motor and generator:\n6.What is the role of the two conducting stationary brushes in a simple electric motor?\nAnswer.The two stationary conducting brushes make the contact in between the two halves of the split rings and the source battery. Current in the coil enters from the source battery through one conducting brush and flows back to the battery through another brush.\nLong Answer Type Question", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXEMPLER QUESTIONS.txt\nLong Answer Type Question\n1.Why does a magnetic compass needle pointing North and South in the absence of a nearby magnet get deflected when a bar magnet or a current carrying loop is brought near it. Describe some salient features of magnetic lines of field concept.\nAnswer.Both the bar magnet or a current carrying loop have their own magnetic field. These external magnetic fields modify the earth\u2019s existing magnetic field. This resultant magnetic field deflects the magnetic compass needle by exerting a magnetic force on it, from its original north-south direction.\nFeatures of magnetic field lines are: \u2018\nMagnetic field lines represent the magnetic field around the magnetic substance or current carrying conductor. They have both direction and magnitude.\nThe field lines emerge from North pole and merge at South pole. Inside the magnet their direction is from south pole to north pole. Hence magnetic field lines are a continuous closed curve.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-EXEMPLER QUESTIONS.txt\nThe field lines emerge from North pole and merge at South pole. Inside the magnet their direction is from south pole to north pole. Hence magnetic field lines are a continuous closed curve.\nThe degree of closeness of the field lines is relative to the strength of magnetic field. Strong magnetic field is indicated by crowded field lines or vice versa.\nNo two magnetic field lines intersect each other. If they intersect, the compass needle would point towards two different directions at a point, which is impossible.\nThe parallel and equidistant field lines in a given region indicate the presence of a uniform magnetic field in that region whereas converging or diverging field lines represent the non-uniform magnetic field.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nCBSE Class 10 Science Notes Chapter 13 Magnetic Effects of Electric Current\nMagnet: Magnetic field and magnetic field lines, Magnetic field due to a current carrying conductor, Right hand thumb rule, Magnetic field due to current through a circular loop. Magnetic field due to current in a solenoid.\nMagnet is an object that attracts objects made of iron, cobalt and nickle. Magnet comes to rest in North \u2013 South direction, when suspended freely.\nUse of Magnets: Magnets are used\nin refrigerators.\nin radio and stereo speakers.\nin audio and video cassette players.\nin children\u2019s toys and;\non hard discs and floppies of computers.\nProperties of Magnet\nA free suspended magnet always points towards the north and south direction.\nThe pole of a magnet which points toward north direction is called north pole or north-seeking.\nThe pole of a magnet which points toward south direction is called south pole or south seeking.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nThe pole of a magnet which points toward north direction is called north pole or north-seeking.\nThe pole of a magnet which points toward south direction is called south pole or south seeking.\nLike poles of magnets repel each other while unlike poles of magnets attract each other.\nMagnetic field: The area around a magnet where a magnetic force is experienced is called the magnetic field. It is a quantity that has both direction and magnitude, (i.e., Vector quantity).\n\nMagnetic field and field lines: The influence of force surrounding a magnet is called magnetic field. In the magnetic field, the force exerted by a magnet can be detected using a compass or any other magnet.\nThe magnetic field is represented by magnetic field lines.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nThe magnetic field is represented by magnetic field lines.\n\nThe imaginary lines of magnetic field around a magnet are called field line or field line of magnet. When iron fillings are allowed to settle around a bar magnet, they get arranged in a pattern which mimicks the magnetic field lines. Field line of a magnet can also be detected using a compass. Magnetic field is a vector quantity, i.e. it has both direction and magnitude.\nDirection of field line: Outside the magnet, the direction of magnetic field line is taken from North pole to South Pole. Inside the magnet, the direction of magnetic field line is taken from South pole to North pole.\nStrength of magnetic field: The closeness of field lines shows the relative strength of magnetic field, i.e. closer lines show stronger magnetic field and vice \u2013 versa. Crowded field lines near the poles of magnet show more strength.\nProperties of magnetic field lines\n(i) They do not intersect each other.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nProperties of magnetic field lines\n(i) They do not intersect each other.\n(ii) It is taken by convention that magnetic field lines emerge from North pole and merge at the South pole. Inside the magnet, their direction is from South pole to North pole. Therefore magnetic field lines are closed curves.\nMagnetic field lines due to current a current carrying straight conductor\nA current carrying straight conductor has magnetic field in the form of concentric circles, around it. Magnetic field of current carrying straight conductor can be shown by magnetic field lines.\nThe direction of magnetic field through a current carrying conductor depends upon the direction of flow electric current.\n\nLet a current carrying conductor be suspended vertically and the electric current is flowing from south to north. In this case, the direction of magnetic field will be anticlockwise. If the current is flowing from north to south, the direction of magnetic field will be clockwise.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nThe direction of magnetic field, in relation to direction of electric\ncurrent through a straight conductor can be depicted by using the Right Hand Thumb Rule. It is also known as Maxwell\u2019s Corkscrew Rule.\nRight-Hand Thumb Rule: If a current carrying conductor is held by right hand, keeping the thumb straight and if the direction of electric current is in the direction of thumb, then the direction of wrapping of other fingers will show the direction of magnetic field.\n\nMaxwell\u2019s Corkscrew rule: As per Maxwell\u2019s Corkscrew Rule, if the direction of forward movement of screw shows the direction of the current, then the direction of rotation of screw shows the direction of magnetic field.\n\nProperties of magnetic field\nThe magnitude of magnetic field increases with increase in electric current and decreases with decrease in electric current.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nProperties of magnetic field\nThe magnitude of magnetic field increases with increase in electric current and decreases with decrease in electric current.\nThe magnitude of magnetic field produced by electric current decreases with increase in distance and vice \u2013 versa. The size of concentric circles of magnetic field lines increases with distance from the conductor, which shows that magnetic field decreases with distance.\nMagnetic field lines are always parallel to each other.\nNo two field lines cross each other.\nMagnetic field lines due to a current through a circular loop\nIn case of a circular current carrying conductor, the magnetic field is produced in the same manner as it is in case of a straight current carrying conductor.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nIn case of a circular current carrying conductor, the magnetic field is produced in the same manner as it is in case of a straight current carrying conductor.\n\nIn case of a circular current carrying conductor, the magnetic field lines would be in the form of iron concentric circles around every part of the FllmSs periphery of the conductor. Since, magnetic field lines tend to remain closer when near to the conductor, so the magnetic field would be stronger near the periphery of the loop. On the other hand, the magnetic field lines would be distant from each other when we move towards the centre of the current carrying loop. Finally, at the centre, the arcs of big circles would appear as a straight line.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nThe direction of the magnetic field can be identified using Right Hand Thumb\u2019s Rule. Let us assume that the current is moving in anti-clockwise direction in the loop. In that case, the magnetic field would be in clockwise direction, at the top of the loop. Moreover, it would be in an anti-clockwise direction at the bottom of the loop.\nClock Face Rule: A current carrying loop works like a disc magnet. The polarity of this magnet can be easily understood with the help of Clock Face Rule. If the current is flowing in anti \u2013 clockwise direction, then the face of the loop shows north pole. On the other hand, if the current is flowing in clockwise direction, then the face of the loop shows south pole.\nMagnetic field and number of turns of coil: Magnitude of magnetic field gets summed up with increase in the number of turns of coil. If there are \u2018n\u2019 turns of coil, magnitude of magnetic field will be \u2018n\u2019 times of magnetic field in case of a single turn of coil.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nThe strength of the magnetic field at the centre of the loop(coil) depends on :\n(i) The radius of the coil: The strength of the magnetic field is inversely proportional to the radius of the coil. If the radius increases, the magnetic strength at the centre decreases\n(ii) The number of turns in the coil : As the number of turns in the coil increase, the magnetic strength at the centre increases, because the current in each circular turn is having the same direction, thus, the field due to each turn adds up.\n(iii) The strength of the current flowing in the coil: As the strength of the current increases, the strength of three magnetic fields also increases.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\n(iii) The strength of the current flowing in the coil: As the strength of the current increases, the strength of three magnetic fields also increases.\nMagnetic field due to a current in a Solenoid: Solenoid is the coil with many circular turns of insulated copper wire wrapped closely in the shape of a cylinder. A current carrying solenoid produces similar pattern of magnetic field as a bar magnet. One end of solenoid behaves as the north pole and another end behaves as the south pole.\n\nMagnetic field lines are parallel inside the solenoid, similar to a bar magnet, which shows that magnetic field is same at all points inside the solenoid.\nMagnetic field produced by a solenoid is similar to a bar magnet.\nThe strength of magnetic field is proportional to the number of turns and magnitude of current.\nBy producing a strong magnetic field inside the solenoid, magnetic materials can be magnetized. Magnet formed by producing magnetic field inside a solenoid is called electromagnet.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nBy producing a strong magnetic field inside the solenoid, magnetic materials can be magnetized. Magnet formed by producing magnetic field inside a solenoid is called electromagnet.\nElectromagnet, Fleming\u2019s Left-Hand Rule, Electric motor, Electromagnetic induction, Fleming\u2019s right hand rule, Electric generator and domestic electic circuits.\nElectromagnet: An electromagnet consists of a long coil of insulated copper wire wrapped on a soft iron.\nMagnet formed by producing magnetic field inside a solenoid is called electromagnet.\n\nForce on a current carrying conductor in a magnetic field: A current carrying conductor exerts a force when a magnet is placed in its vicinity. Similarly, a magnet also exerts equal and opposite force on the current carrying conductor. This was suggested by Marie Ampere, a French Physicist and considered as founder of science of electromagnetism.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nThe direction of force over the conductor gets reversed with the change in direction of flow of electric current. It is observed that the magnitude of force is highest when the direction of current is at right angles to the magnetic field.\nFleming\u2019s Left-Hand Rule: If the direction of electric current is perpendicular to the magnetic field, the direction of force is also perpendicular to both of them. The Fleming\u2019s Left Hand Rule states that if the left hand is stretched in a way that the index finger, the middle finger and the thumb are in mutually perpendicular directions, then the index finger and middle finger of a stretched left hand show the direction of magnetic field and direction of electric current respectively and the thumb shows the direction of motion or force acting on the conductor. The directions of electric current, magnetic field and force are similar to three mutually perpendicular axes, i.e. x, y, and z-axes.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nMany devices, such as electric motor, electric generator, loudspeaker, etc. work on Fleming\u2019s Left Hand Rule.\n\nElectric motor: A device that converts electrical energy to mechanical energy. It is of two types : AC and DC Motor.\nElectrical energy is converted into mechanical energy by using and electric motor. Electric motor works on the basis of rule suggested by Marie Ampere and Fleming\u2019s Left Hand Rule.\nPrinciple of Electric Motor: When a rectangular coil is placed in a magnetic field and a current is passed through it, force acts on the coil, which rotates it continuously. With the rotation of the coil, the shaft attached to it also rotates.\n\nConstruction: It consists of the following parts :\nArmature: It is a rectangular coil (ABCD) which is suspended between the two poles of a magnetic field.\nThe electric supply to the coil is connected with a commutator.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nArmature: It is a rectangular coil (ABCD) which is suspended between the two poles of a magnetic field.\nThe electric supply to the coil is connected with a commutator.\nCommutator or Split \u2013 ring: Commutator is a device which reverses the direction of flow of electric current through a circuit. It is two halves of the same metallic ring.\nMagnet: Magnetic field is supplied bv a permanent magnet NS.\nSliding contacts or Brushes Q which are fixed.\nBattery: These are consists of few cells.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nMagnet: Magnetic field is supplied bv a permanent magnet NS.\nSliding contacts or Brushes Q which are fixed.\nBattery: These are consists of few cells.\nWorking: When an electric current is supplied to the coil of the electric motor, it gets deflected because of magnetic field. As it reaches the halfway, the split ring which acts as commutator reverses the direction of flow of electric current. Reversal of direction of the current, reverses the direction of forces acting on the coil. The change in direction of force pushes the coil, and it moves another half turn. Thus, the coil completes one rotation around the axle. Continuation of this process keeps the motor in rotation.\nIn commercial motor, electromagnet instead of permanent magnet and armature is used. Armature is a soft iron core with large number of conducting wire turns over it. Large number of turns of conducting wire enhances the magnetic field produced by armature.\nUses of motors :\nUsed in electric fans.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nUses of motors :\nUsed in electric fans.\nUsed for pumping water.\nUsed in various toys.\nElectromagnetic Induction: Michael Faraday, an English Physicist is supposed to have studied the generation of electric current using a magnetic field and a conductor.\nElectricity production as a result of magnetism (induced current) is called Electromagnetic Induction.\n\nWhen a conductor is set to move inside a magnetic field or a magnetic field is set to be changing around a conductor, electric current is induced in the conductor. This is just opposite to the exertion of force by a current carrying conductor inside a magnetic field. In other words, when a conductor is brought in relative motion vis \u2013 a \u2013 vis a magnetic field, a potential difference is induced in it. This is known as electromagnetic induction.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nFleming\u2019s Right-Hand Rule: Electromagnetic induction can be explained with the help of Fleming\u2019s Right Hand Rule. If the right hand is structured in a way that the index (fore ginger) finger, middle finger and thumb are in mutually perpendicular directions, then the thumb shows direction of induced current in the conductor, in conductor The directions of movement of conductor, magnetic field and induced current can be compared to three mutually perpendicular axes, i.e. x, y and z axes.\n\nThe mutually perpendicular directions also point to an important fact that when the magnetic field and movement of conductor are perpendicular, the magnitude of induced current would be maximum.\nElectromagnetic induction is used in the conversion of kinetic energy into electrical energy.\nElectric Generator: A device that converts mechanical energy into electrical energy is called an electric generator.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nElectric Generator: A device that converts mechanical energy into electrical energy is called an electric generator.\nElectric generators are of two types: AC generator and a DC generator. Principle of electric generator: Electric motor works on the basis of electromagnetic induction.\n\nConstruction and Working: The structure of an electric generator is similar to that of an electric motor. In case of an electric generator, a rectangular armature is placed within the magnetic field of a permanent magnet. The armature is attached to wire and is positioned in a way that it can move around an axle. When the armature moves within the magnetic field, an electric current is induced. The direction of induced current changes, when the armature crosses the halfway mark of its rotation.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nThus, the direction of current changes once in every rotation. Due to this, the electric generator usually produces alternate current, i.e. A.C. To convert an A.C generator into a D.C generator, a split ring commutator is used. This helps in producing direct current.\nElectrical generator is used to convert mechanical energy into electrical energy.\nA.C and D.C Current\nA.C \u2013 Alternate Current: Current in which direction is changed periodically is called Alternate Current. In India, most of the power stations generate alternate current. The direction of current changes after every 1/100 second in India, i.e. the frequency of A.C in India is 50 Hz. A.C is transmitted upto a long distance without much loss of energy is advantage of A.C over D.C.\n\nD.C \u2013 Direct Current: Current that flows in one direction only is called Direct current. Electrochemical cells produce direct current.\nAdvantages of A.C over D.C\nCost of generatior of A.C is much less than that of D.C.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nAdvantages of A.C over D.C\nCost of generatior of A.C is much less than that of D.C.\nA.C can be easily converted to D.C.\nA.C can be controlled by the use of choke which involves less loss of power whereas, D.C can be controlled using resistances which involves high energy loss.\nAC can be transmitted over long distances without much loss of energy.\nAC machines are stout and durable and do not need much maintenance.\nDisadvantages of AC\nAC cannot be used for the electrolysis process or showing electromagnetism as it reverses its polarity.\nAC is more dangerous than DC.\nDomestic Electric Circuits: We receive electric supply through mains supported through the poles or cables. In our houses, we receive AC electric power of 220 V with a frequency of 50 Hz.\nThe 3 wires are as follows\nLive wire \u2013 (Red insulated, Positive)\nNeutral wire \u2013 (Black insulated, Negative)", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nThe 3 wires are as follows\nLive wire \u2013 (Red insulated, Positive)\nNeutral wire \u2013 (Black insulated, Negative)\nEarth wire \u2013 (Green insulated) for safety measure to ensure that any leakage of current to a metallic body does not give any serious shock to a user.\nShort Circuit: Short-circuiting is caused by the touching of live wires and neutral wire and sudden a large current flows.\nIt happens due to\ndamage pf insulation in power lines.\na fault in an electrical appliance.\nOverloading of an Electric Circuit: The overheating of electrical wire in any circuit due to the flow of a large current through it is called overloading of the electrical circuit.\nA sudden large amount of current flows through the wire, which causes overheating of wire and may cause fire also.\nElectric Fuse: It is a protective device used for protecting the circuit from short-circuiting and overloading. It is a piece of thin wire of material having a low melting point and high resistance.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nElectric Fuse: It is a protective device used for protecting the circuit from short-circuiting and overloading. It is a piece of thin wire of material having a low melting point and high resistance.\nFuse is always connected to live wire.\nFuse is always connected in series to the electric circuit.\nFuse is always connected to the beginning of an electric circuit.\nFuse works on the heating effect.\n\nMagnetic field: The area around a magnet in which other magnet feels force of attraction or repulsion is called Magnetic field.\nMagnetic field lines: The closed curved imaginary lines in the magnetic field which indicate the direction of motion of north pole in the magnetic field if a magnet is free to do so.\nProperties of magnetic field lines.\nMagnetic Field lines originate from the north pole of a magnet and end at its south pole.\nMagnetic Field lines are denser near the poles but rarer at other places.\nThe Magnetic Field lines do not intersect one another.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nMagnetic Field lines are denser near the poles but rarer at other places.\nThe Magnetic Field lines do not intersect one another.\nOersted\u2019s experiment: According to this experiment \u201cA current carrying wire creates a magnetic field around it. The direction of magnetic field depends on the direction of current in conductor.\u201d\nMagnetic field pattern due to straight current carrying conductor are concentric circles whose center lie on the wire.\nThe direction of magnetic field due to straight current carrying conductor can be determined by Right hand thumb rule.\nRight hand thumb rule: According to this rule \u201cif current carrying conductor is held in the right hand in such a way that thumb indicate the direction of current, then the curled finger indicates the direction of magnetic field lines around conductor.\u201d", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nMagnetic field pattern due to current carrying loop: The Magnetic field lines are circular near the current-carrying loop. As we move away from the loop, field lines form bigger and bigger circles. At the center of the circular loop, the magnetic field lines are straight.\nThe solenoid is an insulated and tightly wound long circular wire having large number of turns whose radius is small in comparison to its length. Magnetic field produced by a solenoid is similar to the magnetic field produced by a bar magnet.\nCurrent carrying solenoid is called an electromagnet.\nProperties of magnetic lines of force or magnetic field lines.\nThese lines originate from the north pole and end at the south pole.\nThe magnetic field lines of a magnet form a continuous closed loop.\nTwo magnetic lines of force do not intersect each other.\nThe tangent at any point on the magnetic line gives the direction of the magnetic field at the point.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nTwo magnetic lines of force do not intersect each other.\nThe tangent at any point on the magnetic line gives the direction of the magnetic field at the point.\nFleming\u2019s left hand rule: According to this rule, \u201cif the thumb, forefinger and middle finger of the left hand are stretched perpendicular to each other and if the fore-finger gives the direction of magnetic field, middle finger gives the direction of current, then the thumb will give the direction of motion or the force acting on the current-carrying conductor.\u201d\nPrinciple of an electric motor: A motor works on the principle that when a rectangular coil is placed in a magnetic field and current passes through it, a force acts on the coil which rotates it continuously.\nWhen the coil rotates, the shaft attached to it also rotates. In this way the electrical energy supplied to the motor is converted into the mechanical energy of rotation.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nWhen the coil rotates, the shaft attached to it also rotates. In this way the electrical energy supplied to the motor is converted into the mechanical energy of rotation.\nPrinciple of an electric generator: It is based on the principle of electromagnetic induction. It states that \u201can induced current is produced in a coil placed in a region where the magnetic field changes with time.\u201d The direction of induced current is given by Fleming\u2019s right-hand rule. An electric generator converts mechanical energy into electrical energy.\nElectromagnetic induction: The phenomenon of setting up of an electric current or an induced e.mi. by changing the magnetic lines of force by a moving conductor is called electromagnetic induction.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nElectromagnetic induction: The phenomenon of setting up of an electric current or an induced e.mi. by changing the magnetic lines of force by a moving conductor is called electromagnetic induction.\nMaxwell\u2019s right hand thumb rule: The direction of the current is given by Maxwell\u2019s right-hand thumb rule, \u201cIf the current carrying conductor is gripped with the right hand in such a way that the thumb gives the direction of the current, then the direction of the fingers gives the direction of the magnetic field produced around the conductor.\nFleming\u2019s left-hand rule: The direction of motion of a conductor in a magnetic field is given by Fleming\u2019s left-hand rule. According to this rule, if the thumb, forefinger and middle finger of the left hand are stretched perpendicular to each other and if fore-finger gives the direction of the magnetic field and the middle finger gives the direction of current then, the thumb will give the direction of the motion of the conductor carrying the current.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-NOTES.txt\nFleming\u2019s right-hand rule: The direction of the induced current is given by Fleming\u2019s right-hand rule. According to this rule if the thumb, forefinger and middle finger of the right hand are stretched perpendicular to each other and if the fore-finger gives the direction of the magnetic field and the thumb gives the direction of motion, then the middle finger will give the direction of the induced current in the conductor.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nImportant Questions of Magnetic Effects of Electric Current Class 10 Science Chapter 13\nQuestion 1.\nWhat is meant by magnetic field?\nAnswer:\nMagnetic field : It is defined as the space surrounding the magnet in which magnetic force can be experienced.\nQuestion 2.\nDraw magnetic field lines around a bar magnet. Name the device which is used to draw magnetic field lines. (Board Term I, 2015)\nAnswer:\n\nCompass needle is used to draw magnetic field lines.\nQuestion 3.\nDesign an activity to demonstrate that a bar magnet has a magnetic field around it. (Board Term I, 2017)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nAnswer:\n\nCompass needle is used to draw magnetic field lines.\nQuestion 3.\nDesign an activity to demonstrate that a bar magnet has a magnetic field around it. (Board Term I, 2017)\nAnswer:\nOne can easily demonstrate the presence of field lines around a bar magnet using compass needles. Place the magnet on a white sheet and mark its boundaries on sheet. Place the compass near the north pole of magnet and mark the position of needle. Now move the compass such that its south pole occupies the position previously occupied by its north pole. Repeat this step several times and you will have pattern as shown in the figure.\n\nRepeat the above procedure and draw as many lines as you can. These lines represent the magnetic field around the magnet. These are known as magnetic field lines.\nQuestion 4.\nWhat are magnetic field lines? Justify the following statements:\n(a) Two magnetic field lines never intersect each other.\n\n(b) Magnetic field are closed curves. (Board Term I, 2016)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nWhat are magnetic field lines? Justify the following statements:\n(a) Two magnetic field lines never intersect each other.\n\n(b) Magnetic field are closed curves. (Board Term I, 2016)\nAnswer:\nImaginary continuous closed curves used to represent the magnetic field in a region is known as magnetic field lines. It is directed from north pole to south pole outside the magnet and south pole to north pole inside the magnet.\n\n(a) The direction of magnetic field (B) at any point is obtained by drawing a tangent to the magnetic field line at that point. In case, two magnetic field lines intersect each other at the point P as shown in figure, magnetic field at P will have two directions, shown by two arrows, one drawn to each magnetic field line at P, which is not possible.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(b) It is taken by convention that the field lines emerges from north pole and merge at the south pole. Inside the magnet, the direction of field lines is from its south pole to its north pole. Thus, the magnetic field lines are closed curves.\nQuestion 5.\n(a) What is meant by a magnetic field? Mention two parameters that are necessary to describe it completely.\n(b) If field lines of a magnetic field are crossed at a point, what does it indicate? (Board Term I, 2013)\nAnswer:\n(a) Refer to answer 1.\nNecessary parameters are:\nMagnitude of magnetic field.\nDirection of field lines\n(b) If field lines of a magnetic field are crossed at a point, it indicates that there are two direction of magnetic field at a point which is not possible.\nQuestion 6.\nA compass needle is placed near a current carrying straight conductor. State your observation for the following cases and give reasons for the same in each case.\n(a) Magnitude of electric current is increased.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(a) Magnitude of electric current is increased.\n(b) The compass needle is displaced away from the conductor. (AI 2019)\nAnswer:\n(a) As the amount of magnetic field strength is directly proportional to the amount of current, so the deflection of compass needle increases.\n(b) Since magnetic field strength at a point is inversely proportional to the distance from the wire. Hence deflection of compass decreases when it is displaced away from the conductor.\nQuestion 7.\nState how the magnetic field produced by a straight current carrying conductor at a point depends on\n(a) current through the conductor\n(b) distance of point from conductor. (Board Term I, 2014)\nAnswer:\nStrength of magnetic field produced by a straight current-carrying wire at a given point is\n(a) directly proportional to the current passing through it.\n(b) inversely proportional to the distance of that point from the wire.\n\nQuestion 8.\nGive reason for the following", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(a) directly proportional to the current passing through it.\n(b) inversely proportional to the distance of that point from the wire.\n\nQuestion 8.\nGive reason for the following\n(i) There is either a convergence or a divergence of magnetic field lines near the ends of a current carrying straight solenoid.\n(ii) The current carrying solenoid when suspended freely rests along a particular direction. (2/3, 2020)\nAnswer:\n(i) There is either a convergence or a divergence of magnetic field lines near the ends of a current carrying straight solenoid because it behaves similar to that of a bar magnet and has a magnetic field line pattern similar to that of a bar magnet. Thus the ends of the straight solenoid behaves like poles of the magnet, where the converging end is the south pole and the diverging end is the north pole.\n(ii) The current carrying solenoid behaves similar to that of a bar magnet and when freely suspended aligns itself in the north-south direction.\nQuestion 9.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(ii) The current carrying solenoid behaves similar to that of a bar magnet and when freely suspended aligns itself in the north-south direction.\nQuestion 9.\nFind the direction of magnetic field due to a current carrying circular coil held:\n(i) vertically in North \u2013 South plane and an observer looking it from east sees the current to flow in anticlockwise direction,\n\n(ii) vertically in East \u2013 West plane and an observer looking it from south sees the current to flow in anticlockwise direction,\n(iii) horizontally and an observer looking at it from below sees current to flow in clockwise direction .(Board Term I, 2017)\nAnswer:\nAccording to right hand rule, the direction of magnetic field is\n(i) west to east\n(ii) north to south\n(iii) into the paper.\nQuestion 10.\n(a) State three factors on which the strength of magnetic field produced by a current carrying solenoid depends.\n(b) Draw circuit diagram of a solenoid to prepare an electromagnet. (Board Term I, 2016)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(b) Draw circuit diagram of a solenoid to prepare an electromagnet. (Board Term I, 2016)\nAnswer:\n(a) Strength of magnetic field produced by a current carrying solenoid depends upon the following factors:\nnumber of turns in the coil\namount of current flowing through it\nradius of coil\nMaterial of core of the solenoid.\n(b) A strong magnetic field produced inside a solenoid can be used to magnetise a piece of magnetic material, like soft iron, when placed inside the coil. The magnet so formed is called an electromagnet.\n\nQuestion 11.\n(a) State Right Hand Thumb rule to find the direction of the magnetic field around a current carrying straight conductor.\n(b) How will the magnetic field be affected on:\n(i) increasing the current through the conductor\n(ii) reversing the direction of flow of current in the conductor? (Board Term I, 2015)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(b) How will the magnetic field be affected on:\n(i) increasing the current through the conductor\n(ii) reversing the direction of flow of current in the conductor? (Board Term I, 2015)\nAnswer:\n(a) It states that you are holding a current carrying straight conductor in your right hand such that the thumb points towards the direction of current. Then your finger will wrap around the conductor in the direction of the field lines of the magnetic field.\n(b) (i) If the current is increased, the magnetic field strength also increases.\n(ii) If the direction of current is reversed, the direction of magnetic field also get reversed.\nQuestion 12.\nDiagram shows the lengthwise section of a current carrying solenoid. \u29bb indicates current entering into the page, \u2a00 indicates current emerging out of the page. Decide which end of the solenoid A or B, will behave as north pole. Give reason for your answer. Also draw field lines inside the solenoid.\n\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nAnswer:\n\nUsing right hand thumb rube we can draw the magnetic field lines around the conductor as shown. From figure, end A of solenoid act as north pole and end B will act as south pole. Inside the solenoid field lines are in the form of parallel straight lines.\nQuestion 13.\nWrite one application of right hand thumb rule. (1/3, Board Term I, 2013)\nAnswer:\nIt is used to find the direction of magnetic field around a current carrying conductor.\nQuestion 14.\n(a) What is an electromagnet? List any two uses.\n(b) Draw a labelled diagram to show how an electromagnet is made.\n(c) State the purpose of soft iron core used in making an electromagnet.\n(d) List two ways of increasing the strength of an electromagnet if the material of the electromagnet is fixed. (2020)\nAnswer:\n(a, b) Refer to answer 10(b).\nElectromagnets are used in electric motors and generators, electric bells and buzzers, loudspeakers and headphones etc.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nAnswer:\n(a, b) Refer to answer 10(b).\nElectromagnets are used in electric motors and generators, electric bells and buzzers, loudspeakers and headphones etc.\n(c) The soft iron core placed in an electromagnet increases the strength of the magnetic field produced. Thus increasing the strength of electromagnet.\n(d) The strength of electromagnet can be increased by\n(i) Increasing the current passing through the coil.\n(ii) Increasing the number of turns in the coil.\nQuestion 15.\nWhat is solenoid? Draw the pattern of magnetic field lines of\n(i) a current carrying solenoid and\n(ii) a bar magnet.\nList two distinguishing features between the two fields. (Delhi 2019)\nAnswer:\n(i) Solenoid : A coil of many circular turns of insulated copper wire wrapped in the shape of cylinder is called solenoid.\n\nThe pattern of magnetic field lines inside the solenoid indicates that the magnetic field is the same at all points inside the solenoid. That is, the field is uniform inside the solenoid.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nThe pattern of magnetic field lines inside the solenoid indicates that the magnetic field is the same at all points inside the solenoid. That is, the field is uniform inside the solenoid.\n(ii) Magnetic field lines around a bar magnet.\n\nFollowing are the distinguishing features between the two fields.\n(a) A bar magnet is a permanent magnet whereas solenoid is an electromagnet, therefore field produced by solenoid is temporary and stay till current flows through it.\n(b) Magnetic field produced by solenoid is more stronger than magnetic field of a bar magnet.\nQuestion 16.\nWhat are magnetic field lines? List three characteristics of these lines. Describe in brief an activity to study the magnetic field lines due to a current carrying circular oil. (Board Term I, 2017, 2016)\nAnswer:\nMagnetic field lines : These are the imaginary close curves which are used to represent the magnetic field around the magnet.\nThe properties of the magnetic field lines are listed below:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nAnswer:\nMagnetic field lines : These are the imaginary close curves which are used to represent the magnetic field around the magnet.\nThe properties of the magnetic field lines are listed below:\nMagnetic field lines start at the north pole and end at the south pole.\nMagnetic field lines do not intersect each other, because there cant be two directions of the magnetic field at any one point.\nThe degree of closeness of the field lines depends upon the strength of the magnetic field. Stronger the field, closer are the field lines.\nIn order to find the magnetic field due to a coil, it is held in a vertical plane and is made to pass through a smooth cardboard in such a way that the centre (O) of the coil lies at the cardboard. A current is passed through the coil and iron filings are sprinkled on the cardboard. These iron filings arrange themselves in a pattern similar to one shown in the figure. This pattern represents the magnetic field lines due to the coil.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nIn order to find the direction of magnetic field lines, we plot the magnetic field with the help of a compass needle. The pattern of magnetic field lines so obtained is shown in figure (b). From this pattern, the following important conclusion have been drawn.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nThe magnetic field lines near the coil are nearly circular and concentric. This is due to the reason that the segments of the coil in contact with the board at the points A and B are almost like straight conductors. The direction of the field lines can also be found by applying right-hand thumb rule.\nThe field lines are in the same direction in the space enclosed by the coil.\nNear the centre of the coil, the field lines are nearly straight and parallel. As such the magnetic field at the centre of the coil can be taken to be uniform.\nThe direction of the magnetic field at the centre is perpendicular to the plane of the coil.\nAs we move towards the centre of the coil, the strength of magnetic field increases. Magnetic field is maximum at its centre. This is due to the reason that the two magnetic field (one due to the semicircular segment of the coil through A and the other due to the semicircular segment through B) assist each other.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nThe magnitude of the magnetic field at the centre of the coil is directly proportional to the current flowing through it and total number of turns and inversely proportional to the radius of the coil. This is due to the reason that the current in all the circular turns of the coil is in the same direction. As such, the resultant magnetic field due to the coil is equal to the sum of the field due to all these turns.\nQuestion 17.\nDraw the magnetic field lines through and around a single loop of wire carrying electric current. (2/5, Board Term I, 2016)\nAnswer:\n\nQuestion 18.\nWhat is a solenoid? Draw a diagram to show field lines of the magnetic field through and around a current carrying solenoid. State the use of magnetic field produced inside a solenoid. List two properties of magnetic lines of force. (Board Term I, 2015)\nAnswer:\nRefer to answer 15(i).", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nAnswer:\nRefer to answer 15(i).\nSolenoid is used to form strong but temporary magnet called electromagnets. These electromagnets are used in wide variety of instruments and used to lift heavy iron, objects.\nProperties of magnetic filed lines :\nRefer to answer 16.\nQuestion 19.\nState the effect of a magnetic field on the path of a moving charged particle. (Board Term I, 2014)\nAnswer:\nA charged particle moving in a magnetic field may experience a force in the direction perpendicular to direction of magnetic field and direction of motion of particle. This force deflects the charged particle from its path.\nQuestion 20.\nState the direction of magnetic field in the following case.\n\nAnswer:\nUsing Fleming\u2019s left hand rule, the direction of magnetic field is out of the plane of paper.\nQuestion 21.\nWrite one application of Flemings left hand rule. (1/3, Board Term I, 2013)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nAnswer:\nUsing Fleming\u2019s left hand rule, the direction of magnetic field is out of the plane of paper.\nQuestion 21.\nWrite one application of Flemings left hand rule. (1/3, Board Term I, 2013)\nAnswer:\nFlemings left hand rule is used to find the direction of force on a current carrying conductor placed in a magnetic field acting perpendicular to the direction of current.\nQuestion 22.\nA current carrying conductor is placed in a magnetic field. Now answer the following.\n(i) List the factors on which the magnitude of force experienced by conductor depends.\n(ii) When is the magnitude of this force maximum?\n(iii) State the rule which helps, in finding the direction of motion of conductor.\n(iv) If initially this force was acting from right to left, how will the direction of force change if:\n(a) direction of magnetic field is reversed?\n(b) direction of current is reversed? (Board Term I, 2017)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(a) direction of magnetic field is reversed?\n(b) direction of current is reversed? (Board Term I, 2017)\nAnswer:\n(i) When a current carrying wire is placed in a magnetic field, it experiences a magnetic force that depends on\n(a) current flowing in the conductor\n(b) strength of magnetic field\n(c) length of the conductor\n(d) angle between the element of length and the magnetic field.\n(ii) Force experienced by a current carrying conductor placed in a magnetic field is largest when the direction of current is perpendicular to the direction of magnetic field.\n(iii) The rule used in finding the direction of motion of the conductor placed in a magnetic field is Flemings left hand rule.\nFleming\u2019s left hand rule is as follows:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(iii) The rule used in finding the direction of motion of the conductor placed in a magnetic field is Flemings left hand rule.\nFleming\u2019s left hand rule is as follows:\nStretch out the thumb, the forefinger, and the second (middle) finger of the left hand so that these are at right angles to each other. If the forefinger gives the direction of the magnetic field (N to S), the second (middle) finger the direction of current then the thumb gives the direction of the force acting on the conductor.\n(iv) (a) Direction of force will be reversed when direction of magnetic field is reversed, i.e., now force on conductor will act from left to right.\n(b) Direction of force will be reversed, if the direction of current is reversed, i.e., the force on the conductor will act from left to right.\nQuestion 23.\nState whether an alpha particle will experience any force in a magnetic field if (alpha particles are positively charged particles)\n(i) it is placed in the field at rest.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nQuestion 23.\nState whether an alpha particle will experience any force in a magnetic field if (alpha particles are positively charged particles)\n(i) it is placed in the field at rest.\n(ii) it moves in the magnetic field parallel to field lines.\n(iii) it moves in the magnetic field perpendicular to field lines.\nJustify your answer in each case. (Board Term I, 2016)\nAnswer:\n(i) No, alpha particle will not experience any force if it is at rest, because only moving charge particle can experience force when placed in a magnetic field.\n(ii) No, alpha particle will not experience any force if it moves in the magnetic field parallel to field lines because charge particle experiences force only when it moves at an angle other than 0\u00b0 with magnetic field.\n(iii) Alpha particle will experience a force in the direction perpendicular to the direction of magnetic field and direction of motion of alpha particle.\nQuestion 24.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(iii) Alpha particle will experience a force in the direction perpendicular to the direction of magnetic field and direction of motion of alpha particle.\nQuestion 24.\nDescribe an activity with labelled diagram to show that a force acts on current carrying conductor placed in a magnetic field and its direction of current through conductor. Name the rule which determines the direction of this force. (Board Term I, 2016)\nAnswer:\nA small aluminium rod suspended horizontally from a stand using two connecting wires. Place a strong horseshoe magnet in such a way that the rod lies between the two poles with the magnetic field directed upwards. For this, put the north pole of the magnet vertically below and south pole vertically above the aluminium rod.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nConnect the aluminium rod in series with a battery, a key and a rheostat. Pass a current through the aluminium rod from one end to other (B to A). The rod is displaced towards left. When the direction of current flowing through the rod is reversed, the displacement of rod will be towards right. Direction of force on a current carrying conductor is determined by Fleming\u2019s left hand rule.\nQuestion 25.\n(a) State Flemings left hand rule.\n(b) Write the principle of working of an electric motor.\n(c) Explain the function of the following parts of an electric motor.\n(i) Armature (ii) Brushes (iii) Split ring (2018)\nAnswer:\n(a) Refer to answer 22 (iii).\n(b) Principle : Current carrying conductor when placed at right angle to a magnetic field, experiences a force due to which we get motion. The direction of the force is given by Flemings left hand rule.\n(c) (i) Armature is a conductive part of motor which generates torque in the motor.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(c) (i) Armature is a conductive part of motor which generates torque in the motor.\n(ii) The two stationary brushes in a simple electric motor draw current from the battery and supply it to the armature of motor.\n(iii) The role of split ring is to change the direction of current flowing through the coil after each half-rotation of coil.\nQuestion 26.\nThe change in magnetic field lines in a coil is the cause of induced electric current it. Name the underlying phenomenon. (2020)\nAnswer:\nThe phenomenon in which electric current is generated by varying magnetic fields around a coil is called electromagnetic induction.\nQuestion 27.\nDefine the term induced electric current. (2020)\nAnswer:\nThe current induced in a conductor when the magnetic field around it changes is known as induced electric current.\nQuestion 28.\nFlemings Right-hand rule gives\n(a) magnitude of the induced current.\n(b) magnitude of the magnetic field.\n(c) direction of the induced current.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nQuestion 28.\nFlemings Right-hand rule gives\n(a) magnitude of the induced current.\n(b) magnitude of the magnetic field.\n(c) direction of the induced current.\n(d) both, direction and magnitude of the induced current. (2020)\nAnswer:\n(c) Flemings Right-hand rule gives the direction of induced current.\nQuestion 29.\nWhat is the function of a galvanometer in a circuit? (Delhi 2019)\nAnswer:\nGalvanometer is an instrument that can detect the presence of electric current in a circuit.\nQuestion 30.\nWrite any one method to induce current in a coil. (Board Term I, 2016)\nAnswer:\nBy keeping the magnet in a fixed position and moving the coil towards and away from the magnet, we can induce current in the coil.\nQuestion 31.\nTwo coils of insulated copper wire are wound over a non-conducting cylinder as shown. Coil 1 has comparatively large number of turns. State your observations, when\n\n(i) Key K is closed\n(ii) Key K is opened\nGive reason for each of your observations. (2020)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(i) Key K is closed\n(ii) Key K is opened\nGive reason for each of your observations. (2020)\nAnswer:\n(i) When key is closed, after setting up the circuit as shown, one can observe a deflection on the galvanometer connected to the second coil. This is because, a potential difference and thus a current is induced in coil 2 as there is change in the current and the magnetic field associated with it in coil 1. When the magnetic field changes in coil 1, the magnetic field lines around coil 2 also changes. This induces a current in coil 2.\n(ii) When key K is opened, after closing it for sometime it can be observed that the galvanometer show a deflection, but this time in the opposite direction. This is because, when the current stops flowing in coil 1, the magnetic field associated with it changes in the opposite direction as in the first case, thus inducing a current in the opposite direction.\nQuestion 32.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nQuestion 32.\nTwo circular coils P and Q are kept close to each other, of which coil P carries a current. What will you observe in the galvanometer connected across the coil Q\n(a) if current in the coil P is changed?\n(b) if both the coils are moved in the same direction with the same speed?\nGive reason to justify your answer in each\nAnswer:\n(a) When the amount of current in the coil P is changed, an induced current will induce in the coil Q due to change in magnetic field lines i.e., magnetic flux.\n(b) If both the coils are moved in the same direction with the same speed, then there is no net change in magnetic flux. Hence there will be no deflection in the galvanometer.\nQuestion 33.\nIn Faradays experiment if instead of moving the magnet towards the coil we move the coil towards the magnet. Will there be any induced current? Justify your answer. Compare the two cases. (Board Term I, 2017)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nAnswer:\nYes, these will be an induced current in both the cases as there is a change in the number of magnetic field line associated with the coil or we can say that there is a motion of a magnet with respect to the coil.\nSame of current will be induced and the direction of flow of current will also be the same in the two cases.\nQuestion 34.\nWrite one application of Fleming\u2019s right hand rule. (1/3, Board Term I, 2013)\nAnswer:\nFlemings right hand rule is used to find the direction of induced current.\nQuestion 35.\n(a) A coil of insulated copper wire is connected to a galvanometer. With the help of a labelled diagram state what would be seen if a bar magnet with its south pole towards one face of this coil is\n(i) moved quickly towards it,\n(ii) moved quickly away from it,\n(iii) placed near its one face?\n(b) Name the phenomena involved in the above cases.\n(c) State Fleming\u2019s right hand rule. (Board Term I, 2017)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(ii) moved quickly away from it,\n(iii) placed near its one face?\n(b) Name the phenomena involved in the above cases.\n(c) State Fleming\u2019s right hand rule. (Board Term I, 2017)\nAnswer:\n(a) If a coil of insulated wire is connected to a galvanometer and a bar magnet with south pole is moved towards one face of the coil then, given situation is shown in the figure.\n\n(i) Moved quickly towards the coil : A current is induced in clockwise direction in the coil with respect to the side facing the north pole of the magnet and needle of galvanometer will deflect in one direction from zero position.\n\n(ii) Moved quickly away from coil : A current is induced in anti-clockwise direction in the coil with respect to the side facing the north pole of the magnet and the needle of the galvanometer will deflect in opposite direction from (i).\n\n(iii) Placed near its one face : No deflection of the needle of galvanometer is observed.\n(b) The phenomena involved is called electromag-netic induction.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(iii) Placed near its one face : No deflection of the needle of galvanometer is observed.\n(b) The phenomena involved is called electromag-netic induction.\n(c) Fleming\u2019s right hand rule: Stretch the right hand such that the first finger, the central finger and the thumb are mutually perpendicular to each other.\nIf the first finger points along the direction of the field (magnetic field) and the thumb points along the direction of motion of the conductor, then the direction of induced current is given by the direction of the central finger.\nQuestion 36.\nWrite the frequency of alternating current (AC) in India. How many times per second it changes its direction? (Board Term I, 2015)\nAnswer:\nThe frequency of A.C. in India is 50 Hz and it changes direction twice in each cycle. Therefore, it changes direction 2 \u00d7 50 = 100 times in one second.\nQuestion 37.\nHow is the type of current that we receive in domestic circuit different from the one that runs a clock? (Board Term I, 2014)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nQuestion 37.\nHow is the type of current that we receive in domestic circuit different from the one that runs a clock? (Board Term I, 2014)\nAnswer:\nThe current that we receive from domestic circuit is alternating current (A.C.) and the current that issuse to run clock is direct current (D.C.). Direct current always flow in one direction whereas the alternating current reverses its direction periodically.\nQuestion 38.\nDefine alternating current and direct current.\nExplain why alternating current is preferred over direct current for transmission over long distances. (Board Term I, 2014)\nAnswer:\nAlternating current (A.C.) : An electric current whose magnitude changes with time and direction reverses periodically is called alternating current.\nDirect current (D.C.) : An electric current whose magnitude is either constant or variable but the direction of flow in a conductor remains the same is called direct current.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nDirect current (D.C.) : An electric current whose magnitude is either constant or variable but the direction of flow in a conductor remains the same is called direct current.\nA.C. can be transmitted to distant places without much loss of electric power than D.C. That is why A.C. is preferred over D.C. for transmission of current over a long distances.\nQuestion 39.\n(i) Alternating current has a frequency of 50 Hz. What is meant by this statement? How many times does it change its direction in one second? Give reason for your answer.\n(ii) Mention the frequency of D.C that is given by a cell. (Board Term I, 2013)\nAnswer:\n(i) The frequency of household supply of A.C. in India is 50 Hz. This means, A.C. completes 50 cycles in one second. Thus, A.C. changes direction 2 \u00d7 50 = 100 times in one second.\n(ii) Frequency of D.C. is zero as its direction does not change with time.\nQuestion 40.\nAt the time of short circuit, the electric current in the circuit.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(ii) Frequency of D.C. is zero as its direction does not change with time.\nQuestion 40.\nAt the time of short circuit, the electric current in the circuit.\n(a) vary continuously (b) does not change\n(c) reduces substantially\n(d) increases heavily. (2020)\nAnswer:\n(d) At the time of short circuit, the live and neutral wire come in direct contact, thus increasing the current in the circuit abruptly.\nQuestion 41.\nMention and explain the function of an earth wire. Why it is necessary to earth metallic appliances? (Board Term I, 2013)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nQuestion 41.\nMention and explain the function of an earth wire. Why it is necessary to earth metallic appliances? (Board Term I, 2013)\nAnswer:\nMany electric appliances of daily use like electric press, heater, toaster, refrigerator, table fan etc. have a metallic body. If the insulation of any of these appliances melts and makes contact with the metallic casing, the person touching it is likely to receive a severe electric shock. This is due to the reason that the metallic casing will be at the same potential as the applied one. Obviously, the electric current will flow through the body of the person who touches the appliance. To avoid such serious accidents, the metal casing of the electric appliance is earthed. Since the earth does not offer any resistance, the current flows to the earth through the earth wire instead of flowing through the body of the person.\nQuestion 42.\nGive reason for the following :", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nQuestion 42.\nGive reason for the following :\nThe burnt out fuse should be replaced by another fuse of identical rating. (1/3, 2020)\nAnswer:\nA burnt out fuse should be replaced with identical rating because it helps in protecting the circuit from overloading and short circuiting. If a fuse of higher rating is used then it may not melt and cut off the supply during overloading. Similarly a fuse of lower rating may melt frequently even for a normal flow of current. This results in decreasing the efficiency of the circuit.\nQuestion 43.\nGive reasons for the following:\n(a) It is dangerous to touch the live wire of the main supply rather than neutral wire.\n(b) In household circuit, parallel combination of resistances is used.\n(c) Using fuse in a household electric circuit is important. (Board Term I, 2017)\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(b) In household circuit, parallel combination of resistances is used.\n(c) Using fuse in a household electric circuit is important. (Board Term I, 2017)\nAnswer:\n(a) Live wire is at 220V and neutral wire is at zero volt since the electric current flows from higher potential to lower potential, we can get an electric shock by touching live wire but that is not the case with neutral wire.\n(b) In parallel combination, each resistor gets same potential from the source. We can use separate on/off switches with each appliance. Also in case if any one resistor fails then the circuit will not break. So, it is safe and convenient to connect household circuit in parallel combination of resistors\n(c) Fuse is an important safety device. It is used in series with any electrical appliance and protects it from short-circuiting and overloading.\nQuestion 44.\n(a) Fuse acts like a watchman in an electric circuit. Justify this statement.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nQuestion 44.\n(a) Fuse acts like a watchman in an electric circuit. Justify this statement.\n(b) Mention the usual current rating of the fuse wire in the line to (i) lights and fans (ii) appliance of 2 kW or more power. (Board Term I, 2014)\nAnswer:\n(a) When an unduly high electric current flows through the circuit, the fuse wire melts due to joule heating effect and breaks the circuit. Hence, it keeps an eye on the amount of current flowing and also stops the current if exceeds the maximum value. So, fuse acts like a watchman in an electric circuit.\n(b) (i) A fuse of rating 5A is usually used for lights and fans.\n(ii) A fuse of rating 15 A is usually used for appliance of 2 kW or more power.\nQuestion 45.\n(a) State Fleming\u2019s Left-hand rule.\n(b) List three characteristic features of the electric current used in our homes.\n(c) What is a fuse? Why is it called a safety device?\n(d) Why is it necessary to earth metallic electric appliances?", + "Question 45.\n(a) State Fleming\u2019s Left-hand rule.\n(b) List three characteristic features of the electric current used in our homes.\n(c) What is a fuse? Why is it called a safety device?\n(d) Why is it necessary to earth metallic electric appliances? (2020)\nAnswer:\n(a) Refer to answer 22(iii).", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\n(c) What is a fuse? Why is it called a safety device?\n(d) Why is it necessary to earth metallic electric appliances? (2020)\nAnswer:\n(a) Refer to answer 22(iii).\n(b) (i) The electric current which we receive in our homes are alternating current with a frequency of 50 Hz.\n(ii) There are mainly two wires used in the power supply provided to us. The one usually with a red insulation is called the live wire (or positive wire) and the one with black insulation is called the neutral wire (or negative wire). The potential difference between these wires is 220 V.\n(iii) Often, there are two separate circuits used in our homes, one of 15 A current rating for appliances with higher power rating and the other circuit of 5 A rating for appliances such as fans, bulbs, etc.\n(c) A fuse is safety device use to limit the current in an electric circuit.\nRefer to answer 44(a).\n(d) Refer to answer 41.\nQuestion 46.\n(a) Name two safety measures commonly used in an electric circuit and appliances.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nRefer to answer 44(a).\n(d) Refer to answer 41.\nQuestion 46.\n(a) Name two safety measures commonly used in an electric circuit and appliances.\n(b) What precaution should be taken to avoid the overloading of domestic electric circuits? (Board Term I, 2017)\nAnswer:\n(a) Fuse and the connection of earthing wire are the two safety measure commonly used in electric circuit and appliances.\n(b) Provide fuses/MCBs of proper rating.\nQuestion 47.\n(a) Draw a schematic diagram of a common domestic circuit showing provision of\n(i) Earth wire, (ii) Main fuse\n(iii) Electricity meter and\n(iv) Distribution box.\n(b) Distinguish between short circuiting and overloading. (Board Term I, 2015)\nAnswer:\n\n(b) Overloading : The condition in which a high current flows through the circuit and at the same time too many appliances are switched on then the total current drawn through the circuit may exceed its rated value.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-IMPORTANT QUESTIONS.txt\nShort circuiting: The condition when the live wire comes in direct contact with the neutral wire, due to which a high current flows in the circuit.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 13 Intext Questions\nPage Number: 224\nQuestion 1\nWhy does a compass needle get deflected when brought near a bar magnet ?\nAnswer:\nThe magnetic field of the magnet exerts force on both the poles of the compass needle. The forces experienced by the two poles are equal and opposite. These two forces form a couple which deflects the compass needle.\nPage Number: 228\nQuestion 1\nDraw magnetic field lines around a bar magnet.\nAnswer:\n\nQuestion 2\nList the properties of magnetic lines of force.\nAnswer:\nProperties of magnetic lines of force :\nThe magnetic field lines originate from the north pole of a magnet and end at its south pole.\nThe magnetic field lines become closer to each other near the poles of a magnet but they are widely separated at other places.\nTwo magnetic field lines do not intersect one another.\nQuestion 3\nWhy don\u2019t two magnetic lines of force intersect each other ?\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nTwo magnetic field lines do not intersect one another.\nQuestion 3\nWhy don\u2019t two magnetic lines of force intersect each other ?\nAnswer:\nThis is due to the fact that the resultant force on a north pole at any point can be only in one direction. But if the two magnetic field lines intersect one another, then the resultant force on north pole placed at the point of intersection will be along two directions, which is not possible.\nPage Number: 229 \u2013 230\nQuestion 1\nConsider a circular loop of wire lying on the plane of the table. Let the current pass through the loop clockwise. Apply the right hand rule to find out the direction of the magnetic field inside and outside the loop.\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nAnswer:\nAs shown in figure alongside, each section of wire produces its concentric set of lines of force. By applying right hand thumb rule, we find that all the sections produce magnetic field downwards at all points inside the loop while at the outside points, the field is directed upwards. Therefore, the magnetic field acts normally into the plane of the paper at the points inside the loop and normally out of the plane of paper at points outside the loop.\n\nQuestion 2\nThe magnetic field in a given region is uniform. Draw a diagram to represent it. [CBSE 2013, 2014]\nAnswer:\nA uniform magnetic field in a region is represented by drawing parallel straight lines, ail pointing in the same direction.\nFor example, the uniform magnetic field which exists inside a current-carrying solenoid can be represented by parallel straight lines pointing from its S-pole to N-pole (as shown in figure).\n\nQuestion 3\nChoose the correct option.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 3\nChoose the correct option.\nThe magnetic field inside a long straight solenoid-carrying current\n(i) is zero\n(ii) decreases as we move towards its end\n(iii) increases as we move towards its end\n(iv) is the same at all points\nAnswer:\n(iv) Is the same at all points.\nPage Number: 231 \u2013 232\nQuestion 1\nWhich of the following property of a proton can change while it moves freely in a magnetic field. (There may be more than one correct answer.)\n(i) Mass\n(ii) Speed\n(iii) Velocity\n(iv) Momentum\nAnswer:\nThe correct options are (iii) velocity, (iv) momentum.\nQuestion 2\nIn Activity 13.7 how do we think the displacement of rod AB will be affected if (i) current in rod AB is increased (ii) a stronger horse-shoe magnet is used; and (iii) length of the rod AB is increased ?\nAnswer:\n(i) When the current in the rod AB is increased, force exerted on the conductor increases, so the displacement of the rod increases.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nAnswer:\n(i) When the current in the rod AB is increased, force exerted on the conductor increases, so the displacement of the rod increases.\n(ii) When a stronger horse-shoe magnet is used, the magnitude of the magnetic field increases. This increases the force exerted on the rod and the displacement of the rod.\n(iii) When the length of the rod AB is increased, force exerted on the conductor increases, so the displacement of the rod increases.\nQuestion 3\nA positively-charged particle (alpha particle) projected towards west is deflected towards north by a magnetic field.\nThe direction of magnetic field is :\n(i) towards south\n(ii) towards east\n(iii) downward\n(iv) upward\nAnswer:\n(iv) Upward.\nHere, the positively charged alpha particles are moving towards west, so the direction of current is towards east. The deflection is towards north, so the force is towards north, so, we are given that\n(i) direction of current is towards west\n(ii) direction of force is towards north.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(i) direction of current is towards west\n(ii) direction of force is towards north.\nLet us now hold the forefinger, middle finger and thumb of our left-hand at right angles to one another. Adjust the hand in such a way that our mid finger points towards west (in the direction of current) and thumb points towards north (in the direction of force). Now, if we look at our forefinger, it will be pointing upward. Because the direction of forefinger gives the direction of magnetic field, therefore, the magnetic field is in the upward direction.\nPage Number: 233\nQuestion 1\nState Fleming\u2019s left hand rule. [CBSE 2018]\nAnswer:\nFleming\u2019s left hand rule : Stretch the first finger, the middle finger and the thumb of your left hand mutually perpendicular to each other in such a way that the first finger represents the direction of the magnetic field, the middle finger represents the direction of the current in the conductor, then the thumb will represent the direction of motion of the conductor.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 2\nWhat is the principle of an electric motor ? [CBSE 2018]\nAnswer:\nA motor works on the principle of magnetic effect of current. When a rectangular coil is placed in a magnetic field and current is passed through it, a force acts on the coil which rotates it continuously.\nWhen the coil rotates, the shaft attached to it also rotates. In this way the electrical energy supplied to the motor is converted into the mechanical energy of rotation.\nQuestion 3\nWhat is the role of the split ring in an electric motor ?\nAnswer:\nThe split ring reverses the direction of current in the armature coil after every half rotation, i.e., it acts as a commutator. The reversed current reverses the direction of the forces acting on the two arms of the armature after every half rotation. This allows the armature coil to rotate continuously in the same direction.\nPage Number: 236\nQuestion 1\nExplain different ways to induce current in a coil.\nAnswer:\nDifferent ways to induce current in a coil are :", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nPage Number: 236\nQuestion 1\nExplain different ways to induce current in a coil.\nAnswer:\nDifferent ways to induce current in a coil are :\nmoving a magnet towards or away from the coil or vice-versa, and\nchanging current in the neighbouring coil.\nPage Number: 237\nQuestion 1\nState the principle of an electric generator.\nAnswer:\nThe electric generator works on the principle that when a straight conductor is moved in a magnetic field, then current is induced in the conductor.\nIn an electric generator, a rectangular coil is made to rotate rapidly in the magnetic field between the poles of a horse-shoe type magnet. When the coil rotates, it cuts the magnetic field lines due to which a current is produced in the coil.\nQuestion 2\nName some sources of direct current.\nAnswer:\nSome of the sources of direct current are dry cells, button cells, lead accumulators.\nQuestion 3\nWhich sources produce alternating current ?\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nName some sources of direct current.\nAnswer:\nSome of the sources of direct current are dry cells, button cells, lead accumulators.\nQuestion 3\nWhich sources produce alternating current ?\nAnswer:\nAlternating current is produced by AC generators of nuclear power plants, thermal power plants, hydroelectric power stations, etc.\nQuestion 4\nChoose the correct option : A rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each:\n(i) two revolution\n(ii) one revolution\n(iii) half revolution\n(iv) one-fourth revolution\nAnswer:\n(iii) Half revolution.\nPage Number: 238\nQuestion 1\nName two safety measures commonly used in electric circuits and appliances.\nAnswer:\n(i) Earthing and\n(ii) Electric fuse.\nQuestion 2\nAn electric oven of 2 kW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect ? Explain.\nAnswer:\nThe electric oven draws a current given by", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nAnswer:\nThe electric oven draws a current given by\n\nThus the electric oven draws current much more than the current rating 5 A. That is the circuit is overloaded. Due to excessive current, the fuse wire will blow and the circuit will break.\nWhat precautions should be taken to avoid the overloading of domestic electric circuits ?\nTo avoid the overloading of domestic electric circuits, the following precautions should be taken :\n(i) The wires used in the circuit must be coated with good insulating materials like PVC, etc.\n(ii) The circuit must be divided into different sections and a safety fuse must be used in each section.\n(iii) High power appliances like air-conditioner, refrigerator, a water heater, etc. should not be used simultaneously.\nNCERT Solutions for Class 10 Science Chapter 13 Textbook Chapter End Questions\nQuestion 1\nWhich of the following correctly describes the magnetic field near a long straight wire ?\n(i) the field consists of straight lines perpendicular to the wire", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 1\nWhich of the following correctly describes the magnetic field near a long straight wire ?\n(i) the field consists of straight lines perpendicular to the wire\n(ii) the field consists of straight lines parallel to the wire\n(iii) the field consists of radial lines originating from the wire\n(iv) the field consists of concentric circles centred on the wire\nAnswer:\n(iv) The field consists of concentric circles centred on the wire\nQuestion 2\nThe phenomenon of electromagnetic induction is\n(i) the process of charging a body\n(ii) the process of generating magnetic field due to a current passing through a coil\n(iii) producing induced current in a coil due to relative motion between a magnet and the coil\n(iv) the process of rotating a coil of an electric motor\nAnswer:\n(iii) Producing induced current in a coil due to relative motion between a magnet and the coil\nQuestion 3\nThe device used for producing electric current is called a\n(i) generator\n(ii) galvanometer\n(iii) ammeter\n(iv) motor", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 3\nThe device used for producing electric current is called a\n(i) generator\n(ii) galvanometer\n(iii) ammeter\n(iv) motor\nAnswer:\n(i) Generator.\nQuestion 4\nThe essential difference between an AC generator and a DC generator is that\n(i) AC generator has an electromagnet while a DC generator has permanent magnet\n(ii) DC generator will generate a higher voltage\n(iii) AC generator will generate a higher voltage\n(iv) AC generator has slip rings while the DC generator has a commutator\nAnswer:\n(iv) AC generator has slip rings while the DC generator has a commutator\nQuestion 5\nAt the time of short circuit, the current in the circuit\n(i) reduces substantially\n(ii) does not change\n(iii) increases heavily\n(iv) varies continuously\nAnswer:\n(iii) Increases heavily.\nQuestion 6\nState whether the following statements are True or False.\n(i) An electric motor converts mechanical energy into electrical energy.\n(ii) An electric generator works on the principle of electromagnetic induction.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(i) An electric motor converts mechanical energy into electrical energy.\n(ii) An electric generator works on the principle of electromagnetic induction.\n(iii) The field at the centre a long circular coil carrying current will be parallel straight lines.\n(iv) A wire with a green insulation is usually the live wire of an electric supply.\nAnswer:\n(i) False\n(ii) True\n(iii) True\n(iv) False.\nQuestion 7\nList three sources of magnetic fields.\nAnswer:\n(i) Current carrying conductor\n(ii) Electromagnets\n(iii) Permanent magnets\nQuestion 8\nHow docs a solenoid behave like a magnet ? Can you determine the north and south poles of a current-carrying solenoid with the help of a bar magnet? Explain.\nAnswer:\nA solenoid behaves like a magnet in the following ways.\nThe magnetic field produced by a current carrying solenoid is very much similar to that of a bar magnet.\nLike a bar magnet, one end of the solenoid has N-polarity while the other end has S-polarity.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nThe magnetic field produced by a current carrying solenoid is very much similar to that of a bar magnet.\nLike a bar magnet, one end of the solenoid has N-polarity while the other end has S-polarity.\nTo determine the north and south poles, we bring N-pole of the bar magnet near one end of the solenoid. If there is an attraction, then that end of the solenoid has south polarity and the other has north polarity. If there is a repulsion, then that end of the solenoid has north polarity and the other end has south polarity because similar poles repel each other.\nQuestion 9\nWhen is the force experienced by a current-carrying conductor placed in a magnetic field largest ?\nAnswer:\nWhen the conductor carries current in a direction perpendicular to the direction of the magnetic field, the force experienced by the conductor is largest.\nQuestion 10", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nAnswer:\nWhen the conductor carries current in a direction perpendicular to the direction of the magnetic field, the force experienced by the conductor is largest.\nQuestion 10\nImagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field ?\nAnswer:\nHere the electron beam is moving from our back wall to the front wall, so the direction of current will be in the opposite direction, from front wall towards back wall or towards us. The direction of deflection (or force) is towards our right side.\nWe now know two things :\ndirection of current is from front towards us, and\ndirection of force is towards our right side.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nWe now know two things :\ndirection of current is from front towards us, and\ndirection of force is towards our right side.\nLet us now hold the forefinger, middle finger and thumb of our left hand at right angles to one another. We now adjust the hand in such a way that our centre finger points towards us (in the direction of current) and thumb points towards right side (in the direction of force). Now, if we look at our forefinger, it will be pointing vertically downwards. Since the direction of forefinger gives the direction of magnetic field, therefore, the magnetic field is in the vertically downward direction.\nQuestion 11\nDraw a labelled diagram of an electric motor. Explain its principle and working. What is the function of a split ring in an electric motor ?\nAnswer:\nElectric Motor : The device used to convert electrical energy to mechanical energy is called Electric Motor. It is used in fans, machines, etc.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nAnswer:\nElectric Motor : The device used to convert electrical energy to mechanical energy is called Electric Motor. It is used in fans, machines, etc.\nPrinciple : NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric CurrentElectric motor works on the principle of force experienced by a current carrying conductor in a magnetic field. The two forces in the opposite sides are equal and opposite. Since they act in different lines they bring rotational motion.\nWorking of an electric motor :\nWhen current starts to flow, the coil ABCD is in horizontal position. The direction of current through armature coil has the direction from A to B in the arm AB and from C to D in the arm CD. The direction of force exerted on the coil can be found through Fleming\u2019s left hand law.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nAccording to this law, it is found that the force exerted on the part AB, pushes the coil downwards. While the force exerted on the part CD pushes it upwards. In this way, these two forces being equal and opposite form a couple that rotates the coil in anticlockwise direction.\n\nWhen the coil is in vertical position, the brushes X and Y would touch the centre of the commutator and the current in the coil is stopped. Though current is stopped but the coil comes back in horizontal state due to momentum.\nAfter half rotation, the polarity of the commutator also changes, because now Q makes contact with brush X and P with brush Y. Therefore, now the force exerts downwards on the arm AB and upwards on the arm CD and thus again a couple of forces is formed that rotates the coil in clockwise direction. This process is repeated again and again and the coil rotates til! the current flows across it.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nFunction of split ring : Split ring in a motor acts as a commutator, i.e., it reverses the flow of current in the circuit due to which the direction of the forces acting on the arms also reverses.\nQuestion 12\nName some devices in which electric motors are used.\nAnswer:\nElectric motor is used in the appliances like electric fans, washing machine, mixers, grinders, blenders, computers, MP3 players, etc.\nQuestion 13\nA coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is (t) pushed into the coil (ii) withdrawn from inside the coil (iii) held stationary inside the coil ? [CBSE (Delhi) 2017, AICBSE 2016]\nAnswer:\n(i) As a bar magnet is pushed into the coil, a momentary deflection is observed in the galvanometer indicating the production of a momentary current in the coil.\n(ii) When the bar magnet is withdrawn from the coil, the deflection of galvanometer is in opposite direction showing the production of an opposite current.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(ii) When the bar magnet is withdrawn from the coil, the deflection of galvanometer is in opposite direction showing the production of an opposite current.\n(iii) When the bar magnet is held stationary inside the coil, there is no deflection in galvanometer indicating that no current is produced in the coil.\nQuestion 14\nTwo circular coils A and B are placed closed to each other. If the current in the coil A is changed, will some current be induced in the coil B ? Give reason.\nAnswer:\nYes, some current will be induced in the coil B. When the current in coil A is changed, some current is induced in the coil B. Due to change in current in coil A, the magnetic field lines linked with coil A and with coil B get changed. This sets up induced current in coil B.\nQuestion 15", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 15\nState the rule to determine the direction of a (i) magnetic field produced around a straight conductor-carrying current (it) force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and (in) current induced in a coil due to its rotation in a magnetic field.\nAnswer:\n(i) Right hand thumb rule : If the current carrying conductor is held in the right hand such that the thumb points in the direction of the current, then the direction of the curl of the fingers will give the direction of the magnetic field.\n(ii) Fleming\u2019s left hand rule : NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current Stretch the forefinger, the central finger and the thumb of the left hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field, the middle finger in the direction of current, then the thumb points in the direction of force in the conductor.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(iii) Fleming\u2019s right hand rule : Stretch the thumb, forefinger and the central finger of the right hand mutually perpendicular to each other. If the forefinger points in the direction of magnetic field, thumb in the direction of motion of the conductor, then the middle finger points in the direction of current induced in the conductor.\nQuestion 16\nExplain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of brushes ?\nAnswer:\nPrinciple : The electric generator is based on the principle of electromagnetic induction. When a coil is rotated with respect to a magnetic field, the number of magnetic field lines through the coil changes. Due to this a current is induced in the coil whose direction can be found by Fleming\u2019s right hand rule.\n\nWorking : When the armature coil ABCD rotates in a magnetic field produced by the permanent magnets, it cuts through the magnetic lines of force.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nWorking : When the armature coil ABCD rotates in a magnetic field produced by the permanent magnets, it cuts through the magnetic lines of force.\nDue to the rotation of armature coil, the associated magnetic field changes and an induced electromagnetic force is produced in it. The direction of this induced electromotive force or current can be determined by using Fleming\u2019s right hand rule.\nIn first half cycle the current flows in one direction by brush B1 and in second it flows in opposite direction by brush B2. This process continues. So the current produced is alternating in nature.\nFunctions of Brushes : Brushes in contact with rings provide the current for external use.\nQuestion 17\nWhen does an electric short circuit occur ?\nAnswer:\nIn a domestic circuit, short-circuit occurs when live and neutral wire come in direct contact with each other without any resistance. The resistance of the circuit becomes zero and excessive current starts to flow through it.\nQuestion 18", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 18\nWhat is the function of an earth wire ? Why is it necessary to earth metallic appliances ?\nAnswer:\nEarth wire is a safety measure that provides a low resistance conducting path to the current. Sometimes due to excess heat or wear and tear, the live wire comes in direct contact with the metallic cover of the appliances, which can give an electric shock on touching them. To prevent from the shock the metallic part is connected to the earth through a three-pin plug due to which the current flows to the earth at the instant there is a short circuit.\nIt is necessary to earth metallic appliances because it ensures that if there is any current leakage in the metallic cover, the potential of the appliance becomes equal to that of the earth. The potential of the earth is zero. As a result, the person handling the appliance will not get an electric shock.\nNCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nNCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current\nMagnetic effects of current: Magnetic field, field lines, field due to a current carrying conductor, field due to current carrying coil or solenoid; Force on current carrying conductor, Fleming\u2019s left hand rule. Electromagnetic induction, Induced potential difference, Induced current, Fleming\u2019s right hand rule, Direct current, Alternating current, frequency of AC, Advantage of AC over DC, Domestic electric circuits.\nFormulae Handbook for Class 10 Maths and Science\nQuestion 1\nWhy does a compass needle gets deflected when brought near a bar magnet?\nSolution:\nA compass needle is, in fact, a small bar magnet. If this is brought near another bar magnet, the like poles repel and the needle gets deflected.\nQuestion 2\nDraw magnetic field lines around a bar magnet?\nSolution:\n\nQuestion 3\nList the properties of magnetic lines of force.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 2\nDraw magnetic field lines around a bar magnet?\nSolution:\n\nQuestion 3\nList the properties of magnetic lines of force.\nSolution:\na) Magnetic lines are directed from the north pole towards the south pole.\nb) They do not cross each other.\nc) They are more crowded near the poles than at any other region in the field.\nd) They are closed curves.\ne) In the uniform magnetic field, the lines of force are parallel to one another.\nDownload NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current PDF\nQuestion 4\nWhy don\u2019t two magnetic lines of force intersect each other?\nSolution:\nNo two field-lines are found to cross each other. If they did, it would mean that at the point of intersection, the compass needle would point towards two directions, which is not possible.\nQuestion 5", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 5\nConsider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right-hand rule to find out the magnetic field inside and outside the loop.\nSolution:\nAt every point of a current \u2013carrying loop, the concentric circles representing the magnetic field around it would become larger and larger as we move away from the wire. By the time we reach at the center of the circular loop, the arc of these big circles would appear as straight lines.\nQuestion 6\nThe magnetic field in a given region is uniform. Draw a diagram to represent it.\nSolution:\n\nQuestion 7\nThe magnetic field inside a long straight solenoid-carrying current\na) is zero\nb) decreases as we move towards its end\nc) increases as we move towards its end\nd) is the same at all points\nSolution:\nd) is the same at all points\nQuestion 8\nWhich of the following property of a proton can change while it moves freely in a magnetic field?\na) Mass\nb) Speed\nc) Velocity", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nSolution:\nd) is the same at all points\nQuestion 8\nWhich of the following property of a proton can change while it moves freely in a magnetic field?\na) Mass\nb) Speed\nc) Velocity\nd) Momentum\nSolution:\nc) Velocity\nd) Momentum.\nQuestion 9\n(a) Current in rod AB is increased?\n(b) A stronger horseshoe magnet is used?\n(c) Length of the rod AB is increased?\nSolution:\n(a) If the current in rod AB is increased, the displacement of rod AB will not be affected.\n(b) If a stronger horseshoe magnet is used, force is exerted and hence the displacement increases.\n(c) If the length of the rod AB is increased there is no change in the displacement of the rod AB.\nQuestion 10\nA positively-charged particle projected towards west is deflected towards north by a magnetic field. The direction of the magnetic field is\na) Towards south\nb) Towards east\nc) Downward\nd) Upward\nSolution:\nb) Towards east.\nQuestion 11\nState Fleming\u2019s left-hand rule.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\na) Towards south\nb) Towards east\nc) Downward\nd) Upward\nSolution:\nb) Towards east.\nQuestion 11\nState Fleming\u2019s left-hand rule.\nSolution:\nFleming\u2019s left-hand rule states that, stretch the thumb, fore finger and middle finger of the left hand such that they are mutually perpendicular. If the first finger points in the direction of magnetic field and the second finger in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.\nQuestion 12\nWhat is the principle of an electric motor?\nSolution:\nPrinciple of an electric motor:\nThe working of the electric motor is based on the mechanical effect of an electric current. A conductor carrying a current placed in a magnetic field experiences a mechanical force.\nIn the motor, when a current is passed through a rectangular coil of wire placed in a magnetic field, the coil rotates continuously.\nQuestion 13\nWhat is the role of the split ring in an electric motor?\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 13\nWhat is the role of the split ring in an electric motor?\nSolution:\nIn electric motor, the split ring acts as a commutator. A device that reverses the direction of flow of current through a circuit is called a commutator. The reversal of current also reverses the direction of force acting on the two arms AB and CD.\nQuestion 14\nExplain different ways to induce current in a coil.\nSolution:\nCurrent can be induced in a coil either by moving it in a magnetic field or by changing the magnetic field around it. The induced current is found to be the highest when the direction of motion of the coil is at right angles to the magnetic field. The process, by which a changing magnetic field in a conductor induces a current in another conductor, is called electromagnetic induction.\nQuestion 15\nState the principle of an electric generator.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 15\nState the principle of an electric generator.\nSolution:\nA generator is also known as a dynamo. It is a device used to convert mechanical energy in to electrical energy. The mechanical energy is used to rotate a conductor in a magnetic field to produce electricity. It is an application of electromagnetic induction.\nAn A.C generator generates an alternating current.\nA D.C generator is used to deliver a current, which flows in the same direction.\nQuestion 16\nName some source of direct current.\nSolution:\nThe source of direct current is a split-ring type commutator, one brush is at all times in contact with the arm moving up in the field, while the other is in contact with the arm moving down. Thus a unidirectional current is produced.\nQuestion 17\nWhich sources produce alternating current?\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 17\nWhich sources produce alternating current?\nSolution:\nThe sources which produce alternating current is a permanent magnet called the field magnet, armature, slip ring and carbon brushes. After every half rotation the polarity of the current in the respective arms changes. Such a current, Which changes direction after equal intervals of time, is called an alternating current.\nQuestion 18\nA rectangular coil of copper wires is rotated in a magnetic field. The direction of the induced current changes once in each:\na) Two revolutions\nb) One revolution\nc) Half revolutions\nd) One-fourth revolutions.\nSolution:\nb) One revolution.\nQuestion 19\nName two safety measures commonly used in electric circuits and appliances.\nSolution:\nThe use of an electric fuse prevents the electric circuits and appliance from a possible damage by passing the flow of unduly high electric current. The Joule heating that takes place in the fuse melts it to break the electric circuit.\nQuestion 20", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 20\nAn electric oven of 2 KW power rating is operated in a domestic electric circuit (220 V) that has a current rating of 5 A. What result do you expect? Explain.\nSolution:\nV = 220 V, I = 5 A\nPower, P = VI\nP = 220 \u00d7 5\nP = 1100 W\nTherefore, power P = 1100 W = 1.1 KW\nTherefore, an electric oven of 2 KW power rating cannot be operated in a domestic electric circuit (220 V) that has a current rating of 5 A because electric oven has higher power than the power of the electric circuit.\nQuestion 21\nWhat precaution should be taken to avoid the overloading of domestic electric circuits?\nSolution:\nFuse is the most important safety device, to avoid the overloading of domestic electric circuits.\nToo many appliances should not be connected to a single socket.\nQuestion 22\nWhich of the following correctly describes the magnetic field near a long\nstraight wire?\n(a) The field consists of straight lines perpendicular to the wire.\n(b) The field consists of straight lines parallel to the wire.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nstraight wire?\n(a) The field consists of straight lines perpendicular to the wire.\n(b) The field consists of straight lines parallel to the wire.\n(c) The field consists of radial lines originating from the wire.\n(d) The field consists of concentric circles centred on the wire.\nSolution:\n(d) The field consists of concentric circles centred on the wire.\nQuestion 23\nThe phenomenon of electromagnetic induction is\n(a) the process of charging a body.\n(b) the process of generating magnetic field due to a current passing through a coil.\n(c) producing induced current in a coil due to relative motion between a magnet and the coil.\n(d) the process of rotating a coil of an electric motor.\nSolution:\n(c) producing induced current in a coil due to relative motion between a magnet and the coil.\nQuestion 24\nThe device used for producing electric current is called a\n(a) generator.\n(b) galvanometer.\n(c) ammeter.\n(d) motor.\nSolution:\n(a) generator.\nQuestion 25", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 24\nThe device used for producing electric current is called a\n(a) generator.\n(b) galvanometer.\n(c) ammeter.\n(d) motor.\nSolution:\n(a) generator.\nQuestion 25\nThe essential difference between an AC generator and a DC generator is that:\n(a) AC generator has an electromagnet while a DC generator has permanent magnet.\n(b) DC generator will generate a higher voltage.\n(c) AC generator will generate a higher voltage.\n(d) AC generator has slip rings while the DC generator has a commutator\nSolution:\n(d) AC generator has slip rings while the DC generator has a commutator.\nQuestion 26\nAt the time of short circuit, the current in the circuit\n(a) reduces substantially.\n(b) does not change.\n(c) increases heavily.\n(d) vary continuously.\nSolution:\n(c) increases heavily.\nQuestion 27\nState whether the following statements are true or false.\nSolution:\n(a) An electric motor converts mechanical energy into electrical energy \u2013 false.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nSolution:\n(c) increases heavily.\nQuestion 27\nState whether the following statements are true or false.\nSolution:\n(a) An electric motor converts mechanical energy into electrical energy \u2013 false.\n(b) An electric generator works on the principle of electromagnetic induction \u2013 true\n(c) The field at the centre of a long circular coil carrying current will be parallel straight lines \u2013 true.\n(d) A wire with a green insulation is usually the live wire of an electric supply \u2013 true.\nQuestion 28\nList three sources of magnetic fields.\nSolution:\na) Magnetic field due to a current through a straight conductor.\nb) Magnetic field due to a current in a solenoid.\nc) Magnetic field due to a current through a circular loop.\nQuestion 29\nHow does a solenoid behave like a magnet? Can you determine the north and the south poles of a current-carrying solenoid with the help of a bar magnet? Explain.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 29\nHow does a solenoid behave like a magnet? Can you determine the north and the south poles of a current-carrying solenoid with the help of a bar magnet? Explain.\nSolution:\n\nA coil of many circular turns of insulated copper wire wrapped closely in the shape of the cylinder is called a solenoid. The pattern of the magnetic field lines around a current- carrying solenoid is shown in this figure. In fact, one end of the solenoid behaves as a magnetic north pole, while the other behaves as the south pole. The field lines inside the solenoid are in the form of parallel straight lines. This indicates that the magnetic field is the same at all points inside the solenoid. That is, the field is uniform inside the solenoid.\nA strong magnetic field produced inside a solenoid can be used to magnetise a piece of magnetic material, like soft iron, when placed inside the coil. The magnet so formed is called an electromagnet.\nQuestion 30", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 30\nWhen is the force experienced by a current-carrying conductor placed in a magnetic field the largest?\nSolution:\nThe force experienced by a current-carrying conductor placed in a magnetic field is largest provided when the direction of current is at right angles to the direction of the magnetic field.\nQuestion 31\nImagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?\nSolution:\nThe direction of magnetic field is towards west.\nQuestion 32\nDraw a labelled diagram of an electric motor. Explain its principle and working.What is the function of a split ring in an electric motor?\nSolution:\n\nA motor is a device that converts the electrical energy into mechanical energy.\nPrinciple", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nSolution:\n\nA motor is a device that converts the electrical energy into mechanical energy.\nPrinciple\nAn electric motor is based on the fact that when a current carrying conductor is placed in a magnetic field the conductor experiences a force which is given by Fleming\u2019s Left Hand Rule. For example, when a rectangular coil is placed in the magnetic field and current is passed through it, a torque acts on the coil, which rotates it continuously. When the coil rotates, the shaft attached to it also rotates and therefore the electrical energy supplied to the motor is converted into the mechanical energy of rotation.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nAn electrical motor consists of a rectangular coil ABCD of insulated copper wire, wound on a soft iron core called armature. The coil is mounted between the poles of a magnet in such a way that it can rotate between the poles N and S. The two ends of the coil are soldered to the ends of a commutator whose main function is to reverse the direction of the current flowing through the coil every time the coil just passes the vertical position during its revolution.\nWorking\nSuppose the coil ABCD is initially at a horizontal position. When the switch is in ON position the current enters the coil through the carbon brushes and the half ring \u2018A\u2019 of the commutator.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nWorking\nSuppose the coil ABCD is initially at a horizontal position. When the switch is in ON position the current enters the coil through the carbon brushes and the half ring \u2018A\u2019 of the commutator.\nThe current flows in the direction DCBA and leaves via the half ring \u2018B\u2019. In the side PQ of the coil, the direction is from Q to P towards the south and the direction of the magnetic field is from the N to S pole towards the east. So, by applying Fleming\u2019s left hand rule, we find that it will experience a force in upward direction. Similarly, the side SR of the coil will experience a downward force. Thus we have two parallel wires experiencing forces in opposite directions. They form a couple tending to rotate the coil in the anticlockwise direction.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nWhen the coil goes beyond the vertical position, the two commutator half rings automatically changes contact from one brush to the other. This reverses the direction of current through the coil which, in turn, reverses the direction of forces acting on the two sides of the coil. The sides of the coil are interchanged, but rotate in the same anticlockwise direction. This process is repeated again and again and the coil continues to rotate as long as the current is passing.\nQuestion 33\nName some devices in which electric motors are used.\nSolution:\nElectric fans, refrigerators, mixers, washing machines, computers, MP3 players etc are some devices in which electric motors are used.\nQuestion 34\nA coil of insulated copper wire is connected to a galvanometer. What will happen if a bar magnet is\n(i) pushed into the coil,\n(ii) withdrawn from inside the coil,\n(iii) held stationary inside the coil?\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(i) pushed into the coil,\n(ii) withdrawn from inside the coil,\n(iii) held stationary inside the coil?\nSolution:\n(i) A deflection is observed in the galvanometer due to the induced current because of the changing magnetic flux (increasing) through the turns of the coil connected to the galvanometer.\n(ii) A deflection is again observed in the galvanometer, as when it is pulled out, the flux linked with the coil due to the bar magnet decreases. Hence a current flows in the coil to reduce the change in flux. The deflection can be observed in the opposite direction as compared with the previous case.\n(iii) No deflection is observed in the galvanometer. The flux linked with the coil due to the magnetic field is at a constant. Hence no current is induced due to the bar magnet.\nQuestion 35\nTwo circular coils A and B are placed closed to each other. If the current in the coil A is changed, will some current be induced in the coil B? Give reason.\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 35\nTwo circular coils A and B are placed closed to each other. If the current in the coil A is changed, will some current be induced in the coil B? Give reason.\nSolution:\nYes, if the current in the coil A is changed, then some current will be induced in the coil B because due to the change in the magnetic field effect around the coils.\nQuestion 36\nState the rule to determine the direction of a\n(i) magnetic field produced around a straight conductor-carrying current,\n(ii) force experienced by a current-carrying straight conductor placed in a magnetic field which is perpendicular to it, and\n(iii) current induced in a coil due to its rotation in a magnetic field.\nSolution:\n(i) Right-hand thumb rule", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(iii) current induced in a coil due to its rotation in a magnetic field.\nSolution:\n(i) Right-hand thumb rule\nImagine that we are holding a current carrying straight conductor in the right hand such that the thumb points towards the direction of current. Then our fingers will wrap around the conductor in the direction of the field lines of the magnetic field. This is known as Right-hand thumb rule.\n(ii) Fleming\u2019s left-hand rule\nFleming\u2019s left-hand rule states that, stretch the thumb, fore finger and middle finger of the left hand such that they are mutually perpendicular. If the first finger points in the direction of magnetic field and the second finger in the direction of current, then the thumb will point in the direction of motion or the force acting on the conductor.\n(iii) Fleming\u2019s right-hand rule", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(iii) Fleming\u2019s right-hand rule\nIf the thumb and the first two fingers of right hand are held at right angles to each other, with the Forefinger held in the direction of the field, and the thumb in the direction of motion, the induced current I flows in the direction of the middle finger.\nQuestion 37\nExplain the underlying principle and working of an electric generator by drawing a labelled diagram. What is the function of brushes?\nSolution:\nA C. generator\n\u201cA C. generator\u201d means \u201cAlternating Current generator\u201d. That is, an A. C. generator produces alternating current, which alternates (changes) in polarity continuously. We will now describe the construction an working of the A. C. generator or A. C. dynamo.\nConstruction of an A. C. generator", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nConstruction of an A. C. generator\nA simple A. C. generator consists of a rectangular coil ABCD that can be rotated rapidly between the poles N and S of a strong horseshoe type magnet M. The coil is made of a large number of turns of insulated copper wire. The ends A and D of the rectangular coil are connected to two circular pieces of copper metal called slip rings R1 and R2. As the slip rings R1 and R2 rotate with the coil, the two pieces of carbon called brushes, B1 and B2, keep contact with them. So, the current produced in the rotating coil can be tapped out through slip rings into the carbon brushes. From the carbon brushes B1 and B2 we take the current into various electrical appliances like radio, T. V., electric iron, bulbs, etc. But in this figure, we have shown only a galvanometer G connected the two carbon brushes.\n\nWorking of an A. C. generator", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nWorking of an A. C. generator\nSuppose that the generator coil ABCD is initially in the horizontal position. Again suppose that he coil ABCD is being rotated in the anticlockwise direction between the poles N and S of a horseshoe type magnet.\n(i) As the coil rotates in the anticlockwise direction, the side AB of the coil moves down cutting the magnetic lines of force near the N-pole of the magnet, and side CD moves up, cutting the lines of force near the S-pole of the magnet. Due to this, induced current is produced in the sides AB and DC of the coil. On applying Fleming\u2019s right hand rule to the side AB and DC of the coil, we find that the currents are in the direction B to A and D to C respectively. Thus, the induced currents in the two sides of the coil are in the same direction, and we get an effective induced current in the direction BADC.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(ii) After half revolution, the sides AB and DC of the coil will interchange their positions. The side AB will come on the right hand side and DC will come on the left side. So, after half a revolution, side AB starts moving up and side DC starts coming down. As a result of this, the direction of induced current in each side of the coil is reversed after half a revolution. Since the direction of induced current in the coil is reversed after half revolution so the polarity (positive and negative) of the two ends of the coil also changes after half revolution. The end of coil which was positive in the first half of rotation becomes negative in the second in the second half. And the end which was negative in the first half revolution becomes positive in the second half of revolution. Thus, in 1 revolution of the coil, the current changes its direction 2 times.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nThe alternating current (A. C.) produced in India has a frequency of 50 Hz. That is, the coil is rotated at the rate of 50 revolutions per second. Since in 1 revolution of coil, the current changes its direction 2 times, so in 50 revolutions of coil, the current changes its direction 2 \u00d7 50 = 100 times. Thus, the A. C. supply in India changes its direction 100 times in 1 second. Another way of saying this is that the alternating current produced in India changes its direction every 1/100 second. That is, each terminal of the coil is positive (+) for 1/100 of a second and negative (-) for the next 1/100 of a second. This process is repeated again and again with the result that there is actually no positive and negative in an A. C. generator. We will now describe why the direction of induced current in the coil of an A. C. generator changes after every half revolution of the coil.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nAfter every half revolution, each side of the generator coil starts moving in the opposite direction in the magnetic field. The side of the coil which was initially moving downwards in a magnetic field, after half revolution, it starts moving in opposite direction \u2013 upwards. Similarly the side of coil which was initially moving upwards, after half revolution, it starts moving downwards. Due to the change in the direction of motion of the two sides of the coil in the magnetic field after every half revolution, the direction of current produced in them also changes after every half revolution.\nD. C. generator\n\u201cD. C. generator\u201d means \u201cDirect Current generator\u201d. That is, a D. C. generator produces direct current and not alternating current. We will now describe the construction and working of D. C. generator or D. C. Dynamo.\nConstruction of a D. C. generator", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nA simple D. C. generator consists of a rectangular coil ABCD which cab be rotated rapidly between the poles N and S of a strong horse-shoe type magnet M. The generator coil is made of a large number of turns of insulated copper wire. The two ends of the coil are connected to the two copper half rings (or split rings) R1 and R2 of a commutator. There are two carbon brushes B1 and B2 which press lightly against the two half rings. When the coil is rotated, the two half rings R1 and R2 touch the two carbon brushes B1 and B2 one by one. So the current produced in the rotating coil can be tapped out through the commutator half rings into the carbon brushes. From the carbon brushes B1 and B2, we can take the current into the various electrical appliances like radio, T. V., electric iron, bulbs, etc. But in this figure, we have shown only a galvanometer G connected between the two carbon brushes. The galvanometer is a current detecting and current measuring instrument.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nWorking of a D. C. generator\nSuppose that the generator coil ABCD is initially in the horizontal position. Again suppose that he coil ABCD is being rotated in the anticlockwise direction between the poles N and S of a horseshoe type magnet.\n(iii) As the coil rotates in the anticlockwise direction, the side AB of the coil moves down cutting the magnetic lines of force near the N-pole of the magnet, and side DC moves up, cutting the lines of force near the S-pole of the magnet. Due to this, induced current is produced in the sides AB and DC of the coil. On applying Fleming\u2019s right hand rule to the side AB and DC of the coil we find that the currents in them are in the direction B to A and D to C respectively. Thus, the induced currents in the two sides of the coil are in the same direction, and we get an effective induced current in the direction BADC. Due to this the brush B1 becomes a positive (+) pole and brush B2 becomes negative (-) pole of the generator.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(iv) After half revolution, the sides AB and DC of the coil will interchange their positions. The side AB will come on the right hand side and start moving up whereas side DC will come on then the two commutator half rings R1 and R2 automatically change their contacts from one carbon brush to the other. Due to this change, the current keeps flowing in the same direction in the other circuits. The brush B1 always remaining positive terminal and brush B2 always remaining negative terminal of the generator. Thus, a D. C. generator supplies a current in one direction by the use of a commutator consisting of two, half-rings of copper. In the above discussion we have used the word D. C. generator everywhere. Please note that we can also write D. C. dynamo in place of D. C. generator.\nQuestion 38\nWhen does an electric short circuit occur?\nSolution:\nShort circuiting", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 38\nWhen does an electric short circuit occur?\nSolution:\nShort circuiting\nIf the plastic insulation of the live wire and neutral wire gets torn, then the two wires touch each other. This touching of the live wire and neutral wire directly is known as short-circuiting. The current passing through the circuit formed by these wires is very large and consequently a high heating effect is created which may lead to fire.\nQuestion 39\nWhat is the function of an earth wire? Why is it necessary to earth metallic appliances?\nSolution:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 39\nWhat is the function of an earth wire? Why is it necessary to earth metallic appliances?\nSolution:\nTo avoid electric shocks, the metal body of an electrical device is \u2018earthed\u2019. A wire called \u2018earth wire\u2019 is used to connect the metal body of the electrical device to the earth, which is at zero potential. In household circuits, we have three wires, the live wire, the neutral wire and the earth wire. One end of the earth wire is connected to the device and the other end of the wire is connected to the earth. We now say that the device is \u201cearthed\u201d or \u201cgrounded\u201d. Usually the three wires are connected to a three-pin plug. The neutral wire or the earth connection carries the high current to the earth from the device and prevents an electric shock.\nMultiple Choice Questions (MCQs) [1 Mark each]\nQuestion 1.\nA compass is to be placed near a bar magnet with unknown poles. Outside the magnetic field, the compass needle is pointing towards North as shown below:\n\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 1.\nA compass is to be placed near a bar magnet with unknown poles. Outside the magnetic field, the compass needle is pointing towards North as shown below:\n\nAnswer:\n(c) Magnetic field lines always point from North pole to South pole around the magnet. Thus, compass needle follows the path difference of magnetic field lines.\nQuestion 2.\nA bar magnet is broken into three parts X, Y and Z.\n\nWhich diagram show the poles in X, Y and Z?\n\nAnswer:\n(d) When a magnet is broken into three parts X, Y and Z, each part is still a magnet and the strength of the magnetic force remains the same.\nQuestion 3.\nAn unmagnetised iron bar is placed near the end of a bar magnet.\n\nWhich of the following diagram is correct?\n\nAnswer:\n(b) The end of the iron bar nearer to the South pole of the bar magnet becomes induced North pole while the other end is South pole.\nQuestion 4.\nThe diagram shows a current-carrying wire passing through the centre of a square cardboard.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 4.\nThe diagram shows a current-carrying wire passing through the centre of a square cardboard.\n\nHow do the strengths of the magnetic field at points X, Y and Z compare?\n(a) equal at X, Y and Z\n(b) stronger at Y than X, equal at Y and Z\n(c) weaker at Y than Z, stronger at Y than Z\n(d) stronger at Y than X, weaker at Z than X\nAnswer:\n(d) The closer to the wire, the stronger is the magnetic field strength. Since, the magnetic field is circular. Y is the closest followed by X than Z.\nQuestion 5.\nA circular loop placed in a plane perpendicular to the plane of paper carries a current when the key is on.\n\nThe current as seen from points A and B (in the plane of paper and on the axis of the coil) is anti-clockwise and clockwise, respectively.\nThe magnetic field lines point from B to A. The N-pole of the resultant magnet is on the face close to\n(a) A\n(b) B\n(c) A if the current is small and B if the current is large", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nThe magnetic field lines point from B to A. The N-pole of the resultant magnet is on the face close to\n(a) A\n(b) B\n(c) A if the current is small and B if the current is large\n(d) B if the current is small and A if the current is large [NCERT Exemplar]\nAnswer:\n(a) The N-pole of the resultant magnet is on the face close to A because, the magnetic field lines enter in loop from B and come out from A. Also, as a matter of fact magnetic lines come out of the N-pole of magnet. Therefore, face close to A represents N-pole. The currents in A and B are same.\nQuestion 6.\nA bar magnet is used to pick up an iron nail.\n\nAt which parts X, Y and Z is the easiest for the magnet to pick up the iron nail?\n(a) At X\n(b) AtY\n(c) At Z\n(d) It makes no difference\nAnswer:\n(c) The region with the highest density of magnetic field lines has the greatest strength.\nQuestion 7.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(a) At X\n(b) AtY\n(c) At Z\n(d) It makes no difference\nAnswer:\n(c) The region with the highest density of magnetic field lines has the greatest strength.\nQuestion 7.\nIf the key in the arrangement as shown below is taken out (the circuit is made open) and magnetic field lines are drawn over the horizontal plane ABCD, the lines are [NCERT Exemplar]\n\n(a) concentric circles\n(b) elliptical in shape\n(c) straight lines parallel to each other\n(d) concentric circles near the point O but of elliptical shapes as we go away from it\nAnswer:\n(c) When the key is taken out (the circuit is made open), no current flows through the wire, hence no magnetic field exists due to the conductor.\nThe only magnetic field is due to Earth\u2019s magnetic field and are straight lines parallel to each other. The horizontal component is directed from geographical South to geographical North on the horizontal plane ABCD.\nQuestion 8.\nFour metal rods are placed in turn inside the solenoid to attract paper clips.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nQuestion 8.\nFour metal rods are placed in turn inside the solenoid to attract paper clips.\nThe table below gives the results of the experiment when current is switched on and off.\nWhich rod would be the most suitable to use for the core of the solenoid in a circuit breaker?\nAnswer:\n(c) The core of the solenoid in a circuit breaker must be made of a soft type magnetic material which can be strongly magnetised but does not retain induced magnetism.\nQuestion 9.\nPermanent magnets can be made using hard magnetic materials.\nWhich of the following is not the correct method to make permanent magnets?\n(a) Using a bar magnet to stroke a steel bar\n(b) Using two bar magnets to stroke a steel bar\n(c) Placing a steel bar in a solenoid that connects to a DC supply\n(d) Placing a steel bar in a solenoid that connects to an AC supply, then slowly withdrawing the steel bar away from the solenoid in West-East direction\nAnswer:", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(d) Placing a steel bar in a solenoid that connects to an AC supply, then slowly withdrawing the steel bar away from the solenoid in West-East direction\nAnswer:\n(d) The AC supply will mix up the direction of the magnetic domains. In fact, this is one of the methods to demagnetise magnets.\nQuestion 10.\nIn the arrangement shown in figure, there are two coils wound on a non-conducting cylindrical rod. Initially, the key is not inserted. Then, the key is inserted and later removed. Then,\n\n(a) the deflection in the galvanometer remains zero throughout.\n(b) there is a momentary deflection in the galvanometer but it dies out shortly and there is no effect when the key is removed.\n(c) there are momentary galvanometer deflections that die out shortly, the deflections are in the same direction.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\n(c) there are momentary galvanometer deflections that die out shortly, the deflections are in the same direction.\n(d) there are momentary galvanometer deflections that die out shortly, the deflections are in opposite directions. Thus, the galvanometer shows momentary deflections in opposite directions. [NCERT Exemplar]\nAnswer:\n(d) In the given arrangement, whenever an electric current through the first coil is changed, an emf is induced in the coil due to change in magnetic field lines which pass through the neighbouring second coil. When key is inserted and removed, then the magnetic field lines passing through second coil increases and decreases in two cases respectively. Therefore, the direction of current in two cases is opposite. Thus, the galvanometer shows momentary deflections in opposite directions.\n Magnetic Effects of Electric Current Mind Map\nProperties of Magnets\nAttractive property: Magnets attract magnetic materials like \u2013 iron, cobalt, nickel, etc", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nMagnetic Effects of Electric Current Mind Map\nProperties of Magnets\nAttractive property: Magnets attract magnetic materials like \u2013 iron, cobalt, nickel, etc\nDirective property: A freely suspended magnet always aligns in north- south direction\nOpposite poles attract and like poles repel.\nPoles exist in pairs North and South\nRepulsion is a sure test of magnet\nMagnetic Field\nSpace around a magnet in which magnetic effect is experienced\nMagnetic Field Lines\nA line such that the tangent at any point on it gives the direction of the magnetic field at that point.\nProperties of Magnetic Field Lines\nAll field lines are closed curves.\nField lines are close together near the poles.\nTwo field lines never intersect each other.\nMagnetic Field Due to a Current Carrying Conductor\nThe magnetic field around a straight conductor carrying current is in the form of closed circular loops, in a plane perpendicular to the conductor.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nMagnetic Field Due to a Current Carrying Conductor\nThe magnetic field around a straight conductor carrying current is in the form of closed circular loops, in a plane perpendicular to the conductor.\nDirection of magnetic field can be determined by using Right hand thumb rule\n\nSolenoid\nA solenoid is a long cylindrical helix, which produces a magnetic field when an electric current is passed through it.\nThe magnetic field within the solenoid is uniform and parallel to the axis of solenoid.\n\nThe magnetic field due to solenoid depends upon\n(a) number of turns i.e., B \u221d n\n(b) strength of current i.e, B \u221d I\n(c) Nature of material inside solenoid i.e., B \u221d \u03bc\nMagnetic Field Due to a Circular Current Carrying Loop\nAt every\u2019 point of a current carrying loop, the concentric circles representing the magnetic field around it would become larger as we move away.\n\nDirection of magnetic field can be determined by using right hand rule\nForce on a Current Carrying Conductor", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nDirection of magnetic field can be determined by using right hand rule\nForce on a Current Carrying Conductor\nThe force experienced by the conductor\n\nF\n\u2192\n=IL\u00d7\nB\n\u2192\n\nDirection of force can be determined by Fleming\u2019s left hand rule, right hand palm or screw rule\n\nElectromagnet\nA solenoid with a soft iron core. A magnetic field is produced then an electric current flows through a coil of wire.\nUses\nFor lifting and transporting large masses of iron scrap.\nFor loading furnaces with iron.\nElectric bell, telegraph, electric motor, relay, loud speaker, microphone\nFor separating magnetic substances such as iron from other debris\nIn scientific research to study the magnetic properties of a substance in a magnetic field\nElectric Motor\nIt converts electrical energy to mechanical energy.\nIt works on the principle of electromagnetic induction\nParts of Electric Motor\nArmature\nField magnet\nSplit-ring\nCommutator converts alternating voltage into direct voltage across the brushes.\nBrushes\nBattery\nUses", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nParts of Electric Motor\nArmature\nField magnet\nSplit-ring\nCommutator converts alternating voltage into direct voltage across the brushes.\nBrushes\nBattery\nUses\nThe d.c. motors are used in d.c. fans\nThey are used for pumping water\nBig d.c. motors are used for running tramcars and even trains\nElectric Generator\nIt converts mechanical energy to electrical energy.\nIt works on the principle of electromagnetic induction\nParts of Electric Generator\nArmature\nMagnet\nSlip-rings\nBrushes\nSplit ring type commutator for direct current generator\n\nTypes of Electric Generator\nD.C. Generator\nIt is a type of generator which is used to produce induced current which flows in one direction\nA.C. Generator\nIt generates alternating current that changes its polarity after every half rotation\nElectromagnetic Induction\nVoltage is induced by the relative motion between a wire and a magnetic field.\nThe amount of voltage induced depends on how fast the magnetic field lines are entering or leaving the coil.", + "CHAPTER 12-MAGNETIC EFFECTS OF ELECTRIC CURRENT-INTEXT QUESTIONS.txt\nVoltage is induced by the relative motion between a wire and a magnetic field.\nThe amount of voltage induced depends on how fast the magnetic field lines are entering or leaving the coil.\nSafety Devices\nWhen too much current flows or short circuit these devices breaks the circuit.\nThey have less melting point.\nFuses\nMiniature circuit breakers (MCBs)", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\nNCERT Exemplar Class 10 Science Chapter 3 Metals And Non-Metals\nShort Answer Type Questions\n1.Iqbal treated a lustrous, divalent element M with sodium hydroxide. He observed the formation of bubbles in reaction mixture. He made the same observations when this element was treated with hydrochloric acid. Suggest how can he identify the produced gas. Write chemical equations for both the reactions.\nAnswer:\nM + 2NaOH \u2014>Na2M02 + H2(g)\nM + 2HCl \u2014> MCl2 + H2(g)\nBring a burning candle near the gas. If it burns with pop sound, the gas is hydrogen and the element is a metal.\n2 During extraction of metals, electolytic refining is used to obtain pure metals.\n(a) Which material will be used as anode and cathode for refining of silver metal by this process?\n(b) Suggest a suitable electrolyte also.\n(c) In this electrolytic cell, where do we get pure silver after passing electric current?", + "(a) Which material will be used as anode and cathode for refining of silver metal by this process?\n(b) Suggest a suitable electrolyte also.\n(c) In this electrolytic cell, where do we get pure silver after passing electric current?\nAnswer:\n(a) Pure silver rod will be used as cathode and impure silver rod will be used as anode.", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\n(c) In this electrolytic cell, where do we get pure silver after passing electric current?\nAnswer:\n(a) Pure silver rod will be used as cathode and impure silver rod will be used as anode.\n(b) Ag NO3 (aq) can be used as electrolyte.\n(c) Pure silver will be formed at cathode.\nAt anode : Ag \u2014> Ag+ +e\u2013\nAt cathode : Ag+ +e\u2013\u2014> Ag\n3.Generally, when metals are treated with mineral acids, hydrogen gas is liberated but when metals (except Mn and Mg), treated with HNOs, hydrogen is not liberated. Why?\nAnswer.It is because HNO3 is an oxidising agent and it gets reduced to NO if it is dilute and NO2if HNOs is concentrated, it oxidises H2 to H2O.\n4.Compound X and aluminium are used to join railway tracks.\n(a) Identify the compound X.\n(b) Name the reaction.\n(c) Write down its reaction.\nAnswer.\n\n5.What are the constituents of solder alloy? Which property of solder makes it suitable for welding electrical wires?", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\n(b) Name the reaction.\n(c) Write down its reaction.\nAnswer.\n\n5.What are the constituents of solder alloy? Which property of solder makes it suitable for welding electrical wires?\nAnswer. Solder is made up of lead and tin. It has low melting point, therefore, it is used for soldering (welding) electrical wires.\n6.A metal A, which is used in thermite process, when heated with oxygen gives\nan oxide B, which is amphoteric in nature? Identify A and B. Write down the reactions of oxide B with HC1 and NaOH.\nAnswer: \u2018A\u2019 is aluminium.\n\n7.What happens when\n(a) ZnCOs is heated in the absence of oxygen?\n(b) a mixture of Cu2O and Cu2S is heated?\nAnswer:\n\n8.A non-metal A is an important constituent of our food and forms two oxides B and C. Oxide B is toxic whereas oxide C causes global warming.\n(a) Identify A, B and C.\n(b) To which group of periodic table does A belong?\nAnswer:\n(a) A is carbon.", + "(a) Identify A, B and C.\n(b) To which group of periodic table does A belong?\nAnswer:\n(a) A is carbon. It forms two oxides: CO (B) is toxic whereas CO2 (C) causes", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\n(a) Identify A, B and C.\n(b) To which group of periodic table does A belong?\nAnswer:\n(a) A is carbon. It forms two oxides: CO (B) is toxic whereas CO2 (C) causes\nglobal warming as it absorbs heat radiations from atmosphere.\n(b) A belongs to group 14 of periodic table.\n9.An element A reacts with water to form a compound B which is used in white washing. The compound B on heating forms an oxide C which on treatment with water gives back B. Identify A, B and C and give the reactions involved.\n\u2018A\u2019 is calcium metal, \u2018B\u2019 is calcium hydroxide, \u2018C\u2019 is calcium oxide.\nAnswer:\n\n10. A metal M does not liberate hydrogen from acids but reacts reacts with oxygen to give a black colour product. Identify M and black coloured product and also explain the reaction of M with oxygen.\nAnswer:\nM is \u2018Cu\u2019. It does not liberate hydrogen with dilute acid as it is less reactive than hydrogen gas.", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\nAnswer:\nM is \u2018Cu\u2019. It does not liberate hydrogen with dilute acid as it is less reactive than hydrogen gas.\n\n11. A solution of CuS04 was kept in an iron pot. After few days, the iron pot was found to have a number of holes in it. Explain the reason in terms of reactivity. Write the equation of the reaction involved.\nAnswer: Iron is more reactive than copper. Therefore, it displaces copper from copper sulphate solution.\n\n12. When a metal \u2018X\u2019 is treated with cold water, It gives base salt \u2018X\u2019 with molecular formula XOH (Molecular mass = 40) and liberate a gas Z which easily catches fire. Identify X, Y and Z and also write the reaction involved.\nAnswer.\n\u2018X\u2019 is Na \u2018Y\u2019 is NaOH \u2018Z\u2019 is H2 gas\n\n13. A non-metal \u2018X\u2019 exists in two different forms \u2018Y\u2019 and \u2018Z\\ \u2018Y\u2019 is hardest natural substance, whereas \u2018Z\u2019 is good conductor of electricity.", + "Answer.\n\u2018X\u2019 is Na \u2018Y\u2019 is NaOH \u2018Z\u2019 is H2 gas\n\n13. A non-metal \u2018X\u2019 exists in two different forms \u2018Y\u2019 and \u2018Z\\ \u2018Y\u2019 is hardest natural substance, whereas \u2018Z\u2019 is good conductor of electricity. Identify X, Y and Z.\nAnswer: \u2018X\u2019 is carbon \u2018Y\u2019 is diamond \u2018Z\u2019 is graphite.\n\u2018Y\u2019 (diamond) is hardest natural substance.", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\nAnswer: \u2018X\u2019 is carbon \u2018Y\u2019 is diamond \u2018Z\u2019 is graphite.\n\u2018Y\u2019 (diamond) is hardest natural substance.\n\u2018Z\u2019 (graphite) is good conductor of electricity.\n14. The following reaction takes place when aluminium powder is heated with MnO2 :\n3MnO2(s) + 4Al(s)\u2014\u2013> 3Mn(l) + 2Al2O3(l) + Heat\nMolten Molten\n(a) Is aluminium getting reduced ?\n(b) Is MnO2 getting oxidised ?\nAnswer:\n(a) No, aluminium is getting oxidised.\n(b) No, MnO2 is getting reduced.\n15. A metal that exists as liquid at room temperature is obtained by heating its sulphide in the presence of air. Identify the metal and its ore and give the reaction involved.\nAnswer: The metal is mercury.", + "(b) No, MnO2 is getting reduced.\n15. A metal that exists as liquid at room temperature is obtained by heating its sulphide in the presence of air. Identify the metal and its ore and give the reaction involved.\nAnswer: The metal is mercury.\n2HgS(s) + 3O2(g)\u2014\u2013> 2HgO(s) + 2SO2(g)\nHgS(s) + 2HgO(s)\u2014\u2013> 3Hg(l) + SO2(g)\n16. Give the formulae of the stable binary compounds that would be formed by the combination of following pairs of elements :\n(a) Mg and N2 (b) Li and 02\n(c) A1 and Cl2 (d) K and O2\nAnswer:\n(a) 3Mg + N2\u2014\u2013> Mg3N2\n(b) 4Li + O2\u2014\u2013> 2Li20\n(c) 2Al + 3Cl2\u2014\u2013> 2AlCl3", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\n(a) Mg and N2 (b) Li and 02\n(c) A1 and Cl2 (d) K and O2\nAnswer:\n(a) 3Mg + N2\u2014\u2013> Mg3N2\n(b) 4Li + O2\u2014\u2013> 2Li20\n(c) 2Al + 3Cl2\u2014\u2013> 2AlCl3\n(d) 4K + O2 \u2014\u2013> 2K2O\n17.Give two examples each of the metals that are good conductors and poor conductors of heat respectively.\nAnswer: Good conductors of heat are copper and silver. Poor conductors of heat are lead and mercury.\n18. Name-one metal and one non-metal that exist in liquid state at room temperature. Also name two metals having melting point less than 310 K (37 \u00b0C).\nAnswer: Metal in liquid state is mercury, non-metal in liquid state is bromine.\nGallium (Ga) and Cesium (Cs) have melting point below 310 K.\n19.", + "Also name two metals having melting point less than 310 K (37 \u00b0C).\nAnswer: Metal in liquid state is mercury, non-metal in liquid state is bromine.\nGallium (Ga) and Cesium (Cs) have melting point below 310 K.\n19. An alkali metal A gives a compound B (molecular mass = 40) on reacting with water. The compound B gives a soluble compound C on treatment with aluminium oxide. Identify A, B and C and give the reaction involved.\nAnswer:\n\n\u2018C\u2019 is sodium meta aluminate, soluble in water.", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\nAnswer:\n\n\u2018C\u2019 is sodium meta aluminate, soluble in water.\n20. Give the reaction involved during extraction of zinc from its ore by\n(a) roasting of zinc ore.\n(b) calcination of zinc ore.\nAnswer:\n\n21. An element forms an oxide, A2Os which is acidic in nature. Identify A as a metal or non-metal.\nAnswer: A is non-metal as non-metallic oxides are acidic in nature.\nLong Answer Type Questions\n22. A non-metal A which is the largest constituent of air, when heated with H2 in 1 : 3 ratio in the presence of catalyst (Fe) gives a gas B. On heating with O2, it gives an oxide C. If this oxide is passed into water in the presence of air, it gives an acid D which acts as a strong oxidising agent.\n(a) Identify A, B, C and D.\n(b) To which group of periodic table does this non-metal belong?\nAnswer:\n(a) N2 is largest constituent of air, when heated with H2 in the ratio of 1 : 3 .", + "(a) Identify A, B, C and D.\n(b) To which group of periodic table does this non-metal belong?\nAnswer:\n(a) N2 is largest constituent of air, when heated with H2 in the ratio of 1 : 3 .\nin the presence of catalyst (Fe) gives a gas NH3(B).", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\nAnswer:\n(a) N2 is largest constituent of air, when heated with H2 in the ratio of 1 : 3 .\nin the presence of catalyst (Fe) gives a gas NH3(B).\n\nNitrogen reacts with oxygen on heating to form nitrogen monoxide \u2018C\u2019, which gets oxidised in the presence of O2 to nitrogen dioxide. Nitrogen dioxide dissolves in water in the presence of oxygen to form nitric acid which is an oxidising agent.\n\n\u2018A\u2019 is N2, \u2018B\u2019 is NH3, \u2018C\u2019 is NO and \u2018D\u2019 is HNO3.\n(b) Group 15\n23. Explain the following:\n(a) Reactivity of A1 decreases if it is dipped in HNO3.\n(b) Carbon cannot reduce the oxides of Na or Mg.\n(c) NaCl is not a conductor of electricity in solid state whereas it does conduct electricity in aqueous solution as well as in molten state.\n(d) Iron articles are galvanised.\n(e) Metals like Na, K, Ca and Mg are never found in their free state in nature.", + "(c) NaCl is not a conductor of electricity in solid state whereas it does conduct electricity in aqueous solution as well as in molten state.\n(d) Iron articles are galvanised.\n(e) Metals like Na, K, Ca and Mg are never found in their free state in nature.\nAnswer:\n(a) It is due to the formation of oxide layer on its surface.", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\n(d) Iron articles are galvanised.\n(e) Metals like Na, K, Ca and Mg are never found in their free state in nature.\nAnswer:\n(a) It is due to the formation of oxide layer on its surface.\n(b) It is because Na or Mg are strong reducing agents, therefore, carbon cannot reduce their oxides.\n(c) NaCl does not conduct electricity in solid state as it does not have free ions to move in solid state but in aqueous solution and in molten states ions are free to move.\n(d) Iron articles are galvanised to protect them from rusting.\n(e) It is because these metals are highly reactive and occur in the form of their compounds.\n24.An element A burns with golden flame in air. It reacts with another element B, atomic number 17 to give a product C. An aqueous solution of product C on electrolysis gives a compound D and liberates hydrogen. Identify A, B, C and D. Also write down the equations for the reactions involved, HOTS;\nAnswer: A is sodium as it burns with golden flame in air.", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\nAnswer: A is sodium as it burns with golden flame in air.\n\u2018B\u2019 (17) has electronic configuration 2, 8, 7 so \u2018B\u2019 is chlorine\n\nA is Na, \u2018B\u2019 is Cl2 , \u2018C\u2019 is NaCl and \u2018D\u2019 is NaOH.\n25.Give the steps involved in the extraction of metals of low and medium reactivity from their respective sulphide ores.\nAnswer:\n\nMetal sulphide like HgS reacts with 02 to form mercuric oxide and sulphur dioxide. Mercuric oxide is unstable changes to Hg(l) which can be refined by distillation.\n2HgS 4- 3O2 \u2014\u2013> 2HgO + 2SO2\nHgS + 2HgO \u2014\u2013> 3Hg + SO2\nSulphide ore like Zn, on roasting (in presence of O2 ) gives metal oxide which on reduction with metal sulphide (ZnS) or suitable reducing agent gives metal which is refined by electrolyte refining.", + "2ZnS + 3O2 \u2014\u2013> 2ZnO + 2SO2\n2ZnO + ZnS \u2014\u2013> 3Zn + SO2\nAlternative method of reduction :\nZnO + C \u2014\u2013> Zn + CO\nPure zinc is obtained by electrolytic refining.", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\n2ZnS + 3O2 \u2014\u2013> 2ZnO + 2SO2\n2ZnO + ZnS \u2014\u2013> 3Zn + SO2\nAlternative method of reduction :\nZnO + C \u2014\u2013> Zn + CO\nPure zinc is obtained by electrolytic refining.\n26.Of the three metals X, Y and Z, X reacts with cold water, Y with hot water and Z with steam only. Identify X, Y and Z and also arrange them in order of increasing reactivity.\nAnswer: X\u2019 is sodium or potassium metal.\n2Na(s) + 2HsO(l) \u2014\u2013> 2NaOH(aq) + H2(g)\n\u2018X\u2019 \u2018cold\u2019\nOr\n2K(s) + 2H2O(l) \u2014\u2013> 2KOH(aq) + H2(g)\n\u2018X\u2019 \u2018cold\u2019\n\nFe < Mg < Na < K is an increasing order of reactivity.\n27.Two ores A and B were taken.", + "27.Two ores A and B were taken. On heating, ore A gives CO2 whereas, ore B gives SO2 . What steps will you take to convert them into metals ?\nAnswer:Since ore \u2018A\u2019 gives CO2 and ore \u2018B\u2019 gives SO2 . Therefore, A is metal carbonate ore and \u2018B\u2019 is metal sulphide ore.\nMetal carbonate on heating gives \u2018MO\u2019.\n\n\u2018MO\u2019, on reduction with coke gives metal and carbon monoxide.\n\nMetal \u2018M\u2019 is obtained which can be purified by electrolytic refining.", + "CHAPTER 3- METALS AND NON METALS- EXEMPLER QUESTIONS.txt\nMetal carbonate on heating gives \u2018MO\u2019.\n\n\u2018MO\u2019, on reduction with coke gives metal and carbon monoxide.\n\nMetal \u2018M\u2019 is obtained which can be purified by electrolytic refining.\nMetal sulphide is heated in presence of oxygen to form metal oxide and S02.\n\nMetal sulphide reacts with metal oxide to form impure metal and S02.\n\nMO can also be reduced with carbon to form metal. The metal \u2018M\u2019 from ore \u2018B\u2019 is obtained.\n\n28.Impure metal is purified by electrolytic refining\n(i) Given below are steps for extraction of copper from its ore. Write the reaction involved.\n(a) Roasting of copper (I) sulphide\n(b) Reduction of copper (I) oxide with copper (I) sulphide\n(c) Electrolytic refining\n(ii) Draw the neat and labelled diagram for electrolytic refining of copper.", + "Write the reaction involved.\n(a) Roasting of copper (I) sulphide\n(b) Reduction of copper (I) oxide with copper (I) sulphide\n(c) Electrolytic refining\n(ii) Draw the neat and labelled diagram for electrolytic refining of copper.\nAnswer:\n(i)\n(a) 2CU2S(S) + 3O2(g) \u2014\u2014\u2013> 2Cu2O(s) + 2SO2(g)\n(b) 2CU2O(S) + Cu2S(s) \u2014\u2014\u2013> 6CU(s) + SO2(g)", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nQuestion-1\na) Due to which property of metals are metallic foils prepared?\nb) Write uses of aluminium foils\nSolution:\na) Metallic foils are prepared, making use of the malleable property of metals.\nb) Aluminium foils are used for\n(i) wrapping chocolates and food stuff\n(ii) to prepare hydrogen.\nQuestion-2\nName one non-metal and one metal, which are in liquid state at room temperature.\nSolution:\nBromine is the non\u2013metal and mercury is the metal which are in liquid state at room temperature.\nQuestion-3\nGive reasons for the following\ni) Zinc oxide is considered an amphoteric oxide.\nii) Non-metals in general do not displace hydrogen from dilute acids.\niii) Metals conduct electricity.\nSolution:\ni) Zinc oxide is considered an amphoteric oxide because it shows both acidic and basic behaviour.\nii) Non-metals do not react with dilute acids hence do not displace hydrogen from dilute acids.", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\ni) Zinc oxide is considered an amphoteric oxide because it shows both acidic and basic behaviour.\nii) Non-metals do not react with dilute acids hence do not displace hydrogen from dilute acids.\n\niii) thatMetals conduct electricity because they contain free electrons which can move easily through the metal and conduct electric current.\nQuestion-4\nIdentify the following oxides with neutral, acidic, basic and amphoteric oxides.\ni) Al2O3 ii) MgO iii) SO2 iv) CO; v) Na2O vi) N2O vii) ZnO viii) CO2 ix) SO3 x) CaO.", + "i) Al2O3 ii) MgO iii) SO2 iv) CO; v) Na2O vi) N2O vii) ZnO viii) CO2 ix) SO3 x) CaO.\nSolution:\nNeutral oxides: CO; N2O\nAcidic oxides: SO2; CO2; SO3\nBasic oxides: MgO; Na2O; CaO\nAmphoteric oxides: Al2O3 ; ZnO\nQuestion-5\nBy what processes do you concentrate the following ores:\ni) PbS(galena)\nii) Fe2O3(haematite)\niii) Al2O3.2H2O (bauxite).\nSolution:\ni) PbS (galena) is concentrated by froth floatation method.\nii) Fe2O3 (haematite) is concentrated by magnetic separation.\niii) Al2O3.2H2O (bauxite) is concentrated by leaching process (chemical method).\nQuestion-6", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nii) Fe2O3 (haematite) is concentrated by magnetic separation.\niii) Al2O3.2H2O (bauxite) is concentrated by leaching process (chemical method).\nQuestion-6\nWhy is that iron does not occur as a metal in the crust of the earth? What are the common compound forms in which iron occurs? Which of these is more often used for extracting iron? Why are carbon and limestone mixed with iron ore before feeding it into the blast furnace? Write the chemical equations for the reduction step and the slag formation step.\nSolution:", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nSolution:\nIron is quite reactive and so it does not occur as a metal in the crust of the earth. The common compounds formed are its oxide, a carbonate and a sulphide. The oxide ore of iron called haematite is more often used for extracting iron. The carbon and limestone which is mixed with iron ore is called the charge. This charge is responsible for the formation of iron metal. Coke acts the reducing agent and addition of limestone helps to remove the earthy impurities like sand from the blast furnace by forming fusible slag.\nReduction of Iron(III)oxide or haematite to Iron: carbon monoxide reduces iron(III)oxide to iron metal\n\nQuestion-7\nBy what chemical processes do you obtain the following metals?\ni) Mercury from HgO ii) Iron from Fe2O3 iii) Manganese from MnO2 iv) Sodium form NaCl v) Aluminium from Al2O3.\n\nQuestion-8\ni) What is the composition of molten slag, formed in the extraction of iron in the blast furnace?\nii) What is the use of slag?\nSolution:", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nQuestion-8\ni) What is the composition of molten slag, formed in the extraction of iron in the blast furnace?\nii) What is the use of slag?\nSolution:\ni) Molten slag in the extraction of iron is calcium silicate(CaSiO3).\nii) Slag is used for constructions of roads.\nQuestion-9\nWrite the name of the compound, when iron is corroded?\nSolution:\nWhen iron is corroded rust is formed which is hydrated iron (III) oxide (Fe2O3.x.H2O)\nQuestion-10\nGive reasons for the following\ni. Na, K and Ca metals form hydrides by combination with hydrogen gas, but most other metals do not.\nii. Aluminium easily combines with oxygen but still it can be used for making kitchen utensils.\nSolution:\ni. Metals like Na, K and Ca can force the hydrogen atoms to accept electrons given by them and form hydrides.\nii. Aluminium forms a protective oxide layer when it combines with oxygen and hence it is used for making kitchen utensils.\nQuestion-11\ni) Among iron and aluminium which rusts easily?", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nii. Aluminium forms a protective oxide layer when it combines with oxygen and hence it is used for making kitchen utensils.\nQuestion-11\ni) Among iron and aluminium which rusts easily?\nii) What are the conditions, necessary for rusting of iron?\nSolution:\ni) Among iron and aluminium, iron rusts easily. It forms Al2O3. That is why iron pipes are wrapped with aluminium metal to prevent rusting of iron.\nii) 1) Air or oxygen 2) Water, containing dissolved air.\nThese are necessary conditions for rusting of iron.\nQuestion-12\nName an ore of zinc other than zinc oxide. By what process can this ore be converted to zinc oxide.\nSolution:\nZinc blende(ZnS) is an ore of zinc other than zinc oxide. Roasting is the process by which ZnS can be converted into zinc oxide.\nQuestion-13\nWhat property of hydrogen is observed in the following experiments?\ni) Balloon, filled with hydrogen, floats.\nii) Pop sound is heard when a burning splint is introduced at the mouth of hydrogen filled jar.", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\ni) Balloon, filled with hydrogen, floats.\nii) Pop sound is heard when a burning splint is introduced at the mouth of hydrogen filled jar.\niii) When black cupric oxide is heated in a stream of hydrogen, reddish colour is obtained.\nSolution:\ni) Hydrogen is light. So hydrogen filled balloons float.\nii) Pop sound is due to the combustible property of hydrogen. Hydrogen burns with oxygen at the mouth of the hydrogen filled jar.\niii) Hydrogen has reducing property. Hence it reduces black cupric oxide to red copper metal.\nQuestion-14\nSelect the non-metals amongst the following\ni) Bismuth ii) Sulphur iii) Arsenic iv) Hydrogen.\nSolution:\nHydrogen and Sulphur are the non-metals.\nQuestion-15\nHow do you test ammonia gas?\nSolution:\nAmmonia gas is tested, by showing a drop of conc. HCl by means of a glass rod to it, when dense white fumes of ammonium chloride are formed.\nAmmonium chloride (dense white fumes).\nQuestion-16\nWrite chemical equations for reactions taking place when", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nAmmonium chloride (dense white fumes).\nQuestion-16\nWrite chemical equations for reactions taking place when\ni) Cinnabar is heated in air.\nii) Concentrated sulphuric acid and sulphur are heated together. Calcium metal reacts with water.\nSolution:\n\nQuestion-17\nWhat are the uses of ammonia?\nSolution:\ni) Ammonia is used in the manufacture of fertilizers like ammonium sulphate and ammonium nitrate.\nii) To manufacture nitric acid by Ostwald\u2019s process.\niii) To prepare dyes, explosives, cellulose acetate etc.\nQuestion-18\nWhy do some metals like, Na, K, Ca, Mg not occur in nature as free elements?\nSolution:", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\niii) To prepare dyes, explosives, cellulose acetate etc.\nQuestion-18\nWhy do some metals like, Na, K, Ca, Mg not occur in nature as free elements?\nSolution:\nMetals like Na, K, etc (alkali metals) and Ca, Mg etc (alkaline earth metals) are very reactive and hence they react with atmospheric oxygen and carbon dioxide and also with other non-metals like sulphur present in the earth\u2019s crust to form compounds like oxides, carbonates, sulphides, sulphates and chlorides. So they do not occur in free state, but are found in the form of the above compounds.\nQuestion-19\nWrite chemical equations for reactions taking place when:\ni) Manganese dioxide is heated with aluminium powder.\nii) Steam is passed over hot iron.\niii) A mixture of ammonia and oxygen is passed over platinum heated to 1073 K.\nSolution:\n\nQuestion-20\nWhy are metals like sodium, calcium, magnesium, aluminium etc. obtained by electrolytic reduction?\nSolution:", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nSolution:\n\nQuestion-20\nWhy are metals like sodium, calcium, magnesium, aluminium etc. obtained by electrolytic reduction?\nSolution:\nMetals like Na, Ca, Mg, Al etc, which are high up in the activity series are electropositive elements and possess a high affinity for oxygen. So their oxides are very much stable and cannot be reduced with common reducing agents like carbon. So carbon reduction process cannot be applied to prepare these metals. They can only be prepared by electrolytic reduction of their fused chlorides or oxides.\nQuestion-21\nState briefly how you will extract iron from its ore. Draw a neat and labelled diagram. Write all the chemical reactions involved in it.\nSolution:\nIron is usually extracted from its chief ore, Haemitite.\nThe various steps involved in the production of iron metal from haematite are as follows:\n1. Concentration of ore", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nSolution:\nIron is usually extracted from its chief ore, Haemitite.\nThe various steps involved in the production of iron metal from haematite are as follows:\n1. Concentration of ore\nThe concentration of haematite or is done by the method of hydraulic washing. The ore is spilt into small pieces and then washed in a stream of water to remove sand, clay. In this way ,a fairly concentrated ore is obtained and usually there is no need of any further concentration.\n2. Calcination\nThe washed iron ore is then strongly heated in the absence of air to expel water sticking to it.\n3. Reduction\nThe washed and dried ore is mixed with weighed quantities of coke and limestone and put into a blast furnace from the top .A blast of hot air is blown into the furnace from near its bottom. This air is to supply for the burning of coke.\nThe two reactions which take place in the blast furnace leading to the formation of iron metal are: formation of carbon monoxide and reduction of haematite.\n\n\nQuestion-22", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nThe two reactions which take place in the blast furnace leading to the formation of iron metal are: formation of carbon monoxide and reduction of haematite.\n\n\nQuestion-22\nWrite the chemical formula for the following.\na) Alumina b) Silica c) Rust d) Oleum e) Slag f) Sodium amalgam g) Bauxite h) Haematite.\nSolution:\nQuestion-23\nIn Frasch process of digging out sulphur from a deep underground, how has it made possible to life sulphur from the bed to surface of each? What is the basic principle on which extraction of sulphur is based? Draw a neat and labelled diagram. Besides elemental form, in what compound forms is sulphur found. Give two physical properties of sulphur.\nSolution:", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nSolution:\nIn the Frasch process in order to bring out sulphur from the underground deposits, a hole of about 30 cm diameter is bored up to the sulphur bed. It is based on the principle that sulphur has a comparatively low melting point of 1150 or 3880 due to which it can be melted by introducing super-heated water in the underground sulphur deposits. This mixture of molten sulphur and water is then brought out by the force of hot, compressed air.\nThree concentric pipes A, B, C of different diameters are put down in the boring.", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nThree concentric pipes A, B, C of different diameters are put down in the boring.\n\nThe three pipes are arranged one inside the other as shown in the above figure. Super-heated water under pressure is forced down through the outermost pipe A. The heat given out by hot water melts the underground sulphur. A jet of hot, compressed air is then sent down through the innermost pipe C. The force of compressed air brings the mixture of molten sulphur and water through the central pipe B. The sulphur water emulsion is kept in settling tanks when sulphur gets cooled and separates from water in the form of a yellow solid. The sulphur obtained by this method is 99.5% pure.\nQuestion-24\nThe copper vessels used in homes are found to be coated with green colour. Answer the following.\na) What is the chemical composition of the green coating?\nb) Why do copper vessels form such a green coating?\nc) What is the name of the phenomenon, responsible for the green coating?", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\na) What is the chemical composition of the green coating?\nb) Why do copper vessels form such a green coating?\nc) What is the name of the phenomenon, responsible for the green coating?\nd) Copper vessels are best cleaned with tamarind or lemon juice, but not with soap. Why?\nSolution:\na) Green coloured coating on copper is due to the formation of basic copper carbonate, CuCO3.Cu(OH)2.\nb) Copper forms green coating of basic copper carbonate as it reacts with water, atmospheric oxygen and carbon dioxide.\nc) Corrosion phenomenon is responsible for the formation of basic copper carbonate (green).\nd) Since basic copper carbonate is basic in nature, it can be removed by tartaric acid present in tamarind or citric acid, present in lemon. Soap is basic in nature. So copper vessels are best cleaned with tamarind or lemon.\nQuestion-25\nGive reasons for the following\ni. Silicon counts among metalloids\nii. Carbon is not used for making aluminium from aluminium oxide.", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nQuestion-25\nGive reasons for the following\ni. Silicon counts among metalloids\nii. Carbon is not used for making aluminium from aluminium oxide.\niii. For making hydrogen by reaction with hydrochloric acid, granulated zinc is preferred to a block of zinc.\nSolution:\ni. Silicon shows properties of both metals and non-metals and hence silicon counts among metalloids.\nii. Carbon has less affinity for oxygen than Aluminium and hence it cannot eliminate oxygen from Aluminium oxide.\niii. Zinc granules are preferred to a block of zinc because it offers a large surface area for the reaction with the acid.\nQuestion-26\nExplain\ni) Why sugar chars (turns black) when conc. H2SO4 is added to it?\nii) What is the blue colour of crystalline copper sulphate due to?\n\nii) Blue colour of crystalline copper sulphate is due to five water molecules of hydration in CuSO4 5H2O.\nQuestion-27\nWrite chemical equations for reactions taking place when\ni. Zinc carbonate is calcined.", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nQuestion-27\nWrite chemical equations for reactions taking place when\ni. Zinc carbonate is calcined.\nii. Ammonia gas is passed over heated copper (II) oxide.\nSolution:\n\nQuestion-28\n(i) What is the colour of potassium dichromate solution?\n(ii) What is the colour of Cr3+ ion?\n(iii) What property of SO2 is responsible for bringing change of colour of potassium dichromate?\nSolution:\n(i) Potassium dichromate solution is orange yellow colour.\n(ii) The colour of Cr3+ ion is green.\n(iii) Reducing property of sulphur dioxide is responsible to change the orange yellow colour of K2Cr2O7 to green colour.\nQuestion-29\nHow is chloride of lime chemically different from calcium chloride? Why does chloride of lime gradually lose its chlorine when kept exposed to air?\nSolution:", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nQuestion-29\nHow is chloride of lime chemically different from calcium chloride? Why does chloride of lime gradually lose its chlorine when kept exposed to air?\nSolution:\nChemically, chloride of lime is calcium oxy chloride, CaOCl2, also called bleaching powder. When exposed to air it gradually lose its chlorine because it reacts with carbon dioxide present in air to produce calcium carbonate and chlorine gas.\nQuestion-30\nWrite short notes on the bleaching action of sulphur dioxide.\nSolution:\na) Sulphur dioxide bleaches the vegetable colours in the presence of moisture.\nb) It bleaches the vegetable colours by reduction. It removes oxygen from coloured material.\nc) Its bleaching action is temporary, because atmospheric oxygen again oxidises the bleached article back to its original colour.\nd) SO2 is used to bleach delicate fabrics like wool, silk etc.\nQuestion-31\n(a) What is the atomicity of sulphur?", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nd) SO2 is used to bleach delicate fabrics like wool, silk etc.\nQuestion-31\n(a) What is the atomicity of sulphur?\n(b) How do you bring the change of rhombic sulphur to monoclinic sulphur and vice versa?\n(c) What is the action of conc. HNO3 on sulphur?\nSolution:\n(a) Atomicity of sulphur is 8. Its formula is S8.\n\nQuestion-32\nName two metals used in making fuse wires.\nSolution:\nCopper and Aluminium are the two metals used in making fuse wires.\nQuestion-33\n(i) How can ammonia be converted to NO2 and nitric acid?\n(ii) What is liquor ammonia?\nSolution:\n\nQuestion-34\nComplete the following reactions\n\nQuestion-35\nDefine the following terms\ni) Mineral\nii) Ore\niii) gangue.\nSolution:\ni) The inorganic elements or compounds which occur naturally in the earth\u2019s crust are known as minerals.\nii) The minerals from which a metal can be profitably extracted are called ores.\niii) The impurity of sand and rocky materials present in the ore is known as gangue.", + "Solution:\ni) The inorganic elements or compounds which occur naturally in the earth\u2019s crust are known as minerals.\nii) The minerals from which a metal can be profitably extracted are called ores.\niii) The impurity of sand and rocky materials present in the ore is known as gangue.\nQuestion-36", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nii) The minerals from which a metal can be profitably extracted are called ores.\niii) The impurity of sand and rocky materials present in the ore is known as gangue.\nQuestion-36\na) State why hydrogen is said to be a clean fuel?\nb) What property of hydrogen is involved in being a good fuel?\nc) What is hydrogenation of oils? Write one application.\nSolution:\na) Hydrogen is a clean fuel because its reaction with oxygen produces only water. It does not form gases like CO2, NO2 etc, which pollute air.\nb) Hydrogen, on combustion with oxygen, produces water and a lot of heat. This is an exothermic reaction.\nc) Oils are unsaturated molecules, containing double bond. So hydrogen atoms are added at the double bonds of the molecules of oil. The reaction of hydrogen with oils is known as hydrogenation.\nThe hydrogenation reaction takes place at 473 K and in the presence of nickel catalyst.\nVanaspati or dalda ghee is prepared by hydrogenation of vegetable oils.\nQuestion-37", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nThe hydrogenation reaction takes place at 473 K and in the presence of nickel catalyst.\nVanaspati or dalda ghee is prepared by hydrogenation of vegetable oils.\nQuestion-37\nWhat is the formula of Ammonia? Write the flow diagram for the manufacture of ammonia by Haber\u2019s process. State any two of its physical properties. Write the equation for its reaction with HCl.\nSolution:\nThe formula of Ammonia is NH3.\nThe flow diagram of the manufacture of ammonia by Haber\u2019s process is given below:\n\nPhysical Properties\n1. Ammonia is a colourless gas having a characteristic pungent smell.\n2. Ammonia is highly soluble in water.\nQuestion-38\nHow do metals react with hydrogen? Explain with an example.\nSolution:\nOnly very few reactive metals like sodium, potassium, calcium and magnesium react with hydrogen to form metal hydrides. Example when hydrogen gas is passed overheated sodium, sodium hydride is formed.\nQuestion-39\nName a metal which is stored in kerosene oil.\nSolution:", + "CHAPTER 3- METALS AND NON METALS- EXTRA QUESTIONS.txt\nQuestion-39\nName a metal which is stored in kerosene oil.\nSolution:\nSodium is a metal which is stored in kerosene oil.\nQuestion-40\nMetals are electropositive in nature. Explain.\nSolution:\nMetals are electropositive in nature because they can form positive ions by the loss of electrons.\nQuestion-41\nExplain the displacement of hydrogen by a metal from an acid with an example. Give the equation to support the same.\nSolution:\n\nQuestion-42\nWhy is sodium kept in kerosene?\nSolution:\nSodium reacts so vigorously that they catch fir if it is kept in the open air and hence sodium is kept in kerosene.\nQuestion-43\nName the element or compound which is associated with Hall\u2019s process.\nSolution:\nAluminium is associated with Hall\u2019s process.\nQuestion-44\nWhich metal is used in nuclear reactors and aerospace projects?\nSolution:\nAluminium is used in nuclear reactors and aerospace projects.", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nNCERT Solutions for Class 10 Science Chapter 2 Intext Questions\nPage Number: 18\nQuestion 1\nYou have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution, respectively. If you are given only red litmus paper, how will you identify the contents of each test tube ?\nAnswer:\n(i) Put the red litmus paper in all the test tubes, turn by turn. The solution which turns red litmus to blue will be a basic solution. The blue litmus paper formed here can now be used to test the acidic solution.\n(ii) Put the blue litmus paper obtained above in the remaining two test-tubes, turn-by-turn. The solution which turns the blue litmus paper to red will be the acidic solution.\n\n(iii) The solution which has no effect on any litmus paper will be neutral and hence it will be distilled water.\nPage Number: 22\nQuestion 1\nWhy should curd and sour substances not be kept in brass and copper vessels ?\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nPage Number: 22\nQuestion 1\nWhy should curd and sour substances not be kept in brass and copper vessels ?\nAnswer:\nCurd and sour substances should not be kept in brass and copper vessels because these and other sour food-stuffs contain acids which can react with the metal of the vessel to form poisonous metal compounds which can cause food poisoning and affect our health adversely.\nQuestion 2\nWhich gas is usually liberated when an acid reacts with a metal ? Illustrate with an example. How will you test for the presence of this gas ?\nAnswer:\n(i) Hydrogen (H2) gas is liberated when an acid reacts with a metal.\n(ii) Illustration : Set up the apparatus as shown in the given figure. Take some zinc granules in the test tube. Add about 5 mL dilute hydrochloric acid slowly. Soon the reaction between zinc and hydrochloric acid starts and hydrogen gas is evolved.\n\n(iii) Test for H2 gas :\nH2 gas is not soluble in water. When passed through soap solution, it gets trapped into bubbles.", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\n(iii) Test for H2 gas :\nH2 gas is not soluble in water. When passed through soap solution, it gets trapped into bubbles.\nBring a burning candle near the soap bubble filled with gas. The soap bubble bursts and hydrogen gas burns with a pop sound.\nQuestion 3\nMetal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.\nAnswer:\nAs the end product is calcium chloride and the gas formed is carbon dioxide, the metal compound A must be calcium carbonate. Therefore, the reaction between calcium carbonate and hydrochloric acid is\n\nPage Number: 25\nQuestion 1\nWhy do HCl, HNO3, etc show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character ?\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nPage Number: 25\nQuestion 1\nWhy do HCl, HNO3, etc show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character ?\nAnswer:\nH+ ions in aqueous solution are responsible for acidic character. HCl, HNO3, etc. give H+ ions in water while alcohol and glucose do not give H+ ion in water. Therefore, alcohol and glucose do not show acidic character.\nQuestion 2\nWhy does an aqueous solution of an acid conduct electricity ?\nAnswer:\nThe aqueous solution of an acid conducts electricity due to the presence of charged particles called ions in it.\nQuestion 3\nWhy does dry HCl gas not change the colour of the dry litmus paper ?\nAnswer:\nDry HCl gas does not give H+ ions and therefore does not change the colour of dry litmus paper.\nQuestion 4\nWhile diluting an acid, why is it recommended that the acid should be added to water and not water to the acid ?\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 4\nWhile diluting an acid, why is it recommended that the acid should be added to water and not water to the acid ?\nAnswer:\nWhile diluting an acid it is recommended that the acid should be added to water and not water to the acid because if water is added to concentrated acid to dilute it, then a large amount of heat is evolved at once. This heat changes some of the water to steam explosively which can splash the acid on one\u2019s face or clothes and cause acid burns.\nQuestion 5\nHow is the concentration of hydronium ions (H3O+) affected when a solution of an acid is diluted ?\nAnswer:\nWhen a given amount of an acid is added to water, there is a fixed number of hydronium ions per volume of the solution. On dilution, the number of hydronium ions per volume decreases and concentration decreases.\nQuestion 6\nHow is the concentration of hydroxide ions (OH\u2013) affected when excess base is dissolved in a solution of sodium hydroxide ?\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 6\nHow is the concentration of hydroxide ions (OH\u2013) affected when excess base is dissolved in a solution of sodium hydroxide ?\nAnswer:\nThe concentration of hydroxide ions will increase when excess base is dissolved in a solution of sodium hydroxide, but it happens to a limited extent only after which the concentration becomes almost constant.\nPage Number: 28\nQuestion 1\nYou have two solutions A and B. The pH of solution A is 6 and pH of solution B is 8. Which solution has more hydrogen ion concentration? Which of this is acidic and which one is basic ?\nAnswer:\nA pH value of less than 7 indicates an acidic solution, while greater than 7 indicates a basic solution. Since solution A has more hydrogen ion concentration, solution A is acidic and solution B is basic.\nQuestion 2\nWhat effect does the concentration of H+ (aq) ions have on the nature of the solution ?\nAnswer:\nMore the concentration of H+ ions, higher the acidic nature of the solution.\nQuestion 3", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 2\nWhat effect does the concentration of H+ (aq) ions have on the nature of the solution ?\nAnswer:\nMore the concentration of H+ ions, higher the acidic nature of the solution.\nQuestion 3\nDo basic solutions also have H+ (aq) ions ? If yes, then why are these basic ?\nAnswer:\nBasic solutions have H+ (aq) ions. But these are far less in number than OH\u2013 ions that is responsible for their basic nature.\nQuestion 4\nUnder what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate) ?\nAnswer:\nIf the soil is too acidic (having low pH) then it is treated with materials like quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate).\nPage Number: 33\nQuestion 1\nWhat is the common name of the compound CaOCl2 ?\nAnswer:\nBleaching powder.", + "Page Number: 33\nQuestion 1\nWhat is the common name of the compound CaOCl2 ?\nAnswer:\nBleaching powder.\nQuestion 2\nName the substance which on treatment with chlorine yields bleaching powder.\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nPage Number: 33\nQuestion 1\nWhat is the common name of the compound CaOCl2 ?\nAnswer:\nBleaching powder.\nQuestion 2\nName the substance which on treatment with chlorine yields bleaching powder.\nAnswer:\nSlaked lime Ca (OH)2.\nQuestion 3\nName the sodium compound which is used for softening hard water.\nAnswer:\nSodium carbonate.\nQuestion 4\nWhat will happen if a solution of sodium hydrogen carbonate is heated. Give the equation of the reaction involved ?\nAnswer:\nSolution of sodium hydrogen carbonate on heating gives sodium carbonate and carbon dioxide gas is evolved.\n\nQuestion 5\nWrite an equation to show the reaction between plaster of Paris and water.\nAnswer:\n\nNCERT Solutions for Class 10 Science Chapter 2 Textbook Chapter End Questions\nQuestion 1\nA solution turns red litmus blue, its pH is likely to be\n(a) 1\n(b) 4\n(c) 5\n(d) 10\nAnswer:\n(d) 10\nQuestion 2\nA solution reacts with crushed-egg shells to give a gas that turns lime water milky.", + "The solution contains\n(a) NaCl\n(b) HCl\n(c) LiCl\n(d) KCl", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\n(a) 1\n(b) 4\n(c) 5\n(d) 10\nAnswer:\n(d) 10\nQuestion 2\nA solution reacts with crushed-egg shells to give a gas that turns lime water milky. The solution contains\n(a) NaCl\n(b) HCl\n(c) LiCl\n(d) KCl\nAnswer:\n(b) HCl\nQuestion 3\n10 mL of a solution of NaOH is found to be completely neutralised by 8 mL of a given solution of HC1. If we take 20 mL of the same solution of NaOH, the amount of HC1 solution (the same solution as before) required to neutralise it will be\n(a) 4 mL\n(b) 8 mL\n(c) 12 mL\n(d) 16 mL\nAnswer:\n(d) 16 mL\nQuestion 4\nWhich one of the following types of medicines is used for treating indigestion?", + "(a) Antibiotic\n(b) Analgesic\n(c) Antacid\n(d) Antiseptic\nAnswer:\n(c) Antacid\nQuestion 5\nWrite word equations and then balanced equations for the reaction taking place when\n(a) dilute sulphuric acid reacts with zinc granules\n(b) dilute hydrochloric acid reacts with magnesium ribbon\n(c) dilute sulphuric acid reacts with aluminium powder\n(d) dilute hydrochloric acid reacts with iron filing", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\n(b) dilute hydrochloric acid reacts with magnesium ribbon\n(c) dilute sulphuric acid reacts with aluminium powder\n(d) dilute hydrochloric acid reacts with iron filing\nAnswer:\n(a) Zinc + dilute sulphuric acid \u2192 Zinc sulphate + Hydrogen\nZn (s) + H2SO4 (aq) \u2192 ZnSO4 (aq) + H2 (g)\n(b) Magnesium ribbon + dil. Hydrochloric acid \u2192 Magnesium chloride + Hydrogen\nMg (s) + 2 HCl (aq) \u2192 MgCl2 (aq) + H2 (g)\n(c) Aluminium powder + dil.", + "Hydrochloric acid \u2192 Magnesium chloride + Hydrogen\nMg (s) + 2 HCl (aq) \u2192 MgCl2 (aq) + H2 (g)\n(c) Aluminium powder + dil. Sulphuric acid > Aluminium sulphate + Hydrogen\n2Al (s) + 3H2SO4 (aq) \u2192 Al2 (SO4)3 (aq) + 3H2 (g)\n(d) Iron filings + Dilute hydrochloric acid > Ferric chloride + Hydrogen\n2Fe (s) + 6HCl (aq) \u2192 2FeCl3 (aq) + 3H2 (g)\nQuestion 6\nCompounds such as alcohol and glucose also contain hydrogen but are not categorised as acids. Describe an activity to prove it.\nAnswer:\nThough compounds like alcohol and glucose contain hydrogen but they do not ionise in the solution to produce H+ ions on passing current through them.\n(i) Take solutions of alcohols and glucose.", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nThough compounds like alcohol and glucose contain hydrogen but they do not ionise in the solution to produce H+ ions on passing current through them.\n(i) Take solutions of alcohols and glucose.\n(ii) Fix two nails on a cork, and place the cork in 100 mL beaker.\n(iii) Connect the nails to the two terminals of a 6 volt battery through a bulb and a switch, as shown in the given Figure.\n\n(iv) Now pour alcohol in the beaker and switch on the current.\n(v) The bulb does not glow.\n(vi) Repeat the experiment with glucose. The bulb does not glow in this case also.\n(vii) This means no ions or H+ ions are present in the solution.\nThis shows that alcohols and glucose are not acids.\nQuestion 7\nWhy does distilled water not conduct electricity, whereas rainwater does ?\nAnswer:\nDistilled water does not conduct electricity because it does not contain any ionic compound (like acids, bases or salts) dissolved in it.", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nAnswer:\nDistilled water does not conduct electricity because it does not contain any ionic compound (like acids, bases or salts) dissolved in it.\nRainwater, while falling to the earth through the atmosphere, dissolves an acidic gas carbon dioxide from the air and forms carbonic acid (H2CO3). Carbonic acid provides hydrogen ions, H+ (aq) and carbonate ions, CO(aq)32to rainwater. Hence, due to the presence of carbonic acid which provides ions to rainwater, the rainwater conducts electricity.\nQuestion 8\nWhy do acids not show acidic behaviour in the absence of water ?\nAnswer:\nThe acidic behaviour of acids is due to the presence of hydrogen ions, [H+ (aq) ions], in them. The acid produces hydrogen ions only in the presence of water. So in the absence of water, an acid will not form hydrogen ions and hence will not show its acidic behaviour.\nQuestion 9\nFive solutions A, B, C, D and E when tested with universal indicator showed pH as 4, 1, 11, 7 and 9 respectively. Which solution is", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 9\nFive solutions A, B, C, D and E when tested with universal indicator showed pH as 4, 1, 11, 7 and 9 respectively. Which solution is\n(a) Neutral\n(b) Strongly alkaline\n(c) Strongly acidic\n(d) Weakly acidic\n(e) Weakly alkaline\nArrange the pH in increasing order of hydrogen ion concentration.\nAnswer:\n(a) D\n(b) C\n(c) B\n(d) A\n(e) E\nIncreasing order of hydrogen ion concentration\n11 < 9 < 7 < 4 < 1\ni. e., C < E < D < A < B\nQuestion 10\nEqual lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test tube A, while acetic acid (CH3COOH) is added to test tube B. In which test tube will the fizzing occur more vigorously and why ?\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nAnswer:\nFizzing will occur more vigorously in test tube A. Hydrochloric acid (HCl) is a strong acid whereas acetic acid (CH3COOH) is a weak acid. Being strong acid, the hydrochloric acid solution contains a much greater amount of hydrogen ions in it due to which the fizzing will occur more vigorously in test tube A (containing hydrochloric acid). The fizzing is due to the evolution of hydrogen gas which is formed by the action of acid on the magnesium metal of magnesium ribbon.\nQuestion 11\nfresh milk has a pH of 6. How do you think the pH will change as it turns into curd ? Explain your answer.\nAnswer:\npH of milk falls below 6 as it turns into curd due to the formation of lactic acid during this process. Lactic acid present in it reduces its pH value.\nQuestion 12\nA milkman adds a very small amount of baking soda to fresh milk.\n(a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline ?", + "Lactic acid present in it reduces its pH value.\nQuestion 12\nA milkman adds a very small amount of baking soda to fresh milk.\n(a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline ?\n(b) Why does this milk take a long time to set as curd ?\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\n(a) Why does he shift the pH of the fresh milk from 6 to slightly alkaline ?\n(b) Why does this milk take a long time to set as curd ?\nAnswer:\n(a) Milk is made slightly alkaline so that it may not get sour easily due to the formation of lactic acid in it.\n(b) The alkaline milk takes a longer time to set into curd because the lactic acid being formed has to first neutralise the alkali present in it.\nQuestion 13\nPlaster of Paris should be stored in a moisture proof container. Explain why?\nAnswer:\nPlaster of Paris should be stored in a moisture proof container because the presence of moisture can cause slow setting of plaster of Paris by bringing about its hydration. This will make the plaster of Paris useless after sometime.\nQuestion 14\nWhat is a neutralisation reaction ? Give two examples.\nAnswer:\nThe reaction between an acid and a base to form salt and water is called a neutralisation reaction.\nExamples:\n\nQuestion 15\nGive two important uses of washing soda and baking soda.\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nAnswer:\nThe reaction between an acid and a base to form salt and water is called a neutralisation reaction.\nExamples:\n\nQuestion 15\nGive two important uses of washing soda and baking soda.\nAnswer:\nUses of washing soda :\n(i) Washing soda is used in glass, soap and paper industries.\n(ii) It is used for removing permanent hardness of water.\nUses of baking soda :\n(i) Baking soda is used as an antacid in medicines to remove acidity of the stomach.\n(ii) Baking soda is used for making baking powder (used in making cakes, bread, etc.).\nNCERT Solutions for Class 10 Science Chapter 2 Acids, Bases, and Salts\nNCERT Solutions for Class 10 Science (Chemistry) Chapter 2 Acids, bases, and salts are part of NCERT Solutions for Class 10 Science. Here we have given Class 10 Science NCERT Solutions Chapter 2.\nQuestion 1", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 1\nYou have been provided with three test tubes. One of them contains distilled water and the other two contain an acidic solution and a basic solution respectively. If you are given only red litmus paper, how will you identify the contents of each test tube?\nSolution:\nThe contents of each test tube would be identified by change in colour of red litmus paper. For example, when we wet the red litmus paper with the basic solution, it changes into blue colour. Put the changed blue litmus paper in the solution which turns the blue to red will be the acidic solution. The solution, which has no effect on any litmus paper, will be neutral and hence it will be distilled water.\nDownload NCERT Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts PDF\nQuestion 2\nWhy should curd and sour substances not be kept in brass and copper vessels?\nSolution:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nDownload NCERT Solutions for Class 10 Science Chapter 2 Acids, Bases and Salts PDF\nQuestion 2\nWhy should curd and sour substances not be kept in brass and copper vessels?\nSolution:\nCurd and other sour foodstuffs contain acids, which can react with the metal of the vessel to form poisonous metal compounds which can cause food poisoning and damage our health.\nQuestion 3\nWhich gas is usually liberated when an acid reacts with a metal?\nSolution:\nWhen an acid reacts with metal, a salt and hydrogen gas is formed. i.e\n\nQuestion 4\nMetal compound A reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.\nSolution:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nSolution:\nThe gas that extinguishes a burning candle is carbon dioxide, which is formed by the action of dilute hydrochloric acid on a metal carbonate and produces effervescence. Now, since one of the compounds formed is calcium chloride, it shows that the metal compound is calcium carbonate. Thus, the metal compound A is calcium carbonate (CaCO3). Calcium carbonate reacts with dilute hydrochloric acid to form calcium chloride, carbon dioxide and water. This can be written as:\nQuestion 5\nWhy do HCl, HNO3, etc., show acidic characters in aqueous solutions while solutions of compounds like alcohol and glucose do not show acidic character?\nSolution:\nAn acid is a substance, which dissociates on dissolving in water to produce hydrogen ions [H+(aq) ions]. The acids like HCl, H2SO4, HNO3 and CH3COOH, etc., show acidic character because they dissociate in aqueous solutions to produce hydrogen ions, H+(aq) ions.", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nThe compounds such as glucose and alcohol also contain hydrogen but they do not show acidic character. The aqueous solutions of glucose and alcohol do not show acidic character because the hydrogen in them does not separate out as hydrogen ions [H+ (aq) ions] on dissolving in water.\nQuestion 6\nWhy does an aqueous solution of acid conduct electricity?\nSolution:\nThe aqueous solution of an acid conducts electricity due to the presence of charged particles called ions in it.\nQuestion 7\nWhy does dry HCl gas not change the colour of the dry litmus paper?\nSolution:\nDry HCl gas does not contain any hydrogen ions in it, so it does not show acidic behaviour. In fact, dry HCl gas does not change the colour of dry litmus paper as it has no hydrogen ions [H+(aq) ions] in it.\nQuestion 8\nWhile diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?\nSolution:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 8\nWhile diluting an acid, why is it recommended that the acid should be added to water and not water to the acid?\nSolution:\nDiluting an acid should be done by adding concentrated acid to water gradually with stirring and not by adding water to concentrated acid.\nThe heat is evolved gradually when a concentrated acid is added to water for diluting an acid and the large amount of water is easily absorbed.\nIf, however, water is added to concentrated acid to dilute it, a large amount of heat is evolved at once. The heat generated may cause the mixture to splash the acid on our face or clothes and cause acid burns.\nQuestion 9\nHow is the concentration of hydronium ions (H3O+) affected when a solution of an acid is diluted?\nSolution:\nWhen the concentrated solution of an acid is diluted by mixing water, the concentration of hydronium ions H3O+ per unit volume decreases.\nQuestion 10\nHow is the concentration of hydroxide ions (OH-) affected when excess base is dissolved in water?", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 10\nHow is the concentration of hydroxide ions (OH-) affected when excess base is dissolved in water?\nSolution:\nWhen the solution of a base is diluted by mixing more water in it, the concentration of hydroxide ions (OH\u2013 ions) per unit volume decreases.\nQuestion 11\nYou have two solutions A and B. The pH of solution A is 6 and pH of solution B is 8.\ni. Which solution has more hydrogen ion concentration?\nii. Which of this is acidic and which one is basic?\nSolution:\nThe pH of a solution is inversely proportional to its hydrogen ion concentration. This means that the solution having lower pH will have more hydrogen ion concentration. In this case, solution A(having a lower pH of 6) will have more hydrogen ion concentration.Solution A is acidic and solution B is basic.\nQuestion 12\nWhat effect does the concentration of H+(aq) ions have on the nature of the solution?\nSolution:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 12\nWhat effect does the concentration of H+(aq) ions have on the nature of the solution?\nSolution:\nAcids produce hydrogen ions in water. So, when an acid is added to water, the concentration of hydrogen ions in water increases. The solution of acid thus formed will have more of hydrogen ions and it will be acidic in nature.\nQuestion 13\nDo basic solutions also have H+(aq) ions? If yes, why are these basic?\nSolution:\nNo, the basic solution doesn\u2019t have H+ ions as the solution has excess of hydroxide ions.\nQuestion 14\nUnder what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?\nSolution:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nUnder what soil condition do you think a farmer would treat the soil of his fields with quick lime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate)?\nSolution:\nMost often the soil in the fields is too acidic. If the soil is too acidic (having low pH), it is treated with materials like quicklime (calcium oxide) or slaked lime (calcium hydroxide) or chalk (calcium carbonate). Thus, a farmer should add lime or slaked lime in his fields when the soil is too acidic.\nQuestion 15\nWhat is the common name of the compound CaOCl2?\nSolution:\nThe common name of the compound CaOCl2 is bleaching powder.\nQuestion 16\nName the substance that on treatment with chlorine yields bleaching powder.\nSolution:\nCalcium hydroxide is the substance that on treatment with chlorine yields bleaching powder.\nQuestion 17\nName the sodium compound, which is used, for softening hard water.\nSolution:\nSodium carbonate (washing soda) is used for softening hard water.\nQuestion 18", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 17\nName the sodium compound, which is used, for softening hard water.\nSolution:\nSodium carbonate (washing soda) is used for softening hard water.\nQuestion 18\nWhat will happen if a solution of sodium hydro carbonate is heated? Give the equation of the reaction involved.\nSodium carbonate and carbon dioxide are evolved when sodium hydro carbonate is heated.\n\nQuestion 19\nWrite an equation to show the reaction between plaster of Paris and water.\nSolution:\nPlaster of Paris has a very remarkable property of setting into a hard mass on wetting with water. So, when water is added to plaster of Paris, it sets into a hard mass in about half an hour. The setting of plaster of Paris is due to the hydration crystals of gypsum, which set to form a hard, solid mass.\nQuestion 20\nWhy does distilled water not conduct electricity, whereas rainwater does?\nSolution:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 20\nWhy does distilled water not conduct electricity, whereas rainwater does?\nSolution:\nDistilled water does not conduct electricity because it does not contain any ionic compound (like acids, bases or salts) dissolved in it. On the other hand, rain water conducts electricity. This can be explained as follows: Rain water, while falling to the earth through the atmosphere, dissolves an acidic gas carbon dioxide from the air and forms carbonic acid (H2CO3). Carbonic acid provides hydrogen ions, H+(aq), and carbonate ions, CO2-3 (aq), to rain water. So, due to the presence of carbonic acid (which provides ions to rain water), the rain water conducts electricity.\nQuestion 21\nWhy do acids not show acidic behaviour in the absence of water?\nSolution:\nThe acidic behaviour of acid is due to the presence of hydrogen ions. The acids will not show its acidic behaviour in the absence of water, this is because the acids produce hydrogen ions only in the presence of water.\nQuestion 22", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 22\nFive solutions A, B, C, D and E when tested with universal indicator showed pH as 4, 1, 11, 7 and 9 respectively. Which solution is?\n(i) Neutral?\n(ii) Strongly alkaline?\n(iii) Strongly acidic?\n(iv) Weakly acidic?\n(v) Weakly alkaline? Arrange the pH in increasing order of hydrogen-ion concentration.\nSolution:\n\nQuestion 23\nEqual lengths of magnesium ribbons are taken in test tubes A and B. Hydrochloric acid (HCl) is added to test-tube A while acetic acid (CH3COOH) is added to test-tube B. In which test-tube will fizzing occur more vigorously and why?\nSolution:\nAcetic acid (CH3COOH) is a weak acid whereas hydrochloric acid (HCl) is a strong acid. Fizzing occurs in the test tube due to the evolution of hydrogen gas by the action of acid on magnesium ribbon.", + "Solution:\nAcetic acid (CH3COOH) is a weak acid whereas hydrochloric acid (HCl) is a strong acid. Fizzing occurs in the test tube due to the evolution of hydrogen gas by the action of acid on magnesium ribbon. Since hydrochloric acid is a strong acid a large amount of hydrogen gas is liberated in the test tube A. So fizzing occurs more vigorously in test tube A .\nQuestion 24", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 24\nFresh milk has a pH of 6. How do you think the pH will change as it turns into curd? Explain.\nSolution:\nThe pH will change to below 6, as lactic acid is formed when milk turns into curd.\nQuestion 25\nPlaster of Paris should be stored in a moisture-proof container. Why?\nSolution:\nThe presence of moisture can affect the slow setting of plaster of Paris by bringing about its hydration. This will make the plaster of Paris useless after some time.\nQuestion 26\nWhat is a neutralization reaction? Give two examples.\nSolution:\nThe reaction of an acid and a base, giving rise to the corresponding salt and water is called neutralization reaction.\nExamples:\nNaOH + HCl \u2192 NaCl + H2O\nMg(OH)2 + H2CO3 \u2192 MgCO3 + 2H2O\n\nQuestion 27\nGive two important uses of washing soda and baking soda.\nSolution:\nWashing soda\n1. It is often used as an electrolyte.\n2. Domestically it is used as a water softener during laundry.\nBaking soda", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nGive two important uses of washing soda and baking soda.\nSolution:\nWashing soda\n1. It is often used as an electrolyte.\n2. Domestically it is used as a water softener during laundry.\nBaking soda\n1. It is used to test garden soil for acidity. If it develops bubbles, the soil is too acidic.\n2. Washing a car with it can remove dead bug bodies without damaging the paint.\nNCERT Solutions for Class 10 Science Chapter 2 (MCQs) [1 Mark each]\nQuestion 1.\nEqual pieces of zinc granules are dropped in four test tubes. Following substances are poured in all the four test tubes. The reaction will be vigorous with [CCE 2014]\n(a) CH3COOH\n(b) HCl\n(c) sodium bicarbonate solution\n(d) lemon juice\nAnswer:\n(b) Strong adds like HCl react vigorously with active metals like Zn and form metal salt and evolve H2 gas.\n\nQuestion 2.\nWhich of the following statements shows the property of an acid?", + "Question 2.\nWhich of the following statements shows the property of an acid? [CCE 2014]\n(a) It turns blue litmus to red\n(b) It is sour in taste\n(c) It has no effect on red litmus", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nQuestion 2.\nWhich of the following statements shows the property of an acid? [CCE 2014]\n(a) It turns blue litmus to red\n(b) It is sour in taste\n(c) It has no effect on red litmus\n(d) All of the above\nAnswer:\n(d) An acid turns blue litmus red. Thus, it has no effect on red litmus and acids are sour in taste.\nQuestion 3.\nA drop of a liquid sample was put on the pH paper. It was observed that the colour of the pH paper turned blue. The liquid sample is [CCE 2014]\n(a) lemon juice\n(b) sodium bicarbonate solution\n(c) distilled water\n(d) hydrochloric acid\nAnswer:\n(b) The liquid sample is of sodium bicarbonate (NaHC3) solution. It is a basic solution. And we know that a basic solution turns pH paper blue.\nQuestion 4.\nTwo solutions X and Y were found to have pH value of 4 and 10 respectively.", + "It is a basic solution. And we know that a basic solution turns pH paper blue.\nQuestion 4.\nTwo solutions X and Y were found to have pH value of 4 and 10 respectively. The inference that can be drawn is [CCE 2014]\n(a) X is a base and Y is an acid\n(b) Both X and Y are acidic solutions\n(c) X is an acid and Yis a base\n(d) Both X and Y are bases\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\n(a) X is a base and Y is an acid\n(b) Both X and Y are acidic solutions\n(c) X is an acid and Yis a base\n(d) Both X and Y are bases\nAnswer:\n(c) Any solution having pH > 7 will be a base while the solution having pH < 7 will surely be an acid. Hence, it can be concluded that X is an acid (pH=4, i.e. < 7) and Yis a base (pH =10, i.e. > 7).\nQuestion 5.\nA student was asked to collect apparatus from lab store, for doing experiment of pH of given sample. Identify the article which he is not supposed to pick. [CCE 2014]\n(a) pH paper\n(b) Dropper\n(c) Litmus paper\n(d) Petri dish\nAnswer:\n(d) Petri dish is not required for doing experiment of pH.\nQuestion 6.\nWhich one of the following would you need to identify the gas that evolve when you heat NaOH solution with zinc metal?", + "Question 6.\nWhich one of the following would you need to identify the gas that evolve when you heat NaOH solution with zinc metal? [CCE 2014]\n(a) Red litmus solution\n(b) Blue litmus solution\n(c) A burning splinter / matchstick\n(d) Lime water\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\n(a) Red litmus solution\n(b) Blue litmus solution\n(c) A burning splinter / matchstick\n(d) Lime water\nAnswer:\n(c) When a base like NaOH is treated with any active metal like Zn, it produces H2 gas. And the presence of the hydrogen gas can be tested by bringing a burning splinter/ matchstick near the gas produced. The gas will burn with a pop sound confirming the presence of hydrogen gas.\n\nQuestion 7.\nA solution has pH value of 5. On adding 10 mL of NaCl to it, what will be the pH of the new solution?\n(a) More than 5\n(b) Less than 5\n(c) Only seven\n(d) No change in pH\nAnswer:\n(d) As NaCl is a neutral solution due to its complete ionization in H2O and has pH = 7. Thus, it will not affect the pH of the solution. Hence, the pH of the solution will remain unaffected.\nQuestion 8.\nA salt is dissolved in water.", + "Thus, it will not affect the pH of the solution. Hence, the pH of the solution will remain unaffected.\nQuestion 8.\nA salt is dissolved in water. The pH of this salt solution was found to be 7 by measuring the pH with a universal indicator paper. The salt is most likely to be\n(a) Na2CO3\n(b) KCl\n(c) NH4Cl\n(d) CH3COONa\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\n(a) Na2CO3\n(b) KCl\n(c) NH4Cl\n(d) CH3COONa\nAnswer:\n(b) Salts which are completely ionisable in water are said to be neutral salts and their pH is equal to 7. Among the given salts, only KCl is completely ionisable in the following manner: KCl (aq) \u2192 K+ + Cl\nHence, the pH of KCl will be 7.\nQuestion 9.\nA student performed an experiment using zinc granules and sodium carbonate with sodium hydroxide and hydrochloric acid under different\nconditions as shown below.\n\nIn which set up, no gas is evolved?\n(a) I\n(b) II\n(c) III\n(d) IV\nAnswer:\n(d) Gas will not evolve in the case of IV set up as NaOH does not react with sodium carbonate.\nQuestion 10.\nFour set ups as given below are arranged to identify the gas evolved when dilute hydrochloric acid was added to zinc granules. Which is the most appropriate set up?", + "Question 10.\nFour set ups as given below are arranged to identify the gas evolved when dilute hydrochloric acid was added to zinc granules. Which is the most appropriate set up?\n\n(a) I\n(b) II\n(c) IV\n(d) III\nAnswer:\n(c) The gas evolved can be tested as shown in IV set up because to evolve H2 gas delivery tube, should not dip in the acid.\nQuestion 11.", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\n(a) I\n(b) II\n(c) IV\n(d) III\nAnswer:\n(c) The gas evolved can be tested as shown in IV set up because to evolve H2 gas delivery tube, should not dip in the acid.\nQuestion 11.\nFour students I, II, III and IV were asked to examine the changes for blue and red litmus paper strips with dil. HCl (Solution A) and dil. NaOH (Solutions). The following observations were reported by the 4 students. The sign \u2026\u2026\u2026\u2026. indicates no colour change. [CCE 2015 ]\nThe correct observation would be of student\n(a) I\n(b) II\n(c) III\n(d) IV\nAnswer:\n(c) The correct observation is taken by student III because HCl (solution A) is an acid which turns blue litmus red and dil. NaOH (solution B) is a base which turns red litmus blue.\nQuestion 12.\nA sample of soil is mixed with water and allowed to settle. The clear supernatant solution turns the pH paper yellowish orange.", + "NaOH (solution B) is a base which turns red litmus blue.\nQuestion 12.\nA sample of soil is mixed with water and allowed to settle. The clear supernatant solution turns the pH paper yellowish orange. Which of the following would change the colour of this pH paper to greenish blue? [NCERT Exemplar]\n(a) Lemon juice\n(b) Vinegar\n(c) Common salt", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\n(a) Lemon juice\n(b) Vinegar\n(c) Common salt\n(d) An antacid\nAnswer:\n(d) As pH paper turns greenish blue for weakly basic compound and antacids contain weak base like Mg(OH)2. So, an antacid would change the colour of this pH paper to greenish blue. Other options (a) and (b) contain acids and option (c) is a neutral salt.\nClass 10 Science Acid, Bases and Salts Mind Map\nAcids\nProduce [H+] in H2O\nSour taste\nTurns blue litmus red\nActs as electrolyte in solution\nChemical Properties Of Acids\nAcids + Metal \u2192 Salt + H2\ne.g.", + "H2SO4 + Zn \u2192 ZnSO4 + H2\nAcids + Metal Carbonate/Metal hydrogen\nCarbonate = Salt + CO2\ne.g NaHCO3 + HCl \u2192 NaCl + H2O + CO2\nAcids + Bases \u2192 Salt + water\nNaOH + HCl \u2192 NaCl + H2O\nAcids + Metal oxide \u2192 Salt + Water\nH2SO4 + CuO \u2192 CuSO4 + H2O\nBases\nProduce [OH ] in H2O\nBitter taste\nTurns red litmus blue\nActs as electrolyte in solution\nWater soluble bases are known as alkali\nChemical Properties Of Bases\nBases + Metals \u2192 Salt + H2\n2NaOH + Zn \u2192 Na2ZnO2 + H2", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nBitter taste\nTurns red litmus blue\nActs as electrolyte in solution\nWater soluble bases are known as alkali\nChemical Properties Of Bases\nBases + Metals \u2192 Salt + H2\n2NaOH + Zn \u2192 Na2ZnO2 + H2\nBases + Acids \u2192 Salt + Water\nKOH + HCl \u2192 KCl + H20\nBase + Non-metallic oxide \u2192 Salt + water\n2NaOH + CO2 \u2192 Na2CO3 + H2O\nSalt\nSalt is formed by combination of acid and base through neutralization reaction.\nwater of Crystallization :\nIt is the fixed number of water molecules present in one formula unit of a salt e.g.\nCuSO4.5H2O, Na2CO3.10H2O\nTypes of Salts\nThe acidic & basic nature of salts depends on the acid and base combined in neutralization reaction.\nSome Important Salts\nPlaster of Paris (CaSO4. \n1\n2\n H2O)\n\nUsed for making toys, material for decor action, smooth surfaces.", + "Some Important Salts\nPlaster of Paris (CaSO4. \n1\n2\n H2O)\n\nUsed for making toys, material for decor action, smooth surfaces.\nBleaching Powder (CaOCl2):\nCa(OH)2 + Cl2 \u2192 CaOCl2 + H2O\ndry slaked lime\nUsed\nas an oxidising agent in chemical industry.\nIn disinfecting water.\nCommon Salt NaCl:\nMain source is sea water.", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nBleaching Powder (CaOCl2):\nCa(OH)2 + Cl2 \u2192 CaOCl2 + H2O\ndry slaked lime\nUsed\nas an oxidising agent in chemical industry.\nIn disinfecting water.\nCommon Salt NaCl:\nMain source is sea water.\nAlso exists in the form of rock hence also known as rock salt.\nImportant component of food.\nUsed in preparation of sodium hydroxide, baking soda & washing soda.\nBaking Soda (NaHCO3)\nNaCl + H2O + CO2 + NH23 \u2192 NH4Cl + NaHCO3\nMild non \u2013 corrosive\n\nUsed as in baking cakes as antacid\nFire extinguisher\nSodium Hydroxide (NaOH)\nPrepared by chlor alkali process .\n\nWashing Soda (Na2CO3. 10H2O) :\n\u2022 Na2CO3 + 10H2O \u2192 4 Na2CO3.", + "Washing Soda (Na2CO3. 10H2O) :\n\u2022 Na2CO3 + 10H2O \u2192 4 Na2CO3. 10H2O\n\u2022 Used in glass, soap & paper indusby, removing Permanent hardness of water and cleaning agent\nIndicators\nThese are the substances which indicate the acidic or basic nature of the solution by their colour change.\nStrength of Acids and Bases in Solution\nSome animals like bee and plants like nettle secretes highly acidic substance for self defense.", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nStrength of Acids and Bases in Solution\nSome animals like bee and plants like nettle secretes highly acidic substance for self defense.\nLower pH of sour & sweet food can cause tooth decay. The pH of mouth should be more than 5.5\nThe inner lining of stomach protects vital cells from the acidic pH which is developed by HCl secreted by stomach.\nThe optimum pH range for human body is 7 to 7.8\nAddition of Acids or Bases to Water\nAlways add acid to water and not water to acid because this process is highly exothermic. The acid must be added slowly to water by constant shirring on. Adding water to a concentrated acid, the heat generated may cause the mixture to splash out and cause burns.", + "CHAPTER 2-ACIDS,BASES AND SALTS - INTEXT QUESTIONS.pdf.txt\nNow that you are provided all the necessary information regarding NCERT Solutions for Class 10 Science Chapter 2 Activities and we hope this detailed article on NCERT Solutions For Class 10 Science Chapter 2 Acids And Bases is helpful. If you have any query regarding this article or Science Class 10 Chapter 2 Acids And Bases, drop your comments in the comment box below and we will get back to you as soon as possible.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nImportant Questions of Acids Bases and Salts Class 10 Science Chapter 2\nQuestion 1.\nWith the help of an example explain what happens when a base reacts with a non- metallic oxide. What do you infer about the nature of non-metal oxide? (Board Term I, 2017)\nAnswer:\nOxides of non-metals react with bases to form salt and water. For example, the reaction between carbon dioxide and calcium hydroxide. Calcium hydroxide, which is a base, reacts with carbon dioxide to produce salt and water.\n\nHence, oxides of non-metals are acidic in nature.\nQuestion 2.\nWhat is observed when carbon dioxide gas is passed through lime water\n(i) for a short duration?\n(ii) for a long duration? Also write the chemical equations for the reactions involved. (Board Term I, 2016)\nAnswer:\n(i) When CO2 is passed through lime water for short interval of time, it turns milky due to the formation of insoluble calcium carbonate.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nAnswer:\n(i) When CO2 is passed through lime water for short interval of time, it turns milky due to the formation of insoluble calcium carbonate.\n\n(ii) If CO2 is passed for long duration through lime water, the white precipitate formed dissolves due to the formation of soluble calcium hydrogen carbonate and the solution becomes clear.\n\nQuestion 3.\n2 mL of sodium hydroxide solution is added to a few pieces of granulated zinc metal taken in a test tube. When the content are warmed, a gas evolves which is bubbled through a soap solution before testing. Write the equation of the chemical reaction involved and the test to detect the gas. Name the gas which will be evolved when the same metal reacts with dilute solution of a strong acid.\nAnswer:\nIt is observed that active metals like zinc react with strong bases like NaOH, KOH etc. to liberate hydrogen gas and corresponding salt.\n\nThe evolution of gas is confirmed by the bubble formation in soap solution.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nThe evolution of gas is confirmed by the bubble formation in soap solution.\nTest to detect H2 gas: When burning matchstick is kept on the mouth of this test tube, pop sound is heard which confirms the presence of H2 gas. When Zn metal reacts with dilute solution of strong acid, H2 gas is evolved.\n\nQuestion 4.\nWrite the names of the product formed when zinc reacts with NaOH. Also write the balanced chemical equation for the reaction involved. Write a test to confirm the presence of the gas evolved during this reaction. (Board Term I, 2015)\nAnswer:\nRefer to answer 3.\nQuestion 5.\nTo. a solution of sodium hydroxide in a test tube, two drops of phenolphthalein are added.\n(i) State the colour change observed.\n(ii) If dil HCl is added dropwise to the solution, what will be the colour change?\n(iii) On adding few drops of NaOH solution to the above mixture the colour of the solution reappears. Why? (Board Term I, 2013)\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(iii) On adding few drops of NaOH solution to the above mixture the colour of the solution reappears. Why? (Board Term I, 2013)\nAnswer:\n(i) On adding phenolphthalein to NaOH solution, the colour becomes pink.\n(ii) On adding dilute HCl solution dropwise to the same test tube, the pink colour disappears and the solution again becomes colourless.\n(iii) On again adding NaOH to the above mixture, pink colour reappears because the medium becomes basic again.\nQuestion 6.\nA cloth\u2019strip dipped in onion juice is used for testing a liquid \u2018X. The liquid \u2018X changes its\nodour. Which type of an indicator is onion juice? The liquid \u2018X turns blue litmus red. List the observations the liquid \u2018X will show on reacting with the following :\n(a) Zinc granules\n(b) Solid sodium carbonate\nWrite the chemical equations for the reactions involved.\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(a) Zinc granules\n(b) Solid sodium carbonate\nWrite the chemical equations for the reactions involved.\nAnswer:\nOnion juice is an olfactory indicator. Olfactory indicators give one type of odour in acidic medium and a different odour in basic medium. As the liquid \u2018X\u2019 turns blue litmus red, hence it is an acidic solution.\n(a) Acids react with active metals such as zinc, magnesium etc. and evolve hydrogen gas, for example,\nZn(s) dil.H2SO4(aq) \u2192 ZnSO4H2(g)\n(b) Acids react with metal carbonates to give carbon dioxide with brisk effervescence.\nFor example, Na2CO3 + H2SO4 \u2192 Na2SO4 + CO2 + H2O\nQuestion 7.\n(a) Write the chemical name and formula of marble.\n(b) It has been found that marbles of Taj are getting corroded due to development of industrial areas around it. Explain this fact giving a chemical equation.\n(c) (i) What happens when CO2 is passed through lime water?", + "(a) Write the chemical name and formula of marble.\n(b) It has been found that marbles of Taj are getting corroded due to development of industrial areas around it. Explain this fact giving a chemical equation.\n(c) (i) What happens when CO2 is passed through lime water?\n(ii) What happens when CO2 is passed in excess through lime? (Board Term I, 2013)\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(c) (i) What happens when CO2 is passed through lime water?\n(ii) What happens when CO2 is passed in excess through lime? (Board Term I, 2013)\nAnswer:\n(a) The chemical formula of marble (lime stone) is CaCO3. Its chemical name is calcium carbonate.\n(b) Taj Mahal, one of the seven wonders of the world situated at Agra, is continuously losing its luster day by day due to rapid industrialisation which causes acid rain.\nThe sulphuric acid present in the acid rain causes the marble (CaCO3) to be washed off as calcium sulphate (CaSO4), leading to the deterioration of such a splendid piece of architecture.\nCaCO3(s) + H2SO4(aq) \u2192 CaSO4(aq) + H2Ol + CO2(g)\n(c) Refer to answer 2.\nQuestion 8.\nOn diluting an acid, it is advised to add acid to water and not water to acid. Explain why it is so advised? (Board Term I, 2014)\nDraw a labelled diagram to show the preparation of hydrogen chloride gas in laboratory.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nDraw a labelled diagram to show the preparation of hydrogen chloride gas in laboratory.\n(ii) Test the gas evolved first with dry and then with wet litmus paper. In which of the two cases, does the litmus paper show change in colour?\n(iii) State the reason of exhibiting acidic character by dry HCl gas/HCl solution.\nAnswer:\nDiluting a concentrated acid with water is a highly exothermic process. So, when water is added to concentrated acid, large amounts of heat is liberated which changes some water to steam explosively which can splash the acid and even the glass apparatus may break due to excessive heating.\nQuestion 9.\n(i) Draw a labelled diagram to show the preparation of hydrogen chloride gas in laboratory.\n(ii) Test the gas evolved first with dry and then with wet litmus paper. In which of the two cases, does the litmus paper show change in colour?\n(iii) State the reason of exhibiting acidic character by dry HCl gas/HCl solution. (2020)\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(iii) State the reason of exhibiting acidic character by dry HCl gas/HCl solution. (2020)\nAnswer:\n\n(ii) There is no change in the colour of \u2018dry\u2019 blue litmus paper but \u2018moist\u2019 blue litmus paper turns red if brought near the mouth of the test tube.\nThis shows that HCl gas does not show acidic behaviour in absence of water but it shows acidic behaviour in presence of water.\n(iii) When HC1 gas dissolves in water, forms hydrochloric acid solution i.e., HCl(aq) which then produces H+(aq) or H3O+(aq) ions.\nHCl + H2O \u2192 H3O+ + Cl\u2013\nDue to the presence of H+ or H3O+ it shows acidic behaviour.\nQuestion 10.", + "HCl + H2O \u2192 H3O+ + Cl\u2013\nDue to the presence of H+ or H3O+ it shows acidic behaviour.\nQuestion 10.\nComplete and balance the following chemical equations :\n(i) NaOH(aq) + Zn(s) \u2192\n(ii) CaCO3(s) + H2O(l) + CO2(g) \u2192\n(iii) HCl(aq) + H2O(l) \u2192\nAnswer:\n(i) 2NaOH(aq) + Zn(s) \u2192 Na2ZnO2(aq) + H2(g)\n(ii) CaCO3(s) + CO2(g) + H2Ol \u2192 Ca(HCO3)2(aq)\n(iii) HCl(aq) + H2Ol > H3O+ Cl\u2013(aq)\nQuestion 11.\nHow the following substances will dissociate to produce ions in their solutions?\n(i) Hydrochloric acid", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(ii) CaCO3(s) + CO2(g) + H2Ol \u2192 Ca(HCO3)2(aq)\n(iii) HCl(aq) + H2Ol > H3O+ Cl\u2013(aq)\nQuestion 11.\nHow the following substances will dissociate to produce ions in their solutions?\n(i) Hydrochloric acid\n(ii) Nitric acid\n(iii) Sulphuric acid\n(iv) Sodium hydroxide\n(v) Potassium hydroxide\n(vi) Magnesium hydroxide (Board Term 1,", + "(i) Hydrochloric acid\n(ii) Nitric acid\n(iii) Sulphuric acid\n(iv) Sodium hydroxide\n(v) Potassium hydroxide\n(vi) Magnesium hydroxide (Board Term 1, 2017)\nAnswer:\nDissociation of various substances to produce ions in their solutions are :\n(i) Hydrochloric acid (HCl):\nHCl(aq) \u21cc H+(aq) + Cl\u2013(aq)\n(ii) Nitric acid (HNO3\nHNO3(aq) \u21cc + H+aq + NO\u20133(aq)\n(iii) Sulphuric acid (H2SO4):\nH2SO4(aq) \u21cc 2H+(aq) + SO2-4(aq)\n(iv) Sodium hydroxide (NaOH):\nNaOH(aq) \u21cc Na+(aq) + OH\u2013(aq)\n(v) Potassium hydroxide (KOH) :\nKOH(aq)\u21cc K+(aq) + OH\u2013(aq)\n(vi) Magnesium hydroxide [Mg(OH)2] :\nMg(OH)2(aq) \u21cc Mg2++(aq) + 2OH\u2013 (aq)\nQuestion 12.", + "Sugandha prepares HCl gas in her school laboratory using certain chemicals. She puts both dry and wet blue litmus papers in contact with the gas.\n(i) Name the reagents used by Sugandha to prepare HCl gas.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(i) Name the reagents used by Sugandha to prepare HCl gas.\n(ii) State the colour changes observed with the dry and wet blue litmus papers.\n(iii) Show the formation of ions when HCl gas combines with water. (Board Term I, 2013)\nAnswer:\n(i) Dense white fumes of hydrogen chloride gas are evolved on heating solid sodium chloride with concentrated sulphuric acid.\n\n(ii) Refer to answer 9(ii).\n(iii) Refer to answer 9(iii).\nQuestion 13.\n(a) Illustrate an activity to investigate whether all compounds containing hydrogen are acidic.\n(b) What happens when hydrochloric acid and sodium hydroxide are dissolved in water. Explain by giving equation of each. (Board Term 1, 2016)\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(b) What happens when hydrochloric acid and sodium hydroxide are dissolved in water. Explain by giving equation of each. (Board Term 1, 2016)\nAnswer:\n(a) Take two beakers, one containing HCl acid and other containing alcohol which is not an acid but contains hydrogen. Now, fix two iron nails on a rubber cork and insert in a beaker and connect the nail to the two terminal of 6V battery through a switch and a bulb. Pour some dilute HCl solution in beaker and switch on the current. The bulb starts glowing. This shows that acids get dissociated as H+ and Cl\u2013 ions and these ions are responsible for conducting electricity.\n\nLet us now take alcohol solution in the beaker and switch on the current. The bulb does not glow in this case. This shows that alcohol does not conduct electricity.\n\nSo, all acids have hydrogen but all hydrogen containing compounds are not acid.\n(b) HCl dissociates in aqueous solution to give hydrogen ions (or hydronium ions) and chloride ions.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nSo, all acids have hydrogen but all hydrogen containing compounds are not acid.\n(b) HCl dissociates in aqueous solution to give hydrogen ions (or hydronium ions) and chloride ions.\n\nNaOH when dissolved in water produces sodium ions and hydroxide ions in the solution.\n\nQuestion 14.\nAn aqueous solution \u2018A\u2019 turns phenolphthalein solution pink. On addition of an aqueous solution \u2018B\u2019 to \u2018A\u2019 the pink colour disappears. The following statement is true for solution \u2018A\u2019 and \u2018B\u2019:\n(a) A is strongly basic and B is a weak base.\n(b) A is strongly acidic and B is a weak acid.\n(c) A has pH greater than 7 and B has pH less than 7.\n(d) A has pH less than 7 and B has pH greater than 7. (2020)\nAnswer:\n(c) As the aqueous solution of A turns phenolphthalein solution pink, hence A is basic in nature. On adding an acidic solution, the pink colour will disappear. Hence, B is an acid.\nQuestion 15.", + "(2020)\nAnswer:\n(c) As the aqueous solution of A turns phenolphthalein solution pink, hence A is basic in nature. On adding an acidic solution, the pink colour will disappear. Hence, B is an acid.\nQuestion 15.\nOut of HCl and CH3COOH, which one is a weak acid and why? Explain with the help of an example. (AI 2019)\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nQuestion 15.\nOut of HCl and CH3COOH, which one is a weak acid and why? Explain with the help of an example. (AI 2019)\nAnswer:\nOut of HCl and CH3COOH, CH3COOH is a weak acid because it dissociates partially in the solution. This can be proved with the help of following example.\nIf 1 M HCl and 1 M CH3COOH are taken in the beaker as shown in the figure, greater deflection is observed in case of HCl which shows that more ions are produced by HCl in solution which produce more current.\n\nQuestion 16.\nExplain how an antacid works. (Board Term I, 2017)\nAnswer:\nThe acidity produced due to excess hydrochloric acid in the stomach which cause indigestion, produce pain and irritation. Milk of magnesia (chemically magnesium hydroxide) is used as an antacid. Since, it is basic in nature, reacts with the excess hydrochloric acid present in the stomach and neutralises it.\n\nQuestion 17.", + "Milk of magnesia (chemically magnesium hydroxide) is used as an antacid. Since, it is basic in nature, reacts with the excess hydrochloric acid present in the stomach and neutralises it.\n\nQuestion 17.\n(a) Three acidic solutions A, B and C have pH = 0, 3 and 5 respectively.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nQuestion 17.\n(a) Three acidic solutions A, B and C have pH = 0, 3 and 5 respectively.\n(i) Which solution has highest concentration of H+ ions?\n(ii) Which solution has the lowest concentration of H+ ions?\n(b) How concentrated sulphuric acid can be diluted? Describe the process. (Board Term I, 2014)\nAnswer:\n(a) (i) The solution having lower pH will have more hydrogen ion concentration. Hence, solution \u2018A\u2019 will have highest H+ ion concentration.\n(ii) Solution CC\u2019 i.e., pH = 5 has the lowest concentration of H+ ions.\n(b) Mixing of an acid with water is called dilution. This process is highly exothermic and therefore, acid is always added to the water not water to acid. The process for diluting concentrated sulphuric acid is :\n(i) Take about 10 mL of water in a beaker.\n(ii) Add concentrated sulphuric acid dropwise to water and swirl the beaker slowly.\nQuestion 18.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(i) Take about 10 mL of water in a beaker.\n(ii) Add concentrated sulphuric acid dropwise to water and swirl the beaker slowly.\nQuestion 18.\nA compound P forms the enamel of teeth. It is the hardest substance of the body. It doesn\u2019t dissolve in water but gets corroded when the pH is lowered below 5.5.\n(a) Identify the compound P.\n(b) How does it undergo damage due to eating chocolate and sweets? What should we do to prevent tooth decay? (Board Term I, 2014, 2013)\nAnswer:\n(a) The compound P is calcium phosphate, (b) Eating chocolates and sweets produce large amount of acid in the mouth which is not completely neutralised by the saliva produced in the mouth. Excess acid attacks the enamel and tooth decay starts as pH of the mouth falls below 5.5. The best way to prevent tooth decay is to clean the teeth by using toothpastes after eating food. Toothpastes which are generally basic neutralise the excess acid in the mouth.\nQuestion 19.\nBaking soda is a mixture of", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nQuestion 19.\nBaking soda is a mixture of\n(a) sodium carbonate and acetic acid\n(b) sodium carbonate and tartaric acid\n(c) sodium hydrogen carbonate and tartaric acid\n(d) sodium hydrogen carbonate and acetic acid.\nAnswer:\n(c) : Baking soda is a mixture of sodium hydrogen carbonate and a mild edible acid like tartaric acid or citric acid.\nQuestion 20.\nThe chemical formula for plaster of Paris is\n(a) CaSO4.2H2O\n(b) CaSO4.2H2O\n(c) CaSO4.\n1\n2\nH2O\n(d) 2CaSO4.2H2O\nAnswer:\n(c, d) : Plaster of Paris is calcium sulphate hemihydrate which can be represented as,\nCaSO4. \n1\n2\n H2O and 2CaSO4.H20.\nQuestion 21.\n\u201cSodium hydrogen carbonate is a basic salt\u201d. Justify this statement. How is it converted into washing soda?", + "1\n2\n H2O and 2CaSO4.H20.\nQuestion 21.\n\u201cSodium hydrogen carbonate is a basic salt\u201d. Justify this statement. How is it converted into washing soda? (AI2019)\nAnswer:\nSodium hydrogen carbonate (NaHCO3) is basic in nature as on hydrolysis it gives a mixture of strong base (NaOH) and weak acid (H2CO3). Sodium hydrogen carbonate is converted to washing soda in the following way:\n(i) Thermal decomposition of NaHCO3:", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(i) Thermal decomposition of NaHCO3:\n\n(ii) Recrystallisation of sodium carbonate:\n\nQuestion 22.\nWrite the chemical formula of Bleaching powder. How is bleaching powder prepared? For what purpose is it used in drinking water? (Board Term I, 2016)\nAnswer:\nThe chemical formula of bleaching powder is CaOCl2.\nIt is prepared by the action of chlorine gas on dry slaked lime Ca(OH)2.\n\nThe chlorine used in the above reaction is the by-product produced during the electrolysis of brine. It is used in disinfecting drinking water as chlorine liberated by it, kills the germs.\nQuestion 23.\nA student collected common names and formulae of some substances but he forgot to note which formula is for which compound. Help him to match the correct formula. (Board Term I, 2013)\nAnswer:\n(i) Caustic soda \u2192 NaOH\n(ii) Slaked lime \u2192 Ca(OH)2\n(iii) Baking soda \u2192 NaHCO3\n(iv) Lime \u2192 CaO\nQuestion 24.", + "Help him to match the correct formula. (Board Term I, 2013)\nAnswer:\n(i) Caustic soda \u2192 NaOH\n(ii) Slaked lime \u2192 Ca(OH)2\n(iii) Baking soda \u2192 NaHCO3\n(iv) Lime \u2192 CaO\nQuestion 24.\nList the important products of the Chlor-alkali process. Write one important use of each. (2020)\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(ii) Slaked lime \u2192 Ca(OH)2\n(iii) Baking soda \u2192 NaHCO3\n(iv) Lime \u2192 CaO\nQuestion 24.\nList the important products of the Chlor-alkali process. Write one important use of each. (2020)\nAnswer:\nSodium hydroxide is prepared by electrolysis of an aqueous solution of sodium chloride (brine). The complete reaction can be represented as:\n\nThe process of electrolysis of sodium chloride solution is called chlor-alkali process because of the products formed : chlor for chlorine and alkali for sodium hydroxide. The three very useful products obtained by the electrolysis of sodium chloride solution are sodium hydroxide, chlorine and hydrogen.\nAt anode : Cl2 gas is liberated At cathode : H2 gas is liberated.\nUses of sodium hydroxide: In the manufacture of soaps and detergents.\nUses of chlorine : As a germicide and disinfectant for sterilisation of drinking water and for water of swimming pools.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nUses of sodium hydroxide: In the manufacture of soaps and detergents.\nUses of chlorine : As a germicide and disinfectant for sterilisation of drinking water and for water of swimming pools.\nUses of hydrogen: In the manufacture of ammonia which is used for the preparation of various fertilizers like urea, ammonium sulphate etc.\nQuestion 25.\nHow is washing soda prepared from sodium carbonate? Give its chemical equation. State the type of this salt. Name the type of hardness of water which can be removed by it? (2020)\nAnswer:\nWashing soda is prepared by recrystallisation of sodium carbonate:\n\nIt is used to remove the permanent hardness of water. Hard water is treated with a calculated amount of washing soda when chlorides and sulphates of calcium and magnesium present in hard water get precipitated as insoluble calcium and magnesium carbonates which can be easily filtered off. The water thus becomes soft.\nCaCl2 + Na2CO3 \u2192 CaCO3\u2193 + 2NaCl\nMgSO4 + Na2CO3 \u2192 MgCOsub>3\u2193 + Na2SO4\nQuestion 26.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nCaCl2 + Na2CO3 \u2192 CaCO3\u2193 + 2NaCl\nMgSO4 + Na2CO3 \u2192 MgCOsub>3\u2193 + Na2SO4\nQuestion 26.\nGive reasons for the following:\n(i) Only one half of water molecule is shown in the formula of plaster of Paris.\n(ii) Sodium hydrogen carbonate is used as an antacid.\n(iii) On strong heating, blue coloured copper sulphate crystals turn white. (2020)\nAnswer:\n(i) Only one half of water molecule is shown in the formula of plaster of Paris (CaSO4. \n1\n2\n H2O) as one molecule of water is being shared by two molecules of calcium sulphate (CaSO4). So the effective water of crystallisation for one CaSO4 unit comes to half molecule of water.\n(ii) Acidity can be neutralised by a base. Sodium hydrogen carbonate can be used as an antacid solution because it is a weak base and will react\nwith excess acid produced in the stomach due to hyperacidity and will neutralise it.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nwith excess acid produced in the stomach due to hyperacidity and will neutralise it.\n(iii) Blue coloured copper sulphate crystals are hydrated copper sulphate, CuSO4.5H2O. On heating blue copper sulphate crystals looses its water of crystallisation and turns into anhydrous copper sulphate which is white in colour.\n\nQuestion 27.\nDuring electrolysis of brine, a gas \u2018G\u2019 is liberated at anode. When this gas \u2018G\u2019 is passed through slaked lime, a compound \u2018C\u2019 is formed, which is used for disinfecting drinking water.\n(i) Write formula of \u2018G\u2019 and \u2018C\u2019.\n(ii) State the chemical equations involved.\n(iii) What is common name of compound \u2018C\u2019 ? Give its chemical name. (2020)\nAnswer:\n(i) During electrolysis of brine, chlorine is obtained at anode. When chlorine is passed through slaked lime, bleaching powder is formed which is used for disinfecting drinking water. Hence, G is Cl2 and C is CaOCl2.", + "(2020)\nAnswer:\n(i) During electrolysis of brine, chlorine is obtained at anode. When chlorine is passed through slaked lime, bleaching powder is formed which is used for disinfecting drinking water. Hence, G is Cl2 and C is CaOCl2.\n\n(iii) Common name of C is bleaching powder. Its chemical name is calcium hypochlorite.\nQuestion 28.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(iii) Common name of C is bleaching powder. Its chemical name is calcium hypochlorite.\nQuestion 28.\nIdentify the acid and the base from which sodium chloride is obtained. Which type of salt is it? When is it called rock salt? How is rock salt formed? (Delhi 2019)\nAnswer:\nSodium chloride is obtained by the neutralisation of sodium hydroxide (base) with hydrochloric acid (acid). It is a neutral salt. Common salt found in the form of solid deposits is often brown in colour due to presence of impurities which is called rock salt. Rock salt is formed by evaporation of salty water of inland lakes.\nQuestion 29.\nA white powder is added while baking cakes to make it soft and spongy. Name its main ingredients. Explain the function of each ingredient. Write the chemical reaction taking place when the powder is heated during baking. (AI2019)\nAnswer:", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nAnswer:\nThe white powder added while baking cakes to make it soft and spongy is baking powder. Its main ingredients are sodium hydrogen carbonate and a mild edible acid like tartaric acid or citric acid. NaHCO3 decomposes to give out CO2 which causes the cake to rise and makes it soft and spongy. The function of tartaric acid or citric acid is to neutralise sodium carbonate formed during heating which can otherwise make the cake bitter. Reaction taking place when the powder is heated:\n\nQuestion 30.\nThe pH of a salt used to make tasty and crispy pakoras is 14. Identify the salt and write a chemical equation for its formation. List its two uses. (2018)\nAnswer:\nSalt used to make tasty and crispy pakoras is sodium bicarbonate (NaHCO3), pH = 9. On large scale, sodium bicarbonate is prepared as:\n\nTwo uses of sodium bicarbonate are as follows :\n(i) It is used as an antacid in medicines.\n(ii) It is used as an additive in food and drinks.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nTwo uses of sodium bicarbonate are as follows :\n(i) It is used as an antacid in medicines.\n(ii) It is used as an additive in food and drinks.\nNote : In the question paper, the given pH is 14 which should be 9.\nQuestion 31.\nWrite one point of difference between each of the following:\n(i) A hydrated salt and an anhydrous salt.\n(ii) Washing soda and soda ash.\n(iii) Baking soda and baking powder. (Board Term 1,2017)\nAnswer:\n(i)\n(ii)\n(iii)\nQuestion 32.\nComplete the following table:\nAnswer:\nQuestion 33.\nA white coloured powder is used by doctors for supporting fractured bones.\n(a) Write chemical name and formula of the powder.\n(b) When this white powder is mixed with water a hard solid mass is obtained. Write balanced chemical equation for this change. (Board Term I, 2016)\nAnswer:\n(a) Chemical name of the powder is calcium sulphate hemihydrate. Chemical formula of the powder is CaSO4. \n1\n2\n ApH2O.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nAnswer:\n(a) Chemical name of the powder is calcium sulphate hemihydrate. Chemical formula of the powder is CaSO4. \n1\n2\n ApH2O.\n(b) When water is added to plaster of Paris, it sets into a hard mass in about half an hour. The setting of plaster of Paris is due to its hydration to form crystals of gypsum which set to form a hard, solid mass.\n\nQuestion 34.\n(a) Define an acid-base indicator. Mention one synthetic acid-base indicator.\n(b) If someone in the family is suffering from a problem of acidity after overeating, which of the following substances would you suggest as a remedy?\nLemon juice, vinegar or baking soda solution. Mention the property on the basis of which you will choose the remedy. (Board Term I, 2014)\nAnswer:\n(a) Acid \u2013 base indicators : The indicators which show different colours in acidic and basic medium are called acid-base indicators. Phenolphthalein is a synthetic indicator.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nAnswer:\n(a) Acid \u2013 base indicators : The indicators which show different colours in acidic and basic medium are called acid-base indicators. Phenolphthalein is a synthetic indicator.\n(b) Acidity can be neutralised by a base. Hence, we should choose baking soda solution because it is a weak base and will react with excess acid produced in the stomach due to hyperacidity and will neutralise it.\nQuestion 35.\nDefine water of crystallisation. Give the chemical formula for two compounds as examples. How can it be proved that the water of crystallisation makes a difference in the state and colour of the compounds? (2020)\nAnswer:\nWater of crystallisation : It is the fixed number of water molecules present in one formula unit of a salt, e.g., Gypsum (CaSO4.2H2O) has two molecules of water of crystallisation.\nIn hydrated copper sulphate (CuSO4.5H2O), there are five molecules of water of crystallisation.\nActivity:", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nIn hydrated copper sulphate (CuSO4.5H2O), there are five molecules of water of crystallisation.\nActivity:\n\u2013 Take few crystals of copper sulphate in a dry boiling tube. These are blue in colour.\n\u2013 Heat the boiling tube by holding it with a test tube holder on the flame of the burner.\n\nObservations : You will observe that the colour of copper sulphate crystals after heating becomes white. You may also notice water droplets on the mouth side of the boiling tube which are obtained from water of crystallisation.After adding 2-3 drops of water on the white sample of copper sulphate (obtained after heating) you will observe that the blue colour of copper sulphate crystals is restored.\n\nQuestion 36.\n(a) A student dropped a few pieces of marble in dilute hydrochloric acid contained in a test tube. The evolved gas was passed through lime water. What change would be observed in lime water? Write balanced chemical equations for both the changes observed.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(b) State the chemical property in each case on which the following uses of baking soda are based:\n(i) as an antacid\n(ii) as a constituent of baking powder. (Board Term I, 2017)\nAnswer:\n(a) When marble reacts with dilute HCl carbon dioxide gas is liberated.\n\nWhen CO2 gas is passed through lime water, insoluble calcium carbonate is formed which appears milky.\n\n(b) (i) The excess acid formed in the stomach due to various reasons (one being overeating) is neutralised by sodium hydrogen carbonate. Hence, it is used as an ingredient of antacid.\n(ii) Baking soda (sodium hydrogen carbonate) is a constituent of baking power. On heating it gives out CO2 which causes the cake to rise and make it soft and spongy.\nQuestion 37.\n(a) What are anhydrous and hydrated salts? Explain with a suitable example of each]\n(b) How is plaster of Paris prepared? What reaction takes place when it sets to a hard mass? (Board Term I, 2017)\nAnswer:\n(a) Refer to answer 31(i).", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(b) How is plaster of Paris prepared? What reaction takes place when it sets to a hard mass? (Board Term I, 2017)\nAnswer:\n(a) Refer to answer 31(i).\n(b) It is prepared from gypsum which is calcium sulphate dihydrate (CaSO4.2H2O). Gypsum is heated in a kiln to a temperature of 100\u00b0C (373 K). At this temperature, it loses three-fourth of its water of crystallisation forming plaster of Paris.\n\nRefer to answer 33(b).\nQuestion 38.\n(a) Write the chemical formula of hydrated copper sulphate and anhydrous copper sulphate. Giving an activity illustrate how these two are interconvertible.\n(b) Write chemical names and formulae of plaster of Paris and gypsum. (Board Term 1, 2016)\nAnswer:\n(a) The chemical formula of hydrated copper sulphate is CuSO4.5H2O(s) and anhydrous copper sulphate is CuSO4(s).\nFor activity refer to answer 35.", + "(Board Term 1, 2016)\nAnswer:\n(a) The chemical formula of hydrated copper sulphate is CuSO4.5H2O(s) and anhydrous copper sulphate is CuSO4(s).\nFor activity refer to answer 35.\n(b) Plaster of Paris is calcium sulphate hemihydrate; CaSO4.\n1\n2\n H2O and Gypsum is calcium sulphate dihydrate;\n(CaSO4.2H2O).\nQuestion 39.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nFor activity refer to answer 35.\n(b) Plaster of Paris is calcium sulphate hemihydrate; CaSO4.\n1\n2\n H2O and Gypsum is calcium sulphate dihydrate;\n(CaSO4.2H2O).\nQuestion 39.\nHow is sodium hydroxide produced? Write the balanced chemical equation also. Why is this process called as chlor-alkali process? In this process name the products given off at:\n(a) anode\n(b) cathode\nWrite one use of each of these products. (Board Term I, 2015)\nAnswer:\nRefer to answer 24.\nQuestion 40.\nWhat is water of crystallization? Write the common name and chemical formula of a commercially important compound which has ten water molecules as water of crystallization. How is this compound obtained? Write the chemical equation also. List any two uses of this compound. (Board Term I, 2015)\nAnswer:\nWater of crystallization : Crystals of some salts contain certain amount of associated water.\nThe water associated with the crystal (or molecule) of any salt is called water of crystallisation.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\nWater of crystallization : Crystals of some salts contain certain amount of associated water.\nThe water associated with the crystal (or molecule) of any salt is called water of crystallisation.\nThe hydrated salt is known as washing soda which is sodium carbonate containing 10 molecules of water of crystallization, i.e., it is sodium carbonate decahydrate. Its molecular formula is Na2CO3.10H2O.\nIt can be obtained by heating baking soda followed by recrystallisation from its aqueous solution.\n\nUses of sodium carbonate:\n(i) For the manufacture of glass, soap, papers and chemicals like caustic soda (NaOH), borax, etc.\n(ii) For washing purposes (laundry works).\nQuestion 41.\n(a) Name and describe giving chemical equation the process used for producing sodium hydroxide. Why is this process so named?\n(b) Give one use of each of any two products obtained in this process. (Board Term I, 2014)\nAnswer:\nRefer to answer 24.\nQuestion 42.", + "CHAPTER 2-ACIDS,BASES AND SALTS -IMPORTANT QUESTIONSS.pdf.txt\n(b) Give one use of each of any two products obtained in this process. (Board Term I, 2014)\nAnswer:\nRefer to answer 24.\nQuestion 42.\n(a) You have three solutions \u2013 A, B and C having a pH of 6, 2 and 9 respectively. Arrange these solutions in increasing order of hydrogen ion concentration. Which of the three is most acidic? What happens to the hydrogen ion concentration in A as it is diluted?\n(b) If someone is suffering from a stomach problem called acidity, why is a solution of baking soda offered as a remedy?\n(c) Write chemical name and formula of baking soda. (Board Term I, 2013)\nAnswer:\n(a) The solution having lower pH will have more hydrogen ion concentration. Hence, solution B (i.e., pH = 2) will have more hydrogen ion concentration.\n\nSolution B is most acidic.\nAdding water to solution A, will reduce the concentration of hydrogen ions in the solution.\n(b) Refer to answer 26(ii).\n(c) Refer to answer 31(iii).\nNCERT Solutions for Class 10", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nCBSE Class 10 Science Notes Chapter 2 Acids Bases and Salts\nIndicators: Indicators are substances which indicate the acidic or basic nature of the solution by the colour change.\nTypes of Indicator: There are many types of indicators. Some common types of indicators are:\n1. Natural Indicators: Indicators obtained from natural sources are called Natural Indicators. Litmus, turmeric, red cabbage, China rose, etc., are some common natural indicators used widely to show the acidic or basic character of substances.\nLitmus: Litmus is obtained from lichens. The solution of litmus is purple in colour. Litmus paper comes in two colours- blue and red.\nAn acid turns blue litmus paper red.\nA base turns red litmus paper blue.\nTurmeric: Turmeric is another natural indicator. Turmeric is yellow in colour. Turmeric solution or paper turns reddish brown with base. Turmeric does not change colour with acid.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nTurmeric: Turmeric is another natural indicator. Turmeric is yellow in colour. Turmeric solution or paper turns reddish brown with base. Turmeric does not change colour with acid.\nRed Cabbage: The juice of red cabbage is originally purple in colour. Juice of red cabbage turns reddish with acid and turns greenish with base.\n2. Olfactory Indicator: Substances which change their smell when mixed with acid or base are known as Olfactory Indicators. For example; Onion, vanilla etc.\nOnion: Paste or juice of onion loses its smell when added with base. It does not change its smell with acid.\n\nVanilla: The smell of vanilla vanishes with base, but its smell does not vanish with an acid.\n\nOlfactory Indicators are used to ensure the participation of visually impaired students in the laboratory.\n3. Synthetic Indicator: Indicators that are synthesized in the laboratory are known as Synthetic Indicators. For example; Phenolphthalein, methyl orange, etc.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\n3. Synthetic Indicator: Indicators that are synthesized in the laboratory are known as Synthetic Indicators. For example; Phenolphthalein, methyl orange, etc.\nPhenolphthalein is a colourless liquid. It remains colourless with acid but turns into pink with a base.\nMethyl orange is originally orange in colour. It turns into the red with acid and turns into yellow with base.\nAcids: Acids are sour in taste, turn blue litmus red, and dissolve in water to release H+ ions.\nExample: Sulphuric acid (H2SO4), Acetic Acid (CH3COOH), Nitric Acid (HNO3) etc.\nProperties of Acids:\nAcids have a sour taste.\nTurns blue litmus red.\nAcid solution conducts electricity.\nRelease H+ ions in aqueous solution.\nTypes of Acids: Acids are divided into two types on the basis of their occurrence i.e., Natural acids and Mineral acids.\n(i) Natural Acids: Acids which are obtained from natural sources are called Natural Acids or Organic Acids.", + "Release H+ ions in aqueous solution.\nTypes of Acids: Acids are divided into two types on the basis of their occurrence i.e., Natural acids and Mineral acids.\n(i) Natural Acids: Acids which are obtained from natural sources are called Natural Acids or Organic Acids.\nExamples:\nMethanoic acid (HCOOH)\nAcetic acid (CH3COOH)", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\n(i) Natural Acids: Acids which are obtained from natural sources are called Natural Acids or Organic Acids.\nExamples:\nMethanoic acid (HCOOH)\nAcetic acid (CH3COOH)\nOxalic acid (C2H2O4) etc.\n(ii) Mineral Acids: Acids that are prepared from minerals are known as Mineral Acids Example; Inorganic acids, man-made acids or synthetic acid are also known as Mineral Acids.\nExample:\nHydrochloric acid (HCl)\nSulphuric acid (H2SO4)\nNitric acid (HNO3)\nCarbonic acid (H2CO3)\nPhosphoric acid (H3PO4) etc.\nChemical Properties of Acid:\n(i) Reaction of acids with metal: Acids give hydrogen gas along with respective salt when they react with a metal.\nMetal + Acid \u2192 Salt + Hydrogen\nExamples:\nHydrogen gas and zinc chloride are formed when hydrochloric acid reacts with zinc metal.\n\nHydrogen gas and sodium sulphate are formed when sulphuric acid reacts with sodium metal.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nExamples:\nHydrogen gas and zinc chloride are formed when hydrochloric acid reacts with zinc metal.\n\nHydrogen gas and sodium sulphate are formed when sulphuric acid reacts with sodium metal.\n\nTest For Hydrogen Gas: The gas evolved after reaction of acid with metal can be tested by bringing a lighted candle near it. If the gas bums with a pop sound, then it confirms the evolution of hydrogen gas. Burning with pop sound is the characteristic test for hydrogen gas.\n(ii) Reaction of acids with metal carbonate: Acids give carbon dioxide gas and respective salts along with water when they react with metal carbonates.\nMetal carbonate + Acid \u2192 Salt + Carbon dioxide + Water\nExamples:\nHydrochloric acid gives carbon dioxide gas, sodium chloride along with water when reacts with sodium carbonate.\n\nSulphuric acid gives calcium sulphate, carbon dioxide gas, calcium sulphate and water when it reacts with calcium carbonate", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nSulphuric acid gives calcium sulphate, carbon dioxide gas, calcium sulphate and water when it reacts with calcium carbonate\n\nNitric acid gives sodium nitrate, water and carbon dioxide gas when it reacts with sodium carbonate.\n\n(iii) Reaction of acid with hydrogen carbonates (bicarbonates): Acids give carbon dioxide gas, respective salt and water when they react with metal hydrogen carbonate.\nAcid + Metal hydrogen carbonate \u2192 Salt + Carbon dioxide + Water\nExample:\nSulphuric acid gives sodium sulphate, Carbon dioxide gas and water when it reacts with sodium bicarbonate.\n\nTest For Evolution of Carbon Dioxide Gas: Carbon dioxide turns lime water milky when passed through it. This is the characteristic test for carbon dioxide gas.\nThe gas evolved because of reaction of the acid with metal carbonate or metal hydrogen carbonate turns lime water milky. This shows that the gas is carbon dioxide gas. This happens because of the formation of a white precipitate of calcium carbonate.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nBut when excess of carbon dioxide is passed through lime water, it makes milky colour of lime water disappear. This happens because of formation of calcium hydrogen carbonate. As calcium hydrogen carbonate is soluble in water, thus, the milky colour of solution mixture disappears.\nCommon in Acids: Acids give hydrogen gas when they react with metal. This shows that all acids contains hydrogen. For example; Hydrochloric acid (HCl), sulphuric acid (H2SO4), nitric acid (HNO3), etc.\nWhen an acid is dissolved in water, it dissociates hydrogen. The dissociation of hydrogen ion in aqueous solution is the common property in all acids. Because of the dissociation of hydrogen ion in aqueous solution, an acid shows acidic behaviour.\nExamples:\nHydrochloric acid (HCl) gives hydrogen ion (H+) and chloride ion (Cl\u2013) when it is dissolved in water.\n\nAcetic acid (CH3COOH) gives acetate ion (CH3COO\u2013) and hydrogen ion (H+).\n\nAcids\nStrong Acids", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nHydrochloric acid (HCl) gives hydrogen ion (H+) and chloride ion (Cl\u2013) when it is dissolved in water.\n\nAcetic acid (CH3COOH) gives acetate ion (CH3COO\u2013) and hydrogen ion (H+).\n\nAcids\nStrong Acids\nAn acid which is completely ionised in water and produces (H+) is called Strong Acid.\nExamples: Hydrochloric acid (HCl), Sulphuric acid (H2SO4), Nitric acid (HNO3)\nWeak Acids\nAn acid which is partially ionised in water and thus produces a small amount of hydrogen ions (H+) is called a Weak Acid.\nExample: Acetic acid (CH3COOH), Carbonic acid (H2CO3)\nWhen a concentrated solution of acid is diluted by mixing water, then the concentration of Hydrogen ions (H+) or hydronium ion (H3O\u2013) per unit volume decreases.\nBases: Bases are bitter in taste, have soapy touch, turn red litmus blue and give hydroxide ions (OH\u2013) in aqueous solution.", + "Bases: Bases are bitter in taste, have soapy touch, turn red litmus blue and give hydroxide ions (OH\u2013) in aqueous solution.\nExamples: Sodium hydroxide (caustic soda) \u2013 NaOH\nCalcium hydroxide \u2013 Ca(OH)2\nPotassium hydroxide (caustic potash) \u2013 (KOH)\nProperties of Bases:\nHave a bitter taste.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nExamples: Sodium hydroxide (caustic soda) \u2013 NaOH\nCalcium hydroxide \u2013 Ca(OH)2\nPotassium hydroxide (caustic potash) \u2013 (KOH)\nProperties of Bases:\nHave a bitter taste.\nSoapy to touch.\nTurns red litmus blue.\nConducts electricity in solution.\nRelease OH\u2013 ions in Aqueous Solution\nTypes of bases: Bases can be divided in two types \u2013 Water soluble and Water-insoluble.\nThe hydroxide of alkali and alkaline earth metals are soluble in water. These are also known as alkali.\nFor example; sodium hydroxide, magnesium hydroxide, calcium hydroxide, etc. Alkali is considered a strong base.\nChemical properties of bases:\n(i) Reaction of Base with Metals: When alkali (base) reacts with metal, it produces salt and hydrogen gas.\nAlkali + Metal \u2192 Salt + Hydrogen\nExamples: Sodium hydroxide gives hydrogen gas and sodium zincate when reacts with zinc metal.\n\nSodium aluminate and hydrogen gas are formed when sodium hydroxide reacts with aluminium metal.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nExamples: Sodium hydroxide gives hydrogen gas and sodium zincate when reacts with zinc metal.\n\nSodium aluminate and hydrogen gas are formed when sodium hydroxide reacts with aluminium metal.\n\n(ii) Reaction of Base with Oxides of Non-metals: Non-metal oxides are acidic in nature. For example; carbon dioxide is a non-metal oxide. When carbon dioxide is dissolved in water it produces carbonic acid.\nTherefore, when a base reacts with non-metal oxide, both neutralize each other resulting respective salt and water.\nBase + Non-metal oxide \u2192 Salt + Water\n(Non-metal oxides are acidic in nature)\nExamples:\nSodium hydroxide gives sodium carbonate and water when it reacts with carbon dioxide.\n\nCalcium hydroxide gives calcium carbonate and water when it reacts with carbon dioxide.\n\n(iii) Neutralisation Reaction: An acid neutralizes a base when they react with each other and respective salt and water are formed.\nAcid + Base \u2192 Salt + Water", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\n(iii) Neutralisation Reaction: An acid neutralizes a base when they react with each other and respective salt and water are formed.\nAcid + Base \u2192 Salt + Water\nSince, the reaction between acid and base both neutralize each other, hence, it is also known as Neutralization Reaction.\nExamples: Sodium chloride and water are formed when hydrochloric acid reacts with sodium hydroxide (a strong base).\n\nIn a similar way, calcium chloride is formed along with water when hydrochloric acid reacts with calcium hydroxide (a base).\n\n(iv) Reaction of Acid with Metal Oxides: Metal oxides are basic in nature. Thus, when an acid reacts with a metal oxide both neutralize each other. In this reaction, the respective salt and water are formed.\nAcid + Metal Oxide \u2192 Salt + Water\n(Metal oxides are basic in nature)\nExamples:", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nAcid + Metal Oxide \u2192 Salt + Water\n(Metal oxides are basic in nature)\nExamples:\nCalcium is a metal, thus, calcium oxide is a metallic oxide which is basic in nature. When an acid, such as hydrochloric acid, reacts with calcium oxide, neutralization reaction takes place and calcium chloride, along with water is formed.\n\nSimilarly, when sulphuric acid reacts with zinc oxide, zinc sulphate and water are formed.\n\nCommon in all bases: A base dissociates hydroxide ion in water, which is responsible for the basic behaviour of a compound.\nExample: When sodium hydroxide is dissolved in water, it dissociates hydroxide ion and sodium ion.\n\nSimilarly, when potassium hydroxide is dissolved in water, it dissociates hydroxide ion and potassium ion.\n\nThus, the base shows its basic character because of dissociation of hydroxide ion.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nSimilarly, when potassium hydroxide is dissolved in water, it dissociates hydroxide ion and potassium ion.\n\nThus, the base shows its basic character because of dissociation of hydroxide ion.\nNeutralisation Reaction: When an acid reacts with a base, the hydrogen ion of acid combines with the hydroxide ion of base and forms water. As these ions combine together and form water instead of remaining free, thus, both neutralize each other.\n\nExample: When sodium hydroxide (a base) reacts with hydrochloric acid, sodium hydroxide breaks into a sodium ion and hydroxide ion and hydrochloric acid breaks into hydrogen ion and chloride ion.\nHydrogen ion and hydroxide ion combine together and form water, while sodium ion and chloride ion combine together and form sodium chloride.\n\nDilution of Acid and Base: The concentration of hydrogen ion in an acid and hydroxide ion in a base, per unit volume, shows the concentration of acid or base.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nDilution of Acid and Base: The concentration of hydrogen ion in an acid and hydroxide ion in a base, per unit volume, shows the concentration of acid or base.\nBy mixing of acid to water, the concentration of hydrogen ion per unit volume decreases. Similarly, by addition of base to water, the concentration of hydroxide ion per unit volume decreases. This process of addition of acid or base to water is called Dilution and the acid or base is called Diluted.\nThe dilution of acid or base is exothermic. Thus, acid or base is always added to water and water is never added to acid or base. If water is added to a concentrated acid or base, a lot of heat is generated, which may cause splashing out of acid or base and may cause severe damage as concentrated acid and base are highly corrosive.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nStrength of Acid and Base: Acids in which complete dissociation of hydrogen ion takes place are called Strong Acids. Similarly, bases in which complete dissociation of hydroxide ion takes place are called Strong Bases.\nIn mineral acid, such as hydrochloric acid, sulphuric acid, nitric acid, etc. hydrogen ion dissociates completely and hence, they are considered as strong acids. Since inorganic acids hydrogen ions do not dissociate completely, so they are weak acids.\n\nFor water or neutral solutions : pH = 7\nFor acidic solutions : pH < 7\nFor basic solution : pH > 7\nUniversal Indicator: Using a litmus paper, phenolphthalein, methyl orange, etc. only the acidic or basic character of a solution can be determined, but the use of these indicators does not give the idea about the strength of acid or base. So, to get the strength as well as acidic and basic nature of a given solution universal indicator is used.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nUniversal indicator shows different colour over the range of pH value from 1 to 14 for a given solution. Universal indicator is available both in the form of strips and solution. Universal indicator is the combination of many indicators, such as water, propanol, phenolphthalein, sodium salt, sodium hydroxide, methyl red, bromothymol blue monosodium salt, and thymol blue monosodium salt. The colour matching chart is supplied with a universal indicator which shows the different colours for different values of pH.\n\npH value shown by different colours role of pH everyday life:\n(i) pH in our digestive system: Dilute HCl (Hydrochloric acid) helps in digestion of food (proteins) in our stomach. Excess acid in stomach causes acidity (indigestion). Antacids like magnesium hydroxide [Mg(OH)2] also known as milk of magnesia and sodium hydrogen carbonate (baking soda) are used to neutralize excess acid.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\n(ii) Tooth decay caused by acids: The bacteria present in our mouth converts the sugar into acids. When the pH of acid formed in the mouth falls below 5.5, tooth-decaying starts. The excess acid has to be removed by cleaning the teeth with a good quality toothpaste because these kinds of toothpaste are alkaline in nature.\n(iii) Soil of pH and plant growth: Most of the plants have a healthy growth when the soil has a specific pH (close to 7) range which should be neither alkaline nor highly acidic. Therefore,\nCompound \u2018X\u2019 is Sodium hydroxide (NaOH).\nCompound \u2018A\u2019 is Zinc sulphate (ZnSO4).\nCompound \u2018B\u2019 is Sodium chloride (NaCl).\nCompound \u2018C\u2019 is Sodium acetate (CH3COONa)", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nCompound \u2018X\u2019 is Sodium hydroxide (NaOH).\nCompound \u2018A\u2019 is Zinc sulphate (ZnSO4).\nCompound \u2018B\u2019 is Sodium chloride (NaCl).\nCompound \u2018C\u2019 is Sodium acetate (CH3COONa)\nSalts: Salts are the ionic compounds which are produced after the neutralization reaction between acid and base. Salts are electrically neutral. There are number of salts but sodium chloride is the most common among them. Sodium chloride is also known as table salt or common salt. Sodium chloride is used to enhance the taste of food.\nCharacteristics of salt:\nMost of the salts are crystalline soild.\nSalts may be transparent or opaque.\nMost of the salts are soluble in water.\nSolution of the salts conducts electricity in their molten state also.\nThe salt may be salty, sour, sweet, bitter and umami (savoury).\nNeutral salts are odourless.\nSalts can be colourless or coloured.\nFamily of Salt: Salts having common acidic or basic radicals are said to belong to the same family.\nExample:", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nNeutral salts are odourless.\nSalts can be colourless or coloured.\nFamily of Salt: Salts having common acidic or basic radicals are said to belong to the same family.\nExample:\n(i) Sodium chloride (NaCl) and Calcium chloride (CaCl2) belongs to chloride family.\n(ii) Calcium chloride (CaCl2) and Calcium sulphate (CaSO4) belongs to calcium family.\n(iii) Zinc chloride (ZnCl2) and Zinc sulphate (ZnSO4) belongs to the zinc family.\nNeutral, Acidic and Basic Salts:\n(i) Neutral Salt: Salts produced because of reaction between a strong acid and strong base are neutral in nature. The pH value of such salts is equal to 7, i.e. neutral.\nExample : Sodium chloride, Sodium sulphate. Postassium chloride, etc.\nSodium chloride (NaCl): It is formed after the reaction between hydrochloric acid (a strong acid) and sodium hydroxide (a strong base).", + "neutral.\nExample : Sodium chloride, Sodium sulphate. Postassium chloride, etc.\nSodium chloride (NaCl): It is formed after the reaction between hydrochloric acid (a strong acid) and sodium hydroxide (a strong base).\n\nSodium Sulphate (Na2SO4): It is formed after the reaction between sodium hydroxide (a strong base) and sulphuric acid ( a strong acid).", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nSodium Sulphate (Na2SO4): It is formed after the reaction between sodium hydroxide (a strong base) and sulphuric acid ( a strong acid).\n\nPotassium Chloride (KCl): It is formed after the reaction between potassium hydroxide (a strong base) and hydrochloric acid (a strong acid).\n\n(ii) Acidic Salts: Salts which are formed after the reaction between a strong acid and weak base are called Acidic salts. The pH value of acidic salt is lower than 7. For example Ammonium sulphate, Ammonium chloride, etc.\nAmmonium chloride is formed after reaction between hydrochloric acid (a strong acid) and ammonium hydroxide (a weak base).\n\nAmmonium sulphate is formed after reaction between ammonium hydroxide (a weak base) and sulphuric acid (a strong acid).\n\n(iii) Basic Salts: Salts which are formed after the reaction between a weak acid and strong base are called Basic Salts. For example; Sodium carbonate, Sodium acetate, etc.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\n(iii) Basic Salts: Salts which are formed after the reaction between a weak acid and strong base are called Basic Salts. For example; Sodium carbonate, Sodium acetate, etc.\nSodium carbonate is formed after the reaction between sodium hydroxide (a strong base) and carbonic acid (a weak acid).\n\nSodium acetate is formed after the reaction between a strong base, sodium hydroxide (a strong base) and acetic acid, (a weak acid).\n\nCause of formation of acidic, basic and neutral salts:\nWhen a strong acid reacts with a weak base, the base is unable to fully neutralize the acid. Due to this, an acidic salt is formed.\nWhen a strong base reacts with a weak acid, the acid is unable to fully neutralize the base. Due to this, a basic salt is formed.\nWhen equally strong acid and a base react, they fully neutralize each other. Due to this, a neutral salt is formed.\npH value of salt:\nNeutral salt: The pH value of a neutral salt is almost equal to 7.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\npH value of salt:\nNeutral salt: The pH value of a neutral salt is almost equal to 7.\nAcidic salt: The pH value of an acidic salt is less than 7.\nBasic salt: The pH value of a basic salt is more than 7.\nSome Important Chemical Compounds\n1. Common Salt (Sodium Chloride): Sodium chloride (NaCl) is also known as Common or Table Salt. It is formed after the reaction between sodium hydroxide and hydrochloric acid. It is a neutral salt. The pH value of sodium chloride is about 7. Sodium chloride is used to enhance the taste of food. Sodium chloride is used in the manufacturing of many chemicals.\nImportant chemical from sodium chloride", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nImportant chemical from sodium chloride\nSodium Hydroxide (NaOH): Sodium hydroxide is a strong base. It is also known as caustic soda. It is obtained by the electrolytic decomposition of solution of sodium chloride (brine). In the process of electrolytic decomposition of brine (aqueous solution of sodium chloride), brine decomposes to form sodium hydroxide. In this process, chlorine is obtained at anode and hydrogen gas is obtained at cathode as by products. This whole process is known as Chlor \u2013 Alkali process.\n\nUse of products after the electrolysis of brine:\nHydrogen gas is used as fuel, margarine, in making of ammonia for fertilizer, etc.\nChlorine gas is used in water treatment, manufacturing of PVC, disinfectants, CFC, pesticides. It is also used in the manufacturing of bleaching powder and hydrochloric acid.\nSodium hydroxide is used for degreasing of metals, manufacturing of paper, soap, detergents, artificial fibres, bleach, etc.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nSodium hydroxide is used for degreasing of metals, manufacturing of paper, soap, detergents, artificial fibres, bleach, etc.\n2. Bleaching Powder (CaOCl2): Bleaching powder is also known as chloride of lime. It is a solid and yellowish white in colour. Bleaching powder can be easily identified by the strong smell of chlorine.\nWhen calcium hydroxide (slaked lime) reacts with chlorine, it gives calcium oxychloride (bleaching powder) and water is formed.\n\nAqueous solution of bleaching powder is basic in nature. The term bleach means removal of colour. Bleaching powder is often used as bleaching agent. It works because of oxidation. Chlorine in the bleaching powder is responsible for bleaching effect.\nUse of Bleaching Powder:\nBleaching powder is used as disinfectant to clean water, moss remover, weed killers, etc.\nBleaching powder is used for bleaching of cotton in textile industry, bleaching of wood pulp in paper industry.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nBleaching powder is used for bleaching of cotton in textile industry, bleaching of wood pulp in paper industry.\nBleaching powder is used as oxidizing agent in many industries, such as textiles industry, paper industry, etc.\n3. Baking Soda (NaHCO3): Baking soda is another important product which can be obtained using byproducts of chlor \u2013 alkali process. The chemical name of baking soda is sodium hydrogen carbonate (NaHCO3) or sodium bicarbonate. Bread soda, cooking soda, bicarbonate of soda, sodium bicarb, bicarb of soda or simply bicarb, etc. are some other names of baking soda.\nPreparation Method: Baking soda is obtained by the reaction of brine with carbon dioxide and ammonia. This is known as Solvay process.\n\nIn this process, calcium carbonate is used as the source of CO2 and the resultant calcium oxide is used to recover ammonia from ammonium chloride.\nProperties of Sodium Bicarbonate:\nSodium bicarbonate is white crystalline solid, but it appears as fine powder.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nProperties of Sodium Bicarbonate:\nSodium bicarbonate is white crystalline solid, but it appears as fine powder.\nSodium hydrogen carbonate is amphoteric in nature.\nSodium hydrogen carbonate is sparingly soluble in water.\nThermal decomposition of sodium hydrogen carbonate (baking soda).\nWhen baking soda is heated, it decomposes into sodium carbonate, carbon dioxide and water.\n2NaHCO3 + heat \u2192 Na2CO3 + CO2 + H2O\nSodium carbonate formed after thermal decomposition of sodium hydrogen carbonate decomposes into sodium oxide and carbon dioxide on further heating.\nNa2CO3 \u2192 Na2O + CO2\nThis reaction is known as Dehydration reaction.\nUse of Baking Soda:\nBaking soda is used in making of baking powder, which is used in cooking as it produces carbon dioxide which makes the batter soft and spongy.\nBaking soda is used as an antacid.\nBaking soda is used in toothpaste which makes the teeth white and plaque free.\nBaking soda is used in cleansing of ornaments made of silver.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nBaking soda is used as an antacid.\nBaking soda is used in toothpaste which makes the teeth white and plaque free.\nBaking soda is used in cleansing of ornaments made of silver.\nSince sodium hydrogen carbonate gives carbon dioxide and sodium oxide on strong heating, thus, it, is used as a fire extinguisher.\nBaking Powder: Baking powder produces carbon dioxide on heating, so it is used in cooking to make the batter spongy. Although, baking soda also produces carbon dioxide on heating, but it is not used in cooking because on heating, baking soda produces sodium carbonate along with carbon dioxide. The sodium carbonate, thus, produced, makes the taste bitter.\n\nBaking powder is the mixture of baking soda and a mild edible acid. Generally, tartaric acid is mixed with baking soda to make baking powder.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nBaking powder is the mixture of baking soda and a mild edible acid. Generally, tartaric acid is mixed with baking soda to make baking powder.\n\nWhen baking powder is heated, sodium hydrogen carbonate (NaHCO3) decomposes to give CO2 and sodium carbonate (Na2CO3). CO2 causes bread and cake fluffy. Tartaric acid helps to remove bitter taste due to formation of Na2CO3.\n4. Washing Soda (Sodium Carbonate)\nPreparation Method: Sodium carbonate is manufactured by the thermal decomposition of sodium hydrogen carbonate obtained by Solvay process.\n\nThe sodium carbonate obtained in this process is dry. It is called Soda ash or Anhydrous sodium carbonate. Washing soda is obtained by rehydration of anhydrous sodium carbonate.\n\nSince there are 10 water molecules in washing soda, hence, it is known as Sodium Bicarbonate Decahydrate.\nSodium carbonate is a crystalline solid and it is soluble in water when most of the carbonates are insoluble in water.\nUse of sodium carbonate:", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nSodium carbonate is a crystalline solid and it is soluble in water when most of the carbonates are insoluble in water.\nUse of sodium carbonate:\nIt is used in the cleaning of cloths, especially in rural areas.\nIn the making of detergent cake and powder.\nIn removing the permanent hardness of water.\nIt is used in glass and paper industries.\nThe water of Crystallization: Many salts contain water molecule and are known as Hydrated Salts. The water molecule present in salt is known as Water of crystallization.\nExamples:\nCopper sulphate pentahydrate (CuSO4.5H2O): Blue colour of copper sulphate is due to presence of 5 molecules of water. When copper sulphate is heated, it loses water molecules and turns: into grey \u2013 white colour, which is known as anhydrous copper sulphate. After adding water, anhydrous copper sulphate becomes blue again.\n\nAcids: Substances which turn blue litmus solution red are called acids. Acids are sour in taste.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nAcids: Substances which turn blue litmus solution red are called acids. Acids are sour in taste.\nBases: Substances which change red litmus solution blue are called bases. They are bitter in taste.\nMineral Acids: Acids which are obtained from minerals like sulphates, nitrates, chlorides etc. are called mineral acids, example, H2SO4 (Sulphuric acid), HNO3 (Nitric acid) and HCl (Hydrochloric acid).\nOrganic Acids: Acids which are obtained from plants and animals are called organic acids. Example citric acid, ascorbic acid, tartaric acid, lactic acid, acetic acid.\nHydronium Ions: They are formed by the reaction of H+ (from acid) and H2O. It is because H+ is unstable.\nUniversal Indicator: A universal indicator is a mixture of indicators which shows a gradual but well-marked series of colour changes over a very wide range of change in concentration of H+ ions.", + "It is because H+ is unstable.\nUniversal Indicator: A universal indicator is a mixture of indicators which shows a gradual but well-marked series of colour changes over a very wide range of change in concentration of H+ ions.\nStrong Acids: Acids which dissociate into ions completely are called strong acids. Example, H2SO4, HCl.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nStrong Acids: Acids which dissociate into ions completely are called strong acids. Example, H2SO4, HCl.\nWeak Acids: Acids which do not dissociate into ions completely are called weak acids. Example, citric acid, acetic acid.\nChemical Properties of Acids:\nAcids react with active metals to give salt and hydrogen gas.\nAcids react with metal carbonates and metal hydrogen carbonates to give salt, water and carbon dioxide.\nAcids react with bases to give salt and water. This reaction is called a neutralization reaction.\nAcids react with metal oxides to give salt and water.\nChemical Properties of Bases:\nReaction with metals: Certain metals such as zinc, aluminium and tin react with alkali solutions on heating and hydrogen gas is evolved.\nReaction with acids: Bases react with acids to form salt and water.\nIndicators: Indicators are substances which indicate the acidic or basic nature of the solution by their colour change.", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nReaction with acids: Bases react with acids to form salt and water.\nIndicators: Indicators are substances which indicate the acidic or basic nature of the solution by their colour change.\npH Scale: A scale for measuring hydrogen ion concentration in a solution.\nThe pH of a solution is defined as the negative logarithm of hydrogen ion concentration in moles per litre.\npH = -log [H+]\npH = -log [H3O+]\nwhere [H+] or [H3O+] represents concentrations of hydrogen ions in a solution.\nThe pH of a neutral solution is 7.\nThe pH of an acidic solution is < 7.\nThe pH of a basic solution is > 7.", + "The pH of a neutral solution is 7.\nThe pH of an acidic solution is < 7.\nThe pH of a basic solution is > 7.\nSome Important Compounds and their Uses:\nEquations of Acids, Bases and Salts:\nAcid + Metal \u2192 Salt + Hydrogen gas\nH2SO4 + Zn \u2192 ZnSO4 + H2\nBase + Metal \u2192 Salt + Hydrogen gas\n2NaOH + Zn \u2192 Na2ZnO2 (Sodium zincate) + H2\nBase + Acid \u2192 Salt + Water\nNaOH (aq) + HCl (aq) \u2192 NaCl (aq) + H2O (l)\nAcids give hydronium ions in water\nHCl + H2O \u2192 H3O+ + Cl\u2013\nBases generate OH- ions in water", + "CHAPTER 2-ACIDS,BASES AND SALTS - NOTES pdf.txt\nBase + Acid \u2192 Salt + Water\nNaOH (aq) + HCl (aq) \u2192 NaCl (aq) + H2O (l)\nAcids give hydronium ions in water\nHCl + H2O \u2192 H3O+ + Cl\u2013\nBases generate OH- ions in water\nNaOH (aq) + H2O \u2192 Na+ (aq) + O\u2013 (aq)\nReactions Of Important Chemical Compounds:\nPreparation of Bleaching powder: By the action of chlorine on dry slaked lime\nCa(OH)2 + Cl2 \u2192 CaOCl2 + H2O\nOn heating, baking soda liberates CO2\n\nPreparation of Plaster of Paris:", + "CHAPTER 2-ACIDS,BASES AND SALTS -EXEMPLER QUESTIONS.pdf.txt\nNCERT Exemplar Class 10 Science Chapter 2 Acids, Bases And Salts\nShort Answer Type Questions\n1.Match the acids given in Column (A) with their correct source given in Column (B).\n\nAns.(a) \u2014> (iv) (b) \u2014> (iii) (c) \u2014> (ii) (d) \u2014> (i)\n\nLactic acid is present in curd, acetic acid is present in vinegar, citric acid is present in lemon and oxalic acid is present in tomato.\n2.What will be the action of the following substances on litmus paper?\nDry.HCl gas, Moistened NH3 gas, Lemon juice, Carbonated soft drink, Curd, Soap solution.\nAns.Dry HC1 gas will not have any effect on litmus paper. Moistened NH3 gas will turn red litmus blue. Curd, lemon juice, carbonated soft drink will turn blue litmus red. Soap solution will turn red litmus blue.\n3.Name the acid present in ant sting and give its chemical formula.", + "Moistened NH3 gas will turn red litmus blue. Curd, lemon juice, carbonated soft drink will turn blue litmus red. Soap solution will turn red litmus blue.\n3.Name the acid present in ant sting and give its chemical formula. Also give the common method to get relief from the discomfort caused by the ant sting.", + "CHAPTER 2-ACIDS,BASES AND SALTS -EXEMPLER QUESTIONS.pdf.txt\n3.Name the acid present in ant sting and give its chemical formula. Also give the common method to get relief from the discomfort caused by the ant sting.\nAns. Ant sting contains methanoic acid (formic acid). Its chemical formula is HCOOH. The common method to get relief is to apply paste of \nNaC\nO\n3\n (baking soda) on it.\n4.What happens when nitric acid is added to egg-shell?\nAns. Egg-shell is made up of calcium carbonate which will react with HNO3 to form \nC\n2\n(g) \nand\nH\n2\n(l) along with calcium nitrate.\n\nCa\nCO\n3\n+2HN\nO\n3\n\u27f6Ca(N\nO\n3\n)\n2\n+C\nO\n2\n+\nH\n2\nO\n5.A student prepared solutions of (i) an acid and (ii) a base in two separate\nbeakers. She forgot to label the solutions and litmus paper is not available in the laboratory.", + "She forgot to label the solutions and litmus paper is not available in the laboratory. Since, both the solutions are colourless, how will she distinguish between the two?\nAns.Add phenolphthalein to a portion of each solution in separate test tube. If it turns pink, the beaker contains base whereas if it remains colourless, it is an acid.", + "CHAPTER 2-ACIDS,BASES AND SALTS -EXEMPLER QUESTIONS.pdf.txt\nAns.Add phenolphthalein to a portion of each solution in separate test tube. If it turns pink, the beaker contains base whereas if it remains colourless, it is an acid.\n\nIf phenolphthalein is not available, pH paper can be used. Acid will turn pH paper red, base will turn pH paper blue.\n6.How would you distinguish between baking powder and washing soda by heating?\nAns.\n\n7.Salt A commonly used in bakery products on heating gets converted into another salt B which itself is used for removal of hardness of water and a gas C is evolved. The gas C when passed through lime water, turns it milky. Identify A, B and C.\nAns.\n\n8.In one of the industrial processes used for manufacture of sodium hydroxide, a gas X is formed as by-product. The gas X reacts with lime water to give a compound Y which is used as a bleaching agent in chemical industry. Identify X and Y giving the chemical equation of the reactions involved.\nAns.", + "CHAPTER 2-ACIDS,BASES AND SALTS -EXEMPLER QUESTIONS.pdf.txt\nAns.\n\n9.What are strong and weak acids? In the following list of acids, separate strong acids from weak acids.\nHydrochloric acid, citric acid, acetic acid, nitric acid, formic acid, sulphuric acid.\nAns. Strong acids are those acids which are completely ionised in aqueous solution. Weak acids are those which do not ionise completely in aqueous solution. Strong adds: HC1,\nHN\nO\n3\n , \nH\n2\nS\nO\n4\n Weak acids: Citric acid, acetic acid, formic acid\n1O.A metal carbonate X on reacting with an acid gives a gas which when passed through a solution Y gives the carbonate back. On the other hand, a gas G that is obtained at anode during electrolysis of brine is passed on dry Y, it gives a compound Z, used for disinfecting drinking water. Identify X, Y, G and Z.\nAns.", + "CHAPTER 2-ACIDS,BASES AND SALTS -EXEMPLER QUESTIONS.pdf.txt\nAns.\n\n11.A dry pellet of a common base B, when kept in open absorbs moisture and turns sticky. The compound is also a by-product of chlor-alkali process. Identify B. What type of reaction occurs when B is treated with an acidic oxide? Write a balanced chemical equation for one such solution.\nAns.\u2018B\u2019 is NaOH. It absorbs moisture from atmosphere because it is hygroscopic in nature. It is obtained by product of chlor-alkali process.\n\n\u2018B\u2019 reacts with acidic oxide \nC\nO\n2\n to form salt and water. This reaction is called neutralisation reaction.\n12.A sulphate salt of Group 2 element of the Periodic Table is a white, soft substance, which can be moulded into different shapes by making its dough. When this compound is left in open for sometime, it becomes a solid mass and cannot be used for moulding purposes. Identify the sulphate salt. Why does it show such a behaviour? Give the reaction involved.\nAns. Salt is CaS04 \n1\n2\n H20, Plaster of Paris, white soft substance. It can be dough,", + "CHAPTER 2-ACIDS,BASES AND SALTS -EXEMPLER QUESTIONS.pdf.txt\nAns. Salt is CaS04 \n1\n2\n H20, Plaster of Paris, white soft substance. It can be dough,\nmoulded into different shapes, as 2 moles of CaS04 share 1 mole of HgO molecule.\n\nCa\nSO\n4\n1\n2\nH\n2\nO+\n3\n2\nH\n2\nO\u27f6Ca\nSO\n4\n2H\n2\nO\n\nWhen it is left in open, it becomes solid mass CaS042H20 (Gypsum) which cannot be used for moulding purposes as it is hard solid mass.\n13.Identify the compound X on the basis of the reactions given below. Also, write the name and chemical formulae of A, B and C.\n\nAns.\n\n14.Match the important chemicals given in Column (A) with the chemical formulae given in Column (B).\n\nAns.\n\n15.Fill in the missing data in the following data.\n\nAns.\n\n16.When Zinc metal is treated with dilute solution of strong acid, a gas is involved which is utilized in hydrogenation of oil.", + "Ans.\n\n15.Fill in the missing data in the following data.\n\nAns.\n\n16.When Zinc metal is treated with dilute solution of strong acid, a gas is involved which is utilized in hydrogenation of oil. Name the gas involved. Write the chemical equation for the reaction involved and also write a test to detect the gas formed.\nHydrogen gas is liberated\nAns.\n\nZn(s)+\nH\n2\nS\nO\n4\n(dil.)\u27f6ZnS\nO\n4\n(aq)+\nH\n2\n(g)", + "CHAPTER 2-ACIDS,BASES AND SALTS -EXEMPLER QUESTIONS.pdf.txt\nHydrogen gas is liberated\nAns.\n\nZn(s)+\nH\n2\nS\nO\n4\n(dil.)\u27f6ZnS\nO\n4\n(aq)+\nH\n2\n(g)\n\nTest: Bring a burning splinter near the gas. The gas will burn with \u2018Pop\u2019 sound which shows gas is H2.\n17.In the following schematic diagram for the preparation of hydrogen gas as shown in the given figure, what would happen if following changes are made ?\n\n(\u0430)In place of zinc granules, same amount of zinc dust is taken in the test tube.\n(b)Instead of dilute sulphuric acid, dilute hydrochloric acid is taken.\n(c)In place of zinc, copper turnings are taken.\n(d)Sodium hydroxide is taken in place of dilute sulphuric acid and the tube\nis heated.\nAns.(a)The reaction will become faster because zinc dust has more surface area.\n(b)Nearly same amount of hydrogen gas will be evolved.\n(c)No reaction will take place as copper is less reactive than hydrogen.", + "Ans.(a)The reaction will become faster because zinc dust has more surface area.\n(b)Nearly same amount of hydrogen gas will be evolved.\n(c)No reaction will take place as copper is less reactive than hydrogen.\n(d)The reaction will take place and hydrogen gas will be evolved.", + "CHAPTER 2-ACIDS,BASES AND SALTS -EXEMPLER QUESTIONS.pdf.txt\n(b)Nearly same amount of hydrogen gas will be evolved.\n(c)No reaction will take place as copper is less reactive than hydrogen.\n(d)The reaction will take place and hydrogen gas will be evolved.\n\n18.For making cake, baking powder is taken. If at home your mother uses baking soda instead of baking powder in cake.\n(a)How will it affect the taste of the cake and why?\n(b)How can baking soda be converted into baking powder?\n(c)What is role of tartaric acid added to baking soda?\nAns.(a)The cake will taste bitter due to formation of sodium carbonate.\n(b)Tartaric acid should be added to baking soda to convert it into baking powder.\n(c)Tartaric acid neutralizes sodium carbonate formed and will not make the taste bitter.", + "CHAPTER 2-ACIDS,BASES AND SALTS - EXTRA QUESTIONS.pdf.txt\nQuestion-1\nWhat is an acid?\nSolution:\nAn acid is a hydrogen-containing chemical compound which, when dissolved in water, gives hydrogen ion (H+) or hydrated hydrogen ion (H2O. H+) or hydronium ion (H3O+).\n\nQuestion-2\nWhat are bases and alkalies?\nSolution:\nOxides and hydroxides of metals and metal like radicals (e.g., NH4+ ions) are called bases. Bases ionise to give OH\u2013 ions in aqueous solution. Bases may be soluble or insoluble in water. The soluble bases are called alkalies. Thus all alkalies are bases but all bases are not alkalies.\nExamples\nNaOH and Cu (OH)2 both are bases, but, since NaOH is soluble in water, it is an alkali. On the other hand, since Cu (OH)2 is insoluble in water, it is not an alkali. Other examples of alkalies are KOH, Ca (OH)2 and NH4OH.", + "On the other hand, since Cu (OH)2 is insoluble in water, it is not an alkali. Other examples of alkalies are KOH, Ca (OH)2 and NH4OH.\nMore Resources\nNCERT Solutions\nNCERT Solutions for Class 10 Science\nQuestion-3\nDefine pH.\nSolution:", + "CHAPTER 2-ACIDS,BASES AND SALTS - EXTRA QUESTIONS.pdf.txt\nMore Resources\nNCERT Solutions\nNCERT Solutions for Class 10 Science\nQuestion-3\nDefine pH.\nSolution:\npH of a given solution is the negative logarithm to the base 10 of the hydrogen ion concentration, [H+] expressed in g ions/lit or moles/lit. Thus\npH=- log1- [H+].\nQuestion-4\nWhat are the practical applications of neutralisation reactions?\nSolution:\nBeing alkaline in nature, cold milk is used to neutralise the acidity produced by HCl present in the gastric juice in the stomach.\nAstronauts in space ships use this reaction to neutralise the dangerous levels of CO2.\n\nFarmers add slaked lime (calcium hydroxide) to reduce acidity of soil.\n\nSting of ants and bees contains formic acid. This can be neutralised by rubbing soap, which contains free sodium hydroxide.\nPersons suffering from acidity are given antacid tablets, containing magnesium hydroxide which neutralises excess HCl produced, in stomach. Alternately, they are advised to sip cold milk, which neutralises HCl.\nQuestion-5", + "CHAPTER 2-ACIDS,BASES AND SALTS - EXTRA QUESTIONS.pdf.txt\nQuestion-5\nWhy the salts solutions of strong acid and strong alkali are neutral?\nSolution:\nLet us take the example of potassium sulphate, which is a salt of strong acid [sulphuric acid] and strong base [potassium hydroxide solution].\nFrom the above equation, it is clear that water is always feebly ionised and hence solution of potassium sulphate is neutral in nature.\n\nQuestion-6\nWhat is an universal indicator?\nSolution:\nUniversal indicator is a solution, which undergoes several colour, changes over a wide range of pH. The colour is used to \u2018indicate\u2019 pH directly. Universal indicators are usually mixtures of several indicators.\nQuestion-7\nWhy common indicators cannot determine pH value of a solution?\nSolution:", + "CHAPTER 2-ACIDS,BASES AND SALTS - EXTRA QUESTIONS.pdf.txt\nQuestion-7\nWhy common indicators cannot determine pH value of a solution?\nSolution:\nCommon indicators, such as litmus; methyl orange and phenolphthalein can easily tell, us whether a solution is acidic or alkaline, but they cannot easily tell how much a given acidic solution is stronger than another acidic solution. It means they cannot really tell us the pH value of different acidic or alkaline solutions.\nIn case of litmus, we cannot tell about pH values of 5,6 and 7. Similarly, in case of methyl orange pH values between 3 and 8 will not be very accurate.\nQuestion-8\nWhat are the general characteristics of acids?\nSolution:\nThey have a sour taste\nThey turn blue litmus to red and methyl orange to red.\nThey react with active metals to give hydrogen\nThey decompose carbonates to produced carbon dioxide and water.\n\nQuestion-9\nWhat is called deliquescent? Give examples.\nSolution:", + "CHAPTER 2-ACIDS,BASES AND SALTS - EXTRA QUESTIONS.pdf.txt\nThey react with active metals to give hydrogen\nThey decompose carbonates to produced carbon dioxide and water.\n\nQuestion-9\nWhat is called deliquescent? Give examples.\nSolution:\nCompounds that take up enough water from the air to dissolve in the water they have taken up are called deliquescent. Calcium chloride (CaCl2) and Sodium hydroxide (NaOH) are the examples of deliquescent.\nQuestion-10\nWrite the uses of chlorines.\nSolution:\nChlorine is used in the production of bleaching powder.\nChlorine is used for making solvents for dry cleaning.\nChlorine is used to sterilise drinking water supply, and the water in swimming pools.\nIt is used in the production of hydrochloric acid.\nWe hope given extra questions for class 10 science helpful to you. In case if you need any assistance, please comment below.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nCBSE Class 10 Science Notes Chapter 1 Chemical Reactions and Equations\nChemical Reactions and Equations: Balanced and unbalanced chemical equations and balancing of chemical equations.\nWhat is a chemical reaction Class 10?\nChemical Reaction: The transformation of chemical substance into another chemical substance is known as Chemical Reaction. For example: Rusting of iron, the setting of milk into curd, digestion of food, respiration, etc.\nIn a chemical reaction, a new substance is formed which is completely different in properties from the original substance, so in a chemical reaction, a chemical change takes place.\nOnly a rearrangement of atoms takes place in a chemical reaction.\nThe substances which take part in a chemical reaction are called reactants.\nThe new substances produced as a result of a chemical reaction are called products.\nExample: The burning of magnesium in the air to form magnesium oxide is an example of a chemical reaction.\n2Mg(s) + O2(g) \n\u25b3\n\u2192\n 2MgO(s)", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nExample: The burning of magnesium in the air to form magnesium oxide is an example of a chemical reaction.\n2Mg(s) + O2(g) \n\u25b3\n\u2192\n 2MgO(s)\nBefore burning in air, the magnesium ribbon is cleaned by rubbing with sandpaper.\nThis is done to remove the protective layer of basic magnesium carbonate from the surface of the magnesium ribbon.\nReactant: Substances which take part in a chemical reaction are called reactants.\nExample: Mg and O2.\nProduct: New substance formed after a chemical reaction is called a product.\nExample: MgO.\nCharacteristics of Chemical Reactions :\n(i) Evolution of gas: The chemical reaction between zinc and dilute sulphuric acid is characterised by the evolution of hydrogen gas.\nZn(s) + H2SO4(aq) \u2192 ZnSO4(aq) + H2(g) \u2191\n(ii) Change in Colour: The chemical reaction between citric acid and purple coloured potassium permanganate solution is characterised by a change in colour from purple to colourless.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\n(ii) Change in Colour: The chemical reaction between citric acid and purple coloured potassium permanganate solution is characterised by a change in colour from purple to colourless.\nThe chemical reaction between sulphur dioxide gas and acidified potassium dichromate solution is characterized by a change in colour from orange to green.\n(iii) Change in state of substance: The combustion reaction of candle wax is characterised by a change in state from solid to liquid and gas (because the wax is a solid, water formed by the combustion of wax is a liquid at room temperature whereas, carbon dioxide produced by the combustion of wax is a gas). There are some chemical reactions which can show more than one characteristics.\n(iv) Change in temperature: The chemical reaction between quick lime water to form slaked lime is characterized by a change in temperature (which is a rise in temperature).", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\n(iv) Change in temperature: The chemical reaction between quick lime water to form slaked lime is characterized by a change in temperature (which is a rise in temperature).\nThe chemical reaction between zinc granules and dilute sulphuric acid is also characterised by a change in temperature (which is a rise in temperature).\n(v) Formation of precipitate: The chemical reaction between sulphuric acid and barium chloride solution is characterised by the formation of a white precipitate of barium sulphate.\nBaCl2(aq) + H2SO4(aq) \u2192 BaSO4(s) (ppt) + 2HCl(aq)\nWhat is a chemical Equation Class 10?\nChemical Equation: Representation of chemical reaction using symbols and formulae of the substances is called Chemical Equation.\nExample: A + B \u2192 C + D\nIn this equation, A and B are called reactants and C and D are called the products. The arrow shows the direction of the chemical reaction. Condition, if any, is written generally above the arrow.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nIn this equation, A and B are called reactants and C and D are called the products. The arrow shows the direction of the chemical reaction. Condition, if any, is written generally above the arrow.\nWhen hydrogen reacts with oxygen, it gives water. This reaction can be represented by the following chemical equation:\nHydrogen + Oxygen \u2192 Water\nH2 + O2 \u2192 H2O\nIn the first equation, words are used and in second, symbols of substances are used to write the chemical equation. For convenience, the symbol of substance is used to represent chemical equations.\nA chemical equation is a way to represent the chemical reaction in a concise and informative way.\nA chemical equation can be divided into two types: Balanced Chemical Equation and Unbalanced Chemical Equation.\n(a) Balanced Chemical Equation: A balanced chemical equation has the number of atoms of each element equal on both sides.\nExample: Zn + H2SO4 \u2192 ZnSO4 + H2", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\n(a) Balanced Chemical Equation: A balanced chemical equation has the number of atoms of each element equal on both sides.\nExample: Zn + H2SO4 \u2192 ZnSO4 + H2\nIn this equation, numbers of zinc, hydrogen and sulphate are equal on both sides, so it is a Balanced Chemical Equation.\nAccording to the Law of Conservation of Mass, mass can neither be created nor destroyed in a chemical reaction. To obey this law, the total mass of elements present in reactants must be equal to the total mass of elements present in products.\n(b) Unbalanced Chemical Equation: If the number of atoms of each element in reactants is not equal to the number of atoms of each element present in the product, then the chemical equation is called Unbalanced Chemical Equation.\nExample: Fe + H2O \u2192 Fe3O4 + H2", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nExample: Fe + H2O \u2192 Fe3O4 + H2\nIn this example, a number of atoms of elements are not equal on two sides of the reaction. For example; on the left-hand side only one iron atom is present, while three iron atoms are present on the right-hand side. Therefore, it is an unbalanced chemical equation.\nBalancing a Chemical Equation: To balance the given or any chemical equation, follow these steps:\nFe + H2O \u2192 Fe3O4 + H2\nWrite the number of atoms of elements present in reactants and in products in a table as shown here.\nBalance the atom which is maximum in number on either side of a chemical equation.\nIn this equation, the number of oxygen atom is the maximum on the RHS.\nTo balance the oxygen, one needs to multiply the oxygen on the LHS by 4, so that, the number of oxygen atoms becomes equal on both sides.\nFe + 4 \u00d7 H2O \u2192 Fe3O4 + H2\nNow, the number of hydrogen atoms becomes 8 on the LHS, which is more than that on the RHS.", + "Fe + 4 \u00d7 H2O \u2192 Fe3O4 + H2\nNow, the number of hydrogen atoms becomes 8 on the LHS, which is more than that on the RHS. To balance it, one needs to multiply the hydrogen on the RHS by 4.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nFe + 4 \u00d7 H2O \u2192 Fe3O4 + H2\nNow, the number of hydrogen atoms becomes 8 on the LHS, which is more than that on the RHS. To balance it, one needs to multiply the hydrogen on the RHS by 4.\nFe + 4 \u00d7 H2O \u2192 Fe3O4 + 4 \u00d7 H2\nAfter that, the number of oxygen and hydrogen atoms becomes equal on both sides. The number of iron is one on the LHS, while it is three on the RHS. To balance it, multiply the iron on the LHS by 3.\n3 \u00d7 Fe + 4 \u00d7 H2O \u2192 Fe3O4 + 4 \u00d7 H2\nNow the number of atoms of each element becomes equal on both sides. Thus, this equation becomes a balanced equation.\nAfter balancing, the above equation can be written as follows:\n3Fe + 4H2O \u2192 Fe3O4 + 4H2.", + "Thus, this equation becomes a balanced equation.\nAfter balancing, the above equation can be written as follows:\n3Fe + 4H2O \u2192 Fe3O4 + 4H2.\nTo Make Equations More Informative:\nWriting the symbols of physical states of substances in a chemical equation:\nBy writing the physical states of substances, a chemical equation becomes more informative.\nGaseous state is represented by symbol (g).\nLiquid state is represented by symbol (l).\nSolid state is written by symbol (s).", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nGaseous state is represented by symbol (g).\nLiquid state is represented by symbol (l).\nSolid state is written by symbol (s).\nAqueous solution is written by symbol (aq).\nWriting the condition in which reaction takes place: The condition is generally written above and/or below the arrow of a chemical equation.\nThus, by writing the symbols of the physical state of substances and condition under which reaction takes place, a chemical equation can be made more informative.\nWhat are the types of a chemical reaction Class 10?\nTypes of Chemical Reactions: Combination Reaction, Decomposition Reaction, Displacement Reaction, Double Displacement Reaction, Neutralization Reactions, Exothermic \u2013 Endothermic Reactions and Oxidation-Reduction Reactions.\nTypes of Chemical Reactions:\nChemical reactions can be classified in following types:\n(i) Combination Reaction: Reactions in which two or more reactants combine to form one product are called Combination Reactions.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nChemical reactions can be classified in following types:\n(i) Combination Reaction: Reactions in which two or more reactants combine to form one product are called Combination Reactions.\nA general combination reaction can be represented by the chemical equation given here:\nA + B \u2192 AB\nExamples:\nWhen magnesium is burnt in the air (oxygen), magnesium oxide is formed. In this reaction, magnesium is combined with oxygen.\nMg(s) + O2(g) \u2192 2MgO(s)\nMagnesium + Oxygen \u2192 Magnesium Oxide\nWhen carbon is burnt in oxygen (air), carbon dioxide is formed. In this reaction, carbon is combined with oxygen.\nC (s) + O2(g) \u2192 CO2(g)\nCarbon + Oxygen \u2192 Carbon dioxide\n(ii) Decomposition Reaction: Reactions in which one compound decomposes in two or more compounds or elements are known as Decomposition Reaction. A decomposition reaction is just the opposite of combination reaction.\nA general decomposition reaction can be represented as follows :\nAB \u2192 A + B\nExamples:", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nA general decomposition reaction can be represented as follows :\nAB \u2192 A + B\nExamples:\nWhen calcium carbonate is heated, it decomposes into calcium oxide and carbon dioxide.\nCaCO3(s) \nheat\n\u2212\n\u2192\n\u2212\n CaO(s) + CO2(g)\nCalcium carbonate \u2192 Calcium oxide + Carbon dioxide\nWhen ferric hydroxide is heated, it decomposes into ferric oxide and water\n2Fe(OH)3(s) \n\u25b3\n\u2192\n Fe2O3(s) + 3H2O(l)\nThermal Decomposition: The decomposition of a substance on heating is known as Thermal Decomposition.\nExample: 2Pb(NO3)2(s) \nheat\n\u2212\n\u2192\n\u2212\n 2PbO(s) + 4NO2(g) + O2(g)\nElectrolytic Decomposition: Reactions in which compounds decompose into simpler compounds because of passing of electricity, are known as Electrolytic Decomposition. This is also known as Electrolysis.\nExample: When electricity is passed in water, it decomposes into hydrogen and oxygen.", + "This is also known as Electrolysis.\nExample: When electricity is passed in water, it decomposes into hydrogen and oxygen.\n2H2O(l) \nUndefined control sequence \\xrightarrow\n 2H2(g) + O2(g)", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nExample: When electricity is passed in water, it decomposes into hydrogen and oxygen.\n2H2O(l) \nUndefined control sequence \\xrightarrow\n 2H2(g) + O2(g)\nPhotolysis or Photo Decomposition Reaction: Reactions in which a compound decomposes because of sunlight are known as Photolysis or Photo Decomposition Reaction.\nExample: When silver chloride is put in sunlight, it decomposes into silver metal and chlorine gas.\n2AgCl(s) (white) \nSunlight\n\u2212\n\u2192\n\u2212\n\u2212\n\u2212\n\u2212\n\u2212\n 2Ag(s) (grey) + Cl2(g)\nPhotographic paper has a coat of silver chloride, which turns into grey when exposed to sunlight. It happens because silver chloride is colourless while silver is a grey metal.\n(iii) Displacement Reaction: The chemical reactions in which a more reactive element displaces a less reactive element from a compound is known as Displacement Reactions. Displacement reactions are also known as Substitution Reaction or Single Displacement/ replacement reactions.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nA general displacement reaction can be represented by using a chemical equation as follows :\nA + BC \u2192 AC + B\nDisplacement reaction takes place only when \u2018A\u2019 is more reactive than B. If \u2018B\u2019 is more reactive than \u2018A\u2019, then \u2018A\u2019 will not displace \u2018C\u2019 from \u2018BC\u2019 and reaction will not be taking place.\nExamples:\nWhen zinc reacts with hydrochloric acid, it gives hydrogen gas and zinc chloride.\nZn(s) + 2HCl(aq) \u2192 ZnCl2(aq) + H2(g)\nWhen zinc reacts with copper sulphate, it forms zinc sulphate and copper metal.\nZn(s) + CuSO4(aq) \u2192 ZnSO4(aq) + Cu(s)\n(iv) Double Displacement Reaction: Reactions in which ions are exchanged between two reactants forming new compounds are called Double Displacement Reactions.\nAB + CD \u2192 AC + BD\nExamples:\nWhen the solution of barium chloride reacts with the solution of sodium sulphate, white precipitate of barium sulphate is formed along with sodium chloride.", + "AB + CD \u2192 AC + BD\nExamples:\nWhen the solution of barium chloride reacts with the solution of sodium sulphate, white precipitate of barium sulphate is formed along with sodium chloride.\nBaCl2(aq) + Na2SO4(aq) \u2192 BaSO4(s) (Precipitate) + 2NaCl(aq)", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nBaCl2(aq) + Na2SO4(aq) \u2192 BaSO4(s) (Precipitate) + 2NaCl(aq)\nWhen sodium hydroxide (a base) reacts with hydrochloric acid, sodium chloride and water are formed.\nNaOH(aq) + HCl(aq) \u2192 NaCl(aq) + H2O(l)\nNote: Double Displacement Reaction, in which precipitate is formed, is also known as precipitation reaction. Neutralisation reactions are also examples of double displacement reaction.\nPrecipitation Reaction: The reaction in which precipitate is formed by the mixing of the aqueous solution of two salts is called Precipitation Reaction.\nExample:\n\nNeutralization Reaction: The reaction in which an acid reacts with a base to form salt and water by an exchange of ions is called Neutralization Reaction.\nExample:\n\n(v) Oxidation and Reduction Reactions:\nOxidation: Addition of oxygen or non-metallic element or removal of hydrogen or metallic element from a compound is known as Oxidation.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nExample:\n\n(v) Oxidation and Reduction Reactions:\nOxidation: Addition of oxygen or non-metallic element or removal of hydrogen or metallic element from a compound is known as Oxidation.\nElements or compounds in which oxygen or non-metallic element is added or hydrogen or metallic element is removed are called to be Oxidized.\nReduction: Addition of hydrogen or metallic element or removal of oxygen or non-metallic element from a compound is called Reduction.\nThe compound or element which goes under reduction in called to be Reduced.\nOxidation and Reduction take place together.\nOxidizing agent:\nThe substance which gives oxygen for oxidation is called an Oxidizing agent.\nThe substance which removes hydrogen is also called an Oxidizing agent.\nReducing agent:\nThe substance which gives hydrogen for reduction is called a Reducing agent.\nThe substance which removes oxygen is also called a Reducing agent.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nReducing agent:\nThe substance which gives hydrogen for reduction is called a Reducing agent.\nThe substance which removes oxygen is also called a Reducing agent.\nThe reaction in which oxidation and reduction both take place simultaneously is called Redox reaction.\nWhen copper oxide is heated with hydrogen, then copper metal and hydrogen are formed.\nCuO + H2 \u2192 Cu + H2O\n(i) In this reaction, CuO is changing into Cu. Oxygen is being removed from copper oxide. Removal of oxygen from a substance is called Reduction, so copper oxide is being reduced to copper.\n(ii) In this reaction, H2 is changing to H2O. Oxygen is being added to hydrogen. Addition of oxygen to a substance is called Oxidation, so hydrogen is being oxidised to water.\nThe substance which gets oxidised is the reducing agent.\nThe substance which gets reduced is the oxidizing agent.\n(vi) Exothermic and Endothermic Reactions:", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nThe substance which gets oxidised is the reducing agent.\nThe substance which gets reduced is the oxidizing agent.\n(vi) Exothermic and Endothermic Reactions:\nExothermic Reaction: Reaction which produces energy is called Exothermic Reaction. Most of the decomposition reactions are exothermic.\nExample:\nRespiration is a decomposition reaction in which energy is released.\n\nWhen quick lime (CaO) is added to water, it releases energy.\n\nEndothermic Reaction: A chemical reaction in which heat energy is absorbed is called Endothermic Reaction.\nExample: Decomposition of calcium carbonate.\n\nEffects of Oxidation Reactions in Everyday life: Corrosion and Rancidity.\nCorrosion: The process of slow conversion of metals into their undesirable compounds due to their reaction with oxygen, water, acids, gases etc. present in the atmosphere is called Corrosion.\nExample: Rusting of iron.\nRusting: Iron when reacts with oxygen and moisture forms red substance which is called Rust.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nExample: Rusting of iron.\nRusting: Iron when reacts with oxygen and moisture forms red substance which is called Rust.\n\nThe rusting of iron is a redox reaction.\nCorrosion (rusting) weakens the iron and steel objects and structures such as railings, car bodies, bridges and ships etc. and cuts short their life.\nMethods to Prevent Rusting\nBy painting.\nBy greasing and oiling.\nBy galvanisation.\nCorrosion of Copper: Copper objects lose their lustre and shine after some time because the surface of these objects acquires a green coating of basic copper carbonate, CuCO3.Cu(OH)2 when exposed to air.\n\nCorrosion of Silver Metal: The surface of silver metal gets tarnished (becomes dull) on exposure to air, due to the formation of a coating of black silver sulphide(Ag2S) on its surface by the action of H2S gas present in the air.", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nRancidity: The taste and odour of food materials containing fat and oil changes when they are left exposed to air for a long time. This is called Rancidity. It is caused due to the oxidation of fat and oil present in food materials.\nMethods to prevent rancidity:\nBy adding anti-oxidant.\nVacuum packing.\nReplacing air by nitrogen.\nRefrigeration of foodstuff.\n1. Chemical Reaction: During chemical reactions, the chemical composition of substances changes or new substances are formed.\n2. Chemical Equation: Chemical reactions can be written in chemical equation form which should always be balanced.\n3. Types of Chemical Reactions:\nCombination reaction: A single product is formed from two or more reactants.\n2Mg + O2 \u2192 2MgO\nDecomposition reaction: A single reactant breaks down to yield two or more products.", + "3. Types of Chemical Reactions:\nCombination reaction: A single product is formed from two or more reactants.\n2Mg + O2 \u2192 2MgO\nDecomposition reaction: A single reactant breaks down to yield two or more products.\nThermal decomposition: 2Pb(NO2)2 \u2192 2PbO + 4NO2 + O2\nElectrolysis: 2H20 \u2192 2H2 + O2\nPhotochemical reaction: 2AgBr \u2192 2Ag + Br2", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nThermal decomposition: 2Pb(NO2)2 \u2192 2PbO + 4NO2 + O2\nElectrolysis: 2H20 \u2192 2H2 + O2\nPhotochemical reaction: 2AgBr \u2192 2Ag + Br2\nDisplacement reaction: One element is displaced by another element.\nZn + CuSO4 \u2192 ZnSO4 + Cu\nDouble displacement reaction: Exchange of ions between reactants.\nAgNO3 + NaCl \u2192 AgCl + NaNO3\nRedox reaction: Both oxidation and reduction take place simultaneously.\nCuO + H2 \u2192 Cu + H2O\nExothermic reaction: A chemical reaction in which heat energy is evolved.\nC + O2 \u2192 CO2 (g) + heat\nEndothermic reaction: A chemical reaction in which heat energy is absorbed.\nZnCO3 + Heat \u2192 ZnO + CO2\nRedox reaction: Chemical reaction in which both oxidation and reduction take place simultaneously.\n4. Oxidation: Reaction that involves the gain of oxygen or loss of hydrogen.\n5.", + "ZnCO3 + Heat \u2192 ZnO + CO2\nRedox reaction: Chemical reaction in which both oxidation and reduction take place simultaneously.\n4. Oxidation: Reaction that involves the gain of oxygen or loss of hydrogen.\n5. Reduction: Reaction that shows the loss of oxygen or gain of hydrogen.\nZnO + C \u2192 Zn + CO\nZnO is reduced to Zn\u2014reduction. C is oxidized to CO\u2014Oxidation.\n6. Effects of Oxidation Reactions in Our Daily Life:", + "CHAPTER 1- CHEMICAL REACTIONS AND EQUATIONS-NOTES pdf.txt\nZnO + C \u2192 Zn + CO\nZnO is reduced to Zn\u2014reduction. C is oxidized to CO\u2014Oxidation.\n6. Effects of Oxidation Reactions in Our Daily Life:\nCorrosion: It is an undesirable change that occurs in metals when they are attacked by moisture, air, acids and bases.\nExample, Corrosion (rusting) of Iron: Fe2O3. nH2O (Hydrated iron oxide)\nRancidity: Undesirable change that takes place in oil containing food items due to the oxidation of fatty acids.\nPreventive methods of rancidity: Adding antioxidants to the food materials, storing food in the airtight container, flushing out air with nitrogen gas and refrigeration.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-NOTES.txt\nD.A.V. PUBLIC SCHOOL , VELACHERY, CHENNAI - 42 \n EXPERIENTIAL LEARNING ACTIVITIES \nCLASS : X SUBJECT : Physics \nCHAPTER No .: 10 TOPIC : Light \u2013 Reflection and Refraction \nSUB -TOPIC : Image formation of a Concave mirror and C onvex lens \nNAME OF THE ACTIVITY : U-V method \nTYPE OF ACTIVITY : Group PLACE OF THE ACTIVITY : School (Physics Laboratory) \nMATERIALS USED : Light source, concave mirror, convex lens, white screen, Lens/mirror stand, \nmetre scale. \nMETHODOLOGY : \nStudents are divided into two groups. \nGROUP 1 : \n1) The approximate focal length of the concave mirror is found by distant object method. \n2) A met re scale is placed on a flat surface and the concave mirror is fixed at \u20180\u2019 (zero) of the \nmeter scale. \n3) Distances \u2018F\u2019 and \u2018C\u2019 are marked fro m the concave mirror.", + "2) A met re scale is placed on a flat surface and the concave mirror is fixed at \u20180\u2019 (zero) of the \nmeter scale. \n3) Distances \u2018F\u2019 and \u2018C\u2019 are marked fro m the concave mirror. \n4) The light source is placed at different positions (say beyond C, at C, between F and C, at F \nand between F and C)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-NOTES.txt\nmeter scale. \n3) Distances \u2018F\u2019 and \u2018C\u2019 are marked fro m the concave mirror. \n4) The light source is placed at different positions (say beyond C, at C, between F and C, at F \nand between F and C) \n5) The screen is moved in front of the mirror to get a clear image. \n6) Position and nature of the image is noted. \nGRO UP 2 : \n1) The approximate focal length of the convex lens is found by distant object method. \n2) A metre scale is placed on a flat surface and the convex lens is placed at \u201850\u2019 mark of the \nmetre scale. \n3) Distances \u2018F\u2019 and \u20182F\u2019 are marked on both the sides of the convex lens. \n4) The light source is placed at different positions (say beyond 2F, at 2F, between F and 2F, at F, \nbetween F and O) on one side of the lens. \n5) The screen is moved on the other side to get a clear image.", + "4) The light source is placed at different positions (say beyond 2F, at 2F, between F and 2F, at F, \nbetween F and O) on one side of the lens. \n5) The screen is moved on the other side to get a clear image. \n6) Position and nature of the image formed is noted. \nQUESTIONS : \n1) In what direction is the screen moved when the light source is brought closer to the convex \nlens?", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-NOTES.txt\n6) Position and nature of the image formed is noted. \nQUESTIONS : \n1) In what direction is the screen moved when the light source is brought closer to the convex \nlens? \n2) Write two characteristics of real image. \n3) How does the size of the image change as the object is brought closer to t he convex lens/ \nconcave mirror ? \nCURIOSITY QUESTIONS : \n1) Why is a concave mirror used in operation theatres ? \nAnswer Guide: \nThink about: \n1) What type of mirror is used as converging mirrors? \n2) At what position of the object do we get a powerful parallel beam of light?", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nLight Reflection and Refraction Previous \nyears Questions \nSHORT ANSWER TYPE QUESTIONS [II] [3 Marks] \u2013 Year 2007 \n1. A concave lens has focal length of 20 cm. At what distance from the lens a 5 \ncm tall object be placed so that it forms an image at 15 cm from the lens? Also \ncalculate the size of the image formed. [Delhi] \nAnswer. \n \n2. An object 50 cm tall is placed on the principal axis of a convex lens. Its 20 cm \ntall image is formed on the screen placed at a distance of 10 cm from the lens. \nCalculate the focal length of the lens. [All India] \nAnswer. \nLONG ANSWER TYPE QUESTION [5 Marks] \u2013 Year 2007 \n3.", + "Its 20 cm \ntall image is formed on the screen placed at a distance of 10 cm from the lens. \nCalculate the focal length of the lens. [All India] \nAnswer. \nLONG ANSWER TYPE QUESTION [5 Marks] \u2013 Year 2007 \n3. Draw the ray diagram in each case to show the position and nature of the \nimage formed when the object is placed: \n(i) at the centre of curvature of a concave mi rror \n(ii) between the pole P and focus F of a concave mirror \n(iii) in front of a convex mirror \n(iv) at 2F of a convex lens \n(v) in front of a concave lens \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nNature of image: Real, inverted and same size image is formed at the centre of \ncurvature. \n \nNature of image: Virtual, enlarged and erect image is formed behind the mirror. \n \nNature of image: Virtual, erect and diminished, image is formed behind the mirror. \n \nNature of image: Real, inverted and size to size, image is formed at 2F on the other side \nof lens. \n \nNature of image: Virtual, erect and diminished image is formed between O and F on the \nsame side of object. \nVERY SHORT ANSWER TYPE QUESTIONS [1 Mark] \u2013 Year 2008 \n4. If a light ray IM is incident on the surface AB as shown, identify the correct \nemergent ray. [Delhi (C)] \n \nAnswer. Ray NQ, as it has to be parallel to ray OS. \n5. The refractive indices of four media A, B, C and D are given in the following \ntable:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nAnswer. Ray NQ, as it has to be parallel to ray OS. \n5. The refractive indices of four media A, B, C and D are given in the following \ntable: \n \nIf light, travels from one medium to another, in which case the change in speed \nwill be (i) minimum, (ii) maximum? [Delhi(C)] \nAnswer. (i) Minimum change is seen as light moves between 1.50 and 1.52, i.e. B and \nC. \n(ii) Maximum change when light moves between 1.33 and 2.40, i.e. A and D. \n6. \u201cThe refractive index of diamond is 2.42\u201d. What is th e meaning of this \nstatement in relation to speed of light? [Delhi] \nAnswer. \n \n7. Redraw the given diagram and show the path of the refracted ray: [All India(C)] \n \nAnswer. \n \n8.", + "What is th e meaning of this \nstatement in relation to speed of light? [Delhi] \nAnswer. \n \n7. Redraw the given diagram and show the path of the refracted ray: [All India(C)] \n \nAnswer. \n \n8. Redraw the given diagram and show the path of the refracted ray: [All India(C)] \n \nAnswer. \n \n9. Draw the following diagram in your answer book and show the formation of \nimage of the object AB with the help of suitable rays. [All India] \n \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nAnswer. \n \n9. Draw the following diagram in your answer book and show the formation of \nimage of the object AB with the help of suitable rays. [All India] \n \nAnswer. \n \n10. Which kind of mirrors are used in the headlights of a motor -car and why? \n[Foreign] \nAnswer. Concave mirror, to get the parallel beam of light. \nSHORT ANSWER TYPE QUESTIONS [I] [2 Marks] \u2013 Year 2008 \n11. Explain with the help of a diagram, why a pencil partly immersed in water \nappears to be bent at the water surface. [Delhi] \nAnswer. Light from different points on the pencil, immersed in water refracts and \nappears to come from a point above the original position. \n \n12. Draw ray diagrams to represent the nature, position and relative size of the \nimage formed by a convex lens for the object placed: \n(a) at 2F 1 \n(b) between F 1 and the optical centre O of lens: [All India] \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n13. A ray of light, incident obliquely on a face of a rectangular glass slab placed in \nair, emerges from the opposite face parallel to the incident ray. State two factors \non which the lateral displacement of the emergent ray depends. [Foreign] \nAnswer. Lateral displacement depends on the: \n1. angle of incidence, \n2. thickness of slab, and \n3. refractive index of the material. (any two) \n \n \nSHORT ANSWER TYPE QUESTIONS [II] [3 Marks] \u2013 Year 2008 \n14. An object 2 cm in size is placed 30 cm in front of a concave mirror of focal \nlength 15 cm. At what distance from the mirror should a screen be placed in order \nto obtain a sharp image? What will be the nature and the size of the image \nformed? Draw a ray diagram to show the formation of the image in this case. \n[Delhi(C)] \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n15. An object 2 cm high is placed at a distance of 64 cm from a white screen. On \nplacing a convex lens at a distance of 32 cm from the object it is found that a \ndistinct im age of the object is formed on the screen. What is the focal length of \nthe convex lens and size of the image formed on the screen? Draw a ray diagram \nto show the formation of the image in this position of the object with respect to \nthe lens. [Delhi(C)] \nAnswer. Since, object -screen distance is double of object -lens separation, the object is \nat a distance of 2f from the lens and the image should be of the same size of the object. \n \n16. A convex lens has a focal length of 10 cm. At what distance from the lens \nshould the object be placed so that it forms a real and inverted image 20 cm away \nfrom the lens? What would be the size of the image formed if the object is 2 cm \nhigh?", + "A convex lens has a focal length of 10 cm. At what distance from the lens \nshould the object be placed so that it forms a real and inverted image 20 cm away \nfrom the lens? What would be the size of the image formed if the object is 2 cm \nhigh? With the help of a ray diagram show the formation of the image by the lens \nin this case. [All India (C)] \n \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nAnswer. \n \nLONG ANSWER TYPE QUESTION [5 Marks] \u2013 Year 2008 \n17. (a) It is desired to obtain an erect image of an object, using a concave mirror \nof focal length 20 cm. \n(i) What should be the range of distance of the object from the mirror? \n(ii) Will the image be bigger or smaller than the object? \n(iii) Draw a ray diagram to show the image formation in this case. \n(b) One half a convex lens of focal length 20 cm is covered with a black paper. \n(i)Will the lens produce a complete image of the obj ect? \n(ii)Show the formation of image of an object placed at 2F1 of such covered lens \nwith the help of a ray diagram. \n(iii)How will the intensity of the image formed by half -covered lens compare with \nnon-covered lens? \u2018 [Foreign] \nAnswer. \n \nVERY SHORT ANSWER TYPE QUESTIONS [1 Mark] \u2013 Year 2009 \n18.", + "(iii)How will the intensity of the image formed by half -covered lens compare with \nnon-covered lens? \u2018 [Foreign] \nAnswer. \n \nVERY SHORT ANSWER TYPE QUESTIONS [1 Mark] \u2013 Year 2009 \n18. Redraw the diagram given below in your answer book and show the direction \nof the light ray after reflection from the mirror. [Delhi (C)] \n \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nAnswer. \n \n19. Redraw the diagram given below in your answer book and show the direction \nof the light ray after reflection from the mirror. [Delhi (C)] \n \nAnswer. \n \n20. Redraw the diagram given below in your answer book and show the direction \nof the light ray after reflection from the mirror. [Delhi (C)] \n \nAnswer. \n \n21. Redraw the diagram given below in your answer book and show the direction \nof the light ray after refraction from the lens. [Delhi, All India(C)] \n \nAnswer. \n \n22. Why does a ray of light bend when it travels from one medium into another? \n[Delhi] \nAnswer. Due to change in velocity in the medium and to reduce the time taken to travel \nthe same. \n23. Redraw the diagram given below in your answer book and show the direction \nof the light ray after refraction from the lens. [All India (C)] \n \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nAnswer. \n \nSHORT ANSWER TYPE QUESTION S [I] [2 Marks] \u2013 Year 2009 \n24. What is the minimum number of rays required for locating the image formed \nby a concave mirror for an object. Draw a ray diagram to show the formation of a \nvirtual image by a concave mirror. [Delhi] \nAnswer. \n \n25. The refractive index of water is 1.33 and the speed of light in air is 3 x 108 ms -\n1. Calculate the speed of light in water. [Foreign] \nAnswer. \n \n26. The refractive index of glass is 1.50 and the speed of light in air is 3 x 108 ms -\n1. Calculate the speed of light in glass. [Foreign] \nAnswer. \n \nSHORT ANSWER TYPE QUESTIONS [II] [3 Marks] \u2013 Year 2009 \n27. For which position of the object does a convex lens form a virtual and erect \nimage? Explain with the help of a ray diagram. [All India] \nAnswer.", + "SHORT ANSWER TYPE QUESTIONS [II] [3 Marks] \u2013 Year 2009 \n27. For which position of the object does a convex lens form a virtual and erect \nimage? Explain with the help of a ray diagram. [All India] \nAnswer. When the object i s placed between the focus and the optical centre, a virtual \nand erect image is formed.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nVERY SHORT ANSWER TYPE QUESTIONS [1 Mark] \u2013 Year 2010 \n28. Explain why a ray of light passing through the centre of curvature of a \nconcave mirror, gets reflected along the same path. [Delhi] \nAnswer. The ray passing through the centre of curvature incident to the mirror along its \nnormal so \u2220i= \u2220r = 0. Therefore, the ray retraces its path. \n29. What is the nature of the image formed by a concave mirror if the \nmagnification produced by the mirror is +3? [Delhi] \nAnswer. Positive sign of magnification indicates that image is virtual, erect and \nenlarged. \n30. Between which two points of a concave mirror should an object be placed to \nobtain a magnification of -3? [Delhi] \nAnswer. Negative sign of magnification indicates that image is real and inverted. Also \nsize of image is enlarged. So, object must be positioned between F and 2F, i.e. C.", + "[Delhi] \nAnswer. Negative sign of magnification indicates that image is real and inverted. Also \nsize of image is enlarged. So, object must be positioned between F and 2F, i.e. C. \n31. \u201cThe refractive index of carbon disulphide is 1.63.\u201d What is the meaning of", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nsize of image is enlarged. So, object must be positioned between F and 2F, i.e. C. \n31. \u201cThe refractive index of carbon disulphide is 1.63.\u201d What is the meaning of \nthis statement in relation to speed of light? [Delhi] \nAnswer. Speed of light in carbon disulphide is 1/1.63 times the speed of light in free \nspace. \n32. The outer surface of a hollow sphere of aluminium of radius 50 cm is to be \nused as a mirror. What will be the focal le ngth of this mirror? Which type of \nspherical mirror will it provide? [Delhi] \nAnswer. f = R/2 = 25 cm. It will form a convex mirror. \n33. Between which two points related to a concave mirror should an object be \nplaced to obtain on a screen an image twice the size of the object? [All India] \nAnswer. Real Image: Between F and C. \n34.", + "It will form a convex mirror. \n33. Between which two points related to a concave mirror should an object be \nplaced to obtain on a screen an image twice the size of the object? [All India] \nAnswer. Real Image: Between F and C. \n34. How should a ray of light be incident on a rectangular glass slab so that it \ncomes out from the opposite side of the slab without being displaced? [Foreign]", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n34. How should a ray of light be incident on a rectangular glass slab so that it \ncomes out from the opposite side of the slab without being displaced? [Foreign] \nAnswer. Along the normal to the surface or at an incident angle of i = 0\u00b0. \n35. A girl was playing with a thin beam of light from her laser torch by directing it \nfrom different directions on a convex lens held vertically. She was surprised to \nsee that in a particular direction the beam of light continues to move along the \nsame direction after passing through the lens. State the reason for this \nobservation. [Foreign] \nAnswer. A ray of light passing through the optical centre of the convex lens will \ncontinue to move along the same dire ction after refracting through the lens. 36. A ray of light enters a rectangular glass slab of refractive index 1.5. It is found \nthat the ray emerges from the opposite face of the slab without being displaced. If", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nthat the ray emerges from the opposite face of the slab without being displaced. If \nits speed in air is 3 x 108 ms -1 then what is its speed in glass? [Foreign] \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n37. The speed of light in a transparent medium is 0.6 times that of its speed in \nvacuum. What is the refractive index of the medium? [Foreign] \nAnswer. \n \nSHORT ANSWER TYPE QUESTION [I] [2 Marks] \u2013 Year 2010 \n38. In an experiment with a rectangular glass slab, a student observed that a ray \nof light incident at an angle of 55\u00b0 with the normal on one face of the slab, after \nrefraction strikes the opposite face of the slab before emerging out into air \nmaking an angle of 40\u00b0 with the normal. Draw a labelled diagram to show the path \nof this ray. What value would you assign to the angle of refraction and angle of \nemergence? [All India] \nAnswer. \n \n \nSHORT ANSWER TYPE QUESTION [II] [3 Marks] \u2013 Year 2010 \n39. At what distance sho uld an object be placed from a convex lens of focal \nlength 18 cm to obtain an image at 24 cm from it on the other side.", + "SHORT ANSWER TYPE QUESTION [II] [3 Marks] \u2013 Year 2010 \n39. At what distance sho uld an object be placed from a convex lens of focal \nlength 18 cm to obtain an image at 24 cm from it on the other side. What will be \nthe magnification produced in this case? [Delhi] \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n40. How far should an object be placed from a .convex lens of fo cal length 20 cm \nto obtain its image at a distance of 30 cm from the lens? What will be the height \nof the image if the object is 6 cm tall? [All India] \nAnswer. \n \n \n41. The image of an object placed at 60 cm in front of a lens is obtained on a \nscreen at a d istance of 120 cm from it. Find the focal length of the lens. What \nwould be the height of the image if the object is 5 cm high? [Foreign] \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nSHORT ANSWER TYPE QUESTION [I] [2 Marks] \u2013 Year 2011 \n42. State the two laws of reflection of light. [Delhi] \nAnswer. Laws of reflection of light are: \n1. The angle of incidence is equal to the angle of reflection. \n2. The incident ray, the normal to the reflecting surface at the point of incidence and \nreflected ray from that point, all lies in the same plane. \n43. Define and show on a diagram, the following terms relating to a concave \nmirror: \n(i) Aperture \n(ii) Radius of curvature [Foreign] \nAnswer. (i) The diameter of the reflecting surface of the mirror is called aperture.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n(ii) The radius of the sphere of which t he reflecting surface of the spherical mirror forms \na part is called the radius of curvature of the mirror. \n44. Define the focus of a concave mirror. If the radius of curvature of a convex \nmirror is 30 cm, what would be its focal length? [Foreign] \nAnswer. The point on the principal axis where all the rays parallel to it meet after \nreflection is called focus. Since, R = 30 cm and f=R/2 we have, f=+15 cm for a convex \nmirror. \n45. Distinguish between a real and a virtual image of an object. What type of \nimage i s formed (i) by a plane mirror, (ii) on a cinema screen? [Foreign] \nAnswer. If light rays after reflection converge to a point to form an image on its own, it \nis called a real image. If they are diverging, then they form a virtual image. Real image \ncan be o btained on a screen, while a virtual image cannot be.", + "If light rays after reflection converge to a point to form an image on its own, it \nis called a real image. If they are diverging, then they form a virtual image. Real image \ncan be o btained on a screen, while a virtual image cannot be. \n(i) Plane mirror forms virtual image. \n(ii) On cinema screen, real image is formed.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\ncan be o btained on a screen, while a virtual image cannot be. \n(i) Plane mirror forms virtual image. \n(ii) On cinema screen, real image is formed. \nSHORT ANSWER TYPE QUEST10NS [II] [3 Marks] \u2013 Year 2011 \n46. Draw a ray diagram and also state the position, the relative size and the nature \nof image formed by a concave mirror when the object is placed at the centre of \ncurvature of the mirror. [Delhi] \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n47. Define \u2018refractive index of a transparent medium.\u2019 What is its unit? Which has \na higher refractive index, gla ss or water? [Delhi] \nAnswer. The ratio of the speed of light in the free space (c) to the speed of light in given \nmedium (\u0475) is called its refractive index. \nn=c/\u0475 \nIt has no unit. Glass has more refractive index than water. \n48. A ray of light travelling in air enters obliquely into water. Does the light ray \nbend towards or away from the normal ? Why? Draw a ray diagram to show the \nrefraction of light in this situation. [Delhi] \nAnswer. Light bends towards the normal because water is denser than air.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nThe ray bends towards the normal because it has to travel with a lesser speed in water \nbut with shorter time. \n49. (a) \u201cThe refractive index of diamond is 2.42\u201d. What is the meaning of this \nstatement? \n(b) Name a liquid whose mass density is less than that of water but it is optically \ndenser than water. [Delhi] \nAnswer. (a) This means that the ratio of speed of light in air and the speed of light in \ndiamond is equal to 2.42. (b) Kerosene \n50. An object is placed between infinity and the pole of a convex mirror. Draw a \nray diagram and also state the position, the relative size and the nature of the \nimage formed. [All India] \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n51. What is the principle of reversibility of light? Show that the incident ray of \nlight is parallel to the emergent ray of light when light falls obliquely on a side of \na rectangular glass slab. [All India] \nAnswer. The final path of the ray of light after reflections or refractions is reversed; the \nray retraces its entire path. This principle is called reversibility of light. \n \n \nHence i ncident ray PQ is parallel to the emergent ray RS when light falls obliquely on a \nside of rectangular glass slab. \n52. With the help of a ray diagram explain why a convex mirror is preferred for \nrear view mirrors in motor cars. [Foreign] \nAnswer. (i) It alwa ys forms a virtual, erect and diminished image. \n(ii) The field of view increases while using a convex mirror as shown.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n53. What is understood by lateral displacement of light? Illustrate it with the help \nof a diagram. List any two factors on which the la teral displacement of a \nparticular substance depends. [Foreign] \nAnswer. Lateral displacement is the perpendicular distance between the incident ray \nproduced and the emergent ray. Lateral displacement in the diagram is BL. The lateral \ndisplacement depends o n the thickness of the slab, the incident and refraction angles. \n \n54. Draw the ray diagram and also state the position, relative size and nature of \nthe image formed by a concave mirror when the object is placed between its \ncentre of curvature, C and focus , F. [Foreign] \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nLONG ANSWER TYPE QUESTIONS [5 Marks] \u2013 Year 2011 \n55. (a) If the image formed by a lens is diminished in size and erect, for all \npositions of the object, what type of lens is it? \n(b) Name the point on the lens through which a ray of light passes undeviated. \n(c) An object is placed perpendicular to the principal axis of a convex lens of \nfocal length 20 cm. The distance of the object from the lens is 30 cm. Find (i) the \nposition (ii) the magnification and (iii) the nature of the image formed. [Delhi] \nAnswer. (a) Concave lens. \n(b) Optical centre. \n(c) u = \u2013 30 cm, f = 20 cm \n \n(iii) Image is real as v > 0, inverted and enlarged. \n56. (a) What is meant by \u2018power of a lens\u2019? \n(b) State and define the S.I. unit of power of a lens.", + "56. (a) What is meant by \u2018power of a lens\u2019? \n(b) State and define the S.I. unit of power of a lens. \n(c) A convex lens of focal length 25 cm and a concave lens of focal length 10 cm \nare placed in close contact with each other. Calculate the lens power of this \ncombination. [All India] \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n57. (\u0430) Draw a ray diagram to show the formation of image of an object plac ed \nbetween infinity and the optical centre of a concave lens. \n(b) A concave lens of focal length 15 cm forms an image 10 cm from the lens. \nCalculate \n(i) the distance of the object from the lens. \n(ii) the magnification for the image formed. \n(iii) the nature of the image formed. [All India] \nAnswer. (a) Image formation when the object is at any position between infinity and \noptical centre:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n58. (a) Under what condition will a glass lens placed in a transparent liquid \nbecome invisible? \n(b) Describe and illustrate with a diagram, how we should arrange two converging \nlenses so that a parallel beam of light entering one lens emerges as a parallel \nbeam after passing through the second lens. \n(c) An object is placed at a distance of 3 cm from a concave lens of focal length \n12 cm. Find the (i) position and (ii) nature of the image formed. [Foreign] \nAnswer. (a) When the refractive index of glass lens becomes equal to the refractive \nindex of transparent liquid, the glass lens will become invisible. \n(b) Parallel be am converges at focus of the first lens and emerges parallel as it is at the \nfocus of second lens.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n59. (a) With the help of a ray diagram explain why a concave lens diverges the \nrays of a parallel beam of light. \n(b) A 2.0 cm tall object is placed perpend icular to the principal axis of a concave \nlens of focal length 15 cm. At what distance from the lens, should the object be \nplaced so that it forms an image 10 cm from the lens? Also find the nature and \nthe size of image formed. [Foreign] \nAnswer. (a) When a parallel beam of light incident on a front face of concave lens, each \nray of light will refract towards the normal to the surface as it moves from rarer to denser \nmedium and travels in a straight line inside the lens until it reaches the \u2018 back face of th e \nlens. At the back face boundary, each ray of light will again refract and bend away from \nthe normal to the surface as it moves from denser to rarer medium. The course of ray of \nlight is shown in the following figure.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nThus, because of the concave shape of both the faces, the double concave lens diverge \nthe rays of parallel beam of incident light.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nSHORT ANSWER TYPE QUESTION [I] [2 Marks] \u2013 Year 2012 \n60. List four properties of the image formed by a plane mirror. [Delhi] \nAnswer. Properties of image formed by a plane mirror: \n1. It is always virtual and erect. \n2. Its size is equal to that of the object. \n3. It is formed at the same distance behind the mirror as the object is in front of the \nmirror. \n4. It is laterally inverted. \n61. List four properties of the image formed by a convex mirror. [Delhi] \nAnswer. Properties of image formed by a convex mirror: \n1. It is always formed behind the mirror, between the pole and its focus. \n2. It is always virtual and erect. \n3. Its size is always smaller than the object. \n4. Magnification is a lways positive. \n62. List four properties of the image formed by a concave mirror, when object is \nplaced between focus and pole of the mirror.", + "It is always virtual and erect. \n3. Its size is always smaller than the object. \n4. Magnification is a lways positive. \n62. List four properties of the image formed by a concave mirror, when object is \nplaced between focus and pole of the mirror. [Delhi] \nAnswer. \n1. The image is formed behind the mirror. \n2. It is enlarged, he. magnified. \n3. It is virtual.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nplaced between focus and pole of the mirror. [Delhi] \nAnswer. \n1. The image is formed behind the mirror. \n2. It is enlarged, he. magnified. \n3. It is virtual. \n4. It is erect. \n63. To instruct a ray diagram, we use two light rays which are so chosen that it is \neasy to know their directions after refraction from the lens. List these two rays \nand state the path of these rays after refraction. Use these two rays to locate the \nimage of an object placed between \u2018F and \u20182F of a convex lens. [Foreign] \nAnswer. Two rays choose for refraction: \n(i) A ray of light parallel to the principal axis. \n(ii) A ray of light passing through the optical centre of a lens. \nPath of these rays after refract ion: \nIn case of convex lens, the first ray will pass through the principal focus on the other \nside of the lens. \nIn case of concave lens, the first ray will \n(a) appear to diverge or", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nIn case of convex lens, the first ray will pass through the principal focus on the other \nside of the lens. \nIn case of concave lens, the first ray will \n(a) appear to diverge or \n(b) appear to come from the principal focus positioned on the same side of the object. The second ray emerge from the lens without any deviation in the path.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nSHORT ANSWER TYPE QUESTION [II] [3 Marks] \u2013 Year 2012 \n64. State the type of mirror preferred as (i) rear view mirror in vehicles, (i0 shaving \nmirror. Justify your answer g iving two reasons in each case.[Delhi] ? \nAnswer. (i) Convex mirror (ii) Concave mirror \nJustification: \n(i) Convex mirror is used as a rear view mirror because: \n(a) it gives a wider field of view as it is curved outwards and \n(b) it produces erect and diminis hed image of the traffic behind the driver of the vehicle. \n(ii) Concave mirror is used as a shaving mirror to see a large size image of the face. \nWhen the object lies in between pole and principal focus of a concave mirror, it forms a \nvirtual, erect and en larged image behind it. \n65. The image of a candle flame placed at a distance of 45 cm from a spherical \nlens is formed on a screen placed at a distance of 90 cm from the lens.", + "65. The image of a candle flame placed at a distance of 45 cm from a spherical \nlens is formed on a screen placed at a distance of 90 cm from the lens. \nIdentify the type of lens and calculate its focal length. If the height of the fl ame is", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nlens is formed on a screen placed at a distance of 90 cm from the lens. \nIdentify the type of lens and calculate its focal length. If the height of the fl ame is \n2 cm, find the height of its image. [Delhi] \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nSo, height of image is 4 cm. Negative sign indicates that it is formed below the principal \naxis. \n66. A 4 cm tall object is placed perpendicular to the principal axis of a convex \nlens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find \nthe position, size and nature of the image formed, using the lens formula. [All \nIndia] \nAnswer.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nPosition of image: Image is formed at a distance of 48 cm from the optical centre of the \nlens on the same side of the object. It is indicated by the negative sign. \nSize of image: It is three times the size of object, i.e. 12 cm. \nNature of image: Positive sign in the image height indicates that image is virtual and \nerect. \n67. Name the type of m irror used in the following situations: \n(i) Headlights of a car (ii) Rear -view mirror of vehicles (iii) Solar furnace Support \nyour answer with reason. [All India][Foreign] \nAnswer. Type of mirror used in \n(i) Headlights of a car: Concave mirror Concave mirro r is used because light from the \nbulb placed at the focus of it gets reflected and produces a powerful parallel beam of \nlight to illuminate the road. \n(ii) Rear view mirror of vehicles: Convex mirror \nConvex mirror is used because it always produces a virtua l, and erect image whose", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nlight to illuminate the road. \n(ii) Rear view mirror of vehicles: Convex mirror \nConvex mirror is used because it always produces a virtua l, and erect image whose \nsize is smaller than the object. Therefore it enables the driver to see wide field view of \nthe traffic behind the vehicle in a small mirror. \n(iii) Solar furnace: Concave mirror \nConcave mirror has the property to concentrate the sun light coming from sun along with \nheat radiation at its focus. As a result, temperature at its focus increases and the \nsubstance placed at the focal point gets heated to a high temperature. \n68. A 6 cm tall object is placed perpendicular to the principal axi s of a convex \nlens of focal length 15 cm. The distance of the object from the lens is 10 cm. Find \nthe position, size and nature of the image formed, using the lens formula. \n[Foreign] \nAnswer. For convex lens \nf = +15 cm, u = \u2013 10 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nLONG ANSWER TYPE QUESTI ONS [5 Marks] \u2013 Year 2012 \n69. List the sign conventions for reflection of light by spherical mirrors. Draw a \ndiagram and apply these conventions in the determination of focal length of a \nspherical mirror which forms a three times magnified real image of an object \nplaced 16 cm in front of it. [Delhi ] \nAnswer. Sign conventions for reflection of light by spherical mirror are: \n1. The object is always placed to the left of the mirror. \n2. All the distances parallel to the principal axis are always measured from the po le \nof the spherical mirror. \n3. All the distances measured along the direction of incident light (along +ve x -axis), \nare considered to be positive. \n4. Those distances measured opposite to the direction of incidence light (i.e. along -\nve x-axis), are taken as neg ative. \n5. The distances measured in upward direction, i.e. perpendicular to and above the \nprincipal axis (along +ve y -axis), are taken as positive.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nve x-axis), are taken as neg ative. \n5. The distances measured in upward direction, i.e. perpendicular to and above the \nprincipal axis (along +ve y -axis), are taken as positive. \n6. The distances measured in the downward direction, (along -ve y-axis), i.e. \nperpendicular to and below the princ ipal axis are taken as negative.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nSo, focal length of spherical mirror is 12 cm. Negative sign of focal length indicates that \nmirror is concave in nature. \n \n70. State the law of refraction of light that defines the refractive index of a \nmedium with respect to the other. Express it mathematically. How is refractive \nindex of any medium \u2018A\u2019 with respect to a medium \u2018B\u2019 related to the speed of \npropagation of light in two media A and B? State the name of this constant when \none medium is vacuum or air. \nThe refractive indices of glass and water with respect to vacuum are 3/2 and 4/3 \nrespectively. If the speed of light in glass is 2 x 108 m/s, find the speed of light in \n(i) vacuum, (ii) water. [Delhi] \nAnswer. Snell\u2019s law: The ratio of sine of angle of inciden ce (i.e. sin i) to the sine of \nangle of refraction (i.e. sin r) is always constant for the light of given colour and for the \ngiven pair of media.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n71. To construct a ray diagram, we use two light rays which are so chosen that it \nis easy to know their directions after reflection from the mirror. List these two \nrays and state the path of these rays after reflection. Use these rays to locate the \nimage of an object placed between centre of curvature and focus of a concave \nmirror. [All India] \nAnswer. Rays w hich are chosen to construct ray diagram for reflection are: \n(i) A ray parallel to the principal axis and \n(ii) A ray passing through the centre of curvature of a concave mirror or appear to pass \nthrough the centre of curvature of convex mirror. \nPath of the se rays after reflections is: \n(i) After reflection, the first ray will pass through the principal focus of a concave mirror \nor appear to diverge in case of a convex mirror. \n(ii) After reflection, the second ray is reflected back along the same path.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\n72. List the sign conventions that are followed in case of refraction of light \nthrough spherical lenses. Draw a diagram and apply these conventions in \ndetermining the nature and focal length of a spherical lens which forms three \ntimes magnified real image of a n object placed 16 cm from the lens. [Foreign] \nAnswer. Sign conventions for refraction of light through spherical lens are: \n1. The object is always placed to the left of the lens so that incident light moves from \nleft to right. \n2. All distances are to be measure d from the optical centre of the lens. \n3. The distances measured in the direction of incident light (along +ve x -axis) will be \ntaken as positive. while those measured to the left of the origin (along -ve x-axis) \nwill be taken as negative. \n4. All measurements of heights above the principal axis (along +ve y -axis) will be \nconsidered as positive while below it (along -ve y-axis) will be taken as negative.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTION WIT ANSWERS.txt\nSo, focal length of the given spherical lens is 12 cm. The positive sign of focal length \nshows that the nature of spherical lens is convex. \n \n******", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nCLASS \u2013 X \nLESSON PLAN \nCh-10 LIGHT-REFLECTION AND REFRACTION \n \n Unit LIGHT-REFLECTION AND REFRACTION \nClass \nTransaction Total: 12 periods (approx. 40 min each) \nPre-requisite \nfor the course Students will be able to recall \n\u2022 Names of some common sources of light \n\u2022 What are luminous and non-luminous objects? \n\u2022 What is reflection of light? \n\u2022 Laws of reflection of Light. \n\u2022 What type of image is formed by spherical mirror? \n \nAssessment \nof qualifying \nknowledge 1. Written test \n2. Lab activity \n3. Group Discussion \n4. HW notebook \n5. Subject enrichment activity", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nObjective To develop scientific temper and inquisit iveness in students. Also students will \nbe able to apply and relate learned concepts in the ir daily lives. \nGiven the content (chapter) the learners will be ab le to describe the reflection \nand refraction by drawing ray diagram and solving n umerical in their notebook \nwith 90% accuracy. \nLearning \nOutcomes KNOWLEDGE- \n\u2022 Students will be able to compare Spherical Mirrors (concave and convex \nmirror) \n\u2022 Students will be able to draw ray diagrams for imag e formation by \nSpherical Mirrors (concave and convex mirror) \n\u2022 Students will be able to solve numerical based on m irror formula. \n\u2022 Students will be able to compare Spherical lenses ( concave and convex \nlens) \n\u2022 Students will be able to draw ray diagrams for imag e formation by \nSpherical lenses (concave and convex lens) \n\u2022 Students will be able to solve numerical based on l ens formula. \n \nSKILLS & COMPTENCIES- Students would be able to", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nSKILLS & COMPTENCIES- Students would be able to \n \n\u2022 Critically analyse difference between (a) Spherical Mirrors (concave \nand convex mirror) (b) Spherical lenses (concave and convex lens) \n\u2022 Collaborate in groups and draw ray diagrams to show image form ation \nby (a) Spherical Mirrors (concave and convex mirror) \n (b) Spherical lenses (concave and convex lens) \n\u2022 Appreciate the importance of mirrors and lenses collaboratively \u2022 Design Creative methods to draw ray diagrams and solve problems \nbased on mirror and lens formula, \n\u2022 Built character amongst themselves by discussing / communicating the \nimportance of spherical mirrors and lenses in their daily life. \n\u2022 Answer knowledge, understanding, application and di agrammatic skill \nbased questions based on the topic. \n \nATTITUDES AND BELIEFS- Students will be able to", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nATTITUDES AND BELIEFS- Students will be able to \n \n\u2022 Develop character by visualizing and observing the given spherical \nmirrors and lenses in day to day lives collaborating with others in \ndrawing ray diagrams and solving numerical and sensitizing others \nabout their usage in vehicles, doctors, microscope, telescope etc \n \nDOL D1,2,3 \nTransaction \nMethodology \n(The teacher \ncan use the \nmentioned \ntechniques, \nwherever \napplicable, \nand can use \nany other \ntoo.) Transaction would proceed in the following manner- \nActive learning- Identifying Concave Mirror/Convex Mirror/Concave le ns/ \nConvex lens from the given set of optical instrumen ts by touching it or without \ntouching it . ( Critical thinking)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nGroup learning- The class will be divided in eight groups. JIGSAW \u2013 \nExpert Group: The group will be given worksheet by the teacher for drawing ray \ndiagrams for given position of object . \nLearning Group: Students will teach each other the ray diagram learned. \n (Communication, Collaboration) \n \nDifferentiation- The students come back in Expert G roup and by Peer \nLearning : one member of the group will present in class the ray diagram \nlearned. ( Character, Citizenship) \n \nQuestioning- Multiple level question - Teacher will prepare a list of question \nabout the ray diagrams and will ask these questions during discussions. ( Critical \nthinking, Communication) \n \nBrain Storming -The class would start with a discussion on what th e students \nhave already learnt in the previous classes and hen ce what is it that they would \nlearn now. They would also be told the significance of the topic that they would \nbe studying.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nIntroduction of the topic - PPT and Digital Content would be shared \n \nGuided practice followed by Independent Practice- NCERT, Exemplar, spiral \nquestions to be discussed in the classroom. \n \nTechniques to be used: \nSet induction Class discussion' \nQuiz \nDaily Practice Problem \nMCQ \nPeer Assessment \nCase Studies \nLab Activities \nGraphic Organizers for different terms \nLesson progress using Smart Class \nThink Pair Share for solving numerical \nMind Mapping for lesson summarization \nMuddiest Point and One Minute Paper for Closure \nAny Other \n \nResources Text Book: \nNCERT text book for Science \nNCERT Exemplar \n \nSelf Study, \nHome Work, \nAssignments Independent Practice : Students would do the questions in their H.W note books \nas mentioned in the monthly planner.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nSelf Study, \nHome Work, \nAssignments Independent Practice : Students would do the questions in their H.W note books \nas mentioned in the monthly planner. \n \nHW notebooks to be marked as per the given plan: \nAssessment Parameters : The total marks for the activity is 5 marks \nOn time submission\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nPresentation/ Neatness\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u20261 mark \nContent\u2026\u2026\u2026.\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026......3 marks \nIt is also advised that the students come to the cl ass with proper background \nknowledge of the topic under discussion. They can r efer to the resources stated \nabove. \n \nAssessments 3 UNIT TESTS (20, 30, 30)\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026.80 Marks \nMID TERM EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u202680 Marks \nHALF YEARLY EXAMINATION \u2026\u2026\u2026\u2026\u2026\u202680 Marks \nPRE BOARD-I EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026..80 Marks \nPRE BOARD-II EXAMINATION \u2026\u2026\u2026\u2026\u2026\u2026. 80 Marks \nANNUAL BOARD EXAMINATION \u2026\u2026\u2026\u2026\u2026.80 Marks", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nPERIODIC TEST \nBy administering pen paper test we assess creativit y and critical thinking. \nAverage of the best two tests to be taken that will have a weightage of 10 marks. \nBest 2 Tests out of: Units tests, Midterm examinat ion, Half Yearly examination. \n \nINTERNAL ASSESSMENT \u2026\u2026\u2026\u2026\u2026\u2026\u2026\u2026\u202620 Marks \nPeriodical Test 10 Marks \n \nBy administering pen paper test we assess creativit y and critical thinking. \nSubject Enrichment 05 M arks \nWith the help of subject enrichment activities we a ssess communication \n,collaboration, character building and citizenship( peer teaching /lab activity \nof drawing ray diagrams \nNote Book Submission 05 Ma rks \nBy writing the answers of three levels of spiral quest ion we can assess their \nknowledge (remembering type questions) critical thinking ( analytical \nquestions) , creativity (HOT questions)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nAddressing \nClassroom \nDiversity Due to various social backgrounds and multiple inte lligences, the classroom \nmight be a diverse arena. The following techniques can be used for various \ngroups: \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Remembering type to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes and formulae, if not writt en \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nFor differently abled students: \n\u2022 Ignore spelling mistakes and formulae, if not writt en \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes \n \n \nMarks The weightage would be given by CBSE. \nQUESTIONS 1. Differentiate between a real and virtual image. Giv e any two points. \n2. What is minimum number of rays required for locatin g the image formed by a \nconcave mirror for an object? \n3. Define principal focus. \n4. Show the formation of the image with the help of a ray diagram when \nobject is placed 6 cm away from the pole of a conve x mirror. \n5. A concave mirror produces three times enlarged imag e of an object \nplaced at 10cm. in front of it. Calculate the focal length of the mirror. \n6. State Laws of refraction. \n7. Define power of a lens and give its S.I. unit.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nTopic DAY 1 -Recalling Spherical Mirrors and Terms associated w ith Spherical \nMirrors \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires Basic knowledge of Light, Refl ection, Spherical \nMirrors which students had already learnt in class VII and VIII \nAssessment of \nqualifying \nknowledge Facilitator asks the students to make grid in their notebook and write \ndifference between Real, Virtual image , Characteri stics of image formed \nby a plane mirror and compare their points with other students in group . \nObjective Student will be able to understand the va rious terms associated with \nspherical mirrors and why we should not at any ligh t source directly for a \nlong time,", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nLearning \nOutcomes KNOWLEDGE- Students will know and understand \n\u2022 The importance of terms associated with spherical m irrors. \n\u2022 Appreciate the importance of four different kinds o f rays used for \ndrawing ray diagrams. \nSKILLS AND COMPETENCIES- Students would be able to \n\u2022 Critically analyse various terms associated with spherical mirrors. \n\u2022 Appreciate the importance of of four different kind s of rays used for \ndrawing ray diagrams collaboratively. \nATTITUDES AND BELIEFS- Students will be able to \nDevelop Character by Collaborating in groups to study the terms \nassociated with spherical mirrors and sensitizing others to draw the four \nbasic rays for drawing ray diagrams. \nDOL D 1,2 \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nDifferentiation- The students in a group of 4 to 7 will be study the terms", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nDifferentiation- The students in a group of 4 to 7 will be study the terms \nassociated with spherical mirrors and sensitizing others to draw the four \nbasic rays for drawing ray diagrams. ( Character, Collaboration) \nGuided practice followed by Independent Practice - NCERT questions \nto be discussed in the classroom. \nTechniques to be used: \nPeer Assessment \nStudent - teacher interaction, \nWipro- G.O.s(compare- contrast matrix), \nsilent/ loud reading, \ncollaborative learning \nResources Text Book:", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\ncollaborative learning \nResources Text Book: \nNCERT text book for Science. \n NCERT Exemplar", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments Independent Practice : Different ability questions Assessments Assessment of work done by various grou ps on the basis of content, \ncommunication, collaboration, time management and m essage given by \neach group to sensitize others. \nAddressing \nClassroom \nDiversity Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups: \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Simple remembering type to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes. \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nFor differently abled students: \n\u2022 Ignore spelling mistakes. \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes \n \n \n \nTopic DAY 2 -Image formation by Spherical Mirrors (Concave Mirr or and \nConvex Mirror) \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires Basic knowledge of Terms assoc iated with Spherical \nMirrors and the four basic rays for drawing ray diagrams which students \nhad already learnt in previous class . \nAssessment of \nqualifying \nknowledge Facilitator asks the students to recall Focal Lengt h, Focus, Radius of \nCurvature, Centre of Curvature, Aperture etc \nObjective Student will be able to draw ray diagrams for various position of object \nplaced in front of a concave mirror and convex mirr or.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nLearning \nOutcomes KNOWLEDGE- Students will know and understand \n\u2022 The importance of correct way of drawing ray diagra ms with \nspherical mirrors. \n\u2022 Appreciate the importance of four different kinds o f rays used for \ndrawing ray diagrams. \nSKILLS AND COMPETENCIES- Students would be able to \n\u2022 Creatively draw the ray diagrams with object at different positions \nin front of spherical mirrors. \n\u2022 Appreciate the importance of of four different kind s of rays used for \ndrawing ray diagrams collaboratively. ATTITUDES AND BELIEFS- Students will be able to \nDevelop Character by Collaborating in groups JIGSAW to draw the \nray diagrams with object at different positions in front of sphe rical mirrors. \nand sensitizing others to draw the four basic rays for drawing ray \ndiagrams. \nDOL D 1,2 \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner-", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nDOL D 1,2 \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nActive learning- The class will be divided in eight groups. JIGSAW \u2013 \nExpert Group: The group will be given worksheet by the teacher for \ndrawing ray diagrams for given position of object . \nLearning Group: Students will teach each other the ray diagram learned. \n (Communication, Collaboration) \nDifferentiation- The students come back in Expert Group and by Peer \nLearning : one member of the group will present in class the ray diagram \nlearned. ( Character, Citizenship)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nGuided practice followed by Independent Practice - NCERT questions \nto be discussed in the classroom. \nTechniques to be used: \nPeer Assessment \nStudent - teacher interaction, \nWipro- G.O.s(compare- contrast matrix), \ncollaborative learning \nResources Text Book: \nNCERT text book for Science. \n NCERT Exemplar", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments Independent Practice : Different ability questions \nAssessments Assessment of work done by various grou ps on the basis of content, \ncommunication, collaboration, time management and m essage given by \neach group to sensitize others. \nAddressing \nClassroom \nDiversity Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups: \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Remembering type to be completed \n\u2022 Buddy help to be provided \u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes. \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nFor differently abled students: \n\u2022 Ignore spelling mistakes. \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes \n \n \n \nTopic DAY 3 \u2013 Cartesian Convention and Mirror Formula \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires Basic knowledge of ray diagram s which students had \nalready learnt in previous class. \nAssessment of \nqualifying \nknowledge Facilitator asks the students to recall ray diagram s and number line. \nObjective Student will be able to solve numerical b ased on mirror formula for various \nposition of object placed in front of a concave mir ror and convex mirror.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nLearning \nOutcomes KNOWLEDGE- Students will know and understand \n\u2022 The importance of cartesian sign convention for sph erical mirrors. \n\u2022 Appreciate the importance of mirror formula and mag nification. \nSKILLS AND COMPETENCIES- Students would be able to \n\u2022 Critically solve numerical based on mirror formula and \nmagnification. \n\u2022 Appreciate the importance of these by solving numer ical in groups \ncollaboratively. \nATTITUDES AND BELIEFS- Students will be able to \n\u2022 Develop Character by Collaborating in groups Think Pair \nShare in solving numerical and sensitizing others while \ncollaborating in groups. \nDOL D 1,2 \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nActive learning- The class will be divided in smaller groups of two \nstudents each. THINK PAIR SHARE \u2013 \n: The group will be given worksheet by the teacher for solving numerical", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nActive learning- The class will be divided in smaller groups of two \nstudents each. THINK PAIR SHARE \u2013 \n: The group will be given worksheet by the teacher for solving numerical \nbased on mirror formula and magnification. \nPear Learning: Students will teach each other by so lving numerical in pair. \nFacilitator will help in problem solving and doubt clarification. \n (Communication, Collaboration) \nDifferentiation- The students solve numerical in pairs with buddy h elp and \npeer learning, students will learn to help each oth er and work in groups. \n(Character, Citizenship)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nGuided practice followed by Independent Practice - NCERT questions to be discussed in the classroom. \nTechniques to be used: \nPeer Assessment \nStudent -teacher interaction, \nQuiz \nDaily Practice Problem \ncollaborative learning \nResources Text Book: \nNCERT text book for Science. \n NCERT Exemplar", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments Independent Practice : Different ability questions \nAssessments Assessment of work done by various grou ps on the basis of content, \ncommunication, collaboration, time management and m essage given by \neach group to sensitize others. \nAddressing \nClassroom \nDiversity Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups: \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Simple remembering type to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes. \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nTopic DAY 4 -Recalling Spherical lenses and Terms associated wi th Spherical \nlenses \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires Basic knowledge of Light, Refra ction, Spherical lenses \nwhich students had already learnt in class VII and VIII Assessment of \nqualifying \nknowledge Facilitator asks the students to make grid in their notebook and write \ndifference between mirror and lens/ reflection and refraction and compare \ntheir points with other students in group. \nObjective Student will be able to understand the va rious terms associated with \nspherical lenses and why we should not at any light source directly for a \nlong time . \nLearning \nOutcomes KNOWLEDGE- Students will know and understand \n\u2022 The importance of terms associated with spherical l enses. \n\u2022 Appreciate the importance of four different kinds o f rays used for \ndrawing ray diagrams. \nSKILLS AND COMPETENCIES- Students would be able to", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\n\u2022 Appreciate the importance of four different kinds o f rays used for \ndrawing ray diagrams. \nSKILLS AND COMPETENCIES- Students would be able to \n\u2022 Critically analyse various terms associated with spherical lenses. \n\u2022 Appreciate the importance of of four different kind s of rays used for \ndrawing ray diagrams collaboratively. \nATTITUDES AND BELIEFS- Students will be able to \nDevelop Character by Collaborating in groups to study the terms \nassociated with spherical lenses and sensitizing others to draw the four \nbasic rays for drawing ray diagrams. \nDOL D 1,2,3 \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nDifferentiation- The students in a group of 4 to 7 will be study the terms \nassociated with spherical lenses and sensitizing others to draw the four \nbasic rays for drawing ray diagrams. ( Character, Collaboration)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nassociated with spherical lenses and sensitizing others to draw the four \nbasic rays for drawing ray diagrams. ( Character, Collaboration) \nGuided practice followed by Independent Practice - NCERT questions \nto be discussed in the classroom. \nTechniques to be used: \nPeer Assessment \nStudent - teacher interaction \nWipro- G.O.s(compare- contrast matrix), \nsilent/ loud reading, \ncollaborative learning \n \nResources Text Book: \nNCERT text book for Science. \n NCERT Exemplar", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments Independent Practice : Different ability questions \nAssessments Assessment of work done by various grou ps on the basis of content, \ncommunication, collaboration, time management and m essage given by \neach group to sensitize others. \nAddressing \nClassroom Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used Diversity for various groups: \n \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Simple remembering type to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes. \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nFor differently abled students: \n\u2022 Ignore spelling mistakes. \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes \n \n \n \nTopic DAY 5 -Image formation by Spherical Lenses (Concave and Co nvex Lens) \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires Basic knowledge of Terms associ ated with Spherical \nLenses and the four basic rays for drawing ray diagrams which students \nhad already learnt in previous class. \nAssessment of \nqualifying \nknowledge Facilitator asks the students to recall Focal Lengt h, Focus, Radius of \nCurvature, Centre of Curvature, Aperture etc \nObjective Student will be able to draw ray diagrams for various position of object \nplaced in front of a concave and convex lens", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nLearning \nOutcomes KNOWLEDGE- Students will know and understand \n\u2022 The importance of correct way of drawing ray diagra ms with \nspherical lenses. \n\u2022 Appreciate the importance of four different kinds o f rays used for \ndrawing ray diagrams. \nSKILLS AND COMPETENCIES- Students would be able to \n\u2022 Creatively draw the ray diagrams with object at different positions \nin front of spherical lenses. \n\u2022 Appreciate the importance of of four different kind s of rays used for \ndrawing ray diagrams collaboratively. \nATTITUDES AND BELIEFS- Students will be able to \nDevelop Character by Collaborating in groups JIGSAW to draw the \nray diagrams with object at different positions in front of sphe rical lenses. \nand sensitizing others to draw the four basic rays for drawing ray \ndiagrams. DOL D 1,2,3 \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner-", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nActive learning- The class will be divided in eight groups. JIGSAW \u2013 \nExpert Group: The group will be given worksheet by the teacher for \ndrawing ray diagrams for given position of object . \nLearning Group: Students will teach each other the ray diagram learned. \n (Communication, Collaboration) \nDifferentiation- The students come back in Expert Group and by Peer \nLearning : one member of the group will present in class the ray diagram \nlearned. ( Character, Citizenship)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nGuided practice followed by Independent Practice - NCERT questions \nto be discussed in the classroom. \nTechniques to be used: \nPeer Assessment \nStudent - teacher interaction, \nWipro- G.O.s(compare- contrast matrix), \ncollaborative learning \nResources Text Book: \nNCERT text book for Science. \n NCERT Exemplar", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments Independent Practice : Different ability Questions \nAssessments Assessment of work done by various grou ps on the basis of content, \ncommunication, collaboration, time management and m essage given by \neach group to sensitize others. \nAddressing \nClassroom \nDiversity Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups: \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Simple remembering type to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes. \n\u2022 Call parents at regular intervals \u2022 Provide grade-up classes", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nFor differently abled students: \n\u2022 Ignore spelling mistakes. \n\u2022 Call parents at regular intervals \u2022 Provide grade-up classes \n \n \n \nTopic DAY 6 \u2013 Lens Formula and Magnification \nClass Transaction 40 min each \nPre-requisite for \nthe course This topic requires Basic knowledge of ray diagrams which students had \nalready learnt in previous class. \nAssessment of \nqualifying \nknowledge Facilitator asks the students to recall ray diagram s and number line. \nObjective Student will be able to solve numerical b ased on lens formula for various \nposition of object placed in front of a concave and convex lens.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nLearning \nOutcomes KNOWLEDGE- Students will know and understand \n\u2022 The importance of cartesian sign convention for sph erical lenses. \n\u2022 Appreciate the importance of lens formula and magni fication. \nSKILLS AND COMPETENCIES- Students would be able to \n\u2022 Critically solve numerical based on lens formula and \nmagnification. \n\u2022 Appreciate the importance of these by solving numer ical in groups \ncollaboratively. \nATTITUDES AND BELIEFS- Students will be able to \n\u2022 Develop Character by Collaborating in groups Think Pair \nShare in solving numerical and sensitizing others while \ncollaborating in groups. \nDOL D 1,2,3 \nTransaction \nMethodology \n(The teacher can \nuse the mentioned \ntechniques, \nwherever \napplicable, and can \nuse any other too.) Transaction would proceed in the following manner- \nActive learning- The class will be divided in smaller groups of two \nstudents each. THINK PAIR SHARE \u2013 \n: The group will be given worksheet by the teacher for solving numerical", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nActive learning- The class will be divided in smaller groups of two \nstudents each. THINK PAIR SHARE \u2013 \n: The group will be given worksheet by the teacher for solving numerical \nbased on lens formula and magnification. \nPear Learning: Students will teach each other by so lving numerical in pair. \nFacilitator will help in problem solving and doubt clarification. \n (Communication, Collaboration) \nDifferentiation- The students solve numerical in pairs with buddy h elp and \npeer learning, students will learn to help each oth er and work in groups. \n(Character, Citizenship)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nGuided practice followed by Independent Practice - NCERT questions \nto be discussed in the classroom. \nTechniques to be used: \nPeer Assessment \nStudent -teacher interaction, \nQuiz \nDaily Practice Problem collaborative learning \nResources Text Book: \nNCERT text book for Science. \n NCERT Exemplar", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nClosure Involving students in summarizing by asking them to prepare questions \nbased on the topic. \nSelf Study, Home \nWork, \nAssignments Independent Practice : Different ability Questions \nAssessments Assessment of work done by various grou ps on the basis of content, \ncommunication, collaboration, time management and m essage given by \neach group to sensitize others. \nAddressing \nClassroom \nDiversity Due to various social backgrounds and multiple inte lligences, the \nclassroom might be a diverse arena. The following t echniques can be used \nfor various groups: \n \nFor gifted students: \n\u2022 HOT questions to be done \n\u2022 Encouragement for referring other resources \n \nFor weak students: \n\u2022 Simple remembering type questions to be completed \n\u2022 Buddy help to be provided \n\u2022 Provide grade-up classes \n \nFor differently abled students: \n\u2022 Ignore spelling mistakes. \n\u2022 Call parents at regular intervals \n\u2022 Provide grade-up classes", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nDAY 7& 8 \nDAY 7 \nPr. 4: To estimate the focal length of concave mirr or and convex lens. \nAim: To determine the focal length of a concave mirror b y obtaining image of a distant object. \nMaterial required: A concave mirror, a mirror holder, a screen fixed o n a stand, and a scale. \nTheory: When parallel rays of light fall on a concave mirr or the rays meet at a point in front of \nthe mirror after reflection from it. This point is the focus of the mirror. \n For a parallel beam of light coming from a distant object, a real, inverted and very small image \nsize is formed at the focus of the mirror Since the image formed by the mirror is real, it can be \nobtained on a screen. The distance between the pole O of the concave mirror and the focus F is \nthe focal length of the concave mirror. Thus, the f ocal length of a concave mirror can be \nestimated by obtaining a real image of a distant ob ject at its focus.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nthe focal length of the concave mirror. Thus, the f ocal length of a concave mirror can be \nestimated by obtaining a real image of a distant ob ject at its focus. \nProcedure: 1. Fix a concave mirror in the mirror holder and place it on the table near an open window. \n2. Turn the face of mirror towards a distant objec t eg.a tree ,an electricity pole ,a building. \n3. Place the screen fitted to a stand in front of the concave mirror. \n4. Move the screen back and forth until a sharp, clear and inverted image of the distant object \n is formed on it \n5. Mark the position of the centre of the stand h olding the mirror and the screen when a sharp \n image of the distant object has been obtained on the screen. \n6. Measure the horizontal distance between the ce ntre of the concave mirror and the screen \n with the help of a measuring scale.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\n6. Measure the horizontal distance between the ce ntre of the concave mirror and the screen \n with the help of a measuring scale. \n7. Repeat the experiment two more times by obtain ing the images of two different distant \n objects. Measure the distances between the co ncave mirror a nd the screen in each case. \n Record them in the observation table.Find the mean value of the focal length. \n \n \nObservation and Calculation: \nS.NO. Name of the \ndistant object Distance between concave mirror and screen f Mean focal length of \na concave mirror f \n (cm) (m) (m) \n1 \n2 \n3", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nResult: The approximate value of focal length of the given concave mirror is ____ m. \nPrecautions: \n1. Concave mirror should be placed near window thr ough which sufficient sunlight enters. \n2. There should be no obstacle in the path of ligh t from the distant object, on the concave mirror. \n3. The image of the sun should be focused only on the screen. \n4. To obtain a sharp and clear image of the distan t object it must well illuminated so. \n5. The base of the stands of the concave mirror an d screen should be parallel to the scale. \n6. The mirror holder along with the mirror should be kept perpendicular to the measuring scale.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nDAY 8 \nAim: To determine the focal length of a thin convex lens by obtaining image of a distant object. \nMaterial required: A thin convex lens, a lens holder, screen fixed to a stand, scale. \nTheory: The light rays coming from a distant object can be considered to be parallel to each \nother. When a parallel beam of light falls on a con vex lens, the rays, after refraction, converge at \na point on its other side. This point is one of the two foci of the lens and form real, inverted \ndim inished image. \nProcedure: \n(i) Fix a convex lens on a lens holder and place it on the table or platform near an open window through which sufficient sunlight enters. T urn the face of lens towards a distant object. \n(ii) Place the screen fixed to a stand on the oth er side of the lens. \n(iii) Adjust the position of screen to get a sharp , clear and inverted image of the object on it. \n(iv) Mark the position of the lens and that of the screen", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\n(iii) Adjust the position of screen to get a sharp , clear and inverted image of the object on it. \n(iv) Mark the position of the lens and that of the screen \n(v) Measure the distance between the centre of th e convex lens and the screen. \n(vi) Repeat the experiment two times by obtaining the images of two different distant objects. \n(vii) Measure the distance between the convex lens and the screen in each case. Record them in \n the observation table. \n(viii) Find the average or mean value of the focal length.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nObservation and Calculation: \nS.NO. Name of the \ndistant object Distance between convex lens and screen f Mean focal length of \na convex lens f \n (cm) (m) (m) \n1 \n2 \n3 \n \n \nResult: The approximate value of focal length of the given convex lens is ____ m. \nPrecautions and sources of error: \n1. The principal axis of the lens should be horizon tal (the lens should be placed vertically). \n2. There should be no obstacle in the path of light from the distant object on the convex lens. \n3. The image of the sun formed by the lens should b e focused only on the screen. The image of \n sun should never be seen directly with the nak ed eye. \n4. Adjust the position of convex lens such that the light rays coming from the distant object fall \n on the lens without any obstruction. \n5. The distant object should be well illuminated.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nDAY 9& 10 \nPr.5: To study refraction through glass slab \nAim: To trace the path of a ray of light passing obliqu ely through a rectangular glass slab for \ndifferent angles of incidence and to measure the an gle of incidence, angle of refraction, the angle \nof emergence. \nMaterial required: A rectangular glass slab, drawing board, white she et of paper, protractor, \na measuring scale, pins, and drawing pins. \nTheory: When a ray of light passes from air to glass throug h a rectangular glass slab, it bends towards the normal at the surface of the air-glass boundary (AD). The phenomenon of change in \nthe direction of a ray of light when it enters from one medium to the other is known as refraction. \nThen, the refracted ray strikes the face BC of the glass slab that forms the glass-air boundary at \nthe opposite face of the glass slab ABCD. It underg oes refraction again. The deviation of the ray", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nThen, the refracted ray strikes the face BC of the glass slab that forms the glass-air boundary at \nthe opposite face of the glass slab ABCD. It underg oes refraction again. The deviation of the ray \nof light this time is away from the normal at the p oint of incidence .The refracted ray is known as \nthe emergent ray with respect to the incident ray i ncident at the face AD. The angle between the \nemergent ray and the normal to the face BC is known as angle of emergence ( e). \n \n \nProcedure: \n1. Fix a white sheet of paper on a drawing board. \n2. Place the rectangular glass slab in the middle of the paper and mark its boundary ABCD \n with a pencil. \n3. Remove the rectangular glass slab. Draw a thin line XO (with an arrow) inclined to the face \n AD of the glass slab at any angle preferably between 30\u00ba and 60\u00ba \n4. Replace the glass slab exactly over the bounda ry mark on the paper.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nAD of the glass slab at any angle preferably between 30\u00ba and 60\u00ba \n4. Replace the glass slab exactly over the bounda ry mark on the paper. \n5. Fix two pins P1 and P2 on the line XO. \n6. Observe the images of pins P1 and P2 through t he face BC of the rectangular glass slab. \n7. Fix pins P3 and P4 such that feet of all the p ins appear to be in a straight line. \n8. Remove the pins and the glass slab and mark th e positions of the pins. Join points P3 and P4 \n and extend the line up to point O\u2019 where it m eets the face BC. Also join the points O and O \n9. Draw the normal at the point of incidence O an d point O \u2032. \n10. Measure and record the angles i, r, e. \n11. Repeat the experiment for two more angles of i ncidence in the range 30\u00ba to 60\u00ba and record \n the values of angles i, r, and e in each case.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nObservation: \nS \nNo. Angle of incidence \ni Angle of refraction \nr Angle of emergence \ne Deviation \ni-e", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nResult: \nThe paths of different rays of light through a glas s slab are shown in figure As i = e , the emergent ray emerging out of the rectangular glass slab, is parallel to, but laterally displaced \nwith respect to the incident ray. Angle of refracti on r increases with increase in angle of \nincidence i. \nPrecautions and sources of error: \n1. The glass slab should be perfectly rectangular with all its faces smooth. \n2. The tips of pins P1, P2, P3, and P4 should be sharp. \n3. The angle of incidence should preferably be be tween 30\u00ba and 60\u00ba. \n4. Thin lines should be drawn, using a sharp penc il. \n5. The angles should be measured accurately, usin g a good quality protractor having clear \n markings, by keeping the eye above the markin g. \n6.", + "4. Thin lines should be drawn, using a sharp penc il. \n5. The angles should be measured accurately, usin g a good quality protractor having clear \n markings, by keeping the eye above the markin g. \n6. While viewing the collinearity of pins , the e ye should be kept at some distance from the pins \n so that the feet of all of them can be seen simulta neously in the same straight line.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nDAY 11 & 12 \nSUBJECT ENRICHMENT ACTIVITY \n \nPr. 3: To draw the images of an object formed by a convex lens, when placed at various \npositions. \n \n SUBJECT ENRICHMENT ACTIVITY \nAIM \nTo draw the images of an object formed by a convex lens, when placed at various positions. \nTHEORY \nThe light rays when refracted through a convex lens obey the laws of refraction. The formation \nof images by a convex lens can be studied by drawin g ray diagrams. Neat ray diagrams can be \ndrawn for various positions of an object in front o f a convex lens, using the New Cartesian \nSign Convention and convenient rays for locating th e image. It may be considered that the \nconvex lens is thin and that it has a small apertur e. After locating the position of the image, its \nnature, and size can be determined.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nMATERIALS REQUIRED \nMeasuring scale, graph paper, sheets of white paper , protractor, and drawing pins or adhesive \ntape. \nStudents will be individually following the given p rocedure , analyzing and drawing the ray \ndiagrams. \nCritical Thinking, Problem Solving, Creative Thinki ng \nPROCEDURE 1. Take a white sheet of paper or a graph paper . \n2. Draw a thin line of about 15 - 18 cm length in the middle of the paper. \n3. Mark a point O at the centre of this line. \n4. Make a convex lens LL about this point O. Assu me O as the optical centre of the lens. \n5. Mark points F1 and F2 on either side of the le ns such that OF 1 = OF 2. Also mark points \n2F 1 \n and 2F 2 on the line at double the distances OF 1 and OF 2 \n6. Draw an object AB of suitable height h, shown to be placed at infinity \n7.", + "Also mark points \n2F 1 \n and 2F 2 on the line at double the distances OF 1 and OF 2 \n6. Draw an object AB of suitable height h, shown to be placed at infinity \n7. Draw thin lines, representing incident rays co ming from the object parallel to the principal", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\n6. Draw an object AB of suitable height h, shown to be placed at infinity \n7. Draw thin lines, representing incident rays co ming from the object parallel to the principal \n axis of the Convex lens. These rays after ref raction intersect at the focus F 2 a diminished \n image of the distant object is formed at the point F 2 \n8. Repeat the above steps, to draw ray diagrams f or each position of the object from Fig (a) -\n(f). \n9. Measure the heights of the object, its image i n all cases Figs. (a) - (f)]. \n10. Describe the nature, position and relative size of the image, of the object placed at various \n positions. \n11. Tabulate the results in a convenient format or observation table. \n12. The ray diagrams for the image formation of the object by a convex lens can also be drawn \non a graph paper and measurement of the lengths h and h\u2032\uf020 of the size of the object and its \nimage taken more conveniently.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-LESSON PLAN.txt\nOBSERVATIONS, RESULT AND CONCLUSION \nCommunication Skill \nThe students will analyze the results and comment about the characteristics of image formed \nby a convex lens for various positions of the objec t. \n \nPRECAUTIONS \nCritical Thinking \n\u2022 Use a very sharp tipped pencil to draw thin lines to represent incident and refracted rays. \n\u2022 The convex lens drawn should be thin and of small aperture. (This is required for obtaining \n the distinct image.)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-IMPORTANT POINTS.txt\nFocus Spot: \nImage formation by a concave lens \nWhy: \nStudents are unable to understand the image formed by a concave lens as they find the \nconcept of drawing ray diagram of virtual images abstract due to the effect of concave lens. \nImpact: \nStudents are unable to visualise and draw a correct ray diagram displaying the passage of \nlight, object point, image point and magnification.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-Very short ANSWER type questions.txt\nVery short answer type questions \n1. The power of a lens is \u2013 40, what is its focal length? \n2. State laws of reflection. \n3. What are the properties of image formed by a plane mirror? \n4. Give two uses of convex mirror. \n5. What is refractive index? \n6. Draw a ray diagram to show the image formed by a concave lens for the object placed at \ninfinity. \n7. Draw a ray diagram to show the path of light when it travels through glass slab. \n8. State the lens formula and mirror formula. \n9. The refractive index of water is 1.33 and kerosene is 1.4 4 calculate the refractive index of \n kerosene with respect to water. \n10. Draw and explain the ray diagram formed by a convex mirror when: \n(a) object is at infinity. \n(b) object is at finite distance from the mirror. \n11. Differentiate between concave lens and convex lens.", + "10. Draw and explain the ray diagram formed by a convex mirror when: \n(a) object is at infinity. \n(b) object is at finite distance from the mirror. \n11. Differentiate between concave lens and convex lens. \n(a) Name two mirrors used in this fair shop. \n(b) Name the mirror in which the size of image is small. \n12. Name two types of spherical mirror.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-Very short ANSWER type questions.txt\n(a) Name two mirrors used in this fair shop. \n(b) Name the mirror in which the size of image is small. \n12. Name two types of spherical mirror. \n13. Give uses of concave mirror. \n14. Give uses of convex mirror. \n15. What are the two types of refractive index? \n16. Why do we prefer a convex mirror as a rear view mirror in vehicles? \n17. Name the type of mirror used in the following situations. \n(a) Headlights of a car. \n(b) Side/rear -view mirror of a vehicle. \n(c) Solar furnace. \n18. The magnification produced by a plane mirror is +1. What does this mean? \n19. Find the focal length of a lens of power \u20132.0 D. What type of lens is this? \n20. An incident ray makes an angle of 60\u00b0 with the mirror. What is the angle of reflection? \n21. Why does a ray of light bend when it travels from one medium into another? \n22.", + "20. An incident ray makes an angle of 60\u00b0 with the mirror. What is the angle of reflection? \n21. Why does a ray of light bend when it travels from one medium into another? \n22. The po wer of a lens is \u20134.0 D. What is the nature of this lens? \n23. Which type of mirrors are used to give an erect and enlarged image of an object?", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-Very short ANSWER type questions.txt\n22. The po wer of a lens is \u20134.0 D. What is the nature of this lens? \n23. Which type of mirrors are used to give an erect and enlarged image of an object? \n24. Which kind of mirrors are used in the headlights of a motor -car and why? \n25. What is refractive index?", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nScience \uf031\uf036\uf030\uf04c\uf069\uf067\uf068\uf074\uf020\uf096\uf020\uf052\uf065\uf066\uf06c\uf065\uf063\uf074\uf069\uf06f\uf06e\uf020\uf061\uf06e\uf064\n\uf052\uf065\uf066\uf072\uf061\uf063\uf074\uf069\uf06f\uf06e\uf031\uf030 \uf043\uf048\uf041\uf050\uf054\uf045\uf052\nWe see a variety of objects in the world around us. However , we ar e\nunable to see anything in a dark room. On lighting up the room,\nthings become visible. What makes things visible? During the day, the\nsunlight helps us to see objects. An object reflects light that falls on it.This reflected light, when received by our eyes, enables us to see things.\nWe are able to see thr ough a transparent medium as light is transmitted\nthrough it.", + "During the day, the\nsunlight helps us to see objects. An object reflects light that falls on it.This reflected light, when received by our eyes, enables us to see things.\nWe are able to see thr ough a transparent medium as light is transmitted\nthrough it. There are a number of common wonderful phenomena\nassociated with light such as image formation by mirrors, the twinklingof stars, the beautiful colours of a rainbow, bending of light by a medium\nand so on. A study of the properties of light helps us to explore them.\nBy observing the common optical phenomena around us, we may\nconclude that light seems to travel in straight lines. The fact that a small\nsource of light casts a sharp shadow of an opaque object points to this", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nconclude that light seems to travel in straight lines. The fact that a small\nsource of light casts a sharp shadow of an opaque object points to this\nstraight-line path of light, usually indicated as a ray of light.\uf04d\uf06f\uf072\uf065\uf020\uf074\uf06f\uf020\uf04b\uf06e\uf06f\uf077\uf021If an opaque object on the path of light becomes very small , light has a tendency to\nbend around it and not walk in a straight line \u2013 an effect known as the diffraction oflight. Then the straight-line treatment of optics using rays fails. To explain phenomena\nsuch as diffraction, light is thought of as a wave, the details of which you will study\nin higher classes. Again, at the beginning of the 20\nth century, it became known that\nthe wave theory of light often becomes inadequate for treatment of the interaction oflight with matter , and light often behaves somewhat like a stream of particles . This", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nthe wave theory of light often becomes inadequate for treatment of the interaction oflight with matter , and light often behaves somewhat like a stream of particles . This\nconfusion about the true nature of light continued for some years till a modernquantum theory of light emerged in which light is neither a \u2018wave\u2019 nor a \u2018particle\u2019 \u2013the new theory reconciles the particle properties of light with the wave nature.\nIn this Chapter , we shall study the phenomena of r eflection and\nrefraction of light using the straight-line propagation of light. These basicconcepts will help us in the study of some of the optical phenomena innatur e. We shall try to understand in this Chapter the reflection of light\nby spherical mirrors and refraction of light and their application in real\nlife situations.\n\uf031\uf030\uf02e\uf031\uf020\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030", + "\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf031\uf020\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf031\uf020\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf031\uf020\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046", + "\uf047\uf048\uf054\uf031\uf030\uf02e\uf031\uf020\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf031\uf020\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\nA highly polished sur face, such as a mirror , reflects most of the light", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n\uf031\uf030\uf02e\uf031\uf020\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf031\uf020\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf031\uf020\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf031\uf020\uf052\uf045", + "\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf031\uf020\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf031\uf020\uf052\uf045\uf046\uf04c\uf045\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\nA highly polished sur face, such as a mirror , reflects most of the light\nfalling on it. Y ou ar e already familiar with the laws of r eflection of light.", + "\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\nA highly polished sur face, such as a mirror , reflects most of the light\nfalling on it. Y ou ar e already familiar with the laws of r eflection of light.\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf036\uf031Let us recall these laws \u2013\n(i)The angle of incidence is equal to the angle of reflection, and\n(ii)The incident ray, the normal to the mirror at the point of incidence\nand the reflected ray, all lie in the same plane.\nThese laws of reflection are applicable to all types of reflecting surfaces\nincluding spherical sur faces. Y ou ar e familiar with the formation of image\nby a plane mirr or. What ar e the pr operties of the image? Image for med\nby a plane mirror is always virtual and erect. The size of the image is\nequal to that of the object. The image formed is as far behind the mirror\nas the object is in fr ont of it. Further , the image is laterally inverted.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nequal to that of the object. The image formed is as far behind the mirror\nas the object is in fr ont of it. Further , the image is laterally inverted.\nHow would the images be when the reflecting surfaces are curved? Letus explore.", + "The image formed is as far behind the mirror\nas the object is in fr ont of it. Further , the image is laterally inverted.\nHow would the images be when the reflecting surfaces are curved? Letus explore.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\n\uf06eTake a large shining spoon.", + "Try to view your face in its curved\nsurface.\n\uf06eDo you get the image? Is it smaller or larger?\n\uf06eMove the spoon slowly away from your face. Observe the image.\nHow does it change?\n\uf06eReverse the spoon and repeat the Activity. How does the imagelook like now?\n\uf06eCompare the characteristics of the image on the two surfaces.\nThe curved surface of a shining spoon could be considered as a curved\nmirror. The most commonly used type of curved mirror is the spherical\nmirror. The reflecting surface of such mirrors can be considered to for m\na part of the surface of a sphere. Such mirrors, whose reflecting surfaces", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nmirror. The reflecting surface of such mirrors can be considered to for m\na part of the surface of a sphere. Such mirrors, whose reflecting surfaces\nare spherical, are called spherical mirr ors. W e shall now study about\nspherical mirrors in some detail.\n\uf031\uf030\uf02e\uf032\uf020\uf053\uf050\uf048\uf045\uf052\uf049\uf043\uf031\uf030\uf02e\uf032\uf020\uf053\uf050\uf048\uf045\uf052\uf049\uf043\uf031\uf030\uf02e\uf032\uf020\uf053\uf050\uf048\uf045\uf052\uf049\uf043\uf031\uf030\uf02e\uf032\uf020\uf053\uf050\uf048\uf045\uf052\uf049\uf043\uf031\uf030\uf02e\uf032\uf020\uf053\uf050\uf048\uf045\uf052\uf049\uf043", + "\uf041\uf04c\uf020\uf04d\uf049\uf052\uf052\uf04f\uf052\uf053\uf041\uf04c\uf020\uf04d\uf049\uf052\uf052\uf04f\uf052\uf053\uf041\uf04c\uf020\uf04d\uf049\uf052\uf052\uf04f\uf052\uf053\uf041\uf04c\uf020\uf04d\uf049\uf052\uf052\uf04f\uf052\uf053\uf041\uf04c\uf020\uf04d\uf049\uf052\uf052\uf04f\uf052\uf053\nThe reflecting surface of a spherical mirror may be curved inwards or\noutwar ds. A spherical mirr or, whose r eflecting sur face is curved inwards,\nthat is, faces towar ds the centr e of the sphere, is called a concave mirr or.\nA spherical mirror whose reflecting surface is curved outwards, is called\na convex mirr or. The schematic r epresentation of these mirrors is shown\nin Fig. 10.1.", + "A spherical mirror whose reflecting surface is curved outwards, is called\na convex mirr or. The schematic r epresentation of these mirrors is shown\nin Fig. 10.1. You may note in these diagrams that the backof the mirror is shaded.\nYou may now understand that the surface of the spoon\ncurved inwards can be approximated to a concave mirror\nand the surface of the spoon bulged outwards can be", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nYou may now understand that the surface of the spoon\ncurved inwards can be approximated to a concave mirror\nand the surface of the spoon bulged outwards can be\nappr oximated to a convex mirr or.\nBefore we move further on spherical mirrors, we need to\nrecognise and understand the meaning of a few terms. These\nterms are commonly used in discussions about spherical\nmirrors. The centre of the reflecting surface of a spherical\nmirror is a point called the pole. It lies on the surface of the\nmirror. The pole is usually r epresented by the letter P .Figure 10.1Figure 10.1Figure 10.1Figure 10.1Figure 10.1\nSchematic representation of sphericalmirrors; the shaded side is non-reflecting.(a) Concave mirror (b) Convex mirror\n2019-20\nScience \uf031\uf036\uf032The reflecting surface of a spherical mirror forms a part of a sphere.\nThis sphere has a centre. This point is called the centre of curvature of\nthe spherical mirr or.", + "This sphere has a centre. This point is called the centre of curvature of\nthe spherical mirr or. It is r epresented by the letter C. Please note that the", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nThis sphere has a centre. This point is called the centre of curvature of\nthe spherical mirr or. It is r epresented by the letter C. Please note that the\ncentr e of curvature is not a part of the mirr or. It lies outside its r eflecting\nsurface. The centre of curvature of a concave mirror lies in front of it.\nHowever, it lies behind the mirr or in case of a convex mirr or. You may\nnote this in Fig.10.2 (a) and (b). The radius of the sphere of which thereflecting surface of a spherical mirror forms a part, is called the radius\nof curvatur e of the mirr or. It is r epresented by the letter R. You may note\nthat the distance PC is equal to the radius of curvature. Imagine a straightline passing through the pole and the centre of curvature of a spherical\nmirror. This line is called the principal axis. Remember that principal\naxis is normal to the mirror at its pole. Let us understand an importantterm related to mirrors, through an Activity.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nmirror. This line is called the principal axis. Remember that principal\naxis is normal to the mirror at its pole. Let us understand an importantterm related to mirrors, through an Activity.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nmirror. This line is called the principal axis. Remember that principal\naxis is normal to the mirror at its pole. Let us understand an importantterm related to mirrors, through an Activity.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf032\nCAUTION : Do not look at the Sun directly or even into a mirror\nreflecting sunlight.", + "It may damage your eyes.\n\uf06eHold a concave mirror in your hand and direct its reflecting surface\ntowards the Sun.\n\uf06eDirect the light reflected by the mirror on to a sheet of paper heldclose to the mirr or.\n\uf06eMove the sheet of paper back and forth gradually until you findon the paper sheet a bright, sharp spot of light.\n\uf06eHold the mirror and the paper in the same position for a fewminutes. What do you observe? Why?\nThe paper at first begins to burn producing smoke. Eventually it\nmay even catch fire. Why does it burn? The light from the Sun is converged\nat a point, as a sharp, bright spot by the mirr or. In fact, this spot of light", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nmay even catch fire. Why does it burn? The light from the Sun is converged\nat a point, as a sharp, bright spot by the mirr or. In fact, this spot of light\nis the image of the Sun on the sheet of paper. This point isthe focus of the concave mirr or. The heat pr oduced due to\nthe concentration of sunlight ignites the paper. The distanceof this image from the position of the mirror gives the\nappr oximate value of focal length of the mirr or.\nLet us try to understand this observation with the help\nof a ray diagram.\nObserve Fig.10.2 (a) closely. A number of rays parallel\nto the principal axis are falling on a concave mirr or. Observe\nthe reflected rays. They are all meeting/intersecting at apoint on the principal axis of the mirr or. This point is called\nthe principal focus of the concave mirr or. Similarly, observe\nFig. 10.2 (b). How are the rays parallel to the principal axis,\nreflected by a convex mirror? The reflected rays appear to", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nthe principal focus of the concave mirr or. Similarly, observe\nFig. 10.2 (b). How are the rays parallel to the principal axis,\nreflected by a convex mirror? The reflected rays appear to\ncome from a point on the principal axis. This point is called\nthe principal focus of the convex mirr or. The principal focus\nis represented by the letter F . The distance between the\npole and the principal focus of a spherical mirror is called\nthe focal length. It is represented by the letter f.\nFigure 10.2Figure 10.2Figure 10.2Figure 10.2Figure 10.2\n(a) Concave mirror(b) Convex mirror(b)(a)\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf036\uf033The reflecting surface of a spherical mirror is by-and-large spherical.\nThe surface, then, has a circular outline. The diameter of the reflecting\nsurface of spherical mirror is called its aperture. In Fig.10.2, distance\nMN represents the apertur e. We shall consider in our discussion only", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nsurface of spherical mirror is called its aperture. In Fig.10.2, distance\nMN represents the apertur e. We shall consider in our discussion only\nsuch spherical mirrors whose aperture is much smaller than its radiusof curvature.\nIs there a relationship between the radius of curvature R, and focal\nlength f, of a spherical mirror? For spherical mirrors of small apertures,\nthe radius of curvature is found to be equal to twice the focal length. W e\nput this as R = 2f . This implies that the principal focus of a spherical\nmirror lies midway between the pole and centre of curvature.", + "For spherical mirrors of small apertures,\nthe radius of curvature is found to be equal to twice the focal length. W e\nput this as R = 2f . This implies that the principal focus of a spherical\nmirror lies midway between the pole and centre of curvature.\n\uf031\uf030\uf02e\uf032\uf02e\uf031\uf020\uf020\uf049\uf06d\uf061\uf067\uf065\uf020\uf046 \uf06f\uf072\uf06d\uf061\uf074\uf069\uf06f\uf06e\uf020\uf062\uf079\uf020\uf053\uf070\uf068\uf065\uf072\uf069\uf063\uf061\uf06c\uf020\uf04d\uf069\uf072\uf072\uf06f\uf072\uf073\nYou have studied about the image formation by plane mirrors. Y ou also\nknow the nature, position and relative size of the images formed by them.How about the images formed by spherical mirrors? How can we locate\nthe image formed by a concave mirror for different positions of the object?", + "Y ou also\nknow the nature, position and relative size of the images formed by them.How about the images formed by spherical mirrors? How can we locate\nthe image formed by a concave mirror for different positions of the object?\nAre the images real or virtual? Are they enlarged, diminished or have", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nthe image formed by a concave mirror for different positions of the object?\nAre the images real or virtual? Are they enlarged, diminished or have\nthe same size? We shall explore this with an Activity.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nthe image formed by a concave mirror for different positions of the object?\nAre the images real or virtual? Are they enlarged, diminished or have\nthe same size? We shall explore this with an Activity.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf033\nYou have already lear nt a way of determining the focal length of a\nconcave mirror .", + "In Activity 10.2, you have seen that the sharp bright\nspot of light you got on the paper is, in fact, the image of the Sun. It\nwas a tiny, real, inverted image. You got the approximate focal lengthof the concave mirror by measuring the distance of the image fromthe mirr or.\n\uf06eTake a concave mirr or. Find out its approximate focal length in\nthe way described above. Note down the value of focal length. (Y ou\ncan also find it out by obtaining image of a distant object on asheet of paper .)\n\uf06eMark a line on a Table with a chalk. Place the concave mirror ona stand. Place the stand over the line such that its pole lies overthe line.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n\uf06eMark a line on a Table with a chalk. Place the concave mirror ona stand. Place the stand over the line such that its pole lies overthe line.\n\uf06eDraw with a chalk two more lines parallel to the previous line\nsuch that the distance between any two successive lines is equal\nto the focal length of the mirr or. These lines will now corr espond\nto the positions of the points P , F and C, r espectively. Remember \u2013\nFor a spherical mirror of small aperture, the principal focus F lies\nmid-way between the pole P and the centre of curvature C.\n\uf06eKeep a bright object, say a burning candle, at a position far beyondC. Place a paper screen and move it in front of the mirror till youobtain a sharp bright image of the candle flame on it.\n\uf06eObserve the image carefully. Note down its nature, position andrelative size with respect to the object size.", + "Place a paper screen and move it in front of the mirror till youobtain a sharp bright image of the candle flame on it.\n\uf06eObserve the image carefully. Note down its nature, position andrelative size with respect to the object size.\n\uf06eRepeat the activity by placing the candle \u2013 (a) just beyond C,\n(b) at C, (c) between F and C, (d) at F , and (e) between P and F.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n\uf06eRepeat the activity by placing the candle \u2013 (a) just beyond C,\n(b) at C, (c) between F and C, (d) at F , and (e) between P and F.\n\uf06eIn one of the cases, you may not get the image on the screen.Identify the position of the object in such a case. Then, look for itsvirtual image in the mirror itself.\n\uf06eNote down and tabulate your observations.\n2019-20\nScience \uf031\uf036\uf034You will see in the above Activity that the nature, position and size of\nthe image formed by a concave mirror depends on the position of the\nobject in relation to points P , F and C. The image for med is r eal for some\npositions of the object. It is found to be a virtual image for a certain other\nposition. The image is either magnified, reduced or has the same size,\ndepending on the position of the object. A summary of these observationsis given for your reference in Table 10.1.", + "It is found to be a virtual image for a certain other\nposition. The image is either magnified, reduced or has the same size,\ndepending on the position of the object. A summary of these observationsis given for your reference in Table 10.1.\nTable 10.1 Image formation by a concave mirror for different positions of the object\nPosition of the Position of the Size of the Nature of the", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nTable 10.1 Image formation by a concave mirror for different positions of the object\nPosition of the Position of the Size of the Nature of the\nobject image image image\nAt infinity At the focus F Highly diminished, Real and inverted\npoint-sized\nBeyond C Between F and C Diminished Real and inverted\nAt C At C Same size Real and inverted\nBetween C and F Beyond C Enlarged Real and inverted\nAt F At infinity Highly enlarged Real and inverted\nBetween P and F Behind the mirror Enlarged Virtual and erect\n\uf031\uf030\uf02e\uf032\uf02e\uf032\uf052\uf065\uf070\uf072\uf065\uf073\uf065\uf06e\uf074\uf061\uf074\uf069\uf06f\uf06e\uf020\uf06f\uf066\uf020\uf049\uf06d\uf061\uf067\uf065\uf073\uf020\uf046", + "\uf06f\uf072\uf06d\uf065\uf064\uf020\uf062\uf079\uf020\uf053\uf070\uf068\uf065\uf072\uf069\uf063\uf061\uf06c\n\uf04d\uf069\uf072\uf072\uf06f\uf072\uf073\uf020\uf055\uf073\uf069\uf06e\uf067\uf020\uf052\uf061\uf079\uf020\uf044\uf069\uf061\uf067\uf072\uf061\uf06d\uf073\nWe can also study the formation of images by spherical mirrors by\ndrawing ray diagrams. Consider an extended object, of finite size, placed\nin front of a spherical mirr or. Each small portion of the extended object\nacts like a point source. An infinite number of rays originate from each", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nin front of a spherical mirr or. Each small portion of the extended object\nacts like a point source. An infinite number of rays originate from each\nof these points. To construct the ray diagrams, in order to locate theimage of an object, an arbitrarily large number of rays emanating from a\npoint could be considered. However , it is mor e convenient to consider\nonly two rays, for the sake of clarity of the ray diagram. These rays are\nso chosen that it is easy to know their directions after reflection from themirror.\nThe intersection of at least two reflected rays give the position of image\nof the point object. Any two of the following rays can be considered forlocating the image.\n(i)A ray parallel to the\nprincipal axis , after\nreflection, will pass through\nthe principal focus in case of\na concave mirror or appearto diverge from the principal\nfocus in case of a convex\nmirror. This is illustrated in\nFig.10.3 (a) and (b).\n(a) (b)\nFigure FigureFigure FigureFigure 10.3 10.310.3 10.310.3", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\na concave mirror or appearto diverge from the principal\nfocus in case of a convex\nmirror. This is illustrated in\nFig.10.3 (a) and (b).\n(a) (b)\nFigure FigureFigure FigureFigure 10.3 10.310.3 10.310.3\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf036\uf035(ii)A ray passing through the\nprincipal focus of a concave\nmirror or a ray which is\ndirected towards the\nprincipal focus of a convex\nmirror, after reflection, will\nemerge parallel to theprincipal axis. This isillustrated in Fig.10.4 (a)and (b).\n(iii) A ray passing through thecentre of curvature of a\nconcave mirror or directedin the direction of the centreof curvature of a convexmirr or, after r eflection, is\nreflected back along thesame path. This isillustrated in Fig.10.5 (a)and (b).", + "This isillustrated in Fig.10.5 (a)and (b). The light rays comeback along the same pathbecause the incident rays\nfall on the mirror along the\nnormal to the reflectingsurface.\n(iv) A ray incident obliquely tothe principal axis , towards", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nfall on the mirror along the\nnormal to the reflectingsurface.\n(iv) A ray incident obliquely tothe principal axis , towards\na point P (pole of the mirror),on the concave mirror[Fig. 10.6 (a)] or a convexmirror [Fig. 10.6 (b)], isreflected obliquely. Theincident and reflected raysfollow the laws of reflectionat the point of incidence\n(point P), making equal\nangles with the principal axis.(a) (b)\nFigure 10.4Figure 10.4Figure 10.4Figure 10.4Figure 10.4\nRemember that in all the above cases the laws of reflection are followed.\nAt the point of incidence, the incident ray is reflected in such a way that\nthe angle of reflection equals the angle of incidence.\n(a)Image formation by Concave Mirror\nFigure 10.7 illustrates the ray diagrams for the formation of imageby a concave mirror for various positions of the object.", + "At the point of incidence, the incident ray is reflected in such a way that\nthe angle of reflection equals the angle of incidence.\n(a)Image formation by Concave Mirror\nFigure 10.7 illustrates the ray diagrams for the formation of imageby a concave mirror for various positions of the object.(b) (a)\nFigure 10.5Figure 10.5Figure 10.5Figure 10.5Figure 10.5\n(a) (b)\nFigure 10.6Figure 10.6Figure 10.6Figure 10.6Figure 10.6\n2019-20", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nFigure 10.5Figure 10.5Figure 10.5Figure 10.5Figure 10.5\n(a) (b)\nFigure 10.6Figure 10.6Figure 10.6Figure 10.6Figure 10.6\n2019-20\nScience \uf031\uf036\uf036Figure 10.7Figure 10.7Figure 10.7Figure 10.7Figure 10.", + "6Figure 10.6Figure 10.6Figure 10.6Figure 10.6\n2019-20\nScience \uf031\uf036\uf036Figure 10.7Figure 10.7Figure 10.7Figure 10.7Figure 10.7 Ray diagrams for the image formation by a concave mirror\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf034\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf034\n\uf06eDraw neat ray diagrams for each position of the object shown in\nTable 10.", + "1.\n\uf06eYou may take any two of the rays mentioned in the previous sectionfor locating the image.\n\uf06eCompare your diagram with those given in Fig. 10.7.\n\uf06eDescribe the nature, position and relative size of the image formed\nin each case.\n\uf06eTabulate the results in a convenient format.\nUses of concave mirrors\nConcave mirrors are commonly used in torches, search-lights and\nvehicles headlights to get powerful parallel beams of light. They are\noften used as shaving mirrors to see a larger image of the face. The\ndentists use concave mirrors to see large images of the teeth of patients.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\noften used as shaving mirrors to see a larger image of the face. The\ndentists use concave mirrors to see large images of the teeth of patients.\nLarge concave mirrors are used to concentrate sunlight to produce\nheat in solar furnaces.\n(b)Image formation by a Convex MirrorWe studied the image formation by a concave mirr or. Now we shall\nstudy the formation of image by a convex mirr or.\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf036\uf037We consider two positions of the object for studying the image for med\nby a convex mirr or. First is when the object is at infinity and the second\nposition is when the object is at a finite distance from the mirr or. The ray\ndiagrams for the formation of image by a convex mirror for these two\npositions of the object are shown in Fig.10.8 (a) and (b), respectively.", + "The ray\ndiagrams for the formation of image by a convex mirror for these two\npositions of the object are shown in Fig.10.8 (a) and (b), respectively.\nThe results are summarised in Table 10.2.\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf035\n\uf06eTake a convex mirr or.", + "Hold it in one hand.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nThe results are summarised in Table 10.2.\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf035\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf035\n\uf06eTake a convex mirr or. Hold it in one hand.\n\uf06eHold a pencil in the upright position in the other hand.", + "Hold it in one hand.\n\uf06eHold a pencil in the upright position in the other hand.\n\uf06eObserve the image of the pencil in the mirr or. Is the image er ect or\ninverted? Is it diminished or enlarged?\n\uf06eMove the pencil away from the mirror slowly. Does the image\nbecome smaller or larger?\n\uf06eRepeat this Activity carefully. State whether the image will movecloser to or farther away from the focus as the object is moved\naway from the mirror?\nFigure 10.8 Figure 10.8 Figure 10.8 Figure 10.8 Figure 10.8 Formation of image by a convex mirror\nYou have so far studied the image for mation by a plane mirr or, a\nconcave mirror and a convex mirr or. Which of these mirrors will give the\nfull image of a large object? Let us explore through an Activity.", + "Which of these mirrors will give the\nfull image of a large object? Let us explore through an Activity.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf036", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nfull image of a large object? Let us explore through an Activity.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf036\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf036\n\uf06eObserve the image of a distant object, say a distant tree, in aplane mirr or.", + "\uf06eCould you see a full-length image?Table 10.2 Nature, position and relative size of the image formed by a convex mirror\nPosition of the Position of the Size of the Nature of the\nobject image image image\nAt infinity At the focus F , Highly diminished, Virtual and erect\nbehind the mirror point-sized\nBetween infinity Between P and F , Diminished Virtual and erect\nand the pole P of behind the mirror\nthe mirror\n2019-20\nScience \uf031\uf036\uf038\uf06eTry with plane mirrors of different sizes. Did you see the entire\nobject in the image?\n\uf06eRepeat this Activity with a concave mirr or. Did the mirr or show\nfull length image of the object?\n\uf06eNow try using a convex mirror . Did you succeed? Explain your\nobservations with reason.\nYou can see a full-length image of a tall building/tr ee in a small", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nfull length image of the object?\n\uf06eNow try using a convex mirror . Did you succeed? Explain your\nobservations with reason.\nYou can see a full-length image of a tall building/tr ee in a small\nconvex mirr or. One such mirr or is fitted in a wall of Agra Fort facing T aj\nMahal. If you visit the Agra Fort, try to observe the full image of Taj\nMahal. To view distinctly, you should stand suitably at the terrace\nadjoining the wall.\nUses of convex mirrors\nConvex mirr ors ar e commonly used as rear -view (wing) mirrors in\nvehicles. These mirrors are fitted on the sides of the vehicle, enabling the\ndriver to see traffic behind him/her to facilitate safe driving. Convex\nmirrors are preferred because they always give an erect, though\ndiminished, image. Also, they have a wider field of view as they are curved\noutwards. Thus, convex mirrors enable the driver to view much larger\narea than would be possible with a plane mirr or.", + "Convex\nmirrors are preferred because they always give an erect, though\ndiminished, image. Also, they have a wider field of view as they are curved\noutwards. Thus, convex mirrors enable the driver to view much larger\narea than would be possible with a plane mirr or.\n\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. Define the principal focus of a concave mirr or.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\noutwards. Thus, convex mirrors enable the driver to view much larger\narea than would be possible with a plane mirr or.\n\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. Define the principal focus of a concave mirr or.\n2. The radius of curvature of a spherical mirror is 20 cm. What is its focal\nlength?\n3. Name a mirror that can give an erect and enlarged image of an object.\n4. Why do we prefer a convex mirror as a r ear-view mirror in vehicles?", + "2. The radius of curvature of a spherical mirror is 20 cm. What is its focal\nlength?\n3. Name a mirror that can give an erect and enlarged image of an object.\n4. Why do we prefer a convex mirror as a r ear-view mirror in vehicles?\n\uf031\uf030\uf02e\uf032\uf02e\uf033\uf020\uf053\uf069\uf067\uf06e\uf020\uf043\uf06f\uf06e\uf076\uf065\uf06e\uf074\uf069\uf06f\uf06e\uf020\uf066\uf06f\uf072\uf020\uf052\uf065\uf066\uf06c\uf065\uf063\uf074\uf069\uf06f\uf06e\uf020\uf062\uf079\uf020\uf053\uf070\uf068\uf065\uf072\uf069\uf063\uf061\uf06c\uf020\uf04d\uf069\uf072\uf072\uf06f\uf072\uf073\nWhile dealing with the reflection of light by spherical mirrors, we shall\nfollow a set of sign conventions called the New Cartesian Sign\nConvention .", + "In this convention, the pole (P) of the mirror is taken as the\norigin (Fig. 10.9). The principal axis of the mirror is taken as the x-axis(X\u2019X) of the coordinate system. The conventions are as follows \u2013\n(i)The object is always placed to the left of the mirr or. This implies\nthat the light from the object falls on the mirror from the left-handside.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n(i)The object is always placed to the left of the mirr or. This implies\nthat the light from the object falls on the mirror from the left-handside.\n(ii) All distances parallel to the principal axis are measured from thepole of the mirr or.\n(iii) All the distances measured to the right of the origin (along+ x-axis) are taken as positive while those measured to the left ofthe origin (along \u2013 x-axis) are taken as negative.\n(iv) Distances measured perpendicular to and above the principal axis\n(along + y-axis) are taken as positive.\n(v) Distances measured perpendicular to and below the principal axis\n(along \u2013y-axis) are taken as negative.\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf036\uf039The New Cartesian Sign Convention described above is illustrated in\nFig.10.9 for your reference. These sign conventions are applied to obtain\nthe mirror formula and solve related numerical problems.", + "2019-20\nLight \u2013 Reflection and Refraction \uf031\uf036\uf039The New Cartesian Sign Convention described above is illustrated in\nFig.10.9 for your reference. These sign conventions are applied to obtain\nthe mirror formula and solve related numerical problems.\n\uf031\uf030\uf02e\uf032\uf02e\uf034\uf020\uf020\uf04d\uf069\uf072 \uf072\uf06f\uf072\uf020\uf046\uf06f\uf072\uf06d\uf075\uf06c\uf061\uf020\uf061\uf06e\uf064\uf020\uf020\uf04d\uf061\uf067\uf06e\uf069\uf066\uf069\uf063\uf061\uf074\uf069\uf06f\uf06e\nIn a spherical mirror , the distance of the", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nthe mirror formula and solve related numerical problems.\n\uf031\uf030\uf02e\uf032\uf02e\uf034\uf020\uf020\uf04d\uf069\uf072 \uf072\uf06f\uf072\uf020\uf046\uf06f\uf072\uf06d\uf075\uf06c\uf061\uf020\uf061\uf06e\uf064\uf020\uf020\uf04d\uf061\uf067\uf06e\uf069\uf066\uf069\uf063\uf061\uf074\uf069\uf06f\uf06e\nIn a spherical mirror , the distance of the\nobject from its pole is called the object\ndistance ( u). The distance of the image from\nthe pole of the mirror is called the imagedistance ( v). You alr eady know that the\ndistance of the principal focus from the poleis called the focal length ( f).", + "The distance of the image from\nthe pole of the mirror is called the imagedistance ( v). You alr eady know that the\ndistance of the principal focus from the poleis called the focal length ( f). There is a\nrelationship between these three quantitiesgiven by the mirr or for mula which is\nexpressed as\n11 1\nvu f+= (10.1)\nThis formula is valid in all situations for all\nspherical mirrors for all positions of the\nobject. You must use the New Cartesian SignConvention while substituting numerical\nvalues for u, v, f , and R in the mirror formula\nfor solving problems.\nMagnification\nMagnification produced by a spherical mirror gives the relative extent to\nwhich the image of an object is magnified with respect to the object size.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nfor solving problems.\nMagnification\nMagnification produced by a spherical mirror gives the relative extent to\nwhich the image of an object is magnified with respect to the object size.\nIt is expressed as the ratio of the height of the image to the height of the\nobject. It is usually represented by the letter m.\nIf h is the height of the object and h\n\u2032 is the height of the image, then\nthe magnification m produced by a spherical mirror is given by\nm = Height of the image ( )\nHeight of the object ( )\u2032h\nh\nm = \u2032h\nh(10.2)\nThe magnification m is also related to the object distance ( u) and\nimage distance ( v). It can be expressed as:\nMagnification ( m) = \u2032=\u2212h\nhv\nu(10.3)\nYou may note that the height of the object is taken to be positive as\nthe object is usually placed above the principal axis. The height of theimage should be taken as positive for virtual images. However, it is to betaken as negative for real images.", + "The height of theimage should be taken as positive for virtual images. However, it is to betaken as negative for real images. A negative sign in the value of the", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nmagnification indicates that the image is real. A positive sign in the value\nof the magnification indicates that the image is virtual.Figure 10.9Figure 10.9Figure 10.9Figure 10.9Figure 10.9\nThe New Cartesian Sign Convention for spherical mirrors\n2019-20\nScience \uf031\uf037\uf030Example 10.1\nA convex mirror used for r ear-view on an automobile has a radius of\ncurvatur e of 3.00 m. If a bus is located at 5.00 m from this mirr or,\nfind the position, nature and size of the image.\nSolution\nRadius of curvature, R=+ 3.00 m;\nObject-distance, u=\u2013 5.00 m;\nImage-distance, v=?\nHeight of the image, h\u2032=?", + "Solution\nRadius of curvature, R=+ 3.00 m;\nObject-distance, u=\u2013 5.00 m;\nImage-distance, v=?\nHeight of the image, h\u2032=?\nFocal length, f = R/2 = + 3.00 m\n2= + 1.50 m (as the principal focus of\n a convex mirror is behind the mirror)\nSince 11 1+=vu f\nor, 111\nvfu=\u2212 = + 1\n1.50 \u2013 1\n( 5.00)\u2212 = 1\n1.50+ 1\n5.00\n= 5.00 1.50\n7.50+\nv = 7.50\n6.50+ = + 1.15 m\nThe image is 1.15 m at the back of the mirror .\nMagnification, m = 'hv\nhu=\u2212 = \u2013 1.15 m\n5.00 m\u2212\n = + 0.23\nThe image is virtual, erect and smaller in size by a factor of 0.23.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n6.50+ = + 1.15 m\nThe image is 1.15 m at the back of the mirror .\nMagnification, m = 'hv\nhu=\u2212 = \u2013 1.15 m\n5.00 m\u2212\n = + 0.23\nThe image is virtual, erect and smaller in size by a factor of 0.23.\nExample 10.2An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave\nmirror of focal length 15.0 cm. At what distance from the mirror\nshould a screen be placed in order to obtain a sharp image? Findthe nature and the size of the image.\nSolution\nObject-size, h = + 4.0 cm;\nObject-distance, u = \u2013 25.0 cm;\nFocal length, f = \u201315.0 cm;\nImage-distance, v = ?\nImage-size, h\n\u2032 = ?\nFrom Eq.", + "Solution\nObject-size, h = + 4.0 cm;\nObject-distance, u = \u2013 25.0 cm;\nFocal length, f = \u201315.0 cm;\nImage-distance, v = ?\nImage-size, h\n\u2032 = ?\nFrom Eq. (10.1):\n11 1\nvu f+=\nor, 11 1\nvf u=\u2212 =1 1 11\n15.0 25.0 15.0 25.0\u2212 =\u2212 +\u2212\u2212\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf037\uf031or, 1 5.0 3.0 2.0\n75.0 75.0 v\u2212+ \u2212== or, v = \u2013 37.5 cm\nThe scr een should be placed at 37.5 cm in front of the mirror . The\nimage is real.", + "The\nimage is real.\nAlso, magnification, m = 'hv\nhu=\u2212\nor, h\u2032 = \u2013vh\nu = ( 37.5 cm) ( 4.0 cm)\n( 25.0 cm)\u2212+\u2212\u2212\nHeight of the image, h\u2032 = \u2013 6.0 cm", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nimage is real.\nAlso, magnification, m = 'hv\nhu=\u2212\nor, h\u2032 = \u2013vh\nu = ( 37.5 cm) ( 4.0 cm)\n( 25.0 cm)\u2212+\u2212\u2212\nHeight of the image, h\u2032 = \u2013 6.0 cm\nThe image is inverted and enlarged.\n\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. Find the focal length of a convex mirror whose radius of curvature is\n32 cm.\n2. A concave mirror produces three times magnified (enlarged) real image\nof an object placed at 10 cm in front of it. Where is the image located?\n\uf031\uf030\uf02e\uf033\uf020\uf052\uf045\uf046\uf052\uf041\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054", + "\uf033\uf020\uf052\uf045\uf046\uf052\uf041\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf033\uf020\uf052\uf045\uf046\uf052\uf041\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf033\uf020\uf052\uf045\uf046\uf052\uf041\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf033\uf020\uf052\uf045\uf046\uf052\uf041\uf043\uf054\uf049\uf04f\uf04e\uf020", + "\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf033\uf020\uf052\uf045\uf046\uf052\uf041\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\uf031\uf030\uf02e\uf033\uf020\uf052\uf045\uf046\uf052\uf041\uf043\uf054\uf049\uf04f\uf04e\uf020\uf04f\uf046\uf020\uf04c\uf049\uf047\uf048\uf054\nLight seems to travel along straight-line paths in a transparent medium.\nWhat happens when light enters from one transparent medium to\nanother? Does it still move along a straight-line path or change its\ndirection? W e shall r ecall some of our day-to-day experiences.\nYou might have observed that the bottom of a tank or a pond\ncontaining water appears to be raised. Similarly, when a thick glass slab", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\ndirection? W e shall r ecall some of our day-to-day experiences.\nYou might have observed that the bottom of a tank or a pond\ncontaining water appears to be raised. Similarly, when a thick glass slab\nis placed over some printed matter, the letters appear raised when viewed\nthrough the glass slab. Why does it happen? Have you seen a pencilpartly immersed in water in a glass tumbler? It appears to be displaced\nat the interface of air and water . You might have observed that a lemon\nkept in water in a glass tumbler appears to be bigger than its actualsize, when viewed from the sides. How can you account for such\nexperiences?\nLet us consider the case of the apparent displacement of a pencil,\npartly immersed in water. The light reaching you from the portion of the\npencil inside water seems to come from a different direction, compared\nto the part above water. This makes the pencil appear to be displaced at\nthe interface. For similar reasons, the letters appear to be raised, when", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nto the part above water. This makes the pencil appear to be displaced at\nthe interface. For similar reasons, the letters appear to be raised, when\nseen through a glass slab placed over it.\nDoes a pencil appear to be displaced to the same extent, if instead of\nwater , we use liquids like kerosene or turpentine? Will the letters appear\nto rise to the same height if we replace a glass slab with a transparent\nplastic slab? Y ou will find that the extent of the effect is different for\ndifferent pair of media. These observations indicate that light does not\n2019-20\nScience \uf031\uf037\uf032travel in the same direction in all media. It appears that when travelling\nobliquely from one medium to another , the dir ection of propagation of\nlight in the second medium changes. This phenomenon is known asrefraction of light. Let us understand this phenomenon further by doing\na few activities .", + "It appears that when travelling\nobliquely from one medium to another , the dir ection of propagation of\nlight in the second medium changes. This phenomenon is known asrefraction of light. Let us understand this phenomenon further by doing\na few activities .\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf037", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\na few activities .\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf037\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf037\n\uf06ePlace a coin at the bottom of a bucket filled with water.\n\uf06eWith your eye to a side above water, try to pick up the coin in one\ngo.", + "\uf06eWith your eye to a side above water, try to pick up the coin in one\ngo. Did you succeed in picking up the coin?\n\uf06eRepeat the Activity. Why did you not succeed in doing it in one go?\n\uf06eAsk your friends to do this. Compare your experience with theirs.", + "Did you succeed in picking up the coin?\n\uf06eRepeat the Activity. Why did you not succeed in doing it in one go?\n\uf06eAsk your friends to do this. Compare your experience with theirs.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf038\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf038\n\uf06ePlace a large shallow bowl on a Table and put a coin in it.", + "\uf06eMove away slowly from the bowl. Stop when the coin justdisappears from your sight.\n\uf06eAsk a friend to pour water gently into the bowl without disturbingthe coin.\n\uf06eKeep looking for the coin from your position. Does the coin becomesvisible again from your position? How could this happen?\nThe coin becomes visible again on pouring water into the bowl. The\ncoin appears slightly raised above its actual position due to refraction of\nlight.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nThe coin becomes visible again on pouring water into the bowl. The\ncoin appears slightly raised above its actual position due to refraction of\nlight.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nThe coin becomes visible again on pouring water into the bowl. The\ncoin appears slightly raised above its actual position due to refraction of\nlight.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf039\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf039\n\uf06eDraw a thick straight line in ink, over a sheet of white paper placed\non a Table.", + "\uf06ePlace a glass slab over the line in such a way that one of its edgesmakes an angle with the line.\n\uf06eLook at the portion of the line under the slab from the sides. Whatdo you observe? Does the line under the glass slab appear to be\nbent at the edges?\n\uf06eNext, place the glass slab such that it is normal to the line. Whatdo you observe now? Does the part of the line under the glass slab\nappear bent?\n\uf06eLook at the line from the top of the glass slab. Does the part of theline, beneath the slab, appear to be raised? Why does this happen?", + "Whatdo you observe now? Does the part of the line under the glass slab\nappear bent?\n\uf06eLook at the line from the top of the glass slab. Does the part of theline, beneath the slab, appear to be raised? Why does this happen?\n\uf031\uf030\uf02e\uf033\uf02e\uf031\uf020\uf052\uf065\uf066\uf072\uf061\uf063\uf074\uf069\uf06f\uf06e\uf020\uf074\uf068\uf072\uf06f\uf075\uf067\uf068\uf020\uf061\uf020\uf052\uf065\uf063\uf074\uf061\uf06e\uf067\uf075\uf06c\uf061\uf072\uf020\uf047\uf06c\uf061\uf073\uf073\uf020\uf053\uf06c\uf061\uf062\nTo understand the phenomenon of refraction of light through a glass\nslab, let us do an Activity.\n2019-20", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n\uf031\uf030\uf02e\uf033\uf02e\uf031\uf020\uf052\uf065\uf066\uf072\uf061\uf063\uf074\uf069\uf06f\uf06e\uf020\uf074\uf068\uf072\uf06f\uf075\uf067\uf068\uf020\uf061\uf020\uf052\uf065\uf063\uf074\uf061\uf06e\uf067\uf075\uf06c\uf061\uf072\uf020\uf047\uf06c\uf061\uf073\uf073\uf020\uf053\uf06c\uf061\uf062\nTo understand the phenomenon of refraction of light through a glass\nslab, let us do an Activity.\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf037\uf033In this Activity, you will note, the light ray has changed its direction\nat points O and O \u2032. Note that both the points O and O \u2032 lie on surfaces\nseparating two transparent media.", + "Note that both the points O and O \u2032 lie on surfaces\nseparating two transparent media. Draw a perpendicular NN\u2019 to AB at O\nand another perpendicular MM \u2032 to CD at O \u2032. The light ray at point O has\nentered from a rarer medium to a denser medium, that is, from air toglass. Note that the light ray has bent towardsthe normal. At O\n\u2032, the light ray has entered\nfrom glass to air , that is, fr om a denser\nmedium to a rarer medium. The light herehas bent away from the normal. Compare the\nangle of incidence with the angle of refraction\nat both refracting surfaces AB and CD.\nIn Fig. 10.10, a ray EO is obliquely\nincident on surface AB, called incident ray.OO\n\u2032 is the refracted ray and O\u2032 H is the", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nat both refracting surfaces AB and CD.\nIn Fig. 10.10, a ray EO is obliquely\nincident on surface AB, called incident ray.OO\n\u2032 is the refracted ray and O\u2032 H is the\nemer gent ray. You may observe that the\nemergent ray is parallel to the direction ofthe incident ray. Why does it happen so? Theextent of bending of the ray of light at theopposite parallel faces AB (air -glass interface)\nand CD (glass-air interface) of the rectangularglass slab is equal and opposite. This is why\nthe ray emerges parallel to the incident ray.\nHowever, the light ray is shifted sideward\nslightly. What happens when a light ray isincident normally to the interface of twomedia? Try and find out.\nNow you are familiar with the refraction of light. Refraction is due to\nchange in the speed of light as it enters from one transparent medium toanother .", + "What happens when a light ray isincident normally to the interface of twomedia? Try and find out.\nNow you are familiar with the refraction of light. Refraction is due to\nchange in the speed of light as it enters from one transparent medium toanother . Experiments show that r efraction of light occurs according to\ncertain laws.\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf030", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\ncertain laws.\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf030\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf030\n\uf06eFix a sheet of white paper on a drawing board using drawing pins.", + "\uf06ePlace a rectangular glass slab over the sheet in the middle.\n\uf06eDraw the outline of the slab with a pencil. Let us name the outline\nas ABCD.\n\uf06eTake four identical pins.\n\uf06eFix two pins, say E and F, vertically such that the line joining the\npins is inclined to the edge AB.\n\uf06eLook for the images of the pins E and F through the opposite edge.Fix two other pins, say G and H, such that these pins and the\nimages of E and F lie on a straight line.\n\uf06eRemove the pins and the slab.\n\uf06eJoin the positions of tip of the pins E and F and produce the line\nup to AB. Let EF meet AB at O. Similarly, join the positions of tip\nof the pins G and H and produce it up to the edge CD. Let HG\nmeet CD at O \u2032.\n\uf06eJoin O and O \u2032. Also produce EF up to P, as shown by a dotted line\nin Fig. 10.10.\nFigure 10.10Figure 10.10Figure 10.10Figure 10.10Figure 10.10", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nmeet CD at O \u2032.\n\uf06eJoin O and O \u2032. Also produce EF up to P, as shown by a dotted line\nin Fig. 10.10.\nFigure 10.10Figure 10.10Figure 10.10Figure 10.10Figure 10.10\nRefraction of light through a rectangular glass slab\n2019-20\nScience \uf031\uf037\uf034The following are the laws of refraction of light.\n (i)The incident ray, the refracted ray and the normal to the interface\nof two transparent media at the point of incidence, all lie in the\nsame plane.\n(ii)The ratio of sine of angle of incidence to the sine of angle ofrefraction is a constant, for the light of a given colour and for\nthe given pair of media. This law is also known as Snell\u2019s law of\nrefraction.", + "(ii)The ratio of sine of angle of incidence to the sine of angle ofrefraction is a constant, for the light of a given colour and for\nthe given pair of media. This law is also known as Snell\u2019s law of\nrefraction. (This is true for angle 0 < i < 90\no)\nIf i is the angle of incidence and r is the angle of refraction, then,\nsin\nsini\nr = constant (10.4)\nThis constant value is called the refractive index of the second medium\nwith respect to the first. Let us study about refractive index in some detail.\n\uf031\uf030\uf02e\uf033\uf02e\uf032\uf020\uf054\uf068\uf065\uf020\uf052\uf065\uf066\uf072\uf061\uf063\uf074\uf069\uf076\uf065\uf020\uf049\uf06e\uf064\uf065\uf078", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nr = constant (10.4)\nThis constant value is called the refractive index of the second medium\nwith respect to the first. Let us study about refractive index in some detail.\n\uf031\uf030\uf02e\uf033\uf02e\uf032\uf020\uf054\uf068\uf065\uf020\uf052\uf065\uf066\uf072\uf061\uf063\uf074\uf069\uf076\uf065\uf020\uf049\uf06e\uf064\uf065\uf078\nYou have alr eady studied that a ray of light that travels obliquely from\none transparent medium into another will change its direction in the\nsecond medium. The extent of the change in direction that takes placein a given pair of media may be expressed in terms of the refractive index,the \u201cconstant\u201d appearing on the right-hand side of Eq.(10.4).\nThe refractive index can be linked to an important physical quantity,\nthe relative speed of propagation of light in different media. It turnsout that light propagates with different speeds in different media.", + "The refractive index can be linked to an important physical quantity,\nthe relative speed of propagation of light in different media. It turnsout that light propagates with different speeds in different media. Lighttravels fastest in vacuum with speed of 3\u00d710\n8 m s\u20131. In air , the speed of\nlight is only marginally less, compared to that in vacuum. It reducesconsiderably in glass or water . The value of the r efractive index for a", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n8 m s\u20131. In air , the speed of\nlight is only marginally less, compared to that in vacuum. It reducesconsiderably in glass or water . The value of the r efractive index for a\ngiven pair of media depends upon the speed of light in the two media, asgiven below.\nConsider a ray of light travelling from medium 1 into medium 2, as\nshown in Fig.10.11. Let v\n1 be the speed of light in medium 1 and v2 be\nthe speed of light in medium 2. The refractive index of medium 2 withrespect to medium 1 is given by the ratio of the speed of light in medium\n1 and the speed of light in medium 2. This is usually represented by the\nsymbol n\n21. This can be expressed in an equation form as\nn21=Speed of light in medium 1 \nSpeed of light in medium 2=v\nv1\n2(10.5)\nBy the same argument, the refractive index of medium\n1 with respect to medium 2 is represented as n12.", + "It is given\nby\nn12=Speed of light in medium 2 \nSpeed of light in medium 1=v\nv21 (10.6)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n2(10.5)\nBy the same argument, the refractive index of medium\n1 with respect to medium 2 is represented as n12. It is given\nby\nn12=Speed of light in medium 2 \nSpeed of light in medium 1=v\nv21 (10.6)\nIf medium 1 is vacuum or air , then the r efractive index of medium 2\nis considered with respect to vacuum. This is called the absolute refractive\nindex of the medium. It is simply represented as n2. If c is the speed ofFigure 10.11Figure 10.11Figure 10.11Figure 10.11Figure 10.11\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf037\uf035The ability of a medium to refract light is also expressed in terms of its optical density.\nOptical density has a definite connotation. It is not the same as mass density. We have\nbeen using the ter ms \u2018rar er medium\u2019 and \u2018denser medium\u2019 in this Chapter.", + "Optical density has a definite connotation. It is not the same as mass density. We have\nbeen using the ter ms \u2018rar er medium\u2019 and \u2018denser medium\u2019 in this Chapter. It actually\nmeans \u2018optically rarer medium\u2019 and \u2018optically denser medium\u2019, respectively. When canwe say that a medium is optically denser than the other? In comparing two media, the", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nmeans \u2018optically rarer medium\u2019 and \u2018optically denser medium\u2019, respectively. When canwe say that a medium is optically denser than the other? In comparing two media, the\none with the lar ger refractive index is optically denser medium than the other . The other\nmedium of lower refractive index is optically rarer. The speed of light is higher in a rarer\nmedium than a denser medium. Thus, a ray of light travelling from a rarer medium to a\ndenser medium slows down and bends towards the normal. When it travels from a\ndenser medium to a rarer medium, it speeds up and bends away from the normal.Note from Table 10.3 that an optically denser medium may not\npossess greater mass density.", + "When it travels from a\ndenser medium to a rarer medium, it speeds up and bends away from the normal.Note from Table 10.3 that an optically denser medium may not\npossess greater mass density. For example, kerosene having higher\nrefractive index, is optically denser than water, although its mass density\nis less than water .Material Refractive Material Refractive\nmedium index medium index\nAir 1.0003 Canada 1.53\nBalsam\nIce 1.31\nWater 1.33 Rock salt 1.54\nAlcohol 1.36\nKerosene 1.44 Carbon 1.63", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nis less than water .Material Refractive Material Refractive\nmedium index medium index\nAir 1.0003 Canada 1.53\nBalsam\nIce 1.31\nWater 1.33 Rock salt 1.54\nAlcohol 1.36\nKerosene 1.44 Carbon 1.63\ndisulphide\nFused 1.46\nquartz Dense 1.65\nflint glass\nTurpentine 1.47\noil Ruby 1.71\nBenzene 1.50\nSapphire 1.77\nCrown 1.52\nglass Diamond 2.42light in air and v is the speed of light in the medium, then, the refractive\nindex of the medium nm is given by\nnm = Speed of light in air \nSpeed of light in the medium =c\nv(10.7)\nThe absolute refractive index of a medium is simply called its refractive\nindex. The refractive index of several media is given in Table 10.3.", + "The refractive index of several media is given in Table 10.3. Fromthe T able you can know that the refractive index of water , n\nw = 1.33.\nThis means that the ratio of the speed of light in air and the speed of\nlight in water is equal to 1.33. Similarly, the refractive index of crown\nglass, ng =1.52. Such data ar e helpful in many places. However , you\nneed not memorise the data.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nlight in water is equal to 1.33. Similarly, the refractive index of crown\nglass, ng =1.52. Such data ar e helpful in many places. However , you\nneed not memorise the data.", + "Similarly, the refractive index of crown\nglass, ng =1.52. Such data ar e helpful in many places. However , you\nneed not memorise the data.\nTable 10.3 Absolute refractive index of some material media\uf04d\uf06f\uf072\uf065\uf020\uf074\uf06f\uf020\uf04b\uf06e\uf06f\uf077\uf021\n2019-20\nScience \uf031\uf037\uf036\uf031\uf030\uf02e\uf033\uf02e\uf033\uf020\uf052\uf065\uf066\uf072\uf061\uf063\uf074\uf069\uf06f\uf06e\uf020\uf062\uf079\uf020\uf053\uf070\uf068\uf065\uf072\uf069\uf063\uf061\uf06c\uf020\uf04c\uf065\uf06e\uf073\uf065\uf073\nYou might have seen watchmakers using a small magnifying glass to\nsee tiny parts. Have you ever touched the surface of a magnifying glass\nwith your hand?", + "Have you ever touched the surface of a magnifying glass\nwith your hand? Is it plane surface or curved? Is it thicker in the middle\nor at the edges? The glasses used in spectacles and that by a watchmaker\nare examples of lenses. What is a lens? How does it bend light rays? W e\nshall discuss these in this section.\nA transparent material bound by two surfaces, of which one or both\nsurfaces are spherical, forms a lens. This meansthat a lens is bound by at least one spherical\nsurface. In such lenses, the other surface would\nbe plane. A lens may have two spherical", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nsurfaces are spherical, forms a lens. This meansthat a lens is bound by at least one spherical\nsurface. In such lenses, the other surface would\nbe plane. A lens may have two spherical\nsurfaces, bulging outwards. Such a lens is called\na double convex lens. It is simply called a convex\nlens. It is thicker at the middle as compared to\nthe edges. Convex lens converges light rays as\nshown in Fig. 10.12 (a). Hence convex lenses are\nalso called converging lenses. Similarly, a double\nconcave lens is bounded by two spherical\nsurfaces, curved inwards. It is thicker at the\nedges than at the middle. Such lenses diverge\nlight rays as shown in Fig. 10.12 (b). Such lenses\nare also called diverging lenses. A double concave\nlens is simply called a concave lens.\nA lens, either a convex lens or a concave lens,\nhas two spherical surfaces. Each of these surfaces\nforms a part of a sphere.", + "Such lenses\nare also called diverging lenses. A double concave\nlens is simply called a concave lens.\nA lens, either a convex lens or a concave lens,\nhas two spherical surfaces. Each of these surfaces\nforms a part of a sphere. The centres of these\nspheres are called centres of curvature of the lens.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nA lens, either a convex lens or a concave lens,\nhas two spherical surfaces. Each of these surfaces\nforms a part of a sphere. The centres of these\nspheres are called centres of curvature of the lens.\nThe centre of curvature of a lens is usually\nrepresented by the letter C. Since there are two\ncentres of curvature, we may represent them as C\n1 and C2. An imaginary\nstraight line passing through the two centres of curvature of a lens is\ncalled its principal axis. The central point of a lens is its optical centre. It is\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. A ray of light travelling in air enters obliquely into water . Does the light\nray bend towards the normal or away from the normal? Why?\n2. Light enters from air to glass having refractive index 1.50. What is the\nspeed of light in the glass? The speed of light in vacuum is 3 \u00d7 108 m s\u20131.", + "Why?\n2. Light enters from air to glass having refractive index 1.50. What is the\nspeed of light in the glass? The speed of light in vacuum is 3 \u00d7 108 m s\u20131.\n3. Find out, from Table 10.3, the medium having highest optical density.\nAlso find the medium with lowest optical density.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nspeed of light in the glass? The speed of light in vacuum is 3 \u00d7 108 m s\u20131.\n3. Find out, from Table 10.3, the medium having highest optical density.\nAlso find the medium with lowest optical density.\n4. You ar e given kerosene, turpentine and water . In which of these does\nthe light travel fastest? Use the information given in Table 10.3.\n5. The refractive index of diamond is 2.42. What is the meaning of thisstatement?\n(a)\nFigure 10.12Figure 10.12Figure 10.12Figure 10.12Figure 10.12\n(a) Converging action of a convex lens, (b) divergingaction of a concave lens(b)\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf037\uf037usually represented by the letter O. A ray of light through the optical\ncentre of a lens passes without suffering any deviation. The effective\ndiameter of the circular outline of a spherical lens is called its aperture.\nWe shall confine our discussion in this Chapter to such lenses whose", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\ndiameter of the circular outline of a spherical lens is called its aperture.\nWe shall confine our discussion in this Chapter to such lenses whose\naperture is much less than its radius of curvature and the two centres ofcurvatures are equidistant from the optical centre O. Such lenses are\ncalled thin lenses with small apertures. What happens when parallel rays\nof light are incident on a lens? Let us do an Activity to understand this.", + "What happens when parallel rays\nof light are incident on a lens? Let us do an Activity to understand this.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf031\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf031\nCAUTION : Do not look at the Sun directly or through a lens while\ndoing this Activity or otherwise.", + "You may damage your eyes if you\ndo so.\n\uf06eHold a convex lens in your hand. Direct it towards the Sun.\n\uf06eFocus the light from the Sun on a sheet of paper . Obtain a sharp\nbright image of the Sun.\n\uf06eHold the paper and the lens in the same position for a while. Keep\nobserving the paper . What happened? Why? Recall your experience\nin Activity 10.2.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nbright image of the Sun.\n\uf06eHold the paper and the lens in the same position for a while. Keep\nobserving the paper . What happened? Why? Recall your experience\nin Activity 10.2.\nThe paper begins to burn producing smoke. It may even catch fire\nafter a while. Why does this happen? The light from the Sun constitutes\nparallel rays of light. These rays were converged by the lens at the sharp\nbright spot for med on the paper . In fact, the bright spot you got on the\npaper is a real image of the Sun. The concentration of the sunlight at apoint generated heat. This caused the paper to burn.\nNow, we shall consider rays of light parallel to the principal axis of a\nlens. What happens when you pass such rays of light through a lens?\nThis is illustrated for a convex lens in Fig.10.12 (a) and for a concave\nlens in Fig.10.12 (b).\nObserve Fig.10.12 (a) carefully.", + "What happens when you pass such rays of light through a lens?\nThis is illustrated for a convex lens in Fig.10.12 (a) and for a concave\nlens in Fig.10.12 (b).\nObserve Fig.10.12 (a) carefully. Several rays of light parallel to the\nprincipal axis are falling on a convex lens. These rays, after refraction", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nlens in Fig.10.12 (b).\nObserve Fig.10.12 (a) carefully. Several rays of light parallel to the\nprincipal axis are falling on a convex lens. These rays, after refraction\nfrom the lens, are converging to a point on the principal axis. This point\non the principal axis is called the principal focus of the lens. Let us see\nnow the action of a concave lens.\nObserve Fig.10.12 (b) carefully. Several rays of light parallel to the\nprincipal axis are falling on a concave lens. These rays, after refraction\nfrom the lens, are appearing to diverge from a point on the principal\naxis. This point on the principal axis is called the principal focus of the\nconcave lens.\nIf you pass parallel rays from the opposite surface of the lens, you\nget another principal focus on the opposite side. Letter F is usually used\nto represent principal focus. However , a lens has two principal foci. They\nare represented by F\n1 and F2. The distance of the principal focus from", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nto represent principal focus. However , a lens has two principal foci. They\nare represented by F\n1 and F2. The distance of the principal focus from\nthe optical centre of a lens is called its focal length. The letter f is used to\nrepresent the focal length. How can you find the focal length of a convexlens? Recall the Activity 10.11. In this Activity, the distance between the\nposition of the lens and the position of the image of the Sun gives the\napproximate focal length of the lens.", + "How can you find the focal length of a convexlens? Recall the Activity 10.11. In this Activity, the distance between the\nposition of the lens and the position of the image of the Sun gives the\napproximate focal length of the lens.\n2019-20\nScience \uf031\uf037\uf038\uf031\uf030\uf02e\uf033\uf02e\uf034\uf020\uf049\uf06d\uf061\uf067\uf065\uf020\uf046 \uf06f\uf072\uf06d\uf061\uf074\uf069\uf06f\uf06e\uf020\uf062\uf079\uf020\uf04c\uf065\uf06e\uf073\uf065\uf073\nLenses form images by refracting light. How do lenses form images?\nWhat is their nature? Let us study this for a convex lens first.", + "How do lenses form images?\nWhat is their nature? Let us study this for a convex lens first.\n\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf032\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf032\n\uf06eTake a convex lens.", + "Find its approximate focal length in a way\ndescribed in Activity 10.11.\n\uf06eDraw five parallel straight lines, using chalk, on a long Table suchthat the distance between the successive lines is equal to thefocal length of the lens.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\ndescribed in Activity 10.11.\n\uf06eDraw five parallel straight lines, using chalk, on a long Table suchthat the distance between the successive lines is equal to thefocal length of the lens.\n\uf06ePlace the lens on a lens stand. Place it on the central line such\nthat the optical centre of the lens lies just over the line.\n\uf06eThe two lines on either side of the lens correspond to F and 2F of\nthe lens respectively. Mark them with appropriate letters such as2F\n1, F1, F2 and 2F2, respectively.\n\uf06ePlace a burning candle, far beyond 2F1 to the left. Obtain a clear\nsharp image on a screen on the opposite side of the lens.\n\uf06eNote down the nature, position and relative size of the image.\n\uf06eRepeat this Activity by placing object just behind 2F1, between F1\nand 2F1 at F1, between F1 and O. Note down and tabulate your\nobservations.", + "\uf06eNote down the nature, position and relative size of the image.\n\uf06eRepeat this Activity by placing object just behind 2F1, between F1\nand 2F1 at F1, between F1 and O. Note down and tabulate your\nobservations.\nThe nature, position and relative size of the image formed by convex\nlens for various positions of the object is summarised in Table 10.4.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nobservations.\nThe nature, position and relative size of the image formed by convex\nlens for various positions of the object is summarised in Table 10.4.\nLet us now do an Activity to study the nature, position and relative\nsize of the image formed by a concave lens.Table 10.4 Nature, position and relative size of the image formed by a convex lens for various\npositions of the object\nPosition of the Position of Relative size of Nature of\nobject the image the image the image\nAt infinity At focus F2Highly diminished, Real and inverted\npoint-sized\nBeyond 2F1Between F2 and 2F2Diminished Real and inverted\nAt 2F1At 2F2Same size Real and inverted\nBetween F1 and 2F1Beyond 2F2Enlarged Real and inverted\nAt focus F1At infinity Infinitely large or Real and inverted\nhighly enlarged\nBetween focus F1On the same side Enlarged Virtual and erect\nand of the lens as the\noptical centre O object\n2019-20", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nAt focus F1At infinity Infinitely large or Real and inverted\nhighly enlarged\nBetween focus F1On the same side Enlarged Virtual and erect\nand of the lens as the\noptical centre O object\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf037\uf039The summary of the above Activity is given in Table 10.5 below.", + "5 below.\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf033\uf041\uf063\uf074\uf069\uf076\uf069\uf074\uf079\uf020\uf031\uf030\uf02e\uf031\uf033\n\uf06eTake a concave lens. Place it on a lens stand.\n\uf06ePlace a burning candle on one side of the lens.", + "Place it on a lens stand.\n\uf06ePlace a burning candle on one side of the lens.\n\uf06eLook through the lens from the other side and observe the image.\nTry to get the image on a screen, if possible. If not, observe the\nimage directly through the lens.\n\uf06eNote down the nature, relative size and approximate position ofthe image.\n\uf06eMove the candle away from the lens. Note the change in the sizeof the image. What happens to the size of the image when the\ncandle is placed too far away from the lens.\nWhat conclusion can you draw from this Activity? A concave lens", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\ncandle is placed too far away from the lens.\nWhat conclusion can you draw from this Activity? A concave lens\nwill always give a virtual, erect and diminished image, irrespective of the\nposition of the object.\n\uf031\uf030\uf02e\uf033\uf02e\uf035\uf020\uf049\uf06d\uf061\uf067\uf065\uf020\uf046 \uf06f\uf072\uf06d\uf061\uf074\uf069\uf06f\uf06e\uf020\uf069\uf06e\uf020\uf04c\uf065\uf06e\uf073\uf065\uf073\uf020\uf055\uf073\uf069\uf06e\uf067\uf020\uf052\uf061\uf079\uf020\uf044\uf069\uf061\uf067\uf072\uf061\uf06d\uf073\nWe can r epresent image for mation by lenses using ray diagrams.", + "Ray\ndiagrams will also help us to study the nature, position and relative size\nof the image formed by lenses. For drawing ray diagrams in lenses, alike\nof spherical mirrors, we consider any two of the following rays \u2013\n(i)A ray of light from the object, parallel to the principal axis, afterrefraction from a convex lens, passes through the principal focus\non the other side of the lens, as shown in Fig. 10.13 (a). In case of\na concave lens,\nthe ray appears\nto diverge from\nthe principal\nfocus located\non the same\nside of the lens,\nas shown in\nFig. 10.13 (b). (a) (b)Table 10.5 Nature, position and relative size of the image formed by a concave lens for various", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nthe principal\nfocus located\non the same\nside of the lens,\nas shown in\nFig. 10.13 (b). (a) (b)Table 10.5 Nature, position and relative size of the image formed by a concave lens for various\npositions of the object\nPosition of the Position of Relative size of Nature of\nobject the image the image the image\nAt infinity At focus F1Highly diminished, Virtual and erect\npoint-sized\nBetween infinity and Between focus F1Diminished Virtual and erect\noptical centre O and optical centre O\nof the lens\nFigure 10.13Figure 10.13Figure 10.13Figure 10.13Figure 10.13\n2019-20\nScience \uf031\uf038\uf030(ii)A ray of light passing\nthrough a principal\nfocus, after refraction\nfrom a convex lens, will\nemerge parallel to the\nprincipal axis. This is\nshown in Fig. 10.14 (a).", + "This is\nshown in Fig. 10.14 (a).\nA ray of light appearing\nto meet at the principal\nfocus of a concave lens,\nafter refraction, will\nemerge parallel to the\nprincipal axis. This is\nshown in Fig.10.14 (b).\n(iii) A ray of light passingthrough the optical\ncentre of a lens will", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nfocus of a concave lens,\nafter refraction, will\nemerge parallel to the\nprincipal axis. This is\nshown in Fig.10.14 (b).\n(iii) A ray of light passingthrough the optical\ncentre of a lens will\nemerge without any\ndeviation. This is\nillustrated in Fig.10.15(a)\nand Fig.10.15 (b).(b) (a)\n(b) (a)\nThe ray diagrams for the image formation in a convex lens for a few\npositions of the object are shown in Fig. 10.16. The ray diagrams\nrepresenting the image formation in a concave lens for various positions\nof the object are shown in Fig.", + "10.16. The ray diagrams\nrepresenting the image formation in a concave lens for various positions\nof the object are shown in Fig. 10.17.Figure 10.14Figure 10.14Figure 10.14Figure 10.14Figure 10.14\nFigure 10.15Figure 10.15Figure 10.15Figure 10.15Figure 10.15\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf038\uf031Figure 10.16Figure 10.16Figure 10.16Figure 10.16Figure 10.16 The position, size and the nature of the image formed by\na convex lens for various positions of the object", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\na convex lens for various positions of the object\nFigure 10.17 Figure 10.17 Figure 10.17 Figure 10.17 Figure 10.17 Nature, position and relative size of the image formed by a concave lens\n\uf031\uf030\uf02e\uf033\uf02e\uf036\uf020\uf053\uf069\uf067\uf06e\uf020\uf043\uf06f\uf06e\uf076\uf065\uf06e\uf074\uf069\uf06f\uf06e\uf020\uf066\uf06f\uf072\uf020\uf053\uf070\uf068\uf065\uf072\uf069\uf063\uf061\uf06c\uf020\uf04c\uf065\uf06e\uf073\uf065\uf073\nFor lenses, we follow sign convention, similar to the one used for spherical\nmirr ors. W e apply the rules for signs of distances, except that all\nmeasurements are taken from the optical centre of the lens.", + "W e apply the rules for signs of distances, except that all\nmeasurements are taken from the optical centre of the lens. Accordingto the convention, the focal length of a convex lens is positive and that ofa concave lens is negative. Y ou must take car e to apply appropriate\nsigns for the values of u, v, f, object height h and image height h\u2032.\n\uf031\uf030\uf02e\uf033\uf02e\uf037\uf020\uf04c\uf065\uf06e\uf073\uf020\uf046 \uf06f\uf072\uf06d\uf075\uf06c\uf061\uf020\uf061\uf06e\uf064\uf020\uf04d\uf061\uf067\uf06e\uf069\uf066\uf069\uf063\uf061\uf074\uf069\uf06f\uf06e\nAs we have a formula for spherical mirrors, we also have formula for\nspherical lenses. This formula gives the relationship between object-\ndistance ( u), image-distance ( v) and the focal length ( f).", + "This formula gives the relationship between object-\ndistance ( u), image-distance ( v) and the focal length ( f). The lens formula\nis expressed as\n11 1\nvu f\u2212= (10.8)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nspherical lenses. This formula gives the relationship between object-\ndistance ( u), image-distance ( v) and the focal length ( f). The lens formula\nis expressed as\n11 1\nvu f\u2212= (10.8)\nThe lens formula given above is general and is valid in all situations\nfor any spherical lens. Take proper care of the signs of different quantities,\nwhile putting numerical values for solving problems relating to lenses.\n2019-20\nScience \uf031\uf038\uf032Magnification\nThe magnification produced by a lens, similar to that for spherical\nmirrors, is defined as the ratio of the height of the image and the heightof the object. Magnification is represented by the letter m. If h is the\nheight of the object and h\n\u2032 is the height of the image given by a lens, then\nthe magnification produced by the lens is given by,\nm = Height of the Image\nHeight of the object=h\nh\u2032(10.9)\nMagnification produced by a lens is also related to the object-distance\nu, and the image-distance v. This relationship is given by", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nm = Height of the Image\nHeight of the object=h\nh\u2032(10.9)\nMagnification produced by a lens is also related to the object-distance\nu, and the image-distance v. This relationship is given by\nMagnification (m ) = h\u2032/h = v/u (10.10)\nExample 10.3\nA concave lens has focal length of 15 cm. At what distance should\nthe object from the lens be placed so that it forms an image at 10 cmfrom the lens? Also, find the magnification produced by the lens.\nSolution\nA concave lens always forms a virtual, erect image on the same sideof the object.\nImage-distance v= \u201310 cm;\nFocal length f = \u201315 cm;\nObject-distance u=?\nSince \n11 1\nvu f\u2212=\nor,111\u2013uv f=\n()11 1 11\u2013\u2013\u201310 \u201315 10 15u= =+\n1 32 1\n30 30 u\u2212+==\u2212\nor, u = \u2013 30 cm\nThus, the object-distance is 30 cm.", + "Magnification m = v/u\nm = \u2212\n\u2212=+10 cm\n30 cm1\n30.33/asymptequal\nThe positive sign shows that the image is erect and virtual. The image\nis one-third of the size of the object.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nMagnification m = v/u\nm = \u2212\n\u2212=+10 cm\n30 cm1\n30.33/asymptequal\nThe positive sign shows that the image is erect and virtual. The image\nis one-third of the size of the object.\nExample 10.4A 2.0 cm tall object is placed perpendicular to the principal axis of aconvex lens of focal length 10 cm. The distance of the object from the\nlens is 15 cm. Find the nature, position and size of the image. Also\nfind its magnification.\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf038\uf033Solution\nHeight of the object h= + 2.0 cm;\nFocal length f =+ 10 cm;\nobject-distance u = \u201315 cm;\nImage-distance v =?\nHeight of the image h\u2032=?", + "Height of the image h\u2032=?\nSince11 1\nvu f\u2212=\nor,11 1\nvu f=+\n1 1 1 11\n( 15) 10 15 10v= + =\u2212 +\u2212\n1 23 1\n30 30 v\u2212+==\nor, v = + 30 cm\nThe positive sign of v shows that the image is formed at a distance of\n30 cm on the other side of the optical centre. The image is real and\ninverted.\nMagnification m = 'hv\nhu=\nor, h\u2032 = h (v/u)\nHeight of the image, h\u2032 = (2.0) (+30/\u201315) = \u2013 4.0 cm\nMagnification m = v/u\nor, 30cm215cmm+= =\u2212\u2212", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\ninverted.\nMagnification m = 'hv\nhu=\nor, h\u2032 = h (v/u)\nHeight of the image, h\u2032 = (2.0) (+30/\u201315) = \u2013 4.0 cm\nMagnification m = v/u\nor, 30cm215cmm+= =\u2212\u2212\nThe negative signs of m and h\u2032 show that the image is inverted and\nreal. It is formed below the principal axis. Thus, a real, inverted image,\n4 cm tall, is formed at a distance of 30 cm on the other side of the\nlens. The image is two times enlarged.\n\uf031\uf030\uf02e\uf033\uf02e\uf038\uf020\uf050\uf06f\uf077\uf065\uf072\uf020\uf06f\uf066\uf020\uf061\uf020\uf04c\uf065\uf06e\uf073\nYou have alr eady learnt that the ability of a lens to converge or diver ge\nlight rays depends on its focal length.", + "For example, a convex lens ofshort focal length bends the light rays through large angles, by focussing\nthem closer to the optical centre. Similarly, concave lens of very short\nfocal length causes higher divergence than the one with longer focallength. The degree of convergence or divergence of light rays achievedby a lens is expressed in terms of its power. The power of a lens is defined", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nas the reciprocal of its focal length. It is r epresented by the letter P. The\npower P of a lens of focal length f is given by\nP = \n1\nf(10.11)\n2019-20\nScience \uf031\uf038\uf034The SI unit of power of a lens is \u2018dioptre\u2019. It is denoted by the letter D.\nIf f is expressed in metres, then, power is expressed in dioptres. Thus,\n1 dioptre is the power of a lens whose focal length is 1 metre. 1D = 1m\u20131.\nYou may note that the power of a convex lens is positive and that of a\nconcave lens is negative.\nOpticians prescribe corrective lenses indicating their powers. Let us\nsay the lens prescribed has power equal to + 2.0 D. This means the lens\nprescribed is convex.", + "You may note that the power of a convex lens is positive and that of a\nconcave lens is negative.\nOpticians prescribe corrective lenses indicating their powers. Let us\nsay the lens prescribed has power equal to + 2.0 D. This means the lens\nprescribed is convex. The focal length of the lens is + 0.50 m. Similarly,\na lens of power \u2013 2.5 D has a focal length of \u2013 0.40 m. The lens is concave.\nMany optical instruments consist of a number of lenses. They are combined to increasethe magnification and sharpness of the image. The net power ( P) of the lenses placed", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nMany optical instruments consist of a number of lenses. They are combined to increasethe magnification and sharpness of the image. The net power ( P) of the lenses placed\nin contact is given by the algebraic sum of the individual powers P\n1, P2, P3, \u2026 as\nP = P1 + P2 + P3 + \u2026\nThe use of powers, instead of focal lengths, for lenses is quite convenient for opticians.During eye-testing, an optician puts several different combinations of corrective lenses\nof known power, in contact, inside the testing spectacles\u2019 frame. The optician calculates\nthe power of the lens required by simple algebraic addition. For example, a combination\nof two lenses of power + 2.0 D and + 0.25 D is equivalent to a single lens of power + 2.25 D.\nThe simple additive property of the powers of lenses can be used to design lens systems\nto minimise certain defects in images produced by a single lens. Such a lens system,\nconsisting of several lenses, in contact, is commonly used in the design of lenses of", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nto minimise certain defects in images produced by a single lens. Such a lens system,\nconsisting of several lenses, in contact, is commonly used in the design of lenses of\ncamera, microscopes and telescopes.\n\uf051\uf055\uf045\uf053\uf054\uf049\uf04f\uf04e\uf053\n\uf03f1. Define 1 dioptre of power of a lens.\n2. A convex lens forms a real and inverted image of a needle at a distance\nof 50 cm from it. Where is the needle placed in front of the convex lens\nif the image is equal to the size of the object? Also, find the power of the\nlens.\n3. Find the power of a concave lens of focal length 2 m.", + "Where is the needle placed in front of the convex lens\nif the image is equal to the size of the object? Also, find the power of the\nlens.\n3. Find the power of a concave lens of focal length 2 m.\n\uf057\uf068\uf061\uf074\uf020\uf079 \uf06f\uf075\uf020\uf068\uf061\uf076\uf065\uf020\uf06c\uf065\uf061\uf072\uf06e\uf074\n\uf06e Light seems to travel in straight lines.\n\uf06e Mirrors and lenses form images of objects. Images can be either real or virtual,\ndepending on the position of the object.\n\uf06e The reflecting surfaces, of all types, obey the laws of reflection. The refractingsurfaces obey the laws of refraction.\n\uf06e New Cartesian Sign Conventions are followed for spherical mirrors and lenses.\uf04d\uf06f\uf072\uf065\uf020\uf074\uf06f\uf020\uf04b\uf06e\uf06f\uf077\uf021\n2019-20", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n\uf06e New Cartesian Sign Conventions are followed for spherical mirrors and lenses.\uf04d\uf06f\uf072\uf065\uf020\uf074\uf06f\uf020\uf04b\uf06e\uf06f\uf077\uf021\n2019-20\nLight \u2013 Reflection and Refraction \uf031\uf038\uf035\uf06e Mirror formula, 111+=vu f, gives the relationship between the object-distance (u ),\nimage-distance ( v), and focal length (f) of a spherical mirr or.\n\uf06e The focal length of a spherical mirror is equal to half its radius of curvature.\n\uf06e The magnification produced by a spherical mirror is the ratio of the height of the\nimage to the height of the object.\n\uf06e A light ray travelling obliquely from a denser medium to a rarer medium bendsaway from the normal. A light ray bends towards the normal when it travels obliquely\nfrom a rarer to a denser medium.\n\uf06e Light travels in vacuum with an enormous speed of 3\u00d7108 m s-1.", + "A light ray bends towards the normal when it travels obliquely\nfrom a rarer to a denser medium.\n\uf06e Light travels in vacuum with an enormous speed of 3\u00d7108 m s-1. The speed of light\nis different in different media.\n\uf06e The refractive index of a transparent medium is the ratio of the speed of light invacuum to that in the medium.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nis different in different media.\n\uf06e The refractive index of a transparent medium is the ratio of the speed of light invacuum to that in the medium.\n\uf06e In case of a rectangular glass slab, the refraction takes place at both air -glass\ninterface and glass-air interface. The emergent ray is parallel to the direction of\nincident ray.\n\uf06e Lens formula, 111\u2013=vu f, gives the relationship between the object-distance ( u),\nimage-distance ( v), and the focal length ( f) of a spherical lens.\n\uf06e Power of a lens is the reciprocal of its focal length. The SI unit of power of a lens isdioptre .\n\uf045\uf058\uf045\uf052\uf043\uf049\uf053\uf045\uf053\n1.Which one of the following materials cannot be used to make a lens?\n(a) Water (b) Glass (c) Plastic (d) Clay\n2.The image formed by a concave mirror is observed to be virtual, erect and largerthan the object. Where should be the position of the object?", + "(a) Water (b) Glass (c) Plastic (d) Clay\n2.The image formed by a concave mirror is observed to be virtual, erect and largerthan the object. Where should be the position of the object?\n(a) Between the principal focus and the centre of curvature\n(b) At the centre of curvature\n(c) Beyond the centre of curvature", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n(a) Between the principal focus and the centre of curvature\n(b) At the centre of curvature\n(c) Beyond the centre of curvature\n(d) Between the pole of the mirror and its principal focus.\n3.Where should an object be placed in front of a convex lens to get a real image of thesize of the object?\n(a) At the principal focus of the lens\n(b) At twice the focal length\n(c) At infinity\n(d) Between the optical centre of the lens and its principal focus.\n4.A spherical mirror and a thin spherical lens have each a focal length of \u201315 cm. The\nmirror and the lens are likely to be\n(a) both concave.\n(b) both convex.\n2019-20\nScience \uf031\uf038\uf036(c) the mirror is concave and the lens is convex.\n(d) the mirror is convex, but the lens is concave.\n5.No matter how far you stand fr om a mirr or, your image appears erect. The mirr or\nis likely to be\n(a) only plane.\n(b) only concave.\n(c) only convex.", + "(d) the mirror is convex, but the lens is concave.\n5.No matter how far you stand fr om a mirr or, your image appears erect. The mirr or\nis likely to be\n(a) only plane.\n(b) only concave.\n(c) only convex.\n(d) either plane or convex.\n6.Which of the following lenses would you prefer to use while reading small letters", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\nis likely to be\n(a) only plane.\n(b) only concave.\n(c) only convex.\n(d) either plane or convex.\n6.Which of the following lenses would you prefer to use while reading small letters\nfound in a dictionary?\n(a) A convex lens of focal length 50 cm.\n(b) A concave lens of focal length 50 cm.\n(c) A convex lens of focal length 5 cm.\n(d) A concave lens of focal length 5 cm.\n7.We wish to obtain an erect image of an object, using a concave mirr or of focal\nlength 15 cm. What should be the range of distance of the object from the mirror?\nWhat is the nature of the image? Is the image larger or smaller than the object?Draw a ray diagram to show the image formation in this case.\n8.Name the type of mirror used in the following situations.\n(a) Headlights of a car .\n(b) Side/r ear-view mirr or of a vehicle.\n(c) Solar furnace.\nSupport your answer with reason.\n9. One-half of a convex lens is covered with a black paper . Will this lens pr oduce a", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n(b) Side/r ear-view mirr or of a vehicle.\n(c) Solar furnace.\nSupport your answer with reason.\n9. One-half of a convex lens is covered with a black paper . Will this lens pr oduce a\ncomplete image of the object? Verify your answer experimentally. Explain yourobservations.\n10.An object 5 cm in length is held 25 cm away from a converging lens of focal length10 cm. Draw the ray diagram and find the position, size and the nature of theimage formed.\n11. A concave lens of focal length 15 cm forms an image 10 cm from the lens. How faris the object placed from the lens? Draw the ray diagram.\n12. An object is placed at a distance of 10 cm from a convex mirror of focal length\n15 cm. Find the position and nature of the image.\n13. The magnification produced by a plane mirror is +1. What does this mean?\n14.", + "Draw the ray diagram.\n12. An object is placed at a distance of 10 cm from a convex mirror of focal length\n15 cm. Find the position and nature of the image.\n13. The magnification produced by a plane mirror is +1. What does this mean?\n14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex\nmirror of radius of curvature 30 cm. Find the position of the image, its natureand size.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEXT BOOK.txt\n14. An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex\nmirror of radius of curvature 30 cm. Find the position of the image, its natureand size.\n15. An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal\nlength 18 cm. At what distance from the mirror should a screen be placed, so that\na sharp focussed image can be obtained? Find the size and the nature of the image.\n16. Find the focal length of a lens of power \u2013 2.0 D. What type of lens is this?\n17. A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of\nthe lens. Is the prescribed lens diverging or converging?\n2019-20", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-Long answer type questions.txt\nLong answer questions \n1. A convex lens has a focal length of 12 cm. At what distance from the lens should an \nobject of height 6 cm be placed so that on the other side of the lens its real and inverted \nimage is formed 24 cm away from the lens? What would be the size of the image formed? \nDraw a ray diagram to show the image formed in this case.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-Long answer type questions.txt\n2. A convex lens has a focal length of 15 cm. At what distance from the lens should the \nobject be placed so that is for ms on its other side a real and inverted image 30 cm away \nfrom the lens? What would be the size of image formed if the object is 5 cm high? With \nthe help of a ray diagram show the formation of the image by the lens in this case. \n3. Draw and explain the ray di agram formed by a convex mirror when \n(a) object is at infinity. (b) object is at finite distance from the mirror. \n4. Name the type of mirror used in the following situations: \n(i)Rear view mirror in vehicles \n(ii) Solar furnace \n(iii) Torch \n(iv) Solar cooker \n(v) To get the full length image of tall building. \n5. With the help of a ray diagram show the position, size and the nature of the image \nformed by a convex lens for various positions of the object. \n6. With the help of a ray diagram show the position, size and the natur e of the image", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-Long answer type questions.txt\nformed by a convex lens for various positions of the object. \n6. With the help of a ray diagram show the position, size and the natur e of the image \nformed by a concave mirror for various positions of the object. \n7. Explain refraction through glass slab with necessary diagrams \n8. It is desired to obtain an erect image of an object using a concave mirror of focal \nlength 20 cm. \n(i) What should be the range of distance of the object from the mirror? \n(ii) Will the image be bigger or smaller than the object? \n(iii) Draw a ray diagram to show the image formation in this case. \n9. One-half of a convex lens of focal length 20 cm is covered with a black pap er. \n(i) Will the lens produce a complete image of the object? \n(ii) Show the formation of image of an object placed at 2F1 of such covered lens \nwith the help of a ray diagram. \n(iii) How will the intensity of the image formed by half covered lens compare wit h \nnon-covered lens?", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-Long answer type questions.txt\nwith the help of a ray diagram. \n(iii) How will the intensity of the image formed by half covered lens compare wit h \nnon-covered lens? \n10. Draw ray diagrams to represent the nature, position and relative size of the image formed \nby a convex lens for the object placed: \n(a) At 2F \n(b) Between F1 and the optical centre O of lens \n11. Comment on the size, position of the image forme d by a concave mirror of focal length \n18 cm when an object is placed: \n(i) at 22 cm (ii) 14 cm (ii) 40 cm. \n12. in front of mirror without calculations. \n13. Two medium with refractive index 1.31 and 1.50 are given. In which case (i) bending of light is more? \n(ii) sp eed of light is more? \n14. Give the uses of concave mirror and convex lens . \n15. a)Name a mirror that can give an erect and enlarged image of an object.", + "In which case (i) bending of light is more? \n(ii) sp eed of light is more? \n14. Give the uses of concave mirror and convex lens . \n15. a)Name a mirror that can give an erect and enlarged image of an object. \nb)Why do we prefer a convex mirror as a rear -view mirror in vehicles? \nc)Find the focal length of a convex mirror whose radius of curvature is 32 cm.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-Long answer type questions.txt\nb)Why do we prefer a convex mirror as a rear -view mirror in vehicles? \nc)Find the focal length of a convex mirror whose radius of curvature is 32 cm. \nd) concave mirror produces three times magnified (enlarged) real image of an object placed \nat 10 cm in front of it. Where is the image located?", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-ACTIVITIES.txt\nPhysics \nActivity 3 \u2013 Refractive Index \nChapter \u2013 Light \nClass : X Duration : 15 minutes \nNature of activity : Group \nLearning Outcome : Students will be able to illustrate that there is no refraction taking place when a ray \nof light passes from one medium to another having similar refractive indices. \nWhat we have to do? \nTo demonstrate that a ray of light passes straight through two media having similar refractive indices \nmaking one of the media invisible. \nWhat do we require? \n Vegetable oil, large borosil beaker, smaller borosil beaker. \nWhat do we need to know? \nRefraction is the phenomenon of bending of light as it passes thr ough one medium to another where \nboth the media have different optical densities which is measured in terms of Refractive index. A ray of", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-ACTIVITIES.txt\nboth the media have different optical densities which is measured in terms of Refractive index. A ray of \nlight bends towards the normal as it passes from a medium of low refractive index (rarer) to a medium \nof higher refrac tive index (denser). It further means that if two media have similar or equal refractive \nindices then the light ray will pass straight through them. Hence the media will seemingly merge or one \nof the medium will actually become invisible!! \nHow will you p roceed? \n\u27a2 Teacher\u2019s Activity \nTeacher will ask students to recall the phenomenon of refraction and also guide students on how to \nperform the activity. \n\u27a2 Student\u2019s Activity \n\u2022 Place the smaller beaker inside the larger beaker and observe if both the beakers are visible. \n\u2022 Now remove the smaller beaker and fill the larger beaker with vegetable oil up to 3/4th of its \nheight. \n\u2022 Immerse the smaller beaker inside the larger beaker fi lled with vegetable oil and observe if the", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-ACTIVITIES.txt\nheight. \n\u2022 Immerse the smaller beaker inside the larger beaker fi lled with vegetable oil and observe if the \nsmaller beaker is visible. \nWhat have you observed? \nS.No Smaller beaker is visible when oil \nwas not filled in larger beaker \n( YES/NO) Smaller beaker when oil \nwas filled in larger beaker \n ( YES/NO) 1", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-ACTIVITIES.txt\nWhat have you learned? \nRefraction of a ray of light is due to the difference in ______________________ of two media and if this \ndifference is negligible then the ray of light passes _________________ through both the media. \n \nWhat more can we do? \n1. Suggest an experiment to fi nd out the refractive index of vegetable oil as well as glass beaker. \n2. Experimentally find out \"if vegetable oil is replaced by glycerin then will you see the same result \nas in the above activity\"? \nExtended Worksheet \n1. Define absolute refractive index o f a medium. \n2. How is refractive index of a medium related to optical density of that medium? \n3. Is optical density same as mass density? Support your answer by giving example. \n4. If the speed of light is 2X108 m/s in a medium then calculate the absolute refr active index of that \nmedium. Given that speed of light in vacuum/air is 3X108 m/s. \n5. Draw the path of a ray of light as it passes from AIR \u2013 WATER - DIAMOND - GLASS -AIR.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEACHING METHODS.txt\nPhysics \nActivity 3 \u2013 Refractive Index \nChapter \u2013 Light \nClass : X Duration : 15 minutes \nNature of activity : Group \nLearning Outcome : Students will be able to illustrate that there is no refraction taking place when a ray \nof light passes from one medium to another having similar refractive indices. \nWhat we have to do? \nTo demonstrate that a ray of light passes straight through two media having similar refractive indices \nmaking one of the media invisible. \nWhat do we require? \n Vegetable oil, large borosil beaker, smaller borosil beaker. \nWhat do we need to know? \nRefraction is the phenomenon of bending of light as it passes thr ough one medium to another where \nboth the media have different optical densities which is measured in terms of Refractive index. A ray of", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEACHING METHODS.txt\nboth the media have different optical densities which is measured in terms of Refractive index. A ray of \nlight bends towards the normal as it passes from a medium of low refractive index (rarer) to a medium \nof higher refrac tive index (denser). It further means that if two media have similar or equal refractive \nindices then the light ray will pass straight through them. Hence the media will seemingly merge or one \nof the medium will actually become invisible!! \nHow will you p roceed? \n\u27a2 Teacher\u2019s Activity \nTeacher will ask students to recall the phenomenon of refraction and also guide students on how to \nperform the activity. \n\u27a2 Student\u2019s Activity \n\u2022 Place the smaller beaker inside the larger beaker and observe if both the beakers are visible. \n\u2022 Now remove the smaller beaker and fill the larger beaker with vegetable oil up to 3/4th of its \nheight. \n\u2022 Immerse the smaller beaker inside the larger beaker fi lled with vegetable oil and observe if the", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEACHING METHODS.txt\nheight. \n\u2022 Immerse the smaller beaker inside the larger beaker fi lled with vegetable oil and observe if the \nsmaller beaker is visible. \nWhat have you observed? \nS.No Smaller beaker is visible when oil \nwas not filled in larger beaker \n( YES/NO) Smaller beaker when oil \nwas filled in larger beaker \n ( YES/NO) 1", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-TEACHING METHODS.txt\nWhat have you learned? \nRefraction of a ray of light is due to the difference in ______________________ of two media and if this \ndifference is negligible then the ray of light passes _________________ through both the media. \n \nWhat more can we do? \n1. Suggest an experiment to fi nd out the refractive index of vegetable oil as well as glass beaker. \n2. Experimentally find out \"if vegetable oil is replaced by glycerin then will you see the same result \nas in the above activity\"? \nExtended Worksheet \n1. Define absolute refractive index o f a medium. \n2. How is refractive index of a medium related to optical density of that medium? \n3. Is optical density same as mass density? Support your answer by giving example. \n4. If the speed of light is 2X108 m/s in a medium then calculate the absolute refr active index of that \nmedium. Given that speed of light in vacuum/air is 3X108 m/s. \n5. Draw the path of a ray of light as it passes from AIR \u2013 WATER - DIAMOND - GLASS -AIR.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nwww.ncrtsolutions.in \n \nwww.ncrtsolutions.in \n Light - Reflection and Refraction \nQuestion 1: Define the principal focus of a concave mirror. \n \nAnswer: Light rays that are parallel to the principal axis of a concave mirror converge at a \nspecific point on its principal axis after reflecting from the mirror. This point is known as the \nprincipal focus of the concave mirror. \n \nQuestion 2: The radius of curvat ure of a spherical mirror is 20 cm. What is its focal length? \n \nAnswer: Radius of curvature, R = 20 cm \n \nRadius of curvature of a spherical mirror = 2 \u00d7 Focal length ( f ) \n \nR = 2 f", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nHence, the focal length of the given spherical mirror is 10 cm. \n \nQuestion 3: Name the mirror that can give an erect and enlarged image of an object. \n \nAnswer: When an object is placed between the pole and the principal focus of a concave mirror, \nthe image formed is virtual, erect, and enlarged. \n \nQuestion 4: Why do we prefer a convex mirror as a rear -view mirror in vehicles? \n \nAnswer: Convex mirrors give a virtual, erect, and diminished image of the objects placed in front \nof them. They are preferred as a rear -view mirror in vehicles because they give a wider field of \nview, which allows the driver to see most of the traffic behind him. \n \nQuestion 5: Find the focal length of a convex mirror whose radius of curvature is 32 cm. \n \nAnswer: Radius of curvature, R = 32 cm \n \nRadius of curvature = 2 \u00d7 Focal length ( f ) \n \nR = 2 f", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nHence, the focal length of the given convex mirror is 16 cm. \n \nQuestion 6: A concave mirror produces three times magnified (enlarged) real image of object \nplaced at 10 cm in front of it. Where is the image located? \n \nAnswer: Magnification produced by a spherical mirror is given by the relation, \n \nwww.ncrtsolutions.in \n \nwww.ncrtsolutions.in \n \n \nLet the height of the object, ho = h \n \nThen, height of the image, hI = \u22123h (Image formed is real)", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nObject distance, u = \u221210 cm \n \nv = 3 \u00d7 (\u221210) = \u221230 cm \n \nHere, the negative sign indicates that an inverted image is formed at a distance of 30 \ncm in front of the given concave mirror. \n \nQuestion 7: A ray of light travelling in air enters obliquely into water. Does the light ray bend \ntowards the normal or away f rom the normal? Why? \n \nAnswer: The light ray bends towards the normal. \n \nWhen a ray of light travels from an optically rarer medium to an optically denser medium, it gets \nbent towards the normal. Since water is optically denser than air, a ray of light tra velling from air \ninto the water will bend towards the normal. \n \nQuestion 8: Light enters from air to glass having refractive index 1.50. What is the speed of light \nin the glass? The speed of light in vacuum is 3 \u00d7 108 m s \u22121. \n \nRefractive index of a medium nm is given by,", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nSpeed of light in vacuum, c = 3 \u00d7 108 m s \u22121 \n \nRefractive index of glass, ng = 1.50 \n \nSpeed of light in the glass, \n \nQuestion 9: Find out, from Table, the medium having highest optical density. Also find the \nmedium with lowest optical density. \n \nMaterial Refractive index Material medium Refractive \nwww.ncrtsolutions.in", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nwww.ncrtsolutions.in \n \nmedium \n \nindex \n \nAir \n 1.0003 \n Canada Balsam \n 1.53 \n \nIce \n 1.31 \n - \n - \n \nWater \n 1.33 \n Rock salt \n 1.54 \n \nAlcohol \n 1.36 \n - \n - \n \nKerosene \n 1.44 \n Carbon disulphide \n 1.63 \n \nFused \n \nquartz \n 1.46 \n Dense \n \nflint glass \n 1.65 \n \nTurpentine oil \n 1.47 \n Ruby \n 1.71 \n \nBenzene \n \nCrown 1.50 \n Sapphire \n 1.77 \n \n \nglass 1.52 \n Diamond \n 2.42 \n \n \n \nAnswer: Highest optical density = Diamond \n \nLowest optical density = Air \n \nOptical density of a medium is directly related with the refractive index of that medium.", + "A \nmedium which has the highest refractive index will have the highest optical density and vice -\nversa. \n \nIt can be observed from table 10.3 that diamond and air respec tively have the highest and lowest \nrefractive index. Therefore, diamond has the highest optical density and air has the lowest optical \ndensity.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nrefractive index. Therefore, diamond has the highest optical density and air has the lowest optical \ndensity. \n \nQuestion 10: You are given kerosene, turpentine and water. In which of these does the light travel \nfastest? Us e the information given in Table. \n \nMaterial \n \nmedium \n Refractive indexMaterial medium Refractive \n \nindex \n \nAir \n 1.0003 \n Canada Balsam \n 1.53 \n www.ncrtsolutions.in", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nwww.ncrtsolutions.in \n Ice \n 1.31 \n - \n - \n \nWater \n 1.33 \n Rock salt \n 1.54 \n \nAlcohol \n 1.36 \n - \n - \n \nKerosene \n 1.44 \n Carbon disulphide \n 1.63 \n \nFused \n \nquartz \n 1.46 \n Dense \n \nflint glass \n 1.65 \n \nTurpentine oil \n 1.47 \n Ruby \n 1.71 \n \nBenzene \n \nCrown 1.50 \n Sapphire \n 1.77 \n \n \nglass 1.52 \n Diamond \n 2.42 \n \n \n \nAnswer: Speed of light in a medium is given by the relation for refractive index (nm). The \nrelation is given as", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nIt can be inferred from the relation that light will travel the slowest in the material which has the \nhighest refractive index and travel the fast est in the material which has the lowest refractive index. \n \nIt can be observed from table 10.3 that the refractive indices of kerosene, turpentine, and water are \n1.44, 1.47, and 1.33 respectively. Therefore, light travels the fastest in water. \n \nQuestion 11: The refractive index of diamond is 2.42. What is the meaning of this statement? \n \nAnswer: Refractive index of a medium nm is related to the speed of light in that medium v by the \nrelation: \n \n \n \nWhere, c is the speed of light in vacuum/air \n \nwww.ncrtsolutions.in", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nWhere, c is the speed of light in vacuum/air \n \nwww.ncrtsolutions.in \n \nwww.ncrtsolutions.in \n The refractive index of diamond is 2.42. This suggests that the speed of light in diamond will \nreduce by a factor 2.42 compared to its speed in air. \n \nQuestion 12: Define 1 dioptre of power of a lens. \n \nAnswer: Power of lens is defined as the reciprocal of its focal length. If P is the power of a lens of \nfocal length Fin metres, then", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nThe S.I. unit of power of a lens is Dioptre. It is denoted by D. \n \n1 dioptre is defined as the power of a lens of focal length 1 metre. \n \n1 D = 1 m\u22121 \n \nQuestion 13: A convex lens forms a real and inverted image of a needle at a distance of 50 cm \nfrom it. Where is the needle placed in front of the convex lens if the image is equal to the size of \nthe object? Also, find the power of the lens. \n \nAnswer: When an object is placed at the centre of curvature, 2F1, of a convex lens, its image is \nformed at the centre of curvature, 2F2, on the other side of the lens. The image formed is inverted \nand of the same size as the object, as shown in the given figure.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nIt is given that the image of the needle is formed at a distance of 50 cm from the convex lens. \nHence, the needle is placed in front of the lens at a distance of \n \n50 cm. \n \nObject distance, u = \u221250 cm \n \nImage distance, v = 50 cm \n \nFocal length = f \n \nAccording to the lens formula, \n \nwww.ncrtsolutions.in \n \nwww.ncrtsolutions.in \n \nHence, the power of the given lens is +4 D. \n \n \n \nQuestion 14: Find the power of a concave lens of focal length 2 m. \n \nAnswer: Focal length of concave lens, f = 2 m", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nHere, negative sign arises due to the divergent nature of concave lens. \n \nHence, the power of the given concave lens is \u22120.5 D. \n \nQuestion 15: Which one of the following materials cannot be used to make a lens? \n \n(a) Water \n \n(b) Glass \n \n(c) Plastic \n \n(d) Clay \n \nAnswer: (d) A lens allows light to pass through it. Since clay does not show such property, it \ncannot be used to make a lens. \n \nQuestion 16: The image formed by a concave mirror is observed to be virtual, erect and larger \nthan the object. Where should be the posit ion of the object? \n \n(a) Between the principal focus and the centre of curvature \n \n(b) At the centre of curvature \n \n(c) Beyond the centre of curvature \n \n(d) Between the pole of the mirror and its principal focus \n \nQuestion 17: Where should an object be placed in front of a convex lens to get a real image of the \nsize of the object? \n \n www.ncrtsolutions.in", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nwww.ncrtsolutions.in \n \n(a) At the principal focus of the lens \n \n(b) At twice the focal length \n \n(c) At infinity \n \n(d) Between the optical centre of the lens and its principal focus. \n \nAnswer: (b) When an object is placed at the centre of curvature in front of a convex lens, its \nimage is formed at the centre of curvature on the other side of the lens. The image formed is real, \ninverted, and of the same size as the object. \n \nQuestion 18: A spherical mirror and a thin spherical lens have each a focal length of \u221215 cm. The \nmirror and the lens are likely to be \n \n(a) both concave \n \n(b) both convex \n \n(c) the mirror is concave and the lens is convex \n \n(d) the mirror is convex, but the lens is concave \n \nAnswer: (a) By convention, the focal length of a concave mirror and a concave lens are taken as \nnegative. Hence, both the spherical mirror and the thin spherical lens are concave in nature.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nAnswer: (a) By convention, the focal length of a concave mirror and a concave lens are taken as \nnegative. Hence, both the spherical mirror and the thin spherical lens are concave in nature. \n \nQuestion 19: No matter how far you stand from a mirror, your image appea rs erect. The mirror is \nlikely to be \n \n(a) plane \n \n(b) concave \n \n(c) convex \n \n(d) either plane or convex \n \nAnswer: (d) A convex mirror always gives a virtual and erect image of smaller size of the object \nplaced in front of it. Similarly, a plane mirror will always give a virtual and erect image of same \nsize as that of the object placed in front of it. Therefore, the gi ven mirror could be either plane or \nconvex. \n \nQuestion 20: Which of the following lenses would you prefer to use while reading small letters \nfound in a dictionary? \n \n(a) A convex lens of focal length 50 cm \n \n(b) A concave lens of focal length 50 cm \n \n(c) A convex lens of focal length 5 cm \n www.ncrtsolutions.in", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nwww.ncrtsolutions.in \n (d) A concave lens of focal length 5 cm \n \nAnswer: (c) A convex lens gives a magnified image of an object when it is placed between the \nradius of curvature and focal length. Also, magnification is more for convex lenses havi ng shorter \nfocal length. Therefore, for reading small letters, a convex lens of focal length 5 cm should be \nused. \n \nQuestion 21: We wish to obtain an erect image of an object, using a concave mirror of focal \nlength 15 cm. What should be the range of distan ce of the object from the mirror? What is the \nnature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show \nthe image formation in this case. \n \nAnswer: Range of object distance = 0 cm to15 cm \n \nA concave mirror gives an er ect image when an object is placed between its pole (P) and the \nprincipal focus (F). \n \nHence, to obtain an erect image of an object from a concave mirror of focal length 15 cm, the", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nprincipal focus (F). \n \nHence, to obtain an erect image of an object from a concave mirror of focal length 15 cm, the \nobject must be placed anywhere between the pole and the focus. The image fo rmed will be virtual, \nerect, and magnified in nature, as shown in the given figure.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nQuestion 19: Name the type of mirror used in the following situations. \n \n(a) Headlights of a car \n \n(b) Side/rear -view mirror of a vehicle \n \n(c) Solar furnace \n \nSupport your answer with reason. \n \nAnswer: (a) Concave (b) Convex (c) Concave \n \nExplanation \n \n(a) Concave mirror is used in the headlights of a car. This is because concave mirrors can produce \npowerful parallel beam of light when the light source is placed at their principal focus. \n \n(b) Convex mirror is used in side/rear view mirror of a vehicle. Convex mirrors give a virtual, \nerect, and diminished image of the objects placed in front of it. Because of this, they have a \nwide field of view. It enables the driver to see most of the traffic behind him/her. \n \nwww.ncrtsolutions.in", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nwww.ncrtsolutions.in \n (c) Concave mirrors are convergent mirrors. That is why they are used to construct solar furnaces. \nConcave mirrors converge the light incident on them at a single point known as principal focus. \nHence, they can be used to produce a large amou nt of heat at that point. \n \nQuestion 20: One-half of a convex lens is covered with a black paper. Will this lens produce a \ncomplete image of the object? Verify your answer experimentally. Explain your observations. \n \nAnswer: The convex lens will form compl ete image of an object, even if its one half is covered \nwith black paper. It can be understood by the following two cases. \n \nCase I \n \nWhen the upper half of the lens is covered \n \nIn this case, a ray of light coming from the object will be refracted by the lower half of the lens. \nThese rays meet at the other side of the lens to form the image of the given object, as shown in the \nfollowing figure.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nCase II \n \nWhen the lower half of the lens is covered \n \nIn this case, a ray of light coming from the object is refracted by the upper half of the lens. These \nrays meet at the other side of the lens to form the image of the given object, as shown in the \nfollowing figure. \n \n \n \nQuestion 21: An object 5 cm in leng th is held 25 cm away from a converging lens of focal length \n10 cm. Draw the ray diagram and find the position, size and the nature of the image formed. \n \nAnswer: Object distance, u = \u221225 cm \n \nObject height, ho = 5 cm \n \nFocal length, f = +10 cm \nwww.ncrtsolutions.in \n \nwww.ncrtsolutions.in \n \nAccording to the lens formula, \n \n \n \nThe positive value of v shows that the image is formed at the other side of the lens. \n \n \n \nThe negative sign shows that the image is real and formed behind the lens.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nThe positive value of v shows that the image is formed at the other side of the lens. \n \n \n \nThe negative sign shows that the image is real and formed behind the lens. \n \n \n \nThe negative value of image height indicates that the image formed is inverted. \n \nThe position, size, and nature of image are shown in the following ray diagram. \n \n \n \nQuestion 22: A concave lens of focal length 15 cm forms an image 10 cm from the lens. Ho w far \nis the object placed from the lens? Draw the ray diagram. \n \nAnswer: Focal length of concave lens (OF1), f = \u221215 cm \n \nImage distance, v = \u221210 cm \n \nAccording to the lens formula, \n \n \n \nwww.ncrtsolutions.in \n \nwww.ncrtsolutions.in \n The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown \nin the following ray diagram.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nwww.ncrtsolutions.in \n \nwww.ncrtsolutions.in \n The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown \nin the following ray diagram. \n \n \n \nQuestion 23: An object is placed at a distance of 10 cm from a convex mirror of focal length 15 \ncm. Find the position and nature of the image. \n \nAnswer: Focal length of convex mirror, f = +15 cm \n \nObject distance, u = \u221210 cm \n \nAccording to the mirror formula, \n \nThe positive value of v indicates that the image is formed behind the mirror. \n \n \n \nThe positive value of magnification indicates that the image formed is virtual and erect. \n \nThe magnification produced by a plane mirror is +1. What does this mean? \n \nMagnification produced by a mirror is given by the relation", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nThe magnification produced by a plane mirror is +1. It shows that the image formed by the plane \nmirror is of the same size as that of the object. The positive sign shows that the image formed is \nvirtual and erect. \n \nQuestion 24: An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex \nmirror of radius of curvature 30 cm. Find the position of the image, its nature and size. \n \nAnswer: Object distance, u = \u221220 cm \n \nObject height, h = 5 cm \n \n \n www.ncrtsolutions.in \n \nwww.ncrtsolutions.in \n \nRadius of curvature, R = 30 cm \n \nRadius of cur vature = 2 \u00d7 Focal length \n \nR = 2 f \n \nf = 15 cm \n \nAccording to the mirror formula, \n \nThe positive value of v indicates that the image is formed behind the mirror.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nThe positive value of v indicates that the image is formed behind the mirror. \n \nThe positive value of image height indicates that the image formed is erect. \n \nTherefore, the image formed is virtual, erect, and smaller in size. \n \nQuestion 25: An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal \nlength 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused \nimage can be obtained? Find the size and the nature of the image. \n \nAnswer: Object distance, u = \u221227 cm \n \nObject height, h = 7 cm \n \nFocal length, f = \u221218 cm \n \nAccording to the mirror formula, \n \n \n \n \n \n \n \n \n www.ncrtsolutions.in \n \nwww.ncrtsolutions.in \n \n \nThe screen should be placed at a distance of 54 cm in front of the given mirror. \n \n \n \nThe negative value of magnification indica tes that the image formed is real.", + "CHAPTER 9- LIGHT REFLECTION AND REFRACTION-QUESTIONS.txt\nThe negative value of magnification indica tes that the image formed is real. \n \n \n \nThe negative value of image height indicates that the image formed is inverted. \n \nQuestion 26: Find the focal length of a lens of power \u22122.0 D. What type of lens is this? \n \nAnswer: A concave lens has a negative focal length. Hence, it is a concave lens. \n \nQuestion 27: A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of \nthe lens. Is the prescribed lens diverging or converging? \n \nAnswer: A convex lens has a positive focal length. Hence, it is a convex lens or a converging \nlens." ] \ No newline at end of file