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CBSE Term II, , 2022, , Contents, CHAPTER, Carbon and its Compounds, , -, , CHAPTER, Periodic Classification of Elements, , -, , CHAPTER, How do Organisms Reproduce, , -, , CHAPTER, Heredity and Evolution, , -, , CHAPTER, Electricity, , -, , CHAPTER, Magnetic Effects of Electric Current, , -, , CHAPTER, Our Environment, , -, , Practice Papers, , -, , -, , Watch Free Learning Videos, Subscribe arihant, , Channel, , þ Video Solutions of CBSE Sample Papers, þ Chapterwise Important MCQs, þ CBSE Updates

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CBSE Term II, , 2022, , Syllabus, CBSE Term II Class XII, One Paper, No., , Max Marks:, Units, , Marks, , I., , Chemical Substances-Nature and Behaviour: Chapter and, , II., , World of Living: Chapter and, , IV., , Effects of Current: Chapter, , V., , Natural Resources: Chapter, , and, , THEME MATERIALS, UNIT - I, , Chemical Substances - Nature and Behaviour, , Chapter, , Carbon and its Compounds, Carbon compounds: Covalent bonding in carbon compounds. Versatile nature of, carbon.Homologous series., , Chapter, , Periodic Classification of Elements, Periodic Classification of Elements: Need for classification, early attempts at, classification of elements Dobereiner s Triads, Newland s Law of Octaves, Mendeleev s, Periodic Table , Modern periodic table, gradation in properties, valency, atomic number,, metallic and non-metallic properties., , THEME THE WORLD OF THE LIVING, UNIT - II, , World of Living, , Chapter, , How do Organisms Reproduce, Reproduction: Reproduction in animals and plants asexual and sexual reproductive, health-need and methods of family planning. Safe sex vs HIV AIDS.Child bearing and, women s health., , Chapter, , Heredity and Evolution, Heredity: Heredity; Mendel s contribution- Laws for inheritance of traits: Sex, determination: brief introduction;

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CBSE Term II, , 2022, , THEME NATURAL PHENOMENA, UNIT - IV, , Effects of Current, , Chapter, , Electricity, Ohm s law; Resistance, Resistivity, Factors on which the resistance of a conductor, depends. Series combination of resistors, parallel combination of resistors and its, applications in daily life. Heating effect of electric current and its applications in daily, life. Electric power, Interrelation between P, V, I and R., , Chapter, , Magnetic Effects of Current, Magnetic 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 s Left Hand Rule, Electric Motor, Electromagnetic induction., Induced potential difference, Induced current. Fleming s Right Hand Rule., , THEME NATURAL RESOURCES, UNIT - IV, , Natural Resources, , Chapter, , Our Environment, Our Environment: Eco-system, Environmental problems, Ozone depletion, waste, production and their solutions. Biodegradable and non-biodegradable substances.

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CHAPTER 01, , Carbon and It's, Compounds, In this Chapter..., !, , Covalent Bonding in Carbon, Compounds, , !, , Hydrocarbons, , !, , Isomerism, , !, , Allotropes of Carbon, , !, , Functional Groups, , !, , Organic Compounds, , !, , Homologous Series, , Carbon is the third most important element after oxygen and, hydrogen, for the existence of life on the Earth. The Earth, crust has only 0.02% carbon which is present in the form of, minerals (like carbonates, hydrogen-carbonates, coal,, petroleum, etc.) and the atmosphere has 0.03% of carbon, dioxide., Fuels (like wood, kerosene, coal, LPG, CNG, petrol, etc.), clothing material (like cotton, nylon, polyester, etc.), paper,, rubber, plastics, leather, drugs and dyes are all made up of, carbon., , Covalent Bonding in Carbon Compounds, The bonds which are formed by the sharing of an electron pair, between the atoms (either same or different atoms) are known, as covalent bonds., Atomic number of carbon (C) is 6., K L, , So, its electronic configuration = 2 , 4 ., Thus, there are 4 electrons in its outermost shell and its octet, can be completed by the following two ways, It could gain 4 electrons and form C 4− anion. But for a, nucleus having 6 protons, it would be difficult to hold on 10, electrons, i.e. 4 extra electrons., !, , It could lose 4 electrons and form C 4+ cation. But a large, amount of energy is required to remove 4 electrons, leaving behind a carbon cation with 6 protons in its, nucleus holding on just two electrons together, which is, not possible., In order to overcome this problem, carbon shares its, valence electrons with other atoms of carbon or with atoms of, other elements., , !, , These shared electrons belong to the outermost shells of, both atoms and in this way, both atoms attain the nearest, noble gas configuration. This type of bonding is called, covalent bonding., Compounds having covalent bonds are called covalent, compounds, these are generally poor conductor of, electricity., , Examples of Covalent Bonding, Some example depicting of covalent bonding are as follows, 1. Formation of Methane (CH 4 ), , In the formation of a methane molecule, one carbon atom, shares its 4 electrons with four hydrogen atoms (one electron, of each hydrogen atom). It shows carbon is tetravalent

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02, , CBSE Term II Science X, , because it possesses 4 valence electrons and hydrogen is, monovalent because it has only 1 valence electron., Before, After, Shared, combination Electrons, combination, H, , H, H, , C, , H, , H, , H, , C, , H, H or H—C—H, H, , H, Methane (CH 4 ) molecule, , 2. Formation of Water Molecule (H 2O), , Atomic number of O = 8, K L, Electronic configuration = 2, 6, Atomic number of H = 1, Electronic configuration = K, 1, To attain the stable electronic configuration of the nearest, noble gas, hydrogen needs 1 electron and oxygen needs, 2 electrons. So, two hydrogen atoms share an electron pair, with the oxygen atom such that hydrogen acquires a duplet, configuration and oxygen an octet, resulting in the, formation of two single covalent bonds., Before, combination, , H, , O, , H, , After, combination Shared electrons, , H, , O, , H or H — O —H, , Water (H 2 O) molecule, , Properties of Covalent Compounds, The compounds containing covalent bonds are called, covalent compounds. They have following properties i.e., Covalent compounds have low melting and boiling points, due to small intermolecular forces of attraction between the, atoms., Covalent compounds are generally poor conductors of, electricity. This is because the electrons are shared, between atoms and no charged particles are formed in these, compounds., Covalent compounds are generally volatile in nature., !, , !, , !, , Allotropes of Carbon, Allotropy is the property by virtue of which an element exists, in more than one form and each form has different physical, properties but identical chemical properties. These different, forms are called allotropes. Carbon exists in different, allotropic forms; some of them are diamond, graphite and, fullerene., Diamond is a colourless transparent substance and very, hard whereas graphite is an opaque substance which have, smooth surface and slippery to touch., !, , !, , Fullerenes are recently discovered allotropic forms of, carbon which were prepared for the first time by HW, Kroto, Smalley and Robert Curt by the action of laser beam, on the vapours of graphite., , Versatile Nature of Carbon, Main factors that are responsible for the formation of large, number of carbon compounds are, (i) Catenation The property of self linking of elements mainly, C-atoms through covalent bonds to form long, straight or, branched chains and rings of different sizes is called, catenation. Carbon shows maximum catenation in the, periodic table due to its small size and strong C ⎯ C bond., (ii) Tetravalency of carbon The valency of carbon is four, i.e., it is capable of bonding or pairing with four other carbon, atoms or with the atoms of some other monovalent, elements like hydrogen, halogen (chlorine, bromine), etc., (iii) Tendency to form multiple bonds Carbon has a strong, tendency to form multiple bonds due to it’s small size. It, shares more than one electron pair with its own atoms or, with the atoms of elements like oxygen, nitrogen,, sulphur, etc., , Organic Compounds, The compounds of carbon except its oxides, carbonates and, hydrogen carbonate salts, are known as organic compounds., In 1828, German chemist Friedrich Wohler accidently, prepared urea from ammonium cyanate when he was trying to, prepare ammonium cyanate by heating ammonium sulphate, and potassium cyanate. Thus, synthesis of urea discarded the, vital force theory., , Hydrocarbons, Organic compounds made up of carbon and hydrogen are, called hydrocarbons. These are of two types, i.e., 1. Aliphatic Saturated Hydrocarbons, Saturated aliphatic hydrocarbons are called alkane or paraffin., All the carbon atoms in these are bonded through single bonds., The general formula of these compounds is CnH 2n+ 2 and suffix, “-ane” is used in their nomenclature., e.g. CH 4 (methane), C 2H 6 (ethane), etc., 2. Aliphatic Unsaturated Hydrocarbons, Those compounds in which at least one double or triple bond, is present between two carbon atoms are called unsaturated, hdyrocarbons., Aliphatic unsaturated hydrocarbons are of two types, (i) Alkene Those hydrocarbons which have at least one, carbon-carbon double bond are called alkenes or olefins., The general formula of these compounds is CnH 2n ,, e.g. C 2H 4 (ethene).

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03, , CBSE Term II Science X, , (ii) Alkyne Those hydrocarbons which must have at least, one carbon-carbon triple bond are called alkynes. The, general common formula of these compounds is, CnH 2n− 2 , e.g. C 2H 2 (ethyne)., , Structure of Saturated and Unsaturated, Compounds, Steps to draw the structure of carbon compound are, Step I. First connect all the carbon atoms together with a, single bond., Step II. After that use the hydrogen atoms to satisfy the, remaining valencies of carbon (as carbon forms 4, bonds due to its 4 valency)., Step III. If number of available H-atoms are less than what is, required, satisfy the remaining valency by using, double or triple bond., 1. Structure of Propane (C 3H 8 ), , Same rules are followed here as in case of ethane. Here, the, three carbon atoms are linked together with a single bond., [Step I], , C—C—C, , To satisfy the remaining valencies of carbon atoms, hydrogen, atoms are linked with them., H H H, ⏐ ⏐ ⏐, [Step II], H — C— C— C— H, ⏐ ⏐ ⏐, H H H, 2 carbon atoms are bonded to 3 hydrogen atoms and 1 carbon, atom is bonded to 2 hydrogen atoms., Electron dot structure of propane, H, , H, , H, , C, , C, , C, , H, , H, , H, , To satisfy it, a double bond is used between the two carbon, atoms., H, H, [Step III], C == C, H, H, Now, all the four valencies of carbon are satisfied., Electron dot structure of ethene, , H, C, , H, , 3. Structure of Ethyne (C 2H 2 ), , Link the two carbon atoms by single bond., C—C, , [Step I], , Link the two hydrogen atoms with unsatisfied valencies of, carbon., [Step II], H ⎯C ⎯ C ⎯H, But in this case even after linking the available hydrogen, atoms with carbon atoms, still two valencies of each carbon is, unsatisfy. To satisfy it, a triple bond is used between the two, carbon atoms., [Step III], H ⎯C ≡≡ C ⎯H, In ethyne, the two carbon atoms share three pairs of, electrons among themselves to form a carbon-carbon triple, bond., Each carbon atom shares one electron with each hydrogen, atom to form two carbon-hydrogen single bonds., Electron dot structure of ethyne, , H, , C, , C, , H, , Structure of Cyclic Compounds, In some compounds, carbon atoms are arranged in the form, of ring. e.g. cyclohexane (C 6H12 ) and benzene (C 6H 6 )., , 2. Structure of Ethene (C 2H 4 ), , Link the two carbon atoms by single bond., C—C, , C, , H, , H, H, , H, , Link the four hydrogen atoms with carbon atom to satisfy the, unsatisfied valencies of carbon., H, H, ", !, [Step II], —, C, C, !, ", H, H, But in this case, even after linking the available hydrogen, atoms with carbon atoms, still one valency of each carbon, remains unsatisfy., , H, , HH, , [Step I], H, H C, H, C, H, , C, , C, , C, H, C H, H, C, H, , C—H, , H—C, H—C, , C—H, C, , H H, , H, , C6H12, (Cyclohexane), , C6H6, (Benzene)

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04, , CBSE Term II Science X, , Isomerism, , Names and Formulae of Functional Group, , Organic compounds with same molecular formula but, different chemical and physical properties are called, isomers. This phenomenon is called isomerism., The difference in properties of these compounds is due to, the difference in their structures. These compounds have, identical molecular formula but different structures. Hence,, they are called structural isomers and phenomenon is, called structural isomerism., e.g. Two structural isomers are possible for butane (C 4H10 )., H, H, , C, H, , H, C, H, , H, C, H, , C, , H, , H, , H, H, , H, , H, , Straight chain structure, , C, H, , C, H, , H, , C, , H, , C, , H, , H, , H, , Branched chain structure, , Functional Groups, Groups which combine with a carbon chain and decide its, chemical properties are called functional groups., e.g. — OH in R—OH will decide the chemical properties, of this compound, thus it is a functional group., , Name of functional group, , Formula of functional group, , Alcohol, , — OH, , Aldehyde, , O, ⏐⏐, — CHO or ⎯ C ⎯ H, O, ⏐⏐, ⎯C⎯, O, ⏐⏐, ⎯ C ⎯O ⎯H, , Ketone, , Carboxylic acid, , Thus, — OH in methyl alcohol (CH 3OH) and —COOH, (carboxylic acid) in formic acid (H ⋅ COOH) are present as, functional groups., , Homologous Series, A series of similarly constituted compounds in which the, members present have the same functional group and similar, chemical properties and any two successive members in a, particular series differ in their molecular formula by a, ⎯ CH 2 — unit, is called a homologous series., e.g. CH 4 , C 2H 6 , C 3H 8 , C 4H10 are the members of alkane, family.

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Chapter, Practice, PART 1, Objective Questions, , 8. Which of the following elements does not show, tetravalency?, (a) Ge, , (b) Si, , (c) C, , (d) O, , 9. ......... is the first synthesised organic compound., !, , Multiple Choice Questions, , 1. Carbon exists in the atmosphere in the form of, (NCERT Exemplar), (a) only carbon monoxide, (b) carbon monoxide in traces and carbon dioxide, (c) only carbon dioxide, (d) coal, , 2. Which of the following will contain covalent double, bond between its atoms?, (a) H2, (c) NaCl, , 3. Which of the following is the correct representation, of electron dot structure of nitrogen? (NCERT Exemplar), (a), , •• • •, , (c) •• N, , •, •, , N ••, , (b), , •, •, •• N • • N •, •• •, ••, , (d) •• N •••• N ••, , 4. Carbon can use four hydrogen atoms to form, methane (CH 4 ), because, (a) valency of carbon is four, (b) valency of hydrogen is one, (c) Both (a) and (b), (d) carbon gets noble gas configuration by making four, covalent bonds with hydrogen, , 5. A molecule of ammonia (NH 3 ) has, , (NCERT Exemplar), , (a) only single bonds, (b) only double bonds, (c) only triple bonds, (d) two double bonds and one single bond, , 6. The structure of S 8 molecule is ........ shaped., (a) ring, (c) circle, , (b) Urea, (d) Benzene, , 10. Pentane has the molecular formula C 5 H12 . It has, (NCERT Exemplar), , (a) 5 covalent bonds, (c) 16 covalent bonds, , (b) 12 covalent bonds, (d) 17 covalent bonds, , 11. Match the following :, Column I, , (b) O 2, (d) Cl 2, , •• ••, •• N • N •, •, •, , (a) Alcohol, (c) Vinegar, , (b) crown, (d) rectangle, , 7. Which of the following is not the use of graphite?, (a) It is used as lubricant, (b) It is used in manufacturing of lead-pencils, (c) It is used in manufacturing of artificial diamond, (d) It is used for making insulated plates, , Column II, , A. C 3H8, , (i) Cyclic compound, , B. C 6H6, , (ii) Alkyne, , C. C 2H2, , (iii) Alkene, , D. C 4H8, , (iv) Alkane, , Codes, A, (a) (iv), (b) (iii), (c) (ii), (d) (ii), , B, (i), (iv), (i), (iii), , C, (ii), (i), (iv), (i), , D, (iii), (ii), (iii), (iv), , 12. Structural formula of ethyne is, (a) H ⎯ C ≡≡ C ⎯ H, H, H, !, ", (c), C ==C, !, ", H, H, , (NCERT Exemplar), , (b) H3C ⎯C ≡≡ C ⎯ H, H, H, !, ", (d) H ⎯ C ⎯C ⎯ H, !, ", H, H, , 13. Which among the following are unsaturated, hydrocarbons?, (i) CH3 ⎯ CH2 ⎯ CH3, CH3, ⏐, (iii) CH3 ⎯ C⎯ CH3, ⏐, CH3, (a) (i) and (iii), (c) (ii) and (iv), , (ii) CH3 ⎯ CH == CH ⎯ CH3, , (iv) CH3 ⎯ C == CH2, ⏐, CH3, (b) (ii) and (iii) (NCERT Exemplar), (d) (ii), (iii) and (iv)

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06, , CBSE Term II Science X, , 14. Structural formula of benzene is, , (NCERT Exemplar), , H, C, (a), , H—C, H, C, H, , H, , C—H, , H, H, (b) H, H, , C, , C, , C, , C, , H, H, H, H, , 19. Assertion Carbon shows maximum catenation, property in the periodic table., Reason Carbon has small size and thus, forms, strong C⎯ C bond., , C, , C, H, , H, , H, , 20. Assertion Following are the members of a, , H, , homologous series:, CH3 OH, CH3 CH2 OH, CH3 CH2 CH2 OH, Reason A series of compounds with same, functional group but differing by ⎯ CH2 ⎯ unit is, (CBSE 2020), called a homologous series., , C, , H—C, , C—H, , H—C, , C—H, , C, , H—C, (d), , H—C, , C, , C, , C, , H, , H, , H, H, H, H, !, , 15. Which of the following are correct structural isomers of, H, , H, , H, , ⏐, , ⏐, , ⏐, , ⏐ ⏐, , (ii) H⎯ C ⎯ C ⎯ C ⎯ H, ⏐, C, H, ⏐ ⏐, , butane?, H H H H, ⏐ ⏐ ⏐ ⏐, (i) H⎯C ⎯C ⎯C ⎯C ⎯H, ⏐ ⏐ ⏐ ⏐, H H H H, , H, H, , H, , H, H, H, H, ⏐, ⏐, ⏐, ⏐, ⏐, (iii) H⎯C ⎯ ⎯C ⎯⎯C ⎯H (iv) H ⎯ C ⎯ C ⎯ H, |, |, ⏐, ⏐, ⏐, H H⎯C ⎯H H, H ⎯C ⎯C ⎯H, |, |, ⏐, H H, H, (NCERT Exemplar), , (a) (i) and (iii) (b) (ii) and (iv) (c) (i) and (ii), !, , Reason The various layers of carbon atoms in, graphite are held together by weak van der Waals’, forces., , H, C, , C—H, , C, (c), , 18. Assertion Graphite is slippery to touch., , (d) (iii) and (iv), , Assertion-Reasoning MCQs, Direction (Q. Nos. 16-20) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative choices,, any one of which is the correct answer. You have to select, one of the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true., , 16. Assertion Covalent compounds are poor conductor of, electricity., Reason The electrons are shared between atoms and no, charged particles are present., 17. Assertion Diamond does not conduct electricity., Reason Diamond has high refractive index., , Case Based MCQs, , 21. Read the following and answer the questions from, (i) to (v) given below, The bonds which are formed by the sharing of an, electron pair between the atoms (either same or, different atoms) are known as covalent bonds., As neutral atom carbon has electronic configuration, K L. To gain inert gas configuration carbon can, 2 4, , either donate 4 valence electrons (helium gas, configuration) or gain 4 electrons (neon gas, configuration), but it cannot do so. To acquire inert, gas configuration carbon can only share its 4 valence, electrons with other atoms forming covalent bonds., The concept of covalent bonds was given by, Langmuir and Lewis to explain bonding in non-ionic, compounds. The covalent bonds are of three types. If, each atom contributes one electron, the covalent, bond formed is called a single covalent bond and is, represented by a single line (—) and if each atom, contributes two electrons, the covalent bond formed, is called a double bond and is represented by a double, line (==) and if each atom contributes three electrons,, the covalent bond formed is called a triple bond and is, represented by a triple line ( ≡≡ )., The electrons in a covalent bond are simultaneously, attracted by the two atomic nuclei. A covalent bond, forms when the difference between the, electronegativities of two atoms is too small for an, electron transfer to occur to form ions., (i) Which of the following do not contain a triple, bond?, I. SO2, II. N2, (a) I and II, (b) I, III and IV, (c) III and IV, (d) I and IV, , III. HCl, , IV. NH3

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07, , CBSE Term II Science X, , (ii) Which of the following contains a double bond?, (a) O2, , (b) N2, , (c) CH4, , (d) H2O, , (iii) Chlorine forms a diatomic molecule, Cl 2 . The, electron dot structure for this molecule is, (a), , (c), , Cl, , Cl, , Cl, , Cl, , (b), , (d), , Cl, , Cl, , Cl, , Cl, , (iv) What is the covalency of nitrogen?, (a) 1, , (b) 2, , (c) 3, , (d) 4, , Which statements about substance X are correct?, I. It is a covalent compound., II. It has a giant molecular structure., III. It has the same structure as graphite., IV. It has the same structure as diamond., (a) I and III, (b) II and III, (c) II and IV, (d) I, II and IV, , (ii) Which of the following is correct about the structure, of diamond?, (a) Carbon atoms are held together by single covalent, bonds, (b) Electrons move freely through the structure, (c) Layers of atoms slide easily over each other, (d) Carbon atoms conduct electricity in the molten state, , (iii) Which three allotropes of carbon, do the given figures, represents?, , (v) The shared pair of electrons is said to constitute a, ……… bond between two hydrogen atoms., (a) single, (c) triple, , (b) double, (d) ionic, , 22. Read the following and answer the questions from (i), to (v) given below, Allotropy is the property by virtue of which an, element exists in more than one form and these, different forms of an element are called allotropes., Allotropes have similar chemical properties but they, differ in their physical properties. Carbon exists in, crystalline and amorphous forms., In crystalline form, it occur as diamond, graphite and, fullerenes. Diamond is a colourless, transparent, substance having extraordinary brilliance. It is the, hardest natural substance known. It is used for cutting, marble, granite and glass. Graphite is a greyish-black, opaque substance. It is lighter than diamond, i.e., it, has lower density. It has sheet like structure having, hexagonal layers., One layer slides over the other layer which makes it, soft to touch. It is the reason that graphite is used as a, lubricant. The amorphous form of carbon is also, known as micro-crystalline form which consists of, coal, lampblack and charcoal., (i) Substance X is a moderate conductor of electricity., Substance X has the structure shown below :, , (I), , (II), , I, (a), , (III), , II, , III, , Diamond Graphite, , Buckminster fullerene, , (b) Graphite Buckminster, fullerene, , Diamond, , (c), , Graphite, , Diamond Buckminster, fullerene, , (d) Graphite Diamond, , Buckminster fullerene, , (iv) Identify the incorrect statement(s)., I. Diamond is the hardest substance known while, graphite is smooth and slippery., II. Diamond is made up of billions of carbon atoms., Each carbon atom is bonded to four other carbon, atoms in a tetrahedral manner to form a giant, lattice. All carbon atoms are bonded by strong, covalent bonds., III. Graphite is a poor conductor of electricity unlike, other non-metals., IV. Graphite has a giant covalent structure that is, made up of layers of carbon atoms. In each layer,, each carbon atom is bonded to three other carbon, atoms to form hexagonal rings of carbon atoms., (a) I and III, (c) II and IV, , (b) Only III, (d) I, II and IV, , (v) Which of the following is an example of amorphous, form of carbon ?, (a) Wood, (c) Chalk, , (b) Oil, (d) Coke

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08, , CBSE Term II Science X, , PART 2, Subjective Questions, !, , 15. Write the molecular formula of the following, compounds and draw their electron dot structures, (i) Ethane, (ii) Ethene, (iii) Ethyne, , 16. What is meant by isomers? Draw the structures of two, isomers of butane, C 4 H10 . Explain, why we cannot, have isomers of first three members of alkane series?, , Short Answer Type Questions, , 1. What do you mean by covalent bonding?, 2. Explain the nature of the covalent bond using the, bond formation in CH 3Cl ., , 17. Draw the possible isomers of the compound with, molecular formula C 3 H6 O and also give their, (NCERT Exemplar), electron dot structures., , (NCERT), , 3. What would be the electron dot structure of carbon, dioxide which has molecular formula CO 2 ? (NCERT), , 18. What is meant by functional group in carbon, compounds? Write in tabular form the structural, formula and the functional group present in the, following compounds:, (i) Ethanol, (ii) Ethanoic acid, , 4. (i) Explain the formation of calcium chloride with the, help of electron dot structure., (Atomic numbers of Ca = 20; Cl =17), (ii) Why do ionic compounds not conduct electricity in, solid state but conduct electricity in molten and, aqueous state?, , 19. Describe the applications of homologous series., 20. Why 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., , 5. Carbon a group (14) element in the periodic, table, is known to form compounds with many, elements. Write an example of a compound formed, with, (i) chlorine (group 17 of periodic table), (ii) oxygen (group 16 of periodic table)(NCERT Exemplar), , 6. Why covalent compounds are volatile in nature with, low boiling and low melting point?, (ii) an alkyne, , 8. Covalent compounds are generally poor conductors, of electricity. Why?, , (CBSE 2020), , 9. Diamond is a poor conductor of electricity while, graphite is a good conductor. Assign reason., , 10. Why diamond has high melting point?, 11. What are the main factors that enables carbon to, form large number of compounds?, , 12. Select saturated hydrocarbons from the following:, , !, , Long Answer Type Questions, , cations nor C 4 − anions, but forms covalent, compound. Also state reasons to explain why, covalent compounds, (i) Are bad conductors of electricity?, (ii) Have low melting and boiling point?, 23. Explain the formation of oxygen (O 2 ) molecule and, sulphur (S 8 ) molecule., , 24. What are covalent compounds? Why are they, different from ionic compounds? List their three, characteristics properties., , 25. Why organic compounds are called as hydrocarbons?, , C 3H 6 ; C 5H 10 ; C 4 H 10 ; C 2H 4 ; C 6H 14, , 13. What will be the formula and electron dot structure, of cyclopentane?, , C 2 H2 and C 2 H5 COOH., , 22. State the reason why carbon can neither form C 4 +, , 7. State the valency of each carbon atom in, (i) an alkane and, , 21. Write the next higher order homologous of CH2 O,, , (NCERT), , 14. Answer the following, (i) Carbon is a versatile element. Give reason., (ii) Explain the structural difference between saturated, and unsaturated hydrocarbons with two examples, each., (iii) What is a functional group? Write examples of four, different functional groups., , Write the general formula for homologous series of, alkanes, alkenes and alkynes and also draw the, structure of the first member of each series., , 26. Identify the functional group present in the, following compounds, H O, ⏐ ⏐⏐, (i) H ⎯ C ⎯C ⎯OH, ⏐, H, , H, ⏐, (ii) H ⎯ C ⎯, ⏐, H, , H H, ⏐, ⏐, C ⎯C ⎯H, ⏐, ⏐, OH H

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09, , CBSE Term II Science X, , H, H H, ⏐, ⏐ ⏐, (iii) H ⎯ C ⎯CHO, (iv) H ⎯ C ⎯ C ⎯ Br, ⏐, ⏐ ⏐, H, H H, H H O H, ⏐ ⏐ ⏐⏐ ⏐, (v) H ⎯ C ⎯ C ⎯ C ⎯ C ⎯ H, ⏐ ⏐, ⏐, H H, H, !, , Case Based Questions, , 27. Read the following and answer the questions from (i), to (v) given below, Compounds which contain only carbon and, hydrogen are called hydrocarbon. Among these, the, compounds containing all single covalent bonds are, called saturated hydrocarbons while the compounds, containing atleast one double or triple bond are, called unsaturated hydrocarbons., Saturated hydrocarbons after combustion give a, clean flame while unsaturated hydrocarbons given a, yellow sooty flame. Unsaturated hydrocarbons are, more reactive than saturated hydrocarbons., Unsaturated hydrocarbons add hydrogen in the, presence of catalysts such as palladium or nickel to, give saturated hydrocarbons., Study the table related to three hydrocarbons, A , B , C and answer the questions that follows, Organic compound, , Molecular formula, , A, , C 3H8, , B, , C5H10, , C, , C 4H6, , (i) What is the name of compound B?, (ii) Write two differences between saturated and, unsaturated hydrocarbons., (iii) In unsaturated compounds, what is the minimum, number of carbon atoms and why?, (iv) Among compounds A, B and C, which of the, following is saturated hydrocarbon?, (v) Compound C belongs to which category of, hydrocarbon and what is it’s general formula ?, , 28. Read the following and answer the questions from (i), to (v) given below, Organic compounds with same molecular formula, but different chemical and physical properties are, called isomers. This phenomenon is called isomerism., When the isomerism is due to difference in the, arrangement of atoms within the molecule, without, , any reference to space, the phenomenon is called, structural isomerism. In other words, structural, isomers are compounds that have the same molecular, formula but different structural formulas, i.e. they are, different in the order in which different atoms are, linked or they have different connectivities depending, upon the order they are put together. In these, compounds, carbon atoms can be linked together in the, form of straight chains, branched chains or even rings., (i) Name any set of compound that have same molecular, formula but different structural formula., (ii) Which property of carbon leads to formation of, branched chains?, (iii) How many isomers of pentane are possible?, (iv) Name two compounds that contains six carbon atoms, and have cyclic structure., (v) What is the minimum number of carbon atoms, required to form an isomer?, , 29. Read the following and answer the questions from (i) to, (v) given below, Hydrocarbons are the simplest organic compounds and, are regarded as parent organic compounds. All other, compounds are considered to be derived from them by, the replacement of one or more hydrogen atoms by, other atoms or group of atoms., Functional groups is an atom or group of atoms which, makes a carbon compound (or organic compound), reactive and decide it’s properties. A series of organic, compounds having same functional group with similar, or almost identical chemical characteristics in which, all the members can be represented by the same, general formula and two consecutive members of series, differ by ⎯CH 2 group in their molecular formula is, called a homologous series., H, H, H H, ⏐, ⏐, ⏐ ⏐, H ⎯ C ≡≡ C ⎯ C ⎯ H, H ⎯ C ⎯ Br, H ⎯ C ⎯ C ⎯ OH,, ⏐, ⏐, ⏐ ⏐, H, H, H H, (P), , (Q ), , (R), , H H, ⏐ ⏐, H", !H, H ⎯ C ⎯ C ⎯ H, H ⎯ C ≡≡ C ⎯ H, ! C ==C ", H, H, ( T), ⏐ ⏐, (U), H H, ( S), , (i), (ii), (iii), (iv), , Which compounds belongs to same homologous series?, What is the functional group of compound R ?, Compound T belongs to which homologous series ?, Among the P, S, T and U compounds which belongs to, alkane series?, (v) With respect to Q, R, S and U compounds which one is, an unsaturated hydrocarbon?

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EXPLANATIONS, Objective Questions, , 10. (c) The structural formula of pentane C5H12 is, , 1. (c) Carbon exists in the atmosphere in the form of carbon, dioxide gas (CO2 ) in air (only 0.03%). Carbon also occurs in, the earth’s crust in the form of minerals like carbonates. It also, occurs in the form of fossil fuels, organic compounds, wood,, cotton and wool, etc., 2. (b) Oxygen atom has six (6) valence electrons. Thus, to, complete its octet, it forms double bond with another oxygen, atom to get O2 molecule as, O, , O, , or, , O==O, , Shared electrons, (covalent bond, formation), , 3. (d) Electronic configuration of N (atomic number 7) is K L ., 2 5, Therefore, it needs three more electrons to complete its, octet. Each nitrogen atom shares three electrons to form a, molecule of N 2 as, N, , N, , 4. (c) Carbon has 4 electron in its valence shell, while hydrogen, has one electron in its valence shell., To complete their octet and duplet respectively, they form, covalent bonds. Carbon utilises its 4 valence electron and, forms 4 covalent bonds with 4 hydrogen atoms, using one, valence electron with each hydrogen atom., 5. (a) A molecule of ammonia (NH3) has only single bonds and, these are covalent bonds., Lone pair, H×, , N, , ×, , H, , or, , ×, , H, , H—N—H, |, H, , 6. (b) 8 covalent bonds are formed in S molecule., , S, , S, S, , S, , S, , S, S, S, , Crown shaped (S 8 ) molecule, , 7. (d) Graphite can not be used for making insulated plates, as, it is a good conductor of electricity., 8. (d) C, Si and Ge belongs to group 14 and their valency is 4., But oxygen has electronic configuration 2, 6. So, its valency, is 2. Hence, it does not show tetravalency., 9. (b) Friedrich Wohler accidently prepared urea from, ammonium cyanate and the synthesis of urea discarded the, vital force theory., , H H H H H, ⏐ ⏐ ⏐ ⏐ ⏐, H ⎯ C ⎯ C ⎯ C ⎯ C ⎯ C ⎯H, ⏐ ⏐ ⏐ ⏐ ⏐, H H H H H, It contains 16 covalent bonds., 11. (a) C3H8 is an alkane because it resembles with the general, formula of alkane, i.e. CnH2n + 2 . C 6H6 is benzene which is a, cyclic ring having double bonds in alternate carbon atoms, C 2H2 resembles with the general formula of alkyne and, C 4H8 is an alkene because number of H-atoms are double of, that of carbon atoms., 12. (a) ‘Eth’ represents 2 carbon atoms and ‘yne’ shows presence, of a triple bond. Thus, ethyne has the structural formula as, H ⎯ C ≡≡ C ⎯ H. It is also known as acetylene., 13. (c) Unsaturated hydrocarbons have double or triple bond in, their structure. Both (ii) and (iv) have double carbon-carbon, bonds in their structures., 14. (c) Benzene molecule contains alternate single and double, bonds. Its formula is C 6H6 ., In structure (a), double bonds are not at alternate positions., In structure (b), the formula is C 6H12 and in structure (d),, the formula is C 6H8 ., 15. (a) Structure (i) is n-butane., Structure (iii) is iso-butane., Since, molecular formula is same, only structures are, different. So, (i) and (iii) are isomers while structure (ii) and, (iv) have molecular formula C 4H8 ., 16. (a) Both A and R are true and R is the correct explanation of A., Covalent compounds consist of molecules and not ions, which can transfer charge., 17. (b) Both A and R are true but R is not the correct explanation, of A., Diamond is not good conductor of electricity because of the, absence of free electrons., 18. (a) Both A and R are true and R is the correct explanation of A., A graphite crystal consists of various layers of carbon atoms, in which each carbon atom is joined to three other atoms by, strong covalent bonds. The various layers of carbon atoms in, graphite are held together by weak van der Waals’ forces, making it slippery to touch., 19. (a) Both A and R are true and R is the correct explanation of A., Catenation is the bonding of atoms of the same element into, a series, called as chain., Catenation occurs more readily with carbon, which forms, strong covalent bond with other C-atoms to form long chains, and structures., 20. (a) Both A and R are true and R is the correct explanation of A., The alchohols have general formula of CnH2n + 1OH. So, the, alcohols have the series of formula from the different, compounds with different between the succeeding and, preceding molecules being a — CH2 — unit.

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11, , CBSE Term II Science X, , 21., , (i) (b) The structures of the following given compounds are:, I. SO2, II. N2, O == S == O, N ≡≡ N, III. HCl, IV. NH3, ••, , H ⎯ Cl, , H ⎯N ⎯H, ⏐, H, ∴ I, III and IV do not contain a triple bond. Hence,, option (b) is correct., (ii) (a) O2 contains a double bond between it’s atoms., The structures of the given compounds are :, (a) O 2 ⇒ O == O, (b) N 2 ⇒ N ≡≡ N, H, ⏐, (c) CH4 ⇒ H ⎯ C ⎯ H, ⏐, H, O, , (d) H2O ⇒, H, , H, , (iii) (c) In chlorine molecule, both chlorine atoms, contribute one electron and thus share single electron, pair to form single covalent bond. As electrons are, shared by both atoms, they acquire inert gas, configuration of argon atom in valence shell., , Cl, , Cl, , Sharing of, electrons, , Cl, , Cl, , or Cl Cl or Cl — Cl, , One shared electron pair, , (iv) (c) The number of electrons shared between two atoms, to complete their octet is known as the covalency of, that atom. Therefore, the covalency of nitrogen is, three because it needs three electrons to complete it’s, octet., (v) (a) The shared pair of electrons constitute a single bond, between the two H-atoms, which is represented by a, single line between two H-atoms., , H, , ×, ×, , H, , or, , H—H, , Single bond showing H2 molecule, , 22., , (i) (c) Each atom is covalently bonded to four other atoms,, which in turn, are bonded to four more atoms. Thus, X, is a giant molecule and has a structure similar to that of, diamond. Substance X is not a compound as it consists, of only one type of atoms., (ii) (a) In the structure of diamond, carbon atoms are held, together by single covalent bonds as this is a rigid, three-dimensional network structure because each, carbon atom is bonded to four other carbon atoms., (iii) (d) (I) is the structure of graphite crystal which, consists of layers of carbon atoms or sheets of carbon, atoms., , (II) is the structure of diamond and (III) is the structure, of Buckminster fullerene as their structure resembles, with geodesic domes., (iv) (b) In graphite, only three valence electrons are used, for bond formation and hence fourth electron is free to, move which makes it a good conductor of electricity., (v) (d) Coke is an example of amorphous form of carbon, which is obtained as a residue in destructive distillation, of coal., Subjective Questions, 1. Carbon shares it’s valence electrons with other atoms of, carbon or with atoms of other elements in order to complete, it’s octet. These shared electrons belong to the outermost, shells of both atoms and in this way, both atoms attain the, nearest noble gas configuration. This type of bonding is, called covalent bonding., 2. The bonds that are formed by sharing electrons are known, as covalent bond. In covalent bonding, both atoms share the, valence electrons, i.e. the shared electrons belong to the, valence shells of both the atoms. CH3Cl is called, chloromethane, which contains 1 carbon atom, 3 hydrogen, atoms and 1 chlorine atom., K L, Electronic configuration of carbon, 6 = 2 , 4, K, Electronic configuration of hydrogen, 1 = 1, K L M, Electronic configuration of chlorine,17= 2 , 8 , 7, Carbon atom has four outermost electrons, each hydrogen, atom has one electron and chlorine has seven outermost, electrons. Carbon shares its four outermost electrons with, 3 hydrogen atoms and 1 chlorine atom to form CH3Cl as, follows, H, 3H + C + Cl, , H, , C, , H, Cl or H—C—Cl, H, , H, , 3. Atomic number of C = 6, Electronic configuration = K , L, 2 4, Atomic number of O = 8, Electronic configuration = K , L, 2 6, To attain the stable electronic configuration, carbon needs, 4 electrons, while oxygen needs 2 electrons. So, in CO 2 , each, oxygen atom share two electrons from carbon. Thus, oxygen, and carbon both complete their octet., , Before, combination, , O, , C, , O, , After, combination Shared electrons, , O, , C, , O, , or, O ==C == O, Carbon dioxide (CO2 ) molecule

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12, , CBSE Term II Science X, , 4. (i) The formation of calcium chloride with the help of, electron dot structure., Element, Atomic, Electronic, number, configuration, Calcium (Ca), , 20, , 2, 8, 8, 2, , Chlorine (Cl), , 17, , 2, 8, 7, , Cl, Ca, , Ca2+, , +, , Cl, , –, , 2, , Cl, , Ca, , +2Cl − ⎯→ CaCl 2, , Two valence electrons of calcium attack the valency of, two chlorine to attain the noble gas configuration., (ii) Ionic compounds do not conduct electricity in solid state, but conduct electricity in molten and aqueous state, because in solid state, there is no free ion to move and, pass electricity. Whereas in the molten and aqueous, state, there is free ions to move and pass electricity., K L, (i) Electronic configuration of carbon, C(6) is 2 , 4., K L M, Electronic configuration of chlorine, Cl (17 ) is 2 , 8 , 7 ., To attain the electronic configuration of the nearest noble, gas, carbon needs 4 electron and chlorine needs 1 electron., So, with chlorine, carbon forms carbon tetrachloride., Electron dot structure and structural formula of CCl 4 is as, follows, , Cl, , Cl, , —, , Cl, , C, , Cl, , or Cl — C — Cl, , —, Cl, , Cl, , Carbon tetrachloride (CCl 4 ), , K L, (ii) Electronic configuration of oxygen, O(8) is 2 , 6 ., With oxygen, carbon forms carbon dioxide. To attain the, electronic configuration of the nearest noble gas, carbon, needs 4 electrons and oxygen needs 2 electrons., Therefore, in CO 2 , each oxygen atom shares 2 electrons, with carbon. Electronic configuration of carbon (6) is, K L., 2, 4, The electron dot structure and structural formula of CO2, is as follows:, , O, , C, , O, , or O, , — —, , 5., , 2+, , Carbon dioxide (CO 2 ), , C, , O, , 6. Covalent compounds have low melting and boiling points, due to small intermolecular forces of attraction between the, atoms., 7. (i) Valency of each carbon atom in an alkane is four., (ii) Valency of each carbon atom in an alkyne is four., 8. Covalent compounds are poor conductors of electricity, because covalent bonds are formed by sharing of electrons, between atoms., So, they don’t have a free electron that is required for, electricity transfer (electricity is the flow of free electrons)., Thus, they are bad conductors., 9. In the structure of diamond, all the four valence electrons of, carbon are involved in the formation of covalent bonds., Thus, no free electrons are available., Whereas, in the structure of graphite, three electrons in the, valence shell of carbon are involved in covalent bond, formation and the fourth electron is free to move., Therefore, graphite is a good conductor of electricity., 10. Diamond has a giant structure that consists of carbon atoms, in which each carbon atom is bonded to four other carbon, atoms forming a rigid three-dimensional network structure,, which is responsible for it’s hardness., So, a lot of energy is required to break the network, of strong covalent bonds. That’s why it has high melting, point., 11. The main factors that enables carbon to form large number, of compounds are, (i) Catenation The tendency of carbon to form chains of, identical atoms is known as catenation. Carbon forms, long chains by combining with other carbon atoms, through covalent bonds., (ii) Tetravalency It has 4 valence electrons, so it can form, 4 covalent bonds with four different atoms, or two, double bonds or a single and a triple bond with other, atoms. This tendency helps carbon to form a large, number of compounds., (iii) Tetravalency Carbon forms strong bonds with most of, other elements like H, O, N, S, Cl etc., due to it’s, small size which helps it to attract more number of, electrons., 12. The hydrocarbons in which all the carbon atoms are, connected by only single bonds are called saturated, hydrocarbons or alkanes or paraffins. The general formula of, these compounds is CnH2n + 2 ., where, n = number of carbon atoms in one molecule of a, hydrocarbon., Amongst, the given compounds, only C 4H10 and C 6H14, belongs to the formula of CnH2n + 2 .Therefore, C 4H10 and, C 6H14 are saturated hydrocarbons., 13. General formula of cycloalkane = CnH2n, In cyclopentane, n = 5, ∴ Formula of cyclopentane, C5H5 × 2 = C5H10

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13, , CBSE Term II Science X, , Electron dot structure of cyclopentane, H, H, H, , ×, , ×, , ×, C, , × C, , H, , H H, , H, ×, C ×, H, , H× C, ×, H, , or, , C ×H, ×, H, , H, H, , C, C, C, H H, , C, C, , H, H, H, H, , H, , The structures of possible isomers of butane ( C4H10 ) are, H H H, H H H H, ⏐ ⏐ ⏐ ⏐, ⏐ ⏐ ⏐, H ⎯ C ⎯ C ⎯ C ⎯ C ⎯H H ⎯ C ⎯ C ⎯ C ⎯H, ⏐ ⏐ ⏐ ⏐, ⏐ ⏐ ⏐, H ⏐ H, H H H H, n- butane, H ⎯C ⎯ H, (I), ⏐, H, Iso -butane, , (i) Carbon is a versatile element because of its properties., It shows the property of catenation due to which it, forms a large number of compounds. Carbon is, tetravalent. Due to this, it forms covalent compounds, only., (ii) Saturated hydrocarbons contain carbon-carbon single, bonds. e.g. Methane (CH4 ), ethane (C 2H6 )., Unsaturated hydrocarbons contain atleast one, carbon-carbon double or triple bond. e.g. Propene, (C 3H6 ), butyne (C 4H6 )., (iii) Functional group is an atom or group of atoms joined in, a specific manner which is responsible for the, characteristics chemical properties of the organic, compounds., Examples are alcohols ( ⎯OH) , aldehyde group, ( ⎯ CHO), carboxylic group ( ⎯COOH) , ketone ( ⎯CO), etc., 15. (i) Molecular formula of ethane is C 2 H6. Its electron dot, structure is, , (II), , 14., , H, , H, , H, , C, , C, , H, , H, , The first three members of alkane series are:, (i) CH4 (methane), (ii) C 2H6 (ethane), (iii) C 3H8 (propane), In the above members of alkane series, it is not possible to, have different arrangements of carbon atoms, because, branching is not possible from either first or last carbon., Thus, we cannot have isomers of first three members of, alkane series., 17. There are four isomers possible for the molecular formula, C 3H6O. These are as follows:, H, H H H, ⏐, , ••, , ••, , (ii) CH3 ⎯ C ⎯ CH3 → H •• C •• C, ••, ••, ⏐⏐, H • ••, O, •O, , •, •, , ••, ••, , H, , H, , H, , H, , ••, , ••, , H, , H, (iv) CH2 == CH ⎯CH2 ⎯OH →, , H ••, , C, , ••, , C, , C, , ••, ••, , ••, , ••, , ••, , ••, , C C •• O •• H, ••, , •, •, , H H, , 18. An atom or a group of atoms present in a molecule which, largely determines it’s chemical properties is called functional, group., Compound, , (iii) Molecular formula of ethyne is C 2H2. Its electron dot, structure is, , H, , ••, , ••, , H H, , H, , H, , ••, , (iii) CH3 ⎯ CH == CH ⎯ OH → H •• C •• C •• •• C •• O •• H, , H, C, , ••, , C •• H, , ••, , C, , ••, , ••, , (ii) Molecular formula of ethene is C 2H4. Its electron dot, structure is, , H, , ••, , (i) CH3CH2CHO or CH3CH2 — C → H •• C •• C •• C •• •• O ••, •• ••, ⏐⏐, H, H, O, H, H, , (i) Ethanol, (C 2H5OH), , H H, ⏐ ⏐, H ⎯ C ⎯ C ⎯OH, ⏐ ⏐, H H, , (ii) Ethanoic acid, (CH3COOH), , H O, ⏐ ⏐⏐, H ⎯ C ⎯ C ⎯ OH, ⏐, H, , H, , 16. Isomers are those molecules which have same molecular, formula but different structural formula, i.e. show different, properties., , Structural formula, , Functional group, ⎯OH, ( Alcoholic ), , O, ⏐⏐, ⎯ C ⎯ OH, (Carboxylic acid)

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14, , CBSE Term II Science X, , Compounds, CH 2 O, , O, , 22. Atomic number of carbon is six. This means that it has four, electrons in its outermost shell and it needs four more, electrons to attain noble gas electronic configuration. It, does not form C 4+ cation, as the removal of four valence, electrons will require a huge amount of energy., The cation formed will have six protons and two electrons., This makes it highly unstable. Carbon is unable to form, C 4 − anion as its nucleus with six protons will not be able to, hold ten electrons due to its small size. Thus, carbon, achieves noble gas electronic configuration by sharing its, four electrons either with same or different other atoms,, i.e. it forms covalent compounds., (i) Covalent compounds does not have free ions, due to, this they are bad conductors of electricity in solid,, molten or aqueous state., (ii) Covalent compounds are formed by covalent bonds and, it has been found that the intermolecular force of, attraction in covalent compounds are weak., Thus, low amount of energy is required to break these, force of attraction. Hence, their melting and boiling, points are quite low., 23. Formation of Oxygen Molecule (O 2 ), The atomic number of oxygen is 8 and electronic, configuration is 2, 6, i.e. has 6 electrons in it’s outermost, , or O ⎯, ⎯O, , Oxygen molecule, , The atomic number of sulphur is 16 and electronic, configuration is 2, 8, 6. It also has 6 electrons in it’s outermost, shell and requires 2 electrons to complete it’s octet state., So, each sulphur atom shares two electrons, 1 with each, adjoining sulphur atom by single covalent bonds and thus,, complete it’s octet., , Before, combination, + S +, , S, , S, +, , After, combination, S, +, , S, , S, +, , S, +, S +, , S, , S, , +, , S, , C 3H7COOH, C 4H9 COOH,, C5H11COOH..., , O, , S, , C 2H5COOH, , O, , Formation of Sulphur Molecule (S8 ), , C 2H4O, C 3H6O, C 4H8O..., C 3H4 , C 4H6 , C5H8 ..., , O, , Oxygen atoms, , Homologue compounds, , C 2H2, , +, , S, , The physical properties are determined by alkyl group/, hydrocarbon part/part other than the functional group., The chemical properties are determined by functional, group such as ⎯ OH group., 21. Add —CH 2 group to each compound to obtain next, homologous., , Shared electrons, , S, , !, , shell. So, it requires 2 electrons to complete it’s octet for, attaining noble gas configuration. Hence, it shares two, electrons with another atom of oxygen to make a molecule of, oxygen., By doing so, both the atoms of oxygen get 8 electrons in their, outermost shell. Thus, a double bond is formed between two, oxygen atoms which consists of four electrons., Before, After, combination, combination, , S, , !, , S, , 19. Application of homologous series are as follows, All members of homologous series shows similar chemical, properties and generally prepared through one common, method, e.g. all alkenes are prepared by dehydration of, corresponding alcohols., The physical properties of the members change gradually,, i.e. show gradation in properties as the number of carbon, atom per molecule increased., 20. A homologous series is the family of organic compounds, having the same functional group, similar chemical, properties but the successive (adjacent) members of the, series are differ by a CH2 unit or 14 mass units., Consecutive members of the homologous series of alcohols, are, CH3OH ⎤, They differ by ⎯ CH2 unit., C2H5OH⎥⎦, , 8 covalent bonds are formed, , S, , S, S, , S, , S, , S, S, S, , Crown shaped (S 8 ) molecule, , Eight sulphur atoms form a puckered ring or crown structure, to form an eight atom molecule., 24. Covalent 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., They 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 sodiu m gets completely transferred to outer shell, of chlorine atom. The characteristic properties of covalent, compounds are, (i) They are generally insoluble or less soluble in water but, soluble in organic solvents., (ii) They have low melting and boiling points.

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15, , CBSE Term II Science X, , (iii) They do not conduct electricity as they do not contain, ions., (iv) They are volatile in nature., 25. Organic compounds are called as hydrocarbons because, they are made up of only the elements-carbon (C) and, hydrogen (H)., General formula for the homologous series of alkanes is, C n H2n + 2 which are classified as saturated hydrocarbons or, alkanes. First member of the alkane family is “methane”., H, ⏐, H ⎯ C ⎯H, ⏐, H, General formula for the homologous series of alkenes is, Cn H2n . which are classified in the category of unsaturated, hydrocarbons. They are known as alkenes or olefins. First, member of the alkene family is “ethene”., H", !H, C ==C ", H!, H, General formula for the homologous series of alkynes is, C n H2n − 2 and they are also in the category of unsaturated, hydrocarbons. First member of the alkyne family is “ethyne”., H ⎯ C ≡≡ C ⎯ H, 26. The functional group present in the following compounds are, H O, ⏐ ⏐⏐, (i) H ⎯ C ⎯ C ⎯ OH, ⏐, H, , O, ⏐⏐, This compound contains ⎯ C ⎯ OH functional group, which is the formula of carboxylic acid., , H, H, H, ⏐, ⏐, ⏐, (ii) H ⎯ C ⎯ C ⎯ C ⎯ H, ⏐, ⏐, ⏐, H OH H, In this compound, ⎯OH functional group is present, which is the formula of alcohol., H, ⏐, (iii) H ⎯ C ⎯ CHO, ⏐, H, This compound contains ⎯ CHO functional group, which belongs to the formula of aldehyde., H H, ⏐ ⏐, (iv) H ⎯ C ⎯ C ⎯ Br, ⏐ ⏐, H H, This compound contains halo (bromo) functional group, as ⎯Br., , H H O H, ⏐ ⏐ ⏐⏐ ⏐, (v) H ⎯ C ⎯ C ⎯ C ⎯ C ⎯ H, ⏐ ⏐, ⏐, H H, H, , 27., , O, ⏐⏐, This compound contains ⎯ C ⎯ functional group, which belongs to ketone., (i) Compound B has molecular formula as C5H 10 and, contains five number of carbon atoms, i.e. n = 5., It resembles with the general formula of alkene which, is Cn H2n ., So, the name of this compound is pent + ene = pentene., When five number of carbon atoms are present, it is, named as “pent”., , (ii) Saturated, hydrocarbons, , Unsaturated hydrocarbons, , These hydrocarbons, are linked by only, single covalent, bond., , These hydrocarbons contains at, least one double or triple bond, along with single bonds., They are divided into two, categories, → Alkenes or Olefins, → Alkynes., , General formula of, these compounds is, Cn H2n + 2 ., , General formula of alkene is, Cn H2n and general formula of, alkyne is Cn H2n − 2 ., , (iii) The minimum number of carbon atoms present in an, unsaturated compound is two because formation of, double or triple bonds is possible only between two, carbon atoms., (iv) The molecular formula of compound A is C3H8 , i.e., contains three number of carbon atoms and resembles, with the general formula of alkanes which is Cn H2n + 2 ., So, A is saturated hydrocarbon., While the molecular formula of compounds B and C is, C5H10 and C4H6 which resembles with the general, formula of alkene and alkyne. So, B and C are, unsaturated hydrocarbon., (v) The molecular formula of compound C is C 4H6 which, resembles with alkyne because there is four number of, carbon atoms and 6H-atoms, i.e. number of H-atoms, are only increased by 2. So, the general formula of, alkyne is CnH2n − 2 ., 28. (i) Butane and iso-butane are the compounds that, have same molecular formula but different structural, formula., CH3CH2CH2CH3, CH3 CH C H3, Butane, ⏐, CH3, Iso - butane

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16, , CBSE Term II Science X, , (v) Minimum four carbon atoms are required to show, isomerism because branching is not possible with, carbon-1, 2 and 3., i.e., C —C ( 3 -carbon atoms), C—C—C, |, C 3, 14442444, , (ii) Due to catenation property of carbon, it forms long,, straight or branched chains and rings of different sizes., (iii) Pentane (C5H12 ) has three structural isomers:, CH3 ⎯ CH2 ⎯ CH2 ⎯ CH2 ⎯ CH3, n-pentane, , CH3 ⎯ CH ⎯ CH2 ⎯ CH3, ⏐, CH3, , Same, , C—C —C ( 4 -carbon atoms), |, C 3, 144444244444, , C—C—C—C, , Iso -pentane, , CH3, ⏐, CH3 ⎯ C ⎯ CH3, ⏐, CH3, , Isomers, , 29., , Neo -pentane, , (iv) Cyclohexane (C 6H14 ) and benzene (C 6H6 ) are two, compounds that contain six carbon atoms and have, cyclic structure., , H, , H H, H, H, H, H, , C, C, , C, , C, , C, C, , H H, C6H12, (Cyclohexane), , H, H, H, H, , H, C, C, H, , C, , C, , H, C, C, , H, C6H6, (Benzene), , H, , (i) P and T are the compounds that belongs to same, homologous series. Both these compounds are alkynes, and differ by ⎯CH2 unit in their molecular formula., (ii) The functional group of compounds (R) is ⎯OH which, is the formula of alcohol., (iii) (T) is an alkyne having general formula of Cn H2n − 2 ., H H, ⏐ ⏐, (iv) Compound (S), i.e. H ⎯ C ⎯ C ⎯ H belongs to an alkane, ⏐ ⏐, H H, series having general formula of Cn H2n + 2 ., H, H, (v) Compound (U), i.e., is unsaturated, C == C, H, H, hydrocarbon becuase it contain double bond, i.e., belongs to alkene.

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Chapter Test, , 8. Assertion n-butane and iso-butane are examples of, structural isomers., Reason Isomerism is possible only with, hydrocarbons having 4 or more carbon atoms., , Multiple Choice Questions, , 1. Which of the following is not a property of carbon?, (a) Carbon compounds are good conductor of heat and, electricity, (b) Carbon compounds are poor conductor of heat and electricity, (c) Most of the carbon compounds are covalent compounds, (d) Boiling and melting point of carbon compounds are relatively, lower than those of ionic compounds, , 2. Which of the following is purest form of carbon?, (a) Charcoal, , (b) Coal, , (c) Diamond, , Short Answer Type Questions, , 9. What are the two properties of carbon which lead, to the huge number of carbon compounds., , 10. What is covalent bond? What type of bond exists in, (i) CCl4, , electricity.” Give reason to justify this statement., , 12. Give answers to the following statements., , (d) Graphite, , (i) An allotrope of carbon which has a, two dimensional layered structure consisting, of fused benzene rings., (ii) An allotrope of carbon which looks like a, soccer ball., (iii) An allotrope of carbon which contains both, single and double bonds., , 3. Buckminster fullerene is an allotropic form of, (a) phosphorus (b) sulphur, , (c) carbon, , (d) tin, , 4. Which of the following is not a straight chain hydrocarbon?, (a) H 3C ⎯CH 2 ⎯CH 2 ⎯CH 2 ⎯C H 2, ⏐, CH3, (b) H 3C ⎯CH 2 ⎯CH 2 ⎯CH 2 ⎯CH 2 ⎯CH 3, CH 3, ⏐, (c) H 2 C ⎯H 2C ⎯H 2C ⎯CH 2, ⏐, CH3, H 3C, ", (d), CH ⎯CH 2 ⎯CH 2 ⎯CH 3, !, H 3C, , 13. Diamond and graphite show different physical, properties although they are made up of carbon., Name this relationship between diamond, and graphite. Give the basis of this relationship also., , 14., 15., , 5. Which of the following is not the property of homologous, series?, (a) They differ by ⎯ CH 2 units, (b) They differ by − 14 units by mass, (c) They all contain double bond, (d) They can be represented by a general formula, , (a) Both A and R are true and R is the correct explanation of A., (b) Both A and R are true, but R is not the correct explanation of A., (c) A is true, but R is false., (d) A is false, but R is true., , 6. Assertion Carbon has a tendency to form multiple bonds., Reason Carbon has small size., , 7. Assertion The only element that can form large number, of compounds is carbon., Reason Carbon is tetravalent in nature and shows the, property of catenation., , How diamond can be prepared artificially?, 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., , 16. A compound has the formula H2Y, (Y=non-metal). State the following :, (i) The outer electronic configuration of Y., (ii) The valency of Y., (iii) The bonding present in H2Y., , Assertion-Reasoning MCQs, Direction (Q. Nos. 6-8) Each of these questions contains, two statements Assertion (A) and Reason (R). Each of, these questions also has four alternative choices, any one, of which is the correct answer. You have to select one of, the codes (a), (b), (c) and (d) given below., , (ii) CaCl2, , 11. “Carbon tetrachloride is not a good conductor of, , Long Answer Type Questions, , 17., , (i) What are covalent bonds ?, (ii) How many covalent bonds are present in, ethane with molecular formula C 2H6 ?, (iii) Write the formula and draw electron dot, structure of carbon tetrachloride., (iv) In electron dot structure, the valence shell, electrons are represented by crosses or dots., (a) The atomic number of chlorine is 17. Write, its electronic configuration., (b) Draw the electron dot structure of chlorine, molecule., , 18. Write the structural formulae of all the isomers of, an alkane with six C-atoms (C 6H14 ) ., , 19. State five characteristics of a homologous series., , Answers, Multiple Choice Questions, 1. (a), , 2. (c), , 3. (c), , 4. (d), , Assertion-Reasoning MCQs, 6. (a), , 7. (d), , 8. (b), , 5. (c), , For Detailed Solutions, Scan the code

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CHAPTER 02, , Periodic Classification, of Elements, In this Chapter..., !, , Earlier Attempts at the Classification of Elements, , !, , Mendeleev’s Periodic Table, , !, , Modern Periodic Table, , All substances are made up of elements. At present, there are, 118 elements known, out of which 98 are naturally occurring., In order to study the properties of all these elements, separately, scientists felt the necessity to group elements, having similar characteristics together., , elements in the same slot and also put some unlike elements, under the same column which have very different properties, than other elements., , Earlier Attempts at the, Classification of Elements, , According to this, the physical and chemical properties of the, elements are periodic function of their atomic masses, i.e. on, arranging the elements in increasing order of their atomic, masses, the similar properties were repeated after regular, intervals., , Several attempts have been made to classify the elements, according to their properties. Later, many classifications were, tried. Some important of them are discussed below, , Mendeleev’s Periodic Table, , He arranged three elements with similar properties into, groups which are known as triads and showed that when, three elements in a triad were arranged in order of increasing, atomic masses, the atomic mass of middle element was, roughly the average of atomic masses of other two elements., He could identify only three triads from the elements known, at that time which are, Li, Na, K; Ca, Sr, Ba; Cl, Br, I, , He took the formulae of the hydrides and oxides formed by, an element as one of the basic properties of an element for its, classification. e.g. Hydride of carbon, CH 4 as RH 4 and its, oxides, CO 2 as RO 2 ., He then arranged 63 elements in the increasing order of, their atomic masses and found that there was a periodic, recurrence of elements with similar physical and chemical, properties. He observed that elements with similar properties, fall in the same vertical column. These vertical column are, called groups and horizontal rows of elements are called, periods., , Newland’s Law of Octaves, , Features of Mendeleev’s Periodic Table, , Dobereiner’s Triads, , Newland arranged the known elements in order of increasing, atomic masses and found that every eighth element had, properties similar to that of the first. This law was applicable, only upto calcium and he assumed that there were only 56, elements. To fit elements into his table, he adjusted two, , !, , !, , It consists of 8 vertical columns, called groups and 6 horizontal, rows, called periods., In every period, elements are arranged in increasing order of, their atomic masses.

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19, , CBSE Term II Science X, , He left gaps for the elements not discovered at that time and, named such elements by prefixing a Sanskrit numeral, Eka (one), divi (two) to the name of the preceding similar, element in the same group. e.g. Eka-boron, Eka-aluminium,, which after their discovery were named as scandium, gallium., He also predicted the atomic masses and properties of several, elements that were not known at that time., , !, , !, , !, , Limitations of Mendeleev’s Periodic Table, !, , Properties of Eka-aluminium and Gallium, !, , Eka-aluminium, , Property, , Gallium, !, , Atomic mass, , One of the strengths of Mendeleev’s periodic table was that,, when noble gases like helium, neon were discovered, they, could be placed in a new group without disturbing the existing, order., , 68, , 69.7, , Formula of oxide, , E 2O3, , Ga 2O3, , !, , Formula of chloride, , ECl 3, , GaCl 3, , !, , Elements with dissimilar properties were kept in same, group., Position of hydrogen was not fixed in periodic table., Elements with similar properties were kept in different, groups., Heavier elements were kept before the lighter elements., Position of isotopes and isobars could not be explained., , Mendeleev’s Periodic Table (Published in a German journal in 1872), In the formula of oxides and hydrides at the top of the columns, the letter ‘R’ is used to represent any of the elements in the, group., Group →, , I, , II, , III, , IV, , V, , VI, , VII, , VIII, , Oxide, Hydride, , R2O, RH, , RO, RH 2, , R 2O3, RH3, , RO2, , R 2O5, RH 3, , RO 3, , R 2O7, RH, , RO4, , RH 4, , RH 2, , Periods ↓, , A, , 1, , H, 1.008, , 2, , Li, 6.939, , Be, 9.012, , B, 10.81, , C, 12.011, , N, 14.007, , O, 15.999, , F, 18.998, , 3, , Na, 22.99, , Mg, 24.31, , Al, 29.98, , Si, 28.09, , P, 30.974, , S, 32.06, , Cl, 35.453, , 4 First series, , K, 39.102, , Ca, 40.08, , Second series, 5 First series, , Second series, , B A, , Rb, 85.47, , B A, , Zn Ga, 65.37 69.72, , Cs, 132.90, , Ba, 137.34, Au, 196.97, , La, 138.91, , Hg Tl, 200.59 204.37, , V, 50.94, As, 74.92, , Y, 88.91, Cd In, 112.40 114.82, , B A, , Ti, 47.90, Ge, 72.59, , Sr, 87.62, Ag, 107.87, , B A, , Sc, 44.96, , Cu, 63.54, , Second series, 6 First series, , B A, , Pb, 207.19, , Modern Periodic Table, In 1913, Henry Moseley showed that the atomic number of, an element is a more fundamental property. On the basis of, this, he modified Mendeleev’s periodic law as “physical and, chemical properties of the elements are a periodic function of, their atomic number”. This is called modern periodic law., , Nb, 92.91, Sb, 121.75, , Hf, 178.49, , Cr, 50.20, Se, 78.96, , Zr, 91.22, Sn, 118.69, , B A, , Ta, 180.95, , B, , Transition series, , Mn Fe, Co, Ni, 54.94 55.85 58.93 58.71, Br, 79.909, , Mo, 95.94, Te, 127.60, , Tc Ru, Rh, Pd, 99 101.07 102.91 106.4, I, 126.90, , W, 183.85, , Os, Ir, Pt, 190.2 192.2 195.09, , Bi, 208.98, , When the elements were arranged in the increasing order of, their atomic number, the obtained table is called modern, periodic table., In this periodic table, hydrogen is kept at the top left corner, because of its unique characteristics. The position of cobalt, and nickel is also justified.

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21, , CBSE Term II Science X, , Features of Modern Periodic Table, This table has 18 vertical columns, known as groups and, 7 horizontal rows, known as periods., A few important features of the elements present in groups, and periods are as follows, !, , !, , !, , !, , !, , !, , !, , !, , The groups are not divided into sub-groups., The elements present in a group have the same number of, valence electrons and valency., The number of shells increases as we go down the group., The elements present in a group have identical chemical, properties and their physical properties like density, melting, point vary gradually., Elements of a period have the same number of shells but they, do not contain the same number of valence electrons. So, their, chemical properties are also different., The number of valence shell electrons increases by one unit as, the atomic number increases by one unit on moving from left, to right in a period., In this table, elements of group 13, 14, 15, 16 and 17 are called, normal elements which includes metals, non-metals and, metalloids and elements of group 3, 4, 5, 6, 7, 8, 9, 10, 11 and, 12 are called transition elements., In this periodic table, elements from atomic number 58 to 71, called as lanthanides and elements from 91 to 103 called as, actinoids are kept out of the table., , Position of Elements in the Modern, Periodic Table, For this, first of all write electronic configuration of the given, element. Number of shells present in the electronic, configuration shows the period number of that element., Number of valence electrons present in the electronic, configuration show the group number of the element., e.g. Electronic configuration of the element with atomic, number 19 is 2, 8, 8, 1, since it has four shells, thus it is, element of fourth period. Due to presence of one electron in, the last shell, its group number is 1. After donating one, , electron, it acquires a stable configuration, hence its valency, is also 1., , Trends in Modern Periodic Table, (i) Valency In a period, it increases with respect to, hydrogen from 1 to 4 after that it decreases. On the other, hand with respect to oxygen, valency increases from 1, to 7. In a group, valency remains same as outer, electronic configuration is same., (ii) Atomic size Atomic size decreases on moving from left, to right in a period due to increase in nuclear charge., It increases down the group as new shells are being, added., (iii) Metallic and non-metallic properties Effective nuclear, charge acting on the valence shell electrons increases, across a period and decreases down the group., Therefore, metallic character decreases across a period, and increases down a group. Non-metallic character,, however increases across a period and decreases down a, group., Metals like Na, Mg are present on left side of periodic, table, whereas non-metals like S, Cl are present on right, side of periodic table., There are some metals which exhibits both the, properties of metal and non-metals. These are called, metalloids like Po, Te, Sb, etc., (iv) Electronegativity The electronegativity of the elements, increases along a period, since the non-metallic, character increases. Similarly, it decreases down the, group, since the non-metallic character decreases., (v) Nature of oxides On moving from left to right in a, period, due to increase in non-metallic character,, basic nature of oxides decreases while acidic nature, increases., Na 2O MgO , Al 2O 3 , SiO 2 , P2O 5 SO 2 , Cl 2O 7, 144244, 3 Amphoteric Weakly 14243 Strongly, Strongly basic, , acidic, , Acidic, , On going down the group, the order is reversed., , acidic

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Chapter, Practice, PART 1, Objective Questions, !, , Multiple Choice Questions, , 1. Dobereiner could identify how many triads from the, elements known at that time ?, (a) One, , (b) Two, , 9. Which of the following statements is not a correct, statement about the trends when going from left to, (NCERT), right across the periods of periodic table?, (a) The elements become less metallic in nature, (b) The number of valence electrons increases, (c) The atoms lose their electrons more easily, (d) The oxides become more acidic, , 10. Match the following columns :, , (c) Three, , (d) Four, , Column I, (Elements), , 2. Upto which element, the law of octaves was found to, be applicable?, (a) Oxygen, , (NCERT Exemplar), , (b) Calcium, , (c) Cobalt, , (d) Potassium, , 3. What type of oxide would Eka-aluminium form?, (NCERT Exemplar), , (a) EO3, , (b) E 3 O2, , (c) E 2 O3, , (d) EO, , 4. Which one of the following elements exhibit maximum, number of valence electrons?, (a) Na, , (b) Al, , (NCERT Exemplar), , (c) Si, , (d) P, , 5. Choose the correct statement., (a) Valency and valence electrons are always equal, (b) Valency is determined by using the number of valence, electrons, (c) Valence electrons are always equal to 8-valency, (d) Valency is always equal to 8-valence electrons, , 6. Elements with code letters Q and R occupy the, positions shown in the outline of the periodic table., , Column II, (Groups), , A. Be, , (i) Group 17, , B. F, , (ii) Group 15, , C. P, , (iii) Group 2, , D. Ar, , (iv) Group 18, , Codes, A B C D, (a) (iii) (i) (ii) (iv), (c) (ii) (iv) (iii) (i), , A B C D, (b) (i) (ii) (iii) (iv), (d) (iv) (ii) (i) (iii), , 11. Which metal is among the first ten elements in, modern periodic table?, (a) Lithium (Li), (c) Carbon (C), , (b) Boron (B), (d) Potassium (K), , 12. An element X is forming an acidic oxide. Its position, in modern periodic table will be, (a) group 1 and period 3, (c) group 13 and period 3, , (CBSE 2020), , (b) group 2 and period 3, (d) group 16 and period 3, , 13. An oxide of a metal shows no action with blue-litmus, but turns red-litmus to blue in its aqueous solution., The possible formula of the metal oxide can be, , R, Q, , (a) MgO, , What is the formula of the compound formed, between them?, (a) QR2, , (b) Q2R, , (c) Q2R3, , (d) Q3R2, , (b) CO2, , (c) SO2, , (d) H2O, , 14. The position of four elements A, B, C and D in the, modern periodic table are shown below. The, element which is most likely to form a basic oxide is, , 7. Which of the following statements is incorrect for, atomic size ?, (a) Atomic size of B > Be, (c) Atomic size of N > O, , (b) Atomic size of Be > B, (d) Atomic size of C > N, , B, , C, D, , 8. Which of the following gives the correct increasing, , A, , order of the atomic radii of O, F and N?, (NCERT Exemplar), , (a) O, F, N, , (b) N, F, O, , (c) O, N, F, , (d) F, O, N, , (a) B, , (b) C, , (c) A, , (d) D

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23, , CBSE Term II Science X, , h, , k, , Modern periodic law states that ‘‘the physical and, chemical properties of all the elements are the, periodic function of their atomic radius., The atomic radii of group 1 and group 2 elements, are given in the table., , i, , l, , Group 1 Atomic radii (nm) Group 2 Atomic radii (nm), , 15. Consider the part of periodic table given below:, Group/Period 1, I, , a, , II, , b, , III, , c, , IV, , d, , 2 13, , 14, , 15, , 16 17, , 18, j, , e, , g, f, , Li, , 0.152, , Be, , 0.112, , Na, , 0.186, , Mg, , 0.160, , k, , 0.231, , Ca, , 0.197, , Assertion-Reasoning MCQs, , Rb, , 0.244, , Sr, , 0.215, , Direction (Q. Nos. 16-20) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative choices,, any one of which is the correct answer. You have to select, one of the codes (a), (b), (c) and (d) given below., , Cs, , 0.262, , Ba, , 0.217, , Fr, , 0.270, , Ra, , 0.220, , The most electropositive element is, (a) c, !, , (b) d, , (c) l, , (d) k, , (a) Both A and R are true and R is the correct explanation, of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true., , 16. Assertion Be, Mg and Ca can be classified as, , 17., , 18., 19., , 20., , Dobereiner’s triads., Reason Atomic mass of Mg is approximately the, average of the sum of atomic masses of Be and Ca., Assertion Sodium is an element of group 1 of the, modern periodic table., Reason All the elements of group 1 of the modern, periodic table contain two electrons in their, outermost shell., Assertion Noble gases are also called inert gases., Reason Noble gases have a complete octet., Assertion Atomic size increases on going down the, group in the modern periodic table., Reason New shells are being added as we go down, the group in the modern periodic table., Assertion Electronegativity of flourine is greater than, that of oxygen., Reason The electronegativity of the elements increases, along a period since the metallic character increases., , Case Based MCQs, 21. Read the following and answer the questions from (i), , !, , to (v) given below, The modern periodic table is based on the increasing, order of atomic radius. The atomic size increases, down the group. This is because new shell are being, added as we go down the group. The atomic radius, decreases in moving from left to right along a period., , (i) Who discovered modern periodic table?, (a) Henry Moseley, (b) Johann Wolfgang Dobereiner, (c) John Newlands, (d) Dmitri Ivanovich Mendeleev, , (ii) Along the period 3, which atom can loose electron, easily?, (a) Li, , (b) Na, , (c) Be, , (d) Mg, , (iii) Which metal has the highest metallic character in, group 1?, (a) Li, , (b) Na, , (c) K, , (d) Rb, , (iv) Which of the following properties increases down, the group?, (a) Electronegativity, (b) Electropositive nature of element, (c) Non-metallic character, (d) Acidic oxide character, , (v) Which atom is most electronegative along the, period 4?, (a) K, , (b) Rb, , (c) Sr, , (d) Ca, , 22. Read the following and answer the questions from, (i) to (v) given below, The noble gases make up a class of chemical, elements with similar properties under standard, conditions, they are all odourless, colourless,, monoatomic gases with very low chemical reactivity., These gases are also called as inert gases or, aerogens. The properties of the noble gases can be, well explained by modern theories of atomic, structure. Their outer shell of valence electrons is, considered to be full, giving them little tendency to, participate in chemical reactions., The linkage formed from one electrostatic attraction, between oppositely charged ions in a chemical, compound is called ionic or electrocovalent bond., Such a bond forms when the valence electrons of, one atom are transferred permanently to another, atom.

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24, , CBSE Term II Science X, , The table given below refers to the elements of the, periodic table with atomic number from 3 to 18., These elements are shown by letters., (Not by the usual symbols of the elements)., 3, , 4, , 5, , 6, , 7, , 8, , 9, , 10, , A, , B, , C, , D, , E, , F, , G, , H, , 11, , 12, , 13, , 14, , 15, , 16, , 17, , 18, , I, , J, , K, , L, , M, , N, , O, , P, , (i) Which of the following are noble gases ?, (a) H and P, (c) D and L, , (b) C and M, (d) E and P, , (iii) Which of the following elements have valency 4 ?, (a) F and N, (c) D and L, , (b) C and K, (d) H and P, , (iv) Which of the following can form ionic bond?, (a) B and H, (c) E and P, , (b) J and P, (d) A and G, , (d) C and K, , PART 2, Subjective Questions, !, , (NCERT), , 7. In Mendeleev’s periodic table, the elements were, arranged in the increasing order of their atomic, masses. However, cobalt with atomic mass of, 58.93 amu was placed before nickel having an, atomic mass of 58.71 amu. Give reason for the same., (NCERT Exemplar), , 8. Write the formulae of chlorides of Eka-silicon, , 10. How it can be proved that the basic structure of the, ‘modern periodic table’ is based on the electronic, configuration of atoms of different elements?, (CBSE 2019), , 11. ‘Hydrogen occupies a unique position in modern, periodic table’. Justify the statement.(NCERT Exemplar), , (v) Which of the following elements have 2 valence, electrons?, (a) D and L (b) B and J (c) A and I, , creating his periodic table?, , and Eka-aluminium, the elements predicted by, Mendeleev., 9. Write two main characteristics of Mendeleev’s, periodic table and write name of elements of second, period., , (b) G and O, (d) A and I, , (ii) Which are halogens ?, (a) H and L, (c) G and O, , 6. What were the criteria used by Mendeleev in, , element A (atomic number 19) combines with the, element B (atomic number 17). Draw its electronic, dot structure. What is the nature of the bond, (NCERT Exemplar), formed?, , 1. What were the limitations of Dobereiner’s, (NCERT), , 2. Can the following groups of elements be classified as, Dobereiner’s triad?, (i) Na, Si, Cl, (ii) Be, Mg, Ca, Atomic mass of Be-9; Na-23; Mg-24; Si-28; Cl-35; Ca-40, Justify your answer in each case., (CBSE 2019), , 3. Did Dobereiner’s triads also exist in the columns of, Newlands’ octaves? Compare and find out., , Mendeleev’s periodic table and the modern, periodic table., (ii) With the help of an example, explain, Dobereiner’s Triads., (iii) State modern periodic law., (CBSE 2020), , 13. Write the formula of the product formed when the, , Short Answer Type Questions, classification?, , 12. (i) List any two distinguishing features between, , (NCERT), , 4. Elements have been arranged in the following, sequence on the basis of their increasing atomic, masses. F, Na, Mg, Al, Si, P, S, Cl, Ar, K., (i) Pick two sets of elements which have similar, properties., (ii) The given sequence represents which law of, classification of elements?, 5. What were the limitations of Newlands’ law of, (NCERT), octaves?, , 14. Compare the radii of two species X and Y. Give, reasons for your answer., (a) X has 12 protons and 12 electrons, (b) Y has 12, protons and 10 electrons, (NCERT Exemplar), , 15. Arrange the following elements in increasing order of, their atomic radii., (i) Li, Be, F and N, , (ii) Cl, At, Br and I, (NCERT Exemplar), , 16. An element X of group 15 exists as diatomic, molecule and combines with hydrogen at 773 K in, presence of the catalyst to form a compound,, ammonia which has a characteristic pungent smell., (i) Identify the element X. How many valence, electrons does it have?, (ii) Draw the electron dot structure of the diatomic, molecule of X. What type of bond is formed in it?, (iii) Draw the electron dot structure for ammonia and, what type of bond is formed in it?

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25, , CBSE Term II Science X, , Answer the following question in relation to the, above group of elements., (i) Which element has the most metallic character?, (ii) Which element would be expected to have the, highest electronegativity?, (iii) Will the elements in the group to the right of this, boron group be more metallic or less metallic in, character? Justify your answer., , 17. A salt when dissolved in water dissociates, into cations and anions as follows:, + H2 O, , AB, , !, – H2 O, , A+ + B−, , If both the ions consist same number of electrons, and the molecular weight of salt is 74.5, then identify, the position of A and B in the periodic table., , 18. Three elements A, B and C have 3, 4 and 2 electrons, respectively in their outermost shell. Give the group, number to which they belong in the modern periodic, (NCERT Exemplar), table. Also, give their valencies., , 19. Based on the group valency of elements, write the, molecular formula of the following compounds, giving justification for each., (i) Oxide of first group elements, (ii) Halide of the elements of group thirteen, (iii) Compound formed when an element A of group, 2 combines with an element, B of group seventeen., (CBSE 2019), , 20. From the elements Li, K, Mg, C, Al, S identify the, (i) elements belonging to the same group., (CBSE 2020), (ii) element which has the tendency to lose two electrons., (iii) element which prefers sharing of electrons to, complete its octet., (iv) most metallic element., (v) element that forms acidic oxide., (vi) element that belongs to group 13., , 21. The following table shows the position of five, 1, , 2, , A, , 3, , 2, , 3 to 12, , 13, , 14, , 15, , 16, , 17 18, B, , D, , C, , E, , Answer the following giving reasons:, (i) Which element is a metal with valency two?, (ii) Which element is least reactive?, (iii) Out of D and E which element has a smaller atomic, (CBSE Sample Paper), radius?, , 22. (i) What term can be used for the elements separating, metal from non-metals and why?, (ii) Give the names of the metalloids in the periodic, table along with their atomic number., (iii) In which groups of the periodic table are they, located?, , 23. A group of elements in the periodic table are given, below (boron is the first member of the group and, thallium is the last)., Boron, aluminium, gallium, indium, thallium, , Long Answer Type Questions, , 24. Use Mendeleev’s periodic table to predict the, formulae for the oxides of the following elements., (NCERT), K, C, Al, Si and Ba., , 25. Compare and constrast the arrangement of element in, Mendeleev’s periodic table and the modern periodic, (NCERT), table., , 26. An element is placed in 2nd group and 3rd period of, the periodic table, burns in presence of oxygen to, (NCERT Exemplar), form a basic oxide., (i) Identify the element., (ii) Write the electronic configuration., (iii) Write a balanced equation when it burns in the, presence of air., (iv) Write a balanced equation when this oxide is, dissolved in water., (v) Draw the electron dot structure for the formation of, this oxide., , 27. Which elements has, , elements A, B, C, D and E in the modern periodic, table., Group →, Period ↓, , !, , (i) two shells, both of which are completely filled with, electrons?, (ii) the electronic configuration 2, 8, 2?, (iii) a total of three shells, with four electrons in its, valence shell?, (iv) a total of two shells, with three electrons in its, valence shell?, (v) twice as many electrons in its second shell as in its, first shell?, (NCERT), , 28. An element X (atomic number =17 ) reacts with an, element Y (atomic number = 20) to form a divalent, halide., (i) Where in the periodic table are elements X and Y, placed?, (ii) Classify X and Y as metal(s), non-metal(s) or, metalloid(s)., (iii) What will be the nature of oxide of element Y?, Identify the nature of bonding in the compound, formed., (iv) Draw the electron dot structure of the divalent, halide., (NCERT Exemplar)

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26, , CBSE Term II Science X, , 29. (i) Using the part of the periodic table given below, answer the questions that follows:, Groups → 1, , 2, , 13, , 14, , 15, , 16, , 17, , 18, , (iii), , Periods ↓, 1, , H, , 2, , Li, , B, , C, , N, , O, , F, , He, Ne, , 3, , Na Mg Al, , Si, , P, , S, , Cl, , Ar, , 4, , K, , Be, Ca, , (a) Na has physical and chemical properties similar to, which element(s)., (b) Write the electronic configuration of N and P., Which one of these will be more electronegative, and why?, (c) State a chemical property common to fluorine and, chlorine., , (ii) The neutral atom of an element E consists 12, electrons in its atoms., (a) In which period and group is E placed?, (b) Name the element E., (c) How many electrons it needs to lose or gain to, achieve noble gas configuration ?, (d) What will be the nature of oxide (acidic/basic) of E?, Justify your answer., (e) Write the formulae of chloride of E., !, , (ii), , Case Based Questions, , 30. Read the following and answer the questions from (i), to (v) given below, Modern periodic law states that the physical and, chemical properties of the elements are periodic, function of their atomic radius. When these elements, were arranged in the increasing order of their atomic, number, the obtained table is called modern periodic, table., Numerous forms of the periodic table have been, devised form time to time. A new version, which is, most convinient and widely used is the modern, periodic table. This table consist of 18 vertical, columns called groups and 7 horizontal rows,, known as periods. The first period consists of two, elements., The subsequent period consist of 8, 8, 18, 18 and 32, elements respectively. The seventh period is, incomplete and like the sixth period would have, maximum of 32 electrons., (i) An element belongs to group 17. It is present in, third period and its atomic number is 17. What is, , (iv), (v), , the atomic number of the element belonging to, same group and present in fifth period ?, Name the elements present in the first period of the, modern periodic table., Atoms of different elements with same number of, shells are placed in the same period. Explain., What is the electronic configuration of the element, present in third period and belongs to group 15 ?, How many elements are present in mdoern periodic, table ?, , 31. Read the following and answer the questions from (i), to (v) given below, Valency is the combining capacity of an atom of an, element to acquire noble gas configuration. If, depends upon the number of valence electrons, present in outermost shell of its atom. For the, elements of group 1, 2, 13 and 14 valency = numbers, of valency electrons(s), whereas for the elements of, group 15 onwards valency = 8 − valence electrons., The concept of valency is simple and rationalise the, atomic composition of a large number of compounds., Yet, in many chemistry courses, it is sidelined in, favour of electronic theories of bonding, which are, more difficult. When the theory of valency was, devised, chemists thought that all compounds were, molecular. We now known that many are, non-molecular, i.e. they comprise a large number of, atoms bound together in a continuous framework., The theory can however be adapted to include, non-molecular compounds., The atomic number and valence of element A, B, C,, D and E are given in below table., Elements Atomic number Number of valence electron, A, , 3, , 1, , B, , 9, , 7, , C, , 17, , 7, , D, , 20, , 2, , E, , 36, , 0, , (i) What is the valency of element B ?, (ii) If element C reacts with an element D, which type, of compound they will form., (iii) What is the name of element E ?, (iv) Element A belongs to which period in the modern, periodic table ?, (v) Among A, C, D and E, which element belongs to, group 17 ?

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EXPLANATIONS, Objective Questions, 1. (c) Dobereiner could identify only three triads from the, elements known at that time. These are, Li, Na, K; Ca, Sr, Ba; Cl, Br, I, 2. (b) Newlands’ law of octaves was applicable only to lighter, elements having atomic masses upto 40 u, i.e. upto calcium., After calcium, every eighth element did not possess, properties similar to that of the first element., 3. (c) Gallium has a valency of 3. Hence, it forms an oxide, having molecular formula E 2O3 . In other options, valency of, E is not 3., 4. (d) Na (group 1) has one, Al (group 13) has three (13-10), Si, (group 14) has four (14-10) and P (group 15) has five (15-10), valence electrons. Therefore, P has maximum number of, valence electrons, i.e. 5 (maximum among the given)., 5. (b) The number of electrons present in the outermost orbit, of the element are known as valence electrons, while, number of electrons used by any element in any chemical, reaction is called its valency., Both are co-related as follows, (i) For first four elements in any period,, Valence electrons = Valency, (ii) For last four elements in any period,, Valency = 8 − valence electrons, 6. (d) Q belongs to group II, so its valency is +2. R belongs to, 15 or VA group, so its valency is −3 (as it requires 3 electrons, to complete its octet)., Q, +2, , Q3R2, , –3 Charge, , 10. (a) The correct match for the given item is, A-(iii), B-(i), C-(ii), D-(iv)., Be belongs to group 2 because it contains 2 valence, electrons., F belongs to group 17 because it contains 7 valence, electrons., P contains 5 electrons in it’s outermost shell, so it belongs, to group 15., Ar belongs to group 18 because it has full-filled electronic, configuration., !, , !, , !, , Elements Be, , R, , 7. (a) In long form of periodic table, atomic size decreases, along the period due to increase in effective nuclear charge., Thus, atomic size of B is less than of Be., 8. (d) Because along the period, atomic radii decreases as the, atomic number increases. Thus, the correct increasing order, is F < O < N., 9. (c) On moving from left to right, the atomic number, increases and hence, the nuclear charge increases. With the, increase of nuclear charge, the force binding the electron, increases, so the atom lose the electrons with more difficulty, and not easily., , !, , 11. (a) Lithium is the metal among the first ten elements as, boron is a metalloid, carbon is a non-metal and potassium is, not in first ten elements because it’s atomic number is 19., 12. (d) Oxides of metals are of basic in nature while those of, non-metals are acidic., Group 1 and group 2 consists of metals. Therefore, all the, elements of these groups form basic oxides. The element, present in group 13 and period 3 is aluminium, whose oxides, is amphoteric in nature., Group 16 consists of non-metals. Therefore, all the elements, of this group forms acidic oxide. Thus, an element X which, form an acidic oxides belongs to group 16 and period 3., 13. (a)Q The oxide turns the red-litmus to blue but shows no, effect on blue litmus. Also, the oxide is a metal oxide, thus, one of its component must be a metal. Hence, it is MgO., 14. (c) The element A will form a basic oxide while B, C and D, will form an acidic oxide. This is because on moving along a, period, the acidic character of the oxide increases., 15. (b) The most electropositive element is ‘d’ as electropositive, character increases down the group and decreases along a, period., 16. (a) Both A and R are true and R is the correct explanation, of A. According to the Dobereiner’s triads, the three, elements in a traid were arranged in the order of increasing, atomic masses, the atomic mass of middle element was, roughly the average of the atomic masses of the other two, elements. So, taking Be, Mg and Ca as a triad., , 17., , 18., , 19., 20., , 21., , Mg, , Ca, , Atomic mass 9, 24, 40, Average atomic mass of first and third element, 9 + 40, = 24.5, 2, (c) A is true but R is false. All the elements of group I contains, one valence electron and valency as the electronic, configuration of Li = 2, 1, Na = 2, 8, 1 and K = 2, 8, 8, 1. All the, elements contain only one valence electron., (a) Both A and R are true and R is the correct explanation of A., Noble gases are also called inert gases because they don’t need, to react with other elements to fill their outer shell octet, as, they already posses full valence shell., (a) Both A and R are true and R is the correct explanation of A., (c) A is true but R is false. Electronegativity of fluorine is, greater than that of oxygen, since the non-metallic character, increases along a period from left to right in the modern, periodic table., (i) (a) In 1913, Henry Moseley discovered the modern, periodic table by modifying Mendeleev’s periodic law., (ii) (b) Along the period 3, (Na) sodium can loose electron, easily because moving from left to right along a period,, atomic radius decreases due to increase in effective, nuclear charge which tends to pull the valence, electrons closer to the nucleus and hence, tendency to, lose the electrons decreases.

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28, , CBSE Term II Science X, , (iii) (d) Rb has the highest metallic character because down, the group, the effective nuclear charge decreases as the, atomic radius increases. So, the outermost electrons are, farther away from nucleus which can be lost easily., Hence, metallic character increases down the group., (iv) (b) Electropositive nature increases down the group, due to increase in atomic size., (v) (d) Along the period 4, Ca is the most electronegative, atom because along a period, non-metallic character, increases and so electronegativity also increases., 22. (i) (a) H and P have complete octet. So, they are noble gases., (ii) (c) G and O have 7 electrons in their outermost shell., So, they belongs to halogen group., (iii) (c) D has 6 electrons. So, its electronic configuration is 2, 4., L has 14 electrons. So, its electronic configuration is, 2, 8, 4., ∴Both have 4 valency., (iv) (d) A has one valence electron. So, its valency is 1., It can form A + ion., G has seven valence electron. It needs one electron to, complete its octet. So, A can make ionic bond with G, and form stable compound, AG., (v) (b) B has 4 electrons. So, its electronic configuration is, 2, 2 and hence, have 2 valence electrons., J has 12 electrons. So, it’s electronic configuration is 2,, 8, 2 so, it also contains 2 valence electrons., ∴ Both B and J contains 2 valence electrons., Subjective Questions, , 3. Yes, Dobereiner’s triads also exist in the columns of, Newlands’ octaves, e.g. lithium (Li), sodium (Na) and, potassium (K) constitute a Dobereiner’s triads. Now, if we, consider Li as the first element, then the eighth element, from it is Na and if we consider Na as the first element, then, the eight element from it is K., Similarly, Dobereiner’s triad consisting of the elements, beryllium (Be), magnesium (Mg) and calcium (Ca) is also, included in the column of Newlands’ octaves., Thus, Dobereiner’s triads are included in the columns of, Newlands’ octaves., 4. (i) Here, the elements are arranged in the order of, increasing atomic masses, so according to Newlands’, law of octaves there is a repetition of every eighth, element as compared to the given element. The two, sets of elements which have similar properties are, Set I → F, Cl, Set II → Na, K, F and Cl are first and eighth element in the above, sequence, therefore, they have similar properties., Although Na and K have similar properties but they are, not related as first and eighth element in the above, sequence., (ii) The given sequence is according to Newlands’ law of, octaves represented as, F Na Mg Al Si P S Cl Ar K, 5. This law was applicable only upto calcium. After calcium,, every eighth element did not possess the same properties, similar to that of the first., Newland assumed that there were only 56 elements existed, in nature and no more elements would be discovered in the, future. But, later on, several new elements were discovered,, whose properties did not fit into the law of octaves., In order to fit elements into his table, Newlands’ adjusted two, elements in the same slot and also put some unlike elements, under the same column., e.g. Cobalt and nickel are in the same slot and these are, placed in the same column as fluorine, chlorine and bromine, which have very different properties than these elements., Iron, which resembles cobalt and nickel in properties, has, been placed far away from these elements. Hence, Newlands’, law of octaves worked well with lighter elements only., 6. The criteria used by Mendeleev were:, (i) The arrangement of elements in increasing order of, atomic masses., (ii) Similarity in chemical properties of the elements., 7. In Mendeleev’s periodic table, cobalt (Co) with a higher, atomic mass of 58.93 u is placed before nickel (Ni) due to the, following reasons :, (i) The properties of cobalt are similar to those of rhodium, (Rh) and iridium (Ir) (same group) and, (ii) The properties of nickel are similar to those of, palladium (Pd) and platinum (Pt) (same group)., 8. Eka-silicon is germanium (Ge). It lies in group 4 of the, Mendeleev’s periodic table and thus, has a valency of 4., ∴ The formula of its chloride is GeCl 4 ., !, , !, , 1. All the elements discovered at that time could not be, classified into triads, only a limited number of elements, could be arranged in such triads., e.g. The three elements nitrogen (N), phosphorus (P) and, arsenic (As) have similar properties. Therefore, they should, be regarded to form a triad., However, the actual mass of the middle element P (31.04) is, much lower than the average (44.454) of the atomic masses, of nitrogen (14.4) and arsenic (74.94). Therefore, these three, elements do not constitute a Dobereiner’s triad inspite of, their similar chemical properties., 2. (i) Na, Si and Cl have different properties, therefore, they, do not form Dobereiner’s triad even though the atomic, mass of the middle atom (Si) is approximately the, average of the atomic masses of Na and Cl, i.e., Na (23); Si (28); Cl (35), 23 + 35 58, =, = 29, 2, 2, (ii) Be, Mg and Ca have many similar properties and also, the atomic mass of the middle element Mg is, approximately the average of the atomic masses of Be, and Ca, i.e., Be (9); Mg (24); Ca (40), 9 + 40 49, Atomic mass of Mg =, =, = 24.5, 2, 2, Therefore, they form Dobereiner’s triad., Atomic mass of Si =, , !

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29, , CBSE Term II Science X, , Eka-aluminium is gallium (Ga). It lies in group 3 of the, Mendeleev’s periodic table and thus, has a valency of 3., ∴ The formula of its chloride is GaCl 3 ., 9. Two main characteristics of Mendeleev’s periodic table are :, (i) It consists of 8 vertical columns, called groups and, 6 horizontal rows, called period., (ii) In every period, elements are arranged in increasing, order of their atomic masses., Name of elements of second period are lithium, beryllium,, boron, carbon, nitrogen, oxygen, fluorine., 10. Electronic configuration of an element decides its position, in modern periodic table., If we take an example of sodium (Na), which has, atomic number = 11, i.e. it’s electronic configuration = 2 , 8, 1, As Na contains 1 electron in its outermost shell, this means, that it belongs to group 1 and sodium contains 3 shells so, it, belongs to period number 3., ∴ We can conclude that,, Group number = Number of valence electrons, (When valence electrons are 1 and 2), and group number = 10 + valence electrons, (When valence electrons are 3 and above), Period number = Number of shells in which electrons are filled., 11. Hydrogen occupies a unique position in the modern, periodic table due to the following reasons, (i) Both hydrogen and alkali metals have similar outer, electronic configuration as both have one electron in, the valence shell. Therefore, some of the properties of, hydrogen are similar to those of alkali metals and hence,, it can be placed in group 1 alongwith alkali metals., (ii) Both hydrogen and halogens have similar outer, electronic configuration (both have one electron less, than the nearest inert gas configuration). Therefore,, some of the properties of hydrogen are similar to those, of halogens and hence, it can be placed in group 17, alongwith halogens., (iii) In some properties, it differs from both hydrogen and, halogens, e.g. the oxide of hydrogen, i.e. H2O is neutral, but the oxides of alkali metals (i.e. Na 2O, K 2O etc.) are, basic while those of halogens (i.e. Cl 2O7 , Br2O5 , I2O5, etc.) are acidic., 12. (i), Mendeleev’s periodic table, , In the Mendeleev’s, periodic table, the, elements were arranged, in increasing order of, their atomic masses., This table consists of 8, groups and 6 periods., , Modern periodic table, , In modern periodic table,, the elements are arranged in, the increasing order of their, atomic number., This contains 18 groups and, 7 periods., , (ii) Dobereiner arrange the elements with similar, properties into groups having three elements each and, named these groups as triads., He showed that when the three elements in a triad, were arranged in the order of increasing atomic masses,, the atomic mass of the middle element was roughly the, , average of the atomic masses of other two elements,, e.g., Elements, Cl, Br, I, Atomic mass, 35.5, 80, 127, Average atomic mass, 35.5 + 127 = 81.25, of first and that elements, 2, (iii) Modern periodic law The physical and chemical, properties are a periodic function of their atomic number., 13. Atomic number of A = 19, Electronic configuration is 2, 8, 8, 1., Hence, element A is metal potassium (K) and, Atomic number of B = 17., Electronic configuration is 2, 8, 7., It is a non-metal, chlorine (Cl)., So, the electron dot structure of KCl is, K, 2, 8, 8, 1, , +, , Potassium, , ××, ×Cl×, ×, ××, , 2, 8, 7, Chlorine, , K+, , ××, ×Cl×, ×, ××, , –, , 2, 8, 8 2, 8, 8, Potassium chloride, , The bond formed between K+ and Cl − is ionic bond and, formula of the product formed K + Cl − or KCl., 14. Since, species X has 12 protons and 12 electrons, it is, electrically neutral. Since, species Y has 12 protons and, 10 electrons, therefore, it has two units positive charge., The electronic configuration of the two species are, Species X, Species Y, K L M, K L, 2 8 2, 2 8, Since, species X has three shells while species Y has two, shells, therefore, species Y has smaller radius than species X., 15. (i) Atomic radii decreases along a period from left to right, due to increase in nuclear charge. Li, Be, F and N, belong to same period. Thus, the atomic radii of Li, Be,, F and N increases in the order:, F < N < Be < Li, (ii) Atomic radii increase in a group from top to bottom due, to the corresponding increase in the number of filled, electronic shells. Cl, At, Br, I belong to same group. Thus,, atomic radii of Cl, At, Br and I increase in the order:, Cl < Br < I < At, 16. Since, the element ‘X’ of group 15 exists as a, diatomic molecule and combines with hydrogen at 773 K, in presence of a catalyst to form ammonia which has a, characteristic smell, therefore, the element ‘X’ is, nitrogen (N)., N2, Nitrogen, (diatomic molecule), , + 3H2, Hydrogen, , K, ⎯773, ⎯⎯, →, Catalyst, , 2NH3, Ammonia, (pungent smell), , (i) The atomic number of nitrogen is 7. So, its electronic, configuration is 2, 5. Thus, it has five valence electrons., (ii) Nitrogen has 5 valence electrons. Therefore, it needs 3, more electrons to complete its octet., To do so, it shares three of its electrons with three, electrons of the other nitrogen atom to form a diatomic, molecule of N 2 gas.

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30, , CBSE Term II Science X, , (Q Valency of group 2 element ( A ) is 2 and that of group, 17 element (B) is 1)., ∴, B, A, , Thus, three covalent bonds are formed between two, nitrogen atoms and each nitrogen atom is left with one, lone pair of electrons., N, , + N, , N, , Two nitrogen atoms, , N, , or N, , 2, , N, , Nitrogen molecule, , 20. The electronic configurations of the given elements, , (iii) Electron dot structure for ammonia is as follows :, H, , are as follows, 2, 1, → 2, 8, 8, 1, 12 Mg → 2, 8, 2, 3 Li →, , H, , 19 K, , ×, , N, , +, , 3H×, , N, , ×, , H, Nitrogen atom, , Three hydrogen, atoms, , ×, , H, , or, , N, , H, , H, , 17. Since, both the ions consists of same number of, electrons and has +1 and −1 charges, hence the ions, should belong to group 1A (cation, i.e. A + ) and group VII A, (anion, i.e. B− )., Same number of electrons indicates that their electronic, configuration is same as that of a noble gas whose, atomic number lie between that of the two elements A, and B., Dividing the molecular weight (which is sum of atomic, masses of A and B), we get the rough idea about the atomic, 74.5, mass of the noble gas which is, = 37.25, i.e. nearest to, 2, argon (Ar - 40)., Hence, A is K (group IA, 4th period) and element B is Cl, (group VII A, 3rd period)., 18. (i) Element A has 3 valence electrons, therefore, its, valency is 3 and thus belongs to group 13 (3 + 10 ). As, such, it could be any one of the following elements : B,, Al, Ga, In or Tl., (ii) Element B has 4 valence electrons, therefore, its, valency is 4 and it belongs to group 14 (4 + 10). The, element B could be any one of the following : C, Si, Ge,, Sn or Pb., (iii) Element C has two valence electrons, therefore, its, valency is 2 and it belongs to group 2. The element C, could be any one of the following : Be, Mg, Ca, Sr, Ba, or Ra., 19. (i) The valency of the group 1 elements is 1 and that of, oxygen is 2., O, , 1, 2, ∴ Molecular formula of oxide = Na 2O, (ii) The valency of group 13 element is 3 and that of halide, is 1., Al, Cl, , 3, 1, ∴ Molecular formula of halide = AlCl 3, (iii) Molecular formula of compound formed = AB2, , 2, 4, 2, 8, 3, 16 S→ 2, 8, 6, Element belong to same group are Li and K as they, both contain one electron in their outermost shell., Element which has tendency lose two electrons is, magnesium as it cantains 2 electrons its outermost, shell., Element which prefer sharing of electron to complete, its octet is carbon due to its small size and strong C ⎯ C, bond., Most metallic element is potassium. Elements of group, 1 are metallic in nature as they readily loose their one, valence electron., Non-metals form acidic oxides. Among the given, elements, S is a non-metal. Thus, it forms most acidic, oxide., Aluminium belongs to group 13 as it contains, 3 elements in its outermost shell., 6C →, , 13 Al →, , Ammonia molecule, , In NH3 molecule, there are three N—H single, covalent bonds and one lone pair of electrons on the, nitrogen atom., , Na, , 1, , ⇒ AB2, , (i), (ii), , (iii), , (iv), , (v), , (vi), , (i) Element ‘D’ is a metal with valency two., Because, the group number of an element having upto, two valence electrons is equal to the number of valence, electrons., (ii) ‘C’ is the least reactive element. Because, it belongs to, group 18. Group 18 (Noble gases) are least reactive due, zero valency of group 18 elements., (iii) ‘E’ has a smaller atomic radius than ‘D’ because ‘E’ is, on the right side of the modern periodic table., Across the period, atomic size/radius decreases on, moving left to right. This is due to an increase in, nuclear charge which tends to pull the valence, electrons closer to the nucleus and reduces the size of, the atoms., 22. (i) Metalloids as these elements show the properties of, both the metals and non-metals., (ii) The list of metalloids alongwith their atomic number is, as follows:, B (5), Si(14), Ge (32), As (33), Sb (51), Te (52) and Po (84)., (iii) These elements are located in groups 13, 14, 15 and 16., 23. (i) Thallium has the most metallic character. Metallic, character increases down in a group., (ii) Boron has the highest electronegativity because, electronegativity decreases down a group., (iii) Less metallic in character, because on moving across a, period, metallic nature decreases., , 21.

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31, , CBSE Term II Science X, , 24. Oxygen is a member of group VI A in Mendeleev’s periodic, table. Its valency is 2. Similarly, the valencies of all the, elements given can be predicted from their respective group., This can help in writing the formula of their oxides., (i) Potassium (K) is a member of group IA. Its valency, is 1. Therefore, the formula of its oxide is K 2O., (ii) Carbon (C) is a member of group IVA. Its valency is 4., Therefore, the formula of its oxide is C 2O 4 or CO 2 ., (iii) Aluminium (Al) belongs to group IIIA and its valency is, 3. Therefore, the formula of its oxide is Al 2O 3 ., (iv) Silicon (Si) is present in group IVA after carbon. Its, valency is also 4. Therefore, the formula of its oxide is, Si 2O 4 or SiO 2 ., (v) Barium (Ba) belongs to group IIA and its valency is 2., Therefore, the formula of its oxide is Ba 2O 2 or BaO., 25., , Mendeleev’s periodic table, , Modern periodic table, , The properties of elements The properties of elements are, are the periodic functions of the periodic functions of their, atomic number., their atomic mass., It has 8 groups., , It has 18 groups., , There is no place for, isotopes of an element., , Isotopes of an element are, assigned the same place with, their respective elements as, they have the same atomic, number., , No fixed position was given Hydrogen is given a special, to hydrogen in this periodic position in modern periodic, table., table., Inert gases were not known Inert gases have been placed at, at the time of Mendeleev., the end of period in group 18., 26., , (i) Since, the element lies in group 2, it must be an, alkaline earth metal. Since, it lies in the third period, it, must be magnesium (Mg)., (ii) Atomic number of Mg is 12, therefore, its electronic, K L M, configuration is 2 , 8, 2 ., (iii) When Mg burns in the presence of air, it forms a basic, oxide, MgO., ⎯, ⎯→ 2MgO (s), 2Mg (s) + O2(g ) ⎯Heat, Magnesium, , Magnesium oxide, , Oxygen, , (iv) When MgO is dissolved in water, it forms magnesium, hydroxide., ⎯→ 2Mg(OH)2(aq ), 2MgO (s) + 2 H2O (l ) ⎯, , Magnesium oxide, , Magnesium hydroxide, , Water, , (v) Mg has 2 valence electrons [as electronic configuration, of 12 Mg = [2 , 8, 2]. Oxygen has 6 valence electrons [as, electronic configuration of 8 O = [2, 6]. Electron dot, structure for the formation of magnesium oxide., Mg, , +, , ××, O×, ×, ××, , 2, 8, 2, , 2, 6, , Magnesium, , Oxygen, , Heat, , ××, , Mg2+ O ×, ×, , 2–, , ××, , Magnesium oxide, , (i) Noble gases are the elements which have completely, filled shells. The noble gas with two shells (K, L) is Ne, having atomic number 10 and electronic configuration, K L both of the shells are completely filled., 28, (ii) Electronic configuration 2, 8, 2 suggests that atomic, number is 12 (2 + 8 + 2). Magnesium (Mg) has atomic, number 12., (iii) The element with three shells and four electrons in, the valence shell will have electronic configuration, KL M, . The atomic number of this element is, 28 4, 14 (2+8+4) so it will belong to group 14. Hence, it is, silicon (Si)., (iv) Element with two shells and 3 electrons in the valence, shell will exist in second period and will have the, electronic configuration K2 L3 ., The atomic number of this element will be 5 (2, 3). So,, it will be boron (B)., (v) The element has two shells. We know that first shell, can have only 2 electrons, so according to the question, there will be 4 electrons (double the number of, electrons in the first shell) in the valence shell. The, electronic configuration will be K2 L4 , so the atomic, number is 6. Hence, the element is carbon (C)., 28. (i) The electronic configuration of element X with atomic, number 17 is 2, 8, 7. Since, it has 7 valence electrons., Therefore, it lies in group 17 (10 + 7 ). Further, since in, element X, third shell is being filled, it lies in third, period. In other words, X is chlorine., , 27., , The electronic configuration of element Y with atomic, number 20 is 2, 8, 8, 2. Since, it has 2 valence electrons,, it lies in group 2. Further, since in element Y, fourth, shell is being filled, it lies in 4th period. In other words,, Y is calcium., (ii) Since, element X (i.e. Cl) has seven electrons in the, valence shell and needs one more electron to complete, its octet. Therefore, it is a non-metal. Further, the, element Y has two electrons in the valence shell that, can be easily lost to achieve the stable electronic, configuration of the nearest inert gas, therefore, it is a, metal., (iii) Since, element Y (i.e. Ca) is a metal, therefore, its oxide, (i.e. CaO) must be basic in nature. Further, metals and, non-metals form ionic compounds, therefore, the, nature of bonding in calcium oxide is ionic., (iv) Electronic configuration of 20 Ca = 2 , 8, 8, 2 [valence, electrons = 2], electronic configuration of 17 Cl = 2 , 8, 7, [valence electrons = 7]. The electron dot structure of, divalent metal halide,, ×××, –, × Cl×, ××, ××, +, Ca2+ × Cl ×, ×××, i.e. CaCl 2 Ca, ×× × 2, × Cl×, –, ××, ××, or Y 2+ × X ×, 2, 8, 8, 2, 2, 8, 7, ×, ×× 2, Calcium, Chlorine, Calcium chloride, X, Y

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32, 29., , CBSE Term II Science X, , (i) (a) Lithium and potassium, due to presence of same, number of valence electrons., (b) N → 2, 5, P → 2, 8, 5, N is more electronegative element as, electronegativity decrease on moving down the, group., (c) Both fluorine and chlorine form their hydrides on, reacting with hydrogen., H2 + F2 ⎯→ 2HF, H2 + Cl2 ⎯→ 2HCl, K L M, (ii) (a) The electronic configuration of E = 2 , 8, 2, Hence, it should be placed in 3rd period and, group II A., (b) The element E is magnesium (Mg)., (c) Mg loses 2 electrons to form noble gas (Ne) with, K L, configuration as 2 , 8., , (d) Since, Mg is electropositive, it will form basic, oxide., (e) The formula of chloride is ECl 2 or MgCl 2 ., 30. (i) The element whose atomic number is 17 and belongs to, third period is chlorine. If we go down the group, i.e. in, fifth period, number of shells increases as18e− are, increased in consecutive periods., Therefore, the atomic number of the element, belonging to same group and present in fifth period is, 17 + 18 + 18 = 53., (ii) Hydrogen (H) and helium (He)., (iii) The number of valence shell electrons increases by one, unit as the atomic number increases by one unit on, moving from left to right in a period. Therefore, the, atoms of different elements with same number of shells, are placed in the same period., , (iv) As the element belongs to group-15, i.e., it’s valence, electrons are 5 and valency is 3, so it has the electronic, configuration = 2, 8, 5., (v) At present, 118 elements are known to us. All these, have different properties. Out of these 118 elements,, only 94 are naturally occurring., 31., , (i) The atomic number of element B is 9 and number of, valence electrons are 7, i.e. it belongs to group 17., Valency = 8 − valence electrons, = 8 −7 =1, Hence, the valency of B is 1., (ii) Electronic configuration of element C is 2, 8, 7, i.e. it, has 7 valence electrons and valency is −1 (as it requires, 1 electron to complete it’s octet) and electronic, configuration of element D is 2, 8, 8, 2, i.e. it has 2, valence electrons and valency., C, , D, , –1, , +2 Charge, C2D, , This forms a divalent compound (C 2D)., (iii) The atomic number of element E is 36, so the, electronic configuration is [Ar] 3d 10 4s 2 4p 6 . Therefore,, this element belongs to 4th period as last electron goes, in 4th shell and have zero number of valence electrons., Hence, the element is krypton., (iv) The atomic number of element A is 3 and electronic, configuration is1s 2 2s1 . As last electron enters in 2nd, shell, so, it belongs to period 2., (v) The element that belongs to group 17 is ‘C’ because, this group contain those elements which have 7, electrons in their outermost shell and element C also, contains 7 valence electrons. Therefore, option (b) is, correct.

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Chapter Test, , Short Answer Type Questions, , 9. Give an account of the process adopted by Mendeleev, for the classification of elements. How did he arrive at, “periodic law”?, , Multiple Choice Questions, , 1. Dobereiner grouped the elements into ....... and, Newlands’ gave the ........, (a) periods, triads, (c) octaves, periods, , (b) triads, law of octaves, (d) triads, periodic law, , 2. In Mendeleev’s periodic table, gaps were left for the, elements to be discovered later. Which of the following, elements found a place in the periodic table later?, (a) Germanium, (c) Oxygen, , (b) Chlorine, (d) Silicon, , 3. An element which is an essential constituent of all, organic compounds belongs to, (a) group 1, (c) group 15, , (b) group 14, (d) group 16, , 4. Which of the following is the outermost shell for, elements of period 2?, (a) K-shell, (c) M-shell, , (b) L-shell, (d) N-shell, , 5. Which group of element will form an acidic oxide?, (a) Group 2, (c) Group 15, , (b) Group 13, (d) Group 18, , Assertion-Reasoning MCQs, Direction (Q. Nos. 6-8) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d), given below., (a) Both A and R are true and R is the correct explanation of A., (b) Both A and R are true, but R is not the correct explanation, of A., (c) A is true, but R is false., (d) A is false, but R is true., , 6. Assertion According to Mendeleev, periodic properties, of elements are functions of their atomic number., Reason Atomic number is equal to the number, ofprotons., , 7. Assertion The elements of the same group have, similar chemical properties., Reason The elements of the same group have the, different number of valence electrons., , 8. Assertion Silicon, germanium and arsenic are the, metalloids in the modern periodic table., Reason Silicon, germanium and arsenic has properties, of metal as well as non-metals., , 10. What are valence electrons ? Does the number of, valence electrons increase or decrease on moving from, left to right in a period ? How does valency of elements, very in the period ?, , 11. (i) How is the valency of an element determine if, its electronic configuration is known? Determine the, valency of an element of atomic number 9., (ii) Which one of the above elements belonging to the, fourth period has bigger atomic radius and why?, , 12. Name, (i) three elements that have a single electron in their, outermost shells., (ii) two elements that have two electrons in their, outermost shells., (iii) three elements with filled outermost shells., , 13. Identify and name the metals out of the following, elements whose electronic configurations are given, below., (a) 2, 8, 2, , (b) 2, 8, 1, , 1. (b), , 2. (a), , 3. (b), , 4. (b), , Assertion-Reasoning MCQs, 6. (d), , 7. (c), , 8. (b), , 5. (c), , (d) 2, 1, , the periodic table as boron have in common?, (ii) What property do all elements in the same column of, the periodic table as fluorine have in common?, , 15. Arrange the following elements in the increasing order, of their metallic character Mg, Ca, K, Ge, Ga., , Long Answer Type Questions, , 16. What are the limitations of Mendeleev’s periodic table?, 17. (i) The modern periodic table has been evolved through, the early attempts of Dobereiner, Newland and, Mendeleev. List one advantage and one limitation of, all the three attempts., (ii) Name the scientist who first of all showed that atomic, number of an element is a more fundamental, property than its atomic mass., (iii) State modern periodic law., , 18. (i) List two criteria Mendeleev’s used in his periodic table, to classify the elements. State Mendeleev’s periodic, law and explain why no fixed position was assigned to, hydrogen in Mendeleev's periodic table ?, (ii) How and why does the atomic size of elements vary as, we move, (a) from left to right in a period, and, (b) down a group in the modern periodic, , 19. Write five main characteristics of group in modern, periodic table., , Answers, Multiple Choice Questions, , (c) 2, 8, 7, , 14. (i) What property do all elements in the same column of, , For Detailed Solutions, Scan the code

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34, , CHAPTER 03, , How do Organisms, Reproduce ?, In this Chapter..., !, , The Fundamentals of Reproduction, , !, , Types of Reproduction, , !, , Human Reproductive System, , !, , Fertilisation and Post-fertilisation Changes, , !, , Menstruation, , !, , Reproductive Health, , !, , Sexually Transmitted Diseases, , The process by which living organisms produce its own kind, of individuals to maintain the continuity of species is, called reproduction. Like other essential life processes,, reproduction is not essential to maintain life of an individual., But is a fundamental feature of all known life, each individual, organism exists as the result of reproduction., , The Fundamentals of Reproduction, The process of reproduction involves the formation of DNA, copy and other cellular apparatus required by the cells of an, individual. DNA is the blueprint of all the basic design of, organisms. It is present in the nucleus of a cell as a condensed, structure called chromosome. It acts as the information source, and helps in making different proteins and cellular machinery, of cell, which makes up the different body designs., , Variations, DNA copying during cell division always causes some or, other type of variations in newly formed cell. This brings the, differences found in the morphological and physiological, features of an organism., Since no biochemical reaction is absolutely reliable,, DNA copies generated are similar, but not absolutely, identical. Variations lead to evolution by increasing the, chances of survival of some individuals. Hence, important for, the survival of species., , Types of Reproduction, Reproduction is mainly of two types, i.e. asexual reproduction, and sexual reproduction., Asexual Reproduction It is a rapid mode of multiplication, in which one parent (either male or female) is involved. The, new individuals produced are identical to their parents., Sexual Reproduction In this process, the gametes from, parents of opposite sex (male and female) fuse together to, form a zygote. This zygote develops further and gives rise to, new offspring. The individual produced by this method, exhibits variation., , !, , !, , I. Modes of Asexual Reproduction, Asexual reproduction occurs in unicellular organisms by, fission, budding, spore formation, fragmentation, regeneration, (in animals) and vegetative propagation (in plants). It occurs in, multicellular organisms by budding and regeneration., These are as follows, (i) Fission The process where a unicellular organism splits, itself into two or more daughter cells. It is of two types,, i.e., (a) Binary Fission In this process, parent cell divides, into two identical daughter cells, e.g. Amoeba,, Leishmania.

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35, , CBSE Term II Science X, , In this process, nuclear division is followed by, cytoplasmic division., Pseudopodia, Parent cell, , (iv) Budding A daughter organism is formed from a small, projection known as bud. It develops as an outgrowth due, to repeated cell divisions of the parent body. When fully, grown, it detaches to grow into a new independent, individual, e.g. Hydra., , Nucleus, , Tentacles, , Pseudopodia, withdrawn, Parent, Hydra, , Nuclear, division, Cytoplasmic, division, Two daughter, cells formed, Binary fission in Amoeba, , (b) Multiple Fission In this process, parent cell, divides into many identical daughter organisms, simultaneously, e.g. Plasmodium., , Growing, bud, , New Hydra, , Budding in Hydra, , (v) Vegetative Propagation The process of obtaining complete, plant from any vegetative part of plant is called vegetative, propagation. This is mainly of two types, (a) Natural Vegetative Propagation The vegetative, propagation that occurs automatically in plants is called, natural vegetative propagation. It can be achieved by, root, stem, leaf, etc, e.g. Bryophyllum., Bud, , Many daughter cells produced, Nucleus, by multiple fission, Daughter cells, Cytoplasm, released, , Leaf of Bryophyllum with buds, , Protective wall, , Cell, , Multiple fission in Plasmodium, , (ii) Fragmentation The parent body on maturation breaks, up into two or more small fragments, which later grow, into a complete new organism, e.g. Spirogyra., (iii) Regeneration In this process, all fragments or parts, that are separated from the body develop in new, animals, e.g. sponge, Planaria, Hydra, etc., , (b) Artificial Vegetative Propagation The artificially made, vegetative propagules in plants by humans, is called, artificial vegetative reproduction, e.g. cutting, layering,, grafting, etc., Grafting A small part of stem from one plant without, roots (scion) is attached to the part with root (stock) of, another plant., Layering The development of roots on a stem, while, the stem is still attached to the parent plant is called, layering., Tissue culture It is a technique used for growing new, plants using living tissues (like flower buds, stems,, growing tips, leaves, etc.) in vitro in an artificial culture, medium., Using this technique, large number of plants can be, developed from a single parent., (vi) Spore Formation It is a type of asexual reproduction, where blob-like structures called sporangia are, involved. These cells or spores have the ability to, germinate under favourable conditions forming new, plants, e.g. Rhizopus., !, , !, , !, , !, , Regeneration in Planaria

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36, , CBSE Term II Science X, , Sporangium, , !, , Ovary It is the swollen bottom part of carpel. It contains, ovules having an egg cell (female gamete)., Stigma, , Sporangiophore, , Anther, Filament, , Style, Carpel, , Hypha, (Singular-hypha, plural-hyphae), , v, , II. Modes of Sexual Reproduction, In this type of reproduction, both sexes, i.e. male and, female are involved. Sex cell or gamete of one parent, (male) fuses with the sex cell or gamete of another, parent (female). This results in production of a new cell, called zygote., Thus, the sexual mode of reproduction involves two, major processes, (i) Formation of gametes by meiosis, (ii) Fusion of gametes, , 1. Sexual Reproduction in Flowering Plants, Angiosperms bear the reproductive parts within the, flower and their seeds are enclosed in a fruit. Most plants, have both male and female reproductive, organs in the same flower and are known as bisexual, flowers. While others have either male or female, reproductive parts in a flower known as unisexual flowers., A flower comprises of four main parts, i.e. sepals, petals,, stamens and carpels. Stamens and carpels are the, reproductive parts of a flower., (i) Stamen It is the male reproductive part of the, flower., (ii) Anther It is a bilobed structure containing two, pollen sacs present at tip of stamen. These produce, pollen grains that are yellowish in colour., (iii) Carpel (Pistil) It is the female reproductive part,, which is present in the centre of the flower. It, comprises of three parts, Stigma It is the terminal part of carpel which may, be sticky. It helps in receiving the pollen grains, during pollination., Style It is the middle elongated part of carpel. It, helps in the attachment of stigma to the ovary., , Petal, Ovary, , Spore formation in Rhizopus, , Sepal, , Longitudinal section of flower, , Pollination, The transfer of pollen grains from the anther of the stamen to the, stigma of a flower is termed as pollination. The pollen grains can, be transferred by various agents like wind, water, insects and, animals., Pollination usually occurs in two ways, (i) Self-pollination The pollen from the stamen of a flower is, transferred to the stigma of the same flower or another flower, of same plant., (ii) Cross-pollination The pollen from the stamen of a flower is, transferred to the stigma of another flower of different plant, of the same species., Fertilisation, It is the process of fusion of male and female gametes. It gives rise, to a zygote. As soon as the pollen lands on suitable stigma, it, reaches the female germ cells in ovary. This occurs via pollen, tube. The pollen tube grows out of the pollen grain, travels, through the style and finally reaches the ovary., After fertilisation, ovule develops a rough coat around itself and gets, converted to seeds and ovary ripens as fruit., The seed contains future embryo that grows under suitable, conditions (germination). The fertilisation in the flowering plant is, shown in the given figure., Pollen grain, Stigma, Male gamete (Male germ cell), Pollen tube, , !, , !, , Stamen, , Spores, , Ovary, Female gamete, , Germination of pollen grain on stigma

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37, , CBSE Term II Science X, , (ii) Female Reproductive System It includes internal and, external sex organs that function in reproduction of new, offspring. In human, female reproductive system is, immature at birth and develops to maturity at puberty, to be able to produce gametes, and to carry a foetus., , 2. Sexual Reproduction in Human Beings, Human beings can reproduce sexually after attaining puberty. It, represents period of adolescence when reproductive organs, start developing and sexual maturity is attained. Some changes, during puberty are common in males and females such as hair, growth in armpits and genitals, oily skin, acne, etc., Specific changes in boys include facial hair growth, hoarse, voice, etc. In girls, these changes are enlargement of breast size,, begining of menstruation, etc., , Oviduct or, Fallopian tube, Ovary, Uterus, , Human Reproductive System, , Cervix, , The system of organs required by males and females for the, process of sexual reproduction is called reproductive system., (i) Male Reproductive System It includes parts which, produce the germ cells and those which deliver these cells, at the site of fertilisation., Seminal vesicle, , Vagina, Female reproductive system, , Parts and Details of the Female Reproductive System, Parts, , Ureter, Bladder, , Ovaries, , Prostate gland, , !, , Penis, Urethra, , Vas deferens, , Scrotum, , Details, !, , !, , Oviduct, (Fallopian tube), , !, , !, , Testis, Male reproductive system, , !, , !, , Parts and Details of the Male Reproductive System, Parts, Testes, , Details, !, , !, , !, , Scrotum, , !, , !, , Vas, deferens, Urethra, Prostate, gland and, seminal, vesicles, Penis, , !, , !, , !, , !, , !, , !, , Sperms, , !, , Paired, oval-shaped male sex organs., Consist of seminiferous tubules, where the sperms are, produced., Produce a male sex hormone called testosterone, which, brings about changes in appearance of boys at puberty., Small pouch that contains testis., Present outside the abdominal cavity. As sperms are, formed here, this requires a lower temperature than the, normal body temperature., Tube-like structure which connects testis to the urethra, in order to allow the passage of semen., , Uterus (womb), , !, , Cervix, , !, , Vagina, , !, , !, , External male genital organ., Transfers sperms into the vagina of the female during, copulation., Tiny and motile bodies that use their long tail to move, through the female reproductive tract., , It has a funnel-shaped opening near the, ovary., Carries ova or egg from ovary to the uterus., It is the site of fertilisation., These open into the uterus from both, the sides., Hollow, pear-shaped, bag-like structure., The growth and development of foetus takes, place., It is the lower and the narrower portion of, uterus which opens into the vagina., Receives the sperms from the male partner., Serves as a birth canal., , Fertilisation and Post-Fertilisation, Changes, !, , Common passage for both the sperms and urine. It, never carries both of them at the same time., Secretes seminal fluid and nutrients., Fluid and nutrients combine with sperm to form semen., Milky, viscous fluid contains fructose, proteins and other, chemicals for nourishing and stimulating sperms., , !, , Paired, oval-shaped organs located in the, abdominal cavity near the kidney., Produce thousands of ova or egg cells., Secrete female sex hormones like oestrogen, and progesterone., , !, , !, , Fusion of sperm with ovum is called fertilisation. It, results in the formation of diploid zygote. This process, takes place in the oviduct or Fallopian tube. The, formation of embryo is the result of cleavage and growth, in zygote., The embryo sinks downward, reaches into the soft, uterine lining and gets embedded. This process is known, as implantation., A disc-like structure called placenta grows between the, uterine wall and embryo. It has finger-like projections, called villi, which provide surface area for the exchange, of nutrients, oxygen and waste products between the, embryo and the mother.

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38, !, , CBSE Term II Science X, , Childbirth (after a gestation period of approximately, 9 months) occurs by strong rhythmic contractions of, uterine muscles., , Methods, , Diaphragm, Intra-Uterine, Contraceptive, Device (IUCD), , Menstruation, In the absence of fertilisation, the uterine lining which, becomes thick and spongy to receive a fertilised egg, is no, longer required. It sheds out as blood and mucus which, lasts for about 2-8 days and occurs every month. This, phase is known as menstruation., , Examples, , Hormonal, , Oral, contraceptive, pills, , Details, !, , !, , !, , !, , !, , Reproductive Health, It can be defined as the state of physical, mental and, social fitness to lead a healthy reproductive life. Good, reproductive health provides both male and female with, the fertility control methods., awareness about how to limit their family size., protection from infection and sexually transmitted, diseases., , !, , Chemical, , Spermicide, , !, , !, , !, , !, , !, , Rubber cup that is placed in the vagina, over the cervix., Copper-T placed in uterus by doctor., Used to prevent pregnancy., Can cause side effects due to the, irritation of uterus., Contain hormones, which prevent, release of ovum, so that fertilisation, cannot occur. These disturb the, hormonal balance (levels of FSH and, LH) of the body., Can cause side effects also., Applied in vagina., Kills sperms., Can only be used with condoms or, diaphragm., , !, , Surgical, , Vasectomy, , !, , Sex Ratio, The ratio of the number of females to the number of, males in a population is known as sex ratio. A balanced, female-male sex ratio is necessary for a healthy society., , !, , !, , Tubectomy, , !, , Population Size, The rates of birth and death in a given population, determine its size. The population size increases if the, birth rate is higher than the death rate and vice-versa., !, , Methods of Family Planning, The sexual act always carries the risk of potential, pregnancy. In order to avoid unplanned pregnancies,, many ways have been devised, which are called, contraception or birth control methods., Methods of Family Planning, Methods, , Examples, , Barrier, , Condom, , Details, !, , !, , Rubber sheath worn over the, penis to stop sperm from, entering the vagina., Prevents transmission of, Sexually Transmitted, Diseases (STDs) and has no, side effect., , !, , Small portion of the sperm duct is cut or, tied properly. Therefore, the sperm, transfer will be prevented., Prevents sperms from coming out of, urethra., An irreversible process., Small portion of oviduct is cut or, tied properly. The Fallopian tube in, the female gets blocked. The egg, will not be able to reach the uterus, and thus,fertilisation will not take, place., Prevents the egg from meeting the, sperms., An irreversible process., , Female Foeticide, The killing of unborn girl child is called female foeticide. It is, happening because of misuse of ultrasound technique by which, people get to know the sex of the child. If it is female, they get it, removed by surgery., , Sexually Transmitted Diseases (STDs), Sexually Transmitted Diseases (STDs) are caused by different, pathogens transmitted by an intimate contact between healthy, person and an infected person.

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39, , CBSE Term II Science X, , Some Common STDs, Infections, Bacterial, infections, , Examples, , Causative Organisms, , Gonorrhoea, , Neisseria gonorrhoeae, , Comments, Contracted on during unprotected sexual intercourse with an infected person., Also passed by an infected mother to the developing foetus., Infects ureter in men and cervix in women., Treatment with antibiotics is effective., Symptoms of gonorrhoea, Discharge of pus from penis and vagina., Burning sensation on urinating., , !, , !, , !, , !, , !, , !, , Syphilis, , Treponema pallidum, , !, , !, , !, , !, , Syphilis is transmitted from person to person by direct contact with syphilis sores, These occur mainly on the external genitals, vagina, anus or in the rectum, can also, occur on lips and mouth., Syphilis can be transmitted during vaginal, anal or oral sexual contact., Pregnant women with the disease can pass it to their unborn children., Can be cured by antibiotics., , Symptoms of syphilis, !, , !, , Viral, infections, , AIDS, (Acquired, Immuno, Deficiency, Syndrome), , HIV, (Human, Immunodeficiency, Virus), , Appearance of sores on body parts., Fever, ulcers, bone pain, liver disease and anaemia. These symptoms slow up, during the tertiary stage of syphilis., , Incurable and fatal as it suppresses the immune system of the body. It can be, transferred in following ways,, during unprotected sexual intercourse with an infected person., sharing needles and transfusion of HIV unscreened blood., from the mother to the child via placenta during pregnancy., , !, , !, , !, , Genital, warts, , HPV (Human, Papilloma Virus), , !, , !, , Causes warts over external genitalia and perianal area., Podophyllum preparations are effective in treatment.

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Chapter, Practice, cut pieces of the two Planaria worms could, regenerate to form the complete worm ?, , PART 1, Objective Questions, , R, , P, !, , Multiple Choice Questions, 1. Which of the following statement is correct, regarding DNA copying?, I. It is a basic event in reproduction., II. Two copies of DNA are produced in a reproducing, cell., III. Copies generated are always identical., IV. It is accompanied by the creation of an additional, cellular apparatus., Codes, (a) I, II and IV, (c) I and II, , S, , Q, (I), , (II), , (a) Only P, (b) R and S, (c) P and Q, (d) P, Q, R and S, , 6. Identify the organisms A and B and mode of asexual, reproduction exhibited by them., , (b) I, II and III, (d) II and IV, , 2. The diagram given below shows, A, , (a) A—Bryophyllum, Vegetative propagation, B—Plasmodium, Multiple fission, (b) A—Plasmodium, Multiple fission, B—Bryophyllum, Vegetative propagation, (c) A—Planaria, Budding, B—Plasmodium, Binary fission, (d) A—Hydra, Budding, B—Rhizopus, Spore formation, , (a) binary fission in Amoeba, (b) multiple fission in Plasmodium, (c) budding in Hydra, (d) binary fission in Leishmania, , 3. Asexual reproduction takes place through budding, in, (a) Amoeba, (c) Plasmodium, , B, , 7. Observe the diagram of pistil of a flower., , (NCERT), , (b) Yeast, (d) Leishmania, , 4. An organism which can completely regenerate from, , A, , B, , its cut body parts is, (a) Paramecium, (c) Planaria, , (b) Amoeba, (d) Rhizopus, , 5. A Planaria worm is cut horizontally from the middle, into two halves P and Q. Another Planaria worm is, cut vertically into two halves R and S. Which of the, , C, D

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41, , CBSE Term II Science X, , Match the labelling referred in Column I and, correlate with the function in Column II., , Match the labelling referred in Column I and, correlate with the function in Column II., Column I, , Codes, A, (a) 2, (b) 3, (c) 1, (d) 4, , Column II, , Column I, , Column II, , A, , 1. Produce ovum, , 2. Receptor of pollen grains., , B, , 2. Site of fertilisation, , C, , 3. Pollen grain travels through it., , C, , 3. Site of implantation, , D, , 4. It ripens as fruit., , D, , 4. Blood and mucus comes out, , A, , 1. Female germ cell., , B, , B, 3, 4, 2, 1, , C, 4, 1, 3, 2, , Codes, A, (a) 1, (c) 2, , D, 1, 2, 4, 3, , B, 3, 4, , C, 2, 1, , 12., , D, 4, 3, , A, (b) 2, (d) 2, , B, 1, 1, , C, 4, 3, , D, 3, 4, , Eggs are produced in ovary, , 8. Carefully study the diagram of flower with labels A to, , Released in Fallopian tube, , D. Select the option which gives correct identification, and main function and/or characteristic., , A, B, D, C, , (a) A—Anther–Formation of ovules, (b) B—Stigma–Receives pollen grains, (c) C—Ovary–Contains pollens which develop into seeds, (d) D—Filament–Lifts anther to disperse pollen grains, , 9. In a flower, the parts that produce male and female, gametes (germ cells) are, , (NCERT Exemplar), , (a) stamen and anther, (b) filament and stigma, (c) anther and ovary, (d) stamen and style, , Fertilised, , Unfertilised, , X, , Y, , Identify the correct option from the given table which, represents the correct fate of egg occurring at X and Y,, respectively., Pregnancy Menstruation, (a) !, , ", , (b) ", , !, , (c) !, , !, , (d) ", , ", , 13. The figure given below shows male reproductive, system with labels (i) to (iv)., , Identify the correct label with its functions., , (i), , 10. Which of the following is not a secondary, , (ii), , reproductive organ?, (a) Fallopian tube, (c) Ovary, , (iii), , (b) Uterus, (d) Vagina, , 11. The figure given below shows a female reproductive, system in humans with labels A to D., A, , D, B, , C, , (iv), , (a) (i) Bladder–It is essential for the mobility of sperms, (b) (ii) Scrotum–It transfers sperm into the vagina of female, (c) (iii) Urethra–It is a common passage for both the sperm, and urine, (d) (iv) Testis–Its secretion form 20-30% of semen, , 14. Which of the following option contains male, reproductive organs of humans?, (a) Seminal vesicle, uterus and penis, (b) Prostate gland, penis and vas deferens, (c) Vagina, testis and penis, (d) Cervix, scrotum and seminal vesicle

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42, , CBSE Term II Science X, , 15. Which of the following is not the function of, , 21. Assertion The uterus prepares itself every month to, , testosterone?, (a) Regulation of characters of puberty in boys, (b) Regulation of production of sperms, (c) Development of bones and muscles, (d) Regulation of metabolism for body growth, , receive a fertilised egg., Reason The ovary releases one egg every month., , 22. Assertion AIDS is an incurable and a fatal bacterial, infection., Reason It suppresses the immune system of the, body., , 16. What happens to a man, when the labelled part X is, cut?, !, , Case Based MCQs, , 23. Read the following and answer the questions from (i), , X, , (a) Sperm ejaculation stops, (b) Testosterone production stops, (c) Urine formation stops, (d) Semen formation stops, , 17. Which among the following diseases is not sexual, transmitted?, , (NCERT Exemplar), , (a) Syphilis, (b) Hepatitis, (c) HIV-AIDS, (d) Gonorrhoea, !, , Assertion-Reasoning MCQs, Direction (Q. Nos. 18-22) For given questions, two, statements are given, one labelled Assertion (A) and, the other labelled Reason (R). Select the correct, answer to these questions from the codes (a), (b), (c), and (d) as given below., (a) Both A and R are true and R is the correct explanation, of A, (b) Both A and R are true, but R is not the correct, explanation of A, (c) A is true, but R is false, (d) A is false, but R is true, , 18. Assertion Individuals produced by asexual, reproduction are known as clones., Reason They are known as clones because they are, genetically identical., , 19. Assertion Ureter forms the common passage for both, the sperms and urine., Reason It never carries both of them at the same, time., , 20. Assertion Vagina is also called as birth canal., Reason During birth, the baby passes through the, vagina., , to (v) given below, Farmers, gardeners and horticulturists have, developed various artificial methods of vegetative, propagation for growing plants in gardens and, nurseries. A very simple method of propagation, involves a piece of the parent plants stem with nodes, and internodes is placed in moist soil., This grows into a new plant. In grafting, the cutting, of a plants are attached to the stem of a rooted plant., The attached cutting becomes a part of the rooted, plant, draws nutrition from it and grows roots at the, joint. Now if it is separated, it grows into a new plant., In layering, one or more branches of the parent plant, are bent close to the ground and covered with moist, soil. The covered portions grow roots and develop, into new plants., (i) Which part of plant is more suitable for vegetative, propagation?, (a) Stem, (c) Root, , (b) Leaves, (d) Bulbils, , (ii) Choose the odd one out., (a) Potato, (c) Bryophyllum, , (b) Sugarcane, (d) Wheat, , (iii) In which type of artificial propagation, stock and, scion are involved?, (a) Tissue culture, (b) Cuttings, (c) Grafting, (d) Layering, , (iv) Which of the following statement is correct about, artificial vegetative propagation of plants?, (a) We get seedless plants by this method, (b) The new plants produced by this method are similar, to the parent plants, (c) Many plants can be grown from just one parent plant, (d) All of the above, , (v) Which of the following method is suitable for, combining the desirable characters of two plants, together in a single plant?, (a) Layering, (c) Grafting, , (b) Cutting, (d) All of these

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43, , CBSE Term II Science X, , 24. Read the following and answer the questions from (i), to (v) given below, The uterus or womb is a hollow, pear-shaped organ, in a women’s lower stomach between the bladder and, rectum. It sheds the lining each month during, menstruation. A fertilised egg becomes implanted in, the uterus and the foetus develops., A, , 3. What is a clone? Why do offspring formed by asexual, reproduction exhibit remarkable similarity?, (NCERT Exemplar), , 4. Colonies of yeast fail to multiply in water, but multiply, in sugar solution. Give one reason for this., (NCERT Exemplar), , 5. Illustrate with example, the division and fragmentation, method of reproduction in living organisms., , 6. List two advantages of vegetative propagation over other, modes of reproduction., , 7. Name a plant in which layering produces a new plant., 8. Write one main difference between asexual and sexual, , B, , mode of reproduction. Which species is likely to have, comparatively better chances of survival, the one, reproducing asexually or the one reproducing sexually?, Give reason to justify your answer., (CBSE 2018), , C, , (i) Name the parts labelled as A and C., , 9. In tobacco plant, the male gametes have twenty four, , (a) Umbilical cord and placenta, (b) Umbilical cord and uterine wall, (c) Fallopian tube and ovum, (d) Cytoplasm and umbilical cord, , chromosomes. What is the number of chromosomes in, the female gamete? What is the number of, (NCERT Exemplar), chromosomes in the zygote?, , 10. Differentiate between self-pollination and cross-, , (ii) What determines the sex of a child?, (a) Chromosome content of the ovum, (b) Chromosome content of the sperm, (c) Number of days between ovulation and fertilisation, (d) Number of days between fertilisation and implantation, , (iii) Which of the following is embedded in the uterine, wall ?, (a) Zygote, (c) Placenta, , (b) Embryo’s head, (d) Eggs, , (iv) Which of the following is a temporary method of, family planning ?, (a) Vasectomy, (c) Copper-T, , (b) Tubectomy, (d) Both (a) and (b), , pollination., , 11. In a bisexual flower inspite of the young stamens being, removed artificially, the flower produces fruit. Provide, a suitable explanation for the above situation., (NCERT Exemplar), , 12. Differentiate between unisexual and bisexual flowers, and give one example of each., , 13. (i) List two reasons for the appearance of, variations among the progeny formed by sexual, reproduction., (ii), , A, B, , (v) Union of male and female gametes take places in, (a) uterus, (c) vagina, , (b) ovary, (d) oviduct, , C, , PART 2, Subjective Questions, !, , D, (CBSE 2016), , Short Answer Type Questions, , 1. What is the importance of DNA copying in, reproduction?, , (NCERT), , 2. Why is variation beneficial to the species, but not, necessary for the individual?, , (a), (b), (c), (d), , Name the part marked as A in the diagram., How does A reach part B?, State the importance of the part C., What happens to the part marked as D after, fertilisation is over?

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44, , CBSE Term II Science X, , 14. (i) Draw a diagram showing germination of pollen on, stigma of a flower and mark on it the following, (CBSE 2020), organs/parts, (a) Pollen grain, (c) Stigma, , (b) Pollen tube, (d) Female germ cell, , 28. If a woman is using a copper-T, will it help in, protecting her from sexually transmitted diseases ?, (NCERT), , 29. Write a short note on family planning., 30. (i) ‘Use of a condom is beneficial for both the sexes, involved in a sexual act.’ Justify this statement, giving two reasons., (ii) How do oral contraceptives help in avoiding, pregnancies?, (iii) What is sex selective abortion? How does it, affect a healthy society? (State any one, (CBSE 2020), consequence)., , (ii) State the significance of pollen tube., (iii) Name the parts of flower that develop after, fertilisation into, (a) Seed, (b) Fruit, , 15. Why cannot fertilisation take place in flowers if, pollination does not occur?, , (NCERT Exemplar), , 16. How are general growth and sexual maturation different, from each other?, , (NCERT Exemplar), , 31. What are the various ways to avoid pregnancy?, Elaborate any one method., , 17. Draw the human female reproductive system and label, the following parts, (i) Which organ produces ovum?, (ii) Where does fertilisation take place?, (iii) Where does implantation of embryo take place?, (CBSE 2015, 2019), , 18. List two functions of ovary of female reproductive, system., , (CBSE 2016), , 19. A newly married couple wants to conceive as quickly, as possible. What is the first sign of pregnancy shown, by the woman ?, , 20. What changes are observed in the uterus if fertilisation, does not occur?, , (NCERT Exemplar), , 21. How does the embryo get nourishment inside the, mother’s body?, , !, , (NCERT Exemplar), , Long Answer Type Questions, , 32. Reproduction is essentially a phenomenon that is, not for the survival of an individual, but for the, (NCERT Exemplar), stability of a species. Justify., , 33. ‘Reproduction helps in providing stability to, population of a species’. Justify this statement., , 34. (i) Name the mode of reproduction of the, following organisms and state the important, feature of each mode, (a) Planaria, (b) Hydra, (c) Rhizopus, (ii) We can develop new plants from the leaves of, Bryophyllum. Comment., , (NCERT, CBSE 2015), , 35. Explain the fertilisation process in plant with the, , 22. What is the function of the umbilical cord ?, 23. Why are testes located outside the abdominal cavity?, 24. Trace the path of sperm during ejaculation and mention, , help of a labelled diagram of a longitudinal section of, a flower., , the glands associated with the male reproductive system, (NCERT Exemplar), and their functions., , 25. What would be the ratio of chromosome number, between an egg and its zygote? How is the sperm, genetically different from the egg?, , 26. State any two methods of contracting an STD other, than the sexual contact., , 27. How can people practice safe sex to avoid contracting an, STD ?, , 36. Define pollination. Explain the different types of, pollination. List two agents of pollination. How does, suitable pollination lead of fertilisation? (CBSE 2019), , 37. Distinguish between pollination and fertilisation., Mention the site and product, of fertilisation in a, flower. Draw a neat, labelled diagram of a pistil, showing pollen tube growth and its entry into the, (NCERT Exemplar), ovule., , 38. Trace the change that takes place in a flower from, gamete formation to fruit formation., , (CBSE 2020)

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45, , CBSE Term II Science X, , 39. Based on the given diagram answer the questions, , 43. Give reasons., , given below., , (i) Placenta is extremely essential for foetal, development., (ii) Blocking of vas deferens prevents pregnancy., (iii) Wind acts as a pollinating agent., (iv) Use of condoms prevents pregnancy., (v) Blocking of Fallopian tubes prevents pregnancy., , B, , A, , Bladder, , 44. List four points of significance of reproductive health, in a society. Name any two areas related to the, reproductive health which have improved over the, past 50 years in our country., !, , D, C, , Testis, , (i) Label the parts A, B, C and D., (ii) Name the hormone secreted by testis and, mention its role., (iii) State the functions of B and C in the process of, (CBSE 2020), reproduction., , Case Based Questions, , 45. Read the following and answer the questions from, (i) to (v) given below, Salman soaked a few seeds of Bengal gram (chana), and kept them overnight. Next morning, he drained, the excess water and covered the seeds with a wet, cloth. He then left the seeds for a day., After one day, he opened the seeds into two parts, and carefully observe the different parts., , 40. (i) Identify the given diagram. Name the parts, labelled as A to E., , (CBSE 2019), , A, , C, D, , B, , E, , (ii) What is contraception? List three advantages of, adopting contraceptive measures., , 41. (i) Write the function of following parts in human, female reproductive system., (a) Ovary (b) Oviduct (c) Uterus, (ii) Describe in brief the structure and function of, placenta., (CBSE 2018), , 42. Trace out the movement and fate of egg in female, body., , (i) Which part of the flower develops into seed?, (ii) Where do the energy required for germination of, seeds come from?, (iii) Which part of the seed emerges out the first?, (iv) Define germination., (v) What are the necessary conditions for seed, germination?, , 46. Read the following and answer the questions from (i), to (v) given below, As soon as boys and girls reach adolescent age,, certain changes start happening in their bodies, under the influence of sex hormones produced in, their bodies. These changes are mostly related to, height, size, voice pitch, physical attributes, etc., The table below shows the average height of boys, and girls upto the age of 18 years.

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46, , CBSE Term II Science X, , Age/Years, , Average Height / cm, Boys, , Testicles are also part of the endocrine system., A, , Girls, I, , 0 (at birth), , 52, , 51, , 1, , 76, , 75, , 2, , 88, , 88, , C, , 3, , 97, , 97, , 4, , 103, , 103, , D, E, , 5, , 110, , 110, , 6, , 118, , 117, , 7, , 125, , 122, , 8, , 131, , 128, , 9, , 135, , 133, , 10, , 141, , 140, , 11, , 145, , 146, , 12, , 150, , 153, , 13, , 156, , 158, , 14, , 164, , 161, , 15, , 169, , 162, , 16, , 172, , 162, , 17, , 174, , 162, , 18, , 175, , 162, , (i) State the changes happening in adolescent boys, and girls., (ii) When does the most rapid growth take place?, (iii) The increase in height in girls almost ceases at, what age?, (iv) Significant spurt in increase of height of boys, occurs at what the age?, (v) What are the changes that are common in both, boys and girls at the age of adolescence?, , 47. Read the following and answer the questions from, (i) to (v) given below., The male has reproductive organs or genitals, that are both inside and outside the pelvis. The, male genitals include the testicles, the duct, system, accessory glands and the penis., In male who has reached sexual maturity, the, two oval-shaped testicles make and store, millions of tiny sperm cells., , B, , H, , G, , F, , (i) Name the organ that acts as both endocrine and, exocrine gland?, (ii) How is the sperm genetically different from the egg?, (iii) What is semen?, (iv) A man wants a surgical operation for family planning., Which part of his reproductive system needs to be, operate?, (v) What would be the ratio of chromosome number, between an egg and its zygote?, , 48. Read the following and answer the questions from, (i) to (v) given below, The term Sexually Transmitted Disease (STD) refers to a, condition passed from one person to another through, sexual contact. However, it is not the only way STDs can, be transmitted., An STD develops without any symptoms early on, or if, any symptoms appear they are often dismissed as regular, infections., At present, there are several type of STDs known which, are caused by different type of pathogens., Some of these STDs are curable, while other are not. The, only full proof way of avoiding an STD is to practice safe, sex., (i) Give two examples of STDs., (ii) Do you think like viruses, bacteria can also cause an, STD? Give an example., (iii) Name a method of contraception which protects us from, acquiring sexually transmitted diseases?, (iv) What are IUCD? Given one example., (v) Emergency contraceptives may prevent pregnancy if, used within 72 hrs of …, … .

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EXPLANATIONS, Objective Questions, 1. (a) Statements I, II and IV are correct, whereas statement, III is incorrect. Incorrect statement can be corrected as, The process of copying of DNA will have some variations, each time. As a result, the DNA copies generated will be, similar, but may not be identical to the original., 2. (d) The given diagram shows how Leishmania reproduces, through binary fission to produce two daughter cells., 3. (b) Asexual reproduction in Hydra and yeast, takes place by, budding., 4. (c) Planaria is a flatworm which possesses high regeneration, ability. If its body somehow gets cut into pieces, then each, piece can regenerate into a complete organism by growing, the missing parts., 5. (d) The Planaria reproduces through regeneration method,, therefore all the cut parts (i.e. P , Q, R and S) of the Planaria, will regenerate to form complete worms., 6. (a) A is Bryophyllum, it reproduces by vegetative, propagation., B is Plasmodium, it reproduces by multiple fission., 7. (a) The given diagram shows various parts of pistil., A is stigma, it helps in receiving the pollen grains during, pollination., B is style, the pollen tube grows out of the pollen grain and, travels through the style., C is ovary, it ripens as fruit after the process of fertilisation., D is ovule, it is the female gamete or female germ-cell of, the flower., 8. (d) Option (d) is correct labelled part with its, functions/characteristics., D is filament, it lifts anther to disperse pollen grains., Other options are incorrect labelled parts with their, functions characteristics and can be corrected as, A is anther, it is a site of pollen formation., B is style, it lifts stigma to receive pollen., C is ovary, it contains ovule which develops into seeds while, ovary forms the fruit., 9. (c) The stamen (male part) contains a swollen topmost part, called anther which contains male gametes, i.e. pollen, grains. Ovary (female part) makes ovules (female gametes), and stores them., 10. (c) The primary reproductive organs or gonads consist of the, ovaries and testes. All other organs, ducts and glands in the, reproductive system are considered as secondary or, accessory reproductive organs. So, ovary is not a secondary, reproductive organ., !, , !, , 12. (c) The given flow chart shows the movement and fate of egg, in female body., When egg gets fertilised it forms zygote which develops, into embryo. Early embryo gets implanted leading to, pregnancy (X)., When egg does not get fertilised, it degenerates and passes, out as menstrual discharge through vagina therefore leads to, menstruation (Y)., 13. (c) Option (c) is correct labelled part with its function., (iii) is urethra, it is a common passage for both the sperm, and urine., Other options are incorrect labelled parts with their, functions and can be corrected as, (i) is prostate gland, its secretion form 20-30% of, semen which is essential for the mobility of sperms., (ii) is penis, it transfers sperm into the vagina of female, during copulation., (iv) is testis, it produces sperms and a male sex hormone, called testosterone., 14. (b) Option (b) contains male reproductive organs of humans., The male human reproductive system consists of seminal, vesicle, prostate gland, testis, vas deferens, urethra, penis,, scrotum, etc., whereas uterus, vagina and cervix are parts of, female reproductive system., 15. (d) Regulation of metabolism for body growth is not a, function of testosterone (but by thyroid hormones)., , !, , !, , 11. (b) The correct matches are as follows, A Oviduct or Fallopian tubes are the site of fertilisation., B Ovary releases one mature egg (ovum) every month., C Vagina, in absence of fertilisation, the lining of uterus, slowly breaks and comes out through the vagina as, blood and mucus., D. Lining of uterus is site for implantation of embryo., Thus, option (b) is correct., , 16. (a) Vasectomy is a minor surgery to block sperm, from reaching the semen that is ejaculated from the, penis., The labelled part X in the given diagram is vas deferens., Once vas deferens are cut, sperms cannot get into the, semen or out of the body. Hence, ejaculation of sperms, stops., 17. (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. It is not a sexual, transmitted disease., 18. (a) Both A and R are true and R is the correct explanation of A., The new individuals produced by asexual reproduction are, always genetically identical to each other and their parents,, hence are known as clones., 19. (d) A is false, but R is true because, In males, urethra forms the common passage for both the, sperms and urine. whereas ureters are tubes that propel, urine from the kidneys to the urinary bladder. It never, carries sperms., 20. (a) Both A and R are true and R is the correct explanation of A., Vagina is called as birth canal, because the fully matured baby, passes through the vagina during birth., 21. (a) Both A and R are true and R is the correct explanation of A., Since, the ovary releases one egg every month, the uterus, also prepares itself every month to receive a fertilised egg., Its lining become thick and spongy to nourish the, developing embryo.

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48, 22. (d) A is false, but R is true because, AIDS (Acquired Immuno Deficiency Syndrome) is caused, by HIV (Human Immunodeficiency Virus), so it is a viral, disease. The virus attacks the body’s immune system and, suppresses it., (i) (a) Stems are more suitable for vegetative propagation, due to the presence of nodes in them., (ii) (d) Some plants undergo vegetative propagation such as, potato (through nodes), sugarcane (through stem),, Bryophyllum (through leaves). Wheat plant does not, undergo vegetative propagation and is grown by sowing, seeds., (iii) (c) In grafting, the cut stem of a rooted plant is called, stock and the cut stem of another plant without roots is, called scion., (iv) (d) The process of growing many plants from one plant, by man-made methods is called artificial vegetative, propagation of plants. The new plants produced by this, method have similar characterstics as their parent, plant. Many seedless plants can grow by this method., Hence, all the given statements are correct about, artificial vegetative propagation., Thus, option (d) is correct., (v) (c) Grafting is a most suitable method of artificial, propagation for combining the desirable characters of, two plants together in a single plant., 24. (i) (a) A is umbilical cord, it is a narrow tube-like, structure that connects the developing baby to the, placenta., C is placenta, it is an organ attached to the lining of, the womb that delivers oxygen and nutrients to the, growing baby., (ii) (b) Sex of a child is always determined by the type of, sex- chromosomes received by the father. Male, produces two types of sperms, either having, X-chromosome or Y-chromosome., (iii) (c) Placenta is embedded in the uterine wall., (iv) (c) Copper-T is an IUD (Intrauterine Device). It, releases copper ions which are toxic to the egg and, sperms. It stops sperm from fertilising the egg., Therefore, it prevents pregnancy for upto 10 years., (v) (d) Oviduct also known as Fallopian tube, is the site of, fertilisation, where male and female gametes fuse to, form zygote. So, union of male and female takes place, in oviducts., 23., , Subjective Questions, 1. The importance of DNA copying during reproduction are, (i) It is responsible for the transmission of parental, characteristics to the offsprings., (ii) During DNA copying in reproduction, the changes, occur due to the inheritance of traits from both the, parents. This leads to certain genetic variations, which, are useful for the evolution of species over a period of, time., 2. Variations allow organisms to exist in diverse habitats or, niches. In its absence, a species may remain restricted to a, particular area. If this area gets drastically altered due to various, natural or man-made causes, the species may be wiped out., , CBSE Term II Science X, , However, if some variations are present in few individuals, it, would help them to colonise other habitats and survive. But, if, variations are present in a single organism, there would be a, very little chance for it to survive and species is lost forever., Hence, variation is beneficial to the species, but not, necessary for the individuals., 3. Clones are the offsprings produced by one parent through, asexual reproduction. These are genetically identical to the, parent. The clones possess exact copies of the DNA of their, parent and hence show remarkable similarity to the parent, and to one another., 4. When the colony of yeast is in water, it does not get, nutrition. Sugar solution, on the contrary provides nutrition., As the yeast gets nutrition and thus energy, it grows and, begins to produce buds. This is why colonies of yeast fail to, multiply in water, but multiply in sugar solution., 5. Multicellular organisms like filamentous algae (Spirogyra), and sea animal called sea anemone on maturation breakup, into two or more small fragments or pieces. Each fragment, subsequently grows to form a complete new organism. This, type of asexual reproduction is known as fragmentation., 6. Two advantages of vegetative propagation are as follows, (i) Vegetative reproduction is easier and faster methods of, reproduction., (ii) It is useful in those plants/animals, which cannot, reproduce sexually., 7. Layering is a type of vegetative propagation, e.g. lemon,, rose, jasmine, strawberry, etc., can produce new plant by the, process of layering., 8. The main difference between sexual and asexual, reproduction involves the production and union of gametes, in the process of fertilisation in sexually reproducing, organisms which do not occur in asexual mode of, reproduction., Sexual reproduction is considered to be superior over, asexual reproduction as it leads to variations, while asexual, reproduction does not induce variations among progeny, individuals., Advantages of variations in individuals are, (i) It brings adaptation in individuals., (ii) It helps in the survival of species., (iii) It is the basis of evolution., Hence, the species that reproduce through sexual, reproduction have better chances of survival., 9. The number of chromosome in the female gamete would, be same as that in the male gamete, i.e. it will have 24, chromosomes. The number of chromosome in the zygote, would be double the number present in the gamete and, hence, it would be 48., 10. Pollination is transfer of pollen grains from anther to the, stigma of a flower. It is of two types, (i) Self-pollination Transfer of pollen from the stamens of, a flower to the stigma of the same flower or on the, stigma of other flower of the same plant., (ii) Cross-pollination Transfer of pollen from the stamens of, a flower to the stigma of different flower of different plant, of same species.

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49, , CBSE Term II Science X, , 11. A bisexual flower has the male as well as female, reproductive organs. If the young stamen (i.e. male unit) is, removed artificially, the flower still has its pistil (i.e. female, unit) intact. Therefore, cross-pollination can occur., When the pollen grains from the anther of another flower, are transferred to the stigma of this flower with the help of, pollinating agents such as insects, bees, wind and water, it, causes cross-pollination. After the pollen grains fall on, stigma, the next step is fertilisation, followed by formation of, fruits and seeds., 12. Stamens and carpels (pistils) are the reproductive organs of a, flower, i.e. organs by which sexual reproduction in floral, plants takes place., Most plants have both male and female reproductive organs, in the same flower and are known as bisexual flowers, e.g., lily, rose, etc., while others have either male or female, reproductive parts in a flower known as unisexual flowers,, e.g. papaya, watermelon, etc., (i) Variations appear among the progeny formed by sexual, reproduction due to the following reasons, (a) Sexual reproduction results in new combinations of, genes that are brought together during the formation, of gametes by meiotic divisions (I and II)., (b) The combination of two sets of chromosomes, one, between the homologous chromosome arms set from, each parent during zygote formation, leads to, variation within a species., (ii) (a) A–Pollen grain, (b) Pollen grain reaches part B, i.e. stigma by, pollinating agents such as insects, wind, water, etc., This process is known as pollination., (c) Part C is pollen tube. It allows the passage for the, male gametes to reach the ovary having female, gamete for fertilisation., (d) Part D, i.e. female gamete or egg cell that forms, zygote after fertilisation., 14. (i), , 13., , (a) Pollen grain, (c) Stigma, , (b) Pollen tube, , (d) Female, germ cell, , (ii) The pollen tube takes its origin from intine of pollen, grains. It grows through the style and reaches the, micropyle of ovule. It carries male nuclei to the ovule, for fertilisation., (iii) (a) Ovule develops into seed., (b) Mature ovary develops into fruit., 15. In a flower, fertilisation requires both male and female, gametes. So, it is necessary that the male gamete reaches the, female gamete. This can happen when the pollen grains are, transferred to the stigma through any means of pollination., , Hence, fertilisation cannot take place in flowers if pollination, does not occur due to the absence of pollen tube (i.e. the male, gamete)., 16. General growth refers to different types of developmental, process in the body like increase in height, weight gain,, changes in shape and size of the body. During this phase, the, reproductive organs develop at a slower rate., During sexual maturation, the changes that occur prepare, the body for sexual reproduction. These are specific changes, reflected at puberty like cracking of voice, new hair patterns,, development of breast in female, etc., 17., , Oviduct or, Fallopian tube, Ovary, , Uterus, Cervix, Vagina, Female reproductive system, , (i) Ovum is produced by ovaries which are paired,, oval-shaped organs., (ii) Oviduct or Fallopian tubes are the site of fertilisation., They have funnel-shaped opening near ovary and carry, ova or egg from ovary to uterus., (iii) Implantation refers to embedding of the embryo in the, thick lining of uterus., 18. Ovary in females is responsible for the production of female, gametes (ova) and also produces female sex hormones, i.e., oestrogen and progesterone., 19. The absence of menstrual cycle may be the first indication of, pregnancy in a woman., 20. If the egg is not fertilised, it lives for about one day. Since,, the ovary releases one egg every month, the uterus also, prepares itself every month to receive a fertilised egg. Its, lining becomes thick and spongy, which is required for, nourishing the embryo. If fertilisation, however, does not take, place this lining is not needed in the absence of fertilisation, and it slowly breaks and comes out through the vagina as, blood and mucus. This cycle takes place roughly every, month and is known as menstruation cycle and usually lasts, for about 2-8 days., 21. The embryo gets nutrition from the mother’s blood with the, help of a special tissue called placenta. This is a disc-like, tissue which develops between the uterine wall and embryo., As mother eats, the food passes through the digestive system, where it breaks down into small particles. These nutrients, travel through the mother’s bloodstream and get exchanged, with the bloodstream of foetus through placenta., 22. The umbilical cord contains blood vessels which supply, blood between the foetus and the placenta., 23. Testes are located outside the abdominal cavity because, sperm formation requires a lower temperature than the, normal body temperature., 24. Path of sperm during ejaculation Formation of sperms take, place in testis. Sperms come out from testis into the vas, deferens. It then unites with another tube called urethra

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50, , CBSE Term II Science X, , (iv) Surgical method for permanent contraception., , coming from the urinary bladder. Along the path of vas, deferens, glands like the prostate and the seminal vesicle, add their secretion, so that sperms are in fluid medium to, make their transport easier. This fluid also provides nutrition., Glands associated with male reproductive system are, (i) Testis It secretes the male sex hormone, testosterone., (ii) Prostate Gland It makes the semen medium alkaline., (iii) Cowper’s Gland It secretion of this gland lubricates, the urethra before ejaculation., (iv) Seminal Vesicle It adds fluid content to semen., , 26. Two methods of contracting an STD other than the sexual, contact are as follows, (i) Sharing needles with an infected person., (ii) Transfusion of STD unscreened blood., 27. People can practice safe sex by using condoms as it acts as, barrier method of contraception and does not allow entry of, semen into vagina. Therefore, prevent STDs and avoid, chances of pregnancy., 28. No, copper-T does not prevent the transmission of sexually, transmitted diseases. Copper-T only prevents implantation., The only safe method that can be used to prevent the, transmission of sexually transmitted diseases is condoms., 29. Family planning refers to the regulation of conception by the, use of contraceptive methods or devices to limit the number, of offspring., The methods used to prevent the occurrence of pregnancy, are called contraceptive methods. These can be barrier,, hormonal, chemical and surgical methods., 30., , (i) Use of a condom is beneficial for both the sexes involved, in a sexual act. It is because of the following facts, (a) It prevents pregnancy which is not desired by a, couple., (b) It saves both the partners from sexually transmitted, diseases like AIDS, etc., (ii) Oral contraceptives are the hormonal pills which are, taken by the females after their menstruation ends up., It is taken for 21 days daily. It changes the cyclic events, of ovulation, etc. So, mature ovum is not available for, fertilisation., (iii) Sex selective abortion means if the foetus is female, it is, killed and extracted. This creates an imbalanced in the, society by disturbing the sex ratio., , 31. Ways to avoid pregnancy are called contraceptive methods., It includes a number of ways such as, (i) Mechanical barrier, e.g. condom., (ii) Drugs (oral pills for females)., (iii) IUCD, e.g. copper-T., , 32. All the living organisms need energy for their survival and, growth. This energy is obtained from various life processes, such as nutrition, excretion and respiration., Thus, these phenomena are essential for the survival of an, individual. Compared to these life processes, reproduction, is not essential for survival of an individual., It is basically important for continuity of the generation of an, organism or species as DNA copying during reproduction, helps to produce similar individuals as their parents to, maintain stability of a species., 33. A species occupies a well-defined niche in an ecosystem,, using its ability to reproduce. During reproduction, copies of, DNA pass from one generation to the next. This copying of, DNA takes place with consistency in reproducing organisms, and this is important for the maintenance of body design, features (physiological as well as structural) which allows the, organism to use that particular niche. Reproduction is, therefore, linked to the stability of population of a species., (i), , (a) Planaria—Regeneration, (b) Hydra—Budding, (c) Rhizopus—Sporulation, (ii) The leaves of Bryophyllum bear vegetative adventitious, buds which on separation can give rise to new plants., 35. Stamens and carpels are the reproductive parts of a flower., Stamen is the male reproductive part of the flower., Anther is a bilobed structure containing two pollen sacs, present at tip of stamen. These produce pollen grains that, are yellowish in colour., Carpel (Pistil) is the female reproductive part, which is, present in the centre of the flower., It comprises of three parts, (i) Stigma It is the terminal part of carpel which may be, sticky. It helps in receiving the pollen grains during, pollination., (ii) Style It is the middle elongated part of carpel. It helps in, the attachment of stigma to the ovary., (iii) Ovary It is the swollen bottom part of carpel. It contains, ovules having an egg cell (female gamete)., 34., , !, , !, , !, , Stigma, Style, Pistil, , Anther, Filament, Petal, , Ovary, , Sepals, , Longitudinal section of flower, , Stamen, , 25. The ratio of chromosome number between egg and its zygote, is 1 : 2. An egg is a female gamete and it has haploid number of, chromosomes. During fertilisation, it fuses with male gamete, (also having haploid number of chromosomes) to form a zygote, which now has diploid number of chromosomes., Sperms and eggs are genetically different in terms of nature, of sex chromosome. The sperm contains either X or, Y-chromosome, whereas an egg will always have an, X-chromosome., , Mechanical Barrier There are a number of methods that, create barrier between sperm and egg, Some of them are as follows, Condom It is a fine rubber balloon-like structure worn over, the penis during sexual intercourse. Semen is collected in it, and not discharged into the vagina. This method also, prevents the spread of STDs like AIDS, syphilis, etc., Diaphragms or Caps It can be fitted in the cervix of a, woman to prevent semen from reaching the Fallopian, tube.

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51, , CBSE Term II Science X, , The pollen grain is transferred from the stamen to the stigma. It, is transferred 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. After the pollen, lands on a suitable stigma, it has to reach the female germ cells, which are present in the ovary., For thus, a tube grows out of the pollen grain and travels, through the style to reach the ovary., , Fertilisation is the process of fusion of male germ cells with, the female gametes. It gives rise to a zygote. As soon as the, pollen lands on suitable stigma, it reaches the female germ, cells in ovary. This occurs via pollen tube. The pollen tube, grows out of the pollen grain, travels through the style and, finally reaches the ovary., 36. The transfer of pollen grains from the anther of the stamen, to the stigma of a flower is termed as pollination. There are, two types of pollination, (i) Self-pollination The pollen from the stamen of a, flower is transferred to the stigma of the same flower, or another flower on the same plant., (ii) Cross-pollination The pollen from the stamen of a, flower is transferred to the stigma of another flower of, different individual of the same species., , Pollen grains, Stigma, Male germ cell, Pollen tube, , The pollen grains can be transferred by various agents like, wind, water, insects and animals. As soon as the pollen, lands on suitable stigma, it reaches the female germ cells in, ovary. This occurs via pollen tube., The pollen tube grows out of the pollen grain, travels, through the style and finally reaches the ovary where it, fuses with female gamete (ovule) to give rise to zygote., Hence, pollination is followed by fertilisation in plants., , Ovary, Female, germ cell, , Germination of pollen on stigma The male germ cell, produced by pollen grain fuses with the female gamete precut in, the ovule. This fusion of germ cells is called fertilisation and, gives rise to the zygote., After the fertilisation, the zygote divides several times to form an, embryo within the ovule. The ovule develops a hard coat and is, gradually converted into a seed. The ovary grows rapidly and, ripens to form fruit. Meanwhile the petals, sepals, stamens, style, and stigma may shrivel and fall off., , 37. Distinguishes between pollination and fertilisation are as, follows, , Pollination, , Fertilisation, , It is the transfer of pollen grains It is the fusion of male and, from anther to the stigma., female gametes., It is a physical process., , It is a biological process., , The site of fertilisation is ovule in the ovary., The product of fertilisation is a zygote., Pollen grain, Stigma, Male germ cells, Pollen tube, , Ovary, , Female, germ cell, , Pistil showing pollen tube growth, and its entry into ovule, , 38. Stamen is the male reproductive part and it produces, pollen grains. The ovary contains ovules and each ovule has, egg cell., , 39., , (i) A–Ureter, , C–Urethra, , B–Seminal vesicle, , D–Vas deferens, , (ii) Testosterone hormone is secreted by testis. It controls, spermatogenesis (formation of sperm) and secondary, sexual characters in male adolescents., (iii) Seminal vesicle B temporarily stores sperms., Urethra (C ) It transports and releases urine and sperms, outside the body., 40. (i) The given figure represents the female reproductive, system. The parts labelled as A-E are, A. Oviduct or Fallopian tube B. Ovary, C. Uterus, D. Cervix, E. Vagina, (ii) The prevention of pregnancies by using artificial method is, called as contraception., Advantages of using contraceptive measures are, (a) To control family size, population rise or birth rate., This is done by creating awareness about small, families using contraceptive measures., (b) To prevent chances of meeting female egg and male, sperm, thus preventing future unwanted pregnancies., (c) Use of barrier methods of contraception protects both, the partners from contracting sexually transmitted, diseases like AIDS.

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52, , CBSE Term II Science X, , (i) Prevention of unwanted pregnancies by using, contraceptives have shown the development of health in, women., (ii) Awareness of advantages of small families by using, contraceptives has led to economic growth of the family., , 41. (i), , Parts of female, reproductive system, , Functions, , (a) Ovaries, , Produce thousands of ova or egg, cells. Secrete female sex hormones, like oestrogen and progesterone., , (b) Oviduct, (Fallopian tube), , Carries ova or egg from ovary to the, uterus. It is the site of fertilisation., , (c) Uterus (Womb), , Here, the growth and development, of foetus (embryo) take place., Rhythmic contractions of the, muscles in the uterus cause labour, pain and childbirth., , (ii) Structure of Placenta It is a disc between uterine wall, and embryo which is embedded in the uterine wall. It, contains villi on the embryo's side of the tissue. On the, mother's side, blood spaces are present, which, surrounded the villi., Functions of Placenta It provides a large surface area, for glucose and oxygen to pass from the mother to the, embryo. It also removes the waste generated by embryo,, transferring it to mother's blood., 42. Movement and fate of egg in female body, Eggs are produced in ovary, Released in Fallopian tube, , 43., , If fertilised,, form zygote, , If unfertilised,, degenerate, , Early embryo gets implanted, leading to pregnancy and, foetus starts to develop, , Passes out as, menstrual discharge, , (i) Placenta is extremely essential for foetal development, because it helps in nutrition, respiration, excretion, etc.,, of the foetus through the maternal supply., (ii) Blocking of vas deferens prevents passage of sperms,, hence, there is no fertilisation so it prevents pregnancy., (iii) Wind acts as a pollinating agent because it helps in, transfer of light weighted pollen grains from anther to, stigma of a flower., (iv) Condoms prevent entry of sperms into vagina, hence, prevents pregnancy., (v) If Fallopian tube is blocked, sperm and egg do not meet or, fuse and fertilisation does not take place., , 44. Reproductive health is a state of physical, emotional, mental, and social well-being in relation to sexuality., Significance of reproductive health in a society, (i) It prevents the spread of various Sexually Transmitted, Diseases (STDs)., (ii) Proper medication and checkups will help in the, production of healthy children., (iii) Better sex education and awareness help in maintaining, the population and prevent the population explosion., (iv) Unwanted pregnancies are avoided., The reproductive health in India has improved, tremendously over the past 50 years. The areas in which, reproductive health has improved includes, , (i) The ovules present in ovary of a flower develops into, seeds., (ii) Cotyledons of seed store food which fulfil the energy, requirement for seed germination., (iii) Radicle or future root of the plant emerges out the first, from the germinating seed., (iv) The process of developing seed into a seedling under, appropriate conditions is known as germination., (v) All seeds need water, oxygen and proper temperature in, order to germinate., 46. (i) The changes happening in adolescent boys and girls are, The moustach starts appearing in boys and their voice, becomes hoarse. There is onset of menstrual cycle in, girls and their mammary glands starts developing., (ii) Most rapid growth takes place within one year after the, birth of a baby., (iii) The increase in height of girls ceases at the age of 15 years., (iv) In boys, significant spurt in height occurs at the age of, 11-12 yrs., (v) Both boys and girls grow body hair in their pubic area,, as well as under the arms and on the legs at the age of, adolescence., , 45., , (i) The two main functions of the testes (F) are to produce, sperm and to produce the male sex hormones, (testosterone). This makes the testis both an endocrine, and exocrine gland., (ii) The sperm and eggs are genetically different terms of, nature of sex chromosome. The sperm contains either X, or Y-chromosome whereas an egg will always have an, X-chromosome., (iii) Semen is a fluid which contains sperm cells and, secretion of accessory glands. It is a milky, viscuous fluid, contains fructose, proteins and other chemicals for, nourishing sperms., (iv) Vasectomy is a form of male birth control that cuts the, supply of sperm to semen. It is done by cutting and, sealing the tubes that carry sperm, i.e. vas deferens (H)., (v) Fusion of a sperm and an egg lead to the formation of, a zygote. Therefore, a zygote is diploid in nature,, i.e. 2n, whereas a sperm and an egg are haploid in, nature, i.e. n. Hence, the ratio of chromosome number, between an egg and its zygote is 1 : 2., 48. (i) (a) AIDS (Acquired Immuno Deficiency Syndrome), (b) Genital warts, (ii) Yes, bacteria are also known to cause STDs., For example, Syphilis is an STD caused by bacteria,, Treponema, pallidum., (iii) Condoms protect us from acquiring STDs. It also helps, in avoiding pregnancy., (iv) IUCD stands for Intra-Uterine Contraceptive Device. It, is used to prevent pregnancy, e.g. Copper-T, (v) Emergency contraceptives may prevent pregnancy if, used within 72 hrs of coitus/intercourse., 47.

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Chapter Test, Multiple Choice Questions, , 1. Exchange of genetic material takes place in, (a) vegetative reproduction, (c) sexual reproduction, , (b) asexual reproduction, (d) budding, , 2. Identify the option that indicates the correct function or, purpose served by A, B and C., B, A, C, , (a) Both A and R are true and R is the correct explanation, of A, (b) Both A and R are true, but R is not the correct, explanation of A, (c) A is true, but R is false, (d) A is false, but R is true, , 6. Assertion Pollen grains from the carpel stick to the, stigma of stamen., Reason The fertilised egg cells grow inside the, ovary and become seeds., , 7. Assertion Menstruation is the regular discharge of, blood from the thick uterine lining., Reason It occurs when egg is fertilised by sperm., , 8. Assertion HIV-AIDS is a bacterial disease., (a) A-Future shoot, B-Future root, C-Store food, (b) A-Future root, B-Future shoot, C-Store food, (c) A-Store food, B-Future shoot, C-Future root, (d) A-Future root, B-Store food, C-Future shoot, , 3. Which of the following option shows the correct, arrangement of events in logical sequence in the life cycle, of a human?, (a) Menarche → Gestation → Menopause → Parturition →, Insemination, (b) Parturition → Gestation → Insemination → Menarche →, Menopause, (c) Menarche → Insemination → Gestation → Parturition →, Menopause, (d) Menarche → Insemination → Parturition → Gestation →, Menopause, , 4. Select the barrier methods of contraception among the, following., , Reason It spreads through sharing of infected, needles., , Short Answer Type Questions, , 9. List out the basic features of reproduction., 10. List two advantages of vegetative reproduction, practiced in case of an orange., , 11. Mention any one disadvantage of producing new, plants by vegetative propagation., , 12. Write the name of gamete producing parts of a, flower., , 13. When does pollen tube develop in a flower?, 14. Write the dual purposes served by urethra in males., 15. What is placenta? Describe its structure. State its, functions in case of pregnant human female., , 16. What do you understand by reproductive health, and sex ratio?, , (a) Oral pill, (b) Femidom, (c) Copper-T, (d) Vasectomy, , Long Answer Type Questions, , 17. Given below is the diagram of a reproductive system., A, , 5. Identify the disease which is caused by a virus., I. Syphilis, III. Gonorrhoea, , I, B, , II. Genital warts, IV. AIDS, , C, D, , Codes, (a) I and II, (c) II and IV, , (b) II and III, (d) I and III, , G, , Assertion-Reasoning MCQs, , F, , Direction (Q. Nos. 6-7) Each of these questions contains, two statements, Assertion (A) and Reason (R). Each of, these questions also has four alternative choices, any one, of which is the correct answer. You have to select one of, the codes (a), (b), (c) and (d) given below., , (i) Name the system., (ii) Name the parts labelled as A-I., (iii) Describe the functions of partsC,D,E and F., , 18. What are the different methods of contraception?, , Answers, Multiple Choice Questions, 1. (c), , 2. (c), , 3. (c), , 4. (b), , Assertion-Reasoning MCQs, 6. (d), , 7. (c), , 8. (d), , E, , H, , 5. (c), , For Detailed Solutions, Scan the code

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CHAPTER 04, , Heredity and, Evolution, In this Chapter..., !, , Accumulation of Variations During Reproduction, , !, , Heredity (Inheritance of Traits), , !, , Mendel’s Contribution towards the Inheritance of Traits, , !, , Experiment Conducted by Mendel, , !, , Sex-Determination, , Through the process of reproduction individuals give rise to, new individuals that are similar (not same) to the parents., This similarity in progeny or offspring or child is due to, transmission of characters or traits from parents to their, progeny., The transfer of characters from parents to offspring is known, as heredity and the process through which characters or traits, pass from one generation to another is called inheritance., , Heredity (Inheritance of Traits), Traits or characteristics, which are passed on from parents to, their offspring (generation to generation) are controlled by, genes., A gene is a unit of DNA which governs the synthesis of one, protein that constants a specific character of an organism., e.g. Inheritance of free or attached earlobes., , Accumulation of Variations During, Reproduction, The difference in the characters among the individuals of a, species is termed as variations. These variations, are accumulated by the process of sexual reproduction., Depending upon the nature of variations, different, individuals would have different advantages, the most, important advantage of variation to a species is that it, increases the chances of its survival in a changing, environment., , (a) Free earlobe, , (b) Attached earlobe, , Rules for inheritance of traits Inheritance of a trait is related, to the fact that both father and mother contribute equally, towards the genetic makeup of their offspring, i.e. for each, trait two versions are available in the child.

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55, , CBSE Term II Science X, , Some Important Terms and Definitions Used in Heredity, Terms, , Definitions, , Chromosome A long thread-like structure in the nucleus. It appears, during cell division and carries genes., Gene, , A functional unit of heredity. It is present on chromosome., It is a piece of DNA that codes for one protein that inturn, determines a particular character (phenotype)., , Character, , The feature or characteristic of an individual like height,, colour, shape, etc., , Trait, , An inherited character, i.e. feature, which is normally, inherited and has its detectable variant too, e.g. tall and, dwarf are traits of a character, i.e. height., , Allele, , One of the different forms of a particular gene, occupying, the same position on a chromosome., , Hybrid, , An individual having two different alleles for the same trait., , Dominant, allele, , An allele, whose phenotype will be expressed even in the, presence of another allele of that gene. It is represented by, a capital letter, e.g. T., , Recessive, allele, , An allele, which gets masked in the presence of a dominant, allele and can only affect the phenotype in the absence of a, dominant gene. It is represented by a small letter, e.g. t., , Genotype, , Genetic composition of an individual., , Phenotype, , The expression of the genotype, which is an observable or, measurable characteristic., , Back cross, , Crossing F1 hybrid with one of its parents,, e.g. Tt × tt or Tt × TT., , Experiments Conducted By Mendel, More than a century ago, Mendel worked out the main, rules for inheritance. He performed following two, experiments, 1. Monohybrid Cross : Inheritance of Traits for, One Contrasting Character, Mendel took pea plants with different characteristics, such as height (tall and short plants)., The progeny produced from them (F1 -generation, plants) were all tall. Mendel then allowed F1 progeny, plants to undergo self-pollination., In the F2 -generation, he found that all plants were not, tall, three quarter were tall and one quarter of them, were short. This observation indicated that both the, traits of shortness and tallness were inherited in, F1 -generation. But, only the tallness trait was, expressed in F1 -generation., Two copies of the traits are inherited in each sexually, reproducing organism., , !, , !, , !, , !, , Monohybrid A hybridisation cross in which inheritance of only one pair, cross, of contrasting characters is studied., Dihybrid, cross, , Tall, Short, (TT) × (tt), , A cross in which inheritance of two pairs of contrasting, characters is simultaneously studied., , Homozygous A condition in which an individual possesses a pair of, identical alleles controlling a given character and will, breed true for this character (e.g. occurrence of two, identical alleles for tallness in a P1 tall pea plant)., , P×P, , Reproductive cells containing only one set (haploid) of, dissimilar chromosome., , Test cross, , Crossing F1 heterozygote with homozygous recessive, parent, e.g. F1 hybrid tall plant (Tt) with pure dwarf, plant (tt)., , Mendel’s Contribution towards the, Inheritance of Traits, The Austrian monk, Gregor Johann Mendel is known as Father of, Genetics. He performed many experiments on pea (Pisum sativum), plant related to hybridisation., He studied seven pairs of contrasting characters in pea plants and, only one character at a time., , F1, , Mendel’s experiment showing law of dominance, !, , Heterozygous A condition in which an individual has a pair of contrasting, alleles for any one character and will not breed true for this, character (e.g. simultaneous existence of dominant and, recessive alleles in F1 - hybrid tall pea plant)., Gametes, , All tall offsprings, (Tt), , TT and Tt are phenotypically tall plants, whereas tt is, a short plant. For a plant to be tall, the single copy of, ‘T’ is enough. Therefore, in traits Tt, ‘T’ is a dominant, trait, while ‘t’ is a recessive trait., , Tall, Tall, Tall, Tall, Tall, Short, (Tt) × (Tt) F1, (TT), (Tt), (Tt), (tt) F2, Mendel’s experiment showing law of segregation, !, , In F2 -generation, both the characters are recovered,, though one of these is not seen in F1 stage. During, gamete formation, the factor or allele of a pair segregate, from each other., Thus, the phenotypic ratio is 3 : 1 and the genotypic ratio, is 1 : 2 : 1 for the inheritance of traits for one contrasting, character, i.e. monohybrid cross.

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56, , CBSE Term II Science X, , 2. Dihybrid Cross : Inheritance of Traits for Two, Visible Contrasting Characters, Mendel took pea plants with two contrasting characters, i.e., one with a green round seed and the other one with a yellow, wrinkled seed., When the F1 progeny was obtained, they had round and, yellow seeds, thus establishing that round and yellow are, dominant traits., Mendel then allowed the F1 progeny to be self-crossed, (self-pollination) to obtain F2 progeny. He found that seeds, were round yellow, round green, wrinkled yellow and some, were wrinkled green., The ratio of plants with above characteristics was 9 : 3 : 3 :1,, respectively (Mendel observed that two new combinations, had appeared in F2 )., !, , !, , sufficient protein will be produced for normal body functions., If the gene for a specific protein is altered, the protein will be, less efficient or will not be functional at all., , Mechanism of Inheritance, !, , !, , !, , In F2 -generation, all the four characters were assorted out, independent of the others. Therefore, he said that a pair of, alternating or contrasting characters behaves independently, of the other pair. For example, seed colour is independent of, seed coat. The independent inheritance of two separate traits, shape and colour of seeds is schematically shown below, Parents, RRyy, rrYY, ×, (Round green), (Wrinkled yellow), , ↓, , ↓, , Ry, , rY, , ↓, , ↓, , Gametes, F2-generation, , RrYy, (Round yellow), ×, F1, F2-generation, , F1, , RY, , Ry, , rY, , ry, , RRYY, , RRYy, , RrYY, , RrYy, , RRYy, , RRyy, , RryY, , Rryy, , RrYY, , RrYy, , rrYY, , rrYy, , RrYy, , Rryy, , rrYy, , rryy, , RY, , Ry, , !, , Both the parents contribute a copy of the same gene to their, progeny. Each germ cell thus, has one set of gene, present as, chromosome. Each cell of the body will have two copies of, each chromosome, one inherited from each parent., When two germ cells combine, they restore the normal, number of chromosomes in the progeny. This ensures the, stability of the DNA of species. Such mechanism of, inheritance explains the result of Mendel’s experiments. It, is used by all sexually and asexually reproducing organisms., , Sex-Determination, It is the process by which sex of a newborn individual is, determined. Different strategies can determine sex in, different species. For example, in reptiles environment, factors such as temperature at which fertilised eggs are kept, determine sex of the offspring. The determination of sex, occurs largely by genetic control in human beings. In human, beings, there are 23 pairs of chromosomes, out of which, 22 pairs are autosomes and one pair is sex-chromosomes., Females have a perfect pair of sex chromosome, (homogametic), but males have a mismatched pair, (heterogametic) in which one is X (normal sized) and the, other is Y-chromosome (short in size)., Hence, an egg fertilised by X-chromosome carrying sperm, results in a zygote with XX, which becomes a female and if an, egg is fertilised by Y-chromosome carrying sperm, it results in a, XY zygote that becomes male., Thus, the sex of the children will be determined by what they, inherit from their father. A child who inherits an, X-chromosome will be a girl and one who inherits a, Y-chromosome will be a boy. The inheritance of sex in, humans is diagrammatically shown below, , rY, , ry, , XY, , F2 -generation 315 round yellow, , Ratio, 9, , Male, , 108 round green, , 3, , 101 wrinkled yellow, , 3, , Gametes, , 32 wrinkled green, , 1, , X, , 556 seeds, , XX, , Female, , Y, , X, , 16, , Expression of Traits, Cellular DNA is the source of information for making, proteins in the cell. A section of DNA that provides, information for one particular protein is called a gene for that, protein. Expression of trait in body depends on the, functioning of a gene. If the gene is working normally,, , Zygote, XX, , XY, , ↓, , ↓, , Offsprings, Female, , Male, , Sex-determination in human beings

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Chapter, Practice, PART 1, Objective Questions, !, , Multiple Choice Questions, , 1. The process through which characters pass from one, generation to another is called, (a) inheritance, (c) variation, , (b) heredity, (d) evolution, , 2. Which of the following statements is not true with, respect to variation?, (a) All variations in a species have equal chance of survival, (b) Change in genetic composition results in variation, (c) Selection of variants by environment factors forms the, basis of evolutionary processes., (d) Variation is minimum in asexual reproduction., , 3. Free earlobes are ...(i)... traits and attached earlobes, are ...(ii)... traits., (a) (i) recessive, (ii) dominant, (b) (i) dominant, (ii) recessive, (c) (i) inherited, (ii) dominant, (d) None of the above, , 4. A trait in an organism is influenced by, (a) paternal DNA only, (NCERT Exemplar), (b) maternal DNA only, (c) both maternal and paternal DNA, (d) neither by paternal nor by maternal DNA, , 5. Which amongst the listed tools was used to study the, law of inheritance in pea plant by Gregor J Mendel?, (a) Family tree, (c) Punnett square, , (b) Pedigree chart, (d) Herbarium sheet, , 6. A cross between a tall plant (TT) and short pea plant, (tt) resulted in progeny that were all tall plants, (NCERT Exemplar), because, (a) tallness is the dominant trait, (b) shortness is the dominant trait, (c) tallness is the recessive trait, (d) height of pea plant is not governed by gene ‘T’ or ‘t’, , 7. If a pea plant with round green seed (RRyy) is crossed, pea plant with wrinkled yellow seed (rrYY), the seeds, produced in F 1 -generation are, (a) wrinkled and green, (c) round and yellow, , (b) wrinkled and yellow, (d) round and green, , 8. The genotype of the height of an organism is written, as Tt. What conclusion may be drawn?, (a) The allele for height has at least two different genes, (b) There are atleast two different alleles for the gene for, height, (c) There are two different genes for height, each having a, single allele, (d) There is one allele for height with two different forms, , 9. The first generation cross of pure tall and short pea, plant gives tall pea plants. F2 -generation will give, (a) dwarf pea plant, (b) tall pea plant, (c) Both tall and dwarf plants (d) None of these, , 10. A cross between a pure-breed pea plant A and B is, shown below., Parents plants : A × B, F1 -generation : A, F2 -generation : A, A, A, B, Choose the correct option for A and B., (a) A are tall and B are round, (b) A are tall and B are dwarf, (c) A are dwarf and B are tall, (d) A are round and B are tall, , 11. If pea plants having round green seeds and wrinkled, yellow seeds are crossed, what phenotypic ratio will, be obtained in F2 progeny plants?, (a) 1 : 2 : 1, , (b) 3 : 1, , (c) 9 : 3 : 3 : 1, , (d) 9 : 3 : 4, , 12. The result of a dihybrid cross between two, individuals is recorded as, Phenotypes of, progeny, , Numbers of seeds, obtained, , Round, A, , 315, , Round, B, , 108, , C, Yellow, , 101, , Wrinkled, D, , 32

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58, , CBSE Term II Science X, , Choose the correct option for A , B , C and D., (a), (b), (c), (d), , A, Green, Yellow, Yellow, Green, , B, Yellow, Green, Yellow, Green, , C, Round, Wrinkled, Wrinkled, Round, , 15. Choose the correct statement from the following., , D, Yellow, Green, Green, Yellow, , I. Variation in plants are much lesser than human, beings., II. Each trait in child is influenced by only paternal, DNA., III. An individual having two different alleles for the, same trait is called hybrid., IV. Traits that are passed on from parents to their, offspring are controlled by genes., , 13. Carefully study the cross shown below with labels (i),, (ii) and (iii). Identify the option that indicates the, correct labellings., , Codes, (a) I, II and III, (c) II, III and IV, Purple, , (i) White, , !, , (ii), (iii), , (a), (b), (c), (d), , Purple Purple, , White, , (i) Cross-fertilisation, (ii) Self-fertilisation, (iii) F1 -generation, (i) Self-fertilisation, (ii) Cross-fertilisation, (iii) F 2 -generation, (i) Cross-fertilisation, (ii) Self-fertilisation, (iii) F 2 -generation, (i) Self-fertilisation, (ii) F 2 -generation (iii) Cross-fertilisation, , 14. Observe the diagram given below., , 16. Assertion Dominant allele is an allele whose, , 1. Y sperm, , B., , 2. Ovum, , 19. Assertion All the human female gemetes will have, , C., , 3. X sperm, , D., , 4. Female zygote, , (C), , Male zygote, , Match the labelling referred in Column I and correlate, with Column II., Column I, , C, 2, 1, 3, 4, , (d) A is false, but R is true, , A., , (B), , (D), , B, 1, 3, 1, 2, , (a) Both A and R are true and R is the correct explanation, of A, (b) Both A and R are true, but R is not the correct, explanation of A, (c) A is true, but R is false, , phenotype expresses even in the presence of, another allele of that gene., Reason It is represented by a capital letter, e.g. T., 17. Assertion Mendel self-crossed F 1 progeny to, obtain F2 -generation., Reason F1 progeny of a tall plant with round seeds, and a dwarf plant with wrinkled seeds are all dwarf, plants having wrinkled seeds., 18. Assertion The ratio of F2 plants when Mendel took, pea plants with two contrasting characters was, 9 : 3 : 3 : 1., Reason The ratio of F2 plants when Mendel took, pea plants with one contrasting character was 1 : 1., , (A), , Codes, A, (a) 3, (b) 2, (c) 2, (d) 1, , Assertion-Reasoning MCQs, Direction (Q. Nos. 16-20) Each of these questions, contains two statements, Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given, below., , Purple, , Purple, , (b) I, III and IV, (d) I, II and IV, , D, 4, 4, 4, 3, , Column II, , only X-chromosome., Reason Females are homogametic with two, X-chromosomes., , 20. Assertion The sex of a child will be determined by, chromosome received from the father., Reason A human male has one X and one, Y-chromosome.

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59, , CBSE Term II Science X, , !, , His experiments with garden pea along with the, inferences drawn together constitute, the foundation, of modern genetics., Mendel’s contributions were unique because of the, use of distinct variables and application of, mathematics to the problem. He kept the record of, each generation separately and studied the, inheritance of only one pair of characters at a time., (i) Mendel took ......... contrasting characteristics of pea, plants., , Case Based MCQs, , 21. Read the following and answer the questions from (i), to (v) given below, Mendel’s experiment on sweet pea plants having, axial flowers with round seeds (AARR) and, terminal flowers with wrinkled seeds (aarr) is shown, below., Axial, Round, AARR, , Parents, , Gametes, , AR, AaRr, , Terminal, wrinkled, aarr, , ar, , AR, AaRr, , AaRr, , (a) eight, (c) six, , (ii) After cross pollination of true-breeding tall and, dwarf plants, the F 1 -generation was self-fertilised., The resultant plants have genotype in the ratio, , ar, AaRr, , (a)1 : 2 : 1 (homozygous tall : heterozygous tall :, homozygous dwarf), (b)1 : 2 : 1 (homozygous tall : dwarf : heterozygous tall), (c) 3 : 1 (tall : dwarf), (d) 3 : 1 (dwarf : tall), , (i) Phenotype of F1 progeny, (a) axial round, (b) axial wrinkled, (c) terminal wrinked, (d) terminal round, , (iii) Which Mendelian law states that inheritance of one, character is always independent to the inheritance, of other character within the same individual?, , (ii) Phenotype of F 2 progeny produced upon by the, self-pollination of F1 progeny, (a) axial round and axial winkled, (b) terminal round, (c) terminal wrinkled, (d) All of the above, , (a) Law of dominance, (b) Law of segregation, (c) Law of independent assortment, (d) Both (b) and (c), , (iii) The phenotypic ratio of the F 2 -generation will be, (a) 3 : 1, (c)1 : 2 : 1, , (iv) Which one is the possible progeny in, F 2 -generation of pure breed tall plant with round, seed and short plant with wrinkled seed?, , (b) 9 : 3 : 3 : 1, (d)1 : 3, , (iv) A cross between two individuals results in 9 : 3 : 3 :1, for four possible phenotypes of progeny. This is an, example of a, (a) dihybrid cross, (c) monohyorid cross, , (b) seven, (d) five, , (a) Tall plant with round seed, (b) Tall plant with wrinkled seed, (c) Short plant with round seed, (d) All of the above, , (b) F1 -generation, (d) test cross, , (v) Round and yellow seed is, , (v) Which of the following law is mainly explained by, the given cross?, (a) Law of dominance, (b) Law of segregation, (c) Law of independent assortment, (d) None of the above, , 23. Read the following and answer the questions from (i), to (v) given below, Gregor Johann Mendel is known as a ‘Father of, Modern Genetics’ for his work in the field of the, genetics. He worked out the main rules for, inheritance patterns. The heredity in most of the, living organisms is found to be regulated by certain, definite principles., Mendel opted for garden pea (Pisum sativum) to, conduct his experiments. He performed, self-pollination and cross-pollination to understand, the inheritance patterns of traits., , (a) recessive, (c) hybrid, , (b) dominant, (d) incomplete dominance, , PART 2, Subjective Questions, !, , Short Answer Type Questions, , 1. What do you mean by heredity? Who is its founder?, (CBSE 2020), , 2. If a trait ‘A’ exists in 10% of the 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?, , 3. How does the creation of variations in a species, promote survival?

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60, , CBSE Term II Science X, , Recompile the above observations and explain the, law of inheritance associated with them., , 4. In any population, no two individuals are absolutely, similar. Why?, , 5. A child questioned his teacher that why do organisms, , 16. Name the plant Mendel used for his experiments., 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, (CBSE Delhi 2019), F2 -generation plants., , resemble their parents more as compared to, grandparents. In which way, will the teacher explain, (CBSE 2015), to the child?, , 6. Why did Mendel choose pea plant for his experiments?, 7. How do Mendel’s experiments show that traits may, dominant or recessive?, , 17. Study the following cross showing self-pollination in, F1 progeny. Fill in the blank and answer the, questions that follows., , (CBSE 2016), , 8. What do you mean by dominant and recessive, 9. Explain difference between phenotype and genotype., 10. In a pea plant, find the contrasting trait if, (i) the position of flower is terminal, (ii) the flower is white in colour, (iii) shape of pod is constricted, , 13. A man with blood group A married a woman with blood, group O and their daughter has blood group O. Is this, information enough to tell you which of the traits blood, group A or O is dominant? Why or why not? (NCERT), , 14. Write the phenotypic ratio of progeny of, (2018, 15), F2 -generation of a dihybrid cross., 15. Consider the cross between two parents with, contrasting characteristics given below., Parents, RRyy, rrYY, ×, (Round green), (Wrinkled yellow), Gametes, , Ry, , beings?, , RrYy, (Round yellow), ×, , F2 -generation 315 round yellow, 108 round green, 101 wrinkled yellow, 32 wrinkled green, 556 seeds, , significant role in determining the sex of a newborn?, (NCERT Exemplar), , Long Answer Type Questions, 20. ‘A trait may be inherited, but may not be expressed’., , !, , Justify this statement with the help of a suitable, example., , 21. Discuss Mendelian laws with examples., 22. What do you understand by Mendel’s dihybrid, cross? When pea plants having round and yellow, seeds were crossed with plants having wrinkled and, green seeds and then all the plants of F 1 -generation, had round and yellow seeds. Find out the, phenotypic ratio of F2 -generation., yellow seeds, following types of seeds were, obtained by Mendel, , F1-generation, , F1, , 19. Does genetic combination of mother play a, , 23. After self-pollination in pea plants with round,, , rY, , F1, , F1-generation, , 18. How is the sex of a child determined in human, , 12. In a certain species of animal black fur (B) is dominant, over brown fur (b). Predict a genotype and phenotype, of the offspring when both parents are Bb or have, heterozygous black fur., , × ……, , (i) In above question, what is the combination of, characters in the F 2 progeny? What are the ratios?, (ii) Give reasons for the appearance of new combination, of characters in the F1 progeny. (NCERT Exemplar), , 11. How do Mendel’s experiments show that traits are, (CBSE 2016), , RrYy, (Round yellow), , (CBSE 2015), , inherited independently?, , (Round green), , Parents, × rrYY, (Wrinkled yellow), , RRyy, , characteristics?, , Ratio, 9, , Seeds, , Numbers, , Round, yellow, , 630, , Round, green, , 216, , 3, , Wrinkled, yellow, , 202, , 3, , Wrinkled, green, , 64, , 1, 16, , Analyse the result and describe the mechanism of, inheritance which explains these results. (CBSE 2020)

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61, , CBSE Term II Science X, , 24. ‘In humans, there is a 50% probability of the birth of a, boy and 50% probability that a girl will be born’. Justify, the statement on the basis of the mechanism of, (CBSE 2020), sex-determination in human beings., !, , Case Based Questions, , 25. Read the following and answer the questions from (i), to (v) given below, Inheritance from the previous generation provides, both, a common body design and subtle changes in it,, for the next generation. Now, when the new, generation reproduces, the second generation, produced will have variations that they inherit from, the first generation, as well as newly created, differences.], For example, if one bacterium divides and gives rise, to two individuals each of them will divided again, and give rise to two other individuals in the next, generation., The four individual bacteria generated would be very, similar with minor differences that occurred due to, small inaccuracies to copying of DNA., However, in sexual reproduction, even greater, diversity will be generated. Depending upon the, nature of variations, different individuals would, have different advantages, the most important, advantage of variation to a species is that it, increases the chance of its survival in a changing, environment., (i) What do you understand by the term ‘gene’ ?, (ii) In which type of reproduction exchange of genetic, material takes place ?, (iii) What is the cause of variation in asexual reproducing, organisms ?, (iv) Differentiate between genotype and phenotype of an, organism., (v) What is the chemical composition of chromosome ?, , 26. Read the following and answer the questions from (i), to (v) given below, A person first crossed pure-breed pea plants having, round-yellow seeds with pure-breed pea plants having, wrinkled-green seeds and found that only A-B type of, seeds were produced in the F1 -generation., When F1 -generation pea plants having A-B type, of seeds were cross-breed by self-pollination,, then in addition to the original round yellow and, , wrinkled-green seeds, two new varieties A-D and, C-B types of seeds were also obtained., (i) What are A-B type of seeds?, (ii) State whether A and B are dominant traits or, recessive traits., (iii) What are A-D type of seeds?, (iv) What are C-B type of seeds?, (v) Out of A-B and A-D types of seeds, which one will, be produced in (a) minimum number and (b), maximum number in the F2 -generation?, , 27. Read the following and answer the questions from, (i) to (v) given below, Mendel selected garden pea for his experiments, because he discovered for the first time the, occurrence of two types of seeds in pea plants, growing in the garden of his monastery. Mendel then, used a number of contrasting visible characters of, garden peas like., Dominant, , Recessive, , 1. Plant height, , Characters, , Tall, , Dwarf, , 2. Flower position, , Axial, , Terminal, , 3. Pod colour, , Green, , Yellow, , 4. Pod shape, , Full, , Constricted, , 5. Flower colour, , Violet, , White, , 6. Seed shape, , Round, , Wrinkled, , 7. Seed colour, , Yellow, , Green, , Mendel’s experiments were performed in three, stages in selection of pure or true breeding, parents, hybridisation and obtaining of, F1 -generation of plants and self-pollination of, hybrid plants and raising of subsequent generations, like F2 , F3 , F4 , etc., (i) How many contrasting traits were taken by, Mendel in his monohybrid crosses?, (ii) Give a monohybrid cross to explain the, F1 -generation formed by a plant with green pod, colour and yellow pod colour., (iii) State the first law of Mendel., (iv) What do understand by homozygous tall plant and, heterozygous tall plant?, (v) A Mendelian experiment consisted of breeding, pea plant bearing violet flowers with pea plants, bearing while flowers. What will be the results in, F1 progeny?

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62, , CBSE Term II Science X, , 28. Read the following and answer the questions from, (i) to (v) given below, Observe the following cross between tall plants, having round seeds and dwarf plants having, wrinkled seeds., The individuals obtained in the F1 -generation were, thereafter self-crossed., , 29. Read the following and answer the questions from (i) to, (v) given below, A person crossed pure-breed tall pea plants with, pure-breed dwarf pea plants and obtained pea plants of, F1 -generation. He then performed two types of, experiments., In the first, he self-crossed the plants of F1 -generation, (experiment A) and in the second, he crossed the plants of, TTRR, ttrr, ×, F, 1 -generation with the pure-breed dwarf parent plants, (Tall, Round), (Dwarf, Wrinkled), (experiment, B)., TtRr, × TtRr (F1 -generation self-crossed), Experiment, A : F1 progeny×F1 progeny, (Tall, Round), (Tall, Round), Experiment, B, : F1 progeny×Homozygous dwarf plant, (i) What would be the phenotypes of the individuals, obtained in the F2 -generation? Give their ratios., (i) What would be the phenotype of plants in the, F1 -generation?, (ii) Why do you think all the individuals of the, (ii), What would be the phenotype and genotype ratio of, F1 -generation were tall with round seeds?, F, 2 -generation in experiment ‘A’?, (iii) What will be the number of progeny obtained in, (iii) How would the genotypic ratio in F 2 generation, F 2 -generation in a dihybrid cross with pure, different in experiment ‘B’?, dominant traits?, (iv) How do we describe the phenotypic character that is, (iv) What would be the phenotype of the individual, experssed in F1 -generation? What is the term given to, obtained in F 1 -generation?, the contrasting character?, (v) According to you, what is responsible for the, (v) Name the type of cross shown in experiment B., inheritance of traits?

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EXPLANATIONS, Objective Questions, 1. (a) The process through which characters or traits pass from, one generation to another is called inheritance., 2. (a) Statement in option (a) is not true with respect to, variation because, 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 cellular apparatus it, inherits. Such, a newborn cell dies soon., 3. (b) Free earlobes are dominant (i) traits and attached, earlobes are recessive (ii) traits., 4. (c) A trait in an organism is influenced by both maternal and, paternal DNA., 5. (c) Punnett square was used by Gregor J Mendel to determine, the law of inheritance in his experiments with pea plant., 6. (a) In F1 -generation, the cross between TT and tt will result, into all tall plants, because tallness is the dominant trait., RRyy, rrYY, 7. (c), Parents, ×, (Round, green), , (Wrinkled, yellow), , Ry, , rY, , RrYy, , Gametes, , F 1 -generation, , (Round, yellow), , 8. (b), , An allele is a variant form of a gene, which are located, at the same position on a chromosomes. The heterozygous, organism (Tt) shows that it has two different alleles (T-tall, and t-dwarf) for the gene of the height., , 9. (c) When a pure tall (TT) plant is hybridised with dwarf, plant (tt), it will always give tall plants in F1 -generation., But, when they are self-pollinated, it will give both tall and, dwarf plants in F2 -generation., 10. (b) When pure breed pea plant A is crossed with pure breed, pea plant B. It is found that the plants which look like B do not, appear in F1 -generation but re-emerge in F2 -generation,, which means A is dominant over B trait and both of them are, contrasting traits. (i.e. A are tall and B are dwarf)., 11. (c) If pea plants having round green seeds and wrinkled, yellow seeds are crossed, phenotypic ratio of 9 : 3 : 3 : 1 will, be obtained in F2 progeny., 12. (b) The dihybrid cross between round green plant and, wrinkled yellow plant is shown as, Parents, , RRyy, , ×, , (Round, green), , Gametes, F1 -generation, , Ry, , Selfing, , F1, , F2 -generation, , (Wrinkled, yellow), , rY, (RrYy), (Round, yellow), , 315 round, yellow ( A ), 108 round, green (B), 101 wrinkled (C ), yellow, 32 wrinkled, green (D), 556 seeds, , 14. (a) The given diagram shows sex-determination in human., There are two types of sperms in male. Sperms having, X-chromosome (A) and sperms having Y-chromosome (B),, whereas the female ovum contains only one type of, chromosome (C), i.e. X., When X sperm fuses with ovum, the female zygote (D) is, formed, whereas when Y sperm fuses with ovum, the male, zygote is formed. Thus, option (a) is correct., 15. (b) Statements I, III and IV are correct, whereas statement, II is incorrect and can be corrected as, Each trait in child is influenced by both paternal as well as, maternal DNA., 16. (b) Both A and R are true, but R is not the correct explanation, of A., Dominant allele is an allele whose phenotype will be, expressed even in the presence of another allele of that gene. It, is represented by a capital letter, e.g. T. It can be expressed, itself in both homozygous and heterozygous conditions., 17. (c) A is true, but R is false because, When Mendel self-crossed F1 -generation to obtain, F2 -generation. F1 progeny of a tall plant with round seeds, and a dwarf plant with wrinkled seeds are all tall plants with, round seeds. This is because tallness and round shape are, both dominant traits, while dwarfness and wrinkled shapes, of seeds are recessive traits., 18. (c) A is true, but R is false., The ratio of F2 plants when Mendel took pea plant with one, contrasting characters, i.e. monohybrid cross was 3 : 1., 19. (a) Both A and R are true and R is the correct explanation of A., Females are homogametic with two X-chromosomes. That is, why, all human female gametes will have only X-chromosomes., 20. (a) Both A and R are true and R is the correct explanation of A., A human male has one X and one Y-chromosome. When a child, inherits X-chromosome from the father, it will be a girl and one, who inherits a Y-chromosome will be a boy., 21., , rrYY, , ×, ↓, , 13. (c) In the given cross, a purple flowered plant (dominant) is, cross fertilised with white flowered plant (recessive). The, F1 -progeny obtained contain all plants with purple flowers., The progeny of F1 -generation is self-fertilised to give rise to, F 2 -generation having plants with purple flowers and white, flowers in ratio 3 : 1., Hence, (i) represent cross-fertilisation, (ii) represent, self-fertilisation and (iii) represent F2 -generation., , F1, Ratio, 9, 3, 3, 1, 16, , (i) (a) Axial round seed is the phenotype of F 1 -generation., (ii) (d) Phenotype of F 2 progeny produced upon by the, self-pollination of F1 progeny will be axial round,, axial-wrinkled, terminal round and terminal wrinkled, plants., Thus, option (d) is correct., (iii) (b) The phenotypic ratio of the F 2 -generation will be, 9 : 3 : 3 : 1., (iv) (a) A cross between two individuals results in 9 : 3 : 3 : 1, for four possible phenotypes of progeny. This is an, example of a dihybrid cross., (v) (c) The given cross mainly explains the law of, independent assortment.

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64, 22., , CBSE Term II Science X, , (i) (b) Mendel studied seven pairs of contrasting characteristics, of pea plants., (ii) (a) After cross pollination of true breeding tall and dwarf, plants, the F1 -generation was self-fertilised., The resultant plants have genotype in the ratio1 : 2 : 1, (homozygous tall : heterozygous tall : homozygous dwarf), ×, , TT, T, , Tt, , ×, , Tt, , tt, , Parents, , t, , Gametes, , F1 -generation, , Tt Selfing, , 5. The two parents involved in sexual reproduction produce, gametes which fuse together forming a zygote. It gradually, develops into a young child showing certain similarities, with the parents. Since, a child inherits its characters from, both the parents the resemblance with them is very close., The grandparents and the child resemble less closely, because a gap of gene pool is created by the parents of the, child, Variations of two generations mixing together and addition, of new variations from parents, increases the difference, between them to a greater extent. Hence, a child resembles, more closely to its parents than the grandparents., 6. Reasons for selecting pea plant for experiment by Mendel, are as follows, Pea is an annual plant with short life cycle. So, several, generations can be studied in short period., It produces bisexual flowers, which are mainly, self-pollinating., It can be cross-pollinated., A number of contrasting characters were available in it., !, , !, , TT, Tt, Tt, tt, , Homozygous F2 -generation, tall, Heterozygous Tall, Homozygous dwarf, , Hence, the genotypic ratio is 1 : 2 : 1., (iii) (c) Law of independent assortment states that inheritance of, one character is always independent to the inheritance of, other characters within a same individual., (iv) (d) In F 2 -generation of pure breed tall plant with round, seeds and short plant with wrinkled seed, the possible, progeny will be observed are tall plant with round seeds,, tall plants with wrinkled seeds, short plant with round, seeds and short plant with wrinkled seeds., Thus, option (d) is correct., (v) (b) Round and yellow seeds show dominant character., Subjective Questions, 1. The transmission of characters from the parents to their, offspring is called heredity. Heredity was discovered by Gregor, Mendel through his work on pea plant., 2. In a population of asexually reproducing species, the chances of, appearance of new traits due to variations are very low and the, trait which is already present in the population is likely to be in, higher percentage and would have been arisen earlier., Therefore, the trait B present in 60% of the population is the, trait which have arisen earlier., 3. During reproduction, copying of DNA takes place, which is not, 100% accurate, thereby causing variations. If these variations, are favourable, they help the individuals to survive and pass, these variations to their progeny., Depending upon the nature of variations, different individuals, have different advantages, which promotes their survival like, bacteria which can withstand heat will survive better in a heat, wave., 4. Variations occur in the genes of the organisms produced due to, the mutations, reshuffling of genes and inheritance of acquired, traits during the evolutionary process which make all, individuals different from one another. Thus, in any population,, no two individuals are absolutely similar., , !, , !, , 7. Mendel crossed a pure tall pea plant (TT) with pure dwarf, pea plant (tt) and observed that all the progeny were hybrid, tall (Tt), i.e. only one of the traits was able to express itself in, the F1 -generation, which is the dominant trait., The other trait is called the recessive trait which remains, suppressed., Parents, , TT, (Tall parent), Homozygous, , ×, , tt, (Dwarf parent), Homozygous, , Tt, , F1-generation, , (All hybrid tall), Heterozygous, , However, when he self-crossed plants of F1 -generation, he, observed that one-fourth of the plants were dwarf and, three-fourth were tall., F1-parents, , Tt, , Gametes, , T, , F2-generation TT, , Tt, t, , T, , Tt, , Tt, , Tall, Tall, (Homozygous) (Heterozygous), , t, , tt, Dwarf, (Homozygous), , The expressed trait T for the tallness is dominant trait,, while the trait ‘t’ of dwarfness is recessive. Thus, Mendel’s, experiments show that trait may be dominant or recessive., 8. Dominant character The character which will express in, F1 -generation in both homozygous and heterozygous, conditions are dominant characters., e.g. Tallness of plant, purple flower colour, etc., Recessive character The character which will express only, in homozygous condition, but not in heterozygous conditions, or in F1 -generation is known as recessive character., e.g. Dwarfness, white flower colour, etc.

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65, , CBSE Term II Science X, , 9., , Phenotype, , It is the genetic makeup of an, individual for a character., , It represents the external, morphology of an organism, for a particular character,, , Same phenotype may or may Same genotype produces, not belong to same genotype. same phenotype., 10. Contrasting traits of pea plant were used by Mendel and, were classified as dominant or recessive., , Characters, , Given traits Contrasting traits, , Position of flower, , Terminal, , Axial, , (ii) Colour of flower, , White, , Violet, , (iii) Shape of pod, , Constricted, , Full, , (i), , 11. Mendel performed a dihybrid cross between pure pea plants, to show that traits are inherited independently. He selected a, pea plant with round green (RRyy) and wrinkled yellow, (rrYY) seeds., In the F1 progeny, it was found that all plants were round, yellow. But in F 2 progeny, some plants were round green and, some were wrinkled yellow., However, there were plants which showed new, combinations. Some of them were round with yellow seeds,, while others were wrinkled with green seeds. Thus, the, round/wrinkled trait and green/yellow seed traits are, independently inherited., 12., Parents :, , (Black Fur), Bb, , Gametes :, , Phenotype, Genotype, , (Black Fur), Bb, , b, , B, , ×, , b, , B, , B, , b, , B, , BB, (Pure black), , Bb, (Hybrid black), , b, , Bb, (Hybrid black), , bb, (Pure brown), , F1 -generation, , In case II Let assume that ‘O’ is dominant. In this case, the, child may have blood group ‘O’. Since in both the, assumptions, the child can have blood group ‘O’, so it, cannot infer which trait is dominant., , Genotype, , 3 : 1, Black, Brown, 1, :, 2 :, Pure Hybrid, black, black, , 1, Pure brown, , Therefore, the phenotypic ratio of black fur and brown fur, offspring is 3 : 1. The genotypic ratio of offspring is 1 : 2 : 1., 13. The information is insufficient to tell whether the trait ‘A’ or ‘O’, is dominant. It can find out by assuming the following cases., In case I Let assume that trait ‘A’ is dominant. Father that have, IAIO and mother having IOIO type of gene for blood group. In, this case 50% of the progeny will have blood group ‘A’ and 50%, of the progenies will have blood group ‘O’, when father’s blood, group is IAIO and mother is IO IO ., Whereas, in case of father having IAIA type of gene and mother, having IOIO type of gene, all of the progeny (100%) will have, blood group A., , 14. When pea plants with two contrasting characters, i.e. one, with a green round seeds and the other with a yellow, wrinkled seeds are crossed, all the F1 progeny obtained had, round and yellow seeds., When the F1 progeny is self-crossed to obtain F2 progeny,, four types of seeds were obtained as round yellow, round, green, wrinkled yellow and wrinkled green in ratio, 9 : 3 : 3 : 1 respectively., Hence, the phenotypic ratio of F2 progeny is 9 : 3 : 3 : 1., 15. Mendel took pea plants with two contrasting characters, i.e., one with a green round seeds and the other with a yellow, wrinkled seeds. When the F1 progeny was obtained, they had, round and yellow seeds. Mendel then allowed the F1 progeny, to be self-crossed to obtain F2 progeny., He found that seeds were round yellow, round green,, wrinkled yellow and some were wrinkled green. The ratio of, plants with above characteristics was 9 : 3 : 3 : 1, respectively., In F1-generation, all the characters were asserted out, independently of each other. Therefore, he stated that a pair, of contrasting or alternating characters behave, independently of the other pair., 16. Mendel used the pea plant for his experiments. He took pea, plants with different characteristics such as height (tall and, short plants). The progeny produced from them, (F1 -generation) plants were all tall., Mendel then allowed F1 progeny plants to undergo selfpollination. In the F 2 -generation, he found that all plants, were not tall, three quarter were tall and one quarter of, them were short. The ratio he obtained in F 2 -generation, plants is 9 : 3 : 3 : 1., 17., RRyy, (Round green), , Parents, × rrYY, (Wrinkled yellow), , RrYy, × RrYy, (Round yellow), (Round yellow), , F1 -generation, , (i) In F2 -generation, the combination of characters is, Round yellow = 9, Round green = 3, Wrinkled yellow = 3, Wrinkled green = 1, Thus, the ratio is 9 : 3 : 3 : 1, (ii) In F1 -generation, the production of all round yellow, (RrYy) seeds explains that the round shape and yellow, colour of the seeds were dominant traits over the, wrinkled shape and green colour of the seeds which, segregated during F2 -generation., 18. A male gamete carries either one X or one Y-chromosome,, while a female gamete carries only X-chromosomes., Therefore, sex of the child depends upon what happens, during fertilisation., (i) If a sperm carrying X-chromosome fertilises the egg,, the child born will be a female (XX)., (ii) If a sperm carrying Y-chromosome fertilises the egg,, the child born will be a male (XY).

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66, , CBSE Term II Science X, , Thus, the sperm (the male gamete) determines the sex of the, child., XX, (Female), ↓, X, , XY, (Male), , X, , XX, (Female child), , Gametes, , Y, , XY Offsprings, (Male child), , 19. No, the genetic combination of mother does not play any, significant role in determining the sex of a newborn. This is, because the female cell carries two X-chromosomes (XX)., While, the male cell carries one X and one Y-chromosome., The fusion of X-chromosome bearing sperm (of male cell), with X-chromosome of female egg produces a female child,, while the fusion of Y-chromosome bearing sperm (of male, cell) with X-chromosome of female egg produces a male, child. Therefore, it is the contribution of father which, determines the sex of a newborn., 20. ‘A trait may be inherited, but may not be expressed’. Mendel, crossed tall pea plants with dwarf pea plants., , Parents, , Pure tall plant Pure dwarf plant, (TT), (tt), , (Tt), , (Tt), , (Tt), , (Tt), , (ii) Law of Segregation or Law of Purity of Gametes, (Second Law) This law states that the factors or alleles of, a pair segregate from each other during gamete formation, such that a gamete receives only one of the two factors., They do not show any blending., e.g. The reappearance of wrinkled seed character in, F2 -generation, which was suppressed in F1 -generation by, the round seed character., This law is also be proven by monohybrid cross., Parents, , Tt, , Gametes, , T, , ´, , Tt, , t, , T, , F2-generation TT, , Tt, , Tt, , tt, , Tall, Tall, (Homozygous) (Heterozygous), , Dwarf, (Homozygous), , (iii) Law of Independent Assortment (Third Law) This law, states that two factors of each character assort or separate, out independent of the factors of other characters at the, time of gamete formation and get randomly rearranged in, the offspring., e.g. In dihybrid cross, between pure round and yellow, seed pea plants with plants having wrinkled and green, seeds, the F2 -generation produced two parental and two, hybrid combinations., This law can be proven by dihybrid cross, , F1-generation, , Parents, RRyy, rr YY, ×, (Round green), (Wrinkled yellow), , Selfing, of F1, , ↓, , ↓, , Ry, , rY, , (TT), , (Tt), , (Tt), , (tt), , F2-generation, , Mendel’s observation F1 -generation contained all tall plants, with genotype Tt, where ‘T’ represents a dominant trait and ‘t’, represents a recessive trait. This means when F1 -generation, underwent selfing, the trait that was unexpressed inF1 (dwarf), was observed in some F2 -progeny. Thus, both traits, tall and, dwarf, were expressed in F2 - generation in the ratio of 3 : 1., 21. The three major laws of inheritance proposed by Mendel are, as follows, (i) Law of Dominance (First Law) According to this law,, when two homozygous individuals with one or more, contrasting characters are crossed, the F1 hybrid have, both the contrasting alleles of a pair, but only one (i.e., dominant trait) allele expresses and it does not allow, the other one (recessive trait) to appear., Pure tall plant, (Dominant), (TT), , ×, , Pure dwarf plant, (Recessive), (tt), Parents, , ↓, , ↓, , Gametes, , Tall plant Tall plant, (Tt), (Tt), , RrYy, (Round yellow), , F1-generation, , ×, F1, , F1, , RY, , Ry, , rY, , ry, , RRYY, , RRYy, , RrYY, , RrYy, , RRYy, , RRyy, , RryY, , Rryy, , RrYY, , RrYy, , rrYY, , rrYy, , RrYy, , Rryy, , rrYy, , rryy, , F2-generation, RY, , Ry, , rY, , ry, , F2 -generation 315 round yellow, , (Tt), All plants tall (Hybrid), , F1-generation, , Explanation of law of dominance, , t, , Ratio, 9, , 108 round green, , 3, , 101 wrinkled yellow, , 3, , 32 wrinkled green, 556 seeds, , 1, 16

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67, , CBSE Term II Science X, , All children will inherit an X-chromosome from their mother, regardless of whether they are boys or girls., , 22. Dihybrid Cross When cross between two varieties having, two contrasting characters takes place, then it is called, dihybrid cross., In F1-generation, on the basis of law of dominance, only, dominant characters appear. During gamete formation, genes, of contrasting characters separate and only one gene enters, each gamete., On self-fertilisation, the F1-generation develops, F2 -generation in which 9 : 3 : 3 : 1 phenotypic ratio appears., In F2 -generation, new combination also develops., Parents :, , (, , ), , (, , ), , Gametes :, , F1-generation, , F2-generation, , Thus, the sex of the children will be determined by what they, inherit from their father. A child who inherits an, X-chromosome from her father will be a girl and one who, inherits a Y-chromosome from him will be a boy., (i) A gene is a unit of DNA which governs the synthesis of, one protein that provides a specific character of the, organisms., (ii) Exchange of genetic material takes place in sexual, reproduction., (iii) Environmental factors and mutations cause variation in, asexual reproducing organisms., (iv) The genetic constitution of an organisms is called, genotype, whereas the phenotype is the physical, appearance or characteristics of the organism., (v) The chemical composition of a chromosome is DNA and, histone proteins., 26. (i) A-B type of seeds are round in shape and yellow in colour, as round and yellow both constitute the dominant, character, hence expressed in F1-generation., (ii) A(round) and B(yellow) are dominant traits., (iii) Round-green (A-D)., (iv) Wrinkled-yellow (C-B)., (v) (a) A-D in minimum number., (b) A-B in maximum number., 27. (i) One contrasting trait, i.e. tall and dwarf plants were, taken by Mendel in his monohybrid crosses., (ii), 25., , GG, (Green pod colour), , ×, , Parents, gg, (Yellow pod colour), , g, , G, Gg, , Gametes, , F1-generation, , (Green pod colour), , (iii) The first law of Mendel is law of dominance, which, states that the when two alleles of an inherited pair is, heterozygous, then the allele that is expressed is, dominant, whereas the allele that is not expressed is, recessive., (iv) Homozygous tall plant will have two identical copies of, single gene, i.e. TT., Heterozygous tall plant will have two different copies of, single gene, i.e. Tt., (v) Heterozygous pea plants with violet flowers will result, in F1 -progeny., , In dihybrid cross, Phenotype ratio, , 23. In dihybrid cross, the ratio of phenotype of F2 -generation, obtained by selfing of F1 -generation is 9 : 3 : 3 : 1, in which, parental as well as new combination are observed. This, shows law of independent assortment in which two, characters under consideration assort in an independent, manner to give rise to different combination., It means that the genes of all the characters are, independent from each other and combined to make new, varieties., 24. Humans have 23 pairs of chromosome. Both male and, female carry two sets of sex chromosome., Male (XY) has one X and one Y sex chromosome. Female, (XX) has both X sex chromosome., , 28., , (i) In F2 -generation, the phenotypes of the individuals, obtained would be, Tall and round = 9; Tall and wrinkled = 3;, Dwarf and round = 3; Dwarf and wrinkled = 1, Thus, the ratio is 9 : 3 : 3 : 1.

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68, , CBSE Term II Science X, , (ii) The appearance of all tall plants with round seeds, shows that the tallness and round-shaped seeds are, dominant traits over the dwarfness and wrinkled, shape of the seeds., (iii) The number of progeny obtained in F2 -generation, in a dihybrid cross with pure dominant traits, i.e., TTRR will be 1 which is formed by the fertilisation, of gametes TR and TR., (iv) In F1 -generation1 , all tall and round plants will be, obtained., (v) Gene is the carrier that leads to the inheritance of, traits. It is the part of a chromosome that controls, the appearance of a set of hereditary, characteristics., , 29., , (i) The phenotype of all the plants in the F1 -generation would, be tall., (ii) In experiment ‘A’, the phenotypic ratio of tall and dwarf, plants would be 3 : 1 : : Tall : Dwarf, whereas the genotypic, ratio would be 1 : 2 : 1 for TT, Tt, tt genotype., (iii) When crossed with homozygous recessive parent the, genotypic ratio would be 1 : 1 for Tt, tt genotype., (iv) The phenotypic character that a capable of expressing in, the F1 -generation is described as ‘dominant’. The, contrasting character, i.e. dwarfness is the recessive, character., (v) In experiment B, test cross is done between the F1, heterozygote with homozygous recessive parent.

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Chapter Test, , (b) Both A and R are true, but R is not the correct explanation, of A, (c) A is true, but R is false, (d) A is false, but R is true, , 6. Assertion Genes present in every cell of an organism, control the traits of the organisms., , Multiple Choice Questions, , 1. The number of pair(s) of sex chromosomes in the, zygote of humans is, (a) one, , (b) two, , (c) three, , (d) four, , 2. The following results were obtained by a scientist, who crossed the F 1 -generation of pure breeding, parents for round and wrinkled seeds., Dominants, trait, Round seeds, , Recessive trait, , Number of F 2, offspring, , Wrinkled seeds, , 7524, , From these results, it can be concluded that the, actual number of round seeds he obtained was, (a) 1881, (c) 2508, , 3., , (b) 22572, (d) 5643, , A recessive homozygote is crossed with a, heterozygote of the same gene. What will be the, phenotype of the F1 -generation?, (a), (b), (c), (d), , All dominant, 75% dominant, 25% recessive, 50% dominant, 50% recessive, 25% dominant, 50% heterozygous, 25% recessive, , 4. Identify the genotype of parent plants by observing, the result of the cross given below., F1-generation, Aa, Aa, Aa, Aa, (a) Both parents are homozygous, (b) Both parents are heterozygous, (c) One parent is homozygous and other parent is, heterozygous, (d) Cannot say, , 5. Choose the correctly matched pair., (a) X and Y-chromosomes—Autosomes, (b) Father of Genetics—Gregor Johann Mendel, (c) Monohybrid cross—Two pairs of contrasting characters, (d) Dihybrid cross—One pair of contrasting characters, , Assertion-Reasoning MCQs, , Reason Gene is specific segment of DNA occupying, specific position on chromosome., 7. Assertion Traits like tallness and dwarfness in pea plant are, inherited independently., Reason When a homozygous tall pea plant is crossed with, dwarf pea plant, medium-sized pea plant is obtained in, F 1 -generation., 8. Assertion In human females, all the chromosomes are, perfectly paired except, sex chromosomes., Reason X and Y are sex chromosomes in males., , Short Answer Type Questions, , 9. ‘Only variations that confer an advantage to an individual, organisms will survive in a population’. Do you agree with, this statement? Why or why not?, , 10. Differentiate between homozygote and heterozygote., 11. Outline a project which aims to find the dominant coat, colour in dogs., , 12. If we cross pure breed tall (dominant) pea plants with pure, breed dwarf (recessive) pea plants, we get pea plants of, F1-generation. If we now self-cross the pea plant of, F1-generation, then we obtain pea plants of F2-generation., (i) What do the plants of F1-generation look like?, (ii) What is the ratio of tall plants to dwarf plants in, F2 -generation?, (iii) State the type of plants not found inF1-generation, but, appeared inF2-generation mentioning the reason for the, same., , 13. How is the equal genetic contribution of male and female, parents ensured in the progeny?, , Long Answer Type Questions, , 14. Ovum and sperm are both female and male gametes,, respectively. But, what is so intricate in sperm which makes, it solely responsible for determining the sex of the child?, Explain., , 15. In the following crosses, write the characteristics of the, , Direction (Q. Nos. 6-8) Each of these questions, contains two statements, Assertion (A) and Reason, (R). Each of these questions also has four, alternative choices, any one of which is the correct, answer. You have to select one of the codes (a), (b),, (c) and (d) given below., (a) Both A and R are true and R is the correct explanation, of A, , progeny obtained., (i) RRYY, RRYY, ×, (Round yellow) (Round yellow), , (ii) RrYy, , ×, , RrYy, , (Round yellow) (Round yellow), (iii) rryy, rryy, ×, (Wrinkled green) (Wrinkled green), , (iv) RRYY, rryy, ×, (Round yellow) (Wrinkled green), , Answers, Multiple Choice Questions, 1. (a), , 2. (d), , 3. (c), , 4. (a), , Assertion-Reasoning MCQs, 6. (b), , 7. (c), , 8. (d), , 5. (b), , For Detailed Solutions, Scan the code

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CHAPTER 05, , Electricity, In this Chapter..., !, , Ohm’s Law, , !, , Resistance, , !, , Combination of Resistors, , !, , Heating Effect of Electric Current, , !, , Electric Power, , or, , V = IR, , V, R, where, R is the constant of proportionality called resistance of, the conductor at a given temperature., The conductors which obey Ohm’s law are called ohmic, conductors while the conductors which do not obey Ohm’s, law are called non-ohmic conductors., , or, , I=, , Y, A, , O, , pe, , It gives a relationship between current I, flowing in a, metallic wire and potential difference V, across its terminals., According to this law, the electric current flowing through a, conductor is directly proportional to the potential difference, applied across its ends, providing the physical conditions, (such as temperature) remain unchanged., If V is the potential difference applied across the ends of a, conductor through which current I flows, then according to, Ohm’s law,, [at constant temperature], V ∝I, , The graph between the potential difference V and the, corresponding current I is found to be a straight line passing, through the origin for ohmic (metallic) conductors., , Sl, o, , Ohm’s Law, , V-I Graph, , Potential, difference, V, , Electricity is one of the most convenient and widely used, forms of energy in today’s world. It is a controllable and, convenient form of energy., , θ, Current, I, , X, , V-I graph for metallic conductor, , Resistance, It is that property of a conductor by virtue of which it, opposes/resists the flow of charges/current through it. Its SI, unit is ohm and it is represented by Ω., V, Resistance of a conductor is given by, R = ., I, It is said to be 1 ohm, if a potential difference of 1 volt across, the ends of the conductor makes a current of 1 ampere to flow, through it.

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71, , CBSE Term II Science X, , i.e., ⇒, , 1 volt, 1 ampere, 1V, 1Ω =, = 1 VA –1, 1A, , 1 ohm =, , Factors on which the Resistance of a, Conductor Depends, , Combination of Resistors, There are two methods of joining the resistors together which, are as given below., 1. Series Combination, When two or more resistors are connected end to end to each, other, then they are said to be connected in series., V1, R1, , The electrical resistance of a conductor depends on the, following factors, (i) Length of the Conductor The resistance of a conductor, R is directly proportional to its length l., i.e., …(i), R∝l, (ii) Area of Cross-section of the Conductor The resistance, of a conductor R is inversely proportional to its area of, cross-section A ., 1, …(ii), i.e., R∝, A, (iii) Nature of the Material of the Conductor The resistance, of a conductor depends on the nature of the material of, which it is made. Some materials have low resistance,, whereas others have high resistance., Therefore, from Eqs. (i) and (ii), we can write, l, l, or R = ρ, R∝, A, A, where, ρ is the constant of proportionality and is called, electric resistivity or specific resistance of the material, of the conductor., , V3, R3, , I, , !, , !, , A, , –, + –, K, , V, , The equivalent resistance is equal to the sum of the all, individual resistances., i.e.,, R = R1 + R 2 + R 3, The equivalent resistance is thus greater than the resistances, of either resistor. This is also known as maximum effective, resistance., The current through each resistor is same. The potential, difference across each resistor is different., 2. Parallel Combination, When two or more resistors are connected simultaneously, between two points to each other, then they are said to be, connected in parallel combination., R1, I1, I3, , R2, I2, , R3, , !, , !, , I, +, , + V –, , Resistivity, It is defined as the resistance of a conductor of unit length and, unit area of cross-section. Its SI unit is ohm-metre (Ω-m)., The resistivity of a material does not depend on its length, or thickness but depends on the nature of the substance, and temperature. It is a characteristic property of the, material of the conductor and varies only, if its, temperature changes., Insulators such as glass, rubber, ebonite, etc., have a very, high resistivity (10 12 to1017 Ω -m), while conductors have a, very low resistivity (10 −8 to10 − 6 Ω-m)., Alloys have higher resistivity than that of their constituent, metals. They do not oxidise easily at high temperatures,, this is why they are used to make heating elements of, devices such as electric iron, heaters, etc., Tungsten is almost used exclusively for filaments of, electric bulbs, whereas copper and aluminium are, generally used for electrical transmission lines., , V2, R2, , I, , +, , K, , +, , V, , V, , –, , I, +, A, –, , –, , The reciprocal of equivalent resistance is equal to the sum of, the reciprocal of individual resistances., 1, 1, 1, 1, i.e., =, +, +, R R1 R 2 R 3, The equivalent resistance is less than the resistance of, either resistor. This is also known as minimum effective, resistance., The current from the source is greater than the current, through either resistor. The potential difference across each, resistor is same.

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72, , Heating Effect of Electric Current, When an electric current is passed through a high resistance, wire like nichrome wire, then the wire becomes very hot and, produces heat., In purely resistive circuits, the source of energy continuously, gets dissipated entirely in the form of heat. This is called the, heating effect of current., This is obtained by the transformation of electrical energy, into heat energy. e.g. electric heater, electric iron, etc., Heat produced is expressed as, H = I 2 × R × t, It is known as Joule’s law of heating., This law implies that heat produced in a resistor is, (i) directly proportional to the square of current for a given, resistance., (ii) directly proportional to the resistance for a given current., (iii) directly proportional to the time for which the current, flows through the resistor., , Practical Applications of Heating Effect of, Electric Current, There are two applications of heating effect of electric, current which are given below, 1. Electric Bulb, It has a filament made of tungsten. So, most of the power, consumed by this, is dissipated in the form of heat and some, part is converted into light because it has high resistivity and, high melting point., , CBSE Term II Science X, , The filament is thermally isolated and the bulb is filled with, chemically inactive nitrogen and argon gas to prolong the life, of filament., 2. Electric Fuse, It is used as a safety device in household circuits. It protects, the circuits, by stopping the flow of any unduly high electric, current. It is connected in series with the mains supply. It, consists of an alloy of lead and tin which has appropriate, melting point., When the current flowing through the circuit exceeds the safe, limit, then the fuse wire melts and breaks the circuit. This, helps to protect the other circuit elements from heavy current., Fuses are always rated for different current values such as 1 A,, 2 A, 5 A, 10 A, 15 A, etc., , Electric Power, It is defined as the amount of electric energy consumed in a, circuit per unit time., Electric power is expressed as, P = VI, V2, or, P=, R, The SI unit of electric power is watt (W)., It is said to be 1 watt, if 1 ampere current flows through a, circuit having 1 volt potential difference., i.e., 1 watt = 1 volt × 1 ampere = 1 VA, Commercial unit of electrical energy is kilowatt-hour., 1 kWh = 3.6 × 10 6 J

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Solved, Examples, RA R π d2, =, 4l, l, 30 × π × (6 × 10−4 )2, =, 4×2, , We know that, ρ =, , Example 1. The potential difference between the, terminals of an electric heater is 75 V when it draws, a current of 5 A from the source. What current will the, heater draw, if the potential difference is increased to, 150 V ?, , = 4.24 × 10−6 Ω-m, The resistivity of the metal at 25°C is 4.24 × 10−6 Ω-m., , Sol. Given, potential difference, V = 75 V, Current,, , ⎡, πd 2 ⎤, ⎢Q A = 4 ⎥, ⎣, ⎦, , I= 5A, V, We know that, R =, I, 75, ⇒ R=, = 15 Ω, 5, , Example 4. Three resistors of 5 Ω, 10 Ω and 15 Ω are, , When potential difference is increased to150 V, then, current is, Vʹ 150, Iʹ =, =, = 10 A, R, 15, , Sol., , connected in series with a 12 V power supply. Calculate, their combined resistance, the current that flows in the, circuit and in each resistor and the potential difference, across each resistor., Given, R1 = 5 Ω, R 2 = 10 Ω, R 3 = 15 Ω, V = 12 V,, , R = ?, I = ? and V1 , V2 , V3 = ?, 5Ω, , 10 Ω, , 15 Ω, , So, the current through the heater becomes 10 A., , Example 2. A wire of given material having length l and, area of cross-section A has a resistance of10 Ω. What would, be the resistance of another wire of the same material, having length l/4 and area of cross-section 2. 5 A?, and resistance, R1 = 10 Ω,, , ⇒, , ρl, = 10 Ω, A, 10A, ρ=, l, , R1 =, , l / 4 10A, l, [from Eq. (i)], =, ⋅, l 4 × 2.5A, 2.5A, , = 1Ω, So, the resistance of that wire is 1 Ω., , Example 3. Resistance of a metal wire of length 2 m is, 30 Ω at temperature 25°C. If the diameter of the wire is, 0.6 mm, then what will be the resistivity of the metal at that, temperature?, Sol. Given, length of wire, l = 2 m, Resistance, R = 30 Ω , Temperature, T = 25°C, Diameter of wire, d = 0 .6 mm = 6 × 10−4 m, Resistivity of the wire, ρ = ?, , –, , 12 V, , ∴ The current flowing through the circuit (I), …(i), , For second wire, length = l/ 4, area of cross-section = 2.5 A, ∴ Resistance, R 2 = ρ, , +, , According to question, the three resistors are connected in, series combination, then equivalent resistance,, R = R1 + R 2 + R 3, = 5 + 10 + 15 = 30 Ω, , Sol. For first wire, length = l, area of cross-section = A, , i.e., , I, , I, , Potential of power supply ( V ), Total resistance of the circuit (R ), 12, =, 30, 2, = = 0.4 A, 5, In series combination, the current flowing through each, resistor is equal to total current flowing through the circuit., Therefore, current flowing through each resistor is 0.4 A., ∴Potential difference across first resistor,, V1 = IR1 = 0.4 × 5 = 2 V, Potential difference across second resistor,, V2 = IR 2 = 0.4 ×10 = 4 V, and potential difference across third resistor,, V3 = IR 3 = 0.4 ×15 = 6 V, =

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74, , CBSE Term II Science X, , Example 5. Study the following electric circuit. Find the, readings of (i) the ammeter and (ii) the voltmeter., +, 4Ω, , V, , B, 2Ω, –, , ( ), K, , 3V, , Current (I) =, , –, , A, , +, , We know that,, , – A+, , Sol. In the given circuit, the resistance of 4 Ω and bulb resistance, of 2 Ω are connected in series, so equivalent resistance of the, circuit,, R = R1 + R 2 = 4 Ω + 2 Ω = 6 Ω, (i) Total current flowing in the circuit, (I), Potential difference ( V ) 3, =, = = 0.5A, Total resistance (R ), 6, In series combination, current flowing through each, component of the circuit is same and is equal to the total, current flowing in the circuit. So, 0.5 A current will flow, through the ammeter, so its reading will be 0.5 A., (ii) Reading of voltmeter = Potential difference across 2 Ω, bulb, ∴, V = IR = 0.5 × 2 = 1 V, , Potential difference ( V ), Resistance (R ), , We get the following results for the current, 12, Current through 40 Ω resistor, I1 =, = 0.3 A, 40, Also, I2 = 0.3 A, 12, Current through 20 Ω resistor, I3 =, = 06, . A, 20, ∴Current, I = I1 + I2 + I3 = 0.3 A + 0.3 A + 06, . A, = 1.2 A, , Example 7. In the given figure, R 1 = 5 Ω, R 2 = 10 Ω, R 3 =, 15 Ω, R 4 = 20 Ω, R 5 = 25 Ω and a 15 V battery is, connected to the arrangement. Calculate, (i) the total resistance in the circuit, and, (ii) the total current flowing in the circuit., R1, R2, I, , [Q current flowing through the bulb is 0.5 A], , R3, , Example 6. Two 40 Ω resistors and a 20 Ω resistor are all, connected in parallel with a 12 V power supply. Calculate, their effective resistance and the current through each, resistor. What is the current flowing through the supply?, Sol. Given, R1 = 40 Ω, R 2 = 40 Ω, R 3 = 20 Ω,, V = 12 V,R = ?, I, I1 , I2 , I3 = ?, 12 V, +, –, I, , I1, , R1=40 Ω, , I2, , R2=40 Ω, , I3, , R3=20 Ω, , I, , According to circuit, the three resistors are connected in, parallel combination, then effective resistance,, 1, 1, 1, 1, =, +, +, R R1 R 2 R 3, 1, 1, 1, +, +, 40 40 20, 1+1+ 2, 4, 1, =, =, =, 40, 40 10, R = 10 Ω, =, , ⇒, , So, the three resistors together have an effective resistance, of10 Ω. Each resistor has a potential difference of12 V across it., Because in parallel combination, the potential difference, across each resistance is equal to the total potential, difference applied on the combination., , +, A, –, , 15 V, , R4, , ( ), , R5, , Sol. Resistors R1 and R 2 are in parallel., 1, 1, 1, 1 1 1, 10, So,, =, +, ⇒, = +, ⇒ Rʹ = Ω, Rʹ R1 R 2, Rʹ 5 10, 3, Similarly, R 3 ,R 4 and R5 are in parallel, 1, 1, 1, 1, 1, 1, 1, ⇒, =, +, +, = +, +, Rʹʹ R 3 R 4 R5 15 20 25, 1, 20 + 15 + 12, 300, =, ⇒ Rʹʹ =, Rʹʹ, 300, 47, Thus, the total resistance,, 10 300, R = Rʹ + Rʹʹ = +, = 3.33 + 6.38 = 971, . Ω, 3, 47, The total current, I =, , V 15, =, = 1.54 A, R 971, ., , Example 8. Find the equivalent resistance of the following, circuit. Also, find the current and potential at each resistor., 1Ω, R2, 2Ω, , R3, 3Ω, , A 2Ω, R1, , 2Ω B, R5, , R4, +, , –, , 9V, , K

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75, , CBSE Term II Science X, , Sol. In the given circuit, R 2 , R 3 and R 4 are in parallel, combination. As, currents through R 2 , R 3 and R 4 are, different. So, their equivalent resistance R is, , I3, , 1Ω, I = I1, , R2, 2Ω, , A, , 2Ω, , 2Ω, , R1, , R4, , R5, K, , 1, 1, 1, 1, =, +, +, Rʹ R 2 R 3 R 4, 1 1 1 6 + 3 + 2 11, = + + =, =, 1 2 3, 6, 6, 6, Rʹ =, Ω, ⇒, 11, Now, the given circuit can be redrawn as shown below, A, , 2Ω, , 6/11 Ω, , 2Ω, , R1, , Rʹ, , R5, , 9V, , +, , B, , K, , –, , Now, R1 , Rʹ and R5 are in series combination. As, current, through R1 , Rʹ and R5 is same., So, equivalent resistance of the whole circuit is, 6, R = R1 + Rʹ + R5 = 2 + + 2, 11, 22 + 6 + 22 50, =, =, Ω, 11, 11, Now, total current flowing through the circuit,, V, 9, 99, I= =, =, ≈2 A, R 50 / 11 50, Current through R1 and R5 will be same as these are in, series combination and will be equal to the total current, flowing through the circuit., ∴, I = I1 = I5 = 2 A, Potential drop at R1 , V1 = I1R1 = 2 × 2 = 4 V, Potential drop at R5 , V5 = I5R5 = 2 × 2 = 4 V, Now, potential drop at Rʹ , Vʹ can be calculated as, V = V1 + V5 + Vʹ, ⇒, ⇒, , 9 = 4 + 4 + Vʹ, Vʹ = 1 V, , I, , B, , –, , 9V, , R1, I4, , R3, 3Ω, , +, , R2, , I2, , R3, , I5 = I, , R4, , R5, , + –, , K, , As R 2 , R 3 and R 4 are in parallel combination, so potential, drop at all resistances will be same as 1 V., V2 = V3 = V4 = Vʹ = 1V, Current through R 2 , I2 =, , V2, Vʹ 1, =, = =1 A, R2 R2 1, , Similarly,, , V3, Vʹ 1, =, = = 0.5 A, R3 R3 2, , and, , I3 =, I4 =, , V4, Vʹ 1, =, = = 0.33 A, R4 R4 3, , Example 9. An electric iron consumes energy at a rate, of 880 W, when heating is at the maximum rate and 340 W,, when the heating is at the minimum, the voltage is 220 V., What are the current and the resistance in each case?, Sol. Power, P = VI, Current, I = P / V, (i) When heating is at the maximum rate, I =, Resistance of the electric iron, R =, , V 220, =, = 55 Ω, I, 4, , (ii) When heating is at the minimum rate, I =, Resistance of the electric iron, R =, , 880, = 4A, 220, , 340, = 1. 54 A, 220, , V 220, =, = 142 . 85 Ω, I 1. 54, , Example 10. An electric heater of resistance 500 Ω is, connected to a mains supply for 30 min. If 15 A current, flows through the filament of the heater, then calculate the, heat energy produced in the heater., Sol. Given, resistance of the filament, R = 500 Ω, Current, I = 15 A, Time, t = 30 min = 30 × 60 = 1800 s, Heat produced, H = I2Rt, = 152 × 500 × 1800, = 2025 × 10 5 J = 20.25 × 107 J

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Chapter, Practice, PART 1, Objective Questions, Multiple Choice Questions, , 1. Which one of the following is the correct set-up for, studying the dependence of the current on the, potential difference across a resistor?, (CBSE 2019), R, +, , R, +, , V –, , –, (a) A, + +, , (b), –, , A –, , –, V, + +, , – V +, , –, , +, , V –, , +, , A –, , increasing its temperature., (a) True, (c) Can’t say, (d) Information insufficient, , (b) False, , 5. A cylindrical conductor of length l and uniform area, of cross-section A has resistance R. Another, conductor of length 2l and resistance R of the same, (CBSE 2020), material has area of cross-section, (a) A / 2, (c) 2A, , (b) 3A / 2, (d) 3A, , 6. The substance which have a higher value of, resistance and small number of free electrons in it, is, called, (a) resistor, (c) good conductor, , (b) poor conductor, (d) insulator, , 7. What is the maximum resistance which can be made, using four resistors each of, , –, , (c) A, , R, , + +, , R, , (d), +, , –, , –, , 2. A student carries out an experiment and plots the V-I, graph of three samples of nichrome wire with, resistances R1 , R 2 and R 3 respectively. Which of the, following is true?, I (ampere), , !, , 4. Specific resistance of a conductor increases on, , R1, , (a) 2 Ω, (c) 2.5 Ω, , 1, Ω?, 2, , (b)1Ω, (d) 8 Ω, , 8. Three 2 Ω resistances are connected so as to make a, triangle. The resistance between any two vertices is, (a) 6 Ω, 3, (c) Ω, 4, , (b) 2 Ω, 4, (d) Ω, 3, , 9. Determine the equivalent resistance between points, X and Y in the following circuit., , R2, , 3!, , R3, 3!, , 3!, , 3!, , V(volt), , (a) R1 = R 2 = R 3, (c) R 3 > R 2 > R1, , (CBSE 2020), , (b) R1 > R 2 > R 3, (d) R 2 > R 3 > R1, , 3. The resistivity does not change, if, , (NCERT Exemplar), , (a) the material is changed, (b) the temperature is changed, (c) the shape of the resistor is changed, (d) Both material and temperature are changed, , X, , (a) 5 Ω, (c) 9 Ω, , Y, , (b)12 Ω, (d) 6 Ω, , 10. The filament of bulb is made up of, (a) copper, (c) tin, , (b) tungsten, (d) lead

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77, , CBSE Term II Science X, , 11. The current flowing through a wire of resistance 2 Ω, varies with time as shown in the given figure. The, amount of heat produced (in J) in 3 s would be, , I (A), , 3, , B, , C, , A, , D, , 17. Assertion The connecting wires are made of copper., Reason The electrical conductivity of copper is high., , 18. Assertion When the resistances are connected, end-to-end consecutively, they are said to be in series., Reason In case the total resistance is to be, decreased, then the individual resistances are, connected in parallel., , 2, H, , 1, 0, –1, –2, , G, , I, J, , t (s), , 19. Assertion The fuse is placed in series with the device., E, , Reason Fuse consists of a piece of wire made of a, metal or an alloy of appropriate melting point., , F, , –3, , (a) 2 J, (c) 28 J, , 20. Assertion The 200 W bulbs glow with more, , (b) 18 J, (d) 10 J, , brightness than 100 W bulbs., Reason A 100 W bulb has more resistance than, 200 W bulb., , 12. The rate at which energy is delivered by a current is, determined by, (a) heat, (c) potential difference, , (b) electric power, (d) work, , 13. In an electrical circuit three incandescent bulbs A , B, and C of rating 40 W, 60 W and 100 W, respectively, are connected in parallel to an electric source. Which, of the following is likely to happen regarding their, brightness?, (a) Brightness of all the bulbs will be the same, (b) Brightness of bulb A will be the maximum, (c) Brightness of bulb B will be more than that of A, (d) Brightness of bulb C will be less than that of B, , 14. An electric kettle consumes 1 kW of electric power, , !, , Case Based MCQs, , 21. Read the following and answer questions from (i) to (v), given below, The relationship between potential difference and, current was first established by George Simon Ohm, called Ohm’s law. An electric circuit is shown below, to verify Ohm’s law., +, , (b) 2 A, , (c) 4 A, , 0.75 V and 100 mA. Then its power is, , !, , Voltmeter, , Conductor, , +, , –, , C, , B, S, Battery Switch, , Rh, Rheostat, , Sliding contact, of rheostat, , (d) 5 A, , 15. LED indicator of a TV in your house operates at, (a) 75 mW, (c) 0.75 mW, , –, , Connecting, wires, , when operated at 220 V. A fuse wire of what rating, must be used for it?, (NCERT Exemplar), (a) 1 A, , V, , A, , (b) 1.00 mW, (d) 7.5 mW, , Assertion-Reasoning MCQs, Direction (Q. Nos. 16-20) Each of these questions, contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You have to, select one of the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., , Although Ohm’s law has been found valid over a, large class of materials, there do exist metals and, devices used in electric circuits where the, proportionality of V and I does not hold., (i) Materials which follow Ohm’s law are, (a) ohmic conductors, (c) insulators, , (b) non-ohmic conductors, (d) superconductors, , (ii) For insulator at room temperature, the graph, between V and I is given. Which one is correct?, I, , I, , (a), , (b), O, , (d) A is false, but R is true., , V, V, O, I, , I, , 16. Assertion Longer wires have greater resistance and, the smaller wires have lesser resistance., Reason Resistance is inversely proportional to the, length of the wire., , (c), , (d), O, , V, , O, , V

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78, , CBSE Term II Science X, , (iii) The slope of V - I graph (V on x-axis and I on, y-axis) gives, (a) charge, (b) reciprocal of resistance, (c) resistance, (d) reciprocal of charge, , !, , (iv) By increasing the voltage across a conductor, the, (a) current will decrease, (b) resistance will increase, (c) resistance will decrease, (d) current will increase, , Short Answer Type Questions, , 1. State the law which gives the relationship between, current and potential difference. Define the unit of, resistance., , (v) When a 9 V battery is connected across a, conductor and the current flows is 01, . A, the, resistance is, (a) 90 Ω, (c) 09, . Ω, , PART 2, Subjective Questions, , 2. Define resistance. Give its SI unit., (CBSE 2019), 3. The values of current I flowing in a given resistor for, the corresponding values of potential difference V, across the resistor are as given below, , (b) 9 Ω, (d) 900 Ω, , I, , 2.5, , 3.0, , 4.0, , 5.0, , 0.4, , 0.5, , 0.6, , 0.8, , 1.0, , Plot a graph between V and I and also calculate the, (CBSE 2018), resistance of that resistor., , 4. Why are coils of electric toasters and electric irons, made of alloy rather than a pure metal?, , (NCERT), , 5. (i) List the factors on which the resistance of a, , +, Voltage, , 2.0, , V (volts), , 22. Read the following and answer the questions from (i), to (v) given below, The electrical energy consumed by an electrical, appliance is given by the product of its power rating, and the time for which it is used. The SI unit of, electrical energy is Joule (as shown in figure)., , 0.5 1.0 1.5, , I (amperes) 0.1 0.2 0.3, , R, , conductor in the shape of a wire depends., (ii) Why are metals good conductors of electricity,, whereas glass is a bad conductor of electricity?, Give reason., (iii) What type of material is used in electrical heating, devices? Give reason., (CBSE 2018), , Power, , –, , Actually, Joule represents a very small quantity of, energy and therefore it is inconvenient to use where, a large quantity of energy is involved., , I, , 6., A, , (i) The SI unit of electric energy per unit time is, (a) joule, (c) watt, , (b) joule-second, (d) watt-second, , 12 V, , R1=4 Ω, , R2, , (ii) Kilowatt-hour is equal to, (a) 36, . × 104 J, 6, , (c) 36 × 10 J, , (b) 36, . × 106 J, (d) 36 × 104 J, , (iii) The energy dissipated by the heater is E. When, the time of operating the heater is doubled, the, energy dissipated is, (a) doubled, (c) remains same, , (b) half, (d) four times, , (iv) The power of a lamp is 60 W. The energy, consumed in 1 minute is, (a) 360 J, (c) 3600 J, , (b) 36 J, (d) 3.6 J, , (v) Calculate the energy transformed by a 5 A current, flowing through a resistor of 2 Ω for 30 minutes., (a) 40 kJ, (c) 10 kJ, , (b) 60 kJ, (d) 90 kJ, , B, , A student has two resistors 2 Ω and 3 Ω. She has to put, one of them in place of R 2 as shown in the circuit., The current that she needs in the entire circuit is, exactly 9 A., Show by calculation which of the two resistors she, should choose., , 7. A metal wire has diameter of 0.25 mm and electrical, resistivity of 0. 8 × 10 −8 Ω-m., (i) What will be the length of this wire to make a, resistance 5 Ω?, (ii) How much will the resistance change, if the, diameter of the wire is doubled?

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79, , CBSE Term II Science X, , 8. Redraw the given circuit, putting in an ammeter to, measure the current through the resistors and a, voltmeter to measure the potential difference across, 12 Ω resistor. What would be the readings in, (NCERT), ammeter and voltmeter?, +, , 6V, , 15. In an electrical circuit, two resistors of 2 Ω and 4 Ω, are connected in series to a 6 V battery. Find the heat, dissipated by the 4 Ω resistor in 5 s. (NCERT Exemplar), , 16. Why does the cord of an electric heater not glow, while heating element does?, , commercial unit of electrical energy?, , K, 8Ω, , 5Ω, , 12 Ω, , 19. The electric power consumed by a device may be, , 9. What is the reason of connecting electrical appliances, in parallel combination in household circuit?, , 10. What is (i) the highest and (ii) the lowest total, resistance which can be secured by combinations, of four coils of resistances 4 Ω, 8 Ω, 12 Ω and 24 Ω?, (NCERT), , 11. Consider the circuit diagram as given below, R5 = 3 Ω, , –, , +, , A, , R3 = 3 Ω, R4 = 3 Ω, , ( ), , K, , (NCERT), , 17. Why fuse are used in electrical circuits?, 18. Define the SI unit of electric power. What is the, , –, , calculated by using either of the two expressions, P = I 2R or P = V 2 / R ., The first expression indicates that the power is, directly proportional to R, whereas the second, expression indicates inverse proportionality. How, can seemingly different dependence of P on R in, these expressions be explained?, 20. Three 2 Ω 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, (NCERT Exemplar), flow through the three resistors?, 2Ω, , R1 = 3 Ω, , I, , R2 = 3 Ω, , A, , If R1 = R 2 = R 3 = R 4 = R 5 = 3 Ω, then find the, equivalent resistance of the circuit., , 12. With the help of suitable circuit diagram prove that, , +, E, –, , Q, , I, , C, 2Ω, , to a 250 V line mains. Solve, (i) the electric current drawn by it., (ii) energy consumed by it in 50 h., (iii) cost of energy consumed, if each unit costs ` 6., , 13. A battery E is connected to three identical lamps P, Q, , P, , B, , 21. An electric geyser rated 1500 W, 250 V is connected, , 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.(CBSE 2019), and R as shown in figure. Initially, the switch S is kept, open and the lamp P and Q are observed to glow with, same brightness. Then, switch S is closed., , 2Ω, , Long Answer Type Questions, 22. In the given circuit, A , B , C and D are four lamps, , !, , connected with a battery of 60 V., 60 V, , S, R, , How will the brightness of glow of bulbs P and Q, change? Justify your answer., , 14. (i) Write Joule’s law of heating., (ii) Two lamps, one rated 100 W, 220 V and the other, 60 W, 220 V are connected in parallel to electric, main supply., Find the current drawn by two bulbs from the line,, if the supply voltage is 220 V., (CBSE 2018), , 3A A 4A B 5A C 3A D, , Analyse the circuit to answer the following questions., (i) What kind of combination are the lamps arranged in, (series or parallel)?, (ii) Explain with reference to your above answer, what, are the advantages (any two) of this combination of, lamps?, (iii) Explain with proper calculations which lamp glows, the brightest., (iv) Find out the total resistance of the circuit.

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80, , CBSE Term II Science X, , 23. Find the equivalent resistance of the following circuit., Also, find the current and potential at each resistor., 1Ω, R2, 2Ω, , R3, 3Ω, , A2Ω, R1, , 2Ω B, R5, , R4, , +, , –, , 9V, , K, , 24. (i) State Ohm’s law., (ii) How is an ammeter connected in an electric circuit?, (iii) The power of a lamp is 100 W. Find the energy, consumed by it in 1 min., (iv) A wire of resistance 5Ω is bent in the form of a, closed circle. Find the resistance between two, points at the ends of any diameter of the circle., , 25. An electric lamp of resistance 20 Ω and a conductor, , !, , Case Based Questions, , 28. Read the following and answer the questions from (i), to (v) given below, Several resistors may be combined to form a, network. The combination should have two end, points to connect it with a battery or other circuit, elements. When the resistors are connected in series,, the current in each resistor is same but the potential, difference is different in each resistor., When the resistors are connected in parallel, the, voltage drop across each resistor is same but the, current is different in each resistor., (i) What do you mean by complex circuit?, (ii) In the graph, it shows the resistance in series and, parallel for two identical wires. Which of the, following denotes series combination and parallel, combination, individually?, B, , I, , of resistance 4 Ω are connected to a 6 V battery as, shown in the circuit given below., , A, , 4Ω, I, , V, , +, , K, , –, 6V, , –, , A, , (iii) What is the equivalent resistance of the circuit, given below?, , +, , Calculate, (i) the total resistance of the circuit., (ii) the current through the circuit., (iii) the potential difference across the (a) electric, lamp and (b) conductor., (iv) the power of the lamp., (CBSE 2019), , 26. (i) How should two resistors, with resistances R1 ( Ω ), and R 2 ( Ω ) be connected to a battery of e.m.f. V volts, so that the electrical power consumed is minimum?, (ii) In a house, 3 bulbs of 100 watt each are lighted for, 5 hours daily, 2 fans of 50 watt each are used for, 10 hours daily and an electric heater of 1.00 kWh, 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 ` 3.60 per kWh., , 27. (a) Define power and state its SI unit., (b) A torch bulb is rated 5V and 500 mA. Calculate its, (i) power, (ii) resistances, 1, (iii) energy consumed when it is lighted for 2 hours., 2, , RΩ, , RΩ, , RΩ, , RΩ, , A, , B, , (iv) What is minimum effective resistance?, (v) When three resistors of resistances R, 2R and 3R, are connected in series then, how will be the, value of current gets affected in each resistor by, applying a voltage V across the circuit?, , 29. Read the following and answer the questions from (i), to (v) given below, A cell or a battery is the source of electrical energy., Due to the chemical reactions inside them a, potential difference is setup which is responsible for, the flow of current through any electrical circuit., So, to maintain this flow, the source continuously has, to provide the energy. But only a part of this energy, helps in maintaining the current consumed into, useful work., Rest of it may be consumed in the form of heat by, raising the temperature of the appliances.

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81, , CBSE Term II Science X, , (i) How heat produced in a resistor is related to current, flowing in that resistor?, (ii) Give two practical application of heating effect of, current., (iii) Which type of energy is transformed into heat energy?, (iv) An electric iron of resistance 25 Ω takes a current of, 7 A. Calculate the heat developed in 0.5 min., (v) 200 J of heat is produced in 10 s in a 5 Ω resistance., Find the potential difference across the resistor., , 30. Read the following and answer questions from (i) to (v), given below, In resistance for a system of the resistor, there are two, methods of joining the resistors together as shown below, R1, , P, , R1, , M, , R2, , V, , S, , A, , +, , Rh I, +, , R3, , X, , R3, , R2, , L, , V, , Y, Q, , It showed an electric circuit in which 3 resistors, having resistor R 1 , R 2 and R 3 respectively are, joined end to end i.e in series., While the combination of the resistors in which, 3 resistors connected together which point X, and Y are said to be parallel., (i) Calculate the potential difference across a, series combination of resistors., (ii) What is the value of current in a series, combination?, (iii) Write the formula of electrical energy, dissipated in the resistor., (iv) If 200 resistors, each of value 0.2 Ω are, connected in series, what will be the resultant, resistance?, (v) What will be the effective resistance shown in, figure below?, 6Ω, , T, –, , I, , 6Ω, , K, –, +, , K, –, , 2Ω, –, , A+

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EXPLANATIONS, Objective Questions, , 8. (d) Resistance between terminals A and B, C, , 1. (c) Option (c) represents correct set-up for studying the, dependence of the current on the potential difference across, a resistor because ammeter A is connected in series while, voltmeter V is connected across parallel of resistor R., , 2Ω, , 2. (c) At given voltage, current I is inversely proportional to, 1, resistance, i.e. I ∝ . At given voltage I is maximum for R1 ,, R, therefore, R1 is minimum., , B, , 2Ω, , = ( 2 Ω , 2 Ω in series) parallel to 2 Ω, , I1, I (ampere), , A, , 2Ω, , = ( 2 + 2 )||2 Ω = 4 Ω || 2 Ω =, , 4×2 4, = Ω, 6, 3, , 9. (a) As, R eq, I2, , 3!, , 3!, , X, , Y, , I3, 3!, , V(volt), 3!, , At given voltage,, , I1 > I2 > I3, , ∴, , R1 < R 2 < R 3, , or, , R 3 > R 2 > R1, , 3!, , 3!, , ⇒, 6!, , 3. (c) Resistivity depends on the nature of material and the, temperature but does not depend on the shape of the, resistor., 4. (a) True; Specific resistance of a conductor increases on, increasing its temperature., , ⇒, , 3!, , 2!, , ⇒, , 3×6, =2 !, 3+6, , 5!, , 10. (b) The filament of electric bulb is made up of tungsten, because it has a very high resistance., , 5. (c) In first case,, , 11. (c), As heat produced, H = I2Rt, , RAʹ, Resistivity of the conductor, ρ =, l, , …(i), , Here, R = 2 Ω, For AD, H1 = 32 × 2 × 1 = 18 J, , In second case,, Resistivity of the conductor, ρ =, , RAʹ, 2l, , …(ii), , Q For same material, resistivity of the conductor will be, same., So, on substituting the value of ρ in Eq. (ii), we have, RA RAʹ, =, ⇒ Aʹ = 2 A, l, 2l, 6. (b) A material that offers higher resistance as compared to, conductors to the flow of electron, so has small number of, free electrons is called a poor conductor., 7. (a) The maximum resistance is obtained when resistors are, connected in series combination., Thus equivalent resistance, R eq = R1 + R 2 + R 3 + R 4, =, , 1 1 1 1, + + + =2 Ω, 2 2 2 2, , For DG , H2 = ( −2 )2 × 2 × 1 = 8 J, For GJ, H3 = (1)2 × 2 × 1 = 2 J, ∴ Total amount of heat produced in 3 s is,, H = H1 + H2 + H3, ⇒, , H = 18 + 8 + 2 = 28J, , 12. (b) Electric power determines the rate at which energy is, delivered by a current., 13. (c) Bulbs are rated presuming that they all are connected, with the same voltage supply. In parallel combination,, voltage across each bulb is same, so greater the watt, greater, the brightness., ∴ Brightness (B) of 3 bulbs are as follows, B100 > B60 > B40, ⇒, , BC > BB > BA

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83, , CBSE Term II Science X, , 14. (d) Given, power, P = 1 kW = 1000 W, Voltage,, Current,, , V = 220 V, I= ?, P 1000, I= =, = 4. 5 A, V 220, Thus, the rating of fuse-wire is 5 A which is greater than 4.5 A., , 15. (a) Given, V = 075, . V, I = 100 mA = 100 × 10−3 A = 01, . A, ∴ Power, P = VI = 075, . × 01, ., ⇒, , P = 75 mW, , 16. (c) As we know, resistance of conductor depends on the, length of conductor as R ∝ l, As the length of wire is more, resistance of conductor is, more., ∴A is true but R is false., 17. (a) Copper conducts the current without offering much, resistance due to high electrical conductivity. Hence,, conducting wires are made of copper., ∴Both A and R are true and R is the correct explanation of A., 18. (b) Resistors are connected in series, when they join, end-to-end consecutively with each other. In parallel, combination, the total resistance is always less than the least, resistance in combination., ∴ Both A and R are true but R is not the correct explanation, of A., 19. (b) Fuse is a safety device that is used in household circuits., It is connected in series with the main supply., Fuse consists of an alloy of lead and tin which has, appropriate melting point., ∴ Both A and R are true but R is not the correct explanation, of A., V2, P, 1, R∝, P, i.e. Higher the wattage of a bulb, lesser is the resistance and, so it will glow brighter., , 20. (a) The resistance, R =, , 22. (i) (c) The SI unit of electric energy per unit time is watt., electric energy J, P=, = = Js −1 = W, Q, time, s, (ii) (b) 1 kilowatt-hour is equal to 36, . × 106 joule., (iii) (a) As E ∝ t, ∴ When the time of operating the heater is doubled, the, energy dissipated is doubled., (iv) (c) Given, P = 60 W, t =1 min = 1 × 60 s, ∴, E = P × t = 60 × 60, ⇒, E = 3600 J, (v) (d) Given, I = 5A, R = 2 Ω , t = 30 min = 1800 s, ∴, , E = I2Rt = 5 × 5 × 2 × 1800, , ⇒, , E = 90000 = 90 kJ, , Subjective Questions, 1. Ohm’s law gives the relationship between current I flowing, in a metallic wire and potential difference V, across its, terminals. According to this law, the electric current flowing, through a conductor is directly proportional to the potential, difference applied across its ends, providing physical, conditions remains unchanged., i.e., V ∝ I or V = IR, where R is constant of proportionality called resistance of, conductor at a given temperature., Unit of resistance is ohm. It is said to be 1 ohm, if potential, difference of 1 volt across the ends of the conductor makes a, current of 1 A to flow through it., 1V, i.e., 1 ohm ( Ω ) =, = VA −1, 1A, 2. Resistance is the property of a conductor by virtue of which, it opposes/resists the flow of charges/flow of current through, it. Its SI unit is ohm and is represented by the Greek letter, V, Ω (ohm). Resistance of a conductor is given by R = ., I, 3. Scale : At x-axis, 1 div (1 cm) = 0.1 A, At y-axis, 1 div (1 cm) = 0.5 V, The graph between current and potential difference is, shown below., , ∴Both A and R are true and R is the correct explanation of A., 21. (i) (a) Materials that follow Ohm’s law are called ohmic, conductor. V - I graph is a straight line passing from, origin for ohmic conductors., (ii) (c) V - I curve for insulator is straight line that lies on, voltage axis at room temperature., I 1, (iii) (b) Slope of V - I graph = = ., V R, (iv) (d) On increasing the voltage, the resistance remains, same but current increases., (v) (a) Given, V = 9 V, I = 0.1 A, V 9, R= =, = 90 Ω, ∴, I 01, ., , Y, 5.0, 4.0, 3.0, 2.5, V, (Volts) 2.0, 1.5, 1.0, 0.5, 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8, I (Ampere), , X, , Resistance (R) of the resistor is determined by the slope of, V-I graph.

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84, , CBSE Term II Science X, , V, I, y 2 − y1 1.5 − 10, ., R=, =, x 2 − x1 0.3 − 0.2, 0. 5, =, = 5Ω, 0.1, , ∴ Resistance = R = Slope of graph =, ∴, , 5 × π × 1. 56 × 10−8, = 3062, . m, 0. 8 × 10−8, ρl, (ii) ∴Resistance, R =, A, ρl, ρl, 4, =, =, ×, 2, π D2, ⎛ D⎞, π⎜ ⎟, ⎝ 2⎠, D, [Q A = πr 2 and r = ], 2, ρl, ρl, [Q D has become 2D], R' =, =, 2, A, ⎛ 2D ⎞, π⎜, ⎟, ⎝ 2 ⎠, ρl, π D2, ρl, R', ρl × 4, Now,, =, ÷, R πD 2, πD 2, ρl, πD 2, 1, =, ×, =, 2, ρl × 4 4, πD, R, R' =, ∴, 4, Thus, resistance will decrease by 4 times., =, , I, A, , R2, , 8., +, , 6V, , –, , B, , Let R be the resistance of the entire circuit., Given, overall current needed, I = 9 A, Voltage,, V = 12 V, Using Ohm's law, V = IR, V 12 4, ⇒, R= =, = Ω, I, 9 3, Now, the resistors R1 and R 2 are in parallel combination., 1, 1, 1, 1 1, 1, =, +, ⇒ = +, ∴, R R1 R 2, R 4 R2, 4, 3 1, 1, [QR = Ω], ⇒, = +, 3, 4 4 R2, 1, 3 1 2 1, = − = =, ⇒, R2 4 4 4 2, ⇒, R2 = 2 Ω, So, the student should choose 2 Ω resistor., 7. Given, diameter = 0.25 mm, Resistivity, ρ = 0. 8 × 10−8 Ω-m, (i) Resistance, R = 5 Ω, ρl, We know that, R =, A, , D, ], 2, , =, , 5. (i) Following factors on which resistance of a wire depends, (a) length of wire : R ∝ l, 1, (b) area of cross-section of wire : R ∝, A, (c) resistivity of material of wire : R ∝ ρ, l, R =ρ, ∴, A, (ii) Metals are good conductors as their resistivity is very low, whereas glass is a bad conductor having high resistivity., (iii) Alloys are used as heating elements as their resistivity, and melting points both are very high., 6. The given circuit is shown below, , R1=4 Ω, , RA, =, ρ, , 2, , [Q A = πr 2 and r =, , 4. Alloys have a higher resistivity than their constituent metals., They do not oxidise or burn at higher temperatures as they, have high melting point. Thus, they are used to make coils of, electrical toasters and electric irons rather than pure metals., , 12 V, , l=, , ⇒, , ⎤, ⎡0. 25, × 10−3, ⎥⎦, ⎢⎣ 2, −8, 0. 8 × 10, , 5×π ×, , K, 5Ω, , 8Ω, , R1, , R2, , 12 Ω, +, , V, , –, A, +, , –, , Equivalent resistance of the circuit,, R = R1 + R 2 + R 3 = 5 + 8 + 12 = 25 Ω, [QR1 , R 2 and R 3 are connected in series], In series combination, current flowing through all the, resistances is same and equal to the total current flowing, through the circuit., ∴ Current in the resistors,, V 6, I= =, = 0. 24 A, R 25, ∴ Ammeter reading = 0.24 A, Potential across12 Ω resistance,, V = IR = 0.24 × 12 = 2.88 V, ∴ Voltmeter reading is 2 . 88 V., 9. Parallel combination of resistances is highly useful in, circuits used in daily life, as the circuits used have, components of different resistances requiring different, amounts of current.

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85, , CBSE Term II Science X, , This type of combination in a circuit divides the current, among the components (electrical gadgets), so that they can, have necessary amount of current to operate properly. This, is the reason of connecting electrical appliances in parallel, combination in household circuit., , Now, the given circuit can be redrawn as shown below, +, , A, R'' =2 Ω, , 8Ω, , 12 Ω, , ), , 10. (i) Resistance is maximum when resistors are connected in, series., 4Ω, , (, , K, , 24 Ω, B, , A, , R max = 4 + 8 + 12 + 24 = 48 Ω, (ii) Resistance is minimum when resistors are connected in, parallel., 4Ω, , R1 =3 Ω, , Now, it is clear from the above circuit that all the resistances, R5 , Rʹʹ and R1 are in series combination., As, current through R1 ,Rʹʹ and R5 is same., ∴ Equivalent resistance of the circuit is, , B, , A, , R5 =3 Ω, , –, , 8Ω, , R = R5 + Rʹʹ + R1 = 3 Ω + 2 Ω + 3 Ω = 8 Ω, 12. The following figure shows the connection of resistors in, parallel, , 12 Ω, , R1, 24 Ω, , Rmin. = 1 /, , I1, , 1 ⎤ 24, ⎡1 1 1, + +, +, =, Ω=2Ω, ⎢⎣ 4 8 12 24 ⎥⎦ 12, , I3, , 11. From the combination, it can be observed that R 2 and R 3 are, in series order., , I, , +, , K, R3 = 3 Ω, , R4, , ( ), , =, , 3Ω, , A, K, , R3, V, , R1= 3 Ω, , –, , +, , V, , –, , An applied potential difference V produces current I1 in, R1 , I2 in R 2 and I3 in R 3 ., Total current, I = I1 + I2 + I3, By Ohm’s law, I1 =, , R2 = 3 Ω, , Now, the given circuit can be redrawn as shown below, , Thus,, , V V, V, V, =, +, +, R R1 R 2 R 3, , ⇒, , ⎛ 1, V, 1, 1 ⎞, = V⎜, +, +, ⎟, ⎝ R1 R 2 R 3 ⎠, R, , ⇒, , 1, 1, 1, 1, =, +, +, R R1 R 2 R 3, , ⇒, , If R is the equivalent resistance, then I =, , –, , ), , R4, =, , 3Ω, , A, , R5 = 3 Ω, , (, , K, , R'= 6 Ω, , R1= 3 Ω, , Now, it can be seen that R 4 and Rʹ are in parallel, combination. As, currents through R 4 and Rʹ are, different. So, their equivalent resistance can be calculated, as below, 1 1 1+ 2 3 1, 1, 1, 1, =, +, = + =, = =, Rʹʹ Rʹ R 4, 6 3, 6, 6 2, ∴, , Rʹʹ = 2 Ω, , …(i), , V, V, V, and I3 =, , I2 =, R1, R2, R3, , As current through R 2 and R 3 is same. So, their equivalent, resistance is Rʹ = R 2 + R 3 = 3 Ω + 3 Ω = 6 Ω, , +, , I, +, A, –, , R5 = 3 Ω, , –, , +, , R2, I2, , V, R, [from Eq. (i)], , 13. The brightness of glow of bulb P will increase and brightness, of glow of bulb Q will decrease., This is because, on closing S, bulbs Q and R will be in parallel, and the combination will be in series with bulb P. Hence, the, total resistance of the circuit will decrease and the current, flowing in the circuit will increase. Therefore, the glow of, bulb P will increase., Also, bulbs Q and R will be in parallel in this case. So, the, current gets divided and lesser current flows through Q and, hence the glow of bulb Q decreases.

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86, , CBSE Term II Science X, , 14. (i) According to Joule’s law of heating, amount of heat, produced in a resistor is, (a) directly proportional to square of current flowing, through the resistor., ∴, H ∝ I2, (b) directly proportional to resistance of the resistor., ∴, H ∝R, (c) directly proportional to time for which the current, flows through the resistor., ∴, H∝ t, Hence,, H = I2Rt, (ii), , L1, , 100 W, 220 V, , L2, , 60 W, 220 V, , 220 V, , Here, potential, V = 220 V, Power, P1 = 100 W, P2 = 60 W, As, current drawn is given by, Power (P ), [From P = VI], I=, Voltage ( V ), 100, 60, So,, I1 =, = 0.45 A and I2 =, = 0.27 A, 220, 220, 15. Given, R1 = 2 Ω, R 2 = 4 Ω, t = 5s, V = 6 V, ∴Net resistance, R = R1 + R 2 = 2 Ω + 4 Ω = 6 Ω, V 6V, =1 A, ∴ Current, I = =, R 6Ω, In series, same 1 A current passes through both resistors., ∴ Heat dissipated, H = I2R1 × t = (1)2 × 4 × 5 = 20 J, 16. The heating element of heater is made up of an alloy that has, very high resistance. So when the current flows through it, it, becomes very hot and glows red. But the resistance of cord, is less because it is made up of copper or aluminium, so it, does not glow., 17. Electric fuse is used to protect the electrical circuit from, overloading and short circuit. When the current flowing, through a circuit exceeds the safe limit, the temperature of, fuse wire increases and due to heating effect, it gets melt, and breaks the circuit., 18. The 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 1 V. Thus,, 1W = 1 volt × 1 ampere = 1 VA, The commercial unit of electrical energy is kilowatt hour, (kWh), commonly known as ‘unit’., 1 kWh = 36, . × 106 J, 19. P = I2R is used when current flowing in every component of, the circuit is constant. This is the case of series combination, of the devices in the circuit., , P = V 2 / R is used when potential difference ( V ) across every, component of the circuit is constant. This expression is used, in case of parallel combination in the circuit. In series, combination, R is greater than the value of R in parallel, combination., 20. Resistance, R = 2 Ω, Maximum power, Pmax = 18 W, Maximum current, I max = ?, P = I 2R, P, 18, =, = 3 A = I max, R, 2, Maximum current that can flow through 2 Ω resistor is 3 A., This current divides along B and C because they are in, parallel combination. Voltage across B and C remains same, 1, and hence I ∝ . Since, B and C have same resistance 2 Ω, R, 3, each, and has same current. i.e. I = = 1.5 A flowing through, 2, B and C., 21. Given, power, P = 1500 W,, ⇒, , I =, , Voltage, V = 250 V, (i) Electric current drawn,, P 1500, I= =, = 6A, V 250, (ii) Energy consumed, E = Power × Time, [Q t = 50 h], = 1500 × 50, = 75000 Wh = 75 kWh, [Q 1 kW = 1000 W], [Q 1 unit = 1 kWh], = 75 unit, (iii)Q Cost of energy consumed = 75 × 6 = ` 450, 22. The given circuit is shown below, , 60 V, 3A A 4A B 5A C 3A D, , (i) In the circuit, all the lamps have same voltage, i.e. 60 V, but each lamp is having different current. So, the lamps, are arranged in parallel combination., (ii) The two advantages of lamps in parallel combination are, (a) if one lamp gets faulty, it will not affect the working, of other lamps., (b) in parallel combination of lamps, each lamp will use, the full potential of the battery., (iii) The lamp with the highest power will glow the brightest., As, power = Voltage × Current, In this case, all the lamps have same voltage i.e., 60 V.

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87, , CBSE Term II Science X, , For lamp A , current = 3 A, Power = 60 × 3 = 180 W, ∴, For lamp B, current = 4 A, Power = 60 × 4 = 240 W, ∴, For lamp C, current = 5 A, Power = 60 × 5 = 300 W, ∴, For lamp D, current = 3 A, Power = 60 × 3 = 180 W, ∴, As, the lamp C is having the maximum power, so it will, glow the brightest., (iv) Let R be the total resistance of the circuit., Total current in the circuit, I = 3 + 4 + 5 + 3 = 15 A, Voltage, V = 60 V, Using Ohm's law, V = IR, V 60, ⇒, R= =, =4Ω, I 15, 23. In the given circuit, R 2 , R 3 and R 4 are in parallel, combination. As, currents through R 2 , R 3 and R 4 are, different. So, their equivalent resistance R is, 1Ω, R2, 2Ω, , A, , 2Ω, , R3, 3Ω, , 2Ω, , R1, , R4, , R5, , B, , –, , +, , 9V, , K, , 1, 1, 1, 1, =, +, +, Rʹ R 2 R 3 R 4, 1 1 1 6 + 3 + 2 11, = + + =, =, 1 2 3, 6, 6, 6, Ω, 11, Now, the given circuit can be redrawn as shown below, Rʹ =, , ⇒, , A, , 2Ω, , 6/11 Ω, , 2Ω, , R1, , Rʹ, , R5, , 9V, , +, , B, , K, , R2, , I2, I3, I = I1, , R1, I4, I, , R3, , I5 = I, , R4, , R5, , + –, , K, , As R 2 , R 3 and R 4 are in parallel combination, so potential, drop at all resistances will be same as 1 V., V2 = V3 = V4 = Vʹ = 1V, Current through R 2 ,, I2 =, , V2, Vʹ 1, =, = =1 A, R2 R2 1, , Similarly,, , I3 =, , V3, Vʹ 1, =, = = 0. 5 A, R3 R3 2, , and, , I4 =, , V4, Vʹ, 1, =, =, = 0.33 A, R4 R4, 3, , 24. (i) A states that the electric current flowing through a, conductor directly proportional to the potential, difference applied across its ends provided the physical, conditions such as temperature remains unchanged., (ii) An ammeter is connected in series in a circuit., (iii) Given, P = 100 W and time t = 1 min = 60 s., As, energy E = Pt = 100 × 60 = 6000 J, (iv) Given, a wire with resistance 5 Ω., Now, wire is converted into a ring as shown below, A, , B, , –, , Now, R1 , Rʹ and R5 are in series combination. As, current, through R1 , Rʹ and R5 is same., So, equivalent resistance of the whole circuit is, R = R1 + Rʹ + R5, 6, 22 + 6 + 22 50, = 2+ + 2 =, =, Ω, 11, 11, 11, Now, total current flowing through the circuit,, V 9 99, =, = ≈ 2A, R 50 50, 11, Current through R1 and R5 will be same as these are in, series combination and will be equal to the total current, flowing through the circuit., ∴, I = I1 = I5 = 2 A, I=, , Potential drop at R1 ,, V1 = I1R1 = 2 × 2 = 4 V, Potential drop at R5 ,, V5 = I5R5 = 2 × 2 = 4 V, Now, potential drop at Rʹ , Vʹ can be calculated as, V = V1 + V5 + Vʹ, ⇒, 9 = 4 + 4 + Vʹ, ⇒, Vʹ = 1 V, , The equivalent circuit,, A, , Hence, R eq = R AB =, , B, , 2.5 × 2.5, = 1.25W, 2.5 + 2.5, , 25. Given, resistance of lamp (R1 ) = 20 Ω, Resistance of conductor (R 2 ) = 4 Ω, Potential difference of battery ( V ) = 6 V, (i) Total resistance,, R = R1 + R 2 = 20 + 4 = 24 Ω

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88, , CBSE Term II Science X, , (ii) Current through the circuit, V 6, I= =, = 0.25 A, R 24, (iii) (a) Potential difference across electric lamp = IR1, = 0.25 × 20 = 5 V, (b) Potential difference across conductor, = IR 2 = 0.25 × 4 = 1 V, (iv) Power of lamp = I2R1, = (0.25)2 × 20, = 00625, ., × 20 = 1.25 W, 26. (i) Power is given as, P = VI where V is the voltage and I is, the current. Power consumed is minimum when the, current passing is minimum. So, the resistors should be, connected in series., (ii) Power of each bulb= 100 W, Total power of 3 bulbs, PB = 3 × 100 = 300 W, Energy consumed by 3 bulbs in a day,, E B = PB × t B, [Q t B = 5 h], = 300W × 5 h = 1. 5 kWh, Similarly, power of 2 fans, PF = 2 × 50 W = 100 W, Energy consumed by 2 fans, E F = PF × t F, [Q t F = 10 h], = 100 W × 10 h, = 1 kWh, Now, energy consumed by heater in a day, E H = PH × t H, 1, 1, = 1kWh × h [∴ PH = 1 kWh, t H = h], 2, 2, = 0. 5 kWh, Total energy consumed in a day, E = EB + EF + EH, = (1. 5 + 1 + 0. 5) kWh = 3 kWh, Energy consumed in a month of 31 days, = E × 31 = 93 kWh, ∴The cost of energy consumed, = (` 3.60/kWh) × (93 kWh), = ` 334.80, 27. (a) Power is defined as the rate at which electric energy is, dissipated or consumed in an electric circuit., P = VI = I2R = V 2 /R, The 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 1 V., 1 W =1 volt × 1 ampere =1 VA, (b) Given, voltage rating, V = 5 V and current rating I= 500 mA, (i) As, we know power of bulb, [Q A, P = VI = 5 × 500 × 10−3, −3, 1mA = 10 A], = 2.5 W, = 2.5 × 10− 3 kW, , (ii) Resistance of bulb,, V, R=, I, , (Ohm’s law), , 5, = 10 Ω, 500 × 10− 3, 1, (iii) Energy consumed in 2 hour, 2, 2.5 × 2.5, 1, [2 hour = 2.5 hour], E = P.t. =, 1000, 2, 6.25, =, = 0.00625 kWh, 1000, 28. (i) The electrical circuit in which some resistances are, connected in series combination and some in parallel, combination, this type of combination is called complex, circuit., (ii) Resistance in series is always greater than resistance in, parallel. The slope of V-I graph gives resistance., Hence, line B denotes resistance in series combination, and line A denotes resistance in parallel combination., (iii) Equivalent circuit diagram is, ⇒, , R=, , RS1, , A, , B, , RS2, , R S 1 = R + R = 2R, R S 2 = 2R + 2R = 4R, Both of them are in parallel., R S2 × R S1, 4, = R, R eq =, ∴, R S1 + R S2 3, (iv) When the equivalent resistance is less than the resistance, of least resistor. This is known as minimum effective, resistance., (v) In series combination, the value of current is same, through each resistor., So, it will not be affected at all., 29. (i) According to Joule’s law of heating, the heat produced in, a resistor is directly proportional to the square of current, flowing through that resistor., (ii) Practical application of heating effect of current is, electric heater, electric iron etc., (iii) Electrical energy is transformed into heat energy., (iv) Given, resistance R = 25 Ω; current I = 7 A;, Time t = 0.5 min = 0.5 × 60 = 30 s. Heat H = ?, We know that, Heat H = I2Rt, H = (7 )2 × 25 × 30 = 36750 J, . × 104 J, = 368, So, the heat developed is 368, . × 104 J.

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89, , CBSE Term II Science X, , (v) Given, Heat H = 200 J, Resistance, R = 5 Ω, Time t = 10 s, Potential difference V = ?, We know that,, Heat, H = I2Rt, ⇒ Current, I =, , H, Rt, , R1 × R 2, R1 + R 2, 6×2, =, 6+ 2, 3, RP = Ω, 2, Equivalent circuit is shown below,, ∴, , 200, =2A, 5 × 10, So, the potential difference across the resistor is, 7=, , V = IR, , (iii) (c) Electrical energy = Power × Time, ⇒, W = P × T = VIT, (iv) (b) Resistance of each resistor = 0.2 Ω, Total resistance in series R S = n × R = 200 × 0.2, ⇒, R S = 40 Ω, (v) (d) Resistor of 6 Ω and 2 Ω are in parallel., , [by Ohm’s law], , = 2 × 5 = 10 V, 30. (i) (a) An applied potential V produces current I in the, resistors and R1 , R 2 and R 3 causing a potential drop, V1 , V2 and V3 respectively, through each resistor., Total potential, V = V1 + V2 + V3, (ii) (a) In series combination of resistors, the current is same, at every point of the circuit, i.e. the current through each, resistor is same., , RP =, , 3/2 Ω, , Both of them are in series, ⎛3, ⎞, R eq = ⎜ + 6⎟ Ω, ∴, ⎝2, ⎠, ⇒, , R eq =, , 15, Ω, 2, , 6Ω

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Chapter Test, , 8. Assertion Heater wire must have high resistance, and high melting point., Reason If resistance is high, the electric conductivity, will be less., , Multiple Choice Questions, , Short Answer Type Questions, , 1. Which of the following obeys Ohm’s law?, (a) Filament of a bulb, (c) Nichrome, , 9. Draw the V-I graph for ohmic and non-ohmic, , (b) LED, (d) Transistor, , conductors., , 2. What is the minimum resistance which can be made using, five resistors each of 1/5 Ω?, (a) 1/5 Ω, , (b) 1/25 Ω, , (c) 1/10 Ω, , (d) 25 Ω, , 3. Two resistors of resistance 2 Ω and 4 Ω when connected to, a battery will have, (a) same current flowing through them when connected in parallel, (b) same current flowing through them when connected in series, (c) same potential difference across them when connected in series, (d) different potential difference across them when connected in, parallel, , 4. Two bulbs have the following ratings, (i) 40 W, 220 V, (ii) 20 W, 110 V, The ratio of their resistances is, (a) 1 : 2, , (b) 2 : 1, , (c) 1 : 1, , (d) 1 : 3, , 5. At the time of short circuit, the electric current in the, circuit, (a) varies continuously, (b) reduces substantially, (c) does not change, (d) increases heavily, , 10. A wire of resistance 5 Ω is bent in the form of a, closed circle. Find the resistance between two, points at the ends of any diameter of the circle., , 11. Give two characteristics properties of copper wire, which makes it suitable for use as fuse wire., , 12. A copper wire has diameter 0.5 mm and resistivity, ρ = 1.6 × 10 −8 Ω-m. What will be the length of its, wire to make its resistance 10 Ω ? How much does, the resistance change, if diameter is doubled?, , 13. Two conducting wires of 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, what is the ratio of heat, produced in series and parallel combinations ?, , 14. Compare the power used in 2 Ω resistor in each of, the following circuits (i) a 6 V battery in series with 1, Ω and resistors, (ii) a 4 V battery in parallel with, 12 Ω and 2 Ω resistors., , Long Answer Type Questions, , 15. Two metallic wires A and B are of same material, , Assertion-Reasoning MCQs, Direction (Q. Nos. 6-8) Each of these questions contains, two statements Assertion (A) and Reason (R). Each of, these questions also has four alternative choices, any one, of which is the correct answer. You have to select one of, the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct explanation of A., (b) Both A and R are true, but R is not the correct explanation, of A., (c) A is true, but R is false., (d) A is false, but R is true., , 6. Assertion At high temperatures, metal wires have a, greater chance of short circuiting, Reason Both resistance and resistivity of a material vary, with temperature., , 7. Assertion Incandescent bulb passes an electric current, through a metal filament to produce heat., Reason The higher the wattage, the more energy is, consumed., , where length of wire A is l and wire B is 2l and radius, of wire A is r and wire B is 2r., (i) When both of them are connected in series,, find the ratio of total resistance in series, combination with wire A ., (ii) When both of them are connected in parallel,, Find the ratio of total resistance in parallel, combination with wire A., , 16. (i) Two lamps rated 100 W, 220 V and 10 W, 220 V are, connected in parallel to 220 V supply. Calculate the, total current through the circuit., (ii) Two resistors X and Y of resistances 2 Ω and, 3 Ω respectively are first joined in parallel and, then in series. In each case, the supplied voltage is, 5 V., (a) Draw circuit diagrams to show the combination, of resistors in each case., (b) Calculate the voltage across the 3 Ω resistor in, the series combination of resistors., , Answers, Multiple Choice Questions, 1. (c), , 2. (b), , 3. (b), , 4. (b), , Assertion-Reasoning MCQs, 6. (a), , 7. (b), , 8. (d), , 5. (d), , For Detailed Solutions, Scan the code

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CHAPTER 6, , Magnetic Effects, of Electric Current, In this Chapter..., !, , Magnetic Field, , !, , Electric Motor, , !, , Electromagnetic Induction, , In 1820, Christian Oersted discovered that a compass needle, get deflected when a current carrying metallic conductor is, placed nearby it. He concluded that the deflection of, compass needle was due to the magnetic field produced by, the electric current., Hence, it was deduced that electricity and magnetism are, related to each other., , Magnetic Field, It is the space around a magnet in which its effect can be, experienced i.e. its force can be detected. It is a vector, quantity. The SI unit of magnetic field is Tesla., Magnetic Lines of Force, They are the imaginary lines representing magnetic field, around a magnet. When iron fillings are kept near a magnet,, they get arranged in a pattern which represents the magnetic, field lines., S, , N, , Magnetic field lines around a bar magnet, , A compass needle behaves as a small bar magnet whose one, end points towards North and other end towards South., , Properties of Magnetic Field Lines, The magnetic field lines have the following properties, They originate from North pole of a magnet and end at its, South pole, by convention., These lines are closed and continuous curves., They are crowded near the poles, where the magnetic field, is strong and separated far from the poles, where the, magnetic field is weak., Field lines never intersect with each other. If they do, that, would mean that there are two directions of the magnetic, field at the point of intersection, which is impossible., !, , !, , !, , !, , Magnetic Field due to a Current, Carrying Conductor, When electric current flows through a metallic conductor, a, magnetic field is produced around it., Different magnetic field patterns are produced by current, carrying conductors of different shapes., , Magnetic Field due to a Current through, a Straight Conductor, The magnetic field lines around a current carrying, straight conductor are concentric circles whose centres lie, on the wire.

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92, , CBSE Term II Science X, , The magnitude of magnetic field B produced by a straight, current carrying wire at a given point is, (i) Directly proportional to the current I passing through, the wire,, i.e., …(i), B ∝I, (ii) Inversely proportional to the distance r from the current, carrying conductor,, 1, …(ii), i.e., B∝, r, I, , I, , Concentric field lines around a straight conductor, , I, r, If the direction of current in a straight wire is known, then the, direction of magnetic field produced by it is obtained by, Maxwell’s right hand thumb rule., , By using Eqs. (i) and (ii), we get, B ∝, , Maxwell’s Right Hand Thumb Rule, It states that, if you hold the current carrying straight wire in, the grip of your right hand in such a way that the stretched, thumb points in the direction of current, then the direction of, the curl of the fingers will give the direction of the magnetic, field. This rule is also called Maxwell’s corkscrew rule., , At every point on a current carrying circular loop, the, magnetic field is in the form of concentric circles around it., As we move away from it, the circles would become larger, and larger., When we reach the centre of loop, the field appears to be a, straight line. The magnetic field produced by current, carrying circular wire at a given point is, (i) Directly proportional to the amount of current (I ), passing through it,, i.e., ...(i), B ∝I, (ii) Directly proportional to the number of turns ( N ) of the, wire,, i.e., ...(ii), B∝N, This is because the current in each turn is in the same, direction. Therefore, the field due to these turns get, added up., Thus, the strength of magnetic field produced by a, current carrying circular coil can be increased by, increasing the number of turns of the coil., increasing the current flowing through the coil., !, , !, , Magnetic Field due to a Current, in a Solenoid, A solenoid is defined as a coil consisting of a large number, of circular turns of insulated copper wire. These turns are, wrapped closely to form a cylinder., , S, , N, +, , –, , K, , Magnetic field, , Current, , Magnetic field lines of force due to a current, carrying solenoid, , Maxwell’s right hand thumb rule, , Magnetic Field due to a Current, through a Circular Loop, The magnetic field lines due to a circular coil are shown in the, given figure., , N, , S, , –, +, , Magnetic field lines due to a, current through a circular loop, , The field lines around a current carrying solenoid are, similar to that produced by a bar magnet. This means that a, current carrying solenoid behaves as if it has North pole and, South pole. The field lines inside the solenoid are parallel to, each other., Thus, the strength of magnetic field is the same, i.e. uniform, at all points inside a solenoid., Electromagnet, The 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, electromagnet., The magnetic effect remains only till the current is flowing, through the solenoid.

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93, , CBSE Term II Science X, , An electromagnet is used in electric bells, electric, motors, telephone diaphragms, loudspeakers and for, sorting scrap metal., , Electric Motor, It is a rotating device used for converting electric energy into, mechanical energy., Principle, , –, , +, , It is based on the principle that when a rectangular coil is placed, in a magnetic field and current is passed through it, two equal and, opposite forces act on the coil which rotate it continuously., , K, , Construction, , An electromagnet, , Force on a Current Carrying, Conductor in a Magnetic Field, When a current carrying conductor is placed in a, magnetic field, it experiences a force. The force acting is, due to interaction between magnetic field produced by, the current carrying conductor and external magnetic, field in which the conductor is placed., The direction of force on the conductor depends on the, following factors, (i) Direction of current The direction of force on the, conductor can be reversed by reversing the, direction of current., (ii) Direction of magnetic field The direction of force, on the conductor can be reversed by reversing the, direction of magnetic field by interchanging the, position of poles., Force on the conductor is maximum when the direction, of current is at right angles to the direction of magnetic, field., , Fleming’s Left Hand Rule, , It consists of a rectangular coil, connected to a source of current, and a switch., The commutators R1 and R 2 are fixed to the coil and pressed, tightly against the brushes X and Y., The function of commutator is to reverse the direction of current, flowing through the coil, after every half rotation. In an electric, motor, split rings act as commutator., Rectangular coil, B, , N, , Field, , D, , A, Permanent, magnet, , R1 R2, , Split rings, (R1 and R2), , Y, Brushes, (X and Y ), , X, Axle, K, +, , –, , Working, !, , !, , Field, , Force, or, motion, , S, , A simple electric motor, , The direction of force which acts on a current carrying, conductor placed in a magnetic field is given by Fleming’s, left hand rule., , Current, , C, , !, , Force, , Current, , Fleming’s left hand rule, , It states that, if the forefinger, thumb and middle, finger of left hand are stretched mutually perpendicular, to each other, such that the forefinger points along, the direction of external magnetic field, middle, finger indicates the direction of current, then the, thumb points towards the direction of force acting on, the conductor., , !, , !, , !, , !, , Let coil ABCD be in horizontal position. When the key is, closed, the current flows in the coil ABCD through brush X and, flows back to the battery through the brush Y via ring R 2 ., No force acts on arms BC and ADas they are parallel to magnetic, field. Arm AB experiences a force in downward direction and arm, CD experiences an equal force in upward direction., The direction of force is obtained by applying Fleming’s left, hand rule. This causes the coil to rotate in the anti-clockwise, direction., When the rotating coil is in the vertical position, the brushes, lose contact with the rings and current stops flowing. But the, coil does not stop due to inertia of motion., When the coil rotates, the rings change their positions and, come in contact with opposite brushes., This reverses the direction of current through the coil but the, direction of current on right hand side of the coil remains the, same., So, the force on right hand side is always upwards and a force on, left hand side is always in downward direction. Thus, the coil, continues to rotate in anti-clockwise direction.

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94, , CBSE Term II Science X, , The speed of rotation of the motor can be increased by, increasing the strength of the current in the coil., increasing the number of turns in the coil., increasing the area of the coil., increasing the strength of magnetic field., , Ways to Induce Current in a Circuit, , !, , !, , !, , !, , Commercial Electric Motor, It has the following components, An electromagnet in the place of permanent magnet., A large number of turns of conducting wire in the current, carrying coil., A soft iron core on which the coil is wound. The combination, of soft iron core and coil is called armature. It enhances the, power of motor., Electric motor is used in electric fans, refrigerators, mixers,, washing machines, computers, MP3 players, etc., , There are different methods by which current can be, induced in a circuit, (i) By Moving a Coil in a Magnetic Field Current can, be induced in coil either by moving it in a magnetic, field or by changing the magnetic field around it as, indicated by deflection in galvanometer needle., Motion of coil, , !, , Stationary, magnet, N, S, , !, , Coil, , !, , !, , Electromagnetic Induction, Production of an electric current in a closed circuit by a, changing magnetic field is called an induced current. This, phenomenon is called electromagnetic induction., Direction of motion, Coil, , N, , S, A bar magnet, , G, , Galvanometer, , G, , Moving coil towards stationary magnet, , The induced current is found to be maximum when, the direction of motion of the coil is at right angle, to the magnetic field., The direction of induced current can be reversed by, reversing the direction of magnetic field. If the coil as, well as the magnet are stationary, then no current is, induced in the coil., (ii) By Changing the Magnetic Field Around a Nearby, Coil Consider two coils, where coil 1 is called as, primary coil and coil 2 as secondary coil. Primary coil, is connected to a battery., Coil 1, , Coil 2, , A bar magnet is pushed inside a fixed current carrying coil, , Fleming’s Right Hand Rule, The direction of induced current is given by Fleming’s right hand, rule. It states that, if the forefinger, middle finger and thumb of the, right hand are stretched at right angles to each other, with the, forefinger in the direction of the magnetic field and the thumb in, the direction of the motion of the wire, then the induced current in, the wire is in the direction of the middle finger., Galvanometer, It is an instrument that can detect the presence of current in a, circuit. The pointer remains at zero (the centre of the scale) for, zero current flowing through it., Depending upon the direction of current, it deflects either to, the left or to the right of the zero mark., , +, , –, K, , G, , Set up of two stationary coils, where current is, induced in coil 2 when current in coil 1 is changed, , When the key (K) is closed, the current in primary coil, takes a little time to rise from zero to a maximum, value. This causes a momentary change in the, magnetic field around this coil. This induces a, momentary current in the secondary coil., The same happens in the reverse direction when the, key (K) is opened. Current is induced in coil 2 when, current in coil 1 is changed which is indicated by the, deflection in galvanometer needle.

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95, , Chapter, Practice, PART 1, Objective Questions, , (a) Magnetic field lines are closed and continuous curves, (b) Magnetic field lines never intersect with each other, (c) Magnetic field lines are crowded near the poles, (d) All of the above, , 5. Four students A, B, C and D plotted the sketch of, the patterns of magnetic field lines representing the, magnetic field around a current carrying straight, wire as shown. Whose sketch is correctly, represented?, – A+, , 2. Which of the following is the correct representation, of uniform magnetic field?, (a) N, , (b) S, , S, , – A+, +, , 1. Which of the following is the property of magnetic, field lines ?, , +, , Multiple Choice Questions, , –, , –, , K, Student B, , K, Student A, , N, , – A+, , S, , magnetic field near a long straight wire?, , (NCERT), , 4. 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 the magnetic field at a point, will be North to South, N, , –, , –, , K, Student C, , 3. Which of the following correctly describes the, (a) The field consists of straight lines perpendicular to the, wire, (b) The field consists of straight lines parallel to the wire, (c) The field consists of radial lines originating from the, wire, (d) The field consists of concentric circles centred on the, wire, , +, , (d) N, , N, , – A+, , +, , (c) N, , (a) A, , K, Student D, , (b) B, , (c) C, , plane of paper carries a current when the key is ON., The 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., The magnetic field lines point from B to A., The N-pole of the resultant magnet is on the face, (NCERT Exemplar), close to, A, , B, Variable, resistance, , W, , E, , S, W, , E, , (NCERT), , (d) D, , 6. A circular loop placed in a plane perpendicular to the, , +, , !, , (a) directly above the wire, (b) directly below the wire, (c) at a point located in the plane of the paper, on the, North side of the wire, (d) at a point located in the plane of the paper, on the, South side of the wire, , +, , A, , –, K, , –

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96, , CBSE Term II Science X, , (a) B, (b) A, (c) A, if the current is small and B, if the current is large, (d) B, if the current is small and A, if the current is large, , 7. Which of the following properties of a proton can, change when it moves freely in a magnetic field?, (a) Mass, (c) Velocity, , (b) Speed, (d) Momentum, , (NCERT), , 8. A positively charged particle (α-particle) projected, towards West is deflected towards North by a, magnetic field. The direction of magnetic field is, (NCERT), , (a) towards South, (c) downwards, , (a) forces both pointing into the plane of paper, (b) forces both pointing out of the plane of paper, (c) forces pointing into the plane of paper and out of the, plane of paper, respectively, (d) force pointing opposite and along the direction of, the uniform magnetic field, respectively, , 12. A rectangular loop carrying a current i is situated, near a long straight wire such that the wire is, parallel to one of the sides of the loop and is in, the plane of the loop. If a steady current i is, created in wire as shown in figure below, then, the loop will, i, , (b) towards East, (d) upwards, , i, , 9. The strength of magnetic field inside a long current, carrying straight solenoid is, (a) more at the ends than at the centre, (b) minimum in the middle, (c) same at all point, (d) found to increase from one end to the other, , 10. Match the items in Column I with the items in, Column II and choose the correct codes given below., Column I, , Column II, , A. SI unit of magnetic field (i) Small bar magnet that rotates, B. Magnetic field inside, solenoid, , (ii) Tesla, , C. Compass needle, , (iii) Temporary magnet, , D. Solenoid, , (iv) Uniform value, , Codes, A, (a) (ii), (b) (iii), (c) (i), (d) (iv), , B, (iv), (ii), (iii), (i), , C, (i), (iv), (ii), (iii), , D, (iii), (i), (iv), (ii), , 13. The direction of the induced electric current in a, conductor, when placed in a varying magnetic, (CBSE 2020), field can be assessed by, (a) Maxwell’s right hand-thumb rule, (b) Ohm’s law, (c) Fleming’s left hand rule, (d) Fleming’s right hand rule, , 14. In the arrangement shown in figure below, 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., (NCERT Exemplar), Then,, Coil I, , 11. A uniform magnetic field exists in the plane of paper, pointing from left to right as shown in the figure. In, the field, an electron and a proton move as, shown in the figure. The electron and the proton, (NCERT Exemplar), experience, Proton, , Uniform, magnetic, field, , Electron, , (a) rotate about an axis parallel to the wire, (b) move towards the wire, (c) move away from the wire or towards right, (d) remain stationary, , +, , –, , Coil II, , 9, , 10, , 8 7, , 6, 5, , e, , (a) The deflection in the galvanometer remains zero, throughout., (b) There is a momentary deflection in the, galvanometer only when the key is removed., (c) There are momentary galvanometer deflections that, die out shortly; the deflections are in the same, direction., (d) The galvanometer shows momentary deflections in, opposite directions.

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97, , CBSE Term II Science X, , 15. Match the terms of Column I with Column II and, , 19. Assertion The magnetic field produced by a current, , choose the correct option from the codes given below., (i) Direction of magnetic, force on a North pole, , B. Direction of induced, current, , (ii) Fleming’s left hand rule, , C. Direction of magnetic, field produced by, straight current, carrying conductor, , (iii) Maxwell’s right hand, thumb rule, , D. Direction of magnetic, field lines at a point in, a magnet, , (iv) Fleming’s right hand, rule, , Codes, A, (a) (ii), (b) (i), (c) (ii), (d) (i), !, , Column II, , A. Direction of force, , B, (iv), (ii), (iii), (iii), , C, (iii), (iii), (i), (ii), , D, (i), (iv), (iv), (iv), , Assertion-Reasoning MCQs, Direction (Q. Nos. 16-20) For given questions two, statements are given-one labelled Assertion (A) and the, other labelled Reason (R). Select the correct answer to, these questions from the codes (a), (b), (c) and (d) as, given below., (a) Both A and R are true and R is the correct, explanation of A., (b) Both A and R are true, but R is not the correct, explanation of A., (c) A is true, but R is false., (d) A is false, but R is true., , 16. 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 B produced by a straight, current carrying wire is inversely proportional to, the distance from the wire., , 17. Assertion A current carrying conductor experiences, a force in a magnetic field., Reason The force acting on a current carrying, conductor in a magnetic field is due to interaction, between magnetic field produced by the conductor, and external magnetic field., , 18. Assertion If an electron, moving vertically from outer, space, enters the earth’s magnetic field, then it gets, deflected towards West., Reason Electron has negative charge., , 20. Assertion Production of an electric current in a, closed circuit by a changing electric field is called, an induced current., Reason The direction of induced current is given, by Fleming’s right hand rule., !, , Case Based MCQs, 21. Read the following and answer the questions from, (i) to (v) given below, A solenoid is a long helical coil of wire through, which a current is running in order to create a, magnetic field., The magnetic field of a solenoid is the superposition, of the fields due to the current through each coil. It, is nearly uniform inside the solenoid and close to, zero outside and is similar to the field of a bar, magnet., The following graph is obtained by a researcher,, while doing an experiment to see the variation of the, magnetic field with respect to the current in the, solenoid. The unit of magnetic field as given in the, graph attached is in (mT) and the current is given, in (A)., 18, 16, Magnetic field B (mT), , Column I, , carrying solenoid is independent of its length and, cross-sectional area., Reason The magnetic field inside the solenoid has, variable value., , 14, 12, 10, 8, 6, 4, 2, 0, 0.0, , 0.2, , 0.4, , 0.6, 0.8, Current (A), , 1.0, , 1.2, , (CBSE Sample Paper), , (i) What type of energy conversion is observed in a, linear solenoid?, (a) Mechanical to magnetic, (b) Electrical to magnetic, (c) Electrical to mechanical, (d) Magnetic to mechanical

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98, , CBSE Term II Science X, , (ii) What will happen, if a soft iron bar is placed inside the, solenoid?, (a) A high amount of electric charge flows in the bar, resulting in short-circuit., (b) The bar will be magnetised as long as there is current, in the circuit, (c) The bar will be magnetised permanently, (d) The bar will not be affected by any means, (iii) The magnetic field lines produced inside the solenoid, are similar to that of, (a) a bar magnet, (b) a straight current carrying conductor, (c) a circular current carrying loop, (d) electromagnet of any shape, (iv) After analysing the graph, a student writes the following, statements., I. The magnetic field produced by the solenoid is, inversely proportional to the current., II. The magnetic field produced by the solenoid is, directly proportional to the current., III. The magnetic field produced by the solenoid is, directly proportional to square of the current., IV. The magnetic field produced by the solenoid is, independent of the current., Which of the following would be the correct statement(s)., (a) Only IV, (b) I, III and IV, (c) Both I and II, (d) Only II, (v) From the graph, deduce which of the following, statements is correct?, (a) For a current of 0.8 A, the magnetic field is 13 mT., (b) For larger currents, the magnetic field increases, non-linearly., (c) For a current of 0.8 A, the magnetic field is 1.3 mT., (d) There is not enough information to find the magnetic, field corresponding to 0.8 A current., , 22. Read the following and answer the questions from (i), to (v) given below, A student wants to study the working of electric motor., He used a model of DC motor for electromagnetism as, shown in figure., , (i) The direction of rotation of the coil, when viewed, from the front by the student is, (a) clockwise, (b) anti clockwise, (c) First half clockwise and other half anti-clockwise, (d) First half anti-clockwise and other half clockwise, (ii) The student is still testing on the feasibility of using, metal strips in the model. His observations are given, below., I. As the current reverses in the coil for every half, turn, the coil rotates in one direction., II. The speed of rotation of the motor is increased, if, the value of current is increased., III. The direction of force, acting on the coil is given by, Fleming’s left hand rule., IV. The coil continues its rotation in magnetic field, even if there is no current in circuit., , The correct observations made by him are, (a) I, II and IV, (b) II, III and IV, (c) I, II and III, (d) II and III, (iii) Commercial electric motors do not use, (a) an electromagnet to rotate the armature., (b) effectively large number of turns of conducing, wire in the current carrying coil, (c) a permanent magnet to rotate the armature, (d) a soft iron core on which the coil is wound, (iv) Which one of the following is true about electric, motor?, (a) It converts electrical energy into mechanical, energy, (b) It converts mechanical energy into electrical, energy, (c) It converts magnetic energy into electric energy, (d) It converts electric energy into magnetic energy, (v) The direction of magnetic field at a place is coming, out of the paper. A wire whose direction of current, flow is as shown in the figure is placed there. In, which direction is the force due to the magnetic, field experienced by the wire?, N, , i, , Metal S, strips, , 45°, , N, W, , O, , Rheostat, – +, K, , He fixed the two ends of the coil to a pair of curved, elastic metal strips. The metal strips are connected to, the power supply with a rheostat., , S, ., , (a) North-West direction, (b) North direction, (c) South-West direction, (d) South-East direction, , 45°, , E

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PART 2, Subjective Questions, !, , Short Answer Type Questions, , 1. List the properties of magnetic lines of force. Why, do two magnetic lines of force not intersect with, (NCERT), each other?, , 2. The adjoining diagram shows two, parallel straight conductors carrying, same current. Copy the diagram and draw, the pattern of the magnetic field lines, X, around them showing their directions., What is the magnitude of magnetic field at, a point X which is equidistant from the, conductors? Give justification for your, (CBSE 2019), answer., , 3. How will the strength of the magnetic field change, when the point where magnetic field is to be, determined is moved away from the straight wire, carrying constant current? Justify your answer., (CBSE 2019), , 4. 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?, A, r1, , P, , I, Q, , (i) Vertically in North-South plane and an observer, looking it from East sees the current to flow in, anti-clockwise direction., (ii) Vertically in East-West plane and an observer looking it, from South sees the current to flow in anti-clockwise, direction., (iii) Horizontally and an observer looking at it from below, sees current to flow in clockwise direction., , 8. How does a solenoid behave like a magnet? Can, you determine the North and South poles of a, current carrying solenoid using a bar magnet?, (NCERT), Explain., 9. Give reasons for the following, (i) There is either a convergence or a divergence of, magnetic field lines near the ends of a current, carrying straight solenoid., (ii) The current carrying solenoid when suspended freely, rests along a particular direction., (iii) The strength of magnetic field is uniform inside a, solenoid., (CBSE 2020), , 10. When is the force experienced by a current, carrying conductor placed in a magnetic field, largest? Which rule determines the direction of, force on current carrying conductor?, , 11. A magnetic field is non-uniform but its direction is, constant (East to West) is set-up in a chamber. A, charged particle enters the chamber and travels, undeflected along a straight path with constant, speed., What do you say about the initial velocity of the, particle?, , 12. An α-particle (positive charge) enters, a uniform, magnetic field at right angles to it as shown below., , r2, B, , Given r1 > r2 , where will the strength of the, (NCERT Exemplar), magnetic field be larger?, , α-particle, , Magnetic, field, , 5. A horizontal power line carries a current from East, to West direction. What is the direction of the, magnetic field due to the current in the power, line at a point above and at a point below the power, line?, , 6. A circular loop carrying a current is placed on a, horizontal surface (current is in the clockwise, direction)., What is the direction of its magnetic field at, the centre? What is the direction of the, magnetic field at a point outside the surface of the, loop?, , 7. Find the direction of magnetic field due to a, current carrying circular coil held, , In which direction α-particle moves, if the direction, of magnetic field gets reversed?, , 13. State whether an α-particle will experience any, force in a magnetic field, if (α-particles are, positively charged particles)., (i) It is placed in the field at rest., (ii) It moves in the magnetic field parallel to field, lines., (iii) It moves in the magnetic field perpendicular to, field lines., Justify your answer in each case.

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100, , CBSE Term II Science X, , What are the two observations that can be noted, from the galvanometer reading?, , 14. The electron enters in uniform magnetic field with, three different ways as shown below., e–, B, , 20. (i) A coil of insulated wire is connected to a, , e–, , galvanometer. What would be observed if a, strong bar magnet with its south pole towards, one face of the coil is, (a) moved quickly towards it?, (b) moved quickly away from it?, (c) held stationary near it?, (ii) Name the phenomenon involved., , B, , B, e–, (a), , (b), , (c), , Identify the case in which the force on electron will, be maximum and minimum, respectively. Give, reasons for your answer. Find the direction of, maximum force acting on electron., , 15. The figure shows the split ring commutator and, the two carbon brushes in their respective positions., Carbon brush, , (CBSE 2020, NCERT), !, , Long Answer Type Questions, , 21. 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, (NCERT Exemplar), concept., , 22. PQ is a current carrying conductor in the plane of, , Split ring, commutator, + –, , the paper as shown in the figure below., P, , What can you say about carbon brush and split ring, commutator?, , r1, , R, , 16. Sketch the schematic diagram of electric motor. What, is the role of split rings in an electric motor? (NCERT), , 17. (i) In what ways the speed to rotation of an electric, motor is increased?, (ii) Name some devices in which electric motors are, used., (NCERT), , 18. A circular metallic loop is kept above the wire AB as, shown below., , A, , B, , What is the direction of induced current produced in, the loop, if the current flowing in the straight wire, (i) is steady, i.e. does not vary?, (ii) is increasing in magnitude?, Justify your answer in each case., , 19. In the arrangement shown in figure there are two, coils wound on a non-conducting cylindrical rod., Initially the key is not inserted in the circuit. Later, the key is inserted and then removed shortly after., Coil I, , Coil II, , K, G, , S, , r2, Q, , (i) Find the directions of the magnetic fields, produced by it at points R and S., (ii) Given r1 > r2 , where will the strength of the, magnetic field be larger? Give reasons., (iii) If the polarity of the battery connected to the, wire is reversed, how would the direction of the, magnetic field be changed?, (iv) Explain the rule that is used to find the direction, of the magnetic field for a straight current, carrying conductor., (CBSE Sample Paper), , 23. What is solenoid? Draw the pattern of magnetic, field lines of (i) a current carrying solenoid and, (ii) a bar magnet., List two distinguishing features between the two, (CBSE Delhi, 2019), fields., , 24. (i) State Fleming’s left hand rule., (ii) Write the principle of working of an electric, motor., (iii) Explain the function of the following parts of, an electric motor., (a) Armature, (b) Brushes, (c) Split ring, (CBSE 2018, NCERT)

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101, , CBSE Term II Science X, , 25. (i) What is meant by electromagnetic induction? Name, one device which works on the principle of, electromagnetic induction., (ii) Describe three different ways to produce induced, current in a coil of wire., !, , Case Based Questions, , 26. Read the following and answer the questions from (i) to, (v) given below, A bar magnet is moved in and out of a coil, i.e., connected to a sensitive centre zero meter as shown in, the figure given below., , (i) What type of material is placed on white paper?, (ii) Why do the iron fillings arrange in such a, pattern?, (iii) What should we call to the region in which, magnetic force can be detected?, (iv) What do the lines on pattern demonstrate?, (v) Does degree of closeness of the field lines relate, something?, , 28. Read the following and answer the questions, from (i) to (v) given below, Jay performs an experiment by using two, different coils of copper wire having different, number of turns., He inserted them over a non conducting, cylindrical roll as shown., Coil 1, , Sensitive, centre zero, meter, , Coil 2, , N, , Coil, , +, , G, K, Setup of two stationary coils, , Bar magnet, , The meter needle swings to the left when the magnet is, moving towards the coil., (i) How the needle behaves when the bar magnet is at rest?, (ii) How the needle behaves when the bar magnet is at rest, and the coil is moving away from the magnet?, (iii) In which condition, the meter needle swings to the left?, (iv) Which phenomenon is involved in this?, (v) Give two applications based on the given phenomenon., , 27. Read the following and answer the questions from (i) to, (v) given below, A child performs an activity with a special material. He, fixes a sheet of white paper on a drawing board using, some adhesive material and places that material in the, centre of it., White paper, , –, , Then he connects the coil-1, having large, number of turns in series with a battery and plug, key and other coil-2 with galvanometer. As, he, plug the key, he observes some deflection in, galvanometer., (i) Why there is deflection in the galvanometer, when a key is inserted in the circuit ?, (ii) When there is flow of induced current through a, coil ?, (iii) Jay disconnected coil-1 from the battery and, noted the following observation., ‘‘The needle momentarily moves but to the, opposite direction’’. Justify the statement., (iv) When will the induced current is found to be, highest?, (v), Y, N, , S, , Iron fillings, X, , V, , Material, G, , Then he sprinkles some iron fillings uniformly around it, with a salt-sprinkler and taps the board gently., , A conducting rod XY moves across two magnets, as shown and the needle in galvanometer gets, deflected momentarily. What is the name of this, physical phenomenon?

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EXPLANATIONS, Objective Questions, 1. (d) Properties of magnetic field lines are as following, (a) Magnetic field lines are closed and continuous curves., (b) Magnetic field lines never intersect with each other., (c) Magnetic field lines are crowded near the poles., Hence, option (d) is correct., 2. (a) In a uniform magnetic field, the magnetic field lines of, force are parallel and equidistant from each other., The correct representation of uniform magnetic field is, shown below., N, , S, , 3. (d) The magnetic field lines due to a straight current, carrying wire are concentric circles with centre on the wire., 4. (b) According to right hand thumb rule, when conductor is, held in right hand, keeping thumb from East to West. The, curve of the finger will be from North to South at a point, lying directly below the wire., 5. (d) The magnetic field lines around a straight current, carrying conductor are concentric circles and the direction, of magnetic field is given by right hand thumb rule., 6. (b) 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., 7. (d) Proton is a charged particle. When it moves in a, magnetic field, a magnetic force is applied due to its, velocity and hence the momentum changes., 8. (d) The positively charged particle is moving towards West,, i.e. the direction of current is towards West (current flows in, the direction of the motion of positive charge)., The particle is deflected towards North, so the direction of, force is towards North., Thus, from Fleming’s left hand rule, the direction of, magnetic field is in upward direction., 9. (c) The strength of magnetic field inside a long current, carrying straight solenoid is same at all points., Therefore, correct option is (c)., 10. The correct match for the given items is, A – (ii) The SI unit of magnetic field is Tesla (T)., B – (iv) The magnetic field inside solenoid has a uniform value., C – (i) A compass needle is a small bar magnet that can rotate., D– (iii), Solenoid is a temporary magnet., 11. (a) In the given figure, the proton and electron are moving, in opposite direction to each other and in perpendicular to, the direction of magnetic field. Now, we know that the, direction of current is taken opposite to the direction of, motion of electron., , So, both electron and proton has current in same direction., Therefore, the forces acting on it are given by Fleming’s, left hand rule and they are pointing into the plane of the paper., 12. (b) The force on the left side of loop is attractive because the, direction of current in wire and loop is same, while on right, side of loop, force is repulsive., The force on perpendicular sides is zero. Hence, the, attractive force is greater than repulsive force, so, the loop, will move towards the wire., 13. (d) The direction of the induced electric current in a, conductor, when placed in a varying magnetic field can be, assessed by Fleming’s right-hand rule., 14. (d) When key is inserted and removed, then the magnetic, field lines pass through second coil increase and decrease in, two cases respectively., Therefore, the direction of current in two cases is in, opposite directions and the galvanometer shows momentary, deflections in opposite directions., 15. (a) The correct match for the given term is, (A)–(ii) The direction of force on a current carrying is given, by Fleming’s left hand rule., (B)–(iv) The direction of induced current is given by, Fleming’s right hand rule., (C)–(iii) The direction of magnetic field produced by a, straight current carrying conductor is given by Maxwell’s, right hand thumb rule., (D)–(i) The direction of magnetic field lines at a point in a, magnet is the direction of magnetic force on a North pole., 16. (a) The magnitude of magnetic field is, (i) directly proportional to the current I passing through, the wire., (ii) inversely proportional to the distance r from the wire., The magnetic field is stronger at a point which is nearer to the, conductor and goes on decreasing on moving away from the, conductor., ∴Both A and R are true and R is the correct explanation of A., 17. (b) When a current carrying conductor is placed in a, magnetic field, it experiences a force except when it is, placed parallel to the magnetic field. The force acting, on a current carrying conductor depends on magnetic field, produced by the current carrying conductor and external, magnetic field., ∴Both A and R are true but R is not correct explanation of A., 18. (b) The Earth’s magnetic field is towards North and velocity, of electron is downwards. By applying, Fleming’s left hand, rule, the direction of force is towards West. Also, electron, has a negative charge., ∴Both A and R are true but R is not correct explanation of A., 19. (c) The magnetic field is independent of length and area of, solenoid, it only depends on the number of turns and current, flowing through the solenoid. It is uniform inside the solenoid., ∴A is true but R is false.

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103, , CBSE Term II Science X, , 20. (d) When a closed circuit is placed in a varying magnetic, field, an electric current is produced in it which is known as, induced current., The direction of induced current is given by Fleming’s right, hand rule., ∴ A is false but R is true., 21. (i) (c) A linear solenoid is an electromagnetic device that, converts electrical energy into mechanical energy., (ii) (b) When a soft iron bar is placed inside the solenoid, it, will magnetise the iron bar as long as there is current in, the circuit. Hence, the strength of the magnetic field, inside the solenoid will also increase., (iii) (a) The magnetic field lines produced inside the solenoid, are similar to that produced by a bar magnet. The field, lines inside the solenoid are parallel to each other. The, magnetic field lines due to a current carrying solenoid, and bar magnet are shown below, , S, , N, +, , S, , N, , Subjective Questions, 1. The properties of magnetic field lines are, (i) They originate from North pole of a magnet and end at, its South pole., (ii) They form closed and continuous curves., (iii) They never intersect each other., (iv) They are uniform inside the magnet., If two magnetic field lines intersect each other, then at the, point of intersection, there will be two directions of, magnetic field lines, which is not possible. Hence, the, magnetic field lines never intersect with each other., 2. Magnetic field lines due to parallel current carrying, conductors are shown in figure., Magnetic field at X will be zero as both conductors are equal, in magnitude and are opposite in direction., , X, , –, , K, , Field lines produced by a, current carrying solenoid, , Field lines produced by, a bar magnet, , (iv) (d) The given magnetic field versus current graph is, linear. It is clear that magnetic field increases with, increase in current. Hence, the magnetic field produced, by the solenoid is directly proportional to the current., (v) (a) From graph, when current = 0.8 A, the magnetic field, is 13 mT., Upto current of 1.0 mA, the magnetic field increases, linearly with current., For larger values of current (>1A), the graph does not, depict any information., 22. (i) (b) The direction of rotation of the coil, when viewed, from the front is anti-clockwise., (ii) (c) Observations (I), (II) and (III) are correct while (IV), can be corrected as, when there is no current in the coil,, it will stop rotating because force is produced only when, there is flowing current in coil., (iii) (c) Commercial electric motors do not use permanent, magnet to rotate the armature because permanent, magnets are weak and do not produce strong magnetic, field in the region., (iv) (a) Electric motor is a device which converts electrical, energy into mechanical energy., (v) (d) By applying, Fleming’s left hand rule, the direction of, force experienced by wire due to magnetic field is, South-East direction., , 3. The strength of the magnetic field is inversely proportional, 1, to the distance i.e. B ∝ ., r, As, the point where magnetic field is to be determined is, moved away from the straight wire carrying constant, current, the magnetic field strength decreases., 4. According to the right hand thumb rule, magnetic field at P, is directed into the plane of paper and at Q, it is out of the, plane of paper. The strength of the magnetic field at Q will, 1, be larger as strength of the field ∝, ., r (distance), Here, r1 > r2, ∴ B1 < B2 i.e. B2 has larger field., 5. According to right hand thumb rule,, N, , Power line, W, , E, S, , (i) The direction of magnetic field at a point above the, power line is from South to North., (ii) The direction of magnetic field at a point below the, power line is from North to South., 6. The magnetic field lines are concentric circles at every point, on a current carrying circular loop., The direction of magnetic field is determined by right hand, thumb rule. At the centre of the circular loop, the magnetic, field lines are straight and points towards North.

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104, , CBSE Term II Science X, , The direction of magnetic field lines at the point outside the, surface of the loop is shown below., , particle enters East to West or West to East as shown, below., , Circular loop, carrying current, , B, , B, , q, , q, Magnetic, field lines, (anti-clockwise), , Magnetic, field lines, (clockwise), +, , –, , Key, ( ), , Battery, , 7. (i) When the coil is kept in the North-South plane and the, current is flowing in the anti-clockwise through the loop,, then the magnetic field is in the East to West direction., (ii) When the coil is in vertically East-West plane and current, through the coil is in anti-clockwise direction, then the, magnetic field is in the South to North direction., (iii) When a circular coil carrying current is placed horizontally, and the direction of the current is clockwise, then the, direction of the field for the observer positioned below the, coil is in the downward direction., 8. A solenoid behaves like a magnet when electric current passes, through it., One end of a solenoid behaves as a North pole and the other, end behaves as a South pole. We can use a bar magnet to, determine the North and South poles of a current carrying, solenoid by using the property, i.e. like poles repel and unlike, poles attract each other., The end of solenoid which attracts North pole of a bar magnet, is magnetic South pole of the solenoid. The end of solenoid, which repels the North pole of a bar magnet is the magnetic, North pole of the solenoid., 9. (i) The magnetic field lines are crowded (convergent) near the, poles of solenoid. Hence, the magnetic field is strong and, divergent, where the magnetic field is weak., (ii) A freely suspended current carrying solenoid always points, in the North-South direction even in the absence of any, other magnet. Because the earth itself behaves as a magnet, or solenoid to point always in a particular direction., (iii) The field lines around a current carrying solenoid are, similar to field lines of a bar magnet. So, inside the solenoid, field lines are parallel to each other and the strength of, magnetic field is same i.e. uniform at all points inside a, solenoid., 10. The force experienced by a current carrying conductor placed, in a magnetic field is the largest, when conductor is kept, perpendicular to the direction of the magnetic field., Fleming’s left hand rule determines the direction of force on a, current carrying conductor. It states that, if the forefinger, (magnetic field), middle finger (current and thumb are, stretched mutually, then the direction of force acting on, conductor is given by thumb., 11. If a charged particle moves parallel or anti-parallel to the, magnetic field, no magnetic force will act on it and remains, undeflected. So, in the given condition either the charged, , 12. Here, the force acting on the α-particle is directed, perpendicular to the plane of paper in inward direction by, Fleming’s left hand rule. If the direction of magnetic field, gets reversed then the force will also act in opposite, direction of α-particle i.e. the force experienced by, α-particle is now in outward direction to the plane of, paper., 13. (i) No, it will not experience any force. As, magnetic field, exerts force on a moving charged particle only., (ii) No, it will not experience any force because magnetic, field exerts a force in perpendicular direction to, motion of the particle., (iii) Yes, it will experience a force in a direction, perpendicular to the direction of its own motion and, the direction of magnetic field can be determined by, Fleming’s left hand rule., 14. Force on electron is maximum in case (a), because here, the direction of motion of electron is perpendicular to the, direction of magnetic field B., Similarly, the force on electron is minimum in case (c), because, in this case the direction of motion of electron is, along the direction of magnetic field B, as electron is, moving along B., Hence, the direction of maximum force acting on electron, is perpendicular to the plane of paper and directed into it., 15. When the gaps of the split ring commutator are aligned, with the carbon brushes, then contacts are broken and the, current is temporarily cut-off., However, the coil keeps on rotating in the same direction, due to its inertia until the split ring commutator and the, carbon brushes are in contact again., 16. The schematic diagram of electric motor is shown below,, Rotating of coil, (Clockwise) Rectangular coil, B, , C, , N, , S, D, , A, Permanent, magnet, , R1 R2, , Split rings, (R1 and R2), , Y, Brushes, (X and Y ), , X, Axle, , Shaft, , K, +, , –, , Split rings act as commutator and its function is to reverse, the direction of current flowing through the coil.

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105, , CBSE Term II Science X, , 17. (i) The speed of rotation of the motor can be increased by, increasing the strength of the current in the coil., increasing the number of turns in the coil., increasing the area of the coil., increasing the strength of magnetic field., (ii) Electric motor is used in electric fans, refrigerators,, mixers, washing machine, computers, MP3 player, etc., 18. (i) No induced current will be produced in the loop as the, constant current flowing in the straight wire produces a, constant magnetic field. Hence, no induced current is, produced in the loop., (ii) Since, current in the straight wire is changing, hence, induced current will be produced in it. According to, Fleming’s right hand rule, the current flowing in the, loop will be in clockwise direction., 19. The observation that can be noted from the galvanometer, reading are, (i) There are momentary galvanometer deflections that die, out shortly., (ii) The deflections are in opposite directions., 20. (i) (a) Galvanometer gets deflection in one direction., (b) Galvanometer gets deflection in opposite direction of, the first one., (c) Galvanometer shows no deflection., (ii) Phenomenon involved is electromagnetic induction., !, , !, , !, , !, , 21. When a magnetic compass needle pointing North and, South in the absence of a nearly magnet or a current loop,, it is acting upon by the earth’s magnetic field only. But in, the presence of a magnet or a current loop (which also has, a magnetic field) the earth’s magnetic field near the, compass is modified and the needle is deflected from, North and South directions., The salient features of magnetic field lines are, (i) A magnetic field line is directed from North-pole to, South-pole outside the magnet., (ii) A magnetic field line is a closed and continuous curve., (iii) The magnetic field lines never intersect each other,, because if two lines meet, it means that force is acting, in two directions at that point which is not possible., (iv) Closer the field lines, stronger is the magnetic field, and vice-versa is also true., (v) Magnetic field lines are parallel and equidistant, those, represent a uniform magnetic field., 22. (i) According to right hand thumb rule, the magnetic field, produced by PQ at point R is into the plane of the paper, and at point S is out of the plane of paper., (ii) Here, r1 > r2, The magnetic field will be larger at point S as compared, to that at point R., This is because the magnetic field produced by a, straight current-carrying conductor is inversely, proportional to the distance from the wire. So, the, magnetic field will be larger at the point which is, nearer to the conductor., As, point S is nearer to the conductor as compared to, point R. So, field at S > field at R., , (iii) If the polarity of the battery is reversed, the current will, be going from top to bottom in the wire and the magnetic, field lines will now be in the clockwise direction on the, plane, which is perpendicular to the wire carrying, current., (iv) Maxwell's right hand thumb rule is used to find the, direction of the magnetic field for a straight current, carrying conductor., This law states that, if you hold the current carrying straight, wire in the grip of your right hand in such a way that the, stretched thumb points in the direction of current, then the, direction of the curl of the fingers will give the direction of, the magnetic field., , Magnetic field, Current, , 23. A solenoid is defined as a coil consisting of a large number of, circular turns of insulated copper wire. These turns are, wrapped closely to form a cylinder., , S, , N, –, , +, , K, , (i) Magnetic field lines of force due to a, current carrying solenoid, , S, , N, , (ii) Field lines around a bar magnet, , Distinguishing features are as follow, Magnetic field outside the solenoid is negligible as, compared to the bar magnet., Magnetic field of solenoid can be varied as per our, requirement just by changing current or core of, solenoid but in bar magnet it is fixed., !, , !, , 24. (i) Fleming’s left hand rule states that, if the forefinger,, thumb and middle finger of left hand are stretched, mutually perpendicular to each other, such that the, forefinger points along the direction of external magnetic, field, middle finger indicates the direction of current,, then the thumb points towards the direction of force, acting on the conductor., (ii) Electric motor is based on the principle that, when a, rectangular coil is placed in a magnetic field and current, is passed through it, two equal and opposite forces act on, the coil which rotate it continuously.

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106, , CBSE Term II Science X, , (iii), , (a) Armature It is a coil wound over a soft iron core., It rotates in magnetic field, when a current flows, through it., (b) Brushes They provide a sliding contact and, facilitate current through armature while it, rotates in field., (c) Split ring It ensures unidirectional current, through armature as it rotates in field., , 25. (i) An electric current produced in a closed circuit by a, changing magnetic field is called an induced current., This phenomenon is called electromagnetic induction., An electric generator works on the basis of, electromagnetic induction., (ii) Three different ways to produce induced current in a, coil of wire are as following, (a) If a magnetic field is changed around a coil, then, an induced current is set up in it. It can be done, by taking a bar magnet and bringing it closer to, the coil or taking it away from the coil., (b) If a coil is moved in a magnetic field, then, an induced current is set up in the coil., (c) By changing the magnitude of current flowing, through another coil kept close to the coil., 26. (i) Since, the magnet is at rest, so no current is induced in, the coil., (ii) Since, the bar magnet is at rest and the coil is moving, away from the magnet, needle of the galvanometer coil, swing towards right., (iii) When the magnet will move towards the coil, the, meter needle swings to the left., (iv) The phenomenon involved in this is electromagnetic, induction (EMI). In this phenomenon, current is, induced in the circuit by changing the magnetic field., , (v) Electric generators, transformers etc., are the applications, of electromagnetic induction., 27. (i) Bar magnet is placed on white paper which attracts the, sprinkled iron fillings., (ii) The bar magnet exerts its influence in the region, surrounding it. Therefore, the iron fillings experience a, force and this force makes iron fillings to arrange in a, pattern., (iii) The region surrounding a magnet, in which the force of the, magnet can be detected, is said to have a magnetic field., (iv) The iron fillings arrange themselves in a particular, alignment that are called magnetic field lines. These field, lines represent the region in which force of the magnet, can be detected., (v) The relative strength of the magnetic field is shown by the, degree of closeness of the field lines. The field is stronger,, as the force acting on the pole of the magnet is greater,, where the field lines are crowded., 28. (i) As the current in first coil changes, the magnetic field, associated with it also changes. Hence, the current is, induced in coil-2., (ii) By Faraday’s law of electromagnetic induction, the, induced current flows in a coil only when there is a change, in the magnetic field within the coil., (iii) A current carrying coil has a magnetic field just like bar, magnet. In experiment, if coil-1 is disconnected from, battery, then the needle momentarily moves, but to the, opposite side. It means the current flows in the opposite, direction in coil-2., (iv) The induced current is found to be highest when the, direction of motion of the coil is at right angles to the, magnetic field., (v) The phenomenon associated with this is electromagnetic, induction.

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Chapter Test, Multiple Choice Questions, , 1. Which of the following is the correct representation of, magnetic field lines due to a bar magnet?, N, , N, , (a), , (d), , (c), , (b), S, , S, , 6. Assertion When current passes through a solenoid,, N, , N, , S, , S, , 2. Choose the incorrect statement from the following, regarding magnetic lines of field., (a) The direction of magnetic field at a point is taken to be the, direction in which the North pole of a magnetic compass, needle points., (b) Magnetic field lines are closed curves., (c) If magnetic field lines are parallel and equidistant, they, represent zero field strength., (d) Relative strength of magnetic field is shown by the degree, of closeness of the field lines., , 3. The magnetic field at a distance r from a long wire, carrying current I is 0.4 T. The value of magnetic field, at a distance 2r is, (a) 0.2 T, , (b) 0.1 T, , (c) 0.15 T, , (a) Both A and R are true and R is the correct explanation of A., (b) Both A and R are true, but R is not the correct explanation, of A., (c) A is true, but R is false., (d) A is false, but R is true., , (d) 1 T, , 4. A coil, carrying a current is arranged within a magnetic, field. The coil experiences forces that can make it move., Y, Current, , then it tends to contract., Reason The current flowing through two parallel wires, in the same direction gives rise to force of attraction on, each other., , 7. Assertion Galvanometer is used to measure polarity., Reason Galvanometer is an instrument which is used to, detect current in any circuit., , 8. Assertion A pump operated by electric motor starts, pumping liquid., Reason Motor converts mechanical energy to electrical, energy., Short Answer Type Questions, , 9. What are magnetic field lines? Justify the following, statements., (i) Two magnetic field lines never intersect each other., (ii) Magnetic field lines are closed curves., , 10. List some methods of producing magnetic fields., 11. It is established that an electric current through a, 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 in each case., , 12. How can it be shown that a magnetic field exists around a, wire through which a direct current is passing?, , 13. Meena draws magnetic field lines of field close to the, , Magnetic field, Current, X, , In which direction does the coil move?, (a) Along the magnetic field, (c) Out of the paper, , (b) Turns about the axis XY, (d) From Y to X, , 5. Which of the following is not the application of electric, motor?, (a) Washing machines, (c) Electric drills, , (b) Mixers and grinders, (d) Transformers, , Assertion-Reasoning MCQs, , axis of a current carrying circular loop. As she moves, away from the centre of the circular loop, she, observes that the lines keep on diverging. How will you, explain her observation?, , 14. (i) Two circular coils P and Q are kept close to each, other, of which coil P carries a current. If coil P is, moved towards Q, then will some current be, induced in coil Q? Give reason for your answer and, name the phenomenon involved., (ii) What happens, if coil P is moved away from Q?, (iii) State few methods of inducing current in a coil., , Direction (Q. Nos. 6-8) Each of these questions, , Long Answer Type Questions, , contains two statements Assertion (A) and Reason (R)., Each of these questions also has four alternative, choices, any one of which is the correct answer. You, have to select one of the codes (a), (b), (c) and (d) given, below., , 15. With the help of a diagram of experimental set-up, , Answers, , describe an activity to show that the force acting on a, current carrying conductor placed in a magnetic field, increases with increase in field strength., , 16. How electric motor works?, , Multiple Choice Questions, 1. (d), , 2. (c), , 3. (a), , 4. (b), , Assertion-Reasoning MCQs, 6. (a), , 7. (b), , 8. (c), , 5. (d), , For Detailed Solutions, Scan the code

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108, , CHAPTER 7, , Our Environment, In this Chapter..., !, , Ecosystem, , !, , Food Chain, Trophic Levels,, Energy flow and Food Web, , !, , Environmental Problems, , All those things and set of conditions that influence the life of, an organism like their growth, survival, development and, reproduction, constitute the environment., It is composed of physical surrounding (e.g. air, water), living, beings (e.g. plants, animals) and climatic conditions (e.g., rainfall, temperature) of the region., , Ecosystem, It is the structural and functional unit of biosphere. It is a, stable ecological unit where regular input of energy and, circulation of matter takes place. The term ecosystem was, coined by AG Tansley (in 1935)., All the interacting or living organisms (biotic components) in, an area together with the non-living constituents (abiotic, components) of the environment form an ecosystem, e.g. a, lake, a field or a forest., , Types of Ecosystem, There are two types of ecosystem, i.e. natural and artificial, ecosystem., 1. Natural Ecosystem, The naturally existing ecosystem without any human support is, called as natural ecosystem. Depending upon the habitats,, natural ecosystem may be terrestrial (desert, grassland and, forest) and aquatic (ponds, lakes, estuaries and marine)., , 2. Artificial Ecosystem, An ecosystem which is created and maintained by humans, is called as artificial or man-made ecosystem. It rely on, human efforts to sustain. It does not possess a self-regulating, mechanism., e.g. Aquariums, botanical gardens, field crops, etc., Agro-ecosystem is the largest man-made ecosystem., , Components of Ecosystem, The ecosystem encompasses both living (biotic) and, non-living (abiotic) components of the earth., 1. Biotic Components, These include all the living organisms present in the, ecosystem, i.e. plants, animals and microorganisms., On the basis of food, they consume the different living, organisms can be categorised into three groups. These are as, follows, (i) Producers All green plants and certain blue-green, algae which can produce food by the process of, photosynthesis are producers. These are also called, autotrophs. These are the source of nutrition for rest of, ecosystem., (ii) Consumers These are dependent on producers for, their nutritional requirement and consume food prepared, by producers. These are also called heterotrophs.

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109, , CBSE Term II Science X, , Consumers can be further divided into the following three, categories, Herbivores These are primary consumers which feed, directly on the producers, i.e. plants. e.g. Grazing, animals., Carnivores The carnivores which feed on herbivores, are called secondary consumers., Some are predators which attack and kill their prey and, feed on their bodies. Some are scavengers that feed on, dead animals that they find. These are called tertiary, consumers., Omnivores These are animals that feed on both plants, and animals, e.g. humans and bears., (iii) Decomposers These are microorganisms which feed on, decaying and dead organic matter. They breakdown the, remains of dead animals and plants, to release simpler, inorganic materials, making nutrients available to, producers., e.g. Bacteria and fungi., !, , !, , !, , Trophic Levels, The transfer of food or energy takes place through various, steps or levels in the food chain known as trophic levels., The producers (autotrophs) are present at the first trophic, level. They fix solar energy, making it available for, consumers (heterotrophs). The herbivores or the primary, consumers are found at the second trophic level., Small carnivores or secondary consumers are present at the, third trophic level. The large or the tertiary consumers, form the fourth trophic level., , !, , 2. Abiotic Components, The abiotic components of an ecosystem are the non-living, components on which living organisms depends. These, components are light, temperature, water, atmospheric gases,, wind, etc., , Trophic, levels, IV, , Secondary, consumers, , III, , Primary consumers, , II, , Producers, , I, , Trophic levels in an ecosystem, , Energy Flow, !, , Food Chain, It is a linear network of living organisms in a community, through which energy is transferred in the form of food., It describes relationship of organisms about ‘who eats whom’., , Tertiary, consumers, , !, , !, , The green plants in a terrestrial ecosystem capture about, 1% of the energy of sunlight (light energy). They convert, it into food (chemical energy)., When green plants are eaten by primary consumers, a, great deal of energy is lost as heat and an average of 10%, of the energy of food eaten by an organism is turned back, into its own body and made available for the next level, of consumers. This is known as the 10% law (Lindemann, in 1942)., Therefore, 10% can be taken as the average value for the, amount of organic matter that is present at each step and, reaches to the next trophic level., The loss of energy at each step is very large. Only a little, energy is available for the next level of consumers, food, chains generally consist of three or four steps., Top carnivores, , (a), , (b), , Carnivores, , (c), , Food chain in nature; (a) in forest,, (b) in grassland and (c) in a pond, , Herbivores, , On the basis of choice of habitat, food chains are of two types, (i) Terrestrial food chain It is the food chain present on, land., e.g. Grass → Insects → Snake → Hawk., (ii) Aquatic food chain It is the food chain present in, different water bodies., e.g. Phytoplankton → Zooplankton → Fish → Shark, , Producers, Sunlight, , Flow of energy in an ecosystem, !, , The flow of energy is unidirectional, i.e. the energy which, passes to the next trophic level, does not come back to the, lower trophic level.

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110, , CBSE Term II Science X, , Food Web, It is the interconnection of different food chains, which correlate at, various trophic levels operating in an ecosystem. Each organism is, generally eaten by two or more other kinds of organisms. They in, turn are eaten by several other organisms., , In the atmosphere, UV radiations breakdown CFCs, molecules and release chlorine atom. These atoms on, reacting with ozone, dissociate ozone molecules into, oxygen., Thus, it leads to depletion of ozone layer., , Managing the Garbage We Produce, , Food web consisting of many interlinked food chains, , Environmental Problems, Human are an integral part of the environment. Various human, activities pollute the environment in various ways and pose., Serious environmental problems such as biomagnification, ozone, layer depletion, waste accumulation are as follows, , Biological Magnification or Biomagnification, It is the phenomenon of progressive increase in the concentration, of non-biodegradable toxicants in organisms at each successive, trophic level. It is also called bioconcentration. These toxic, chemicals enter the food chain through pesticides like DDT., The maximum concentration of these chemicals gets accumulated, in humans as they occupy the topmost place in any food chain., , Depletion of Ozone Layer, Ozone (O 3 ) is a molecule formed by three atoms of oxygen. It is, found in the stratosphere. It shields the surface of the Earth from, harmful ultraviolet (UV) radiations of the Sun., Formation of Ozone Layer, Ozone is a product of UV radiations acting on oxygen (O 2 ) molecule., The high energy UV radiations split apart some molecular oxygen, (O 2 ) into free oxygen (O) atoms. These atoms are very reactive and, combine with the molecular oxygen to form ozone., UV, , O 2 ⎯→ O + O; O + O 2 ⎯→, , O3, , (Ozone), , Ozone Depleting Substance, The increasing use of synthetic chemicals like Chlorofluorocarbons, (CFCs) which are used in refrigerants as coolant and in fire, extinguishers are responsible for ozone layer depletion., , The household waste is called garbage. Every, household produces a lot of garbage on daily basis., Improvements in our lifestyle have resulted in, greater amounts of waste material generation., These waste substances can be divided into two main, groups, (i) Biodegradable substances The substances which, can be disposed off naturally by the action of, microorganisms like bacteria, fungi, etc., are called, biodegradable substances., e.g. Tea leaves, waste paper, left over food, etc., (ii) Non-Biodegradable substances The substances, that cannot be converted into harmless simpler, forms by the action of microorganisms are called, non-biodegradable substances. These are toxic,, harmful, may be inert and accumulate in the, environment or food chain., e.g. Plastics, insecticides, heavy metals, etc., , Methods of Waste Disposal, Various methods of waste disposal are as given below, (i) Recycling It is the processing of waste materials to, form new products., (ii) Composting Biodegradable domestic wastes, can be buried in a pit, dug into ground and, are converted into compost and used as manure., (iii) Incineration It is burning of a substance at high, temperature to form ash. It is commonly used to, dispose hospital waste., (iv) Landfills Solid waste is dumped into a low lying, area and covered with soil., (v) Sewage treatment Organic material in the sewage, is allowed to settle down and decompose in large, tanks. The water from these tanks is cleaned and is, released into waterbodies., (vi) Biogas production In some places, sewage is, decomposed anaerobically to yield biogas and, manure., Note Disposable paper cups which are biodegradable in nature, are now being used in trains instead of plastic cups and, kulhads., As the production of kulhads at a large scale resulted in the, loss of top fertile layer of soil which leads to soil erosion,, whereas, plastic cups are non-biodegradable in nature,, therefore cause environmental problems.

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111, , Chapter, Practice, PART 1, Objective Questions, !, , 5. In the given food chain, organisms are labelled as A, to D. Match the labelling referred in Column I, with their most suitable feature in Column II., , Multiple Choice Questions, , 1. Which of the following is an incorrect pair?, , D, , (a) River—Natural ecosystem, (b) Air—Biotic component, (c) Crop field—Artificial ecosystem, (d) Water—Abiotic component, , C, B, , 2. Which of the following is not a functional component of, , A, , an ecosystem?, (a) Communities, (b) Decomposers, (c) Sunlight, (d) Energy flow, , Column I, , 3. Flow of energy in an ecosystem is always, (a) unidirectional, (b) bidirectional, (c) multidirectional, (d) no specific direction, , (NCERT Exemplar), , 4. The diagram represents the flow of energy within a, balanced ecosystem., The boxes represents various trophic levels. Select the, option which gives correct identification and main, function or characteristic, P, , Q, , R, , S, , (a) P-Producers : They are heterotrophic in nature, (b) Q-Primary consumers : They eat only plants and their products, (c) R-Herbivores : They are secondary consumers, (d) S-Decomposers : They act as tertiary consumers in a food chain, , Codes, A, (a) 3, (c) 3, , Column II, , A, , 1. Primary carnivore, , B, , 2. Secondary carnivore, , C, , 3. Autotrophs, , D, , 4. Primary consumer, , B, 4, 1, , C, 1, 4, , D, 2, 2, , A, (b) 4, (d) 3, , B, 3, 2, , C, 2, 1, , D, 1, 4, , 6. What will happen if deer is missing in the food, chain given below?, Grass → Deer → Tiger, , (NCERT Exemplar), , (a) The population of tiger increases, (b) The population of grass decreases, (c) Tiger will start eating grass, (d) The population of tiger decreases and the population of, grass increases, , 7. The diagram shows the flow of energy through an, ecosystem., 3, Sun, , Producers, , 4, 1, , Primary, consumers, , 2, , Secondary, consumers

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112, , CBSE Term II Science X, , The smallest amount of energy transferred between, organisms and the largest amount of energy lost to, the ecosystem is represented by which arrows?, Smallest energy transfer, , Largest energy loss, , (a), , 4, , 3, , (b), , 2, , 1, , (c), , 2, , 3, , (d), , 1, , 4, , 11. Identify the option that indicates the same trophic, level of different organisms in the given food web., , 8. Which of the following represents the labelled parts, X and Y correctly?, Phytoplankton, Zooplankton, , Y, , X, , X, , Y, , (a) Primary consumer, , Producer, , (b) Secondary consumer, , Decomposer, , (c) Primary consumer, , Secondary consumer, , (d) Carnivore, , Herbivore, , 9. Carefully study the given pyramid of energy flow in, an ecosystem. Select the option which gives correct, identification and their examples., T4, , (a) Goat, Snake, Owl, (c) Frog, Fish, Bird, , (b) Rabbit, Fox, Peacock, (d) Tiger, Fox, Eagle, , 12. What happens in biological magnification?, (a) There is a progressive increase in the level of harmful, substances through trophic levels, (b) There is a progressive increase in the body weight, through trophic levels, (c) There is a progressive increase in number of organisms, through trophic levels, (d) There is a progressive increase in biological activities, through trophic levels, , 13. Which of the following chemicals cause the thinning, , T3, , of layer P?, , T2, , Sunlight, , T1, , (a) T1 —Herbivores—Deer (b) T 2 —Producers—Fungi, (c) T 3 —Omnivores—Bear (d) T 4 —Decomposers—Grass, , Layer P, , 10. Which of the following pyramid given below, , Earth, , correctly represents the food chain?, (a) Sulphur dioxide, (c) Chlorofluorocarbon, Plants, , (a), , Caterpillars, , Plants, , (b), , Toads, , Toads, Caterpillars, , Snakes, , Snakes, , Snakes, , Snakes, , (b) Nitrogen dioxide, (d) Carbon dioxide, , 14. Which of the following is the correct reaction, involved in formation of ozone layer?, (a) O2 ⎯UV, ⎯, ⎯→ O + O; O + O2 ⎯→ O3, (b) O + O ⎯→ O 2; O 2 + O ⎯→ O 3, (c) O 2 ⎯Infrared, ⎯⎯⎯→ O + O; O + O + O ⎯→ O 3, (d) O 2 + O ⎯→ O 3; O 3 ⎯→ O 2 + O, , (c), , Toads, , (d), , Caterpillars, , Toads, Plants, , Plants, , Caterpillars, , 15. Burning of waste products at high temperature to, form ash, reduces waste considerably. This method, of waste disposal is called, (a) composting, (c) recycling, , (b) sewage treatment, (d) incineration

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113, , CBSE Term II Science X, , !, , Assertion-Reasoning MCQs, Direction (Q. Nos. 16-20) Each of these questions, contains two statements, Assertion (A) and Reason (R)., Each of these questions also has four alternative choices, any, one of which is the correct answer. You have to select one of, the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct explanation of A, (b) Both A and R are true, but R is not the correct explanation, of A, (c) A is true, but R is false, (d) A is false, but R is true, , 16. Assertion Biotic components of an ecosystem include all, the living organisms present in that ecosystem., Reason Biotic components also include wind, gases,, light, etc., 17. Assertion Food chains generally consist of more than, four trophic levels., , Reason There is loss of energy at each trophic level, and very little usable energy remains after four trophic, levels., , 18. Assertion Producers are present at the first trophic level., Reason Consumers or heterotrophs fix energy and make, it available for autotrophs., 19. Assertion Certain pesticides and other chemicals used to, protect our crops from diseases and pests are, non-biodegradable., , Reason They do not get accumulated at various trophic, levels., , 20. Assertion Ozone layer shields the surface of Earth from, UV radiations., Reason The UV radiations are highly damaging to, organisms., !, , Case Based MCQs, , 21. Read the following and answer the questions from (i) to, (v) given below, A group of scientists analysed samples of four different, animals from a river for possible accumulation of DDT in, their body due to biomagnification. The result obtained, is shown in the given graph., Z, Amount, of DDT, (in ppm), , (a) W → X → Y → Z, (c) Z → X → W → Y, , (b) Y → W → X → Z, (d) X → W → Z → Y, , (ii) If concentration of DDT in W is estimated to, be 0.4 ppm, then amount of DDT in Y and Z, would respectively be, (a) 0.02 ppm and 2 ppm, (b) 2 ppm and 0.02 ppm, (c) 0.3 ppm and 0.5 ppm, (d) 0.04 ppm and 0.02 ppm, , (iii). Due to biomagnification, which of the following, substances accumulate and go on concentrating, at each trophic level?, (a) Biodegradable substances, (b) Non-biodegradable substances, (c) Biotic components, (d) Abiotic components, , (iv) The decomposers are not included in the food, chain. The correct reason for the same is, because decomposers :, (a) act at every trophic level of the food chain, (b) do not breakdown organic compounds, (c) convert organic material to inorganic forms, (d) release enzymes outside their body to convert, organic material to inorganic forms, , (v) Which amongst the following are hazardous for, the environment?, I. Pesticides, II. Fertilisers, III. Heavy metals, IV. Manures, Codes, (a) I, II and IV, (c) I, II and III, , (b) I and II, (d) Only IV, , 22. Read the following and answer the questions, from (i) to (v) given below, Sheenu took three different types of solid wastes, R, S and T. She buried them under the soil in a, pit, as she wanted to study their rate of, decomposition. Her findings are shown in the, given graph., , Amount, of solid in pit, (grams), , S, , T, R, Number of days, , X, W, Y, , O, , (i) The correct order of animals in food chain, operating in a river is, , Animals, , (i) Which of the following waste does not, decompose at all?, (a) R, (c) T, , (b) S, (d) Both (a) and (c)

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114, , CBSE Term II Science X, , (ii) Choose the correct statements regarding, non-biodegradable wastes., I. Easily decomposed by fungi., II. Enter into food chain and get magnified., III. Biological in origin., IV. Cause soil pollution., Codes, (a) I and II, (c) I and IV, , (NCERT), , 8. Number of trophic levels is limited to 3-4 in a food, , trophic level?, , 10. State 10% law. Explain with an example how energy, flows through different trophic levels., , (b) Leather shoes, (d) Electronic waste, , (a) By recycling, (b) By burning, (c) By disposing them in water bodies, (d) By mixing with biodegradable waste, , 11. Indicate the flow of energy in an ecosystem. Why it, is unidirectional?, , (NCERT Exemplar), , 12. What do you mean by biological magnification?, 13. What is biological magnification? Will the levels of, this magnification be different at different levels of the, (NCERT Exemplar), ecosystem?, , (v) Incineration is another method of waste disposal, which is used for, (a) kitchen waste, (b) electronic waste, (c) biomedical waste, (d) plastic waste, , 14. Write the major cause of ozone depletion. What steps, should be taken to limit the damage to ozone layer?, 15. Explain how ozone being a deadly poison can still, perform an essential function for our environment., (CBSE 2020), , 16. Give reason why a food chain cannot have more than, (CBSE 2020), four trophic levels?, , PART 2, Subjective Questions, , 17. (i) Explain the role of UV radiation in producing ozone, layer., (ii) Mention the reaction involved., (iii) Why is excessive use of CFCs a cause of concern?, , Short Answer Type Questions, , 1. Define an ecosystem. Draw a block diagram to show, the flow of energy in an ecosystem., , food chain and state the different trophic levels in it., , 9. What will happen if we kill all the organisms in one, , (iv) Which of the following is the safe method of, disposal of the non-biodegradable waste?, , !, , 7. What are trophic levels? Give an example of a, , chain. Give reason., , (b) II and IV, (d) II and III, , (iii) Which of the following solid waste comes under the, category of R?, (a) Cow dung, (c) Plastic bottles, , 6. Based on their feeding habits, differentiate between, parasites and decomposers., , (CBSE 2019), , 18. How is ozone formed in the upper atmosphere? State, its importance. What is responsible for its depletion?, Write one harmful effect of ozone depletion., (CBSE 2019), , 2. (i) What do you understand by the term ecosystem?, (ii) Autotrophs are at the first level of food chain. Give, reason., (iii) In a food chain of frogs, grass, insects and snakes,, assign trophic level to frogs. To which category of, (CBSE 2020), consumers do they belong to?, , 3. What is the role of decomposers in the ecosystem?, (NCERT), , 4. Natural water bodies are not regularly cleaned, whereas an aquarium needs regular cleaning. Why?, (CBSE 2019), , 5. How does study of food chain in an area or habitat, help us? Give an example of four steps of food chain, operating in a large lake., , 19. Give some methods that could be applied to reduce, accumulation of pesticides in our body., 20. How can you help in reducing the problem of waste, disposal? Give any two methods., , 21. If all the waste we generate is biodegradable, will this, have no impact on the environment?, , 22. Give any two ways in which non-biodegradable, substances would affect the environment., , 23. Why should biodegradable and non-biodegradable, wastes be discarded in two separate dustbins?, , 24. ‘Effective segregation of wastes at the point of, generation is very important’. Justify this statement.

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115, , CBSE Term II Science X, , 25. Sheenu went to a picnic where she saw a pond. Her, teacher gives her an assignment to draw the pond, ecosystem and show its different components. Help her, to complete the assignment., , 26. Name the wastes which are generated in your house, daily. What measures would you take for their disposal?, (NCERT Exemplar), , 27. A modern insecticide has been introduced with, certain new properties like, accumulation in the bodies, of predators, broken down by soil bacteria, easily, washed into lakes and rivers and taken up by plant roots., Among all these properties, which one will help in, reducing or keeping the level of environment pollution, to lowest?, , 28. Mona eats curd and yogurt and follows vegetarian diet., For this food intake, what should be the trophic level, occupied by her in a food chain ?, !, , Long Answer Type Questions, , 29. What are decomposers? What will be the consequence of, their absence in an ecosystem?, , (NCERT Exemplar), , 30. (i) What are consumers? What will be the consequence of, the absence of primary consumers in an ecosystem?, (ii) What will be the direction of energy transfer in each, of the following cases?, (a) Grasshopper eaten by a frog, (b) Deer feeds on grass, (c) Deer eaten by a lion, , 31. (i) Will the impact of removing all the organisms in, a trophic level be different for different trophic, levels?, (ii) Can the organisms of any trophic level be removed, without causing any damage to the ecosystem?, (NCERT Exemplar), , 32. (i) ‘Energy flow in a food chain is unidirectional’. Justify, this statement., (ii) Explain how the pesticides enter a food chain and, (CBSE 2014), subsequently get into our body., 33. (i) How do food chains get shortened? How does the, shortening of food chain affect the biosphere?, (ii) How will you justify that vegetarian food habits give, us more calories?, , 34. Explain some harmful effects of agricultural practices on, the environment., , (NCERT Exemplar), , 35. What is ozone and how does it affect any, ecosystem?, , (NCERT Exemplar), , 36. (i) What is ‘environmental pollution’?, (ii) Distinguish between biodegradable and, non-biodegradable pollutants., (iii) Choose the non-biodegradable pollutants, from the list given below, Paper, DDT, Radioactive waste, Plastic,, Insecticides., , 37. Suggest any five activities in daily life, which are, eco-friendly., , (NCERT Exemplar), , 38. What are the reasons for the shift from plastic to, kulhads and then finally to paper cups?(CBSE 2020), !, , Case Based Questions, , 39. Read the following and answer the questions, from (i) to (v) given below, Food web is the interconnection of different food, chains, which correlate at various trohpic levels, operating in an ecosystem., Foxes, Snake, Mice, Seeds, , Rabbits, Young trees, , Grass, , Food web, , (i) How many food chains are present in the given, food web?, (ii) Name the primary consumer in the given food, web., (iii) How much percentage of energy is less at each, trophic level?, (iv) If all the foxes are killed due to a disease, what, will be your observations about food web?, (v) Name the organism in which accumulation, of toxic non-biodegradable substances is the, lowest., , 40. Read the following and answer the questions from, (i) to (v) given below, Food chains are very important for the survival of, most species. When only one element is removed, from the food chain it can result in extinction of a, species in some cases.

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116, , CBSE Term II Science X, , The foundation of the food chain consists of, primary producers.Primary producers or, autotrophs can use either solar energy or chemical, energy to create complex organic compounds,, whereas species at higher trophic levels cannot, and so must consume producers or other life that, itself consumes producers., Because the Sun's light is necessary for, photosynthesis, most life could not exist if the Sun, disappeared., Even so, it has recently been discovered that, there are some forms of life, chemotrophs that, appear to gain all their metabolic energy from, chemosynthesis driven by hydrothermal vents,, thus showing that some life may not require solar, energy to thrive., , Analyse the populations and their effects on each other., , Food webs, Secondary, carnivores, , Primary, carnivores, , Herbivores, , Producers, , Grass, , (i) If 10,000 J solar energy falls on green plants in a, terrestrial ecosystem, what percentage of solar, energy will be converted into food energy?, (ii) Why decomposers are not included in the food, chain?, (iii) How primary consumers are different from, primary carnivores?, (iv) State one reason that limits the number of trophic, levels in a food chain., (v) In the given food web, name the organism in, which highest accumulation of toxic substance is, found., , 41. Read the following and answer the questions from, (i) to (v) given below, Each organism is generally eaten by two or more, other kinds of organisms. They in turn are eaten by, several other organisms., Study the figure which shows the similar, relationship between various organisms of, ecosystem through a food web., , Food web, , (i) How many trophic levels are there in the longest food, chain?, (ii) What is the role of rabbit in the given food web?, (iii) Give the significance of food web., (iv) Why do all food chain start with plants?, (v) What leads an organism to increase in number?, 42. Read the following and answer the questions from (i) to, (v) given below, To study the ozone layer depletion, Mrs. Sharma, a, science teacher draw the given flow chart on the, blackboard. She asked students to fill the bubbles, labelled as 1-5 in the flow chart., Ozone depleting, substances, 1, , Formation of ozone, , Effects of ozone, depletion, , 2, Ozone Layer, , Where is it found?, 3, , 5, , What is ozone?, 4, , (i) What are ozone depleting substance?, (ii) How is ozone formed?, (iii) Ravi want to drawn a ozone layer in a diagram, showing different layers of atmosphere, where he, should label it?, (iv) What is the difference between good ozone and bad, ozone?, (v) What could be the effect of ozone depletion?

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EXPLANATIONS, Objective Questions, 1. (b) Option (b) is incorrect pair and can be corrected as, Air is abiotic (non-living) component of ecosystem., 2. (d) The flow of energy is not a functional component of an, ecosystem., 3. (a) Flow of energy in an ecosystem is always unidirectional., 4. (b) In the given diagram, P represents producers as every food, chain begins with plants which are autotrophic in nature. Q is, primary consumer that depends on plants directly, i.e. they are, herbivores., R could be carnivores or omnivores and acts as secondary, consumer in the food chain as it depends upon plant eating, animals., S are decomposers that help in cleaning our environment by, acting on dead and decaying organic matter and decompose, them into soil., Thus, option (b) gives correct identification and main function, or characteristic., 5. (a) The given food chain is as follows, Grass → Grasshopper → Frog → Snake → Hawk, A, B, C, D, E, A. Grass acts as producer in the given food chain as it is, autotrophic in nature and make its own food in the, presence of sunlight. All living organisms depend upon, plants directly or indirectly., B. Grasshoppers are primary consumers as they feed on, plants (grass) directly., C. Frogs are primary carnivores as they feed on plant-eating, animals (grasshoppers)., D. Snakes are secondary carnivores as they eat flesh of other, animals and feed on primary carnivores majorly., Thus, option (a) is correct., 6. (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 when deer is missing., 7. (c) In a food chain, only around 10 % of the available energy is, passed on to the next trophic level. The rest of the energy is lost, to the ecosystem in form of heat., As the trophic level increases, the amount of energy transfer, decreases., Secondary consumers receive the smallest amount of energy, from primary consumers., Hence, arrow 2 shows smallest energy transfer, whereas arrow, 3 shows largest energy loss as 90% of energy at producer level, is lost to the ecosystem, whereas only 10% of energy is, transferred to the primary consumers., , 8. (b) In the given food chain, X could be small fish who feeds on, zooplankton., Hence, X is a secondary consumer. Y acts on every trophic, level which means it is a decomposer who acts on dead remains, of living organisms., 9. (c) In the given pyramid of energy, T1 , T2 , T3 and T4 represent, the trophic levels. T1 represents plants which are producers,, e.g. grass, T2 represents herbivores who are plant-eating, animals, e.g. deer. T3 represents both carnivores or omnivores, who feeds on plant-eating animals, e.g. bear. T4 represents, tertiary consumers who feed on other animals, e.g. lion,, Hence, option (c) is correct., 10. (c) C represents the correct pyramid in a food chain., Population of producers (plants) are maximum in a food chain, to support other animals. As the trophic level increases, the, number of organisms decreases., 11. (c) The given food web contains various food chains, interconnected with each other., Frog, bird and fish act as secondary consumers in their, respective food chains, therefore acquire same trophic level ,, i.e. third trophic level in their respective food chain., 12. (a) Biomagnification is the phenomenon of progressive, increase in the concentration of non-biodegradable toxicants, in organisms at each successive trophic level. It is also called, bioconcentration., 13. (c) The layer P is ozone layer that protects us from harmful UV, rays of Sun. Chlorofluorocarbon is responsible for depletion of, ozone layer., 14. (a) High energy UV radiations split apart molecular, oxygen ( O2 ) into free oxygen atoms (O) which are highly, reactive and combine with molecular oxygen to form ozone, layer., 15. (d) Incineration involve degradation of wastes by burning, them at high temperature., 16. (c) A is true, but R is false., All living organisms are biotic components of ecosystem, whereas all non-living things such as wind, gases, light, water,, etc., are abiotic components of ecosystem., 17. (d) A is false, but R is true., Food chains generally consist up to three or four trophic levels, because there is loss of energy at each trophic level and very, little usable energy remains after three or four trophic levels., 18. (c) A is true, but R is false., Autotrophs (producers) are present at the first trophic, level because they fix solar energy, making it available for, consumers or heterotrophs.

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118, , CBSE Term II Science X, , 19. (c) A is true, but R is false., Certain pesticides and other chemicals used to protect, our crops from diseases and pests do not get degraded, (i.e. non-biodegradable). So, they get accumulated, progressively at each trophic level., 20. (a) Both A and R are true and R is the correct explanation, of A. Ozone layer is present in the stratosphere region of, our atmosphere. It shields the surface of Earth from, harmful UV radiations of Sun which are highly damaging, and can cause various health issues and diseases such as, skin cancer, cataract, etc., 21. (i) (b) Due to biomagnification, the concentration of, harmful chemicals such as DDT increases at each, successive trophic level., The maximum concentration of DDT gets, accumulated in Z and least is found in Y., Therefore, the correct order of animals in food chain, , (iii) (a) Solid waste R completely decomposes in very few days, which means it is an easily decomposable biodegradable waste., e.g. Cow dung, fruit pulp, etc., (iv) (a) Non-biodegradable waste can be decomposed, by recycling or by dumping underground into landfills., (v) (c) Incineration is a method of waste disposal in which, burning of substances take place at high temperature to form, ash. It is used to dispose off hospital or harmful wastes of, biomedical industries., Subjective Questions, 1. Ecosystem is defined as the structural and functional unit of, biosphere. It is a stable ecological unit where continuous input of, energy and circulation of matter occurs., Top carnivores, Carnivores, , is Y → W → X → Z., (ii) (a) The concentration of DDT is found to be less in, organisms of lower trophic level and higher in, organisms of higher trophic levels., If the amount of DDT in W is estimated to be 0.4, ppm then in Y, it has to be less than 0.4 ppm, i.e., 0.02 ppm and in Z, it has to be more the 0.4 ppm, i.e, 2 ppm., Hence, option (a) is correct., (iii) (b) The progressive increase in the concentration of, non-biodegradable toxicants in organisms at each, successive trophic level is known as, biomagnification., (iv) (a) Decomposers are microorganisms that, breakdown dead and decayed organisms into, simpler inorganic materials, making nutrients, available to primary producers. They act at every, trophic level of the food chain., Hence, they do not have a fix position in the food, chain., (v) (c) Toxic chemicals and non-biodegradable, substances such as pesticides, fertilisers and heavy, metals are hazardous to the environment, whereas, manures are organic substance made by, decomposing dead and decayed living organisms, buried under the soil., 22. (i) (b) S is the solid waste that does not decompose at, all. According to the graph, there is no change in, number of days for the amount of decomposition, take place in S waste., Hence, it is a non-biodegradable waste., (ii) (b) Substances which do not decompose by the, action of microorganisms present in the soil are, known as non-biodegradable substances., They enter into food chain through soil and get, magnified into higher trophic levels. In soil, they, also cause pollution which decreases the soil, fertility., , Herbivores, Producers, Sunlight, , Flow of energy in an ecosystem, , 2. (i) Ecosystem is the structural and functional unit of biosphere, and is a stable ecological unit where regular input of energy, and circulation matter takes place., (ii) Autotrophs can make their own food in the presence of, sunlight. They are the ultimate source of energy for each and, every organism of a food chain. Hence, every chain always, starts with producers (autotrophs) that is why they are placed, at the first trophic level of food chain., (iii) In given food chain, Grass→ Insects→ Frogs→ Snakes, Frogs assign at third trophic level. They acts as secondary, consumers who feeds of primary consumers (i.e. insects)., 3. Organisms that feed on dead plants and animals are called, decomposers, e.g. bacteria, fungi, etc. They breakdown the, complex organic compounds present in the dead remains into, simpler substances and obtain nutrition from them. These, substances are released into the soil and to the atmosphere., Thus, they play the following roles, (i) They help in recycling of materials, replenishment of the, soil’s nutrients, etc., (ii) They clean our surroundings by decomposing dead organisms, and organic wastes., 4. Natural water bodies are example of natural ecosystem. They, exist naturally without any human support, whereas aquarium is, an artificial ecosystem which is created and maintained by, humans. It rely on human efforts to sustain. It does not possess a, self-regulating mechanism., 5. The study of food chain in an area or habitat helps in, (i) understanding the energy transfer through organisms., (ii) understanding the ecological balance in a habitat or, ecosystem.

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119, , CBSE Term II Science X, , (iii) understanding harmful human activities and disruption, of ecological balance, if any., An example of four steps of food chain operating in a large lake, is as follows, Algae → Protozoan → Small fish → Big fish., 6. Parasites are organisms (animals or plants) that live in or an, other organism (host) and take benefits by deriving nutrients, from it, i.e. they get food or protection from host organism, e.g., Cuscuta., On other hands, decomposers are organisms that breakdown, dead or decaying organisms into simple inorganic, substances. e.g. Fungi., 7. The transfer of food or energy takes place through various, levels in the food chain, which are known as trophic levels. e.g., Trees, Rabbit →, Snake →, Hawk, →, (First trophic, level), [Producers], , (Second trophic, level), [I consumer], , (Third trophic, level), [II consumer], , (Fourth trophic, level), [III consumer], , 8. In a food chain, about 80-90% of the energy available at a, trophic level is lost during its transfer to next trophic level., Hence, amount of energy available goes on decreasing at each, successive trophic level., , 11. The flow of energy in an ecosystem occurs in the following, sequence, Sun ⎯→ Producer ⎯→ Herbivore ⎯→ Carnivore, (Primary, consumer), , (Secondary, consumer), , The flow of energy is unidirectional because of the reasons, given below, (i) Energy flows progressively from one trophic level to, another and cannot revert back. Energy given out as, heat is lost to the environment and does not return to be, used again., (ii) The available energy decreases at higher trophic level., Out of the total energy available at a particular trophic, level, only 10% is passed on to the next trophic level,, making it impossible for energy to flow in the reverse, direction., 12. Biological magnification also known as biomagnification is the, phenomenon of progressive increase in the concentration of, non-biodegradable toxicants in organisms at each successive, trophic level., 13. Biological magnification refers to the increase in the, concentration of certain toxicants at each successive trophic, level., , If a plant fixes 4000 J energy, then next three successive, trophic levels will get 400 J, 40 J and 4 J, respectively, (according to 10% law). If another level is added in a food chain, then it will get only 0.4 J energy. Thus, usually food chains, remain shorter and limited to 3-4 trophic levels only., , No, the levels of magnification will not be same in all trophic, levels. When the chemicals do not get degraded and get, accumulated progressively at each trophic level, it leads to, biomagnification. Biomagnification is more in organisms of, higher trophic levels., , 9. If we kill all the organisms in one trophic level, the lower trophic, level will grow more in number and the higher trophic level will, not survive., , 14. Depletion of ozone is mainly caused due to the excessive use of, Chlorofluorocarbons (CFCs). These are synthetic chemicals, which are used as refrigerants and in fire extinguishers., , Hence, flow of energy from one trophic level to other will not, take place., , Steps which should be taken to limit the damage to ozone layer, include, (a) Minimising the use of aerosol spray propellants containing, fluorocarbons and chlorofluorocarbons., (b) Exercising control over large scale nuclear explosions, and limited use of supersonic planes., , 10. 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., The flow of energy through a food chain is unidirectional and it, moves progressively through various trophic levels as follows, (i) Green plants capture 1% of energy of the sunlight that, falls on their leaves and convert it into food energy., (ii) When green plants are eaten by primary consumers, a, great deal of energy is lost as heat to the environment., On an average only 10% of food eaten is turned into its, own body and made available for the next level of, consumers., (iii) Thus, 10% can be taken as average value of the amount, of organic matter present at each step and reaches the, next level of consumers., Sun, , Leaf, (producer), , 10% energy, transferred, , Caterpillar, (primary consumer), , 10% energy, Bird, Snake, (tertiary consumer ) transferred (secondary consumer), Energy lost by energy transfer, , 10% energy, transferred, , 15. Ozone layer filter the Sun's ultraviolet radiation (UV-B),, thereby protecting the environment from its harmful effects, and also play key role in regulating the temperature., 16. The energy and biomass decrease from lower to higher trophic, levels, so the length of food chain is restricted and cannot have, more than four trophic levels., 17. (i) UV radiation acts on the O 2 molecule. Higher energy of, UV split apart O 2 ., (ii) O2 ⎯UV, ⎯, ⎯→ O + O, O + O2 ⎯→ O3, , (Ozone), , (iii) CFCs rise up in stratosphere where UV radiation splits, them releasing molecular chlorine (Cl − ). It reacts with, O 3 and releases oxygen, so O 3 gets depleted., 18. Ozone at the higher levels of the atmosphere is a product of UV, radiations acting on oxygen (O 2 ) molecule. The high energy UV, radiations split apart some molecular oxygen (O 2 ) into free

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120, , CBSE Term II Science X, , oxygen (O) atoms. These atoms are very reactive and combine, with the molecular oxygen to form ozone., UV, O2 ⎯ → O + O, O + O2 ⎯→, , O3, (Ozone), , It shields the surface of the Earth from harmful ultraviolet, (UV) radiations of the Sun., Due to environmental pollution, ozone layer has began to, deplete in the 1980s., This was mainly due to the increasing use synthetic, chemicals like Chlorofluorocarbons (CFCs). These are used, in refrigerants as coolant and in fire extinguishers., Due to depletion of ozone layer harmful UV rays can, penetrate or enter our atmosphere.These radiations are, highly damaging to organisms. They can cause skin cancer in, human beings, damage eyes (cause disease called cataract),, decrease crop yield, disturb global rainfall, etc., 19. Following methods could be applied to reduce the, accumulation of pesticides in our body., (i) Minimise the use of pesticides, (ii) Consuming washed fruits and vegetables, (iii) Developing vegetarian feeding habits, (i.e. feed upon plants as plants belong to lower trophic level so,, they have less accumulation of insecticides, whereas, organisms of higher trophic level have higher concentration of, insecticides and pesticides)., 20. We can reduce the problem of waste disposal by the following, methods, Recycling The solid wastes such as paper, plastics, glass, and metals, etc., are recyclable. So, waste paper can be sent, to paper mills for reprocessing to form newspaper. The, plastic articles can be melted and remoulded again to make, new articles., Biodegradable waste The waste such as left over food,, fruits, animal dung, peels of vegetables can be converted, into compost by burying them in a pit dug in ground and, can be used as manure., !, , 24. Effective segregation of waste as biodegradable and, non-biodegradable is much easier to recycle. Biodegradable, waste used to make manure can out of compost, whereas, non-biodegradable waste could be recycled and reused for, various purposes. Also effective segregation of wastes means, that less waste goes to landfill, which makes it cheaper and, better for people and environment., 25. A pond ecosystem refers to freshwater ecosystem where, there are various organisms dependent on each other with, the prevailing water environment for their nutrients and, survival., Phytoplankton → Zooplankton → Small fish → Big fish, There are two components of the pond ecosystem, (i) Abiotic It includes water, dissolved minerals, oxygen, and carbon dioxide. Sunlight is the main source of, energy., (ii) Biotic It consists of phytoplanktons, zooplanktons,, aquatic insects, fishes and other aquatic animals. These, organisms are classified as producers, consumers and, decomposers., 26. The waste generated in our house and measures for its disposal, are given in the table below, , Household wastes, , Kitchen waste like bottles, Prepare a compost pit, plastics, food, etc., Paper wastes like, Should be recycled, newspaper, envelopes, etc., Plastic bags, , Should be safely dumped in, garbage bins for, non-biodegradable wastes, , Vegetable/fruit peels/rind, , Can be placed near trees/plants,, so that on decomposition enrich, the soil with nutrients, , All other wastes, , Segregation into biodegradable, and non-biodegradable wastes, , !, , 21. If all the waste is biodegradable, then there will be no, accumulation of waste on the Earth and it would be a cleaner, place to live., But if, this biodegradable waste is too large in amount then its, slow degradation may lead to air pollution (due to release of, gases) as well as water and land pollution., 22. The two ways in which non-biodegradable substances would, affect the environment are, (i) They make the environment poisonous and unfit for, survival of living forms of life., (ii) They block the transfer of energy and minerals in the, ecosystem., 23. Biodegradable materials are broken down by microorganisms, present in nature into simple harmless substances., Non-biodegradable materials need a different treatment like, heat and temperature for disposal and hence, both should be, discarded in two different dustbins., , Measures for disposal, , 27. Insecticides are non-biodegradable chemicals added to crop, fields to stop the growth of insects infecting the crops., Modern insecticides are being developed keeping in, mind, the harm they cause to the environment and its, components., Biodegradable insecticides can be decomposed into harmless, substances, which will subsequently be dispersed in their, specific pathways and cause no pollution., Non-biodegradable insecticides build up in the fat tissues of, the body and pass on to organisms that feed on them., Hence, they accumulate along the food chain resulting in, significant amounts in the tissues of consumers at the, highest trophic level., The property of newly developed insecticide includes that it, can easily get decomposed into simpler components by soil, bacteria.

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121, , CBSE Term II Science X, , 28. As Mona follows vegetarian diet along with milk products., She should be considered as occupying third trophic level, because the curd is prepared by the milk of cow/buffalo by, the action of bacteria, but the energy of the milk is getting, transferred from cow to Mona, so Mona is at the third, trophic level., Grass, , ⎯⎯⎯→, , (First trophic level), , Cow (milk) ⎯⎯⎯→ Human, (Second trophic level), , (Third trophic level), , 29. Organisms which breakdown the complex organic compounds, present in dead and decaying matter into simpler inorganic, materials are called decomposers, e.g. certain bacteria and, fungi., Decomposers act as cleaning agents of environment by, decomposing dead bodies of plants and animals. They also, help in recycling of materials, replenishment of soil’s, nutrients, etc., The consequence of their absence in an ecosystem can be, disastrous. The dead bodies would persist for long, leading, to their accumulation and thus, polluting the environment., The biogenetic nutrients associated with these remains will, not be returned back to the environment., As a result, all the nutrients present in soil, air and water, would soon be exhausted and the whole life cycle of, organisms will be disrupted., 30. (i) Consumers are the organisms who derive energy by, eating plants or other organisms as they cannot produce, food on their own. Absence of primary consumers in, nature would lead to enormous growth of plants and, decline in the population of carnivore animals, who eat, them. Hence, the whole food web will get distrupted., (ii) The direction of energy transfer in following cases are as, follows, (a) Primary consumer (grasshopper) to secondary, consumer (frog)., (b) Producer (grass) to primary consumer (deer)., (c) Primary consumer (deer) to secondary consumer, (lion)., 31. (i) Yes, the impact of removing all the organisms in a, trophic level will be different for different trophic levels., The lower trophic level of an ecosystem has a greater, number of individuals than the higher trophic levels., Removal of producers will affect all the organisms of, successive trophic levels and it will threat their survival., The removal of higher trophic level will lead to increase, in organisms of lower trophic level and the organisms of, higher trophic level will die due to the shortage of food., (ii) No, removal of all organisms of a trophic level will, disturb the ecosystem. Killing of higher trophic level, organisms will cause explosion in the population of lower, trophic level organisms. This will adversely affect the, ecosystem and thus environment., 32. (i) The producers convert solar energy into chemical energy, in the form of organic compounds. The primary consumers, (herbivores) derive their nutrition from the producers., According to the energy transfer law, only 10% of energy is, transferred from one trophic level to the other., , So, the energy that is captured by the producers does, not revert back to the Sun and the energy transferred to, the herbivores does not come back to the producers. It, just keeps on moving to the next trophic level in one, direction. That is why the flow of energy in the food, chain is always unidirectional., (ii) A large number of pesticides and chemicals are used to, protect our crops from pests and diseases. Some of these, chemicals are washed down from the soil, while some, enter the water bodies. From the soil, they are absorbed, by plants along with water and minerals and from the, water bodies, they are taken up by aquatic plants and, animals. This is how these chemicals enter the food, chain., As these chemicals cannot decompose, they accumulate, progressively at each trophic level. This increase in the, concentration of harmful chemicals with each step of the, food chain is called biomagnification. As human beings, occupy the top level in any food chain, these chemicals, get accumulated in our bodies in considerably high, amount causing diseases., 33. (i) Undesirable activities of man eliminate growth of, organisms belonging to one or more trophic levels in a, food chain. Thus, the food chain gets shortened, e.g., hunting tigers for their skin, etc., It causes imbalance in the functioning of ecosystem and, biosphere. If organisms of one trophic level are, eliminated, the organisms prior to that trophic level will, flourish and increase in number. Also, the organisms of, the subsequent trophic level will sharply decrease,, thereby creating an imbalance., (ii) Vegetarian food chain is advantageous in terms of energy, because it has less number of trophic levels. As we know,, only 10% of the energy is transferred to the next trophic, level in a food chain, so if a person is vegetarian then, he, would have maximum amount of energy by consuming, producers or plants in a food chain., 34. Some harmful effects of agricultural practices on the, environment are as follows, (i) Soil degradation Extensive cropping causes loss of soil, fertility. Also, over the time it can lead to soil erosion, and finally to desertification., (ii) Pollution Use of synthetic chemical fertilisers and, pesticides leads to soil, water and air pollution., (iii) Water shortage Excess use of groundwater for, agriculture lowers the water level. This results in acute, water shortage at many places., (iv) Biomagnification The chemical pesticides, being, non-biodegradable accumulate in organisms in, increasing amounts at each trophic level., (v) Deforestation Indiscriminate cutting of trees for, agriculture has resulted in loss of habitat for wildlife., Thus, it also causes damage to natural ecosystem., 35. Ozone is a triatomic molecule, i.e. made up of three atoms of, oxygen joined together. Its molecular formula is O3 . It can, affect any ecosystem in the following ways, (i) It protects against ultraviolet rays if, present in, stratosphere.

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122, , CBSE Term II Science X, , (ii) Ozone dissipates the energy of UV rays by undergoing, dissociation followed by reassociation., UV, O 2 ⎯→ O + O ; O + O 2 ⎯→ O 3, (Ozone), (iii) In atmosphere, it is highly toxic and causes injury to, mucous membranes, eye irritation and internal, haemorrhages in animals and humans., 36. (i) Environmental pollution is an undesirable change in the, physical, chemical or biological characteristics of the, natural environment, brought about by man’s activities., This pollution may affect the soil, water or air., (ii) Differences between biodegradable and, non-biodegradable pollutants are as follows, , Biodegradable, pollutants, , Non-biodegradable, pollutants, , These pollutants cannot be, These pollutants can be, broken down into non-toxic broken down into non-toxic, substances in nature by the substances by microorganisms., action of microorganisms., They get recycled thus, do They cannot be recycled thus,, not need any dumping sites. require dumping sites., They cause minimum, environmental pollution., , They cause maximum, environmental pollution., , (iii) Non-biodegradable pollutants include DDT, radioactive, waste, plastic, insecticides., 37. Some daily life eco-friendly activities are, (i) Save a tree, use less paper You can buy ‘tree-free’ 100%, post-consumer recycled paper for everything from, greeting cards to toilet paper. Paper with a high, post-consumer waste content uses less pulp and keeps, more waste paper out of landfills., (ii) Opt bamboo for hardwood floors Bamboo is considered, as an environmental-friendly flooring material due to its, high yield and the relatively fast rate at which it, replenishes itself. It takes just 4-6 years for bamboo to, mature, compared to 50-100 years for typical hardwoods., Also look for sources that use formaldehyde-free glues., (iii) Reduce plastics, reduce global warming Unfortunately,, plastics are made from petroleum, the processing and, burning of which is considered one of the, main contributors to global warming, according to the, EPA. In addition, sending plastics to the landfill also, increases greenhouse gases. Reduce, reuse and recycle, our plastics are one of the best ways to combat global, warming., (iv) Use healthier paints Conventional paints contain, solvents, toxic metals and Volatile Organic Compounds, (VOCs) that can cause smog, ozone pollution and indoor, air quality problems with negative health effects,, according to the EPA. These unhealthy ingredients are, released into the air, while we are painting, drying of paint, and even after the paints are completely dry., , (v) Use compost Instead of using synthetic fertilisers,, compost provides a full complement of soil organisms, and the balance of nutrients needed to maintain the, soil’s health. Healthy soil minimises the population of, weeds., 38. Materials that remain for a long time in the environment,, without getting decomposed by any natural agents, also, causing harm to the environment are called, non-biodegradable. Plastic cups are non-biodegradable and, raised the concern towards hygiene, thus they were, replaced by kulhads., Kulhads are made up of clay on a large scale resulted in the, loss of top fertile soil. It is replaced by disposable paper, cups because the paper can be recycled, it is biodegradable, and is eco-friendly material which does not cause, environment pollution., 39. (i) There are 6 food chains that constitute to form the, given food web. These are as follows, 1. Seeds → Mice → Foxes, 2. Young trees → Mice → Foxes, 3. Young trees → Rabbits → Foxes, 4. Grass → Rabbits → Foxes, 5. Grass → Rabbits → Snake, 6. Young trees → Rabbits → Snake, (ii) The primary consumers are the organisms who directly, feed on the producers. In the given food web, rabbits, and mice are the primary consumers, whereas foxes and, snake are the secondary consumers., (iii) About 90% of energy is loss at each trophic level., (iv) The foxes feed on the rabbits and mice. If all the foxes, are killed then there will be no direct predator of, rabbits and mice, hence the number of rabbits and mice, (i.e. both are primary consumers) will increase in the, given ecosystem, which will disturb its balance., (v) Accumulation of toxic non-biodegradable substances, increases at each trophic level. It is least in organisms, of first trophic level (i.e. seeds), young trees and grass, and highest in organisms of third trophic level (i.e. foxes, and snake)., 40. (i) The green plants in a terrestrial ecosystem capture, about 1% of the energy of sunlight (light energy). They, convert it into food energy (chemical energy)., Therefore, if 10,000 J solar energy falls on green plants in, a terrestrial ecosystem, only 1% of solar energy, i.e. 100 J, will be converted into food energy., (ii) Decomposers are not included in the food chain as they, act at every trophic level of the food chain., (iii) Primary consumers are those organisms who directly, feed on plants, e.g. herbivores., Primary carnivores are those organisms who feeds on, plant-eating animals, e.g. snake., (iv) Decrease in energy at higher trophic levels (according to, 10% law) is one of the main reason that limits the number, of trophic levels in a food chain.

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123, , CBSE Term II Science X, , (v) Owl is placed at highest trophic level in the given food, web, therefore the accumulation of non-biodegradable, toxic materials are found to be highest in them due to, biomagnification., 41. (i) The longest food chain consist of maximum five trophic, levels., (ii) Rabbit is a herbivore as it feeds on plants directly. Its, role in the given food web is to transfer the energy from, producers to carnivores., (iii) The significance of food web is as follows, (a) Food chain provides pathways for availability of, food., (b) It allows endangered populations to grow in size., (iv) Sun or solar energy is the ultimate source of energy for, the Earth. Only plants can utilise this energy to make, their on food., (v) When predator for a particular organism decreases in, number, the organisms start increasing in number., As the animals which used to feed on them decrease in, number, therefore, the population of those organisms, increases., , 42. (i) Substances that are responsible for depletion ozone layer, or breakdown of ozone molecules are known as ozone, depleting substances., e.g. CFCs, halogens, nitrous oxide, CCl 4 and CH4 are ozone, depleting substances responsible for ozone layer depletion., (ii) Atomic oxygen is highly reactive. It combines with, molecular oxygen to form ozone., UV, , O2 ⎯→ [O] + [O], 2O2 + 2 [O] ⎯→ 2O3 (Ozone), (iii) Ozone layer is found in the stratosphere around 15-30 km, above the Earth’s surface., (iv) Ozone is a triatomic molecule made up of three atoms of, oxygen (O3 ). It is present in atmosphere as an ozone layer, shield that protects us from high energy UV radiations., So, it is known as good ozone whereas near the surface of, earth ozone act as a highly poisonous gas. Hence, known, bad ozone., (v) Cancers, mutations, effect on eyesight, global warming,, weakening of immune system, etc., are some adverse, effects of ozone depletion.

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Chapter Test, Multiple Choice Questions, , 1. Which of the following option correctly represents the, biotic and abiotic components of ecosystem?, Biotic components, , Abiotic components, , (a), , Soil, , Water, , (b), , Plants, , Animals, , (c), , Animals, , Decomposers, , (d), , Decomposers, , Water, , 2. Biomagnification is highest in, (a) producer, (b) primary consumer, (c) secondary consumer, (d) decomposer, , 7. Assertion (A) The green plants capture about 10%, of the energy of sunlight that falls on their leaves., Reason (R) When green plants are eaten by primary, consumers, a great deal of energy is lost as heat to, the environment., , 8. Assertion (A) Biodegradable domestic wastes can, be buried in a pit, dug into ground., Reason (R) They are converted into compost and, used as manure., , Short Answer Type Questions, 9. Suggest one word for each of the following, statements., I. The physical and biological world where we live., II. The physical factors like rainfall, wind, soil are, termed as., III. Trophic level of plant-eating animals., IV. Each level of food chain where transfer of energy, takes place., , 10. What percentage of solar energy is trapped and, , 3. Organisms which synthesise carbohydrates from inorganic, , utilised in terrestrial and aquatic ecosystem?, , compounds using radiant energy are called, , 11. Which chemical is used in fire extinguishers? How is, , (a) decomposers, (c) herbivores, , 12. Among all four types of animals, i.e. carnivores,, , (b) producers, (d) carnivores, , 4. In a food chain, the snake an predates as rabbit which fed, on fresh green bushes. What percentage of the energy, accumulated by rabbit, would be acquired by snakes?, (a) 90%, (c) 50%, , (b) 10%, (d) 25%, , it harmful?, decomposers, herbivores and producers, how, does energy flow in an ecosystem, occur through, these organisms? Explain with the help of example., , 13. How are water bodies affected by sewage disposal?, Long Answer Type Questions, 14. ‘Number of vultures are decreasing remarkably, , 5. Bad ozone is formed in, (a) atmosphere, (b) ionosphere, (c) stratosphere, (d) troposphere, , Assertion-Reasoning MCQs, Direction (Q. Nos. 6-8) Each of these questions contains, two statements, Assertion (A) and Reason (R). Each of, these questions also has four alternative choices, any one, of which is the correct answer. You have to select one of, the codes (a), (b), (c) and (d) given below., (a) Both A and R are true and R is the correct explanation of A, (b) Both A and R are true, but R is not the correct explanation, of A, (c) A is true, but R is false, (d) A is false, but R is true, , 6. Assertion (A) The autotrophs are at the first trophic level., Reason (R) Autotrophs fix up the solar energy and make it, available for consumers., , now-a-days, which is a matter of concern’., Answer the following questions related to the, above given statement., (i) Vultures belong to which category of animals?, (ii) What is their role in nature to maintain ecological, balance?, (iii) At which trophic level, will you place vultures in a, food chain?, (iv) How much energy is passed on at each trophic, level in a food chain? Give the energy flow, diagram of an ecosystem., (v) Mention one of the main cause for this decline in, the number of vultures., , 15. Salman wants to volunteer for Swachh Bharat, Abhiyan intership. He got a responsibility to device, various methods for garbage management. Help, him to write various ways to control and manage, garbage., , Answers, Multiple Choice Questions, 1. (d), , 2. (c), , 3. (b), , 4. (b), , Assertion-Reasoning MCQs, 6. (a), , 7. (d), , 8. (a), , 5. (d), , For Detailed Solutions, Scan the code

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125, , CBSE Term II Science X, , Science, , Class 10th ( Term II), , Practice Paper 1, , *, , (Solved), Time : 2 Hours, Max. Marks : 40, , General Instructions, , 1. There are 11 questions in the question paper. All questions are compulsory., 2. Section A, Question no. 1 to 3 is a Case Based Questions, which has four MCQs/Questions. Each question carries one mark., 3. Section B, Question no. 4 to 8 are Short Answer Type Questions. Each question carries 2/3 marks., 4. Section C, Question no. 9-11 are Long Answer Type Questions. Each question carries 5 marks., 5. There is no overall choice. However, internal choices have been provided in some questions. Students have to attempt, only one of the alternatives in such questions., , * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this, paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised, not to consider the pattern of this paper as official, it is just for practice purpose., , Section A, 1. Read the following and answer the questions from (i) to (iv) given below, A student wants to study the dependence of potential difference ( V ) on current (I) flowing across a resistor., He plotted a graph by taking I along X-axis and V along Y-axis of all his observations., Potential difference (v), , Y, 1.0, 0.8, 0.6, 0.4, 0.2, X, 0.2 0.4 0.6 0.8 1.0, , (i) The value of resistance used in the circuit is, (a) 1 Ω, , (b) 2 Ω, , Current (I), , (c) 4 Ω, , (d) 0.5 Ω, , (c) reactance, , (d) resistivity, , (c) V ∝ I, , (d) I ∝ V, , (ii) The inverse of the slope of V-I curve gives, (a) resistance, , (b) conductance, , (iii) The correct representation of Ohm’s law is, (a) V ∝, , 1, I, , (b) I ∝, , 1, V, , (iv) His friend noticed some observations from this demonstration as follows, 1. The slope of the line gives the resistance of a conductor., 2. Motion of electrons through a conductor is retarded by its resistance., 3. All conductors do not obey Ohm’s law., , Which of the following observation(s) is/are correct?, (a) 1 and 2, , (b) 2 and 3, , (c) 1 and 3, , (d) 1, 2 and 3

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126, , CBSE Term II Science X, , Or, What is the use of the variable resistor in the circuit in, the Ohm’s law experiment shown below?, , A, , 3. Read the following and answer the questions from, (i) to (iv) given below, Sex-determination is the method by which, distinction between males and females is, established in a species. The sex of an individual is, determined by specific chromosomes. In human, beings, it is determined as follows, , V, , Sperms, and, , X, , I. To change the voltage across the fixed resistor., II. To change the resistance across the fixed resistor., III. To change the current flowing through the fixed, resistor., (a) Only I, (b) Only III, (c) I and III only, (d) I, II and III, , 2. Read the following and answer the questions from, (i) to (iv) given below, In modern periodic table, some properties show a, regular trend when we move along a period from, left to right or in a group from top to bottom., Atomic size refers to the radius of an atom. It may, be visualised as the distance between the centre of, the nucleus and the outermost shell of an isolated, atom. It is measured in picometres., About 75% of all elements in one periodic table are, metals. The most common metal found in the, Earth’s crust is aluminium. The non-metals also are, some of the most abundant elements in the, universe including the Earth’s crust, the, atmosphere and the human body., The position of three elements A, B and C in the, periodic table are shown below, Group 16, , Group 17, , —, , —, , —, , A, , —, , —, , B, , C, , (i) State whether A is a metal or non-metal., (ii) State whether C is more reactive or less reactive, than A., (iii) Will C be larger or smaller in size than B?, (iv) Which type of ion- cation or anion, will be formed, by A?, , Or, What are metalloids? Give examples., , Egg, Y, , X, X Y, , X X, , (i) In XX-XY type of sex-determination,, (a) females produce two different types of gametes, (b) males produce two different types of gametes, (c) female produces gametes with Y-chromosome, (d) male produces gametes with X-chromosome only, , (ii) A couple has six daughters. What is the possibility, of them having a girl child next time?, (a) 10%, , (b) 90%, , (c) 100%, , (d) 50%, , (iii) Number of chromosome present in the egg cell of a, human female is, (a) 23 pairs, (c) 22, , (b) 23, (d) 22 pairs, , (iv) Choose the incorrect statement from the following, set of statements., I. XX-XY type of sex-determination is the example of, male heterogamety., II. XX-XY type of sex-determination is the example of, female heterogamety., III. There are always 50% chances of having a baby girl, child., IV. Changes in the non-reproductive tissues can be, passed on to the DNA of the germ cells., Codes, (a) I and III, (c) III and IV, , (b) II and IV, (d) I and III, , Or, Sex of human child is determined by, (a) size of the egg, (c) type of the sperm, , (b) size of the sperm, (d) type of the egg, , Section B, 4. Choose the kind of chemical bonding (ionic bond,, covalent bond, both ionic and covalent bonds), present in the following compounds. Potassium, chloride, magnesium oxide, sulphuric acid,, ammonium hydroxide, zinc sulphide, phosphorus, [2 M], trichloride (PCl 3 ).

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127, , CBSE Term II Science X, , Or, 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., , Section C, 9. (i) Which is the better way to connect lights and other, appliances in domestic circuit, series connection or, parallel connection? Justify your answer., (ii) An electrician has made electric circuit of a house in, such a way that, if a lamp gets fused in a room of the, house, then all the lamps in other rooms of the, house stop working. What is the defect in this type, of circuit wiring? Give reason., , [2 M], , 5. Differentiate between pollen tube and style. [2 M], 6. For the current carrying solenoid as shown below,, draw magnetic field lines and give reason to, explain that out of the three points A , B and C at, which point, the field strength is maximum and at, which point, it is minimum., , Or, Three incandescent bulbs of 100 W each are, connected in series in an electric circuit. In another, set of three bulbs of the same wattage are connected, in parallel to the source., (i) Will the bulb in the two circuits glow with the same, brightness? Justify your answer., (ii) Now, let one bulb in both the circuits get fused. Will, the rest of the bulbs continue to glow in each circuit?, Give reason., , B, , A, , C, , [3 M], , I, , I, , Or, Diagram shows the lengthwise section of a current, carrying solenoid., ⊗ Indicates current entering into the page,, • Indicates current emerging out of the page., A, , 10. Differentiate between monohybrid and dihybrid, crosses with the help of an example., Or, Explain the term ‘regeneration’ as used in relation to, reproduction of organisms. Describe briefly how, regeneration is carried out in multicellular organisms, like Hydra?, , B, , 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, [3 M], solenoid., , 11. (i) How can pregnancy be prevented surgically?, (ii) Study the diagram given below, , 7. Give an example of each of the following, (i) A carbon compound containing two double, bonds., (ii) A molecule in which central atom is linked to, three other atoms., (iii) A compound containing both ionic and covalent, bonds., [3 M], , Or, Define homologous series. Examine it with an, example. Will there be any change in their physical, [3 M], properties? Give reason for your answer., , 8. Give any three points of differentiation between, acquired and inherited characters with one, example each., , [3 M], , (a) Identify the process., (b) Which organism uses the above method for, reproduction?, (c) How is the above method different from the, process of fragmentation?, , Or, , (i) Why should biodegradable and non-biodegradable, wastes be discarded in two separate dustbins?, (ii) Why do we exemplify crop fields as artificial, ecosystem?

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Explanations, 1., , (i) (a) The slope of V-I graph gives resistance., 0.4 − 0 0.4, ∴ Resistance =, =, = 1Ω, 0.4 − 0 0.4, (ii) (b) The inverse of the slope of V-I curve gives, conductance. i.e., 1, 1, =, = Conductance ( σ ), Slope Resistance (R ), (iii) (d) The correct representation of Ohm’s law is I ∝ V. It, states that current flowing through the conductor is, directly proportional to the potential difference, applied across the conductor., , (iv) (d) All the observations are correct. The slope of the, line in V-I graph gives the resistance of a conductor, whereas motion of electrons through a conductor is, retarded by its resistance., All conductors do not obey Ohm’s law i.e., Ohm’s, law is not universal., Or (c) Ohm’s law states that the resistance (ratio of the voltage, across the fixed resistor to the current flowing through, it) is constant. The aim of Ohm’s law experiment is to, find the resistance of a fixed resistor. By adjusting the, variable resistor., • the voltage across the variable resistor and fixed, resistor will change., • the total resistance in the circuit will change, leading to a change in current flowing through the, fixed resistor., 2. (i) Since, A belongs to group 17 and has 7 valence, electrons so it is a non-metal because it will gain one, electron to complete its octet., (ii) C lies below A and belong to the same group. As we, move down a group, the size increases and, electronegative character decreases. With the decrease, in electronegative character, the electron accepting, tendency and hence the reactivity of non-metals, decreases, so C is less reactive than A., (iii) C and B both belongs to the same period therefore, C is, smaller than B in size because as we move left to right, in a period, atomic size decreases due to increased, effective nuclear charge., (iv) As discussed in that element A has a tendency to gain, electron to complete its octet. It needs to take up one, electron, so it will form anion (A − )., Or These are some elements which exhibit the properties, of both metals and non-metals. These are called, metalloids. In modern periodic table, a zig-zag line, separates metals from non-metals., The border line elements—boron, silicon, antimony are, intermediate in properties, so they are called mtalloids, or semi-metals., 3. (i) (b) In XX-XY type of sex-determination (human’s sexdetermination), the males produce two different types, of gametes, one contains X-chromosome whereas the, other contains Y-chromosome., , (ii) (d) When a sperm having an X-chromosome fuses with, the egg, the zygote formed will develop into a female, baby whereas when a sperm having Y-chromosome, fuses with the egg, the zygote formed will develop into, a male baby. Thus, there are always 50% chances of, having a girl child by the couple., (iii) (d) Gametes are haploid in nature, i.e they contain half, the number of chromosomes as compared to other, somatic cells. Human beings have 23 pairs of, chromosomes. Therefore, the egg cell (gamete) contains, 23 chromosomes only., (iv) (b) Statements II and IV are incorrect and can be, corrected as, XX-XY type of sex-determination shows male, heterogamety and female homogamety., Changes in the non-reproductive tissues cannot be, passed on to the DNA of the germ cells., Or (c) Sex of a human child is determined by the type of sperm, (X or Y) that fuses with egg during fertilisation., A sperm with an X-chromosome will produce a baby girl and, that with Y-chromosome will produce a baby boy on, fertilisation., 4., !, , !, , Ionic bond, (i) Potassium chloride, (ii) Magnesium oxide, (iii) Zinc sulphide, , Covalent, bond, , Both ionic and, covalent bond, , Phosphorus, trichloride, , Sulphuric acid, Ammonium, hydroxide, , Or Cyclopropane, H, , H, C, , H, C, , C, , H, , H, H, , Cyclopropane contains three C ⎯ C single bond and six C ⎯ H, single bond., Total 9 single bonds are present in cyclopropane., 5. Differences between pollen tube and style are as follows, , Pollen Tube, , Style, , A tube growing out of pollen The middle elongated, grain when it reaches stigma. part of the carpel, i.e., female part of a flower., The attachment of stigma, to the ovary., , It transports male gametes, from pollen grains to ovules., , 6. Magnetic field lines due to a solenoid, B, A, C, I, , I

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129, , CBSE Term II Science X, , In case of an ideal solenoid, magnetic field strength is, maximum at point A and is minimum or zero at point B. This, is because the magnetic field is strong, where magnetic field, lines are crowded and is weak, where magnetic field lines, are far apart. At the point C, the density of the field lines is, less than that of point A but greater than that of point B. So,, the order of magnetic field at points A , B and C is, BB < BC < BA, Or From diagram, we can see that current is entering from A, and emerging out from B., , A, , B, , Thus, using right hand thumb rule, direction of magnetic, field lines is from B to A. We know that, magnetic field lines, move from North to South direction. Thus, B represents, North pole or A represents South pole., 7. (i) Carbon dioxide (CO2), , 9., , (i) Parallel connection is a better way to connect lights and, other appliances in domestic circuit., It is because, (a) when we connect a number of devices in parallel, , combination, each device gets the same potential, as provided by the battery and it keeps on, working even, if other devices stop working., (b) parallel connection is helpful when each device, has different resistances and requires different, current for its operation as in this case the, current divides itself through different devices, unlike series connection., (ii) Electrician has made series connection of all the lamps, in electric circuit of house because of which, if one, lamp gets fused, all the other lamps stop working., This is due to the fact that when devices are connected, in series, then if one device fails, the circuit gets broken, and all the devices in that circuit stop working., Or (i) Let us assume that the resistance of each bulb be R., The circuit diagram in two cases may be drawn as given, below, B1, , ••, , (ii) Ammonia molecule (NH3 ) H — N— H, ⏐, H, (iii) Ammonium chloride (NH4Cl) contains both ionic and, covalent bonds., , A1, , A, , B1, +, , B2, –, , B2, , B3, , A2, B3, , K, , A3, K, , V, , +, , H, ⎡, ⎤, |, ⎢, ⎥, −, ⎢ H ⎯ N ⎯ H⎥ Cl, |, ⎢, ⎥, ⎢⎣, ⎥⎦, H, Or A series of similarly constituted compounds in which the, members present have the same functional group and, similar chemical properties and any two successive members, in a particular series differ in their molecular formula by, ( ⎯ CH2 ) unit, is called a homologous series, e.g. alkane, series Cn H2n + 2 ., CH4, , Methane, , C 2H6, , Ethane, , C 3H8, , Propane, , C 4H 10, , Butane, , C5 H 12, , Pentane, , With increase in the molecular mass, a gradual change in, their physical properties is seen. e.g. The melting and, boiling points increase with increasing molecular mass., 8. Differences between acquired and inherited traits are as, follows, Acquired Characters, , Inherited Characters, , They develop in the organism, during their lifetime., , They are received by the, organisms from their parents., , They do not bring any change, in the genes of organisms., , They bring about certain changes, in the genes of the organisms., , They are lost with the death of They are transferred to the next, the individual, e.g. intelligence. generation, e.g. free and fused, earlobes., , +, , –, , Equivalent resistance in series combination, R S = R + R + R = 3R , voltage = V, Let current through each bulb in series combination be I1 ., V, By Ohm’s law, V = I1 × 3R ⇒ I1 =, 3R, ∴Power consumption of each bulb in series combination,, 2, , V2, V2, ⎛ V⎞, ...(i), P1 = I12 ( 3R ) = ⎜ ⎟ × 3R = 2 × 3R =, ⎝ 3R ⎠, 3R, 9R, For parallel circuit, the resistance of each bulb = R, Voltage across each bulb = V, [Q same voltage in parallel combination], ∴ Power consumption of each bulb in parallel combination, is given by, V2, …(ii), P2 =, R, From Eqs. (i) and (ii), we get, ( V2 /R ), P2, =, P1 ( V 2 / 3R ), 2, V, 3R, × 2 =3, ⇒, P2 = 3P1, ⇒, R, V, Therefore, each bulb in parallel combination glows 3 times, brighter than that of each bulb in series combination., (ii) When one bulb gets fused then in series combination, the, circuit gets broken and current stops flowing, whereas in, parallel combination, same voltage continues to act on the, remaining bulbs and hence other bulbs continues to glow, with same brightness.

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130, , CBSE Term II Science X, , 10. Differences between monohybrid and dihybrid crosses are, as follows, Monohybrid cross between homozygous tall plant and, homozygous short plant is shown below, Monohybrid cross, , A hybridisation cross in which, inheritance of only one pair of, contrasting characters is studied., , Dihybrid cross, , A cross in which inheritance of two, pairs of contrasting characters is, simultaneously studied., , Tall, Short, (TT) × (tt), , All tall offsprings, (Tt), , P×P, , Tall, Tall, (Tt) × (Tt), , F1, , Tall, (TT), , F1, , Tall, (Tt), , Tall, (Tt), , Short, (tt) F2, , Or, Regeneration is used in relation to reproduction because, reproduction is the process by which a living organism is able, to produce new individuals of its own kind likewise, regeneration is the ability of some organisms to give rise to, new organisms when the individual is cut or broken up into, many pieces., It is seen in Hydra and Planaria., Regeneration in Multicellular Organism like Hydra, (i) It is carried out by specialised cells., (ii) When Hydra is cut or broken up into many pieces, these, specialised cells proliferate and make large number of, cells., (iii) From this mass of cells, different cells undergo changes, to become various cell types and tissues., (iv) These changes take place in an organised sequence, referred to as development thereby making each piece, to grow into a separate individual., 11. (i) When vas deferens in males are blocked surgically,, sperm transfer is be prevented. Similarly, when, Fallopian tubes are blocked in females the egg will, not be able to reach the uterus therby preventing, pregnancy., (ii) (a) The process in the figure depicts binary fission in, Amoeba, a method of asexual reproduction., (b) Binary fission also occurs in Euglena and Paramecium,, etc., (c) Differences between fission and fragmentation are as, follows, Fission, , Dihybrid cross between pure breed of plants having Round, and Green seed and Wrinkled and Yellow seeds is shown, below, Parents, RRyy, rrYY, ×, (Round green), (Wrinkled yellow), , ↓, , Ry, , rY, , ↓, , ↓, , Gametes, , ↓, F1-generation, RrYy, (Round yellow), ×, F1, , F1, , Or, , RY, , Ry, , rY, , ry, , RRYY, , RRYy, , RrYY, , RrYy, , RRYy, , RRyy, , RrYy, , Rryy, , RrYY, , RrYy, , rrYY, , rrYy, , RrYy, , Rryy, , rrYy, , rryy, , F2, RY, , Ry, , rY, , ry, , F2 -generation 315 round yellow, , Ratio, 9, , 108 round green, , 3, , 101 wrinkled yellow, , 3, , 32 wrinkled green, 556 seeds, , 1, 16, , Fragmentation, , It is the division of parent, body into two identical, daughter cells., , It is the division of, parent body into two, or more small fragments., , It occurs in unicellular, organisms or multicellular, organisms with simple body, organisation., , It occurs only in, multicellular organisms, with complex cellular, organisations., , e.g. Amoeba, Plasmodium, (protozoan), , e.g. Spirogyra (algae), , (i) Biodegradable materials are broken down by, microorganisms in nature into simple harmless, substances. Non-biodegradable materials need a, different treatment like heat and temperature for, disposal and hence, both should be discarded in two, different dustbins., (ii) Artificial ecosystems are those ecosystems which are, modified and managed by human beings., Crop fields are man-made. Here plants do not grow, naturally rather most of the plants are grown by humans, according to the season, type of soil, etc., In crop fields, the land is managed, soil is prepared, for sowing seeds, then irrigated and further, progress is also kept under observation for getting good, yield., This is why, crop fields are known as artificial ecosystem.

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131, , CBSE Term II Science X, , Science, , Class 10th ( Term II), , Practice Paper 2, , *, , (UnSolved), Time : 2 Hours, Max. Marks : 40, , General Instructions, , 1. There are 11 questions in the question paper. All questions are compulsory., 2. Section A, Question no. 1 to 3 is a Case Based Questions, which has four MCQs/Questions. Each question carries one mark., 3. Section B, Question no. 4 to 8 are Short Answer Type Questions. Each question carries 2/3 marks., 4. Section C, Question no. 9-11 are Long Answer Type Questions. Each question carries 5 marks., 5. There is no overall choice. However, internal choices have been provided in some questions. Students have to attempt, only one of the alternatives in such questions., , * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this, paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised, not to consider the pattern of this paper as official, it is just for practice purpose., , Section A, 1. Read the following and answer the questions from (i) to (iv) given below., After the discovery of large number of elements it became necessary to classify them and arrange them in a, regular manner in order of their periodic properties. In order to study the properties of all these elements, separately, scientists felt the necessity to group elements having similar characteristics together. So, all the, elements have been divided into few groups in such a way that elements in the same group have similar, properties of elements., In 1817, Johann Wolfgang Dobereiner’ triad to arrange the elements with similar properties into groups. He, identified some groups fo three elements having similar physical and chemical properties, known as, Dobereiner’s triads., In 1865, John Newlands arranged all known elements in the order of increasing atomic masses and found, that the properties of every eighth element are similar to the properties of the first element., (i) If Cl, Br, I is a Dobereiner’s triad and the atomic masses of Cl and I are 35.5 and 127 respectively, then the, atomic mass of Br is, (a) 162.5, , (b) 91.5, , (c) 81.25, , (d) 45.625, , (c) Li, K, Na, , (d) K, Al, Ca, , (ii) Example of Dobereiner’s triad is, (a) Li, Al, Ca, , (b) Li, Na, K, , (iii) A and B are two elements having similar properties which obey Newlands’ law of octaves. How many elements, are there in between A and B ?, (a) 7, , (b) 8, , (c) 5, , (d) 6, , (iv) According to the Newland’s law of octaves, the properties of magnesium are similar to those of, (a) beryllium, , (b) lithium, , (c) sodium, , Or, On what basis the elements are arranged in Dobereiner’s triad?, (a) Atomic number, (c) Number of neutrons, , (b) Atomic mass, (d) Number of electrons, , (d) potassium

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132, , CBSE Term II Science X, , 2. Read the following and answer the questions from, (i) to (iv) given below., Devices in series and parallel., You must have seen tiny bulbs strung together for, decorating buildings during festivals like Diwali,, and occasions like marriages, etc. These bulbs are, connected in series and the mains voltage is applied, to the combination. (as shown in figure), I, , R1, , R2, , R3, , (b), , (a), , (c), , (d), , R4, , + V–, AC mains, , The potential difference ( V ) of the mains gets, divided across the bulbs ( V = V1 + V2 + V3 + K ), So, a small potential difference exists across each, bulb, close to that required to make the bulb work., (i) The equivalent resistor of series combination of, four equal resistors is S. If they are joined in, parallel, the total resistance is P. The relation, between S and P is given by S = nP, then the, minimum possible value of n is (as shown in, figure), R, , Or Which of the following option(s) is/are correct for, the given statements about series combination?, (i) Current flowing in series combination is same in, all the resistors of the circuit., (ii) Potential difference across each resistor in the, circuit is same., (iii) Equivalent resistance of the circuit is the product, of all the resistances., (a) Only (i), (c) Only (i) and (ii), , (b) Only (ii), (d) Only (i), (ii) and (iii), , 3. Read the following and answer the questions from, (i) to (iv) given below., The amount of ozone in the atmosphere began to, drop sharply in the 1980s. This decrease has been, linked to synthetic chemicals which are used as, refrigerants and in fire extinguishers., , R, R, , R, , R, , R, R, , R, Parallel, , Series, , (a) 12, , Solar radiation, , (b) 14, , (c) 16, , (d) 10, , (ii) In a series combination when high resistors are, connected with each other then, the value of, current for the potential difference V will be, (a) increased, (b) decreased, (c) increased first then decreased, (d) None of the above, , Ozone Layer, , Harmful UV rays, CFCs etc, , (iii) In the circuit shows below, the value of current I, flowing will be, , 1!, , 2V, , 2!, , 1!, , 1, (a) A, 2, , 2, (b)1 A, 3, , (c) 2A, , (d), , 3, A, 2, , (iv) Which arrangement of 3Ω resistors will give a total, resistance of 7Ω?, , (i) Which layer of the atmosphere consists of ozone, layer?, (ii) Why ozone is important for us?, (iii) Give one reason for damaging ozone layer., (iv) What are the effects due to depletion of ozone, layer?, Or, , Where was the first ozone hole discovered?

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133, , CBSE Term II Science X, , 7. It is the responsibility of the government to arrange, , Section B, 4. ‘Males are heterogametic’ Explain., [2 M], 5. What are the advantages of connecting electrical, appliances in parallel with the battery instead of, [2 M], connecting them in series?, Or, I, , 8. A magnetic compass needle is placed in, , A, , 12 V, , the plane of paper near point A as, shown in the figure. In which plane, A, should a straight current carrying, S, conductor be placed, so that it passes, through A and there is no change in the, deflection of the compass? Under what condition is, [3 M], the deflection maximum and why?, , R2, , R1=4 Ω, , B, , A student has two resistors 2 Ω and 3 Ω. She has to, put one of them in place of R 2 as shown in the, circuit. The current that she needs in the entire, circuit is exactly 9 A. Show by calculation which of, [2 M], the two resistors she should choose., , 6. An element X belongs to 3rd period and group 16 of, the modern periodic table., (i) Determine the number of valence electrons and, the valency of X., (ii) Molecular formula of the compound when X, reacts with hydrogen and write its electron dot, structure., (iii) Name the element X and state whether it is, [3 M], metallic or non-metallic., Or Consider the part of periodic table given below and, answer the following questions., Group →, Period↓, , 1, , 1, , a, , 2, , b, , 3, , c, , 4, , d, , for the management and disposal of waste. As an, individual you have no role to play. Do you agree., [3 M], Support your answers with two reasons., Or “The chromosome number of the sexually, producing parents and their offsprings is the same”., [3 M], Justify this statement., , 2, , 13, , 14, , 15, , 16, , 17, , 18, , g, , h, , k, , i, , l, , j, e, f, , (i) The atom of which element is smaller in size e or h?, (ii) Which element is the most electropositive in, nature?, (iii) Which element has only one proton in its atom?, (iv) What is the valency of g?, (v) How many valence electrons does g have?, (vi) Name the element which is a metalloid., [3 M], , Section C, 9. Define all the events in correct sequence that lead, to pregnancy in a female., Or, What is meant by the word contraception? Discuss, the types of surgical method of contraception., , 10. (i) Describe an activity to obtain magnetic field line, around current carrying straight conductor., (ii) State the rule used to find the direction of this, magnetic field., (iii) How does magnitude of magnetic field depend on, current through a conductor?, , Or (i) Prachi draws magnetic field lines of field close to, the axis of a current carrying circular loop. As she, moves away from the centre of the circular loop., As she moves away from centre of the circular, loop, she observes that the lines keep on, diverging. How will you explain her observation?, (ii) How will you use a solenoid to magnetise a steel, bar?, , 11. (i) What are hydrocarbons? Give examples., (ii) Give the structural differences between saturated, and unsaturated hydrocarbons with two examples, each., (iii) What is functional group? Give examples of four, different functional groups., , Or Define structural isomer and draw the isomeric, structures of butane. Compare the structure of, benzene and cyclohexane by drawing them., , Answers, 1. (i) - (c); (ii) - (b); (iii) - (d); (iv) - (a, b), , 2. (i) - (c); (ii) - (b); (iii) - (c); (iv) - (d, a)

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Science, Class 10th ( Term II), , Practice Paper 3, , *, , (UnSolved), Time : 2 Hours, , General Instructions, , Max. Marks : 40, 1. There are 11 questions in the question paper. All questions are compulsory., 2. Section A, Question no. 1 to 3 is a Case Based Questions, which has four MCQs/Questions. Each question carries one mark., 3. Section B, Question no. 4 to 8 are Short Answer Type Questions. Each question carries 2/3 marks., 4. Section C, Question no. 9-11 are Long Answer Type Questions. Each question carries 5 marks., 5. There is no overall choice. However, internal choices have been provided in some questions. Students have to attempt, only one of the alternatives in such questions., * As exact Blue-print and Pattern for CBSE Term II exams is not released yet. So the pattern of this, paper is designed by the author on the basis of trend of past CBSE Papers. Students are advised, not to consider the pattern of this paper as official, it is just for practice purpose., , Section A, 1. Read the following and answer the questions from (i) to (iv) given below, , LH, , FSH, , Ovarian, homone levels, , Pituitary, homone levels, , Menstrual cycle is the cycle of events taking place in female reproductive organs, under the control of sex, hormones, in every 28 days., Diagrammatic representation of various events occurring during menstrual cycle is shown below, , Oestrogen, Progesterone, 1, , 3, , 5, , Menstruation, (1-4 days), , 7, , 9, , 11, , 13 15 17 19 21 23 25 27 29/1, Luteal phase Next cycle, (Secretory phase) begins, , Follicular phase, (Proliferative phase), (5-14 days), , (15-28 days), , (i) What changes take place in the uterus of a menstruating female?, (a) Release of egg into uterus, (c) Breakdown of uterine lining, , (b) Fusion of male and female gametes, (d) Building up of uterine lining\, , (ii) Zygote differs from embryo in a manner that, (a) it is larger in size, (c) it is multicellular, , (b) it is single-celled, (d) it is obtained as a result of fusion, , (iii) The average duration of human pregnancy is, (a) 7 months, , (b) 6 months, , (c) 12 months, , (d) 9 months

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135, , CBSE Term II Science X, , (iv) The release of eggs from the ovary of a human female is, called, (a) ovulation, (c) pregnancy, , (b) menstruation, (d) fertilisation, , Or Uterus is also called, (a) ureter, (c) womb, , (b) fallopian tube, (d) ovary, , 2. Read the following and answer the questions from (i) to, (iv) given below, Food, clothes, medicines, books, or many of the things, are all based on this versatile element carbon. In, addition, all living structures are carbon based. The, earth’s crust has only 0.02% carbon in the form of, minerals. The element carbon occurs in different forms, in nature will widely varying physical properties. Both, diamond and graphite are formed by carbon atoms, the, difference lies in the manner in which the carbon atoms, are bonded to one another. Carbon has the unique ability, to form bonds with other atoms of carbon, giving rise to, large molecules. This property is called catenation., (i) The various layers of carbon atoms in graphite are held, together by ......... force?, (a) H-bonding, (c) London, , (b) weak van der Waals’, (d) electrostatic, , (ii) Which of the following statements is not correct?, (a) Graphite is much less dense than diamond, (b) Graphite is black and soft, (c) Graphite has low melting point, (d) Graphite feels smooth and slippery, , (iii) Which of the following are isomers?, (a) Butane and isobutene, (c) Propane and propyne, , (b) Ethane and ethene, (d) Butane and isobutane, , (iv) Which one of the following is not an allotrope of carbon?, (a) Soot, (c) Diamond, , (b) Graphite, (d) Carborundum, , Or, Propane has the molecular formula C 3 H 8 . It has, (a) 10 covalent bonds, (c) 8 covalent bonds, , (b) 12 covalent bonds, (d) 3 covalent bonds, , 3. Read the following and answer the questions from (i) to, (iv) given below, , A galvanometer pointer gives a momentary, deflection to the right of the zero position, when it is connected to a small coil and a, magnet is allowed to fall towards the coil., (i) Where will the galvanometer pointer move, if, the magnet moves through the coil and it, falls away from the coil?, (ii) Predict the direction of induced current, using Fleming’s right hand rule., (iii) What is the magnitude of the induced, current in the circular loop KLMN of radius, r, if the straight wire PQ carries a steady, current of magnitude 1 ampere?, K, , N, , L, M, , P, , Q, , (iv) How can you show that the magnetic field, produced by a given electric current in the, wire decreases as the distance from the wire, increases?, , Or, State the phenomenon of “electromagnetic, induction”., , Section B, 4. A plant is heterozygous for a pair of alleles., This plant is self-pollinated and the resulting, seeds are germinated and allow to grow., Write the ratios of phenotypes and genotypes, [2 M], of expected offsprings., , 5. Draw the electron dot structure of, (a) H2 S molecule, (b) Ethene, , [2 M], , Or, Write any two difference between covalent, [2 M], and ionic compounds., , 6. What is meant by pollination? Differentiate, , S, , between the two modes of pollination in, flowering plants., , N, Galvanometer, , [3 M], , 7. What is meant by non-biodegradable waste?, Why is the government stressing more on, the use of jute/paper bags instead of plastic, [3 M], bags?

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136, , CBSE Term II Science X, , Or, The sperms are tiny bodies that consist of mainly, genetic material and a long tail, (i) Where are the sperms produced?, (ii) What is the role of the long tail?, (iii) How are the sperms delivered from the site of, their production?, [3 M], , 8. An electric iron consumes energy at a rate of 840, W when heating is at the maximum rate and 360, W when the heating is at the minimum rate. The, applied voltage is 220 V., What is the value of current and the resistance, [3 M], in each case?, Or, Derive an expression for equivalent resistance in, the following case, R1, , R2, , B, R4, , R5, , Decide which resistances are in series and, parallel. Solve for series and then for parallel., Combine both the results to get the equivalent, [3 M], resistance., , Section C, 9. (i) State any three characteristics of sexual, reproduction., (ii) Explain what happens to the egg, once it gets, fertilised in human female., , 1. (i) - (c); (ii) - (b); (iii) - (d); (iv) - (a, c), , X, , 10. A small valued resistance XY is, connected across the ends of a, coil. Predict the direction of, induced current in the resistance, XY, when, , S, , Y, , A, , B, , (i) South pole of a magnet moves towards end A of coil., (ii) South pole of magnet moves away from end A of the, coil., , Or A coil made of insulated copper wire is connected to a, galvanometer. What will happen to the deflection of, the galvanometer if this coil is moved towards a, stationary bar magnet and then moved away from it?, Give reason for your answer and name the, phenomenon involved., , 11. Consider two elements A (atomic number =17) and B, , R3, , A, , Or, What harmful effects do agricultural practices have on, the environment?, , (atomic number =19)., (i) Write the positions of these elements in the modern, periodic table giving justification., (ii) Write the formula of the compound formed when A, combines with B., (iii) Draw the electron dot structure of the compound and, state the nature of the bond formed between the two, elements., Or (i) How is the valency of an element determine if its, electronic configuration is known? Determine the, valency of an element of atomic number 9., (ii) Which one of the above elements belonging to the, fourth period has bigger atomic radius and why?, , Answers, , 2. (i) - (b); (ii) - (c); (iii) - (d); (iv) - (d, a)