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CBSE Class 10 Science 2026 — Paper 31/1/3

80 marks · 3 hours · 39 questions (45 with the OR choices)

33 of its 45 questions also appear in 31/1/1 or 31/1/2; 12 are only in this paper.

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Questions 1–9 · 1 mark each

  1. Q1 1 mark

    Also in 31/1/1 (Q3), 31/1/2 (Q6)

    Identify the correct statement for spirogyra, leishmania and hydra :
    1. (a)they reproduce sexually.
    2. (b)they are unicellular.
    3. (c)they are multicellular.
    4. (d)they reproduce asexually.
    Show answer
    (d) they reproduce asexually.
    Step-by-step solution →
  2. Q2 1 mark

    Also in 31/1/1 (Q1), 31/1/2 (Q2)

    Which structure in a leaf is mainly responsible for gaseous exchange ?
    1. (a)Xylem
    2. (b)Stomata
    3. (c)Phloem
    4. (d)Cuticle
    Show answer
    (b) Stomata
    Step-by-step solution →
  3. Q3 1 mark

    Also in 31/1/1 (Q5), 31/1/2 (Q7)

    Pancreas secretes pancreatic juice which contain certain enzyme that helps in digestion of food. Choose the correct option from the following :
    1. (a)Trypsin digests emulsified fats and lipase digests proteins.
    2. (b)Trypsin digests proteins and lipase digests emulsified fats.
    3. (c)Trypsin and lipase both digests fats.
    4. (d)Trypsin digests proteins and lipase digests carbohydrates.
    Show answer
    (b) Trypsin digests proteins and lipase digests emulsified fats.
    Step-by-step solution →
  4. Q4 1 mark
    From the given situations, identify ‘Chemotropic’ and ‘Geotropic’ movements in parts of plants, respectively :
    1. (i)Growth of pollen tube towards ovule.
    2. (ii)Movement of sunflower towards sunlight.
    3. (iii)Movement of root towards Earth/Gravity.
    4. (iv)Movement of leaves due to breeze. Choose the correct option :
    1. (a)(i) and (iii) respectively
    2. (b)(iii) and (i) respectively
    3. (c)(i), (ii) and (iii), (iv) respectively
    4. (d)(i), (iii) and (ii), (iv) respectively
    Show answer
    (a) (i) and (iii) respectively
    Step-by-step solution →
  5. Q5 1 mark

    Also in 31/1/1 (Q7), 31/1/2 (Q4)

    Which of the following group is not ‘biodegradable’ ?
    1. (a)Vegetable peels, dead leaves, paper
    2. (b)Cow dung, leather bag, water
    3. (c)Polythene bag, rubber band, ball pen
    4. (d)Paper, fruits, bones
    Show answer
    (c) Polythene bag, rubber band, ball pen
    Step-by-step solution →
  6. Q6 1 mark
    Study the given diagram of the heart, with blood vessels marked ‘1’ and ‘2’. From the following statements, establish the relationship between heart and /or the two blood vessels :
    1. (i)Blood vessel 1 - It carries carbon dioxide rich blood to the lungs.
    2. (ii)Blood vessel 2 - It carries oxygen rich blood from the lungs.
    3. (iii)Blood vessel 2 - Left atrium relaxes as it receives blood from this blood vessel.
    4. (iv)Blood vessel 1 - Right atrium has thick wall as it has to pump blood to this vessel. The option with correct statements is :
    1 2
    1. (a)(i) and (ii)
    2. (b)(ii) and (iii)
    3. (c)(ii), (iii) and (iv)
    4. (d)(i), (ii) and (iii)
    Show answer
    (d) (i), (ii) and (iii)
    Step-by-step solution →
  7. Q7 1 mark
    Plants use variety of techniques to get rid of their waste materials. Some are mentioned below. Identify the incorrect one :
    1. (a)Excess water is given out by transpiration.
    2. (b)Gums and Resins are wastes that are stored.
    3. (c)Roots secrete some wastes into the soil.
    4. (d)Flowers can store some waste products.
    Show answer
    (d) Flowers can store some waste products.
    Step-by-step solution →
  8. Q8 1 mark

    Also in 31/1/1 (Q8)

    Directions : Question numbers 8 and 9 are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below :

    Assertion (A): Reflex actions do not involve thinking.

    Reason (R): Most reflex actions are controlled by the spinal cord.

    1. (a)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    2. (b)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
    3. (c)Assertion (A) is true, but Reason (R) is false.
    4. (d)Assertion (A) is false, but Reason (R) is true.
    Show answer
    (a) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    Step-by-step solution →
  9. Q9 1 mark
    Directions : Question numbers 8 and 9 are Assertion and Reason based questions. Two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below :

    Assertion (A): Ozone at higher levels of atmosphere is a product of UV radiation acting on oxygen molecule.

    Reason (R): The higher energy of UV splits apart some molecular .

    1. (a)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    2. (b)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
    3. (c)Assertion (A) is true, but Reason (R) is false.
    4. (d)Assertion (A) is false, but Reason (R) is true.
    Show answer
    (a) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    Step-by-step solution →

Questions 10–12 · 2 marks each

  1. Q10 2 marks
    1. (a)Why is bile juice considered to be very important for digestion, even though it doesn’t contain any digestive enzymes ?
    2. (b)Name some substances present in initial filtrate and subsequently selectively reabsorbed in the tubules of nephron.
    Show answer

    Bile makes the food alkaline and emulsifies fats; glucose, amino acids, salts and water are selectively reabsorbed in the tubules.

    (a) Bile juice has no enzymes, but it is essential for digestion because it makes the acidic food coming from the stomach alkaline, so that the pancreatic and small-intestinal enzymes can act, and its bile salts emulsify fats, breaking large fat globules into small droplets and so increasing the efficiency of the fat-digesting enzyme (lipase). (b) Glucose, amino acids, salts and a major amount of water are present in the initial filtrate and are selectively reabsorbed in the tubules of the nephron.
    Step-by-step solution →
  2. Q11 2 marks

    Also in 31/1/1 (Q12)

    Draw a neat diagram to show germination of pollen on the female reproductive part of the flower. Name and label only the following parts :
    1. (a)The part that receives the pollen grain.
    2. (b)The structure that carries the male germ cell to reach the female germ cell.
    Show answer

    The stigma receives the pollen grain, and the pollen tube carries the male germ cell down to the female germ cell in the ovule.

    (a) Stigma: the sticky top of the pistil, which receives the pollen grain. (b) Pollen tube: it grows out of the germinating pollen grain, down through the style, to the ovule, and carries the male germ cell to the female germ cell. The diagram shows the pistil (stigma, style and ovary with an ovule) with pollen grains on the stigma and a pollen tube growing from one of them down to the ovule; only Stigma and Pollen tube are labelled.
    Step-by-step solution →
  3. Q12(a) 2 marks

    Also in 31/1/1 (Q11(a)), 31/1/2 (Q10(a))

    State two differences between the act of chewing food and salivation on sight of food.
    Show answer

    Chewing is voluntary and controlled by the fore-brain; salivation on sight of food is involuntary and controlled by the medulla.

    1. Chewing food is a voluntary action: we decide to chew and can start or stop it at will. Salivation on sight of food is an involuntary action: the mouth waters automatically and we cannot stop it by thinking. 2. Chewing is controlled by the fore-brain (cerebrum) and is carried out by the muscles of the jaw. Salivation is controlled by the medulla in the hind-brain and is carried out by the salivary glands.
    Step-by-step solution →
    OR
    Q12(b) 2 marks

    Also in 31/1/1 (Q11(b)), 31/1/2 (Q10(b))

    State two differences between pollination and fertilization.
    Show answer

    Pollination transfers pollen to the stigma, whereas fertilisation fuses the male and female germ cells; pollination occurs only in plants.

    1. Pollination is the transfer of pollen grains from the anther to the stigma of a flower, whereas fertilisation is the fusion of the male germ cell with the female germ cell (egg) to form a zygote. 2. Pollination occurs only in plants, whereas fertilisation occurs in both plants and animals. Another acceptable difference: pollination takes place on the stigma, outside the ovary, and transfer between flowers needs agents such as wind, water or insects, whereas fertilisation takes place inside the ovule, after the pollen tube has carried the male germ cell there, and needs no outside agent.
    Step-by-step solution →

Questions 13–14 · 3 marks each

  1. Q13 3 marks

    Also in 31/1/1 (Q14)

    Give differences between the following :
    1. (a)Nephron and neuron
    2. (b)Sensory nerve and motor nerve
    3. (c)Consumers and decomposers
    Show answer

    Nephron forms urine, neuron conducts impulses; sensory nerves carry impulses to the CNS, motor nerves from it; consumers eat, decomposers break down dead remains.

    (a) Nephron: the structural and functional unit of the kidney; it filters the blood and forms urine. Neuron: the structural and functional unit of the nervous system; it receives and conducts information as electrical impulses. (b) Sensory nerve: carries impulses from the receptors of the sense organs to the brain or spinal cord. Motor nerve: carries impulses from the brain or spinal cord to the effectors, that is, muscles and glands. (c) Consumers: get their food directly or indirectly from the producers by eating plants or other animals, and pass food energy along the food chain. Decomposers: micro-organisms such as bacteria and fungi that feed on dead remains and wastes, breaking them down into simple substances that return to the soil.
    Step-by-step solution →
  2. Q14 3 marks

    Also in 31/1/1 (Q13), 31/1/2 (Q14)

    Given below is a pyramid showing various trophic levels in an ecosystem :
    1. (a)From the organisms listed below, identify which one is to be placed at which trophic level ? Deer, Grass, Lion, Snake, Rabbit
    2. (b)Discuss the reason why primary consumers will have more energy as compared to secondary consumers ?
    3. (c)Why is the base of the pyramid broad ?
    (iv) (iii) (ii) (i) Tertiary consumers Secondary consumers Primary consumers Producers
    Show answer

    Grass is the producer, deer and rabbit are primary consumers, snake secondary, lion tertiary; each level passes on only about 10% of its energy.

    (a) (i) Producers: Grass; (ii) Primary consumers: Deer and Rabbit; (iii) Secondary consumers: Snake (the lion also feeds at this level when it eats deer); (iv) Tertiary consumers: Lion. (b) At each trophic level most of the food energy is used up in life processes and lost as heat; only about is passed on to the next level. Primary consumers get energy directly from producers, whereas secondary consumers get only about of the energy of the primary consumers, so primary consumers have more energy. (c) The base is formed by the producers, which trap solar energy directly and so hold the maximum energy (and, in a grassland, the largest number of organisms); because only about of the energy passes to each higher level, every level above receives less energy and supports less life, so the base is the broadest.
    Step-by-step solution →

Question 15 · 4 marks

  1. Q15 4 marks

    Also in 31/1/1 (Q15), 31/1/2 (Q15)

    Mendel took garden pea plants with different characteristics, such as height to study the inheritance pattern of factors (genes). He crossed tall pea plant with short pea plant and obtained all the tall plants in the generation. Answer the following questions :
    1. (a)Why only tall pea plants were observed in progeny ?[1]
    2. (b)By which method did Mendel obtain progeny ?[1]
    3. (c)(i)Write one difference between dominant and recessive trait.[2]
    4. (c)(ii)OR : Write two observations made by Mendel about progeny.[2]
    Show answer

    Tallness is dominant, so all are and tall; self-pollination gives in the ratio ; a recessive trait needs two factors.

    (a) Tallness is dominant over shortness. Every plant received from the tall parent and from the short parent, so all were ; in the dominant is expressed and is masked, so all were tall. (b) By self-pollination of the tall plants (), which gave tall and short plants in the ratio in the . (c)(i) A dominant trait is expressed even when only one factor for it is present (), so it appears in the ; a recessive trait is expressed only when both factors are recessive (), so it is hidden in the and reappears in the . OR (c)(ii) All the plants were tall, showing the trait of only one parent; and none was short or of medium height, so the parental traits did not blend.
    Step-by-step solution →

Question 16 · 5 marks

  1. Q16(a) 5 marks

    Also in 31/1/1 (Q16(a)), 31/1/2 (Q16(b))

    Given below are certain situations. Analyse each and describe its possible impact :
    1. (i)A population of bacteria living in temperate waters whose temperature increased by global warming.
    2. (ii)The sperm encounters the egg when it reaches the oviduct in human females.
    3. (iii)Self pollination does not occur in a flower that contains only pistil.
    4. (iv)Egg does not get fertilised in a human female.
    5. (v)When the seed is placed under appropriate condition of water and air in the soil ?
    Show answer

    Heat-resistant bacteria survive, fertilisation leads to pregnancy, a pistil-only flower needs cross-pollination, an unfertilised egg causes menstruation, and a seed germinates.

    (i) Most of the bacteria would die in the warmer water, but a few variants that are resistant to heat would survive and multiply, so the species survives; this shows the importance of variation. (ii) Fertilisation occurs in the oviduct and a zygote is formed; it divides, gets implanted in the thick lining of the uterus and develops into an embryo, nourished through the placenta, so the woman becomes pregnant. (iii) The flower is unisexual (female) and has no pollen of its own, so it can be pollinated only by cross-pollination, with pollen brought from another flower by wind, water or animals; without that pollen no seeds or fruit form. (iv) The egg lives for about a day; the thick, blood-rich lining of the uterus is not needed, so it breaks down and comes out through the vagina as blood and mucus: menstruation, lasting about two to eight days. (v) The seed germinates: the embryo grows using food stored in the cotyledons, the radicle forms the root and the plumule the shoot, and a seedling develops.
    Step-by-step solution →
    OR
    Q16(b) 5 marks

    Also in 31/1/1 (Q16(b)), 31/1/2 (Q16(a))

    Given below are certain situations. Analyse and describe what would happen when :
    1. (i)Spores are liberated from blob-like structures of the bread mould?
    2. (ii)Leaves of bryophyllum fall on wet soil ?
    3. (iii)A pollen from different species land on the stigma of totally unrelated species ?
    4. (iv)Copper-T is placed in the uterus of a human female ?
    5. (v)Spirogyra breaks into smaller fragments upon maturation ?
    Show answer

    Spores, Bryophyllum buds and Spirogyra fragments each grow into new individuals; unrelated pollen fails to fertilise, and Copper-T prevents pregnancy.

    (i) The blob-like structures are sporangia; the spores released from them are carried by air, and on reaching a moist surface such as bread each spore germinates into a new bread mould (spore formation). (ii) Buds in the notches along the leaf margin develop roots and shoots and grow into new Bryophyllum plants (vegetative propagation). (iii) The pollen does not germinate, or its pollen tube does not grow down to the ovule, on the stigma of an unrelated species, so fertilisation does not occur and no seed forms. (iv) The copper-T, a contraceptive device, prevents pregnancy; it may cause side effects due to irritation of the uterus. (v) Each fragment grows into a new Spirogyra filament, a new individual (fragmentation).
    Step-by-step solution →

Questions 17–24 · 1 mark each

  1. Q17 1 mark

    Also in 31/1/1 (Q19), 31/1/2 (Q21)

    The gases evolved on heating lead (II) nitrate crystals are :
    1. (a) and
    2. (b) and
    3. (c) and
    4. (d) and
    Show answer
    (d) and
    Step-by-step solution →
  2. Q18 1 mark

    Also in 31/1/1 (Q17), 31/1/2 (Q23)

    1. (i)
    2. (ii) Which of the following options clearly describes both the reactions ?
    1. (a)(i) is double displacement, (ii) is displacement reaction.
    2. (b)Both, (i) and (ii) are displacement reactions and precipitation reactions.
    3. (c)Both, (i) and (ii) are double displacement reactions and precipitation reactions.
    4. (d)(i) is displacement, (ii) is double displacement reaction.
    Show answer
    (c) Both, (i) and (ii) are double displacement reactions and precipitation reactions.
    Step-by-step solution →
  3. Q19 1 mark

    Also in 31/1/1 (Q18), 31/1/2 (Q22)

    Which one of the following can be used as an acid-base indicator by a visually impaired (blind) student ?
    1. (a)Turmeric
    2. (b)Vanilla essence
    3. (c)Methyl orange
    4. (d)Litmus
    Show answer
    (b) Vanilla essence
    Step-by-step solution →
  4. Q20 1 mark
    The hydrocarbons with general formula represents :
    1. (a)alkane
    2. (b)alkene
    3. (c)alkyne
    4. (d)cyclic compounds
    Show answer
    (b) alkene
    Step-by-step solution →
  5. Q21 1 mark

    Also in 31/1/1 (Q23), 31/1/2 (Q17)

    When an element ‘X’ reacts with water, it starts floating. Identify the element ‘X’ :
    1. (a)Potassium
    2. (b)Calcium
    3. (c)Sodium
    4. (d)Iron
    Show answer
    (b) Calcium
    Step-by-step solution →
  6. Q22 1 mark

    Also in 31/1/1 (Q21), 31/1/2 (Q19)

    Which of the following is a poor conductor of electricity ?
    1. (a)Pb
    2. (b)Cu
    3. (c)Ag
    4. (d)Al
    Show answer
    (a) Pb
    Step-by-step solution →
  7. Q23 1 mark

    Also in 31/1/1 (Q22), 31/1/2 (Q18)

    The natural sources of oxalic acid, lactic acid and methanoic acid respectively are :
    1. (a)tomato, curd, ant-sting
    2. (b)tomato, orange, nettle-sting
    3. (c)orange, milk, ant-sting
    4. (d)orange, sour milk, nettle-sting
    Show answer
    (a) tomato, curd, ant-sting
    Step-by-step solution →
  8. Q24 1 mark

    Also in 31/1/1 (Q24), 31/1/2 (Q24)

    Directions : For question number 24, two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to this question from the codes (a), (b), (c) and (d) as given below :

    Assertion (A): Carbon shares its valence electrons with other atoms of carbon or with atoms of other elements.

    Reason (R): The shared electrons belong to the outermost shells of both the atoms and lead to both atoms attaining the noble gas configuration.

    1. (a)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    2. (b)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
    3. (c)Assertion (A) is true, but Reason (R) is false.
    4. (d)Assertion (A) is false, but Reason (R) is true.
    Show answer
    (a) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    Step-by-step solution →

Question 25 · 2 marks

  1. Q25 2 marks
    Why does one feels pain and irritation when stung by honey-bee ? Rubbing of baking soda on the stung area gives relief. How ?
    Show answer

    The sting leaves methanoic acid, which causes the pain; baking soda, a mild base, neutralises the acid and gives relief.

    When a honey-bee stings, it leaves an acid (methanoic acid, ) in the skin, and this acid causes the pain and irritation. Baking soda (sodium hydrogencarbonate, ) is a mild base, so rubbing it on the stung area neutralises the acid and gives relief:
    Step-by-step solution →

Questions 26–27 · 3 marks each

  1. Q26 3 marks
    What happens when :
    1. (a)An iron nail is dipped in copper (II) sulphate solution ?
    2. (b)Potassium iodide solution is mixed with lead nitrate solution ?
    3. (c)Silver chloride is exposed to sunlight ? Write balanced chemical equations to support your answer.
    Show answer

    The iron nail gets a copper coating, lead iodide forms as a yellow precipitate, and white silver chloride turns grey in sunlight.

    (a) Iron displaces copper from copper(II) sulphate solution: the nail gets a reddish-brown coating of copper, and the blue colour of the solution fades to pale green (displacement reaction). (b) A yellow precipitate of lead iodide forms (double displacement, or precipitation, reaction). (c) White silver chloride turns grey as sunlight decomposes it into silver and chlorine gas (decomposition by light).
    Step-by-step solution →
  2. Q27(a) 3 marks

    Also in 31/1/1 (Q27(a)), 31/1/2 (Q26(a))

    Explain chlor-alkali process with chemical equation. Name the products formed at anode and cathode.
    Show answer

    Brine electrolysis gives chlorine at the anode, hydrogen at the cathode and sodium hydroxide near the cathode.

    The chlor-alkali process is the electrolysis of brine, a concentrated aqueous solution of sodium chloride. The electric current decomposes it into sodium hydroxide, chlorine and hydrogen; the name comes from chlorine ("chlor") and sodium hydroxide ("alkali"). Anode: chlorine gas, . Cathode: hydrogen gas, ; sodium hydroxide solution is formed near the cathode.
    Step-by-step solution →
    OR
    Q27(b) 3 marks

    Also in 31/1/1 (Q27(b)), 31/1/2 (Q26(b))

    Write the preparation of following compounds with balanced chemical equation :
    1. (i)Baking soda
    2. (ii)Bleaching powder
    3. (iii)Plaster of Paris
    Show answer

    Baking soda from brine, ammonia and carbon dioxide; bleaching powder from chlorine and dry slaked lime; Plaster of Paris by heating gypsum at 373 K.

    (i) Baking soda () is prepared by passing carbon dioxide and ammonia into brine: . (ii) Bleaching powder () is prepared by the action of chlorine on dry slaked lime: . (iii) Plaster of Paris is prepared by heating gypsum at K: .
    Step-by-step solution →

Question 28 · 4 marks

  1. Q28 4 marks

    Also in 31/1/1 (Q28), 31/1/2 (Q28)

    Read the following passage and answer the questions given below : Most of metals occur in combined state in form of ores. Carbonate ores are converted into oxides by calcination and sulphide ores by roasting. Oxides are reduced with suitable reducing agent like carbon to get free metal. Highly reactive metals like – Al, Mg are also used as reducing agents to obtain metal from their oxides. Most reactive metals are obtained by electrolytic reduction of their molten ores. Alloying is a very good method of improving the properties of a metal. We can get desired properties by this method. The electrical conductivity and melting point of an alloy is less than that of pure metals.
    1. (a)Why carbonate or sulphide ores are converted to oxides before extraction of metal from it ?[1]
    2. (b)Write a reaction in which Aluminium is used as a reducing agent to obtain metal from its oxide.[1]
    3. (c)(i)How is copper obtained from its ore () ? Give equations of the reactions.[2]
    4. (c)(ii)(I)OR : Why highly reactive metals cannot be obtained from their oxides by using carbon as a reducing agent ?[1]
    5. (c)(ii)(II)Why solder, an alloy of lead and tin, is used for welding electrical wires together ?[1]
    Show answer

    Oxides are easier to reduce; aluminium reduces iron oxide; copper sulphide is heated in air; carbon cannot reduce reactive metals; solder melts easily.

    (a) A metal is obtained much more easily from its oxide than from its carbonate or sulphide, because carbon readily reduces the oxide. So carbonate ores are calcined and sulphide ores roasted to the oxide first, e.g. and , followed by . (b) Thermit reaction: . (c)(i) Copper is low in the reactivity series, so is simply heated in air (self-reduction): , then . OR (c)(ii) (I) Highly reactive metals have a greater affinity for oxygen than carbon, so carbon cannot remove oxygen from their oxides; they are obtained by electrolytic reduction of their molten compounds. (II) Solder has a low melting point (lower than that of lead or of tin), so it melts easily and joins the wires without melting them, and the joint still conducts electricity.
    Step-by-step solution →

Question 29 · 5 marks

  1. Q29(a) 5 marks

    Also in 31/1/1 (Q29(a)), 31/1/2 (Q29(a))

    1. (i)Give reasons for the following : (I) Covalent compounds are poor conductor of electricity. (II) Soap does not form lather in hard water. (III) Carbon shows catenation but silicon does not.
    2. (ii)Write chemical equations for the following : (I) Oxidation of ethanol by acidified . (II) Hydrogenation of ethene.
    Show answer

    No free ions in covalent compounds; scum forms in hard water; strong C-C bonds give catenation; ethanol gives ethanoic acid, ethene gives ethane.

    (i) (I) In covalent compounds the atoms share electrons, so no ions are formed and there are no free charged particles to carry a current; hence they are poor conductors of electricity. (II) Hard water contains calcium and magnesium salts. Soap reacts with the and ions to form an insoluble precipitate (scum), so the soap is used up and no lather forms. (III) The carbon atom is small, so carbon–carbon bonds are very strong and stable and carbon atoms can join into long chains and rings. Silicon atoms are larger and silicon–silicon bonds are weak, so silicon forms only short, very reactive chains (up to seven or eight atoms). (ii) (I) (ethanol to ethanoic acid). (II) (ethene to ethane).
    Step-by-step solution →
    OR
    Q29(b) 5 marks

    Also in 31/1/1 (Q29(b)), 31/1/2 (Q29(b))

    Mohan heated ethanol with a compound ‘X’ in the presence of a few drops of conc. and observed a sweet smelling compound ‘Y’ is formed. When ‘Y’ is treated with sodium hydroxide it gives back ethanol and a compound ‘Z’.
    1. (i)Identify ‘X’, ‘Y’ and ‘Z’.
    2. (ii)Write the role of conc. in the reaction.
    3. (iii)Write the chemical equations involved and name the reactions.
    Show answer

    X is ethanoic acid, Y is ethyl ethanoate and Z is sodium ethanoate; concentrated sulphuric acid is a catalyst and a dehydrating agent.

    (i) X is ethanoic acid (acetic acid), ; Y is ethyl ethanoate (ethyl acetate), , a sweet-smelling ester; Z is sodium ethanoate (sodium acetate), . (ii) Concentrated acts as a catalyst for the esterification; as a dehydrating agent it also takes up the water formed. (iii) Esterification: . Saponification: .
    Step-by-step solution →

Questions 30–32 · 1 mark each

  1. Q30 1 mark

    Also in 31/1/1 (Q30), 31/1/2 (Q31)

    A convex lens of focal length cm, is forming a real image. If the size of image is same as the size of object, then position of object and position of image will be, respectively :
    1. (a) cm and cm from lens
    2. (b) cm and cm from lens
    3. (c) cm and cm from lens
    4. (d) cm and cm from lens
    Show answer
    (c) cm and cm from lens
    Step-by-step solution →
  2. Q31 1 mark

    Also in 31/1/1 (Q31), 31/1/2 (Q30)

    When you look at an object very close to your eyes, the :
    1. (a)Ciliary muscles of your eye contract and the eye lens becomes thick.
    2. (b)Ciliary muscles of your eye get relaxed and the eye lens becomes thick.
    3. (c)Ciliary muscles of your eye contract and the eye lens becomes thin.
    4. (d)Ciliary muscles of your eye get relaxed and the eye lens becomes thin.
    Show answer
    (a) Ciliary muscles of your eye contract and the eye lens becomes thick.
    Step-by-step solution →
  3. Q32 1 mark

    Also in 31/1/1 (Q32), 31/1/2 (Q32)

    Directions : For question number 32, two statements are given, one labelled as Assertion (A) and the other is labelled as Reason (R). Select the correct answer to this question from the codes (a), (b), (c) and (d) as given below :

    Assertion (A): When rays of white light pass through a prism, on emerging they give spectrum of seven colours.

    Reason (R): It is due to the scattering of light that red light bends minimum and violet light bends the maximum.

    1. (a)Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    2. (b)Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
    3. (c)Assertion (A) is true, but Reason (R) is false.
    4. (d)Assertion (A) is false, but Reason (R) is true.
    Show answer
    (c) Assertion (A) is true, but Reason (R) is false.
    Step-by-step solution →

Questions 33–34 · 2 marks each

  1. Q33 2 marks
    Draw the ray diagram for the image formation by a lens which shows a magnification of .
    Show answer

    A convex lens with the object midway between O and forms a virtual, erect image with .

    Use a convex lens with the object placed between its optical centre and its principal focus - exactly midway, at . In the ray diagram a ray from the top of the object parallel to the principal axis is refracted through , and a ray through passes undeviated, with arrowheads on the rays showing the direction of light; the refracted rays diverge, and their dashed backward extensions meet on the object's side, at . The image A′B′ formed there () is virtual, erect and twice the size of the object: .
    Step-by-step solution →
  2. Q34(a) 2 marks

    Also in 31/1/1 (Q34(a)), 31/1/2 (Q33(a))

    The resistance of a wire of 0.01 cm radius and 1.0 cm length is 7 Ω. Calculate its resistivity.
    Show answer

    The resistivity of the wire's material is .

    The resistivity of the material of the wire is m (that is, cm).
    Step-by-step solution →
    OR
    Q34(b) 2 marks

    Also in 31/1/1 (Q34(b)), 31/1/2 (Q33(b))

    An electric heater is rated 220 V; 11 A. Calculate the power consumed if the heater is operated at 200 V.
    Show answer

    W: the heater consumes 2 kW when operated at 200 V.

    The heater consumes W ( kW) at V. Its resistance is , so W.
    Step-by-step solution →

Questions 35–37 · 3 marks each

  1. Q35 3 marks

    Also in 31/1/1 (Q37), 31/1/2 (Q36)

    1. (a)The pattern of magnetic field due to a current carrying wire depends upon the shape made by that wire. Justify.
    2. (b)A current carrying straight wire AB is shown in the given diagram. Out of X, Y and Z on which point will the strength of magnetic field be maximum and why ?
    X Y Z A B Electric Current i
    Show answer

    Wire shape decides the field pattern; the field is maximum at X, the point nearest the wire.

    (a) Each small part of a current-carrying wire produces circular field lines around itself, so the overall field depends on how the wire is laid out. For a straight wire the field lines are concentric circles round the wire; for a circular loop they are small circles near the wire that become nearly straight lines at the centre, at right angles to the plane of the loop; for a solenoid they are parallel straight lines inside (a uniform field) and like those of a bar magnet outside. The same current thus gives different patterns for different shapes. (b) The field is maximum at X, because X is nearest the wire and the magnetic field of a straight current-carrying wire decreases as the distance from the wire increases (its field-line circles are closest together near the wire).
    Step-by-step solution →
  2. Q36 3 marks

    Also in 31/1/1 (Q36), 31/1/2 (Q35)

    1. (a)Describe an activity to show that a current carrying conductor, placed in an external magnetic field experiences a force.
    2. (b)Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally towards the front wall from the back wall, is deflected by a strong magnetic field to your right side. Find the direction of the magnetic field.
    Show answer

    A rod between magnet poles is displaced when current flows, showing a force; the electron-beam field is vertically downwards.

    (a) Suspend a small aluminium rod horizontally from a stand by two connecting wires so that it lies between the poles of a strong horseshoe magnet (north pole below, south pole above), and connect it in series with a battery, a key and a rheostat. When a current is passed, the rod is displaced sideways; when the direction of the current is reversed, or the poles of the magnet are interchanged, it is displaced in the opposite direction. This shows that a current-carrying conductor placed in a magnetic field experiences a force, whose direction depends on the directions of the current and the field (Fleming's left-hand rule). (b) The magnetic field is directed vertically downwards.
    Step-by-step solution →
  3. Q37 3 marks
    1. (a)What is hypermetropia ?
    2. (b)Write any one cause of hypermetropia.
    3. (c)With the help of a suitable ray diagram, explain how hypermetropia is corrected ?
    Show answer

    Hypermetropia: nearby objects look blurred (near point beyond 25 cm); eye-lens focal length too long is one cause; a convex lens corrects it.

    (a) Hypermetropia (far-sightedness) is the defect of vision in which a person can see distant objects clearly but cannot see nearby objects distinctly: the near point of the eye lies farther than the normal cm, because light from a nearby object is focused behind the retina. (b) Cause (any one): the focal length of the eye lens becomes too long (the eye lens is not converging enough), or the eyeball becomes too small. (c) It is corrected with spectacles having a convex (converging) lens of suitable power. The convex lens bends the rays from an object at N, cm away, so that they appear to come from the near point N′ of the defective eye; the eye can focus light from N′, so the image is formed on the retina. In the ray diagram, rays from N pass through the convex lens, their backward extensions meet at N′, and the eye lens brings them to a focus on the retina.
    Step-by-step solution →

Question 38 · 4 marks

  1. Q38 4 marks

    Also in 31/1/1 (Q38), 31/1/2 (Q38)

    Read the following passage and answer the questions given : Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.
    1. (a)A convex lens of focal length 20 cm is used to form an image. If an object is placed at 40 cm from the lens, what will be the position and nature of image ?[1]
    2. (b)Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.[1]
    3. (c) (i)A lens combination consists of a convex lens of focal length 30 cm and a concave lens of focal length 15 cm placed together. Find the equivalent focal length and power of this lens combination.[2]
    4. (c) (ii)OR : Two lenses are placed in contact. One is a concave lens with focal length 2 m and the other is a convex lens with focal length 1.5 m. What type of lens will the combination behave as (convex or concave) ? Give reason.[2]
    Show answer

    Image at cm is real, inverted, same size; a concave lens gives a virtual, diminished image; combination cm, D.

    (a) The image is formed at cm, that is cm on the other side of the lens (at ); it is real, inverted and of the same size as the object (). (b) The ray parallel to the axis diverges as if it came from and the ray through goes straight on; they meet (by backward extension) between and , so the image is formed between the optical centre and the focus on the object's side, nearer the lens than the object - virtual, erect and diminished, as shown in the ray diagram. (c) (i) , so the equivalent focal length is cm and the power is D; the combination acts as a concave (diverging) lens. OR (c) (ii) D (focal length m). The total power is positive, so the combination behaves as a convex (converging) lens, because the convex lens has the shorter focal length and hence the greater power.
    Step-by-step solution →

Question 39 · 5 marks

  1. Q39(a) 5 marks
    Study the given electric circuit in which 2 A electric current is flowing between points X and Y.
    1. (i)Using the battery, key, voltmeter and ammeter in this given electric circuit, redraw a circuit diagram in which and resistors are connected between X and Y in parallel combination.
    2. (ii)Circuit drawn by you, in which resistors are connected in parallel combination, calculate the electric current flowing through resistor.
    V A X 2 Ω 4 Ω Y − + 2A 2A 2A − +
    Show answer

    Connect the and resistors in parallel between X and Y; the resistor then carries A.

    (i) Both the and the resistor are connected between X and Y, side by side; the battery, the key and the ammeter stay in series in the main line (ammeter + towards the battery's + terminal), and the voltmeter is connected across X and Y (+ at Y). (ii) From the given series circuit the battery's potential difference is V. In the parallel circuit the resistor has the whole V across it, so the current through it is A. (The resistor carries A and the ammeter reads A.)
    Step-by-step solution →
    OR
    Q39(b) 5 marks
    1. (i)Two lamps ‘A’ and ‘B’ of rating 50 W; 220 V and 100 W, 220 V are connected in series combination. Find out the ratio of the resistances () of these lamps.
    2. (ii)Derive the expression for the equivalent resistance of three resistors , and connected in parallel combination.
    Show answer

    , and the equivalent resistance of three parallel resistors follows .

    (i) gives and , so (connecting the lamps in series does not change their resistances). (ii) Let , and be connected in parallel across a potential difference . Each resistor has the same across it, and the main current divides as . By Ohm's law , , and , so ; dividing by gives .
    Step-by-step solution →

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