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CBSE Class 10 Science 2026 — Paper 31/8/2/SB

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

28 of its 45 questions also appear in 31/8/1/SB, 31/8/3/SB and other papers; 17 are only in this paper.

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Questions 1–9 · 1 mark each

  1. Q1 1 mark

    Also in 31/8/1/SB (Q7), 31/8/3/SB (Q4)

    Two pea plants one with round green seeds (RRyy) and another with wrinkled yellow seeds (rrYY) were crossed. All the seeds of progeny have round, yellow seeds (RrYy). When plants are self-pollinated, the progeny will have new combination of characters.
    1. (i)Round, yellow
    2. (ii)Round, green
    3. (iii)Wrinkled, yellow
    4. (iv)Wrinkled, green Choose the option that shows new combination of characters in progeny.
    1. (a)(i) and (ii)
    2. (b)(i) and (iv)
    3. (c)(iii) and (iv)
    4. (d)(i) and (iii)
    Show answer
    (b) (i) and (iv)
    Step-by-step solution →
  2. Q2 1 mark

    Also in 31/8/1/SB (Q4), 31/8/3/SB (Q7)

    Examine the given food chains to establish the correct one in an ecosystem :
    1. (a)Seed mouse snake deer
    2. (b)Algae small fish large fish hawk
    3. (c)Grasshopper frog snake elephant
    4. (d)Plants rabbit deer wolf
    Show answer
    (b) Algae small fish large fish hawk
    Step-by-step solution →
  3. Q3 1 mark
    ‘The lymph is colourless tissue fluid.’ This indicates that it lacks ____(i)____ cells and hence cannot carry ____(ii)____ . From the following options, choose the one that correctly identifies
    1. (i)and
    2. (ii).
    1. (a)(i) – Platelets ; (ii) – excess fluid
    2. (b)(i) – Red blood ; (ii) – oxygen
    3. (c)(i) – Red blood ; (ii) – fats
    4. (d)(i) – White blood ; (ii) – salts
    Show answer
    (b) (i) – Red blood ; (ii) – oxygen
    Step-by-step solution →
  4. Q4 1 mark

    Also in 31/8/1/SB (Q2), 31/8/3/SB (Q5)

    Which of the following groups consists entirely of substances that are broken down by an enzyme, lipase ?
    1. (a)Carbohydrates, proteins and vitamins
    2. (b)Fats, oils and ghee
    3. (c)Minerals, starch and vitamins
    4. (d)Proteins, sugars and esters
    Show answer
    (b) Fats, oils and ghee
    Step-by-step solution →
  5. Q5 1 mark
    Select the incorrect statement from the following :
    1. (a)Iodine is necessary for production of thyroxin hormone by pituitary gland.
    2. (b)Thyroxin regulates metabolism of carbohydrates, fats and protein.
    3. (c)Swelling of neck may indicate goitre due to iodine deficiency.
    4. (d)Thyroid is located in the neck region in our body.
    Show answer
    (a) Iodine is necessary for production of thyroxin hormone by pituitary gland.
    Step-by-step solution →
  6. Q6 1 mark

    Also in 31/8/1/SB (Q1), 31/8/3/SB (Q3)

    Select the correct option that shows the path of flow of only the deoxygenated blood through the human circulatory system.
    1. (a)Body left atrium left ventricle lungs right atrium right ventricle body.
    2. (b)Right atrium right ventricle lungs left atrium left ventricle body
    3. (c)Body left atrium left ventricle lungs
    4. (d)Body right atrium right ventricle lungs
    Show answer
    (d) Body right atrium right ventricle lungs
    Step-by-step solution →
  7. Q7 1 mark

    Also in 31/8/1/SB (Q5), 31/8/3/SB (Q1)

    A man underwent MRI of brain which shows some defects in the cerebellum of his brain. Which activity of that person will be affected ?
    1. (a)Heart beat
    2. (b)Feeling of hunger and thirst
    3. (c)Reflexes
    4. (d)Walking in straight line
    Show answer
    (d) Walking in straight line
    Step-by-step solution →
  8. Q8 1 mark

    Also in 31/8/1/SB (Q9), 31/8/3/SB (Q8)

    Question No. 8 & 9 consists of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option from (a), (b), (c) and (d) given below :

    Assertion (A): The energy transferred to a lion when it eats a deer is only a small fraction of the deer’s total energy content.

    Reason (R): Energy transfer is efficient at all trophic levels with nearly 90% of energy passed on to the next level.

    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 →
  9. Q9 1 mark
    Question No. 8 & 9 consists of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option from (a), (b), (c) and (d) given below :

    Assertion (A): A reflex arc provides an immediate and automatic response to a stimulus.

    Reason (R): The nerve pathway in a reflex arc involves the brain for the fastest possible processing.

    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 10–12 · 2 marks each

  1. Q10(a) 2 marks
    Transport of water and minerals in plants takes place through xylem tissue where vessels and tracheids of roots, stems and leaves are interconnected. This forms a continuous system of channels reaching all plant parts. Simple physical forces are involved in the process. If the soil has adequate water, name and explain which force will be acting effectively at :
    1. (i)Night
    2. (ii)Day
    Show answer

    At night root pressure acts, with stomata closed; during the day transpiration pull acts, with stomata open.

    (i) Night — root pressure. With the stomata closed there is almost no evaporation from the leaves, so no suction is produced above. Meanwhile the root cells go on actively taking up ions from the soil and passing them into the xylem; this raises the concentration inside the root above that of the soil water, water enters the root by osmosis, and its continuous entry at the base pushes the column of water up the xylem. That push is root pressure. (ii) Day — transpiration pull. The stomata are open, water evaporates from the leaf cells and diffuses out, and each molecule lost is replaced from the xylem. This creates a suction at the top of the water column, and because the vessels and tracheids form one continuous channel holding an unbroken thread of water, the suction is transmitted all the way down and pulls water up from the roots. Transpiration pull is by far the stronger of the two and is the main force during the day.
    Step-by-step solution →
    OR
    Q10(b) 2 marks
    Although plants do not have excretory system still they are able to get rid-off their waste products. How ?
    Show answer

    Plants release gaseous wastes by diffusion, lose excess water by transpiration, store wastes, shed them in leaves, or release them into the soil.

    Plants do not need an excretory organ because every kind of waste they make already has a way out of the body, and their slow rate of metabolism means the waste is never produced faster than these routes can clear it. (i) Gaseous wastes leave by diffusion. Oxygen made in photosynthesis and carbon dioxide made in respiration pass out through the stomata of the leaves and the lenticels of the stem, and excess water is lost as water vapour through the stomata during transpiration. (ii) Wastes that cannot diffuse out are stored where they do no harm. Many are kept in the vacuoles of the cells, and others are laid down as resins and gums, chiefly in the old xylem, which is dead tissue. (iii) Some wastes are moved into leaves, and the leaves are then shed, so the waste goes with them. (iv) Some waste substances are released into the soil around the roots.
    Step-by-step solution →
  2. Q11 2 marks

    Also in 31/8/1/SB (Q12), 31/8/3/SB (Q10)

    The sensitive plant shows movement of leaves in response to touch. Explain such movement briefly.
    Show answer

    On touch, cells at the leaflet base lose water and shrink, so the leaflets fold; this nastic movement does not depend on growth.

    The folding of the leaves of the sensitive plant on being touched is a movement that is independent of growth, that is, a nastic movement. The stimulus is touch. The plant has no nervous tissue, so the information that a leaf has been touched is passed on from one cell to the next as an electrical and chemical signal until it reaches the cells at the base of the leaflets. Plants have no muscle cells; instead, their cells change shape by changing the amount of water in them. The cells in the swelling at the base of each leaflet lose water, become flaccid and shrink, so the leaflets fold up in pairs and the leaf droops. When these cells take up water again they swell, and the leaflets spread out as before, so the movement is reversible.
    Step-by-step solution →
  3. Q12 2 marks
    (A) Name the reproductive and non-reproductive parts of bread mould (Rhizopus). (B) Give any two advantages of sexual reproduction in plants.
    Show answer

    (A) Sporangia are reproductive, hyphae non-reproductive; (B) sexual reproduction gives variation and seeds that spread and survive.

    (A) The reproductive parts of bread mould are the tiny blob-on-a-stick structures: each is a sporangium — a knob-like case full of spores — borne on an erect stalk, and each spore it releases can grow into a new Rhizopus. The non-reproductive parts are the thread-like hyphae that make up the mycelium: the stolons that spread over the bread and the rhizoids that grow into it, which anchor the mould and absorb food from the bread. (B) Two advantages of sexual reproduction in plants are: (i) It produces variation. The offspring gets genes from two different parents, so it is not a copy of either; these variations help the species survive a changing environment and provide the raw material for evolution. (ii) It produces seeds. A seed carries the embryo with a store of food, can be dispersed by wind, water or animals so that new plants grow away from the parent, and can remain dormant through unfavourable conditions, then develop into a seedling when conditions are suitable.
    Step-by-step solution →

Questions 13–14 · 3 marks each

  1. Q13 3 marks

    Also in 31/8/1/SB (Q14), 31/8/3/SB (Q13)

    “Each organism has two sets of all genes, one inherited from each parent i.e. both parents contribute equally to the DNA of the progeny during sexual reproduction”. If such is the case, then how is the stability of DNA of the species ensured ? Discuss.
    Show answer

    Gametes get half the chromosomes (one set), and fertilisation restores the full number (), so the DNA of the species stays stable.

    If both parents really passed on the two sets of genes each of them carries, the progeny would have four sets, the next generation eight, and the amount of DNA would double in every generation. This does not happen, because sexual reproduction includes a halving step. Each body cell has two copies of each chromosome, one from the male parent and the other from the female parent. When germ cells are formed, each germ cell (gamete) takes only one chromosome from each pair, which may be of either maternal or paternal origin, so a gamete carries one set of genes instead of two and half the number of chromosomes; in human beings a body cell has 46 chromosomes (23 pairs) while a sperm or an egg has 23. When two germ cells unite at fertilisation, the zygote once again has two copies of each chromosome, 23 plus 23 giving 46, so the normal number of chromosomes is restored in the progeny. This re-establishment of the number of chromosomes, and so of the amount of DNA, in every generation ensures the stability of the DNA of the species.
    Step-by-step solution →
  2. Q14 3 marks
    Consider a simple aquatic ecosystem like a lake which consists of algae, small fish and large fishes. It was found that the lake water was contaminated with a low concentration of a non-biodegradable pesticide, DDT. Based on this ecosystem, answer the following questions :
    1. (a)In which organism will the concentration of the pesticide DDT be the highest ? Name and explain the phenomenon.
    2. (b)Why is the flow of energy in a food chain always unidirectional ?
    Show answer

    Large fishes have the most DDT through biological magnification; energy flow is one-way as it enters only at producers and is lost as heat.

    (a) The concentration of DDT will be highest in the large fishes. The lake's food chain is algae, then small fish, then large fish, and the large fish stand at the highest trophic level. The phenomenon is biological magnification (biomagnification): DDT is non-biodegradable, so it is neither broken down nor excreted and stays stored in the body. The algae absorb it from the water and retain it; each small fish eats a great many algae and keeps the DDT of all of them; each large fish eats many small fish and keeps all of theirs. Since the amount of pesticide taken in grows far faster than body mass, its concentration increases at every successive trophic level and is greatest at the top. (b) The flow of energy in a food chain is unidirectional for two reasons. First, the energy enters the chain only once — sunlight is captured only by the green plants, the producers — and thereafter it passes from producer to herbivore to carnivore and never back; a carnivore returns no energy to the herbivore it ate. Second, at each transfer only about ten per cent of the energy passes on, while about ninety per cent is spent in the life processes of that organism and escapes into the surroundings as heat. No organism can take that heat back and convert it into food, so the energy leaves the chain permanently at every step and cannot be reused or returned.
    Step-by-step solution →

Question 15 · 4 marks

  1. Q15 4 marks

    Also in 31/8/1/SB (Q15), 31/8/3/SB (Q15)

    The given diagram shows the human kidney and its associated structures. Study the Table I and Table II that shows the concentration of certain substances found within the structure P and Q.
    Table I (structure P) and Table II (structure Q) - concentration of components, in per cent
    ComponentsTable I - Structure P, concentration %Table II - Structure Q, concentration %
    Urea0.051.00
    Amino acids0.060.00
    Salts1.721.60
    Glucose0.100.00
    P Q
    1. (A)Where are kidneys located in the human body ? Name the filtration unit of them.[1]
    2. (B)List two factors on which reabsorption of water from the filtrate depends.[1]
    3. (C)Why is there variation in the concentration of given substance in structure P and Q listed in table I and II respectively ?[2]
    4. (C)OR : If a person is suffering from diabetes, concentration of which substance present in structure Q (listed in table II) will be modified and why ?[2]
    Show answer

    Kidneys lie in the abdomen beside the backbone, with the nephron as filtration unit; filtration then selective reabsorption changes blood into urine.

    (A) The kidneys are a pair of bean-shaped organs lying in the abdomen, one on either side of the backbone, towards the back of the abdominal cavity. Their filtration unit is the nephron. (B) The amount of water reabsorbed from the filtrate depends on (i) how much excess water there is in the body, and (ii) how much dissolved waste there is to be excreted. (C) P is the blood vessel carrying blood into the kidney, so Table I is the composition of blood, and Q is the ureter carrying urine away, so Table II is the composition of urine. In the nephron the blood is first filtered at the glomerulus, so water and all the small dissolved substances enter the tubule, and the filtrate is then selectively reabsorbed as it flows along the tubule. All the glucose and all the amino acids are useful and are taken back completely, so they fall from 0.10% and 0.06% in P to 0.00% in Q. Most of the salts are taken back along with most of the water, so their concentration hardly changes, 1.72% to 1.60%. Urea is a nitrogenous waste and is not reabsorbed, while the water around it is, so it is left behind and becomes concentrated, rising from 0.05% in P to 1.00% in Q. OR (C) The concentration of glucose in structure Q will be modified: there will be more glucose in Q. In a healthy person it is 0.00% because all the filtered glucose is reabsorbed by the tubule. A person suffering from diabetes does not have sufficient insulin, so the glucose level of the blood rises; more glucose is then filtered than the tubule is able to reabsorb, the excess is not taken back and passes out in the urine, so glucose now appears in structure Q at a value above 0.00%.
    Step-by-step solution →

Question 16 · 5 marks

  1. Q16(a) 5 marks

    Also in 31/8/1/SB (Q16(b)), 31/8/3/SB (Q16(b))

    1. (i)Name the special tissue inside the uterus that provides nutrition to the developing embryo. Describe its structure. State its two functions besides nutrition.
    2. (ii)Write any two sexually transmitted diseases in humans and their causative microbial agents.
    Show answer

    The placenta nourishes the embryo, exchanges gases and removes wastes; gonorrhoea (bacterium) and AIDS (virus) are two sexually transmitted diseases.

    (i) The tissue is the placenta. It is a disc-shaped tissue embedded in the wall of the uterus; villi grow out from the embryo's side of the disc into blood spaces (sinuses) filled with the mother's blood, which gives a very large surface for exchange, and the embryo is joined to the disc by the umbilical cord. Two functions besides nutrition: it passes oxygen from the mother's blood to the embryo and carries carbon dioxide back; and it removes the embryo's nitrogenous wastes, chiefly urea, into the mother's blood. (The placenta also secretes hormones that help to maintain the pregnancy.) (ii) Gonorrhoea is caused by the bacterium Neisseria gonorrhoeae, and AIDS is caused by a virus, the Human Immunodeficiency Virus (HIV). Syphilis (bacterium Treponema pallidum) and warts (human papilloma virus) are equally acceptable pairs.
    Step-by-step solution →
    OR
    Q16(b) 5 marks

    Also in 31/8/1/SB (Q16(a)), 31/8/3/SB (Q16(a))

    In flowering plants, the process of pollination leads to fertilization.
    1. (i)Explain with the help of a well-labelled diagram how the male germ cell from the pollen grain reaches the female germ cell located in the ovule.
    2. (ii)Differentiate between self-pollination and cross-pollination.
    3. (iii)Identify and name the parts ‘R’ and ‘S’ in the given diagram of seed.
    R S
    Show answer

    Pollen tube carries the male germ cell to the egg; self-pollination is within a flower, cross-pollination between plants; R is plumule, S radicle.

    (i) Pollination brings the pollen grain to the stigma of the carpel, while the female germ cell lies inside an ovule in the ovary. The pollen grain that lands on a suitable stigma germinates and puts out a pollen tube, which grows through the tissue of the stigma and downward through the style until it reaches the ovary and enters the ovule. The male germ cell travels down this pollen tube and is released inside the ovule, where it fuses with the female germ cell, that is, with the egg. This fusion is fertilisation and it forms the zygote, which divides to form the embryo; the ovule then becomes the seed and the ovary becomes the fruit. The diagram should show the stigma with a germinating pollen grain, the pollen tube growing down the style with the male germ cell in it, the ovary, the ovule and the female germ cell inside it, all labelled. (ii) Self-pollination: pollen grains are transferred from the anther to the stigma of the same flower; cross-pollination: pollen grains are transferred from the anther of one flower to the stigma of a different flower, usually a flower on another plant of the same species. Self-pollination usually needs no outside agent; cross-pollination needs an agent such as wind, water, insects or birds. The progeny of self-pollination show little variation; cross-pollination between different plants brings together genes from two individuals, so its progeny show more variation. (iii) R is the plumule, the part of the embryo above the attachment of the cotyledons, which grows into the shoot; S is the radicle, the part below it, which grows into the root.
    Step-by-step solution →

Questions 17–24 · 1 mark each

  1. Q17 1 mark
    The number of isomers of butane is
    1. (a)
    2. (b)
    3. (c)
    4. (d)
    Show answer
    (b)
    Step-by-step solution →
  2. Q18 1 mark
    Which of the following compound is formed when gas is passed through ammoniacal brine solution ?
    1. (a)
    2. (b)
    3. (c)
    4. (d)
    Show answer
    (d)
    Step-by-step solution →
  3. Q19 1 mark
    Which of the following compounds will undergo addition reaction ?
    1. (a)
    2. (b)
    3. (c)
    4. (d)
    Show answer
    (b)
    Step-by-step solution →
  4. Q20 1 mark
    In the compound , which of the following functional group is present ?
    1. (a)Carboxylic acid
    2. (b)Alcohol
    3. (c)Aldehyde
    4. (d)Ketone
    Show answer
    (a) Carboxylic acid
    Step-by-step solution →
  5. Q21 1 mark

    Also in 31/8/1/SB (Q23), 31/8/3/SB (Q21)

    Which of the following acid-base indicators will be used by a visually challenged student to detect the presence of acid in a given solution ?
    1. (a)Blue litmus
    2. (b)Clove oil
    3. (c)Red cabbage extract
    4. (d)Methyl orange
    Show answer
    (b) Clove oil
    Step-by-step solution →
  6. Q22 1 mark
    Which of the following aqueous solution reacts with crushed egg-shells to give a gas that turns lime water milky ?
    1. (a)
    2. (b)
    3. (c)
    4. (d)
    Show answer
    (b)
    Step-by-step solution →
  7. Q23 1 mark

    Also in 31/8/1/SB (Q20), 31/8/3/SB (Q22)

    During the preparation of hydrogen chloride gas on a humid day, the gas is usually passed through a guard tube containing calcium chloride. The role of taken in guard tube is to –
    1. (a)Absorb the evolved gas
    2. (b)Moisten the gas
    3. (c)Absorb moisture from the gas
    4. (d)Absorb chloride ions from the evolved gas
    Show answer
    (c) Absorb moisture from the gas
    Step-by-step solution →
  8. Q24 1 mark

    Also in 31/8/1/SB (Q24), 31/8/3/SB (Q24)

    For question number 24, two statements are given-one labelled as Assertion (A) and the other labelled as Reason (R). Answer this question selecting the appropriate option from (a), (b), (c) and (d) given below :

    Assertion (A): Sodium acetate is a basic salt.

    Reason (R): It is because it is a salt of weak acid and strong base.

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

Question 25 · 2 marks

  1. Q25 2 marks
    Write chemical equations for the reactions taking place when : (A) is heated with aluminium powder. (B) Sodium metal reacts with water.
    Show answer

    Heated with aluminium, iron oxide gives aluminium oxide and molten iron; sodium and water give sodium hydroxide and hydrogen.

    (A) - the thermite reaction; so much heat is released that the iron is obtained molten. (B) - the reaction is so exothermic that the hydrogen evolved catches fire.
    Step-by-step solution →

Questions 26–27 · 3 marks each

  1. Q26 3 marks

    Also in 31/8/1/SB (Q27), 31/8/3/SB (Q27)

    (A) Name the acid present in ant sting and give its chemical formula. Give the common method to get relief from the pain and irritation caused by ant sting. (B) Two solutions ‘X’ and ‘Y’ have pH values and respectively. Which of these will turn blue litmus to red and which will turn phenolphthalein from colourless to pink ?
    Show answer

    Ant-sting acid is methanoic acid, relieved by baking soda; X (pH 3.0) turns blue litmus red; Y (pH 9.5) turns phenolphthalein pink.

    (A) The acid present in an ant sting is methanoic acid (formic acid), . Relief is obtained by rubbing the stung area with a mild base such as moist baking soda (sodium hydrogencarbonate, ), which neutralises the acid; calamine solution, containing zinc carbonate, may be used in the same way. (B) ‘X’, of pH , is acidic and will turn blue litmus red. ‘Y’, of pH , is basic and will turn phenolphthalein from colourless to pink.
    Step-by-step solution →
  2. Q27(a) 3 marks

    Also in 31/8/1/SB (Q26(a)), 31/8/3/SB (Q26(a))

    1. (i)State any two reasons for carbon forming a large number of compounds. Why does carbon form compounds mainly by covalent bonds ?
    2. (ii)Identify the compound which contains aldehyde as its functional group : , , ,
    Show answer

    Catenation and tetravalency let carbon form many compounds; carbon shares its four valence electrons, forming covalent bonds; has the aldehyde group.

    (i) Two reasons: (1) Catenation — carbon atoms link to one another by strong covalent bonds and build long straight chains, branched chains and closed rings; (2) Tetravalency — carbon has four valence electrons, so one carbon atom can bond with four other atoms of carbon or of other monovalent elements, and because the carbon atom is small these bonds are strong and stable. Carbon forms compounds mainly by covalent bonds because it has four electrons in its outermost shell: it can neither gain four electrons (a nucleus of six protons cannot hold ten electrons) nor lose four electrons (that would need a very large amount of energy), so it attains a noble-gas arrangement by **sharing** its four valence electrons, and a shared pair of electrons is a covalent bond. (ii) (propanal) — it carries the aldehyde group .
    Step-by-step solution →
    OR
    Q27(b) 3 marks

    Also in 31/8/1/SB (Q26(b)), 31/8/3/SB (Q26(b))

    An organic compound ‘X’ is a constituent of wine. ‘X’ on reacting with acidified forms another compound ‘Y’ with molecular formula . ‘Y’ produces brisk effervescence with sodium carbonate.
    1. (i)Identify ‘X’ and ‘Y’.
    2. (ii)Name the gas evolved when ‘Y’ reacts with sodium carbonate.
    3. (iii)Write the chemical equations involved for the chemical reactions taking place.
    Show answer

    X is ethanol and Y is ethanoic acid; the gas evolved is carbon dioxide.

    (i) ‘X’ is ethanol, — the alcohol present in wine; ‘Y’ is ethanoic acid (acetic acid), , whose molecular formula is . (ii) The gas evolved is carbon dioxide, . (iii) and
    Step-by-step solution →

Question 28 · 4 marks

  1. Q28 4 marks

    Also in 31/8/1/SB (Q28), 31/8/3/SB (Q28)

    The Statue of Liberty is one of the seven wonders. It is present on Liberty island in New York harbour in U.S.A. It is made up of a metal. Over a period of time, the metal’s colour has changed and statue gives greenish look. Answer the following questions based on the above information :
    1. (A)What is corrosion ?[1]
    2. (B)Why do silver articles turn black after some time ?[1]
    3. (C)Copper reacts with moist gas ‘A’ present in the air and slowly loses its shiny surface. It turns green. Identify ‘A’ and the substance of green coating formed on the surface of copper.[2]
    4. (C)OR : What is an alloy ? How is an alloy made ?[2]
    Show answer

    Corrosion eats away a metal's surface; silver turns black due to silver sulphide; gas A is carbon dioxide, the green coating basic copper carbonate.

    (A) Corrosion is the process in which the surface of a metal is slowly eaten away by the action of air, moisture, acids or other chemicals around it, the metal being converted into a surface compound such as an oxide, a carbonate or a sulphide. (B) Silver articles turn black because silver reacts with the sulphur present in the air, as hydrogen sulphide, to form a black coating of silver sulphide, : . (C) ‘A’ is carbon dioxide, , and the green coating formed on copper is basic copper carbonate, : . (C) OR : An alloy is a homogeneous mixture of two or more metals, or of a metal and a non-metal. It is made by first melting the main metal, then dissolving the other metals or non-metals in it in a fixed proportion, and then cooling the mixture to room temperature.
    Step-by-step solution →

Question 29 · 5 marks

  1. Q29(a) 5 marks

    Also in 31/8/1/SB (Q29(a)), 31/8/3/SB (Q29(a))

    Write chemical equations for the following when
    1. (a)Silver bromide is exposed to sunlight.
    2. (b)A piece of lead metal is dropped into copper (II) chloride solution.
    3. (c)Hydrogen gas is passed over heated copper oxide.
    4. (d)Methane gas is burnt in air.
    5. (e)Quick lime is mixed with water.
    Show answer

    Silver bromide decomposes in sunlight, lead displaces copper, hydrogen reduces copper oxide, methane burns, and quick lime combines with water.

    (a) — photolytic decomposition. (b) — displacement. (c) — reduction of copper oxide. (d) — combustion. (e) — combination.
    Step-by-step solution →
    OR
    Q29(b) 5 marks

    Also in 31/8/1/SB (Q29(b)), 31/8/3/SB (Q29(b))

    1. (a)Study the experimental setup shown in the diagram and write chemical equation for the same. Name and define the type of reaction. Mention the colour of the products formed.
    2. (b)What type of reaction takes place when aqueous solution of lead (II) nitrate and potassium iodide are mixed together ? Write chemical equation also.
    A pair of tongs Boiling tube Lead nitrate Burner
    Show answer

    Heated lead nitrate decomposes into yellow lead oxide, brown nitrogen dioxide and oxygen; lead nitrate and potassium iodide solutions give a double displacement precipitate.

    (a) . It is a thermal decomposition reaction - a reaction in which a single compound breaks down on heating into two or more simpler substances. Colours of the products: lead oxide is a yellow solid, nitrogen dioxide is a reddish brown gas and oxygen is colourless. (b) A double displacement reaction, and since an insoluble solid separates out it is a precipitation reaction: , the precipitate being yellow lead iodide.
    Step-by-step solution →

Questions 30–32 · 1 mark each

  1. Q30 1 mark

    Also in 31/8/1/SB (Q31), 31/8/3/SB (Q31)

    In the given ray diagram the values of the angle of incidence and the angle of reflection are :
    N 40°
    1. (a),
    2. (b),
    3. (c),
    4. (d),
    Show answer
    (d) ,
    Step-by-step solution →
  2. Q31 1 mark

    Also in 31/8/1/SB (Q30)

    When a convergent beam of light is incident on a convex mirror, the reflected rays appear to be coming from a point 20 cm behind the mirror and retrace their path. The focal length of the mirror is
    1. (a)20 cm
    2. (b)10 cm
    3. (c)40 cm
    4. (d)30 cm
    Show answer
    (b) 10 cm
    Step-by-step solution →
  3. Q32 1 mark

    Also in 31/8/1/SB (Q32), 31/8/3/SB (Q32)

    For question number 32, two statements are given – one labelled as Assertion (A) and the other labelled as Reason (R). Answer this question selecting the appropriate option from (a), (b), (c) and (d) given below :

    Assertion (A): The bending of a straight wire in the form of current carrying circular coil will change the pattern of magnetic field around it.

    Reason (R): Strength of magnetic field depends upon number of turns of coil and current flowing through the coil.

    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
    (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
    Step-by-step solution →

Questions 33–34 · 2 marks each

  1. Q33 2 marks
    (A) In the given electric circuit of Ohm's law experiment, identify P and X. (B) Write the uses of Q and X in the given circuit.
    P Q R (Resistor) X Z Y
    Show answer

    P is a voltmeter and X is a rheostat; Q (ammeter) measures the current, and X varies it.

    (A) P is a voltmeter - it is connected in parallel across the resistor R, so it reads the potential difference across R. X is a rheostat, that is, a variable resistance, shown by the slanting arrow drawn across the zig-zag. (B) Q is the ammeter: it is connected in series, so the whole circuit current passes through it and it measures the current flowing in the circuit, through R. X, the rheostat, is used to change (regulate) the current in the circuit without changing the voltage source, so that several pairs of readings of and can be taken with the same resistor and shown to be constant; it also keeps the current within safe limits.
    Step-by-step solution →
  2. Q34(a) 2 marks

    Also in 31/8/1/SB (Q33(a)), 31/8/3/SB (Q34(a))

    What is meant by scattering of light ? How do our observations differ when white light (sun light) is scattered from
    1. (i)very fine particles of the medium and
    2. (ii)very large particles of the medium ?
    Show answer

    Scattering sends light off in different directions; very fine particles give blue scattered light, very large particles give white.

    Scattering of light is the phenomenon in which a part of the light falling on the particles of a medium - molecules, dust, smoke or water droplets - is redirected (sent out again) in different directions, away from its original straight path. (i) Very fine particles: they scatter mainly the shorter wavelengths, so the scattered light appears **blue**. This is why the clear sky looks blue. (ii) Very large particles: they scatter all wavelengths almost equally, so the scattered light appears **white**. This is why clouds, fog and mist look white.
    Step-by-step solution →
    OR
    Q34(b) 2 marks

    Also in 31/3/1 (Q34(b)), 31/3/2 (Q33(b)), 31/3/3 (Q34(b)), 31/8/1/SB (Q33(b)), 31/8/3/SB (Q34(b))

    With the help of a ray diagram, explain any phenomenon of formation of spectrum in nature.
    Show answer

    A rainbow forms when raindrops disperse sunlight by refraction, internal reflection and refraction, giving a spectrum with red outside and violet inside.

    The rainbow is a natural spectrum formed after rain by the dispersion of sunlight by tiny water droplets in the atmosphere. Each droplet acts like a small prism: sunlight is refracted and dispersed as it enters the drop, reflected internally at the back surface, and refracted again as it comes out (see the ray diagram). The colours emerge at slightly different angles - red at about , violet at about to the direction opposite the Sun - so an observer standing with the Sun behind sees a band of colours in the sky, red on the outside and violet on the inside, always on the side opposite the Sun.
    Step-by-step solution →

Questions 35–37 · 3 marks each

  1. Q35 3 marks

    Also in 31/8/1/SB (Q37), 31/8/3/SB (Q36)

    (A) Write any two precautions to be taken to avoid electrical accidents. (B) State how these precautions prevent possible damage to the electric circuit.
    Show answer

    Fit a fuse in the live wire and earth metal-bodied appliances; the fuse breaks the circuit, earthing gives a low-resistance path.

    (A) Two precautions: (i) Fit an electric fuse (or an MCB) of the proper rating in series in the **live** wire of every circuit and of every high-power appliance. (ii) Earth every metal-bodied appliance - connect its metal casing through the green earth wire and the third pin of a three-pin plug to a metal plate buried deep in the ground. (B) How they prevent damage to the circuit: The fuse is a short piece of wire of high resistivity and low melting point. If the current rises above the rated value because of overloading or a short circuit, the heat produced in it, , melts the fuse wire. The circuit is broken at once and the current stops, so the household wiring, the switches and the appliances are never heated to the point where their insulation melts, they are damaged or a fire starts. Earthing provides the metal body of the appliance with a path of very low resistance to the ground. If the live wire touches the casing, the body is held at the potential of the earth (about V), so nobody gets a shock; and because that path has a very low resistance the leakage current is large, so it melts the fuse and disconnects the faulty appliance before its damaged wiring can overheat. (Other acceptable precautions, with the same kind of explanation: do not overload one socket through a multi-plug; use only ISI-marked, properly insulated wires of adequate thickness; switch off the mains before repairs and never touch switches with wet hands.)
    Step-by-step solution →
  2. Q36 3 marks

    Also in 31/8/1/SB (Q35), 31/8/3/SB (Q37)

    (A) A vertical wire is carrying a current in the upward direction. It is placed in a magnetic field, pointing towards the East direction. Find out the direction of force on the wire. (B) Out of the two current carrying circular coils, coil ‘A’ has 10 turns while coil B has 15 turns. Which of the two coils will produce a stronger magnetic field at its centre if same current is flowing through both the coils ? Explain with reason. (C) In the given figure, AB and CD are two magnetic field lines intersecting each other at point O. If a compass is placed at point O, what will happen ? State your conclusions.
    O A B C D
    Show answer

    Force is towards North; coil B (15 turns) gives the stronger field; the compass cannot point two ways, so field lines never intersect.

    (A) The force on the wire is horizontal and directed towards the **north**. By Fleming's left-hand rule, with the forefinger along the field (East) and the middle finger along the current (vertically up), the thumb points North. (B) Coil B, with turns, produces the stronger field. At the centre of a circular coil , so with the same current and the same radius the field is proportional to the number of turns - each turn adds its own field in the same direction - and , that is times as strong. (C) The figure shows the field at O in two directions at once, so a compass needle placed at O would have to point in two directions (along OB and along OD) at the same time. That is not possible, since a needle can settle in only one direction - the situation shown is unreal. Conclusions: the magnetic field at any point has a single definite direction, and therefore two magnetic field lines never intersect each other.
    Step-by-step solution →
  3. Q37 3 marks
    A person cannot read newspaper placed nearer than 50 cm from his eyes. Name the defect of vision he is suffering from. Draw a ray diagram to illustrate the defect. List two possible causes of this defect also.
    Show answer

    Hypermetropia: the near point is 50 cm, so closer objects focus behind the retina; a long focal length or short eyeball causes it.

    The defect is hypermetropia, also called far-sightedness or long-sightedness. The nearest point this eye can focus on has moved out from the normal cm to cm, so light coming from anything closer than cm is brought to a focus behind the retina and the image on the retina is blurred; distant objects are still seen clearly. The ray diagram shows light from an object at the normal near point of cm meeting behind the retina, and light from an object at cm meeting exactly on it. Two possible causes: (i) The focal length of the eye lens is too long, so its converging power is too low; (ii) The eyeball is too short, so the retina lies in front of the point where the refracted rays meet.
    Step-by-step solution →

Question 38 · 4 marks

  1. Q38 4 marks

    Also in 31/8/1/SB (Q38), 31/8/3/SB (Q38)

    Four cells, a thin wire AB, a voltmeter, an ammeter, a plug key and some thick connecting wires are taken. The circuit is set up as shown below. The key is inserted into the plug and readings of ammeter (I) and voltmeter (V) are noted. Then the plug key is removed. Now two cells are connected in the circuit in series. Plugging in the key the new values of ‘V’ and ‘I’ are noted. The experiment is repeated by connecting three cells in series and then four cells. In each case ‘V’ and ‘I’ are noted.
    V + − A − + + − A B
    1. (A)Why was the plug key removed after each measurement of ‘V’ and ‘I’ and inserted again for the next step ?
    2. (B)What change takes place in the reading of the voltmeter when a new cell is added in series in the circuit ? Give reason.
    3. (C)In which case will the readings of voltmeter and ammeter be maximum in the above experiment ? What conclusion can be drawn from the straight line graph between ‘V’ and ‘I’ ?
    4. (C)OR : Suppose this experiment is carried out using only one cell but wires of same length & thickness but made of different materials like nichrome wire, fuse wire, copper wire and iron wire which are replaced one by one. Out of V and I which will change and why ?
    Show answer

    Key removed so wire AB stays cool; each added cell raises the voltmeter reading; four cells give maximum readings; the line shows Ohm's law.

    (A) So that current flows through the circuit only while a reading is being taken. A current left flowing through the thin wire AB would heat it (); the resistance of a metal rises with temperature, so would drift and the V-I graph would not come out straight, since Ohm's law holds only at constant temperature. Removing the key lets AB cool to room temperature between readings and also saves the cells. (B) The voltmeter reading increases. Cells joined in series have their emfs added, so each extra cell applies a larger potential difference across AB; as the wire is unchanged its resistance is unchanged, so the current increases in the same proportion and stays constant. (C) Both readings are maximum when all four cells are connected in series, since that gives the greatest total emf. The straight line through the origin shows that , i.e. is constant for the conductor at constant temperature - this is Ohm's law - and that constant, the slope of the line, is the resistance of the wire AB. (C) OR: will change and will remain the same. The same single cell applies the same potential difference in every trial, while the wires have the same length and the same area of cross-section , so by their resistances differ only through the resistivity of the material; hence is different for each wire - largest for copper, which has the lowest resistivity of the four, and smallest for nichrome, which has the highest. This shows that resistance depends on the nature of the material.
    Step-by-step solution →

Question 39 · 5 marks

  1. Q39(a) 5 marks
    1. (i)With the help of given ray diagram, calculate the refractive index of medium 2 with respect to medium 1.
    2. (ii)If speed of light in medium-1 is m/s, then calculate the speed of light in medium-2.
    3. (iii)Compare the optical density of medium-1 and medium-2 and justify your answer.
    N 30° 60° Medium-1 Medium-2
    Show answer

    ; m/s; medium-2 is optically denser than medium-1.

    The arcs in the figure are drawn from the boundary, so and . (i) . (ii) m/s. (iii) Medium-2 is optically denser than medium-1, because : the ray bends towards the normal on entering medium-2 () and light travels more slowly in it ( m/s against m/s).
    Step-by-step solution →
    OR
    Q39(b) 5 marks
    1. (i)Define the focus of a convex lens.
    2. (ii)An object placed on a meter scale at 8 cm mark, was focused on a white screen placed at 92 cm mark, by using a converging lens placed on the scale at 50 cm mark. (I) Find the focal length of converging lens. (II) Find the position of object if the image forms at 71 cm mark on meter scale. (III) State the nature of image formed if the object is kept on 36 cm mark of meter scale.
    Show answer

    cm, the object for the 71 cm image is at infinity, and the 36 cm mark gives a virtual, erect, enlarged image.

    (i) The principal focus of a convex lens is the point on its principal axis at which rays of light coming parallel to the principal axis meet after refraction through the lens. A convex lens has two such foci, one on each side, each at a distance from the optical centre. (ii)(I) With the lens at the cm mark, cm and cm, so and cm. (II) An image at the cm mark is cm from the lens, which is exactly , so and the object must be at infinity; no mark on the metre scale will give this image, because only a beam of light arriving parallel to the principal axis is brought to a focus at . (III) An object at the cm mark is cm from the lens, which is less than cm, so it lies between the optical centre and the focus: , giving cm and . The image is virtual, erect and enlarged three times, formed cm from the lens on the same side as the object, that is, at the cm mark.
    Step-by-step solution →

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