CBSE Class 10 Science Board Paper 2026 (31/1/1): Solutions

Which part of a leaf mainly handles the exchange of gases?

1. Stomata are the tiny pores found on the leaf surface that allow gases like carbon dioxide and oxygen to enter and exit.
2. Therefore, the part of the leaf that mainly handles the exchange of gases is the stomata, which corresponds to option (B).

Answer: Stomata, the tiny pores on the leaf surface — option (B).

A diagram shows a single amoeba-like cell whose nucleus divides and then the cell splits into two daughter cells. Name this mode of reproduction.

Uses a figure printed in CBSE's paper — what the figure shows: The figure shows one irregular amoeba-like cell; its nucleus elongates and divides, the cell constricts in the middle and separates into two similar daughter cells.

1. The figure shows a single cell dividing into two similar daughter cells, which is the defining characteristic of binary fission.
2. Option (D) is correct because it identifies this process, whereas other options like budding or fragmentation do not match the simple splitting shown.

Answer: Binary fission — option (D).

What is true of Spirogyra, Leishmania and Hydra together?

1. Spirogyra reproduces by fragmentation, Leishmania by binary fission, and Hydra by budding (it can also regenerate from pieces).
2. Leishmania is single-celled while the other two are multicellular, so only the statement about mode of reproduction fits all three.
3. None of these methods involves fusion of gametes, so all three reproduce asexually — option (D).

Answer: All three reproduce asexually (fragmentation, binary fission and budding/regeneration) — option (D).

Which part of the human brain keeps voluntary movements precise?

1. The cerebellum, located in the hindbrain, is responsible for coordinating voluntary movements and maintaining balance.
2. Therefore, the part of the brain that keeps voluntary movements precise is the cerebellum, which corresponds to option (B).

Answer: The cerebellum (hind brain) — option (B).

Which pairing correctly states the job of trypsin and lipase in pancreatic juice?

1. Trypsin is a proteolytic enzyme that breaks down proteins into peptides and amino acids.
2. Lipase is a lipolytic enzyme that breaks down emulsified fats into fatty acids and glycerol.
3. Therefore, the pairing stating trypsin acts on proteins and lipase acts on fats is correct, which is option (B).

Answer: Trypsin breaks down proteins and lipase breaks down emulsified fats — option (B).

For humans, which hold: (i) X-bearing gamete from the mother, (ii) X-bearing gamete from the father, (iii) Y-bearing gamete from the father? Pick the option naming the correct ones.

1. In humans the mother has XX, so every egg carries an X chromosome; the father has XY, so half his sperms carry X and half carry Y.
2. The child's sex depends on which sperm fertilises the egg: an X-bearing sperm gives a girl (XX) and a Y-bearing sperm gives a boy (XY).
3. So the father's X-bearing and Y-bearing gametes, statements (ii) and (iii), are the ones that decide sex — option (A).

Answer: The marking scheme accepts (ii) and (iii): the father gives either an X or a Y gamete, and this decides the sex of the child — option (A).

Note: Statement (i) is also biologically true (every egg carries an X), but no option lists all three; CBSE's marking scheme accepts (A), since the father's gamete is what decides the child's sex.

Which group of items is non-biodegradable?

1. Polythene, rubber bands and ball pens are made of plastics and synthetic materials that microbes cannot break down quickly.
2. They stay in the environment for a very long time, so this group is non-biodegradable — option (C).

Answer: Polythene bag, rubber band and ball pen — option (C).

Assertion: a reflex action happens without thinking. Reason: the spinal cord controls most reflex actions.

1. The assertion is true: a reflex is a quick, automatic response that happens without conscious thinking.
2. The reason is true: in most reflexes the sensory and motor nerves meet in the spinal cord, forming a reflex arc there.
3. Because the response is decided in the spinal cord before the message reaches the thinking part of the brain, the reason explains the assertion — option (A).

Answer: Both are true and the reason explains the assertion, since the reflex arc passes through the spinal cord without the brain's thinking — option (A).

Assertion: haemoglobin in red blood cells is the human respiratory pigment. Reason: haemoglobin binds carbon dioxide very strongly.

1. Haemoglobin is the respiratory pigment in red blood cells, so the assertion is true.
2. Haemoglobin has a high affinity for oxygen, not carbon dioxide, so the reason is false.
3. Since the assertion is true but the reason is false, the correct option is (C).

Answer: The assertion is true but the reason is false — haemoglobin has a high affinity for oxygen, not carbon dioxide — option (C).

What does the diaphragm do in breathing, and where is it located?

1. The diaphragm is a muscle located at the floor or base of the chest cavity.
2. During breathing in, it flattens to enlarge the chest cavity, which draws air into the lungs.

Answer: It flattens when we breathe in, enlarging the chest cavity so air is drawn into the lungs (it helps breathing). It lies at the floor (base) of the chest cavity.

Give two differences between chewing food and salivating when food is seen.

1. Chewing is a voluntary action controlled by the forebrain, whereas salivation upon seeing food is an involuntary reflex action.
2. Salivation is controlled by the medulla in the hind brain, which differs from the forebrain control of chewing.

Answer: (1) Chewing is voluntary; salivation on seeing food is a reflex (involuntary) action. (2) Chewing is controlled by the forebrain; salivation is controlled by the medulla in the hind brain.

Give two differences between pollination and fertilisation.

1. State that pollination is the transfer of pollen grains from the anther to the stigma, whereas fertilisation is the fusion of male and female gametes.
2. State that pollination occurs only in plants, while fertilisation occurs in both plants and animals.

Answer: (1) Pollination is the transfer of pollen from anther to stigma; fertilisation is the fusion of the male and female gametes. (2) Pollination happens only in plants; fertilisation happens in both plants and animals.

Draw pollen germinating on the pistil of a flower and label (a) the part that receives pollen and (b) the structure that takes the male germ cell to the female germ cell.

1. Draw a pistil consisting of the stigma at the top, the style in the middle, and the ovary at the base.
2. Draw a pollen grain on the stigma and a pollen tube growing down through the style to reach the ovule.
3. Label the stigma as the part that receives pollen and the pollen tube as the structure that carries the male germ cell to the female germ cell.

Answer: Diagram: a pistil with stigma, style and ovary; a pollen grain on the stigma sends a pollen tube down through the style to the ovule. Labels: (a) stigma; (b) pollen tube.

A pyramid has levels (i) producers, (ii) primary, (iii) secondary and (iv) tertiary consumers. (a) Place deer, grass, lion, snake and rabbit on these levels. (b) Why do primary consumers get more energy than secondary consumers? (c) Why is the pyramid's base wide?

Uses a figure printed in CBSE's paper — what the figure shows: A four-step pyramid: (i) producers at the base, (ii) primary consumers, (iii) secondary consumers, (iv) tertiary consumers at the apex.

1. (a) Grass makes its own food, so it is the producer at level (i).
2. Deer and rabbit eat plants, so they are primary consumers at level (ii).
3. Snake eats rabbits and lion eats deer, so both feed on herbivores and can sit at level (iii) as secondary consumers.
4. Lion can also eat animals that are themselves flesh-eaters, so it is placed at the top, level (iv), as the tertiary consumer.
5. (b) Primary consumers eat producers directly. Only about 10% of the energy at one level passes to the next, so secondary consumers, one step further, receive much less.
6. (c) Producers are the most numerous and hold the most biomass and energy of all levels, so the pyramid's base is the widest.

Answer: (a) (i) grass; (ii) deer and rabbit; (iii) snake (and lion); (iv) lion. (b) Primary consumers eat plants directly, and only about 10% of the energy at one level passes on to the next, so secondary consumers get less. (c) Producers are the most in number, mass and energy of all levels, so the base is broadest.

Note: CBSE's marking scheme places lion at both level (iii) and level (iv), because a lion can feed at more than one level.

Differentiate: (a) nephron and neuron, (b) sensory and motor nerve, (c) consumers and decomposers.

1. (a) A nephron is the structural and functional unit of the kidney that filters blood to remove nitrogenous waste, whereas a neuron is the basic unit of the nervous system that transmits electrical impulses.
2. (b) A sensory nerve carries impulses from sense organs (receptors) to the brain or spinal cord, while a motor nerve carries impulses from the brain or spinal cord to muscles or glands (effectors).
3. (c) Consumers are organisms that feed on producers or other consumers to obtain energy, whereas decomposers break down dead and decaying organic matter into simple inorganic substances, returning nutrients to the soil.

Answer: (a) A nephron is the filtering unit of the kidney that removes nitrogenous waste from blood; a neuron is the unit of the nervous system that carries messages. (b) A sensory nerve carries impulses from receptors to the brain/spinal cord; a motor nerve carries them from the brain/spinal cord to muscles or glands (effectors). (c) Consumers feed on producers or other consumers and pass energy along the food chain; decomposers break dead matter into simple inorganic substances and return nutrients to the environment.

Mendel crossed tall and short pea plants and got only tall plants in F1. Why?

1. The tall factor (T) is dominant over the short factor (t), so the presence of T masks the expression of t.
2. Since every F1 plant has the genotype Tt, they all exhibit the tall phenotype.

Answer: Every F1 plant is Tt, and the tall factor T is dominant over short t, so all of them are tall.

In the tall × short pea cross, how did Mendel get the F2 generation?

1. Mendel obtained the F2 generation by self-pollinating the F1 hybrid plants.

Answer: By self-pollinating the F1 plants.

Give one difference between a dominant and a recessive trait.

1. A dominant trait expresses itself in the phenotype even when only one allele is present, such as in the genotypes TT or Tt.
2. A recessive trait is masked by the dominant allele and only appears in the phenotype when both alleles are recessive, which corresponds to the genotype tt.

Answer: A dominant trait shows itself even when one copy is present (TT or Tt); a recessive trait shows only when both copies are recessive (tt), and is hidden in the presence of the dominant one.

State two things Mendel noticed about the F1 plants of the tall × short cross.

1. Observe the phenotype of the F1 generation resulting from the cross between tall and short pea plants.
2. Note that all F1 plants exhibited the tall trait, matching the phenotype of only one parent.
3. Observe that no short or medium-height plants appeared in the F1 generation.

Answer: Any two: all F1 plants were tall; no short or medium-height plants appeared; the F1 plants looked like only one parent.

Say what would follow in each case: (i) water holding a bacterial population warms up due to global warming; (ii) a sperm meets the egg in the oviduct; (iii) a flower has only a pistil (no stamens), so it cannot self-pollinate; (iv) the egg of a human female is not fertilised; (v) a seed in soil gets enough water and air.

1. (i) As the water warms up, most bacteria die, but the few heat-resistant variants survive and multiply.
2. (ii) When the sperm meets the egg in the oviduct, fertilisation takes place and a zygote forms.
3. (iii) Since the flower has only a pistil and no stamens, it cannot self-pollinate; only cross-pollination can occur, which may lead to fruit formation.
4. (iv) If the egg is not fertilised, the thick, spongy uterine lining breaks down and leaves the body as blood and mucus, a process called menstruation.
5. (v) When the seed in soil gets enough water and air, it germinates and grows into a seedling.

Answer: (i) Most bacteria die, but the few heat-resistant variants survive and multiply. (ii) Fertilisation takes place and a zygote forms. (iii) Only cross-pollination can occur, which may lead to fruit formation (without it there is no fertilisation or fruit). (iv) The thick, spongy uterine lining breaks down and leaves the body as blood and mucus — menstruation. (v) The seed germinates and grows into a seedling.

Say what happens when: (i) bread-mould spores are released from their sporangia; (ii) Bryophyllum leaves drop onto moist soil; (iii) pollen lands on the stigma of an unrelated species; (iv) a Copper-T is fitted in the uterus; (v) Spirogyra breaks into pieces after maturing.

1. (i) When bread-mould spores are released from sporangia, they land on a moist, warm food surface and grow into new Rhizopus.
2. (ii) When Bryophyllum leaves drop onto moist soil, buds present in the notches of the leaf margin grow into new plants.
3. (iii) When pollen lands on the stigma of an unrelated species, no pollen tube forms, so fertilisation does not happen.
4. (iv) When a Copper-T is fitted in the uterus, it prevents fertilisation or pregnancy.
5. (v) When Spirogyra breaks into pieces after maturing, each fragment grows into a new Spirogyra filament.

Answer: (i) Spores landing on a moist, warm food surface grow into new Rhizopus. (ii) Buds in the notches of the leaf margin grow into new plants. (iii) No pollen tube forms, so fertilisation does not happen. (iv) Fertilisation/pregnancy is prevented. (v) Each fragment grows into a new Spirogyra filament.

Classify both: (i) AgNO₃ + NaCl → NaNO₃ + AgCl and (ii) K₂SO₄ + BaCl₂ → BaSO₄ + 2KCl.

1. Both reactions involve the exchange of ions between two reactants, which defines them as double displacement reactions.
2. In both cases, an insoluble product (AgCl in the first and BaSO₄ in the second) is formed, which is also a characteristic of precipitation reactions.
3. Therefore, both reactions are classified as double displacement and precipitation reactions, corresponding to option (C).

Answer: Both swap ions (double displacement) and both form an insoluble precipitate (AgCl, BaSO₄) — option (C).

Which indicator could a blind student use to test acids and bases?

1. A blind student cannot see color changes, so they must use an olfactory indicator that changes smell.
2. Vanilla essence is an olfactory indicator that loses its characteristic smell when a base is added, making it the correct choice.

Answer: Vanilla essence, an olfactory indicator whose smell changes with a base — option (B).

Which gases come off when lead(II) nitrate is heated?

1. Lead(II) nitrate decomposes on heating to form lead(II) oxide, nitrogen dioxide, and oxygen gas.
2. The balanced chemical equation is 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂, identifying the gases evolved as NO₂ and O₂.

Answer: 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂, so the gases are NO₂ and O₂ — option (D).

Give the functional group of an aldehyde.

1. An aldehyde is an organic compound characterized by a carbonyl group (C=O) bonded to at least one hydrogen atom.
2. The functional group representing this structure is written as –CHO, which corresponds to option (B).

Answer: –CHO — option (B).

Among Pb, Cu, Ag and Al, which conducts electricity poorly?

1. Lead (Pb) has a lower electrical conductivity compared to Copper (Cu), Silver (Ag), and Aluminium (Al).
2. Therefore, Lead (Pb) conducts electricity poorly among the given options.

Answer: Lead (Pb) — option (A).

Name natural sources of oxalic acid, lactic acid and methanoic acid, in that order.

1. Oxalic acid is naturally found in tomatoes, lactic acid is present in curd, and methanoic acid is found in ant stings.
2. Since the order matches Tomato, curd, and ant sting, option (A) is the correct answer.

Answer: Tomato, curd and ant sting — option (A).

Element X begins to float when it reacts with water. Which element is it?

1. Calcium reacts with water less violently than sodium, forming calcium hydroxide and hydrogen gas.
2. The hydrogen bubbles stick to the surface of the metal and make it float, so X is calcium — option (B).

Answer: Calcium — hydrogen bubbles stick to it and make it float — option (B).

Assertion: carbon shares its valence electrons with carbon or other atoms. Reason: shared electrons count in the outer shells of both atoms, so each attains a noble-gas configuration.

1. Assertion is true because carbon forms covalent bonds by sharing its four valence electrons with other atoms.
2. Reason is true because the shared electrons are counted in the outer shells of both atoms, allowing each to attain a stable noble-gas configuration.
3. Since the reason correctly explains the mechanism behind the assertion, option (A) is the correct answer.

Answer: Both are true and the reason explains the assertion — option (A).

Define amphoteric oxides.

1. Amphoteric oxides are metal oxides that react with both acids and bases to produce salt and water.
2. Examples of amphoteric oxides include zinc oxide (ZnO) and aluminium oxide (Al₂O₃).

Answer: Metal oxides that react with both acids and bases to give salt and water (e.g. ZnO, Al₂O₃).

Classify ZnO, Na₂O and CO₂ by nature.

1. ZnO reacts with both acids and bases, so it is classified as amphoteric.
2. Na₂O is a metal oxide that reacts with acids, so it is classified as basic.
3. CO₂ is a non-metal oxide that reacts with bases, so it is classified as acidic.

Answer: ZnO is amphoteric, Na₂O is basic and CO₂ is acidic.

In ZnO + C → Zn + CO, name what is oxidised and what is reduced.

1. Carbon (C) gains oxygen to form CO, so it is oxidised.
2. Zinc oxide (ZnO) loses oxygen to form Zn, so it is reduced.

Answer: C is oxidised (to CO) and ZnO is reduced (to Zn).

Balance Pb(NO₃)₂ + KI → PbI₂ + KNO₃.

1. Identify the unbalanced atoms: there are 2 nitrogen and 6 oxygen atoms on the left, but only 1 nitrogen and 3 oxygen atoms on the right.
2. Place a coefficient of 2 before KNO₃ to balance nitrogen and oxygen, which also balances potassium to 2 on the right.
3. Place a coefficient of 2 before KI to balance the 2 iodine atoms on the right and match the 2 potassium atoms.
4. Verify that lead (1), potassium (2), nitrogen (2), oxygen (6), and iodine (2) are balanced on both sides.

Answer: Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃

Give one example each of decomposition by electricity and by sunlight.

1. Decomposition by electricity (electrolysis) example: 2H₂O → 2H₂ + O₂.
2. Decomposition by sunlight (photolysis) example: 2AgCl → 2Ag + Cl₂.

Answer: Electrolytic: 2H₂O → 2H₂ + O₂ (electricity). By sunlight: 2AgCl → 2Ag + Cl₂ (or 2AgBr → 2Ag + Br₂).

Explain the chlor-alkali process with its equation and name the products at the anode and the cathode.

1. The chlor-alkali process involves the electrolysis of brine, which is an aqueous solution of sodium chloride (NaCl).
2. When electricity is passed through the brine, it decomposes to form sodium hydroxide, chlorine gas, and hydrogen gas.
3. The balanced chemical equation for this reaction is: 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g).
4. Chlorine gas (Cl₂) is produced at the anode, while hydrogen gas (H₂) is produced at the cathode.
5. Sodium hydroxide (NaOH) accumulates in the solution near the cathode, completing the process that yields both chlorine and alkali.

Answer: Electricity passed through brine (aqueous NaCl) gives sodium hydroxide (alkali) and chlorine, hence the name: 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g). Chlorine gas forms at the anode, hydrogen gas at the cathode (NaOH solution near the cathode).

Give the preparation, with balanced equations, of (i) baking soda, (ii) bleaching powder, (iii) plaster of Paris.

1. Preparation of baking soda: Sodium chloride, water, carbon dioxide, and ammonia react to form ammonium chloride and sodium hydrogen carbonate.
2. Balanced equation: NaCl + H₂O + CO₂ + NH₃ → NH₄Cl + NaHCO₃.
3. Preparation of bleaching powder: Chlorine gas is passed over dry slaked lime (calcium hydroxide).
4. Balanced equation: Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O.
5. Preparation of plaster of Paris: Gypsum (calcium sulfate dihydrate) is heated at 373 K.
6. Balanced equation: CaSO₄·2H₂O → CaSO₄·½H₂O + 1½H₂O.

Answer: (i) NaCl + H₂O + CO₂ + NH₃ → NH₄Cl + NaHCO₃. (ii) Chlorine over dry slaked lime: Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O. (iii) Heat gypsum at 373 K: CaSO₄·2H₂O → CaSO₄·½H₂O + 1½H₂O.

Why are carbonate and sulphide ores first changed into oxides before extracting the metal?

1. Carbonate and sulphide ores are first converted into oxides through processes like calcination or roasting.
2. This is done because metals are obtained far more easily by reducing their oxides than by reducing carbonates or sulphides directly.

Answer: Because a metal is obtained far more easily by reducing its oxide than from its carbonate or sulphide.

Give a reaction where aluminium reduces a metal oxide to the metal.

1. Write the balanced chemical equation for the thermite reaction where aluminium reduces iron(III) oxide: Fe₂O₃(s) + 2Al(s) → 2Fe(l) + Al₂O₃(s) + heat.

Answer: Fe₂O₃(s) + 2Al(s) → 2Fe(l) + Al₂O₃(s) + heat (thermit reaction); or 3MnO₂ + 4Al → 3Mn + 2Al₂O₃ + heat.

How is copper extracted from Cu₂S? Give the equations.

1. Roast the ore in air to convert cuprous sulfide into cuprous oxide and sulfur dioxide: 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂.
2. Heat the cuprous oxide with the remaining cuprous sulfide to undergo self-reduction, yielding copper metal and sulfur dioxide: 2Cu₂O + Cu₂S → 6Cu + SO₂.

Answer: Roast the ore in air: 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂. Then Cu₂O reacts with remaining Cu₂S on heating (self-reduction): 2Cu₂O + Cu₂S → 6Cu + SO₂.

(I) Why can't carbon reduce the oxides of highly reactive metals? (II) Why is solder (lead–tin alloy) used to join electrical wires?

1. (I) Highly reactive metals have a stronger affinity for oxygen than carbon does.
2. (I) Therefore, carbon cannot displace the oxygen from their oxides to reduce them.
3. (II) Solder is an alloy of lead and tin which has a low melting point.
4. (II) This low melting point allows it to melt easily and join electrical wires without damaging them.

Answer: (I) Highly reactive metals hold oxygen more strongly than carbon does, so carbon cannot remove it. (II) Solder has a low melting point.

(i) Explain: (I) covalent compounds conduct electricity poorly; (II) soap gives no lather in hard water; (III) carbon shows catenation but silicon hardly does. (ii) Write equations for (I) oxidation of ethanol with acidified K₂Cr₂O₇ and (II) hydrogenation of ethene.

1. (i)(I) Covalent compounds do not form ions, so there are no charged particles to carry current.
2. (i)(II) Soap reacts with calcium and magnesium salts in hard water to form an insoluble scum, so little lather forms.
3. (i)(III) C–C bonds are strong and stable, so carbon forms long chains; Si–Si bonds are weak, so silicon catenates only a little.
4. (ii)(I) CH₃CH₂OH → CH₃COOH on heating with acidified K₂Cr₂O₇ (ethanol is oxidised to ethanoic acid).
5. (ii)(II) CH₂=CH₂ + H₂ → CH₃–CH₃ in the presence of a nickel catalyst.

Answer: (i)(I) They do not form ions (charged particles). (II) Soap reacts with calcium and magnesium salts in hard water to form insoluble scum. (III) C–C bonds are strong and stable, while Si–Si bonds are weak. (ii)(I) CH₃CH₂OH → CH₃COOH (acidified K₂Cr₂O₇, heat). (II) CH₂=CH₂ + H₂ → CH₃–CH₃ (nickel catalyst).

Ethanol heated with X and a little conc. H₂SO₄ gives a sweet-smelling Y; Y with NaOH gives back ethanol and Z. (i) Identify X, Y, Z. (ii) Role of conc. H₂SO₄? (iii) Equations and names of the reactions.

1. Identify X as ethanoic acid (CH₃COOH) because it reacts with ethanol to form a sweet-smelling ester.
2. Identify Y as ethyl ethanoate (CH₃COOC₂H₅), which is the ester produced in the reaction.
3. Identify Z as sodium ethanoate (CH₃COONa), formed when the ester Y reacts with NaOH.
4. State that conc. H₂SO₄ acts as a catalyst and dehydrating agent to facilitate the reaction.
5. Write the equation for esterification: CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O.
6. Write the equation for saponification: CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH.

Answer: (i) X = ethanoic (acetic) acid, CH₃COOH; Y = ethyl ethanoate, CH₃COOC₂H₅ (an ester); Z = sodium ethanoate, CH₃COONa. (ii) It acts as a catalyst (dehydrating agent). (iii) CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O (acid, heat) — esterification; CH₃COOC₂H₅ + NaOH → CH₃COONa + C₂H₅OH — saponification.

A convex lens with f = 15 cm makes a real image equal in size to the object. Give the object and image positions.

1. A real image of the same size (m = −1) forms only when the object is at 2F of a convex lens.
2. With f = 15 cm, u = −2 × 15 = −30 cm, and the image forms at 2F on the other side, v = +30 cm.
3. Check: 1/v − 1/u = 1/30 + 1/30 = 1/15 = 1/f. So u = −30 cm, v = +30 cm — option (C).

Answer: Same-size real image needs the object at 2F: u = -30 cm and v = +30 cm — option (C).

What happens in the eye when you look at something very near?

1. To see a nearby object, the ciliary muscles contract.
2. The eye lens becomes thicker, its focal length decreases and the image of the near object falls on the retina — option (A).

Answer: The ciliary muscles contract and the eye lens becomes thicker (shorter focal length) — option (A).

Assertion: white light leaving a prism splits into a seven-colour spectrum. Reason: scattering makes red bend least and violet bend most.

1. The assertion is true because white light splits into seven colours when passing through a prism.
2. The reason is false because the separation is due to dispersion (different refractive indices for different colours), not scattering.
3. Therefore, the assertion is true but the reason is false, which corresponds to option (C).

Answer: The assertion is true but the reason is false — the colours separate because they refract by different amounts (dispersion), not by scattering — option (C).

(a) From the ray diagram, which type of lens is used? (b) A real image forms 30 cm from the lens and is twice the object's size. Where was the object?

Uses a figure printed in CBSE's paper — what the figure shows: Object AB stands on the axis close to a lens; the rays from B leave the lens diverging, and their backward extensions (dashed) meet at B' on the same side as the object, giving an enlarged, erect, virtual image A'B' farther from the lens — the magnifier case of a convex lens.

1. (a) The rays diverge after refraction and their backward extensions meet to form an erect, magnified virtual image, which is the characteristic behavior of a convex (converging) lens acting as a magnifier.
2. (b) Given the real image is twice the object's size, the magnification m = -2 (negative sign indicates inversion), and the image distance v = +30 cm.
3. Using the magnification formula m = v/u, substitute the values: -2 = 30/u.
4. Solving for u gives u = 30/(-2) = -15 cm, meaning the object was placed 15 cm in front of the lens.

Answer: (a) Convex (converging) lens. (b) For a real, inverted image m = -2 and v = +30 cm; m = v/u gives u = 30/(-2) = -15 cm, i.e. 15 cm in front of the lens.

Note: The figure shows a convex lens used as a magnifier (virtual image), while part (b) is a separate situation with a real image. A real, inverted image has negative magnification, so m = −2 is used.

A wire of radius 0.01 cm and length 1.0 cm has resistance 7 Ω. Find its resistivity.

1. Convert the given dimensions to SI units: radius r = 0.01 cm = 1 × 10⁻⁴ m and length l = 1.0 cm = 0.01 m.
2. Calculate the cross-sectional area using A = πr², which gives A = (22/7) × (1 × 10⁻⁴)² = (22/7) × 10⁻⁸ m².
3. Substitute the values into the resistivity formula ρ = RA/l to get ρ = (7 × (22/7) × 10⁻⁸) / 0.01.
4. Simplify the expression to find the final resistivity: ρ = 22 × 10⁻⁸ / 10⁻² = 2.2 × 10⁻⁵ Ω m.

Answer: r = 1 × 10⁻⁴ m, l = 0.01 m, A = πr² = (22/7) × 10⁻⁸ m². ρ = RA/l = 7 × (22/7) × 10⁻⁸ / 0.01 = 2.2 × 10⁻⁵ Ω m.

A heater rated 220 V, 11 A is run on 200 V. Find the power it uses.

1. Calculate the resistance of the heater using its rated values: R = V/I = 220/11 = 20 Ω.
2. Calculate the power consumed at the new voltage: P = V²/R = 200²/20 = 40000/20 = 2000 W.

Answer: R = V/I = 220/11 = 20 Ω. At 200 V, P = V²/R = 200 × 200 / 20 = 2000 W (2 kW).

A person cannot read a book held nearer than 1 m. Name the defect and draw ray diagrams of the defect and its correction.

1. The person sees distant objects but cannot read nearer than 1 m, so the defect is hypermetropia (far-sightedness).
2. The near point has moved from the normal 25 cm to 1 m.
3. Defect diagram: rays from an object at 25 cm enter the eye and would meet behind the retina, so the image is blurred.
4. Correction diagram: a convex (converging) lens is placed in front of the eye.
5. The convex lens bends the rays from the object at 25 cm so they appear to come from the near point at 1 m, and the eye then focuses them on the retina.

Answer: Hypermetropia (far-sightedness); the near point has moved away to 1 m. Diagram 1: rays from an object at the normal near point (25 cm) would meet behind the retina. Diagram 2: a convex (converging) lens placed before the eye makes those rays appear to come from the far near point N', so the image forms on the retina.

(a) Describe an activity showing that a current-carrying conductor in a magnetic field feels a force. (b) You sit facing the front wall; an electron beam travelling horizontally from the back wall to the front wall is pushed to your right by a strong magnetic field. Find the field direction.

1. Suspend a small aluminium rod horizontally using two connecting wires from a stand.
2. Place the rod between the poles of a strong horseshoe magnet so it is perpendicular to the magnetic field.
3. Connect the rod in series with a battery and a key to complete the circuit.
4. On closing the key, the rod moves or is displaced, demonstrating that a force acts on the current-carrying conductor.
5. Identify that the electron beam travels from the back wall to the front wall, meaning the conventional current flows from the front wall to the back wall.
6. Apply Fleming's left-hand rule with the thumb pointing to the right (force) and the middle finger pointing backward (current).
7. Determine that the forefinger points vertically downwards, indicating the direction of the magnetic field.

Answer: (a) Hang a small aluminium rod horizontally from a stand by two wires, place it between the poles of a strong horseshoe magnet at right angles to the field, and connect it in series with a battery and key. On closing the key the rod moves (is displaced), showing a force acts on it. (b) The electron flow is forward, so the current is backward; with the force to the right, Fleming's left-hand rule gives a magnetic field acting vertically downwards.

(a) Show that the field pattern of a current-carrying wire depends on the wire's shape. (b) A straight wire AB carries current upward; points X, Y, Z lie on a line to one side at increasing distance. Where is the field strongest, and why?

Uses a figure printed in CBSE's paper — what the figure shows: Vertical wire AB with current i flowing from B up to A; a horizontal dashed line from the wire passes through X, then Y, then Z, X being closest to the wire.

1. For part (a), state that a straight current-carrying wire produces magnetic field lines in the form of concentric circles around it.
2. Mention that a circular loop produces field lines that are circular near the wire but become nearly straight and parallel at the center.
3. Note that a solenoid produces a field pattern similar to a bar magnet, with almost uniform parallel lines inside.
4. For part (b), identify that point X is the closest to the wire among X, Y, and Z.
5. State that the magnetic field strength of a straight wire decreases as the distance from the wire increases.
6. Conclude that the field is strongest at X because it is at the minimum distance from the current-carrying wire.

Answer: (a) Any two: a straight wire gives concentric circles around it; a circular loop gives circles near the wire that become nearly straight and parallel at the centre; a solenoid gives a pattern like a bar magnet with almost uniform parallel lines inside. (b) At X, because the field of a straight wire weakens as the distance from the wire increases, and X is nearest.

An object is 40 cm from a convex lens of focal length 20 cm. Give the image position and nature.

1. For a convex lens f = +20 cm and the object distance u = −40 cm.
2. Lens formula: 1/v = 1/f + 1/u = 1/20 − 1/40 = 2/40 − 1/40 = 1/40, so v = +40 cm.
3. The image is 40 cm on the other side of the lens (at 2F); it is real, inverted and the same size as the object.

Answer: f = +20 cm, u = -40 cm; 1/v = 1/f + 1/u = 1/20 - 1/40 = 1/40, so v = +40 cm (at 2F on the other side). Real, inverted and the same size.

Draw the ray diagram for an object between the optical centre and the focus of a concave lens.

1. Draw the principal axis, optical centre O, and focus F on the left side of the concave lens.
2. Place the object between O and F on the left side of the lens.
3. Draw a ray from the top of the object parallel to the principal axis; after refraction, it diverges such that its backward extension passes through F.
4. Draw a second ray from the top of the object passing through the optical centre O; it continues straight without deviation.
5. Extend the refracted rays backwards until they meet at a point between F and O on the left side.
6. Mark the image at this intersection point; it is virtual, erect, and diminished.

Answer: A ray parallel to the axis diverges as if coming from F on the object side; a ray through the optical centre goes straight. Their backward extensions meet between F and O: the image is virtual, erect and diminished, on the same side as the object.

A convex lens (f = 30 cm) and a concave lens (f = 15 cm) are in contact. Find the combined focal length and power.

1. Take signs: convex lens f₁ = +30 cm = +0.30 m; concave lens f₂ = −15 cm = −0.15 m.
2. P₁ = 1/0.30 = +3.33 D and P₂ = 1/(−0.15) = −6.67 D.
3. Lenses in contact: P = P₁ + P₂ = 3.33 − 6.67 = −3.33 D.
4. f = 1/P = 1/(−3.33) ≈ −0.30 m = −30 cm, so the combination acts as a diverging lens.

Answer: P₁ = 1/0.30 m = +3.33 D, P₂ = 1/(-0.15 m) = -6.67 D; P = P₁ + P₂. Result: f = -30 cm, P = -3.33 D (the combination is diverging).

A concave lens of focal length 2 m touches a convex lens of focal length 1.5 m. Is the pair convex or concave in effect? Why?

1. Concave lens: P₁ = −1/2 = −0.5 D. Convex lens: P₂ = +1/1.5 ≈ +0.67 D.
2. Total power P = −0.5 + 0.67 = +1/6 D ≈ +0.17 D, so f = 1/P = +6 m.
3. The power is positive because the convex lens has the shorter focal length and is the stronger lens, so the pair acts as a convex lens.

Answer: Convex. P₁ = -1/2 = -0.5 D and P₂ = +1/1.5 = +0.67 D, so P = +1/6 D, giving f = +6 m. The total power is positive because the convex lens, with the shorter focal length, is the stronger lens.

(i) Changing a conductor's length or cross-section alters its resistance but not its resistivity — why? (ii) Why are heating elements of toasters and irons made of alloys and not pure metals? (iii) Define the SI unit of electric current.

1. (i) Resistance is given by R = ρl/A, which shows it depends on length and area, but resistivity ρ is an intrinsic property of the material, so it remains unchanged.
2. (ii) Alloys are used because they have higher resistivity than pure metals, generating more heat, and they do not oxidise or burn easily at high temperatures.
3. (iii) The SI unit of electric current is the ampere, defined as the current when one coulomb of charge flows through a conductor in one second.

Answer: (i) R = ρl/A depends on length and area, but resistivity ρ is a property of the material itself, so it stays the same. (ii) Alloys have higher resistivity than their constituent metals and do not oxidise (burn) easily at high temperature. (iii) One ampere is the current when one coulomb of charge flows per second (1 A = 1 C/1 s).

(i) How many 8 Ω bulbs in parallel make a 4 V battery give 2 A? (ii) Which device measures current and how is it connected? (iii) State Joule's law of heating.

1. (i) Calculate the total resistance required using Ohm's law: R = V/I = 4/2 = 2 Ω.
2. For n bulbs of 8 Ω in parallel, the equivalent resistance is R = 8/n.
3. Equate the required resistance to the parallel formula and solve for n: 8/n = 2, so n = 4 bulbs.
4. (ii) The device used to measure current is an ammeter.
5. An ammeter is connected in series with the circuit.
6. (iii) Joule's law states that heat produced is directly proportional to the square of current, resistance, and time.
7. The formula for heat produced is H = I²Rt.

Answer: (i) Needed R = V/I = 4/2 = 2 Ω. For n equal bulbs in parallel R = 8/n, so 8/n = 2 and n = 4 bulbs. (ii) An ammeter, connected in series. (iii) Heat produced in a conductor is directly proportional to the square of the current, its resistance and the time for which current flows: H = I²Rt.