CBSE Class 10 Science: 2026 board paper, set 31/5/1

1. Each stomatal pore is bordered by a pair of guard cells. When they take in water they swell and the pore opens; when they lose water they shrink and the pore closes. So the correct option is (C) Guard cells.

   1 mark

Answer: (C) Guard cells

1. The pituitary gland secretes growth hormone, which regulates growth and development of the body. Too little of it in childhood causes dwarfism. Answer: (A).

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Answer: (A) Pituitary gland

1. Glycolysis splits glucose into pyruvate in the cytoplasm without using oxygen.

2. When oxygen is short in muscle cells, pyruvate is converted to lactic acid with release of energy. The build-up of lactic acid causes cramps. Answer: (D).

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Answer: (D)

1. Fertilisation happens in the oviduct. The zygote then moves down and implants in the thick, blood-rich lining of the uterus. Answer: (C).

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Answer: (C) Uterus

1. The egg carries 22 + X and the sperm carries 22 + Y.

2. The zygote therefore has 44 + XY, which gives a male child. Answer: (C).

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Answer: (C) 44 + XY

1. Paper, wood and vegetable peel are broken down by microbes. DDT is a synthetic pesticide that microbes cannot decompose, so it persists in the environment. Answer: (B).

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Answer: (B) DDT

1. Decomposers such as bacteria and fungi break down complex organic matter of dead organisms into simple inorganic substances. These return to the soil and are reused by plants. Answer: (B).

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Answer: (B)

1. A is true: heat-resistant variants survive a heat wave better than the others.

2. R is true, and it explains A: the heat tolerance is itself a variation, and variations improve survival in a changing environment. Answer: (A).

   1 mark

Answer: (A)

1. A is true: peristalsis occurs all along the gut, from the oesophagus to the intestines.

2. R is true and is the cause of A: the rhythmic contraction of muscles in the gut lining is exactly what produces these peristaltic movements. Answer: (A).

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Answer: (A)

1. Draw a sequence of 3–4 labelled sketches of Amoeba (irregular cell with nucleus). Stage 1 – Ingestion: the Amoeba senses a food particle and pushes out finger-like pseudopodia that surround it.

   ½ mark

2. Stage 2: the tips of the pseudopodia fuse, enclosing the food with a little water in a food vacuole.

   ½ mark

3. Stage 3 – Digestion and absorption: digestive enzymes enter the food vacuole and break the food into simple soluble substances, which diffuse into the cytoplasm and are assimilated.

   ½ mark

4. Stage 4 – Egestion: the vacuole carrying the undigested residue moves to the cell surface and throws it out. Label in the diagram: pseudopodia, food particle, food vacuole, nucleus.

   ½ mark

Answer: Labelled sequence: pseudopodia surround food → food vacuole forms → digestion and absorption into cytoplasm → egestion of undigested waste.

1. P – Receptor (sense organ in the skin of the hand that detects heat).

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2. Q – Sensory neuron (carries the impulse from the hand to the spinal cord).

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3. R – Relay neuron (in the spinal cord, passes the impulse from sensory to motor neuron).

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4. S – Effector (muscles of the arm that contract to pull the hand away).

   ½ mark

Answer: P – receptor (skin); Q – sensory neuron; R – relay neuron; S – effector (arm muscle)

1. (i) Promoting cell division – Cytokinin.

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2. (ii) Inhibition of growth – Abscisic acid (ABA).

   ½ mark

3. (iii) Response to light (phototropism) – Auxin, which moves to the shaded side and makes cells there grow longer.

   ½ mark

4. (iv) Wilting of leaves – Abscisic acid (ABA).

   ½ mark

Answer: (i) Cytokinin; (ii) Abscisic acid; (iii) Auxin; (iv) Abscisic acid

1. Oxygen-rich blood from the lungs enters the left atrium through the pulmonary veins; the left atrium contracts and sends it into the left ventricle.

   ½ mark

2. The left ventricle contracts and pumps this blood into the aorta, which carries it to all parts of the body.

   ½ mark

3. Deoxygenated blood from the body returns through the vena cava to the right atrium, which passes it into the right ventricle.

   ½ mark

4. The right ventricle pumps it through the pulmonary artery to the lungs for oxygenation. Thus blood passes through the heart twice in one cycle – double circulation.

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Answer: Lungs → left atrium → left ventricle → aorta → body → right atrium → right ventricle → lungs (blood passes twice through the heart).

1. Parents: Round yellow (RRYY) × Wrinkled green (rryy). Gametes: RY and ry. F₁: all RrYy – round, yellow seeds.

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2. F₁ × F₁ (selfing): RrYy × RrYy. Gametes of each: RY, Ry, rY, ry.

   	RY	Ry	rY	ry
   RY	RRYY	RRYy	RrYY	RrYy
   Ry	RRYy	RRyy	RrYy	Rryy
   rY	RrYY	RrYy	rrYY	rrYy
   ry	RrYy	Rryy	rrYy	rryy

   Phenotypic ratio 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.

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3. (i) RRYY – 1 out of 16.

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4. (ii) rryy – 1 out of 16.

   ½ mark

Answer: F₁ all RrYy (round yellow); F₂ 9:3:3:1; (i) RRYY = 1; (ii) rryy = 1 (out of 16)

1. (a) Harmful effect: plastic is non-biodegradable, so bags stay in soil and water for very long, choking drains and harming animals that swallow them.

   ½ mark

2. Better alternatives: jute bags or cloth bags (also paper bags).

   ½ mark

3. (b) The pesticides and fertilisers are non-biodegradable. From soil/water they are absorbed by plants and aquatic organisms, then pass along the food chain, accumulating progressively at each trophic level. Humans, at the top of food chains, get the highest concentration in their bodies.

   1 mark

4. This phenomenon is called biological magnification.

   1 mark

Answer: (a) Non-biodegradable, persists in environment; use jute/cloth bags. (b) Enter food chain and accumulate at each trophic level, maximum in humans – biological magnification.

1. (a) Structure: Bowman's capsule is the cup-shaped beginning of a nephron, enclosing a cluster of capillaries (glomerulus).

   ½ mark

2. Function: it receives the filtrate filtered from the blood in the glomerulus.

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3. (b) As the filtrate flows along the tubule of the nephron, most of the water and useful substances like glucose, amino acids and salts are selectively reabsorbed into the blood; so only a litre or two remains as urine.

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4. (c) Excretion is the biological process of removing harmful metabolic (nitrogenous) wastes such as urea and uric acid from the body.

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5. It is necessary because these wastes are toxic; if they accumulate they harm the body and the organism cannot survive.

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6. (c) OR – Similarity 1: both remove metabolic wastes from the blood (lungs remove CO₂, kidneys remove urea).

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7. Similarity 2: both contain thin-walled blood capillaries – in alveoli for gaseous exchange and in glomeruli for filtration.

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Answer: (a) Cup-shaped start of nephron; collects filtrate. (b) Selective reabsorption of water and useful substances. (c) Removal of toxic metabolic wastes, needed as they are harmful. OR: both remove wastes; both have thin-walled capillaries.

1. (i)(a) Zygote: divides repeatedly to form the embryo inside the ovule.

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2. (b) Ovule: develops a tough coat and gradually becomes the seed.

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3. (c) Ovary: grows rapidly and ripens to form the fruit.

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4. (d) Sepals: shrivel and fall off (along with petals and stamens).

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5. (ii) Germination is the process by which the embryo in a mature seed grows into a seedling under suitable conditions of water, air and temperature.

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Answer: Zygote → embryo; ovule → seed; ovary → fruit; sepals shrivel and fall; germination = seed developing into a seedling under suitable conditions.

1. (i) Sugarcane reproduces by vegetative propagation (through stem cuttings with nodes).

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2. (ii) Advantage 1: plants raised this way bear flowers and fruits earlier than those grown from seeds.

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3. Advantage 2: the new plants are genetically similar to the parent, so desirable characters are preserved; it also lets seedless plants (e.g. banana) reproduce.

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4. (iii) Examples: rose and grapes (also banana, orange, jasmine).

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5. (iv) Regeneration needs specialised cells that can divide and differentiate into all tissues. Only simple organisms like Hydra and Planaria have such cells; complex animals lack them, so they cannot regenerate a whole body.

   1 mark

Answer: (i) Vegetative propagation; (ii) early flowering/fruiting, offspring identical to parent; (iii) rose, grapes; (iv) only organisms with specialised regenerative cells can regenerate.

1. The electrolysis reaction is $2H_{2}O(l)\to 2H_2(g)+O_2(g)$.

2. Two volumes of hydrogen form for every one volume of oxygen, so the ratio is 2 : 1. Answer: (B).

   1 mark

Answer: (B) 2 : 1

1. $C_4{H}_8$ fits $C_n{H}_{2n}$, so it is an alkene. $C_5{H}_{12}$ fits $C_n{H}_{2n+2}$, so it is an alkane.

2. Every other pair contains only alkanes. So (D) is the pair that is not from the same homologous series.

   1 mark

Answer: (D)

1. $A{l}_2{O}_3$ reacts with acids and also with bases to form salt and water, so it is amphoteric.

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2. $N{a}_{2}O$ and $K_{2}O$ are basic oxides, and $C{O}_2$ is acidic. Answer: (D).

Answer: (D) Al₂O₃

1. CH₃COONa comes from a weak acid (CH₃COOH) and a strong base (NaOH), so the salt is basic. Row (D) is correct.

   1 mark

2. The other rows are wrong: NaCl is neutral, Na₂CO₃ is basic and Na₂SO₄ is neutral.

Answer: (D)

1. $C_3{H}_8$ (propane) is a saturated alkane with only single bonds, so it cannot undergo an addition reaction.

   1 mark

2. The other three compounds have double or triple bonds and do undergo addition. Answer: (C).

Answer: (C) C₃H₈

1. HCl and NaOH ionise in water, giving free ions that carry current, so the bulb glows.

   1 mark

2. Fructose and ethanol do not ionise in water, so the bulb does not glow with them. Answer: (D).

Answer: (D)

1. HNO₃ is a strong oxidising agent, so it usually oxidises any H₂ formed to water. Very dilute HNO₃ is an exception for Mg and Mn.

2. With very dilute HNO₃, Mg releases hydrogen gas: $Mg+2HN{O}_3\to Mg(N{O}_3{)}_2+H_2$. Answer: (C).

   1 mark

Answer: (C) Hydrogen

1. The reaction is $CaO(s)+H_{2}O(l)\to Ca(OH{)}_2(aq)$ + heat, so A is true.

2. R is true, and release of a large amount of heat is what makes a reaction exothermic, so R explains A. Answer: (A).

   1 mark

Answer: (A)

1. (i) Olfactory indicators are substances whose odour changes in acidic or basic medium, e.g. onion, vanilla essence, clove oil.

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2. (ii) Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt, e.g. 5 in $\mathrm{CuS}{\mathrm{O}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$.

   1 mark

Answer: (i) Substances whose smell changes in acid/base; (ii) fixed number of water molecules in one formula unit of a salt.

1. (i) Lead nitrate decomposes on heating and brown fumes of nitrogen dioxide are given off; yellow lead oxide remains.
   $2\mathrm{Pb}(\mathrm{N}{\mathrm{O}}_3{)}_2(\mathrm{s})\stackrel{\text{Heat}}{\to }2\mathrm{PbO}(\mathrm{s})+4\mathrm{N}{\mathrm{O}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})$

   1 mark

2. (ii) Green ferrous sulphate crystals change colour, forming reddish-brown ferric oxide with a smell of burning sulphur.
   $2\mathrm{FeS}{\mathrm{O}}_4(\mathrm{s})\stackrel{\text{Heat}}{\to }\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3(\mathrm{s})+\mathrm{S}{\mathrm{O}}_2(\mathrm{g})+\mathrm{S}{\mathrm{O}}_3(\mathrm{g})$

   1 mark

3. (iii) White silver chloride turns grey as silver metal forms and chlorine is released.
   $2\mathrm{AgCl}(\mathrm{s})\stackrel{\text{Sunlight}}{\to }2\mathrm{Ag}(\mathrm{s})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})$

   1 mark

Answer: (i) Brown NO₂ fumes: 2Pb(NO₃)₂ → 2PbO + 4NO₂ + O₂; (ii) burning-sulphur smell, colour change: 2FeSO₄ → Fe₂O₃ + SO₂ + SO₃; (iii) white AgCl turns grey: 2AgCl → 2Ag + Cl₂

1. (a) Chemical name: calcium sulphate hemihydrate; formula $\mathrm{CaS}{\mathrm{O}}_4\cdot \frac{1}{2}{\mathrm{H}}_2\mathrm{O}$.

   1 mark

2. (b) Gypsum is heated at 373 K:
   $\mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}\stackrel{373K}{\to }\mathrm{CaS}{\mathrm{O}}_4\cdot \frac{1}{2}{\mathrm{H}}_2\mathrm{O}+\frac{3}{2}{\mathrm{H}}_2\mathrm{O}$

   1 mark

3. (c) Uses: (1) supporting fractured bones in the right position by doctors; (2) making toys, decorative materials and smooth surfaces.

   1 mark

Answer: (a) Calcium sulphate hemihydrate, CaSO₄·½H₂O; (b) CaSO₄·2H₂O → (373 K) CaSO₄·½H₂O + 3/2 H₂O; (c) plastering fractured bones, making toys.

1. (a) Ant sting contains methanoic acid (formic acid), HCOOH.

   1 mark

2. (b)(i) Mixing acid with water is highly exothermic. Adding acid slowly to a large amount of water spreads the heat, so there is no excessive local heating, splashing or burns; adding water to acid can boil the mixture and break the container.

   1 mark

3. (b)(ii) Baking soda (NaHCO₃) is a mild base; it neutralises the excess acid in the stomach and gives relief from acidity.

   1 mark

Answer: (a) Methanoic (formic) acid; (b)(i) avoids excessive local heating and splashing; (ii) mild alkali neutralises excess stomach acid.

1. (a) Fourth member (n = 4): $\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_2-\mathrm{OH}$ (draw all C–H bonds with –OH on the end carbon).

   ½ mark

2. Name: Butanol (${\mathrm{C}}_4{\mathrm{H}}_9\mathrm{OH}$).

   ½ mark

3. (b) Ethanol is oxidised to ethanoic acid by alkaline KMnO₄ (oxidising agent).

   ½ mark

4. $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{Alkaline }KMn{O}_4+\text{Heat}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$

   ½ mark

5. (c) $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{Conc. }{H}_{2}S{O}_4}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$ (ethyl ethanoate, sweet-smelling ester)

   1 mark

6. This reaction is called esterification.

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7. (c) OR – Ethanol is dehydrated to ethene.

   ½ mark

8. $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{Hot conc. }{H}_{2}S{O}_4,443K}{\to }\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$

   ½ mark

9. Role of conc. H₂SO₄: dehydrating agent – it removes water from ethanol.

   1 mark

Answer: (a) CH₃CH₂CH₂CH₂OH, butanol; (b) ethanoic acid formed; (c) CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O, esterification. OR: ethene formed, conc. H₂SO₄ is dehydrating agent.

1. (i)(I) In ionic compounds oppositely charged ions are held by strong electrostatic (inter-ionic) forces; a large amount of energy is needed to break them, so melting and boiling points are high.

   1 mark

2. (i)(II) Solder has a low melting point, so it melts easily and joins electrical wires without damaging them.

   1 mark

3. (i)(III) Na and Mg are highly reactive and have more affinity for oxygen than carbon, so carbon cannot remove oxygen from their oxides; they are obtained by electrolytic reduction.

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4. (ii)(I) X is iron(III) oxide, Fe₂O₃.

   ½ mark

5. (II) The reaction is the thermit reaction.

   ½ mark

6. (III) $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3(\mathrm{s})+2\mathrm{Al}(\mathrm{s})\to 2\mathrm{Fe}(\mathrm{l})+\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3(\mathrm{s})+\text{Heat}$ – molten iron fills the gap between rails.

   1 mark

Answer: (i) strong inter-ionic forces; low melting point of solder; Na/Mg have greater affinity for oxygen than C. (ii) X = Fe₂O₃; thermit reaction; Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + heat.

1. (i)(I) Self-reduction of copper: $2\mathrm{C}{\mathrm{u}}_2\mathrm{O}+\mathrm{C}{\mathrm{u}}_2\mathrm{S}\stackrel{\text{Heat}}{\to }6\mathrm{Cu}(\mathrm{s})+\mathrm{S}{\mathrm{O}}_2(\mathrm{g})$

   1 mark

2. (i)(II) Roasting of zinc blende: $2\mathrm{ZnS}(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})\stackrel{\text{Heat}}{\to }2\mathrm{ZnO}(\mathrm{s})+2\mathrm{S}{\mathrm{O}}_2(\mathrm{g})$

   1 mark

3. (ii)(I) PVC is a poor conductor of electricity; the coating insulates current-carrying wires and prevents shocks and short circuits.

   1 mark

4. (ii)(II) Copper does not react with cold water, hot water or steam, so it does not corrode, whereas steel (iron) rusts; copper is also a better conductor of heat.

   1 mark

5. (iii) Ca (2, 8, 8, 2) loses 2 electrons; O (2, 6) gains them:
   $\mathrm{Ca}:+\cdot \ddot{\underset{\cdot \cdot }{O}}\cdot \to \mathrm{C}{\mathrm{a}}^{2+}[:\ddot{\underset{\cdot \cdot }{O}}:{]}^{2-}$ i.e. $\mathrm{C}{\mathrm{a}}^{2+}{\mathrm{O}}^{2-}$ (CaO). Show the two electrons transferring by arrows from Ca to O.

   1 mark

Answer: (i) 2Cu₂O + Cu₂S → 6Cu + SO₂; 2ZnS + 3O₂ → 2ZnO + 2SO₂. (ii) PVC insulates; copper does not react with water/steam (steel rusts). (iii) Ca transfers 2 electrons to O → Ca²⁺[O]²⁻.

1. Rays from the sun meet at the principal focus, so f = 25 cm.

2. A convex lens forms a virtual image only when the object is between the lens and F, that is, closer than 25 cm. Of the options, only 20 cm is less than 25 cm. Answer: (A).

   1 mark

Answer: (A) 20 cm

1. When the eyeball is too short, the image of a near object forms behind the retina. This defect is hypermetropia.

2. It is corrected with a convex (converging) lens. Answer: (A).

   1 mark

Answer: (A) Converging lens

1. A is false: in a strong external magnetic field the needle aligns along that field, not along the earth's north-south direction.

2. R is true: a compass needle is a small magnet and behaves like a freely suspended bar magnet. Answer: (D).

   1 mark

Answer: (D)

1. (a) Speed of light is more in an optically rarer medium and less in an optically denser medium, i.e. the greater the optical density, the smaller the speed of light.

   1 mark

2. (b) Speeds: A ($2.25\times 10^8$) > C ($2.08\times 10^8$) > B ($2\times 10^8$ m/s). Lower speed means higher optical density, so ascending order of optical density: A < C < B.

   1 mark

Answer: (a) Higher optical density → lower speed of light; (b) A < C < B

1. (i) For nearby objects the ciliary muscles contract, so the eye lens becomes thicker (curvature increases); its focal length decreases, and the image of the near object is formed sharply on the retina.

   1 mark

2. (ii) A normal eye can see clearly from its near point 25 cm to its far point at infinity.

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Answer: (i) Curvature increases, focal length decreases, image focused on retina; (ii) 25 cm to infinity

1. The man's near point has receded to 1 m (hypermetropia/presbyopia), so the defect is corrected by a convex (converging) lens. (For reference: $u=-25$ cm, $v=-100$ cm; $\frac{1}{f}=\frac{1}{-100}-\frac{1}{-25}=\frac{3}{100}$, $f\approx +33.3$ cm, $P=+3$ D.)

2. Diagram (a) – defective eye: rays from an object N at 25 cm, after passing through the eye lens, meet at a point behind the retina; mark the near point N′ at 1 m.

   1 mark

3. Diagram (b) – corrected eye: place a convex lens in front of the eye; rays from N (25 cm) are converged so they appear to come from N′ (1 m) and the image forms on the retina. Show arrows on all rays to indicate the direction of light.

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Answer: Convex lens (about +3 D) placed before the eye makes rays from 25 cm appear to come from 1 m and focus on the retina.

1. (a) Ray diagram: Draw a triangular glass prism $PQR$. A ray $EF$ strikes face $PQ$ at $F$ and bends towards the normal $N_1{N}_1^{\prime }$ (air → glass), travelling as $FG$ inside the prism. At face $PR$ it emerges along $GH$, bending away from the normal $N_2{N}_2^{\prime }$ (glass → air). Arrows show the direction of light. Produce the incident ray $EF$ forward and the emergent ray $GH$ backward (dotted) so that they meet.

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2. Labelling: mark angle of incidence $\angle i$ between $EF$ and $N_1{N}_1^{\prime }$, **angle of refraction $\angle r$** between $FG$ and the normal inside the prism, and angle of emergence $\angle e$ at $G$.

   ½ mark

3. Mark the **angle of deviation $\angle D$** as the angle between the produced incident ray and the emergent ray.

   ½ mark

4. (b) The angle of deviation does not change.

   ½ mark

5. Reason: by the principle of reversibility of light, the reversed ray follows exactly the same path backwards; the emergent ray becomes the incident ray, so $\angle e$ and $\angle i$ interchange and the ray retraces its original path, giving the same deviation.

   ½ mark

Answer: (a) Labelled prism diagram with $\angle r$ and $\angle D$ marked; (b) angle of deviation remains the same, as the ray retraces its path.

1. (a) Draw a concave mirror with pole $P$, principal focus $F$ and centre of curvature $C$ on the principal axis. Draw a ray from the object passing through (or directed from) $F$ towards the mirror; after reflection it travels parallel to the principal axis. Show arrows on both rays and label $P$, $F$, $C$ and the principal axis.

   1 mark

2. (b)(i) Virtual image by a convex mirror is always diminished and is formed for all positions of the object; virtual image by a concave mirror is enlarged and is formed only when the object lies between $P$ and $F$.

   1 mark

3. (b)(ii) $m=\frac{h^{\prime }}{h}$; image is inverted (since $m$ is negative), so $h^{\prime }=-20$ cm.

   ½ mark

4. $h=\frac{h^{\prime }}{m}=\frac{-20}{-2}=10$ cm. Height of object = 10 cm.

   ½ mark

Answer: (b)(i) Convex: diminished, for all positions; concave: enlarged, only when object between P and F; (b)(ii) h = 10 cm

1. (a) A heater has low resistance and draws a large current when switched on. The total current in the supply line rises sharply, so the voltage drop across the connecting wires increases and less voltage/current is available to the bulb in parallel; hence the bulb glows dim.

   1 mark

2. (b) Diagram: Connect two $8\Omega$ resistors in parallel, and connect this parallel pair in series with the third $8\Omega$ resistor between points A and B.

   1 mark

3. Justification: parallel pair $\frac{1}{R_1}=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}\Rightarrow R_1=4\Omega$. In series with $8\Omega$: $R_{eq}=R_1+8=4+8=12\Omega$.

   1 mark

Answer: (a) Heater draws large current, reducing current/voltage for the bulb; (b) two 8 Ω in parallel (4 Ω) in series with third 8 Ω gives 12 Ω

1. (a) $R\propto l$; doubling the length doubles resistance, so with the same cell $I=V/R$ is halved. Ammeter reading = $\frac{X}{2}$ A.

   1 mark

2. (b) $R\propto \frac{1}{A}$; doubling the area halves the resistance, so current doubles. Ammeter reading = $2X$ A.

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3. (c) Resistivity ($\rho$) of a material is the resistance of a conductor of that material having unit length and unit area of cross-section ($R=\rho \frac{l}{A}$).

   1 mark

4. SI unit: ohm metre ($\Omega$m).

   ½ mark

5. Resistivity of an alloy is generally higher than that of its constituent metals.

   ½ mark

6. (c) OR (i) Tungsten has a very high melting point (about 3380 °C), so it can be heated till it glows (white hot) without melting.

   1 mark

7. (c) OR (ii) An alloy has higher resistivity than pure metals, so it produces more heat, and it does not oxidise (burn) readily at high temperatures.

   1 mark

Answer: (a) X/2 A; (b) 2X A; (c) resistivity = resistance of unit length and unit cross-section area, unit Ω m, alloy has higher resistivity; OR (i) high melting point; (ii) high resistivity and no oxidation at high temperature

1. (i) Diagram: Draw a bar magnet with N and S marked. Draw closed curved field lines coming out of the N pole and entering the S pole outside the magnet, with arrows; lines are crowded near the poles and spread out away from them; inside the magnet arrows go from S to N.

   1 mark

2. Characteristic 1: Field lines emerge from the north pole and merge at the south pole outside the magnet, and go from south to north inside, so they are closed curves.

   1 mark

3. Characteristic 2: No two field lines cross each other (otherwise there would be two directions of the field at one point). [Also: closeness of lines shows relative strength of field.]

   1 mark

4. (ii) Right-hand thumb rule: Hold the current-carrying straight conductor in the right hand with the thumb pointing along the direction of current; the direction in which the fingers curl around the conductor gives the direction of the magnetic field lines.

   1 mark

5. (iii) The magnetic field becomes weaker (decreases) on the larger circles, i.e. it decreases as the distance from the wire increases ($B\propto \frac{1}{r}$).

   1 mark

Answer: (i) Diagram + closed curves N→S outside, lines never cross; (ii) thumb along current, fingers give field direction; (iii) field decreases on larger circles

1. (i) Diagram: Draw a solenoid (coil of many turns) connected to a battery and key. Inside the coil draw parallel straight field lines with arrows; outside, draw lines looping from one end to the other, like a bar magnet; mark the ends N and S as per the current direction.

   1 mark

2. Feature 1: The field pattern of a current-carrying solenoid is similar to that of a bar magnet; one end behaves as north pole and the other as south pole.

   1 mark

3. Feature 2: Inside the solenoid the field lines are parallel straight lines, showing that the magnetic field is uniform there.

   1 mark

4. (ii) An electromagnet is a piece of magnetic material (soft iron) that behaves as a strong magnet only while electric current flows through a coil around it.

   1 mark

5. It is made by winding insulated copper wire as a coil (solenoid) around a soft iron core and passing current through the coil; the soft iron gets magnetised.

   1 mark

Answer: (i) Diagram; field like a bar magnet, parallel straight lines (uniform field) inside; (ii) electromagnet = soft iron magnetised by current in coil, made by winding insulated copper wire around soft iron core