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

1. Gaseous exchange in a leaf (${\text{CO}}_2$ in, ${\text{O}}_2$ out during photosynthesis) happens through tiny pores on the leaf surface. These pores are the stomata, and guard cells open and close them.

2. Xylem carries water and phloem carries food. The cuticle is a waxy layer that cuts down water loss. None of these exchanges gases. Correct option: (B) Stomata.

   1 mark

Answer: (B) Stomata

1. The figure shows a single amoeba-like cell. Its nucleus divides into two, the cell constricts in the middle, and it splits into two equal daughter cells.

2. A parent cell splitting into two equal halves is binary fission. Budding would show a small outgrowth, and spore formation would need sporangia. Correct option: (D).

   1 mark

Answer: (D) Binary Fission

1. Spirogyra reproduces by fragmentation. Leishmania reproduces by binary fission. Hydra reproduces by budding (and regeneration).

2. All three are asexual methods. Leishmania is unicellular while the other two are multicellular, so (B) and (C) are wrong. Correct option: (D) they reproduce asexually.

   1 mark

Answer: (D) they reproduce asexually

1. The cerebellum, a part of the hindbrain, controls the precision of voluntary actions and maintains posture and balance of the body (e.g. walking in a straight line, picking up a pencil). The cerebrum is the seat of thinking and initiates voluntary actions, while the medulla and pons regulate involuntary actions. Hence the answer is option (B) Cerebellum.

   1 mark

Answer: (B) Cerebellum

1. Pancreatic juice contains trypsin, which digests proteins, and lipase, which breaks down emulsified fats.

2. Only option (B) matches both enzymes correctly. Correct option: (B).

   1 mark

Answer: (B) Trypsin digests proteins and lipase digests emulsified fats.

1. A woman is XX, so every egg carries an X chromosome. A man is XY, so half his sperms carry X and half carry Y.

2. Sex of the child depends on the sperm. An X-carrying sperm from the father gives a girl (XX). A Y-carrying sperm from the father gives a boy (XY). So the deciding gametes are (ii) and (iii). Correct option: (A).

   1 mark

Answer: (A) (ii) and (iii)

1. Biodegradable substances are broken down by decomposers such as bacteria and fungi. Polythene, synthetic rubber and the plastic of a ball pen are not broken down by microbes. All the other groups contain natural materials that decay. Hence the answer is option (C).

   1 mark

Answer: (C) Polythene bag, rubber band, ball pen

1. Assertion: a reflex action is a quick, automatic response, so it happens without conscious thinking. True.

2. Reason: most reflex arcs pass through the spinal cord, which sends the response directly without involving the thinking part of the brain. True, and this explains why no thinking is involved. Correct option: (A).

   1 mark

Answer: (A)

1. Assertion: haemoglobin in red blood cells is the respiratory pigment in humans. True.

2. Reason: haemoglobin has a very high affinity for oxygen, not carbon dioxide. Most ${\text{CO}}_2$ travels dissolved in plasma. So R is false. Correct option: (C).

   1 mark

Answer: (C)

1. Function: during inhalation the diaphragm contracts and flattens. This increases the volume of the chest cavity, so air is drawn into the lungs. During exhalation it relaxes and pushes air out. It therefore helps in breathing.

   1 mark

2. Location: it is a dome-shaped muscular sheet at the base (floor) of the chest cavity. It separates the chest cavity from the abdominal cavity.

   1 mark

Answer: It flattens during inhalation to enlarge the chest cavity and draw air into the lungs (helps breathing); it lies at the base of the chest cavity.

1. Difference 1: chewing food is a voluntary action done consciously. Salivation on seeing food is an involuntary (reflex) action.

   1 mark

2. Difference 2: chewing is controlled by the forebrain (cerebrum). Salivation is controlled by the medulla in the hindbrain.

   1 mark

Answer: Chewing: voluntary, controlled by forebrain; Salivation: involuntary/reflex, controlled by medulla (hindbrain).

1. Difference 1: pollination is the transfer of pollen grains from the anther to the stigma. Fertilisation is the fusion of the male gamete with the female gamete to form a zygote.

   1 mark

2. Difference 2: pollination occurs only in plants and happens before fertilisation. Fertilisation occurs in both plants and animals.

   1 mark

Answer: Pollination: transfer of pollen from anther to stigma, only in plants; Fertilisation: fusion of male and female gametes, in plants and animals.

1. Diagram: draw a pistil with the stigma at the top, a long style and a swollen ovary at the base containing an ovule. Show a pollen grain on the stigma with a pollen tube growing out of it, down through the style, entering the ovule.

   1 mark

2. (a) Label the top sticky surface of the pistil Stigma. It receives the pollen grain.

   ½ mark

3. (b) Label the tube running from the pollen grain through the style Pollen tube. It carries the male germ cell to the female germ cell in the ovule.

   ½ mark

Answer: Neat diagram of pollen germinating on the stigma; labels (a) Stigma, (b) Pollen tube.

1. (a) (i) Producers: Grass; (ii) Primary consumers: Deer, Rabbit; (iii) Secondary consumers: Snake, Lion; (iv) Tertiary consumer: Lion.

   1 mark

2. (b) Primary consumers feed directly on green plants, which hold a large amount of energy. By the 10% law, only about 10% of the energy at one level passes to the next. The rest is used in life processes or lost as heat. So secondary consumers get much less energy than primary consumers.

   1 mark

3. (c) The base is broad because producers are the most numerous and contain the maximum energy and biomass of all trophic levels. Energy, and usually number, decreases at each higher level.

   1 mark

Answer: (a) Grass – producers; Deer, Rabbit – primary; Snake, Lion – secondary; Lion – tertiary. (b) Only ~10% energy passes to the next level. (c) Producers have the highest number/energy/biomass.

1. (a) A nephron is the structural and functional (filtration) unit of the kidney; it filters nitrogenous wastes from blood. A neuron is the structural and functional unit of the nervous system; it transmits information (impulses) from one part of the body to another.

   1 mark

2. (b) A sensory nerve carries impulses from receptors to the CNS (brain and spinal cord). A motor nerve carries impulses from the CNS to effectors (muscles or glands).

   1 mark

3. (c) Consumers feed on producers or other consumers and pass energy along the food chain. Decomposers break down dead organic matter into simple inorganic substances and recycle nutrients back into the environment.

   1 mark

Answer: (a) Nephron: unit of kidney, filters blood; Neuron: unit of nervous system, transmits impulses. (b) Sensory: receptor → CNS; Motor: CNS → effector. (c) Consumers feed on other organisms; Decomposers break down dead matter and recycle nutrients.

1. (a) The tall parent (TT) gives T and the short parent (tt) gives t, so every F₁ plant is Tt. Tallness (T) is dominant over shortness (t), so all F₁ plants appear tall.

   1 mark

2. (b) Mendel obtained the F₂ progeny by self-pollination of the F₁ tall plants.

   1 mark

3. (c)(i) A dominant trait expresses itself even in the presence of the recessive factor, so it shows in both TT and Tt. A recessive trait is masked when the dominant factor is present; it shows only when both factors are recessive (tt).

   2 marks

Answer: (a) F₁ is Tt and tall is dominant over short; (b) self-pollination of F₁; (c)(i) dominant trait shows in TT and Tt, recessive shows only in tt.

1. Observation 1: all plants of the F₁ progeny were tall.

   1 mark

2. Observation 2: no short or medium-height plants appeared in F₁; the F₁ plants resembled only one parent (the tall one).

   1 mark

Answer: All F₁ plants were tall; no medium or short plants appeared (F₁ resembled only one parent).

1. (i) Most bacteria in the population would die in the warmer water. A few variants that can tolerate heat would survive and multiply, so the population would come to be dominated by heat-resistant bacteria. Variation helps survival.

   1 mark

2. (ii) The sperm fuses with the egg, so fertilisation takes place and a zygote is formed. The zygote later develops into an embryo.

   1 mark

3. (iii) The flower is unisexual (only pistil), so self-pollination is impossible. It can produce fruit only if cross-pollination occurs; otherwise there is no fertilisation and no fruit or seed formation.

   1 mark

4. (iv) The thick, spongy uterine lining prepared for the embryo is not needed. It breaks down and comes out through the vagina as blood and mucus, which is menstruation.

   1 mark

5. (v) The seed germinates: the embryo grows into a seedling (radicle forms the root, plumule forms the shoot).

   1 mark

Answer: (i) Most die, heat-resistant variants survive; (ii) fertilisation, zygote forms; (iii) only cross-pollination can lead to fruit, else no fertilisation; (iv) menstruation; (v) germination into a seedling.

1. (i) The spores spread through the air. When they land on a suitable moist surface with the right temperature, each spore germinates and grows into a new Rhizopus (bread mould) plant.

   1 mark

2. (ii) Buds in the notches along the leaf margin develop into new plantlets on the wet soil. This is vegetative propagation.

   1 mark

3. (iii) The stigma of an unrelated species does not support the foreign pollen. No pollen tube forms, so fertilisation does not take place.

   1 mark

4. (iv) The copper-T prevents fertilisation and implantation, so pregnancy is prevented. It is a contraceptive device.

   1 mark

5. (v) Each fragment grows into a new Spirogyra filament. This is fragmentation, an asexual method.

   1 mark

Answer: (i) Spores germinate into new Rhizopus; (ii) leaf-notch buds grow into new plants; (iii) no pollen tube, no fertilisation; (iv) pregnancy prevented; (v) each fragment grows into a new individual.

1. In both reactions two compounds exchange their ions:
   (i) ${\text{AgNO}}_3+\text{NaCl}\to {\text{NaNO}}_3+\text{AgCl}$
   (ii) ${\text{K}}_2{\text{SO}}_4+{\text{BaCl}}_2\to {\text{BaSO}}_4+2\text{KCl}$
   So both are double displacement reactions.

2. Each reaction forms an insoluble product: white $\text{AgCl}$ in (i) and white ${\text{BaSO}}_4$ in (ii). So both are also precipitation reactions. Correct option: (C).

   1 mark

Answer: (C) Both are double displacement and precipitation reactions.

1. A blind student cannot see a colour change, so the indicator must signal acids and bases through some other sense, such as smell.

2. Vanilla essence is an olfactory indicator. Its smell can be detected in acidic solution but is lost in basic solution. Turmeric, methyl orange and litmus all work only by colour change.

   1 mark

3. Correct option: (B) Vanilla essence.

Answer: Vanilla essence, option (B)

1. On heating, lead(II) nitrate undergoes thermal decomposition: $2\mathrm{Pb}(\mathrm{N}{\mathrm{O}}_3{)}_2(\mathrm{s})\stackrel{\text{heat}}{\to }2\mathrm{PbO}(\mathrm{s})+4\mathrm{N}{\mathrm{O}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})$

2. The gases given off are brown nitrogen dioxide ($\mathrm{N}{\mathrm{O}}_2$) and oxygen (${\mathrm{O}}_2$). Correct option: (D).

   1 mark

Answer: NO₂ and O₂, option (D)

1. Match each group to its class: –COOH is a carboxylic acid, –C(=O)– inside a chain is a ketone, and –OH is an alcohol.

2. The aldehyde group is –CHO, for example methanal HCHO. Correct option: (B).

   1 mark

Answer: –CHO, option (B)

1. Silver, copper and aluminium are the best conductors of electricity.

2. Lead (Pb) and mercury are comparatively poor conductors. Correct option: (A) Pb.

   1 mark

Answer: Pb, option (A)

1. Oxalic acid is found in tomato, lactic acid in curd (sour milk), and methanoic acid in ant sting (and nettle sting).

2. Only option (A) gives all three sources correctly and in the right order. Option (B) is wrong because orange contains citric acid, not lactic acid. Correct option: (A).

   1 mark

Answer: tomato, curd, ant-sting, option (A)

1. Calcium reacts less violently with water: $\mathrm{Ca}+2{\mathrm{H}}_2\mathrm{O}\to \mathrm{Ca}(\mathrm{OH}{)}_2+{\mathrm{H}}_2$

2. Bubbles of hydrogen gas stick to the surface of the metal, so calcium starts floating. Potassium and sodium react so violently that the hydrogen catches fire, and iron does not react with cold water. Correct option: (B) Calcium.

   1 mark

Answer: Calcium, option (B)

1. Assertion: carbon has 4 valence electrons and forms covalent bonds by sharing them with carbon or other atoms. This is true.

2. Reason: the shared pair belongs to the outer shells of both atoms, so each atom attains a noble gas configuration. This is true, and it explains why carbon shares electrons. Correct option: (A).

   1 mark

Answer: Option (A)

1. (a) Amphoteric oxides are metal oxides that react with both acids and bases to give salt and water. Examples are $\mathrm{ZnO}$ and $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$.

   1 mark

2. (b) $\mathrm{ZnO}$ is an amphoteric oxide, $\mathrm{N}{\mathrm{a}}_2\mathrm{O}$ is a basic oxide (metal oxide), and $\mathrm{C}{\mathrm{O}}_2$ is an acidic oxide (non-metal oxide).

   1 mark

Answer: (a) Metal oxides reacting with both acids and bases to form salt and water; (b) ZnO amphoteric, Na₂O basic, CO₂ acidic

1. (a) In $\mathrm{ZnO}+\mathrm{C}\to \mathrm{Zn}+\mathrm{CO}$, carbon gains oxygen, so C is oxidised.

   ½ mark

2. (a) ZnO loses oxygen, so ZnO is reduced.

   ½ mark

3. (b) Balanced equation: $\mathrm{Pb}(\mathrm{N}{\mathrm{O}}_3{)}_2+2KI\to \mathrm{Pb}{\mathrm{I}}_2+2KN{\mathrm{O}}_3$

   1 mark

4. (c) Electrolytic decomposition: $2{\mathrm{H}}_2\mathrm{O}\stackrel{\text{electricity}}{\to }2{\mathrm{H}}_2+{\mathrm{O}}_2$

   ½ mark

5. (c) Decomposition by sunlight: $2\mathrm{AgCl}\stackrel{\text{sunlight}}{\to }2\mathrm{Ag}+\mathrm{C}{\mathrm{l}}_2$

   ½ mark

Answer: (a) C oxidised, ZnO reduced; (b) Pb(NO₃)₂ + 2KI → PbI₂ + 2KNO₃; (c) electrolysis of water; 2AgCl → 2Ag + Cl₂ in sunlight

1. When electricity is passed through brine (an aqueous solution of sodium chloride), it decomposes to give sodium hydroxide (an alkali) and chlorine. Because the products are chlorine and an alkali, the process is called the chlor-alkali process.

   1 mark

2. Equation: $2\mathrm{NaCl}(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\stackrel{\text{electricity}}{\to }2\mathrm{NaOH}(\mathrm{aq})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})$

   1 mark

3. At the anode: chlorine gas ($\mathrm{C}{\mathrm{l}}_2$) is formed.

   ½ mark

4. At the cathode: hydrogen gas (${\mathrm{H}}_2$) is formed. Sodium hydroxide solution collects near the cathode.

   ½ mark

Answer: Electrolysis of brine gives NaOH, Cl₂ and H₂; Cl₂ at anode, H₂ at cathode

1. (i) Baking soda is made from sodium chloride, water, ammonia and carbon dioxide: $\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2+\mathrm{N}{\mathrm{H}}_3\to \mathrm{N}{\mathrm{H}}_4\mathrm{Cl}+\mathrm{NaHC}{\mathrm{O}}_3$

   1 mark

2. (ii) Bleaching powder is made by passing chlorine over dry slaked lime: $\mathrm{Ca}(\mathrm{OH}{)}_2+\mathrm{C}{\mathrm{l}}_2\to \mathrm{CaOC}{\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}$

   1 mark

3. (iii) Plaster of Paris is made by heating gypsum at 373 K: $\mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}\stackrel{373K}{\to }\mathrm{CaS}{\mathrm{O}}_4\cdot \frac{1}{2}{\mathrm{H}}_2\mathrm{O}+1\frac{1}{2}{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (i) NaCl + H₂O + CO₂ + NH₃ → NH₄Cl + NaHCO₃; (ii) Ca(OH)₂ + Cl₂ → CaOCl₂ + H₂O; (iii) CaSO₄·2H₂O → CaSO₄·½H₂O + 1½H₂O at 373 K

1. (a) It is easier to obtain a metal from its oxide than from its carbonate or sulphide, because a metal oxide can be reduced readily by carbon. So carbonate ores are calcined and sulphide ores are roasted to convert them into oxides first.

   1 mark

2. (b) Thermit reaction: $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3(\mathrm{s})+2Al(\mathrm{s})\to 2Fe(\mathrm{l})+\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3(\mathrm{s})+\text{heat}$

   1 mark

3. (c)(i) Copper glance ($\mathrm{C}{\mathrm{u}}_2\mathrm{S}$) is roasted in air, which converts part of it to copper(I) oxide: $2\mathrm{C}{\mathrm{u}}_2\mathrm{S}+3{\mathrm{O}}_2\stackrel{\text{heat}}{\to }2\mathrm{C}{\mathrm{u}}_2\mathrm{O}+2\mathrm{S}{\mathrm{O}}_2$

   1 mark

4. (c)(i) The $\mathrm{C}{\mathrm{u}}_2\mathrm{O}$ formed then reacts with the remaining $\mathrm{C}{\mathrm{u}}_2\mathrm{S}$ (self-reduction) to give copper metal: $2\mathrm{C}{\mathrm{u}}_2\mathrm{O}+\mathrm{C}{\mathrm{u}}_2\mathrm{S}\stackrel{\text{heat}}{\to }6\mathrm{Cu}+\mathrm{S}{\mathrm{O}}_2$

   1 mark

5. (c)(ii)(I) [OR alternative] Highly reactive metals such as Na, Mg and Al have a greater affinity for oxygen than carbon has. So carbon cannot remove oxygen from their oxides, and these metals are obtained by electrolytic reduction instead.

   1 mark

6. (c)(ii)(II) [OR alternative] Solder has a low melting point, lower than that of the pure metals lead and tin. So it melts easily and can be used to join electrical wires.

   1 mark

Answer: (a) Oxides are easier to reduce to metal; (b) Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + heat; (c)(i) 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂, then 2Cu₂O + Cu₂S → 6Cu + SO₂; OR (c)(ii) (I) these metals have more affinity for oxygen than carbon; (II) solder has a low melting point

1. (i)(I) Covalent compounds are formed by sharing of electrons, so they do not form ions (charged particles). With no ions to carry charge, they are poor conductors of electricity.

   1 mark

2. (i)(II) Hard water contains calcium and magnesium salts. Soap reacts with these salts to form an insoluble precipitate called scum, so the soap is used up and does not form lather easily.

   1 mark

3. (i)(III) C–C bonds are very strong and stable because carbon is a small atom, so carbon can form long chains (catenation). Si–Si bonds are much weaker, so silicon does not form long stable chains.

   1 mark

4. (ii)(I) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{acidified }{K}_{2}C{r}_2{O}_7,\text{heat}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$ (ethanol is oxidised to ethanoic acid)

   1 mark

5. (ii)(II) $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\stackrel{\text{Ni catalyst}}{\to }\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_3$ (ethene is hydrogenated to ethane)

   1 mark

Answer: (i) (I) No ions formed; (II) Ca/Mg salts form scum with soap; (III) C–C bonds strong, Si–Si bonds weak; (ii) CH₃CH₂OH → CH₃COOH; CH₂=CH₂ + H₂ → CH₃CH₃ (Ni)

1. (i) X is ethanoic acid (acetic acid), $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$, because it reacts with ethanol to give a sweet-smelling compound.

   ½ mark

2. (i) Y is ethyl ethanoate, an ester, $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5$.

   ½ mark

3. (i) Z is sodium ethanoate (sodium acetate), $\mathrm{C}{\mathrm{H}}_3\mathrm{COONa}$.

   ½ mark

4. (ii) Concentrated ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ acts as a catalyst and dehydrating agent: it removes the water formed and speeds up the reaction.

   ½ mark

5. (iii) $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{conc. }{H}_{2}S{O}_4}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{O}$

   1 mark

6. (iii) This reaction is called esterification.

   ½ mark

7. (iii) $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaOH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$

   1 mark

8. (iii) This reaction is called saponification.

   ½ mark

Answer: (i) X = CH₃COOH, Y = CH₃COOC₂H₅, Z = CH₃COONa; (ii) catalyst/dehydrating agent; (iii) esterification and saponification (equations shown)

1. A convex lens forms a real image of the same size when the object is at $2F_1$. The image is then at $2F_2$.

2. $2f=2\times 15=30$ cm. By the sign convention, $u=-30$ cm and $v=+30$ cm. Check: $\frac{1}{30}-\frac{1}{-30}=\frac{1}{15}$. Correct option: (C).

   1 mark

Answer: −30 cm and +30 cm, option (C)

1. To focus on a nearby object, the eye needs a shorter focal length, that is, a more converging lens.

2. The ciliary muscles contract, the lens becomes thicker and its focal length decreases. Correct option: (A).

   1 mark

Answer: Option (A)

1. Assertion: white light passing through a prism splits into a spectrum of seven colours. This is true (dispersion).

2. Reason: the splitting happens because different colours are refracted (bent) by different amounts, red the least and violet the most. It is due to refraction, not scattering, so R is false. Correct option: (C).

   1 mark

Answer: Option (C)

1. (a) The image is enlarged and erect and forms on the same side as the object, with the refracted rays traced back to meet at B'. Only a converging lens does this, so the lens is a convex (converging) lens.

   1 mark

2. (b) The image is real and twice the size of the object, so it is inverted: $m=-2$. Also $v=+30$ cm. For a lens, $m=\frac{v}{u}$.

   ½ mark

3. $u=\frac{v}{m}=\frac{30}{-2}=-15$ cm. The object was placed 15 cm in front of the lens (on the left side).

   ½ mark

Answer: (a) Convex lens; (b) u = −15 cm, i.e. 15 cm in front of the lens

1. Given: $r=0.01$ cm $=1\times 10^{-4}$ m, $l=1$ cm $=0.01$ m, $R=7\Omega$. Formula: $R=\rho \frac{l}{A}$, so $\rho =\frac{RA}{l}=\frac{R\pi r^2}{l}$

   ½ mark

2. Substituting: $\rho =\frac{7\times \frac{22}{7}\times (10^{-4}{)}^2}{0.01}=\frac{22\times 10^{-8}}{10^{-2}}$

   1 mark

3. $\rho =22\times 10^{-6}\Omega \text{m}=2.2\times 10^{-5}\Omega \text{m}$

   ½ mark

Answer: ρ = 2.2 × 10⁻⁵ Ω m

1. The rating tells us the heater draws $I=11\text{A}$ at $V=220\text{V}$. By Ohm's law, its resistance is $R=\frac{V}{I}$.

   ½ mark

2. $R=\frac{220}{11}=20\Omega$. The resistance stays the same when the heater is run at a different voltage.

   ½ mark

3. When it is operated at $V^{\prime }=200\text{V}$, the power consumed is $P=\frac{V^{\prime 2}}{R}$.

   ½ mark

4. $P=\frac{200\times 200}{20}=\frac{40000}{20}=2000\text{W}=2\text{kW}$.

   ½ mark

Answer: R = 20 Ω; power consumed at 200 V = 2000 W (2 kW)

1. The person cannot see nearby objects (closer than 1 m) clearly, so his near point has moved far away. This defect is hypermetropia (far-sightedness). It happens when the focal length of the eye lens is too long or the eyeball has become too small.

   1 mark

2. Diagram 1 – Hypermetropic eye: Draw the eye with its lens. Mark the normal near point N (about 25 cm) and the defective near point N′ farther away. Rays from N enter the eye and meet behind the retina. Show arrows on all rays.

   1 mark

3. Diagram 2 – Correction: Place a convex (converging) lens in front of the eye. Rays from N pass through the convex lens and then appear to come from N′, shown by dashed back-extensions. The eye lens then focuses them on the retina. Show arrows on all rays.

   1 mark

Answer: Hypermetropia (far-sightedness); it is corrected with a convex lens of suitable power. Two ray diagrams are drawn: defect and correction.

1. (a) Procedure: Hang a small aluminium rod AB horizontally from a stand using two connecting wires. Place a strong horse-shoe magnet so that the rod lies between its two poles, at right angles to the magnetic field. Connect the rod in series with a battery, a key and a rheostat. Close the key so that current flows through the rod (say from B to A). A labelled diagram of this set-up may also be drawn.

   1 mark

2. (a) Observation: As soon as current flows, the rod is displaced (it moves to one side). If the current is reversed, or the poles of the magnet are interchanged, the rod moves the opposite way. This shows that a current-carrying conductor placed in a magnetic field experiences a force.

   1 mark

3. (b) An electron beam moving from the back wall to the front wall is the same as a conventional current from the front wall towards the back wall. The force on it is towards your right. By Fleming's left-hand rule (middle finger = current pointing backward, thumb = force pointing right), the forefinger points vertically downward. So the magnetic field is vertically downwards.

   1 mark

Answer: (a) Aluminium rod between the poles of a horse-shoe magnet is displaced when current flows. (b) Magnetic field is vertically downwards.

1. (a) Straight wire: the field lines are concentric circles centred on the wire, lying in a plane perpendicular to it. Draw a vertical wire with circles around it and arrows given by the right-hand thumb rule.

   1 mark

2. (a) Circular loop: near the wire the lines are concentric circles. Towards the centre they become nearly straight and parallel. Solenoid: the pattern is like a bar magnet, with parallel straight lines (a uniform field) inside. Drawing any two of these patterns shows that the field pattern changes with the shape of the wire.

   1 mark

3. (b) The field will be maximum at X.

   ½ mark

4. (b) Reason: the strength of the magnetic field of a straight current-carrying wire decreases as the distance from the wire increases. X is the point closest to the wire.

   ½ mark

Answer: (a) Straight wire gives concentric circles; loop gives circles that become straight near the centre; solenoid gives a bar-magnet-like pattern (any two). (b) At X, because the field decreases with distance from the wire.

1. (a) Given $f=+20$ cm and $u=-40$ cm. Using $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$: $\frac{1}{v}=\frac{1}{20}+\frac{1}{-40}=\frac{2-1}{40}=\frac{1}{40}$, so $v=+40$ cm. The image is 40 cm on the other side of the lens, at $2F_2$.

   ½ mark

2. (a) Nature: the image is real and inverted, and the same size as the object. This is because the object is at $2F_1$.

   ½ mark

3. (b) Ray diagram: draw a concave lens with $F_1$ and $F_2$ marked, and object AB between O and $F_1$. Ray 1 from B goes parallel to the axis. After refraction it diverges, and its dashed back-extension passes through $F_1$. Ray 2 from B passes undeviated through O. The extension and ray 2 meet at B′. The image A′B′ lies between O and $F_1$ on the same side as the object, and is virtual, erect and diminished.

   1 mark

4. (c)(i) $f_1=+30$ cm $=+0.3$ m and $f_2=-15$ cm $=-0.15$ m. Using $P=\frac{1}{f}$ (f in metres): $P_1=\frac{1}{0.3}=+3.33$ D.

   ½ mark

5. (c)(i) $P_2=\frac{1}{-0.15}=-6.67$ D.

   ½ mark

6. (c)(i) Equivalent power: $P=P_1+P_2=3.33-6.67=-3.33$ D.

   ½ mark

7. (c)(i) Equivalent focal length: $f=\frac{1}{P}=\frac{1}{-3.33}=-0.3$ m $=-30$ cm. The combination acts as a diverging lens.

   ½ mark

8. (c)(ii) $f_1=-2$ m (concave) gives $P_1=-\frac{1}{2}=-0.5$ D. $f_2=+1.5$ m (convex) gives $P_2=+\frac{1}{1.5}=+0.67$ D. Then $P=P_1+P_2=-\frac{1}{2}+\frac{2}{3}=+\frac{1}{6}$ D, so $f=+6$ m.

   1 mark

9. (c)(ii) The combined power (and focal length) is positive, so the combination behaves as a convex (converging) lens. Reason: the convex lens has the smaller focal length, so its power is greater and outweighs the concave lens.

   1 mark

Answer: (a) 40 cm behind the lens (at 2F₂), real and inverted, same size; (b) virtual, erect, diminished image between O and F₁ on the object side; (c)(i) f = −30 cm, P = −3.33 D; (c)(ii) behaves as a convex lens (P = +1/6 D, f = +6 m)

1. (i) Resistance is $R=\rho \frac{l}{A}$. It is directly proportional to the length $l$ and inversely proportional to the area of cross-section $A$. So changing $l$ or $A$ changes $R$.

   1 mark

2. (i) Resistivity $\rho$ is a characteristic property of the material (at a given temperature). It does not depend on the dimensions of the conductor, so it does not change.

   1 mark

3. (ii) An alloy has a higher resistivity than its constituent pure metals, so it produces more heat for the same current.

   1 mark

4. (ii) Alloys also do not oxidise (burn) readily at high temperatures. This makes them ideal for heating elements such as those in toasters and irons.

   1 mark

5. (iii) The SI unit of electric current is the ampere (A). One ampere is the current when 1 coulomb of charge flows through a cross-section of a conductor in 1 second: $1\text{A}=\frac{1\text{C}}{1\text{s}}$.

   1 mark

Answer: (i) R = ρl/A depends on l and A, while ρ is a property of the material only; (ii) alloys have high resistivity and do not oxidise at high temperature; (iii) 1 A = flow of 1 C of charge per second

1. (i) Required total resistance: $R=\frac{V}{I}$.

   ½ mark

2. (i) $R=\frac{4}{2}=2\Omega$.

   ½ mark

3. (i) Let $n$ bulbs of $8\Omega$ be joined in parallel. Then $\frac{1}{R}=\frac{1}{8}+\frac{1}{8}+\dots$ ($n$ terms) $=\frac{n}{8}$.

   ½ mark

4. (i) $\frac{1}{2}=\frac{n}{8}$, so $n=4$. 4 bulbs should be connected in parallel.

   ½ mark

5. (ii) The device used for measuring electric current is an ammeter.

   1 mark

6. (ii) It is always connected in series in the circuit.

   1 mark

7. (iii) Joule's law of heating: the heat produced in a resistor is directly proportional to the square of the current, to the resistance, and to the time for which the current flows. So $H=I^{2}Rt$.

   1 mark

Answer: (i) 4 bulbs; (ii) ammeter, connected in series; (iii) H = I²Rt (heat ∝ I², R and t)