CBSE Class 10 Science: 2025 board paper, set 31/2/1

1. Natural gas is mainly methane, $\mathrm{C}{\mathrm{H}}_4$. Burning it in air gives carbon dioxide and water, and heat is given out, so the reaction is exothermic.

2. Check the balance. Left side: 1 C, 4 H, 4 O. Right side of $\mathrm{C}{\mathrm{H}}_4+2{\mathrm{O}}_2\to \mathrm{C}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$: 1 C, 4 H, $2+2=4$ O. So it is balanced.

3. (A) is unbalanced. (B) has $2\mathrm{C}{\mathrm{O}}_2$, which gives 2 C on the right, so it is wrong. (C) is balanced but does not show the energy released. Only (D) is balanced and shows the energy, so the answer is (D).

   1 mark

Answer: (D) $\mathrm{C}{\mathrm{H}}_4+2{\mathrm{O}}_2\to \mathrm{C}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}+\mathrm{Energy}$

1. The sign shows liquid dropping onto a hand and eating into it. This is the corrosive hazard symbol.

2. Of the hydroxides listed, sodium hydroxide (NaOH, caustic soda) is a strong alkali that is highly soluble. It is strongly corrosive and burns the skin.

3. The hydroxides of aluminium and magnesium are weak bases that dissolve only slightly, and calcium hydroxide is much milder. So the container must hold sodium hydroxide: (C).

   1 mark

Answer: (C) Sodium

1. Our body works within a narrow pH range that is close to neutral and slightly basic, about 7.0 to 7.8 (as given in NCERT).

2. So the correct option is (C) 7.0 to 7.8.

   1 mark

Answer: (C) 7.0 to 7.8

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

2. Aluminium is more reactive than iron, so it displaces iron from iron(III) oxide. So much heat is released that the iron formed is molten. This molten iron is used to join railway tracks and cracked machine parts.

3. Aluminium does not react with iron itself, so (A) is wrong. Iron, not aluminium oxide, is the joining material, so (C) is wrong. No Fe–Al alloy is formed, so (D) is wrong. Answer: (B).

   1 mark

Answer: (B)

1. Brass is an alloy of copper and zinc.

2. Bronze is an alloy of copper and tin. (Solder is lead and tin, so it does not fit.)

3. The question takes zinc first and tin second, so the order is Brass, Bronze: (C).

   1 mark

Answer: (C) Brass and Bronze

1. Each stoma is surrounded by two guard cells.

2. When water flows into the guard cells, they swell and the pore opens. When they lose water, they shrink and the pore closes.

3. So stomata are regulated by the amount of water in the guard cells: (D).

   1 mark

Answer: (D) Amount of water in guard cells

1. Arteries and veins have thick, layered walls. Alveoli are air sacs in the lungs, not blood vessels.

2. The smallest blood vessels have walls only one cell thick, which lets materials be exchanged with the tissues. These are the capillaries: (B).

   1 mark

Answer: (B) Capillaries

1. The leaves of the sensitive plant fold immediately on touch, with no growth involved.

2. Plant cells pass the stimulus from cell to cell and then change shape by changing the amount of water in them. This causes the leaves to fold.

3. So the answer is (C), change in amount of water in cells.

   1 mark

Answer: (C) Change in amount of water in cells

1. Bryophyllum reproduces vegetatively through its leaves.

2. Buds form in the notches along the leaf margin. When they fall on soil, they grow into new plants: (B).

   1 mark

Answer: (B)

1. The cell division that halves the chromosome number is meiosis. It forms gametes.

2. Gametes are made in both sexes: eggs in the ovary and sperms in the testis. So the answer is (C), ovary and testis both.

   1 mark

Answer: (C) Ovary and testis both

1. Refractive index of Y with respect to X: $n_{YX}=\frac{n_Y}{n_X}$.

2. $n_{YX}=\frac{4/3}{6/5}=\frac{4}{3}\times \frac{5}{6}=\frac{20}{18}=\frac{10}{9}$

3. Answer: **(A) $\frac{10}{9}$**.

   1 mark

Answer: $\frac{10}{9}$, option (A)

1. By the sign convention, the object and the real image are both in front of the mirror: $u=-30$ cm, $v=-60$ cm.

2. Mirror formula: $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=-\frac{1}{60}-\frac{1}{30}=-\frac{3}{60}=-\frac{1}{20}$

3. $f=-20$ cm: (B).

   1 mark

Answer: $f=-20$ cm, option (B)

1. The angle of incidence is the angle between the incident ray and the normal at the first face. From the figure, this is angle 1. (Angle 7 is measured from the face of the prism, not the normal.)

2. The angle of deviation is the angle between the incident ray extended forward and the emergent ray extended backward. In the figure, this is angle 6.

3. So the answer is (D) 1 and 6.

   1 mark

Answer: (D) 1 and 6

1. $I=500$ mA $=0.5$ A, $V=220$ V

2. $P=VI=220\times 0.5=110$ W: (B).

   1 mark

Answer: 110 W, option (B)

1. Points 1 and 2 are joined to the adjacent corners A and B. Between A and B there are two paths:

   • the single resistor AB: $12\Omega$
   • the path A–C–D–B: $12+12+12=36\Omega$

2. These two paths are in parallel: $R=\frac{12\times 36}{12+36}=\frac{432}{48}=9\Omega$

3. Answer: **(C) $9\Omega$**.

   1 mark

Answer: $9\Omega$, option (C)

1. Leather, cotton, wood, silk and wool are natural materials, and microbes can break them down. They are biodegradable.

2. DDT (a pesticide), polyester (a synthetic fibre) and glass cannot be broken down by microorganisms. So every member of group (C) is non-biodegradable: (C).

   1 mark

Answer: (C) DDT, Polyester, Glass

1. Assertion: metals high in the reactivity series (such as Na, Mg, Al) have a greater affinity for oxygen than carbon has. So carbon cannot reduce their oxides, and they are obtained by electrolysis. A is true.

2. Reason: a more reactive metal can displace a less reactive metal from its compound. For example, the thermit reaction gives Fe from $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3$. R is true.

3. However, displacement does not explain why carbon fails for very reactive metals. R is not the explanation of A, so the answer is (B).

   1 mark

Answer: (B)

1. A mango seed always grows into a mango tree. A is true.

2. Heredity is the reliable passing of traits from parents to offspring. R is true.

3. The seed grows into a mango tree because the traits of the parent are passed on through heredity. So R explains A: (A).

   1 mark

Answer: (A)

1. Nichrome is used in the heating elements of irons, toasters and similar devices. A is true.

2. Nichrome does have a high resistivity, but its resistance does not decrease when it is heated. It also does not oxidise easily at high temperatures. So R is false.

3. Answer: (C).

   1 mark

Answer: (C)

1. Animals cannot digest coal, so they get no energy from eating it. A is true.

2. Each kind of food is broken down by its own specific enzymes. R is true.

3. Animals have no enzyme that can break down coal, which is why they get no energy from it. R explains A: (A).

   1 mark

Answer: (A)

1. (a) Copper is more reactive than silver, so it pushes silver out of silver nitrate solution. This is a displacement reaction.

   ½ mark

2. (a) Equation: $\mathrm{Cu}(\mathrm{s})+2AgN{\mathrm{O}}_3(\mathrm{aq})\to \mathrm{Cu}(\mathrm{N}{\mathrm{O}}_3{)}_2(\mathrm{aq})+2Ag(\mathrm{s})$

   ½ mark

3. (b) Silver is refined by electrolytic refining. The impure silver is made the anode and a thin strip of pure silver is the cathode, with a silver salt solution as the electrolyte. Pure silver deposits on the cathode.

   1 mark

Answer: (a) Displacement reaction: $\mathrm{Cu}+2AgN{\mathrm{O}}_3\to \mathrm{Cu}(\mathrm{N}{\mathrm{O}}_3{)}_2+2Ag$; (b) Electrolytic refining

1. (a) Members of a homologous series have the same functional group and differ by one $-\mathrm{C}{\mathrm{H}}_2-$ unit.

   • (1) is propanal, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ (aldehyde)
   • (4) is butanal, $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{CHO}$ (aldehyde)
   • (2) is propanone (ketone) and (3) is propanoic acid (carboxylic acid)

   So (1) and (4) are in the same series; they differ by one $\mathrm{C}{\mathrm{H}}_2$.

   1 mark

2. (b)(i) As molecular mass increases along the series, the melting point increases gradually.

   ½ mark

3. (b)(ii) Solubility changes gradually. In a given solvent such as water, it decreases as molecular mass increases. Chemical properties stay similar because they depend on the functional group.

   ½ mark

Answer: (a) (1) and (4); (b)(i) melting point increases; (ii) solubility shows gradation (decreases)

1. The hind-brain has three parts. The first is the pons.

   ½ mark

2. The second is the medulla (medulla oblongata).

   ½ mark

3. The third is the cerebellum.

   ½ mark

4. Involuntary actions such as blood pressure, salivation and vomiting are controlled by the medulla.

   ½ mark

Answer: Pons, medulla, cerebellum; medulla controls involuntary actions

1. (i) Seed coat: it is the outer cover that protects the embryo and the seed from injury and drying out.

   ½ mark

2. (ii) Cotyledon: it stores food for the embryo, which is used when the seed germinates.

   ½ mark

3. (iii) Radicle: it grows into the root when the seed germinates.

   ½ mark

4. (iv) Plumule: it grows into the shoot when the seed germinates.

   ½ mark

Answer: (i) protection; (ii) food storage; (iii) future root; (iv) future shoot

1. Prepare a sugar solution: put about 10 g of sugar in a test tube (or small flask) and dissolve it in about 100 mL of water. The sugar is the food for the yeast cells.

   1 mark

2. Add a pinch of yeast granules to the sugar solution. Keep the test tube in a warm place for 1–2 hours. The warmth helps the yeast cells grow and multiply. Yeast has now been cultured.

   1 mark

3. (To observe it, put a drop of the culture on a slide and see it under a microscope. Yeast cells with small buds, i.e. budding, are seen.)

Answer: Dissolve ~10 g sugar in 100 mL water in a test tube, add a pinch of yeast granules, and keep it in a warm place for 1–2 hours.

1. A concave lens diverges (spreads out) a parallel beam of light falling on it.

   1 mark

2. Ray diagram (describe/draw): Draw a concave lens with optical centre O on the principal axis. Draw two rays parallel to the principal axis striking the lens. After refraction they spread outwards (arrows pointing away from the axis). Extend the refracted rays backwards with dotted lines. They meet at a point F on the principal axis on the same side as the incident light. Label F as the principal focus.

   1 mark

Answer: Concave lens; the refracted rays, when produced backwards, appear to meet at the principal focus F on the side of the incident light.

1. (i) Ray diagram (describe/draw): Draw a convex lens with optical centre O. Mark foci F₁ (left) and F₂ (right) on the principal axis. Draw an incident ray from the left that passes through F₁ and strikes the lens. After refraction, draw it emerging parallel to the principal axis on the right side. Put arrows on both rays and label O, F₁ and F₂.

   1 mark

2. (ii) The incident ray passes through the principal focus (F₁) of the convex lens. A ray through the focus becomes parallel to the principal axis after refraction.

   1 mark

Answer: (i) Ray through F₁ emerges parallel to the principal axis; (ii) principal focus (focus).

1. Given: $P=750\text{W}$, $V=220\text{V}$. Current drawn: $I=\frac{P}{V}=\frac{750\text{W}}{220\text{V}}\approx 3.4\text{A}$

   ½ mark

2. No, this kettle cannot be used in this circuit.

   ½ mark

3. Reason: the kettle draws 3.4 A, which is more than the fuse rating of 3 A. The fuse wire will heat up and melt, which breaks the circuit, so the kettle will not work.

   1 mark

Answer: I ≈ 3.4 A > 3 A, so No; the fuse will melt and break the circuit.

1. (i) Decomposition reaction: a reaction in which a single reactant breaks down to form two or more simpler products.

   ½ mark

2. Heat (thermal decomposition): $\mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})\stackrel{\mathrm{Heat}}{\to }\mathrm{CaO}(\mathrm{s})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$

   ½ mark

3. Light (photolytic decomposition): $2AgCl(\mathrm{s})\stackrel{\mathrm{Sunlight}}{\to }2Ag(\mathrm{s})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})$

   ½ mark

4. Electricity (electrolytic decomposition): $2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\stackrel{\mathrm{Electric}\mathrm{current}}{\to }2{\mathrm{H}}_2(\mathrm{g})+{\mathrm{O}}_2(\mathrm{g})$

   ½ mark

5. (ii) When vegetable matter decomposes into compost, heat (energy) is released. A compost pit becomes warm. A reaction that releases heat is exothermic.

   1 mark

Answer: (i) Single reactant → two or more products; e.g. CaCO₃ → CaO + CO₂ (heat), 2AgCl → 2Ag + Cl₂ (light), 2H₂O → 2H₂ + O₂ (electricity). (ii) Heat is released during decomposition.

1. In a combination reaction, two or more reactants join to form a single product. In a decomposition reaction, a single reactant breaks down into two or more products. One joins substances and the other splits them, so the two are opposite processes.

   1 mark

2. Combination: $\mathrm{C}(\mathrm{s})+{\mathrm{O}}_2(\mathrm{g})\to \mathrm{C}{\mathrm{O}}_2(\mathrm{g})+\text{Heat}$. Reactants: carbon and oxygen; product: carbon dioxide.

   1 mark

3. Decomposition: $\mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})\stackrel{\mathrm{Heat}}{\to }\mathrm{CaO}(\mathrm{s})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$. Reactant: calcium carbonate; products: calcium oxide and carbon dioxide.

   1 mark

Answer: Combination joins reactants into one product, while decomposition splits one reactant into several products. Examples: C + O₂ → CO₂; CaCO₃ → CaO + CO₂.

1. Electronic configurations: Na (Z = 11): 2, 8, 1, so it has 1 valence electron. O (Z = 8): 2, 6, so it has 6 valence electrons.

2. (i) Electron-dot structure of sodium: $\mathrm{Na}\cdot$ (one dot).

   ½ mark

3. (ii) Electron-dot structure of oxygen: the symbol O with six dots around it (two pairs and two single dots).

   ½ mark

4. Formation of sodium oxide: each of two Na atoms transfers its one electron (shown by arrows) to one O atom. O gains 2 electrons and completes its octet.
   $2\mathrm{Na}\cdot +\mathrm{O}(6\text{ dots})\to 2\mathrm{N}{\mathrm{a}}^{+}+[\mathrm{O}(8\text{ electrons}){]}^{2-}$, i.e. $(\mathrm{N}{\mathrm{a}}^{+}{)}_2[\mathrm{O}{]}^{2-}$ = $\mathrm{N}{\mathrm{a}}_2\mathrm{O}$.

   1 mark

5. Anion: oxide ion, ${\mathrm{O}}^{2-}$.

   ½ mark

6. Cation: sodium ion, $\mathrm{N}{\mathrm{a}}^{+}$.

   ½ mark

Answer: Na· and O with 6 dots; 2Na· + O → (Na⁺)₂[O]²⁻ (Na₂O); anion O²⁻, cation Na⁺.

1. (a) Hormones are chemical messengers secreted by endocrine glands in one part of the body. They travel through the blood to target organs or tissues and regulate body functions.

   1 mark

2. (b) Example, insulin and blood sugar: when the blood sugar level rises (e.g. after a meal), cells of the pancreas detect it and secrete more insulin. Insulin lowers the blood sugar level.

   1 mark

3. When the blood sugar level falls, the pancreas detects this and reduces insulin secretion. So the timing and amount of the hormone are controlled by a feedback mechanism, and the hormone is released in precise quantities.

   1 mark

Answer: (a) Chemical messengers carried by blood to target organs to regulate body functions. (b) Blood sugar–insulin feedback: high sugar → more insulin; low sugar → less insulin.

1. (a) Let F = gene for free earlobe (dominant) and f = gene for attached earlobe (recessive). The attached-earlobe man must be ff. Because half the children have attached earlobes (ff), the free-earlobe woman must carry f, so she is Ff.
   Parents: woman Ff (free) × man ff (attached).

   ½ mark

2. Gametes: woman gives F and f; man gives f and f.

   ½ mark

3. Cross (F₁ progeny): F × f → Ff, F × f → Ff, f × f → ff, f × f → ff. Two children are Ff (free earlobe) and two are ff (attached earlobe).

   ½ mark

4. Progeny: Ff : ff = 1 : 1, i.e. 50% free earlobe and 50% attached earlobe. The trait combinations are free earlobe (Ff) and attached earlobe (ff).

   ½ mark

5. (b) Gene combination of father: ff

   ½ mark

6. Gene combination of mother: Ff

   ½ mark

Answer: (a) Ff × ff gives progeny Ff (free) : ff (attached) = 50% : 50%. (b) Father ff, mother Ff.

1. (i) The image is on the same side as the object and is larger than the object (8.0 cm vs 2.0 cm). So the image is virtual and erect (and magnified).

   1 mark

2. (ii) Given $h=+2.0$ cm, $h^{\prime }=+8.0$ cm (erect), $u=-6.0$ cm.
   $m=\frac{h^{\prime }}{h}=\frac{v}{u}\Rightarrow \frac{8.0}{2.0}=\frac{v}{-6.0}$

   ½ mark

3. $v=4\times (-6.0)=-24$ cm. The image is 24 cm from the lens, on the same side as the object.

   ½ mark

4. (iii) Lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{-24}-\frac{1}{-6}=\frac{1}{f}$

   ½ mark

5. $\frac{1}{f}=-\frac{1}{24}+\frac{4}{24}=\frac{3}{24}=\frac{1}{8}\Rightarrow f=+8$ cm. The focal length of the convex lens is 8 cm.

   ½ mark

Answer: (i) Virtual, erect (magnified); (ii) v = −24 cm (24 cm on the object side); (iii) f = +8 cm

1. (i) Series diagram: the 2 Ω, 3 Ω and 6 Ω resistors are joined end to end in a single line between two terminals.

   ½ mark

2. In series: $R_s=R_1+R_2+R_3$

   ½ mark

3. $R_s=(2+3+6)\Omega =11\Omega$

   ½ mark

4. (ii) Parallel diagram: the 2 Ω, 3 Ω and 6 Ω resistors are each connected between the same two junction points, forming three separate branches.

   ½ mark

5. In parallel: $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$

   ½ mark

6. $\frac{1}{R_p}=\frac{3+2+1}{6}=\frac{6}{6}=1\Rightarrow R_p=1\Omega$

   ½ mark

Answer: (i) Series: 11 Ω; (ii) Parallel: 1 Ω

1. Function: the ozone layer in the upper atmosphere shields the Earth's surface from harmful ultraviolet (UV) radiation from the Sun. UV radiation damages living organisms (e.g. it can cause skin cancer).

   1 mark

2. Formation: high-energy UV radiation splits some oxygen molecules into free oxygen atoms. Each free atom then combines with an oxygen molecule to form ozone.
   ${\mathrm{O}}_2\stackrel{\mathrm{UV}}{\to }\mathrm{O}+\mathrm{O}$; $\mathrm{O}+{\mathrm{O}}_2\to {\mathrm{O}}_3$

   1 mark

3. Chemicals responsible for depletion: chlorofluorocarbons (CFCs), also called freons, used as refrigerants and in fire extinguishers.

   1 mark

Answer: Ozone absorbs harmful UV radiation; it forms when UV splits O₂ into O atoms, which combine with O₂ to give O₃; CFCs deplete it.

1. (i) X has two carbon atoms and is acidic (turns blue litmus red). A 5–8% solution of it in water is vinegar, which is used as a preservative. So X is ethanoic acid (acetic acid).

   ½ mark

2. Structure: $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$. In expanded form, one carbon carries three H atoms and is bonded to a second carbon. The second carbon has a double bond to O and a single bond to an –OH group (H–C(H)(H)–C(=O)–O–H).

   1 mark

3. (ii) Ethanoic acid is a weak acid and ionises only partially. Its pH is therefore higher than that of a mineral acid such as HCl of similar concentration.

   ½ mark

4. (iii) Name of reaction: esterification.

   ½ mark

5. $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\mathrm{Acid}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3+{\mathrm{H}}_2\mathrm{O}$
   (ethanoic acid + ethanol → ethyl ethanoate (ester, sweet smelling) + water)

   1 mark

6. (iv) $2C{\mathrm{H}}_3\mathrm{COOH}+\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3\to 2C{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2$. The CO₂ gas turns lime water milky.

   1 mark

7. Salt produced: sodium ethanoate (sodium acetate), $\mathrm{C}{\mathrm{H}}_3\mathrm{COONa}$.

   ½ mark

Answer: (i) X = ethanoic acid, CH₃COOH; (ii) pH higher than a mineral acid (weaker acid); (iii) esterification: CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O; (iv) 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂, salt = sodium ethanoate

1. (i) Difference: Saturated hydrocarbons have only single covalent bonds between all carbon atoms (general formula $C_n{H}_{2n+2}$, alkanes). Unsaturated hydrocarbons have at least one double or triple bond between two carbon atoms (general formula $C_n{H}_{2n}$ for alkenes, $C_n{H}_{2n-2}$ for alkynes).

   1 mark

2. (i) **Saturated example – Propane, $C{H}_{3}C{H}_{2}C{H}_3$.** Structural formula: H–C–C–C–H in a chain, with every bond single and each carbon carrying enough H atoms to make 4 bonds. That is, the end carbons have 3 H each and the middle carbon has 2 H.

   1 mark

3. (i) **Unsaturated example – Propene, $C{H}_2=CH-C{H}_3$.** Structural formula: H₂C=CH–CH₃, with one C=C double bond between C-1 and C-2. Propyne, $HC\equiv C-C{H}_3$, is also a correct example.

   1 mark

4. (ii) Method: Heat the unsaturated hydrocarbon with hydrogen in the presence of a nickel (or palladium) catalyst. This addition of hydrogen is called hydrogenation, e.g. $C{H}_2=C{H}_2+H_2\stackrel{Ni}{\to }C{H}_3-C{H}_3$.

   1 mark

5. (ii) Industry: This reaction is used in the vegetable oil (vanaspati ghee) industry to hydrogenate vegetable oils.

   ½ mark

6. (iii) **Butene, $C_4{H}_8$.** Structure (but-1-ene): $C{H}_2=CH-C{H}_2-C{H}_3$, i.e. H₂C=CH–CH₂–CH₃.

   ½ mark

Answer: (i) Saturated: only C–C single bonds, e.g. propane CH₃CH₂CH₃. Unsaturated: at least one C=C or C≡C, e.g. propene CH₂=CH–CH₃. (ii) Hydrogenation (H₂ with Ni/Pd catalyst); used in the vegetable oil (vanaspati) industry. (iii) Butene, CH₂=CH–CH₂–CH₃.

1. (i) Herbivores such as cows eat grass, which is rich in cellulose. Cellulose takes a long time to digest, so herbivores have a longer small intestine to allow complete digestion.

   1 mark

2. (i) Carnivores such as tigers eat meat, which is easier to digest. They therefore have a shorter small intestine. So the length of the small intestine depends on the type of food eaten.

   1 mark

3. (ii) Pancreas: It secretes pancreatic juice into the small intestine. This juice contains trypsin, which digests proteins, and lipase, which breaks down emulsified fats.

   1 mark

4. (ii) Bile juice: It is made in the liver and stored in the gall bladder. Bile makes the acidic food coming from the stomach alkaline so that pancreatic enzymes can act. Its bile salts emulsify fats, breaking large fat globules into smaller ones.

   1 mark

5. (iii) The inner lining of the small intestine has numerous finger-like projections called villi, which greatly increase the surface area for absorption. The villi are richly supplied with blood vessels, which carry the absorbed food to every cell of the body.

   1 mark

Answer: (i) Herbivores need a longer small intestine to digest cellulose; carnivores have a shorter one because meat is easy to digest. (ii) Pancreatic juice: trypsin digests proteins, lipase digests emulsified fats. Bile: makes the medium alkaline and emulsifies fats. (iii) Villi increase the surface area and are richly supplied with blood vessels.

1. (i) The rings of cartilage in the throat (trachea) keep the air passage open. They ensure that it does not collapse when there is little or no air in it.

   1 mark

2. (ii) When we breathe in, the muscles between the ribs contract and lift the ribs upward and outward. At the same time, the diaphragm contracts and flattens.

   1 mark

3. (ii) This makes the chest cavity larger and lowers the pressure inside it. As a result, air is sucked into the lungs and fills the expanded alveoli.

   1 mark

4. (iii) During heavy exercise, the muscle cells need energy faster than oxygen can be supplied. They then switch to anaerobic respiration, in which pyruvate is converted into lactic acid: glucose → pyruvate → lactic acid + energy.

   1 mark

5. (iii) The build-up of lactic acid in the muscles causes muscle cramps.

   1 mark

Answer: (i) The rings keep the air passage from collapsing. (ii) Ribs are lifted and the diaphragm flattens, so the chest cavity enlarges and air is sucked into the lungs. (iii) Lack of oxygen leads to anaerobic respiration; pyruvate becomes lactic acid, and its build-up causes cramps.

1. (i) Diagram: Draw several concentric circles on the horizontal plane Q, centred on the wire XY, spaced farther apart as they move away from the wire. Current flows downward from X to Y. Seen from above, the current is going away from the viewer, so the arrows on the field lines point clockwise.

   1½ marks

2. (ii) Name: Right-hand thumb rule.

   ½ mark

3. (ii) Statement: Imagine holding the current-carrying straight conductor in your right hand so that the thumb points in the direction of the current. The direction in which your fingers curl around the conductor gives the direction of the magnetic field lines.

   1 mark

4. (iii) Fleming's left-hand rule: Stretch the thumb, forefinger and middle finger of the left hand so that they are mutually perpendicular. If the forefinger points along the magnetic field and the middle finger along the current, the thumb points in the direction of the force (motion) on the conductor.

   1 mark

5. (iii) The electron moves upward, so the conventional current is downward. The field points to the right. With the forefinger to the right and the middle finger pointing down, the thumb points out of the plane of the paper (towards the reader).

   1 mark

Answer: (i) Concentric circles around XY, clockwise as seen from above. (ii) Right-hand thumb rule. (iii) The force on the electron is perpendicular to the plane of the paper, directed outward (towards the reader).

1. (i) Definition: A solenoid is a coil of many circular turns of insulated copper wire, wrapped closely in the shape of a cylinder.

   1 mark

2. (i) Diagram: Draw a cylindrical coil connected to a battery. Inside it, draw straight, parallel, equally spaced field lines along the axis. Outside, the lines curve from one end round to the other, like those of a bar magnet, with one end acting as N and the other as S.

   1½ marks

3. (i) Marking: (1) Arrows on the turns show the direction of current, entering at one end from the battery. (2) Arrows on the field lines show them coming out of the N end and entering the S end; inside the solenoid they run from S to N. (3) The region inside the solenoid, where the lines are straight and parallel, is labelled as the region of uniform magnetic field.

   1½ marks

4. (ii) Place a piece of soft iron (a magnetic material) inside the solenoid and pass current through it. The strong field inside magnetises the soft iron, which then acts as an electromagnet as long as the current flows.

   1 mark

Answer: (i) A solenoid is a closely wound cylindrical coil of many turns of insulated copper wire. Its field resembles a bar magnet's and is uniform inside. (ii) Insert a soft iron core inside the current-carrying solenoid.

1. (a) P, a strong acid, lies at the low end of the scale, between pH 0 and 4 (around pH 1).

   ½ mark

2. (a) Q, a strong base, lies at the high end of the scale, between pH 12 and 14.

   ½ mark

3. (b)(i) To increase the pH, add a base such as sodium hydroxide (NaOH).

   ½ mark

4. (b)(ii) To decrease the pH, add an acid such as hydrochloric acid (HCl).

   ½ mark

5. (c)(i) A lower pH means a more acidic solution, so the hydronium ion ($H_3{O}^{+}$) concentration increases. A fall of 2 pH units means 100 times more.

   1 mark

6. (c)(i) The colour of the pH paper changes from yellowish-orange (at pH 4) to red (at pH 2).

   1 mark

7. (c)(ii) Indigestion is caused by excess acid in the stomach, so the stomach fluid has a low pH, about 1 to 3.

   1 mark

8. (c)(ii) The medicines used are called antacids.

   ½ mark

9. (c)(ii) Milk of magnesia is magnesium hydroxide, $Mg(OH{)}_2$.

   ½ mark

Answer: (a) P: pH 0–4; Q: pH 12–14. (b)(i) NaOH (a base); (ii) HCl (an acid). (c)(i) H₃O⁺ concentration increases; colour changes from orange/yellow to red. (c)(ii) pH about 1–3; antacids; magnesium hydroxide, Mg(OH)₂.

1. (a) Fertilisation takes place in the oviduct (fallopian tube).

   1 mark

2. (b) Every month, the lining of the uterus thickens and becomes spongy. It is richly supplied with blood so that it can receive and nourish an embryo if fertilisation occurs.

   1 mark

3. (c)(i) If the egg is not fertilised, the thickened, blood-rich uterine lining is not needed. It slowly breaks down.

   1 mark

4. (c)(i) The broken lining comes out through the vagina as blood and mucus, along with the unfertilised egg. This is menstruation, which occurs roughly every month and lasts about 2–8 days.

   1 mark

5. (c)(ii) The embryo gets nutrition through a special disc-like tissue called the placenta, embedded in the uterine wall. Villi on the embryo's side of the placenta lie in blood spaces of the mother and provide a large surface area.

   1 mark

6. (c)(ii) Glucose and oxygen pass from the mother's blood to the embryo through the placenta, while the embryo's wastes pass into the mother's blood to be removed.

   1 mark

Answer: (a) Oviduct (fallopian tube). (b) The uterine lining thickens, becomes spongy and gets a rich blood supply. (c)(i) The lining breaks down and is shed as blood and mucus with the egg; this is menstruation. (c)(ii) Through the placenta, which passes oxygen and nutrients from the mother's blood to the embryo.

1. (a) Dispersion of light.

   1 mark

2. (b) The different colours of white light bend through different angles with respect to the incident ray as they pass through the prism. Red bends the least and violet the most, so the colours separate into a band.

   1 mark

3. (c)(i) Arrangement: Take two identical glass prisms, A and B. Place prism B upside down (inverted) relative to prism A, next to it.

   1 mark

4. (c)(i) Diagram and explanation: Label white light falling on prism A. Show it splitting into a spectrum between the prisms, labelled R (red) to V (violet). Show the colours entering inverted prism B and emerging as a single beam labelled white light falling on a screen. The second prism bends the colours back by equal and opposite amounts, so they recombine into white light.

   1 mark

5. (c)(ii) Rainbow diagram: Draw a spherical raindrop and label incoming sunlight (white light) striking its upper part. Show the light refracting on entering and splitting into colours (dispersion). Show it reflecting internally at the back of the drop and refracting again as it leaves.

   1 mark

6. (c)(ii) Labels: Mark the emerging red and violet rays, and an observer standing with the sun behind them. Label the steps: refraction and dispersion, internal reflection, refraction. Put arrows on all rays.

   1 mark

Answer: (a) Dispersion of light. (b) Different colours bend through different angles in the prism. (c)(i) Two identical prisms, the second inverted, recombine the spectrum into white light. (c)(ii) Raindrop diagram showing refraction with dispersion, internal reflection and refraction, with the observer facing away from the sun.