CBSE Class 10 Science: sample paper 2024-25

1. Write the skeletal equation: $\mathrm{Pb}(\mathrm{N}{\mathrm{O}}_3{)}_2\stackrel{\mathrm{Heat}}{\to }\mathrm{PbO}+\mathrm{N}{\mathrm{O}}_2+{\mathrm{O}}_2$. Balance N and O first, keeping the coefficient of ${\mathrm{O}}_2$ as 1.

2. Take $p=2$. Then Pb gives $q=2$ and N gives $2\times 2=4$, so $r=4$: $2Pb(\mathrm{N}{\mathrm{O}}_3{)}_2\to 2PbO+4N{\mathrm{O}}_2+{\mathrm{O}}_2$.

   ½ mark

3. Check oxygen. Left side: $2\times 6=12$. Right side: $2+8+2=12$. The equation is balanced, so $p,q,r=2,2,4$, which is option (A).

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Answer: p = 2, q = 2, r = 4; option (A)

1. (a) Aluminium forms the amphoteric oxide $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$, so a matches (iii). (b) Gallium has a very low melting point and melts on the palm, so b matches (i).

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2. (c) Manganese reacts with very dilute nitric acid to give ${\mathrm{H}}_2$, so c matches (iv). (d) Gold lies below hydrogen in the activity series and cannot displace it from acids, so d matches (ii). The correct option is (B).

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Answer: a–(iii), b–(i), c–(iv), d–(ii); option (B)

1. The bulb does not glow, so the solution does not conduct electricity. It therefore contains no free ions, which means it is a non-electrolyte.

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2. $\mathrm{HN}{\mathrm{O}}_3$, ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ and $\mathrm{C}{\mathrm{O}}_2$ in water (carbonic acid) all produce ${\mathrm{H}}^{+}$ ions. Ethanol, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$, does not ionise in water. The answer is ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$, option (B). (For the VI alternative, the strongest electrolyte is dilute acid, option A.)

   ½ mark

Answer: C₂H₅OH, option (B) (VI alternative: dilute acid, option A)

1. Phenolphthalein turns pink only in a basic solution, so A is basic and its pH is greater than 7.

   ½ mark

2. The pink colour disappears when B is added, so B has neutralised the base. B must be an acid, with pH less than 7. The answer is option (C).

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Answer: A has pH > 7 and B has pH < 7; option (C)

1. Lead is more reactive than copper, so it displaces copper from copper sulphate: $\mathrm{Pb}+\mathrm{CuS}{\mathrm{O}}_4\to \mathrm{PbS}{\mathrm{O}}_4+\mathrm{Cu}$.

   ½ mark

2. The blue $\mathrm{C}{\mathrm{u}}^{2+}$ ions are removed, so the solution loses its colour. A more reactive metal replacing a less reactive one is a displacement reaction, option (C).

   ½ mark

Answer: Displacement reaction; option (C)

1. X (2, 8, 7) has 7 valence electrons and tends to gain 1 electron, so X is a non-metal (chlorine). Y (2, 8, 2) has 2 valence electrons and tends to lose them, so Y is a metal (magnesium).

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2. Z (2, 8) has a complete octet and is a noble gas (neon). The correct statement is option (C): X is a non-metal and Y is a metal.

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Answer: X is a non-metal and Y is a metal; option (C)

1. Burning a candle, decomposition of vegetable matter and the reaction of sodium with air all release heat, so they are exothermic.

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2. Cooking food needs a continuous supply of heat, so it absorbs energy. It is endothermic, option (B).

   ½ mark

Answer: Cooking of food; option (B)

1. Ethanol formation in yeast and lactic acid formation in muscles are anaerobic. They release very little energy (2 ATP per glucose).

   ½ mark

2. In human cells, aerobic respiration in the mitochondria breaks pyruvate completely into $\mathrm{C}{\mathrm{O}}_2$ and water. This gives the maximum ATP. The answer is option (D).

   ½ mark

Answer: Carbon dioxide in human cells; option (D)

1. Oxygenated blood from the lungs enters the left atrium and then passes into the left ventricle.

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2. When the thick-walled left ventricle contracts, it pumps this blood into the aorta and on to all parts of the body. The answer is option (B).

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Answer: Contraction of left ventricle; option (B)

1. Cellulose is hard to digest, so it needs more time in the gut.

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2. Herbivores have a longer small intestine, which allows cellulose to be digested. The answer is option (D).

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Answer: Longer small intestine; option (D)

1. The cerebellum, part of the hindbrain, controls the precision of voluntary actions and maintains posture and balance.

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2. Salivation, hunger and blood pressure are controlled by other brain regions (medulla and hypothalamus). The answer is option (C).

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Answer: Posture and balance; option (C)

1. In snails, an individual can change sex during its life depending on environmental conditions.

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2. This means sex is not fixed by inherited genes or sex chromosomes. Sex is not genetically determined in snails, option (D).

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Answer: Sex is not genetically determined in snails; option (D)

1. A line through the centre of curvature to any point on a concave mirror is a radius, so it is normal to the mirror at that point.

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2. A ray travelling along the normal makes an angle of incidence of $0^{\circ }$ and retraces its path. The answer is option (B).

   ½ mark

Answer: A ray passing through the centre of curvature; option (B)

1. In dispersion by a prism, red light deviates the least and violet light deviates the most.

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2. Ray A is the least deviated, so it is red. Ray B is the most deviated, so it is violet. The answer is option (C).

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Answer: Ray A: Red; Ray B: Violet; option (C)

1. Producers capture about 1% of sunlight. They use most of this energy for their own life processes and pass only about 10% on to the next trophic level (the 10% law).

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2. Statement (C) says producers use 10% and transfer the rest, which reverses the 10% law. So (C) is the incorrect statement.

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Answer: Option (C)

1. Decomposers break down dead remains and wastes and return nutrients to the soil.

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2. They do not produce oxygen; it is added to the atmosphere by photosynthesis. The answer is option (B).

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Answer: Enrichment of oxygen in atmosphere; option (B)

1. A reactive metal X reacts with dilute HCl and releases hydrogen gas, which burns with a pop sound. So A is true.

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2. In this reaction the metal displaces hydrogen from the acid ($\mathrm{X}+2HCl\to \mathrm{XC}{\mathrm{l}}_2+{\mathrm{H}}_2$). Hydrogen does not displace the metal, so R is false. The answer is option (C).

   ½ mark

Answer: A is true but R is false; option (C)

1. Germ cells are formed by halving the chromosome number, so they carry half the chromosomes of a body cell. A is true.

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2. At fertilisation, two germ cells fuse and restore the full chromosome number. This is exactly why germ cells must carry half the number, so R is true and explains A. The answer is option (A).

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Answer: Both true, R explains A; option (A)

1. For every position of the object, a convex mirror forms a virtual, erect, diminished image behind the mirror. A is true. By the sign convention, the focus lies behind the mirror, so its focal length is positive. R is true.

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2. The positive sign is only a convention and does not explain why the image is always virtual. That happens because the reflected rays diverge. The answer is option (B).

   ½ mark

Answer: Both true, R not the explanation; option (B)

1. Removing the lions (top consumers) would let herbivore numbers rise, which would overgraze the plants. The food chain would be affected, so A is false.

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2. Plants are producers that make food using sunlight, and lions are consumers. R is true. The answer is option (D).

   ½ mark

Answer: A is false but R is true; option (D)

1. A. $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3+2Al\to \mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3+2Fe+\mathrm{heat}$ is an (exothermic) displacement reaction, which is also a redox reaction.

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2. Reason: aluminium is more reactive than iron, so it displaces iron from iron oxide. Al is oxidised and $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3$ is reduced, and heat is released.

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3. B. $\mathrm{Pb}(\mathrm{N}{\mathrm{O}}_3{)}_2+2KI\to \mathrm{Pb}{\mathrm{I}}_2\downarrow +2KN{\mathrm{O}}_3$ is a double displacement (precipitation) reaction.

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4. Reason: the reactants exchange ions ($\mathrm{P}{\mathrm{b}}^{2+}$ pairs with ${\mathrm{I}}^{-}$ and ${\mathrm{K}}^{+}$ pairs with $\mathrm{N}{\mathrm{O}}_3^{-}$), and an insoluble yellow precipitate of $\mathrm{Pb}{\mathrm{I}}_2$ forms.

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Answer: A: exothermic displacement (redox) reaction; B: double displacement (precipitation) reaction

1. Structure and location. Alveoli are tiny balloon-like air sacs at the ends of the bronchioles inside the lungs. A nephron is a long tubular structure (Bowman's capsule, glomerulus and tubule) and is the filtering unit of the kidney.

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2. Function. Alveoli provide a large surface for the exchange of gases: ${\mathrm{O}}_2$ passes into the blood and $\mathrm{C}{\mathrm{O}}_2$ passes out. A nephron filters blood to remove nitrogenous wastes and forms urine.

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Answer: Alveoli: balloon-like sacs at ends of bronchioles in lungs; gaseous exchange. Nephron: tubular unit in kidneys; filters blood to form urine.

1. Step 1: Chlorophyll absorbs light energy.

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2. Step 2: Light energy is converted into chemical energy, and water molecules split into hydrogen and oxygen.

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3. Step 3: Carbon dioxide is reduced to carbohydrates (glucose): $6C{\mathrm{O}}_2+12{\mathrm{H}}_2\mathrm{O}\underset{\text{chlorophyll}}{\overset{\text{sunlight}}{\to }}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+6{\mathrm{O}}_2+6{\mathrm{H}}_2\mathrm{O}$.

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4. Special feature of desert plants: their stomata open at night, when they take in $\mathrm{C}{\mathrm{O}}_2$ and store it as an intermediate. During the day, with stomata closed to save water, this intermediate is used with the light energy absorbed by chlorophyll to make glucose.

   ½ mark

Answer: Absorption of light by chlorophyll → conversion of light to chemical energy and splitting of water → reduction of CO₂ to carbohydrate; desert plants take in CO₂ at night as an intermediate and use it in the day.

1. (i) Salivary amylase, in the mouth, breaks down starch (a complex carbohydrate) into simple sugar (maltose).

   ½ mark

2. (ii) Pepsin, in the stomach and working in an acidic medium, begins the digestion of proteins.

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3. (iii) Trypsin, from pancreatic juice acting in the small intestine, digests proteins further towards amino acids.

   ½ mark

4. (iv) Lipase, from pancreatic and intestinal juice, breaks down emulsified fats into fatty acids and glycerol.

   ½ mark

Answer: Amylase: starch → sugar; Pepsin: proteins (stomach); Trypsin: proteins → amino acids; Lipase: emulsified fats → fatty acids + glycerol

1. A. The image is upright, larger than the object and on the same side as the object. Only a converging lens forms such an image, with the object between $F$ and $O$. So the lens is a convex lens.

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2. B. The image is formed where the backward (dotted) extensions of the refracted rays meet, on the same side as the object. It is therefore virtual (and also erect and enlarged).

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3. C. For a lens, $m=\frac{h_i}{h_o}=\frac{v}{u}$. The image is virtual and erect, so $m=+2$ and $v=-30$ cm (same side as the object).

   ½ mark

4. $2=\frac{-30}{u}\Rightarrow u=\frac{-30}{2}=-15$ cm. The object is placed 15 cm in front of the lens, on the same side as the image.

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5. (For visually impaired students) A. A concave lens always forms a virtual, erect and diminished image. B. Two uses: correcting myopia (short-sightedness) and door peepholes (also some torches and spectacles).

Answer: A. Convex lens; B. Virtual (erect, enlarged); C. u = −15 cm, i.e. 15 cm in front of the lens. (VI: concave lens; used to correct myopia and in door peepholes)

1. (i) Power in heater X: $P=VI=230\text{V}\times 3.5\text{A}=805\text{W}$

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2. (ii) X and Y are in parallel, so both have the same p.d. of 230 V. By Ohm's law $I=\frac{V}{R}$, so at constant $V$ the current is $I\propto \frac{1}{R}$.

   ½ mark

3. $R_X=2R_Y$, so $I_Y=2I_X=2\times 3.5=7.0\text{A}$

   ½ mark

Answer: (i) 805 W; (ii) 7.0 A

1. Series: $R_s=R+2R=3R$

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2. $P_1=\frac{V^2}{R_s}=\frac{V^2}{3R}$

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3. Parallel: $\frac{1}{R_p}=\frac{1}{R}+\frac{1}{2R}=\frac{3}{2R}$, so $R_p=\frac{2R}{3}$

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4. $P_2=\frac{V^2}{2R/3}=\frac{3V^2}{2R}$. Hence $\frac{P_1}{P_2}=\frac{V^2}{3R}\times \frac{2R}{3V^2}=\frac{2}{9}$, i.e. $P_1:P_2=2:9$.

   ½ mark

5. (For visually impaired students) A. The lowest resistance comes from joining all four in parallel: $\frac{1}{R}=\frac{1}{3}+\frac{1}{6}+\frac{1}{9}+\frac{1}{12}=\frac{12+6+4+3}{36}=\frac{25}{36}$, so $R=1.44\Omega$. OR B. $I=P/V=1000/220\approx 4.54$ A. Use fuse B (5 A): it carries 4.54 A safely, whereas fuse A (2 A) would melt.

Answer: P₁ : P₂ = 2 : 9

1. Pesticides sprayed on the cabbage field are washed into the soil and water bodies. They are non-biodegradable, so they persist for a long time and enter the food chain through plants and aquatic organisms.

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2. These chemicals are not broken down or excreted. They accumulate at each trophic level, so their concentration increases at every step. Humans at the last trophic level therefore receive the highest concentration of pesticide. This is called biological magnification, and it can harm human health.

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3. (For visually impaired students) Food chain: cabbage → rabbit → snake → owl, or cabbage → caterpillar → bird → human. The pesticide then magnifies along the chain as explained above, with humans getting the most.

Answer: Biological magnification: non-biodegradable pesticides accumulate at each trophic level, so humans at the last level get the highest concentration.

1. Shake a little of each powder with water and test the liquid with red and blue litmus. Magnesium oxide dissolves slightly to form magnesium hydroxide, which is basic and turns red litmus blue: $\mathrm{MgO}(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to \mathrm{Mg}(\mathrm{OH}{)}_2(\mathrm{aq})$

   1 mark

2. Aluminium oxide is insoluble in water, so its liquid causes no change in either red or blue litmus.

   1 mark

3. Confirm with dilute NaOH. Aluminium oxide is amphoteric and dissolves: $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3+2NaOH\to 2NaAl{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}$. Magnesium oxide is basic and does not react with the alkali. Label the powder that turned red litmus blue (and did not dissolve in NaOH) MgO, and the other Al₂O₃.

   1 mark

Answer: MgO solution turns red litmus blue; Al₂O₃ is insoluble (no litmus change) and, being amphoteric, dissolves in NaOH while MgO does not.

1. Copper is low in the reactivity series, so its compounds can be reduced to the metal by heating alone (self-reduction). First, heat copper sulphide strongly in air (roasting) so that part of it changes to copper(I) oxide: $2C{\mathrm{u}}_2\mathrm{S}(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})\stackrel{\mathrm{\Delta }}{\to }2C{\mathrm{u}}_2\mathrm{O}(\mathrm{s})+2S{\mathrm{O}}_2(\mathrm{g})$

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2. Then heat the copper(I) oxide with the remaining copper sulphide in the absence of air, which reduces it to copper: $2C{\mathrm{u}}_2\mathrm{O}(\mathrm{s})+\mathrm{C}{\mathrm{u}}_2\mathrm{S}(\mathrm{s})\stackrel{\mathrm{\Delta }}{\to }6Cu(\mathrm{s})+\mathrm{S}{\mathrm{O}}_2(\mathrm{g})$

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3. Here $\mathrm{C}{\mathrm{u}}_2\mathrm{O}$ is reduced (it loses oxygen) and the sulphide is oxidised to $\mathrm{S}{\mathrm{O}}_2$, so copper is obtained by heating alone, without any other reducing agent.

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4. (For visually impaired students) i. Metals such as aluminium and copper are used for cooking utensils because they are good conductors of heat and have high melting points. ii. Nitric acid is a strong oxidising agent. It oxidises the ${\mathrm{H}}_2$ produced to water and is itself reduced to nitrogen oxides (N₂O, NO, NO₂). Only Mg and Mn give ${\mathrm{H}}_2$ with very dilute HNO₃.

Answer: 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂ (roasting); 2Cu₂O + Cu₂S → 6Cu + SO₂ (self-reduction).

1. (i) In the Solvay process, $\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2+\mathrm{N}{\mathrm{H}}_3\to \mathrm{N}{\mathrm{H}}_4\mathrm{Cl}+\mathrm{NaHC}{\mathrm{O}}_3$. So **X = sodium hydrogencarbonate, $\mathrm{NaHC}{\mathrm{O}}_3$** (baking soda).

   ½ mark

2. On heating, $2NaHC{\mathrm{O}}_3\stackrel{\mathrm{\Delta }}{\to }\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2$, so **Z = sodium carbonate, $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$**. Its decahydrate Q is washing soda.

   ½ mark

3. (ii) One compound breaks into simpler substances on heating, so X → Z is a (thermal) decomposition reaction.

   ½ mark

4. (iii) The lower the pH, the higher the ${\mathrm{H}}_3{\mathrm{O}}^{+}$ concentration. Solution A (pH 6) has the maximum number of hydronium ions.

   ½ mark

5. Increasing order of ${\mathrm{H}}^{+}$ concentration: C (pH 9.5) < B (pH 8) < A (pH 6)

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Answer: (i) X = NaHCO₃, Z = Na₂CO₃; (ii) thermal decomposition; (iii) A has the most H₃O⁺; C < B < A

1. (i) A bee sting injects an acid (methanoic acid), which the mild base baking soda ($\mathrm{NaHC}{\mathrm{O}}_3$) neutralises. A wasp sting is basic, so a mild acid such as vinegar (ethanoic acid) neutralises it. In both cases neutralisation relieves the pain.

   1 mark

2. (ii) Soil can become too acidic, which harms crop growth. Quicklime (CaO) is basic and neutralises the excess acid, bringing the soil to a neutral or slightly basic pH suitable for plants.

   1 mark

3. (iii) Marble ($\mathrm{CaC}{\mathrm{O}}_3$) reacts with acid rain and acidic gases ($\mathrm{S}{\mathrm{O}}_2$, $\mathrm{N}{\mathrm{O}}_2$) in the air and corrodes: $\mathrm{CaC}{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{CaS}{\mathrm{O}}_4+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2$. Coating sculptures with protective chemicals stops this contact and conserves them.

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Answer: (i) Neutralisation: bee sting acidic (base used), wasp sting basic (acid used); (ii) quicklime neutralises acidic soil; (iii) chemicals protect marble from acid rain/acidic gases.

1. Root cells in contact with the soil take up water (and minerals). The xylem vessels and tracheids of the roots, stem and leaves are interconnected, forming a continuous water-conducting channel from the roots to the leaves.

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2. During the day the stomata are open and water evaporates from the leaves. This loss of water vapour is transpiration, and the transpiration pull it creates is the main driving force for moving water up the xylem.

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3. Evaporation from the leaf cells creates a suction that pulls a continuous column of water from the root xylem, up through the stem, into the leaves. (At night, root pressure helps push water up.)

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Answer: Water absorbed by the roots moves up through continuous xylem channels, pulled by suction (transpiration pull) created by evaporation from the leaves.

1. Let L = long ears (dominant) and l = short ears (recessive). Parents: father $LL$ (pure long) × mother $ll$ (pure short). Gametes: $L$ and $l$. F1: all Ll (long ears).

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2. F1 × same genotype: $Ll\times Ll$. Gametes $L,l$ from each parent. Punnett square: $LL$, $Ll$, $Ll$, $ll$. Genotypic ratio of F2 = 1 LL : 2 Ll : 1 ll (phenotypic ratio 3 long : 1 short).

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3. B. Now father $ll$ × mother $LL$. F1 is again all $Ll$, and $Ll\times Ll$ again gives 1 LL : 2 Ll : 1 ll. There is no change in the ratio: the cross is still between pure dominant and pure recessive parents, and it does not matter which parent carries which trait.

   1 mark

Answer: A. F2 genotypic ratio 1 LL : 2 Ll : 1 ll; B. No change, still 1 : 2 : 1

1. A. A person with hypermetropia sees distant objects clearly but has difficulty seeing near objects. A person with myopia sees near objects clearly, but distant objects appear blurred.

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2. B (i) The rays from the near point N' converge at a point behind the retina, so the defect is hypermetropia (long-sightedness).

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3. (ii) Cause: the eyeball is too short, or the focal length of the eye lens is too long.

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4. (iii) No. A concave lens diverges the rays and would push the image even farther behind the retina. A convex (converging) lens of suitable power is needed: it converges the rays so that the image forms exactly on the retina.

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5. (For visually impaired students) B. Causes of myopia: excessive curvature (high converging power) of the eye lens, and elongation of the eyeball.

Answer: A. Hypermetropia: near objects unclear; myopia: distant objects unclear. B (i) Hypermetropia; (ii) short eyeball / too long focal length; (iii) No, a convex lens is needed.

1. A. The resistance of a wire is directly proportional to its length: $R\propto l$

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2. It is inversely proportional to its area of cross-section: $R\propto \frac{1}{A}$. Combining, $R\propto \frac{l}{A}$.

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3. So $R=\rho \frac{l}{A}$, where $\rho$ is a constant called the resistivity of the material.

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4. B. Rearranging, $\rho =\frac{RA}{l}$. If $l=1$ m and $A=1{\text{m}}^2$, then $\rho =R$.

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5. Resistivity is the resistance of a wire of the material of length 1 m and cross-sectional area 1 m². Its SI unit is Ω m.

   ½ mark

Answer: R = ρl/A; resistivity is the resistance of a 1 m long conductor of the material with 1 m² cross-sectional area (unit Ω m).

1. A. The magnetic field produced by a straight current-carrying wire is inversely proportional to the distance from the wire. As the compass moves away, the field acting on it becomes weaker, so the deflection decreases.

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2. B. The field is directly proportional to the current. Increasing the current (for example, by using a battery of higher voltage or reducing the circuit resistance) would increase the deflection.

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3. C. From the battery's polarity, the current in the vertical wire flows from the top of the board downwards. By the right-hand thumb rule, point the thumb along the current (downwards); the curled fingers then give the field lines as concentric circles that are clockwise when viewed from above the board.

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4. (For visually impaired students) C. If the battery's polarity is reversed, the current in the wire reverses, so the direction of the magnetic field also reverses (clockwise becomes anticlockwise).

Answer: A. Field weakens with distance (B ∝ 1/r); B. Increase the current; C. Current downward, so field is clockwise (seen from above) by the right-hand thumb rule.

1. (i) Keerthi is correct. Saturated hydrocarbons have only single bonds, so they react by substitution: an H atom is replaced by another atom such as Cl, e.g. $\mathrm{C}{\mathrm{H}}_4+\mathrm{C}{\mathrm{l}}_2\stackrel{\mathrm{sunlight}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}+\mathrm{HCl}$. Unsaturated hydrocarbons have double or triple bonds, so they undergo addition across the multiple bond instead.

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2. (ii) Methane, propane and their isomers burn in oxygen with a clean flame, giving $\mathrm{C}{\mathrm{O}}_2$, water and a large amount of heat. This makes them good fuels, e.g. $\mathrm{C}{\mathrm{H}}_4+2{\mathrm{O}}_2\to \mathrm{C}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}+\mathrm{heat}$ (CNG); propane is a component of LPG.

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3. The immediate lower homologue of propane is **ethane, ${\mathrm{C}}_2{\mathrm{H}}_6$**. Electron dot structure: the two C atoms share one pair of electrons (C–C), and each C shares one electron pair with each of three H atoms. Written in a line: ${\mathrm{H}}_3\mathrm{C}:\mathrm{C}{\mathrm{H}}_3$, with a shared pair (:) between every C and each of its H atoms. That gives 7 shared pairs in all, an octet on each C and a duet on each H.

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4. Two characteristics of a homologous series: (1) successive members differ by a $-\mathrm{C}{\mathrm{H}}_2-$ unit (14 u in molecular mass); (2) all members have the same general formula and functional group, and hence similar chemical properties. Physical properties show a gradual change.

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5. (iii) With pure oxygen, ethyne undergoes complete combustion and gives a very hot, clean blue flame suitable for welding. Air contains only about 21% oxygen, so ethyne burns incompletely in air, giving a sooty flame with less heat that is not hot enough for welding.

   1 mark

Answer: (i) Keerthi: substitution in saturated, addition in unsaturated hydrocarbons; (ii) they burn releasing much heat; ethane C₂H₆ dot structure with 7 shared pairs; homologues differ by –CH₂– and have similar chemical properties; (iii) air gives incomplete combustion and a sooty, less hot flame.

1. (i) I would prefer B, the sodium salt of the long-chain sulphonic acid, which is a detergent.

   ½ mark

2. Reason: Hand-pump (underground) water is hard because it contains $\mathrm{C}{\mathrm{a}}^{2+}$ and $\mathrm{M}{\mathrm{g}}^{2+}$ salts. With these ions, the soap A (a carboxylate) forms an insoluble scum, so the soap is wasted. The calcium and magnesium salts of sulphonic acids are soluble, so B forms no scum and still cleans well in hard water.

   ½ mark

3. (ii) Structure of a soap molecule: A soap molecule has two ends that behave differently.

   • The ionic end ($-\mathrm{CO}{\mathrm{O}}^{-}\mathrm{N}{\mathrm{a}}^{+}$) is hydrophilic and dissolves in water.
   • The long hydrocarbon chain is hydrophobic and dissolves in oil or grease.
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4. Cleansing action: Most dirt is oily and does not dissolve in water. In soap solution, the hydrocarbon tails of many soap molecules dissolve in the oily dirt. The ionic heads stay outside, facing the water. In this way the molecules arrange themselves into spherical clusters called micelles, with the dirt trapped at the centre.

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5. Because the outer surfaces of the micelles are charged, the micelles stay suspended in water as an emulsion and do not come together. When the cloth is rinsed with water, the micelles carry the trapped dirt away, and the cloth becomes clean.

   ½ mark

6. Labelled diagram of a micelle (described in words): Draw a circle for the oil/dirt droplet at the centre. Around it, draw many soap molecules arranged like spokes.

   • The zig-zag hydrocarbon tails (hydrophobic) point inwards into the oil droplet.
   • The round ionic heads ($-\mathrm{CO}{\mathrm{O}}^{-}$, hydrophilic) lie on the outer surface, facing the surrounding water.
   • A few $\mathrm{N}{\mathrm{a}}^{+}$ ions are shown outside.

   Labels needed: oil droplet, hydrophobic tail, hydrophilic (ionic) head, $\mathrm{N}{\mathrm{a}}^{+}$, water.

   1 mark

7. (iii) When the ester is heated with sodium hydroxide, it is hydrolysed to the sodium salt of the acid and the alcohol:
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{COOC}{\mathrm{H}}_3+\mathrm{NaOH}\stackrel{\mathrm{\Delta }}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+\mathrm{C}{\mathrm{H}}_3\mathrm{OH}$$
   The products are sodium ethanoate and methanol.

   ½ mark

8. Making a soap by alkaline hydrolysis of an ester in this way is called saponification.

   ½ mark

Answer: (i) B (sulphonate/detergent): its Ca and Mg salts are soluble, so it forms no scum in hard water; (ii) the hydrophobic tails dissolve in oily dirt and the ionic heads face water, forming micelles that are rinsed away (labelled micelle diagram); (iii) CH₃COOCH₃ + NaOH → CH₃COONa + CH₃OH, saponification

1. (i) New banana plants grow from suckers, which are outgrowths of the stem (vegetative parts). No seeds or gametes are involved.

   • Name: vegetative propagation
   • Type: asexual reproduction
   1 mark

2. (ii) Advantage 1: Plants are raised faster. They flower and fruit sooner than seed-grown plants, so the farmer gets more crops in the same time.

   1 mark

3. Advantage 2: All new plants are genetically identical to the parent. Desirable features such as fruit quality are kept unchanged. Also, the farmer need not depend on seeds or pollinators.

   1 mark

4. (iii) The male flowers produce pollen in their anthers. This pollen is carried by agents such as insects, wind or water to the stigma of female flowers of another banana plant. This is cross-pollination, and it allows fertilisation and fruit formation.

   1 mark

5. (iv) Every reproduction involves copying the parent's DNA. DNA copying is not perfectly accurate, so small variations appear. Hence the offspring is very similar to the parent but not absolutely identical.

   1 mark

Answer: (i) Vegetative propagation, asexual; (ii) e.g. faster flowering and fruiting (more crops in less time) and genetically identical plants with no need for pollinators; (iii) its pollen is carried by insects/wind/water to the stigma of female flowers of another plant (cross-pollination); (iv) small errors in DNA copying cause slight variation

1. (i) The umbilical vessels carry materials between mother and foetus.

   • Glucose and oxygen pass from mother to foetus.
   • Carbon dioxide and nitrogenous wastes pass from foetus to mother.

   Any two of these substances, e.g. glucose and oxygen.

   1 mark

2. (ii) Villi give the placenta a large surface area for exchange. With fewer villi, the surface area is smaller. Less glucose and oxygen pass from mother to embryo, and wastes are removed less efficiently. The baby's growth would therefore be slow (retarded).

   1 mark

3. (iii) The embryo develops in the uterus. To nourish the baby, the uterus develops a thick, spongy lining with a rich blood supply. The embryo is attached to this lining through the placenta, so nutrients and oxygen reach it easily.

   1 mark

4. (iv)(a) The cells carry the XY sex chromosomes, i.e. a Y chromosome is present. The foetus is therefore a male child.

   1 mark

5. (iv)(b) Some people use prenatal tests to find the sex of the foetus. If it is female, they get it aborted. This is female foeticide, which is illegal and lowers the sex ratio. For this reason, prenatal sex determination is banned by law.

   1 mark

Answer: (i) e.g. glucose and oxygen (also CO₂/wastes); (ii) smaller surface area for exchange, so slow growth of the baby; (iii) uterus, whose thick lining has a rich blood supply; (iv)(a) male, (iv)(b) misused for female foeticide

1. (i) The 4.0 Ω resistor and $R_2$ are in parallel, so the potential difference across them is the same.

   ½ mark

2. Using $V=IR$ for the 4.0 Ω resistor:
   $V=1.5\text{A}\times 4.0\Omega =6\text{V}$

   So the p.d. across $R_2$ is 6 V.

   ½ mark

3. (ii) The total current of 2.0 A divides between the two parallel branches. The 4.0 Ω resistor takes 1.5 A.

   ½ mark

4. Current through $R_2$:
   $I_2=2.0-1.5=0.5\text{A}$

   Apply Ohm's law to $R_2$: $6\text{V}=0.5\text{A}\times R_2$.

   ½ mark

5. $R_2=\frac{6}{0.5}=12\Omega$

   ½ mark

6. (iii) $R_1$, the 2.0 Ω resistor and the parallel combination are in series with the 12 V battery. So:

   $V_{R_1}=12\text{V}-V_{\text{parallel}}-V_{2\Omega }$

   ½ mark

7. P.d. across the 2.0 Ω resistor $=2.0\text{A}\times 2.0\Omega =4\text{V}$

   P.d. across the parallel part $=6\text{V}$ (from part (i))

   ½ mark

8. $V_{R_1}=12-6-4=2\text{V}$

   ½ mark

9. $R_1$ is in the main line, so the full 2.0 A flows through it.

   ½ mark

10. $R_1=\frac{V}{I}=\frac{2\text{V}}{2\text{A}}=1\Omega$

    *(Check: the total resistance should be $\frac{12}{2}=6\Omega$. Here $R_1+\frac{4\times 12}{4+12}+2=1+3+2=6\Omega$ ✓)*

    ½ mark

Answer: (i) 6 V; (ii) R₂ = 12 Ω; (iii) R₁ = 1 Ω

1. (i) Use $P=VI$, so $I=\frac{P}{V}$.

   ½ mark

2. Lamp A works at its rated values of 12 V and 24 W:
   $I_A=\frac{24\text{W}}{12\text{V}}=2\text{A}$

   ½ mark

3. (ii) The voltmeter reads 12 V.

   ½ mark

4. Reason: The voltmeter is connected across the parallel combination. One branch of this combination is lamp A, which runs at its rated voltage of 12 V. Branches in parallel have the same p.d., so the voltmeter (and the branch containing B and $R_2$) shows 12 V.

   ½ mark

5. (iii) In the branch containing lamp B and $R_2$ (in series):
   $V_{R_2}+V_B=12\text{V}$

   ½ mark

6. Lamp B works at its rated 6.0 V, so:
   $V_{R_2}=12-6=6\text{V}$

   ½ mark

7. $R_2$ is in series with lamp B, so the current through $R_2$ is the same as the current through B, which is 3.0 A.

   ½ mark

8. $R_2=\frac{V}{I}=\frac{6\text{V}}{3\text{A}}=2\Omega$

   ½ mark

9. (iv) $R_1$ is in the main line, so it carries the total current:
   $I=2\text{A}+3\text{A}=5\text{A}$

   P.d. across $R_1=15\text{V}-12\text{V}=3\text{V}$

   ½ mark

10. $R_1=\frac{3\text{V}}{5\text{A}}=0.6\Omega$

    ½ mark

Answer: (i) 2 A; (ii) 12 V, because the voltmeter is across the parallel part and lamp A runs at 12 V; (iii) R₂ = 2 Ω; (iv) R₁ = 0.6 Ω

1. A. In reaction 1, electrolysis of an aqueous salt gives caustic soda (NaOH). This happens with brine, so A = sodium chloride (NaCl).

   ½ mark

2. In reaction 1:
   $$2NaCl(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\stackrel{\text{electrolysis}}{\to }2NaOH(\mathrm{aq})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})$$
   Chlorine is released at the anode, so **Y = chlorine gas ($\mathrm{C}{\mathrm{l}}_2$)**.

   ½ mark

3. Hydrogen is released at the cathode, so **Z = hydrogen gas (${\mathrm{H}}_2$)**.

   ½ mark

4. In reaction 2, electrolysis of molten NaCl gives sodium at the cathode (and $\mathrm{C}{\mathrm{l}}_2$ at the anode), so M = sodium metal (Na).

   ½ mark

5. In reaction 3, sodium reacts with ethanoic acid:
   $$2Na+2C{\mathrm{H}}_3\mathrm{COOH}\to 2C{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{H}}_2$$
   So **P = sodium ethanoate ($\mathrm{C}{\mathrm{H}}_3\mathrm{COONa}$)**.

   ½ mark

6. In reaction 4, an acid reacts with an alcohol Q (ethanol) in the presence of conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$. This is esterification:
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{conc.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$$
   So R = ethyl ethanoate (an ester).

   Check with reaction 5: $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaOH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$, which gives P and Q ✓.

   ½ mark

7. B. Activity to test Z (hydrogen):
   1. Put zinc granules with dilute sulphuric acid in a test tube fitted with a cork and a delivery tube.
   2. Pass the gas evolved into soap solution, so that soap bubbles filled with the gas form.
   3. Bring a burning candle near a gas-filled bubble.

   The gas burns with a pop sound, which confirms it is hydrogen:
   $\mathrm{Zn}+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{ZnS}{\mathrm{O}}_4+{\mathrm{H}}_2$

   1 mark

8. C. (Alternative to B) The process is the chlor-alkali process. It is named after its products: chlorine and an alkali, sodium hydroxide (with hydrogen as a by-product). They are formed when electricity is passed through aqueous NaCl (brine).

   1 mark

Answer: A: A = sodium chloride, Y = chlorine, Z = hydrogen, M = sodium, P = sodium ethanoate, R = ethyl ethanoate (ester); B: hydrogen from Zn + dil. H₂SO₄ burns with a pop sound; OR C: chlor-alkali process, named after its products chlorine and alkali (NaOH)

1. A. The bending of the shoot towards light is phototropism. It is a directional movement that depends on growth.

   ½ mark

2. The plant hormone auxin is made at the shoot tip and helps cells grow longer.

   ½ mark

3. When light falls from one side, auxin diffuses towards the shaded side of the shoot.

   ½ mark

4. The cells on the shaded side therefore elongate more than those on the lit side. The shoot bends towards the light.

   ½ mark

5. B. (Alternative to A) The folding of 'Touch me not' leaves is a growth-independent movement and happens quickly.

   ½ mark

6. The touch stimulus is passed from cell to cell by electrical–chemical signals.

   ½ mark

7. The movement occurs at a point different from the point of touch, showing that the information has travelled.

   ½ mark

8. The cells there change shape by changing the amount of water in them. They swell or shrink, which makes the leaflets fold and droop.

   ½ mark

9. C. Both are tropic movements, i.e. directional growth in response to a stimulus.

   • The pollen tube grows towards the ovule in response to chemicals: chemotropism.
   • The shoot in figure A bends towards light: phototropism.
   1 mark

10. D. Plants and animals both pass information by electrical–chemical means, but they differ in two ways.

    • Signal conduction: Plants have no specialised tissue such as nerves for conducting information. Animals have nerve cells.
    • Change of shape: Plant cells move by changing their water content. Animal body parts move because special proteins in muscle cells change the shape of those cells.
    1 mark

Answer: A: phototropism, as auxin moves to the shaded side and those cells elongate more, so the shoot bends towards light; OR B: growth-independent movement, with electrical–chemical signals causing cells to lose or gain water and change shape, so leaves fold; C: pollen tube growth is chemotropism, while shoot bending is phototropism (both directional growth movements); D: plants have no nervous tissue and move by changing water content, while animals use nerves and contractile muscle proteins

1. A. The concave mirror first forms a real, inverted image of the star near its focus. The eyepiece lens then acts like a magnifying glass on this image.

   ½ mark

2. The observer therefore sees a magnified, virtual image (inverted with respect to the star) through the eyepiece.

   ½ mark

3. B. Light from a distant star arrives as parallel rays. These rays must be gathered and brought to a focus, which only a concave (converging) mirror can do. A concave mirror is therefore used.

   1 mark

4. C. The eyepiece is a convex (converging) lens.

   1 mark

5. Reason: It collects the rays reflected by the plane mirror. The image formed by the concave mirror lies within the focal length of this lens, so the lens forms an enlarged image that the eye can view comfortably, just as a reading glass magnifies an object.

   1 mark

6. D. (Alternative to C) The plane mirror is set at an angle in the path of the rays converging from the concave mirror. It reflects these rays sideways out of the tube towards the eyepiece. The observer can then view the image without blocking the light entering the telescope.

   1 mark

7. The plane mirror does not change the size of the image. It only shifts the position of the image formed by the curved mirror, and it laterally inverts that image.

   1 mark

Answer: A: a magnified virtual image (inverted w.r.t. the star); the intermediate image by the mirror is real; B: concave mirror; C: convex (converging) lens, which magnifies the image formed by the mirror; OR D: the plane mirror redirects the rays out of the tube to the eyepiece, shifting (and laterally inverting) the image