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

1. Balanced equation for electrolysis: $2{\mathrm{H}}_2\mathrm{O}\to 2{\mathrm{H}}_2+{\mathrm{O}}_2$

2. Mass of hydrogen $=2\times 2=4$ g. Mass of oxygen $=1\times 32=32$ g.

3. $M_H:M_O=4:32=1:8$, so the answer is option (D).

   1 mark

Answer: 1 : 8, option (D)

1. Al has valency 3 and O has valency 2. Crossing the valencies gives $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$: $4\mathrm{Al}+3{\mathrm{O}}_2\to 2\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$

2. Mg has valency 2, so its oxide is $\mathrm{MgO}$: $2\mathrm{Mg}+{\mathrm{O}}_2\to 2\mathrm{MgO}$. The answer is option (B).

   1 mark

Answer: Al₂O₃ and MgO, option (B)

1. A has pH 5. This is slightly below 7, so A is a weak acid. B has pH 7, so B is neutral.

2. C has pH 13, which is far above 7, so C is a strong base. D has pH 3, which is far below 7, so D is a strong acid.

3. The order is weak acid, neutral, strong base, strong acid, which is option (D).

   1 mark

Answer: Option (D)

1. (i) $\mathrm{HCl}+\mathrm{NaOH}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}$

2. (ii) $\mathrm{MgO}+2HCl\to \mathrm{MgC}{\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}$

3. (iii) $\mathrm{C}{\mathrm{O}}_2+2NaOH\to \mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{O}$

4. Every reaction gives a salt and water, so the answer is option (A). Na₂CO₃ is basic, not neutral, so (B) is wrong.

   1 mark

Answer: Salt and water is formed, option (A)

1. A loses electrons and B gains them, so C is an ionic compound.

2. Ionic compounds have a high melting point, are soluble in water and conduct electricity when molten or in solution. These match (A), (B) and (D).

3. The attraction between their ions is strong, not weak. So property (C) is not shown, and the answer is option (C).

   1 mark

Answer: Option (C)

1. Metals at the top of the reactivity series (Na, K, Ca, Mg, Al) cannot be reduced by carbon. They are obtained by electrolysis of their molten compounds.

2. Calcium and magnesium are extracted by electrolysis of their molten chlorides. Gold, silver and iron are less reactive and are extracted by other methods. The answer is option (B).

   1 mark

Answer: Calcium and magnesium, option (B)

1. Mg (2, 8, 2) loses its two valence electrons to O (2, 6).

2. This forms $\mathrm{M}{\mathrm{g}}^{2+}$ and ${\mathrm{O}}^{2-}$ (an octet), giving $\mathrm{M}{\mathrm{g}}^{2+}[\mathrm{O}{]}^{2-}$ in a 1 : 1 ratio.

3. Option (A) shows two electrons moving from Mg to O and the ions $\mathrm{M}{\mathrm{g}}^{2+}$ and ${\mathrm{O}}^{2-}$, so it is correct.

   1 mark

Answer: Option (A)

1. Saliva contains the enzyme salivary amylase.

2. Salivary amylase breaks down starch into simple sugars. Less saliva therefore affects this conversion, so the answer is option (C).

   1 mark

Answer: Starch into simple sugars, option (C)

1. When light falls on one side of a shoot, auxin diffuses to the shaded side.

2. Auxin makes the cells on the shaded side grow longer, so the shoot bends towards light. The answer is option (D).

   1 mark

Answer: Auxins, option (D)

1. A bisexual flower has both stamens and pistil, so (i) is true and (ii) is false.

2. It can be self-pollinated or cross-pollinated, so (iii) is true. It can produce fruits on its own, so (iv) is false.

3. The answer is option (C), (i) and (iii).

   1 mark

Answer: (i) and (iii), option (C)

1. Gametes: RRyy gives only Ry and rrYY gives only rY.

2. Every F₁ plant is RrYy. R is dominant over r and Y is dominant over y.

3. So all F₁ seeds are round and yellow (100%), which is option (C).

   1 mark

Answer: 100% round and yellow, option (C)

1. Glucose is broken down to pyruvate (glycolysis) in the cytoplasm of all cells. So site (a) is the cytoplasm.

2. Pyruvate changes to lactic acid in muscle cells when oxygen is lacking. So site (b) is oxygen-deficient muscle cells. The answer is option (D).

   1 mark

Answer: Cytoplasm and oxygen deficient muscle cells, option (D)

1. X is a concave mirror. Sunlight comes from a very distant object, so the image is real, highly diminished and at the focus. Statement (i) is true.

2. Y is a convex mirror. It always forms a virtual, erect and diminished image. Statement (ii) is true.

3. The answer is option (A), (i) and (ii).

   1 mark

Answer: (i) and (ii), option (A)

1. Weak ciliary muscles and a less flexible eye lens in old age cause presbyopia.

2. Such a person often has both near and far defects, which is corrected by a bifocal lens. The answer is option (B).

   1 mark

Answer: Presbyopia and bifocal lens, option (B)

1. (i) Grass → rabbit → wolf → lion is a valid chain.

2. (iii) Grass → grasshopper → frog → snake → hawk is a valid chain.

3. (ii) has no proper sequence: man does not eat plankton, and grasshoppers do not eat man. (iv) fails because a snake does not eat grass.

4. So (ii) and (iv) are not food chains, and the answer is option (D).

   1 mark

Answer: (ii) and (iv), option (D)

1. Green plants convert only about 1% of the sunlight falling on their leaves into food energy.

2. So about $100\%-1\%=99\%$ is not converted. The answer is option (D).

   1 mark

Answer: 99%, option (D)

1. Assertion: Decomposition reactions need energy (heat, light or electricity) to break bonds, so they are generally endothermic. The Assertion is true.

2. Reason: Composting releases heat, so it is exothermic. The Reason is true.

3. The Reason is an exception and does not explain why decomposition is generally endothermic. The answer is option (B).

   1 mark

Answer: Option (B)

1. Assertion: A child inherits the basic design of the human body. The Assertion is true.

2. Reason: A child does not look exactly like its parents. Sexual reproduction produces noticeable variations, so the Reason is false.

3. The answer is option (C).

   1 mark

Answer: Option (C)

1. Field lines never cross each other, so the Assertion is true.

2. If two lines crossed, the field would have two directions at that point. A compass needle cannot point two ways at once, so the Reason is true.

3. The Reason correctly explains the Assertion. The answer is option (A).

   1 mark

Answer: Option (A)

1. Ozone levels fell sharply in the 1980s, mainly because of CFCs. The Assertion is true.

2. $\mathrm{O}+{\mathrm{O}}_2\to {\mathrm{O}}_3$ is how ozone forms, so the Reason is true.

3. The Reason describes how ozone forms, not why it decreased. The answer is option (B).

   1 mark

Answer: Option (B)

1. Reaction: $\mathrm{Zn}(\mathrm{s})+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})\to \mathrm{ZnS}{\mathrm{O}}_4(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$

2. Observation 1: Bubbles of a gas (hydrogen) form on the zinc granules, which shows that a gas is evolved.

   1 mark

3. Observation 2: The conical flask becomes warm, which shows a change (rise) in temperature.

   1 mark

Answer: Evolution of gas (H₂ bubbles) and rise in temperature

1. Draw three Hydra figures in sequence, each a tubular body with tentacles at the top. Figure 1: the parent Hydra with a small outgrowth (bud) on its side. Figure 2: the bud grows larger and develops tentacles at its tip. Figure 3: a fully formed young Hydra still attached at the base, about to detach.

   1 mark

2. Labels: tentacles, bud (growing bud / young Hydra) and parent body.

   1 mark

Answer: Labelled diagram of the stages of budding in Hydra (bud forms, grows tentacles, detaches)

1. Plugging the leak stops the blood pressure from falling, which keeps the pumping system working efficiently.

   1 mark

2. Component: blood platelets.

   ½ mark

3. Function: platelets help the blood clot at the site of injury, which plugs the leak.

   ½ mark

Answer: Prevents fall in blood pressure / maintains pumping efficiency; platelets, which clot blood at the injury site

1. (i) Plants do not move, and many of their tissues are made of dead cells. So their energy needs are low, and a slow transport system is enough.

   1 mark

2. (ii) Phloem translocates the soluble products of photosynthesis (sugars) from the leaves to other parts of the plant. It also carries amino acids and other substances to storage organs (roots, fruits, seeds) and growing parts.

   1 mark

Answer: (i) Low energy needs (no movement, many dead cells); (ii) translocation of food from leaves to other parts

1. Using the sign convention for a concave lens: object distance $u=-60$ cm and focal length $f=-30$ cm. A concave lens is diverging, so $f$ is negative.

2. Lens formula: $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

   ½ mark

3. Substituting the values: $\frac{1}{-30}=\frac{1}{v}-\frac{1}{-60}$, so $\frac{1}{v}=-\frac{1}{30}-\frac{1}{60}$

   ½ mark

4. $\frac{1}{v}=\frac{-2-1}{60}=\frac{-3}{60}=-\frac{1}{20}$, so $v=-20$ cm.

   ½ mark

5. The image is formed 20 cm from the lens, on the same side as the object. The negative sign shows this. The image is virtual and erect.

   ½ mark

Answer: v = -20 cm: image 20 cm from the lens, on the same side as the object (virtual, erect)

1. Resistance is proportional to length ($R=\rho l/A$). Each of the three equal parts therefore has resistance $\frac{R}{3}$.

   ½ mark

2. For resistors in parallel: $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$

   ½ mark

3. Substituting: $\frac{1}{R_p}=\frac{1}{R/3}+\frac{1}{R/3}+\frac{1}{R/3}=\frac{3}{R}+\frac{3}{R}+\frac{3}{R}=\frac{9}{R}$

   ½ mark

4. Therefore the total resistance of the combination is $R_p=\frac{R}{9}$.

   ½ mark

Answer: R_p = R/9

1. Electric power is the rate at which electrical energy is consumed or dissipated in a circuit, that is, electrical energy used per unit time: $P=\frac{W}{t}=VI$.

   1 mark

2. Since $P=VI$, the power consumed is 1 watt when a current of 1 ampere flows through a circuit across a potential difference of 1 volt. Equivalently, 1 joule of electrical energy is consumed in 1 second: $1\text{ W}=1\text{ V}\times 1\text{ A}=1\text{ J/s}$.

   1 mark

Answer: Power = rate of consumption of electrical energy; 1 W when 1 A flows under 1 V (1 J consumed per second)

1. Chemical fertilisers and pesticides sprayed on crops get washed by rain into the soil and water bodies. There they harm both the biotic components (plants, animals, microbes) and the abiotic components (soil, water) of the ecosystem.

   1 mark

2. Most of these chemicals are non-biodegradable. They enter food chains and build up in increasing concentration at each higher trophic level. This is called biological magnification. Organisms at all trophic levels, including humans at the top, therefore suffer harmful health effects. So the statement is justified.

   1 mark

Answer: Chemicals wash into soil and water, are non-biodegradable, and accumulate at each trophic level (biological magnification), harming all organisms.

1. (a) Oxides of moderately reactive metals such as Mn and Fe can be reduced by a more reactive metal like aluminium, which displaces them. These reactions give out so much heat that the metal is obtained in the molten state.

2. (a) Example 1: $3\mathrm{Mn}{\mathrm{O}}_2(\mathrm{s})+4\mathrm{Al}(\mathrm{s})\to 3\mathrm{Mn}(\mathrm{l})+2\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3(\mathrm{s})+\text{heat}$

   1 mark

3. (a) Example 2 (thermit reaction, used to join railway tracks): $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3(\mathrm{s})+2\mathrm{Al}(\mathrm{s})\to 2\mathrm{Fe}(\mathrm{l})+\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3(\mathrm{s})+\text{heat}$

   1 mark

4. (b) Metals high in the reactivity series, such as Na, Ca, Mg and Al, have a greater affinity for oxygen than carbon has. Carbon cannot remove oxygen from their oxides, so these metals are extracted by electrolytic reduction instead.

   1 mark

Answer: (a) 3MnO₂ + 4Al → 3Mn + 2Al₂O₃ + heat; Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + heat; (b) these metals have more affinity for oxygen than carbon

1. Set-up: Take three test tubes labelled A, B and C and place clean iron nails in each. In A, pour some ordinary tap water and cork it, so the nails get both air and water.

   1 mark

2. In B, pour boiled distilled water, which has no dissolved air, add about 1 mL of oil to stop air dissolving, and cork it. This gives water only. In C, put anhydrous calcium chloride, which absorbs moisture, and cork it. This gives dry air only. (Diagram: three corked tubes. A: nails partly in water with air above. B: nails in boiled water under an oil layer. C: nails above anhydrous $\mathrm{CaC}{\mathrm{l}}_2$ in dry air.)

   1 mark

3. Observation: After a few days, the nails in A rust, but the nails in B and C do not. Conclusion: Iron rusts only when it is exposed to both air (oxygen) and water (moisture) together.

   1 mark

Answer: Iron rusts only when both air (oxygen) and water (moisture) are present (nails rust in tube A only).

1. (i) Two metals that react violently with cold water: sodium and potassium (lithium is also acceptable).

   1 mark

2. (i) Observations: (1) the reaction is violent and vigorous, and the metal moves about on the water; (2) a large amount of heat is evolved (exothermic); (3) the hydrogen gas evolved catches fire. Reaction: $2\mathrm{Na}+2{\mathrm{H}}_2\mathrm{O}\to 2\mathrm{NaOH}+{\mathrm{H}}_2+\text{heat}$

   1½ marks

3. (ii) The gas evolved is hydrogen. Bring a burning matchstick or splinter near the gas: it burns with a pop sound, which confirms hydrogen.

   ½ mark

Answer: (i) Sodium, potassium; violent reaction, much heat evolved, gas catches fire; (ii) gas burns with a pop sound (hydrogen)

1. (a) The plant has no nervous tissue. The information that touch has occurred is passed from cell to cell by electrical-chemical means.

   1 mark

2. (b) Plant cells change shape by changing the amount of water in them. They swell or shrink as water moves in or out, and this makes the leaves fold up and droop.

   1 mark

3. (c) Movement in touch-me-not is growth-independent, does not depend on the direction of the stimulus, and is a nastic movement. Movement of pea tendrils is growth-dependent and directional: the tendril grows towards and coils around the support. It is a tropic movement (thigmotropism).

   1 mark

Answer: (a) Electrical-chemical signals between cells; (b) change in water content (swelling/shrinking) of cells; (c) touch-me-not: growth-independent, non-directional (nastic); tendril: growth-dependent, directional (tropic)

1. (a) Chromosomes are thread-like structures in the nucleus. They are made of DNA, which carries genes, the information for inheritance of features from parents to the next generation.

   1 mark

2. (b) Each body cell has two copies of each chromosome, one from each parent. During gamete (germ cell) formation, the chromosome number is halved, so each gamete gets only one set.

   1 mark

3. (b) When the male and female gametes fuse during fertilisation, the zygote formed gets one set from each. This restores the original number of chromosomes, so the DNA content of the species stays the same from generation to generation.

   1 mark

Answer: (a) DNA-containing structures in the nucleus carrying genes; (b) gametes have half the chromosomes, and fusion restores the full number in the zygote, keeping DNA content constant

1. (i) Object at infinity – ray diagram: Draw a convex mirror with its pole P, and mark F and C behind it on the principal axis. Draw two rays parallel to the principal axis striking the mirror. After reflection they diverge. Extend them backwards with dotted lines; they appear to meet at F behind the mirror. Mark arrows on all rays.

   1 mark

2. (i) Image: at the focus F, behind the mirror; highly diminished (point-sized); virtual and erect.

   ½ mark

3. (ii) Object between infinity and P – ray diagram: Place object AB in front of the mirror. From A, draw (1) a ray parallel to the axis, which reflects as if coming from F (dotted backwards), and (2) a ray directed towards C, which reflects back along the same path. The dotted extensions meet behind the mirror between P and F at A'. Draw the image A'B' (small, upright). Mark arrows on all rays.

   1 mark

4. (ii) Image: between P and F, behind the mirror; diminished; virtual and erect.

   ½ mark

Answer: (i) At F behind mirror, highly diminished, virtual, erect; (ii) between P and F behind mirror, diminished, virtual, erect

1. (a) Top parallel group: $\frac{1}{R_1}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{1}{6}$, so $R_1=6\Omega$. Bottom group: $\frac{1}{R_2}=\frac{1}{60}+\frac{1}{40}=\frac{2+3}{120}=\frac{1}{24}$, so $R_2=24\Omega$.

   ½ mark

2. (a) $R_1$ and $R_2$ are in series: $R_{total}=6+24=30\Omega$.

   ½ mark

3. (b) By Ohm's law, $I=\frac{V}{R}=\frac{15\text{ V}}{30\Omega }=0.5\text{ A}$.

   1 mark

4. (c) The whole current flows through $R_1$, so the p.d. across the $10\Omega \parallel 15\Omega$ combination is $V_1=I{R}_1=0.5\text{ A}\times 6\Omega =3.0\text{ V}$.

   1 mark

Answer: (a) 30 Ω; (b) 0.5 A; (c) 3.0 V

1. (a) Resistance of a uniform cylindrical conductor: $R=\rho \frac{l}{A}$, where $\rho$ is the resistivity of the material.

   1 mark

2. (a) Rearranging: $\rho =\frac{RA}{l}$

   ½ mark

3. (a) SI unit of $\rho =\frac{\Omega \times {\text{m}}^2}{\text{m}}=\Omega \text{m}$ (ohm metre).

   ½ mark

4. (b) Alloys are used in heating devices because their resistivity is higher than that of pure metals, so they produce more heat. They also do not oxidise (burn) readily at high temperatures.

   1 mark

Answer: (a) R = ρl/A, so ρ = RA/l; SI unit Ω m; (b) high resistivity and they do not oxidise readily at high temperatures

1. (i) Isomer 1, but-1-ene: $\mathrm{C}{\mathrm{H}}_2=\mathrm{CH}-\mathrm{C}{\mathrm{H}}_2-\mathrm{C}{\mathrm{H}}_3$ (H–C=C double bond at the end of the 4-carbon chain, all H shown).

   ½ mark

2. (i) Isomer 2, but-2-ene: $\mathrm{C}{\mathrm{H}}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{C}{\mathrm{H}}_3$. (2-Methylpropene, $(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{C}=\mathrm{C}{\mathrm{H}}_2$, is also acceptable.)

   ½ mark

3. (ii) (I) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Cl}$: chloropropane (1-chloropropane).

   ½ mark

4. (ii) (II) $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$: butanone (butan-2-one). The C=O is on carbon 2 when the chain is numbered from the nearer end.

   ½ mark

5. (iii) (I) Oxidation of ethanol: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{alkaline }KMn{O}_4+\text{heat}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$. Condition: heating with an oxidising agent (alkaline $\mathrm{KMn}{\mathrm{O}}_4$ or acidified ${\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7$).

   1 mark

6. (iii) (II) Hydrogenation of propene: $\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\stackrel{\text{Ni/Pd}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$. Condition: presence of a nickel or palladium catalyst.

   1 mark

7. (iii) (III) Esterification: $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{\text{conc. }{H}_{2}S{O}_4}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$. Condition: warming in the presence of an acid catalyst (conc. sulphuric acid).

   1 mark

Answer: (i) But-1-ene CH₂=CHCH₂CH₃ and but-2-ene CH₃CH=CHCH₃; (ii) (I) chloropropane, (II) butanone (butan-2-one); (iii) ethanol → ethanoic acid (alk. KMnO₄, heat); propene + H₂ → propane (Ni/Pd); ethanoic acid + ethanol → ethyl ethanoate + water (acid catalyst)

1. (i) X is a good solvent and reacts with sodium, so it must be an alcohol. **X = Ethanol (ethyl alcohol), ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$**.

   ½ mark

2. When sodium reacts with ethanol it forms the salt **Y = Sodium ethoxide, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{ONa}$**.

   ½ mark

3. The gas given off is **Z = Hydrogen, ${\mathrm{H}}_2$**. Hydrogen is used, with a nickel catalyst, to hydrogenate vegetable oil into vegetable ghee. This confirms Z.

   ½ mark

4. Equation to justify: $2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+2\mathrm{Na}\to 2{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{ONa}+{\mathrm{H}}_2\uparrow$ (X → Y + Z).

   ½ mark

5. (ii)(I) Combustion: ethanol burns in oxygen with a clean blue flame. ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+3{\mathrm{O}}_2\to 2\mathrm{C}{\mathrm{O}}_2+3{\mathrm{H}}_2\mathrm{O}$ + heat + light.

   1 mark

6. (ii)(II) Dehydration: hot concentrated sulphuric acid at 443 K removes water from ethanol and gives ethene. ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\underset{\text{443 K}}{\overset{\text{hot conc. }{H}_{2}S{O}_4}{\to }}\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$.

   1 mark

7. (ii)(III) Oxidation: acidified potassium dichromate oxidises ethanol to ethanoic acid. $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+2[\mathrm{O}]\stackrel{\text{acidified }{K}_{2}C{r}_2{O}_7}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{H}}_2\mathrm{O}$. The orange dichromate turns green.

   1 mark

Answer: (i) X = ethanol (C₂H₅OH), Y = sodium ethoxide (C₂H₅ONa), Z = hydrogen (H₂); 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂. (ii)(I) C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O; (II) C₂H₅OH → C₂H₄ + H₂O (hot conc. H₂SO₄, 443 K); (III) C₂H₅OH + 2[O] → CH₃COOH + H₂O (acidified K₂Cr₂O₇)

1. (i)(I) Ovary: produces the female gametes (eggs/ova). One egg is released each month.

   ½ mark

2. The ovary also secretes the female hormones oestrogen and progesterone.

   ½ mark

3. (i)(II) Fallopian tube (oviduct): carries the egg from the ovary towards the uterus. It is the site where the sperm meets the egg, so fertilisation takes place here.

   1 mark

4. (i)(III) Uterus: the zygote/embryo gets implanted in its thick, blood-rich lining. The embryo grows and develops there, getting nourishment through the placenta.

   1 mark

5. (ii) Method 1 – Mechanical barrier (condom): a sheath worn over the penis during intercourse. It stops sperm from reaching the egg and also protects against sexually transmitted infections.

   1 mark

6. (ii) Method 2 – Surgical method (vasectomy): the vas deferens is cut or blocked surgically, so sperms cannot be transferred.

   1 mark

Answer: (i) Ovary – produces eggs and female hormones (oestrogen); Fallopian tube – carries egg, site of fertilisation; Uterus – site of implantation and development of embryo. (ii) Condom (mechanical barrier) and vasectomy (blocking vas deferens).

1. (i) Self-pollination: pollen grains are transferred from the anther (stamen) to the stigma of the same flower, or of another flower on the same plant.

   1 mark

2. Cross-pollination: pollen grains are transferred from the anther of one flower to the stigma of a flower on a different plant of the same species. Agents such as wind, water, insects or birds carry the pollen.

   1 mark

3. (ii) A – Stigma: the sticky top of the pistil. It receives the pollen grains and provides a suitable surface for them to germinate.

   1 mark

4. B – Pollen tube: grows from the germinated pollen grain down through the style. It carries the male germ cells to the female germ cell in the ovule inside the ovary. (If B is read as the style itself: the style connects stigma and ovary and provides the passage through which the pollen tube grows.)

   1 mark

5. C – Female germ cell (egg cell) in the ovule: it fuses with the male germ cell (fertilisation) to form the zygote, which develops into the embryo.

   1 mark

Answer: (i) Self-pollination: pollen to stigma of same flower; cross-pollination: pollen to stigma of another flower of same species. (ii) A – stigma (receives pollen, helps it germinate); B – pollen tube (carries male gametes to the ovule); C – egg/female germ cell (fuses with male gamete to form zygote).

1. (i) The power is negative, so lens X is a concave (diverging) lens.

   ½ mark

2. Use $P=\frac{1}{f(\text{in m})}$, so $-2.5=\frac{1}{f}$.

   ½ mark

3. $f=\frac{1}{-2.5}\text{m}=-0.4\text{m}=-40\text{cm}$.

   ½ mark

4. A concave lens corrects myopia (short-sightedness / near-sightedness).

   ½ mark

5. (ii) Here $m=-2$ and $u=-20$ cm. Since $m=\frac{v}{u}$, we get $v=mu=(-2)(-20)=+40$ cm. A negative magnification means the image is inverted, so the lens is convex.

6. (I) Nature: real and inverted, because $m$ is negative.

   ½ mark

7. (II) Size: magnified, twice the size of the object, since $|m|=2$.

   ½ mark

8. (III) Position: $v=+40$ cm, i.e. 40 cm on the other side of the lens. Here $\frac{1}{f}=\frac{1}{40}+\frac{1}{20}$, so $f\approx 13.3$ cm and $2F\approx 26.7$ cm. The image is therefore **beyond $2F_2$**.

   ½ mark

9. (IV) Sign of image height: negative, because the image is inverted and lies below the principal axis.

   ½ mark

10. (iii) Lens A (focal length 10 cm) shows the higher degree of convergence or divergence.

    ½ mark

11. Reason: $P=\frac{1}{f}$, so the shorter the focal length, the greater the power. $P_A=\frac{100}{10}=10$ D, while $P_B=\frac{100}{20}=5$ D in magnitude.

    ½ mark

Answer: (i) Concave lens, f = −40 cm, corrects myopia; (ii) (I) real and inverted, (II) magnified – twice the object size, (III) 40 cm on the other side of the lens (beyond 2F), (IV) negative; (iii) Lens A (10 cm), because smaller focal length means greater power.

1. (i) Ray diagram (describe and draw): Draw a rectangular glass slab ABCD with air on both sides. Draw an incident ray EO striking face AB obliquely at O, and the normal NN′ at O. In glass the ray bends towards the normal along OO′, so angle $r$ is less than angle $i$.

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2. At face CD, draw the normal MM′ at O′. The ray goes from glass to air and bends away from the normal along O′H. The emergent ray O′H is parallel to the incident ray EO but laterally displaced; show the extended incident ray as a dotted line. Mark $i$, $r$, $e$ (with $e=i$) and put arrows on all rays.

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3. (ii) Snell's law: for light of a given colour and a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant: $\frac{\sin i}{\sin r}=\text{constant}=n_{21}$, the refractive index of the second medium with respect to the first.

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4. (iii)(I) Object distance: a convex lens forms a virtual image only when the object is placed between the optical centre O and the focus F. A concave lens forms a virtual image for the object placed anywhere in front of it.

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5. (iii)(II) Magnification: the virtual image formed by a convex lens is enlarged ($m>1$). The virtual image formed by a concave lens is diminished ($m<1$).

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Answer: (i) Ray bends towards normal entering glass, away from normal on leaving; emergent ray parallel to incident ray (laterally displaced). (ii) sin i / sin r = constant for a given colour and pair of media. (iii) Convex: object between O and F, image magnified; Concave: object anywhere, image diminished.

1. (a) Electrolysis of brine (chlor-alkali process): $2\mathrm{NaCl}(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\stackrel{\text{electricity}}{\to }2\mathrm{NaOH}(\mathrm{aq})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})$. Chlorine forms at the anode, hydrogen at the cathode, and NaOH near the cathode.

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2. (b) Two uses of sodium hydroxide: (1) making soaps and detergents; (2) paper making / artificial fibres / degreasing metals. (Hydrogen: fuel, making margarine or ammonia. Chlorine: water treatment, PVC, bleaching powder.)

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3. (c)(i) A mild, non-corrosive base used for faster cooking is **A = sodium hydrogencarbonate (baking soda), $\mathrm{NaHC}{\mathrm{O}}_3$**.

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4. On strong heating it gives **B = sodium carbonate, $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3$**. Sodium carbonate (washing soda after recrystallisation) removes permanent hardness of water.

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5. Equation: $2\mathrm{NaHC}{\mathrm{O}}_3\stackrel{\text{heat}}{\to }\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2$.

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6. (c)(ii) Water of crystallisation is the fixed number of water molecules chemically present in one formula unit of a salt in its crystals.

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7. Examples: copper sulphate pentahydrate (blue vitriol), $\mathrm{CuS}{\mathrm{O}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$; washing soda, $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3\cdot 10{\mathrm{H}}_2\mathrm{O}$. (Gypsum $\mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}$ is also correct.)

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Answer: (a) 2NaCl + 2H₂O → 2NaOH + Cl₂ + H₂; (b) NaOH: soaps/detergents, paper making; (c)(i) A = NaHCO₃, B = Na₂CO₃; 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂; (c)(ii) fixed number of water molecules in one formula unit of a salt, e.g. CuSO₄·5H₂O, Na₂CO₃·10H₂O

1. (a) The process is photosynthesis.

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2. Definition: green plants take in carbon dioxide and water and, using sunlight absorbed by chlorophyll, convert them into carbohydrates (food), releasing oxygen.

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3. (b) $6\mathrm{C}{\mathrm{O}}_2+12{\mathrm{H}}_2\mathrm{O}\underset{\text{sunlight}}{\overset{\text{chlorophyll}}{\to }}{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+6{\mathrm{O}}_2+6{\mathrm{H}}_2\mathrm{O}$

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4. (c)(i) Step 1 (night): desert plants open their stomata only at night, to reduce water loss. They take in $\mathrm{C}{\mathrm{O}}_2$ and store it as an intermediate (an acid).

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5. Step 2 (day, stomata closed): chlorophyll absorbs light energy.

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6. Step 3: the light energy is converted into chemical energy, and water molecules are split into hydrogen and oxygen.

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7. Step 4: the $\mathrm{C}{\mathrm{O}}_2$ released from the stored intermediate is reduced to carbohydrates using this energy.

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8. (c)(ii)(I) Cloudy weather: the rate of food preparation decreases. Less sunlight reaches the leaves, so chlorophyll absorbs less light energy for photosynthesis.

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9. (c)(ii)(II) Stomata blocked by dust: the rate decreases. Gaseous exchange is reduced, so less $\mathrm{C}{\mathrm{O}}_2$, a raw material, enters the leaf.

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Answer: (a) Photosynthesis – making carbohydrates from CO₂ and water using sunlight and chlorophyll; (b) 6CO₂ + 12H₂O → C₆H₁₂O₆ + 6O₂ + 6H₂O; (c)(i) CO₂ taken in at night and stored as intermediate → light absorbed by chlorophyll in day → light energy converted to chemical energy and water split → CO₂ reduced to carbohydrate; (c)(ii) rate decreases in both: (I) less light, (II) less CO₂ intake due to reduced gaseous exchange

1. (a) Live wire: red insulation (brown in the newer convention).

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2. Neutral wire: black insulation (light blue in the newer convention).

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3. (b) $P=1\text{kW}=1000$ W and $V=220$ V. Using $P=VI$: $I=\frac{P}{V}=\frac{1000}{220}\approx 4.54$ A.

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4. The circuit must carry at least 4.54 A, so its current rating should be 5 A.

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5. (c)(i) Function: the earth wire (green insulation) joins the metal body of the appliance to a metal plate buried deep in the earth. It gives any leakage current a low-resistance path to the ground, which keeps the body at the earth's potential.

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6. Advantage: if the live wire touches the metal body of the electric iron, the current flows to earth. The user does not get a severe electric shock.

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7. (c)(ii) Precaution 1 – Use a fuse (or MCB) of proper rating in the circuit.

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8. Precaution 2 – Earth all metal-bodied appliances using the earth wire.

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9. How the fuse helps: during overloading or a short circuit the fuse wire melts and breaks the circuit. This prevents damage to the circuit and appliances.

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10. How earthing helps: leakage current flows safely to earth, preventing electric shock.

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Answer: (a) Live – red, Neutral – black; (b) I ≈ 4.54 A, so rating 5 A; (c)(i) earth wire gives low-resistance path for leakage current to earth and keeps body at earth potential, so user does not get a shock; (c)(ii) fuse (prevents damage from overloading/short circuit) and earthing (prevents shock from leakage current)