CBSE Class 10 Science Board Paper 2025 (31/1/1): Solutions

When water is electrolysed, what is the mass ratio of hydrogen to oxygen gas set free (MH : MO)?

1. Electrolysis splits water as 2H₂O → 2H₂ + O₂, so 2 molecules of H₂ form for every 1 molecule of O₂.
2. Mass of 2H₂ = 2 × 2 = 4 g and mass of O₂ = 32 g, so MH : MO = 4 : 32 = 1 : 8 — option (D).

Answer: 2H₂O → 2H₂ + O₂ gives 4 g hydrogen for every 32 g oxygen, so MH : MO = 1 : 8 — option (D).

Note: By volume hydrogen and oxygen come off in a 2 : 1 ratio, but by mass the ratio is 1 : 8.

Which oxides form when aluminium and magnesium (in that order) burn in air?

1. Aluminium reacts with oxygen in air to form aluminium oxide, Al₂O₃.
2. Magnesium reacts with oxygen in air to form magnesium oxide, MgO.
3. Thus, the oxides formed are Al₂O₃ and MgO, which corresponds to option (B).

Answer: Aluminium forms Al₂O₃ and magnesium forms MgO — option (B).

On a pH scale, solution A is at pH 5, B at pH 7, C at pH 13 and D at pH 3. Classify A, B, C and D in that order.

Uses a figure printed in CBSE's paper — what the figure shows: A pH strip 1 to 14 is drawn with arrows: D at 3, A at 5, B at 7, C at 13.

1. pH values below 7 indicate acidic solutions, where lower values represent stronger acids, so D (pH 3) is a strong acid and A (pH 5) is a weak acid.
2. pH 7 is neutral, so B is neutral, while pH values above 7 indicate basic solutions, so C (pH 13) is a strong base.
3. Matching the classifications in the order A, B, C, D gives weak acid, neutral, strong base, and strong acid, which corresponds to option (D).

Answer: A (pH 5) weak acid, B (pH 7) neutral, C (pH 13) strong base, D (pH 3) strong acid — option (D).

What is common to these three reactions: dilute HCl with NaOH; MgO with dilute HCl; CO₂ with NaOH?

1. Identify that the reaction between dilute HCl and NaOH is a neutralization reaction producing NaCl and H₂O.
2. Note that MgO reacting with dilute HCl and CO₂ reacting with NaOH also produce a salt and water, confirming that all three reactions yield a salt and water.

Answer: Each one gives a salt and water (NaCl + H₂O, MgCl₂ + H₂O, Na₂CO₃ + H₂O) — option (A).

Element A gives electrons to element B to make compound C. Which property does C NOT have: high melting point, good solubility in water, weak attraction between its ions, or conduction when molten/dissolved?

1. Element A gives electrons to B, forming an ionic compound C with strong electrostatic forces between ions.
2. Ionic compounds have high melting points, dissolve in water, and conduct electricity when molten or dissolved, so 'weak attraction between ions' is the property C does not have.

Answer: C is ionic, so its ions are held by strong electrostatic forces; 'weak forces between ions' is the property it does not show — option (C).

Which pair of metals is got by electrolysing their molten chlorides?

1. Highly reactive metals like calcium and magnesium cannot be extracted by carbon reduction because they are more reactive than carbon.
2. Therefore, they are obtained by the electrolytic reduction of their molten chlorides, making option (B) the correct answer.

Answer: Highly reactive metals such as calcium and magnesium are obtained by electrolytic reduction of molten chlorides — option (B).

Pick the electron-dot scheme that correctly shows how magnesium oxide forms.

Uses a figure printed in CBSE's paper — what the figure shows: Four dot schemes: (A) Mg transfers two electrons to one O, giving Mg²⁺[O]²⁻; (B) one electron transferred, giving Mg⁺[O]⁻; (C) one electron to each of two O atoms, giving Mg²⁺[O⁻]₂; (D) 2Mg with one O giving [Mg²⁺]₂[O]²⁻.

1. Magnesium has 2 valence electrons and Oxygen has 6, so Mg transfers both electrons to one O atom to complete its octet.
2. This forms Mg²⁺ and O²⁻ ions, which corresponds to the electron-dot scheme in option (A).

Answer: Mg gives both its valence electrons to one O atom, forming Mg²⁺ and O²⁻ (O with a complete octet) — option (A).

If less saliva is secreted in the mouth, which conversion is affected?

1. Saliva contains the enzyme salivary amylase, which breaks down starch into simple sugars.
2. Therefore, if less saliva is secreted, the conversion of starch into simple sugars is affected, making option (C) correct.

Answer: Saliva has salivary amylase, which changes starch into simple sugar, so that step suffers — option (C).

Which plant hormone makes shoot cells on the shaded side (away from light) elongate more?

1. Auxin is the plant hormone responsible for phototropism, where shoots bend towards light.
2. Auxin accumulates on the shaded side, causing those cells to elongate more than the cells on the lit side, which is option (D).

Answer: Auxin, which diffuses to the shaded side and makes those cells grow longer — option (D).

Which are true for a bisexual flower: (i) has both stamens and pistil; (ii) has only stamens or only pistil; (iii) can be self- or cross-pollinated; (iv) cannot form fruit by itself?

1. A bisexual flower contains both male (stamens) and female (pistil) reproductive organs, so statement (i) is true and statement (ii) is false.
2. Because it has both organs, it can undergo self-pollination or cross-pollination, making statement (iii) true and statement (iv) false.

Answer: Statements (i) and (iii) are true — option (C).

Pea plants RRyy (round, green) are crossed with rrYY (wrinkled, yellow). What seeds do F1 plants show?

1. Cross the parents RRyy and rrYY, where R (round) and Y (yellow) are dominant over r (wrinkled) and y (green).
2. Each parent produces only one type of gamete: RRyy produces Ry and rrYY produces rY.
3. Fertilization gives the F1 genotype RrYy, which expresses the dominant traits, so 100% of seeds are round and yellow (option C).

Answer: All F1 are RrYy, so 100% of seeds are round and yellow (both dominant traits) — option (C).

Glucose → pyruvate + energy happens at site 'a'; pyruvate → lactic acid + energy happens at site 'b'. Name 'a' and 'b'.

1. Step 1: The breakdown of glucose into pyruvate (glycolysis) occurs in the cytoplasm, so site 'a' is the cytoplasm.
2. Step 2: The conversion of pyruvate into lactic acid (anaerobic respiration) occurs in muscle cells when oxygen is scarce, so site 'b' is muscle cells short of oxygen.
3. Step 3: Therefore, the correct option is (D).

Answer: 'a' is the cytoplasm and 'b' is muscle cells short of oxygen — option (D).

Mirror X focuses sunlight in a solar furnace; mirror Y is a vehicle's side (rear-view) mirror. Which statements are true: (i) X gives a real, small image at its focus; (ii) Y gives a virtual, small, erect image; (iii) X gives a virtual, small, erect image; (iv) Y gives a real, small image at its focus?

1. Mirror X is a concave mirror used in solar furnaces to concentrate sunlight at its focus, forming a real and diminished image of the sun.
2. Mirror Y is a convex mirror used as a rear-view mirror, which always forms a virtual, erect, and diminished image.
3. Therefore, statements (i) and (ii) are correct, corresponding to option (A).

Answer: X is concave and Y is convex, so (i) and (ii) are true — option (A).

An elderly person's ciliary muscles have weakened and the eye lens has become less flexible. Name the defect 'a' and the corrective lens 'b'.

1. The weakening of ciliary muscles and loss of lens flexibility in elderly people causes the defect known as presbyopia.
2. Presbyopia is corrected using a bifocal lens, which combines a convex lens for near vision and a concave lens for far vision.

Answer: The defect is presbyopia, corrected with a bifocal lens — option (B).

Which of these sets (members listed in any order) cannot be arranged as a chain of eater and eaten: (i) lion, grass, wolf, rabbit; (ii) man, fish, plankton, grasshopper; (iii) frog, hawk, grasshopper, snake, grass; (iv) tiger, grass, wolf, snake?

1. Set (i) links up: grass → rabbit → wolf → lion.
2. Set (iii) links up: grass → grasshopper → frog → snake → hawk.
3. Set (ii) has plankton and grasshopper as two separate starting points (one aquatic, one land), so all four cannot sit in one chain; set (iv) has no animal that eats grass and is eaten by the others in a single line (snake does not feed on grass or on wolves).
4. So sets (ii) and (iv) cannot form a single eater–eaten chain — option (D).

Answer: Groups (ii) and (iv) cannot be linked as eater and eaten in a chain — option (D).

About what percentage of sunlight falling on green leaves on land is NOT turned into food energy?

1. Green leaves on land convert only about 1% of the sunlight energy into food energy.
2. Since 1% is converted, the remaining 99% is not turned into food energy, making option (D) correct.

Answer: Leaves store only about 1% of the solar energy, so about 99% is not converted — option (D).

Assertion: most decomposition reactions absorb heat. Reason: the breakdown of organic matter into compost gives out heat.

1. The assertion is true because most decomposition reactions require energy input to break chemical bonds, making them endothermic.
2. The reason is true because the breakdown of organic material releases thermal energy during the formation of compost.
3. Since the reason describes an exothermic process, it does not explain why decomposition reactions are generally endothermic, so option (B) is correct.

Answer: Both statements are true, but the compost example does not explain why decomposition reactions are generally endothermic — option (B).

Assertion: a human baby has all the basic human features. Reason: the baby looks exactly like its parents with very little variation.

1. The assertion is true because a human baby possesses all the basic features characteristic of the human species.
2. The reason is false because a baby does not look exactly like its parents; it shows clear variations due to genetic recombination.

Answer: The assertion is true but the reason is false — a child is not an exact copy of its parents and shows clear variations — option (C).

Assertion: magnetic field lines never intersect. Reason: at a crossing point a compass needle would have to point in two directions at once.

1. Both the assertion and the reason are true statements.
2. The reason correctly explains the assertion because if lines intersected, a compass needle would point in two directions at once, which is impossible.

Answer: Both are true and the reason correctly explains the assertion — option (A).

Assertion: atmospheric ozone started falling sharply during the 1980s. Reason: ozone forms when an oxygen atom joins an O₂ molecule.

1. The assertion is true: ozone in the atmosphere began to drop sharply in the 1980s.
2. The reason is true: ozone (O₃) forms when an oxygen atom (O) joins an oxygen molecule (O₂).
3. How ozone forms does not explain why it fell (the fall was caused by CFCs), so option (B) is correct.

Answer: Both are true, but how ozone forms does not explain why it fell (that was due to CFCs) — option (B).

Zinc granules are put in dilute sulphuric acid in a corked conical flask. Give two observations showing a chemical change has happened.

Uses a figure printed in CBSE's paper — what the figure shows: The figure shows a conical flask closed with a cork and a tube, with dilute sulphuric acid and Zn granules at the bottom.

1. Observe that bubbles of a gas (hydrogen) are evolved from the zinc granules, indicating a chemical reaction.
2. Note that the conical flask becomes warm to the touch, showing that heat is evolved during the reaction.
3. Write the balanced chemical equation: Zn + H₂SO₄ → ZnSO₄ + H₂.

Answer: (1) Bubbles of a gas (hydrogen) are given off. (2) The flask becomes warm, i.e. the temperature rises. (Zn + H₂SO₄ → ZnSO₄ + H₂)

Draw labelled sketches of the stages of budding in Hydra.

1. Draw the parent Hydra with its tentacles and basal disc clearly labelled.
2. Show a small outgrowth or bud forming on the body wall and label it as the bud.
3. Draw the bud growing larger with its own tentacles and mouth developing.
4. Illustrate the mature bud detaching from the parent to form a new Hydra.

Answer: Diagram: (1) parent Hydra with tentacles; (2) a small bud (outgrowth) forms on the body wall by repeated cell division at one spot; (3) the bud grows and develops its own tentacles and mouth; (4) the mature bud detaches as a new Hydra. Label: tentacles, bud, parent body/basal disc.

Apart from cutting blood loss, why must a leak in a blood vessel be sealed? Which blood component helps, and how?

1. A leak in a blood vessel must be sealed because it would lower blood pressure and reduce the efficiency of the heart's pumping system.
2. Platelets help by moving through the blood to the site of injury, where they form a clot to plug the leak.

Answer: A leak would lower blood pressure and reduce the efficiency of the pumping system. Platelets help — they move through the blood and make it clot at the site of injury, plugging the leak.

(i) Why is transport in plants slower than in animals? (ii) What does phloem do in plants?

1. Plants are non-motile organisms with many dead tissues, resulting in lower metabolic energy requirements compared to animals.
2. Because of these low energy needs, plants can afford a slower transport system without compromising survival.
3. Phloem is responsible for the translocation of soluble products of photosynthesis, such as sugars and amino acids.
4. This transport moves nutrients from source organs like leaves to sink organs such as roots, fruits, seeds, and growing regions.

Answer: (i) Plants do not move and many of their tissues are dead cells, so their energy needs are low and a slow system is enough. (ii) Phloem carries soluble products of photosynthesis (and amino acids etc.) from leaves to other parts such as roots, fruits, seeds and growing regions — translocation.

Using the lens formula, locate the image of an object kept 60 cm from a diverging (concave) lens whose focal length is 30 cm.

1. For a diverging lens, u = −60 cm and f = −30 cm.
2. Lens formula: 1/v − 1/u = 1/f, so 1/v = 1/f + 1/u = −1/30 + (−1/60).
3. 1/v = −2/60 − 1/60 = −3/60 = −1/20, so v = −20 cm.
4. The image is 20 cm from the lens on the same side as the object, so it is virtual and erect.

Answer: u = -60 cm, f = -30 cm; 1/v = 1/f + 1/u = -1/30 - 1/60 = -3/60, so v = -20 cm. The image is 20 cm from the lens on the same side as the object (virtual, erect).

A wire with resistance R is divided into 3 equal lengths, which are then connected in parallel. Find the combined resistance.

1. Resistance is proportional to length, so each of the 3 equal pieces has resistance R/3.
2. In parallel: 1/Rp = 1/(R/3) + 1/(R/3) + 1/(R/3) = 3/R + 3/R + 3/R = 9/R.
3. So Rp = R/9.

Answer: Each piece is R/3. 1/Rp = 3/R + 3/R + 3/R = 9/R, so Rp = R/9.

Define electric power. When is the power used in a circuit equal to 1 watt?

1. Electric power is defined as the rate at which electrical energy is used up or dissipated in a circuit.
2. The power is equal to 1 watt when a current of 1 ampere flows through a conductor under a potential difference of 1 volt, meaning 1 joule of energy is consumed in 1 second.

Answer: Electric power is the rate at which electrical energy is used up (dissipated) in a circuit, P = VI. Power is 1 W when a current of 1 A flows under a potential difference of 1 V, i.e. 1 J of energy is used in 1 s.

Explain how heavy use of farm chemicals and pesticides harms the environment.

1. Heavy use of farm chemicals and pesticides causes them to wash into the soil and water bodies, directly affecting both living and non-living parts of the ecosystem.
2. Since these chemicals are mostly non-biodegradable, they enter food chains and accumulate at each trophic level through biological magnification, which harms the health of organisms at all levels, including humans.

Answer: The chemicals wash into soil and water bodies and affect living and non-living parts of the ecosystem. Being mostly non-biodegradable, they enter food chains and build up at each trophic level (biological magnification), harming the health of organisms at all levels, including humans.

(a) Show with two examples that displacement reactions help extract metals from the middle of the reactivity series. (b) Why can carbon not reduce the oxides of highly reactive metals?

1. Displacement reactions are used to extract metals from the middle of the reactivity series by using a more reactive metal to displace them from their oxides.
2. Example 1: Aluminium displaces manganese from manganese dioxide: 3MnO₂ + 4Al → 3Mn + 2Al₂O₃ + heat.
3. Example 2: Aluminium displaces iron from iron oxide in the thermit reaction: Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + heat.
4. Carbon cannot reduce the oxides of highly reactive metals because these metals (like Na, Ca, Mg, Al) have a stronger affinity for oxygen than carbon does.
5. Since carbon cannot remove oxygen from these oxides, these metals are extracted by electrolysis instead.

Answer: (a) Aluminium displaces such metals from their oxides with much heat: 3MnO₂ + 4Al → 3Mn + 2Al₂O₃ + heat; Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + heat (thermit reaction). (b) Metals high in the series (Na, Ca, Mg, Al) attract oxygen more strongly than carbon does, so carbon cannot take the oxygen away; they are got by electrolysis instead.

Describe an activity that shows what conditions are needed for iron objects to rust.

1. Take three clean, dry test tubes labeled A, B, and C, and place a clean iron nail in each tube.
2. In test tube A, add ordinary water and cork it to allow the nail to be exposed to both air and moisture.
3. In test tube B, add boiled distilled water and cover it with a layer of oil to prevent air from dissolving, then cork it to expose the nail to moisture only.
4. In test tube C, place anhydrous calcium chloride to absorb moisture from the air, then cork it to expose the nail to dry air only.
5. Observe the test tubes after a few days and note that only the nail in test tube A shows rust.
6. Conclude that iron rusts only when both air (oxygen) and water are present simultaneously.

Answer: Put clean iron nails in three test tubes A, B, C. A: ordinary water, corked. B: boiled distilled water with a layer of oil on top (no dissolved air), corked. C: anhydrous calcium chloride (absorbs moisture, so dry air), corked. After a few days only the nails in A rust. So iron rusts only when both air (oxygen) and water are present.

(i) Name two metals that react violently with cold water and give three observations when they are dropped into water. (ii) How would you test the gas given off?

1. (i) Two metals that react violently with cold water are sodium and potassium.
2. Observation 1: The reaction is very violent and rapid.
3. Observation 2: A large amount of heat is released, making the solution hot.
4. Observation 3: The hydrogen gas formed may catch fire and burn.
5. (ii) To test the gas, bring a burning splinter near the mouth of the test tube.
6. The gas burns with a distinct 'pop' sound, confirming it is hydrogen.

Answer: (i) Sodium and potassium (or lithium). Observations: the reaction is violent; much heat is released; the hydrogen formed may catch fire. (ii) Bring a burning splinter near the gas — it burns with a pop sound, showing hydrogen.

Touch-me-not leaves fold and droop when touched, though plants have no nerves or muscles. (a) How is the message passed? (b) What lets plant cells produce the visible movement? (c) How does this movement differ from a pea tendril's movement?

1. The message is passed from cell to cell by electrical-chemical means.
2. Plant cells produce visible movement by changing their shape through swelling or shrinking due to water changes.
3. Touch-me-not movement is nastic, meaning it is not growth-related and does not depend on stimulus direction.
4. Tendril movement is tropic, meaning it is growth-related and directional towards the support.

Answer: (a) The cells pass the information from cell to cell by electrical-chemical means. (b) Cells change shape by changing the amount of water in them (swelling or shrinking). (c) Touch-me-not movement is not growth-related and does not depend on the direction of the stimulus (nastic); tendril movement is growth-related and directional towards the support (tropic).

(a) What are chromosomes? (b) In sexually reproducing organisms, how does the amount of DNA stay constant from one generation to the next?

1. (a) Chromosomes are thread-like structures located in the nucleus of a cell.
2. They are made of DNA and carry genes, which hold the information for inheriting traits from parents.
3. (b) In sexually reproducing organisms, body cells contain two sets of chromosomes, one from each parent.
4. Germ cells (gametes) are formed by halving the chromosome number so they carry only one set.
5. When male and female gametes fuse during fertilization, the zygote restores the normal diploid number.
6. This ensures the amount of DNA remains constant from one generation to the next.

Answer: (a) Thread-like structures in the nucleus made of DNA that carry the genes, i.e. the information for inheriting traits from parents. (b) Body cells have two copies of each chromosome (one from each parent); germ cells get only one set (halved). When male and female gametes fuse, the zygote gets back the normal number of chromosomes, so the DNA content of the species stays the same.

Draw ray diagrams for a convex mirror with the object (i) at infinity and (ii) anywhere between infinity and the pole P; state the image's nature, position and size.

1. For case (i), draw two rays parallel to the principal axis incident on the convex mirror.
2. Draw the reflected rays diverging such that their backward extensions meet at the focus F behind the mirror.
3. State that the image is formed at F, is virtual, erect, and highly diminished (point-sized).
4. For case (ii), draw an object placed between infinity and the pole P.
5. Draw a ray parallel to the axis reflecting as if from F, and a ray aimed at the center of curvature C returning along its own path.
6. Show the backward extensions of these reflected rays meeting between P and F behind the mirror.
7. State that the image is formed between P and F, is virtual, erect, and diminished.

Answer: (i) Parallel rays reflect so that they appear to diverge from F behind the mirror: image at F, highly diminished (point-sized), virtual and erect. (ii) Ray parallel to axis reflects as if from F; ray aimed at C returns along itself: image between P and F behind the mirror, diminished, virtual and erect. Arrows must be marked on rays.

A 15 V battery, key and ammeter are in series with two groups: 10 Ω and 15 Ω in parallel, and 60 Ω and 40 Ω in parallel. Find (a) total resistance, (b) current supplied by the battery, (c) voltage across the 10 Ω–15 Ω group.

Uses a figure printed in CBSE's paper — what the figure shows: Circuit: 15 V battery, plug key K and ammeter A all in series with a parallel pair 10 Ω and 15 Ω and another parallel pair 60 Ω and 40 Ω; the two pairs are in series with each other.

1. 10 Ω and 15 Ω in parallel: (10 × 15)/(10 + 15) = 150/25 = 6 Ω.
2. 60 Ω and 40 Ω in parallel: (60 × 40)/(60 + 40) = 2400/100 = 24 Ω.
3. The two groups are in series, so total resistance = 6 + 24 = 30 Ω.
4. Current from the battery: I = V/R = 15/30 = 0.5 A.
5. Voltage across the 10 Ω–15 Ω group = I × 6 Ω = 0.5 × 6 = 3 V.

Answer: (a) 10 Ω ∥ 15 Ω = 6 Ω; 60 Ω ∥ 40 Ω = 24 Ω; in series total = 30 Ω. (b) I = 15 V / 30 Ω = 0.5 A. (c) V = 0.5 A × 6 Ω = 3 V.

(a) Relate resistance and resistivity for a cylindrical wire of length l and cross-section area A, and obtain the unit of resistivity in SI. (b) Give the reason alloys are used in heating appliances.

1. Write the formula relating resistance (R), resistivity (ρ), length (l), and cross-sectional area (A) as R = ρl/A.
2. Rearrange the formula to solve for resistivity, giving ρ = RA/l.
3. Substitute the SI units for resistance (Ω), area (m²), and length (m) into the expression to get Ω × m² / m.
4. Simplify the units to obtain the SI unit of resistivity as ohm metre (Ω m).
5. State that alloys are used in heating appliances because they have higher resistivity than pure metals, producing more heat for the same current.
6. Mention that alloys do not oxidize or burn easily at high temperatures, making them durable for heating elements.

Answer: (a) R = ρl/A, so ρ = RA/l; unit = Ω × m² / m = ohm metre (Ω m). (b) Alloys have higher resistivity than pure metals and do not oxidise (burn) easily at high temperatures.

(i) Draw two isomers of butene, C₄H₈. (ii) Name: (I) CH₃–CH₂–CH₂–Cl; (II) CH₃–CH₂–CO–CH₃. (iii) Give equations with one needed condition each for: (I) full oxidation of ethanol, (II) hydrogenation of propene, (III) ethanoic acid with ethanol.

Uses a figure printed in CBSE's paper — what the figure shows: The paper gives full structural formulas: (I) three carbons in a chain, each with H atoms, last carbon bonded to Cl; (II) four-carbon chain with C=O on the third carbon (CH₃CH₂COCH₃).

1. Draw but-1-ene as CH₂=CH–CH₂–CH₃ and but-2-ene as CH₃–CH=CH–CH₃ to show two isomers of C₄H₈.
2. Name CH₃–CH₂–CH₂–Cl as 1-chloropropane because the chlorine is on the first carbon of a three-carbon chain.
3. Name CH₃–CH₂–CO–CH₃ as butan-2-one because the ketone group is on the second carbon of a four-carbon chain.
4. Write the oxidation equation CH₃CH₂OH + 2[O] → CH₃COOH + H₂O and state the condition as alkaline KMnO₄ or acidified K₂Cr₂O₇ with heat.
5. Write the hydrogenation equation CH₃CH=CH₂ + H₂ → CH₃CH₂CH₃ and state the condition as nickel or palladium catalyst.
6. Write the esterification equation CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O and state the condition as concentrated H₂SO₄ catalyst.

Answer: (i) Any two: CH₂=CH–CH₂–CH₃ (but-1-ene), CH₃–CH=CH–CH₃ (but-2-ene), CH₂=C(CH₃)–CH₃ (2-methylpropene). (ii) (I) chloropropane (1-chloropropane); (II) butanone (butan-2-one). (iii) (I) CH₃CH₂OH → CH₃COOH using alkaline KMnO₄ or acidified K₂Cr₂O₇ with heat; (II) CH₃CH=CH₂ + H₂ → CH₃CH₂CH₃ in presence of nickel/palladium catalyst; (III) CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O in presence of an acid (conc. H₂SO₄) catalyst — esterification.

(i) Compound X is a good solvent; with sodium it gives Y and Z, and Z is used to turn vegetable oil into ghee. Identify X, Y, Z with the equation. (ii) Write equations for ethanol (I) burning in air, (II) heated to 443 K with excess conc. H₂SO₄, (III) with acidified potassium dichromate.

1. Identify Compound X as ethanol (C₂H₅OH) because it is a good solvent and reacts with sodium to release hydrogen gas.
2. Identify Y as sodium ethoxide (C₂H₅ONa) and Z as hydrogen gas (H₂), noting that Z is used in the hydrogenation of vegetable oil to make ghee.
3. Write the balanced equation for the reaction of ethanol with sodium: 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂.
4. Write the equation for the combustion of ethanol in air: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O + heat and light.
5. Write the equation for the dehydration of ethanol at 443 K with excess conc. H₂SO₄: C₂H₅OH → C₂H₄ + H₂O.
6. Write the equation for the oxidation of ethanol with acidified potassium dichromate: C₂H₅OH → CH₃COOH.

Answer: (i) X = ethanol (C₂H₅OH), Y = sodium ethoxide (C₂H₅ONa), Z = hydrogen (H₂, used in hydrogenation of oils): 2CH₃CH₂OH + 2Na → 2CH₃CH₂ONa + H₂. (ii) (I) C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O + heat and light; (II) C₂H₅OH → C₂H₄ + H₂O (dehydration to ethene, conc. H₂SO₄, 443 K); (III) C₂H₅OH → CH₃COOH (oxidation by acidified K₂Cr₂O₇).

(i) Give the function of the (I) ovary, (II) fallopian tube, (III) uterus in human females. (ii) Briefly describe two birth-control methods used by men.

1. (i) (I) The ovary produces female gametes (eggs) and secretes female hormones like oestrogen.
2. (i) (II) The fallopian tube (oviduct) is the site of fertilisation and transports the egg to the uterus.
3. (i) (III) The uterus provides the site for implantation of the embryo and its subsequent development.
4. (ii) One method is the barrier method, where a condom is used on the penis to prevent sperm from reaching the egg.
5. (ii) Another method is the surgical method, known as vasectomy, where the vas deferens is cut or blocked to stop sperm transfer.

Answer: (i) (I) Ovary makes eggs (female gametes) and female hormones such as oestrogen. (II) Fallopian tube (oviduct) is where fertilisation happens and carries the egg to the uterus. (III) Uterus is where the embryo implants and develops. (ii) Barrier method — condom on the penis stops sperm reaching the egg; surgical method — vasectomy, cutting/blocking the vas deferens so sperm cannot be transferred.

(i) How do self-pollination and cross-pollination differ? (ii) In the diagram of a pistil during fertilisation, identify A, B, C and give one function of each.

Uses a figure printed in CBSE's paper — what the figure shows: Diagram of a pistil: A points to the top sticky surface where pollen grains sit (stigma); B points to a tube growing from a pollen grain down through the style (pollen tube); C points to the female germ cell inside the ovule in the ovary (egg cell).

1. Self-pollination is the transfer of pollen grains from the anther to the stigma of the same flower.
2. Cross-pollination is the transfer of pollen grains from the anther of one flower to the stigma of another flower of the same species.
3. Part A is the stigma, which is the sticky surface that receives pollen grains and aids in their germination.
4. Part B is the pollen tube, which grows down through the style to carry male gametes to the ovule.
5. Part C is the egg cell (female gamete), which fuses with a male gamete during fertilisation to form the zygote.

Answer: (i) Self-pollination: pollen moves from the stamen to the stigma of the same flower. Cross-pollination: pollen moves from the stamen of one flower to the stigma of another flower of the same species. (ii) A = stigma — receives pollen and helps it germinate; B = pollen tube — carries the male gametes down to the ovule in the ovary; C = egg cell (female gamete) — fuses with a male gamete to form the zygote.

(i) Lens X has power -2.5 D. Name it, find its focal length in cm, and say which eye defect it corrects. (ii) A lens gives magnification m = -2 for an object 20 cm from its optical centre. Using the sign convention, give the image's (I) nature, (II) size vs object, (III) position, (IV) sign of height. (iii) Lenses A and B have focal lengths of 10 cm and 20 cm. Which converges/diverges light more, and why?

1. (i) P = −2.5 D is negative, so lens X is concave (diverging); f = 1/P = 1/(−2.5) m = −0.4 m = −40 cm.
2. A concave lens is used to correct myopia (near-sightedness).
3. (ii) m = −2 is negative, so the image is real and inverted (I), and |m| = 2 means it is magnified, twice the object's size (II).
4. With u = −20 cm, v = m × u = (−2) × (−20) = 40 cm, so the image is 40 cm on the other side of the lens (III); here 1/f = 1/40 + 1/20 = 3/40, f ≈ 13.3 cm, so 40 cm lies beyond 2F (≈ 26.7 cm).
5. The image is inverted, so its height is negative (IV).
6. (iii) Lens A (f = 10 cm) bends light more than lens B (f = 20 cm), because a shorter focal length means a greater power P = 1/f.

Answer: (i) Concave (diverging) lens; f = 1/P = 1/(-2.5) m = -0.4 m = -40 cm; prescribed for myopia. (ii) (I) real and inverted; (II) magnified, twice the object's size; (III) v = m × u = (-2)(-20) = 40 cm, i.e. 40 cm on the other side of the lens, beyond 2F; (IV) image height is negative. (iii) Lens A (10 cm), because a shorter focal length means greater power, so more convergence/divergence.

(i) Draw the path of a ray going obliquely from air into a rectangular glass slab and out again. (ii) State Snell's law. (iii) Compare the virtual images of a convex and a concave lens by (I) where the object must be and (II) magnification.

1. Draw a rectangular glass slab and an incident ray striking the first surface at an angle.
2. Show the ray bending towards the normal inside the glass and away from the normal upon exiting.
3. Indicate that the emergent ray is parallel to the incident ray but laterally displaced.
4. State Snell's law: For a given pair of media and wavelength, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.
5. Note that a convex lens forms a virtual image only when the object is between the optical center and the focus.
6. State that a concave lens forms a virtual image for any object position.
7. Conclude that the virtual image from a convex lens is magnified, while that from a concave lens is diminished.

Answer: (i) Diagram: the incident ray bends towards the normal on entering glass and away from the normal on leaving; the emergent ray is parallel to the incident ray but laterally displaced (mark arrows, normals, angles i, r, e). (ii) For light of a given colour and a given pair of media, sin i / sin r is a constant (the refractive index). (iii) (I) Convex lens gives a virtual image only when the object is between O and F; a concave lens gives one for any object position. (II) Convex lens: virtual image is magnified; concave lens: virtual image is diminished.

Case: common salt from seawater and rock salt is a raw material for many chemicals. Write the balanced equation for electrolysis of brine.

1. Write the reactants as aqueous sodium chloride and liquid water: NaCl(aq) + H₂O(l).
2. Write the products as aqueous sodium hydroxide, chlorine gas, and hydrogen gas: NaOH(aq) + Cl₂(g) + H₂(g).
3. Balance the equation by placing a coefficient of 2 before NaCl, H₂O, and NaOH to ensure equal atoms of Na, Cl, H, and O on both sides: 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g).

Answer: 2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g)

Give two uses of any one product of brine electrolysis.

1. Select one product of brine electrolysis, such as sodium hydroxide (NaOH).
2. State two uses of NaOH, for example, degreasing metals and making soaps and detergents.

Answer: Any two uses of one product, e.g. NaOH: degreasing metals, making soaps and detergents, paper, artificial fibres; Cl₂: water treatment/disinfectant, PVC, pesticides, bleach; H₂: fuel, margarine, ammonia for fertilisers.

Salt A is a mild, non-corrosive base used to make cooking faster; heating it strongly gives compound B, which is used to soften permanently hard water. Name A and B and write the equation for heating A.

1. Salt A is sodium hydrogencarbonate (baking soda, NaHCO₃), a mild non-corrosive base used for faster cooking.
2. Compound B is sodium carbonate (Na₂CO₃), used to remove permanent hardness of water.
3. On heating: 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂.

Answer: A = sodium hydrogencarbonate (baking soda, NaHCO₃); B = sodium carbonate (Na₂CO₃). 2NaHCO₃ → Na₂CO₃ + H₂O + CO₂ (on heating).

What is water of crystallisation? Name two salts that contain it.

1. Water of crystallisation is the fixed number of water molecules present in one formula unit of a salt crystal.
2. Two salts that contain water of crystallisation are copper sulphate pentahydrate (CuSO₄·5H₂O) and gypsum (CaSO₄·2H₂O).

Answer: It is the fixed number of water molecules present in one formula unit of a salt crystal. Examples: copper sulphate pentahydrate CuSO₄·5H₂O, gypsum CaSO₄·2H₂O (also washing soda Na₂CO₃·10H₂O, FeSO₄·7H₂O).

Case: living things need energy for maintenance even at rest, and it comes from food made by autotrophs. Give the name and meaning of the process green plants use to make their food.

1. Name the process as Photosynthesis.
2. Define it as the process where green plants use sunlight absorbed by chlorophyll to convert carbon dioxide and water into carbohydrates, changing light energy into chemical energy.

Answer: Photosynthesis — the process in which green plants use sunlight, absorbed by chlorophyll, to turn carbon dioxide and water into carbohydrates (light energy changed to chemical energy).

Write the chemical equation for photosynthesis.

1. Write the reactants as 6CO₂ and 12H₂O, and the products as C₆H₁₂O₆, 6O₂, and 6H₂O.
2. Add the conditions 'in sunlight' and 'with chlorophyll' above the reaction arrow to complete the equation.

Answer: 6CO₂ + 12H₂O → C₆H₁₂O₆ + 6O₂ + 6H₂O (in sunlight, with chlorophyll)

List in order the steps by which desert plants make food.

1. Desert plants open their stomata at night to take in carbon dioxide and store it as an intermediate compound.
2. During the day, chlorophyll absorbs light energy, which is converted into chemical energy while water is split.
3. The stored carbon dioxide is then reduced to carbohydrate using the chemical energy produced.

Answer: Desert plants open stomata at night and take in CO₂, storing it as an intermediate. By day: (1) chlorophyll absorbs light energy; (2) light energy is changed to chemical energy and water is split; (3) the stored CO₂ is reduced to carbohydrate using this energy.

With reasons, say how the rate of food-making by green plants changes (I) on a cloudy day and (II) when dust blocks the stomata.

1. Photosynthesis requires sunlight and carbon dioxide as essential reactants.
2. On a cloudy day, the intensity of sunlight reaching the leaves is reduced.
3. Since light is a limiting factor, the rate of food-making falls on a cloudy day.
4. When dust blocks the stomata, the entry of carbon dioxide into the leaf is restricted.
5. Reduced availability of CO₂ slows down the process, causing the rate of food-making to fall.

Answer: (I) It falls, because less sunlight is available. (II) It falls, because blocked stomata reduce gas exchange, so less CO₂ enters the leaf.

Case: homes get mains power at 220 V between live and neutral wires. What colours are the insulation of the supply wires?

1. The live wire is insulated with red colour.
2. The neutral wire is insulated with black colour.

Answer: Live wire — red; neutral wire — black.

What current rating should a 220 V circuit have to run a 1 kW electric iron?

1. Given power P = 1 kW = 1000 W and voltage V = 220 V.
2. Using the formula I = P/V, we calculate I = 1000/220 ≈ 4.55 A.
3. Since the calculated current is approximately 4.55 A, the circuit should have a current rating of 5 A.

Answer: I = P/V = 1000 W / 220 V ≈ 4.55 A, so the circuit should be rated 5 A.

Note: CBSE's marking scheme prints 4.54 A; 1000/220 = 4.545… ≈ 4.55 A. Either way a 5 A rating is right.

What does the earth wire do, and how does it help with appliances like an electric iron?

1. The earth wire provides a low-resistance path to the ground for any current that leaks to the metal body of an appliance.
2. This ensures the metal body remains at earth's potential, preventing a severe electric shock to the user.

Answer: It gives a low-resistance path so any current leaking to the metal body flows to earth, keeping the body at earth's potential. So the user does not get a severe electric shock.

Give two precautions against electrical accidents and how each prevents damage.

1. Precaution 1: Use a fuse of proper rating in the circuit.
2. The fuse melts and breaks the circuit during overloading or short-circuit, which protects the appliances and wiring from damage.
3. Precaution 2: Earth the metal bodies of electrical appliances.
4. Earthing provides a low-resistance path for leaked current to flow into the ground, preventing electric shock to the user.

Answer: (1) Use a fuse of proper rating — it melts and breaks the circuit on overloading or short-circuit, protecting the circuit and appliances. (2) Earth metal-bodied appliances — leaked current goes to earth, preventing electric shock.