CBSE Class 10 Science Board Paper 2024 (31/1/1): Solutions

Granulated zinc is warmed with about 2 mL of NaOH solution. Which balanced equation describes the reaction?

1. Zinc is one of the metals that react with a strong base such as sodium hydroxide; the products are a salt (sodium zincate) and hydrogen gas.
2. Balanced equation: 2NaOH + Zn → Na₂ZnO₂ + H₂, which is option (b).

Answer: Zinc reacts with sodium hydroxide to give sodium zincate and hydrogen: 2NaOH + Zn → Na₂ZnO₂ + H₂ — option (b).

Which decomposition uses light as its energy source: FeSO₄ heating, electrolysis of water, AgBr breakdown, or CaCO₃ heating?

1. Photolytic decomposition is the process where a compound breaks down in the presence of light.
2. Silver bromide (AgBr) decomposes into silver and bromine when exposed to sunlight, as shown by the equation 2AgBr → 2Ag + Br₂.

Answer: Silver bromide breaks down in sunlight: 2AgBr → 2Ag + Br₂ — option (c).

Name, in order, a metal and a non-metal that are liquids at room temperature.

1. Mercury is the only metal that exists in a liquid state at room temperature.
2. Bromine is the only non-metal that exists in a liquid state at room temperature.
3. Therefore, the correct pair in order is Mercury and Bromine, which corresponds to option (c).

Answer: Mercury (metal) and bromine (non-metal) — option (c).

Carbon compounds: (i) conduct electricity well, (ii) conduct electricity poorly, (iii) have strong intermolecular forces, (iv) have weak intermolecular forces. Which pair is correct?

1. Carbon compounds are covalent, meaning electrons are shared rather than free to move, so they conduct electricity poorly (ii).
2. Covalent molecules typically have weak intermolecular forces, so (iv) is also correct.
3. Thus, the correct pair is (ii) and (iv), which is option (c).

Answer: Carbon compounds are covalent: poor conductors with weak forces between molecules, so (ii) and (iv) — option (c).

Among FeSO₄, CuSO₄, CaSO₄ and Na₂CO₃, which crystal holds the most water of crystallisation per formula unit?

1. Write the usual crystal forms: FeSO₄·7H₂O (green vitriol), CuSO₄·5H₂O (blue vitriol), CaSO₄·2H₂O (gypsum) and Na₂CO₃·10H₂O (washing soda).
2. Washing soda has 10 water molecules per formula unit, more than 7, 5 or 2, so the answer is Na₂CO₃ — option (d).

Answer: Washing soda Na₂CO₃·10H₂O has 10 (vs 7 in FeSO₄·7H₂O, 5 in CuSO₄·5H₂O, 2 in CaSO₄·2H₂O), so Na₂CO₃ — option (d).

What type of oxides are Al₂O₃ and ZnO?

1. Al₂O₃ and ZnO react with both acids and bases to form salts and water.
2. Since they show properties of both acidic and basic oxides, they are classified as amphoteric oxides, which corresponds to option (c).

Answer: They react with both acids and bases, so they are amphoteric — option (c).

Why is MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂ a redox reaction?

1. MnO₂ loses oxygen to form MnCl₂, which is reduction.
2. HCl loses hydrogen to form Cl₂, which is oxidation.
3. Since both oxidation and reduction occur, it is a redox reaction (option d).

Answer: MnO₂ loses oxygen (reduced) while HCl loses hydrogen to give Cl₂ (oxidised) — option (d).

Pick the true statements: a baby's sex is decided by (i) what comes from the mother, (ii) what comes from the father; (iii) a boy is more likely than a girl; (iv) sex is fixed at fertilisation when the zygote forms.

1. The father's sperm determines the baby's sex because it carries either an X or a Y chromosome, while the mother's egg always carries an X chromosome.
2. Sex is fixed at the moment of fertilisation when the sperm fuses with the egg to form the zygote, making statements (ii) and (iv) correct.

Answer: The father's sperm carries X or Y and sex is fixed at fertilisation, so (ii) and (iv) — option (b).

At which site of a nerve cell is the electrical impulse changed into a chemical signal: axon, dendrites, neuromuscular junction, or cell body?

1. The electrical impulse travels along the axon to the end of the nerve cell.
2. At the neuromuscular junction, the electrical signal triggers the release of chemical neurotransmitters, converting the electrical impulse into a chemical signal.

Answer: Neuromuscular junction — option (c).

A stoma closes because: (i) CO₂ is needed, (ii) CO₂ is not needed, (iii) guard cells lose water, (iv) guard cells take in water. Which reasons are right?

1. The stoma closes when the leaf does not need CO₂ for photosynthesis; then water leaves the guard cells, they shrink and the pore shuts.
2. So reasons (ii) and (iii) are right — option (c).

Answer: Stomata close when CO₂ is not needed and the guard cells lose water and shrink, so (ii) and (iii) — option (c).

Where must an object be kept in front of a convex lens so that a real image of equal size forms on a screen?

1. For a convex lens, a real image of the same size as the object is formed when the object is placed at twice the focal length (2F).
2. Therefore, the object must be kept at 2F, which corresponds to option (c).

Answer: At 2F, i.e. twice the focal length from the lens — option (c).

Name the light-sensitive screen of the eye on which the eye lens forms the image.

1. The light-sensitive screen of the eye where the image is formed is called the retina.
2. Therefore, the correct option is (d).

Answer: Retina — option (d).

Choose the diagram that shows the field pattern inside a solenoid carrying current.

Uses a figure printed in CBSE's paper — what the figure shows: Four sketches: (a) parallel equally spaced straight lines with arrows in one direction; (b) concentric circles round a point; (c) straight parallel lines in the middle with curved loops above and below; (d) curved lines bending around two current symbols.

1. Recall that the magnetic field inside a long solenoid is uniform, meaning it has the same strength and direction at every point.
2. A uniform magnetic field is represented by straight, parallel, and equally spaced lines, which matches the pattern in option (a).

Answer: Inside a solenoid the field is uniform, shown by straight, parallel, evenly spaced lines all pointing one way — option (a).

In which listed food chain is the second trophic level (herbivore) absent: grass-goat-lion; zooplankton-phytoplankton-small fish-large fish; tiger-grass-snake-frog; grasshopper-grass-snake-frog-eagle?

1. In a correct food chain the producer (a plant) comes first and a herbivore comes second.
2. Grass-goat-lion has the herbivore goat; the plankton chain has zooplankton feeding on phytoplankton; grasshopper eats grass in the last chain.
3. In tiger-grass-snake-frog a carnivore comes first and the grass is followed by a snake, so no herbivore occupies the second level — option (c).

Answer: Tiger, grass, snake, frog has no herbivore (primary consumer) — option (c).

Which organism reproduces asexually by multiple fission: yeast, Leishmania, Paramoecium or Plasmodium?

1. Identify that multiple fission is a form of asexual reproduction where a single parent cell divides into many daughter cells simultaneously.
2. Recall that Plasmodium, the malaria parasite, reproduces asexually by multiple fission, whereas yeast reproduces by budding and Leishmania and Paramoecium reproduce by binary fission.

Answer: Plasmodium (the malaria parasite) — option (d).

Assertion: zinc and nitric acid do not release hydrogen. Reason: HNO₃ is an oxidising agent that turns the H₂ into water and is itself reduced.

1. Assertion is true: when zinc reacts with nitric acid, hydrogen gas is not given off.
2. Reason is true: nitric acid is a strong oxidising agent; it oxidises the H₂ formed to water and is itself reduced to oxides of nitrogen.
3. The reason explains the assertion, so the answer is option (a).

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

Assertion: harmful chemicals collect most in the top trophic level. Reason: such chemicals are sprayed on crops against pests and diseases.

1. Both the assertion and the reason are factually true statements.
2. The reason correctly identifies the source of the chemicals but does not explain the mechanism of accumulation at the top trophic level.
3. The build-up is due to biomagnification, where non-biodegradable chemicals increase in concentration as they move up the food chain.
4. Therefore, the correct option is (b), as the reason is not the correct explanation for the assertion.

Answer: Both statements are true, but spraying does not explain why the build-up is greatest at the top (that is biomagnification) — option (b).

Assertion: aquatic animals breathe much faster than land animals. Reason: water holds far more dissolved oxygen than air.

1. Air contains about 21% oxygen, while water holds only a small fraction of dissolved oxygen, so the reason is false.
2. Because water has much less dissolved oxygen than air, aquatic animals must breathe faster to extract enough oxygen, making the assertion true but the reason incorrect.

Answer: The assertion is true but the reason is false: water has much less dissolved oxygen than air, which is why breathing is faster — option (c).

Assertion: a rainbow is a natural spectrum of sunlight. Reason: it forms when the sun is overhead and droplets are in the air.

1. A rainbow is indeed a natural spectrum formed by the dispersion of sunlight, so the assertion is true.
2. The reason is false because a rainbow is seen opposite the sun with the sun behind the observer, not when the sun is overhead.
3. Therefore, the assertion is true but the reason is false, which corresponds to option (c).

Answer: The assertion is true but the reason is false: a rainbow is seen opposite the sun (sun behind the observer), not with the sun overhead — option (c).

What kind of reaction occurs between calcium oxide and water? Support it with the balanced equation.

1. The reaction between calcium oxide and water is a combination reaction because two reactants combine to form a single product.
2. The balanced chemical equation is CaO(s) + H₂O(l) → Ca(OH)₂(aq) + heat, showing that quick lime forms slaked lime and releases energy.

Answer: Combination reaction (also exothermic), since two reactants give one product: CaO(s) + H₂O(l) → Ca(OH)₂(aq) + heat. Quick lime becomes slaked lime.

Give one function each in human digestion of: (i) hydrochloric acid, (ii) villi, (iii) anal sphincter, (iv) lipase.

1. (i) Hydrochloric acid makes the stomach acidic so pepsin can act and kills germs.
2. (ii) Villi increase the surface area for absorbing digested food.
3. (iii) The anal sphincter controls the exit of waste.
4. (iv) Lipase digests emulsified fats.

Answer: (i) HCl makes the stomach acidic so pepsin can act, and kills germs. (ii) Villi increase surface area for absorbing digested food. (iii) The anal sphincter controls the exit of waste. (iv) Lipase digests emulsified fats.

Compare the folding of a touch-me-not (sensitive plant) leaf with the downward growth of roots.

1. Leaf folding is a nastic movement triggered by touch, while root growth is a tropic movement triggered by gravity.
2. Leaf folding involves no growth and is non-directional, whereas root growth is a growth movement and is directional (geotropism).

Answer: Leaf folding responds to touch, involves no growth and has no direction; root movement responds to gravity, is a growth movement and is directional (geotropism). Any two differences.

Which hormone controls carbohydrate, protein and fat metabolism, and which gland makes it? Why should our salt be iodised?

1. Thyroxine is the hormone that controls carbohydrate, protein, and fat metabolism.
2. It is secreted by the thyroid gland.
3. Iodine is an essential component required for the synthesis of thyroxine.
4. A deficiency of iodine in the diet leads to the enlargement of the thyroid gland, known as goitre, so salt is iodised to prevent this.

Answer: Thyroxine, made by the thyroid gland. Iodine is needed to make thyroxine; lack of iodine in food causes goitre, so iodised salt is used.

A convex mirror has focal length 15 cm and an object is 10 cm in front of it. Locate the image.

1. Given focal length f = +15 cm and object distance u = -10 cm.
2. Using the mirror formula 1/v = 1/f - 1/u, substitute values to get 1/v = 1/15 - (-1/10).
3. Simplify the equation to 1/v = 1/15 + 1/10 = 5/30, which gives v = +6 cm.
4. The positive value of v indicates the image is formed 6 cm behind the mirror, making it virtual and erect.

Answer: u = -10 cm, f = +15 cm; 1/v = 1/f - 1/u = 1/15 + 1/10 = 5/30, so v = 6 cm (+6 cm). The image is 6 cm behind the mirror (virtual, erect).

Using three 6 Ω resistors, how do you get a total of 9 Ω? Justify.

1. Connect two 6 Ω resistors in parallel to find their equivalent resistance.
2. Calculate the parallel resistance using 1/Rp = 1/6 + 1/6, which gives 1/Rp = 2/6, so Rp = 3 Ω.
3. Connect the third 6 Ω resistor in series with this parallel combination.
4. Calculate the total resistance as R_total = 6 + 3 = 9 Ω.

Answer: Put two 6 Ω resistors in parallel (1/Rp = 1/6 + 1/6, Rp = 3 Ω) and the third 6 Ω in series with that pair: total = 6 + 3 = 9 Ω.

A 6 V battery drives current through 1 Ω and 2 Ω in series. Find the power used in the 2 Ω resistor.

Uses a figure printed in CBSE's paper — what the figure shows: Circuit: a 6 V battery and a plug key in series with a 1 Ω resistor and a 2 Ω resistor.

1. Calculate the total resistance of the series combination: R_total = 1 Ω + 2 Ω = 3 Ω.
2. Determine the current flowing through the circuit using Ohm's law: I = V / R_total = 6 V / 3 Ω = 2 A.
3. Calculate the power dissipated in the 2 Ω resistor using the formula P = I²R: P = (2 A)² × 2 Ω = 4 × 2 = 8 W.

Answer: Total R = 1 + 2 = 3 Ω, I = 6/3 = 2 A, P = I²R = 2² × 2 = 8 W.

(i) Why can two magnetic field lines never cross? (ii) How is a uniform field shown? Sketch it.

1. Two magnetic field lines cannot cross because at the point of intersection, the magnetic field would have two different directions.
2. A compass needle placed at that point would try to align with both directions simultaneously, which is physically impossible.
3. A uniform magnetic field is represented by drawing parallel straight lines that are equally spaced apart.
4. Arrows on these lines must all point in the same direction to indicate the uniformity of the field.

Answer: (i) At a crossing point the field would have two directions and a compass would point two ways at once, which is impossible. (ii) By parallel straight lines equally spaced, all with arrows in the same direction.

Give one equation each for a reaction showing (i) colour change, (ii) temperature change, (iii) a precipitate; state the colour change, rise/fall of temperature, or the precipitate.

1. For the colour change reaction, write the equation Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s) and state that the blue solution turns green.
2. For the temperature change reaction, write the equation NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l) + heat and state that the temperature rises.
3. For the precipitate reaction, write the equation Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq) and state that a yellow precipitate of PbI₂ forms.

Answer: (i) Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s): blue solution turns green. (ii) NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l) + heat: temperature rises. (iii) Pb(NO₃)₂(aq) + 2KI(aq) → PbI₂(s) + 2KNO₃(aq): yellow precipitate of PbI₂. Other correct examples are accepted.

(i) Tomato juice has pH 4.6; how will it taste and why? (ii) Tell a strong acid from a weak base by the ions they form in water. (iii) How does acid rain harm aquatic animals?

1. (i) pH 4.6 is less than 7, so tomato juice is acidic and tastes slightly sour.
2. (ii) A strong acid gives a large number of H⁺ (H₃O⁺) ions in water, while a weak base gives only a few OH⁻ ions.
3. (iii) Aquatic animals survive only in a narrow pH range of about 7.0 to 7.8.
4. Acid rain (pH below 5.6) lowers the pH of river water below this range, making survival difficult.

Answer: (i) Slightly sour, because pH 4.6 (below 7) means it is acidic. (ii) A strong acid gives many H⁺ (H₃O⁺) ions; a weak base gives few OH⁻ ions. (iii) Living things survive only in a narrow pH range (about 7.0 to 7.8); acid rain (pH below 5.6) lowers river water pH, making survival hard.

(i) Why do large multicellular animals need a respiratory pigment? (ii) Give reasons: (a) cartilage rings in the throat, (b) lungs keep some residual air, (c) diaphragm flattens and ribs rise on breathing in, (d) alveoli walls have many blood vessels.

1. (i) In a large body, diffusion alone is too slow to take oxygen to every part, so a respiratory pigment such as haemoglobin carries it.
2. (ii)(a) Rings of cartilage keep the air passage from collapsing. (b) Residual air in the lungs gives time to absorb O₂ and release CO₂.
3. (ii)(c) The diaphragm flattening and ribs rising enlarge the chest cavity, so air rushes into the lungs.
4. (ii)(d) The alveoli are where gases are exchanged between air and blood, so their walls carry a dense network of blood vessels.

Answer: (i) Diffusion alone is too slow to supply oxygen to all parts of a big body, so a pigment like haemoglobin carries it. (ii) (a) They keep the air passage from collapsing. (b) There is time to absorb O₂ and release CO₂. (c) It enlarges the chest cavity so air rushes in. (d) Gas exchange between air and blood happens in the alveoli.

What is a reflex action? Draw a flow chart of the pathway for a reflex like sneezing.

1. A reflex action is a sudden, automatic response to a stimulus that happens without conscious thinking.
2. Receptors in the lining of the nose detect the irritating stimulus (dust, pepper) and start a nerve impulse.
3. A sensory neuron carries the impulse to a reflex centre, where a relay connection passes it straight to a motor neuron; for sneezing this centre is in the medulla of the hind-brain, while for reflexes like pulling the hand from a hot object it is in the spinal cord.
4. The motor neuron carries the impulse to the effectors — muscles of the chest, diaphragm and throat.
5. Flow chart: stimulus → receptor (nose) → sensory neuron → relay centre → motor neuron → effector muscles → response (sneeze).

Answer: A reflex action is a sudden, automatic response to a stimulus without conscious thought. Path: stimulus → receptor (nose) → sensory neuron → spinal cord (relay neuron) → motor neuron → effector (muscles) → response (sneeze).

Note: CBSE's marking scheme draws the relay in the spinal cord, as in the hand-withdrawal reflex arc in NCERT; for sneezing the relay centre is actually the medulla, but the order receptor → sensory neuron → relay → motor neuron → effector is the same.

From the eye diagram: (i) name the defect and give the reason, (ii) give two causes, (iii) draw how it is corrected.

Uses a figure printed in CBSE's paper — what the figure shows: Diagram: point N far from the eye and N' closer; rays from N' enter the eye and meet at a point behind the retina.

1. (i) The defect is hypermetropia (far-sightedness): rays from a nearby object meet behind the retina, so near objects look blurred and the near point N is farther than the normal near point N' at 25 cm.
2. (ii) Causes: the focal length of the eye lens is too long, or the eyeball has become too small (short).
3. (iii) Correction: place a convex (converging) lens of suitable power in front of the eye.
4. The lens bends rays from an object at N' so that they appear to come from N; the eye can focus these, and the image forms on the retina.

Answer: (i) Hypermetropia (far-sightedness): light from the near object focuses behind the retina; the near point N is farther than the normal N' (25 cm). (ii) Eye lens focal length too long; eyeball too short. (iii) A convex lens of suitable power makes rays from N' appear to come from N, so the image falls on the retina.

Name and state the rule for the direction of (i) the field around a straight wire with current, (ii) the force on a straight current-carrying wire kept at right angles to a magnetic field.

1. State that the direction of the magnetic field around a straight current-carrying wire is given by the Right-hand thumb rule.
2. Explain the rule: If the right hand grips the wire with the thumb pointing in the direction of the current, the curled fingers indicate the direction of the magnetic field lines.
3. State that the direction of the force on a current-carrying conductor in a magnetic field is given by Fleming's left-hand rule.
4. Explain the rule: Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular; the forefinger points in the direction of the magnetic field, the middle finger in the direction of the current, and the thumb indicates the direction of the force or motion.

Answer: (i) Right-hand thumb rule: grip the wire in the right hand with the thumb along the current; the curled fingers give the field direction. (ii) Fleming's left-hand rule: hold thumb, forefinger and middle finger of the left hand mutually perpendicular; forefinger = field, middle finger = current, thumb = force (motion).

In Plants → Deer → Lion, what happens to levels 1 and 3 if all deer are removed? Would lions be hit the same way if they were part of a food web? Explain.

1. Deer are the herbivores that eat the plants; with the deer gone, the plants (level 1) are no longer eaten and their number increases.
2. Lions (level 3) lose their only food in this chain, so their number decreases.
3. In a food web, lions can eat other animals, so they would not be hit as badly.
4. A food web has many linked feeding paths, which makes it more stable than a single chain.

Answer: Plants (first level) will increase and lions (third level) will decrease. No — in a food web lions can find other prey, so they will not starve; webs are more stable.

Gas X is poisonous yet does a vital job high in the atmosphere. Name it and its job, the chemical blamed for its decline, and the international step taken to protect its layer.

1. Gas X is ozone (O₃); it is poisonous, but its layer high in the atmosphere does a vital job.
2. The ozone layer shields the earth from the sun's harmful ultraviolet (UV) rays.
3. Chlorofluorocarbons (CFCs), used as refrigerants and in fire extinguishers, are blamed for its decline.
4. In 1987 UNEP brought about an agreement to freeze CFC production at 1986 levels.

Answer: X is ozone (O₃); it shields the earth from the sun's UV rays. Chlorofluorocarbons (CFCs) deplete it. UNEP got an agreement (1987) to freeze CFC production at 1986 levels; CFC-free refrigerators are now made.

(i) Define a homologous series. (ii) Why do C₄H₈ boil and melt higher than C₃H₆ or C₂H₄? (iii) Why is there no gradation in chemical properties within the series? (iv) Name and draw the aldehyde and the ketone of formula C₃H₆O.

1. (i) A homologous series is a series of carbon compounds with the same functional group replacing hydrogen in the chain; members have similar chemical properties and each differs from the next by a –CH₂– unit.
2. (ii) C₂H₄, C₃H₆ and C₄H₈ belong to one series; as the molecular mass rises, melting and boiling points rise, so C₄H₈ has the highest.
3. (iii) Chemical properties are decided by the functional group, which is the same in every member, so they show no gradation.
4. (iv) Aldehyde of formula C₃H₆O: propanal, CH₃CH₂CHO.
5. Ketone of formula C₃H₆O: propanone (acetone), CH₃COCH₃.

Answer: (i) A series of carbon compounds with the same functional group replacing hydrogen in the chain (similar chemical properties, successive members differ by CH₂). (ii) Melting and boiling points rise with molecular mass. (iii) Chemical properties depend on the functional group, which is the same in all members. (iv) Aldehyde: propanal, CH₃CH₂CHO; ketone: propanone (acetone), CH₃COCH₃.

(i) Compound X has two carbons and an '-ol' name; give its name and structure. (ii) What forms when X is heated with excess concentrated H₂SO₄ at 443 K? Give the equation with conditions and the acid's role. (iii) Name and draw the electron dot structure of the hydrocarbon formed.

1. (i) X has two carbons and the -ol ending, so it is ethanol, CH₃CH₂OH (C₂H₅OH).
2. (ii) Heating ethanol at 443 K with excess concentrated H₂SO₄ removes water from it and gives ethene.
3. Equation: CH₃CH₂OH → (hot conc. H₂SO₄, 443 K) CH₂=CH₂ + H₂O; the acid acts as a dehydrating agent.
4. (iii) The hydrocarbon is ethene, C₂H₄: in its electron dot structure the two carbon atoms share two electron pairs (a double bond).
5. Each carbon also shares one electron pair with each of its two hydrogen atoms.

Answer: (i) Ethanol, CH₃CH₂OH (C₂H₅OH). (ii) Ethene forms: CH₃CH₂OH → (hot conc. H₂SO₄, 443 K) CH₂=CH₂ + H₂O; conc. H₂SO₄ acts as a dehydrating agent. (iii) Ethene, C₂H₄: the two carbons share two electron pairs (double bond) and each carbon shares one pair with each of two hydrogens.

(i) Give three methods/devices women use to avoid pregnancy and a side effect of each. (ii) In a woman, what happens if (a) fertilisation occurs, (b) the egg is not fertilised?

1. (i) Oral pills change the hormonal balance so eggs are not released; side effect: they can upset the body's hormonal balance.
2. The loop or copper-T is placed in the uterus to prevent pregnancy; side effect: it may irritate the uterus.
3. Surgical blocking of the fallopian tubes is a permanent method; side effect: if not done properly it may cause infection.
4. (ii)(a) If fertilisation occurs, the zygote implants in the thick uterine lining and keeps dividing to form the embryo.
5. (ii)(b) If the egg is not fertilised, the thick, spongy uterine lining breaks down and leaves through the vagina as blood and mucus — menstruation.

Answer: (i) Oral pills (chemical) – upset hormonal balance; copper-T/loop (barrier) – irritation of the uterus; surgical blocking of fallopian tubes – may cause infection. (ii) (a) The zygote implants in the uterine lining and keeps dividing to form the embryo. (b) The thick, spongy uterine lining breaks down and leaves through the vagina as blood and mucus (menstruation).

(i) Draw spore formation in Rhizopus, labelling its reproductive and non-reproductive parts; why does it not grow on dry bread? (ii) Name and explain how Hydra reproduces.

1. Draw a diagram of Rhizopus showing the sporangium at the tip of the sporangiophore and label it as the reproductive part.
2. Label the thread-like structures supporting the sporangium as hyphae, which are the non-reproductive parts.
3. State that Rhizopus does not grow on dry bread because the spores require moisture to germinate and absorb nutrients.
4. Identify the mode of reproduction in Hydra as budding.
5. Explain that regenerative cells divide repeatedly at one spot to form a bud, which grows into a tiny Hydra and detaches when mature.

Answer: (i) Diagram of Rhizopus: (a) reproductive part – sporangium (with spores); (b) non-reproductive part – hyphae. Dry bread lacks the moisture (and nutrients) spores need to germinate and grow. (ii) Budding: regenerative cells divide repeatedly at one spot to form a bud, which grows into a tiny Hydra and detaches when mature. (Regeneration from cut pieces is also accepted.)

(i) Define electric power and write it using V and R. (ii) A 220 V oven uses 11 units of energy in 5 h. Find (a) its power, (b) its current, (c) its resistance when red hot.

1. Electric power is defined as the rate at which electrical energy is consumed or dissipated in a circuit.
2. Using Ohm's law (V = IR) and the power formula (P = VI), substituting I gives P = V²/R.
3. Given energy E = 11 units (kWh) and time t = 5 h, calculate power as P = E/t = 11/5 = 2.2 kW.
4. Convert power to watts: P = 2.2 × 1000 = 2200 W.
5. Calculate current using I = P/V = 2200/220 = 10 A.
6. Calculate resistance using R = V²/P = 220²/2200 = 48400/2200 = 22 Ω.

Answer: (i) Power is the rate at which electrical energy is used (dissipated); P = V²/R. (ii) (a) P = 11 kWh / 5 h = 2.2 kW = 2200 W. (b) I = P/V = 2200/220 = 10 A. (c) R = V²/P = 220²/2200 = 22 Ω.

(i) Relate R to resistivity ρ for a wire of length l and cross-section A and derive the SI unit of ρ. (ii) A 3 m wire of area 4×10⁻⁷ m² has resistance 60 Ω; find ρ. (iii) How does ρ change if this wire is stretched to double length? Why?

1. Resistance is directly proportional to length l and inversely proportional to area A, so R = ρl/A.
2. Rearranging the formula gives ρ = RA/l.
3. Substituting SI units, ρ = Ω × m² / m = Ω m, so the SI unit of resistivity is ohm metre.
4. Given R = 60 Ω, A = 4×10⁻⁷ m², and l = 3 m, calculate ρ = (60 × 4×10⁻⁷) / 3.
5. Performing the division, ρ = 240×10⁻⁷ / 3 = 80×10⁻⁷ Ω m = 8×10⁻⁶ Ω m.
6. Resistivity does not change when the wire is stretched because it is an intrinsic property depending only on the material, not its dimensions.

Answer: (i) R = ρl/A, so ρ = RA/l with unit Ω·m²/m = Ω m (ohm metre). (ii) ρ = 60 × 4×10⁻⁷ / 3 = 80 × 10⁻⁷ Ω m = 8 × 10⁻⁶ Ω m. (iii) No change: resistivity depends only on the material, not on the wire's size.

Note: Stretching to double length makes the resistance 4 times, but the resistivity stays the same.

In electrolytic refining of copper, what are the cathode and the anode made of?

1. In electrolytic refining, the cathode is made of a thin strip of pure copper to collect the deposited metal.
2. The anode is made of impure copper, which dissolves into the electrolyte during the process.

Answer: Cathode: a thin strip of pure copper. Anode: impure copper.

Which electrolyte solution is used for refining copper electrolytically? Give its formula.

1. The electrolyte used for refining copper is an acidified solution of copper sulphate.
2. The chemical formula of this electrolyte is CuSO₄.

Answer: Acidified copper sulphate solution, CuSO₄.

Explain how copper is purified when current flows through the electrolytic cell.

1. At the anode, impure copper oxidizes and dissolves into the solution as copper ions: Cu → Cu²⁺ + 2e⁻.
2. At the cathode, these copper ions gain electrons and deposit as pure copper metal: Cu²⁺ + 2e⁻ → Cu.
3. Soluble impurities remain in the electrolyte, while insoluble impurities settle below the anode as anode mud.

Answer: Copper from the impure anode dissolves as Cu²⁺ (Cu → Cu²⁺ + 2e⁻) and an equal amount of pure copper deposits on the cathode (Cu²⁺ + 2e⁻ → Cu). Soluble impurities stay in solution; insoluble ones settle below the anode as anode mud.

Copper sulphate solution is in beakers A and B. A zinc strip goes in A and a silver strip in B. What is seen after about 2 hours, and why?

1. In beaker A, the zinc strip displaces copper from the solution because zinc is more reactive than copper, causing the blue colour to fade and copper to deposit on the strip.
2. In beaker B, no change is observed because silver is less reactive than copper and therefore cannot displace it from the copper sulphate solution.

Answer: Beaker A: the blue colour fades (solution becomes colourless) and copper deposits, because zinc is more reactive than copper and displaces it. Beaker B: no change, because silver is less reactive than copper.

Mendel crossed pure tall with pure short pea plants. State two observations he made on the F₁ plants.

1. All F₁ plants were tall, showing that the tall trait is dominant over the short trait.
2. None of the F₁ plants were of medium height, indicating that traits do not blend but remain distinct.

Answer: All F₁ plants were tall (no short ones), and none were of medium height — only the dominant parent's trait appeared, not a blend.

Give one way a dominant trait differs from a recessive trait.

1. A dominant trait is expressed when only one copy of the allele is present in the genotype.
2. A recessive trait is expressed only when two copies of the allele are present.

Answer: A dominant trait shows even with one copy; a recessive trait shows only when two copies are present.

For RRYY (round yellow) × rryy (wrinkled green), F₂ showed 4 kinds of seeds. How was F₂ obtained, what is the ratio of the two parental types, and what was concluded?

1. The F₂ generation was obtained by self-pollinating the F₁ hybrid plants.
2. In the F₂ ratio of 9:3:3:1, the parental types are round yellow (9) and wrinkled green (1), giving a ratio of 9 : 1.
3. The conclusion is that the two traits are inherited independently of each other.

Answer: By self-pollinating the F₁ plants. Parental types round yellow : wrinkled green = 9 : 1 (out of 9:3:3:1). Conclusion: the two traits are inherited independently.

Justify: a trait can be passed on yet stay hidden.

1. Cross pea plants with yellow seeds and plants with green seeds.
2. All F₁ plants have yellow seeds, so the green trait does not show in F₁.
3. Self-pollinate the F₁ plants: both yellow and green seeds appear in F₂.
4. So the green trait was passed on to F₁ but stayed hidden (not expressed).

Answer: Crossing yellow-seeded with green-seeded pea plants gives only yellow seeds in F₁, but selfing F₁ brings back both yellow and green in F₂. So the green trait was inherited by F₁ but not expressed.

Concave mirrors: A (f 20 cm, object at 45 cm), B (f 15 cm, object at 30 cm), C (f 30 cm, object at 20 cm). Which gives a diminished image? Justify.

1. Calculate the position of the centre of curvature (C) for each mirror using C = 2f: Mirror A has C = 40 cm, Mirror B has C = 30 cm, and Mirror C has C = 60 cm.
2. Compare the object distance (u) with C for each mirror: For Mirror A, u = 45 cm which is greater than C (40 cm); for Mirror B, u = 30 cm which equals C; for Mirror C, u = 20 cm which is less than C (60 cm).
3. Apply the rule that a concave mirror forms a diminished image only when the object is placed beyond C, which applies only to Mirror A.

Answer: Mirror A (case 1): the object at 45 cm is beyond C (2f = 40 cm), so the image is diminished (real, inverted, between F and C).

Mirror B has f = 15 cm with the object at 30 cm. Give two properties of the image.

1. The object is placed at 30 cm, which is twice the focal length (2 × 15 cm), so the object is at the centre of curvature (C).
2. When the object is at C, the image is formed at C and is real, inverted, and of the same size as the object.

Answer: The object is at C (2f), so the image is real, inverted and of the same size (formed at C). Any two.

Mirror C has f = 30 cm with the object at 20 cm. Give the image's nature and size and justify with a ray diagram.

1. The object (20 cm) is closer than the focus (30 cm), so it lies between the pole P and the focus F of the concave mirror.
2. Ray diagram: a ray parallel to the principal axis reflects through F; a ray aimed along the line from C through the object strikes the mirror normally and reflects back along itself.
3. The reflected rays diverge; extended backwards they meet behind the mirror, so the image is virtual, erect and magnified (larger than the object).

Answer: The object is between P and F, so the image is virtual, erect and magnified, formed behind the mirror. Ray diagram: a ray parallel to the axis reflects through F, a ray towards C (or through P) reflects back; extended backwards they meet behind the mirror.

An object is 18 cm from a concave mirror of focal length 12 cm. Locate the image.

1. Given: Object distance u = -18 cm and focal length f = -12 cm (using the Cartesian sign convention).
2. Using the mirror formula 1/v + 1/u = 1/f, substitute the values to get 1/v - 1/18 = -1/12.
3. Rearrange to find 1/v = -1/12 + 1/18, which simplifies to 1/v = -3/36 + 2/36 = -1/36.
4. Therefore, v = -36 cm, meaning the image is formed 36 cm in front of the mirror and is real and inverted.

Answer: u = -18 cm, f = -12 cm; 1/v = 1/f - 1/u = -1/12 + 1/18 = -1/36, so v = -36 cm. The image is 36 cm in front of the mirror (real, inverted).