CBSE Class 10 Science: 2023 board paper, set 31/6/1

1. Oxides that react with both acids and bases are called amphoteric oxides.

2. $ZnO$ and $A{l}_2{O}_3$ are amphoteric. $MgO$ and $CaO$ are basic only. Hence the answer is option (B).

   1 mark

Answer: II and III, option (B)

1. Universal indicator turning pink shows the solution is acidic, so its pH is less than 7.

2. $N{H}_{4}Cl$ is formed from $HCl$ (a strong acid) and $N{H}_{4}OH$ (a weak base). Its solution is therefore acidic. Hence the answer is option (C).

   1 mark

Answer: Option (C)

1. Zinc sulphate dissolves in water, so it is written as $(aq)$. Hydrogen is released as a gas, so it is written as $(g)$. Hence the answer is option (B).

   1 mark

Answer: X = (aq), Y = (g), option (B)

1. Phenolphthalein turns pink only in a basic solution. $N{a}_{2}C{O}_3$ is a salt of a strong base and a weak acid, so its solution is basic. This makes it X.

   ½ mark

2. $N{H}_{4}Cl$ gives an acidic solution, so phenolphthalein stays colourless. This makes it Y. Hence the answer is option (A).

   ½ mark

Answer: X = Na2CO3, Y = NH4Cl, option (A)

1. From the figure, (1) is the dark round nucleus and (2) is the green chloroplast.

2. (3) is the large clear vacuole and (4) is the guard cell. Hence the answer is option (B).

   1 mark

Answer: Option (B)

1. Precise voluntary actions, posture and balance, such as walking in a straight line or riding a bicycle, are controlled by the cerebellum.

2. The cerebellum is a part of the hind brain. Hence the answer is option (C).

   1 mark

Answer: Hind brain, Cerebellum, option (C)

1. A concave mirror forms an erect (virtual, enlarged) image only when the object is between the pole P and the focus F.

2. Here $f=10$ cm, so the object must be less than 10 cm from the mirror. Hence the answer is option (A).

   1 mark

Answer: Less than 10 cm, option (A)

1. Bronze is an alloy of copper and tin. Brass, by contrast, is copper and zinc. Hence the answer is option (D).

   1 mark

Answer: Copper and Tin, option (D)

1. The $F_1$ plants are all $Tt$. Selfing them, $Tt\times Tt$, gives $TT:Tt:tt=1:2:1$ in $F_2$.

   ½ mark

2. Pure tall ($TT$) to pure short ($tt$) is therefore $1:1$. Hence the answer is option (C).

   ½ mark

Answer: 1 : 1, option (C)

1. Primary consumers feed directly on producers. In this web, mice eat seeds and plants, and rabbits eat plants and grass.

2. So mice and rabbit are the primary consumers. Hence the answer is option (D).

   1 mark

Answer: Mice and Rabbit, option (D)

1. In Leishmania, binary fission takes place along a definite (longitudinal) plane. Arranging the drawings in sequence gives I → III → II → V → IV. Hence the answer is option (B).

   1 mark

Answer: I, III, II, V, IV, option (B)

1. In I, Mg displaces H from HCl, so it is a displacement reaction.

   ½ mark

2. In II, NaOH and HCl exchange ions to form NaCl and water (neutralisation), so it is a double displacement reaction. Hence the answer is option (B).

   ½ mark

Answer: Option (B)

1. Violet light bends the most and red light bends the least when passing through a prism.

2. P is deviated more, so it is violet; Q is deviated less, so it is red. Hence the answer is option (B).

   1 mark

Answer: Violet and Red, option (B)

1. A fruit develops from the ovary, so only a flower that has a pistil (ovary) can form a fruit.

2. Flowers X and Y bear a pistil, while Z bears only stamens. Hence the answer is option (C).

   1 mark

Answer: 'X' and 'Y' only, option (C)

1. Inside a long solenoid the field lines are parallel straight lines, which shows the field is uniform. It is therefore the same at all points. Hence the answer is option (D).

   1 mark

Answer: Same at all points, option (D)

1. Light enters the eye through the cornea, the thin transparent membrane at the front of the eyeball. Hence the answer is option (D).

   1 mark

Answer: Cornea, option (D)

1. Assertion is false: acid must be added slowly to water, not water to acid.

   ½ mark

2. Reason is true: dissolving an acid in water is highly exothermic, and adding water to acid can make the mixture splash. Hence the answer is option (D).

   ½ mark

Answer: Option (D)

1. Both statements are true. Energy does not return to autotrophs precisely because energy flows in only one direction in a food chain, so R explains A. Hence the answer is option (A).

   1 mark

Answer: Option (A)

1. Assertion is true: Amoeba engulfs food using pseudopodia, which are finger-like extensions of the cell surface.

   ½ mark

2. Reason is false: not all unicellular organisms take in food through the whole surface; Paramecium, for example, takes food in at a specific spot. Hence the answer is option (C).

   ½ mark

Answer: Option (C)

1. Ethanol is a covalent compound with weak intermolecular forces, so it has low melting and boiling points.

2. NaCl has strong electrostatic forces of attraction between its ions, giving it high melting and boiling points. Both statements are true and R explains A. Hence the answer is option (A).

   1 mark

Answer: Option (A)

1. Yes, it is a redox reaction, because oxidation and reduction take place together in it.

   1 mark

2. $\mathrm{HCl}$ loses hydrogen and forms $\mathrm{C}{\mathrm{l}}_2$, so HCl is oxidised to $\mathrm{C}{\mathrm{l}}_2$.

   ½ mark

3. $\mathrm{Mn}{\mathrm{O}}_2$ loses oxygen and forms $\mathrm{MnC}{\mathrm{l}}_2$, so MnO₂ is reduced to $\mathrm{MnC}{\mathrm{l}}_2$.

   ½ mark

Answer: Yes, it is a redox reaction: HCl is oxidised to Cl₂ and MnO₂ is reduced to MnCl₂.

1. Difference 1 – growth: The folding of the leaves of the sensitive plant (touch-me-not) is not a growth movement. It happens when cells change shape by losing or gaining water. The bending of a shoot towards light is a growth movement: the cells on the shaded side grow faster.

   1 mark

2. Difference 2 – stimulus and direction: In the sensitive plant the stimulus is touch. The movement is fast and reversible, and it does not depend on the direction of the stimulus. In the shoot the stimulus is light (phototropism). The movement is slow and irreversible, and it is directional, towards the light.

   1 mark

Answer: Sensitive plant: non-growth, fast, reversible, non-directional, stimulus touch. Shoot: growth movement, slow, irreversible, directional, stimulus light.

1. The electrical impulse reaches the axon ending of the first neuron. There it causes the release of chemicals (neurotransmitters) into the small gap called the synapse.

   1 mark

2. These chemicals cross the gap and reach the dendrite of the next neuron, where they start a new electrical impulse. In this way the impulse passes from one neuron to the next, in one direction only.

   1 mark

Answer: At the synapse the electrical impulse is converted into chemical signals. These cross the gap and set off a fresh electrical impulse in the next neuron.

1. The enzyme in saliva is salivary amylase (ptyalin).

   ½ mark

2. It is produced by the salivary glands.

   ½ mark

3. If this gland stops secreting the enzyme, starch will not be broken down into sugar (maltose) in the mouth, so the digestion of starch will not begin there.

   1 mark

Answer: Salivary amylase (ptyalin); salivary gland; starch will not be broken down into sugar in the mouth.

1. By Ohm's law, $I=\frac{V}{R}$. With $R$ constant, $I^{\prime }=\frac{V/4}{R}=\frac{I}{4}$. So the current becomes one-fourth of its original value.

   1 mark

2. Ohm's law: the potential difference across the ends of a conductor is directly proportional to the current through it, provided the temperature remains the same: $V\propto I$, or $V=IR$.

   1 mark

Answer: Current becomes one-fourth of its initial value; Ohm's law.

1. (a) Medium B is optically denser than medium A.

   ½ mark

2. Reason: on entering medium B, the ray bends towards the normal, so $\angle r<\angle i$. Light slows down in a denser medium.

   ½ mark

3. (b) $n_{BA}=\frac{\text{speed of light in A}}{\text{speed of light in B}}=\frac{v_a}{v_b}$

   1 mark

Answer: (a) Medium B (ray bends towards normal, ∠r < ∠i); (b) n_BA = v_a / v_b

1. (a) Ray diagram (in words): Draw a horizontal interface with diamond below and water above, and a dashed normal at the point of incidence. Draw the incident ray in diamond, with an arrow, meeting the interface at angle $\angle i$. Draw the refracted ray in water bending away from the normal, at angle $\angle r>\angle i$, because water is optically rarer than diamond. Label $\angle i$, $\angle r$, both media and the arrows.

   1 mark

2. (b) $n_{wd}=\frac{n_w}{n_d}=\frac{1.33}{2.42}$

   ½ mark

3. $n_{wd}\approx 0.55$

   ½ mark

Answer: (a) Ray bends away from the normal in water (∠r > ∠i); (b) refractive index of water w.r.t. diamond ≈ 0.55

1. (a) Right-hand thumb rule: Hold the current-carrying straight conductor in the right hand with the thumb pointing in the direction of the current. The curled fingers then show the direction of the magnetic field lines around the conductor.

   1 mark

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

   1 mark

Answer: (a) Right-hand thumb rule; (b) Fleming's left-hand rule

1. Oxygen-rich blood from the lungs comes through the pulmonary veins into the left atrium.

   ½ mark

2. The left atrium contracts and pushes the blood into the left ventricle.

   ½ mark

3. The left ventricle contracts and pumps the oxygenated blood through the aorta to all parts of the body.

   ½ mark

4. Deoxygenated blood from the body comes through the vena cava into the right atrium.

   ½ mark

5. The right atrium contracts and sends the blood into the right ventricle.

   ½ mark

6. The right ventricle contracts and pumps the blood through the pulmonary artery to the lungs for oxygenation. Valves prevent backflow, so the two kinds of blood never mix (double circulation).

   ½ mark

Answer: Lungs → left atrium → left ventricle → body; body → right atrium → right ventricle → lungs.

1. (i) X is calcium carbonate,

   ½ mark

2. formula $\mathrm{CaC}{\mathrm{O}}_3$ (e.g. marble or limestone).

   ½ mark

3. (ii) With dilute HCl: $\mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})+2HCl(\mathrm{aq})\to \mathrm{CaC}{\mathrm{l}}_2(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$

   1 mark

4. The $\mathrm{C}{\mathrm{O}}_2$ turns lime water milky: $\mathrm{Ca}(\mathrm{OH}{)}_2(\mathrm{aq})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})\to \mathrm{CaC}{\mathrm{O}}_3(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$. The milkiness is due to insoluble $\mathrm{CaC}{\mathrm{O}}_3$.

   1 mark

Answer: (i) Calcium carbonate, CaCO₃; (ii) CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂ and Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

1. (i) Gas G is hydrogen (${\mathrm{H}}_2$). Metal M is an amphoteric metal such as zinc.

   ½ mark

2. (ii) Test: bring a burning splinter or matchstick near the gas. It burns with a pop sound, which confirms hydrogen.

   ½ mark

3. (iii)(1) With an acid: $\mathrm{Zn}(\mathrm{s})+2HCl(\mathrm{aq})\to \mathrm{ZnC}{\mathrm{l}}_2(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$

   1 mark

4. (iii)(2) With a base: $\mathrm{Zn}(\mathrm{s})+2NaOH(\mathrm{aq})\to \mathrm{N}{\mathrm{a}}_2\mathrm{Zn}{\mathrm{O}}_2(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$ (sodium zincate).

   1 mark

Answer: (i) Hydrogen; (ii) burns with a pop sound; (iii) Zn + 2HCl → ZnCl₂ + H₂; Zn + 2NaOH → Na₂ZnO₂ + H₂

1. The gland is the adrenal gland,

   ½ mark

2. and the hormone it secretes is adrenaline.

   ½ mark

3. Response 1: the heart beats faster, so more oxygen is supplied to the muscles.

   1 mark

4. Response 2: the breathing rate increases, and blood supply to the digestive system and skin is reduced and diverted to the skeletal muscles.

   1 mark

Answer: Adrenal gland; adrenaline. Responses: faster heartbeat, faster breathing (and blood diverted to skeletal muscles).

1. (i) A – sensory neuron

   ½ mark

2. B – relay neuron (in the spinal cord)

   ½ mark

3. C – effector, i.e. the muscle of the arm

   ½ mark

4. (ii) A carries the impulse from the receptor (skin of the hand) to the spinal cord.

   ½ mark

5. C responds to the stimulus by contracting, so the hand is pulled away from the hot pan.

   ½ mark

6. (iii) Reflex arcs have evolved because the thinking process of the brain is not fast enough. A reflex arc gives a quick response without conscious thought, which protects the body from danger.

   ½ mark

Answer: (i) A sensory neuron, B relay neuron, C effector (muscle); (ii) A carries impulse from receptor to spinal cord, C responds by contracting; (iii) the brain's thinking is too slow, so reflex arcs give quick protective responses.

1. (a) Change in temperature: quicklime reacts vigorously with water and releases a lot of heat. $\mathrm{CaO}(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to \mathrm{Ca}(\mathrm{OH}{)}_2(\mathrm{aq})+\mathrm{Heat}$

   1 mark

2. (b) Evolution of a gas: zinc granules react with dilute sulphuric acid and bubbles of hydrogen are seen. $\mathrm{Zn}(\mathrm{s})+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})\to \mathrm{ZnS}{\mathrm{O}}_4(\mathrm{aq})+{\mathrm{H}}_2(\mathrm{g})$

   1 mark

3. (c) Change in colour: an iron nail kept in blue copper sulphate solution turns the solution light green, and a brown coat of copper forms on the nail. $\mathrm{Fe}(\mathrm{s})+\mathrm{CuS}{\mathrm{O}}_4(\mathrm{aq})\to \mathrm{FeS}{\mathrm{O}}_4(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})$. Another example is blue $\mathrm{CuS}{\mathrm{O}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$ turning white on heating.

   1 mark

Answer: (a) CaO + H₂O → Ca(OH)₂ + heat; (b) Zn + H₂SO₄ → ZnSO₄ + H₂; (c) Fe + CuSO₄ → FeSO₄ + Cu (blue to green)

1. Reason 1: excessive curvature of the eye lens (focal length too short).

   ½ mark

2. Reason 2: elongation of the eyeball (distance between lens and retina too large).

   ½ mark

3. (a) Diagram (in words): Draw parallel rays from a distant object entering the eye lens. They meet at a point in front of the retina, so the image of a distant object is blurred. Mark the far point and the arrows on the rays.

   1 mark

4. (b) Diagram (in words): Place a concave (diverging) lens of suitable power in front of the eye. The parallel rays are first diverged, so they appear to come from the far point. The eye lens then focuses them on the retina. Mark the arrows on the rays.

   1 mark

Answer: Causes: excessive curvature of eye lens, elongated eyeball. Image forms in front of the retina; corrected by a concave lens.

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

   1 mark

2. A solenoid behaves like a bar magnet when an electric current is passed through it. One end acts as a north pole and the other as a south pole.

   1 mark

3. Diagram (in words): Draw the coil with the current direction marked by arrows on the turns. Inside the solenoid, draw straight, parallel, equally spaced field lines, showing a uniform field. Outside, draw lines that curve from one end (N) to the other (S), like a bar magnet. Mark the arrows on the field lines.

   1 mark

Answer: A solenoid is a closely wound cylindrical coil of insulated copper wire; it acts as a magnet when current flows; field inside is uniform (parallel straight lines).

1. (a)(i) Autotrophs capture about 1% of the solar energy falling on their leaves.

   ½ mark

2. (a)(ii) About 10% of the energy is transferred from the autotrophs to the next trophic level.

   ½ mark

3. (b) Trophic levels are the various steps in a food chain at which energy is transferred, e.g. producers, herbivores, carnivores.

   1 mark

4. Only about 10% of the energy passes on at each level, so the usable energy keeps decreasing. After four or five trophic levels, too little energy is left to support another level.

   1 mark

Answer: (a) (i) 1%; (ii) 10%; (b) trophic levels are steps of energy transfer in a food chain; energy decreases at each level, so too little remains beyond 4–5 levels.

1. (i) A is ethanoic acid (acetic acid), $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$.

   ½ mark

2. Its nature is acidic.

   ½ mark

3. Its functional group is the carboxylic acid group, $-\mathrm{COOH}$.

   ½ mark

4. Reaction with a base: $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+\mathrm{NaOH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}$ (sodium ethanoate + water).

   1 mark

5. (ii)(1) B is ethanol, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$.

   ½ mark

6. C is the ester ethyl ethanoate, $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5$.

   ½ mark

7. (ii)(2) The acid (conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$) acts as a catalyst.

   ½ mark

8. (ii)(3) Esterification: $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{\mathrm{acid}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (i) A = ethanoic acid, acidic, –COOH; CH₃COOH + NaOH → CH₃COONa + H₂O. (ii) B = ethanol, C = ethyl ethanoate (ester); acid is a catalyst; CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

1. (i) The compound formed is ethene, ${\mathrm{C}}_2{\mathrm{H}}_4$: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\underset{443\mathrm{K}}{\overset{conc.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }}\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$

   ½ mark

2. Electron dot structure (in words): The two carbon atoms share two pairs of electrons between them, forming a double bond. Each carbon also shares one electron pair with each of two hydrogen atoms: ${\mathrm{H}}_2\mathrm{C}::\mathrm{C}{\mathrm{H}}_2$, with dots for the four C–H pairs.

   1 mark

3. Conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ acts as a dehydrating agent: it removes water from ethanol.

   ½ mark

4. (ii) Hydrogenation is the addition of hydrogen to unsaturated hydrocarbons, in the presence of a catalyst such as nickel or palladium, to form saturated compounds.

   1 mark

5. Equation: ${\mathrm{R}}_2\mathrm{C}=\mathrm{C}{\mathrm{R}}_2+{\mathrm{H}}_2\stackrel{\mathrm{Ni}\mathrm{catalyst}}{\to }{\mathrm{R}}_2\mathrm{CH}-\mathrm{CH}{\mathrm{R}}_2$, e.g. $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\stackrel{\mathrm{Ni}}{\to }\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_3$

   1 mark

6. Role in industry: it is used to hydrogenate vegetable oils into vegetable ghee (fats) with saturated carbon chains.

   1 mark

Answer: (i) Ethene (CH₂=CH₂); conc. H₂SO₄ is a dehydrating agent. (ii) Hydrogenation is addition of H₂ to unsaturated compounds with Ni/Pd catalyst; used to convert vegetable oils into vegetable ghee.

1. (a) DNA is the information source for making proteins. If the DNA copy changes, different proteins are made. Proteins control body structures, so this leads to an altered body design.

   1 mark

2. (b) Pollination brings pollen grains to the stigma. Without it, no male gametes reach the ovule, so fusion with the female gamete (fertilisation) cannot occur.

   1 mark

3. (c) In many multicellular organisms, the body is made of specialised cells organised into tissues and organs at definite positions. A simple broken piece cannot reorganise into a complete organism, so fragmentation or regeneration is not possible for them.

   1 mark

4. (d) Vegetative propagation is used for plants such as banana, orange, rose and jasmine, which have lost the capacity to produce seeds. It also gives flowers and fruits earlier, and plants genetically like the parent. Plants that set seed easily do not need it, so it is used only for some plants.

   1 mark

5. (e) During gamete formation the chromosome number is reduced to half. At fertilisation, the fusion of male and female gametes restores the original number in the zygote, so offspring have the same number of chromosomes as the parents.

   1 mark

Answer: (a) Changed DNA gives different proteins, so altered body design; (b) no male gametes reach ovule; (c) specialised cells organised into tissues/organs cannot regenerate from a piece; (d) used for plants that have lost the ability to produce seeds (banana, rose, etc.); (e) gametes have half the chromosomes and fusion restores the full number.

1. (i) Resistance: Resistance is the property of a conductor by which it opposes the flow of charge (electric current) through it. By Ohm's law, $R=\frac{V}{I}$.

   1 mark

2. **SI unit – ohm ($\Omega$):** A conductor has a resistance of $1\Omega$ if a current of $1$ A flows through it when a potential difference of $1$ V is applied across its ends, i.e. $1\Omega =\frac{1\text{ V}}{1\text{ A}}$.

   1 mark

3. (ii) Resistance of a conductor depends on (any two): (1) its length $l$ ($R\propto l$); (2) its area of cross-section $A$ ($R\propto \frac{1}{A}$); (3) the nature of its material; (4) its temperature.

   1 mark

4. (iii) For a wire of radius $r$: $R=\rho \frac{l}{A}=\rho \frac{l}{\pi r^2}$.

   (1) If only the length is doubled ($l\to 2l$), $R_1=\rho \frac{2l}{\pi r^2}=2R$, so the resistance doubles because $R\propto l$.

   1 mark

5. (2) If the radius is also doubled ($l\to 2l$, $r\to 2r$):
   $R^{\prime }=\rho \frac{2l}{\pi (2r{)}^2}=\rho \frac{2l}{4\pi r^2}=\frac{1}{2}\rho \frac{l}{\pi r^2}=\frac{R}{2}$.

   The resistance becomes half of its original value. Doubling the radius makes the area of cross-section four times larger, which outweighs the doubling of length.

   1 mark

Answer: (i) Opposition to the flow of current, $R=V/I$; $1\Omega$ = 1 V/1 A. (ii) Length and area of cross-section (also material, temperature). (iii) Length doubled alone gives $2R$; length and radius both doubled gives $R^{\prime }=R/2$.

1. (i) No, the bulbs in the two circuits will not glow with the same brightness.

   1 mark

2. Series circuit: The total resistance becomes $3R$, so the current falls. The source voltage is also shared, each bulb getting only $\frac{V}{3}$. Each bulb therefore consumes less power, $P=\frac{V^2}{R}$ becomes $\frac{(V/3{)}^2}{R}=\frac{P}{9}$, and the bulbs glow dimly.

   1 mark

3. Parallel circuit: Each bulb gets the full source voltage $V$ and hence its required current. Each bulb works at its rated 100 W and glows with normal (full) brightness. So the bulbs in parallel glow brighter than those in series.

   1 mark

4. (ii) Series circuit: If one bulb fuses, the only path for the current is broken. The circuit becomes open, no current flows, and none of the remaining bulbs glows.

   1 mark

5. Parallel circuit: Each bulb is on a separate branch across the source. If one bulb fuses, the other two still get the full voltage and their own current. They continue to glow with the same brightness as before.

   1 mark

Answer: (i) No – the series bulbs glow dimly (voltage shared, current reduced) and the parallel bulbs glow at full brightness (full voltage across each). (ii) Series: no bulb glows (circuit broken); parallel: the other two glow with the same brightness.

1. (a) A metal that reacts vigorously with both air and water is highly reactive (e.g. Na, K). It is obtained by electrolytic reduction, i.e. electrolysis of its molten chloride or oxide; the metal is deposited at the cathode.

   1 mark

2. (b) Aluminium is a highly reactive metal and has a greater affinity for oxygen than carbon has. Carbon therefore cannot remove oxygen from ${\text{Al}}_2{\text{O}}_3$, so it cannot reduce aluminium oxide to aluminium.

   1 mark

3. (c) [Option A] Mercury is a metal of low reactivity. Cinnabar (HgS) is heated strongly in air, which converts it into mercuric oxide:
   $2\text{HgS}(s)+3{\text{O}}_2(g)\stackrel{\text{heat}}{\to }2\text{HgO}(s)+2{\text{SO}}_2(g)$

   1 mark

4. On further heating, mercuric oxide is reduced to mercury:
   $2\text{HgO}(s)\stackrel{\text{heat}}{\to }2\text{Hg}(l)+{\text{O}}_2(g)$

   1 mark

5. (c) [Option B] Roasting: The sulphide ore is heated strongly in the presence of excess air, which converts it into the metal oxide.
   $2\text{ZnS}(s)+3{\text{O}}_2(g)\stackrel{\text{heat}}{\to }2\text{ZnO}(s)+2{\text{SO}}_2(g)$

6. Calcination: The carbonate ore is heated strongly in a limited supply of air, which converts it into the metal oxide and releases CO₂.
   ${\text{ZnCO}}_3(s)\stackrel{\text{heat}}{\to }\text{ZnO}(s)+{\text{CO}}_2(g)$

   (Option B carries the same 2 marks as Option A.)

Answer: (a) Electrolytic reduction. (b) Al has more affinity for oxygen than carbon, so carbon cannot reduce Al₂O₃. (c) A: 2HgS + 3O₂ → 2HgO + 2SO₂, then 2HgO → 2Hg + O₂ on heating. (c) B: roasting = sulphide ore heated in excess air (2ZnS + 3O₂ → 2ZnO + 2SO₂); calcination = carbonate ore heated in limited air (ZnCO₃ → ZnO + CO₂).

1. (a) A human zygote has 46 chromosomes (23 pairs). Each gamete (sperm or egg) has 23 chromosomes, which is half the number.

   1 mark

2. (b) In some reptiles sex is not decided genetically. The temperature at which the fertilised eggs are incubated decides whether the young developing in them will be male or female. For example, in turtles a high incubation temperature produces females, while in lizards it produces males.

   1 mark

3. (c) [Option A] Flow chart (in words):
   Father (XY) → two kinds of sperm: 50% carry X, 50% carry Y.
   Mother (XX) → only one kind of egg: all carry X.
   X-sperm + X-egg → XX zygote → girl.
   Y-sperm + X-egg → XY zygote → boy.

   1 mark

4. Which type of sperm fertilises the egg is purely a matter of chance, so each child has a 50 : 50 probability of being a boy or a girl. Neither parent can be held responsible for the sex of the child.

   1 mark

5. (c) [Option B] A human female has a perfect pair of sex chromosomes, XX. During gamete formation the chromosome pairs separate, so each egg receives one chromosome of each pair. Since both sex chromosomes are X, every egg gets an X chromosome.

   (Option B carries the same 2 marks as Option A.)

Answer: (a) Zygote: 46 chromosomes (23 pairs); gamete: 23. (b) The incubation temperature of the eggs decides the sex in some reptiles. (c) A: XY father gives X or Y sperm, XX mother gives only X eggs; XX gives a girl, XY gives a boy, by chance (50%). (c) B: females are XX, so every egg gets one X.

1. (a) In Case I, $f=20$ cm, so $C=2f=40$ cm. The object at 45 cm lies beyond C. The image therefore forms between F and C and is (any two): real, inverted and diminished.

   1 mark

2. (b) Case II. Here $f=15$ cm, so $2f=30$ cm. The object distance is 30 cm, which means the object is placed at the centre of curvature C.

   ½ mark

3. A concave mirror forms a real, inverted image of the same size at C when the object is placed at C.

   ½ mark

4. (c) [Option A] Dentists use concave mirrors.

   1 mark

5. When a tooth is placed very close to a concave mirror (between P and F), the mirror forms an erect and enlarged (magnified) image. This lets the dentist see the tooth clearly.

   1 mark

6. (c) [Option B] Case III (object distance 20 cm, focal length 30 cm) resembles a shaving mirror. Here the object lies between the pole P and the focus F, so the image is virtual, erect and enlarged.

7. Ray diagram (in words): Draw the concave mirror with its principal axis and mark P, F (30 cm) and C (60 cm). Place the object AB at 20 cm, between P and F. Ray 1 from A runs parallel to the axis and reflects through F. Ray 2 from A is directed towards the mirror along the line joining C to A and is reflected back along the same path. The reflected rays diverge, so extend them backwards with dotted lines behind the mirror. They meet at A′, giving an erect, enlarged, virtual image A′B′ behind the mirror. Arrows must show the direction of the rays.

   (Option B carries the same 2 marks as Option A: 1 for identifying Case III, 1 for the diagram.)

Answer: (a) Real, inverted, diminished (any two). (b) Case II – the object is at C (u = 2f = 30 cm). (c) A: concave mirror, as it gives an erect, enlarged image of a tooth held close to it. (c) B: Case III (u = 20 cm, f = 30 cm), object between P and F gives a virtual, erect, enlarged image (ray diagram).