CBSE Class 10 Science: 2023 board paper, set 31/5/1

1. Olfactory indicators change their smell in acidic or basic medium. Clove oil and vanilla essence both lose their odour in a basic solution, so both are olfactory indicators; onion is also olfactory, but turmeric, litmus and methyl orange change colour, not smell. Option (A).

   1 mark

Answer: (A) Clove oil and vanilla essence

1. Quicklime reacts vigorously with water to form slaked lime, releasing heat and no gas: $\mathrm{CaO}(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to \mathrm{Ca}(\mathrm{OH}{)}_2(\mathrm{aq})$ + Heat. This equation is balanced, so the answer is (C).

   1 mark

Answer: (C)

1. Na loses electrons to become $\mathrm{N}{\mathrm{a}}^{+}$ in NaOH, so Na is oxidised. It reduces water to ${\mathrm{H}}_2$, so Na is the reducing agent. Option (A).

   1 mark

Answer: (A) Na

1. A milkman adds a very small amount of baking soda ($\mathrm{NaHC}{\mathrm{O}}_3$) to fresh milk. This shifts the pH slightly to the alkaline side, so the milk takes longer to curdle. Option (D).

   1 mark

Answer: (D) Baking soda

1. Amphoteric oxides such as $\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3$ and ZnO show both acidic and basic behaviour. They react with acids as well as with bases, and in both cases form salt and water. Option (C).

   1 mark

Answer: (C)

1. When HCl gas dissolves in water it ionises, and the ${\mathrm{H}}^{+}$ ion combines with water: $\mathrm{HCl}+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_3{\mathrm{O}}^{+}+\mathrm{C}{\mathrm{l}}^{-}$. This forms hydronium ions. Option (C).

   1 mark

Answer: (C) Hydrogen chloride gas and water

1. Oxygen has 6 valence electrons, so each atom shares 2 electrons and ${\mathrm{O}}_2$ has a double bond. The overlap region must therefore show two pairs (two crosses and two dots), with each atom's complete outer shell drawn. This matches option (A).

   1 mark

Answer: (A)

1. Loss of water as vapour from the aerial parts of a plant is transpiration. Besides helping in the upward movement of water, it also cools the plant, that is, it helps in temperature regulation. Option (B).

   1 mark

Answer: (B)

1. The amount of oxygen dissolved in water is far lower than the amount of oxygen in air. Aquatic organisms therefore have to breathe faster to obtain enough oxygen. Option (B).

   1 mark

Answer: (B)

1. The trait that expresses itself in the ${\mathrm{F}}_1$ generation is the dominant trait. A trait passed on from one generation to the next is an inherited trait. Option (C).

   1 mark

Answer: (C) Dominant trait, Inherited trait

1. Gustatory receptors detect taste, and they are located on the tongue. Option (C).

   1 mark

Answer: (C) tongue

1. Gonorrhoea and syphilis are bacterial infections, while warts and HIV-AIDS are viral infections. The correct 'bacterial, viral' order is therefore syphilis and warts. Option (D).

   1 mark

Answer: (D) Syphilis and Warts

1. Current is charge per unit time, $I=Q/t$, which gives $Q=It$. Potential difference is work done per unit charge, $V=W/Q$, which gives $W=VQ$. Option (B).

   1 mark

Answer: (B)

1. Apply the right-hand thumb rule with the thumb pointing west, the direction of the current. The fingers curl so that the field is north to south at a point directly below the wire, i.e. beneath the plane of the paper. Option (B).

   1 mark

Answer: (B) directly below the wire

1. Current drawn: $I=P/V=1000/220\approx 4.55$ A.

   ½ mark

2. The fuse must be rated just above this current, so the minimum rating is 5 A. Option (D).

   ½ mark

Answer: (D) 5 A

1. The field pattern of a current-carrying solenoid is similar to that of a bar magnet. Statement (C) says the patterns are different, so it is the incorrect statement. Statements (A), (B) and (D) are all true. Option (C).

   1 mark

Answer: (C)

1. ZnO loses oxygen to become Zn, so ZnO is reduced; the Assertion is true. Carbon takes up that oxygen to form CO, so carbon is the reducing agent; the Reason is true and correctly explains the Assertion. Option (A).

   1 mark

Answer: (A)

1. Human populations do show a great deal of variation, so the Assertion is true. Variations do not all have equal chances of survival, because nature selects the variations best suited to the environment. The Reason is false. Option (C).

   1 mark

Answer: (C)

1. Ventricles have thicker walls than atria, so the Assertion is false. The Reason is true: ventricles must pump blood to various organs at high pressure. Option (D).

   1 mark

Answer: (D)

1. Field lines never intersect, because the field would otherwise have two directions at one point; the Assertion is true. The field strength does increase with current, so the Reason is true too. However, this does not explain why the lines do not intersect. Option (B).

   1 mark

Answer: (B)

1. (i) 'X' is gypsum, $\mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}$ (calcium sulphate dihydrate).

   ½ mark

2. 'Y' is Plaster of Paris, $\mathrm{CaS}{\mathrm{O}}_4\cdot \frac{1}{2}{\mathrm{H}}_2\mathrm{O}$ (calcium sulphate hemihydrate). Doctors use it to hold fractured bones in place. On heating at 373 K: $\mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}\stackrel{373\mathrm{K}}{\to }\mathrm{CaS}{\mathrm{O}}_4\cdot \frac{1}{2}{\mathrm{H}}_2\mathrm{O}+1\frac{1}{2}{\mathrm{H}}_2\mathrm{O}$

   ½ mark

3. (ii) 'X' is obtained again by mixing 'Y' with water. Plaster of Paris takes up water and sets into a hard mass of gypsum:
   $\mathrm{CaS}{\mathrm{O}}_4\cdot \frac{1}{2}{\mathrm{H}}_2\mathrm{O}+1\frac{1}{2}{\mathrm{H}}_2\mathrm{O}\to \mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (i) X = Gypsum, CaSO₄·2H₂O; Y = Plaster of Paris, CaSO₄·½H₂O; (ii) by adding water to Y: CaSO₄·½H₂O + 1½H₂O → CaSO₄·2H₂O

1. (i) Solution M has the higher hydrogen ion concentration.

   ½ mark

2. A red colour with universal indicator means a low pH, so M is acidic and contains more ${\mathrm{H}}^{+}$ ions. N gives blue, so it is basic and has fewer ${\mathrm{H}}^{+}$ ions.

   ½ mark

3. (ii) The salt formed is neutral (green with universal indicator means pH ≈ 7).

   ½ mark

4. Justification: M is an acid and N is a base. They neutralise each other to form a salt and water, so the mixture is neutral.

   ½ mark

Answer: (i) M, because it is acidic (red, low pH); (ii) the salt is neutral, formed by neutralisation of acid M with base N

1. (i) Dendrite

   ½ mark

2. Function: it picks up (acquires) information or stimuli from other neurons or receptors and passes it to the cell body.

   ½ mark

3. (ii) Axon

   ½ mark

4. Function: it is the long fibre along which the information travels as an electrical impulse, from the cell body to the axon terminals.

   ½ mark

Answer: (i) Dendrite: acquires information; (ii) Axon: carries information as an electrical impulse away from the cell body

1. (i) Chlorophyll absorbs light energy.

   ½ mark

2. (ii) The light energy is converted into chemical energy, and water molecules are split into hydrogen and oxygen.

   1 mark

3. (iii) Carbon dioxide is reduced to carbohydrates.

   ½ mark

Answer: Absorption of light by chlorophyll → conversion of light energy to chemical energy and splitting of water → reduction of CO₂ to carbohydrates

1. (1) Transpiration: the leaves lose excess water as water vapour through their stomata, which removes the extra water from the plant.

   1 mark

2. (2) Removal of gaseous wastes: oxygen produced as a by-product of photosynthesis (and $\mathrm{C}{\mathrm{O}}_2$ from respiration) diffuses out through the stomata. Also accepted: waste products stored in old leaves are removed when those leaves are shed.

   1 mark

Answer: Leaves remove excess water by transpiration and give out waste oxygen through stomata (wastes are also stored in leaves that are shed)

1. Enzyme 1: Pepsin, which digests proteins in the stomach.

   ½ mark

2. It is secreted by the gastric glands in the wall of the stomach.

   ½ mark

3. Enzyme 2: Trypsin, which digests proteins in the small intestine.

   ½ mark

4. It is secreted by the pancreas as part of pancreatic juice.

   ½ mark

Answer: Pepsin – gastric glands; Trypsin – pancreas

1. (a) The image is formed behind the retina, so the defect is hypermetropia (far-sightedness).

   ½ mark

2. (b) Cause 1: the focal length of the eye lens is too long.

   ½ mark

3. Cause 2: the eyeball has become too small (too short).

   ½ mark

4. Correction: a convex (converging) lens of suitable power, which brings the image back onto the retina.

   ½ mark

Answer: (a) Hypermetropia; (b) focal length of eye lens too long, eyeball too small; corrected with a convex lens

1. Trophic levels: Grass (1st) → Grasshopper (2nd) → Frog (3rd) → Snake (4th) → Hawk (5th). Working back from the hawk, each lower level has 10 times the energy: Hawk 50 J → Snake 500 J → Frog 5000 J → Grasshopper 50,000 J → Grass 5,00,000 J.

2. Energy at the first trophic level (grass) = 5,00,000 J

   ½ mark

3. Energy at the third trophic level (frog) = 5000 J

   ½ mark

4. Justification (10% law): only about 10% of the energy at one trophic level is passed on to the next level. The rest is used for life processes or lost as heat.

   1 mark

Answer: First trophic level: 5,00,000 J; third trophic level: 5000 J (10% law)

1. (a) A double displacement reaction is one in which the reactants exchange ions with each other to form new compounds.

   1 mark

2. (b)(i) Neutralisation: $\mathrm{HCl}(\mathrm{aq})+\mathrm{NaOH}(\mathrm{aq})\to \mathrm{NaCl}(\mathrm{aq})+{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

   ½ mark

3. Justification: an acid reacts with a base to give a salt and water. The ${\mathrm{H}}^{+}$ and $\mathrm{N}{\mathrm{a}}^{+}$ ions exchange partners, so it is also a double displacement.

   ½ mark

4. (ii) Precipitation: $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})+\mathrm{BaC}{\mathrm{l}}_2(\mathrm{aq})\to \mathrm{BaS}{\mathrm{O}}_4(\mathrm{s})+2NaCl(\mathrm{aq})$

   ½ mark

5. Justification: the ions are exchanged and an insoluble white precipitate of $\mathrm{BaS}{\mathrm{O}}_4$ is formed.

   ½ mark

Answer: (a) Reaction with exchange of ions between reactants; (b)(i) HCl + NaOH → NaCl + H₂O; (ii) Na₂SO₄ + BaCl₂ → BaSO₄↓ + 2NaCl

1. (a) Bacteria in the mouth break down sugar and food particles left after eating and produce acids. These acids lower the pH of the mouth below 5.5, which can cause tooth decay.

   1 mark

2. (b) Heating baking soda gives sodium carbonate. Recrystallising it gives 'X' = washing soda, $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3\cdot 10{\mathrm{H}}_2\mathrm{O}$ (sodium carbonate decahydrate).

   ½ mark

3. Industrial use 1: in the glass, soap and paper industries.

   ½ mark

4. Industrial use 2: in the manufacture of sodium compounds such as borax. (Also accepted: removing permanent hardness of water.)

   ½ mark

5. (c) On heating, blue copper sulphate crystals lose their water of crystallisation and become white anhydrous copper sulphate:
   $\mathrm{CuS}{\mathrm{O}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}\stackrel{\mathrm{heat}}{\to }\mathrm{CuS}{\mathrm{O}}_4+5{\mathrm{H}}_2\mathrm{O}$

   ½ mark

Answer: (a) Acids produced by bacteria degrading sugar/food in the mouth; (b) X = washing soda Na₂CO₃·10H₂O, used in glass/soap/paper industries and manufacture of borax; (c) loss of water of crystallisation

1. (a) Activity: take 1 mL of 1% starch solution in each of two test tubes, A and B.

   ½ mark

2. Add 1 mL of saliva to test tube A only. Leave both tubes undisturbed for 20–30 minutes, then add a few drops of dilute iodine solution to each.

   ½ mark

3. Observation in A: the colour of iodine does not change (no blue-black). The salivary amylase in saliva has broken the starch down into sugar.

   ½ mark

4. Observation in B: the solution turns blue-black because the starch is still present. Conclusion: saliva digests starch into sugar.

   ½ mark

5. (b) Bile is important because it makes the acidic food coming from the stomach alkaline, so that pancreatic enzymes can act on it. It also emulsifies fats, breaking large fat globules into smaller ones (either point).

   1 mark

Answer: (a) Starch + saliva gives no blue-black colour with iodine (starch converted to sugar by salivary amylase), starch alone gives blue-black; (b) bile makes the medium alkaline for pancreatic enzymes and emulsifies fats

1. (i) Given: $h=5$ cm, $u=-20$ cm, $f=-18$ cm (concave lens). Lens formula: $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

   ½ mark

2. (1) $\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{-18}+\frac{1}{-20}=-\frac{10+9}{180}=-\frac{19}{180}$, so $v=-\frac{180}{19}\approx -9.47$ cm. The image is virtual and on the same side as the object.

   ½ mark

3. (2) Magnification: $m=\frac{v}{u}$

   ½ mark

4. $m=\frac{-9.47}{-20}\approx 0.47$. This is positive and less than 1, so the image is erect and diminished (image height ≈ 2.37 cm).

   ½ mark

5. (ii) Convex lens forming a virtual image: the image is erect and magnified, so $m$ is positive and **$m>1$**.

   ½ mark

6. Concave lens: the virtual image is always erect and diminished, so $m$ is positive and **$m<1$**.

   ½ mark

Answer: (i) v ≈ −9.47 cm, m ≈ +0.47; (ii) virtual image: convex lens m > 1, concave lens m < 1 (both positive)

1. (i) Real, inverted, magnified image: the object must lie between F and 2F, i.e. 10 cm < object distance < 20 cm.

   ½ mark

2. (ii) Virtual, erect, magnified image: the object must lie between the optical centre O and F, i.e. object distance < 10 cm.

   ½ mark

3. Ray diagram (i): draw a convex lens with F₁, 2F₁ on the left and F₂, 2F₂ on the right. Place object AB between F₁ and 2F₁. A ray parallel to the axis refracts through F₂. A ray through O goes straight on. They meet beyond 2F₂, giving a real, inverted, enlarged image A′B′ below the axis. Mark arrows on all rays.

   1 mark

4. Ray diagram (ii): place object AB between O and F₁. The ray parallel to the axis refracts through F₂, and the ray through O goes straight on. These rays diverge, so extend them backwards as dotted lines. They meet on the same side as the object, giving a virtual, erect, enlarged image A′B′ behind the object. Mark arrows on all rays.

   1 mark

Answer: (i) Between 10 cm and 20 cm (between F and 2F); (ii) less than 10 cm (between O and F)

1. (a)(i) Retina: it acts as a light-sensitive screen on which the image is formed. Its light-sensitive cells send signals to the brain through the optic nerve.

   1 mark

2. (ii) Pupil: it regulates and controls the amount of light entering the eye. The iris adjusts its size.

   1 mark

3. (b) Ciliary muscles relax or contract to change the curvature (thickness) of the eye lens, and hence its focal length. This lets the eye focus both near and distant objects on the retina (power of accommodation).

   1 mark

Answer: (a) Retina: light-sensitive screen where the image forms; Pupil: controls the amount of light entering the eye; (b) ciliary muscles change the curvature and so the focal length of the eye lens

1. (i)(1) The displacement increases, because the force on the conductor increases with current.

   ½ mark

2. (2) The displacement increases, because a stronger magnetic field exerts a larger force.

   ½ mark

3. (3) The direction of displacement is reversed: the conductor now moves towards the right.

   ½ mark

4. (ii) The rule is Fleming's left-hand rule.

   ½ mark

5. Statement: stretch the thumb, forefinger and middle finger of the left hand so that they are mutually perpendicular. If the forefinger points in the direction of the magnetic field and the middle finger in the direction of the current, the thumb points in the direction of the force (motion) on the conductor.

   1 mark

Answer: (i)(1) Increases; (2) increases; (3) reverses (moves right); (ii) Fleming's left-hand rule

1. Diagram: draw a vertical straight wire passing through the centre of a horizontal cardboard, connected to a battery and key. On the cardboard, draw concentric circles centred on the wire, closer together near the wire.

   ½ mark

2. Mark the direction of current with an arrow on the wire (e.g. upward).

   ½ mark

3. Mark the direction of the field lines with arrows on the circles. For an upward current they run anticlockwise when viewed from above (reverse them if the current is downward).

   ½ mark

4. Rule: Right-hand thumb rule.

   ½ mark

5. Statement: imagine holding the current-carrying straight conductor in your right hand with the thumb pointing in the direction of current. The fingers wrapped around the conductor then give the direction of the magnetic field lines.

   1 mark

Answer: Concentric circular field lines around the wire (anticlockwise from above for upward current); Right-hand thumb rule

1. Formation: in the upper atmosphere, high-energy UV radiation splits oxygen molecules into free (nascent) oxygen atoms: ${\mathrm{O}}_2\stackrel{\mathrm{UV}}{\to }\mathrm{O}+\mathrm{O}$

   1 mark

2. These free oxygen atoms combine with oxygen molecules to form ozone: $\mathrm{O}+{\mathrm{O}}_2\to {\mathrm{O}}_3$

   1 mark

3. Justification: the ozone layer absorbs harmful UV radiation from the Sun. If it is damaged (for example by CFCs), more UV radiation reaches the Earth and causes harm such as skin cancer in human beings.

   1 mark

Answer: UV splits O₂ into O atoms, which combine with O₂ to form O₃; damage lets UV reach the Earth, causing skin cancer and other harm

1. (i) X = ethanol, ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$

   ½ mark

2. Y = ethanoic acid (acetic acid), $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$

   ½ mark

3. Z = ethyl ethanoate (ethyl acetate), $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5$, a sweet-smelling ester

   ½ mark

4. (ii)(1) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{acidified }{\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$

   1 mark

5. (2) $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\underset{\text{warm}}{\overset{\text{conc. }{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }}\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$

   1 mark

6. (iii)(1) Acidified ${\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7$ acts as an oxidising agent: it supplies oxygen to convert ethanol into ethanoic acid.

   ½ mark

7. (2) Conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ acts as a dehydrating agent (and catalyst): it removes the water formed and helps the esterification proceed.

   ½ mark

8. (iv) When the ester Z reacts with an alkali (e.g. NaOH) it gives back the alcohol and the sodium salt of the acid. This is called saponification: $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaOH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$

   ½ mark

Answer: (i) X = ethanol, Y = ethanoic acid, Z = ethyl ethanoate; (ii) equations as shown; (iii) K₂Cr₂O₇: oxidising agent; conc. H₂SO₄: dehydrating agent; (iv) saponification

1. (i) Heating ethanol at 443 K with excess conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ (dehydration):
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\underset{443\mathrm{K}}{\overset{\text{hot conc. }{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }}\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\mathrm{O}$

   1 mark

2. (ii) Adding hydrogen in the presence of a nickel or palladium catalyst (hydrogenation):
   $\mathrm{C}{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}_2\stackrel{Ni/Pd}{\to }\mathrm{C}{\mathrm{H}}_3-\mathrm{C}{\mathrm{H}}_3$

   1 mark

3. (iii) Reacting ethane with chlorine in the presence of sunlight (substitution):
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_3+\mathrm{C}{\mathrm{l}}_2\stackrel{\mathrm{sunlight}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{Cl}+\mathrm{HCl}$

   1 mark

4. (iv) Oxidation with alkaline $\mathrm{KMn}{\mathrm{O}}_4$ or acidified ${\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7$:
   $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\stackrel{\text{acidified }{\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$

   1 mark

5. (v) Warming ethanoic acid with ethanol in the presence of conc. ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ (esterification):
   $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{\text{conc. }{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (i) hot conc. H₂SO₄, 443 K; (ii) H₂ with Ni/Pd; (iii) Cl₂ in sunlight; (iv) acidified K₂Cr₂O₇ / alkaline KMnO₄; (v) ethanol + conc. H₂SO₄, warm

1. (i) The testes are located outside the abdominal cavity, in the scrotum.

   ½ mark

2. Reason: sperm formation needs a temperature lower than normal body temperature.

   ½ mark

3. Function 1: formation of sperms (male gametes).

   ½ mark

4. Function 2: secretion of the male sex hormone testosterone.

   ½ mark

5. (ii)(1) If the egg is fertilised, the zygote divides repeatedly to form an embryo. The embryo gets implanted in the thickened, blood-rich lining of the uterus and develops further, nourished through the placenta.

   1 mark

6. (2) If the egg is not fertilised, the thick inner lining of the uterus is no longer needed. It slowly breaks down and comes out through the vagina as blood and mucus (menstruation), roughly every month.

   1 mark

7. (iii) In males, the vas deferens is blocked or cut (vasectomy) so that sperms cannot be transferred.

   ½ mark

8. In females, the fallopian tube (oviduct) is blocked or cut (tubectomy) so that the egg cannot reach the uterus.

   ½ mark

Answer: (i) In the scrotum, outside the abdomen, as sperm formation needs a lower temperature; functions: sperm formation and testosterone secretion; (ii)(1) zygote forms embryo, which implants in the uterus lining; (2) uterine lining breaks down and is shed as menstruation; (iii) vasectomy (vas deferens blocked) and tubectomy (fallopian tube blocked)

1. (i)(1) When a Bryophyllum leaf falls on moist soil, the buds in the notches along its margin sprout and grow into new plantlets. Mode: vegetative propagation (by leaves).

   1 mark

2. (i)(2) When Planaria is cut into many pieces, each piece grows into a complete new organism. Mode: regeneration.

   1 mark

3. (i)(3) When the sporangia of Rhizopus burst, the spores are released. On reaching a moist, suitable surface, each spore germinates into a new Rhizopus plant. Mode: spore formation.

   1 mark

4. (ii) The zygote divides many times and forms an embryo inside the ovule.

   ½ mark

5. The ovule develops a tough coat and becomes the seed.

   ½ mark

6. The ovary grows rapidly and ripens into the fruit.

   ½ mark

7. The petals, sepals, stamens, style and stigma usually shrivel and fall off.

   ½ mark

Answer: (i) (1) Buds on the leaf margin grow into new plants (vegetative propagation); (2) each piece grows into a complete Planaria (regeneration); (3) the released spores germinate into new Rhizopus plants (spore formation). (ii) Zygote → embryo; ovule → seed; ovary → fruit; other floral parts shrivel and fall off.

1. (a) Ohm's law: The potential difference $V$ across the ends of a metallic wire is directly proportional to the current $I$ through it, provided its temperature stays the same. So $V\propto I$, i.e. $V=IR$.

   1 mark

2. (b) The slope of the V–I graph gives the resistance $R$. Resistance is the property of a conductor by which it opposes the flow of charge (current) through it.

   1 mark

3. From the graph, take the origin $(0,0)$ and the point $(0.8\text{ A},6.0\text{ V})$:
   $R=\text{slope}=\frac{V_2-V_1}{I_2-I_1}=\frac{6.0-0}{0.8-0}$

   1 mark

4. $R=\frac{60}{8}=7.5$

   ½ mark

5. SI unit: ohm, so $R=7.5\Omega$.

   ½ mark

6. (c) $1\text{ kWh}=1000\text{ W}\times 3600\text{ s}=1000{\text{J s}}^{-1}\times 3600\text{ s}=3.6\times 10^6\text{ J}$

   1 mark

Answer: (a) V ∝ I at constant temperature (V = IR); (b) resistance, R = 7.5 Ω; (c) 1 kWh = 3.6 × 10⁶ J

1. (a) The ore of mercury is cinnabar, $\mathrm{HgS}$. Mercury is present in it in the sulphide form.

   1 mark

2. (b) Heating strongly in a limited supply of air is called calcination. Zinc carbonate decomposes into zinc oxide and carbon dioxide:
   $\mathrm{ZnC}{\mathrm{O}}_3(\mathrm{s})\stackrel{\mathrm{heat}}{\to }\mathrm{ZnO}(\mathrm{s})+\mathrm{C}{\mathrm{O}}_2(\mathrm{g})$

   1 mark

3. (c)(I) Metal A is aluminium. The reaction is the thermit reaction.

   1 mark

4. (c)(II) $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3(\mathrm{s})+2Al(\mathrm{s})\to 2Fe(\mathrm{l})+\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3(\mathrm{s})+\mathrm{heat}$
   The molten iron produced is used to join railway tracks.

   1 mark

5. (c) OR: Sodium is highly reactive and has a greater affinity for oxygen than carbon has. Carbon therefore cannot reduce sodium oxide to sodium, so sodium is obtained by electrolytic reduction of molten NaCl.

   1 mark

6. At the cathode: $\mathrm{N}{\mathrm{a}}^{+}+{\mathrm{e}}^{-}\to \mathrm{Na}$

   ½ mark

7. At the anode: $2C{\mathrm{l}}^{-}\to \mathrm{C}{\mathrm{l}}_2+2{\mathrm{e}}^{-}$

   ½ mark

Answer: (a) Cinnabar, HgS (sulphide form); (b) it changes to ZnO and CO₂ (calcination); (c) (I) aluminium, thermit reaction; (II) Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + heat. OR (c): Na has more affinity for oxygen than C; cathode Na⁺ + e⁻ → Na; anode 2Cl⁻ → Cl₂ + 2e⁻

1. (a) The mother always gives an X chromosome, while the father's sperm carries either X or Y. If a sperm carrying X fuses with the ovum (X), the child is XX, a girl.

   ½ mark

2. If a sperm carrying Y fuses with the ovum (X), the child is XY, a boy. So the sex of the baby depends on the chromosome inherited from the father.

   ½ mark

3. (b) In males, one sex chromosome is a normal-sized X and the other is a much shorter Y. The two members of the pair are unequal, so it is called a mismatched pair.

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4. (c) During gamete formation, the chromosome number is reduced to half, so each germ cell carries only one set of chromosomes.

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5. When the male and female gametes fuse at fertilisation, the zygote gets one set from each parent. This restores the original (diploid) number of chromosomes in the progeny.

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6. (c) OR, Example 1: Reptiles (e.g. some turtles). The temperature at which the fertilised eggs are incubated decides whether the young developing in them will be male or female.

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7. Example 2: Snails. An individual can change its sex during its lifetime, so sex is not genetically fixed.

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Answer: (a) Father's sperm carrying X gives a girl (XX); sperm carrying Y gives a boy (XY). (b) X is normal-sized and Y is short. (c) Gametes carry half the chromosomes, and fertilisation restores the full number. OR: reptiles (incubation temperature decides sex) and snails (can change sex).