CBSE Class 10 Science: 2025 board paper, set 31/5/1

1. Magnesium ribbon burns in air with a dazzling white flame and forms a white powder, magnesium oxide: $2\mathrm{Mg}(s)+{\mathrm{O}}_2(g)\to 2\mathrm{MgO}(s)$.

2. These two observations are statements (i) and (ii), so option (C) is correct.

   1 mark

Answer: (C) (i) and (ii)

1. M displaces Fe, so M is more reactive than iron. M cannot displace Al, so M is less reactive than aluminium.

2. In the reactivity series, $\mathrm{Al}>\mathrm{Zn}>\mathrm{Fe}$. Magnesium is above Al, while copper and lead are below Fe, so M is zinc: option (D).

   1 mark

Answer: (D) Zinc

1. The formulae are $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3\cdot 10{\mathrm{H}}_2\mathrm{O}$, $\mathrm{CuS}{\mathrm{O}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$, $\mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}$ and $\mathrm{FeS}{\mathrm{O}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}$. Each contains a fixed number of water molecules of crystallisation, so option (C) is correct.

   1 mark

Answer: (C)

1. Zinc reacts with both a base and an acid to give hydrogen: $\mathrm{Zn}+2NaOH\to \mathrm{N}{\mathrm{a}}_2\mathrm{Zn}{\mathrm{O}}_2+{\mathrm{H}}_2$ and $\mathrm{Zn}+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{ZnS}{\mathrm{O}}_4+{\mathrm{H}}_2$.

2. Copper does not react with either reagent, so X is zinc and G is hydrogen: option (C).

   1 mark

Answer: (C) Zinc and Hydrogen

1. The balanced equation is $2\mathrm{Pb}(\mathrm{N}{\mathrm{O}}_3{)}_2\to 2\mathrm{PbO}+4\mathrm{N}{\mathrm{O}}_2+{\mathrm{O}}_2$.

2. Check: Pb = 2 on each side; N = 4 on each side; O = 12 on the left and $2+8+2=12$ on the right. So a, b, c, d are 2, 2, 4, 1: option (D).

   1 mark

Answer: (D) 2, 2, 4, 1

1. In electrolytic refining, impure copper is made the anode, a thin strip of pure copper is the cathode, and acidified copper sulphate solution is the electrolyte. Copper dissolves from the anode and is deposited on the cathode, so option (A) is correct.

   1 mark

Answer: (A)

1. Four carbon atoms give two skeletons: a straight chain C–C–C–C, and a branched chain made of a three-carbon chain with one carbon attached to the middle carbon.

2. According to the marking scheme, option (B) shows these two skeletons.

   1 mark

Answer: (B)

1. Amoeba first forms pseudopodia (iii), which enclose the food to form a food vacuole (ii). Inside the vacuole, complex food is broken down (iv). The simple nutrients diffuse into the cytoplasm (i), and undigested material is thrown out (v). The sequence matches option (B).

   1 mark

Answer: (B) (iii), (ii), (iv), (i), (v)

1. X is the medulla in the hindbrain. It controls salivation, vomiting and blood pressure.

2. The feeling of hunger is controlled by the forebrain, not the medulla. So the answer is option (B).

   1 mark

Answer: (B) Hunger

1. The filament of Spirogyra breaks into pieces, and each piece grows into a new filament (fragmentation). In Planaria, a cut body part grows into a whole organism (regeneration). So the answer is option (C).

   1 mark

Answer: (C)

1. Cytokinins promote cell division and are present in high amounts in seeds and fruits. Abscisic acid inhibits growth and causes wilting of leaves. So the answer is option (B).

   1 mark

Answer: (B)

1. Glucose is broken down to pyruvate in the cytoplasm. In aerobic respiration, pyruvate is then broken down to $\mathrm{C}{\mathrm{O}}_2$ and water in the mitochondria. So the answer is option (D).

   1 mark

Answer: (D) Cytoplasm and Mitochondria

1. A concave mirror forms a virtual, erect and enlarged image only when the object is placed between the pole P and the focus F. So the dentist keeps the teeth within the focal length: option (B).

   1 mark

Answer: (B)

1. An elongated eyeball causes myopia, because the image forms in front of the retina. A concave (diverging) lens of suitable power corrects it, so the answer is option (B).

   1 mark

Answer: (B) Concave lens

1. A lake is a natural ecosystem. A garden is made and maintained by humans, so it is an artificial ecosystem. The order natural, then man-made, matches option (C).

   1 mark

Answer: (C) Lakes and gardens

1. Excessive use of disposable cups and plates produces large amounts of waste and harms the environment. The other three options are eco-friendly practices, so the answer is option (B).

   1 mark

Answer: (B)

1. In sunlight, $2\mathrm{AgCl}(s)\to 2\mathrm{Ag}(s)+\mathrm{C}{\mathrm{l}}_2(g)$.

2. The silver formed is grey, which is why the white salt turns grey. Both statements are true, and R explains A: option (A).

   1 mark

Answer: (A)

1. Both statements are true. However, the placenta supplies nutrition because its villi provide a large surface area for exchange of substances. Its being a disc embedded in the uterine wall does not explain this. So the answer is option (B).

   1 mark

Answer: (B)

1. A is true: a solenoid's field resembles that of a bar magnet. R is false: the field pattern depends on the shape of the conductor (compare a straight wire, a circular loop and a solenoid). So the answer is option (C).

   1 mark

Answer: (C)

1. A is false: only green plants and some bacteria (autotrophs) can make food using sunlight. R is true: such organisms are called producers. So the answer is option (D).

   1 mark

Answer: (D)

1. (a) The green crystals that lose water of crystallisation and then decompose with a smell of burning sulphur are ferrous sulphate crystals: **X = $\mathrm{FeS}{\mathrm{O}}_4\cdot 7{\mathrm{H}}_2\mathrm{O}$** (ferrous sulphate heptahydrate).

   ½ mark

2. (a) On strong heating: $2\mathrm{FeS}{\mathrm{O}}_4(\mathrm{s})\stackrel{\text{heat}}{\to }\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3(\mathrm{s})+\mathrm{S}{\mathrm{O}}_2(\mathrm{g})+\mathrm{S}{\mathrm{O}}_3(\mathrm{g})$. The solid left is **Y = $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3$** (ferric oxide).

   ½ mark

3. (b) The green crystals first turn white as they become anhydrous $\mathrm{FeS}{\mathrm{O}}_4$. On further heating the white solid turns reddish-brown as $\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3$ forms.

   1 mark

Answer: (a) X = FeSO₄·7H₂O (ferrous sulphate crystals), Y = Fe₂O₃ (ferric oxide); (b) green → white → reddish-brown

1. (a) The testes lie in the scrotum, outside the abdominal cavity, because sperm formation needs a temperature about 2–3 °C lower than normal body temperature, which the scrotum provides.

   1 mark

2. (b) The seminal vesicles and prostate gland add their secretions to the sperms. These secretions make the sperms move in a fluid, which helps in their transport, and also provide nutrition to the sperms.

   1 mark

Answer: (a) Testes need a temperature lower than body temperature for sperm formation; (b) gland secretions help transport sperms and nourish them.

1. Formation: Some plasma, proteins and blood cells escape through the pores in the walls of capillaries into the intercellular spaces of tissues. This fluid is the tissue fluid, called lymph, and it drains into lymphatic capillaries.

   1 mark

2. Function: Lymph carries digested and absorbed fat from the intestine. It also drains excess fluid from the intercellular spaces back into the blood.

   1 mark

Answer: Lymph forms when plasma, proteins and blood cells leak through capillary pores into intercellular spaces; it carries absorbed fat from the intestine and returns excess tissue fluid to the blood.

1. (a) X is Bowman's capsule, the cup-shaped start of the nephron that surrounds the glomerulus.

   ½ mark

2. (a) Role: it collects the filtrate (water, glucose, amino acids, salts and urea) filtered from the blood in the glomerulus.

   ½ mark

3. (b) The initial filtrate also contains useful substances such as glucose, amino acids, salts and much water. The nephron reabsorbs selectively according to how much excess water is present in the body and how much dissolved waste has to be removed. In this way useful substances are kept in the body and only wastes and extra water leave as urine.

   1 mark

Answer: (a) X = Bowman's capsule; it collects the filtrate from the glomerulus. (b) Reabsorption depends on how much water and useful substances the body needs to retain and how much waste must be removed.

1. (a) Optical density: kerosene ($n=1.44$) has a higher refractive index than water ($n=1.33$), so kerosene is optically denser than water.

   ½ mark

2. (b) Mass density: kerosene floats on water, so kerosene has a lower mass density than water.

   ½ mark

3. (c) Speed of light: since $v=\frac{c}{n}$, light travels slower in kerosene and faster in water ($v_{kerosene}<v_{water}$).

   ½ mark

4. Inference: A medium that is optically denser need not have a greater mass density. Optical density and mass density are different properties.

   ½ mark

Answer: Kerosene: higher optical density, lower mass density, lower speed of light than water. Inference: an optically denser medium need not be denser in mass.

1. (A) First application of Joule's heating at home: the electric iron / electric heater / room heater, where the heating element becomes hot when current flows.

   ½ mark

2. (A) Second application: the electric fuse, whose wire melts from the heat produced and breaks the circuit when the current is too large. The filament of an electric bulb also glows because of heating.

   ½ mark

3. (B)(a) $1\text{kWh}=1000\text{W}\times 3600\text{s}$

   ½ mark

4. (B)(a) $=3.6\times 10^6\text{W s}=3.6\times 10^6\text{J}$. So $1\text{kWh}=3.6\times 10^6$ joules.

   ½ mark

5. (B)(b) [OR alternative to (B)(a), ½ + ½] For resistors in parallel: $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}$

6. (B)(b) $\frac{1}{R_p}=\frac{6+3+2}{12}=\frac{11}{12}$, so $R_p=\frac{12}{11}\Omega \approx 1.09\Omega$.

Answer: (A) Electric iron/heater and electric fuse (or bulb); (B)(a) 1 kWh = 3.6 × 10⁶ J; (B)(b) Rp = 12/11 Ω ≈ 1.09 Ω

1. (a) First-trophic-level organisms are the producers (green plants). They trap solar energy and convert it into chemical energy in food. This makes energy available to all the consumers (heterotrophs), so no food chain can exist without them.

   1 mark

2. (b) Energy trapped by autotrophs never goes back to the sun, and energy passed to herbivores never returns to the producers. Energy moves only forward from one trophic level to the next and is no longer available to the previous level. Hence the flow of energy is unidirectional.

   1 mark

Answer: (a) Producers fix solar energy and make it available to consumers; (b) energy passes forward only and never returns to the sun or to earlier trophic levels.

1. Chemical formula of washing soda: $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3\cdot 10{\mathrm{H}}_2\mathrm{O}$ (sodium carbonate decahydrate).

   1 mark

2. Heating baking soda gives sodium carbonate: $2\mathrm{NaHC}{\mathrm{O}}_3\stackrel{\text{heat}}{\to }\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2$. Recrystallising this sodium carbonate from water gives washing soda: $\mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3+10{\mathrm{H}}_2\mathrm{O}\to \mathrm{N}{\mathrm{a}}_2\mathrm{C}{\mathrm{O}}_3\cdot 10{\mathrm{H}}_2\mathrm{O}$.

   1 mark

3. Uses (any two): (i) in the glass, soap and paper industries; (ii) for removing permanent hardness of water; (iii) as a cleaning agent for domestic purposes; (iv) in the manufacture of borax.

   1 mark

Answer: Na₂CO₃·10H₂O; heat baking soda to get Na₂CO₃, then recrystallise; uses: glass/soap/paper industry, removing permanent hardness of water.

1. Test tube A: Metal M (iron, the metal usually protected by galvanisation) corrodes partly. The part above the oil is exposed to moist air and rusts. The part dipped in oil is cut off from air and moisture, so it does not rust.

   1 mark

2. Test tube B: Metal M does not corrode, because it is fully inside the oil and not exposed to moist air.

   1 mark

3. Test tube C: Metal M does not corrode, because anhydrous calcium chloride absorbs moisture and the air is dry. Rusting needs both air and water.

   1 mark

Answer: A: partly corroded (only the part outside the oil); B: no corrosion (covered by oil); C: no corrosion (dry air due to anhydrous CaCl₂).

1. (a) Electronic configurations: Al (Z = 13): 2, 8, 3; N (Z = 7): 2, 5.

   1 mark

2. (a) Al loses its 3 valence electrons to N: $\mathrm{Al}\to \mathrm{A}{\mathrm{l}}^{3+}+3e^{-}$ and $\mathrm{N}+3e^{-}\to {\mathrm{N}}^{3-}$. Both ions now have the octet 2, 8. The electrostatic attraction between them gives $\mathrm{A}{\mathrm{l}}^{3+}{\mathrm{N}}^{3-}$, i.e. AlN. (Dot diagram: Al with 3 crosses → arrows to N with 5 dots → [Al]³⁺ [N with 8 electrons]³⁻.)

   1 mark

3. (b) In ionic compounds the positive and negative ions are held by strong electrostatic (interionic) forces of attraction in a rigid arrangement, so they are solids. When pressure is applied, layers of ions shift, like charges come face to face and repel, so the crystal breaks into pieces; hence they are brittle.

   1 mark

Answer: (a) Al (2,8,3) transfers 3 electrons to N (2,5), giving Al³⁺ and N³⁻, which combine as AlN; (b) strong electrostatic forces between ions make ionic compounds hard, brittle solids.

1. (a) The respiratory pigment in humans is haemoglobin, present in red blood cells.

   ½ mark

2. (a) Function: it has a high affinity for oxygen. It takes up oxygen in the lungs and carries it to the tissues that are deficient in oxygen.

   ½ mark

3. (b) The lungs always keep a residual volume of air so that there is enough time for oxygen to be absorbed into the blood and for carbon dioxide to be released, even during exhalation.

   1 mark

4. (c) Energy released during cellular respiration is stored in ATP. Breaking its terminal phosphate bond releases energy that the cell uses for various processes such as muscle contraction, protein synthesis and conduction of nerve impulses. ATP works like currency in this way, so it is called the energy currency of the cell.

   1 mark

Answer: (a) Haemoglobin – carries O₂ from lungs to tissues; (b) to allow sufficient time for O₂ absorption and CO₂ release; (c) ATP stores energy and supplies it for all cellular activities.

1. (a) Fertilisation is the fusion of the male gamete with the female gamete to form a zygote.

   1 mark

2. (b) Zygote: divides repeatedly to form the embryo inside the ovule.

   ½ mark

3. (b) Ovule: develops a tough coat and becomes the seed.

   ½ mark

4. (b) Ovary: grows rapidly and ripens to form the fruit.

   ½ mark

5. (b) Stamens: wither (shrivel) and fall off, along with the petals and sepals.

   ½ mark

Answer: (a) Fusion of male and female gametes to form a zygote; (b) zygote → embryo, ovule → seed, ovary → fruit, stamens wither and fall off.

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

   ½ mark

2. (a) $f=\frac{1}{P}=\frac{1}{-0.25}\text{m}=-4\text{m}$ (i.e. $-400$ cm).

   ½ mark

3. (b) A concave lens is used to correct myopia (near-sightedness).

   1 mark

4. (c) Nature: the image is virtual and erect. A concave lens always forms such an image, wherever the object is placed.

   ½ mark

5. (c) Size: the image is diminished (smaller than the object) and forms between the optical centre and the focus on the same side as the object.

   ½ mark

Answer: (a) Concave lens, f = −4 m; (b) myopia; (c) virtual, erect and diminished

1. Given: $r=0.01\text{cm}=0.01\times 10^{-2}\text{m}=1\times 10^{-4}\text{m}$, $R=7\Omega$, $\rho =44\times 10^{-6}\Omega \text{m}$.

   ½ mark

2. Area of cross-section: $A=\pi r^2=\frac{22}{7}\times (10^{-4}{)}^2{\text{m}}^2=\frac{22}{7}\times 10^{-8}{\text{m}}^2$.

   ½ mark

3. $R=\rho \frac{l}{A}\Rightarrow l=\frac{RA}{\rho }$

   ½ mark

4. $l=\frac{7\times \frac{22}{7}\times 10^{-8}}{44\times 10^{-6}}=\frac{22\times 10^{-8}}{44\times 10^{-6}}$

   1 mark

5. $l=0.5\times 10^{-2}\text{m}=0.005\text{m}$ (= 0.5 cm).

   ½ mark

Answer: l = 0.5 × 10⁻² m = 0.005 m (0.5 cm)

1. (a) The earth wire is connected to the metallic body of appliances such as an iron, toaster or refrigerator. If current leaks to the metal body, the earth wire gives it a low-resistance path to the earth and keeps the body at the earth's potential (zero). So the user does not get a severe electric shock; the earth wire is a safety measure.

   2 marks

2. (b) Precaution 1: use an electric fuse (or MCB) of proper rating in the circuit.

   ½ mark

3. (b) Precaution 2: do not connect too many appliances to a single socket at the same time.

   ½ mark

Answer: (a) The earth wire keeps the metal body at earth potential and carries leaked current to earth, preventing severe shock; (b) use a fuse of proper rating; avoid connecting many appliances to one socket.

1. (a) A homologous series is a series of carbon compounds that have the same functional group substituting for hydrogen in a carbon chain. They have the same general formula and similar chemical properties, and successive members differ by a $-\mathrm{C}{\mathrm{H}}_2-$ unit.

   1 mark

2. (a) Two consecutive ketones: propanone ${\mathrm{C}}_3{\mathrm{H}}_6\mathrm{O}$ ($\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3$) and butanone ${\mathrm{C}}_4{\mathrm{H}}_8\mathrm{O}$ ($\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{C}}_2{\mathrm{H}}_5$).

   1 mark

3. (b)(i) $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+\mathrm{NaOH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{H}}_2\mathrm{O}$. Product: sodium ethanoate (sodium acetate) and water.

   1 mark

4. (b)(ii) $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}\stackrel{\text{acid}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$. Product: ethyl ethanoate (an ester) and water (esterification).

   1 mark

5. (c) Benzene, ${\mathrm{C}}_6{\mathrm{H}}_6$: six carbon atoms joined in a hexagonal ring with alternating single and double bonds (C=C, C–C, C=C, C–C, C=C, C–C). Each carbon carries one hydrogen atom by a single bond.

   1 mark

Answer: (a) Series with same functional group/general formula and similar properties; C₃H₆O and C₄H₈O; (b)(i) sodium ethanoate, (ii) ethyl ethanoate; (c) hexagonal ring of 6 C with alternate double bonds, one H on each C.

1. (a) Molecular formula of ethyne: ${\mathrm{C}}_2{\mathrm{H}}_2$ ($\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}$).

   1 mark

2. (a) Electron dot structure: the two carbon atoms share three pairs of electrons (a triple bond), and each carbon shares one pair with a hydrogen atom: $\mathrm{H}:\mathrm{C}:::\mathrm{C}:\mathrm{H}$. Each C then has 8 electrons and each H has 2.

   1 mark

3. (b)(i) $2Na+2C{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\to 2C{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{ONa}+{\mathrm{H}}_2$. Products: sodium ethoxide and hydrogen gas.

   1 mark

4. (b)(ii) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}\stackrel{\text{acid}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$. Product: ethyl ethanoate (ester).

   1 mark

5. (b)(iii) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\underset{\text{heat}}{\overset{\text{acidified }{\mathrm{K}}_2\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_7}{\to }}\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$. Product: ethanoic acid (oxidation of ethanol).

   1 mark

Answer: (a) C₂H₂, H:C:::C:H; (b)(i) sodium ethoxide + H₂; (ii) ethyl ethanoate; (iii) ethanoic acid

1. (a)(i) Iodine is needed by the thyroid gland to make the hormone thyroxin. When iodine is deficient, thyroxin is not made properly and the thyroid swells, causing goitre (swollen neck).

   1 mark

2. (a)(ii) Growth hormone from the pituitary gland controls growth. Its deficiency in childhood leads to dwarfism.

   1 mark

3. (a)(iii) At puberty the testes secrete the hormone testosterone, which causes changes such as growth of facial hair in boys.

   1 mark

4. (b) Reason 1: Electrical impulses reach only the cells connected to nerve tissue. Hormones released into the blood can reach all cells of the body.

   1 mark

5. (b) Reason 2: Hormones act steadily and persistently, which is needed to coordinate long-term processes like growth, development and responses to the environment. Nerve cells cannot generate impulses continuously.

   1 mark

Answer: (a)(i) Low iodine → less thyroxin → goitre; (ii) growth hormone deficiency → dwarfism; (iii) testosterone at puberty; (b) hormones reach all cells and act steadily, coordinating growth and development.

1. (a) Difference 1: A voluntary action involves thinking and is done according to our will, e.g. writing. An involuntary action does not involve thinking and is not under our will, e.g. heartbeat.

   1 mark

2. (a) Difference 2: Voluntary actions are controlled by the forebrain (cerebrum). Involuntary actions such as salivation and vomiting are controlled by the hindbrain (medulla).

   1 mark

3. (b) Reflex action: a sudden, automatic response to a stimulus in the environment that occurs without conscious thinking, e.g. pulling the hand away from a hot object.

   1 mark

4. (b) Flow diagram of the path of a nerve impulse: Stimulus → Receptor (in sense organ) → Sensory neuron → Spinal cord (relay neuron) → Motor neuron → Effector (muscle/gland) → Response.

   2 marks

Answer: (a) Voluntary: involves thinking, under our will, controlled by forebrain; involuntary: no thinking, not under our will, controlled by hindbrain. (b) Reflex action = sudden automatic response to a stimulus; Stimulus → Receptor → Sensory neuron → Spinal cord → Motor neuron → Effector.

1. (a)(i) Speed of light: The ray goes from A into B without bending, so light travels at the same speed in A and B. On passing from B into C it bends away from the normal, so light travels faster in C than in A and B. Thus $v_A=v_B<v_C$.

   1 mark

2. (a)(ii) Optical density: A and B have the same optical density. C is optically rarer (less optically dense) than A and B.

   1 mark

3. Justification using refractive index: A ray is not deviated between media of equal refractive index, so $n_A=n_B$. A ray bends away from the normal when it goes into a medium of lower refractive index, so $n_C<n_B$. Since $n=\frac{c}{v}$, a lower refractive index means a higher speed of light and a lower optical density. Hence $n_A=n_B>n_C$.

   1 mark

4. (b)(i) Oblique incidence, B denser than A and C: Draw the ray striking the A–B surface at an angle. At the A–B surface it bends towards the normal inside B. At the B–C surface it bends away from the normal into C. Draw normals (dotted) at both points of incidence and mark arrows on the ray.

   1 mark

5. (b)(ii) Normal incidence: Draw the ray falling perpendicular to the A–B surface. It goes straight through B and C without bending (angle of incidence = 0°, so angle of refraction = 0°).

   1 mark

Answer: (a) Speed: $v_A=v_B<v_C$; optical density: A = B > C, because $n_A=n_B>n_C$ and $n=c/v$. (b)(i) The ray bends towards the normal on entering B and away from the normal on entering C. (ii) The ray passes straight through all three media without deviation.

1. (a) Focal length: In Observation 4, $u=-40$ cm and $v=+40$ cm. The object and image distances are equal, so the object is at $2F_1$ and the image is at $2F_2$. Therefore $2f=40$ cm, which gives $f=20$ cm.

   1 mark

2. Reason: For a convex lens, an object placed at $2F_1$ forms a real, inverted image of the same size at $2F_2$ on the other side of the lens. Only then are $|u|$ and $|v|$ equal.

   1 mark

3. (b) Magnification in Observation 3: $m=\frac{v}{u}=\frac{+60\text{ cm}}{-30\text{ cm}}=-2$. The image is real, inverted and twice the size of the object.

   1 mark

4. (c) Difference 1: In Observation 1, $v=-60$ cm (negative), so the image is virtual and erect ($m=+4$). In Observation 2, $v=+100$ cm (positive), so the image is real and inverted ($m=-4$).

   1 mark

5. Difference 2: In Observation 1 the image forms on the same side as the object and cannot be obtained on a screen. In Observation 2 it forms on the other side of the lens and can be obtained on a screen.

   1 mark

Answer: (a) f = 20 cm, since $u=-40$ cm gives $v=+40$ cm (object at 2F gives image at 2F); (b) m = −2; (c) Obs. 1: virtual, erect, same side as the object; Obs. 2: real, inverted, other side of the lens.

1. (a) The electronic configuration of A (Z = 16) is 2, 8, 6, with 6 valence electrons. It gains 2 electrons (forming $A^{2-}$) or shares 2 electrons to reach the configuration 2, 8, 8 of the nearest noble gas, argon.

   1 mark

2. (b) Butene is $C{H}_2=CH-C{H}_2-C{H}_3$. (i) Single covalent bonds = 8 C–H + 2 C–C = 10. (ii) Double covalent bonds = 1 (the C=C bond).

   1 mark

3. (c)(A) Nitrogen (2, 5) has 5 valence electrons and needs 3 more to complete its octet. Each hydrogen has 1 electron and needs 1 more for a duplet. So N shares one electron pair with each of three H atoms, forming three N–H single covalent bonds.

   1 mark

4. Electron dot structure: Write N in the centre with three H atoms around it. Put one shared pair (one dot from N, one cross from H) between N and each H, and one lone pair on N. N now has 8 electrons around it and each H has 2: $H:\stackrel{\cdot \cdot }{N}:H$ with the third H below N.

   1 mark

5. (c)(B) Carbon (2, 4) has 4 valence electrons. It cannot easily gain 4 electrons to form $C^{4-}$, because a nucleus with only 6 protons cannot hold 10 electrons.

   1 mark

6. It also cannot easily lose 4 electrons to form $C^{4+}$, because removing 4 electrons needs a very large amount of energy. Therefore carbon completes its octet by sharing its 4 valence electrons, forming covalent bonds.

   1 mark

Answer: (a) Gains or shares 2 electrons to reach argon's configuration 2, 8, 8; (b) (i) 10 single bonds, (ii) 1 double bond; (c)(A) N shares one electron pair with each of 3 H atoms, giving 3 N–H bonds and 1 lone pair on N; (c)(B) forming $C^{4-}$ or $C^{4+}$ is energetically unfavourable, so carbon shares electrons.

1. (a) Chromosomes are thread-like structures in the nucleus, made of DNA and proteins. They carry genes, the information for inheriting traits from parents to offspring.

   1 mark

2. (b) In males the sex-chromosome pair is XY. The X chromosome is of normal size but the Y chromosome is much shorter. The two partners are unequal, so the pair is called mismatched.

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3. (c)(A) Flow chart: Father (XY) → sperms carry either X or Y (50% each). Mother (XX) → all eggs carry X. Then: X egg + X sperm → XX (girl); X egg + Y sperm → XY (boy).

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4. Each type of sperm is equally likely to fertilise the egg, so the probability of a boy = probability of a girl = $\frac{1}{2}$, i.e. 50 : 50. The sex of the child depends on the chromosome inherited from the father.

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5. (c)(B) Example 1, some reptiles (e.g. turtles, lizards): The temperature at which the fertilised eggs are incubated decides whether the offspring is male or female.

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6. Example 2, snails: An individual snail can change its sex during its life, which shows that its sex is not fixed genetically.

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Answer: (a) Thread-like DNA–protein structures in the nucleus carrying genes; (b) the Y chromosome is much shorter than the X; (c)(A) XY × XX gives XX : XY = 1 : 1 (50 : 50); (c)(B) reptiles (incubation temperature decides sex) and snails (can change sex).

1. (a) The direction of a field line is taken as the direction in which the north pole of a compass needle moves when placed in the field. The needle's north pole is repelled by the magnet's N pole and attracted to its S pole. So outside the magnet the lines emerge from N and merge at S.

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2. (b) The closer together the field lines, the stronger the magnetic field in that region. The field is strongest near the poles, where the lines are most crowded.

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3. (c)(A)(i) If two field lines crossed, there would be two tangents (two field directions) at the point of intersection. A compass needle there would have to point in two directions at once, which is impossible. So field lines never intersect.

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4. (c)(A)(ii) Diagram: Draw several straight, parallel, equally spaced lines, all with arrows pointing in the same direction. This represents a uniform magnetic field (as found inside a solenoid).

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5. (c)(B) Diagram: Draw a solenoid (coil of many turns) with the current direction marked. Inside it, draw straight, parallel lines along the axis. Outside, draw closed loops from one end (N) to the other (S), like the field of a bar magnet. Mark arrows from N to S outside and from S to N inside.

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6. The parallel, straight field lines inside the solenoid show that the magnetic field there is uniform: it has the same strength and direction at all points.

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Answer: (a) By convention, a field line's direction is the direction a compass north pole moves, which is away from N and towards S; (b) closer lines mean a stronger field; (c)(A)(i) intersection would give two field directions at one point, which is impossible; (ii) equally spaced parallel straight lines with arrows in one direction; (c)(B) bar-magnet-like pattern with parallel lines inside, showing a uniform field inside the solenoid.