CBSE Class 10 Science: 2026 board paper, set 31/3/1

1. In humans, anaerobic respiration happens in muscle cells during vigorous exercise when oxygen runs short, so (i) is correct. It does not need oxygen, so (ii) is wrong.

2. Glucose (6C) is broken into pyruvate (3C), which is then converted to lactic acid, also a 3-carbon molecule, so (iii) is correct. Option (B).

   1 mark

Answer: (B) (i) and (iii)

1. The experiment showing that CO${}_2$ is needed for photosynthesis requires light to reach the leaves. The bell-jars must therefore be transparent glass jars.

2. Wooden bell-jars block sunlight, so no photosynthesis can occur in either plant. This is the incorrect step. Option (B).

   1 mark

Answer: (B)

1. The cerebellum controls precision of voluntary actions, posture and balance, as needed in walking straight or cycling. It is part of the hind-brain, so the answer is option (C).

   1 mark

Answer: (C) Hind-Brain – Cerebellum

1. At the axon ending, chemicals are released that cross the synapse to the dendrite of the next neuron. Hence the signal passes from the axonal end of one neuron to the dendritic end of another, option (D).

   1 mark

Answer: (D)

1. When organisms feed on several kinds of prey, many food chains interconnect. These branching chains form a food web, option (A).

   1 mark

Answer: (A) Food web

1. In biological magnification, non-biodegradable toxins concentrate further at each trophic level. Humans occupy the top level, so they accumulate the most, option (B).

   1 mark

Answer: (B) Human beings

1. The side of the tendril away from the support grows faster than the side touching it. This makes the tendril coil around the support, so the answer is option (D).

   1 mark

Answer: (D)

1. Decomposers break dead matter into simple substances that return to the soil and replenish it. Without them this recycling would stop, so A is true.

2. R is true and explains why A happens. Option (A).

   1 mark

Answer: (A)

1. The glomerulus, a cluster of thin-walled capillaries, is the filtering unit, just as the alveolar capillaries are in the lungs. So A is true. R correctly describes Bowman's capsule, so R is also true.

2. R only describes the structure of the nephron; it does not explain why the filtering unit is a capillary cluster. Option (B).

   1 mark

Answer: (B)

1. (i) Under the 1987 UNEP agreement (the Montreal Protocol), countries agreed to freeze the production of CFCs at the 1986 levels.

   1 mark

2. (ii) Ozone (${\mathrm{O}}_3$) in the upper atmosphere absorbs the harmful ultraviolet (UV) radiation from the Sun. It shields living beings on the Earth's surface from UV damage such as skin cancer.

   1 mark

Answer: (i) Freeze CFC production at 1986 levels; (ii) Ozone shields the Earth from harmful UV radiation of the Sun.

1. Aerobic respiration provides more energy.

   1 mark

2. Reason: in aerobic respiration glucose is completely broken down in the presence of oxygen into $\mathrm{C}{\mathrm{O}}_2$ and water. This releases much more energy (ATP) than anaerobic respiration, where glucose is only partly broken down into ethanol and $\mathrm{C}{\mathrm{O}}_2$ or into lactic acid.

   1 mark

Answer: Aerobic respiration, because glucose is completely broken down in the presence of oxygen, releasing more energy.

1. (I)(i) Fish have a 2-chambered heart (one atrium and one ventricle).

   ½ mark

2. (I)(ii) Amphibians / Reptiles have a 3-chambered heart with incomplete double circulation.

   ½ mark

3. (I)(iii) In birds and mammals the circulation is double circulation: blood passes through the heart twice in each cycle.

   ½ mark

4. (II) Double circulation (as in birds and mammals) is the most efficient. Oxygenated and deoxygenated blood never mix, so the body gets a highly efficient supply of oxygen.

   ½ mark

Answer: (I)(i) 2-chambered; (ii) Amphibians/Reptiles; (iii) Double circulation; (II) Double circulation

1. Water moves up through this continuous xylem system by two forces: root pressure and transpiration pull.

2. Root pressure: Root cells in contact with the soil actively take up ions.

   ½ mark

3. This creates a difference in ion concentration between the root and the soil. Water therefore moves from the soil into the root xylem to remove the difference, and a column of water is pushed steadily upward.

   1 mark

4. Transpiration pull: Water evaporates (is lost as vapour) through the stomata of the leaves and other aerial parts.

   ½ mark

5. This loss creates a suction that pulls the continuous water column up from the xylem of the roots through the stem to the leaves. It is the main driving force during the day.

   1 mark

Answer: Water rises through the interconnected xylem by root pressure (active ion uptake by roots draws water in and pushes it up) and by transpiration pull (evaporation from leaves creates suction that pulls the water column up).

1. Parents: pure tall (TT) × pure short (tt).

   ½ mark

2. Gametes: the tall parent gives only T; the short parent gives only t.

   ½ mark

3. F₁ progeny: T × t → Tt. All F₁ plants are tall (hybrid tall).

   1 mark

4. Observation 1: all the F₁ plants were tall, so they resembled only one parent (the tall one).

   ½ mark

5. Observation 2: no medium-height or short plants appeared in the F₁ generation. There was no blending; the trait for shortness was not expressed.

   ½ mark

Answer: TT (tall) × tt (short) → gametes T and t → F₁ all Tt; all F₁ plants were tall, with no medium or short plants (tallness is dominant).

1. (i) Transpiration, the loss of water as vapour, is minimised. These plants keep their stomata closed during the day, which cuts water loss so they can survive in dry conditions.

   2 marks

2. (ii) The stomata, the tiny pores on the leaf surface, are responsible for exchange of gases ($\mathrm{C}{\mathrm{O}}_2$ in, ${\mathrm{O}}_2$ out) during photosynthesis.

   2 marks

Answer: (i) Transpiration (water loss through stomata) is minimised; (ii) Stomata

1. (i) The events in photosynthesis are: (1) chlorophyll absorbs light energy; (2) light energy is converted into chemical energy, and water molecules are split into hydrogen and oxygen; (3) carbon dioxide is reduced to carbohydrates.

   1 mark

2. (ii) Such desert plants take in $\mathrm{C}{\mathrm{O}}_2$ at night, when their stomata open. They store it as an intermediate compound and use it during the day.

   1 mark

3. (iii) Diagram: draw a pair of kidney-shaped guard cells facing each other with a pore between them, inside an outline of the leaf epidermis. The guard cells should be bulging (turgid) so the gap between them is wide open.

   1 mark

4. Labels: Guard cells, Stomatal pore (open), Chloroplast (inside the guard cells), Nucleus, Epidermal cells / cell wall.

   1 mark

Answer: (i) Light absorption by chlorophyll; conversion of light to chemical energy with splitting of water; reduction of CO₂ to carbohydrates. (ii) At night. (iii) Labelled diagram of an open stoma.

1. (i) Structure: Bryophyllum leaf, with buds in the notches on its margin.

   ½ mark

2. Type: vegetative propagation, a form of asexual reproduction.

   ½ mark

3. (ii) Advantage (any one): the new plants are genetically identical to the parent, so desirable traits are kept. They also flower and fruit earlier than seed-grown plants, and the method can raise plants that have lost the capacity to produce seeds.

   1 mark

4. Disadvantage: there is no genetic variation, so all the plants can be harmed equally by a disease or by a change in the environment.

   1 mark

5. (iii) Asexual reproduction involves only one parent and no fusion of gametes. The offspring form by mitosis and receive copies of the same DNA, so they have the same genetic make-up and show no variation.

   1 mark

6. (iv) Sexual reproduction combines DNA from two parents, which creates genetic variation. Variation improves the chances of survival of the species in a changing environment and drives evolution.

   1 mark

Answer: (i) Bryophyllum leaf; vegetative propagation (asexual). (ii) Advantage: genetically identical plants / early flowering; Disadvantage: no variation. (iii) Single parent, no gamete fusion, mitotic division. (iv) It produces variation, which aids survival and evolution.

1. (i) Name: multiple fission.

   ½ mark

2. Type: asexual reproduction.

   ½ mark

3. (ii) In stage P, inside the protective cyst, the nucleus divides repeatedly and each nucleus is then surrounded by cytoplasm. Many daughter cells form at once and are released when the cyst wall breaks.

   1 mark

4. (iii) Plasmodium (the malarial parasite) reproduces by multiple fission.

   1 mark

5. (iv) Method: vegetative propagation (for example layering, grafting or cutting), as used for banana, orange, rose and jasmine.

   1 mark

6. Advantage 1: all the plants produced are genetically similar to the parent plant, so desirable characters are kept.

   ½ mark

7. Advantage 2: such plants bear flowers and fruits earlier than plants raised from seeds.

   ½ mark

Answer: (i) Multiple fission; asexual. (ii) Repeated nuclear division followed by cytoplasmic division inside the cyst forms many daughter cells. (iii) Plasmodium. (iv) Vegetative propagation; plants are genetically identical to the parent and flower/fruit earlier.

1. A reactive metal displaces hydrogen from a dilute acid, for example $\mathrm{Zn}+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to \mathrm{ZnS}{\mathrm{O}}_4+{\mathrm{H}}_2$. This is a displacement reaction, option (B).

   1 mark

Answer: (B)

1. $\mathrm{CuO}+2HCl\to \mathrm{CuC}{\mathrm{l}}_2+{\mathrm{H}}_2\mathrm{O}$. The copper(II) chloride formed gives the solution a blue-green colour, option (C).

   1 mark

Answer: (C)

1. ${\mathrm{I}}_2$ needs 2 I atoms, so x = 2. Then 2 K atoms give y = 2.

2. Check: $2KI+{\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{I}}_2+2KOH$ gives O: 2 = 2 and H: 2 = 2, so it is balanced. Option (A).

   1 mark

Answer: (A) 2, 2

1. Both X and Y are ${\mathrm{C}}_4{\mathrm{H}}_{10}$, but X is a straight chain and Y is branched. So they have the same molecular formula with different structures.

2. Compounds like this are structural isomers, not homologues. So (iii) and (iv) are correct, option (D).

   1 mark

Answer: (D) (iii) and (iv)

1. Calcium reacts less violently than sodium. The reaction is exothermic, but the heat released is not enough to ignite the hydrogen. The hydrogen bubbles stick to the metal and make it float. Option (C).

   1 mark

Answer: (C)

1. Glucose and alcohol do not give ions in solution, so they do not conduct electricity. Dilute HCl produces H${}^{+}$ ions, which carry current, so the bulb glows only in case (iii). Option (D).

   1 mark

Answer: (D) Case (iii) only

1. $\mathrm{Zn}+2NaOH\to \mathrm{N}{\mathrm{a}}_2\mathrm{Zn}{\mathrm{O}}_2+{\mathrm{H}}_2$. The product is sodium zincate, option (C).

   1 mark

Answer: (C) $\mathrm{N}{\mathrm{a}}_2\mathrm{Zn}{\mathrm{O}}_2$

1. Catenation, the ability of carbon to link with many other carbon atoms, is possible because the C–C bond is strong and stable. So A is true and R explains it, option (A).

   1 mark

Answer: (A)

1. (i) Electronic configurations: Na (11) = 2, 8, 1; O (8) = 2, 6. Each sodium atom loses its 1 valence electron, and the oxygen atom gains 2 electrons:
   $2\mathrm{N}{\mathrm{a}}^{\bullet }+\ddot{\underset{\cdot \cdot }{\mathrm{O}}}:\to 2\mathrm{N}{\mathrm{a}}^{+}+[:\ddot{\underset{\cdot \cdot }{\mathrm{O}}}:{]}^{2-}$, i.e. $2\mathrm{Na}\to 2\mathrm{N}{\mathrm{a}}^{+}+2e^{-}$ and $\mathrm{O}+2e^{-}\to {\mathrm{O}}^{2-}$, giving $(\mathrm{N}{\mathrm{a}}^{+}{)}_2{\mathrm{O}}^{2-}$ = $\mathrm{N}{\mathrm{a}}_2\mathrm{O}$.

   1 mark

2. (ii) Electronic configurations: Mg (12) = 2, 8, 2; Cl (17) = 2, 8, 7. The magnesium atom gives one electron to each of two chlorine atoms:
   $\mathrm{Mg}\to \mathrm{M}{\mathrm{g}}^{2+}+2e^{-}$ and $2\mathrm{Cl}+2e^{-}\to 2\mathrm{C}{\mathrm{l}}^{-}$. In electron-dot form: $:\ddot{\underset{\cdot \cdot }{\mathrm{Cl}}}\cdot +{\mathrm{Mg}}^{\bullet \bullet }+\cdot \ddot{\underset{\cdot \cdot }{\mathrm{Cl}}}:\to [:\ddot{\underset{\cdot \cdot }{\mathrm{Cl}}}:{]}^{-}\mathrm{M}{\mathrm{g}}^{2+}[:\ddot{\underset{\cdot \cdot }{\mathrm{Cl}}}:{]}^{-}$, giving $\mathrm{MgC}{\mathrm{l}}_2$.

   1 mark

Answer: (i) 2Na → 2Na⁺ + 2e⁻; O + 2e⁻ → O²⁻; Na₂O. (ii) Mg → Mg²⁺ + 2e⁻; 2Cl + 2e⁻ → 2Cl⁻; MgCl₂.

1. Type of reaction: thermal decomposition reaction. On heating, the green crystals first lose water and turn white, then break down into a brown solid and gases with the smell of burning sulphur.

   1 mark

2. Balanced equation: $2\mathrm{FeS}{\mathrm{O}}_4(\mathrm{s})\stackrel{\text{heat}}{\to }\mathrm{F}{\mathrm{e}}_2{\mathrm{O}}_3(\mathrm{s})+\mathrm{S}{\mathrm{O}}_2(\mathrm{g})+\mathrm{S}{\mathrm{O}}_3(\mathrm{g})$

   1 mark

3. It is an endothermic reaction, because heat must be absorbed continuously for the decomposition to take place.

   1 mark

Answer: Thermal decomposition; 2FeSO₄(s) → Fe₂O₃(s) + SO₂(g) + SO₃(g) on heating; endothermic.

1. (i) Dissolving an acid in water is highly exothermic. If acid is added slowly to a large amount of water, the heat is spread through the water, so there is no excessive local heating. If water is added to acid instead, the heat produced locally can make the mixture splash out and cause burns, or crack the glass container.

   1 mark

2. (ii) Dry HCl gas does not ionise, so it produces no ${\mathrm{H}}^{+}$ ions and cannot change the colour of litmus. In dilute hydrochloric acid, HCl ionises in water to give ${\mathrm{H}}^{+}$ (as ${\mathrm{H}}_3{\mathrm{O}}^{+}$) ions, which turn blue litmus red.

   1 mark

3. (iii) Blue copper sulphate crystals are $\mathrm{CuS}{\mathrm{O}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$. On heating they lose their 5 molecules of water of crystallisation and become anhydrous $\mathrm{CuS}{\mathrm{O}}_4$, which is white: $\mathrm{CuS}{\mathrm{O}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}\stackrel{\text{heat}}{\to }\mathrm{CuS}{\mathrm{O}}_4+5{\mathrm{H}}_2\mathrm{O}$.

   1 mark

Answer: (i) Avoids excessive local heating and splashing; (ii) Dry HCl gives no H⁺ ions, but dilute HCl does; (iii) Loss of 5 molecules of water of crystallisation.

1. (i) Chlor-alkali process: electricity is passed through an aqueous solution of NaCl (brine). $\mathrm{C}{\mathrm{l}}_2$ gas forms at the anode, ${\mathrm{H}}_2$ gas at the cathode, and NaOH solution near the cathode:
   $2\mathrm{NaCl}(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\stackrel{\text{electricity}}{\to }2\mathrm{NaOH}(\mathrm{aq})+\mathrm{C}{\mathrm{l}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})$

   1 mark

2. (ii) Baking soda is made from sodium chloride, water, ammonia and carbon dioxide:
   $\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}+\mathrm{C}{\mathrm{O}}_2+\mathrm{N}{\mathrm{H}}_3\to \mathrm{N}{\mathrm{H}}_4\mathrm{Cl}+\mathrm{NaHC}{\mathrm{O}}_3$

   1 mark

3. (iii) Plaster of Paris is made by heating gypsum at 373 K:
   $\mathrm{CaS}{\mathrm{O}}_4\cdot 2{\mathrm{H}}_2\mathrm{O}\stackrel{373\text{K}}{\to }\mathrm{CaS}{\mathrm{O}}_4\cdot \frac{1}{2}{\mathrm{H}}_2\mathrm{O}+1\frac{1}{2}{\mathrm{H}}_2\mathrm{O}$

   1 mark

Answer: (i) 2NaCl + 2H₂O → 2NaOH + Cl₂ + H₂ (electrolysis); (ii) NaCl + H₂O + CO₂ + NH₃ → NH₄Cl + NaHCO₃; (iii) CaSO₄·2H₂O → CaSO₄·½H₂O + 1½H₂O at 373 K.

1. (a) During electrolytic refining, the insoluble impurities that settle at the bottom, below the anode, are called anode mud.

   1 mark

2. (b) Anode: a thick rod of impure copper.

   ½ mark

3. Cathode: a thin strip of pure copper.

   ½ mark

4. (c)(i) Roasting: the sulphide ore is partly heated in air:
   $2\mathrm{C}{\mathrm{u}}_2\mathrm{S}(\mathrm{s})+3{\mathrm{O}}_2(\mathrm{g})\stackrel{\text{heat}}{\to }2\mathrm{C}{\mathrm{u}}_2\mathrm{O}(\mathrm{s})+2\mathrm{S}{\mathrm{O}}_2(\mathrm{g})$

   1 mark

5. Self-reduction: the copper(I) oxide then reacts with the remaining $\mathrm{C}{\mathrm{u}}_2\mathrm{S}$:
   $2\mathrm{C}{\mathrm{u}}_2\mathrm{O}(\mathrm{s})+\mathrm{C}{\mathrm{u}}_2\mathrm{S}(\mathrm{s})\stackrel{\text{heat}}{\to }6\mathrm{Cu}(\mathrm{s})+\mathrm{S}{\mathrm{O}}_2(\mathrm{g})$

   1 mark

6. (c)(ii) (Alternative to (c)(i), 2 marks: diagram 1, labels 1.) Draw an electrolytic tank containing acidified copper sulphate solution, with two electrodes dipped in it and connected to a battery through a switch. The thick impure copper rod is joined to the positive terminal and the thin pure copper strip to the negative terminal. Anode mud is shown at the bottom below the anode. Labels: Anode (impure copper), Cathode (pure copper), Acidified copper sulphate solution (electrolyte), Anode mud, Battery / + and − terminals, Tank.

Answer: (a) Insoluble impurities settling below the anode; (b) Anode: impure copper, Cathode: pure copper; (c)(i) 2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂, then 2Cu₂O + Cu₂S → 6Cu + SO₂; (c)(ii) labelled electrolytic refining set-up.

1. (i) A homologous series is a series of compounds in which the same functional group replaces hydrogen in a carbon chain. Members have the same general formula, and successive members differ by a $-\mathrm{C}{\mathrm{H}}_2-$ unit (14 u).

   1 mark

2. ${\mathrm{C}}_2{\mathrm{H}}_4$ and ${\mathrm{C}}_4{\mathrm{H}}_8$ belong to the same homologous series. Both fit ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n}$ (alkenes), whereas ${\mathrm{C}}_4{\mathrm{H}}_{10}$ is an alkane and ${\mathrm{C}}_3{\mathrm{H}}_4$ is an alkyne.

   1 mark

3. (ii)(I) Esterification: $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}+\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}\stackrel{\text{Acid}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+{\mathrm{H}}_2\mathrm{O}$ (products: ethyl ethanoate, an ester, and water).

   1 mark

4. (ii)(II) Combustion: ${\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}+3{\mathrm{O}}_2\stackrel{\text{Burning}}{\to }2\mathrm{C}{\mathrm{O}}_2+3{\mathrm{H}}_2\mathrm{O}$ (with heat and light).

   1 mark

5. (iii) A soap molecule has an ionic (hydrophilic) end that interacts with water and a long hydrocarbon (hydrophobic) tail that avoids water and interacts with oil or dirt. In water, the molecules therefore cluster with their tails pointing inward and their ionic ends facing outward towards the water. This cluster is a micelle, and it traps oily dirt at its centre.

   1 mark

Answer: (i) Series with the same functional group/general formula, successive members differing by CH₂; C₂H₄ and C₄H₈. (ii)(I) CH₃COOC₂H₅ + H₂O; (II) 2CO₂ + 3H₂O. (iii) The ionic end faces water and the hydrocarbon tail faces oil, so a micelle forms.

1. (i)(I) $\mathrm{C}{\mathrm{H}}_3\mathrm{COC}{\mathrm{H}}_3$ is propanone (acetone), a ketone with 3 carbon atoms.

   ½ mark

2. (i)(II) $\mathrm{C}{\mathrm{H}}_3\mathrm{C}\equiv \mathrm{CH}$ is propyne, an alkyne with 3 carbon atoms.

   ½ mark

3. (ii) Hydrogenation, an addition reaction: hydrogen is added to unsaturated vegetable oils in the presence of a nickel catalyst, converting them into saturated fats.

   1 mark

4. (iii) ${\mathrm{C}}_4{\mathrm{H}}_{10}$ (butane, 58 u) has a higher molecular mass than ${\mathrm{C}}_3{\mathrm{H}}_8$ (propane, 44 u). In a homologous series, melting point increases with increasing molecular mass, so butane has the higher melting point.

   1 mark

5. (iv) Sodium reacts with ethanol to form sodium ethoxide, and hydrogen gas is evolved with effervescence:
   $2\mathrm{Na}+2\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{OH}\to 2\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{ONa}+{\mathrm{H}}_2$

   1 mark

6. (v) Saponification: $\mathrm{C}{\mathrm{H}}_3\mathrm{COO}{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{NaOH}\to \mathrm{C}{\mathrm{H}}_3\mathrm{COONa}+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$. The products are sodium ethanoate and ethanol.

   1 mark

Answer: (i) Propanone, Propyne; (ii) Hydrogenation (addition); (iii) Higher molecular mass; (iv) Sodium ethoxide + H₂ gas; (v) CH₃COONa + C₂H₅OH

1. An ammeter must carry the full circuit current, so it is connected in series. A voltmeter measures the potential difference across two points, so it is connected in parallel. Option (C).

   1 mark

Answer: (C)

1. Red light deviates least and violet light deviates most in a prism. Ray 1 is less deviated, so it is red; ray 2 bends most towards the base, so it is violet. Option (C).

   1 mark

Answer: (C) Ray 1 Red, Ray 2 Violet

1. In an external magnetic field, the needle aligns along that field, not necessarily along the Earth's North–South direction. So A is false.

2. The compass needle is a small magnet, so it behaves like a freely suspended bar magnet. R is true. Option (D).

   1 mark

Answer: (D)

1. (a) Medium A is optically denser. The ray going from A into B bends away from the normal NN′, and light bends away from the normal only when it passes from a denser to a rarer medium.

   1 mark

2. (b) Refractive index of B with respect to A: $n_{BA}=\frac{\text{speed of light in A}}{\text{speed of light in B}}=\frac{V_A}{V_B}$

   1 mark

Answer: (a) Medium A, since the ray bends away from the normal on entering B; (b) $n_{BA}=V_A/V_B$

1. (i) Very fine particles scatter mainly blue light, because very small particles scatter light of shorter wavelengths more than longer wavelengths. This is why the clear sky looks blue.

   1 mark

2. (ii) Very large particles make the scattered light look white, because particles of large size scatter light of all wavelengths almost equally. This is why clouds look white.

   1 mark

Answer: (i) Mainly blue light is scattered (shorter wavelengths are scattered more); (ii) Scattered light appears white (all wavelengths are scattered).

1. Phenomenon: formation of a rainbow. A rainbow is a natural spectrum seen in the sky after a shower, always in the direction opposite to the Sun. Here the tiny water droplets in the atmosphere act like small prisms. Sunlight entering a droplet is refracted and dispersed into its colours. It then undergoes internal reflection at the back surface of the droplet. It is refracted again as it comes out, and the colours spread further apart so that they reach the observer's eye as a band from red to violet.

   1 mark

2. Ray diagram (described):

   • Draw a circular raindrop with a ray of white sunlight entering its upper part. Mark refraction at entry, with the ray splitting into red and violet rays (dispersion).
   • Show both rays striking the back surface of the drop and undergoing internal reflection.
   • Show both rays refracting out of the lower front part of the drop towards the observer.
   • Label: sunlight, raindrop, refraction and dispersion, internal reflection, refraction, red, violet.
   1 mark

Answer: Rainbow: sunlight is refracted and dispersed, internally reflected and refracted again by raindrops (with a ray diagram of a raindrop).

1. (a) Light from the near point N′ converges at a point beyond the retina, so the defect is hypermetropia (far-sightedness).

   ½ mark

2. (a) In hypermetropia the image of a nearby object is formed behind the retina.

   ½ mark

3. (b) Cause 1: the focal length of the eye lens is too long, i.e. the lens has too little converging power.

   ½ mark

4. (b) Cause 2: the eyeball has become too small, i.e. the distance from the lens to the retina is too short.

   ½ mark

5. (c) Correction with a convex (converging) lens of suitable power.
   Diagram (described): place a convex lens in front of the eye. Rays from an object at N (the normal near point, 25 cm) pass through the convex lens. They emerge less divergent, as if coming from the hypermetropic near point N′. The eye lens then focuses them exactly on the retina.
   Label: N, N′, convex lens, eye lens, retina, image on retina.

   1 mark

Answer: (a) Hypermetropia; image formed behind the retina. (b) Focal length of eye lens too long; eyeball too small. (c) Corrected by a convex lens that makes rays from N appear to come from N′, so the image forms on the retina.

1. (a)(i) A battery / electric cell maintains a constant potential difference in a circuit.

   ½ mark

2. (a)(ii) A rheostat (variable resistance) is used to change the current in a circuit.

   ½ mark

3. (b) By Ohm's law, the resistance of the heater is $R=\frac{V}{I}$.

   ½ mark

4. $R=\frac{110\text{V}}{5\text{A}}=22\Omega$

   ½ mark

5. The resistance stays the same, so the new current is $I^{\prime }=\frac{V^{\prime }}{R}$.

   ½ mark

6. $I^{\prime }=\frac{220\text{V}}{22\Omega }=10\text{A}$
   (Equivalently: $V\propto I$, so doubling the potential difference doubles the current, from 5 A to 10 A.)

   ½ mark

Answer: (a)(i) Battery/cell; (ii) Rheostat. (b) R = 22 Ω, current = 10 A

1. (a) Between R and S: the 4 Ω and 1 Ω resistors are in series, giving $4+1=5\Omega$. This 5 Ω is in parallel with the other 5 Ω:
   $\frac{1}{R_1}=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$, so $R_1=\frac{5}{2}\Omega$.

   Between P and Q: the 2 Ω and 3 Ω resistors are in series, giving $2+3=5\Omega$. This 5 Ω is in parallel with the other 5 Ω:
   $R_2=\frac{5}{2}\Omega$.

   The two groups are in series:
   $R=R_1+R_2=\frac{5}{2}+\frac{5}{2}=5\Omega$.

   1 mark

2. (b) $I=\frac{V}{R}=\frac{10\text{V}}{5\Omega }=2\text{A}$

   1 mark

3. (c) The whole current of 2 A flows through the P–Q combination, so
   $V_{PQ}=I{R}_2=2\text{A}\times \frac{5}{2}\Omega =5\text{V}$.

   1 mark

Answer: (a) 5 Ω; (b) 2 A; (c) 5 V

1. (a) In observation 3, $|u|=|v|=50$ cm. For a convex lens the image distance equals the object distance only when the object is at $2F_1$, so $2f=50$ cm and $f=25$ cm.

   1 mark

2. (b) Observation 6 is not correct.

   ½ mark

3. (b) Reason: with $u=-15$ cm the object lies between the optical centre O and $F_1$ (15 cm < 25 cm). The image is then virtual and on the same side as the object, so $v$ must be negative. The table gives $+37.5$ cm.

   ½ mark

4. (c)(i) Magnification is $m=\frac{v}{u}=\frac{+150}{-30}$.

   ½ mark

5. $m=-5$

   ½ mark

6. Nature of image: real and inverted (and 5 times magnified).

   ½ mark

7. Reason: the magnification is negative, which means the image is inverted, and an inverted image formed by a lens is real.

   ½ mark

8. (c)(ii) [alternative] Ray diagram (described):

   • Draw the principal axis and a convex lens at O. Mark $F_1$ and $2F_1$ (25 cm and 50 cm) on the left, and $F_2$ and $2F_2$ on the right.
   • Place an upright object AB at 30 cm, between $F_1$ and $2F_1$.
   • Ray 1: from A parallel to the axis, refracting through $F_2$.
   • Ray 2: from A through O, going straight undeviated.
   • The rays meet beyond $2F_2$ at 150 cm, forming the image A′B′, which is real, inverted and enlarged.
   • Arrows must show the direction of the rays.
   2 marks

Answer: (a) f = 25 cm (object at 2F, u = v = 50 cm). (b) Observation 6; object within focus, so v should be negative. (c)(i) m = −5; real and inverted, since m is negative. (c)(ii) Object between F₁ and 2F₁ gives a real, inverted, enlarged image beyond 2F₂.

1. (i) A solenoid is a coil of many circular turns of insulated copper wire wound closely in the shape of a cylinder.

   1 mark

2. (i) Diagram (described): draw a cylindrical coil connected to a battery, with the current direction marked in the turns.

   • Inside: draw parallel, equally spaced straight field lines along the axis, showing a uniform field.
   • Outside: draw closed lines running from one end to the other, like a bar magnet's field.
   • Mark the N end (where the lines emerge) and the S end (where they enter), with arrows from N to S outside and from S to N inside.
   2 marks

3. (ii) Factor 1: the magnitude of the current flowing through the solenoid.

   ½ mark

4. (ii) Factor 2: the number of turns of the coil.

   ½ mark

5. (iii) Field around a straight current-carrying wire (described): draw a vertical straight wire passing through a horizontal card, with the current marked upward. Around the wire on the card, draw concentric circles centred on it, with the spacing increasing outward. For upward current, mark the circles with arrows anticlockwise as seen from above, by the right-hand thumb rule. Any current direction is acceptable if the field direction matches it.

   1 mark

Answer: (i) Closely wound cylindrical coil of insulated copper wire; uniform parallel field lines inside, bar-magnet-like field outside. (ii) Current and number of turns. (iii) Concentric circles around the wire, with direction by the right-hand thumb rule.

1. (i) Activity: suspend a small aluminium rod horizontally by two connecting wires between the poles of a horseshoe magnet, and connect it in series with a battery, a key and a rheostat. When the key is closed and current passes, the rod is displaced. This shows that a current-carrying conductor in a magnetic field experiences a force. Reversing the current reverses the displacement.

   1 mark

2. (i) Diagram (described): draw a horseshoe magnet with its N pole above and S pole below, and the rod AB lying between the poles, joined to the circuit.

   ½ mark

3. Mark the current from A to B along the rod (for example, into the page).

   ½ mark

4. Mark the magnetic field vertically downward, from N to S.

   ½ mark

5. Mark the force on the rod, perpendicular to both the field and the current (here, horizontally, as given by Fleming's left-hand rule).

   ½ mark

6. (ii)(I) The electron moves upward, so the conventional current is downward. The field points to the left. By Fleming's left-hand rule (forefinger to the left for the field, middle finger downward for the current), the thumb points into the plane of the paper. So the force on the electron is into the plane of the paper.

   1 mark

7. (ii)(II) Reversing the field alone would reverse the force, and reversing the motion alone would also reverse it. Reversing both cancels out, so the direction of the force remains the same (into the plane of the paper).

   1 mark

Answer: (i) The rod is displaced when current flows, so a current-carrying conductor in a field experiences a force perpendicular to both field and current. (ii)(I) Into the plane of the paper; (II) the direction of force remains the same.