CBSE Class 10 Science Sample Paper 2: marking scheme and solutions
Every answer with its marking scheme, the way an examiner awards the marks. Chapters in this paper: 1 Chemical Reactions and Equations; 2 Acids, Bases and Salts; 3 Metals and Non-metals; 4 Carbon and its Compounds; 5 Life Processes; 6 Control and Coordination; 7 How do Organisms Reproduce?; 8 Heredity; 9 Light: Reflection and Refraction; 10 The Human Eye and the Colourful World; 11 Electricity; 12 Magnetic Effects of Electric Current; 13 Our Environment.
Section A (20 × 1 mark)
Answer: (a) Herbivores eating grass need a longer small intestine to digest cellulose; meat is easier to digest, so carnivores have a shorter one.
Answer: (d) V = W/Q = 360 J / 30 C = 12 J/C = 12 V. Multiplying (10800) or inverting the ratio (0.083) are the usual slips.
Answer: (b) Covalent molecules have strong bonds within the molecule but only weak forces between molecules, so little energy separates them and they melt low. Wax has no ions, and it is made of C–H chains.
Answer: (a) Both are true, and the copying errors in R are exactly what makes the four cells slightly different.
Answer: (d) Jasmine, like banana, orange and rose, has lost the capacity to produce seeds and is propagated vegetatively. Mustard, papaya and watermelon all form flowers that give seeds.
Answer: (a) Respiration is exothermic but evolves no light. Burning magnesium and natural gas give both heat and light; photosynthesis absorbs energy.
Answer: (b) Both are true. Gold occurs free because it is at the bottom of the activity series and least reactive; its ductility has nothing to do with that, so R does not explain A.
Answer: (b) DNA is the information source for making proteins, and the section of DNA that carries the information for one protein is the gene for that protein. A chromosome is a whole piece of DNA carrying many genes.
Answer: (a) n = c/v, so a higher refractive index means a lower speed. Crown glass (1.52) has a higher refractive index than water (1.33), so light is slower in glass and faster in water. R explains A.
Answer: (b) Autotrophs capture the energy of sunlight and convert it into chemical energy stored in food.
Answer: (c) Sexual reproduction generates greater diversity than asexual reproduction. The goat kids come from sexual reproduction; the other three come from a single parent asexually and are very similar.
Answer: (d) The field lines are concentric circles centred on the wire. P and Q lie on the same circle, at opposite ends of a diameter, so the field has the same size there but the tangents to the circle point in opposite directions.
Answer: (d) A normal eye sees clearly from its near point, about 25 cm, out to infinity. The coin at 10 cm is nearer than the near point; the page, the bus and the Moon all lie within the range.
Answer: (b) C is at 2f = 20 cm, so the bulb at 35 cm is beyond C. With the object beyond C a concave mirror forms the image between F and C: diminished, real and inverted.
Answer: (c) Non-metals are solids or gases, the one exception being bromine, which is a liquid. Mercury is a liquid but a metal; iodine and sulphur are solids.
Answer: (b) Both are true. The body is earthed because the earth wire gives a low-resistance path that carries away any leakage of current, keeping the body at the earth's potential; the colour of the insulation does not explain this.
Answer: (c) The direction of the field at a point is the direction in which the north pole of a compass placed there points: east. West is where the south pole points.
Answer: (c) Iodine is a non-metal but it is lustrous. Sulphur is a dull solid, bromine a liquid and oxygen a gas.
Answer: (b) Phenolphthalein is pink in a basic solution and colourless in an acidic one. Only sodium hydroxide solution is basic; the other three are acidic.
Answer: (b) At the axon end the impulse releases chemicals which cross the synapse and start a similar impulse in the next neuron's dendrite. With no chemicals released, the impulse still reaches the axon end but goes no further.
Section B (6 × 2 marks)
The brain sits inside a bony box, and within it is contained in a fluid-filled balloon that absorbs shocks. The spinal cord is protected by the vertebral column, or backbone.
- brain: bony box and fluid-filled balloon (shock absorption) (1)
- spinal cord: vertebral column (backbone) (1)
(a) Chlorine: 3 on the right in FeCl₃, 2 in Cl₂; the lowest common number is 6, so 3Cl₂ and 2FeCl₃, then 2Fe: 2Fe + 3Cl₂ → 2FeCl₃. (b) Chlorine: 3 in AlCl₃, so 6HCl with 2AlCl₃; hydrogen 6 on the left needs 3H₂; aluminium 2: 2Al + 6HCl → 2AlCl₃ + 3H₂.
- 2Fe + 3Cl₂ → 2FeCl₃ (1)
- 2Al + 6HCl → 2AlCl₃ + 3H₂ (1)
(a) Oxygen: 5 in P₂O₅ and 2 in O₂; the lowest common number is 10, so 5O₂ and 2P₂O₅, then 4P: 4P + 5O₂ → 2P₂O₅. (b) Oxygen: 3 in KClO₃, 2 in O₂; 6 atoms means 2KClO₃ and 3O₂, then 2KCl: 2KClO₃ → 2KCl + 3O₂.
- 4P + 5O₂ → 2P₂O₅ (1)
- 2KClO₃ → 2KCl + 3O₂ (1)
X is ethene. CH₃–CH₂OH → (hot conc. H₂SO₄, 443 K) CH₂=CH₂ + H₂O. The concentrated sulphuric acid acts as a dehydrating agent, removing water from ethanol.
- X is ethene, with the equation CH₃CH₂OH → CH₂=CH₂ + H₂O (conditions shown) (1)
- concentrated sulphuric acid is the dehydrating agent (1)
The two wires touching make a short circuit: the current no longer passes through the heater, and it rises sharply. This large current through the fuse wire, which is in series, raises its temperature until it melts and breaks the circuit, before the appliances are damaged.
- short circuit: the current rises sharply (1)
- the large current heats the fuse wire until it melts and breaks the circuit (1)
(a) They filter the air entering the body. (b) The air retained during the breathing cycle gives enough time for oxygen to be absorbed and carbon dioxide to be released.
- (a) filter the incoming air (1)
- (b) allows sufficient time for oxygen absorption and carbon dioxide release (1)
The retina. On illumination its light-sensitive cells generate electrical signals, which travel through the optic nerves to the brain; the brain interprets them and we perceive the object.
- names the retina (1)
- electrical signals travel by the optic nerves to the brain, which interprets them (1)
Section C (7 × 3 marks)
(a) 2NaHCO₃ + H₂SO₄ → Na₂SO₄ + 2H₂O + 2CO₂. (b) Carbon dioxide. Passed through lime water, it turns it milky, because insoluble calcium carbonate forms: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O. (c) The carbon dioxide smothers the fire.
- balanced equation with Na₂SO₄, H₂O and CO₂ (1)
- names CO₂ and the lime water test (1)
- CO₂ smothers the fire (1)
(a) A coil of n turns gives n times the field of a single turn, so the field at the centre of Y is 100/25 = 4 times that at the centre of X. (b) The current in each turn flows in the same direction, so every turn, and every section of each turn, contributes field in the same direction inside the coil; the fields add up. (c) Near the centre the circles round each section of wire have grown so large that the field lines appear as straight lines.
- Y gives four times the field of X (1)
- current in every turn in the same direction, so fields add (1)
- lines appear straight near the centre (1)
(i) Geotropism: the shoot grows away from the earth, against the pull of gravity. (ii) Phototropism: the roots bend away from light. (iii) Chemotropism: the pollen tube grows towards the ovule.
- (i) geotropism, shoot away from gravity (1)
- (ii) phototropism, roots away from light (1)
- (iii) chemotropism, towards the ovule (1)
(a) Exothermic: CH₄ + 2O₂ → CO₂ + 2H₂O gives out heat. (b) Endothermic: the mixture absorbs heat from the beaker, which turns cold. (c) Endothermic: decomposition needs a continuous supply of heat to break the calcium carbonate down.
- (a) exothermic, heat given out (1)
- (b) endothermic, heat absorbed (1)
- (c) endothermic, energy needed to decompose (1)
At the first face the ray goes from air into glass, rarer to denser, and bends towards the normal (angle of refraction less than 40°). At the second face it goes from glass into air and bends away from the normal. The bending at the two parallel faces is equal and opposite, so it emerges at 40°, parallel to the incident ray. The sideways shift is the lateral displacement.
- towards the normal at the first face, away at the second (1)
- emerges at 40°, parallel to the incident ray (1)
- names lateral displacement (1)
(i) One germ-cell is large and contains food stores; the other is smaller and likely to be motile. (ii) The motile one is the male gamete; the one with stored food is the female gamete. (iii) In very simple organisms the two germ-cells may be similar to each other.
- one large with food stores, one small and motile (1)
- motile = male, food-storing = female (1)
- in simple organisms the two germ-cells may be similar (1)
(a) About 10% passes at each step: 30 000 → 3000 kJ (zooplankton) → 300 kJ (small fish) → 30 kJ (kingfisher). (b) Energy diminishes at every level, so only a little reaches the kingfisher; it supports only a small population, and numbers fall at each step up the chain.
- uses 10% at each step (1)
- 30 kJ for the kingfisher (1)
- little energy reaches the top, so few individuals (1)
Section D (3 × 5 marks)
(a) Sepals protect the flower while it is a bud; petals are usually coloured and attract the insects and birds that carry pollen. (b) The stamen, which produces yellowish pollen grains. (c) Pistil: ovary (swollen base), style (middle, elongated), stigma (tip, may be sticky). The ovary contains ovules, each with an egg cell. (d) The zygote divides to form an embryo; the ovule develops a tough coat and becomes a seed; the ovary ripens into a fruit.
- sepals protect the bud, petals attract pollinators (1)
- stamen produces pollen grains (1)
- ovary, style, stigma in order (1)
- ovary holds ovules, each with an egg (1)
- zygote → embryo, ovule → seed, ovary → fruit (1)
(a) Sepals protected the flower in the bud; petals, coloured, attracted insects and birds that carry pollen. (b) The stamens, which produce yellowish pollen grains. (c) The pistil: ovary, style and stigma; the ovary ripened into the pod (fruit). (d) The male germ-cell from the pollen grain fused with the egg in the ovule, giving a zygote; the zygote divided into an embryo, and the ovule with a tough coat became the seed, holding the embryo.
- sepals protect the bud, petals attract pollinators (1)
- pollen from the stamens (1)
- ovary, style, stigma named (1)
- ovary became the pod (fruit) (1)
- male germ-cell fuses with the egg in the ovule; seed holds the embryo (1)
(a) The pH scale measures the hydrogen ion concentration of a solution; it generally runs from 0 (very acidic) to 14 (very alkaline). (b) Q, pH 11, is basic (pH greater than 7), so it turns phenolphthalein pink. P, pH 3, has the higher hydrogen ion concentration, because a lower pH means more H⁺ ions. (c) Bacteria in the mouth degrade sugar and food particles to acids. When the pH falls below 5.5, tooth enamel (calcium hydroxyapatite) is corroded. Brushing with a basic toothpaste after eating neutralises the acid.
- (a) hydrogen ion concentration, 0 to 14 (1)
- (b) Q turns phenolphthalein pink, being basic (1)
- (b) P has more H⁺, lower pH (1)
- (c) bacterial acid corrodes enamel below pH 5.5 (1)
- (c) basic toothpaste neutralises the acid (1)
(a) 'p' stands for potenz, the German word for power; pH measures the hydrogen ion concentration of a solution. (b) R is more acidic: the lower pH means a higher hydrogen ion concentration. R is acidic (pH below 7), so it turns blue litmus red. (c) Enamel is calcium hydroxyapatite, a crystalline form of calcium phosphate. Bacteria degrade sugar left in the mouth into acids; below pH 5.5 the enamel is corroded. Toothpaste is basic and neutralises the acid.
- (a) potenz, power; measures H⁺ concentration (1)
- (b) R more acidic, lower pH means more H⁺ (1)
- (b) R turns blue litmus red (1)
- (c) calcium hydroxyapatite; bacterial acid corrodes it below pH 5.5 (1)
- (c) basic toothpaste neutralises the acid (1)
(a) V = IR, so P = VI = (IR)I = I²R; and I = V/R, so P = V(V/R) = V²/R. (b) Geyser: 2 kW × 0.75 h × 30 = 45 kW h = 45 units. Instant heater: 3 kW × 0.2 h × 30 = 18 units. Saving = 27 units × ₹7 = ₹189. (c) I = 3000/230 ≈ 13 A, so the 15 A fuse, the next rating above the working current.
- P = I²R from V = IR (1)
- P = V²/R from I = V/R (1)
- 45 units and 18 units (1)
- saving 27 units = ₹189 (1)
- 13 A, so the 15 A fuse (1)
(a) P = VI with V = IR gives P = I²R; with I = V/R gives P = V²/R. (b) Iron: 1.1 kW × 2 h × 25 = 55 units. Motors: 2 × 0.1 kW × 6 h × 25 = 30 units. Total 85 units × ₹8 = ₹680. (c) I = 1100/230 ≈ 4.8 A, so the 5 A fuse.
- P = I²R from V = IR (1)
- P = V²/R from I = V/R (1)
- 55 + 30 = 85 units (1)
- bill ₹680 (1)
- 4.8 A, so the 5 A fuse (1)
Section E (3 × 4 marks)
Answer: (b) 22 carat is 22 parts gold in 24: 36 × 22/24 = 33 g. 22 g takes the carat number as grams; 34 g subtracts 2 g instead of 2 parts in 24.
Silver or copper, 2 parts to 22 parts of gold, because pure 24-carat gold is too soft for jewellery.
- silver or copper, because pure gold is too soft (1)
Solder is an alloy of lead and tin; it has a low melting point. Brass, an alloy of copper and zinc, is not a good conductor, as the conductivity of an alloy is lower than that of the pure metal.
- lead and tin, low melting point (1)
- brass a poorer conductor than pure copper (1)
Bronze is an alloy of copper and tin. Both its melting point and its electrical conductivity are lower than those of pure copper, which is why bronze is not used as a conductor.
- copper and tin (1)
- melting point and conductivity both lower (1)
Answer: (a) Right-hand thumb rule, thumb up along the current: seen from above the field lines go anticlockwise, so at P, which is east of the cable, the field points north.
Answer: (a) The field decreases as the distance from the conductor increases; Q is farther from the cable than P.
Reversing the current reverses the field, so the north pole at P now points south. Doubling the current makes the field at P larger, so the cable's effect on the needle is stronger.
- needle at P turns to point south, as the field reverses (1)
- larger current gives a stronger field at P (1)
Right-hand thumb rule: hold the cable in the right hand with the thumb along the current; the fingers wrap round in the direction of the field lines. With the thumb pointing up, out of the shelf, the fingers curl anticlockwise as seen from above.
- right-hand thumb rule stated (1)
- anticlockwise seen from above (1)
| F2 class | Number of plants |
|---|---|
| Tall, round seeds | 452 |
| Tall, wrinkled seeds | 148 |
| Short, round seeds | 152 |
| Short, wrinkled seeds | 48 |
| Total | 800 |
Answer: (b) The parents were tall-round and short-wrinkled. The new combinations are tall-wrinkled and short-round: 148 + 152 = 300.
Answer: (b) Each F1 plant got T and R from the tall-round parent and t and r from the short-wrinkled parent: Tt Rr.
New combinations show that height and seed shape are inherited independently of each other. If a whole gene set were inherited from each parent, T would always travel with R and t with r, the traits would be linked, and only tall-round and short-wrinkled plants could appear.
- height and seed shape inherited independently (1)
- whole sets would link T with R and t with r (1)
800/16 = 50, so the expected numbers are 9 × 50 = 450 tall-round, 3 × 50 = 150 tall-wrinkled, 150 short-round and 50 short-wrinkled. The counts 452, 148, 152, 48 are very close, so they fit the 9 : 3 : 3 : 1 ratio.
- 450, 150, 150, 50 from 800/16 (1)
- counts close, so they fit (1)