CBSE Class 10 Science 2025 Question Paper (Set 31/3/1) with Solutions

1. Balance aluminium: $r = 1$ gives $2$ Al on the right, so $p = 2$. Balance oxygen: $\mathrm{Al_2O_3}$ has $3$ O, so $q = 3$. Then $3\mathrm{H_2O}$ has $6$ H, so $s = 3$.
2. $\mathrm{2Al + 3H_2O \rightarrow Al_2O_3 + 3H_2}$, so $p, q, r, s = 2, 3, 1, 3$.

Answer: (C) 2, 3, 1, 3

1. Salts belong to one family when they have the same negative ion (the same acid radical) or the same positive ion.
2. In option B all three are sulphates (they contain $\mathrm{SO_4^{2-}}$), so they are the sulphate family.

Answer: (B) $\mathrm{K_2SO_4, Na_2SO_4, CaSO_4}$

1. Most metals in the middle of the reactivity series (zinc, iron, lead and so on) are obtained by heating their oxides with carbon (coke), which is cheap and takes the oxygen away.

Answer: (A) Reduction with carbon

1. Related structures here means isomers: same molecular formula, different structure. (ii) is $\mathrm{CH_3CH_2CH_2CH_3}$ and (iii) is $\mathrm{CH_3CH(CH_3)CH_3}$. Both are $\mathrm{C_4H_{10}}$, one a straight chain and one branched.
2. (i) is $\mathrm{C_4H_8}$ and (iv) contains chlorine, so neither is an isomer of the others.

Answer: (C) (ii) and (iii)

1. Hydrogenation adds hydrogen across the double bonds, so it is an addition reaction. It needs a nickel or palladium catalyst, and the product is saturated (only single C–C bonds).
2. An acid catalyst is not used, so statement D is the incorrect one.

Answer: (D) It is an addition reaction which occurs in the presence of an acid catalyst.

1. a. Fertilisation takes place in the oviduct: (iii). b. The embryo is implanted in the uterus: (ii). c. Sperm enter through the vagina: (i). d. Wastes of the embryo are removed through the placenta: (iv).

Answer: (D) a-(iii), b-(ii), c-(i), d-(iv)

1. The cerebellum, part of the hind-brain, keeps the posture and balance of the body and coordinates voluntary movements.

Answer: (C) Cerebellum

1. Cytokinins promote cell division, so they are found in high concentration in fruits, seeds and other areas of rapid cell division.

Answer: (B) Cytokinins

1. A bisexual flower has both stamens and pistil. Hibiscus and mustard are bisexual. Papaya and watermelon have unisexual flowers.

Answer: (B) Hibiscus and mustard

1. Gastric glands release hydrochloric acid, the protein-digesting enzyme pepsin and mucus (which protects the stomach lining). Trypsin comes from the pancreas and amylase from the salivary glands.

Answer: (C) Mucus and Pepsin

1. $n = \dfrac{c}{v}$, so $c = n_g v_g = \dfrac{3}{2} \times 2 \times 10^8 = 3 \times 10^8$ m/s. Then $v_w = \dfrac{c}{n_w} = \dfrac{3 \times 10^8}{4/3} = \dfrac{9}{4} \times 10^8$ m/s $= 2.25 \times 10^8$ m/s.
2. The printed options leave out the factor $10^8$, so the question is faulty. CBSE’s key awards the mark to all candidates. The nearest option to the correct digits is (A), $\dfrac{9}{4}$.

Answer: (A) $\dfrac{9}{4}$ (times $10^8$ m/s); CBSE gave full credit to everyone

About CBSE’s key: CBSE’s marking scheme awards this mark to all candidates because the options omit the factor of 10⁸.

1. Very fine particles scatter light of short wavelength the most. Blue has a shorter wavelength than red, orange and yellow, so it is scattered the most. (This is why the sky looks blue.)

Answer: (C) blue

1. The volume stays the same: $A \times l = A' \times 3l$, so the area falls to $A' = \dfrac{A}{3}$.
2. $R' = \rho\dfrac{3l}{A/3} = 9\rho\dfrac{l}{A} = 9R$.

Answer: (C) 9R

Common mistake: Answering 3R. The wire is also thinner, which triples the resistance again.

1. Field lines are closed loops: outside the magnet they run from north to south, and inside the magnet they run from the south pole to the north pole. So statement D is not true.

Answer: (D) The direction of magnetic field lines inside a bar magnet is from its north pole to its south pole.

1. The field inside a long solenoid depends on the current, the number of turns per unit length and the core material. It is uniform and does not depend on the radius of the coil. (The direction of the current only decides which end is north.)

Answer: (D) radius of the coil of the solenoid

Common mistake: Choosing B. Reversing the current reverses the direction of the field, but its strength is unchanged.

1. A self-sustaining aquarium needs aquatic producers, the small herbivorous fish, and decomposers. Terrestrial plants and animals do not live in water, so (ii) is not needed.
2. Marine consumers such as clown fish and sea urchins are not required either, since the herbivorous fish are already the consumers: (iv) is not needed.

Answer: (D) (ii) and (iv)

1. Assertion: true. Nitric acid is a strong oxidising agent, so hydrogen does not escape (except with very dilute acid and magnesium or manganese).
2. Reason: false. Nitric acid is a strong oxidising agent, not a reducing agent; it oxidises the hydrogen to water.

Answer: (C) A is true, but R is false

1. Assertion: true, the brain sends signals to the muscles through nerves. Reason: true, the nerves from the brain and spinal cord form the peripheral nervous system. But the reason only names parts of the system and does not explain why the nervous system works with the muscles.

Answer: (B) Both A and R are true, but R is not the correct explanation of A

1. Both statements are true. But lines do not cross because at a point the field can have only one direction. Merely being a vector does not by itself explain this, so R is not the correct explanation.

Answer: (B) Both A and R are true, but R is not the correct explanation of A

1. Jute breaks down by decomposers and the bag can be used again, so it does not pile up as non-biodegradable waste like plastic bags. The reason explains the assertion.

Answer: (A) Both A and R are true and R is the correct explanation of A

1. (a) Heat, light and electricity. Example (light): $\mathrm{2AgBr \xrightarrow{sunlight} 2Ag + Br_2}$.
2. (b) Green crystals first lose their water of crystallisation: water droplets appear on the cooler upper part of the tube and the colour changes from green to white. $\mathrm{FeSO_4{\cdot}7H_2O \rightarrow FeSO_4 + 7H_2O}$.
3. On stronger heating the solid turns brown (iron(III) oxide) and gives off gases with the smell of burning sulphur: $\mathrm{2FeSO_4 \xrightarrow{heat} Fe_2O_3 + SO_2 + SO_3}$.

Answer: (a) heat, light, electricity; e.g. $2\mathrm{AgBr} \rightarrow 2\mathrm{Ag} + \mathrm{Br_2}$   (b) green $\to$ white, then brown; pungent $\mathrm{SO_2}$

1. (a) (i) Acids give hydrogen ions, $\mathrm{H^+}$ (in water $\mathrm{H_3O^+}$). (ii) Bases give hydroxide ions, $\mathrm{OH^-}$.
2. (b) Dry HCl does not form hydrogen ions in the absence of water. Only the H⁺ ions change the colour of litmus, so dry HCl gas has no effect on dry litmus paper.

Answer: (a) $\mathrm{H^+}$ and $\mathrm{OH^-}$   (b) no water, so no $\mathrm{H^+}$ ions are released

1. Veins carry deoxygenated blood from the different organs and bring it back to the heart (the pulmonary vein is the exception).
2. They do not need thick walls because the blood flows through them at low pressure.

Answer: Bring blood back to the heart; blood in veins is at low pressure

1. (a) Tallness depends on the amount of a plant hormone (a growth hormone). Making this hormone depends on how efficiently an enzyme (a protein) works, and the enzyme is made from a specific section of DNA. A more efficient enzyme gives more hormone, so the plant is taller.
2. (b) A gene.

Answer: The gene makes the enzyme, which makes the growth hormone, which decides tallness; the section of DNA is a gene

1. (a) He sees far objects clearly but not near ones, so he has hypermetropia (far-sightedness). Reasons: the focal length of the eye lens is too long, or the eyeball has become too small. It is corrected with a convex (converging) lens.
2. (b) The ray bends towards the normal on entering the prism and away from the normal on leaving, so it is deviated towards the base. The angle between the original direction and the emergent ray is the angle of deviation (diagram above).

Answer: (a) hypermetropia; long focal length or small eyeball; convex lens   (b) diagram

1. $P = \dfrac{V^2}{R}$, so $R = \dfrac{V^2}{P} = \dfrac{230^2}{1000} = \dfrac{52900}{1000} = 52.9\ \Omega$.

Answer: $52.9\ \Omega$

1. (a) The reactivity series is a list of metals arranged in decreasing order of their reactivity. It is developed by performing displacement reactions: a more reactive metal displaces a less reactive one from its salt solution.
2. In decreasing order of reactivity: calcium, aluminium, lead, copper.
3. (b) $\mathrm{Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe + heat}$ (the thermite reaction).

Answer: (a) Ca, Al, Pb, Cu   (b) $\mathrm{Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe}$ + heat

1. (a)(i) Electricity is passed through brine (aqueous sodium chloride), the chlor-alkali process: $\mathrm{2NaCl + 2H_2O \rightarrow 2NaOH + Cl_2 + H_2}$.
2. (ii) Ammonia and carbon dioxide are passed through brine: $\mathrm{NaCl + H_2O + CO_2 + NH_3 \rightarrow NaHCO_3 + NH_4Cl}$.
3. (b) Set up a circuit with a bulb, a battery and two electrodes dipped in a beaker. Put solutions of glucose and alcohol in the beaker one at a time, and then dilute acid solutions (such as HCl). The bulb does not glow with glucose or alcohol, but glows with the acids.
4. Reason: acids release hydrogen ions in solution, so they conduct electricity. Glucose and alcohol contain hydrogen but do not release it as ions, so they are not acids.

Answer: (a) chlor-alkali process; brine + $\mathrm{CO_2}$ + $\mathrm{NH_3}$   (b) bulb glows only with acids because only acids give $\mathrm{H^+}$ ions

1. (a) Aerobic respiration: pyruvate is broken down to carbon dioxide and water (with energy). Anaerobic respiration in humans (in muscle): pyruvate is changed to lactic acid.
2. (b) Respiration releases carbon dioxide; photosynthesis releases oxygen.
3. (c) Terrestrial animals breathe through lungs (or the skin in some); fishes breathe through gills.

Answer: (a) CO₂ + H₂O vs lactic acid   (b) CO₂ vs O₂   (c) lungs vs gills

1. (a) The parents are $RRYY$ and $rryy$. All F₁ plants are $RrYy$, and round and yellow are dominant, so all the seeds are round and yellow.
2. (b) F₂: round yellow $9$ : round green $3$ : wrinkled yellow $3$ : wrinkled green $1$.
3. (c) The two traits are inherited independently of each other (independent assortment), so the genes recombine in the gametes and give the new combinations round green and wrinkled yellow.

Answer: (a) round and yellow   (b) 9 : 3 : 3 : 1   (c) independent assortment of the traits

1. A rainbow is a natural spectrum (a band of colours) that appears in the sky after a rain shower, when the Sun is behind the observer.
2. It forms by the dispersion of sunlight by tiny water droplets in the air. In each droplet the light is refracted and split into colours as it enters, reflected internally at the back of the drop, and refracted again as it leaves. Violet bends the most and red the least, so a band of seven colours reaches the eye (diagram above, with arrows).

Answer: A spectrum in the sky after rain, formed by refraction, dispersion and internal reflection in water droplets

1. (a) Near the wire (points 1 and 3) the field lines are circles around the wire, in the plane of the cardboard. At the centre of the loop (point 2) the field line is a straight line perpendicular to the plane of the loop.
2. (b) The strength of the current in the loop, and the number of turns of the loop (the field also falls with the radius of the loop).
3. (c) The right-hand thumb rule.

Answer: (a) circles near the wire, straight at the centre   (b) current, number of turns   (c) right-hand thumb rule

1. (a) Reason 1: the energy captured by the plants never goes back to the Sun, and the energy passed to the herbivores does not go back to the producers. Reason 2: as energy moves from one trophic level to the next, it is no longer available to the previous level; a large part is lost at each level (as heat in respiration), so the energy cannot be passed backwards.
2. (b) Only about $10\%$ of the energy passes on to the next level (the 10% law). Producers $10\,000$ J $\rightarrow$ primary consumers $1000$ J $\rightarrow$ secondary consumers $100$ J. So $100$ J is available to the secondary consumers.

Answer: (a) energy is not returned to the Sun or the earlier level   (b) 100 J (10% law)

1. (a)(i) I. Silver (Ag) does not react with oxygen even when hot. II. Aluminium (Al) forms a protective oxide layer. III. Potassium (K) catches fire in the open. IV. Copper (Cu) does not burn but gets a black coating of copper(II) oxide when heated.
2. (ii) Amphoteric oxides are metal oxides that react with both acids and bases to give salt and water. $\mathrm{Al_2O_3 + 6HCl \rightarrow 2AlCl_3 + 3H_2O}$ (acts as a base) and $\mathrm{Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O}$ (acts as an acid).
3. (iii) Alkalis are bases that dissolve in water, for example sodium hydroxide (NaOH) or potassium hydroxide (KOH).
4. (b)(i)(I) Cinnabar is heated in air: $\mathrm{2HgS + 3O_2 \xrightarrow{heat} 2HgO + 2SO_2}$, and then mercury(II) oxide is heated: $\mathrm{2HgO \xrightarrow{heat} 2Hg + O_2}$.
5. (II) Copper sulphide is heated in air: $\mathrm{2Cu_2S + 3O_2 \xrightarrow{heat} 2Cu_2O + 2SO_2}$, and then $\mathrm{2Cu_2O + Cu_2S \xrightarrow{heat} 6Cu + SO_2}$.
6. (ii) (I) Silver gets a coating of silver sulphide, $\mathrm{Ag_2S}$ (black). (II) Copper gets a green coating of basic copper carbonate, $\mathrm{Cu(OH)_2{\cdot}CuCO_3}$.

Answer: (a) Ag, Al, K, Cu; amphoteric oxides; NaOH   (b) roasting of cinnabar and of $\mathrm{Cu_2S}$; $\mathrm{Ag_2S}$ and basic copper carbonate

1. (a)(i) X is the stigma and Y is the anther. (ii) The anther produces yellowish pollen grains. (iii) The transfer of pollen from the anther to the stigma is pollination.
2. (iv) After pollination, a pollen tube grows out of the pollen grain, goes through the style and reaches the ovary. The male germ cell fuses with the female germ cell (the egg) in the ovule to form a zygote. The zygote divides again and again to form an embryo inside the ovule. The ovule develops a tough coat and changes into a seed.
3. (b)(i) Binary fission. (ii) The unicellular organism is Leishmania (it divides by binary fission along its long axis).
4. (iii) Advantages: many offspring are made in a short time; the offspring are genetically identical to the parent, so useful traits are kept; it needs no gametes or a mate; it helps organisms to survive in unfavourable conditions. (Any two.)
5. (iv) Budding: a bud forms as an outgrowth of the parent from repeated cell division at one site. It grows into a small individual, matures, and detaches to become a new independent Hydra.

Answer: (a) X stigma, Y anther; pollen; pollination; pollen tube and fertilisation make the seed   (b) binary fission; Leishmania; two advantages; budding

1. (a)(i) At the first face the ray bends towards the normal by some angle, and at the opposite parallel face it bends away from the normal by the same angle. The two bends are equal and opposite, so the emergent ray is parallel to the incident ray (it is only shifted sideways). A ray falling normally on a face goes straight through without bending, as the angle of incidence is $0^\circ$.
2. (ii) $u = -30$ cm and $f = -20$ cm (concave lens). $\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = -\dfrac{1}{20} - \dfrac{1}{30} = -\dfrac{5}{60} = -\dfrac{1}{12}$, so $v = -12$ cm. The image is $12$ cm from the lens, on the same side as the object.
3. (b)(i) When the flame is beyond the focus, the image is real and inverted. When it is between the pole and the focus, the image is virtual and erect. As the flame is moved away from the pole the image gets smaller (diminished) and moves nearer the focus. When the object is nearly at the centre of curvature C, the image is also formed at C: real, inverted and the same size.
4. (ii) $u = -6.0$ m and $f = +3.0$ m (convex mirror). $\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{1}{2}$, so $v = +2.0$ m. The image is $2.0$ m behind the mirror (virtual, erect and diminished).

Answer: (a) bends are equal and opposite; $v = -12$ cm   (b) real inverted then virtual erect, diminished as it moves away; $v = +2.0$ m

1. (a) Salt A is a soap (used for bathing) and salt B is a detergent. Tubewell water is usually hard. Soap does not lather in hard water, so the foam is the least in test tube III (tubewell water + salt A).
2. (b) Salt A (soap) is the sodium or potassium salt of long-chain carboxylic acids and does not form lather with hard water. Salt B (detergent) is a sodium salt of sulphonic acids (or an ammonium salt) and forms lather even with hard water.
3. (c)(i) Esters are sweet-smelling compounds, used in perfumes and flavourings, formed when an alcohol reacts with a carboxylic acid in the presence of an acid. With sodium hydroxide an ester is changed back to the alcohol and the sodium salt of the carboxylic acid (saponification): $\mathrm{CH_3COOC_2H_5 + NaOH \rightarrow C_2H_5OH + CH_3COONa}$.
4. (c)(ii) Hardness is caused by dissolved calcium and magnesium salts. The soap reacts with them to form an insoluble substance, scum, which is why foam forms with difficulty.

Answer: (a) test tube III   (b) soap vs detergent   (c) esters give alcohol + salt (OR) calcium and magnesium salts; scum

1. (a)(i) Adrenaline. (ii) The adrenal gland (on top of the kidneys).
2. (b) The heart beats faster to supply more oxygen to the muscles; the breathing rate increases; blood is diverted to the skeletal muscles and its supply to the digestive system and the skin is reduced. (Any two.)
3. (c) A hormone, a chemical signal, travels through the bloodstream and reaches a wide range of target cells all over the body, while an electrical impulse travels along a nerve cell to a particular place, and is quicker.

Answer: (a) adrenaline (OR) adrenal gland   (b) faster heartbeat, faster breathing   (c) blood-borne and wide-ranging vs along nerves

1. The graph of $I$ against $V$ is a straight line of slope $\dfrac{1}{R}$. A larger resistance gives a smaller slope.
2. (a) In series the total resistance is the greatest, so the slope is the least: graph A.
3. (b) In parallel the total resistance is the smallest, so the slope is the greatest: graph D.
4. (c)(i) Join two $10\ \Omega$ resistors in parallel: $\dfrac{10 \times 10}{10 + 10} = 5\ \Omega$. Join this pair in series with the third $10\ \Omega$ resistor: $5 + 10 = 15\ \Omega$ (diagram above).
5. (c)(ii) The total resistance in series is $0.1 + 0.2 + 0.3 + 0.4 + 0.5 = 1.5\ \Omega$. $I = \dfrac{V}{R} = \dfrac{6}{1.5} = 4.0$ A. The same current flows through every resistor in a series circuit, so $4.0$ A flows through the $0.3\ \Omega$ resistor.

Answer: (a) A   (b) D   (c) two 10 Ω in parallel, in series with the third (OR) $4.0$ A