CBSE Class 10 Science 2025 Question Paper (Set 31/5/1) with Solutions

1. $\mathrm{2Mg + O_2 \rightarrow 2MgO}$. The ribbon burns with a dazzling white flame and leaves a white powder, magnesium oxide. Magnesium does not vapourise, and magnesium oxide is basic: its solution turns red litmus blue, not blue litmus red.

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

1. M is more reactive than iron but less reactive than aluminium. In the reactivity series Mg > Al > Zn > Fe > Pb > Cu, only zinc lies between aluminium and iron.

Answer: (D) Zinc

1. They are hydrated salts: $\mathrm{Na_2CO_3{\cdot}10H_2O}$, $\mathrm{CuSO_4{\cdot}5H_2O}$, $\mathrm{CaSO_4{\cdot}2H_2O}$ and $\mathrm{FeSO_4{\cdot}7H_2O}$. Each has a fixed number of water molecules of crystallisation. (Gypsum is white, so “all coloured” is wrong.)

Answer: (C) have fixed number of molecules of water of crystallisation in one formula unit

1. Zinc reacts with both: $\mathrm{Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2}$ and $\mathrm{Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2}$. The gas is hydrogen. Copper does not react with dilute sulphuric acid or with sodium hydroxide.

Answer: (C) Zinc and Hydrogen

1. Take $a = 2$: lead gives $b = 2$; nitrogen $4$ gives $c = 4$; oxygen on the left is $12$, on the right $2 + 8 + 2d$, so $d = 1$.
2. $\mathrm{2Pb(NO_3)_2 \rightarrow 2PbO + 4NO_2 + O_2}$.

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

1. The impure copper is the anode and dissolves, pure copper deposits on the cathode, and the electrolyte is acidified copper sulphate solution.

Answer: (A) Impure copper, pure copper, acidified copper sulphate solution

1. Four carbon atoms can form only two skeletons: a straight chain, C–C–C–C, and a branched chain with three carbons in a row and one carbon attached to the middle one. These are the isomers butane and 2-methylpropane.

Answer: (B) straight chain and the branched chain

1. Amoeba first puts out pseudopodia (iii) to engulf the food, forming a food vacuole (ii). Enzymes then break the complex food into simpler substances (iv), these diffuse into the cytoplasm (i), and the undigested material is thrown out (v).

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

1. X is the medulla at the base of the brain. It controls involuntary actions such as salivation, vomiting and blood pressure. Hunger is controlled by the fore-brain (hypothalamus).

Answer: (B) Hunger

1. Spirogyra breaks into pieces and each grows into a new filament: fragmentation. Planaria, if cut, regrows the missing body part: regeneration.

Answer: (C) Fragmentation and regeneration

1. Cytokinins promote cell division. Abscisic acid inhibits growth and causes wilting of leaves.

Answer: (B) Cytokinins and Abscisic acid

1. Glucose is broken down to pyruvate in the cytoplasm. Pyruvate is then broken down to carbon dioxide and water, with the release of much energy, in the mitochondria.

Answer: (D) Cytoplasm and Mitochondria

1. A concave mirror forms a large, virtual and erect image only when the object is between the pole and the focus.

Answer: (B) between pole and focus of the mirror

1. An elongated eyeball forms the image in front of the retina: myopia (short-sightedness). It is corrected with a concave lens, which diverges the light a little before the eye lens.

Answer: (B) concave lens

1. Lakes, forests and ponds are natural ecosystems. Gardens and crop fields are made by humans. Only “lakes and gardens” puts a natural one first and an artificial one second.

Answer: (C) Lakes and gardens

1. Using fewer CFCs, reusing utensils and segregating waste all help the environment. Excessive use of disposable cups and plates creates waste that harms it.

Answer: (B) excessive use of disposable cups and plates

1. $\mathrm{2AgCl \xrightarrow{sunlight} 2Ag + Cl_2}$. The grey colour is the silver that forms, so R explains A.

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

1. Both are true. But being embedded in the uterine wall describes where the placenta is, not why it passes nutrients to the embryo (that is because of its villi and blood supply). So R does not explain A.

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

1. Assertion: true, a solenoid behaves like a bar magnet. Reason: false. The field pattern depends on the shape of the conductor: it is circles for a straight wire, and a bar-magnet pattern for a solenoid.

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

1. Only green plants, algae and some bacteria can photosynthesise. Animals and fungi cannot. So A is false. R, the definition of producers, is true.

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

1. (a) $X$ is hydrated ferrous sulphate (green vitriol), $\mathrm{FeSO_4{\cdot}7H_2O}$. On strong heating it gives the solid $Y$, iron(III) oxide (ferric oxide), $\mathrm{Fe_2O_3}$, and gives off $\mathrm{SO_2}$ and $\mathrm{SO_3}$ with the smell of burning sulphur: $\mathrm{2FeSO_4 \rightarrow Fe_2O_3 + SO_2 + SO_3}$.
2. (b) The green crystals first turn white (on losing the water) and then brown on further heating.

Answer: (a) $X = \mathrm{FeSO_4{\cdot}7H_2O}$, $Y = \mathrm{Fe_2O_3}$   (b) green $\to$ white, then brown

1. (a) Sperm formation needs a temperature lower than the normal body temperature. The testes are in the scrotum, outside the abdominal cavity, where it is cooler.
2. (b) The secretions of the glands (the prostate and seminal vesicles) make it easier for the sperms to be transported, and they give the sperms nutrition.

Answer: (a) cooler temperature for sperm formation   (b) help transport and nourish the sperms

1. (A) Through the pores in the walls of the capillaries, some plasma, proteins and blood cells escape into the spaces between cells in the tissues. This fluid is lymph. It carries digested and absorbed fat from the intestine, and drains excess fluid from the tissues back into the blood.
2. (B)(a) X is the Bowman’s capsule. It collects the filtrate from the glomerulus (the blood capillaries).
3. (b) The nephron adjusts how much water is lost and how much dissolved waste is removed, so the useful substances such as glucose, amino acids and salts, and the water the body needs, are taken back into the blood.

Answer: (A) escaped plasma forms lymph; carries fat and drains fluid   (B) Bowman’s capsule collects the filtrate; useful substances and water are retained

1. A larger refractive index means a larger optical density and a lower speed of light: $v = \dfrac{c}{n}$. So kerosene is optically denser and light travels slower in it.
2. But kerosene floats on water, so its mass density is less than that of water.
3. Inference: an optically denser medium need not have a greater mass density.

Answer: Kerosene is optically denser, light is slower in it, but it has the lower mass density

1. (A) Electric heater, electric iron (also a toaster, an electric kettle) use the heating effect. The electric fuse also works by it: the wire melts when too much current flows.
2. (B)(a) $1\ \mathrm{kWh} = 1000\ \mathrm{W} \times 3600\ \mathrm{s} = 3.6 \times 10^6\ \mathrm{J}$.
3. (b) $\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{6} = \dfrac{6 + 3 + 2}{12} = \dfrac{11}{12}$, so $R_p = \dfrac{12}{11} \approx 1.09\ \Omega$.

Answer: (A) heater, iron, fuse   (B) $1\ \mathrm{kWh} = 3.6 \times 10^6$ J; $R_p \approx 1.09\ \Omega$

1. (a) The organisms at the first trophic level, the producers, fix the energy of sunlight and make it available to the heterotrophs, the consumers.
2. (b) The energy captured by the autotrophs does not return to the Sun, and the energy passed to the herbivores does not go back to the producers. As it moves from one trophic level to the next it is no longer available to the previous one. So the flow is only in one direction.

Answer: (a) producers fix solar energy for all consumers   (b) energy does not revert to the earlier level or the Sun

1. The formula of washing soda is $\mathrm{Na_2CO_3{\cdot}10H_2O}$.
2. Heating baking soda gives sodium carbonate: $\mathrm{2NaHCO_3 \xrightarrow{heat} Na_2CO_3 + H_2O + CO_2}$. Recrystallising the sodium carbonate from water gives washing soda: $\mathrm{Na_2CO_3 + 10H_2O \rightarrow Na_2CO_3{\cdot}10H_2O}$.
3. Uses: in the glass, soap and paper industries; for removing the permanent hardness of water; as a cleansing agent for domestic use. (Any two.)

Answer: $\mathrm{Na_2CO_3{\cdot}10H_2O}$; heat $\mathrm{NaHCO_3}$ then recrystallise; glass/soap industry and softening water

1. (A) M is iron (galvanisation protects iron). A: the part of the metal outside the oil is in moist air, so it corrodes (rusts); the part inside the oil does not, as it cannot react with moist air. B: the metal is fully covered by oil, so no moist air reaches it and it does not corrode. C: the air is dry, there is no moisture, so it does not corrode.
2. (B)(a) Aluminium (2, 8, 3) gives up its three outer electrons, and nitrogen (2, 5) takes three to complete its octet: $\mathrm{Al \rightarrow Al^{3+} + 3e^-}$ and $\mathrm{N + 3e^- \rightarrow N^{3-}}$, giving $\mathrm{AlN}$.
3. (b) In an ionic solid positive and negative ions are held in a rigid lattice by strong electrostatic forces. When pressure is applied, layers of ions slide and like charges come face to face, so they repel and the crystal breaks.

Answer: (A) A: rusts above the oil; B and C: no corrosion   (B) Al³⁺ and N³⁻; brittle because like ions repel when layers slip

1. (a) Haemoglobin. It takes up oxygen from the lungs and carries it to the tissues that lack oxygen.
2. (b) The residual volume of air stays in the lungs so that there is enough time for oxygen to be absorbed and for carbon dioxide to be released, even between breaths.
3. (c) ATP provides energy for the various cellular processes, and it is the form in which the energy released in respiration is stored and passed on, so it is called the energy currency.

Answer: (a) haemoglobin carries oxygen   (b) more time for gas exchange   (c) ATP supplies energy for cell processes

1. (a) Fertilisation is the fusion of the male and female gametes to form a zygote.
2. (b) The zygote divides and forms the embryo. The ovule forms the seed. The ovary grows and forms the fruit. The stamens wither away (shrivel and fall off).

Answer: (a) fusion of gametes   (b) zygote → embryo, ovule → seed, ovary → fruit, stamens wither

1. (a) A negative power means a concave lens. $f = \dfrac{1}{P} = \dfrac{1}{-0.25} = -4$ m.
2. (b) Myopia (short-sightedness).
3. (c) A concave lens always forms a virtual, erect and diminished image, wherever the object is placed.

Answer: (a) concave, $f = -4$ m   (b) myopia   (c) virtual, erect, diminished

1. $r = 0.01$ cm $= 10^{-4}$ m, so $A = \pi r^2 = \dfrac{22}{7} \times 10^{-8}$ m². $R = 7\ \Omega$ and $\rho = 44 \times 10^{-6}\ \Omega\,\mathrm{m}$.
2. $R = \rho\dfrac{l}{A}$, so $l = \dfrac{RA}{\rho} = \dfrac{7 \times \frac{22}{7} \times 10^{-8}}{44 \times 10^{-6}} = \dfrac{22 \times 10^{-8}}{44 \times 10^{-6}} = 0.5 \times 10^{-2}$ m.

Answer: $l = 0.5 \times 10^{-2}$ m $= 0.5$ cm

Common mistake: Using the radius in cm. Convert 0.01 cm to 10⁻⁴ m before squaring it.

1. (a) The earth wire is a safety measure. If current leaks to the metal body of the appliance, the earth wire keeps the body at the potential of the earth, so the user does not get a severe shock.
2. (b) Use a fuse of the proper rating, and do not connect too many appliances to a single socket.

Answer: (a) leakage current flows to the earth, so no shock   (b) proper fuse; do not overload a socket

1. (A)(a) A homologous series is a series of carbon compounds with the same functional group, the same general formula and similar chemical properties, in which each member differs from the next by a $\mathrm{CH_2}$ unit. Two consecutive ketones: $\mathrm{CH_3COCH_3}$ (propanone) and $\mathrm{CH_3COC_2H_5}$ (butanone).
2. (b)(i) $\mathrm{CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O}$ (sodium ethanoate and water). (ii) $\mathrm{CH_3COOH + C_2H_5OH \xrightarrow{acid} CH_3COOC_2H_5 + H_2O}$ (ethyl ethanoate, an ester, and water).
3. (c) Benzene, C₆H₆, is a ring of six carbon atoms with alternate double bonds and one hydrogen on each carbon (diagram above).
4. (B)(a) Ethyne is $\mathrm{C_2H_2}$. Its electron dot structure is $\mathrm{H{:}C{:::}C{:}H}$: the two carbon atoms share three pairs of electrons (a triple bond) and each shares one pair with a hydrogen.
5. (b)(i) $\mathrm{2C_2H_5OH + 2Na \rightarrow 2C_2H_5ONa + H_2}$ (sodium ethoxide and hydrogen). (ii) $\mathrm{C_2H_5OH + CH_3COOH \xrightarrow{acid} CH_3COOC_2H_5 + H_2O}$ (ethyl ethanoate). (iii) $\mathrm{C_2H_5OH \xrightarrow{acidified\ K_2Cr_2O_7,\ heat} CH_3COOH}$ (ethanoic acid).

Answer: (A) same functional group and general formula; $\mathrm{CH_3COCH_3}$, $\mathrm{CH_3COC_2H_5}$; sodium ethanoate; ethyl ethanoate; benzene ring   (B) $\mathrm{C_2H_2}$; sodium ethoxide; ester; ethanoic acid

1. (A)(a)(i) Iodine is needed by the thyroid gland to make the hormone thyroxin. Without it the gland swells: goitre. (ii) A deficiency of growth hormone in childhood causes dwarfism. (iii) At puberty the testes secrete testosterone, which brings out the male features such as facial hair.
2. (b) In a multicellular body the chemical signals (hormones) can reach all the cells steadily and for a long time, even those not joined by nerves. They also coordinate the growth, development and responses to the environment.
3. (B)(a) Voluntary action: involves thinking, is controlled by the fore-brain and occurs according to our will (e.g. writing). Involuntary action: needs no thinking, is controlled by the hind-brain (medulla) or the spinal cord, and does not depend on our will (e.g. the heartbeat).
4. (b) A reflex action is a sudden, automatic action in response to a stimulus in the environment. Path: receptor → sensory neuron → relay neuron (spinal cord) → motor neuron → effector (muscle or gland).

Answer: (A) thyroxin and goitre; growth hormone; testosterone; hormones reach every cell   (B) voluntary vs involuntary; receptor → sensory → relay → motor → effector

1. (A)(a)(i) The ray goes through A to B without bending, so the speed of light is the same in A and B. At the boundary of B and C it bends away from the normal, so the speed in C is greater than in A and B.
2. (ii) Optical density is the opposite of speed: A and B have the same optical density, and C is optically rarer than A and B. In terms of refractive index: $n = \dfrac{c}{v}$, so $n_A = n_B$ (no bending) and $n_C$ is less than $n_A$ and $n_B$ (bending away from the normal).
3. (b)(i) The ray falling obliquely on the denser B bends towards the normal, and it bends away from the normal on going into C. (ii) The ray falling normally goes straight through, with no bending (diagram above).
4. (B)(a) In observation 4 the object and image distances are equal, $u = -40$ cm and $v = +40$ cm. For a convex lens this happens only when the object is at $2f$ (and the image at $2f$ too). So $2f = 40$ cm and $f = 20$ cm.
5. (b) $m = \dfrac{v}{u} = \dfrac{+60}{-30} = -2$.
6. (c) Observation 1 ($u = -15$, $v = -60$): $m = \dfrac{-60}{-15} = +4$, the image is virtual and erect (the object is inside $f$). Observation 2 ($u = -25$, $v = +100$): $m = \dfrac{100}{-25} = -4$, the image is real and inverted. Both are magnified 4 times, but they differ in nature (virtual vs real) and orientation (erect vs inverted), and the first is on the same side as the object.

Answer: (A) same speed in A and B, faster in C; A = B denser than C; sketches (B) f = 20 cm; m = −2; virtual and erect vs real and inverted

1. (a) Element A has the arrangement 2, 8, 6, so it has six outer electrons. It will gain (or share) two electrons to reach 2, 8, 8, the configuration of argon.
2. (b) Butene is $\mathrm{CH_2{=}CH{-}CH_2{-}CH_3}$. (i) Single bonds: $2$ C–C and $8$ C–H, that is $10$. (ii) Double bonds: $1$.
3. (c)(A) Nitrogen (2, 5) has five outer electrons and needs three more. Each hydrogen has one electron and needs one more. Nitrogen shares one electron with each of three hydrogen atoms, forming three single covalent bonds, and keeps one lone pair (diagram above).
4. (B) Carbon has four valence electrons. To get a noble gas configuration it would have to gain four electrons (forming C⁴⁻, which the nucleus of six protons cannot hold) or lose four (forming C⁴⁺, which needs a huge amount of energy). So it shares its electrons instead.

Answer: (a) gain or share 2 electrons   (b) 10 single and 1 double bond   (c) three N–H bonds, one lone pair (OR) a six-proton nucleus cannot hold or give four electrons easily

1. (a) Chromosomes are thread-like structures in the nucleus that carry the genes controlling the traits of an organism. They carry the information for inheritance of features from parents to the next generation in the form of DNA.
2. (b) A human male has one normal-sized X chromosome and a short Y chromosome, so the two do not match in size.
3. (c)(A) The mother (XX) gives an X in every egg. The father (XY) gives X in half his sperms and Y in the other half. So half the children are XX (girls) and half are XY (boys): $50 : 50$ (diagram above).
4. (B) In some reptiles the temperature at which the fertilised eggs are kept decides the sex of the young. In snails the individual can change sex, so sex is not genetically fixed.

Answer: (a) carriers of genes (DNA)   (b) X is normal, Y is short   (c) X, X × X, Y gives girl : boy = 50 : 50 (OR) reptiles by temperature; snails change sex

1. (a) By convention the direction of the field is the direction in which the north pole of a compass needle moves inside the field. The north pole of the needle is pushed away from the magnet’s north pole and pulled towards its south pole, so the lines go from north to south outside the magnet.
2. (b) The closer together the field lines, the stronger the magnetic field.
3. (c)(A)(i) If two lines crossed, at the point of intersection a compass needle would have to point in two directions at once, which is not possible. (ii) A uniform field is shown by parallel, equally spaced straight lines (diagram above).
4. (B) Inside a current-carrying solenoid the field lines are parallel, equally spaced straight lines along the axis, which represents a uniform magnetic field. Outside, the pattern is like that of a bar magnet.

Answer: (a) the north pole of a needle moves from N to S   (b) closer lines mean a stronger field   (c) a needle cannot point two ways; parallel equally spaced lines (OR) uniform field inside