CBSE Class 10 Science Sample Paper 2026 (with Solutions) — Free PDF
This is a full-length CBSE Class 10 Science paper built to the current board blueprint: 39 questions across five sections for 80 marks in 3 hours, covering
TL;DR: This is a full-length CBSE Class 10 Science paper built to the current board blueprint: 39 questions across five sections for 80 marks in 3 hours, cov…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
This is a full-length CBSE Class 10 Science paper built to the current board blueprint: 39 questions across five sections for 80 marks in 3 hours, covering Physics, Chemistry and Biology in the proportions the board uses. Section A is objective, Sections B to D are descriptive and Section E is case-based. Every question carries a fully worked solution explaining the reasoning, not just the final answer.
Board: CBSE · Class: Class 10 · Subject: Science · Total: 80 marks · Time: 3 hours
Section A — Objective (1 mark each) — 20 marks
Questions 1–18 are multiple choice. Questions 19 and 20 are Assertion–Reason: mark (a) if both A and R are true and R explains A, (b) if both are true but R does not explain A, (c) if A is true and R is false, (d) if A is false and R is true.
Q1. The correct balanced equation for the reaction of iron with steam is (1 mark)
Answer: 3Fe + 4H₂O → Fe₃O₄ + 4H₂. Iron reacts with steam (not cold water) to give iron(II,III) oxide and hydrogen gas.
Q2. When quicklime is added to water, the reaction is (1 mark)
Answer: A combination reaction that is exothermic. CaO + H₂O → Ca(OH)₂, releasing a large amount of heat.
Q3. The pH of a solution changes from 4 to 3. The hydrogen ion concentration becomes (1 mark)
Answer: 10 times greater. The pH scale is logarithmic, so a fall of one unit means a tenfold rise in [H⁺].
Q4. Which of the following is used to test whether a substance is a base: red litmus, blue litmus, or dilute HCl? (1 mark)
Answer: Red litmus. A base turns red litmus blue; blue litmus stays blue and so gives no information.
Q5. The common name of sodium hydrogencarbonate is (1 mark)
Answer: Baking soda, NaHCO₃. It is used in antacids and in baking powder with a mild edible acid.
Q6. The reaction 2PbO + C → 2Pb + CO₂ shows that carbon is acting as (1 mark)
Answer: A reducing agent. Carbon removes oxygen from lead oxide, so PbO is reduced and carbon is itself oxidised.
Q7. Which gas is evolved when a metal reacts with a dilute acid? (1 mark)
Answer: Hydrogen. It burns with a pop sound when a lighted splint is brought near, which is the standard test.
Q8. The functional group present in ethanoic acid is (1 mark)
Answer: The carboxylic acid group, −COOH. Ethanoic acid is CH₃COOH.
Q9. The number of covalent bonds in a molecule of ethane (C₂H₆) is (1 mark)
Answer: Seven — six C–H bonds and one C–C bond.
Q10. The site of complete oxidation of glucose in a cell is the (1 mark)
Answer: Mitochondrion. Glycolysis occurs in the cytoplasm, but the complete breakdown to CO₂ and water occurs in the mitochondria.
Q11. The xylem in plants is responsible for (1 mark)
Answer: Transport of water and dissolved minerals from the roots to the rest of the plant. Phloem transports food.
Q12. Which hormone regulates the level of sugar in the blood? (1 mark)
Answer: Insulin, secreted by the pancreas. Its deficiency causes diabetes.
Q13. In human beings, the sex of a child is determined by (1 mark)
Answer: The chromosome inherited from the father. The mother always contributes an X; the father contributes either X (girl) or Y (boy).
Q14. A concave mirror forms a real, inverted and magnified image when the object is placed (1 mark)
Answer: Between the focus F and the centre of curvature C. The image then forms beyond C.
Q15. The refractive index of a medium in which light travels at 2 × 10⁸ m/s is (speed of light in vacuum = 3 × 10⁸ m/s) (1 mark)
Answer: 1.5. n = c/v = (3 × 10⁸)/(2 × 10⁸) = 1.5.
Q16. The power of a lens of focal length 25 cm is (1 mark)
Answer: +4 D. P = 1/f in metres = 1/0.25 = 4 dioptres, positive because a 25 cm focal length here denotes a converging lens.
Q17. Two resistors of 6 Ω and 3 Ω are connected in parallel. Their equivalent resistance is (1 mark)
Answer: 2 Ω. 1/R = 1/6 + 1/3 = 1/2, so R = 2 Ω.
Q18. The direction of the magnetic field around a straight current-carrying conductor is given by (1 mark)
Answer: The right-hand thumb rule: point the right thumb along the current and the curled fingers give the direction of the field.
Q19. Assertion (A): Photosynthesis is an endothermic process. Reason (R): Energy is absorbed from sunlight during photosynthesis. (1 mark)
Answer: (a) Both true and R explains A. Photosynthesis stores absorbed light energy in glucose, which is precisely why it is endothermic.
Q20. Assertion (A): Ozone in the upper atmosphere is harmful to living organisms. Reason (R): Ozone absorbs ultraviolet radiation from the Sun. (1 mark)
Answer: (d) A is false, R is true. Upper-atmosphere ozone protects life by absorbing UV radiation; it is ozone at ground level that is a pollutant.
Section B — Very Short Answer (2 marks each) — 12 marks
Answer all 6 questions. Each carries 2 marks.
Q1. Why is respiration considered an exothermic reaction? Write the overall equation. (2 marks)
Answer: Respiration breaks down glucose and releases the energy stored in it, which the cell uses for its activities — energy is given out, so the reaction is exothermic. C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + energy.
Q2. What is meant by a rancid food, and how can rancidity be prevented? (2 marks)
Answer: Rancidity is the oxidation of fats and oils in food, which spoils their smell and taste. It is prevented by adding antioxidants, packaging in nitrogen, keeping food in airtight containers and refrigerating it.
Q3. Distinguish between an acid and a base in terms of the ions they produce in aqueous solution, with one example of each. (2 marks)
Answer: An acid produces hydrogen ions, H⁺ (as H₃O⁺), in water — for example HCl → H⁺ + Cl⁻. A base produces hydroxide ions, OH⁻ — for example NaOH → Na⁺ + OH⁻.
Q4. Why do we apply paint on iron articles? (2 marks)
Answer: Paint forms a barrier that keeps out air and moisture, both of which are needed for rusting. Without contact with oxygen and water the iron cannot corrode, so painting prevents rusting.
Q5. State two differences between arteries and veins. (2 marks)
Answer: Arteries carry blood away from the heart, usually oxygenated, under high pressure and have thick elastic walls without valves. Veins carry blood towards the heart, usually deoxygenated, under low pressure and have thinner walls with valves that prevent backflow.
Q6. Define the principal focus of a concave mirror. (2 marks)
Answer: It is the point on the principal axis at which rays of light parallel to the principal axis actually meet after reflection from the concave mirror. The distance of this point from the pole is the focal length.
Section C — Short Answer (3 marks each) — 21 marks
Answer all 7 questions. Each carries 3 marks.
Q1. Balance the equation and name the type of reaction: Fe + CuSO₄ → FeSO₄ + Cu. (3 marks)
Answer: The equation is already balanced: Fe + CuSO₄ → FeSO₄ + Cu, with one atom of each element on both sides. It is a displacement reaction, because the more reactive iron displaces the less reactive copper from its salt solution. The blue colour of the copper sulphate solution fades and a reddish-brown deposit of copper forms.
Q2. What happens when sodium hydrogencarbonate is heated? Give the equation and one use of the product. (3 marks)
Answer: On heating, sodium hydrogencarbonate decomposes: 2NaHCO₃ →(heat) Na₂CO₃ + H₂O + CO₂. The carbon dioxide released is what makes cakes and bread rise, which is why baking soda is used in baking powder; the sodium carbonate formed is used in cleaning agents.
Q3. Explain why carbon forms a very large number of compounds. Give any two reasons. (3 marks)
Answer: First, carbon shows catenation — carbon atoms bond strongly to one another to form long chains, branched chains and rings, which no other element does to the same extent. Second, carbon is tetravalent, so each atom can bond with four other atoms of carbon or of other elements such as hydrogen, oxygen, nitrogen and sulphur. The strength of these covalent bonds makes the resulting compounds stable.
Q4. Draw and label the pathway of blood in the human heart, naming the four chambers in order. (3 marks)
Answer: Deoxygenated blood from the body enters the right atrium through the vena cava, passes to the right ventricle, and is pumped through the pulmonary artery to the lungs. Oxygenated blood returns from the lungs through the pulmonary vein into the left atrium, passes to the left ventricle, and is pumped out through the aorta to the body. The four chambers in order are right atrium, right ventricle, left atrium and left ventricle; valves between them prevent backflow.
Q5. State the laws of refraction of light and define absolute refractive index. (3 marks)
Answer: First law: the incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane. Second law (Snell's law): for a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant, sin i / sin r = n. The absolute refractive index of a medium is the ratio of the speed of light in vacuum to the speed of light in that medium, n = c/v.
Q6. An electric bulb is rated 220 V and 100 W. Calculate its resistance and the current it draws in normal operation. (3 marks)
Answer: Power P = V²/R, so R = V²/P = (220 × 220)/100 = 484 Ω. Current I = P/V = 100/220 = 0.45 A. The bulb therefore has a resistance of 484 Ω and draws about 0.45 A.
Q7. What is a food chain? Explain why the number of trophic levels in a food chain is limited to four or five. (3 marks)
Answer: A food chain is the series of organisms through which energy passes as one feeds on another — for example grass → grasshopper → frog → snake. Only about 10% of the energy at one trophic level is transferred to the next, the rest being lost as heat and used in life processes. After four or five levels so little energy remains that it cannot support another population, so food chains are short.
Section D — Long Answer (5 marks each) — 15 marks
Answer all 3 questions. Each carries 5 marks.
Q1. (a) What is meant by the reactivity series of metals? (b) Explain, with equations, why zinc can displace copper from copper sulphate solution but copper cannot displace zinc from zinc sulphate solution. (c) Name one metal that does not react with dilute acids. (5 marks)
Answer: (a) The reactivity series is the arrangement of metals in decreasing order of their tendency to lose electrons and form positive ions, the most reactive at the top. (b) Zinc lies above copper in the series, so it is more reactive and displaces copper: Zn + CuSO₄ → ZnSO₄ + Cu, the blue solution fading as reddish-brown copper is deposited. Copper lies below zinc, is less reactive, and therefore cannot displace it, so Cu + ZnSO₄ → no reaction. (c) Copper (or silver or gold) does not react with dilute acids to release hydrogen.
Q2. (a) Draw a ray diagram showing image formation by a convex lens when the object is placed beyond 2F₁. (b) State the nature, position and size of the image. (c) An object 4 cm high is placed 30 cm from a convex lens of focal length 10 cm. Find the position and height of the image. (5 marks)
Answer: (a) One ray parallel to the principal axis refracts through F₂; a second ray through the optical centre goes straight on; they meet to form the image between F₂ and 2F₂ on the far side. (b) The image is real, inverted and diminished. (c) Using 1/v − 1/u = 1/f with u = −30 cm and f = +10 cm: 1/v = 1/10 + 1/(−30) = (3 − 1)/30 = 2/30, so v = +15 cm — the image is 15 cm from the lens on the opposite side. Magnification m = v/u = 15/(−30) = −0.5, so the image height = −0.5 × 4 = −2 cm: it is 2 cm tall and inverted.
Q3. (a) State Mendel's law of dominance using his monohybrid cross as an example. (b) A tall pea plant (TT) is crossed with a short one (tt). Give the F₁ and F₂ genotypes and the F₂ phenotypic ratio. (c) Why did Mendel choose the pea plant? (5 marks)
Answer: (a) The law of dominance states that when two contrasting forms of a trait are present together, only one — the dominant one — is expressed, while the recessive form remains hidden. In Mendel's cross of tall with short pea plants, all F₁ plants were tall. (b) F₁ is entirely Tt and all plants are tall. Selfing the F₁ gives F₂ genotypes TT, Tt, Tt and tt, a genotypic ratio of 1 : 2 : 1, and since T is dominant the phenotypic ratio is 3 tall : 1 short. (c) The pea plant has clearly contrasting characters, is self-pollinating so pure lines are easily maintained, can be cross-pollinated by hand, and has a short life cycle producing many seeds — so results appear quickly and in large numbers.
Section E — Case-based questions (4 marks each) — 12 marks
Answer all 3 case-based questions. Each has sub-parts totalling 4 marks.
Q1. CASE STUDY 1 — A student tests four household solutions with universal indicator and records: lemon juice pH 2, milk pH 6, water pH 7, soap solution pH 10. (i) Which solution is most acidic and which is most basic? (ii) Which is neutral? (iii) The student's tooth enamel begins to corrode below pH 5.5 — explain why brushing with a toothpaste like the soap solution helps. (4 marks)
Answer: (i) Lemon juice at pH 2 is the most acidic and soap solution at pH 10 the most basic. (ii) Water at pH 7 is neutral. (iii) Bacteria in the mouth produce acids that lower the pH below 5.5, and tooth enamel — calcium phosphate — is corroded by that acid. Toothpastes are mildly basic, so they neutralise the excess acid and raise the pH above the level at which enamel dissolves.
Q2. CASE STUDY 2 — An electric circuit has a 6 V battery connected to two resistors of 4 Ω and 8 Ω joined in series, with an ammeter connected in series. (i) Find the equivalent resistance. (ii) Find the current recorded by the ammeter. (iii) Find the potential difference across the 8 Ω resistor. (4 marks)
Answer: (i) In series the resistances add: R = 4 + 8 = 12 Ω. (ii) I = V/R = 6/12 = 0.5 A, and since the current is the same everywhere in a series circuit the ammeter reads 0.5 A wherever it is placed. (iii) V = IR = 0.5 × 8 = 4 V across the 8 Ω resistor (the remaining 2 V is across the 4 Ω resistor, and 4 + 2 = 6 V as expected).
Q3. CASE STUDY 3 — A pond ecosystem contains algae, small fish that eat the algae, and large fish that eat the small fish. A factory discharges a non-biodegradable pesticide into the pond. (i) Write the food chain and name the producer. (ii) Which organism will have the highest concentration of the pesticide, and what is this phenomenon called? (iii) Suggest one way to reduce this problem. (4 marks)
Answer: (i) The food chain is algae → small fish → large fish, and the algae are the producers because they make their own food by photosynthesis. (ii) The large fish, at the highest trophic level, will carry the highest concentration. Because the pesticide is non-biodegradable it is not broken down or excreted, so it accumulates and becomes more concentrated at each successive level — this is called biological magnification. (iii) Use biodegradable pesticides, or treat factory effluent before discharge, or switch to biological pest control so the harmful chemical never enters the food chain.
FAQs
Is the Class 10 Science exam practical-based?
Yes, 20 marks are for practical assessment, and the remaining 80 marks are for theory.
Which topics are most important for the board exam?
Electricity, heredity, photosynthesis, and chemical reactions are frequently asked topics.
Can I use a calculator in the exam?
Calculators are generally not allowed; you should be able to perform calculations mentally.
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