CBSE Class 10 Science Question Paper 2024 PDF (Set 2 - 31/5/2) is available for download here. CBSE conducted the Science exam on March 2, 2024, from 10:30 AM to 1:30 PM. The total marks for the theory paper are 80. The question paper contains 20% MCQ-based questions, 40% competency-based questions, and 40% short and long answer type questions. Students reported the paper to be of moderate difficulty level.
CBSE Class 10th 2025 Science Question Paper
CBSE Class 10 Science Question Paper 2024 (Set 2 - 31/5/2) with Answer Key
| CBSE Class 10 Science Question Paper 2024 (Set 2 - 31/5/2) with Answer Key | Check Solutions |
CBSE Science Question Paper (Set 2 – 31/5/2) 2024 Solution
| Question | Answer | Detailed Solution |
|---|---|---|
| 1. Which of the following is not a thermal decomposition reaction? (A) 2FeSO4 → Fe2O3 + SO2 + SO3 (B) CaCO3 → CaO + CO2 (C) 2AgCl → 2Ag + Cl2 (D) Pb(NO3)2 → 2PbO + 4NO2 + O2 |
(C) 2AgCl → 2Ag + Cl2 | Thermal decomposition reactions are triggered by heat. Reaction (C) occurs due to light exposure, making it a photochemical decomposition reaction, not a thermal one. |
| 2. The process in which transport of soluble products of photosynthesis takes place in plants is known as: (A) Transpiration (B) Evaporation (C) Conduction (D) Translocation |
(D) Translocation | Translocation moves sugars and other solutes from leaves (source) to other plant parts (sink). It's distinct from processes like transpiration and conduction. |
| 3. Sense organ in which olfactory receptors are present is: (A) Nose (B) Skin (C) Tongue (D) Inner ear |
(A) Nose | Olfactory receptors detect smells and are located in the nasal cavity's olfactory epithelium. |
| 4. The incorrect statement about placenta is: (A) It is a disc embedded in the uterine wall. (B) It contains villi on the embryo’s side of the tissue. (C) It has a very small surface area for glucose and oxygen to pass from mother to the embryo. (D) The embryo gets nutrition from the mother’s blood through it. |
(C) It has a very small surface area for glucose and oxygen to pass from mother to the embryo. | The placenta has a large surface area due to villi, optimizing nutrient and gas exchange between mother and embryo. |
| 5. Which of the following is a redox reaction, but not a combination reaction? (A) C + O2 → CO2 (B) 2H2 + O2 → 2H2O (C) 2Mg + O2 → 2MgO (D) Fe2O3 + 3CO → 2Fe + 3CO2 |
(D) Fe2O3 + 3CO → 2Fe + 3CO2 | This is a redox reaction as Fe2O3 is reduced to Fe and CO is oxidized to CO2. It's not a combination reaction since two products are formed. |
| 6. The salt present in tooth enamel is: (A) Calcium phosphate (B) Magnesium phosphate (C) Sodium phosphate (D) Aluminium phosphate |
(A) Calcium phosphate | Tooth enamel is primarily made of hydroxyapatite, a crystalline form of calcium phosphate, which provides hardness. |
| 7. Identify an involuntary action from the following: (A) Riding a bicycle (B) Picking up a pencil (C) Regular beating of heart (D) Walking in a straight line |
(C) Regular beating of heart | Involuntary actions like heartbeat are controlled by the autonomic nervous system and don't require conscious effort. |
| 8. An aqueous solution of sodium chloride is prepared in distilled water. The pH of this solution is: (A) 6 (B) 8 (C) 7 (D) 3 |
(C) 7 | NaCl is a neutral salt, dissociating fully in water without altering the pH, keeping it at 7. |
| 9. A metal ‘X’ is used in thermit process. When ‘X’ is heated with oxygen, it gives an oxide ‘Y’, which is amphoteric in nature. ‘X’ and ‘Y’ respectively are: (A) Mn, MnO2 (B) Al, Al2O3 (C) Fe, Fe2O3 (D) Mg, MgO |
(B) Al, Al2O3 | Aluminum (Al) is used in the thermite process. Its oxide, Al2O3, is amphoteric, reacting with both acids and bases. |
| 10. Which one of the following is not a natural ecosystem? (A) Pond ecosystem (B) Grassland ecosystem (C) Forest ecosystem (D) Cropland ecosystem |
(D) Cropland ecosystem | Croplands are artificial ecosystems created and maintained by humans, unlike natural ecosystems that develop independently. |
| 11. A uniform magnetic field exists in the plane of paper as shown in the diagram. In this field, an electron (e−) and a positron (p+) enter as shown. The electron and positron experience forces: (A) both pointing into the plane of the paper. (B) both pointing out of the plane of the paper. (C) pointing into the plane of the paper and out of the plane of the paper respectively. (D) pointing out of the plane of the paper and into the plane of the paper respectively. |
(A) both pointing into the plane of the paper. | Using Fleming’s left-hand rule, both particles experience forces pointing into the plane of the paper due to their opposite charges but same motion and field alignment. |
| 12. The current carrying device which produces a magnetic field similar to that of a bar magnet is: (A) A straight conductor (B) A circular loop (C) A solenoid (D) A circular coil |
(C) A solenoid | A solenoid produces a magnetic field with clear north and south poles, resembling a bar magnet, unlike a loop or straight conductor. |
| 13. Select from the following the conditions responsible for the rapid spread of bread mould on a slice of bread: (i) Formation of large number of spores (ii) Presence of moisture and nutrients in bread (iii) Low temperature (iv) Presence of hyphae (A) (i) and (ii) (B) (ii) and (iv) (C) (ii) and (iii) (D) (iii) and (iv) |
(A) (i) and (ii) | Bread mould spreads rapidly due to spores and moisture. Low temperatures inhibit growth, while hyphae spread after germination. |
| 14. How will the image formed by a convex lens be affected, if the upper half of the lens is wrapped with a black paper? (A) The size of the image formed will be one-half of the size of the image due to complete lens. (B) The image of upper half of the object will not be formed. (C) The brightness of the image will reduce. (D) The lower half of the inverted image will not be formed. |
(C) The brightness of the image will reduce. | Covering part of the lens reduces the light passing through, thus decreasing the brightness. The size and orientation remain unaffected. |
| 15. The phenomena of light involved in the formation of a rainbow are: (A) Refraction, reflection and dispersion (B) Refraction, dispersion and internal reflection (C) Reflection, dispersion and internal reflection (D) Refraction, dispersion, scattering and total internal reflection |
(B) Refraction, dispersion and internal reflection | Rainbow formation involves refraction of light entering water droplets, dispersion of light into colors, and internal reflection within droplets. |
| 16. The colour of light for which the refractive index of glass is minimum, is: (A) Red (B) Yellow (C) Green (D) Violet |
(A) Red | Red light, with the longest wavelength, experiences the least refraction and thus has the minimum refractive index in glass. |
| 17. Assertion (A): Ozone layer protects the surface of the Earth from harmful UV radiations. Reason (R): Chlorofluorocarbons (CFCs) are responsible for depletion of ozone layer. |
(B) Both A and R are true, but R is not the correct explanation of A. | Ozone absorbs harmful UV rays, protecting life. While CFCs deplete ozone, the reason doesn't directly explain the assertion. |
| 18. Assertion (A): Some vegetable oils are healthy. Reason (R): Vegetable oils generally have long unsaturated carbon chains. |
(A) Both A and R are true, and R is the correct explanation of A. | Unsaturated fats in vegetable oils improve heart health. The long chains reduce LDL cholesterol, aligning with the assertion. |
| 19. Assertion (A): Sex of the children will be determined by what they inherit from their mother. Reason (R): Women have XX sex chromosomes. |
(D) A is false, but R is true. | The father's chromosome (X or Y) determines the child's sex. Mothers always pass an X chromosome. |
| 20. Assertion (A): Electrons move from lower potential to higher potential in a conductor. Reason (R): A dry cell maintains electric potential difference across the ends of a conductor. |
(B) Both A and R are true, but R is not the correct explanation of A. | Electrons flow from lower to higher potential due to their negative charge. A dry cell maintains the necessary potential difference. |
| 21. (a) Sometimes while running, athletes suffer from muscle cramps. Why? How is the respiration in this case different from aerobic respiration? | (a) Lactic acid accumulation causes cramps; anaerobic respiration occurs. | (a) During vigorous exercise, oxygen supply is insufficient, and anaerobic respiration produces lactic acid, causing cramps. Aerobic respiration requires oxygen and produces more energy, yielding CO2 and water. |
| OR 21.(b) Write the other name given to lymph. State its two functions. | (b) Other name: Tissue fluid. Functions: Nutrient transport and immunity. |
(b) Lymph collects interstitial fluid and aids in immunity via lymphocytes in lymph nodes. |
| 22. Identify the functional group present in the following compounds and name them: (a) CH3COOH (b) HCHO |
(a) Carboxylic acid group (-COOH); Acetic Acid. (b) Aldehyde group (-CHO); Formaldehyde. |
(a) CH3COOH contains a carboxylic acid group (-COOH), forming Acetic Acid (Ethanoic Acid). (b) HCHO contains an aldehyde group (-CHO), forming Formaldehyde (Methanal). |
| 23. (a) Copper powder is taken in a china dish and heated over a burner. Name the product formed and state its color. Write the chemical equation for the reaction involved. | (a) Product: Copper(II) oxide (CuO); Color: Black. Equation: 2Cu + O2 → 2CuO. |
(a) Heating copper powder forms black Copper(II) oxide by reacting with oxygen. |
| OR 23.(b) Write the chemical equation for the reaction when aqueous solutions of barium chloride and sodium sulfate react together. Write the ions present in the precipitate. | (b) Equation: BaCl2 + Na2SO4 → BaSO4↓ + 2NaCl. Ions: Ba2+ and SO42−. |
(b) Reaction of BaCl2 with Na2SO4 forms insoluble white precipitate BaSO4, containing Ba2+ and SO42− ions. |
| 24. Explain how the original number of chromosomes present in the parents is restored in the progeny. Name the cell division by which chromosome number is maintained in the progeny. | Restored during fertilization; maintained by meiosis. | During meiosis, gametes are haploid (n). Fusion of sperm (n) and egg (n) during fertilization restores the diploid (2n) chromosome number in the zygote. |
| 25. Define power of a lens. Find the power of a lens whose focal length is 50 cm. | Definition: Reciprocal of focal length (in meters). Power: +2D. |
Power (P) is measured in diopters (D), calculated as P = 1/f (meters). For f = 50 cm = 0.5 m, P = 1/0.5 = +2D. A convex lens has positive power. |
| 26. An electric source can supply a charge of 750 coulombs. If the current drawn by a device is 15 mA, find the time in which the electric source will be discharged completely. | Time: 13.89 hours. | Given charge \( Q = 750 \, \text{C} \) and current \( I = 15 \, \text{mA} = 0.015 \, \text{A} \), time \( t \) is calculated as \( t = \frac{Q}{I} = \frac{750}{0.015} = 50,000 \, \text{seconds} \). Converting to hours: \( \frac{50,000}{3600} = 13.89 \, \text{hours} \). |
| 27. (a) Study the diagram given below and answer the questions that follow: (i) Name the defect of vision depicted in this diagram and the part of the eye responsible for this condition. (ii) List two causes of this defect. (iii) Name the type of lens used to correct this defect and state its role. |
(a) (i) Hypermetropia (farsightedness); weak lens. (ii) Causes: Weak lens, short eyeball. (iii) Convex lens increases converging power. |
(a) Hypermetropia occurs when the eyeball is too short or the lens is too weak to focus light on the retina. A convex lens corrects it by converging light before it enters the eye. |
| OR 27.(b) What is dispersion of white light? State its cause. Draw a diagram to show dispersion of a beam of white light by a glass prism. | (b) Dispersion: Splitting of white light into its spectrum; caused by varying refraction of wavelengths. | (b) Dispersion occurs because different wavelengths of light refract at different angles in a prism. Violet bends the most, red the least, forming a spectrum. |
| 28. Suggest an activity to differentiate the acidic or basic character of the products obtained on burning magnesium and sulfur. | Magnesium oxide: Basic. Sulfur dioxide: Acidic. |
Burn magnesium to form magnesium oxide, dissolve in water, and test with litmus (turns red litmus blue). Burn sulfur to form sulfur dioxide, dissolve in water, and test with litmus (turns blue litmus red). |
| 29. Study the experimental set-up shown in the diagram and write the chemical equation for the reaction involved. Name and define the type of reaction. List two other metals which can be used in place of iron to show the same type of reaction with copper sulfate solution. | Equation: Fe + CuSO4 → FeSO4 + Cu. Type: Displacement reaction. Metals: Zinc, Magnesium. |
In displacement reactions, a more reactive metal replaces a less reactive metal from its compound. Iron displaces copper from copper sulfate. Zinc and magnesium can also displace copper due to higher reactivity. |
| 30. A plant with violet flowers (VV) was crossed with a plant with white flowers (vv): (a) What color of flowers was obtained in the plants of F1 generation and why? (b) Write the percentage of plants with white flowers in F2 generation if F1 plants were self-pollinated. Give the reason why this trait was not expressed in F1 generation. (c) In what ratio did we get the plants with (VV) and (Vv) gene combinations in the F2 generation? |
(a) Violet; dominant trait. (b) 25%; masked by dominance in F1. (c) Ratio: 1:2. |
(a) F1 generation had violet flowers due to dominance of violet allele (V). (b) Self-pollination of F1 resulted in 25% white flowers (recessive trait vv) in F2. The dominant allele (V) masked the recessive allele in F1. (c) F2 plants with (VV) and (Vv) genotypes appear in a 1:2 ratio. |
| 31. Taking the example of any two animal hormones along with their gland of secretion, explain how these hormones help: (i) In growth and development. (ii) Regulate metabolism in the body. |
Examples: Growth Hormone (Pituitary), Thyroxine (Thyroid). | Growth Hormone promotes cell division, bone growth, and muscle development. Thyroxine regulates the basal metabolic rate and energy production. Both are crucial for overall growth and maintaining body functions. |
| 32. "Earth wire is a safety measure in domestic electric circuits." Justify this statement explaining its role in case of accidental leakage of electric appliances. | Earth wire prevents electric shocks and protects appliances. | Earth wire provides a low-resistance path for leakage current to flow into the ground, preventing shocks and overheating of appliances during faults. |
| 33. Differentiate between food chain and food web. In a food chain consisting of deer, grass, and tiger, if the population of deer decreases, what will happen to the population of organisms belonging to the first and third trophic levels? | Food chain: Linear sequence. Food web: Interconnected food chains. Grass increases; Tiger decreases. |
Food chains are linear, while food webs are complex networks of multiple chains. A decrease in deer increases grass due to reduced grazing and decreases tigers due to lack of prey. |
| 34. (a) Explain the chlor-alkali process and write balanced chemical equations for the reactions that occur. Name the gases obtained at the anode and cathode respectively. Mention two uses each of the two gases obtained. | (a) Gases: Cl2 (anode), H2 (cathode). | (a) Chlor-alkali process electrolyzes brine to produce NaOH, Cl2, and H2. Reactions: 2Cl- → Cl2 + 2e- (anode), 2H2O + 2e- → H2 + 2OH- (cathode). Uses: Cl2: Water purification, PVC. H2: Fuel, ammonia production. |
| 34. OR (b) Common salt is a very important raw material as many compounds of industrial use can be prepared from it. Explain, giving chemical equations, the method of preparation of washing soda from sodium chloride. List four uses of washing soda. | (b) Process: Solvay process. Uses: Cleaning, water softening, glass making. |
(b) Washing soda is prepared via the Solvay process: NaCl + NH3 + CO2 + H2O → NaHCO3 + NH4Cl. Uses: Cleaning agent, water softening, glass and textile industries. |
| 35. (a) (i) The potential difference across the two ends of a circuit component is decreased to one-third of its initial value, while its resistance remains constant. What change will be observed in the current flowing through it? Name and state the law which helps us to answer this question. (ii) Draw a schematic diagram of a circuit consisting of a battery of four 1.5 V cells, a 5Ω resistor, a 10Ω resistor, and a 15Ω resistor, and a plug key, all connected in series. Calculate (I) the electric current passing through the circuit, and (II) the potential difference across the 10Ω resistor when the plug key is closed. |
(a) (i) Current reduces to one-third (Ohm's Law). (ii) Current: 0.2 A; V across 10Ω: 2 V. |
(a) (i) From Ohm's Law V = IR, reducing V to one-third reduces I to one-third since R is constant. (ii) Total resistance R = 5 + 10 + 15 = 30 Ω. Current I = 6/30 = 0.2 A. Potential difference across 10Ω: V = IR = 0.2 × 10 = 2 V. |
| OR 35.(b) When is the potential difference between two points said to be 1 volt? A copper wire has a diameter of 0.2 mm and resistivity of 1.6 × 10-8 Ωm. What will be the length of this wire to make its resistance 14Ω? How much does the resistance change if the diameter of the wire is doubled? | (b) Definition: 1V = 1J/C. Length: 27.5 m; New Resistance: 3.5Ω. |
1 volt is when 1 joule of work moves 1 coulomb of charge. Using R = ρl/A, l = 27.5 m. Doubling the diameter increases area fourfold, reducing resistance to 1/4 of the original |
| 36(a): Design an experiment to demonstrate that carbon dioxide is essential for photosynthesis. Write the observation and conclusion of the experiment. | Experiment: - Take a healthy potted plant and keep it in darkness for 48 hours to destarch its leaves. - Cover one of the leaves with a conical flask containing potassium hydroxide (KOH) to absorb CO₂. - Expose the plant to sunlight for a few hours. - Test the covered and uncovered leaves for starch using iodine solution. |
Observation: The leaf covered with KOH does not turn blue-black, while the uncovered leaf turns blue-black. Conclusion: Carbon dioxide is essential for photosynthesis, as its absence prevents starch synthesis. |
| 36(b)(i): Atmospheric air and exhaled air are passed through lime water. Explain the difference in time taken for lime water to turn milky in both cases. | Lime water turns milky faster with exhaled air. | Reason: Exhaled air contains a higher concentration of CO₂ (~4%) compared to atmospheric air (~0.03–0.04%). Reaction: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O. Higher CO₂ concentration accelerates the formation of calcium carbonate, causing faster milkiness. |
| 36(b)(ii): Draw a diagram of an open stomatal pore and label (i) Guard cells, and (ii) Chloroplast. Mention two functions of stomata. | Diagram: Open stomatal pore with labeled guard cells and chloroplast. Functions: - Facilitate gas exchange during photosynthesis and respiration. - Regulate transpiration for water loss and cooling. |
Stomata are vital for gas exchange and maintaining water balance in plants. Guard cells regulate the opening and closing of stomata based on environmental conditions. |
|
37.A highly polished surface such as a mirror reflects most of the light falling on it. In our daily life, we use two types of mirrors — plane and spherical. The reflecting surface of spherical mirrors may be curved inwards or outwards. In concave mirrors, reflection takes place from the inner surface, while in convex mirrors reflection takes place from the outer surface. |
(a) Principal axis: Line passing through the pole and center of curvature. (b) Radius of curvature: 20 cm. (c)(i) Image position: 6 cm behind the mirror. |
(a) The principal axis is the straight line passing through the pole and the center of curvature of a spherical mirror. (b) Focal length (f) = 10 cm; Radius of curvature (R) = 2f = 2 × 10 = 20 cm. (c)(i) Using mirror formula: 1/f = 1/v + 1/u, with f = +15 cm and u = -10 cm: 1/15 = 1/v - 1/10; v = +6 cm (behind the mirror). |
| OR 37.(c)(ii) A mirror forms a virtual, erect, and diminished image of an object. Identify the type of this mirror. Draw a ray diagram to show the image formation in this case. | (c)(ii) Mirror type: Convex. Image: Virtual, erect, diminished. | (c)(ii) A convex mirror forms a virtual, erect, and diminished image. The ray diagram shows parallel rays diverging after reflection and appearing to come from a focus behind the mirror. |
| 38. (a) What are hydrocarbons? (b) List two properties by virtue of which carbon can form a large number of compounds. (c)(i) Write the formula of the functional group present in (1) aldehydes and (2) ketones. Write the chemical equation for the reaction that occurs between ethanoic acid and ethanol in the presence of a catalyst. |
(a) Compounds of carbon and hydrogen. (b) Tetravalency and catenation. (c)(i) Aldehydes: -CHO, Ketones: -C=O. Reaction: Esterification. |
(a) Hydrocarbons are organic compounds composed only of carbon and hydrogen. (b) Tetravalency: Carbon can form four covalent bonds. Catenation: Ability to form long chains and rings. (c)(i) Aldehyde group: -CHO. Ketone group: -C=O. Reaction: CH3COOH + C2H5OH → CH3COOC2H5 + H2O (Esterification). |
| OR 38.(c)(ii) What are structural isomers? Write the structures of two isomers of butane (C4H10). | (c)(ii) Definition: Same formula, different structures. Examples: n-butane and isobutane. | (c)(ii) Structural isomers have the same molecular formula but different arrangements. Examples: n-butane (CH3-CH2-CH2-CH3) and isobutane (CH3-CH(CH3)-CH3). |
| 39. (a) Write the main difference between self-pollination and cross-pollination. (b) Name the part of the flower which attracts insects for pollination. What happens to this part after fertilization? (c)(i) Define fertilization. What is the fate of ovules and the ovary in a flower after fertilization? |
(a) Self: Same plant; Cross: Different plants. (b) Petals; they wither after fertilization. (c)(i) Ovules → Seeds, Ovary → Fruit. |
(a) Self-pollination: Pollen transfer within the same flower or plant. Cross-pollination: Pollen transfer between flowers of different plants of the same species. (b) Petals attract insects; they wither and fall off after fertilization. (c)(i) Fertilization: Fusion of male and female gametes. Ovules develop into seeds, and the ovary develops into fruit. |
| OR 39.(c)(ii) In a germinating seed, which parts are known as future shoot and future root? Mention the function of cotyledon. | (c)(ii) Shoot: Plumule, Root: Radicle. Cotyledon stores food. | (c)(ii) Plumule is the future shoot; radicle is the future root. Cotyledon stores and provides food to the embryo during germination. |



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