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CBSE Class 10 Science 2024 — Paper 31/4/2

80 marks · 3 hours · 39 questions (45 with the OR choices)

34 of its 45 questions also appear in 31/4/1 or 31/4/3; 11 are only in this paper.

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Questions 1–20 · 1 mark each

  1. Q1 1 mark
    A chemical reaction in which exchange of ions occurs between the reactants, is known as :
    1. (a)Endothermic Reaction
    2. (b)Exothermic Reaction
    3. (c)Double Displacement Reaction
    4. (d)Displacement Reaction
    Show answer
    (c) Double Displacement Reaction
    Step-by-step solution →
  2. Q2 1 mark

    Also in 31/4/1 (Q10), 31/4/3 (Q3)

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  3. Q3 1 mark

    Also in 31/4/1 (Q9), 31/4/3 (Q4)

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  4. Q4 1 mark

    Also in 31/4/1 (Q8)

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  5. Q5 1 mark

    Also in 31/4/1 (Q7), 31/4/3 (Q11)

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  6. Q6 1 mark

    Also in 31/4/1 (Q2)

    An aqueous solution of a salt turns blue litmus to red. The salt could be the one obtained by the reaction of :
    1. (a) and
    2. (b) and
    3. (c) and
    4. (d) and
    Show answer
    (d) and
    Step-by-step solution →
  7. Q7 1 mark

    Also in 31/4/1 (Q3), 31/4/3 (Q6)

    Four solutions, namely glucose, alcohol, hydrochloric acid and sulphuric acid filled in four separate beakers are connected one by one in an electric circuit with a bulb. The solutions in which the bulb will glow when current is passed are :
    1. (a)Glucose and alcohol
    2. (b)Alcohol and hydrochloric acid
    3. (c)Glucose and sulphuric acid
    4. (d)Hydrochloric acid and sulphuric acid
    Show answer
    (d) Hydrochloric acid and sulphuric acid
    Step-by-step solution →
  8. Q8 1 mark

    Also in 31/4/1 (Q4), 31/4/3 (Q7)

    The metals which are found in both free state as well as combined state are :
    1. (a)Gold and platinum
    2. (b)Platinum and silver
    3. (c)Copper and silver
    4. (d)Gold and silver
    Show answer
    (c) Copper and silver
    Step-by-step solution →
  9. Q9 1 mark

    Also in 31/4/1 (Q5), 31/4/3 (Q10)

    The number of single and double bonds present in a molecule of benzene () respectively, are :
    1. (a)6 and 6
    2. (b)9 and 3
    3. (c)3 and 9
    4. (d)3 and 3
    Show answer
    (b) 9 and 3
    Step-by-step solution →
  10. Q10 1 mark

    Also in 31/4/1 (Q6), 31/4/3 (Q9)

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  11. Q11 1 mark

    Also in 31/4/1 (Q16), 31/4/3 (Q12)

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  12. Q12 1 mark

    Also in 31/4/1 (Q15), 31/4/3 (Q13)

    A rectangular loop ABCD carrying a current is situated near a straight conductor XY, such that the conductor is parallel to the side AB of the loop and is in the plane of the loop. If a steady current is established in the conductor as shown, the conductor XY will
    X Y I B C A D I
    1. (a)remain stationary.
    2. (b)move towards the side AB of the loop.
    3. (c)move away from the side AB of the loop.
    4. (d)rotate about its axis.
    Show answer
    (b) move towards the side AB of the loop.
    Step-by-step solution →
  13. Q13 1 mark

    Also in 31/4/1 (Q11), 31/4/3 (Q14)

    Absolute refractive index of glass and water is and respectively. If the speed of light in glass is m/s, the speed of light in water is :
    1. (a) m/s
    2. (b) m/s
    3. (c) m/s
    4. (d) m/s
    Show answer
    (a) m/s
    Step-by-step solution →
  14. Q14 1 mark

    Also in 31/4/1 (Q12), 31/4/3 (Q15)

    When a beam of white light passes through a region having very fine dust particles, the colour of light mainly scattered in that region is :
    1. (a)Red
    2. (b)Orange
    3. (c)Blue
    4. (d)Yellow
    Show answer
    (c) Blue
    Step-by-step solution →
  15. Q15 1 mark

    Also in 31/4/1 (Q13), 31/4/3 (Q16)

    Consider the following combinations of resistors : The combinations having equivalent resistance is/are :
    2 Ω 2 Ω I 3 Ω 3 Ω 3 Ω II 2 Ω 4 Ω 3 Ω 3 Ω III 0.5 Ω 0.5 Ω IV
    1. (a)I and IV
    2. (b)Only IV
    3. (c)I and II
    4. (d)I, II and III
    Show answer
    (c) I and II
    Step-by-step solution →
  16. Q16 1 mark

    Also in 31/4/1 (Q14), 31/4/3 (Q1)

    An electric iron of resistance draws a current of A. The heat developed in the iron in seconds is :
    1. (a)15000 J
    2. (b)6000 J
    3. (c)1500 J
    4. (d)3000 J
    Show answer
    (a) 15000 J
    Step-by-step solution →
  17. Q17 1 mark

    Also in 31/4/1 (Q20), 31/4/3 (Q17)

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  18. Q18 1 mark

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  19. Q19 1 mark

    Also in 31/4/1 (Q17), 31/4/3 (Q19)

    For Questions number 17 to 20, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.

    Assertion (A): The extraction of metals from their sulphide ores cannot take place without roasting of the ore.

    Reason (R): Roasting converts sulphide ores directly into metals.

    1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
    2. (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
    3. (c)Assertion (A) is true, but Reason (R) is false.
    4. (d)Assertion (A) is false, but Reason (R) is true.
    Show answer
    (c) Assertion (A) is true, but Reason (R) is false.
    Step-by-step solution →
  20. Q20 1 mark

    Also in 31/4/1 (Q19), 31/4/3 (Q18)

    For Questions number 17 to 20, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below.

    Assertion (A): Magnetic field lines never intersect each other.

    Reason (R): If they intersect, then at the point of intersection, the compass needle would point towards two directions, which is not possible.

    1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
    2. (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
    3. (c)Assertion (A) is true, but Reason (R) is false.
    4. (d)Assertion (A) is false, but Reason (R) is true.
    Show answer
    (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
    Step-by-step solution →

Questions 21–26 · 2 marks each

  1. Q21(a) 2 marks

    Also in 31/4/1 (Q23(a)), 31/4/3 (Q21(a))

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    Q21(b) 2 marks

    Also in 31/4/1 (Q23(b)), 31/4/3 (Q21(b))

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  2. Q22 2 marks
    A spatula full of sodium carbonate is taken in a test tube and 2 mL of dilute ethanoic acid is added to it.
    1. (a)Write a chemical equation for the reaction.
    2. (b)Suggest a method of testing the gas liberated in the reaction.
    Show answer

    Ethanoic acid and sodium carbonate give sodium ethanoate, water and carbon dioxide; the gas turns lime water milky.

    (a) (sodium ethanoate, water and carbon dioxide; brisk effervescence is seen). (b) Pass the gas liberated through freshly prepared lime water with a delivery tube dipping into it: the lime water turns milky because insoluble calcium carbonate forms, , which shows that the gas is carbon dioxide.
    Step-by-step solution →
  3. Q23(a) 2 marks

    Also in 31/4/1 (Q21(a)), 31/4/3 (Q23(a))

    1 gram of solid sodium chloride was taken in a clean and dry test tube and concentrated sulphuric acid was added to it.
    1. (i)Name the gas evolved in the reaction.
    2. (ii)What will be observed when this gas is tested with (I) dry, and (II) wet blue litmus paper ? Write your conclusion about the nature (acidic/basic) of this gas.
    Show answer

    Hydrogen chloride gas; dry blue litmus is unchanged, wet blue litmus turns red, so the gas is acidic only with water.

    (i) Hydrogen chloride () gas is evolved. (ii) (I) Dry blue litmus paper shows no change — it stays blue; (II) Wet blue litmus paper turns red. Conclusion: hydrogen chloride gas is acidic in nature, but it shows its acidic character only in the presence of water, which lets it form () ions.
    Step-by-step solution →
    OR
    Q23(b) 2 marks

    Also in 31/4/1 (Q21(b)), 31/4/3 (Q23(b))

    Some metals react with acids to produce salt and hydrogen gas. Illustrate it with an example. How will you test the presence of this gas ?
    Show answer

    Zinc and dilute sulphuric acid give zinc sulphate and hydrogen; hydrogen burns with a pop sound near a flame.

    Example: zinc granules react with dilute sulphuric acid to form zinc sulphate (the salt) and hydrogen gas: . Test: pass the gas through soap solution and bring a burning candle near a gas-filled bubble (or bring a burning matchstick near the gas). The gas burns with a pop sound, which shows that it is hydrogen.
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  4. Q24 2 marks

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  5. Q25 2 marks
    Name the phenomenon of light responsible for Tyndall effect. Write an event where this phenomenon can be observed.
    Show answer

    Scattering of light; a sunlight beam entering a smoke-filled room through a small hole is seen as a bright streak.

    The phenomenon is scattering of light: fine (colloidal) particles in the path of a beam scatter the light in all directions, so the path of the beam becomes visible. Event: when a fine beam of sunlight enters a smoke-filled room through a small hole, its path is seen as a bright streak. The same effect is seen when sunlight passes through the canopy of a dense forest, where tiny water droplets in the mist scatter the light.
    Step-by-step solution →
  6. Q26 2 marks
    State Joule’s law of heating. How is this effect useful in electric circuits where fuse is used as a safety device ?
    Show answer

    Heat produced is ; in a fuse, excess current heats the wire until it melts and breaks the circuit.

    Joule's law of heating: when a current flows through a resistor of resistance for a time , the heat produced is ; that is, the heat produced is directly proportional to the square of the current for a given resistance, to the resistance for a given current, and to the time for which the current flows. Use in a fuse: a fuse is a wire of a metal or alloy of suitable melting point connected in series with the appliance, so it carries the same current. If the current becomes larger than the specified value (overloading or short circuit), the heat produced in the fuse wire, which grows as , raises its temperature until it melts. This breaks the circuit and stops the current, protecting the appliances and the wiring from damage.
    Step-by-step solution →

Questions 27–33 · 3 marks each

  1. Q27 3 marks
    A small amount of copper oxide was taken in a beaker and dilute hydrochloric acid was added with continuous stirring of the solution. Name the compound formed and state the colour of its solution. Write balanced chemical equation for the reaction that occurs. Based on the reaction, state the nature (acidic/basic) of copper oxide.
    Show answer

    Copper(II) chloride forms, giving a blue-green solution: ; copper oxide is basic.

    The compound formed is copper(II) chloride, , and its solution is blue-green (the black copper oxide dissolves). Balanced equation: . Nature: copper oxide reacts with an acid to form a salt and water, as a base does, so copper oxide is basic in nature (a basic oxide).
    Step-by-step solution →
  2. Q28 3 marks

    Also in 31/4/1 (Q31), 31/4/3 (Q27)

    Define the term power of accommodation of human eye. What happens to the image distance in the eye when we increase the distance of an object from the eye ? Name and explain the role of the part of human eye responsible for it in this case.
    Show answer

    Power of accommodation is the eye lens's ability to change its focal length; image distance stays fixed as the ciliary muscles relax.

    Power of accommodation: the ability of the eye lens to adjust its focal length (by changing its curvature) so that we can see both nearby and distant objects clearly. When the object is moved farther from the eye, the image distance does not change — the image is still formed on the retina, whose distance from the eye lens is fixed; instead, the focal length of the eye lens changes. The part responsible is the ciliary muscles, which hold the eye lens and change its curvature. As the object moves away, the ciliary muscles relax, the eye lens becomes thinner and its focal length increases, so the image is still focused sharply on the retina.
    Step-by-step solution →
  3. Q29 3 marks

    Also in 31/4/1 (Q33)

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  4. Q30 3 marks

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  5. Q31 3 marks

    Also in 31/4/1 (Q28)

    Why is the conversion of ethanol to ethanoic acid an oxidation reaction ? Name the oxidising agent used in this conversion. Write chemical equation for this oxidation reaction. How is this reaction different from the reaction in which ethanol burns in the presence of oxygen ?
    Show answer

    Ethanol gains oxygen and loses hydrogen, so it is oxidised to ethanoic acid by alkaline ; unlike burning, it is a controlled partial oxidation.

    It is an oxidation because oxygen is added to ethanol and hydrogen is removed from it: () gains one oxygen atom and loses two hydrogen atoms to become (). Oxidising agent: alkaline potassium permanganate (or acidified potassium dichromate). Equation: , i.e. . Difference: this is a controlled (partial) oxidation by an oxidising agent, without burning — the two-carbon chain is kept and ethanoic acid is formed. When ethanol burns in oxygen it undergoes complete oxidation (combustion) with a flame, forming only carbon dioxide and water with the release of heat and light: .
    Step-by-step solution →
  6. Q32 3 marks
    1. (a)How is a solenoid prepared ? Differentiate between a circular coil and a solenoid.
    2. (b)Draw the pattern of the magnetic field lines inside a current carrying solenoid. What does this pattern indicate ?
    Show answer

    A solenoid is insulated wire wound closely into a cylinder; its field inside is uniform, with straight parallel lines.

    (a) A solenoid is prepared by winding a long insulated copper wire closely, in many circular turns placed side by side, round a cylindrical tube, so that the coil takes the shape of a cylinder; its ends are connected to a battery through a key. Difference: a circular coil is a wire bent into a circle, with one or a few turns lying in one place, and its field lines are nearly circles near the wire and nearly straight only near its centre. A solenoid has many turns spread along the length of a cylinder; the field lines inside it are straight and parallel along its whole length, and outside its pattern is like that of a bar magnet, with one end acting as a north pole and the other as a south pole. (b) Inside a current-carrying solenoid the field lines are parallel, equally spaced straight lines along the axis, running from the S end to the N end; outside they leave the N end and return to the S end. This pattern indicates that the magnetic field is the same at all points inside the solenoid, that is, the field inside a solenoid is uniform.
    Step-by-step solution →
  7. Q33(a) 3 marks

    Also in 31/4/1 (Q29(a)), 31/4/3 (Q32(a))

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    Q33(b) 3 marks

    Also in 31/4/1 (Q29(b)), 31/4/3 (Q32(b))

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Questions 34–36 · 5 marks each

  1. Q34(a) 5 marks
    When lead nitrate is heated strongly in a boiling tube, two gases are liberated and a solid residue is left behind in the test tube.
    1. (i)Name the type of chemical reaction and define it.
    2. (ii)Write the name and formula of the coloured gas liberated.
    3. (iii)Write the balanced chemical equation for the reaction.
    4. (iv)Name the residue left in the test tube and state the method of testing its nature (acidic/basic).
    Show answer

    Thermal decomposition gives brown nitrogen dioxide, oxygen and a basic residue of lead oxide, .

    (i) Thermal decomposition reaction. A reaction in which a single reactant breaks down to give simpler products is a decomposition reaction; when the decomposition is brought about by heating, it is called thermal decomposition. (ii) The coloured (brown) gas is nitrogen dioxide, . (iii) . (iv) The residue is lead oxide (lead(II) oxide), . To test its nature, shake a little of it with water and test with red and blue litmus paper, since a basic oxide turns red litmus blue. Because lead oxide dissolves only slightly in water, confirm by warming it with dilute nitric acid: it dissolves to give lead nitrate and water, as a base does. Lead oxide is therefore basic in nature.
    Step-by-step solution →
    OR
    Q34(b) 5 marks
    1. (i)Write balanced chemical equation for the following word equation. Lead nitrate + Potassium iodide → Lead iodide + Potassium nitrate Is this a double displacement reaction ? Justify your answer. Name the compound precipitated and write the ions present in it.
    2. (ii)Write the method of preparation of . What happens when is passed through it ? Write balanced chemical equation for the reaction involved.
    Show answer

    Lead nitrate and potassium iodide give yellow lead iodide by double displacement; slaked lime solution turns milky with carbon dioxide, forming calcium carbonate.

    (i) . Yes, it is a double displacement reaction, because the reactants exchange their ions: lead, first combined with nitrate, combines with iodide, and potassium, first combined with iodide, combines with nitrate. The compound precipitated is lead iodide, , a yellow precipitate; it contains lead ions, , and iodide ions, (two for each ). (ii) Calcium hydroxide (slaked lime) is prepared by adding water to calcium oxide (quicklime); the reaction is vigorous and gives out a large amount of heat: . When carbon dioxide is passed through its solution (lime water), the solution turns milky because insoluble calcium carbonate is formed: .
    Step-by-step solution →
  2. Q35(a) 5 marks

    Also in 31/4/1 (Q36(a))

    The variation of image distance with object distance for a convex lens is given in the following observation table. Analyse it and answer the questions that follow:
    1. (i)Without calculation, find the focal length of the convex lens. Justify your answer.
    2. (ii)Which observation is not correct? Why? Draw ray diagram to find the position of the image formed for this position of the object.
    3. (iii)Find the approximate value of magnification for cm.
    S. No.Object distance (u) cmImage distance (v) cm
    1
    2
    3
    4
    5
    6
    Show answer

    cm; observation 6 is wrong as should be cm; the magnification for cm is .

    (i) cm. In observation 3 the object and the image are at the same distance, 50 cm, on opposite sides of the lens; this happens only when the object is at 2F₁ and the image at 2F₂, so cm and cm. (ii) Observation 6 ( cm, cm) is not correct. The object at 15 cm is between the optical centre and F₁ (), so the lens forms a virtual, erect, magnified image on the same side as the object and must be negative. By the lens formula the image is at cm and is 2.5 times the size of the object, as the ray diagram shows. (iii) : the image is real, inverted and 5 times the size of the object.
    Step-by-step solution →
    OR
    Q35(b) 5 marks

    Also in 31/4/1 (Q36(b))

    1. (i)Define principal axis of a lens. Draw a ray diagram to show what happens when a ray of light parallel to the principal axis of a concave lens passes through it.
    2. (ii)The focal length of a concave lens is 20 cm. At what distance from the lens should a 5 cm tall object be placed so that its image is formed at a distance of 15 cm from the lens? Also calculate the size of the image formed.
    Show answer

    The object must be 60 cm in front of the lens; the image is 1.25 cm tall, virtual and erect.

    (i) The principal axis of a lens is the imaginary straight line passing through the two centres of curvature of the lens. A ray parallel to the principal axis, after refraction through a concave lens, diverges (bends away from the axis) and appears to come from the principal focus F₁ on the same side of the lens as the incident light, as shown in the ray diagram. (ii) With cm and cm, , so cm: the object must be placed 60 cm in front of the lens. Then and cm: the image is 1.25 cm tall, virtual and erect.
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  3. Q36(a) 5 marks

    Also in 31/4/1 (Q35(a)), 31/4/3 (Q34(a))

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    Q36(b) 5 marks

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Questions 37–39 · 4 marks each

  1. Q37 4 marks

    Also in 31/4/1 (Q39), 31/4/3 (Q37)

    Study the following circuit: On the basis of this circuit, answer the following questions:
    A B C 4 Ω 6 Ω 16 Ω 8 Ω 8 Ω 6 V Key
    1. (a)Find the value of total resistance between the points A and B.[1]
    2. (b)Find the resistance between the points B and C.[1]
    3. (c)(i)Calculate the current drawn from the battery, when the key is closed.[2]
    4. (c)(ii)OR : In the above circuit, the resistor or the parallel combination of two resistors of , which one of the two will have more potential difference across its two ends? Justify your answer.[2]
    Show answer

    and ; A; or the resistor has the larger potential difference, 3.2 V against 0.8 V.

    (a) . (b) , so . (c)(i) Total resistance , so the current drawn from the battery is A. (c)(ii) The resistor. It is in series with the parallel combination, so the same current, 0.2 A, flows through both; by the potential difference is V across the resistor but only V across the combination, whose resistance is .
    Step-by-step solution →
  2. Q38 4 marks

    Also in 31/4/1 (Q37), 31/4/3 (Q38)

    Three metal samples of magnesium, aluminium and iron were taken and rubbed with sand paper. These samples were then put separately in test tubes containing dilute hydrochloric acid. Thermometers were also suspended in each test tube so that their bulbs dipped in the acid. The rate of formation of bubbles was observed. The above activity was repeated with dilute nitric acid and the observations were recorded. Answer the following questions:
    1. (a)When activity was done with dilute hydrochloric acid, then in which one of the test tubes was the rate of formation of bubbles the fastest and the thermometer showed the highest temperature?[1]
    2. (b)Which metal did not react with dilute hydrochloric acid? Give reason.[1]
    3. (c)(i)Why is hydrogen gas not evolved when a metal reacts with dilute nitric acid? Name the ultimate products formed in the reaction.[2]
    4. (c)(ii)OR : Name the type of reaction on the basis of which reactivity of metals is decided. You have two metals X and Y. How would you decide which is more reactive than the other?[2]
    Show answer

    Magnesium reacts fastest with hydrochloric acid; none of the three metals fails to react; nitric acid oxidises hydrogen to water; displacement reactions decide reactivity.

    (a) The test tube with magnesium. Magnesium is the most reactive of the three, so it displaces hydrogen from dilute hydrochloric acid fastest and releases the most heat: . (b) None of the three metals fails to react. Magnesium, aluminium and iron all lie above hydrogen in the activity series, so each displaces hydrogen from dilute hydrochloric acid; iron reacts the most slowly. Only a metal below hydrogen, such as copper, does not react with dilute hydrochloric acid, because it cannot displace hydrogen from the acid. (c)(i) Nitric acid is a strong oxidising agent: it oxidises the hydrogen produced to water and is itself reduced to an oxide of nitrogen (, or ). The ultimate products are the metal nitrate, water and an oxide of nitrogen. (c)(ii) Displacement reactions. Dip a strip of metal X in a solution of a salt of Y, and a strip of Y in a solution of a salt of X. If X displaces Y, so that Y is deposited on X, X is more reactive than Y; if instead Y displaces X, Y is more reactive. For example, iron displaces copper from copper sulphate solution, so iron is more reactive than copper.
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  3. Q39 4 marks

    Also in 31/4/1 (Q38), 31/4/3 (Q39)

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