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CBSE Class 10 Science 2024 — Paper 31/3/2

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

32 of its 45 questions also appear in 31/3/1 or 31/3/3; 13 are only in this paper.

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Questions 1–20 · 1 mark each

  1. Q1 1 mark

    Also in 31/3/1 (Q3), 31/3/3 (Q4)

    A chemical compound used in glass, soap and paper industries is
    1. (a)Washing Soda
    2. (b)Baking Soda
    3. (c)Bleaching Powder
    4. (d)Common Salt
    Show answer
    (a) Washing Soda
    Step-by-step solution →
  2. Q2 1 mark

    Also in 31/3/1 (Q7), 31/3/3 (Q1)

    Which one of the following reactions is different from the remaining three ?
    1. (a)
    2. (b)
    3. (c)
    4. (d)
    Show answer
    (b)
    Step-by-step solution →
  3. Q3 1 mark

    Also in 31/3/1 (Q1), 31/3/3 (Q2)

    Identify the product ‘X’ obtained in the following chemical reaction :
    1. (a)Quick lime
    2. (b)Gypsum
    3. (c)Lime Stone
    4. (d)Plaster of Paris
    Show answer
    (a) Quick lime
    Step-by-step solution →
  4. Q4 1 mark

    Also in 31/3/1 (Q2), 31/3/3 (Q3)

    Select a pair of natural indicator from the following :
    1. (a)Litmus and methyl orange
    2. (b)Turmeric and Litmus
    3. (c)Phenolphthalein and methyl orange
    4. (d)Methyl orange and Turmeric
    Show answer
    (b) Turmeric and Litmus
    Step-by-step solution →
  5. Q5 1 mark

    Also in 31/3/1 (Q6)

    Which one of the following hydrocarbons is different from the others ?
    1. (a)
    2. (b)
    3. (c)
    4. (d)
    Show answer
    (b)
    Step-by-step solution →
  6. Q6 1 mark
    Select saponification reaction from the following :
    1. (a)
    2. (b)
    3. (c)
    4. (d)
    Show answer
    (c)
    Step-by-step solution →
  7. Q7 1 mark

    Also in 31/3/1 (Q5)

    Consider the following Chemical equation : In order to balance this chemical equation, the values of , , and must be
    1. (a)1, 6, 2 and 3
    2. (b)1, 6, 3 and 2
    3. (c)2, 6, 2 and 3
    4. (d)2, 6, 3 and 2
    Show answer
    (a) 1, 6, 2 and 3
    Step-by-step solution →
  8. Q8 1 mark

    Also in 31/3/1 (Q14), 31/3/3 (Q13)

    In case of four wires of same material, the resistance will be minimum if the diameter and length of the wire respectively are
    1. (a) and
    2. (b) and
    3. (c) and
    4. (d) and
    Show answer
    (d) and
    Step-by-step solution →
  9. Q9 1 mark

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  10. Q10 1 mark

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  11. Q11 1 mark

    Also in 31/3/1 (Q12)

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  12. Q12 1 mark

    Also in 31/3/1 (Q11), 31/3/3 (Q8)

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  13. Q13 1 mark

    Also in 31/3/1 (Q9)

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  14. Q14 1 mark

    Also in 31/3/1 (Q13)

    The Phenomena of light involved in the formation of a rainbow in the sky are
    1. (a)Refraction, dispersion and reflection
    2. (b)Refraction, dispersion and total internal reflection
    3. (c)Dispersion, scattering and reflection
    4. (d)Dispersion, refraction and internal reflection
    Show answer
    (d) Dispersion, refraction and internal reflection
    Step-by-step solution →
  15. Q15 1 mark

    Also in 31/3/1 (Q16), 31/3/3 (Q15)

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  16. Q16 1 mark

    Also in 31/3/1 (Q15), 31/3/3 (Q16)

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  17. Q17 1 mark
    Q. 17 to 20 consists of two statements – Assertion (A) and Reason (R), answer these questions selecting the appropriate option given below :

    Assertion (A): Oxides of metals show basic characters.

    Reason (R): Oxides of metals react with acid to form salt and water.

    1. (a)Both (A) and (R) are true and (R) is the correct explanation of (A).
    2. (b)Both (A) and (R) are true and (R) is not correct explanation of (A).
    3. (c)(A) is true, but (R) is false.
    4. (d)(A) is false, but (R) is true.
    Show answer
    (a) Both (A) and (R) are true and (R) is the correct explanation of (A).
    Step-by-step solution →
  18. Q18 1 mark

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

    Q. Nos. 17 to 20 consists of two statements – Assertion (A) and Reason (R), answer these questions selecting the appropriate option given below :

    Assertion (A): Red light signals are used to stop the vehicles on the road.

    Reason (R): Red coloured light is scattered the most so as to be visible from a large distance.

    1. (a)Both (A) and (R) are true and (R) is the correct explanation of (A).
    2. (b)Both (A) and (R) are true and (R) is not correct explanation of (A).
    3. (c)(A) is true, but (R) is false.
    4. (d)(A) is false, but (R) is true.
    Show answer
    (c) (A) is true, but (R) is false.
    Step-by-step solution →
  19. Q19 1 mark

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  20. Q20 1 mark

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Questions 21–26 · 2 marks each

  1. Q21 2 marks
    When a few drops of Barium chloride solution are added to an aqueous solution of Sodium sulphate, a white precipitate is obtained.
    1. (a)Write balanced chemical equation for the reaction involved.
    2. (b)What is the other name of this precipitation reaction ? Why is it called so ?
    Show answer

    Barium sulphate precipitates; the reaction is also called double displacement because the two reactants exchange their ions.

    (a) . The white precipitate is barium sulphate. (b) It is a double displacement reaction. It is called so because the two reactants exchange their ions: barium ions combine with sulphate ions to form barium sulphate, and sodium ions combine with chloride ions to form sodium chloride.
    Step-by-step solution →
  2. Q22 2 marks

    Also in 31/3/1 (Q24), 31/3/3 (Q25)

    When do we say that a particular person is suffering from hypermetropia ? List two causes of this defect. Name the type of lens used to correct this defect.
    Show answer

    In hypermetropia nearby objects are not seen clearly because the near point is beyond 25 cm; a convex lens corrects it.

    A person is said to suffer from hypermetropia (far-sightedness) when distant objects are seen clearly but nearby objects cannot be seen distinctly: the near point of the eye is farther away than the normal near point of 25 cm, because light from a nearby object is focused behind the retina. Two causes: (i) the focal length of the eye lens is too long; (ii) the eyeball has become too small. The defect is corrected with a convex (converging) lens of suitable power.
    Step-by-step solution →
  3. Q23(a) 2 marks

    Also in 31/3/1 (Q25(a)), 31/3/3 (Q24(a))

    Draw a labelled diagram to show the pattern of magnetic field lines produced due to a current carrying straight conductor. Mark on it the direction of current in the conductor and the direction of magnetic field lines.
    Show answer

    The field lines are concentric circles round the wire, anticlockwise as seen from above for an upward current.

    Diagram (see the figure): a straight conductor passes vertically through a horizontal cardboard sheet, with the current flowing upwards, marked by an arrow. The magnetic field lines are concentric circles on the card with the conductor at their centre; by the right-hand thumb rule they run anticlockwise when seen from above, marked by arrows on the circles. The circles are closer together near the wire, where the field is stronger. If the current is reversed, the field lines run clockwise.
    Step-by-step solution →
    OR
    Q23(b) 2 marks

    Also in 31/3/1 (Q25(b)), 31/3/3 (Q24(b))

    Name the device used to magnetise a piece of magnetic material. Draw a labelled diagram to show the arrangement used for the magnetisation of a cylinder made of soft iron.
    Show answer

    A solenoid magnetises the soft iron cylinder placed inside it when current flows, forming an electromagnet.

    Device: a solenoid, a coil of many circular turns of insulated copper wire wrapped closely in the shape of a cylinder. Arrangement (see the figure): the soft iron cylinder is placed inside the solenoid, and the ends of the solenoid are connected to a battery through a key. When the key is closed, current flows through the coil and the strong magnetic field inside the solenoid magnetises the cylinder. With the current shown (flowing down the front of each turn), the left end of the cylinder becomes S and the right end N. The magnet formed is an electromagnet.
    Step-by-step solution →
  4. Q24 2 marks

    Also in 31/3/1 (Q22), 31/3/3 (Q23)

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  5. Q25(a) 2 marks

    Also in 31/3/1 (Q23(a)), 31/3/3 (Q22(a))

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    OR
    Q25(b) 2 marks

    Also in 31/3/1 (Q23(b)), 31/3/3 (Q22(b))

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  6. Q26 2 marks

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Questions 27–33 · 3 marks each

  1. Q27(a) 3 marks

    Also in 31/3/1 (Q28(a))

    State giving reason the reduction process to obtain the following metals from their compounds :
    1. (i)Mercury,
    2. (ii)Copper and
    3. (iii)Sodium
    Show answer

    Mercury and copper are obtained by heating alone as they are low in the activity series; sodium by electrolysis of molten sodium chloride.

    (i) Mercury: reduction by heating alone. Mercury is low in the activity series; its ore cinnabar is heated in air to give mercuric oxide, which on further heating is reduced to mercury: ; . (ii) Copper: reduction by heating alone. Copper is also low in the activity series; its ore is heated in air: , then . (iii) Sodium: electrolytic reduction of molten sodium chloride. Sodium is very high in the activity series and has more affinity for oxygen than carbon, so carbon cannot reduce its compounds; sodium is deposited at the cathode () and chlorine is liberated at the anode.
    Step-by-step solution →
    OR
    Q27(b) 3 marks

    Also in 31/3/1 (Q28(b))

    State giving reason for the change in appearance observed when each of the following metal is exposed to atmospheric air for some time :
    1. (i)Silver,
    2. (ii)Copper and
    3. (iii)Iron
    Show answer

    Silver turns black (silver sulphide), copper gets a green coating (basic copper carbonate) and iron gets reddish-brown rust; all are corrosion.

    (i) Silver: the shiny surface turns black, because silver reacts with sulphur (present in the air as compounds such as hydrogen sulphide) to form a black coating of silver sulphide, . (ii) Copper: it loses its shiny brown surface and gains a green coating, because copper reacts with the moist carbon dioxide in the air to form basic copper carbonate, . (iii) Iron: it gets a coating of a reddish-brown flaky substance called rust, because iron reacts with the oxygen and moisture of the air to form hydrated iron(III) oxide, . All three changes are examples of corrosion.
    Step-by-step solution →
  2. Q28 3 marks
    Write electronic configuration of Sodium (At. No.11) and Oxygen (At. No. 8) and show the formation of the ionic compound obtained when these two elements combine. Name anion and cation present in the compound.
    Show answer

    Two sodium atoms give one electron each to oxygen, forming with cations and anions.

    Electronic configurations: Na (11) = 2, 8, 1; O (8) = 2, 6. Formation: each sodium atom loses its one outer electron, , and the oxygen atom gains two electrons, ; so two sodium atoms give one electron each to one oxygen atom, (sodium oxide). Both ions now have the configuration 2, 8. Cation: sodium ion, ; anion: oxide ion, .
    Step-by-step solution →
  3. Q29 3 marks

    Also in 31/3/1 (Q30)

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  4. Q30 3 marks

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  5. Q31 3 marks

    Also in 31/3/1 (Q32), 31/3/3 (Q31)

    Explain in brief the function of an electric fuse in a domestic circuit. An electric heater of current rating 3 kW; 220 V is to be operated in an electric circuit of rating 5 A. What is likely to happen when the heater is switched ‘ON’ ? Justify your answer with necessary calculation.
    Show answer

    The heater draws about A, more than the 5 A rating, so the fuse melts and breaks the circuit.

    Function of a fuse: an electric fuse is a safety device connected in series in the live wire of a circuit; it is a wire of a metal or an alloy of suitable low melting point. When the current exceeds its rated value, because of overloading or a short circuit, the fuse wire heats up, melts and breaks the circuit, protecting the appliances and the wiring from damage. Calculation: the heater draws A. This is much more than the 5 A rating of the circuit, so switching the heater on overloads the circuit: the fuse wire melts and the circuit breaks, and the heater stops working. It should be used on the 15 A circuit.
    Step-by-step solution →
  6. Q32 3 marks

    Also in 31/3/1 (Q31), 31/3/3 (Q33)

    Name and explain the phenomenon of light due to which the path of a beam of light becomes visible when it enters a smoke filled room through a small hole. Also state the dependence of colour of the light we receive on the size of the particle of the medium through which the beam of light passes.
    Show answer

    Tyndall effect: smoke particles scatter the beam, making its path visible; fine particles scatter mainly blue, larger ones longer wavelengths.

    The phenomenon is the Tyndall effect, the scattering of light by colloidal particles. The smoke in the room consists of very fine particles; when the beam entering through the hole strikes them, they scatter the light in all directions. Part of this scattered light reaches our eyes from every point along the beam, so its path becomes visible. The colour of the scattered light depends on the size of the particles: very fine particles scatter mainly blue light, larger particles scatter light of longer wavelengths, and if the particles are large enough the scattered light may even appear white.
    Step-by-step solution →
  7. Q33 3 marks
    List three advantages of parallel circuits in domestic wiring.
    Show answer

    Each appliance gets the full 220 V, works independently with its own switch, and draws the current it needs.

    (i) Each appliance gets the same, full mains voltage (220 V), because every branch is connected directly across the live and neutral wires. (ii) Each appliance can be switched on or off independently with its own switch, and if one appliance is switched off or fails, the others continue to work. (iii) Each appliance draws the current it needs according to its own resistance (the current divides among the branches, and the total resistance is less than the smallest individual resistance), so gadgets needing very different currents, such as a bulb and a heater, work properly together.
    Step-by-step solution →

Questions 34–36 · 5 marks each

  1. Q34(a) 5 marks

    Also in 31/3/1 (Q36(a)), 31/3/3 (Q34(a))

    (i) Draw a ray diagram to show the path of the refracted ray in each of the following cases : A ray of light incident on a concave lens
    1. (1)parallel to its principal axis, and
    2. (2)is directed towards its principal focus. (ii) A 4 cm tall object is placed perpendicular to the principal axis of convex lens of focal length 24 cm. The distance of object from the lens is 16 cm. Find the position and size of image formed.
    Show answer

    Concave lens: the parallel ray appears to come from ; the ray towards emerges parallel. Image: cm, 12 cm tall, virtual, erect.

    (i) (1) The ray parallel to the principal axis, after refraction through the concave lens, diverges and appears to come from the principal focus on the same side as the incident light. (2) The ray directed towards the principal focus on the other side of the lens, after refraction, emerges parallel to the principal axis. The ray diagrams show both paths. (ii) With cm and cm, , so cm: the image is formed 48 cm from the lens on the same side as the object. , so cm: the image is 12 cm tall, virtual, erect and magnified three times.
    Step-by-step solution →
    OR
    Q34(b) 5 marks

    Also in 31/3/1 (Q36(b)), 31/3/3 (Q34(b))

    (i) Draw a ray diagram to show the path of the reflected ray in each of the following cases : A ray of light incident on a convex mirror
    1. (1)parallel to its principal axis, and
    2. (2)is directed towards its principal focus (ii) A 1.5 cm tall candle flame is placed perpendicular to the principal axis of a concave mirror of focal length 12 cm. If the distance of the flame from the pole of the mirror is 18 cm, use mirror formula to determine the position and size of the image formed.
    Show answer

    Convex mirror: parallel ray appears to come from F, ray towards F returns parallel. Concave mirror: cm, real, inverted, 3 cm tall.

    (i) (1) The ray parallel to the principal axis, after reflection from the convex mirror, diverges and appears to come from the principal focus behind the mirror. (2) The ray directed towards the principal focus behind the mirror, after reflection, travels parallel to the principal axis. The ray diagrams show both paths. (ii) With cm and cm, , so cm: the image is formed 36 cm in front of the mirror, on the same side as the flame. , so cm: the image is 3 cm tall, real, inverted and twice the size of the flame.
    Step-by-step solution →
  2. Q35(a) 5 marks

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    Q35(b) 5 marks

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  3. Q36(a) 5 marks

    Also in 31/3/1 (Q34(a)), 31/3/3 (Q36(a))

    (i) Five solutions A, B, C, D and E when tested with pH paper showed pH as 4, 1, 13, 7 and 10 respectively. Which solution is :
    1. (1)Strongly acidic
    2. (2)Strongly alkaline
    3. (3)Weakly acidic
    4. (4)Neutral and
    5. (5)Weakly alkaline ? Arrange the solutions in increasing order of ion concentration. (ii) Write the name and formula of (1) an acidic salt and (2) a basic salt giving the name of the parent acid and parent base used to form the salt in each case.
    Show answer

    ion concentration increases in the order ; is an acidic salt and a basic salt.

    (i) (1) Strongly acidic: B (pH 1); (2) Strongly alkaline: C (pH 13); (3) Weakly acidic: A (pH 4); (4) Neutral: D (pH 7); (5) Weakly alkaline: E (pH 10). Increasing order of ion concentration: , because the lower the pH, the higher the ion concentration. (ii) (1) Acidic salt: ammonium chloride, ; parent acid hydrochloric acid (, strong) and parent base ammonium hydroxide (, weak). (2) Basic salt: sodium carbonate, ; parent acid carbonic acid (, weak) and parent base sodium hydroxide (, strong).
    Step-by-step solution →
    OR
    Q36(b) 5 marks

    Also in 31/3/1 (Q34(b)), 31/3/3 (Q36(b))

    Name and state in brief the process which is used to prepare sodium hydroxide from sodium chloride. In this process along with the main product two gases ‘X’ and ‘Y’ are also given off at the two electrodes. Name ‘X’ and ‘Y’ specifying the name of their respective electrode at which each gas is obtained. One of these gases when reacts with dry calcium hydroxide produces a compound ‘Z’ which is widely used in water treatment plants and textile industries. Name Z and write chemical equation for the reaction involved in its formation.
    Show answer

    Chlor-alkali process; X is chlorine at the anode, Y is hydrogen at the cathode; Z is bleaching powder, .

    The process is the chlor-alkali process: when electricity is passed through brine (an aqueous solution of sodium chloride), it decomposes to form sodium hydroxide, and chlorine and hydrogen are given off: . X is chlorine gas, given off at the anode (positive electrode), and Y is hydrogen gas, given off at the cathode (negative electrode). Sodium hydroxide solution is formed near the cathode. Z is bleaching powder, , formed when chlorine reacts with dry slaked lime: .
    Step-by-step solution →

Questions 37–39 · 4 marks each

  1. Q37 4 marks

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

    More than three million carbon compounds have been discovered in the field of chemistry. The diversity of these compounds is due to the capacity of carbon atoms for bonding with one another as well as with other atoms. Most of the carbon compounds are poor conductors of electricity and have low melting and boiling points.
    C H O , (i) C O (ii)
    Formulae (i) and (ii) for part (b)
    1. (a)Write the molecular formula of first two members of homologous series having functional group .[1]
    2. (b)Given below are the formulae of some functional groups : Write the name of these functional groups.[1]
    3. (c)(i)What would be observed on adding a 5% alkaline potassium permanganate drop by drop to some warm ethanol taken in a test tube ? State the role of in the reaction and write the chemical equation for the reaction involved.[2]
    4. (c)(ii)OR : Write the name of the compound formed when ethanol is heated at 443 K temperature with excess of conc. . What is the role of conc. in the reaction ? Write the chemical equation for the reaction involved.[2]
    Show answer

    The first two bromo compounds are and ; alkaline permanganate oxidises ethanol to ethanoic acid; hot conc. sulphuric acid gives ethene.

    (a) (bromomethane) and (bromoethane). (b) (i) Aldehyde group, ; (ii) Ketone group, . (c)(i) The purple colour of alkaline potassium permanganate disappears as the first drops are added to the warm ethanol; when it is added in excess, the purple colour persists. Alkaline acts as an oxidising agent: it oxidises ethanol to ethanoic acid. (c)(ii) [OR] The compound formed is ethene, . Concentrated sulphuric acid acts as a dehydrating agent, removing water from ethanol.
    Step-by-step solution →
  2. Q38 4 marks

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

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  3. Q39 4 marks

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

    In a domestic circuit five LED bulbs are arranged as shown. The source voltage is 220 V and the power rating of each bulb is marked in the circuit diagram. Based on the following circuit diagram, answer the following questions :
    + − + − K1 K2 22 W, 220 V A 11 W; 55 V B 11 W; 55 V C 11 W; 55 V D 11 W; 55 V E
    1. (a)State what happens when (i) key is closed. (ii) key is closed.[1]
    2. (b)Find the current drawn by the bulb B when it glows.[1]
    3. (c)Calculate (i) the resistance of bulb B, and (ii) total resistance of the combination of four bulbs B, C, D and E.[2]
    4. (c)OR : What would happen to the glow of all the bulbs in the circuit when keys and both are closed and the bulb C suddenly get fused ? Give reason to justify your answer.[2]
    Show answer

    Only A glows with , B to E with ; A, , total ; if C fuses, A still glows.

    (a) (i) When is closed, only bulb A is connected across the 220 V source; it gets its rated voltage and glows with full brightness (22 W), while B, C, D and E do not glow. (ii) When is closed, bulbs B, C, D and E are connected in series across 220 V; each gets 55 V, its rated voltage, so all four glow normally, while A does not glow. (b) A. (c)(i) . (c)(ii) The four bulbs are in series, so . OR (c) Bulbs B, D and E stop glowing, along with the fused bulb C, because they are in series with C: the break in C opens the only path through them, so no current flows in that branch. Bulb A goes on glowing with the same brightness, because it is on a separate parallel branch that still has the full 220 V across it.
    Step-by-step solution →

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