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CBSE Class 10 Science 2024 — Paper 31/5/1

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

40 of its 45 questions also appear in 31/5/2 or 31/5/3; 5 are only in this paper.

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Questions 1–20 · 1 mark each

  1. Q1 1 mark

    Also in 31/5/3 (Q5)

    To balance the following chemical equation, the values of the coefficients , and must be respectively :
    1. (a)
    2. (b)
    3. (c)
    4. (d)
    Show answer
    (c)
    Step-by-step solution →
  2. Q2 1 mark

    Also in 31/5/2 (Q5), 31/5/3 (Q6)

    Which of the following is a redox reaction, but not a combination reaction ?
    1. (a)
    2. (b)
    3. (c)
    4. (d)
    Show answer
    (d)
    Step-by-step solution →
  3. Q3 1 mark

    Also in 31/5/2 (Q6)

    The salt present in tooth enamel is :
    1. (a)Calcium phosphate
    2. (b)Magnesium phosphate
    3. (c)Sodium phosphate
    4. (d)Aluminium phosphate
    Show answer
    (a) Calcium phosphate
    Step-by-step solution →
  4. Q4 1 mark

    Also in 31/5/2 (Q8), 31/5/3 (Q7)

    An aqueous solution of sodium chloride is prepared in distilled water. The pH of this solution is :
    1. (a)
    2. (b)
    3. (c)
    4. (d)
    Show answer
    (c)
    Step-by-step solution →
  5. Q5 1 mark

    Also in 31/5/2 (Q9), 31/5/3 (Q8)

    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 :
    1. (a)
    2. (b)
    3. (c)
    4. (d)
    Show answer
    (b)
    Step-by-step solution →
  6. Q6 1 mark

    Also in 31/5/2 (Q2), 31/5/3 (Q9)

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  7. Q7 1 mark

    Also in 31/5/3 (Q4)

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  8. Q8 1 mark

    Also in 31/5/2 (Q3), 31/5/3 (Q10)

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  9. Q9 1 mark

    Also in 31/5/2 (Q4), 31/5/3 (Q2)

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  10. Q10 1 mark

    Also in 31/5/2 (Q13), 31/5/3 (Q1)

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  11. Q11 1 mark

    Also in 31/5/2 (Q14), 31/5/3 (Q13)

    How will the image formed by a convex lens be affected, if the upper half of the lens is wrapped with a black paper ?
    1. (a)The size of the image formed will be one-half of the size of the image due to complete lens.
    2. (b)The image of upper half of the object will not be formed.
    3. (c)The brightness of the image will reduce.
    4. (d)The lower half of the inverted image will not be formed.
    Show answer
    (c) The brightness of the image will reduce.
    Step-by-step solution →
  12. Q12 1 mark

    Also in 31/5/2 (Q15), 31/5/3 (Q14)

    The phenomena of light involved in the formation of rainbow are :
    1. (a)Refraction, reflection and dispersion
    2. (b)Refraction, dispersion and internal reflection
    3. (c)Reflection, dispersion and internal reflection
    4. (d)Refraction, dispersion, scattering and total internal reflection
    Show answer
    (b) Refraction, dispersion and internal reflection
    Step-by-step solution →
  13. Q13 1 mark

    Also in 31/5/2 (Q16), 31/5/3 (Q15)

    The colour of light for which the refractive index of glass is minimum, is :
    1. (a)Red
    2. (b)Yellow
    3. (c)Green
    4. (d)Violet
    Show answer
    (a) Red
    Step-by-step solution →
  14. Q14 1 mark

    Also in 31/5/2 (Q12), 31/5/3 (Q16)

    The current carrying device which produces a magnetic field similar to that of a bar magnet is :
    1. (a)A straight conductor
    2. (b)A circular loop
    3. (c)A solenoid
    4. (d)A circular coil
    Show answer
    (c) A solenoid
    Step-by-step solution →
  15. Q15 1 mark

    Also in 31/5/2 (Q11), 31/5/3 (Q12)

    A uniform magnetic field exists in the plane of paper as shown in the diagram. In this field, an electron () and a positron () enter as shown. The electron and positron experience forces :
    e⁻ p⁺ Uniform magnetic field
    1. (a)both pointing into the plane of the paper.
    2. (b)both pointing out of the plane of the paper.
    3. (c)pointing into the plane of the paper and out of the plane of the paper respectively.
    4. (d)pointing out of the plane of the paper and into the plane of the paper respectively.
    Show answer
    (a) both pointing into the plane of the paper.
    Step-by-step solution →
  16. Q16 1 mark

    Also in 31/5/2 (Q10), 31/5/3 (Q11)

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  17. Q17 1 mark

    Also in 31/5/2 (Q18), 31/5/3 (Q18)

    Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to this question from the codes (a), (b), (c) and (d) as given below.

    Assertion (A): Some vegetable oils are healthy.

    Reason (R): Vegetable oils generally have long unsaturated carbon chains.

    1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
    2. (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the 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 the Assertion (A).
    Step-by-step solution →
  18. Q18 1 mark

    Also in 31/5/2 (Q19), 31/5/3 (Q19)

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  19. Q19 1 mark

    Also in 31/5/2 (Q20), 31/5/3 (Q17)

    Two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to this question from the codes (a), (b), (c) and (d) as given below.

    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.

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

    Also in 31/5/2 (Q17)

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Questions 21–26 · 2 marks each

  1. Q21(a) 2 marks

    Also in 31/5/2 (Q23(a)), 31/5/3 (Q23(a))

    Copper powder is taken in a china dish and heated over a burner. Name the product formed and state its colour. Write the chemical equation for the reaction involved.
    Show answer

    Heated copper powder forms black copper(II) oxide, : .

    The product is copper(II) oxide, , which is black: the reddish-brown surface of the copper powder becomes coated with it. Equation: .
    Step-by-step solution →
    OR
    Q21(b) 2 marks

    Also in 31/5/2 (Q23(b)), 31/5/3 (Q23(b))

    Write chemical equation for the chemical reaction which occurs when the aqueous solutions of barium chloride and sodium sulphate react together. Write the symbols of the ions present in the compound precipitated in the reaction.
    Show answer

    The precipitate is barium sulphate, , which contains and ions.

    . The precipitate is barium sulphate, , a white solid; the ions present in it are the barium ion, , and the sulphate ion, .
    Step-by-step solution →
  2. Q22 2 marks
    The melting and boiling points of carbon compounds are generally low and they are largely non-conductors of electricity. State two conclusions based on these two properties.
    Show answer

    Low melting and boiling points show weak forces between molecules; non-conduction shows covalent bonding with no ions.

    Conclusion 1: the low melting and boiling points show that the forces of attraction between the molecules of carbon compounds are not very strong, so little heat is needed to separate the molecules. Conclusion 2: being largely non-conductors of electricity shows that the bonding in carbon compounds does not give rise to any ions; their atoms are held together by covalent bonds, formed by the sharing of electrons.
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  3. Q23(a) 2 marks

    Also in 31/5/2 (Q21(a)), 31/5/3 (Q21(a))

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    OR
    Q23(b) 2 marks

    Also in 31/5/2 (Q21(b)), 31/5/3 (Q21(b))

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  4. Q24 2 marks

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  5. Q25 2 marks
    The heat produced at a point due to concentration of sunlight by a convex lens burns a paper.
    1. (a)Explain why it happens.
    2. (b)Name the term (in the context of the lens used) given to the point at which the paper starts burning. What does the bright spot formed on the paper represent ?
    Show answer

    The lens focuses parallel sunlight at its principal focus; the bright spot there is the real image of the Sun.

    (a) The Sun is very far away, so its rays reach the lens parallel to one another. The convex lens is a converging lens: it refracts these parallel rays so that they meet at one point. All the light energy falling on the lens is concentrated into a tiny spot, and the heat produced there raises the temperature of the paper until it burns. (b) The point is the principal focus () of the convex lens. The bright spot on the paper is the real image of the Sun formed by the lens.
    Step-by-step solution →
  6. Q26 2 marks
    An electric source can supply a charge of 500 coulomb. If the current drawn by a device is 25 mA, find the time in which the electric source will be discharged completely.
    Show answer

    The electric source is discharged completely in s (about 5.56 hours).

    s s, which is about h ( h min s).
    Step-by-step solution →

Questions 27–33 · 3 marks each

  1. Q27 3 marks

    Also in 31/5/2 (Q29)

    Study the experimental set-up shown in the diagram and write chemical equation for the chemical 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 sulphate solution.
    Stand Test tube Thread Copper sulphate solution Iron nail
    Show answer

    Iron displaces copper from copper sulphate, a displacement reaction; zinc and magnesium (or aluminium), being above copper, can replace iron.

    Equation: ; the iron nails become coated with brownish copper and the blue colour of the solution fades. Type: displacement reaction, a reaction in which a more reactive element displaces a less reactive element from its compound (here iron, being more reactive, displaces copper from copper sulphate). Two other metals that can be used in place of iron: zinc and magnesium, both above copper in the reactivity series (; ). Aluminium also lies above copper and displaces it, so any two of magnesium, aluminium and zinc are correct. Lead lies above copper as well, but the lead sulphate it forms is insoluble and coats the lead, so in copper sulphate solution the reaction soon stops and the copper coating and the fading blue colour do not develop.
    Step-by-step solution →
  2. Q28 3 marks

    Also in 31/5/3 (Q30)

    Name the ore of mercury and state the form in which it is found in nature. Write the chemical equations along with the condition required for the reactions involved in the extraction of mercury from its ore.
    Show answer

    The ore is cinnabar, , found as a sulphide; roasting gives , which on heating gives mercury.

    Ore: cinnabar, . Mercury is found in nature in the combined state, as its sulphide (mercury(II) sulphide), so cinnabar is a sulphide ore. Extraction: mercury is low in the activity series, so heating alone is enough. (1) Roasting, in which the ore is heated strongly in air: . (2) On further heating the mercuric oxide is reduced to mercury: .
    Step-by-step solution →
  3. Q29 3 marks

    Also in 31/5/2 (Q31)

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  4. Q30 3 marks

    Also in 31/5/3 (Q31)

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  5. Q31(a) 3 marks

    Also in 31/5/2 (Q27(a)), 31/5/3 (Q33(a))

    Study the diagram given below and answer the questions that follow :
    1. (i)Name the defect of vision depicted in this diagram stating the part of the eye responsible for this condition.
    2. (ii)List two causes of this defect.
    3. (iii)Name the type of lens used to correct this defect and state its role in this case.
    N N′
    Show answer

    Hypermetropia arises when the eye lens has too long a focal length (or the eyeball is too small); a convex lens adds converging power.

    (i) The defect is hypermetropia (far-sightedness): light from the nearby point N′ is brought to a focus behind the retina. The part of the eye responsible is the eye lens, whose focal length has become too long (an eyeball that has become too small has the same effect). (ii) Two causes: (1) the focal length of the eye lens is too long; (2) the eyeball has become too small. (iii) A convex (converging) lens of suitable power. It provides the additional converging power needed: it bends the rays from a nearby object so that they appear to come from a farther point that the eye can focus, so the image is formed on the retina.
    Step-by-step solution →
    OR
    Q31(b) 3 marks

    Also in 31/5/2 (Q27(b)), 31/5/3 (Q33(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.
    Show answer

    Dispersion is the splitting of white light into VIBGYOR, because the refractive index of glass differs for each colour.

    Dispersion is the splitting of white light into its component colours, violet, indigo, blue, green, yellow, orange and red (VIBGYOR), when it passes through a glass prism; the band of colours obtained is called a spectrum. Cause: the different colours of light bend through different angles with respect to the incident ray as they pass through the prism, because the refractive index of glass (and so the speed of light in glass) is different for different colours; red bends the least and violet the most, so the colours emerge along different paths and separate. Diagram: a beam of white light entering one face of the prism bends towards the normal and starts to split; at the second face the colours bend away from the normal and fall on a screen as a spectrum, with red least deviated and violet, nearest the base, most deviated.
    Step-by-step solution →
  6. Q32 3 marks
    1. (a)What happens when a bundle of wires of soft iron is placed inside the coil of a solenoid carrying a steady current ? Name the device obtained. Why is it called so ?
    2. (b)Draw the magnetic field lines inside a current carrying solenoid. What does this pattern of magnetic field lines indicate ?
    Show answer

    The soft iron becomes a strong magnet, forming an electromagnet; the field inside a solenoid is uniform.

    (a) The strong magnetic field inside the current-carrying solenoid magnetises the soft iron, so the bundle becomes a strong magnet, one end north and the other south, as long as the current flows. The device obtained is an electromagnet. It is called so because it is a magnet produced by an electric current: it behaves as a magnet only while the current flows through the coil and loses its magnetism when the current is switched off. (b) Inside the solenoid the magnetic field lines are parallel straight lines running from the south-pole end to the north-pole end (outside, they curve from N to S). This pattern indicates that the magnetic field is the same at all points inside the solenoid, that is, the field inside is uniform.
    Step-by-step solution →
  7. Q33 3 marks

    Also in 31/5/2 (Q33), 31/5/3 (Q28)

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Questions 34–36 · 5 marks each

  1. Q34(a) 5 marks

    Also in 31/5/3 (Q35(a))

    A few crystals of ferrous sulphate were taken in a dry boiling tube and heated. Tiny water droplets were observed in the tube after some time.
    1. (i)From where did these water droplets appear ? Explain.
    2. (ii)What colour change will be observed during heating ?
    3. (iii)How many molecules of water are attached per molecule of crystal ? Write the molecular formula of crystalline forms of (I) Copper sulphate, and (II) Sodium carbonate.
    4. (iv)State how is Plaster of Paris obtained from gypsum. Write two uses of Plaster of Paris.
    Show answer

    Water of crystallisation of condenses as droplets; green crystals turn white; heating gypsum at 373 K gives Plaster of Paris.

    (i) The droplets are the water of crystallisation of ferrous sulphate. The crystals are ; heating drives this water out as vapour, which condenses as droplets on the cooler upper part of the dry tube. (ii) The pale green crystals turn white as they lose their water (anhydrous ); on stronger heating they decompose, , leaving reddish-brown ferric oxide. (iii) Seven molecules of water per formula unit (). (I) Copper sulphate: ; (II) Sodium carbonate: . (iv) Gypsum is heated at K: , and the hemihydrate is Plaster of Paris. Uses: supporting fractured bones in the right position; making toys and materials for decoration (also for making surfaces smooth).
    Step-by-step solution →
    OR
    Q34(b) 5 marks

    Also in 31/5/3 (Q35(b))

    An acid ‘X’ present in tamarind when mixed with ‘Y’, produces a mixture ‘Z’. ‘Z’ on addition to a dough when heated makes cakes soft and spongy. ‘Y’ is prepared from common salt and helps in faster cooking.
    1. (i)Write the common names of ‘X’, ‘Y’ and ‘Z’, and the chemical formula of ‘Y’.
    2. (ii)How is ‘Y’ prepared and how does it help in making cakes soft and spongy ? Illustrate the reaction with suitable chemical equation.
    3. (iii)Write the name and chemical formula of a mild base other than ‘Y’ used as an antacid.
    Show answer

    X is tartaric acid, Y baking soda (), Z baking powder; it releases carbon dioxide that makes cakes spongy; antacid: .

    (i) X: tartaric acid; Y: baking soda (sodium hydrogencarbonate), ; Z: baking powder. (ii) Y is prepared from sodium chloride: . When baking powder in the dough is heated or mixed with water, the baking soda reacts with the acid and releases carbon dioxide: (heated alone, ). The carbon dioxide makes the dough rise, so the cake is soft and spongy. (iii) Magnesium hydroxide (milk of magnesia), .
    Step-by-step solution →
  2. Q35(a) 5 marks

    Also in 31/5/2 (Q36(a)), 31/5/3 (Q34(a))

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    Q35(b) 5 marks

    Also in 31/5/2 (Q36(b)), 31/5/3 (Q34(b))

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  3. Q36(a) 5 marks

    Also in 31/5/2 (Q35(a))

    1. (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.
    2. (ii)Draw a schematic diagram of a circuit consisting of a battery of four V cells, a resistor, a resistor and a resistor and a plug key, all connected in series. Now find (I) the electric current passing through the circuit, and (II) potential difference across the resistor when the plug key is closed.
    Show answer

    The current falls to one-third (Ohm's law); the circuit current is A and the resistor has V across it.

    (i) The current also decreases to one-third of its initial value, . The law is Ohm's law: the potential difference across the ends of a conductor is directly proportional to the current through it, provided its temperature remains the same; . (ii) Circuit: the four V cells in series (battery of V), the , and resistors and the plug key, all in one loop. (I) A. (II) V.
    Step-by-step solution →
    OR
    Q36(b) 5 marks

    Also in 31/5/2 (Q35(b))

    1. (i)When is the potential difference between two points said to be 1 volt ?
    2. (ii)A copper wire has a diameter of mm and resistivity of . What will be the length of this wire to make its resistance ? How much does the resistance change, if the diameter of the wire is doubled ?
    Show answer

    One volt is 1 J of work per coulomb; the wire is about 27.5 m long; doubling its diameter gives .

    (i) The potential difference between two points is volt when joule of work is done to move a charge of coulomb from one point to the other (). (ii) , so m. If the diameter is doubled, the area becomes times as large, so the resistance becomes one-fourth: , a decrease of .
    Step-by-step solution →

Questions 37–39 · 4 marks each

  1. Q37 4 marks

    Also in 31/5/2 (Q38), 31/5/3 (Q38)

    Carbon is a versatile element that forms the basis of all living organisms and many of the things we use. A large variety of compounds is formed because of its tetravalency. Compounds of carbon are formed with oxygen, hydrogen, nitrogen, sulphur, chlorine and many other elements. Answer the following questions :
    1. (a)What are hydrocarbons ?[1]
    2. (b)List two properties by virtue of which carbon can form a large number of compounds.[1]
    3. (c)(i)Write the formula of the functional group present in (1) aldehydes, and (2) ketones. Write chemical equation for the reaction that occurs between ethanoic acid and ethanol in the presence of a catalyst.[2]
    4. (c)(ii)OR : What are structural isomers ? Write the structures of two isomers of butane ().[2]
    Show answer

    Hydrocarbons contain only carbon and hydrogen; carbon shows catenation and tetravalency; aldehydes have and ketones have .

    (a) Hydrocarbons are compounds of carbon and hydrogen only, for example methane, . (b) Catenation (carbon atoms join with one another into long chains, branched chains and rings) and tetravalency (each carbon atom forms four bonds with carbon or other atoms). (c)(i) (1) Aldehydes: ; (2) Ketones: . Equation: (ethyl ethanoate, an ester). (c)(ii) Structural isomers are compounds with the same molecular formula but different structures. The two isomers of butane are n-butane, , and isobutane, .
    Step-by-step solution →
  2. Q38 4 marks

    Also in 31/5/2 (Q39), 31/5/3 (Q39)

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  3. Q39 4 marks

    Also in 31/5/2 (Q37), 31/5/3 (Q37)

    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 a 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.
    1. (a)Define the principal axis of a concave mirror.[1]
    2. (b)A ray of light is incident on a concave mirror, parallel to its principal axis. If this ray after reflection from the mirror passes through the principal axis from a point at a distance of 10 cm from the pole of the mirror, find the radius of curvature of the mirror.[1]
    3. (c)(i)An object is placed at a distance of 10 cm from the pole of a convex mirror of focal length 15 cm. Find the position of the image.[2]
    4. (c)(ii)OR : 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.[2]
    Show answer

    The principal axis joins the pole and centre of curvature; cm; cm for the convex mirror; or the mirror is convex.

    (a) The principal axis of a concave mirror is the straight line passing through its pole and its centre of curvature; it is normal to the mirror at the pole. (b) The reflected ray crosses the axis at the principal focus, so cm and cm. (c)(i) With cm and cm, , so cm: the image is formed cm behind the mirror (virtual, erect and diminished, ). (c)(ii) It is a convex mirror. Ray diagram: a ray parallel to the principal axis reflects as if coming from , and a ray directed towards reflects back along itself; their backward extensions meet behind the mirror between and , where the virtual, erect, diminished image is formed.
    Step-by-step solution →

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