Q14. When we connect too many electrical appliances to a single socket in a circuit having a fixed value of current rating: (a) the total resistance of the circuit increases. (b) the current drawn from the source decreases. (c) the current increases beyond the rated value. (d) each appliance starts drawing more and more current from the source.

(c) the current increases beyond the rated value.

Q19. 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): The commercial unit of electric energy is kilowatt hour (kWh). Reason (R): 1 kWh = 10⁶ joule (J) (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false, but Reason (R) is true.

(c) Assertion (A) is true, but Reason (R) is false.

Q25(b). Name the muscles present in the human eye which enable it to focus on objects at varied distances (i.e., distant as well as nearby objects). Explain how it happens.

The ciliary muscles change the lens curvature: they relax for distant objects (thin lens) and contract for near ones (thick lens).

Q36(a). (i) Write the relation between the electrical resistivity ‘ ρ ’ and the resistance ‘ R ’ of a cylindrical conductor of length ‘ l ’ and area of cross-section ‘ A ’. Use this relationship to determine the SI unit of ‘ ρ ’. (ii) Find the electrical resistivity of the material of a wire of length 120 cm and diameter 0.4 mm. The resistance of this wire is 3 Ω.

R = ρl/A gives the unit of resistivity as the ohm metre; the wire's resistivity is 3.14 × 10⁻⁷ Ω m.

Q36(b). (i) What is meant by heating effect of electric current? Derive an expression for the amount of heat produced in a conductor of resistance R through which a current I flows for a time t. (ii) Calculate the amount of heat produced in a conductor of resistance 10 Ω through which a current of 2 A flows for 100 seconds.

Heat produced is H = I²Rt; for 2 A through 10 Ω for 100 s, H = 4000 J.

Q39. In our laboratories we use compound microscopes to see the magnified image of a microscopic object. A compound microscope is made up of two lenses. The lens nearest to the object to be viewed, called objective lens, forms real, inverted and magnified image of the object. This image serves as an object for the second lens called eyepiece. The eyepiece forms virtual, erect and magnified image of its object. Thus, the resultant image formed by a microscope is virtual, inverted and magnified with respect to the microscopic object viewed. (a) The image of an object formed by a convex lens of focal length 2 cm is real, inverted and magnified. What is the range of object distance in this case? (b) The image of an object formed by a convex lens of focal length 6 cm is virtual, erect and magnified. What is the range of object distance in this case? (c)(i) An object is placed at a distance of 12 cm from the optical centre of a convex lens of focal length 18 cm. Draw a labelled ray diagram to show the formation of image in this case. (c)(ii) OR: The image formed by a convex lens is real, inverted and of the same size as the object. If the distance between the objects and the image is 60 cm…

A real, magnified image needs an object between F₁ and 2F₁, a virtual one inside F₁; for 12 cm, v = -36 cm; or f = 15 cm.

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

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

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

Take this paper as a timed mock

Free with an account: the board's time limit, marking at the end, and the answer to every question.

Questions 1–20 · 1 mark each

Q1 1 mark

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Q2 1 mark

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Q3 1 mark

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Q4 1 mark

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Q5 1 mark

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Q6 1 mark

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Q7 1 mark

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Q8 1 mark

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Q9 1 mark

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Q10 1 mark

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Q11 1 mark

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Q12 1 mark

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Q13 1 mark

Also in 31/S/2 (Q12), 31/S/3 (Q13)
The image formed by a concave mirror of focal length $50$ cm is real and of magnification $- 1$ . In this case the distance between the object from its own image is :
1. (a) $50$ cm
2. (b) $100$ cm
3. (c) $200$ cm
4. (d)Zero
Show answer

(d) Zero

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Q14 1 mark
When we connect too many electrical appliances to a single socket in a circuit having a fixed value of current rating :
1. (a)the total resistance of the circuit increases.
2. (b)the current drawn from the source decreases.
3. (c)the current increases beyond the rated value.
4. (d)each appliance starts drawing more and more current from the source.
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(c) the current increases beyond the rated value.

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Q15 1 mark

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Q16 1 mark

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Q17 1 mark

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Q18 1 mark

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Q19 1 mark

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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): The commercial unit of electric energy is kilowatt hour (kWh).
Reason (R): $1 kWh = 1 0^{6}$ joule (J)
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.

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Q20 1 mark

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Questions 21–26 · 2 marks each

Q21 2 marks

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Q22 2 marks

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Q23(a) 2 marks

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Q23(b) 2 marks

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Q24 2 marks

Also in 31/S/2 (Q25), 31/S/3 (Q22)

Draw a labelled ray diagram to show the formation of image due to a concave mirror when an object is placed between its pole and principal focus.

Show answer

The image is formed behind the mirror and is virtual, erect and enlarged.
Draw the concave mirror with P, F and C on the principal axis and place the object AB between P and F. From the top A draw (1) a ray along the line from C through A, which meets the mirror normally and is reflected back along the same path, and (2) a ray to the pole P, which is reflected at an equal angle on the other side of the principal axis (a ray parallel to the axis, reflected through F, may be used instead). The reflected rays diverge; produced backwards as dashed lines, they meet behind the mirror at A′. The image A′B′ is formed behind the mirror and is virtual, erect and enlarged. For example, with $f = - 10$ cm and $u = - 6$ cm the mirror formula gives $v = + 15$ cm and $m = + 2.5$ .

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Q25(a) 2 marks

Also in 31/S/2 (Q24(a)), 31/S/3 (Q23(a))

Draw a ray diagram to show the formation of a rainbow in the sky. Mark on the diagram three points A, B and C as given below : A – where dispersion of light occurs B – where internal reflection of dispersed light occurs C – where refraction of dispersed light occurs

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At A the ray is refracted and dispersed on entering the drop, at B internally reflected, at C refracted again on leaving.
A ray of sunlight (white light) falls on a spherical raindrop. A is the point where it enters the drop: it is refracted and dispersed into its colours, violet bending most and red least. B is the point on the back surface of the drop where the dispersed light is reflected internally. C is the point on the front surface, lower down, where the dispersed light is refracted again as it passes out of the drop. The colours emerge separated and travel towards the observer, who has the Sun behind, and are seen as a rainbow.

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OR

Q25(b) 2 marks

Also in 31/S/2 (Q24(b)), 31/S/3 (Q23(b))

Name the muscles present in the human eye which enable it to focus on objects at varied distances (i.e., distant as well as nearby objects). Explain how it happens.

Show answer

The ciliary muscles change the lens curvature: they relax for distant objects (thin lens) and contract for near ones (thick lens).
The ciliary muscles. They hold the eye lens, which is made of a flexible, jelly-like material, and change its curvature and hence its focal length. For a distant object the ciliary muscles relax: the lens becomes thin, its focal length increases, and the distant object is focused on the retina. For a nearby object the ciliary muscles contract: the curvature of the lens increases and it becomes thicker, its focal length decreases, and the nearby object is focused on the retina. This ability of the eye lens to adjust its focal length is called accommodation.

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Q26 2 marks

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Questions 27–33 · 3 marks each

Q27 3 marks

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Q28(a) 3 marks

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Q28(b) 3 marks

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Q29 3 marks

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Q30 3 marks

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Q31 3 marks

Also in 31/S/2 (Q32), 31/S/3 (Q29)
In the given circuit when key is closed, determine the following :
1. (i)Total resistance offered by the three resistors
2. (ii)Reading of the ammeter
3. (iii)Reading of the voltmeter
Show answer

$R = 12 Ω$ , the ammeter reads $0.5$ A and the voltmeter across the $4 Ω$ resistor reads $2$ V.
(i) Total resistance $R = 2 + 4 + 6 = 12 Ω$ , as the three resistors are in series. (ii) Reading of the ammeter $I = \frac{V}{R} = \frac{6}{12} = 0.5$ A. (iii) Reading of the voltmeter, which is across the $4 Ω$ resistor, $= I R = 0.5 \times 4 = 2$ V.

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Q32 3 marks

Also in 31/S/2 (Q31)
(a) State
1. (i)right-hand thumb rule, and
2. (ii)Fleming’s left-hand rule. (b) Using Fleming’s left-hand rule determine the direction of force experienced by a proton, which enters vertically downwards in a uniform magnetic field acting horizontally from west to east as shown in the diagram.
Show answer

The force on the proton is perpendicular to the plane of the paper, outwards, towards the reader, that is towards the south.
(a) (i) Right-hand thumb rule: hold a current-carrying straight conductor in the right hand with the thumb pointing in the direction of the current; the fingers wrapped round the conductor then point in the direction of the magnetic field lines. (ii) Fleming's left-hand rule: stretch the thumb, forefinger and middle finger of the left hand so that they are mutually perpendicular; if the forefinger points in the direction of the magnetic field and the middle finger in the direction of the current, the thumb points in the direction of the force on the conductor. (b) A proton is positively charged, so the current is along its motion, vertically downwards. With the forefinger towards the east (the field) and the middle finger downwards (the current), the thumb points out of the plane of the paper. The force on the proton is therefore perpendicular to the plane of the paper, directed outwards, towards the reader, that is, towards the south.

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Q33 3 marks

Also in 31/S/3 (Q33)
Draw a ray diagram to show the refraction of a ray of light through a triangular glass prism and mark on it
1. (i)angle of incidence ( $i$ ),
2. (ii)angle of emergence ( $e$ ), and
3. (iii)angle of deviation ( $D$ ). What would happen, if instead of a ray, a beam of white light falls on the prism ? Give reason for the phenomenon that occurs in this case.
Show answer

Draw the ray through the prism marking $i$ , $e$ and $D$ ; white light disperses into a spectrum because colours bend by different angles.
Ray diagram: a ray PE falls on the first face of the prism at E, bends towards the normal and travels along EF inside the glass, then bends away from the normal at F on the second face and emerges along FS. The angle of incidence $i$ is the angle between PE and the normal at E, the angle of emergence $e$ is the angle between FS and the normal at F, and the angle of deviation $D$ is the angle between the incident ray produced forwards and the emergent ray produced backwards. If a beam of white light falls on the prism, it splits into a band of seven colours, violet, indigo, blue, green, yellow, orange and red (a spectrum): this is dispersion. It happens because the different colours of white light bend through different angles with respect to the incident ray as they pass through the prism, red the least and violet the most, so each colour emerges along a different path.

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Questions 34–36 · 5 marks each

Q34(a) 5 marks

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Q34(b) 5 marks

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Q35(a) 5 marks

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Q35(b) 5 marks

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Q36(a) 5 marks

Also in 31/S/3 (Q34(a))
1. (i)Write the relation between the electrical resistivity ‘ $ρ$ ’ and the resistance ‘ $R$ ’ of a cylindrical conductor of length ‘ $l$ ’ and area of cross-section ‘ $A$ ’. Use this relationship to determine the SI unit of ‘ $ρ$ ’.
2. (ii)Find the electrical resistivity of the material of a wire of length $120$ cm and diameter $0.4$ mm. The resistance of this wire is $3 Ω$ .
Show answer

$R = ρ \frac{l}{A}$ gives the unit of resistivity as the ohm metre; the wire's resistivity is $3.14 \times 1 0^{- 7} Ω m$ .
(i) $R = ρ \frac{l}{A}$ , so $ρ = \frac{R A}{l}$ . Substituting SI units, the unit of $ρ$ is $\frac{Ω \times m ^{2}}{m} = Ω m$ (ohm metre). (ii) With $l = 1.2$ m, $r = 2 \times 1 0^{- 4}$ m and $A = π r^{2} = 1.256 \times 1 0^{- 7} m^{2}$ : $ρ = \frac{3 \times 1.256 \times 1 0 ^{- 7}}{1.2} = 3.14 \times 1 0^{- 7} Ω m$ .

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OR
Q36(b) 5 marks

Also in 31/S/3 (Q34(b))
1. (i)What is meant by heating effect of electric current ? Derive an expression for the amount of heat produced in a conductor of resistance $R$ through which a current $I$ flows for a time $t$ .
2. (ii)Calculate the amount of heat produced in a conductor of resistance $10 Ω$ through which a current of $2$ A flows for $100$ seconds.
Show answer

Heat produced is $H = I^{2} R t$ ; for 2 A through $10 Ω$ for 100 s, $H = 4000$ J.
(i) When a current flows through a purely resistive conductor, the energy supplied by the source is dissipated entirely as heat in the conductor; this is the heating effect of electric current. If a current $I$ flows through a resistance $R$ for a time $t$ with a potential difference $V$ across it, the charge moved is $Q = I t$ and the work done by the source is $W = V Q = V I t$ . This work appears as heat, so $H = V I t$ , and with Ohm's law $V = I R$ this gives $H = I^{2} R t$ (Joule's law of heating). (ii) $H = I^{2} R t = 2^{2} \times 10 \times 100 = 4000$ J, that is $4 \times 1 0^{3}$ J.

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Questions 37–39 · 4 marks each

Q37 4 marks

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Q38 4 marks

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Q39 4 marks

Also in 31/S/2 (Q39), 31/S/3 (Q39)
In our laboratories we use compound microscopes to see the magnified image of a microscopic object. A compound microscope is made up of two lenses. The lens nearest to the object to be viewed, called objective lens, forms real, inverted and magnified image of the object. This image serves as an object for the second lens called eyepiece. The eyepiece forms virtual, erect and magnified image of its object. Thus, the resultant image formed by a microscope is virtual, inverted and magnified with respect to the microscopic object viewed.
1. (a)The image of an object formed by a convex lens of focal length $2$ cm is real, inverted and magnified. What is the range of object distance in this case ?[1]
2. (b)The image of an object formed by a convex lens of focal length $6$ cm is virtual, erect and magnified. What is the range of object distance in this case ?[1]
3. (c)(i)An object is placed at a distance of $12$ cm from the optical centre of a convex lens of focal length $18$ cm. Draw a labelled ray diagram to show the formation of image in this case.[2]
4. (c)(ii)OR : The image formed by a convex lens is real, inverted and of the same size as the object. If the distance between the objects and the image is $60$ cm, determine the focal length of the lens. Give justification for your answer.[2]
Show answer

A real, magnified image needs an object between $F_{1}$ and $2 F_{1}$ , a virtual one inside $F_{1}$ ; for 12 cm, $v = - 36$ cm; or $f = 15$ cm.
(a) The object must be between F₁ and 2F₁: its distance from the lens is more than $2$ cm and less than $4$ cm. (b) The object must be between the optical centre O and F₁: its distance from the lens is less than $6$ cm. (c)(i) The object ( $u = - 12$ cm) lies between O and F₁ of the lens ( $f = 18$ cm). A ray from the top of the object parallel to the axis refracts through F₂, and a ray through O goes on undeviated; the refracted rays diverge, and their backward extensions meet on the object's side at $v = - 36$ cm, at 2F₁. The image is virtual, erect and magnified three times ( $m = + 3$ ), as the ray diagram shows. (c)(ii) A real, inverted image of the same size is formed only when the object is at 2F₁, and the image is then at 2F₂. Object and image are therefore $2 f + 2 f = 4 f = 60$ cm apart, so $f = 15$ cm; check: $\frac{1}{30} - \frac{1}{- 30} = \frac{1}{15}$ .

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