ICSE - X Physics Board Paper 2024.txt

ICSE CLASS 10 PHYSICS BOARD PAPER 2024 - COMPLETE ANSWER KEY

SECTION A (40 MARKS)

Question 1 (Multiple Choice - 15 marks)
(i) (b) kinetic energy to sound energy
(ii) (a) 5 N
(iii) (c) 5 m
(iv) (a) 128
(v) (b) A is true but R is false.
(vi) (c) forced vibrations
(vii) (d) Human forearm
(viii) (b) material
(ix) (a) APPLIANCE Mains Blue wire
(x) (a) terminal voltage
(xi) (b) energy is absorbed and temperature remains constant.
(xii) (a) m < 1
(xiii) (c) Remain unchanged
(xiv) (d) Heat capacity
(xv) (a) Move away from the slab

Question 2 (15 marks)
(i)(a) C¹⁴ is a radioisotope because of the instability of the nucleus. Used for carbon dating.
(i)(b) Class II lever.

(ii)(a) Centripetal
(ii)(b) zero

(iii) Solution:
120 × 0.2 = F × 0.8
F = 30 N

(iv)(a) Mechanical advantage using a block and tackle system of 9 pulleys is 9 whereas the mechanical advantage of a single moveable pulley is 2.
(iv)(b) Increase of number of pulley increases mechanical advantage.

(v) Solution:
Pₛᵤₘᵢₜ = 600 J / 10 min = 60 J/min
Pₐₘᵢₜ = 300 J / 20 min = 15 J/min
Ratio = 60:15 = 4:1

(vi)(a) Resistance decreases
(vi)(b) Current (Intensity of light) increases as resistance decreases

(vii) Solution:
For Oxygen: x = 340 × (0.1/2) = 17 m
For Benzene: y = 200 × (0.1/2) = 10 m
So, x:y = 17:10

Question 3 (10 marks)
(i)(a) In between optical centre and focus
(i)(b) Because concave lens produces reduced size image.

(ii) No.
Justification: The geyser draws I = P/V = 1540/220 = 7A.
Hence, 5A rated fuse will blow off for the geyser.

(iii) Two factors:
1. Increasing the number of turns of the armature
2. Increasing the applied battery e.m.f.

(iv) Solution:
Q = mL = 500 × 330 = 165,000 J

(v) Complete reaction:
₉₂U²³⁵ + ₀n¹ → ₅₆Ba¹⁴¹ + ₃₆Kr⁹² + 3₀n¹

SECTION B (40 MARKS)

Question 4 (10 marks)
(i) (3 marks) Solution:
1/v - 1/u = 1/f
1/72 + 1/36 = 1/f
3/72 = 1/f (1 mark for working)
f = 24 cm (1 mark)
Power = 100/24 = 4.17 D (1 mark)

(ii)(a) Microwave
(ii)(b) X-rays
(ii)(c) Both have same speed in vacuum

(iii)(a) Red colour is used as danger signal because it is scattered least due to its longest wavelength in visible range. Hence, it is visible from a very long distance.
(iii)(b)1. Critical angle is 43°
   2. Angle of prism = 43°
   3. If the blue ray is replaced by indigo or violet, there will be total internal reflection.

Question 5 (10 marks)
(i)(a) 8/9
(i)(b) The principle of reversibility
(i)(c) No

(ii)(a) False
(ii)(b) Solution:
Refractive index = (10x/t₁)/(x/t₂) = 10t₂/t₁

(iii)(a) [Diagram showing completed ray path]
(iii)(b) Two factors:
1. Angle of incidence
2. Refractive index

Question 6 (10 marks)
(i)(a) Centre of gravity is the point where the mass of the whole body appears to be concentrated.
(i)(b)1. 2 cm from base (h/3 = 6/3)
    2. Yes

(ii)(a) Marble A
(ii)(b) Along both the paths
(ii)(c) Along both the paths

(iii)(a) [Diagram showing tackle system with VR=2]
(iii)(b)1. Solution:
    Efficiency = M.A./V.R.
    0.8 = M.A./2
    M.A. = 1.6
    2. Effort = Load/M.A. = 48/1.6 = 30 kgf

Question 7 (10 marks)
(i)(a) Ultrasonic wave
(i)(b) Solution:
t = 2d/v = (2×160)/320 = 1 s

(ii)(a) ²³⁴₉₀Y → ²³⁴₉₁Z + ⁰₋₁e
(ii)(b) U-235

(iii)(a) Resonance
(iii)(b) Natural frequency of the air column matched with the frequency of the tuning fork.
(iii)(c) Length of air column should be decrease. Since frequency is inversely proportional to the air column.

Question 8 (10 marks)
(i)(a) Non-ohmic
(i)(b) The resistance is not constant. It increases as voltage and current increases.
(i)(c) The flow of current through the conductor is directly proportional to the potential difference established across the conductor provide its physical conditions are constant.

(ii)(a) Step down transformer
(ii)(b) Wire of secondary coil is thicker because secondary current is higher than primary current.

(iii)(a) Solution:
R₁ = 10 + 6 = 16Ω
R₂ = 12 + 4 = 16Ω
Rₑq = (16×16)/(16+16) = 8Ω
(iii)(b) I = V/R = 4/8 = 0.5A
(iii)(c) Current in both the resistors are equal because both the arms have same resistance value.

Question 9 (10 marks)
(i) Solution:
85 × 4.2 × 25 = m × 336 + m × 4.2 × 5
8925 = 357m
m = 25g

(ii)(a) When lake freezes, it released latent heat to the atmosphere which warms the atmosphere.
(ii)(b) Since there is no change of temperature there is no change in kinetic energy.
(ii)(c) 1 g of ice at 0°C has more heat energy than 1 g water at 0°C.

(iii)(a) Number of turns in armature may be increased.
(iii)(b)1. Maxwell right hand thumb rule.
    2. Direction of current is from X to Y.
    3. As R increases, current decreases. So, strength of magnetic field decreases.