JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Rotational Mechanics

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A wheel of radius 0.2 m rotates freely about its center when a string that is wrapped over its rim is pulled by force of 10 N as shown in figure. The established torque produces an angular acceleration of

2 rad / s^{2}

. Moment of inertia of the wheel is ______

kg m^{2}

.
(Acceleration due to gravity

= 10 m / s^{2}

)

Answer: 1

Solution

The

10

N pull acts along the rim at radius

0.2

m, so it exerts a torque

τ = F R = 10 \times 0.2 = 2 N m

about the centre. This torque drives the angular acceleration through

τ = I α

, giving

I = \frac{τ}{α} = \frac{2}{2} = 1 kg m^{2}

. The moment of inertia of the wheel is

1 kg m^{2}

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Integer answer

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