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MCQEasyJEE 2025Moment of Inertia & Radius of Gyration

JEE Physics 2025 Question with Solution

A wheel of radius 0.2 m0.2 \, m rotates freely about its center when a string that is wrapped over its rim is pulled by a force of 10 N10 \, N as shown in the figure. The established torque produces an angular acceleration of 2 rad/s22 \, rad/s^2. Moment of inertia of the wheel is............. kg m2kg\, m^2.

A wheel suspended about its center with a string wrapped around the rim, pulled downward tangentially by a force of 10 N.
  • A

    1 kg m21 \, kg \, m^2

  • B

    2 kg m22 \, kg \, m^2

  • C

    3 kg m23 \, kg \, m^2

  • D

    4 kg m24 \, kg \, m^2

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: radius r=0.2 mr = 0.2 \, \text{m}, force F=10 NF = 10 \, \text{N}, angular acceleration α=2 rad/s2\alpha = 2 \, \text{rad/s}^2.

Find: moment of inertia II of the wheel.

The torque produced by the tangential force is

τ=F⋅r\tau = F \cdot r

Substituting the given values,

τ=10 N×0.2 m=2 N m\tau = 10 \, \text{N} \times 0.2 \, \text{m} = 2 \, \text{N m}

Using the rotational relation

τ=Iα\tau = I\alpha

we get

I=τα=2 N m2 rad/s2=1 kg m2I = \frac{\tau}{\alpha} = \frac{2 \, \text{N m}}{2 \, \text{rad/s}^2} = 1 \, \text{kg m}^2

Therefore, the moment of inertia of the wheel is 1 kg m21 \, \text{kg m}^2. The correct option is A.

Step-by-step torque relation

Given: The force is applied tangentially at the rim, so the lever arm is the radius rr.

Find: II.

First calculate torque:

τ=Fr=10×0.2=2 N m\tau = Fr = 10 \times 0.2 = 2 \, \text{N m}

Now use Newton's second law for rotation:

τ=Iα\tau = I\alpha

Hence,

I=Frα=10×0.22=1 kg m2I = \frac{Fr}{\alpha} = \frac{10 \times 0.2}{2} = 1 \, \text{kg m}^2

Thus, the required moment of inertia is 1 kg m21 \, \text{kg m}^2.

Common mistakes

  • Using diameter instead of radius for torque. Torque is τ=Fr\tau = Fr, where rr is the perpendicular distance from the axis to the line of action of force. Here that distance is the wheel radius, not the diameter.

  • Writing I=ταI = \tau \alpha instead of τ=Iα\tau = I\alpha. The correct rearrangement is I=ταI = \frac{\tau}{\alpha}, so angular acceleration must be in the denominator.

  • Ignoring that the force is tangential. Since the string is pulled along the tangent, the angle between radius and force is 90∘90^\circ, so torque magnitude is FrFr. Do not multiply by any extra trigonometric factor incorrectly.

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