JEE Main 2025 — Rotational Motion Question with Solution

Moment of Inertia and Angular Motion

For a solid circular disk, the moment of inertia is given by

$$I=\frac{1}{2}M{R}^2.$$

The angular position is defined as

$$\theta (t)=5t^2-8t.$$

Angular Velocity and Acceleration

Differentiate with respect to time to obtain the angular velocity:

$$\omega (t)=\frac{d\theta }{dt}=10t-8.$$

Differentiating again, the angular acceleration is:

$$\alpha (t)=\frac{d\omega }{dt}=10.$$

At time $$t=2\text{s}$$:

Angular velocity:

$$\omega (2)=10(2)-8=12\text{rad/s}.$$

Angular acceleration:

$$\alpha (2)=10{\text{rad/s}}^2.$$

Torque Calculation

The torque applied by the external force is related to the moment of inertia and angular acceleration:

$$\tau =I\alpha =\frac{1}{2}M{R}^2\times 10=5M{R}^2.$$

Power Delivered

Power delivered by a torque is given by:

$$P=\tau \omega .$$

At $$t=2\text{s}$$, substitute the values:

$$P=5M{R}^2\times 12=60M{R}^2.$$

Thus, the power delivered by the applied torque at $$t=2\text{s}$$ is:

$$\boxed{60M{R}^2}.$$