JEE Physics 2025 Question with Solution

Given: $\theta(t) = 5t^2 - 8t$ for a circular disk of mass $M$ and radius $R$.

Find: Power delivered by the applied torque at $t = 2 \, \text{s}$.

Use the relation

$$P = \tau \cdot \omega$$

First find the angular velocity:

$$\omega(t) = \frac{d\theta(t)}{dt} = \frac{d}{dt}(5t^2 - 8t) = 10t - 8$$

At $t = 2 \, \text{s}$,

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

Now find the angular acceleration:

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

For a circular disk about an axis perpendicular to its plane,

$$I = \frac{1}{2}MR^2$$

Torque is

$$\tau = I\alpha = \frac{1}{2}MR^2 \times 10 = 5MR^2$$

Therefore, power is

$$P = \tau \omega = 5MR^2 \times 12 = 60MR^2$$

Therefore, the power delivered is $60MR^2$ and the correct option is A.

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

Find: Power at $t = 2 \, \text{s}$.

From the angular position,

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

so at $t = 2$,

$$\omega = 12 \, \text{rad/s}$$

Also,

$$\alpha = \frac{d\omega}{dt} = 10$$

For the disk,

$$I = \frac{1}{2}MR^2$$

Hence,

$$\tau = I\alpha = \frac{1}{2}MR^2 \times 10 = 5MR^2$$

Now substitute directly into

$$P = \tau\omega$$

to get

$$P = 5MR^2 \times 12 = 60MR^2$$

This works quickly because power in rotational motion depends directly on the instantaneous values of torque and angular speed. Hence the correct option is A.