Q 11-06-067JEE MainJEE Main 2025 (23 Jan, Shift 2)Medium
A circular disk of radius $R$ metre and mass $M$ kg is rotating around the axis perpendicular to the disk. An external torque is applied to the disk such that $\theta(t) = 5t^2 - 8t$, where $\theta(t)$ is the angular position of the rotating disc as a function of time $t$. How much power is delivered by the applied torque when $t = 2\ \text{s}$?
Answer: (D) $60MR^2$
$\omega = \dfrac{d\theta}{dt} = 10t - 8$, so $\omega(2) = 12\ \text{rad/s}$; $\alpha = \dfrac{d\omega}{dt} = 10\ \text{rad/s}^2$.
Taking the axis through the centre, $I = \tfrac{1}{2}MR^2$:
$$\tau = I\alpha = \frac{1}{2}MR^2\times10 = 5MR^2$$
$$P = \tau\omega = 5MR^2\times12 = 60MR^2$$
Solution by Sreeraj P, M.Sc Physics