JEE PYQ: Rotational Motion Question 47
Question 47 - 2019 (09 Apr Shift 1)
A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of $\theta$, where $\theta$ is the angle by which it has rotated, is given as $k\theta^2$. If its moment of inertia is $I$ then the angular acceleration of the disc is:
(1) $\frac{k}{4I},\theta$
(2) $\frac{k}{I},\theta$
(3) $\frac{k}{2I},\theta$
(4) $\frac{2k}{I},\theta$
BETA
