Q 11-06-163JEE MainJEE Main 2021 (26 Feb, Shift 2)Easy
A cord is wound round the circumference of wheel of radius $r$. The axis of the wheel is horizontal and the moment of inertia about it is $I$. A weight $mg$ is attached to the cord at the end. The weight falls from rest. After falling through a distance $h$, the square of angular velocity of wheel will be
Answer: (A) $\frac{2mgh}{I+mr^2}$
The cord does not slip, so $v = \omega r$. Energy conservation:
$$mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}m\omega^2r^2 \Rightarrow \omega^2 = \frac{2mgh}{I + mr^2}$$
Solution by Sreeraj P, M.Sc Physics