Q 11-06-079JEE MainJEE Main 2024 (1 Feb, Shift 2)Medium
A uniform rod $AB$ of mass $2\ \text{kg}$ and length $30\ \text{cm}$ is at rest on a smooth horizontal surface. An impulse of $0.2\ \text{N s}$ is applied to end B. The time taken by the rod to turn through a right angle will be $\dfrac{\pi}{x}\ \text{s}$, where $x = $ ______.
Numerical value type. Enter your answer.
Answer: 4
The impulse (perpendicular to the rod) gives angular impulse about the centre of mass:
$$J\frac{L}{2} = I_{cm}\,\omega \Rightarrow 0.2\times0.15 = \frac{2\times(0.3)^2}{12}\,\omega = 0.015\,\omega \Rightarrow \omega = 2\ \text{rad s}^{-1}$$
Time to turn through $\dfrac{\pi}{2}$: $t = \dfrac{\pi/2}{2} = \dfrac{\pi}{4}\ \text{s}$, so $x = 4$.
Solution by Sreeraj P, M.Sc Physics