JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Rotational Mechanics

The impulse $J=0.2$ N-s applied at end B, a distance $L/2$ from the centre of mass, delivers an angular impulse about the centre: $M_c=\frac{L}{2}J=\frac{0.3}{2}\times 0.2=0.03$ N-m-s. The moment of inertia of the rod about its centre is $I_{cm}=\frac{M{L}^2}{12}=\frac{2\times (0.3{)}^2}{12}=0.015$ kg·m². The angular velocity produced is $\omega =\frac{M_c}{I_{cm}}=\frac{0.03}{0.015}=2$ rad/s, and it stays constant afterward since no further external torque acts. To turn through $\theta =\pi /2$, the time taken is $t=\frac{\theta }{\omega }=\frac{\pi /2}{2}=\frac{\pi }{4}$ s, so $x=4$.