JEE Main 2024 — 1 February, Shift 2 — Question 59

$\mathrm{I}_{\mathrm{cm}}=\frac{\mathrm{ML}^{2}}{12}=\frac{2 \times(0.3)^{2}}{12}=\frac{0.09}{6}$ $\mathrm{M}=\mathrm{I}_{\mathrm{cm}}\left(\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\right)$ $0.03=\frac{0.09}{6}\left(\omega_{\mathrm{f}}\right)$ $\omega_{\mathrm{f}}=2 \mathrm{rad} / \mathrm{s}$ $\theta=\omega \mathrm{t}$ $\mathrm{t}=\frac{\theta}{\omega}=\frac{\pi}{2 \times 2}=\frac{\pi}{4} \sec$. $\mathrm{X}=4$