A uniform rod \(A B\) of mass \(2 \mathrm{~kg}\) and Length \(30 \mathrm{~cm}\) at rest on a smooth horizontal surface. An impulse of force \(0.2\  \mathrm{Ns}\) is applied to end \(B.\) The time taken by the rod to turn through at right angles will be \(\frac{\pi}{\mathrm{x}}\ \mathrm{s}\), where  X = _______.

\(\mathrm{J}=\int \mathrm{Fdt}=0.2 \mathrm{~N}-\mathrm{s}\)

Angular impuls ( \(\overrightarrow{\mathrm{M}})\)

\(\overrightarrow{\mathrm{M}}_{\mathrm{c}}=\int \tau \mathrm{dt}\)

\(=\int \mathrm{F} \frac{\mathrm{L}}{2} \mathrm{dt}\)

\(=\frac{\mathrm{L}}{2} \int \mathrm{Fdt}=\frac{\mathrm{L}}{2} \times \mathrm{J}v\)

\(=\frac{0.3}{2} \times 0.2\)

\(=0.03\)

\(I_{c m}=\frac{\mathrm{LL}^2}{12}=\frac{2 \times(0.3)^2}{12}=\frac{0.09}{6}\)

\(\mathrm{M}=\mathrm{I}_{\mathrm{cm}}\left(\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\right)\)

\(0.03=\frac{0.09}{6}\left(\omega_{\mathrm{f}}\right)\)

\(\omega_{\mathrm{f}}=2 \mathrm{rad} / \mathrm{s}\)

\(\theta=\omega \mathrm{t}\)

\(\mathrm{t}=\frac{\theta}{\omega}=\frac{\pi}{2 \times 2}=\frac{\pi}{4} \mathrm{sec} .\)

\(\mathrm{X}=4\)