QJEEM24Ph919c: JEE Main 2024 Physics - Class 11 JEEM24Ph919c Solution: A uniform rod of mass and length at rest on a smooth horizontal …

Question

Verified Solution • 2025–26 Session moderate

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}{x} \mathrm{~s}

, where

x=

_______ .

Accepted Answer

Impulse

\mathrm{J}=0.2 \mathrm{~N}-\mathrm{S}

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

Angular impuls

(\vec{M})

\begin{aligned}&amp; \vec{M}_c=\int \tau d t \\\\&amp; =\int F \frac{L}{2} d t \\\\&amp; =\frac{L}{2} \int F d t=\frac{L}{2} \times J \\\\&amp; =\frac{0.3}{2} \times 0.2 \\\\&amp; =0.03\end{aligned}

\begin{aligned} &amp; I_{\mathrm{cm}}=\frac{\mathrm{ML}^2}{12}=\frac{2 \times(0.3)^2}{12}=\frac{0.09}{6} \\\\ &amp; \mathrm{M}=\mathrm{I}_{\mathrm{cm}}\left(\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\right) \\\\ &amp; 0.03=\frac{0.09}{6}\left(\omega_{\mathrm{f}}\right) \\\\ &amp; \omega_{\mathrm{f}}=2 \mathrm{rad} / \mathrm{s}\end{aligned}

\theta=\omega \mathrm{t}

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

X=4

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