JEE MAIN - Physics (2024 - 1st February Evening Shift - No. 29)
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 $x=$ _______ .
Answer
4
Explanation
_1st_February_Evening_Shift_en_29_1.png)
Impulse $\mathrm{J}=0.2 \mathrm{~N}-\mathrm{S}$
$$ \mathrm{J}=\int \mathrm{Fdt}=0.2 \mathrm{~N}-\mathrm{s} $$
Angular impuls $(\vec{M})$
$$ \begin{aligned} & \vec{M}_c=\int \tau d t \\\\ & =\int F \frac{L}{2} d t \\\\ & =\frac{L}{2} \int F d t=\frac{L}{2} \times J \\\\ & =\frac{0.3}{2} \times 0.2 \\\\ & =0.03 \end{aligned} $$
$\begin{aligned} & I_{\mathrm{cm}}=\frac{\mathrm{ML}^2}{12}=\frac{2 \times(0.3)^2}{12}=\frac{0.09}{6} \\\\ & \mathrm{M}=\mathrm{I}_{\mathrm{cm}}\left(\omega_{\mathrm{f}}-\omega_{\mathrm{i}}\right) \\\\ & 0.03=\frac{0.09}{6}\left(\omega_{\mathrm{f}}\right) \\\\ & \omega_{\mathrm{f}}=2 \mathrm{rad} / \mathrm{s}\end{aligned}$
$\theta=\omega \mathrm{t}$
$\mathrm{t}=\frac{\theta}{\omega}=\frac{\pi}{2 \times 2}=\frac{\pi}{4} \mathrm{sec}$
$X=4$
Comments (0)
