JEE MAIN - Physics (2023 - 11th April Morning Shift - No. 24)
A solid sphere of mass $$500 \mathrm{~g}$$ and radius $$5 \mathrm{~cm}$$ is rotated about one of its diameter with angular speed of $$10 ~\mathrm{rad} ~\mathrm{s}^{-1}$$. If the moment of inertia of the sphere about its tangent is $$x \times 10^{-2}$$ times its angular momentum about the diameter. Then the value of $$x$$ will be ___________.
Answer
35
Explanation
$$
\begin{aligned}
& L_{\text {diameter }}=\frac{2}{5} M R^2 \omega ; \quad I_{\text {tangent }}=\frac{7}{5} M R^2 \\\\
& \begin{aligned}
\frac{I_{\text {tangent }}}{L_{\text {diameter }}} & =\frac{7 / 5}{2 / 5} \times \frac{1}{\omega}=\frac{7}{2 \omega} \\\\
& =\frac{7}{2 \times 10}=\frac{7}{20} \\\\
&= \frac{700}{20} \times 10^{-2}=35 \times 10^{-2}
\end{aligned}
\end{aligned}
$$
Comments (0)
