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 (integer)
35
Solution
$$
\begin{aligned}
& L_{\text {diameter }}=\frac{2}{5} M R^2 \omega ; \quad I_{\text {tangent }}=\frac{7}{5} M R^2 \\\\
& \begin{aligned}
\frac{I_{\text {tangent }}}{L_{\text {diameter }}} & =\frac{7 / 5}{2 / 5} \times \frac{1}{\omega}=\frac{7}{2 \omega} \\\\
& =\frac{7}{2 \times 10}=\frac{7}{20} \\\\
&= \frac{700}{20} \times 10^{-2}=35 \times 10^{-2}
\end{aligned}
\end{aligned}
$$
About this question
Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass
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