A solid sphere of radius R has its outer half removed, so that its radius becomes R 2 . Then its moment of inertia about the diameter is

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Aicharya · Accepted Answer

Moment of inertia of a solid sphere is given by, $I = \frac{2}{5} M R^{2} = \frac{2}{5} ρ \frac{4}{3} π R^{3} R^{2} = \frac{8}{15} ρ π R^{5}$ .
Therefore, for constant density, the moment of inertia $I \propto R^{5}$ .
$\Rightarrow \frac{I_{1}}{I_{2}} = {(\frac{R_{1}}{R_{2}})}^{5} \Rightarrow I_{2} = {(\frac{1}{2})}^{5} I_{1} \Rightarrow I_{2} = \frac{1}{32} I_{1}$

A solid sphere of radius R has its outer half removed, so that its radius becomes R 2 . Then its moment of…

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