Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis $\mathrm{AB}$ as shown in figure is $\sqrt{8 / x}$. The value of $x$ is :

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{I}_{	ext {sphere }}=\frac{2}{3} \mathrm{MR}^2=\mathrm{Mk}_1^2 \ & \mathrm{I}_{	ext {cylinder }}=\frac{1}{12} \mathrm{M}\left(4 \mathrm{R}^2ight)+\frac{1}{4} \mathrm{MR}^2+\mathrm{M}(2 \mathrm{R})^2 \ & =\frac{67}{12} \mathrm{MR}^2=\mathrm{Mk}_2^2 \ & \frac{\mathrm{k}_1}{\mathrm{k}_2}=\sqrt{\frac{2}{3} \cdot \frac{12}{67}}=\sqrt{\frac{8}{67}}\end{aligned}$

Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of…

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