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A solid sphere of mass $M$, radius $R$ has moment of inertia ' $I$ ' about its diameter. It is recast into a disc of thickness ' $t$ ' whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains ' $I$ '. Radius of the disc will be
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Verified Answer
The correct answer is:
$\frac{2 \mathrm{R}}{\sqrt{15}}$
For solid sphere, $\mathrm{I}=\frac{2}{5} \mathrm{MR}^2$
For disc, $\mathrm{I}^{\prime}=\frac{\mathrm{MR}^{\prime 2}}{2}+\mathrm{MR}^{\prime 2}=\frac{3}{2} \mathrm{MR}^{\prime 2}$
$$
\begin{aligned}
& \because I^{\prime}=I \\
& \therefore \frac{3}{2} M R^{\prime 2}=\frac{2}{5} M^2 \\
& R^{\prime}=\frac{2}{\sqrt{15}} R
\end{aligned}
$$
For disc, $\mathrm{I}^{\prime}=\frac{\mathrm{MR}^{\prime 2}}{2}+\mathrm{MR}^{\prime 2}=\frac{3}{2} \mathrm{MR}^{\prime 2}$
$$
\begin{aligned}
& \because I^{\prime}=I \\
& \therefore \frac{3}{2} M R^{\prime 2}=\frac{2}{5} M^2 \\
& R^{\prime}=\frac{2}{\sqrt{15}} R
\end{aligned}
$$
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