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$$
\text { A solid sphere of radius } R \text { has moment of inertia } J \text { about its qeometrical axis. If it is }
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melted into a disc of radius $r$ and thickness $t$. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$, then the value of $r$ is equal to
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\text { A solid sphere of radius } R \text { has moment of inertia } J \text { about its qeometrical axis. If it is }
$$
melted into a disc of radius $r$ and thickness $t$. If its moment of inertia about the tangential axis (which is perpendicular to plane of the disc), is also equal to $I$, then the value of $r$ is equal to
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Verified Answer
The correct answer is:
$\frac{2}{\sqrt{15}} R$
$\frac{2}{\sqrt{15}} R$
$\frac{2}{3} M R^2=\frac{1}{2} M r^2+M r^2$ or $\frac{2}{3} M R^2=\frac{3}{2} M r^2 \Rightarrow r=\frac{2}{\sqrt{15}} R$
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