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(D) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is given by $I = \frac{2}{5}MR^2$. Let the radius of the lar

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$\frac{I}{243}$

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(D) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is given by $I = \frac{2}{5}MR^2$. Let the radius of the large sphere be $R$ and its mass be $M$. When it is recast into $27$ small spheres of radius $r$,the volume remains constant: $\frac{4}{3}\pi R^3 = 27 	imes \frac{4}{3}\pi r^3$. This implies $R^3 = 27r^3$,so $R = 3r$ or $r = \frac{R}{3}$. The mass of each small sphere $m$ is $\frac{M}{27}$. The moment of inertia of each small sphere about its diameter is $I' = \frac{2}{5}mr^2$. Substituting $m = \frac{M}{27}$ and $r = \frac{R}{3}$: $I' = \frac{2}{5} \left( \frac{M}{27} ight) \left( \frac{R}{3} ight)^2 = \frac{2}{5} \left( \frac{M}{27} ight) \left( \frac{R^2}{9} ight) = \frac{1}{243} \left( \frac{2}{5}MR^2 ight)$. Since $I = \frac{2}{5}MR^2$,we get $I' = \frac{I}{243}$.

The moment of inertia of a solid sphere about its diameter is $I$. It is then recast into $27$ small spheres of the same diameter. The moment of inertia of each small sphere about its diameter is:

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