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

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$\frac{I}{243}$

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(A) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is given by $I = \frac{2}{5}MR^2$.Since the volume remains constant,the volume of the large sphere equals the sum of the volumes of the $27$ small spheres: $\frac{4}{3}\pi R^3 = 27 	imes \frac{4}{3}\pi r^3$,where $r$ is the radius of each small sphere.This simplifies to $R^3 = 27r^3$,so $R = 3r$ or $r = R/3$.The mass of each small sphere $m$ is $M/27$ because the density is uniform.The moment of inertia of each small sphere $I'$ is $\frac{2}{5}mr^2$.Substituting $m = M/27$ and $r = R/3$:$I' = \frac{2}{5} 	imes (M/27) 	imes (R/3)^2 = \frac{2}{5} 	imes \frac{M}{27} 	imes \frac{R^2}{9} = \frac{2}{5}MR^2 	imes \frac{1}{27 	imes 9} = I 	imes \frac{1}{243} = \frac{I}{243}$.

Moment of inertia of a solid sphere about its diameter is $I$. It is then casted into $27$ small spheres of same diameter. The moment of inertia of each new sphere is

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