Question

$I$ is moment of inertia of a copper solid sphere of radius $R$ about an axis passing through its centre. Eight identical copper solid spheres each of radius $R$ recasted into a single larger solid sphere. The moment of inertia of larger solid sphere about an axis passing through its centre is:

• A. $4I$
• B. $8I$
• C. $16I$
• D. $32I$

Answer 1

The correct option is D $32I$

Let Mass and radius of smaller spheres be $M$ and $R$.

So Moment of Inertia (MI) is $I=2M{R}^2/5$

As $8$ such spheres are recast into bigger sphere, the mass of bigger sphere is $8M$ and radius is $(\frac{8\times \frac{4}{3}\pi R^3}{\frac{4}{3}\pi }{)}^{1/3}=2R$

So MI of bigger sphere is $2\left(8M\right)(2R{)}^2/5=32\times 2M{R}^2/5=32I$