A sphere of radius $R$ is cut from a larger solid sphere of radius $2R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

To find the ratio of the moment of inertia of the smaller sphere ($I_{small}$) to the remaining part ($I_{rest}$) about the Y-axis, we use the principle of superposition.1. Mass and DensityLet the larger sphere have radius $R_L=2R$ and the smaller sphere have radius $R_s=R$.The volume of a sphere is proportional to $R^3$.Mass of larger sphere ($M_L$) $\propto (2R{)}^3=8R^3$.Mass of smaller sphere ($M_s$) $\propto R^3$.Therefore, if $M_s=M$, then $M_L=8M$. The mass of the remaining part is $M_{rest}=8M-M=7M$.2. Moment of Inertia of the Small Sphere ($I_{small}$)The smaller sphere is shifted from the Y-axis. Its center is at $x=R$.$I$ about its own center: $I_{cm}=\frac{2}{5}M{R}^2$.Using the Parallel Axis Theorem ($I=I_{cm}+M{d}^2$):
$$I_{small}=\frac{2}{5}M{R}^2+M(R{)}^2=\frac{7}{5}M{R}^2$$
3. Moment of Inertia of the Rest Part ($I_{rest}$)The total moment of inertia of the large sphere about the Y-axis is:
$$I_{total}=\frac{2}{5}{M}_L(2R{)}^2=\frac{2}{5}(8M)(4R^2)=\frac{64}{5}M{R}^2$$
The moment of inertia of the remaining part is the total minus the removed part:
$$I_{rest}=I_{total}-I_{small}=\frac{64}{5}M{R}^2-\frac{7}{5}M{R}^2=\frac{57}{5}M{R}^2$$
4. Calculate the Ratio
$$\text{Ratio}=\frac{I_{small}}{I_{rest}}=\frac{\frac{7}{5}M{R}^2}{\frac{57}{5}M{R}^2}=\frac{7}{57}$$
Correct Option: (3) 7/57