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:

The correct answer is 7/57.

Let us analyze the problem step-by-step using the principle of superposition for the moment of inertia.

1. Determine the masses of the spheres:
Let the uniform density of the solid sphere be $\rho$. The mass of a sphere is proportional to its volume, which depends on the cube of its radius ($V=\frac{4}{3}\pi r^3$).
Let $M$ be the mass of the smaller sphere of radius $R$:

$M=\rho \cdot \frac{4}{3}\pi R^3$

The larger solid sphere has a radius of $2R$. Its mass $M_1$ is given by:

$M_1=\rho \cdot \frac{4}{3}\pi {\left(2R\right)}^3=8\cdot (\rho \cdot \frac{4}{3}\pi R^3)=8M$

Thus, the mass of the larger sphere is $8M$, and the mass of the smaller sphere is $M$.

2. Calculate the moment of inertia of the larger sphere about the Y-axis:
Since the larger sphere is centered at the origin, the Y-axis is a central diametrical axis. The moment of inertia $I_1$ of a solid sphere of mass $M_1$ and radius $2R$ is:

$I_1=\frac{2}{5}{M}_1{\left(2R\right)}^2$

Substituting $M_1=8M$:

$I_1=\frac{2}{5}\left(8M\right)\left(4R^2\right)=\frac{64}{5}M{R}^2$

3. Calculate the moment of inertia of the smaller sphere about the Y-axis:
As shown in the figure, the smaller sphere has a diameter of $2R$ (radius $R$) and is tangent to the Y-axis. The center of this smaller sphere lies on the X-axis at a distance of $d=R$ from the Y-axis.
Using the parallel axis theorem, the moment of inertia $I_2$ of the smaller sphere about the Y-axis is:

$I_2=I_{\text{cm}}+M{d}^2$

$I_2=\frac{2}{5}M{R}^2+M{R}^2=\frac{7}{5}M{R}^2$

4. Calculate the moment of inertia of the remaining (rest) part of the sphere:
By subtraction, the moment of inertia of the remaining part $I_{\text{rest}}$ about the Y-axis is:

$I_{\text{rest}}=I_1-I_2$

$I_{\text{rest}}=\frac{64}{5}M{R}^2-\frac{7}{5}M{R}^2=\frac{57}{5}M{R}^2$

5. Find the ratio:
The ratio of the moment of inertia of the smaller sphere ($I_2$) to that of the rest part of the sphere ($I_{\text{rest}}$) is:

$\frac{I_2}{I_{\text{rest}}}=\frac{\frac{7}{5}M{R}^2}{\frac{57}{5}M{R}^2}=\frac{7}{57}$