A uniform solid cylinder with radius $$R$$ and length $$L$$ has moment of inertia $$I_1$$, about the axis of cylinder. A concentric solid cylinder of radius $$R' = \frac{R}{2}$$ and length $$L' = \frac{L}{2}$$ is carved out of the original cylinder. If $$I_2$$ is the moment of inertia of the carved out portion of the cylinder then $$\frac{I_1}{I_2} =$$ _____.

A uniform solid cylinder of radius $$R$$ and length $$L$$ has moment of inertia $$I_1$$ about its axis, and a concentric cylinder of radius $$R' = \frac{R}{2}$$ and length $$L' = \frac{L}{2}$$ is carved out.

Since the moment of inertia of a solid cylinder about its axis is given by $$I = \frac{1}{2}MR^2$$, for the original cylinder the mass is $$M = \rho \pi R^2 L$$, so its moment of inertia becomes $$I_1 = \frac{1}{2}MR^2 = \frac{1}{2}\rho \pi R^4 L$$. Substituting the corresponding dimensions for the carved-out portion gives its mass as $$M' = \rho \pi \left(\frac{R}{2}\right)^2 \cdot \frac{L}{2} = \frac{\rho \pi R^2 L}{8} = \frac{M}{8}$$, and therefore its moment of inertia is $$I_2 = \frac{1}{2}M'\left(\frac{R}{2}\right)^2 = \frac{1}{2} \cdot \frac{M}{8} \cdot \frac{R^2}{4} = \frac{MR^2}{64}$$.

From the above, the ratio of the moments of inertia is $$\frac{I_1}{I_2} = \frac{\frac{MR^2}{2}}{\frac{MR^2}{64}} = \frac{64}{2} = 32$$. Therefore, the correct answer is 32.