A uniform solid cylinder of length L and radius R has moment of inertia about its axis equal to $$I_{1}$$. A small co-centric cylinder of length L/2 and radius R/3 carved from this cylinder has moment of inertia about its axis equals to $$I_{2}$$. The ratio $$I_{1}/I_{2}$$ is __________.

We need to find the ratio $$I_1/I_2$$ of moments of inertia of a solid cylinder and a smaller co-centric cylinder carved from it.

Moment of inertia of a solid cylinder about its axis:

$$I = \frac{1}{2}MR^2$$

For the large cylinder:

Mass: $$M_1 = \rho \pi R^2 L$$

$$I_1 = \frac{1}{2}M_1 R^2 = \frac{1}{2}\rho \pi R^2 L \times R^2 = \frac{1}{2}\rho \pi R^4 L$$

For the small cylinder:

Radius = R/3, Length = L/2

Mass: $$M_2 = \rho \pi (R/3)^2 (L/2) = \rho \pi R^2 L / 18$$

$$I_2 = \frac{1}{2}M_2 (R/3)^2 = \frac{1}{2} \times \frac{\rho \pi R^2 L}{18} \times \frac{R^2}{9} = \frac{\rho \pi R^4 L}{324}$$

Ratio:

$$\frac{I_1}{I_2} = \frac{\frac{1}{2}\rho \pi R^4 L}{\frac{\rho \pi R^4 L}{324}} = \frac{324}{2} = 162$$

Therefore, $$I_1/I_2 = $$ 162.