JEE Physics 2026 Question with Solution

Given: A uniform solid cylinder has length $L$, radius $R$, and moment of inertia $I_1$ about its axis. A small co-centric cylinder carved from it has length $L/2$, radius $R/3$, and moment of inertia $I_2$.

Find: The ratio $I_1/I_2$.

For a solid cylinder about its own axis,

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

For bodies made of the same material, mass is proportional to volume.

For the original cylinder,

$$M_1 \propto \pi R^2 L$$

For the carved cylinder,

$$M_2 \propto \pi \left(\frac{R}{3}\right)^2 \left(\frac{L}{2}\right)$$

So,

$$M_2 = \frac{\pi R^2 L}{18} \propto \frac{M_1}{18}$$

Hence,

$$\frac{M_1}{M_2} = 18$$

Now write the two moments of inertia:

$$I_1 = \frac{1}{2} M_1 R^2$$ $$I_2 = \frac{1}{2} M_2 \left(\frac{R}{3}\right)^2$$

Therefore,

$$\frac{I_1}{I_2} = \frac{M_1 R^2}{M_2 (R^2/9)} = 9 \times \frac{M_1}{M_2}$$

Substituting $\frac{M_1}{M_2} = 18$,

$$\frac{I_1}{I_2} = 9 \times 18 = 162$$

Therefore, the required ratio is $162$.

Given: The two cylinders are made of the same material.

Find: The ratio $I_1/I_2$.

Use the scaling relation for a solid cylinder about its own axis:

$$I \propto MR^2$$

and since mass is proportional to volume,

$$M \propto R^2L$$

So,

$$I \propto R^4L$$

Hence,

$$\frac{I_1}{I_2} = \frac{R^4L}{\left(\frac{R}{3}\right)^4 \left(\frac{L}{2}\right)}$$ $$= \frac{R^4L}{\frac{R^4}{81} \cdot \frac{L}{2}} = 162$$

This works because both objects are solid cylinders of the same material and the same axis of rotation is used. Therefore, the required ratio is $162$.