JEE Physics 2025 Question with Solution

Given: The moment of inertia of the original rod is

$$\alpha = \frac{ML^2}{12}$$

where the rod has mass $M$ and length $L$.

Find: The moment of inertia of the cross formed by joining the two equal halves, about an axis passing through its center and normal to the plane of the cross.

When the rod is cut into two equal parts, each smaller rod has mass

$$m = \frac{M}{2}$$

and length

$$l = \frac{L}{2}$$

The axis through the center of the cross and normal to its plane passes through the center of each small rod and is perpendicular to its length.

So, for one small rod,

$$I_{\text{small}} = \frac{ml^2}{12}$$

Substituting the values,

$$I_{\text{small}} = \frac{\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2}{12} = \frac{ML^2}{96}$$

The cross consists of two identical small rods, so by superposition,

$$I_{\text{cross}} = 2I_{\text{small}} = 2 \times \frac{ML^2}{96} = \frac{ML^2}{48}$$

Using

$$\alpha = \frac{ML^2}{12}$$

we get

$$I_{\text{cross}} = \frac{ML^2}{48} = \frac{1}{4}\left(\frac{ML^2}{12}\right) = \frac{\alpha}{4}$$

Therefore, the moment of inertia of the cross is $\frac{\alpha}{4}$. The correct option is B.

Given: The original rod has moment of inertia

$$\alpha = \frac{ML^2}{12}$$

Find: The moment of inertia of the new cross-shaped system.

The original rod is divided into two equal rods. For each part:

$$m = \frac{M}{2}, \qquad l = \frac{L}{2}$$

The moment of inertia of a thin rod about an axis through its center and perpendicular to its length is

$$I = \frac{ml^2}{12}$$

Hence for one part,

$$I = \frac{\left(\frac{M}{2}\right)\left(\frac{L}{2}\right)^2}{12}$$ $$I = \frac{\left(\frac{M}{2}\right)\left(\frac{L^2}{4}\right)}{12}$$ $$I = \frac{ML^2}{96}$$

Since two such rods form the cross,

$$I_{\text{cross}} = \frac{ML^2}{96} + \frac{ML^2}{96} = \frac{ML^2}{48}$$

Now compare with

$$\alpha = \frac{ML^2}{12}$$

Therefore,

$$I_{\text{cross}} = \frac{ML^2}{48} = \frac{\alpha}{4}$$

So the correct option is B.