JEE Physics 2025 Question with Solution

Given: A rod of linear mass density $\lambda$ and length $L$ is bent to form a ring of radius $R$.

Find: The moment of inertia of the ring about any diameter.

The total mass of the rod, and hence of the ring, is

$$M = \lambda L$$

Since the rod is bent into a complete ring, its length equals the circumference:

$$L = 2\pi R$$

So,

$$R = \frac{L}{2\pi}$$

For a ring, the moment of inertia about any diameter is

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

Substituting $M = \lambda L$ and $R = \frac{L}{2\pi}$,

$$I = \frac{1}{2} \cdot \lambda L \cdot \left(\frac{L}{2\pi}\right)^2$$ $$I = \frac{\lambda L}{2} \cdot \frac{L^2}{4\pi^2}$$ $$I = \frac{\lambda L^3}{8\pi^2}$$

Therefore, the moment of inertia is $\frac{\lambda L^3}{8\pi^2}$ and the correct option is A.

Given: Linear mass density $\lambda$, rod length $L$, and the rod is bent into a ring.

Find: Moment of inertia of the ring about a diameter.

The solution uses two facts:

1. Mass of the ring:

$$M = \lambda L$$

2. Circumference-radius relation:

$$L = 2\pi R$$

Hence,

$$R = \frac{L}{2\pi}$$

Now use the formula for a ring about its diameter:

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

Substitute for $M$:

$$I = \frac{1}{2}(\lambda L)R^2$$

Then substitute for $R$:

$$I = \frac{1}{2} \lambda L \left(\frac{L}{2\pi}\right)^2$$

Squaring the bracket,

$$I = \frac{1}{2} \lambda L \cdot \frac{L^2}{4\pi^2}$$

Therefore,

$$I = \frac{\lambda L^3}{8\pi^2}$$

So, the correct option is A, corresponding to $\frac{\lambda L^3}{8\pi^2}$.