JEE Main 2025 — Rotational Motion Question with Solution

First, relate the length of the rod to the circumference of the ring:

$L=2\pi R$

The mass of the ring, denoted by $M$, is the product of the linear mass density and the length of the rod:

$M=\lambda \times L$

The moment of inertia of a ring about a diameter is given by the formula:

$I=\frac{M{R}^2}{2}$

Substituting the expression for mass $M$ and using the relation for $R$ derived from the circumference:

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

Simplifying the expression, we arrive at:

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

This is the moment of inertia of the ring about any of its diameters.