JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Rotational Mechanics

The moment of inertia of a thin uniform rod is $\frac{M{L}^2}{12}$ about a perpendicular axis through its centre and $\frac{M{L}^2}{3}$ about a perpendicular axis through one end. A rod of length $L$ and mass $M$ about its midpoint therefore has $\frac{M{L}^2}{12}$. A rod of length $L$ and mass $2M$ about one end has $\frac{(2M)L^2}{3}=\frac{2M{L}^2}{3}$. A rod of length $2L$ and mass $M$ about its midpoint has $\frac{M(2L{)}^2}{12}=\frac{M{L}^2}{3}$. A rod of length $2L$ and mass $2M$ about one end has $\frac{(2M)(2L{)}^2}{3}=\frac{8M{L}^2}{3}$.