JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Rotational Mechanics

Both the parent solid cylinder and the smaller cylinder carved from it are made of the same material of density $\rho$ and each rotates about its own axis, so the axis moment of inertia $\frac{1}{2}m{r}^2$ applies to each. For the parent, $I_1=\frac{1}{2}M{R}^2$ with $M=\rho \pi R^{2}L$. The smaller cylinder has radius $R/3$ and length $L/2$, so its mass is $m=\rho \pi {\left(\frac{R}{3}\right)}^2\frac{L}{2}=\frac{\rho \pi R^{2}L}{18}=\frac{M}{18}$, and its moment of inertia is $I_2=\frac{1}{2}m{\left(\frac{R}{3}\right)}^2=\frac{1}{2}\cdot \frac{M}{18}\cdot \frac{R^2}{9}=\frac{M{R}^2}{324}$. Dividing, $\frac{I_1}{I_2}=\frac{\frac{1}{2}M{R}^2}{\frac{1}{324}M{R}^2}=162$. The ratio $I_1/I_2$ is $162$.