Q75 / 75 · JEE Main 2026 (24 January Shift 2) · Physics
A uniform solid cylinder of length $L$ and radius $R$ has moment of inertia about its axis equal to $I_{1}$. A small co-centric cylinder of length $L / 2$ and radius $R / 3$ carved from this cylinder has moment of inertia about its axis equals to $I_{2}$. The ratio $I_{1} / I_{2}$ is $\_\_\_\_$.
Correct answer: 162
Solution
For solid cylinder: $I = \frac{1}{2}mR^2$
Large cylinder: mass $m_1 = \rho \pi R^2 L$, so $I_1 = \frac{1}{2}\rho \pi R^2 L \times R^2 = \frac{\rho \pi R^4 L}{2}$
Small cylinder (carved out): length L/2, radius R/3, mass $m_2 = \rho \pi(R/3)^2(L/2) = \frac{\rho \pi R^2 L}{18}$
$I_2 = \frac{1}{2} \times \frac{\rho \pi R^2 L}{18} \times (R/3)^2 = \frac{1}{2} \times \frac{\rho \pi R^2 L}{18} \times \frac{R^2}{9} = \frac{\rho \pi R^4 L}{324}$
$\frac{I_1}{I_2} = \frac{\rho \pi R^4 L/2}{\rho \pi R^4 L/324} = \frac{324}{2} = 162$
Large cylinder: mass $m_1 = \rho \pi R^2 L$, so $I_1 = \frac{1}{2}\rho \pi R^2 L \times R^2 = \frac{\rho \pi R^4 L}{2}$
Small cylinder (carved out): length L/2, radius R/3, mass $m_2 = \rho \pi(R/3)^2(L/2) = \frac{\rho \pi R^2 L}{18}$
$I_2 = \frac{1}{2} \times \frac{\rho \pi R^2 L}{18} \times (R/3)^2 = \frac{1}{2} \times \frac{\rho \pi R^2 L}{18} \times \frac{R^2}{9} = \frac{\rho \pi R^4 L}{324}$
$\frac{I_1}{I_2} = \frac{\rho \pi R^4 L/2}{\rho \pi R^4 L/324} = \frac{324}{2} = 162$
Hence, the correct answer is 162.