Q. A uniform cylinder has a radius $R$ and length $L$. If the moment of inertia of this cylinder about an axis passing through its centre and normal to its circular face is equal to the moment of inertia of the same cylinder about an axis passing through its centre and perpendicular to its length, then
A) $L = R$
B) $L = \sqrt{3}\ R$
C) $L = \frac{R}{\sqrt{3}}$
D) $L = \sqrt{\frac{3}{2}R}$
Explanation
(b) M.I. of a cylinder about its centre and parallel to its length $= \frac{MR^{2}}{2}$
M.I. about its centre and perpendicular to its length $= M\left( \frac{L^{2}}{12} + \frac{R^{2}}{4} \right)$
According to problem, $\frac{ML^{2}}{12} + \frac{MR^{2}}{4} = \frac{MR^{2}}{2}$
By solving we get $L = \sqrt{3}R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.