Q. Moment of inertia of a solid cylinder of length $L$ and diameter $D$ about an axis passing through its centre of gravity and perpendicular to its geometric axis is
A) $M\left( \frac{D^{2}}{4} + \frac{L^{2}}{12} \right)$
B) $M\left( \frac{L^{2}}{16} + \frac{D^{2}}{8} \right)$
C) $M\left( \frac{D^{2}}{4} + \frac{L^{2}}{6} \right)$
D) $M\left( \frac{L^{2}}{12} + \frac{D^{2}}{16} \right)$
Explanation
(d) $I = M\left( \frac{L^{2}}{12} + \frac{r^{2}}{4} \right) = M\left( \frac{L^{2}}{12} + \frac{D^{2}}{16} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.