Q. Moment of inertia of a thin rod of mass $M$ and length $L$ about an axis passing through its centre is $\frac{ML^{2}}{M}$. Its moment of inertia about a parallel axis at a distance of $\frac{L}{4}$ from this axis is given by
A) $\frac{ML^{2}}{48}$
B) $\frac{ML^{3}}{48}$
C) $\frac{ML^{2}}{12}$
D) $\frac{7ML^{2}}{48}$
Explanation
(d) Apply parallel axis theorem
$I = I_{CM} + Mh^{2}$
$= \frac{ML^{2}}{12} + M\left( \frac{L}{4} \right)^{2}$
$= \frac{7ML^{2}}{48}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.