Q. The moment of inertia of a uniform rod about a perpendicular axis passing through one end is $I_{1}$. The same rod is bent into a ring and its moment of inertia about a diameter is $I_{2}$. Then $\frac{I_{1}}{I_{2}}$ is
A) $\frac{\pi^{2}}{3}$
B) $\frac{2\pi^{2}}{3}$
C) $\frac{4\pi^{2}}{3}$
D) $\frac{8\pi^{2}}{3}$
Explanation
(d) Let $M$ be the mass and $L$ be the length of given uniform rod
Moment of inertia of the rod about one end is
$I_{1} = \frac{1}{3}ML^{2}$ …(i)
When it is bent into a ring
$\therefore L = 2\pi R$
Or $R = \frac{L}{2\pi}$ …(ii)
Moment of inertia of a ring about its diameter is
$I_{2} = \frac{MR^{2}}{2} = \frac{ML^{2}}{8\pi^{2}}$ [Using (ii)]
$\therefore\frac{I_{1}}{I_{2}} = \frac{ML^{2}}{3} \times \frac{8\pi^{2}}{ML^{2}} = \frac{8\pi^{2}}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.