Q. A thin wire of mass $M$ and length $L$ is bent to form a circular ring. The moment of inertia about its axis is
A) $\frac{1}{4\pi^{2}}{ML}^{2}\ $
B) $\frac{1}{12}{ML}^{2}$
C) $\frac{1}{{3\pi}^{2}}{ML}^{2}$
D) $\frac{1}{\pi^{2}}{ML}^{2}$
Explanation
(a) Here a thin wire of length $L$ is bent to form
a circular ring.
[image:7_image157.emf]
Then, $2\pi r = L\ \ \ \ \ \ \ (r\ is\ the\ radius\ of\ ring)$
$\Rightarrow$ $r = \frac{L}{2\pi}$
Hence, the moment of inertia of the ring about its axis
$I = Mr^{2} \Rightarrow$ $I = M\left( \frac{L}{2\pi} \right)^{2} \Rightarrow$ $I = \frac{ML^{2}}{4\pi^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.