Q. Moment of inertia of ring of mass $M$ and radius $R\ $about an axis passing through the centre and perpendicular to the plane is $I\ $. What is the moment of inertia about its diameter?
A) $I$
B) $\frac{I}{2}$
C) $\frac{I}{\sqrt{2}}$
D) ${I + MR}^{2}$
Explanation
(b) According to the theorem of perpendicular axes.
$I_{AB} + I_{CD} = MR^{2}$
$I_{d} + I_{d} = I$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\because I_{AB} = I_{CD} = I_{d})$
$2I_{d} = I$
$I_{d} = \frac{I}{2}$
[image:7_image107.emf]
where, $I_{d} =$moment of inertia about diameter of the ring, $I =$moment of inertia about axes passing through to the ring.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.