Q. Moment of inertia of ring about its diameter is $I$. Then, moment of inertia about an axis passing through centre perpendicular to its plane is
A) $2I$
B) $\frac{I}{2}$
C) $\frac{3}{2}I$
D) $I$
Explanation
(a) By the theorem of perpendicular axes, the moment of inertia about the central axis $I_{C}$, will be equal to the sum of its moments of inertia about two mutually perpendicular diameters lying in its plane.
Thus, $I_{d} = I = \frac{1}{2}MR^{2}$
$\therefore$ $I_{C} = I + I$
$= \frac{1}{2}MR^{2} + \frac{1}{2}MR^{2}\ $
$= I + I = 2I$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.