Q. Two rings of the same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass of the ring $= m$ and radius $= r$)
A) $\frac{1}{2}mr^{2}$
B) $mr^{2}$
C) $\frac{3}{2}mr^{2}$
D) $2mr^{2}$
Explanation
(c) $I_{1} =$ M.I. of ring about its diameter $= \frac{1}{2}mr^{2}$
$I_{2} =$ M.I. of ring about the axis normal to plane and passing through centre $= mr^{2}$
Two rings are placed according to figure. Then
$I_{xx'} = I_{1} + I_{2} = \frac{1}{2}mr^{2} + mR^{2} = \frac{3}{2}mr^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.