Q. Two thin uniform circular rings each of radius 10 cm and mass 0.1 kg are arranged such that they have a common centre and their planes are perpendicular to each other. The moment of inertia of this system about an axis passing through their common centre and perpendicular to the plane of one of the rings in $kgm^{- 2}$is
A) $15 \times 10^{- 3}$
B) $5 \times 10^{- 3}$
C) $1.5 \times 10^{- 3}$
D) $18 \times 10^{- 4}$
Explanation
(c) Here, $m_{1} = m_{2} = 0.1\ kg$
$r_{1} = r_{2} = 10\ cm = 0.1\ m$
$I = I_{1} + I_{2} = m_{1}r_{1}^{2} + \frac{1}{2}m_{2}r_{2}^{2} = \frac{3}{2}m_{1}r_{1}^{2}$
$= \frac{3}{2} \times 0.1(0.1)^{2} = 1.5 \times 10^{- 3}\ kg\ m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.