Question

Two thin uniform circular rings each of radius $10cm$ and mass $0.1Kg$ are arranged such that they have 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 $kg{m}^2$ is:

• A. $15\times {10}^{-3}$
• B. $5\times {10}^{-3}$
• C. $15\times {10}^{-4}$
• D. $18\times {10}^{-4}$

Answer 1

Answer: (C)

$15\times {10}^{-4}$

Here $M=0.1kg$, $R=10cm=0.1m$

The MI of system is the sum of MI for first ring and MI for the second,

$M{R}^2/2+M{R}^2=3M{R}^2/2=1.5\times 0.1\times {0.1}^2=15\times {10}^{-4}kg{m}^2$