Question

If $I$ is the moment of inertia of a circular ring about a tangent of ring in its plane then the moment of inertia of the same ring about a tangent of ring perpendicular to its plane is:

• A. $\frac{2I}{3}$
• B. $\frac{3I}{2}$
• C. $\frac{4I}{3}$
• D. $\frac{3I}{4}$

Answer 1

The correct option is C $\frac{4I}{3}$

Moment of Inertia of a ring about any axis perpendicular to its plane is $M{R}^2$
By perpendicular axis theorem , moment of Inertia about any diameter is $\frac{M{R}^2}{2}$.
Moment of Inertia about any tangent in the plane is $\frac{M{R}^2}{2}+M{R}^2=\frac{3M{R}^2}{2}=I$
Moment of Inertia about any tangent and perpendicular plane is $M{R}^2+M{R}^2=2M{R}^2$
$=4\left(\frac{M{R}^2}{2}\right)=\frac{4}{3}\left(\frac{3M{R}^2}{2}\right)=\frac{4I}{3}$