Question

The diagram shows a uniform ring of mass M and radius a. If the moment of inertia of the ring about the axis PQ is I, the moment of inertia of the ring about the axis XY is

• A. $I+\frac{1}{4}M{a}^2$
• B. $i+\frac{1}{2}M{a}^2$
• C. $I-\frac{1}{4}M{a}^2$
• D. $i-\frac{1}{2}M{a}^2$

Answer 1

The correct option is C $I-\frac{1}{4}M{a}^2$

$I=I_{cm}+\frac{M{a}^2}{4}$

$=I_{cm}=I-\frac{M{a}^2}{4}$

Hence$,$ option $\left(C\right)$ is correct$.$