Question

From a ring of mass $M$ and radius $R$, a $30^{\circ }$ sector is removed. The moment of inertia of the remaining portion of the ring, about an axis passing through the centre and perpendicular to the plane of the ring is

• A. $\frac{3}{4}M{R}^2$
• B. $\frac{5}{6}M{R}^2$
• C. $M{R}^2$
• D. $\frac{11}{12}M{R}^2$

Answer 1

Answer: (D)

Mass of remaining portion
$=\frac{360^{\circ }-30^{\circ }}{360^{\circ }}\times M$
$=\frac{11}{12}M$
$l=\int dm{R}^2=R^2\int dm$
$=R^2\left[\frac{11}{12}M\right]$
$=\frac{11}{12}M{R}^2$