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

Answer 1

Answer: (D)

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