Question

From a circular ring of mass $M$ and radius $R$, an arc corresponding to a ${90}^{\circ }$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the center of the ring and perpendicular to the plane of the ring is $K$ times $M{R}^2$. Then the value of $K$ is -

• A. $\frac{1}{8}$
• B. $\frac{3}{4}$
• C. $\frac{7}{8}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{3}{4}$

Given:

Mass of ring $=M$; Radius of ring $=R$

Now ${90}^{\circ }$ arc is removed from circular ring, then mass removed is $=\frac{M}{4}$.


Mass of the remaining portion $=\frac{3M}{4}$

Moment of inertia of remaining part $=\int dm{r}^2$

$\Rightarrow I=R^2\int dm$ $(\because r=R)$

$\Rightarrow I=\frac{3M{R}^2}{4}$.

So, the value of $K$ is $\frac{3}{4}$.

Hence, option $\left(\text{B}\right)$ is the correct answer.

Why this question ? Trick : As long as the mass remains constant, $I$ does not change even if a portion of it is removed provided the axis remains the same.