From a circular ring of mass M and radius R an arc corresponding to a 90 ° sector is removed. The moment of…

mass per unit length of the ring is

$\lambda =\frac{M}{2\pi R}$

$\therefore$ Mass of remaining ring is $M^{\text{'}}=\lambda \times \frac{3}{4}\left(2\pi R\right)$

$\therefore M^{\text{'}}=\frac{3}{4}M$

Moment of inertia of remaining part is

$I^{\text{'}}=M^{\text{'}}{R}^2=\frac{3}{4}M{R}^2=KM{R}^2$

$\therefore K=\frac{3}{4}$