From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a 90° sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is ‘K’ times ‘MR²’. Then the value of ‘K’ is :

The correct option is 3/4.

Let's understand the problem step-by-step.
1. A complete uniform circular ring has a mass $M$ and a radius $R$.
2. The moment of inertia of a complete circular ring about an axis passing through its center and perpendicular to its plane is given by:
$I_{\text{total}}=M{R}^2$

3. Let $\lambda$ be the mass per unit length of the ring. Since the total circumference of the ring is $2\pi R$, we have:
$\lambda =\frac{M}{2\pi R}$

4. Now, an arc corresponding to a 90° sector is removed from the ring.
The angle of a complete circle is 360°. Therefore, the fraction of the ring removed is:
$\frac{{90}^{\circ }}{{360}^{\circ }}=\frac{1}{4}$

5. Consequently, the remaining portion of the ring corresponds to a sector of:
${360}^{\circ }-{90}^{\circ }={270}^{\circ }$
This represents $1-\frac{1}{4}=\frac{3}{4}$ of the entire ring.

6. Since the ring is uniform, the mass of the remaining part, $M^{\prime }$, is:
$M^{\prime }=\frac{3}{4}M$

7. The moment of inertia of any part of a ring about the central perpendicular axis depends only on the mass of that part and its distance from the axis. Since every point on the remaining arc of radius $R$ is at a distance $R$ from the axis of rotation, the moment of inertia $I$ of the remaining part is:
$I=M^{\prime }{R}^2$

8. Substituting the value of $M^{\prime }$ into the equation:
$I=\left(\frac{3}{4}M\right)R^2=\frac{3}{4}M{R}^2$

9. The problem states that the moment of inertia of the remaining part is $K$ times $M{R}^2$:
$I=KM{R}^2$
Comparing the two expressions, we get:
$K=\frac{3}{4}$