From a circular ring of mass ‘M’ and radius ‘R’ an arc corresponding to a 908 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 answer is 3/4.

Step-by-step Explanation:
1. Let the total mass of the original uniform circular ring be $M$ and its radius be $R$.
2. The moment of inertia of the complete circular ring of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is given by the standard formula:

$I_{\text{total}}=M{R}^2$

3. Since the ring is uniform, its mass per unit length (linear mass density, $\lambda$) is constant. The total circumference of the ring is $2\pi R$, which corresponds to a total angle of $360°$.
4. An arc corresponding to a $90°$ sector is removed from the ring.
5. The fraction of the ring removed is:

$\frac{90°}{360°}$

6. Therefore, the fraction of the mass of the ring that is removed is $\frac{1}{4}M$, and the mass of the remaining part of the ring, $M^{\prime }$, is:

$M^{\prime }=M-\frac{1}{4}M=\frac{3}{4}M$

7. Since every element of the remaining part of the ring is still at the same perpendicular distance $R$ from the axis passing through the center, the moment of inertia of the remaining part is simply the product of its mass and the square of the radius:

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

8. We are given that the moment of inertia of the remaining part is $KM{R}^2$. Comparing the two expressions:

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

9. Thus, the value of $K$ is:

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