To solve this problem, we need to find the moment of inertia of the remaining part of the ring after removing a $90^\circ$ sector.$\\$Given:$\\$• Total mass of the ring $M \\$• Radius of the ring $R \\$• A $90^\circ$ sector is removed, which is $\frac{1}{4}$ of the ring.$\\$The moment of inertia of the entire ring about an axis through its center and perpendicular to its plane is:$\\ I_{total} = MR^2 \\$Since a $90^\circ$ sector is removed, the mass of the removed sector is:$\\ M_{removed} = \frac{M}{4} \\$The moment of inertia of the removed sector about the same axis is:$\\ I_{removed} = \frac{M}{4} \cdot R^2 = \frac{MR^2}{4} \\$The moment of inertia of the remaining part of the ring is:$\\ I_{remaining} = I_{total} - I_{removed} \\ I_{remaining} = MR^2 - \frac{MR^2}{4} \\$Simplify the expression:$\\ I_{remaining} = \frac{4MR^2}{4} - \frac{MR^2}{4} \\ I_{remaining} = \frac{3MR^2}{4} \\$The moment of inertia of the remaining part is $K$ times $MR^2,$ so:$\\ I_{remaining} = K \cdot MR^2 \\$Equating the two expressions for $I_{remaining}: \\ \frac{3MR^2}{4} = K \cdot MR^2 \\$Divide both sides by $MR^2: \\ K = \frac{3}{4} \\$Therefore, the value of $K$ is $\frac{3}{4}. \\$This corresponds to Option 1.