To solve this problem, we need to find the moment of inertia of the remaining part of the disc after a circular hole is cut out.$\\$Given:$\\$• Radius of the original disc $= R \\$• Mass of the original disc $= M \\$• Diameter of the hole $= R \\$• Radius of the hole $= \frac{R}{2} \\$The moment of inertia of the original disc about a perpendicular axis through its center is:$\\ I_{original} = \frac{1}{2} M R^2 \\$The area of the original disc is:$\\ A_{original} = \pi R^2 \\$The area of the hole is:$\\ A_{hole} = \pi \left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{4} \\$The mass of the hole (assuming uniform mass distribution) is:$\\ M_{hole} = M \cdot \frac{A_{hole}}{A_{original}} = M \cdot \frac{\frac{\pi R^2}{4}}{\pi R^2} = \frac{M}{4} \\$The moment of inertia of the hole about its own center is:$\\ I_{hole\_center} = \frac{1}{2} M_{hole} \left(\frac{R}{2}\right)^2 = \frac{1}{2} \cdot \frac{M}{4} \cdot \frac{R^2}{4} = \frac{M R^2}{32} \\$The distance from the center of the hole to the center of the original disc is $\frac{R}{2}. \\$Using the parallel axis theorem, the moment of inertia of the hole about the center of the original disc is:$\\ I_{hole} = I_{hole\_center} + M_{hole} \left(\frac{R}{2}\right)^2 \\ I_{hole} = \frac{M R^2}{32} + \frac{M}{4} \cdot \frac{R^2}{4} \\ I_{hole} = \frac{M R^2}{32} + \frac{M R^2}{16} \\ I_{hole} = \frac{M R^2}{32} + \frac{2 M R^2}{32} \\ I_{hole} = \frac{3 M R^2}{32} \\$The moment of inertia of the remaining part of the disc is:$\\ I_{remaining} = I_{original} - I_{hole} \\ I_{remaining} = \frac{1}{2} M R^2 - \frac{3 M R^2}{32} \\ I_{remaining} = \frac{16 M R^2}{32} - \frac{3 M R^2}{32} \\ I_{remaining} = \frac{13 M R^2}{32} \\$Therefore, the moment of inertia of the remaining part of the disc is $\frac{13}{32} M R^2. \\$This corresponds to Option 4.