To solve this problem, we need to find the ratio of the radius of gyration of a thin uniform disc about two different axes.$\\$The radius of gyration $k$ is related to the moment of inertia $I$ and mass $m$ by the formula:$\\ k = \sqrt{\frac{I}{m}} \\$1. Moment of Inertia about an axis through the center and normal to the plane:$\\$For a thin uniform disc of mass $m$ and radius $R, \\ I_{center} = \frac{1}{2} m R^2 \\$The radius of gyration $k_{center}$ is:$\\ k_{center} = \sqrt{\frac{I_{center}}{m}} = \sqrt{\frac{\frac{1}{2} m R^2}{m}} = \sqrt{\frac{1}{2}} R \\$2. Moment of Inertia about a diameter:$\\$Using the perpendicular axis theorem, we have:$\\ I_{diameter} = \frac{1}{4} m R^2 \\$The radius of gyration $k_{diameter}$ is:$\\ k_{diameter} = \sqrt{\frac{I_{diameter}}{m}} = \sqrt{\frac{\frac{1}{4} m R^2}{m}} = \frac{R}{2} \\$3. Ratio of the radii of gyration:$\\$We need to find the ratio $\frac{k_{center}}{k_{diameter}}: \\ \frac{k_{center}}{k_{diameter}} = \frac{\sqrt{\frac{1}{2}} R}{\frac{R}{2}} = \frac{\sqrt{\frac{1}{2}}}{\frac{1}{2}} = \sqrt{2} \\$Therefore, the ratio of the radius of gyration of the disc about its center to that about its diameter is $\sqrt{2} : 1. \\$This corresponds to Option 2.