Solution:
About central normal axis, \(I_1 = \frac{1}{2}MR^2 = MK_1^2 \Rightarrow K_1 = \frac{R}{\sqrt{2}}\). About diameter, \(I_2 = \frac{1}{4}MR^2 = MK_2^2 \Rightarrow K_2 = \frac{R}{2}\). The ratio is \(\frac{K_1}{K_2} = \sqrt{2} : 1\).
The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is
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