Q 11-06-248NEETNEET 2022Top questionEasy
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
Answer: (C) $\sqrt{2} : 1$
About the central normal axis: $I_1 = \dfrac{1}{2}MR^2 = Mk_1^2 \Rightarrow k_1 = \dfrac{R}{\sqrt{2}}$.
About a diameter (perpendicular axis theorem, $I_1 = 2I_d$): $I_d = \dfrac{1}{4}MR^2 \Rightarrow k_2 = \dfrac{R}{2}$.
$$\frac{k_1}{k_2} = \frac{R/\sqrt{2}}{R/2} = \sqrt{2} : 1$$
Solution by Sreeraj P, M.Sc Physics