Q 11-06-082JEE MainJEE Main 2024 (5 Apr, Shift 1)Hard
The ratio of the radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moments of inertia about their diameter axes AB as shown in the figure, is $\sqrt{8/x}$. The value of $x$ is
Answer: (D) $67$
Hollow sphere about a diameter: $k_1^2 = \dfrac23R^2$.
Solid cylinder (radius $R$, length $L = 4R$) about a diameter of its end face (perpendicular to its length):
$$I = M\left(\frac{R^2}{4} + \frac{L^2}{3}\right) = M\left(\frac{R^2}{4} + \frac{16R^2}{3}\right) = \frac{67}{12}MR^2 \Rightarrow k_2^2 = \frac{67}{12}R^2$$
$$\frac{k_1}{k_2} = \sqrt{\frac{2/3}{67/12}} = \sqrt{\frac{8}{67}} \Rightarrow x = 67$$
Solution by Sreeraj P, M.Sc Physics