JEE MAIN - Physics (2024 - 5th April Morning Shift - No. 4)
Ratio of radius of gyration of a hollow sphere to that of a solid cylinder of equal mass, for moment of Inertia about their diameter axis $$A B$$ as shown in figure is $$\sqrt{8 / x}$$. The value of $$x$$ is :
_5th_April_Morning_Shift_en_4_1.png)
34
51
67
17
Explanation
For hollow sphere
$$\begin{aligned} & I_1=\frac{2}{3} m R^2=\left(\sqrt{\frac{2}{3}} R\right)^2 m \\ & \therefore \quad K_1=\sqrt{\frac{2}{3}} R \end{aligned}$$
For solid cylinder
$$\begin{aligned} & I_2=\frac{1}{4} m R^2+\frac{m}{3}(4 R)^2 \\ & =\left(\sqrt{\frac{67}{12}} R\right)^2 m \\ & \therefore \quad K_2=\sqrt{\frac{67}{12}} R \\ & \Rightarrow \frac{K_1}{K_2}=\sqrt{\frac{2}{3} \times \frac{12}{67}} \\ & =\sqrt{\frac{8}{67}} \end{aligned}$$
$$\Rightarrow \quad x=67$$
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