Physics · Rotational Dynamics

JEE Main 2024 — 5 April, Shift 1 — Question 40

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 AB as shown in figure is $\sqrt{\frac{8}{x}}$ . The value of $x$ is:

Answer: C

$\mathrm{I}_{\text {sphere }}=\frac{2}{3} \mathrm{MR}^{2}=\mathrm{Mk}_{1}^{2}$

$\mathrm{I}_{\text {cylinder }}=\frac{1}{12} \mathrm{M}\left(4 \mathrm{R}^{2}\right)+\frac{1}{4} \mathrm{MR}^{2}+\mathrm{M}(2 \mathrm{R})^{2}$

$=\frac{67}{12} \mathrm{MR}^{2}=\mathrm{Mk}_{2}^{2}$

$\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}=\sqrt{\frac{2}{3} \cdot \frac{12}{67}}=\sqrt{\frac{8}{67}}$

Answer key and solution verified before publishing.

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