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JEE Mains · Physics · STD 11 - System of particles and rotational motion
  • JEE Mains 2024
  • Hard
  • MCQ

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 \(\sqrt{\frac{8}{x}}\). The value of \(x\) is _______.

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Answer & Solution

Correct Answer

(C) \(67\)

Step-by-step Solution

Detailed explanation

\(\mathrm{I}_{\text {sphere }}=\frac{2}{3} \mathrm{MR}^2=\mathrm{Mk}_1^2\)

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

\(=\frac{67}{12} \mathrm{MR}^2=\mathrm{Mk}_2^2\)

\(\frac{\mathrm{k}_1}{\mathrm{k}_2}=\sqrt{\frac{2}{3} \cdot \frac{12}{67}}=\sqrt{\frac{8}{67}}\)

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