JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 _______.
- A \(34\)
- B \(17\)
- C \(67\)
- D \(51\)
Answer & Solution
Correct Answer
(C) \(67\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}_{\text {sphere }}=\frac{2}{3} \mathrm{MR}^2=\mathrm{Mk}_1^2\) \(\mathrm{I}_{\text {cylinder }}=\frac{1}{12} \mathrm{M}\left(4 \mathrm{R}^2\right)+\frac{1}{4} \mathrm{MR}^2+\mathrm{M}(2 \mathrm{R})^2\) \(=\frac{67}{12} \mathrm{MR}^2=\mathrm{Mk}_2^2\)…
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