Rotational Motion

JEE Main level, single correct, medium

29 of 280

Question 29 of 280

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\mathrm{AB} as shown in figure is 8/x\sqrt{8 / x}. The value of xx is :

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Answer: (D) 6767

Isphere =23MR2=Mk12Icylinder =112M(4R2)+14MR2+M(2R)2=6712MR2=Mk22k1k2=23⋅1267=867\begin{aligned} & \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}}\end{aligned}

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