Rotational Motion

JEE Main level, numerical answer, medium

35 of 280

Question 35 of 280

A ring and a solid sphere rotating about an axis passing through their centres have same radii of gyration. The axis of rotation is perpendicular to plane of ring. The ratio of radius of ring to that of sphere is 2x\sqrt{\dfrac{2}{x}}. The value of xx is _____.

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Answer: 5

The moment of inertia of the ring about an axis perpendicular to the plane of ring is Ir=mk2{I}_{r}=m{k}^{2}.
The moment of inertia of the sphere is
Is=25mR2s{I}_{s}=\dfrac{2}{5}m{{R}^{2}}_{s}
It is given that
mk2=mR2r=25mRs2...(i)m{k}^{2}=m{{R}^{2}}_{r}=\dfrac{2}{5}m{{R}_{s}}^{2}...(i)
From equation (i)
Rr=25Rs{R}_{r}=\sqrt{\dfrac{2}{5}}{R}_{s}
From the question, it can be written that
RrRs=2x\dfrac{{R}_{r}}{{R}_{s}}=\sqrt{\dfrac{2}{x}}
⇒RrRs=25⇒x=5\Rightarrow \dfrac{{R}_{r}}{{R}_{s}}=\sqrt{\dfrac{2}{5}} \Rightarrow x=5

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