A solid sphere and a solid cylinder of same mass and radius are rolling on a horizontal surface without slipping. The ratio of their radius of gyrations respectively (k_ sph : k_ cyl ) is 2: \sqrtx . The value of x is \underline\hspace2cm .

Rotational Motion PYQ — JEE Main 2023 · 15 Apr · Shift 1 · Q72 with Solution

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Correct answer: 5

The radius of gyration $k$ is defined by $I = Mk^2$ so, $k = \sqrt{\frac{I}{M}}.$
For a solid sphere about its center, $I_{\text{sph}} = \frac{2}{5}MR^2.$ Hence, $k_{\text{sph}} = \sqrt{\frac{I_{\text{sph}}}{M}} = \sqrt{\frac{2}{5}R^2} = R\sqrt{\frac{2}{5}}.$
For a solid cylinder about its axis, $I_{\text{cyl}} = \frac{1}{2}MR^2.$ Hence, $k_{\text{cyl}} = \sqrt{\frac{I_{\text{cyl}}}{M}} = \sqrt{\frac{1}{2}R^2} = \frac{R}{\sqrt{2}}.$
Therefore, $k_{\text{sph}} : k_{\text{cyl}} = R\sqrt{\frac{2}{5}} : \frac{R}{\sqrt{2}}.$ Cancel $R$ : $= \sqrt{\frac{2}{5}} : \frac{1}{\sqrt{2}}.$ Multiply both terms by $\sqrt{10}$ (or simplify directly): $\frac{\sqrt{2/5}}{1/\sqrt{2}} = \sqrt{\frac{2}{5}}\cdot \sqrt{2} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}.$ So the ratio is $k_{\text{sph}} : k_{\text{cyl}} = 2 : \sqrt{5}.$
Comparing with the given form $2 : \sqrt{x},$ we get $\sqrt{x} = \sqrt{5} \Rightarrow x=5.$

Therefore, the required integer is $\boxed{5}.$