A solid sphere and a solid cylinder of same m | vedclass

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(A) The moment of inertia $I$ of a body is given by $I = mk^2$,where $m$ is the mass and $k$ is the radius of gyration.For a solid sphere,the moment o

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$5$

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(A) The moment of inertia $I$ of a body is given by $I = mk^2$,where $m$ is the mass and $k$ is the radius of gyration.For a solid sphere,the moment of inertia about its central axis is $I_{	ext{sph}} = \frac{2}{5}mR^2$.Equating this to $mk_{	ext{sph}}^2$,we get $k_{	ext{sph}} = \sqrt{\frac{2}{5}}R$.For a solid cylinder,the moment of inertia about its central axis is $I_{	ext{cyl}} = \frac{1}{2}mR^2$.Equating this to $mk_{	ext{cyl}}^2$,we get $k_{	ext{cyl}} = \frac{R}{\sqrt{2}}$.The ratio of the radii of gyration is $\frac{k_{	ext{sph}}}{k_{	ext{cyl}}} = \frac{\sqrt{2/5}R}{R/\sqrt{2}} = \sqrt{\frac{2}{5}} 	imes \sqrt{2} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}$.Comparing this with the given ratio $\frac{2}{\sqrt{x}}$,we find that $x = 5$.

$A$ 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 $(k_{\text{sph}} : k_{\text{cyl}})$ is $2 : \sqrt{x}$,then the value of $x$ is .............

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