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(B) The formula for the velocity $v$ of a body rolling down an inclined plane without slipping is given by $v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}

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$2$

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(B) The formula for the velocity $v$ of a body rolling down an inclined plane without slipping is given by $v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}$.Here,$h$ is the vertical height of the inclined plane,$h = L \sin 	heta = 60 	imes 10^{-2} 	imes \sin 30^{\circ} = 0.6 	imes 0.5 = 0.3\,m$.For a solid cylinder,the radius of gyration $k$ satisfies $k^2 = \frac{R^2}{2}$,so $\frac{k^2}{R^2} = \frac{1}{2}$.Substituting these values into the formula:$v = \sqrt{\frac{2 	imes 10 	imes 0.3}{1 + 0.5}} = \sqrt{\frac{6}{1.5}} = \sqrt{4} = 2\,ms^{-1}$.

$A$ solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60\,cm$. If the cylinder rolls without slipping,its speed upon reaching the bottom of the inclined plane is $...........\,ms^{-1}$. (Given $g = 10\,ms^{-2}$)

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