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(C) The linear acceleration $a$ of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{K^2}{R^2}}$,where $K$ is the

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$\sqrt{14} : \sqrt{15}$

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(C) The linear acceleration $a$ of a body rolling down an inclined plane is given by $a = \frac{g \sin 	heta}{1 + \frac{K^2}{R^2}}$,where $K$ is the radius of gyration.For a solid sphere,$K^2 = \frac{2}{5}R^2$,so $a_s = \frac{g \sin 	heta}{1 + 2/5} = \frac{5}{7} g \sin 	heta$.For a disc,$K^2 = \frac{1}{2}R^2$,so $a_d = \frac{g \sin 	heta}{1 + 1/2} = \frac{2}{3} g \sin 	heta$.Since the distance $s$ is the same and they start from rest,$s = \frac{1}{2} a t^2$,which implies $t = \sqrt{\frac{2s}{a}}$.Therefore,the ratio of times is $\frac{t_s}{t_d} = \sqrt{\frac{a_d}{a_s}} = \sqrt{\frac{(2/3) g \sin 	heta}{(5/7) g \sin 	heta}} = \sqrt{\frac{2}{3} 	imes \frac{7}{5}} = \sqrt{\frac{14}{15}}$.Thus,the ratio is $\sqrt{14} : \sqrt{15}$.

When a uniform solid sphere and a disc of the same mass and of the same radius roll down a rough inclined plane from rest to the same distance,then the ratio of the time taken by them is

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