A solid sphere and a disc of same mass and ra | vedclass

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(D) The time taken for an object to roll down an inclined plane of height $h$ and angle $	heta$ is given by the formula: $t = \frac{1}{\sin 	heta} \

Vedclass · Accepted Answer

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

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(D) The time taken for an object to roll down an inclined plane of height $h$ and angle $	heta$ is given by the formula: $t = \frac{1}{\sin 	heta} \sqrt{\frac{2h}{g} \left( 1 + \frac{k^2}{R^2} ight)}$.Since $h$,$g$,and $	heta$ are the same for both,the ratio of times is $t_s / t_d = \sqrt{\frac{1 + k_s^2/R^2}{1 + k_d^2/R^2}}$.For a solid sphere,the moment of inertia $I = \frac{2}{5}MR^2$,so $k_s^2/R^2 = 2/5$.For a disc,the moment of inertia $I = \frac{1}{2}MR^2$,so $k_d^2/R^2 = 1/2$.Substituting these values: $t_s / t_d = \sqrt{\frac{1 + 2/5}{1 + 1/2}} = \sqrt{\frac{7/5}{3/2}} = \sqrt{\frac{7}{5} 	imes \frac{2}{3}} = \sqrt{\frac{14}{15}}$.Thus,the ratio is $\sqrt{14} : \sqrt{15}$.

$A$ solid sphere and a disc of same mass and radius start rolling down a rough inclined plane from the same height. The ratio of the time taken in the two cases is:

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