A solid sphere and a disc,having the same mas | vedclass

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(C) 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} \

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

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(C) The time taken for an object to roll down an inclined plane of height $h$ and angle $	heta$ is given by the formula: $t = \frac{1}{\sin 	heta} \sqrt{\frac{2h}{g} \left( 1 + \frac{K^2}{R^2} ight)}$.For a solid sphere,the radius of gyration squared is $K^2 = \frac{2}{5}R^2$,so $\frac{K^2}{R^2} = \frac{2}{5}$.For a disc,the radius of gyration squared is $K^2 = \frac{1}{2}R^2$,so $\frac{K^2}{R^2} = \frac{1}{2}$.The ratio of the time taken by the sphere $(t_S)$ to the time taken by the disc $(t_D)$ is: $\frac{t_S}{t_D} = \sqrt{\frac{1 + (K^2/R^2)_S}{1 + (K^2/R^2)_D}}$.Substituting the values: $\frac{t_S}{t_D} = \sqrt{\frac{1 + 2/5}{1 + 1/2}} = \sqrt{\frac{7/5}{3/2}} = \sqrt{\frac{7}{5} 	imes \frac{2}{3}} = \sqrt{\frac{14}{15}}$.Thus,the ratio is $\sqrt{14}:\sqrt{15}$.

$A$ solid sphere and a disc,having the same mass and radius,roll down from rest from the same height on a smooth inclined plane. The ratio of the time taken by them is:

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