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(B) For slipping (sliding) down an inclined plane,the acceleration is $a_1 = g \sin 	heta$.Using the equation of motion $L = \frac{1}{2} a_1 t_1^2$,w

Vedclass · Accepted Answer

$1 : \sqrt{2}$

Vedclass · Answer

(B) For slipping (sliding) down an inclined plane,the acceleration is $a_1 = g \sin 	heta$.Using the equation of motion $L = \frac{1}{2} a_1 t_1^2$,we get $t_1 = \sqrt{\frac{2L}{g \sin 	heta}}$.For rolling down the same inclined plane,the acceleration of a ring (moment of inertia $I = MR^2$) is $a_2 = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}} = \frac{g \sin 	heta}{1 + 1} = \frac{g \sin 	heta}{2}$.Using the equation of motion $L = \frac{1}{2} a_2 t_2^2$,we get $t_2 = \sqrt{\frac{2L}{a_2}} = \sqrt{\frac{2L}{\frac{g \sin 	heta}{2}}} = \sqrt{\frac{4L}{g \sin 	heta}} = \sqrt{2} \sqrt{\frac{2L}{g \sin 	heta}} = \sqrt{2} t_1$.Therefore,the ratio $\frac{t_1}{t_2} = \frac{1}{\sqrt{2}}$.Thus,option $B$ is correct.

$A$ ring takes time $t_1$ in slipping down an inclined plane of length $L$ and takes time $t_2$ in rolling down the same plane. The ratio $\frac{t_1}{t_2}$ is

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