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(A) The time taken to reach the bottom of an inclined plane of length $\ell$ is given by $t = \sqrt{\frac{2\ell}{a_{cm}}}$.The acceleration of a body

Vedclass · Accepted Answer

$t_1 < t_2$

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(A) The time taken to reach the bottom of an inclined plane of length $\ell$ is given by $t = \sqrt{\frac{2\ell}{a_{cm}}}$.The acceleration of a body rolling down an inclined plane is $a_{cm} = \frac{g \sin 	heta}{1 + \frac{I_{cm}}{MR^2}}$.For a solid sphere,$I_{cm} = \frac{2}{5}MR^2$,so $a_1 = \frac{g \sin 	heta}{1 + 2/5} = \frac{5g \sin 	heta}{7}$.For a hollow sphere,$I_{cm} = \frac{2}{3}MR^2$,so $a_2 = \frac{g \sin 	heta}{1 + 2/3} = \frac{3g \sin 	heta}{5}$.Comparing the accelerations,$a_1 = \frac{5}{7}g \sin 	heta \approx 0.714 g \sin 	heta$ and $a_2 = \frac{3}{5}g \sin 	heta = 0.6 g \sin 	heta$. Thus,$a_1 > a_2$.Since $t \propto \frac{1}{\sqrt{a_{cm}}}$,a higher acceleration results in a shorter time. Therefore,$t_1 < t_2$.

$A$ solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be $t_1$ and $t_2$,respectively,then

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