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(B) For a solid sphere rolling down an inclined plane of height $h$ and inclination $	heta$,the final speed $v$ at the bottom is given by the conserv

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speed will be same,but time of descent will be different

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(B) For a solid sphere rolling down an inclined plane of height $h$ and inclination $	heta$,the final speed $v$ at the bottom is given by the conservation of mechanical energy: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. Since $I = \frac{2}{5}mr^2$ and $\omega = v/r$,we have $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(v/r)^2 = \frac{1}{2}mv^2 + \frac{1}{5}mv^2 = \frac{7}{10}mv^2$. Thus,$v = \sqrt{\frac{10gh}{7}}$. Since $v$ depends only on $h$ and $g$,the speed at the bottom is the same for both planes.However,the acceleration $a$ down the plane is given by $a = \frac{g \sin 	heta}{1 + I/mr^2} = \frac{5}{7}g \sin 	heta$. Since the inclinations $	heta$ are different,the accelerations are different. The time of descent $t = \sqrt{\frac{2s}{a}}$,where $s = h/\sin 	heta$. Thus,$t = \sqrt{\frac{2h}{a \sin 	heta}} = \sqrt{\frac{2h}{\frac{5}{7}g \sin^2 	heta}} = \frac{1}{\sin 	heta} \sqrt{\frac{14h}{5g}}$. Since $	heta$ is different,the time of descent $t$ will be different.

$A$ solid sphere rolls down two different inclined planes of the same height,but of different inclinations. In both cases:

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