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(D) For a cylinder rolling down an inclined plane of height $h$,the mechanical energy is conserved. The potential energy at the top is converted into

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The speed will be same but time of descent will be different.

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(D) For a cylinder rolling down an inclined plane of height $h$,the mechanical energy is conserved. The potential energy at the top is converted into translational and rotational kinetic energy at the bottom: $mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$. Since $I = \frac{1}{2}mr^2$ and $v = r\omega$,we have $mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v}{r})^2 = \frac{3}{4}mv^2$. Thus,$v = \sqrt{\frac{4gh}{3}}$. Since $h$ is the same,the final speed $v$ is the same in both cases.The acceleration of a body rolling down an incline is $a = \frac{g \sin 	heta}{1 + \frac{I}{mr^2}}$. Since the inclination $	heta$ is different,the acceleration $a$ will be different. The time of descent is given by $t = \sqrt{\frac{2s}{a}}$,where $s = \frac{h}{\sin 	heta}$. Substituting $a$,we get $t = \sqrt{\frac{2h}{g \sin^2 	heta} (1 + \frac{I}{mr^2})}$. Since $	heta$ is different,the time of descent $t$ will be different. Therefore,the speed is the same,but the time of descent is different.

$A$ cylinder rolls down two different inclined planes of the same height but of different inclinations.

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