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A solid sphere, released from rest from the top of an inclined plane of inclination $θ_{1,}$ rolls without sliding and reaches the bottom with speed $v_{1}$ and its time of descent is $t_{1} .$ The same sphere is then released from rest from the top of another inclined plane of inclination $θ_{2}$ but of the same height rolls without sliding and reaches the bottom with speed $v_{2}$ and its time of descent is $t_{2} .$ If $θ_{1} > θ_{2},$ then

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$v_{2} > v_{1}; t_{2} < t_{1}$

$v_{2} = v_{1}; t_{2} > t_{1}$

$v_{2} < v_{1}; t_{2} > t_{1}$

$v_{2} = v_{1}; t_{2} = t_{1}$

answer is B.

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Detailed Solution

L: height of the sphere, $θ$ : angle of inclination

time of descent:

$t = \frac{1}{\sin θ} \sqrt{\frac{2 L (1 + \frac{K^{2}}{R^{2}})}{g}}$

$\Rightarrow θ_{1} > θ_{2} \Rightarrow t_{1} < t_{2}$

velocity at bottom:

$v = 0 + a t = \frac{1}{\sin θ} \sqrt{\frac{2 L (1 + \frac{K^{2}}{R^{2}})}{g}} \times \frac{gsin θ}{(1 + \frac{K^{2}}{R^{2}})}$

$\Rightarrow v = \sqrt{\frac{2 L g}{(1 + \frac{K^{2}}{R^{2}})}}$

Velocity is independent of of angle of inclination.

$\Rightarrow θ_{1} > θ_{2} \Rightarrow v_{1} = v_{2}$

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