A small disc is released from rest at A on an inclined plane AB so that it rolls down without slipping. It reaches the bottom with linear velocity $v_{1}$ in time $t_{1}$ . Next a small ring is released from rest on the inclined plane AC so that it rolls down without slipping. It reaches the bottom with linear velocity $v_{2}$ in time $t_{2}$ . Given $θ_{1} = 30^{0}$ , $θ_{2} = 60^{0}$ and $h = 10 m$ . Then,

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

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

$v_{1} > v_{2}, t_{1} > t_{2}$

$v_{1} < v_{2}, t_{1} < t_{2}$

answer is C.

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Detailed Solution

$a = \frac{g \sin θ}{1 + \frac{c m}{M R^{2}}}$
$a_{1} = \frac{2 g \sin θ_{1}}{3}$
$v_{1} = \sqrt{2 a_{1} S_{1}} = \sqrt{2 \times \frac{2 g \sin θ_{1}}{3} \frac{h}{\sin θ_{1}}} = \sqrt{\frac{4 g h}{3}}$
$t_{1} = \frac{v_{1}}{a_{1}} = \sqrt{\frac{12 h}{g}}$
$a_{2} = \frac{g \sin θ_{2}}{2}$
$v_{2} = \sqrt{2 a_{2} S_{2}} = \sqrt{2 \frac{g \sin θ_{2}}{2} \times \frac{h}{\sin θ_{2}}} = \sqrt{g h}$
$t_{2} = \frac{v_{2}}{a_{2}} = \sqrt{\frac{16 h}{3 g}}$
$v_{1} > v_{2}$ and $t_{1} > t_{2}$