A body of mass $m$ slides down an inclined plane and reaches the bottom with a velocity $v$ . If the same mass were in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been

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$v \sqrt{2}$

$\frac{v}{\sqrt{2}}$

$(\sqrt{\frac{2}{5}}) v$

$v$

answer is C.

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Detailed Solution

$v \propto \sqrt{a} \frac{v}{v_{R}} = \sqrt{\frac{a}{a_{R}}} & a_{R} = \frac{g \sin θ}{1 + \frac{I}{m R^{2}}} = \frac{g \sin θ}{2} a = g \sin θ \Rightarrow v_{R} = \frac{v}{\sqrt{2}}$

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