Three bodies a ring $(R)$ , a solid cylinder $(C)$ and a solid sphere $(S)$ having same mass and same radius roll down the inclined plane without slipping. They start from rest, if $v_{R}$ , $v_{C}$ and $v_{S}$ are velocities of respective bodies on reaching the bottom of the plane, then:

v_{R} = v_{C} = v_{S}

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v_{R} > v_{C} > v_{S}

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v_{R} < v_{C} < v_{S}

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v_{R} = v_{C} > v_{S}

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Solution

The correct option is D $v_{R} < v_{C} < v_{S}$
Let the body starts from rest at a height $h$ to roll down inclined plane.
Using work-energy equation : $W = Δ K . E = K . E_{f}$ $(∵ K . E_{i} = 0)$
$∴$ $m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I w^{2}$
But $v = r w$ (pure rolling) and $I = k m r^{2}$
$∴$ We get $m g h = \frac{1}{2} m v^{2} + \frac{1}{2} k m v^{2}$
Or $m g h = \frac{1}{2} m v^{2} [1 + k]$
$⟹$ $v = \sqrt{\frac{2 g h}{1 + k}}$
As we know, $k_{R} = 1$ $k_{C} = 0.5$ $k_{S} = 0.4$
$∴$ $k_{S} < k_{C} < k_{R}$
$⟹ v_{R} < v_{C} < v_{S}$

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