Two bodies, a ring and a solid cylinder of same material are rolling down without slipping on inclined plane. The radii of the bodies are same. The ratio of velocity of the centre of mass at the bottom of the inclined plane of the ring to that of the cylinder is $\frac{\sqrt{x}}{2}$ . then, the value of x is---------.

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answer is 3.

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Detailed Solution

Velocity at the lowest point of inclined plane

$V = \sqrt{\frac{2 g h}{1 + \frac{K^{2}}{R^{2}}}} {(\frac{K^{2}}{R^{2}})}_{R i n g} = 1 & {(\frac{K^{2}}{R^{2}})}_{S o l i d c y l i n d e r} = \frac{1}{2} \frac{V_{R}}{V_{C} = \sqrt{\frac{2 g h}{1 + 1}}} / \sqrt{\frac{2 g h}{1 + \frac{1}{2}}} = \sqrt{\frac{3}{2} \times \frac{1}{2}} = \frac{\sqrt{3}}{2} ∴ x = 3$

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