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A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of $\frac{3 v^{2}}{4 g}$ with respect to the initial position. The object is

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ring

solid sphere

hollow sphere

disc

answer is D.

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Detailed Solution

.mgh=rotational KE

$mgh = \frac{1}{2} {mv}^{2} (1 + \frac{K^{2}}{R^{2}})$

$m g \frac{3 v^{2}}{4 g} = \frac{1}{2} {mv}^{2} (1 + \frac{K^{2}}{R^{2}}) solving K^{2} = \frac{R^{2}}{2} so it is a disc$

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