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A hollow vertical cylinder of radius r and height h has smooth internal surface. A small particle is placed in contact with the inner side of the upper rim, at a point P. It is given a horizontal speed u, tangent to
rim. It leaves the lower rim at point Q, vertically below P. The number of revolutions made by the particle will be

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$n = \frac{u}{\sqrt{2 gh}}$

$n = \frac{u}{2 Ïr} \sqrt{(\frac{2 h}{g})}$

$n = \frac{2 Ïr}{h}$

$n = \frac{h}{2 Ïr}$

answer is D.

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Detailed Solution

For vertical motion

$h = 0 + \frac{1}{2} {gt}^{2} (âµ s = ut + \frac{1}{2} {gt}^{2})$

Let n be the number of revolutions made by the particle. For horizontal motion,

$\begin{array}{l} s = (2 Ïr) n = ut \\ or n = \frac{ut}{2 Ïr} = \frac{u}{2 Ïr} \sqrt{(\frac{2 h}{g})} \end{array}$

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