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Hollow vertical cylinder with 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. Itt is given a horizontal speed v₀ tangential to rim. It leaves the lower rim at point Q, vertically below P. Find the, number of revolutions made by the particle.

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$\frac{h}{2 π R}$

$\frac{v_{0}}{\sqrt{2 gh}}$

$\frac{2 π R}{h}$

$\frac{v_{0}}{2 πR} (\sqrt{\frac{2 h}{g}})$

answer is D.

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Detailed Solution

$∵ t = \sqrt{\frac{2 h}{g}}$

$∴ \frac{2 π R}{v_{0}} \times n = \sqrt{\frac{2 h}{g}}$

$\Rightarrow n = \frac{v_{0}}{2 π R} \sqrt{\frac{2 h}{g}}$

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