Question

A hollow vertical cylinder of radius $R$ and height $h$ has a 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 $v_{0}$ tangential to the rim. It leaves the lower rim at point $Q$ , vertically below $P$ . The number of revolutions made by the particle will be

• A. $\frac{h}{2 \pi R}$
• B. $\frac{v_{0}}{\sqrt{2 g h}}$
• C. $\frac{2 \pi R}{h}$
• D. $\frac{v_{0}}{2 \pi R}\left(\sqrt{\frac{2 h}{g}}\right)$

Answer 1

Answer: (D)

Time to reach the bottom,
$\because t=\sqrt{\frac{2 h}{g}}$
$\therefore \frac{2 \pi R}{v_{0}}\times n=\sqrt{\frac{2 h}{g}}$
$⇒ n=\frac{v_{0}}{2 \pi R}\sqrt{\frac{2 h}{g}}$