A uniform flexible chain of mass m and length l hangs in equilibrium over a smooth horizontal pin of neglible diameter. One end of the chain is given a small verticle displacement so that the chain slips over the pin. The speed of chain when it leaves pin is

Options

A. `sqrt((gl)/(2))`
B. `sqrt(gl)`
C. `sqrt(2gl)`
D. `sqrt(3gl)`

Answer

A

Explanation

(a) Decrease in gravitational potential energy
=increase in kinetic energy.
Initailly centre of mass of chain was at distance `(l)/(4)` below the pin and in final position, it is at distance `(l)/(2)` below the pin. Hence, centre of mass has descended `(l)/(4)`.

`:." " "mg"(l)/(4)=(1)/(2)mv^(2)" or " v=sqrt((gl)/(2))`