Question

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

• A. $\sqrt{\frac{gl}{2}}$
• B. $\sqrt{gl}$
• C. $\sqrt{2gl}$
• D. $\sqrt{3gl}$

Answer 1

The correct option is A $\sqrt{\frac{gl}{2}}$

Decrease in gravitational potential energy= increase in kinetic energy
Initially centre of mass of chain was at distance $\frac{l}{4}$ below the pin and in final position it is at distance $\frac{l}{2}$ below the pin.
Hence centre of mass has descended $\frac{l}{4}$.

$\therefore mg\frac{l}{4}=\frac{1}{2}m{v}^2$
or $v=\sqrt{\frac{gl}{2}}$