Question

A uniform flexible chain of mass $m$ and length $2l$ 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. $\sqrt{gl}$
• B. $\sqrt{\frac{gl}{2}}$
• C. $\sqrt{2gl}$
• D. $\sqrt{3gl}$

Answer 1

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


Initially centre of mass of chain was at distance $2l/4$ below the pin and in final position it is at distance $l/2$ below the pin.

Here, non- conservative force and external force are absent , so we can apply conservation of mechanical energy.

$\therefore U_1+K_1=U_2+K_2$

Taking pin as a reference point,

$-\frac{mgl}{2}+0=-mgl+\frac{1}{2}m{v}^2$

$\Rightarrow mg\frac{l}{2}=\frac{1}{2}m{v}^2$

$\therefore v=\sqrt{gl}$

Hence, option $\left(a\right)$ is correct.

Why this question ? Key concept : $W_{con}+W_{non-con}+W_{ext}=\left(\Delta K\right)$ $-\Delta U+W_{non-con}+W_{ext}=\left(\Delta K\right)$ $W_{non-con}+W_{ext}=\Delta (U+K)$ If work done by non-conservative and external force acting on the system is zero, then mechanical energy of the system is conserved. $K_i+U_i=K_f+U_f$