A uniform chain of length l is placed on a smooth horizontal table, such that half of its length hangs over one edge. It is released from rest, the velocity with which it leaves the table is

Consider the diagram. Let the mass of chain be m.
$\therefore$ Mass per unit length = $\frac{m}{l}$
Consider an elementary length of the chain at a depth x below the table. The potential energy of their parts is given by

$dU=-\frac{m}{l}gxdx$

Now, potential energy of hanging part,

$U_1=\int dU=-{\int }_0^{\frac{l}{2}}\frac{m}{l}gxdx=-\frac{m}{l}g{\left[\frac{x^2}{2}\right]}_0^{\frac{1}{2}}=-\frac{mgl}{8}$

When chain is completely leaves the table, then potential energy,

$U_2=-{\int }_0^l\frac{m}{l}gxdx=-\frac{mg}{2l}{\left[x^2\right]}_0^l=\frac{-mgl}{2}$

Now, loss in potential energy   $=\frac{-mgl}{8}+\frac{mgl}{2}$

From law of conservation of energy,

Loss in potential energy = Gain in kinetic energy

$\Rightarrow \frac{1}{2}m{v}^2=\frac{3mgl}{8}\Rightarrow v=\sqrt{\frac{3gl}{4}}$