Explanation

(c) When there is equal level of liquid in two arms of U-tube, then height of liquid in each arm of U-tube$= \frac{h_{1} + h_{2}}{2}$. We may consider that a length $h_{1} - \frac{(h_{1} + h_{2})}{2} = \frac{h_{1} - h_{2}}{2}$ of the liquid has been transferred from left arm to right arm of U-tube. The mass of the liquid transferred from left arm to right arm of U-tube$= \left( \frac{h_{1} - h_{2}}{2} \right)A\ \rho$, where $A =$area of cross-section of tube and $\rho =$ density of liquid.
The decrease in height of this liquid $= \left( \frac{n_{1} - n_{2}}{2} \right)$
Loss in potential energy of this liquid
$= \left( \frac{h_{1} - h_{2}}{2} \right)^{2}A\ \rho g$
The mass of the entire liquid in U-tube
$= \left( h_{1} + h_{2} + h \right)\rho A$
If this liquid moves with velocity $v$, then its
$KE = \frac{1}{2}\left( h_{1} + h_{2} + h \right)\rho A\ v^{2}$
Using law of conservation of energy, we gave
$\frac{1}{2}\left( h_{1} + h_{2} + h \right)\rho A\ v^{2} = \left( \frac{h_{1} - h_{2}}{2} \right)^{2}A\ \rho g$
Or $v = \left( h_{1} - h_{2} \right)\sqrt{\frac{g}{2(h_{1} + h_{2} + h)}}$