The U-tube shown in the figure has a uniform cross section. A liquid is filled up to heights $h_1$ and $h_2$ in the two arms and then they are allowed to move. Neglecting viscosity and surface tension, find the state of the liquid when the levels equalize in the two arms. Find the correct statement.

When the levels equalize, the height of liquid in each arm is $\left({\mathrm{h}}_1+{\mathrm{h}}_2\right)/2$

We may then visualize that a length

${\mathrm{h}}_1-\left({\mathrm{h}}_1+{\mathrm{h}}_2\right)/2=\left({\mathrm{h}}_1-{\mathrm{h}}_2\right)/2$ of the liquid has been transferred from the left arm to the right arm.

Mass of liquid transferred $=\left({\mathrm{h}}_1-{\mathrm{h}}_2\right)\mathrm{A}\rho /2$

where A is area of hole and $\rho$is density.
Vertical distance through which it moves = $\left({\mathrm{h}}_1-{\mathrm{h}}_2\right)/2$

Loss in gravitational potential energy = ${\left[\left({\mathrm{h}}_1-{\mathrm{h}}_2\right)/2\right]}^2\mathrm{A}\rho \mathrm{g}$

Mass of the entire liquid $=\left(h_1+h_2+h\right)A\rho$

If this moves with velocity v

Kinetic energy $=\frac{1}{2}\left(h_1+h_2+h\right)A\rho v^2$

Equating energies, we get

$v=\left(h_1-h_2\right)\sqrt{\frac{g}{2\left(h_1+h_2+h\right)}}$