Question

The $U$ - tube shown has a uniform cross-section. A liquid is filled in the two arms up to heights $h_1$ and $h_2$ , and then the liquid is allowed to move. Neglect viscosity and surface tension. When the levels equalize in the two arms, the liquid will

• A. Be at rest
• B. Be moving with an acceleration of g $g\left(\frac{h_1-h_2}{h_1+h_2+h}\right)$
• C. Be moving with a velocity of $\left(h_1-h_2\right)\sqrt{\frac{g}{2\left(h_1+h_2+h\right)}}$
• D. Exert a net force to the right on the tube.

Answer 1

Answer: (C)

When the levels equalize, the height of the liquid in each arm $=\frac{{\text{h}}_1+{\text{h}}_2}{2}$
We may then visualize that a length ${\text{h}}_1-\frac{{\text{h}}_1+{\text{h}}_2}{2}=\frac{{\text{h}}_1-{\text{h}}_2}{2}$ of the liquid has been transferred from the left arm to the right arm.
Then, the mass of this liquid $=\left(\frac{h_1-h_2}{2}\right)A\rho$
where, $A$ = area of tube, $\rho$ = density of the liquid
Distance through which it moves down $=\frac{{\text{h}}_1-{\text{h}}_2}{2}$
Then, the mass of this liquid $=\left(\frac{h_1-h_2}{2}\right)A\rho$
where, $A=$ area of tube, $\rho =$ density of the liquid
Distance through which it moves down $=\frac{h_1-h_2}{2}$
$\therefore$ Loss in the gravitational potential energy $={\left(\frac{h_1-h_2}{2}\right)}^{2}A\rho g$
The mass of the entire liquid $=\left(h_1+h_2+h\right)A\rho$
If this moves with a velocity $v$, its kinetic energy $=\frac{1}{2}\left(h_1+h_2+h\right)A\rho v^2$
Equating energies, we get $v=\left(h_1-h_2\right)\sqrt{\frac{g}{2\left(h_1+h_2+h\right)}}$