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 $=\frac{ h _{1}- h _{2}}{2}$
$\therefore \quad$ 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 $v=\left(h_{1}-h_{2}\right) \sqrt{\frac{ g }{2\left( h _{1}+ h _{2}+ h \right)}}$