Question

A liquid of density $\rho $ is filled in a U-tube is accelerated with an acceleration $a$ so that the height of liquid in its two vertical arms are $h_{1}$ and $ \, h_{2}$ as shown in the figure. If $l$ is the length of horizontal arm of the tube, the acceleration $a$ is

• A. $\frac{g \left(\right. h_{1} - h_{2} \left.\right)}{2 l}$ towards right
• B. $\frac{g \left(\right. h_{1} - h_{2} \left.\right)}{2 l}$ towards left
• C. $\frac{g \left(\right. h_{1} - h_{2} \left.\right)}{l}$ towards right
• D. $\frac{g \left(\right. h_{1} - h_{2} \left.\right)}{l}$ towards left

Answer 1

Answer: (C)

Pressure on left end of horizontal tube,
$p_{1}=p_{0}+h_{1}\rho g$
Pressure on right end of horizontal tube,
$p_{2}=p_{0}+h_{2}\rho \, g$
As $p_{1}>p_{2}$ , so acceleration should be towards right hand side. If $A$ is the area of cross-section of the tube in the horizontal portion of U-tube, then
$p_{1} \, A-p_{2} \, A=\left(\right.l \, A \, \rho \left.\right)a$
Or $\left(h_{1} - h_{2}\right)\rho \, g \, A=l \, A \, \rho \, a$
$a = g \frac{\left(\right. h_{1} - h_{2} \left.\right)}{l}$