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 [image:10_image34.emf]

Question

Dr. Abubacker · Accepted 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
$p_{1}\ A - p_{2}\ A = (l\ A\ \rho)a$
Or $\left( h_{1} - h_{2} \right)\rho\ g\ A = l\ A\ \rho\ a = \frac{g(h_{1} - h_{2})}{l}$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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