First slide

Fluid statics

Question

Fig. Shows a U-tube of uniform cross-sectional area A accelerated with acceleration a as shown. If d is the separation between the limbs, then the difference in the levels of the liquid in the U-tube is

Moderate

A

$\frac{ad}{g}$
B

$\frac{g}{ad}$
C

adg
D

ad+g

Solution

Mass of liquid in horizontal portion of U-tube = A $dρ$
Pseudo force on this mass = Ad $ρa$
Force due to pressure difference in the two limbs
$= (h_{1} ρg - h_{2} ρg) A$
Equating, $(h_{1} - h_{2}) ρgA = Adρa$
or $h_{1} - h_{2} = \frac{Adρa}{ρgA} = \frac{ad}{g}$

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