First slide

Fluid dynamics

Question

There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density $ρ$ . The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is

Difficult

A

$ghρa$
B

$\frac{2 g h}{ρ a}$
C

$2 ρ a g h$
D

$\frac{ρ g h}{a}$

Solution

$Net force (reaction) = F = F_{B} - F_{A} = \frac{d p_{B}}{d t} - \frac{d p_{A}}{d t}$
$= a v_{B} ρ \times v_{B} - a v_{A} ρ \times v_{A}$
$∴ F = a ρ (v_{B}^{2} - v_{A}^{2}) . . . . . . (i)$
According to Bernoulli's theorem
$p_{A} + \frac{1}{2} ρ v_{A}^{2} + ρ g h = p_{B} + \frac{1}{2} ρ v_{B}^{2} + 0$
$\Rightarrow \frac{1}{2} ρ (v_{B}^{2} - v_{A}^{2}) = ρ g h \Rightarrow v_{B}^{2} - v_{A}^{2} = 2 g h$
$From equation (i), F = 2 a ρ g h$

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