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

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$g h ρ a$

$\frac{2 g h}{ρ a}$

$2 ρ a g h$

$\frac{ρ g h}{a}$

answer is C.

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Detailed Solution

Net force (reaction) = $F = F_{B} - F_{A} = \frac{{dp}_{B}}{dt} - \frac{{dp}_{A}}{dt}$
$= {av}_{B} ρ \times v_{B} - {av}_{A} ρ \times v_{A}$
$∴$ $F = aρ (v_{B}^{2} - v_{A}^{2})$ …(i)
According to Bernoulli's theorem
$\begin{array}{l} p_{A} + \frac{1}{2} {ρv}_{A}^{2} + ρgh = p_{B} + \frac{1}{2} {ρv}_{B}^{2} + 0 \\ \Rightarrow \frac{1}{2} ρ (v_{B}^{2} - v_{A}^{2}) = ρgh \Rightarrow v_{B}^{2} - v_{A}^{2} = 2 gh \end{array}$
From equation (i), $F = 2 aρgh .$