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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. The tank is resting on a smooth horizontal surface. The horizontal force which will have to be applied on the tank to kept it in equilibrium is

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$ghρa$

$2 ghρa$

$\frac{ρgh}{a}$

$\frac{2 gh}{ρa}$

answer is C.

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Detailed Solution

Thrust force, $F = F_{1} - F_{2} = {ρav}_{1}^{2} - {ρav}_{2}^{2}$
$\begin{array}{l} = ρa (2 {gh}_{1}) - ρa (2 {gh}_{2}) \\ = 2 ρag (h_{1} - h_{2}) = 2 ρagh \end{array}$

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