There are two identical small holes of area of cross section a on the either sides of a tank containing a liquid of density p (shown in figure). 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 :

\frac{2 g h}{p a}

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\frac{p g h}{a}

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g h p a

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2 p a g h

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Solution

The correct option is C $2 p a g h$
Net force (reaction)
$F = F_{B} - F_{A} = \frac{d p_{B}}{d t} - \frac{d p_{A}}{d t}$
$= a v_{B} p \times v_{B} - a v_{A} p \times v_{A}$
$∴ F = a p (v_{B}^{2} - v_{A}^{2})$
According to Bernaulli's theorem
$P_{A} + \frac{1}{2} p v_{A}^{2} + p g h = P_{B} + \frac{1}{2} p v_{B}^{2} + 0$
$\Rightarrow \frac{1}{2} p (v_{B}^{2} - v_{A}^{2}) = p g h$
$\Rightarrow v_{B}^{2} - v_{A}^{2} = 2 g h$
From Eqs (i) and (ii), we get
$F = a p (2 g h) = 2 a p g h$

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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. Tank is resting on a smooth horizontal surface. Horizontal force which has to be applied on the tank to keep it in equilibrium is