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

$g h ρ a$

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$\frac{2 g h}{ρ a}$

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$2 ρ a g h$

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$\frac{ρ g h}{a}$

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Solution

The correct option is C. $2 ρ a g h$ .

Momentum of the mass of water coming out of the hole per sec is thrust.
Net thrust $= F = F_{B} - F_{A} = \frac{d p_{B}}{d t} - \frac{d p_{A}}{d t}$

$= a v_{B} ρ \times v_{B} - a v_{A} ρ \times v_{A}$

$F = a ρ (v_{B}^{2} - v_{A}^{2}$ )
$F = a ρ ((\sqrt{2 g h_{B}})^{2} - (\sqrt{2 g h_{A}})^{2})$
$F = 2 a ρ g (h_{B} - h_{A}) = 2 a ρ g h$
∴ The horizontal force that has to be applied to the tank to keep it in equilibrium is $F = 2 a ρ g h$ .

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