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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$2pagh$

$ghpa$

$\frac{2 gh}{ρ a}$

$\frac{ρ g h}{a}$

answer is C.

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Detailed Solution

Let the holes be at a height of h₁ and h₂ and speed of liquid coming out be v₁ and v₂

Thrust force,

$F = F_{1} - F_{2} = {ρav}_{1}^{2} - {ρav}_{2}^{2}$

$= ρa (2 {gh}_{1}) - ρa (2 {gh}_{2})$

$= 2 ρag (h_{1} - h_{2}) = 2 ρagh$

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