There are two identical small holes of an 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 o

Question

There are two identical small holes of an 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. The horizontal force which will have to be applied on the tank to keep it in equilibrium is

Options:

Answer:

C

Solution:

Thrust force

$F = F_{1} - F_{2} = ρ {av}_{1}^{2} - ρ {av}_{2}^{2}$

$= a ρ (2 {gh}_{1}) - a ρ (2 {gh}_{2})$

$= 2 ρ a g (h_{1} - h_{2})$

$= 2 ρ a gh$

TestHub · Accepted Answer

Thrust force F = F 1 - F 2 = ρ av 1 2 - ρ av 2 2 = a ρ 2 gh 1 - a ρ 2 gh 2 = 2 ρ a g h 1 - h 2 = 2 ρ a gh

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