Momentum and CollisionVery Hard

Question

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 :-
         

Options

A.ρgh A
B.2gh/ρA
C.2ρgh A
D.ρgh/ A

Solution

         
F = FQ - FP =
= AρuQ × uQ - Aρup × up
= Aρ(uQ2 - up2)       ......(1)
PP + ρup2 + ρgh = PPQ + ρuQ2 + 0(∵ PP = PQ)
oe ρ(uQ2 - up2) = ρgh
∴ F = 2Aρgh     (from eqn. (i))

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