Laws of MotionHard
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 will has to be applied on the tank to keep it in equilibrium is :-
Options
A.ghρa
B.
C.2ρagh
D.
Solution
Net force = F = FB - FA =
= avBρ × vB - avAρ × vA
F = aρ(vB2 - VA2) ....(1)
According to Bernoulli′s theorem
ρA +
ρvA2 + ρgh ⇒ ρB + 
ρvB2 + 0⇒
ρ(vB2 - VA2) = ρghvB2 - vA2 = 2gh ....(2)
from equation (1) and (2) we get F = 2aρgh
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