There are two identical small holes, each 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. Then the horizontal force which will have to be applied on the tank to keep it in equilibrium

2 ρ a g h

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5 ρ a g h

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1 ρ a g h

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3 ρ a g h

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Solution

The correct option is A $2 ρ a g h$

Net force by fluid is, F= $F_{B} - F_{A}$
$= \frac{d p_{B}}{d t} - \frac{d p_{A}}{d t} = (a v_{B} ρ \times v_{B}) - (a v_{A} ρ \times v_{A})$
$F = a ρ (v_{B}^{2} - v_{A}^{2})$ ------ (1)
$A c c o r d i n g t o B e r n o u l l i e^{'} s t h e o r e m P_{A} + \frac{1}{2} ρ v_{A}^{2} + ρ g h = P_{B} + \frac{1}{2} ρ v_{B}^{2} + 0$
$ρ g h (v_{B}^{2} - v_{A}^{2}) = 2 g h$
From equation (1), F= $2 a ρ g h$ .

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The correct option is A 2ρaghNet force by fluid is, F= FB−FA=dpBdt−dpAdt=(avBρ×vB)−(avAρ×vA)F=aρ(v2B−v2A)------ (1)AccordingtoBernoullie′stheoremPA+12ρv2A+ρgh=PB+12ρv2B+0ρgh(v2B−v2A)=2ghFrom equation (1), F= 2aρgh.