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 equilbrium is :

g h ρ a

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\frac{2 g h}{ρ a}

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2 ρ a g h

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\frac{ρ g h}{a}

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Solution

The correct option is D $2 ρ a g h$
Momentum of the mass of water coming out of the hole per sec is thrust.
Resultant thrust $F = F_{1} - F_{2} = - ρ a v_{1}^{2} - ρ a v_{2}^{2} = - ρ a \sqrt{2 g h_{1}} - ρ a \sqrt{2 g h_{2}} = - 2 a ρ g (h_{1} - h_{2})$
$∴$ horizontal force that has to be applied to the tank to keep it in equilibrium is
$2 a ρ g (h_{1} - h_{2})$
or
$2 a ρ g h$

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The correct option is D 2ρaghMomentum of the mass of water coming out of the hole per sec is thrust.Resultant thrust F=F1−F2=−ρav21−ρav22=−ρa√2gh1−ρa√2gh2=−2aρg(h1−h2)∴ horizontal force that has to be applied to the tank to keep it in equilibrium is2aρg(h1−h2)or2aρgh