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

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 <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-l7pi0qeobnjn.png" /> (A) ghρ a (B) (2 g h/ρ a) (C) 2ρ agh (D) (ρ g h/a)

Tardigrade Admin · Accepted Answer

Thrust force <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-l7pi0qeotdcx.png" /> textF = textF1 - textF2 = ρ textav12 - ρ textav22 =aρ (2 ( textgh)1)-aρ (2 ( textgh)2) =2ρ a textg(( texth)1 - ( texth)2) =2ρ a textgh

$g h ρ a$

$\frac{2 g h}{ρ a}$

$2 ρ a g h$

$\frac{ρ g h}{a}$

Thrust force

$F = F_{1} - F_{2} = ρ av_{1}^{2} - ρ av_{2}^{2}$
$= a ρ (2 (gh)_{1}) - a ρ (2 (gh)_{2})$
$= 2 ρ a g ((h)_{1} - (h)_{2})$
$= 2 ρ a gh$

ghρa

ρa2gh​

2ρagh

aρgh​

Thrust force F=F1​−F2​=ρav12​−ρav22​ =aρ(2(gh)1​)−aρ(2(gh)2​) =2ρag((h)1​−(h)2​) =2ρagh

$g h ρ a$

$\frac{2 g h}{ρ a}$

$2 ρ a g h$

$\frac{ρ g h}{a}$

Thrust force

$F = F_{1} - F_{2} = ρ av_{1}^{2} - ρ av_{2}^{2}$
$= a ρ (2 (gh)_{1}) - a ρ (2 (gh)_{2})$
$= 2 ρ a g ((h)_{1} - (h)_{2})$
$= 2 ρ a gh$