There are two identical small holes of the 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 . The 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 the 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 . The 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-wqozvoai4rlk.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-wqozvoaiiwhm.png" /> F = F 1- F 2=ρ av 12-ρ av 22 =ρ a (2 gh 1)-ρ a (2 gh 2) =2 operatornamepag( h 1- h 2) =2 pagh

$g h ρ a$

$\frac{2 g h}{ρ a}$

$2 ρ a g h$

$\frac{ρ g h}{a}$

Thrust force

$F = F_{1} - F_{2} = ρ a v_{1}^{2} - ρ a v_{2}^{2}$
$= ρ a (2 g h_{1}) - ρ a (2 g h_{2})$
$= 2 pag (h_{1} - h_{2})$
$= 2$ pagh

ghρa

ρa2gh​

2ρagh

aρgh​

Thrust force F=F1​−F2​=ρav12​−ρav22​ =ρa(2gh1​)−ρa(2gh2​) =2pag(h1​−h2​) =2 pagh

$g h ρ a$

$\frac{2 g h}{ρ a}$

$2 ρ a g h$

$\frac{ρ g h}{a}$

Thrust force

$F = F_{1} - F_{2} = ρ a v_{1}^{2} - ρ a v_{2}^{2}$
$= ρ a (2 g h_{1}) - ρ a (2 g h_{2})$
$= 2 pag (h_{1} - h_{2})$
$= 2$ pagh