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 $\rho $ . 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 -

• A. $gh\rho a$
• B. $\frac{2 g h}{\rho a}$
• C. $2\rho agh$
• D. $\frac{\rho g h}{a}$

Answer 1

Answer: (C)

Thrust force

$F = F _{1}- F _{2}=\rho av _{1}^{2}-\rho av _{2}^{2}$
$=\rho a \left(2 gh _{1}\right)-\rho a \left(2 gh _{2}\right)$
$=2 \operatorname{pag}\left( h _{1}- h _{2}\right)$
$=2$ pagh