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 $\rho $ . 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

• 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

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