Question

There are two identical small holes of 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. Horizontal force which has to be applied on the tank to keep it in equilibrium is $xa\rho gh$. Find $x$.

Answer 1

Net force (reaction),
$F =F_{B}-F_{A}$
$=\frac{d p_{B}}{d t}-\frac{d p_{A}}{d t}$
$=a v_{B} \rho \times v_{B}-a v_{A} \rho \times v_{A}$
$=a \rho\left(v_{B}^{2}-v_{A}^{2}\right)$
$=a \rho\left[(\sqrt{2 g(x+h)})^{2}-(\sqrt{2 g x})^{2}\right]$
$= a \rho(2 g x +2 gh -2 gx )$
$=2 a \rho gh$
$\Rightarrow x =2$