A large container (with open top) of negligible mass and uniform cross-sectional area A has a small hole of cross-sectional area $a$ in its side wall near the bottom. The container is keep on a smooth horizontal platform and contains a liquid of density $\rho$ and mass $m$. If the liquid starts flowing out of the hole at time $t=0$, the initial acceleration of the container is 

Let h be the initial height of the liquid of density $\rho$ in the container of cross-sectional area A. The mass of the liquid in the container initially is (Fig. 7.37) 

                                                                                                             $m=Ah\rho$

From Torricelli's theorem, the velocity of the liquid flowing out of the hole is 

                                         $v=\sqrt{2gh}$

$\therefore$ Volume of liquid flowing out per unit time = av. 

Hence the mass of liquid flowing out per unit time = $\rho$av. Therefore, the momentum carried per unit time by the liquid flowing out is = (mass per unit time) x velocity = ($\rho$av)v = $\rho$av² .

This is the rate of change of momentum of the liquid flowing out which is the force with which the liquid. flows out at t = 0. 

$$\begin{gathered} \therefore Initial acceleration=\frac{force}{mass}=\frac{\rho a{v}^2}{Ah\rho }=\frac{a{v}^2}{Ah} \\ =\frac{a\times 2gh}{Ah} \left(\because v=\sqrt{2gh}\right) \\ =\frac{2ga}{A}. \end{gathered}$$