# A large open top container of negligible mass: Fluid Mechanics

Problem: A large open top container of negligible mass and uniform cross-sectional area A has a small hole of cross-sectional area A/100 in its side wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density ρ and mass m0. Assuming that the liquid starts flowing out horizontally through the hole at t = 0. Calculate,  (IIT JEE 1997)

1. the acceleration of the container.
2. velocity of efflux when 75% of the liquid has drained out.

Answer: The answer is (a) g/50 (b) $\sqrt{gm_0/2A\rho}$.

Solution: The height of the container is given by \begin{align} H&=\frac{V}{A}=\frac{m_0/\rho}{A}=\frac{m_0}{\rho A}. \end{align} The velocity of efflux when container is completely filled with the liquid is \begin{align} v&=\sqrt{2gH}=\sqrt{{2m_0 g}/{\rho A}}.\nonumber \end{align} The mass of the liquid coming out of the hole in one second and the momentum carried by this liquid in one second are \begin{align} &\Delta m=\rho v ({A}/{100})={\rho v A}/{100},\nonumber\\ &\Delta p=\Delta m\, v={\rho A v^2}/{100}.\nonumber \end{align} The rate of change of momentum of the container is equal to the force acting on the container i.e., \begin{align} F=\frac{\Delta p}{\Delta t}=\frac{\Delta p}{1}=\frac{\rho A v^2}{100}.\nonumber \end{align} Newton's second law gives acceleration of the container as \begin{align} a&=\frac{F}{m_0} \\ &=\frac{1}{m_0}\frac{\rho A v^2}{100} \\ &=\frac{1}{m_0}\frac{\rho A}{100}\frac{2m_0 g}{\rho A}\\ &=\frac{g}{50}. \nonumber \end{align} The height of the liquid, when 75% of the liquid is drained out is, $H^\prime=H/4$ and the velocity of efflux at that instant is \begin{align} v^\prime&=\sqrt{2gH^\prime} \\ &=\sqrt{gH/2}\\ &=\sqrt{{m_0g}/{(2\rho A)}}.\nonumber \end{align}