A large open top container of negligible mass: Fluid Mechanics

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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}

A large open top container of negligible mass

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}

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