# 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 m_{0}. Assuming that the liquid starts flowing out horizontally through the hole at t = 0. Calculate, (IIT JEE 1997)

- the acceleration of the container.
- 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}

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