Question

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 $\rho$ and mass $M_0$. Assuming that the liquid starts flowing out horizontally through the hole at $t=0$, calculate The acceleration of the container and its velocity when $75\%$ of the liquid has drained out is given as $\sqrt{\frac{M_{0}g}{xA\rho }}$. Find $x$:

Answer 1

Liquid is escaping from the container. Let y be the height of the liquid at any instant t after start.

Then velocity of escape is given by

$v=\sqrt{2gy}$

Mass of the liquid flowing in time $dt=av\rho dt$, where $a=area$ of the hole $=A\cdot 100$.

Force is the rate of change of momentum. Hence,

$F=\frac{dP}{dt}=\left(\frac{av\rho dt}{dt}\right)v$

$=a{v}^2\rho =a\left(2gy\right)\rho$

As $F=$ mass of container $\times$ acceleration $=M{a}^{\prime }$

$\therefore a^{\prime }=\frac{F}{M}$

But $M=\text{mass of container}=Ay\rho$

$\therefore a^{\prime }=\frac{a\left(2gy\right)\rho }{M}=\frac{\left(\frac{A}{100}\right)2gy\rho }{Ay\rho }=\frac{g}{50}$

When 75% of the liquid has drained out, the height of the liquid y is given by

$Ay\rho =\frac{25}{100}{M}_0\Rightarrow y=\frac{M_0}{4A\rho }$

But $v=\sqrt{2gy}=\sqrt{2g\left(\frac{M_0}{4A\rho }\right)}=\sqrt{\frac{M_{0}g}{2A\rho }}$