Question

A long cylindrical tank of cross-sectional area $0.5m^2$ is filled with water. It has a small hole at a height $50cm$ from the bottom. A movable piston of cross-sectional area almost equal to $0.5m^2$ is fitted on the top of the tank such that it can slide in the tank freely. A load of $20kg$ is applied on the top of the water by piston, as shown in the figure. Calculate the speed of the water jet with which it hits the surface when piston is $1m$ above the bottom. (Ignore the mass of the piston).

Answer 1

Given

$A=0.5W=20kgH=1mh=50cm$

By the Bernoulli theorem:

$\frac{1}{2}\rho v^2=\rho gh+\frac{mg}{A}$

Here, $h=1.0-0.5=0.5$ m,

A=Area of pistone=0.5 m${}^2$

$\therefore$ $v=\sqrt{2gh+\frac{2mg}{\rho A}}$

$=\sqrt{2\times 9.8\times 0.5+\frac{2\times 20\times 9.8}{{10}^3\times 0.5}}$

$=3.25$ m/s

$\therefore$ Speed with which it hits the surface is

$v^{\prime }=\sqrt{v^2+2g{h}^{\prime }}$

$=\sqrt{{\left(3.25\right)}^2+(2\times 9.8\times 0.5)}$

$=4.51$ m/s