Question

A spout pipe of diameter $0.5cm$ is connected horizontally at the bottom of a cylindrical vessel of diameter $15cm$ as shown in figure below.

When water is poured into the vessel, it leaves the spout in the form of a fountain. Find the height (in $m$ ) to which the vertical stream of water goes, if the water level in the vessel is maintained at a constant height of $0.45m$. (Take $g=10m/s^2$ )

Answer 1

Answer: 0.45

Depth of spout pipe, $h=0.45m$;
$\therefore$ Velocity of efflux through the spout pipe,
$v=\sqrt{2gh}=\sqrt{2\times 10\times 0.45}=3m{s}^{-1}$
According to Bernoulli's theorem,
$\frac{P}{\rho }+gh+\frac{1}{2}{v}^2=$ constant
At the mouth of the spout pipe and the top of fountain, pressure is same (equal to atmospheric pressure).
$\therefore gh+\frac{1}{2}{v}^2=$ constant
At the mouth of the spout pipe, $h=0$ and $v=3m{s}^{-1}$
Also, at the top of fountain, $v=0$
Suppose that stream of water in the form of fountain goes up to a height $h^{\prime }$. Then,
$10\times h^{\prime }+\frac{1}{2}\times (0{)}^2=10\times 0+\frac{1}{2}\times (3{)}^2$
$\therefore h^{\prime }=\frac{1}{2}\times \frac{9}{10}=0.45m$