Question

Consider a cylindrical vessel of the diameter of $15cm$ whose bottom is connected horizontally to a spout pipe of diameter $0.5cm$ as shown in the above figure, such that the water in the cylinder leaves the spout in the form of a fountain. If the water level in the vessel is maintained at a constant height of $0.45m$ , then the height (in $m$ ) to which the vertical stream of water goes is $\alpha$ . Write value of $100\alpha$ .
Take $g=10m/s^2$ .

Answer 1

Answer: 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'. Then,
$10\times h^{\prime }+\frac{1}{2}\times (0=10\times 0+\frac{1}{2}\times (3$
$\therefore h^{{}^{\prime }}=\frac{1}{2}\times \frac{9}{10}=0.45m=\alpha 100\alpha =45$