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

Depth of spout pipe, $h=0.45m$
$\therefore $ Velocity of efflux through the spout pipe,
$v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 0 . 45}=3ms^{- 1}$
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 $v=3ms^{- 1}$
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,
$10 \times h ^{\prime}+\frac{1}{2} \times\left(0=10 \times 0+\frac{1}{2} \times(3\right.$
$\therefore h^{'}=\frac{1}{2}\times \frac{9}{10}=0.45m=\alpha 100\alpha =45$