Water is flowing through a cylinderical pipe, of internal diameter 2 cm, into a cylinderical tank of base radius 40 cm, at the rate of 0.4 m/s. Determine the rise in level of water in the tank in half an hour.
Given: A cylinderical pipe of internal diameter=2cm and a cylinderical tank of base with radius =40cm. Rate of the flow of the water flowing in the pipe=0.4\ m/s$
To do: To determine the rise in level of water in the tank in half an hour.
Solution:
Diameter of circular end of pipe $=2\ cm$
Radius of circular end of pipe $r_{1} =\frac{diameter}{2} =\frac{2}{2} =1\ cm=\frac{1}{100} =0.01\ m$
Area of cross section$=\pi r^{2}_{1}$
$=\pi ( 0.01)^{2}$
$=0.0001\pi\ m^{2}$
Speed of water $=0.4\ m/s=0.4\times 60=24\ meter/min$
Volume of water that flows in 1 minute from pipe$=24\times 0.0001\pi$
$=0.0024\pi m^{3}$
Volume of water that flows in 30 minutes from pipe$=30\times 0.00024\pi$
$=0.072\pi m^{3}$
Radius $r_{2}$ of the base of cylindrical tank$=40\ cm=0.4\ m$
Let the cylindrical tank be filled up to h meter in 30 minutes.
Volume of water filled in tank in 30 minutes is equal to the volume of water flowed out
In 30 minutes from the pipe,
$\pi r^{2}_{2} h=0.072\pi$
$\Rightarrow h=\frac{0.072}{0.4^{2}}$
$\Rightarrow h=0.45\ m=45\ cm$
Therefore, The rise in level of water in the tank in half an hour is 45 cm.
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