Water is flowing through a cylindrical pipe of internal diameter 2cm into a cylindrical tank of base radius 40cm at the rate of 0.7m/sec .By how much will the water rise in the tank in half an hour?
Answers
Step-by-step explanation:
Given diameter of the circular pipe = 2 cm
So, the radius of the circular pipe = 2/2 = 1 cm
Height of the circular pipe = 0.7 m = 0.7 * 100 = 70 cm
Now, volume of the water flows in 1 second = πr2 h
= 3.142 * 12 * 70
= 3.142 * 70
Volume of the water flows in 1/2 hours = 3.142 * 70 * 30 * 60
Now, volume of the water flows = Volume of the cylinder
=> 3.142 * 70 * 30 * 60 = πr2 h
=> 3.142 * 70 * 30 * 60 = 3.142 * (40)2 h
=> 70 * 30 * 60 = 40 * 40 * h
=> h = (70 * 30 * 60)/(40 * 40)
=> h = (70 * 3 * 6)/(4 * 4)
=> h = 1260/16
=> h = 78.85 cm
So, the level of water rise in the tank in half an hour is 78.75 cm
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The volume of water raised after 30 mins in the tank is given as 395.64 litres
Given :
Diameter of cylindrical pipe = 2cm
The rate of water flow = 0.7m/sec
To Find :
Much water rise in the tank in half and hour
Formula Applied :
Solution :
As by given ,
Cross section area of pipe =Area of circle=
Diameter of pipe = 2cm
Radius =
Area of circle =
Hence the volume of water flowing per second ,
Hence of water is flowing per second.
The amount of water flowing for 30 mins is given as,
Hence the volume of water raised after 30 mins in litres is given as 395.64 litres