water is flowing through a cylindrical pipe of internal diameter 2m , into a cylindrical tank of base radius 40 cm at the rate of 0.7m/ sec .By how much will the water rise in the tank in half an hour?
Answers
Answer:
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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Answer:
Step-by-step explanation:
