Water is flowing through a cylindrical pipe of internal diameter 2cm,
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:
H = 78.75 cm
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 = 78.75 cm
So, the level of water rise in the tank in half an hour is 78.75 cm
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Step-by-step explanation:
For pipe, r = 1cm
Length of water flowing in 1 sec, h = 0.7m = 7cm
Cylindrical Tank, R = 40 cm, rise in water level = H
Volume of water flowing in 1 sec = πr2h = π x 1 x 1 x 70
= 70π
Volume of water flowing in 60 sec = 70π x 60
Volume of water flowing in 30 minutes = 70π x 60 x 30
Volume of water in Tank = πr2H = π x 40 x 40 x H
Volume of water in Tank = Volume of water flowing in 30 minutes
π x 40 x 40 x H = 70π x 60 x 30
H = 78.75cm