Question

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​

Answer 1

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