Question

water flows through a circular pipe of internal diameter is 2 cm @ 0.7 M per second into a cylindrical tank the radius of whose base is 40 cm by how much will the level of 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