Question

Q. Water flows through a circular pipe, whose internal diameter is 2 cm, at the rate of 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 in the cylindrical tank use in half an hour?

Answer 1

Solution:

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

Answer 2

Given Diameter of the pipe = 2cm.

Then the radius of the pipe r = d/2

                                                 = 1cm.



Given Rate of flow of water = 0.7m per second

                                              = 70cm.


Given volume of the water flows in 1 sec = Base area of the pipe * Rate of water flow

   = pir^2h

   = 22/7 * 1 * 70

   = 220cm^3.


The volume of water flows in 30 minutes = 220 * 30 * 60

                                                                     = 396000cm^3.


Now,

Given radius = 40cm.

pir^2h = 396000 

22/7 * (40)^2 * h = 396000

35200/7 * h = 396000

35200 * h = 2772000

h = 2772000/35200

h = 78.75.


Therefore the level of water rise = 78.75m.


Hope this helps!