Question

Water is flowing at the rate of 0.7 m/sec through a circular pipe whose internal diameter is 2 cm into a cylindrical tank, the radius of whose base is 40 cm. Determine the increase in the level of water in half hour.

Answer 1

 internal diameter = 2 cm

radius = 2/2 cm = 1cm

 water flows at the rate of 0.7 m/s.

 In 1 second it covers distance = 0.7 m = 70 cm

 volume of water flows in 1 second = πr²d = 3.14 x 12 x 70 = 219.8 m³

Thus volume of water flows in 1 and half hour  = 5400 x 219.8 = 1186920 m³

radius of cylindrical tank (r) =  40 cm

Let the increase level of water = h cm

Volume of cylindrical tank = π(40)²h

Si water flows out at the rate of 0.7 m/s into a cylindrical tank. 

So,volume of water flows in 1and half hour = Volume of cylindrical tank

1186920 m³= π(40)²h

h = 1186920 / ((3.14 x (40)²)

h = 236.25 m

Hence, water rise by 236.25 m.

Answer 2

in half hour vol.of water in pipe=70x30x60=126000
vol.of cylindrical tank=1600h
so h=126000/1600
=78.75cm