Question

What is flowing at the rate of 2.52 km throw cylindrical pipe in syndicate and the radius of whose base is 40 cm if the increased level of water in the tank in half an hour at 3.15 m find internal diameter of the pipe?

Answer 1

Increase in the water level in half an hour = 3.15 m = 315 cm

Radius of the water tank = 40 cm

Volume of the water that falls in the tank in half an hour = πr²h

= 22/7*40*40*315

= 1584000 cu cm

Rate of the water flow = 2.52 km/hr

Length of water column in half an hour = (2.52*30)/60

= 1.26 km = 126000 cm

Let the internal diameter of the cylindrical pipe be d.

Volume of water that flows through the pipe in half an hour = π*(d/2)²*126000

As we know that,

Volume of the water that flows through pipe in half an hour = Volume of water that falls in the cylindrical tank in half an hour

⇒ 22/7*(d/2)²*126000 = 1584000

⇒ 22/7*d²/4*126000 = 1584000

⇒ d² = 16

⇒ d = √16

⇒ d = 4

So, the internal diameter of the pipe is 4 cm

hope this answer helpful u