Question

WATER IS FLOWING AT A RATE OF 2.52KM/HR THROUGH A CYLINDRICAL PIPE INTO A CYLINDRICAL TANK,THE RADIUS OF WHOSE BASE IS 40CM.IF THE INCREASE IN THE LEVEL OF WATER IN THE TANK ,IN THE HALF OF THE HOUR IS 3.15M , FIND THE INTERNAL DIAMETER OF THE PIPE

Answer 1

Solution:-

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
Answer.

Answer 2

Answer:4cm

Step-by-step explanation:

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