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

Given that rate of flow of water = 2.52km/hr and h = 3.15m = 315cm.

Given radius = d/2 = 40cm.Then the diameter d = 2r.    -------- (1) 

The volume of water tank that fills in half an hour(30 minutes) = pir^2h

 = 22/7 * 40 * 40 * 315

= 1584000cm^3.     ----------- (2)


Length of water column in 30 minutes = 2.52 * 30/60

                                                                 = 126000cm^3.


Volume of water through flows through the pipe in 30 minutes

= pir^2l

= 22/7 * (d/2)^2 * 126000.


Now, Volume of the water flown in 30 minutes = volume of the water that fills.

22/7 * d^2/4 * 126000 = 1584000

d^2/4 = 4

d^2 = 16

d = 4 cm.


The internal diameter of the pipe = 4cm.


Hope this helps!

Answer 2

Answer:

Internal Diameter of Pipe =4cm

Step-by-step explanation:✪ ✫ ✬ ✭ ✮ ✯

Increase in level of water in tank in half hour =3.15m

⇒Increase in level of water in 1hour =6.3m

Or increase in level of water in cylindrical tank  =6.3m/h

So Volume of water in tank in 1hour =πr²h

given that radius of cylindrical tank =40cm =0.4m

Volume of water =(22/7)(0.4)²×6.3

=22×(0.16)×(0.9)

=22×1.44

=3.168m³

It means 3.168m³ Volume of water increases per hour.

So Volume of water flowing thorough Pipe =3.168m³/h

Or (2.52km/h)×cross sectional area of pipe =3.168m³/h

Or (2520m/h)×πr² =3.168m³/h

Or 360×22r²=3.168

Or r² =3.168/7920

Or r²=0.0004m²

⇒ r=+0.02m=2cm  (radius is always +ve)

D=2r=4cm  ✪ ✫ ✬ ✭ ✮ ✯