Water flows out through a circular pipe whose internal diameter is 2 cm at the rate of 0.8 m/s into a cylindrical tank, the radius of whose base is 40 cm. By how much will the level rise in one and a half hour?
Answers
Answer:
Internal diameter of the pipe =2cm
So its radius =1cm=
100
1
m
Water that flows out through the pipe in 6ms
−1
So volume of water that flows out through the pipe in 1sec=π×
100
1
2
×6m
3
∴ In 30 minutes, volume of water flow =π
100×100
1
×6×30×60m
3
This must be equal to the volume of water that rises in the cylindrical tank after 30 minutes and height up to which it rises say h.
Radius of tank =60cm=
100
60
m
Volume =π(
100
60
)
2
h
=π(
100
60
)
2
h=π×
100×100
1
×6×30×60
⇒
100×100
60×60
h=
100×100
6×30×60
⇒h=
36
3×36
=3m
So required height will be 3m.
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Answer:
If internal diameter=2cm,
then internal radius of the pipe, r = 1 cm.
length of water flowing in 1 second, h = 0.8 m = 80 cm.
Volume of water flowing in 1 second
πr²h=(π×1×1×80)cm3=(80π)cm³
Time=1 and half hour=90 min
Volume of water flowing in 90 minutes
=(80π×60×90)cm³=(432000π)cm³.
Radius of cylindrical tank, R = 40 cm.
Let the rise in level of water be H cm.
Volume of water in the tank.
=πR²H=(π×40×40×H)cm³
=(1600πH)cm³.
Volume of water in the tank = volume of water in the tank
⇒ 1600πH=432000π⇒H=432000/1600=270
Hence, rise in level = 270 cm.