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Question
Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m per second. Determine the rise in level of water in the tank in half an hour.
Solution
We have,
the internal radius of the cylindrical pipe, r = 2/2 = 1 cm and
the base radius of cylindrical tank, R = 40 cm.
Also, the rate of flow, h = 0.4 m/s = 40 cm/s
Let the rise in level of water be H.
Now,
The volume of water flowing out of the cylindrical pipe in 1 sec =πr2h = π × 1 × 1 × 40 = 40π cm3
As,
Volume of water in the cylindrical tank = Volume of standing water in cylindrical pipe
⇒πR2H = 72000π
⇒R^2H = 72000`
⇒ 40 × 40 × H = 7200H
`rArr H = 72000/(40xx40)`
∴ H = 45 cm
So, the rise in level of water in the tank in half an hour is 45 cm.
Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.
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