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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/s. Determine the rise in level of water in the tank in half an hour ?
Solution
Diameter of circular end of pipe = 2 cm
∴ Radius (r1) of circular end of pipe =`2/200m=0.01m`.
Area of cross-section = `pixxr_1^2=pixx(0.01)^2=0.0001pim^2`
Speed of water = 0.4 m/s = `0.4xx60=24 \text{metre/min}`
Volume of water that flows in 1 minute from pipe = 24 ×`0.0001pi` m3 = 0.0024π m3
Volume of water that flows in 30 minutes from pipe = 30 × 0.0024π m3= 0.072π m3
Radius (r2) of base of cylindrical tank = 40 cm = 0.4 m
Let the cylindrical tank be filled up to h m in 30 minutes.
Volume of water filled in tank in 30 minutes is equal to the volume of water flowed in 30 minutes from the pipe.
∴ π × (r2)2 ×h= 0.072π
⇒ (0.4)2 ×h= 0.072
⇒ 0.16 ×h= 0.072
`rArrh=0.072/0.16`
⇒ h = 0.45 m = 45 cm
Therefore, the rise in level of water in the tank in half an hour is 45 cm.
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