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 ?

Answer 1

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.