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What will be the surface area of the basin for a flow of 2*10^6 litre per day having a surface loading rate 10,000 litre/day/m^2?

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in Environmental Engineering by (81.4k points)
What will be the surface area of the basin for a flow of 2*10^6 litre per day having a surface loading rate 10,000 litre/day/m^2?

(a) 100

(b) 200

(c) 300

(d) 400

I got this question in an online interview.

Question is taken from Design Elements of Sedimentation in chapter Sedimentation of Environmental Engineering

1 Answer

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Best answer
The correct option is (b) 200

The best I can explain: Surface area of basin = flow of water/surface loading rate = 2000000/10000=200m^2.

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