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If a sewer drain carrying a discharge of 2 cumecs, outfalls into a river carrying a discharge of 10 cumecs and having DO equal to 8.4 mg/l, the resultant DO of the mixture will be equal to:


1. 5.0 mg/l
2. 7.0 mg/l
3. 10.5 mg/l
4. 15.0 mg/l
5.

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Best answer
Correct Answer - Option 2 : 7.0 mg/l

Concept:

Resultant DO after mix will be given by,

\({\rm{DO}} = \frac{{{{\rm{Q}}_1}{\rm{D}}{{\rm{O}}_1} + {{\rm{Q}}_2}{\rm{D}}{{\rm{O}}_2}}}{{{{\rm{Q}}_1} + {{\rm{Q}}_2}}}\)

Where,

Q1 = Discharge of river, Q2 = Discharge of sewer,

DO1 = Dissolved oxygen of river, and DO2 = Dissolved oxygen of sewer

Calculation:

Given: 

Q1 = 10 cumec, Q2 = 2 cumec,

DO1 = 8.4 mg/L, and DO2 = 0 mg/L

\(\therefore {\rm{DO}} = \frac{{10 \times 8.4 \ +\ 2 \times 0}}{{10 + 2}} = 7{\rm{\;mg}}/{\rm{L}}\)

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