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