TGPSC HMWSSB Manager 2015 Paper 2 · Question 86

Question 86 of the TSPSC Manager (Engineering), HMWSSB, 2015: Paper 2 (Civil Engineering) (1 November 2015). Try it first, then open the answer and explanation.

1. Manager 2015 Paper 2 (Civil) · Q86 · Environmental Engineering · Sewerage & Wastewater Treatment

   If a sewer carrying a discharge of 3 cumecs outfalls into a river having a discharge of 10 cumecs and DO equal to 9.1mg/lt, the resultant DO of the mix will be equal to

   1. 5mg/lt
   2. 6mg/lt
   3. 7mg/lt
   4. 8mg/lt
   Show answer

   Answer: (3) 7mg/lt

   Taking the DO of raw sewage as zero, the mixed DO = (10 × 9.1 + 3 × 0)/(10 + 3) = 91/13 = 7 mg/lt.

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