TGPSC HMWSSB Manager 2015 Paper 2 · Question 80

Question 80 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) · Q80 · Environmental Engineering · Sewerage & Wastewater Treatment

   A 2% solution of a sewage sample is kept at an incubation temperature of 20°C. If initial DO and final DO values after 5 days of incubation period are 8.5 mg/lit and 5.5 mg/lit respectively, then the BOD will be

   1. 50mg/lit
   2. 150mg/lit
   3. 250mg/lit
   4. 350mg/lit
   Show answer

   Answer: (2) 150mg/lit

   BOD₅ = (DO_initial − DO_final)/dilution fraction = (8.5 − 5.5)/0.02 = 150 mg/lit.

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