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GATE Exam Questions

Q1: In a BOD test using 5% dilution of the sample (15 ML of sample and 285 mL of dilution water), dissolved oxygen values for the sample and dilution water blank bottles after five days incubation at 20°C were 3.80 and 8.80 mg/L. respectively. Dissolved oxygen originally present in the undiluted sample was 0.80 mg/L. The 5-day 20°C BOD of the sample is

A 116 mg/L

B 108 mg/L

C 100 mg/L

D 92 Mg/L

ANS:D - 92 Mg/L

Bod initial = ((15*.8)+(285*8.8))/300 = 8.4.
Bod final= 3.8,
Bod = (8.4-3.8)/(5/100)=4.6*20=90.



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