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UPSC Civil Service Exam Questions

Q1: What is the BOD5 at 20°C of a waste that yields an oxygen consumption of 2 mg/l from a 0.5% diluted sample ?

A 50 mg/l

B 400 mg/l

C 200 mgl

D 250 mgl

ANS:B - 400 mg/l

BOD5= (Initial oxygen level - Final Oxygen level) x Dilution Factor.
Dilution Factor = Volume of raw sewage/Volume of treated(or Diluted) Sample.
ie BOD5 = 2 x (100 /0.5).
BOD5 = 400 mg/l.



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