Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 4 (Q.No. 5)
5.
In a BOD test, 5 ml of waste is added to 295 ml of aerated pure water. Initial dissolved oxygen (D.O) content of the diluted sample is 7.8 mg/l. After 5 days of incubation at 20°C, the D.O. content of the sample is reduced to 4.4 mg/l. The BOD of the waste water is :
Discussion:
14 comments Page 2 of 2.
Sachin s Naik said:
6 years ago
The correct answer is 200.6mg/l.
Abhishek said:
4 years ago
Df = 295÷5
Bod = (intial Do-final DO) * D.F,
= 200.6.
Bod = (intial Do-final DO) * D.F,
= 200.6.
Sharath Manohar said:
4 years ago
295/5 * (7.8-4.4) = 200 mg/l is the actual right answer.
Lencho Edeto said:
3 years ago
BOD = [DO( initial) -DO final)] * Dilution factor.
= (7.8-4.4)mg/lit * (295/5).
= 200.6 mg/lit.
= (7.8-4.4)mg/lit * (295/5).
= 200.6 mg/lit.
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