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.
Priyanka said:
9 years ago
The correct answer is 204.
(1)
Divya Vishnoi said:
9 years ago
BOD = (initial D.O - Final D.O.)* Dilution factor.
= (7.8 - 4.4)*300/5.
= 204 mg/l.
= (7.8 - 4.4)*300/5.
= 204 mg/l.
(1)
Nayan Kulkarni said:
1 decade ago
BOD = (DO initial -DO final)*Dilution factor.
= (7.8-4.4)*(295/5).
= 200.6 mg/lit
= (7.8-4.4)*(295/5).
= 200.6 mg/lit
Jay Shah said:
1 decade ago
BOD = (INITIAL-FINAL)*300/ML.
= (7.8-4.4)*300/5 = 204 MG/L.
= (7.8-4.4)*300/5 = 204 MG/L.
(1)
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