Civil Engineering - GATE Exam Questions - Discussion

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 :

[A]. 196 mg/l
[B]. 200 m/l
[C]. 204 mg/l
[D]. 208 mg/l

Answer: Option B

Explanation:

No answer description available for this question.

Jay Shah said: (Feb 4, 2015)  
BOD = (INITIAL-FINAL)*300/ML.

= (7.8-4.4)*300/5 = 204 MG/L.

Nayan Kulkarni said: (Jun 4, 2015)  
BOD = (DO initial -DO final)*Dilution factor.
= (7.8-4.4)*(295/5).
= 200.6 mg/lit

Divya Vishnoi said: (Apr 29, 2016)  
BOD = (initial D.O - Final D.O.)* Dilution factor.
= (7.8 - 4.4)*300/5.
= 204 mg/l.

Priyanka said: (Mar 25, 2017)  
The correct answer is 204.

Erick said: (Jun 8, 2017)  
The first initial DO of the mixture is required.

DO = 7.8 x (295/300) = 7.67,
BOD = (7.67-4.4)/(5/300) = 196.2 mg/L.

Himanshu said: (Nov 16, 2017)  
BOD = (initial D.O - Final D.O.)/ Dilution factor.
= (7.8 - 4.4)/(5/300).
= 204 mg/l.

Keshav Kaushik said: (Jan 9, 2019)  
Dilution Factor = volume of wastewater sample/(volume of wastewater sample + dilution sample).

B.O.D = Initial D.O - Final D.O / dilution factor = 204mg/l.

Manisha Nimje said: (Jan 9, 2020)  
The correct answer is 204 mg/l.

Sachin S Naik said: (Jan 30, 2020)  
The correct answer is 200.6mg/l.

Post your comments here:

Name *:

Email   : (optional)

» Your comments will be displayed only after manual approval.