Civil Engineering - GATE Exam Questions - Discussion

27. 

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

Answer: Option D

Explanation:

No answer description available for this question.

Gajanan said: (Aug 11, 2017)  
Bod= dissolved oxygen * dilution factor.
D.f= 5%=100/5 =20.
Given D.O as 0.80& 3.80.
Bod= .80 * 20= 16mg/l.
Bod= 3.8 * 20= 76 mg/l.
Bod= 16+76=92mg/l.

Sumit Kumar said: (Dec 11, 2017)  
Should be 60.
Because DO FINAL- DO INITIAL=3.
BOD IS=20*3.

Vaish said: (Feb 23, 2019)  
It Should be option A.

100 * 5/5 + 100 * .8/5 = 116.

Abhiram P said: (Dec 10, 2020)  
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.

Priyanka said: (Jan 12, 2021)  
A is the correct answer.

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