Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 7 (Q.No. 27)
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
Discussion:
5 comments Page 1 of 1.
Gajanan said:
7 years ago
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.
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.
(4)
Sumit kumar said:
7 years ago
Should be 60.
Because DO FINAL- DO INITIAL=3.
BOD IS=20*3.
Because DO FINAL- DO INITIAL=3.
BOD IS=20*3.
Vaish said:
6 years ago
It Should be option A.
100 * 5/5 + 100 * .8/5 = 116.
100 * 5/5 + 100 * .8/5 = 116.
ABHIRAM P said:
4 years ago
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.
Bod final= 3.8,
Bod = (8.4-3.8)/(5/100)=4.6*20=90.
Priyanka said:
4 years ago
A is the correct answer.
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