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If the initial D.O. of sewage sample is 6mg/l and final D.O. is 2 mg/l. The dilution is 1%. Then the BOD of the sample is-
- a)4 mg/l
- b)6 mg/l
- c)40 mg/l
- d)400 mg/l
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
If the initial D.O. of sewage sample is 6mg/l and final D.O. is 2 mg/...
BOD = (Initial D.O. – Final D.O.) × Dilution Factor
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Hence the correct answer is option d.
Most Upvoted Answer
If the initial D.O. of sewage sample is 6mg/l and final D.O. is 2 mg/...
BOD (Biochemical Oxygen Demand) is a measure of the amount of dissolved oxygen required by microorganisms to decompose organic matter in water. It is an important parameter used to assess the level of pollution in water bodies.
Given:
Initial D.O. (Dissolved Oxygen) = 6 mg/l
Final D.O. = 2 mg/l
Dilution = 1%
To calculate the BOD of the sample, we can use the formula:
BOD = (Initial D.O. - Final D.O.) * Dilution
Let's calculate the BOD using the given values:
BOD = (6 mg/l - 2 mg/l) * 1%
Calculating the difference in D.O.:
BOD = 4 mg/l * 1%
Converting the dilution from percentage to decimal:
BOD = 4 mg/l * 0.01
Simplifying the calculation:
BOD = 0.04 mg/l
Therefore, the BOD of the sample is 0.04 mg/l.
However, the correct answer provided is 400 mg/l, which seems to be an error. The correct answer should be 0.04 mg/l based on the given information.
In conclusion, the BOD of the sewage sample is 0.04 mg/l, not 400 mg/l as mentioned in the options.
Given:
Initial D.O. (Dissolved Oxygen) = 6 mg/l
Final D.O. = 2 mg/l
Dilution = 1%
To calculate the BOD of the sample, we can use the formula:
BOD = (Initial D.O. - Final D.O.) * Dilution
Let's calculate the BOD using the given values:
BOD = (6 mg/l - 2 mg/l) * 1%
Calculating the difference in D.O.:
BOD = 4 mg/l * 1%
Converting the dilution from percentage to decimal:
BOD = 4 mg/l * 0.01
Simplifying the calculation:
BOD = 0.04 mg/l
Therefore, the BOD of the sample is 0.04 mg/l.
However, the correct answer provided is 400 mg/l, which seems to be an error. The correct answer should be 0.04 mg/l based on the given information.
In conclusion, the BOD of the sewage sample is 0.04 mg/l, not 400 mg/l as mentioned in the options.
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If the initial D.O. of sewage sample is 6mg/l and final D.O. is 2 mg/l. The dilution is 1%. Then the BOD of the sample is-a)4 mg/lb)6 mg/lc)40 mg/ld)400 mg/lCorrect answer is option 'D'. Can you explain this answer?
Question Description
If the initial D.O. of sewage sample is 6mg/l and final D.O. is 2 mg/l. The dilution is 1%. Then the BOD of the sample is-a)4 mg/lb)6 mg/lc)40 mg/ld)400 mg/lCorrect answer is option 'D'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about If the initial D.O. of sewage sample is 6mg/l and final D.O. is 2 mg/l. The dilution is 1%. Then the BOD of the sample is-a)4 mg/lb)6 mg/lc)40 mg/ld)400 mg/lCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the initial D.O. of sewage sample is 6mg/l and final D.O. is 2 mg/l. The dilution is 1%. Then the BOD of the sample is-a)4 mg/lb)6 mg/lc)40 mg/ld)400 mg/lCorrect answer is option 'D'. Can you explain this answer?.
If the initial D.O. of sewage sample is 6mg/l and final D.O. is 2 mg/l. The dilution is 1%. Then the BOD of the sample is-a)4 mg/lb)6 mg/lc)40 mg/ld)400 mg/lCorrect answer is option 'D'. Can you explain this answer? for SSC 2024 is part of SSC preparation. The Question and answers have been prepared according to the SSC exam syllabus. Information about If the initial D.O. of sewage sample is 6mg/l and final D.O. is 2 mg/l. The dilution is 1%. Then the BOD of the sample is-a)4 mg/lb)6 mg/lc)40 mg/ld)400 mg/lCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for SSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the initial D.O. of sewage sample is 6mg/l and final D.O. is 2 mg/l. The dilution is 1%. Then the BOD of the sample is-a)4 mg/lb)6 mg/lc)40 mg/ld)400 mg/lCorrect answer is option 'D'. Can you explain this answer?.
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