250+ TOP MCQs on Population Equivalent and Relative Stability and Answers

Environmental Engineering Questions for Entrance exams on “Population Equivalent and Relative Stability”.

1. The average standard BOD of domestic sewage is measured in ____________
a) ppm
b) Kg/day
c) Kg per person per day
d) Number of persons per day
Answer: c
Clarification: The average standard BOD of domestic sewage is measured in 0.08Kg per person per day.

2. Which of the following id the correct expression of population equivalent?
a) BOD of industrial sewage in kg/day * 0.08
b) 0.08 * BOD of industrial sewage kg/day
c) BOD of industrial sewage kg/day / 0.08kg/person/day
d) BOD of industrial sewage kg/day + 0.08
Answer: c
Clarification: BOD of industrial sewage kg/day /(0.08kg/person/day) where 0.08 represents the average standard BOD of domestic sewage.

3. The expression of relative stability at 20^oC is given by ___________
a) S = (1-(0.794)^T20)
b) S = 100 (1-(0.794)^T20)
c) S = 1-(0.794)^T20
d) S = 100 (0.794)^T20
Answer: b
Clarification: The expression of relative stability at 20^oC is given by
S=100 (1-(0.794)^T20) where T20 represents the incubation period at 20^oC.

4. The expression of relative stability at 37^oC is given by ___________
a) S = (1-(0.794)^T37)
b) S = 100 (1-(0.794)^T37)
c) S = 1-(0.794)^T20
d) S = 100 (1-(0.630)^T37)
Answer: d
Clarification: The expression of relative stability at 37^oC is given by
S = 100 (1-(0.630)^T37) where T37 represents the incubation period at 37^oC.

5. What is the value of BOD of industrial sewage in kg/day, given population equivalent as 6000 persons?
a) 480
b) 160
c) 270
d) 100
Answer: a
Clarification: BOD of industrial sewage in kg/day = Population equivalent * Average standard BOD of domestic sewage = 6000*0.08 = 480kg/day.

6. Calculate the percentage of relative stability, given 5 day incubation period at 37⁰C.
a) 50%
b) 60%
c) 90%
d) 99%
Answer: c
Clarification: Relative stability at 370C is given by, S = 100 (1-(0.630) ^T37) = 100 (1-(0.630)⁵) = 90.08% which is approximately 90%.

7. _____ is the amount of oxygen required to oxidize both organic and inorganic matter in sewage.
a) Turbidity
b) BOD
c) COD
d) DO
Answer: c
Clarification: COD is the amount of oxygen required to oxidize both organic and inorganic matter in sewage. It is always greater than BOD.

8. COD is abbreviated as ___________
a) Chemical oxygen demand
b) Complex oxygen demand
c) Customary oxygen demand
d) Chemical oxygen deficit
Answer: a
Clarification: COD is abbreviated as chemical oxygen demand. It is measured in ppm or mg/L.

9. The oxidizing agent used in COD test is ___________
a) Potassium chloride
b) Potassium per-manganate
c) Potassium chromate
d) Potassium dichromate
Answer: d
Clarification: In COD test, oxidization of organic matter takes place by potassium dichromate in the presence of sulphuric acid using ferroin as indicator.

10. 5 days BOD at 20^oC is ___ of total demand.
a) 20%
b) 47%
c) 68%
d) 100%
Answer: c
Clarification: 5 days BOD at 20^oC is the standard BOD which is 68% of total demand, whereas 10 days BOD at 20^oC is 90% of total demand.

11. Which of the following expression is correct regarding deoxygenation constant where symbols have their usual meaning?
a) KDT = KD20 + (1.047)^T-20
b) KDT = KD20 / (1.047)^T-20
c) KDT = KD20 (1.047)^T-20
d) KDT = KD20 (1.04^T)
Answer: c
Clarification: KDT = K_D20 (1.047)^T-20 where, KDT is the deoxygenation constant at temperature ‘T’ and KD20 is the deoxygenation constant at 20^oC.

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