Engineering Questions with Answers - Multiple Choice Questions
Environmental Engineering MCQ’s – Population Equivalent and Relative Stability
1 - Question
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
View Answer
Answer: c
Explanation: The average standard BOD of domestic sewage is measured in 0.08Kg per person per day.
2 - Question
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
View Answer
Answer: c
Explanation: BOD of industrial sewage kg/day /(0.08kg/person/day) where 0.08 represents the average standard BOD of domestic sewage.
3 - Question
The expression of relative stability at 20oC 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
View Answer
Answer: b
Explanation: The expression of relative stability at 20oC is given by
S=100 (1-(0.794)T20) where T20 represents the incubation period at 20oC.
4 - Question
The expression of relative stability at 37oC 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)
View Answer
Answer: d
Explanation: The expression of relative stability at 37oC is given by
S = 100 (1-(0.630)T37) where T37 represents the incubation period at 37oC.
5 - Question
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
View Answer
Answer: a
Explanation: BOD of industrial sewage in kg/day = Population equivalent * Average standard BOD of domestic sewage = 6000*0.08 = 480kg/day.
6 - Question
Calculate the percentage of relative stability, given 5 day incubation period at 370C.
a) 50%
b) 60%
c) 90%
d) 99%
View Answer
Answer: c
Explanation: 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 - Question
_____ is the amount of oxygen required to oxidize both organic and inorganic matter in sewage.
a) Turbidity
b) BOD
c) COD
d) DO
View Answer
Answer: c
Explanation: COD is the amount of oxygen required to oxidize both organic and inorganic matter in sewage. It is always greater than BOD.
8 - Question
COD is abbreviated as ___________
a) Chemical oxygen demand
b) Complex oxygen demand
c) Customary oxygen demand
d) Chemical oxygen deficit
View Answer
Answer: a
Explanation: COD is abbreviated as chemical oxygen demand. It is measured in ppm or mg/L.
9 - Question
The oxidizing agent used in COD test is ___________
a) Potassium chloride
b) Potassium per-manganate
c) Potassium chromate
d) Potassium dichromate
View Answer
Answer: d
Explanation: In COD test, oxidization of organic matter takes place by potassium dichromate in the presence of sulphuric acid using ferroin as indicator.
10 - Question
5 days BOD at 20oC is ___ of total demand.
a) 20%
b) 47%
c) 68%
d) 100%
View Answer
Answer: c
Explanation: 5 days BOD at 20oC is the standard BOD which is 68% of total demand, whereas 10 days BOD at 20oC is 90% of total demand.
11 - Question
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 (1.04T)
View Answer
Answer: c
Explanation: KDT = KD20 (1.047)T-20 where, KDT is the deoxygenation constant at temperature ‘T’ and KD20 is the deoxygenation constant at 20oC.