Category: Environmental Engineering Objective Questions

Environmental Engineering Multiple Choice Questions on “Rapid Sand Filter”.

1. The depth of an enclosed tank of a rapid sand filter lies in the range of ______
a) 1-2m
b) 2.5-3.5m
c) 2-5m
d) 3-5m
Answer: b
Clarification: The depth of an enclosed tank of a rapid sand filter lies in the range of 2.5-3.5m and its surface area lies between 20-50m².

2. The effective size of a rapid sand filter is ______
a) 0.45mm
b) 0.7mm
c) 0.8mm
d) 0.9mm
Answer: a
Clarification: The effective size of a rapid sand filter lies in the range of 0.35-0.6mm and its common value is 0.45mm.

3. The finer the suspended impurities to be removed ___________
a) Larger will be the sand size
b) Smaller will be the sand size
c) Smaller will be the surface area of sand grains
d) Narrow will be the sand depth of the bed
Answer: b
Clarification: The finer the suspended impurities to be removed, smaller will be the sand size in order to increase the surface area of sand grains in the sand bed.

4. What is the value of the uniformity coefficient in a rapid sand filter?
a) 1
b) 1.5
c) 2
d) 2.5
Answer: b
Clarification: The value of uniformity coefficient in a rapid sand filter lies between 1.2 and 1.7 and its common value is 1.5.

5. What is the effective size of Anthracite when used as a filter material?
a) 0.1-0.2mm
b) 0.2-0.4mm
c) 0.4-0.6mm
d) 0.7-0.75mm
Answer: d
Clarification: The effective size of Anthracite when used as a filter material lies between 0.7-0.75mm and its uniformity coefficient is 1.75.

6. Consider the following statement regarding the depth of sand of the rapid gravity filter?
i. It should be checked against breakage of floc
ii. The sand depth should lie between 100-200cm
iii. The floc should be attached through the sand bed
Which of the above statement is/are correct?
a) Only i
b) i, ii
c) ii, iii
d) Only ii
Answer: a
Clarification: Sand depth should be such that floc does not break through the sand bed and the depth of sand should vary between 60-90 cm.

7. Which of the following statement is wrong regarding Anthracite?
a) It requires less water
b) It is less inert to caustic solutions than sand
c) It is more costly per tonne than sand
d) It is used in industrial filters.
Answer: b
Clarification: Anthracite is a costlier filter material used in industrial filter which is more inert to hot caustic solutions than sand.

8. The minimum depth of sand of a rapid sand filter is given by __________
a) Stokes formula
b) Clark’s formula
c) Hudson formula
d) Emden formula
Answer: c
Clarification: The minimum depth of sand of a rapid sand filter is given by
Qd³h/l=B*29323, Where Q=filtration rate, d = sand size, H = head loss, l=depth of sand bed, B=break through index.

9. The thickness of the base material of a rapid sand filter on which filter media are supported is __________
a) 45-60cm
b) 20-80cm
c) 30-75cm
d) 10-30cm
Answer: a
Clarification: The base material of a rapid sand filter consists of 45-60 cm thick gravel bed which is laid in layers of 15 cm with a finer topmost layer (2-6mm) and coarser bottom layer (20-50mm).

10. In which type of under drainage system of rapid sand filter, laterals are provided with strainers?
a) Perforated pipe system
b) Pipe and strainer system
c) Wheeler system
d) Wagner system
Answer: b
Clarification: In Perforated pipe system, laterals are provided with strainers. Strainers are made of brass or bronze and are in umbrella shape, having a small perforation. They distribute the wash water without jet action.

11. Which device is used to supply air for the agitation of sand grains during washing of filters?
a) Rate control device
b) Tube settlers
c) Air compressors
d) Flocculator
Answer: c
Clarification: Air compressor is used to supply air for the agitation of sand grains at the rate of 0.6-0.8m³ per minute per m² of filter area for 5 minutes.

Environmental Engineering Interview Questions for freshers on “Removal of Iron, Manganese, Colour, Odour and Taste”.

1. Iron and Manganese are generally present in water supplies as ____________
a) Suspension
b) Solution
c) Colloidal particles
d) Either suspension or solution
Answer: d
Clarification: Iron and Manganese are generally present in water supplies in the form of hydrated oxide as suspension and bicarbonates as a solution.

2. What concentration of iron and manganese will cause unpleasant taste and odor of water?
a) 1ppm
b) 2ppm
c) 3ppm
d) >3ppm
Answer: d
Clarification: When the concentration of iron and manganese is greater than 3, it will result in unpleasant taste and odor, unwanted precipitation and difficulties in manufacturing processes.

3. The presence of Sulfate iron in the water results in ____________
a) Odorless water
b) Acidity in water
c) Colorless water
d) Growth of crenothrix in water mains
Answer: b
Clarification: The presence of Sulfate iron in the water results in acidity in water and corrosiveness on iron and brass.

4. The reddish tinge in the water is due to the presence of ____________
a) Zinc
b) Copper
c) Iron
d) Manganese
Answer: c
Clarification: The reddish tinge in the water is due to the presence of iron. Water contains iron by the reaction of CO₂ in contact with iron ore forming soluble ferrous bicarbonate.

5. The brown tinge in the water is due to the presence of ____________
a) Zinc
b) Copper
c) Iron
d) Manganese
Answer: d
Clarification: The brown tinge in the water is due to the presence of manganese, which occurs when it is present in an amount greater than 0.3ppm.

6. By aeration process, dissolved iron is oxidized into ____________
a) Ferrous oxide
b) Ferric oxide
c) Ferrous hydroxide
d) Ferric hydroxide
Answer: b
Clarification: By aeration process, dissolved iron is oxidized into ferric oxide, which is insoluble in water.

7. By which process dissolved manganese compounds can be converted into insoluble manganese compounds?
a) Aeration
b) Coagulation
c) Sedimentation
d) Flocculation
Answer: a
Clarification: The dissolved manganese compounds can be converted into insoluble manganese compounds by aeration The precipitated floc can be settled down in settling tanks.

8. For removal of iron from water ________ mg of oxygen is required.
a) 0.14
b) 0.50
c) 1
d) 1.5
Answer: a
Clarification: The reaction for removal of iron from water is – 4Fe + O₂ + 10H₂O = 4Fe (OH)₃ + 8H
1mg of Fe requires 0.14 mg of O₂.

9. For removal of manganese from water ________ mg of oxygen is required.
a) 0.14
b) 0.50
c) 0.29
d) 0.90
Answer: c
Clarification: The reaction for removal of manganese from water is – 4Mn + 3O₂ + 6H₂O = 6MnO₂ + 12H
1mg of Mn requires 0.29 mg of O₂.

10. Removal of manganese from water takes place at a pH of ______
a) 1.5
b) 3
c) 6
d) 9.4
Answer: d
Clarification: Removal of manganese from water takes place at a pH between 9.4 to 9.6. Pre- chlorination up to 0.7 to 1 mg/l affect the oxidation and precipitation of manganese.

11. When small amounts of iron and manganese are present in water, they are removed by ____________
a) Aeration
b) Coagulation
c) Manganese zeolite
d) Filtration
Answer: c
Clarification: When small amounts of iron and manganese are present in water, they are removed by manganese zeolite where iron and manganese are oxidized to insoluble hydrated oxides which are removed by filtration action of zeolite bed.

12. Removal of iron from water takes place at a pH of ______
a) 1.5
b) 3
c) 8.2
d) 9.4
Answer: d
Clarification: Removal of iron from water takes place at a pH of 8.2. It can be removed in an oxidized condition by softening water by the excess lime process.

13. The taste produced by a microorganism, can be controlled by ____________
a) Potassium per-manganate
b) Potassium dichromate
c) Sodium chloride
d) Iodine
Answer: a
Clarification: The taste produced by a microorganism, can be controlled by potassium per-manganate. It can be controlled alone or in combination with chlorine.

14. Which of the following regarding ozonation is incorrect?
a) Improves the appearance of water
b) Removes traces of color
c) Remove odor
d) Remove phenolic and algae taste
Answer: d
Clarification: Ozonation improves the appearance of water, removes traces of color, remove odor and is also effective against earthy and moldy tastes and odor.

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Environmental Engineering Multiple Choice Questions on “Storm Water Flow”.

1. Which of the following represents the minimum flow?

Figure: Hourly variation of sewage flow
a) A
b) B
c) C
d) D
Answer: a
Clarification: The figure clearly shows that the minimum flow occurs at point A and its magnitude is 50.

2. Which of the following represents the peak flow?

Figure: Hourly variation of sewage flow
a) A
b) B
c) C
d) D
Answer: b
Clarification: The peak flow is the maximum flow which occurs at point B and its magnitude is 170.

3. The point R in the figure represents the ________

Figure: Hourly variation of sewage flow
a) Minimum flow
b) Maximum flow
c) Average rate of flow
d) Peak flow
Answer: c
Clarification: The average rate of flow lies between the maximum and minimum flow. Here, R represents the average rate of flow which lies between point B and A.

4. At what time the peak flow takes place?

Figure: Hourly variation of sewage flow
a) 3hours
b) 7hours
c) 10hours
d) 14hours
Answer: c
Clarification: The peak flow takes place at point B and the corresponding time is 10 hours.

5. In India, the maximum sewage flow occurs normally in which month?
a) July
b) January
c) February
d) December
Answer: a
Clarification: In India, the maximum sewage flow occurs during June and July as maximum utilization of water takes place.

6. The minimum sewage flow occurs normally in which month in India?
a) March
b) January
c) August
d) July
Answer: b
Clarification: The minimum sewage flow in India is assumed to occur during December or January.

7. The peaking factor is defined by __________
a) Peak flow * Average flow
b) Peak flow – Average flow
c) Peak flow + Average flow
d) Peak flow / Average flow
Answer: d
Clarification: The peaking factor is the ratio of Peak flow to the Average flow and it is highest for domestic sewer as they receive the flow directly from the source.

8. Which of the following has a maximum peaking factor?
a) Domestic sewer
b) Branch sewer
c) Main sewer
d) Trunk sewer
Answer: a
Clarification: The domestic sewer has the maximum peaking factor of 6 while Branch sewer, Main sewer and Truck sewer has a peaking factor of 3, 2.5 and 2 respectively.

9. The Lateral sewer has a peaking factor of ____________
a) 1
b) 2
c) 3
d) 4
Answer: d
Clarification: The peaking factor of lateral sewer lies between 4 to 6 which is greater than Branch sewer, Main sewer and Truck sewer.

10. The population of the city is 30,000. What is the peaking factor?
a) 1
b) 2
c) 2.25
d) 2.5
Answer: d
Clarification: When the population of the city is between 20,000 to 50,000, the peaking factor is 2.5.

11. The peaking factor is directly proportional to the population of the city.
a) True
b) False
Answer: b
Clarification: When the population of the city is less than 20,000, the peaking factor is 3.5 and when the population of the city is above 7,50,000, the peaking factor is 2. So, the peaking factor varies inversely with the population.

12. _________ flow persists for various time duration.
a) Minimum sewage
b) Maximum sewage
c) Sustained
d) Average sewage
Answer: c
Clarification: Sustained flow persists for various time duration. They are important in sewage design when they are higher or lower than than average flow.

13. The wet weather flow is also known by ______
a) Minimum sewage flow
b) Maximum sewage flow
c) Storm water flow
d) Average rate of flow
Answer: c
Clarification: The wet weather is also known by the storm water flow and is commonly abbreviated as W.W.F.

14. Consider the following statement.
The storm water flow depends on
i. Catchment area
ii. Ground slope
iii. Quality of water
vi. Rainfall duration
Which of the following is correct?
a) i, ii
b) i, ii, iii
c) i, ii, iv
d) i, iv
Answer: c
Clarification: The storm water flow depends on the catchment area, ground slope, permeability of ground, extent of vegetative growth, rainfall intensity and duration, condition of the ground and climatic conditions.

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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Environmental Engineering Multiple Choice Questions & Answers on “Mechanical Aeration System, Treatment of Sludge”.

1. The device used for the aeration in the combined aerator is ___________
a) Header pipe
b) Diffuser plate and paddles
c) Cone with spiral vanes
d) Diffuser plate
Answer: b
Clarification: In combined aeration, both diffused air aeration and mechanical aeration take place using both diffuser plate and paddles.

2. The detention period of the Sheffield mechanical aeration tank is ___________
a) 15 hours
b) 1minute
c) 1 hour
d) 4 hours
Answer: a
Clarification: The detention period of a Sheffield mechanical aeration tank is 15 hours. Its return sludge is 15-20% of sewage flow.

3. The minimum detention period of combined aerator is ___________
a) 1 hour
b) 2 hours
c) 3 hours
d) 4 hours
Answer: c
Clarification: The detention period of combined aerator lies between 3 to 4 hours and the minimum detention period is 3 hours.

4. The speed at which the bubbles are formed in the simplex aerator?
a) 15rpm
b) 20rpm
c) 60rpm
d) 30rpm
Answer: c
Clarification: The cones, with spiral vanes rotates at a high speed of 60rpm that results in the formation of bubbles in a simplex aerator resulting in the aeration of sewage.

5. Dorroco aerator is a type of ___________
a) Combined aerator
b) Mechanical aerator
c) Ridge and furrow type aerator
d) Spiral flow type aerator
Answer: a
Clarification: Dorroco aerator is a type of combined aerator which is efficient in comparison to diffused air and mechanical aerator.

6. Simplex aerator is a type of ___________
a) Combined aerator
b) Mechanical aerator
c) Ridge and furrow type aerator
d) Spiral flow type aerator
Answer: b
Clarification: Simplex aerator is a type of mechanical aerator where a square or rectangular tank with a hopper bottom is used.

7. Which gas is produced in aerobic digestion of sludge?
a) NO₂
b) CO₂
c) O₂
d) H₂S
Answer: b
Clarification: Carbon dioxide gas is produced in the aerobic digestion of sludge by the oxidation of organic matter by bacteria into CO₂.

8. Which of the following process is not involved in the treatment of sludge?
a) Dewatering
b) Drying
c) Flocculation
d) Conditioning
Answer: c
Clarification: The process involved in the treatment of sludge are thickening, stabilization, dewatering and dying.

9. Which of the following is the final stage in the sludge treatment process?
a) Digestion
b) Dewatering
c) Drying
d) Thickening
Answer: c
Clarification: Drying of the sludge is the final stage in the sludge treatment process where the most stable form of sludge is obtained.

10. In which sludge treatment process, the organic solids are converted into more stable form?
a) Dewatering
b) Thickening
c) Digestion
d) Conditioning
Answer: c
Clarification: In the sludge digestion process, the organic solids are converted into more stable form. It makes the dewatering and drying process of sludge simpler.

11. The sludge treatment process to increase the solid content of sludge is ___________
a) Dewatering
b) Thickening
c) Conditioning
d) Drying
Answer: b
Clarification: In sludge thickening, the solid content of sewage is increased by gravity flotation and clarifier.

12. The process of removal of water from the sludge is called ______
a) Dewatering
b) Thickening
c) Digestion
d) Drying
Answer: a
Clarification: Dewatering is the process of removal of water from the sludge. It can be done by drying or subjected the sludge with pressure.

13. In which sludge treatment process, the sludge is treated with chemicals?
a) Dewatering
b) Thickening
c) Conditioning
d) Drying
Answer: c
Clarification: In sludge conditioning, the sludge is treated with chemicals in order to separate the water from the sludge and improving the dewatering characteristics of the sludge.

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