Category: Environmental Engineering Objective Questions

Environmental Science Multiple Choice Questions on “Particulates”.

1. What is the size range of respirable suspended particulate matter?
a) Less than 1 micrometre
b) Less than 10 micrometre
c) Less than 100 micrometre
d) Less than 0.1 micrometre
Answer: b
Clarification: Particulate matter of size less than 10 micrometre, fall under the category of suspended particulate matter.

2. Air-borne particulates are classified as group 1 carcinogens.
a) True
b) False
Answer: a
Clarification: Air borne particulate matter is classified as group 1 carcinogen by WHO and IARC.

3. Which of the following is a viable particulate?
a) Smoke
b) Mist
c) Dust
d) Moulds
Answer: d
Clarification: Viable particulate matter consists of miniscule living organisms that float in the atmosphere.

4. Which type of particulate is condensed form of vapours?
a) Mist
b) Dust
c) Fumes
d) Smoke
Answer: c
Clarification: Fumes are condensed vapours, especially metallic vapour and is regarded as a type of particulate matter.

5. What is the composition of photochemical smog?
a) Nitrogen oxides, ketones and ozone
b) VOCs and hydrocarbons
c) Peroxy-acetyl-nitrate, peroxy-benzoyl-nitrate, peroxy fornyl-nitrate
d) All of the mentioned
Answer: d
Clarification: The photochemical smog constitutes of oxides of nitrogen, aldehydes, ketones, ground level ozone as well as PAN, PBN, PFN which are the free radicals formed due to reaction of ketones, nitrogen oxides and oxygen in the presence of sunlight.

6. Which of the following constituent of photochemical smog causes the bronzing of plants?
a) PBN
b) PAN
c) PFN
d) Ketones
Answer: b
Clarification: Peroxy-acetyl-nitrate is known to have sever effects on plants and causes it’s surface to bronze and glaze.

7. What is the reason behind the yellow colour of smog?
a) Nitrogen dioxide
b) Sulphur dioxide
c) Sulphate ions
d) Nitrate ions
Answer: c
Clarification: Sulphate ions formed by the conversion of sulphur dioxide due to high humidity and low temperature, is responsible for the yellow colour of smog.

8. Which of the following aerosols have the best absorbing properties?
a) Carbon black
b) Soot
c) Elemental Carbon
d) All of the mentioned
Answer: d
Clarification: Elemental carbon (also called soot/ carbon black/ black carbon) is a type of aerosol with the best absorbing property and used to absorb poisonous gases.

9. What is the size range of atmospheric particulate matter?
a) 0.1 – 10 microns
b) 0.1 – 1 micron
c) 1 – 10 microns
d) 10 – 100 microns
Answer: a
Clarification: Atmospheric particulate matter comprises of particles of sizes in the range of 0.1 – 10 micrometres.

10. In rural areas, what has contributed significantly to particulate pollution?
a) Incomplete combustion in vehicles
b) Using wood for fire and cooking
c) Fertilisers
d) All of the mentioned
Answer: b
Clarification: In rural areas, due to usage of wood for fire and cooking leading to the emission of soot and other organic gases has significantly contributed to particulate pollution.

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Environmental Engineering Multiple Choice Questions on “Theory of Sedimentation”.

1. When impurities are separated by the gravitation of settling particles, the operation is called _____
a) Sedimentation with coagulant
b) Plain sedimentation
c) Secondary sedimentation
d) Disinfection
Answer: b
Clarification: In plain sedimentation, impurities are separated from a suspending fluid by use of natural forces like gravitational forces.

2. In which type of settling, sedimentation of discrete particles takes place?
a) Zone settling
b) Compression settling
c) Hindered settling
d) Discrete settling
Answer: d
Clarification: In Discrete settling, sedimentation of discrete particles takes place where the particles have little tendency to flocculate.

3. In which settling type, dilute suspension of particles takes place?
a) Zone settling
b) Compression settling
c) Hindered settling
d) Discrete settling
Answer: c
Clarification: In this type of settling, particle mass increase and it settle down due to flocculation.

4. In which type of sedimentation, flocculent suspension of intermediate concentration takes place?
a) Type I sedimentation
b) Type II sedimentation
c) Type III sedimentation
d) Type IV sedimentation
Answer: c
Clarification: In type III sedimentation, inter-particle forces hold the particle together and flocculent suspension of intermediate concentration takes place.

5. In which type of sedimentation, flocculent suspension of high concentration takes place?
a) Type I sedimentation
b) Type II sedimentation
c) Type III sedimentation
d) Type IV sedimentation
Answer: d
Clarification: In type IV sedimentation, flocculent suspension of high concentration takes place and particles comes in contact with each other resulting in the formation of structure.

6. What is the settling velocity of the particle if its diameter is 2 * 10^-3cm. Given G=2.65, viscosity v=8 * 10^-3cm²/Sec?
a) 0.01cm/Sec
b) 0.13cm/Sec
c) 0.24cm/Sec
d) 0.36cm/Sec
Answer: b
Clarification: Settling velocity of the particle is given by vs=g (G-1) d²/18v
Specific gravity G=1, g=981cm²/Sec, viscosity v=8 * 10-3cm²/Sec
vs = 981* (2.65-1) *0.002*0.002/(18*.008) = 0.13cm/Sec.

7. In a rectangular horizontal flow tank, the maximum permissible velocity is ____ m/Sec.
a) 3
b) 0.3
c) 0.03
d) 0.003
Answer: b
Clarification: In the rectangular tank, the ideal conditions are that the flow is horizontal, velocity is uniform in all directions and the maximum permissible limit of velocity is .03m/Sec.

8. The time period for which the water is stored in a sedimentation tank is called _____
a) Time of flow
b) Frequency of flow
c) Settling time
d) Detention period
Answer: d
Clarification: Detention period is the time period at which the water is stored in a tank. It varies for different purposes like in the septic tank or oxidation pond, its value is higher.

9. Settling velocity of a spherical body in a viscous fluid is given by ___________
a) Reynolds law
b) Newton’s law
c) Stokes law
d) Charles law
Answer: c
Clarification: According to Stokes law, settling velocity of the particle is given by
vs=g (G-1) d²/18v where G is the specific gravity, d is the diameter of spherical particles and v is the viscosity.

10. In a fill and draw type sedimentation tank, a detention period of ____ hours is provided.
a) 6
b) 12
c) 18
d) 24
Answer: d
Clarification: In this tank, water is detained for 24 hours. Then cleaning is done for 6-12 hours, so, the whole cycle of operations takes around 30-36 hours.

11. Which of the following represents the correct relation between displacement velocity and diameter of the particle?
a) v1 = (8Bg (G-1) d/f) ^1/2
b) v1 = (8Bg (G-1) d2/f)²
c) v1 = (8Bg (G-1) d/f)
d) v1 = (8Bg (G-1) d/f)²
Answer: a
Clarification: Displacement velocity is given by v1 = (8Bg (G-1) d/f)^1/2 where B is a constant with value range 0.04-0.06, G is specific gravity, d is the diameter of the particle and f is the Darcy Weisbach friction factor.

Environmental Engineering Multiple Choice Questions on “Forms of Chlorination”.

1. The normal dose of chlorine during plain chlorination is ___________
a) 0.5-1ppm
b) 0.1-0.2ppm
c) 0.1-0.5ppm
d) 1-2ppm
Answer: a
Clarification: The normal dose of chlorine during plain chlorination is 0.5-1ppm. It should be applied when the water is clear and its turbidity does not exceed 30 ppm.

2. The normal dose of chlorine during pre chlorination is ___________
a) 0.5-1ppm
b) 0.1-0.2ppm
c) 0.1-0.5ppm
d) 1-2ppm
Answer: c
Clarification: The normal dose of chlorine during pre chlorination is 0.5-1ppm. This results in the oxidization of organic matter and reduce the amount of coagulants required in the water.

3. The normal dose of chlorine during post chlorination is ___________
a) 0.5-1ppm
b) 0.1-0.2ppm
c) 0.1-0.5ppm
d) 1-2ppm
Answer: b
Clarification: The normal dose of chlorine during post chlorination is 0.1-0.2ppm. It is used for protection against contamination.

4. In which form of chlorination, chlorine applies to raw water supply as it enters the distribution system?
a) Plain chlorination
b) Pre chlorination
c) Super chlorination
d) Double chlorination
Answer: a
Clarification: In plain chlorination, chlorine applies to raw water supply as it enters the distribution system. It checks the growth of organic matter, algae and remove taste and odor from water.

5. _________ is the application of chlorine before filtration.
a) Plain chlorination
b) Pre chlorination
c) Super chlorination
d) Double chlorination
Answer: b
Clarification: Pre chlorination is the application of chlorine before filtration. This results in a decrease of coagulant required in the water.

6. __________ is the application of chlorine at two or more points in the purification process.
a) Plain chlorination
b) Pre chlorination
c) Super chlorination
d) Double chlorination
Answer: d
Clarification: Double chlorination is the application of chlorine at two or more points in the purification process. It is applied just before when the water enters the sedimentation tank and after when it leaves the filter plant.

7. The normal dose of chlorine during break point chlorination is ____________
a) 0.5-1ppm
b) 0.1-0.2ppm
c) 3-7ppm
d) 1-2ppm
Answer: b
Clarification: The normal dose of chlorine during post chlorination is 0.1-0.2ppm. It kills bacteria and oxidizes the organic matter present in water.

8. Which of the following regarding break point chlorination is wrong?
a) It will have adequate chlorine residual
b) It will remove manganese
c) It will remove taste and odor
d) It is determined by physical test
Answer: d
Clarification: The break point chlorination is determined by laboratory test. It is represented by instantaneous yellow color if the orthotolidine test when the orthotolidine test is applied.

9. _______ is the application of chlorine beyond the stage of the break point.
a) Plain chlorination
b) Pre chlorination
c) Super chlorination
d) Double chlorination
Answer: c
Clarification: Super chlorination is the application of chlorine beyond the stage of the break point. It is followed by a contact period of 30-60 minutes.

10. The process of removal of excess chlorine from water is ____________
a) Plain chlorination
b) Dechlorination
c) Super chlorination
d) Double chlorination
Answer: b
Clarification: Dechlorination is the process of removal of excess chlorine from water. It can be done aeration or by the use of various chemicals like activated carbon, sodium sulfate in liquid form.

11. Which color is formed when orthotolidine is oxidized by chlorine?
a) Yellow
b) Orange
c) Green
d) Blue
Answer: a
Clarification: Orthotolidine is a colorless organic liquid that is oxidized into a yellow colored compound called Holoquinone.

Environmental Engineering Multiple Choice Questions on “Types of Pipes”.

1. Which of the following is incorrect regarding cast iron pipes?
a) They are non corrosive
b) They are durable
c) They can be produced by centrifugal process
d) Their length varies from 80-100cm
Answer: d
Clarification: Cast iron pipes are resistant to corrosion, has long life, can be produced by the centrifugal process and their length varies from 3-6m.

2. Horizontally cast iron pipes are called _________
a) Mcwane pipes
b) Pit cast pipes
c) Galvanized iron pipes
d) Wrought iron pipes
Answer: a
Clarification: The cast iron pipes which are horizontally cast are called Mcwane pipes while those that are vertically cast are called as Pit cast pipes.

3. Mcwane pipes are ___ stronger in tension than Pit cast pipes.
a) 20%
b) 30%
c) 50%
d) 100%
Answer: d
Clarification: Mcwane pipes are 100% stronger in tension and 50% stronger in rupture than Pit cast pipes.

4. The enlarged end of a cast iron pipe is called ________
a) Lead
b) Socket
c) Hemp
d) Spigot end
Answer: b
Clarification: The enlarged end of a cast iron pipe is called Socket. It accommodates the spigot end.

5. What is the quantity of lead required for a 15cm diameter cast iron pipe?
a) 5kg
b) 10kg
c) 15kg
d) 20kg
Answer: a
Clarification: The quantity of lead required depends on the diameter of cast iron pipe. For 15cm diameter cast iron pipe, it requires 5kg lead, whereas for 120 cm diameter pipe, it requires 55kg lead.

6. Which of the following joint is called run lead joint?
a) Tyton joint
b) Coupled joint
c) Spigot and socket joint
d) Flanged joint
Answer: c
Clarification: Spigot and socket joint is called run lead joint as it contains molten lead, which is heated at 4000C and applied at the joints.

7. In which of the following joint, pipes do not have sockets or spigots?
a) Tyton joint
b) Coupled joint
c) Spigot and socket joint
d) Expansion joint
Answer: b
Clarification: In Coupled joints, pipes do not have sockets or spigots. There are two types of coupling joints- dresser coupling and victaulic coupling joints. These joints are highly flexible.

8. Which of the following joint is highly influenced by temperature?
a) Tyton joint
b) Coupled joint
c) Spigot and socket joint
d) Expansion joint
Answer: d
Clarification: Expansion joints allow the expansion and contraction of the joint due to change in temperature. The space between the socket and spigot in this joint is filled with an elastic gasket.

9. Which of the following is the advantage of cast iron pipes?
a) They are subjected to tuberculation in certain water
b) They are fragile
c) They are heavier
d) They are resistant to corrosion.
Answer: d
Clarification: The cast iron pipes are subjected to tuberculation which reduces their carrying capacity, they are fragile, heavier, resulting in more cost when their diameter exceeds 120cm and they are resistant to corrosion.

10. Which grade of concrete is used in Reinforced concrete pipes?
a) M10
b) M15
c) M20
d) M25
Answer: d
Clarification: M20 grade of concrete is used for making Reinforced concrete pipes with longitudinal reinforcement equal to at least 0.25% of cross sectional area of concrete.

11. The pre-stressed concrete pipes are used for heads ____
a) <8m
b) <61m
c) >60m
d) <35m
Answer: c
Clarification: The pre-stressed concrete pipes are used for heads greater than 60m, plain cement concrete up to 7m and reinforced concrete up to 60m.

12. Which of the following is correct regarding Cement concrete pipes?
a) They are corrosive
b) The maintenance cost is high
c) Unreinforced pipes are liable to tensile crack
d) It is easy to repair them
Answer: c
Clarification: Cement concrete pipes are anti corrosive with low maintenance cost, difficult to repair and the unreinforced pipes are subjected to tensile cracks.

13. The tensile strength of Asbestos cement pipes is _____
a) 200kg/cm²
b) 300kg/cm²
c) 400kg/cm²
d) 600kg/cm²
Answer: a
Clarification: The tensile strength of Asbestos cement pipes is 200kg/cm² and its compressive strength is 600kg/cm².

14. Which of the following is a disadvantage of Asbestos cement pipe?
a) They are light weight
b) They have smooth internal surface
c) They are anti corrosive
d) They are soft and brittle.
Answer: d
Clarification: The disadvantage of Asbestos cement pipe is that it is soft and brittle, so they are weak under impact loading due to moving traffic.

15. Lead pipes are used in water supply mains.
a) True
b) False
Answer: b
Clarification: Lead pipes are used in service connection and in internal plumbing, but are not used in water supply mains due to the poisonous effect of lead.

Environmental Engineering Multiple Choice Questions on “Forces and Loads on Sewer”.

1. The magnitude of hoop stress on sewer is given __________
a) PD/2t
b) PD/t
c) 2PD/t
d) PD/2
Answer: a
Clarification: The magnitude of hoop stress on sewer is given by PD/2t where, P is the internal pressure, D is the diameter of sewer and t is the thickness of the shell.

2. In order to reduce temperature stress ______________ joints are provided.
a) Contraction
b) Expansion
c) Bell and spigot
d) Collar
Answer: b
Clarification: In order to reduce temperature stress, expansion joints are provided at intervals of 20m to 30m.

3. The expression for the temperature stress is given by __________
a) E_Pαt
b) E_Pαt/2
c) 2E_Pαt
d) E_P/αt
Answer: a
Clarification: The expression for the temperature stress is given by E_Pαt where, EP is the modulus of elasticity of pipe, α is the coefficient of thermal expansion and ‘t’ is the change in temperature.

4. The type of backfill is saturated clay. What is its unit weight?
a) 1040kg/m³
b) 2080kg/m³
c) 2000kg/m³
d) 1500kg/m³
Answer: b
Clarification: When the type of backfill is saturated clay, then its unit weight is 2080kg/m³.

5. Which of the following has the minimum unit weight?
a) Dry sand
b) Wet sand
c) Damp clay
d) Ordinary sand
Answer: a
Clarification: The unit weight of wet sand, damp clay, ordinary sand and dry sand are 1920kg/m³, 1920kg/m³, 1840kg/m³ and 1600kg/m³ respectively.

6. The type of soil is loose dry sand. What is the coefficient of cohesion of the soil?
a) 100
b) 200
c) 500
d) 0
Answer: d
Clarification: Loose dry sand is a cohesionless soil. So, the value of the coefficient of cohesion is 0.

7. Which of the following order is correct regarding coefficient of cohesion?
a) Hard clay < Dense sand > Medium clay > Soft clay
b) Hard clay > Dense sand > Medium clay > Soft clay
c) Medium clay > Dense sand > Hard clay > Soft clay
d) Dense sand > Medium clay > Hard clay > Soft clay
Answer: b
Clarification: The coefficient of cohesion of Hard clay, Dense sand, Medium clay and Soft clay are 4700kg/m², 1400kg/m², 1200kg/m² and 200kg/m² respectively.

8. How many types of underground sewer are there on the basis of construction conditions?
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: Depending upon the type of the construction conditions, underground sewers are of 3 types, namely Trench ditch condition, Embankment condition and Tunnel condition.

9. _______ condition exists when the sewer is laid at a depth between 9m to 12m.
a) Trench ditch
b) Trench
c) Tunnel
d) Embankment
Answer: c
Clarification: Tunnel condition exists when the sewer is laid at a depth between 9m to 12m. This process is done by jacking or by tunneling.

10. Negative projecting condition is a type of ______ condition.
a) Trench ditch
b) Trench
c) Tunnel
d) Embankment
Answer: d
Clarification: Negative projecting condition exists when the trench is narrower with respect to the size of the pipe and shallow with respect to the depth of cover.

Environmental Engineering Multiple Choice Questions on “Design of Sedimentation Tank”.

1. Which of the following cannot be the velocity of flow in the horizontal flow sedimentation tank?
a) 12cm/min
b) 39cm/min
c) 5cm/min
d) 20cm/min
Answer: b
Clarification: The maximum value of the velocity of flow is 30cm/min, so it cannot be 39cm/min.

2. The deviation from the ideal flow of wastewater in the sedimentation tank causes ____________
a) Whirlpool
b) Increase in tank efficiency
c) Short circuiting
d) Coagulation
Answer: c
Clarification: The deviation from the ideal flow of wastewater in the sedimentation tank causes short circuiting. It results in variation in the detention period in the top and bottom layer of wastewater in the tank.

3. The overflow rate for a plain sedimentation tank is ______ litre/day/metre².
a) 50000
b) 40000
c) 12000
d) 30000
Answer: c
Clarification: The overflow rate for a plain sedimentation tank lies between 12000-18000 litre/day/metre².

4. The detention period for a sedimentation tank where coagulation takes place is ____________
a) 30 seconds
b) 1 minute
c) 1hour
d) 2hour
Answer: d
Clarification: The detention period for a sedimentation tank without coagulation is 4 hours, whereas with coagulation is 2 – 2.5 hours.

5. The test for finding the optimum coagulant dose in water is ____________
a) Chromatography
b) Jar test
c) Kjeldahl test
d) Dilatancy test
Answer: b
Clarification: Jar test is used to find the optimum coagulant concentration that is required to cause coagulation in water so that the settling of suspended impurities takes place.

6. The average time required by water to pass through the settling tank is called _______
a) Detention time
b) Flowing through period
c) Time of flow
d) Mean time
Answer: b
Clarification: The average time required by water to pass through the settling tank is called ‘flowing through period’ which is always less than the detention period.

7. The quantity of water flowing per hour per unit horizontal area is called ____
a) Detention time
b) Flowing through period
c) Displacement time
d) Overflow rate
Answer: d
Clarification: Overflow rate = Q/A where Q is discharge rate in litre per day, A is the horizontal area in m².

8. In a rectangular horizontal flow tank, the maximum permissible velocity is ____ m/Sec.
a) 3
b) 0.3
c) 0.03
d) 0.003
Answer: b
Clarification: In the rectangular tank, the ideal conditions are that the flow is horizontal, velocity is uniform in all directions and the maximum permissible limit of velocity is 0.03m/Sec.