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All questions of Environmental Engineering for Civil Engineering (CE) Exam

The ultimate BOD value of a waste -
  • a)
    increase with temperature.
  • b)
    decreases with temperature.
  • c)
    remains the same at all temperatures.
  • d)
    doubles with every 10°c rise in temperature.
Correct answer is option 'C'. Can you explain this answer?

Puja Sharma answered
The ultimate BOD value of waste is the amount of biodegradable organic matter presents in the sewage sample. At any temperature the amount of organic matter present does not change, so the ultimate BOD will remain same for any temperature.
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The maximum efficiency of BOD removal is achieved in which of the following?
  • a)
    Oxidation ditch
  • b)
    Oxidation ponds
  • c)
    Aerated lagoons
  • d)
    Trickling filter
Correct answer is option 'A'. Can you explain this answer?

Anirudh Kulkarni answered
The main advantage of oxidation ditch is the ability to achieve removal performance objective with low operational requirements and eminence costs.An oxidation ditch is a modified activated sludge biological treatment process that uses long solids retention times (SRTs) to remove biodegradable organics. The typical oxidation ditch is equipped with aeration rotors or brushes that provide aeration and circulation. The wastewater moves through the ditch at 1 to 2ft/s. The ditch may be designed for continuous or intermittent operation. Because of this feature, this process may be adaptable to the fluctuations in flows and loadings associated with recreation area wastewater production. Several manufacturers have developed modifications to the oxidation ditch design to remove nutrients in conditions cycled or phased between the anoxic and aerobic states.
Oxidation ponds, also called lagoons or stabilization ponds, are large, shallow ponds designed to treat wastewater through the interaction of sunlight bacteria, and algae. Algae grow using energy from the sun and carbon dioxide and inorganic compounds released by bacteria in water. During the process of photosynthesis, the algae release oxygen needed by aerobic bacteria. Mechanical aerators are sometimes installed to supply yet more oxygen, thereby reducing the required size of the pond. Sludge deposits in the pond must eventually be removed by dredging. Algae remaining in the pond effluent can be removed by filtration or by a combination of chemical treatment and settling.

A water sample analysis produces alkalinity and total hardness of 200 mg/L and 250 mg/L as CaCO3 respectively. The carbonate and non-carbonate hardness (as CaCO3) will respectively be
  • a)
    200 mg/L and 50 mg/L
  • b)
    200 mg/L and 0 mg/L
  • c)
    250 mg/L and 50 mg/L
  • d)
    250 mg/L and 0 mg/L
Correct answer is option 'A'. Can you explain this answer?

Lavanya Menon answered
Concept:
Carbonate hardness = minimum of {Alkalinity, Total Hardness}
Total hardness = Carbonate Hardness + Non-carbonate Hardness.
Calculation:
Carbonate Hardness= minimum of {200 mg/L , 250 mg/L}
Carbonate Hardness = 200 mg/L as CaCO3
Non-Carbonate Hardness = Total Hardness – Carbonate Hardness
= 250 – 200
= 50 mg/L ​as CaCO3

Two primary air pollutants are
  • a)
    Sulfur oxide and ozone
  • b)
    Nitrogen oxide and Peroxyacetylnitrate
  • c)
    Sulphur oxide and hydrocarbon
  • d)
    Ozone and Peroxyacetylnitrate
Correct answer is option 'C'. Can you explain this answer?

Aarav Chauhan answered
Sulphur dioxide carbon monoxide, nitrogen oxides, lead, hydrocarbons, allergic agents like pollens and spores and radioactive substances are primary pollutants. Sulphuric acid, ozone, formaldehydes & peroxyacylnitrates (PAN) are secondary pollutants.

The design parameter for flocculation is given by a dimensionless number Gt, where G is the velocity gradient and t is the detention time. Values of Gt ranging from 104 to 105 are commonly used, with t ranging from 10 to 30 min. the most preferred combination of G and t to produce smaller and denser flocs is
  • a)
    Large G values with short t
  • b)
    Large G values with long t
  • c)
    Small G values with short t
  • d)
    Small G values with long t
Correct answer is option 'A'. Can you explain this answer?

Lakshmi Datta answered
Large G values with short times tend to produce small, dense flocs, while low G values and long times produce large, lighter flocs. Since large, dense flocs are more easily removed in the settling basin, it may be advantageous to vary the G values over the length of the flocculation basin. The small dense flocs produced at high G values subsequently combine into larger flocs at the lower G values. Reduction in G values by a factor of 2 from the influent end to the effluent end of the flocculator has been shown to be effective.

A channel-type grit chamber has a flow-through velocity of 0.15 m/s, a depth of 0.8 m and a length of 10 m. For inorganic particles with the specific gravity of 2.7, and considering kinematic viscosity of 8×10-7 m2/s the largest diameter that can be removed with 95% efficiency is
  • a)
    0.0452 mm
  • b)
    0.0786 mm
  • c)
    0.0873 mm
  • d)
    0.0992 mm
Correct answer is option 'D'. Can you explain this answer?

Bhaskar Unni answered
The settling velocity can be determined through the equation,

where H = depth of the tank, L = length of the tank, Vs  = Settling velocity, and VH = Flow through velocity

Now, using Stokes law,

 
where G = specific gravity of the particle, ν =kinematic viscosity of the particle, g = acceleration due to gravity, d = diameter of the particle
For 95% removal efficiency,

d = 9.92 × 10−5m
d = 0.0992.mm

Which of the following engineering system for air pollution control device can collect a particle size of less than one micrometer and can also have an efficiency as high as 99%
  • a)
    Centrifugal Collector
  • b)
    Electrostatic Precipitator
  • c)
    Fabric Filtration
  • d)
    Cyclonic Wet Collector
Correct answer is option 'C'. Can you explain this answer?

Sagnik Sen answered


Fabric Filtration

Fabric filtration is an engineering system for air pollution control that can collect particles as small as less than one micrometer in size with an efficiency as high as 99%. Here's why fabric filtration is the correct choice for this scenario:

Efficiency
- Fabric filtration systems are known for their high efficiency in capturing particles of various sizes, including ultra-fine particles smaller than one micrometer.
- The fabric filters have a high collection efficiency, often reaching up to 99% or more, making them extremely effective at removing pollutants from the air.

Particle Size Collection
- Fabric filters are designed with a dense network of fibers that can capture particles as small as less than one micrometer.
- The small spaces between the fibers create a barrier that effectively traps even the tiniest particles, ensuring thorough air filtration.

Versatility
- Fabric filtration systems can be customized to suit different air pollution control needs and can be used in various industries and applications.
- They can handle a wide range of particle sizes and can be designed to target specific pollutants based on the requirements of the air quality standards.

Conclusion
In conclusion, fabric filtration is a reliable engineering system for air pollution control that can effectively collect particles smaller than one micrometer with a high efficiency of up to 99%. Its versatility, efficiency, and ability to capture ultra-fine particles make it a suitable choice for addressing air pollution challenges.

Pollution potential of domestic sewage generated in a town and its industrial sewage can be compared with reference to
  • a)
    their BOD value
  • b)
    population equivalent
  • c)
    their volume
  • d)
    the relative density
Correct answer is option 'B'. Can you explain this answer?

Jyoti Choudhury answered
Understanding Pollution Potential of Sewage
When comparing the pollution potential of domestic sewage generated in a town to that of industrial sewage, using the population equivalent is the most effective approach. Here's why:
Population Equivalent Explained
- The population equivalent (PE) is a measure that equates the pollutant load of industrial sewage to the load produced by a certain number of people in a community.
- It provides a standardized way to assess the impact of different types of sewage on the environment, allowing for a more accurate comparison.
Why Not Other Options?
- BOD Value: While Biochemical Oxygen Demand (BOD) is important for measuring organic pollutants, it does not account for the volume or specific characteristics of industrial effluents.
- Volume: The volume of sewage does not directly correlate with its pollution potential. Different types of sewage can have vastly different pollutant concentrations, rendering volume alone insufficient for comparison.
- Relative Density: This factor mainly addresses the physical characteristics of sewage rather than its biological or chemical potential to cause pollution.
Conclusion
Using population equivalent allows for a comprehensive and comparable assessment of pollution potential from domestic and industrial sources. It incorporates both the quantity and quality of sewage, making it a crucial metric for environmental engineers and policymakers.

The ultimate BOD of wastewater is 300 mg/L and the reaction rate constant (to the base ‘e’) at 20°C is 0.3585/day, then the 5 days BOD at 20oC will be.
  • a)
    175 mg/L
  • b)
    50 mg/L
  • c)
    250 mg/L
  • d)
    200 mg/L
Correct answer is option 'C'. Can you explain this answer?

Sarthak Kulkarni answered
Concept:
Ultimate BOD = Lo
BOD5­ = 5day BOD, of water/waste water
BOD at any time ‘t’ is (Lt)
Lt = Lo(1 − e−kDt)
Where Lo = ultimate BOD
KD = Rate constant
Calculation:
Lt = 300 × (1 – e-0.3585 × t)
L5 = BOD5 = 300 × (1 – e-0.3585 × 5)
= 250 mg/L

Which sewer is having the maximum fluctuations in sewer discharge?
  • a)
    Outfall Sewer
  • b)
    Branch sewer
  • c)
    Lateral sewer
  • d)
    Main Sewer
Correct answer is option 'C'. Can you explain this answer?

Nitya Nambiar answered
The fluctuation in sewage discharge is found to be maximum in the lateral sewer.
The fluctuation in sewage discharge reduced from lateral sewer to branch sewer, from branch sewer to the main sewer and from the main sewer to outfall sewer. An example of the respective fluctuation values are given below:

For water, the dosage at a breakpoint is 1.5 mg/L and residual chlorine at that time is found to be 0.3 mg/L. If cumulative chlorine added is 2 mg/L, the total residual chlorine will be –
  • a)
    1.2 mg/L
  • b)
    0.8 mg/L
  • c)
    0.5 mg/L
  • d)
    Zero
Correct answer is option 'B'. Can you explain this answer?

Pallabi Tiwari answered
Break point chlorination = Chlorine demand = 1.5 mg/L 
Combined Cl2 residual at that time = 0.3 mg/L
Chlorine dosage = 2 mg/L
∴ Free chlorine residual = Dose – demand
= 2 – 1.5
= 0.5 mg/L
∴ Total chlorine residual = Free chlorine residual + combined chlorine residual
= 0.5 + 0.3
= 0.8 mg/L

For the Sewer of size 550 mm, it is designed to run “x” times full. The value of x is
  • a)
    1/2
  • b)
    3/4
  • c)
    3/2
  • d)
    2/3
Correct answer is option 'D'. Can you explain this answer?

Partho Jain answered
Sewers are never designed to run at full depth. If the size of the Sewer is less than 400 mm, these are designed to run half full. If the Sewer size is in the range of 400 to 900 mm, these are designed to run 2/3rd full. When the size of Sewer is more than 900 mm, these are designed for 3/4th full.

Match the following:-
​
  • a)
    1 - Q, 2 - P, 3 - S, 4 - R
  • b)
    1 - Q, 2 - P, 3 - R, 4 - S
  • c)
    1 - R, 2 - Q, 3 - P, 4 - S
  • d)
    1 - R, 2 - Q, 3 - S, 4 - P
Correct answer is option 'B'. Can you explain this answer?

Diya Dasgupta answered
Free Ammonia – indicate recent pollution
Nitrate – indicate old pollution (fully oxidized)
Nitrite – indicate partly decomposed condition
Organic – indicate quantity of nitrogen before decomposition started.

A large stream has a reoxygenation constant of 0.45 per day and at a point at which organic pollutant is discharged, it is saturated with oxygen at 12 mg/L (Do = 0). Below the outfall, the ultimate demand for oxygen is found to be 19 mg/l and the Deoxygenation Constant is 0.2 per day. Calculate, the D.O at 36 km downstream Assume velocity of the stream to be 1.2 m/s.
  • a)
    10 mg/l
  • b)
    16.65 mg/l
  • c)
    6.65 mg/l
  • d)
    9.65 mg/l
Correct answer is option 'D'. Can you explain this answer?

Shilpa Pillai answered
Given data:
Reoxygenation constant (k1) = 0.45 per day
Initial dissolved oxygen (Do) = 0 mg/L
Ultimate demand for oxygen (U) = 19 mg/L
Deoxygenation constant (k2) = 0.2 per day
Distance downstream (d) = 36 km = 36,000 m
Velocity of the stream (v) = 1.2 m/s

To calculate the dissolved oxygen (DO) at 36 km downstream, we need to consider the reoxygenation and deoxygenation processes in the stream.

1. Reoxygenation Process:
The reoxygenation process replenishes the dissolved oxygen in the stream. The rate of reoxygenation is given by the equation:
dDO/dt = k1 * (U - DO)
where dDO/dt is the rate of change of dissolved oxygen with respect to time.

2. Deoxygenation Process:
The deoxygenation process consumes the dissolved oxygen in the stream. The rate of deoxygenation is given by the equation:
dDO/dt = -k2 * DO

Since the stream is saturated with oxygen at the point of discharge (Do = 0), we can assume the deoxygenation process dominates initially.

3. Calculation of DO at 36 km downstream:
We can solve the differential equation for the deoxygenation process to find the dissolved oxygen at any point downstream.
Integrating the equation, we get:
ln(DO) = -k2 * t + C
where C is the integration constant.

Using the initial condition at the point of discharge (Do = 0), we can determine the integration constant C:
ln(0) = -k2 * 0 + C
C = ln(0)
C = -∞ (negative infinity)

Now, let's substitute the values and calculate the dissolved oxygen at 36 km downstream:
ln(DO) = -k2 * t - ∞
ln(DO) = -k2 * t

Solving for DO:
DO = e^(-k2 * t)

Substituting the values:
DO = e^(-0.2 * 36,000 / 1.2)

Calculating the value:
DO ≈ 9.65 mg/L

Therefore, the dissolved oxygen at 36 km downstream is approximately 9.65 mg/L.
Hence, the correct answer is option 'D'.

The efficiency of sediment removal in a continuous flow sedimentation tank does not depend upon-
  • a)
    depth of the tank.
  • b)
    discharge through the tank.
  • c)
    length of the tank.
  • d)
    width of the tank.
Correct answer is option 'A'. Can you explain this answer?

Pallabi Tiwari answered
For particle removal, setting velocity > Surface loading rate.
Settling velocity is depended upon the particle size & SLR is on the surface area which depends on length and width.
So efficiency of particle removal does not depend upon the depth of the tank.

Arrange the given lakes in the descending order of their productivity level
Mesotrophic Lakes
Oligotrophic Lakes
Eutrophic Lakes
  • a)
    Mesotrophic Lakes, Oligotrophic Lakes and Eutrophic Lakes
  • b)
    Eutrophic Lakes, Mesotrophic Lakes and Oligotrophic Lakes
  • c)
    Oligotrophic Lakes, Eutrophic Lakes and Mesotrophic Lakes
  • d)
    Oligotrophic Lakes, Mesotrophic Lakes and Eutrophic Lakes
Correct answer is option 'B'. Can you explain this answer?

Manasa Sen answered
Eutrophic Lakes: These are the lakes of high productivity level, hence support high growth of algae.
Mesotrophic Lakes: These are the lakes of intermediate productivity level, hence support intermediate growth of algae. 
Oligotrophic Lakes: These are the lakes of least productivity level, hence support negligible growth of algae.

An air parcel having 40° temperature moves from ground level to 500m elevation in dry air following the “adiabatic lapse rate”. The temperature of air parcel at 500 m elevation will be
  • a)
    35°C
  • b)
    38°C
  • c)
    41°C
  • d)
    44°C
Correct answer is option 'A'. Can you explain this answer?

Ashwin Desai answered
Dry air cools at the rate of 9.8°C per km and it is called dry adiabatic lapse rate. In saturated (wet) air, this rate is calculated to be 6°C per km and is known as wet adiabatic lapse rate.
Resulting temperature of air
= 40 − 9.8 × 500/1000 = 40 − 4.9 = 35.1∘C

Two pipe system of providing building drainage consists of
  • a)
    One solid pipe + one waste pipe + one vent pipe + one sullage pipe
  • b)
    One soil pipe + one waste pipe + two vent pipes
  • c)
    Two soil pipes + two waste pipes
  • d)
    Two soil pipes only
Correct answer is option 'B'. Can you explain this answer?

Subhankar Khanna answered
In this system, two sets of vertical pipes are laid, i.e., one for draining night soil, and the other for draining sullage. The pipes of the first set carrying night soil are called soil pipes, and the pipes of the second set carrying sullage from baths etc., are called sullage pipes or waste pipes. The soil pipe as well as the waste pipe, are separately ventilated by providing separate vent pipes or antisiphonage pipes.

Calculate the minimum velocity (m/sec) required to transport Coarse sand through a sewer of 35 cm diameter with sand particles of 0.97 mm diameter and specific gravity 2.59, and organic matter of 4.75 mm average size with a specific gravity of 1.42. The friction factor for the sewer material may be assumed as 0.04 and roughness coefficient of 0.013. Consider k = 0.05 for inorganic solids and 0.07 for organic solids.
    Correct answer is between '0.48,0.56'. Can you explain this answer?

    Arya Menon answered

    Where Vsc = self cleaning velocity or minimum velocity in m/sec
    k = constant depend upon type of Particle present in the Sewage
    f = friction factor
    d = size of particle under consideration(m)
    For inorganic solid i.e. Coarse sand
    VSC1 = 0.389 m/sec
    For organic solid

     
    VSC2 = 0.523 m/s
    Minimum velocity required will be maximum of VSC1 and VSC2
    So, VSC = 0.523 m/sec

    The effective size of the particle (D10) of filter [slow sand filter] is
    • a)
      0.25 mm
    • b)
      0.50 mm
    • c)
      0.11 mm
    • d)
      0.35 mm
    Correct answer is option 'A'. Can you explain this answer?

    Akanksha Mehta answered
    As per Government of India’s Effective size (D10) of filter for slow sand filter is in the range of (0.2 to 0.3) mm and Coefficient of uniformity for this filter media is in the range of (3 – 5)

    Arrange the layers of the atmosphere in order of increasing heights from the surface of the earth 
    • a)
      Mesosphere, Stratosphere, Thermosphere, and Troposphere
    • b)
      Stratosphere, Troposphere, Mesosphere, Thermosphere
    • c)
      Troposphere, Stratosphere, Mesosphere, Thermosphere
    • d)
      Troposphere, Mesosphere, Stratosphere, Thermosphere
    Correct answer is option 'C'. Can you explain this answer?

    Bhaskar Rane answered
    The Layers of atmosphere in order of increasing heights form the surge of earth
    a) Troposphere_____ 0 to 11 km _____ 15 to -56°C
    b) Stratosphere _____ 11 to 50 km _____ 15 to 0.2°C
    c) Mesosphere _____ 50 to 85 km _____ -02° to -92°C
    d) Thermosphere _____ 85 to 500 km _____ -92° to 1200°C

    Answer using the codes given below the list:-
    • a)
      1 – P, 2 – Q , 3 - S , 4 - R
    • b)
      1 - P, 2 – Q , 3 - R , 4 - S
    • c)
      1 - R, 2 –S  , 3 - P , 4 - Q
    • d)
      1 – R , 2 – Q , 3 – S , 4 - P
    Correct answer is option 'C'. Can you explain this answer?

    Kavya Mehta answered
    1) Measurement suspended solid is done by a gravimetric test involving the mass of residual measurement.
    2) Measurement of Turbidity: -
    A) Based on absorption principles
    i) Turbidity Rod
    ii) Jackson’s Turbidity meter
    iii) Baylis Turbidimeter
    B) Based on the scattering principle
    i) Modern Nephelometer
    3) Through Nessler’s tubes on Burgess scale, true colour is measured.

    Chapter doubts & questions for Environmental Engineering - RRB JE for Civil Engineering 2025 is part of Civil Engineering (CE) exam preparation. The chapters have been prepared according to the Civil Engineering (CE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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