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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

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.

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)

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.

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:

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

    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?

    Telecom Tuners answered
    - The pollution potential of sewage is often measured using the concept of "population equivalent."
    - Population equivalent reflects the organic pollution load that domestic sewage would impose if it were discharged untreated.
    - It provides a comparative scale by equating the pollution load of industrial sewage to a certain number of people.
    - This makes it easier to compare different sewage types, as it standardizes the pollution potential and facilitates planning and treatment processes.
    - Hence, the correct answer is B: population equivalent.

    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.

    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.

    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

    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

    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

    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?

    Bhargavi Sarkar answered
    Understanding Sewer Design Capacity
    In civil engineering, the design capacity of a sewer system is crucial for efficient wastewater management. The sewer size of 550 mm indicates its diameter, affecting its volume and flow characteristics.
    Concept of "x" times full
    The term “x times full” refers to the flow capacity that the sewer is designed to handle without risking overflow or blockage. The value of "x" is determined based on various factors including:
    - Peak Flow Rates: Anticipating maximum wastewater discharge during busy periods.
    - Sewer System Characteristics: Design considerations such as slope, material, and intended use affect flow dynamics.
    - Regulatory Standards: Compliance with hydraulic standards and guidelines.
    Choosing the Correct Value of "x"
    The value of "x" can vary based on specific design criteria. For a sewer of size 550 mm, the correct answer being 2/3 suggests:
    - Hydraulic Efficiency: A sewer operating at 2/3 full ensures optimal flow velocity, reducing sedimentation and potential blockages.
    - Safety Margin: This value provides a buffer for peak flows, allowing the system to handle unexpected surges without overflow.
    - Design Practice: It aligns with established engineering norms where sewer systems are often designed to operate below full capacity for reliability.
    Conclusion
    In summary, the value of "x" being 2/3 for a 550 mm sewer size reflects a careful balance between capacity, safety, and efficiency, ensuring the sewer system functions effectively under varying flow conditions.

    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

    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.

    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.

    The technique utilized for the disposal of biomedical waste​, in which it is in contact with steam under controlled pressure and temperature condition with the end goal to complete sterilization is
    • a)
      Thermal Pyrolysis
    • b)
      Shredding
    • c)
      Autoclaving
    • d)
      Pulverization
    Correct answer is option 'C'. Can you explain this answer?

    Nilanjan Chawla answered
    Autoclaving:- The method used for the disposal of biomedical waste is , in which it is brought in closed contact with steam under controlled pressure and temperature condition in order to carry out disinfection.
    Pulverization and Shredding: Basically this is not a method of disposal but in these method heavier solids are broken into the lighter one either by cutting action or by grinding action.
    Pulverization refers to the action of crushing and grinding, whereas shredding refers to the action of cutting and tearing.
    Pyrolysis: is a method of disposal of the refuse in which burning of refuse is generally done at high temperature in the almost absence of oxygen.

    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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