Sewage MCQ Quiz - Objective Question with Answer for Sewage - Download Free PDF

Last updated on May 15, 2025

Latest Sewage MCQ Objective Questions

Sewage Question 1:

If 1% solution of a sewage sample is incubated for 5 days at 20°C and depletion of oxygen was found to be 3 ppm, BOD of the sewage is:

  1. 225 ppm
  2. 200 ppm
  3. 300 ppm
  4. 250 ppm
  5. 600 ppm

Answer (Detailed Solution Below)

Option 3 : 300 ppm

Sewage Question 1 Detailed Solution

Concept:

Biological Oxygen Demand(BOD):

Biological Oxygen Demand or Biochemical oxygen demand (BOD) is the amount of dissolved oxygen used by microorganisms to break down organic matter in water.

A low BOD is an indicator of good quality water, while a high BOD indicates polluted water.

BOD = ( DOi - DFf ) × Dilution factor

Where,

BOD = Biochemical oxygen demand in ppm or mg/lit

DOi = Initial dissolved oxygen in mg/lit.

DOf =Final dissolved oxygen in mg/lit.

Dilution factor \(= \frac{{Volume\ of\ the\ diluted\ sample}}{{{\rm{Volume \ of\ the\ undiluted\ sewage\ sample}}}} \)

Calculation:

Given:

Dilution factor \(= \frac{{100}}{{{\rm{\% \;solution}}}} = \frac{{100}}{1} = 100\)

Depletion of oxygen = 3 ppm

BOD5 = Depletion of oxygen × Dilution factor

BOD= 3 × 100 = 300 ppm

∴ The BOD of the sewage is 300 ppm.

Sewage Question 2:

Imhoff tank results in ______ removal of solids and _______ removal of BOD.

  1. 30 - 40 %, 60 - 65 %
  2. 60 - 65 %, ​ 30 - 40 %
  3.  60 - 70 %, 98 - 99 %
  4. 20 - 30 %, 60 - 70 %
  5. 30 - 40 %, ​ 10 - 20 %

Answer (Detailed Solution Below)

Option 2 : 60 - 65 %, ​ 30 - 40 %

Sewage Question 2 Detailed Solution

Explanation:

Imhoff tank results in 60 - 65 % removal of solids and 30 - 40 % removal of BOD.

Imhoff tank:

  • An Imhoff tank is an improvement over the septic tank in which the incoming sewage is not allowed to get mixed up with the sludge produced and the outgoing effluent is not allowed to carry with it a large amount of organic load, as in the case of the septic tank.
  • Imhoff tanks combine the advantages of both the septic as well as sedimentation tanks.
  • An Imhoff tank is two storey tank in which the upper chamber served as sedimentation (aerobic condition) and the lower chamber served as digestion chamber (anaerobic condition).
  • They find use in case of small treatment plants requiring only primary treatment.
  • They are quite economical and do not require skilled attention during operations.
  • The result obtained is quite good, with 60 to 65 % removal of solids and 30 to 40 % removal of BOD.
  • There is no problem of sludge disposal, as in the case of sedimentation tan

Sewage Question 3:

Waste stabilization ponds can be:-

  1. Aerobic
  2. Anaerobic
  3. Facultative
  4. Any one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 4 : Any one of the above

Sewage Question 3 Detailed Solution

Wastewater Stabilization Ponds (WSPs)/Oxidation pond:

It is 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.

The ponds can be used individually or linked in a series for improved treatment. There are three types of ponds, (1) anaerobic, (2) facultative, and (3) aerobic (maturation), each with different treatment and design characteristics.

RRB JE CE R 24 14Q Environmental Engineering Subject Test Part 2(Hindi) Nitesh Madhushri D3

Sewage Question 4:

P. Trickling filter consists of a bed of crushed stone, gravel or slag of relatively large size.

Q. As sewage passes through the filtering media of the trickling filter, an organic film is formed around the particles of filtering media.

Identify the correct option.

  1. P is correct and Q is incorrect
  2. P is incorrect and Q is correct
  3. Both are incorrect
  4. Both are correct
  5. Insufficient Information

Answer (Detailed Solution Below)

Option 4 : Both are correct

Sewage Question 4 Detailed Solution

Concept:

Trickling Filters:

  • A trickling filter is an attached growth process i.e. process in which microorganisms responsible for treatment are attached to inert packing material.
  • The packing material used in attached growth processes includes rock, gravel, slag, sand, redwood, and a wide range of plastic and other synthetic materials.

F1 A.M Madhu 14.04.20 D12

Figure: Trickling Filter

The process of trickling filter is as follows:

  • The wastewater in the trickling filter is distributed over the top area of a vessel containing non-submerged packing material.
  • Air circulation in the void space, by either natural draft or blowers, provides oxygen for the microorganisms growing as an attached biofilm.
  • During operation, the organic material present in the wastewater is metabolized by the biomass attached to the medium. The biological slime grows in thickness as the organic matter abstracted from the flowing wastewater is synthesized into new cellular material.
  • The thickness of the aerobic layer is limited by the depth of penetration of oxygen into the microbial layer.
  • The micro-organisms near the medium face enter the endogenous phase as the substrate is metabolized before it can reach the micro-organisms near the medium face as a result of increased thickness of the slime layer and lose their ability to cling to the media surface.
  • The liquid then washes the slime off the medium and a new slime layer starts to grow. This phenomenon of losing the slime layer is called sloughing.
  • The sloughed-off film and treated wastewater are collected by an under drainage which also allows circulation of air through the filter.
  • The collected liquid is passed to a settling tank used for solid-liquid separation.

Sewage Question 5:

Which of the following statements is/are correct in relation to sedimentation with coagulation?

1. Coagulation is usually found necessary for water having turbidity greater than 30-50 ppm.

2. Alum produces heavy floc compared with iron salt coagulants.

3. It is difficult to dewater the sludge formed by alum coagulant.

  1. Statements 1 and 2 are correct, while statement 3 is incorrect.
  2. Statement 1 is correct, while statements 2 and 3 are incorrect. 
  3. Statement 1 is incorrect, while statements 2 and 3 are correct.
  4. Statements 1 and 3 are correct, while statement 2 is incorrect.

Answer (Detailed Solution Below)

Option 4 : Statements 1 and 3 are correct, while statement 2 is incorrect.

Sewage Question 5 Detailed Solution

Explanation:

Sedimentation with Coagulation

Sedimentation with coagulation is a water treatment process used to remove suspended solids from water by adding a coagulant to form flocs, which then settle out of the water.

Analyzing the Given Statements

  1. Statement 1: "Coagulation is usually found necessary for water having turbidity greater than 30-50 ppm." (Correct)

    • Coagulation is typically required for water with higher turbidity levels to effectively remove suspended particles.

  2. Statement 2: "Alum produces heavy floc compared with iron salt coagulants." (Incorrect)

    • Iron salt coagulants tend to produce heavier and more compact flocs compared to alum.

  3. Statement 3: "It is difficult to dewater the sludge formed by alum coagulant." (Correct)

    • The sludge formed by alum coagulant has a gelatinous nature, making it more challenging to dewater compared to other coagulants.

Top Sewage MCQ Objective Questions

Sewers must be checked for minimum velocities at their minimum hourly flows which is equal to:

  1. 1/3 average daily flows
  2. 1/4 average daily flows
  3. average daily flows
  4. 1/2 average daily flows

Answer (Detailed Solution Below)

Option 1 : 1/3 average daily flows

Sewage Question 6 Detailed Solution

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Concept:

The peak flow can be considered as 1.5 times the annual average daily flow.

For a design of the treatment facility, the peak factor is considered as 1.5 times the annual average daily flow.

The minimum flow passing through sewers is also important to develop self-cleansing velocity to avoid silting in sewers. This flow will generate in the sewers during late night hours. The effect of this flow is more pronounced on lateral sewers than the main sewers.

Sewers must be checked for minimum velocity as follows:

Minimum daily flow = 2/3 Annual average daily flow

Minimum hourly flow = 1/2 Minimum daily flow

Minimum hourly flow = 1/3 Annual average daily flow

The activated sludge process is an

  1. Aerobic attached growth system
  2. Anaerobic attached growth system
  3. Anaerobic suspended growth system
  4. Aerobic suspended system

Answer (Detailed Solution Below)

Option 4 : Aerobic suspended system

Sewage Question 7 Detailed Solution

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Explanation:

Activated sludge process:

  • The essential features of the activated sludge process are an aeration stage, solids-liquid separation following aeration, and a sludge recycle system.
  • Wastewater after primary treatment enters an aeration tank where the organic matter is brought into intimate contact with the sludge from the secondary clarifier.
  • It requires less space, does not produce obnoxious odor, and requires less time for wastewater treatment.
  • It requires skilled supervision


Followings are the classification of secondary treatment units:

Method

Contact Mechanism

Decomposition

Trickling filter

Attached growth

Aerobic

Rotating biological contactor

Attached growth

Aerobic

Activated sludge process

 

Suspended growth

Aerobic

Oxidation pond

Suspended growth

Aerobic

Septic tank

Suspended growth

Anaerobic

Imhoff tank

Suspended growth

Anaerobic

Match List - I with List-II and select the correct answer using the codes given below the lists:

List - I

(Treatment Unit)

List - II

(Detention Time)

A. Grit chamber

i. Six hours

B. Primary sedimentation

ii. Two minutes

C. Activated sludge

iii. Two hours

D. Sludge digestion

iv. Twenty days

  1. A - iii, B - i, C - iv, D - ii
  2. A - ii, B - iii, C - i, D - iv
  3. A - ii, B - i, C - iii, D - iv
  4. A - i, B - ii, C - iii, D - iv

Answer (Detailed Solution Below)

Option 2 : A - ii, B - iii, C - i, D - iv

Sewage Question 8 Detailed Solution

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Explanation:

Detention time for different type of Treatment unit

Treatment Unit

Detention time

Grit chamber

30-60 second

Primary sedimentation

2-2.5 hour

Sludge digestion

20-30 days

Activated sludge process

2-4 hour

Oxidation pond

2-6 weeks

Septic tank

12-36 hour

 

Note: Septic tank has high detention time whereas Grit chamber has less detention time.

What will be the Sludge Volume Index (SVI) (ml/gram) if 100 ml of sludge collected in 30 mins on drying weighs 800 mg?

  1. 125
  2. 8
  3. 0.008
  4. 0.125

Answer (Detailed Solution Below)

Option 1 : 125

Sewage Question 9 Detailed Solution

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Concept:

Sludge volume index is the volume occupied in mL by one gm of solids in the mixed liquor after settling for 30 minutes.

The recommended value of the Sludge Volume Index (SVI) for municipal sewage is approximately lying in the range of 80 to 150.

\({\rm{Sludge\;volume\;index}} = \frac{{{\rm{Volume\;of\;settled\;sludge\;in\;ml}}}}{{{\rm{MLSS\;present}}}}\)

Calculation:

Given,

Volume of settled sludge = 100 ml, Drying weight (MLSS) = 800 mg

\(\rm SVI = \frac{{100}}{{800 \ \times \ {{10}^{ - 3}}}}\ ml/gram\)

SVI = 125 ml/gram

A ______ tank is an underground chamber made of concrete, fiberglass, or plastic through which domestic wastewater (sewage) flows for basic treatment.

  1. soap
  2. pressure
  3. flush
  4. septic

Answer (Detailed Solution Below)

Option 4 : septic

Sewage Question 10 Detailed Solution

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The correct answer is septic.

Key Points

  • A septic tank is concrete, fibreglass, or plastic subterranean chamber through which domestic wastewater (sewage) flows for basic treatment.
  • Anaerobic action occurs within the septic tank. The term "septic" refers to the anaerobic bacterial environment that forms in the tank and decomposes the waste that is expelled.
  • Domestic sewage, as well as animal and poultry wastes, are examples of N-rich materials that supply nutrients for anaerobic organism development and multiplication.

Important Points

  • A pressure/vacuum vent is installed in pressure tanks to avoid venting loss due to boiling and breathing loss due to daily temperature or barometric pressure variations. 

The moisture content of sewage sludge of two samples was reduced as follows:

Sample A: 97% to 95%.

Sample B: 98% to 96%

Select the correct inference.

  1. The decrease in volume for samples A and B is the same = 50%
  2. The decrease in volume for Samples A and B is the same = 60%
  3. There is an increase in volume of 60% for Sample A and 50% for Sample B
  4. There is a decrease in volume of 60% for Sample A and 50% for Sample B

Answer (Detailed Solution Below)

Option 4 : There is a decrease in volume of 60% for Sample A and 50% for Sample B

Sewage Question 11 Detailed Solution

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Explanation:

Sewage Sludge Sample A:

Assume, The initial volume of sample A is V1 and its moisture content is 97%

The final Volume of sample A is V2 and its moisture content is 95%

Now using the given condition we find the relationship between V1 & V2

V1 (100 - 97) = V2 (100 - 95)

\(V_2 = {3\over 5}V_1\)

So, The volume changes are given by

\(\Delta V_A = {V_1 - V_2\over V_1} \times 100\)

\(\Delta V_A = {V_1 - {3\over 5}V_1\over V_1} \times 100\) = 40%

Hence, there is a decrease in volume in sample A is 60% or (100 - 40), when the moisture content reduces from 97% to 95%.

Sewage Sludge Sample B:

Assume, The initial volume of sample B is V1 and its moisture content is 98%

The final Volume of sample B is V2 and its moisture content is 96%

Now using the given condition we find the relationship between V1 & V2

V1 (100 - 98) = V2 (100 - 96)

\(V_2 = {1\over 2}V_1\)

So, The volume changes are given by

\(\Delta V_B = {V_1 - V_2\over V_1} \times 100\)

\(\Delta V_B = {V_1 - {1\over 2}V_1\over V_1} \times 100\) = 50%

Hence, there is a decrease in volume in sample B is 50% or (100 - 50), when the moisture content reduces from 98% to 96%.

What will be the Sludge Volume Index (SVI) if 100 ml of sludge collected in 30 mins on drying weighs 800 mg?

  1.  115
  2. 78
  3. 125
  4. 100

Answer (Detailed Solution Below)

Option 3 : 125

Sewage Question 12 Detailed Solution

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Concept:

Sludge volume index is the volume occupied in mL by one gm of solids in the mixed liquor after settling for 30 minutes.

The recommended value of the Sludge Volume Index (SVI) for municipal sewage is approximately lies in the range of 80 to 150.

\({\rm{Sludge\;volume\;index}} = \frac{{{\rm{Volume\;of\;settled\;sludge\;in\;ml}}}}{{{\rm{MLSS\;present}}}}\)

Calculation:

Given,

Volume of settled sludge = 100 ml, Drying weight (MLSS) = 800 mg

\(\rm SVI = \frac{{100}}{{800 \ \times \ {{10}^{ - 3}}}}\ ml/gram\)

SVI = 125 ml/gram

In which of the following wastewater treatment units is organic matter destroyed and stabilised by anaerobic bacteria?

  1. Trickling filter
  2. Oxidation pond
  3. Sedimentation tank
  4. Septic tank

Answer (Detailed Solution Below)

Option 4 : Septic tank

Sewage Question 13 Detailed Solution

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Concept:

A septic tank is an underground chamber made of concrete, fiberglass or plastic through which domestic wastewater (sewage) flows for basic treatment. Action within the septic tank is anaerobic in nature.

The term "septic" refers to the anaerobic bacterial environment that develops in the tank which decomposes the waste discharged into the tank.

Followings are the classification of secondary treatment units

S.No.

Method

Contact Mechanism

Decomposition

1

Trickling filter

Attached growth

Aerobic

2

Rotating biological contactor

Attached growth

Aerobic

3

Activated sludge process

Suspended growth

Aerobic

4

Oxidation pond

Suspended growth

Aerobic

5

Septic tank

Suspended growth

Anaerobic

6

Imhoff tank

Suspended growth

Anaerobic

Which type of particle has the least self-cleansing velocity?

  1. Fine gravel
  2. Fine clay and silt
  3. Coarse sand
  4. Find sand and clay

Answer (Detailed Solution Below)

Option 2 : Fine clay and silt

Sewage Question 14 Detailed Solution

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Self-cleansing velocity is defined as the minimum velocity of flow at which no solid is deposited in the sewer even at minimum flow.

It can be calculated as:

\(V = \frac{1}{n}{R^{\frac{1}{6}}}{\left\{ {{K_s}\left( {{G_s} - 1} \right){d_p}} \right\}^{\frac{1}{2}}}\)

Gs = Specific gravity of particle

n = Manning’s Coefficient

dp = Particle size

Ks = Dimensionless constant

R = Hydraulic radius of sewer

It can be concluded that less will be the size of particle, less will be its self-cleansing velocity. Among Fine gravel, Fine clay and silt, Coarse sand & Fine sand and clay, Fine clay and silt has less size of particle, Hence Fine clay and Silt has less self-cleansing velocity.

Aerobic attached growth process is classified under:

  1. Activated sludge process
  2. Chemical clarification 
  3. Preliminary treatment
  4. Biological treatment

Answer (Detailed Solution Below)

Option 4 : Biological treatment

Sewage Question 15 Detailed Solution

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Explanation:

Treatment units Aerobic Anaerobic
Based on an attached growth system

1. Trickling Filter

2. Rotatory Biological Contactor

 
Based on a suspended growth system

1. Activated Sludge Process

2. Oxidation Pond

1. Upflow Anaerobic Sludge Blanket Reactor

2. Septic Tank

3. Imhoff Tank

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