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

Last updated on Jun 11, 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:

In designing Imhoff tanks, the usual retention period is 

  1. 2 hours
  2. 20 hours
  3. 14 hours
  4. 8 hours

Answer (Detailed Solution Below)

Option 1 : 2 hours

Sewage Question 2 Detailed Solution

Explaination:

  • The retention period of 2 hours refers specifically to the settling or sedimentation chamber of the Imhoff tank, where raw sewage is retained to allow suspended solids to settle out by gravity.

  • This retention time is crucial to ensure efficient settling of solids before they enter the digestion chamber below.

  • The digestion chamber, which is separate and located beneath the settling chamber, retains sludge for a much longer time (usually 15-20 days) for anaerobic digestion.

 Additional Information

  • Imhoff tanks are two-story septic tanks designed to combine sedimentation and sludge digestion in one structure but in separate chambers to avoid disturbance.The upper chamber serves as the sedimentation tank where suspended solids settle out of the wastewater.

  • Settled solids fall through slots or openings into the lower digestion chamber, where anaerobic bacteria break down organic matter.Anaerobic digestion in the lower chamber reduces sludge volume and stabilizes it, producing biogas.

  • Imhoff tanks are typically used for primary sewage treatment in small to medium-sized communities.They require minimal mechanical equipment and have low operational costs but need regular desludging to maintain efficiency.

  • Proper design parameters include adequate surface area for sedimentation, flow rates, and ensuring the settling time is sufficient (usually around 2 hours).

F1 N.M M.P 25.07.19 D 6

Sewage Question 3:

For roughing type trickling filters what would be the BOD removal rate?

  1. 50 - 80%
  2. 40 - 70%
  3. 80 - 90% 
  4. 60 - 90%

Answer (Detailed Solution Below)

Option 2 : 40 - 70%

Sewage Question 3 Detailed Solution

Explanation:

Roughing Trickling Filters are designed as pre-treatment units and are typically used before high-rate or standard-rate trickling filters or activated sludge processes.

  • Their primary role is to reduce high organic loads (i.e., high BOD levels) before final treatment.

  • Typical BOD removal efficiency for roughing type trickling filters ranges from 40% to 70%, depending on:

    • Influent BOD concentration

    • Hydraulic and organic loading rates

    • Filter media and depth

    • Retention time

 Additional InformationRoughing Trickling Filters

  • Roughing trickling filters are used as a pre-treatment step to reduce high concentrations of Biochemical Oxygen Demand (BOD) from wastewater. They are typically installed before conventional biological treatment units to prevent overloading and ensure efficient functioning of subsequent stages.
  •  These filters are designed to handle high organic loads without requiring effluent recirculation. They operate effectively even with raw or partially settled sewage, making them suitable for treating industrial or domestic wastewater with high BOD levels.
  • RTFs are deeper than standard trickling filters and use coarse media such as large stones or specially designed plastic modules. The media provides a surface for microbial biofilm growth, which biologically breaks down organic pollutants as the wastewater flows downward.
  • Roughing filters can remove about 50% to 80% of BOD, depending on the influent characteristics and filter design. While they do not achieve final effluent quality, they significantly reduce the load on downstream processes.
  • They require minimal maintenance and have low energy consumption since they rely mostly on gravity flow and natural aeration. Their robust design allows them to withstand fluctuations in flow rate and organic strength, making them ideal for developing regions or facilities with variable wastewater characteristics.

Sewage Question 4:

Which type of bacteria is used in trickling filters? 

  1. Facultative
  2. Blue-green bacteria
  3. Anaerobic
  4. Nitrifying

Answer (Detailed Solution Below)

Option 1 : Facultative

Sewage Question 4 Detailed Solution

Explanation:

  • Facultative bacteria are used in trickling filters because they have the ability to survive and function in both aerobic (with oxygen) and anaerobic (without oxygen) environments.
  • This flexibility allows them to efficiently break down organic matter in the wastewater as it passes through the filter media, which may have zones with varying oxygen levels.
  • Their metabolic versatility helps in stabilizing the waste and reducing pollutants, making them ideal for biological treatment in trickling filters.

 Additional Information

Blue-green bacteria

  • These are photosynthetic bacteria (cyanobacteria) that produce oxygen through photosynthesis.

  • They are mostly found in natural water bodies and are not primarily used for wastewater treatment in trickling filters since trickling filters rely on bacterial decomposition rather than photosynthesis.

Anaerobic bacteria

  • These bacteria thrive in environments without oxygen and are typically used in anaerobic digestion processes to decompose organic matter in sludge treatment.

  • Trickling filters, however, are aerobic systems where oxygen is available, so anaerobic bacteria are not the main microorganisms involved.

Nitrifying bacteria

  • These are specialized aerobic bacteria responsible for converting ammonia into nitrites and then nitrates, a process called nitrification.

  • While nitrifying bacteria can be present in trickling filters, especially in the later stages, the primary bacteria responsible for organic matter breakdown are facultative bacteria.

  • Nitrifying bacteria specifically target nitrogen compounds, not the bulk organic material.

Sewage Question 5:

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 5 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

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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