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:
Answer (Detailed Solution Below)
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
BOD5 = 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.
Answer (Detailed Solution Below)
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:-
Answer (Detailed Solution Below)
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.
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.
Answer (Detailed Solution Below)
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.
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.
Answer (Detailed Solution Below)
Sewage Question 5 Detailed Solution
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
-
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.
-
-
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.
-
-
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.
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Top Sewage MCQ Objective Questions
Sewers must be checked for minimum velocities at their minimum hourly flows which is equal to:
Answer (Detailed Solution Below)
Sewage Question 6 Detailed Solution
Download Solution PDFConcept:
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
Answer (Detailed Solution Below)
Sewage Question 7 Detailed Solution
Download Solution PDFExplanation:
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 |
Answer (Detailed Solution Below)
Sewage Question 8 Detailed Solution
Download Solution PDFExplanation:
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?
Answer (Detailed Solution Below)
Sewage Question 9 Detailed Solution
Download Solution PDFConcept:
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.
Answer (Detailed Solution Below)
Sewage Question 10 Detailed Solution
Download Solution PDFThe 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.
Answer (Detailed Solution Below)
Sewage Question 11 Detailed Solution
Download Solution PDFExplanation:
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?
Answer (Detailed Solution Below)
Sewage Question 12 Detailed Solution
Download Solution PDFConcept:
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?
Answer (Detailed Solution Below)
Sewage Question 13 Detailed Solution
Download Solution PDFConcept:
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?
Answer (Detailed Solution Below)
Sewage Question 14 Detailed Solution
Download Solution PDFSelf-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:
Answer (Detailed Solution Below)
Sewage Question 15 Detailed Solution
Download Solution PDFExplanation:
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 |