Water Treatment MCQ Quiz - Objective Question with Answer for Water Treatment - Download Free PDF
Last updated on Jun 11, 2025
Latest Water Treatment MCQ Objective Questions
Water Treatment Question 1:
A soil is composed of solid spherical grains of identical specific gravity and diameter between 0.075 mm and 0.0075 mm. If the terminal velocity of largest particle falling through water without flocculation is 0.5 mm/sec then, that for the smallest particle would be
Answer (Detailed Solution Below)
Water Treatment Question 1 Detailed Solution
Concept:
The settling velocity of sand particles is given by Stoke's Law,
\({{\rm{V}}_{\rm{s}}} = \frac{{g \times \left( {G - 1} \right) \times {d^2}}}{{18\;ν }}\)...... (1)
G -Specific gravity of sand particles
ν - Kinematic viscosity
d - Diameter of sand particles
Calculation:
Given:
Let V1 and V2 be settling velocities of larger and smaller grain respectively
V1 = 0.5 mm/sec
Diameter of larger grain d1 = 0.075 mm
Diameter of smaller grain d2 = 0.0075 mm
g is constant, specific gravity identical for both grains, both the particles are falling through water hence ν remains same. Hence equation 1 becomes
⇒ \(\frac{{{V_1}}}{{{V_2}}}\; = \;{\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^2}\)
⇒ \(\frac{{{0.5}}}{{{V_2}}}\; = \;{\left( {\frac{{{0.075}}}{{{0.0075}}}} \right)^2}\)
⇒ V2 = 0.005 mm/sec
Water Treatment Question 2:
American Public Health Association formula for aeration time (T) in hours is (where, La = BOD of the aeration tank sewage influent (mg/litre) to be removed)
Answer (Detailed Solution Below)
Water Treatment Question 2 Detailed Solution
Explanation:
American Public Health Association Formula for Aeration Time (T):
The formula for calculating the aeration time (T) in hours is given as:
T = (La / 20) - 1
Where:
- La = BOD (Biochemical Oxygen Demand) of the aeration tank sewage influent (mg/L) to be removed.
- T = Aeration time in hours.
This formula determines the time required for the aeration process in a sewage treatment plant, where the influent water is treated to remove the desired amount of BOD. Biochemical Oxygen Demand is a critical parameter in wastewater treatment, as it indicates the amount of organic matter present in the water. Efficient removal of BOD ensures that the treated water meets environmental discharge standards and reduces the load on receiving water bodies.
Derivation and Application:
The formula T = (La / 20) - 1 is derived based on empirical relationships and operational data from sewage treatment plants. It accounts for the time required for microbial activity in the aeration tank to break down organic matter, as indicated by the BOD value of the influent sewage. The "20" in the denominator represents the average rate of BOD removal efficiency (mg/L per hour) under standard conditions in a typical aeration tank.
Let’s break this down step by step:
- The BOD of the sewage influent (La) represents the amount of oxygen that microorganisms will consume to degrade organic matter.
- The rate of BOD removal is assumed to be 20 mg/L per hour in this formula, which is an average value based on experimental data and operational conditions of aeration tanks.
- Dividing La by 20 gives the base time required to remove the specified amount of BOD.
- Subtracting 1 from the result accounts for additional factors, such as initial conditions and system efficiency adjustments.
Thus, the formula provides a quick and reliable estimate of the aeration time required for a given influent sewage BOD level under typical operating conditions.
Water Treatment Question 3:
Which of the following is false about rapid gravity type filters used for water purification?
Answer (Detailed Solution Below)
Water Treatment Question 3 Detailed Solution
Explanation:
Rapid Gravity Type Filters
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Used for large-scale municipal water treatment to remove suspended solids after coagulation and flocculation.
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Water passes rapidly through a sand bed under gravity (rate: 4,000–12,000 liters/m²/hr).
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Coagulation is essential before filtration to form flocs for efficient removal.
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Requires periodic backwashing (usually every 1–2 days) to clean clogged filter media.
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Needs skilled supervision for operation, backwash management, and chemical handling.
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Initial and operational costs are high compared to slow sand filters.
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Removes turbidity, color, and microorganisms effectively when used with disinfection.
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Typically consists of a filter bed (sand), underdrain system, and wash water troughs.
Additional InformationSlow Sand Filters:
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Operate at a slow filtration rate (100–200 liters/m²/hr), much slower than rapid filters.
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Mainly used in small-scale or rural water treatment plants with low turbidity water.
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Filtration is done through a fine sand bed, forming a biologically active layer called schmutzdecke on top.
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No need for coagulation or chemical treatment before filtration.
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Require large land area due to slow filtration rate and longer filter runs.
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Cleaning is done by scraping the top sand layer, not backwashing.
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Low operational cost and does not require highly skilled supervision.
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Effective in removing bacteria, turbidity, and pathogens through biological and physical processes.
Water Treatment Question 4:
Aeration of water is done to remove
Answer (Detailed Solution Below)
Water Treatment Question 4 Detailed Solution
Explanation:
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Aeration involves exposing water to air, which helps remove dissolved gases like hydrogen sulfide responsible for bad odor.
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It also helps oxidize dissolved iron and manganese, improving taste and smell.
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Aeration does not remove hardness, bacteria, or colour directly.
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It is a common preliminary water treatment step to improve water quality and acceptability.
Additional Information
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Coagulation and Flocculation: Chemicals (like alum) are added to water to cause small suspended particles to clump together into larger particles (flocs) which settle out, clarifying the water.
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Filtration: Water passes through filters (sand, gravel, charcoal) to remove remaining suspended solids, turbidity, and some microorganisms.
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Disinfection: Chemicals like chlorine, ozone, or UV light are used to kill or deactivate harmful pathogens, ensuring water is safe to drink.
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Softening: Removes hardness-causing minerals (calcium, magnesium) using processes like lime-soda treatment or ion exchange.
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Sedimentation: Allows heavy particles to settle at the bottom of tanks before filtration, reducing the load on filters.
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Activated Carbon Treatment: Removes organic compounds, odors, and some chemicals through adsorption on carbon surfaces.
Water Treatment Question 5:
Consider the following statements:
1. Alum coagulation decreses the alkalinity of water.
2. Alum coagulation increases permanent hardness of water
3. Alum coagulation decreses pH of water.
4. Alum coagulation produces aluminium hydroxide flocks in the flocculation process.
Which of the statements given above the correct?
Answer (Detailed Solution Below)
Water Treatment Question 5 Detailed Solution
Explanation:
(i) Alum is the name given to the aluminium sulphate with its chemical formula is Al2(SO4)3.18H2O.
(ii) The alum when added to raw water, reacts with the bicarbonate alkalinities, which are generally present in raw supplies, so as to form a gelatinous precipitate (floc) of aluminium hydroxide.
Al2(SO4)3.18H2O + 3Ca(HCO3)2 → 3CaSO4 + 2Al(OH)3 + 6CO2
(iii) From the above equation, it becomes evident that the addition of alum to water imparts permanent hardness to it, in the form of calcium sulphate.
(iv) The CO2 gas which is evolved, causes corrosiveness.
(v) On addition to water, alum react with HCO3-, CO32-, and OH- and form aquametalllic cation and reduces the alkalinity of solutions, thus decreases the pH of water.
Top Water Treatment MCQ Objective Questions
Which out of the following does not help in disinfecting water?
Answer (Detailed Solution Below)
Water Treatment Question 6 Detailed Solution
Download Solution PDFThe correct answer is Filtration.
Key Points
Filtration
- Filtration is a process used to separate solids from liquids or gases by using a filter medium that allows fluid to pass through but not solid.
- Filtration can be mechanical-biological or physical.
- It does not help in disinfecting water.
- If you want to disinfect the water you can boil it or you can use alum for chlorine tablets.
Boiling
- Boiling is used to kill pathogenic bacteria, viruses, and protozoa.
Chlorination
- Chlorine kills a large variety of microbial waterborne pathogens.
- Chlorination is the process of adding chlorine to the water to disinfect it and kill pathogens.
- Amount of chlorine required for water disinfection around 1-16 milligrams per litre of water.
Coagulation
- Alum acts as an electrolyte that helps in settling the suspended matter in water.
- The process of adding alum to disinfect water is called coagulation.
Zero hardness of water is achieved by
Answer (Detailed Solution Below)
Water Treatment Question 7 Detailed Solution
Download Solution PDFConcept:
- Water softening: It is the process of hardness removal from the water. It is caused by multivalent cation and affects water quality.
- Lime soda method: It is a water softening method in which lime and soda ash are added to the water, which causes the precipitation of multivalent cation as CaCO3.
- Precipitation of CaCO3 occurs only when the pH of water is greater than 9, so in case of less pH alkalinity is added to the water. In this process small amount of Ca2+ and Mg2+ precipitates very late, which will create incrustation in the pipe, so to avoid this recarbonation is done to dissolve back this small amount of cation.
- Due to this, the method does not give zero hardness.
- Ion Exchange Process: Ion-exchange resin, (zeolite) exchanges one ion from the water being treated for another ion that is in the resin (sodium is one component of softening salt, with chlorine being the other). Zeolite resin exchanges sodium for calcium and magnesium. It can produce water with zero hardness.
Which coagulant is widely used for sewage treatment?
Answer (Detailed Solution Below)
Water Treatment Question 8 Detailed Solution
Download Solution PDFConcept:
Coagulation:
- Coagulation is a process used to neutralize charges and form a gelatinous mass to trap or bridge particles thus forming a mass large enough to settle or be trapped in the filter.
- Coagulation destabilizes the charges of the particles. Coagulants with charges opposite to those of the suspended solids are added to the water to neutralize the negative charges on dispersed non-settable solids such as clay and organic substances.
- Coagulant chemicals are substances that promote the coagulation of liquids into solids.
Coagulants used for wastewater treatment:
- Aluminium chloride
- Ferric chloride
- Ferric and ferrous sulphate
- Aluminium chlorohydrate
Coagulants used for water treatment:
- Alum
- Ferrous sulphate
- chlorinated ferrous sulphate
- Sodium aluminate
Ferric chloride is widely used for sewage treatment and Alum is widely used for water treatment.
Select the correct statement with respect to a slow sand filter.
Answer (Detailed Solution Below)
Water Treatment Question 9 Detailed Solution
Download Solution PDFExplanation:
Item |
Slow Sand Filter |
Rapid Sand Filter |
Pre-treatment |
Not required except plain sedimentation |
Coagulation, Flocculation, and Sedimentation |
Base materials |
Gravel base of 30 to 75 cm depth with 3 to 65 mm size graded gravel. |
Gravel base of 45 to 50 cm depth with gravel size varies from 3 to 50 mm in 4 or 5 layers |
Filter sand
|
|
|
Under drainage system |
Open jointed pipes or drains covered with perforated blocks |
Perforated pipe laterals discharging into the main header |
Size of each unit |
50 to 200 sq.m |
10 to 100 sq.m |
Rate of filtration |
100 to 200 Lph/sq.m |
4800 to 7200 Lph/sq.m |
Cost
|
|
|
Efficiency
|
Low; < 30 NTU 98 to 99% |
Any level of turbidity of feed water; (with pre-treatment) 80 to 90% |
Suitability |
For water supply to rural areas and small town |
For public water supply to towns and cities |
Post-treatment |
Slight disinfection |
Complete disinfection is a must |
Ease of constructive |
Simple |
Complicated; |
Skilled supervision |
Not essential |
Essential |
Loss of head
|
|
|
Method of cleaning |
|
|
Quantity of wash water required |
0.2 to 0.5% of total water filtered |
1 to 5% of the total water filtered |
Cleaning Interval |
Three to four months |
One to two days |
For a slow sand filter
The rate of filtration is 100 - 200 lit/hour/sqm
The spacing between two bars in medium size screen ranges from
Answer (Detailed Solution Below)
Water Treatment Question 10 Detailed Solution
Download Solution PDFSome important points about screens in the screening process
- The coarse screen consists of parallel iron rods are placed vertically or at a slight slope at about 25 to 50 mm apart
- The fine screen is usually made of woven wire mesh with an opening not more than 6 mm square
- The spacing between two bars in medium size screen ranges from 15 to 40 mm
- The clear spacing between the bars may be in the range of 15 mm to 75 mm in case of a mechanically cleaned bar screen
- However, for the manually cleaned bar screen, the clear spacing used is in the range 25 mm to 50 mm
A village has a population of 200 with an average rate of water demand of 100 litres per capita per day. A rapid sand filter having an average filtration rate of 100 liter/hour/m2 is to be designed for water treatment. The area of rapid sand filter required is:
(Take the ratio of maximum demand to average demand as 1.5)
Answer (Detailed Solution Below)
Water Treatment Question 11 Detailed Solution
Download Solution PDFExplanation:
Given Data:
Village population = 200
The average rate of water demand = 100 litres per capita per day
average filtration rate = 100 liter/hour/m2 = 100 x 24 liter/day/m2
Ratio of maximum demand to average demand = 1.5
Design dicharge = 1.5 x Village population x average rate of water demand
= 1.5 x 200 x 100 = 30000 liter/day/m2
Area of rapid sand filter = \(\frac{Design \ discharge}{Average \ filtration \ rate}\)
Area of rapid sand filter = \(\frac{30000}{100\ \times 24}\)
Area of rapid sand filter = 12.5 m2
Additional InformationFor Rapid Sand Filter
- Numbers of filters (N) = 1.22 \(\sqrt (\)Q), where Q is in MLD
- Total cross-sectional area of perforations = 0.2% of filtered area
- Cross-sectional area of on lateral is 2 or 4 times the total cross-sectional area of perforation.
For a given discharge, the efficiency of sedimentation tank can be increased by:
Answer (Detailed Solution Below)
Water Treatment Question 12 Detailed Solution
Download Solution PDFExplanation:
- The efficiency of the sedimentation tank indicates the overall percentage removal of suspended matter at a given overflow rate or surface loading rate.
- The efficiency of the sedimentation tank increases if the overflow rate reduces (more time available to particles for settle).
overflow rate, \({V_s} = \frac{Q}{{BL}}\)
From the equation, it is clear that if the surface area (B × L) of the tank increases, the overflow rate reduces, and efficiency increases for a given discharge.
Important Points
Efficiency depends on the following parameter during sedimentation:
- The velocity of flow: Efficiency increases if the velocity of flow reduces
- Viscosity: Efficiency increases if viscosity reduces (Viscosity changes but we can not do)
- Size of particle: Efficiency increases if the size of particle increases.
Temporary hardness in water is due to presence of
Answer (Detailed Solution Below)
Water Treatment Question 13 Detailed Solution
Download Solution PDFThe correct answer is Carbonates of calcium.
Key Points
- Hardness in water is that characteristic which prevents the formation of sufficient leather or foam, when such hard waters are mixed with soap.
- It is usually caused by the presence of calcium and magnesium salts present in water, which form scum by reaction with soap.
- Hard waters are undesirable because they may lead to greater soap consumption, scaling of boilers, causing corrosion and incrustation of pipes, making foods tasteless, etc.
- If bicarbonates and carbonates of calcium and magnesium are present in water, the water is rendered hard temporarily, as this hardness can be removed to some extent by simple boiling or to full extent by adding lime to the water. Such a hardness is known as temporary hardness or carbonate hardness.
- When such waters are boiled carbon dioxide gas escapes out and the insoluble calcium carbonate gets precipitated. Magnesium carbonate being fairly soluble in water won’t get removed by boiling. The temporary hard waters, therefore do cause deposition of calcium scales in boilers.
As per IS 2296-1982, which category of surface water can be used as drinking water source without conventional treatment but after disinfection?
Answer (Detailed Solution Below)
Water Treatment Question 14 Detailed Solution
Download Solution PDFConcepts:
As per IS 2296: 1982, the surface water quality standards are classified into 5 categories which are specified below:
- Class A – Drinking water without conventional treatment but after disinfection.
- Class B – Water for outdoor bathing.
- Class C – Drinking water with conventional treatment followed by disinfection.
- Class D – Water for fish culture and wild life propagation.
- Class E – Water for irrigation, industrial cooling and controlled waste disposal.
If the dosage of Alum is 15 mg/l, then the estimated quantity of Alum required (MT) for 10 million liter of water per day will be:
Answer (Detailed Solution Below)
Water Treatment Question 15 Detailed Solution
Download Solution PDFCalculations:
Given Data:
Alum dose required = 15 mg/ltr
Total water = 10 million litre = 10 × 106 = 107 litres
For 1 litre of water, alum required = 15 mg
For 107 litres of water, alum required = 15 × 107 mg = 150 kg = 0.15 MT
As 1 Metric Ton = 1000 Kg