Water Treatment MCQ Quiz - Objective Question with Answer for Water Treatment - Download Free PDF

Last updated on Jun 20, 2025

Latest Water Treatment MCQ Objective Questions

Water Treatment Question 1:

Which of the following treatments reduce salinity of water?
(i) Flocculation and sedimentation.
(ii) Filtration.
(iii) Reverse osmosis.
(iv) Electrodialysis.
Select the correct answer given below.

  1. (i) and (ii)
  2. (iii) and (iv)
  3. (ii) and (iii)
  4. (i) and (iv)

Answer (Detailed Solution Below)

Option 2 : (iii) and (iv)

Water Treatment Question 1 Detailed Solution

Explanation:

Salinity Reduction (Desalination)

  • Salinity in water refers to the concentration of dissolved salts, primarily chlorides, sulfates, sodium, and other ions. Removing these requires advanced treatment processes capable of separating dissolved ions.
  • Reverse Osmosis (RO) is a membrane-based process where water is forced under pressure through a semi-permeable membrane.

  • It effectively removes dissolved salts, heavy metals, and other impurities.

  • Widely used for desalination of seawater and brackish water.

  • Electrodialysis uses ion-exchange membranes and electric potential to remove dissolved ions from water.

  • Positively and negatively charged ions migrate through membranes to separate chambers, leaving behind desalinated water.

  • Effective for saline groundwater and industrial wastewater.

 Additional InformationPrimary Treatment of Water

  • Removal of Large and Settleable Solids
    Primary treatment involves physical processes like screening and sedimentation to remove floating debris, grit, and suspended solids. Large solids are first removed by bar screens, followed by grit chambers to eliminate heavier inorganic matter such as sand and gravel.

  • Sedimentation for Suspended Solids
    In primary sedimentation tanks, water is held for several hours to allow heavier suspended particles to settle at the bottom as sludge, while lighter materials like oils and grease float to the surface for removal. This reduces the total suspended solids (TSS) load entering secondary treatment.

  • Partial Reduction of Organic Load
    Though not designed to remove dissolved contaminants, primary treatment helps in reducing the organic load (BOD) by about 25–30%, making downstream biological processes more efficient. It also stabilizes the flow and composition of wastewater.

  • Sludge and Scum Handling
    The settled sludge is collected and sent for further treatment like thickening or digestion. The floating scum, consisting of oils and grease, is skimmed off separately. Proper handling of these by-products is crucial to prevent blockages and maintain operational efficiency.

Water Treatment Question 2:

Considering
(i) Screening
(ii) Sedimentation
(iii) Disinfection
(iv) Filtration. . .

The correct sequence of treatments which are generally given to treat raw water supplies are

  1. (i) (ii) (iii) (iv)
  2. (i) (ii) (iv) (iii)
  3. (i) (iii) (iv) (ii)
  4. (iv) (iii) (ii) (i)

Answer (Detailed Solution Below)

Option 2 : (i) (ii) (iv) (iii)

Water Treatment Question 2 Detailed Solution

Explanation:

This sequence — Screening → Sedimentation → Filtration → Disinfection — is the standard flow in municipal and industrial water treatment plants.

Screening 

  • The first step in water treatment.

  • Removes large floating debris such as leaves, twigs, and other large particles from raw water.

  • Prevents clogging and damage to downstream equipment.

Sedimentation

  • Water is allowed to stand in sedimentation tanks so that suspended solids settle at the bottom.

  • Reduces the load on the filtration system by removing heavier particles.

  • Enhances clarity and prepares the water for more refined treatment.

 Additional Information

Filtration 

  • The settled water is passed through filters (commonly sand or multimedia filters) to remove finer suspended particles.

  • Further improves the clarity of the water.

  • Critical for ensuring the effectiveness of the next stage, disinfection.

Disinfection 

  • Final step to kill or inactivate pathogens (bacteria, viruses, protozoa).

  • Usually achieved through chlorination, UV treatment, or ozonation.

  • Ensures the water is safe for human consumption.

Water Treatment Question 3:

What is the minimum residual chlorine desired after the chlorination process?

  1. 0.2 ppm
  2. 2 ppm
  3. 20 ppm
  4. 1 ppm

Answer (Detailed Solution Below)

Option 1 : 0.2 ppm

Water Treatment Question 3 Detailed Solution

Explanation

  • Residual chlorine is the amount of chlorine left in the water after a certain contact time, ensuring ongoing disinfection.

  • The Bureau of Indian Standards (BIS) and WHO guidelines recommend a minimum of 0.2 ppm (or mg/L) of free residual chlorine after 30 minutes of contact time at pH < 8.

  • This ensures that the water remains safe from microbial contamination during storage and distribution.

Additional Information

Types of Chlorine Residual:

  • Free Residual Chlorine: Most effective form, includes hypochlorous acid (HOCl) and hypochlorite ion (OCl⁻).

  • Combined Residual Chlorine: Reacts with ammonia to form chloramines—less effective but more stable.

  • Total Residual Chlorine: Sum of free and combined residual chlorine.

Water Treatment Question 4:

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 

  1. 0.005
  2. 0.05
  3. 5
  4. 50
  5. 0.0005

Answer (Detailed Solution Below)

Option 1 : 0.005

Water Treatment Question 4 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 5:

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)

  1. T = (La/2) - 1
  2. T = (La/4) - 1
  3. T = (La/20) - 1
  4. T = 2La - 1

Answer (Detailed Solution Below)

Option 3 : T = (La/20) - 1

Water Treatment Question 5 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:

  1. The BOD of the sewage influent (La) represents the amount of oxygen that microorganisms will consume to degrade organic matter.
  2. 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.
  3. Dividing La by 20 gives the base time required to remove the specified amount of BOD.
  4. 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.

Top Water Treatment MCQ Objective Questions

Which out of the following does not help in disinfecting water?

  1. Filtration
  2. Chlorine tablets
  3. Alums
  4. Boiling

Answer (Detailed Solution Below)

Option 1 : Filtration

Water Treatment Question 6 Detailed Solution

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The 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

  1. using lime soda process
  2. excess lime treatment
  3. ion exchange method
  4. using excess alum dosage

Answer (Detailed Solution Below)

Option 3 : ion exchange method

Water Treatment Question 7 Detailed Solution

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Concept:

  • 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?

  1. Lime
  2. Alum
  3. Ferric chloride
  4. Ferric sulphate

Answer (Detailed Solution Below)

Option 3 : Ferric chloride

Water Treatment Question 8 Detailed Solution

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Concept:

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.

  1. The rate of filtration is in the range of 100-150 m3/m2/day.
  2. The effective size of sand particles shall be 0.15 to 0.3 mm, and their uniformity coefficient shall be preferably below 3.
  3. Backwashing daily or on alternate days is necessary for cleaning.
  4. The filtering medium has effective size 0.45 to 0.7 mm, uniformity coefficient 1.3 to 1.7, multiple graded layers of sand with bacteria and odour removal.

Answer (Detailed Solution Below)

Option 2 : The effective size of sand particles shall be 0.15 to 0.3 mm, and their uniformity coefficient shall be preferably below 3.

Water Treatment Question 9 Detailed Solution

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Explanation:

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

  • Effective size
  • Uniformity coefficient
  • The thickness of sand bed
  • 0.2 to 0.4 mm
  • 1.8 - 3
  • 90 to 110 cm
  • 0.35 to 0.55 mm
  • 1.3 to 1.7
  • 60 to 75 cm

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

  • Installation
  • O & M
  • High
  • Low
  • Low
  • High

Efficiency

  • Turbidity of feed water
  • Removal of bacteria

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

  • Initial
  • Final
  • 10c m
  • 80 to 120 cm
  • 30 cm
  • 250 to 350 cm

Method of cleaning

  • Scrapping and removing Schmutzedecke and 1.5 to 3 cm thick sand layer
  • Laborious
  • Backwashing with or without compressed air agitation
  • Simple and easy

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

  1. 20 – 30 mm
  2. 20 – 40 mm
  3. 10 – 20 mm
  4. 10 – 30 mm

Answer (Detailed Solution Below)

Option 2 : 20 – 40 mm

Water Treatment Question 10 Detailed Solution

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Some 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)

  1. 14.50 m2
  2. 8.00 m2
  3. 12.50 m2
  4. 10.00 m2

Answer (Detailed Solution Below)

Option 3 : 12.50 m2

Water Treatment Question 11 Detailed Solution

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Explanation:

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:

  1. Increasing the depth of the tank
  2. Increasing the surface area of the tank
  3. Decreasing the depth of the tank
  4. Decreasing the surface area of the tank

Answer (Detailed Solution Below)

Option 2 : Increasing the surface area of the tank

Water Treatment Question 12 Detailed Solution

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Explanation:

  • 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

  1. Carbonates of calcium
  2. Sulphates of calcium
  3. Chlorides of calcium
  4. Nitrates of calcium

Answer (Detailed Solution Below)

Option 1 : Carbonates of calcium

Water Treatment Question 13 Detailed Solution

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The 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?

  1. Category B
  2. Category C
  3. Category D
  4. Category A

Answer (Detailed Solution Below)

Option 4 : Category A

Water Treatment Question 14 Detailed Solution

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Concepts:

As per IS 2296: 1982, the surface water quality standards are classified into 5 categories which are specified below:

  1. Class A – Drinking water without conventional treatment but after disinfection.
  1. Class B – Water for outdoor bathing.
  2. Class C – Drinking water with conventional treatment followed by disinfection.
  3. Class D – Water for fish culture and wild life propagation.
  4. 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:

  1. 0.015
  2. 0.15
  3. 15
  4. 1.5

Answer (Detailed Solution Below)

Option 2 : 0.15

Water Treatment Question 15 Detailed Solution

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Calculations:

Given Data:

Alum dose required = 15 mg/ltr

Total water = 10 million litre = 10 × 106 = 10litres

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

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