Category: WASTE WATER Engineering Objective Questions

250+ TOP MCQs on Self Purification of Natural Streams and Answers

Waste Water Engineering Multiple Choice Questions on “Self Purification of Natural Streams”.

1. Which one of the following is the basic indicator of river health?
a) BOD
b) COD
c) DO
d) ThOD
Answer: c
Clarification: The amount of dissolved Oxygen (DO) in water is one of the most commonly used indicators of a river health. Below 4 or 5 mg/l, the life forms that survive are reduced.

2. What is the minimum amount of DO required for the life survival of aquatic animals?
a) 10 mg/l
b) 5 mg/l
c) 2 mg/l
d) 1 mg/l
Answer: c
Clarification: As DO drops below 4 or 5 mg/L the forms of life that can survive begin to be reduced. A minimum of about 2.0 mg/L of dissolved oxygen is required to maintain higher life forms.

3. Oxygen demanding wastes improves DO.
a) True
b) False
Answer: b
Clarification: Oxygen demanding wastes remove DO. Plants add DO during the day but remove it at night. Respiration of organisms removes oxygen. Temperature is reduced in summer which inturn reduces flow.

4. In the concept of self purification of natural streams, complete the following phrase. Solution to pollution is _________
a) Control
b) Dilution
c) Reuse
d) Recycle
Answer: b
Clarification: Waste water disposal practices are based on the premise that “the solution to pollution is dilution”. In this method relatively small quantities of waste are discharged into large bodies of water.

5. _________ is accomplished by the replenishment of oxygen lost to bacterial degradation of organic waste.
a) Gas transfer
b) Dilution
c) Filtration
d) Re-suspension
Answer: a
Clarification: The transfer of gases into and out of water is an important part of the natural purification process. Gas transfer is accomplished by the replenishment of oxygen lost to bacterial degradation of organic waste.

6. On which of the following does the self purification process does not depend?
a) Volume
b) Flow rate
c) Temperature
d) Aquatic species
Answer: d
Clarification: The speed and completeness of the self purification process depend upon many factors like volume, flow rate, the turbulence of flow, variation in sunlight, etc. Aquatic species are affected by polluted streams.

7. Flowing water bodies recover rapidly.
a) True
b) False
Answer: a
Clarification: Flowing water bodies like streams, canals and rivers can recover rapidly from degradable, oxygen demanding wastes and excess heat through a combination of dilution and bacterial decay.

8. In a flowing stream, the breakdown of degradable wastes by bacteria ________ dissolved oxygen.
a) Increases
b) Depletes
c) Maintains
d) Improves
Answer: b
Clarification: In a flowing stream, the breakdown of degradable wastes by bacteria depletes dissolved oxygen. This eliminates the populations with high oxygen requirements until the stream is cleansed of wastes.

9. What is the objective of water quality management?
a) Control the discharge of pollutants
b) Pollutants are discharged into flowing streams
c) Selective pollutants are released
d) Only highly toxic pollutants are released
Answer: a
Clarification: To control the discharge of pollutants so that the water quality is not degraded to an unacceptable extent below the natural background level is the objective of water quality management.

10. The impact of pollution depends upon nature of the pollutants and the ___________
a) Toxic contaminants
b) Season
c) Contaminants
d) Characteristics of river
Answer: d
Clarification: The impact of pollution depends upon nature of the pollutants and the characteristics of the river like discharge and speed of flowing water, depth of the river, type of bottom, surrounding vegetation, etc.

250+ TOP MCQs on Chemical Clarification – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Chemical Clarification – 1”.

1. Clarification removes most of the turbidity.
a) True
b) False
Answer: a
Clarification: Most of the turbidity is removed by clarification, making the water crystal clear. Disinfection is usually the final step in the treatment of drinking water which destroys pathogenic microbes.

2. What is the first step in clarification?
a) Sedimentation
b) Coagulation
c) Flocculation
d) Screening
Answer: b
Clarification: Finely divided particles suspended in surface water repel each other because most of the surfaces are negatively charged. Coagulation is the first step to neutralize the charged particles and form flocs.

3. The agglomeration of destabilized particles into large particles is called _________
a) Sedimentation
b) Coagulation
c) Flocculation
d) Disinfection
Answer: c
Clarification: Flocculation is the agglomeration of destabilized particles into large particles and can be enhanced by the addition of high-molecular-weight, water-soluble organic polymers. These polymers increase floc size by charged site binding and by molecular bridging.

4. What is the density of alum content used in water treatment?
a) 9.8 lb/gal
b) 10.1 lb/gal
c) 11.1 lb/gal
d) 9.2 lb/gal
Answer: c
Clarification: Alum is the primary coagulant. Alum is widely used because it is cheap and easily available. The chemical formula for alum is AlCl3. It is used in the form of a liquid and density is 11.1 lb/gal.

5. What is the speed of the impeller in a flash mixer?
a) 90 rpm
b) 100 rpm
c) 110 rpm
d) 120 rpm
Answer: c
Clarification: The impeller used in the flash mixer is a propeller type. The average speed of the impeller used in the flash mixer is 110 rpm. Rpm refers to the number of rotations per minute.

6. Microflocculation is brought about by which phenomenon?
a) Velocity gradient
b) Differential settling
c) Brownian movement
d) Differential mixing
Answer: c
Clarification: Microflocculation is brought about by Brownian movement. This is also known as perikinetic flocculation. In this type of flocculation, particle aggregation is brought about by random thermal motion.

7. Macro flocculation is brought about by which phenomenon?
a) Velocity gradient
b) Differential settling
c) Brownian movement
d) Differential mixing
Answer: a
Clarification: Macroflocculation is brought about by velocity gradient. This is also known as orthokinetic flocculation. In this type of flocculation, particles to be flocculated are mixed.

8. Alum precipitation occurs at which pH?
a) 1-3
b) 3-5
c) 5-7
d) 7-9
Answer: c
Clarification: Alum precipitation occurs at 5-7. Minimum solubility occurs at 6. Alum is a coagulating agent.

9. Iron precipitation occurs at which pH?
a) 1-3
b) 3-5
c) 5-7
d) 7-9
Answer: d
Clarification: Iron precipitation occurs at 7-9. Minimum solubility occurs at 8. Water containing iron usually is brown in colour.

10. Out of these which is not used as a chemical coagulant?
a) Alum
b) Calcium chloride
c) Ferric Chloride
d) Poly Iron Chloride
Answer: d
Clarification: Poly Iron chloride is not a chemical coagulant. It is a flocculant. It is used in bringing about flocculation.

11. What is the one disadvantage of adding chemicals to bring about precipitation?
a) The TSS increases
b) The TDS increases
c) BOD increases
d) It imparts colour
Answer: b
Clarification: The disadvantage of adding chemicals to bring about precipitation is that this increases the TDS. There is an increase in the dissolved constituents. Addition of chlorine increases the TDS in the effluent.

12. When alum is added to the waste water containing calcium ions which compound is formed as a precipitate?
a) Al(OH)3
b) Al2 (SO4)3
c) Ca(OH)2
d) CaCO3
Answer: a
Clarification: When alum is added to a waste water containing calcium salts Al (OH)3 is precipitated. This sis a gelatinous floc. This reaction sweeps out suspended particles.

13. Along with ferrous sulphate what other chemical is added in order to bring about precipitation?
a) Alum
b) PAC
c) Lime
d) Polyelectrolyte
Answer: c
Clarification: Along with ferrous sulfate, lime is generally added to bring about precipitation. This is done in order to increase the pH. Ferric hydroxide precipitates are formed only at very high pH.

14. Why is Ferric ammonium sulfate generally not used as a coagulant?
a) It is very expensive
b) It is dependent on the Dissolved oxygen present in the waste water
c) It is a very slow process
d) It works only at high pH
Answer: b
Clarification: Ferric ammonium sulfate is not a preferred coagulant. This is because only if oxygen is present ferrous hydroxide forms ferric hydroxide which is a precipitate. Thus this is dependent on the dissolved oxygen of the waste water.

15. What is the recommended surface loading rate in case of alum floc suspension?
a) 30-70 m3/m2.d
b) 10-30 m3/m2.d
c) 70-100 m3/m2.d
d) <10 m3/m2.d
Answer: a
Clarification: The recommended surface loading rate for alum floc suspension is 30-70 m3/m2.d. The typical value is 70 m3/m2.d. In case of iron floc suspension the recommended surface loading rate is 30-70 m3/m2.d

250+ TOP MCQs on Activated Sludge Treatment Systems – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Activated Sludge Treatment Systems – 1”.

1. Why is the sludge aerated?
a) To avoid bacterial growth
b) To increase bacterial growth
c) To maintain pH
d) To maintain temperature
Answer: b
Clarification: An activated-sludge reactor is a system in which pre-treated sewage (i.e. having passed through primary treatment) is aerated to promote the growth of bacteria (cells) that gradually consume the organics in the sewage.

2. ________ is treated in activated sludge reactor.
a) Pre-treated sludge
b) Treated sludge
c) Macronutrients
d) Micro-organisms
Answer: a
Clarification: Pre-treated sludge is treated in an activated sludge reactor. This helps to reduce a load of treatment on the activated sludge process.

3. After the treatment, the BOD demand ______
a) Remains constant
b) Decreases
c) Increases
d) Alters
Answer: b
Clarification: After the treatment, the BOD demand decreases. BOD stands for biological oxygen demand and COD stands for chemical oxygen demand.

4. Which is the next reactor after activated sludge reactor in the treatment process?
a) Flocculation unit
b) Aeration unit
c) Clarifier
d) Disinfection unit
Answer: c
Clarification: Clarifier (settling tank) is the next reactor where the solids (mostly cells, called sludge at this stage) are separated from the water. The system is commonly operated in continuous mode (as opposed to batch mode).

5. Where is the sludge at the bottom of the clarifier processed to?
a) Settling unit
b) Aerator
c) Flocculation unit
d) Disinfection unit
Answer: b
Clarification: The sludge coming from the bottom of the clarifier is processes to aerator unit and this clearly indicates activated sludge process.

6. Which of these is not an alternative to activated sludge treatment systems?
a) Stabilization ponds
b) Rotating biological reactors
c) Trickling filter
d) Screening units
Answer: d
Clarification: In the trickling filter, pre-treated sewage is sprayed on the substrate. Cells can grow by the available oxygen. Stabilization ponds are similar to aerated lagoons and the process is natural.

7. An activated sludge system consists of two components, an aerator and ________
a) Screening units
b) Disinfection unit
c) Flocculation unit
d) Clarifier
Answer: d
Clarification: Aerator and clarifier are the two components present in activated sludge systems. An activated sludge system consists of two components such as aerator and clarifier.

8. How is air pumped in the aerator unit?
a) Bubbled from bottom
b) Sides
c) Bubbled from top
d) Sprayed
Answer: a
Clarification: As cells need oxygen for their metabolism, air is injected from the bottom of the aerator. Rising bubbles agitate the water well and create good contact between the three ingredients: cells, sewage and oxygen.

9. Which are the three ingredients in activated sludge systems?
a) Cells, sewage and oxygen
b) Cells, sewage and nitrogen
c) Solids, sewage and oxygen
d) Solids, water and oxygen
Answer: a
Clarification: The cells need oxygen for their metabolism, air is injected from the bottom of the aerator. The water is well agitated by the rising bubbles and creates good contact between the three ingredients: cells, sewage and oxygen.

10. Mechanical stirring can be done instead of injection of air from the bottom.
a) True
b) False
Answer: a
Clarification: Activated-sludge aerators are well agitated by mechanical stirring from the top or injection of air from the bottom.

11. The sludge particles concentration is increased by the growth of the organism in aeration tanks.
a) True
b) False
Answer: b
Clarification: The wastewater is first screened and then it is mixed with different amounts of the recycled liquid containing high proportion of organisms and sludge concentration decreases as organism growth increases.

12. The organisms feed on _____ in aeration tanks.
a) Water
b) Air
c) Sludge particles
d) Bacteria
Answer: c
Clarification: The organisms in the reactor multiply by feeding of the organic solids present in the reactor. The reactor consists of wastewater containing organic solids.

250+ TOP MCQs on Oil and Grease Removal Methods and Answers

Waste Water Engineering Multiple Choice Questions on “Oil and Grease Removal Methods”.

1. Oil and grease is the presence of inorganics in wastewater.
a) True
b) False
Answer: b
Clarification: Oil and Grease (O&G) are a common occurrence in wastewater. An EPA commissioned study recently concluded that O&G is an indicator of the presence of numerous other organics in a wastewater, the types that partition into oil.

2. The surfactants have ______ chains.
a) Linear
b) Hydrocarbon
c) Complex
d) Carbon
Answer: b
Clarification: Oil is chemically emulsified in water when emulsifiers such as surfactants or soaps are present. The surfactants have hydrocarbon chains. The simplest ones are sodium laurel sulphate or stearic acid which has a hydrophilic (water loving) and a lipophilic (oil loving) end.

3. Lipophilic end is water liking.
a) True
b) False
Answer: b
Clarification: The surfactants have hydrocarbon chains. The simplest ones are sodium laurel sulphate or stearic acid which has a hydrophilic (water loving) and a lipophilic (oil loving) end. The lipophilic end enters the oil droplet, while the hydrophilic end remains in the water.

4. The lipophilic end ______
a) Remains in water
b) Enters oil droplet
c) Dissolves
d) Stays on top of water surface
Answer: b
Clarification: The lipophilic end enters the oil droplet, while the hydrophilic end remains in the water. Since this creates a charge on the otherwise neutral oil droplet, the droplets will repel each other and disperse.

5. What is the size of the oil droplets?
a) Less than 50 microns
b) Less than 40 microns
c) Less than 30 microns
d) Less than 20 microns
Answer: d
Clarification: The droplets are less than 20 microns, while the colour of the water is white. The white colour is an indicator that the emulsion must be split to allow removal of the oil. The source of such oils is metal working fluids, coolants, lubricants, motor oil, hydraulic fluids, etc.

6. What is the colour of the emulsion?
a) White
b) Grey
c) Black
d) Yellow
Answer: a
Clarification: The white colour is an indicator that the emulsion must be split to allow removal of the oil. The source of such oils is metal working fluids, coolants, lubricants, motor oil, hydraulic fluids, etc.

7. What is the size of dissolved oil droplets?
a) Less than 10 microns
b) Less than 8 microns
c) Less than 7 microns
d) Less than 5 microns
Answer: d
Clarification: These are oils from the light end of the oil spectrum such as benzene, toluene and xylene. The molecules are less than five microns in size. They are removed very effectively by activated carbon.

8. _________ act as a coupling agent between oil oil/water phases.
a) Oil
b) Water
c) Emulsifier
d) Disinfectants
Answer: c
Clarification: Emulsifier act as a coupling agent between the oil/water phases. Because the emulsifier is polar on one end (i.e., it has a charge) and is non-polar at the other end, it prevents the oil from approaching and coalescing.

9. An emulsion is a _____ system.
a) Homogenous
b) Heterogenous
c) Natural
d) Oxidized
Answer: b
Clarification: An emulsion is a heterogeneous system that consists of at least one immiscible liquid intimately dispersed in another liquid in the form of droplets, whose diameter generally exceeds 0.1 microns.

10. Surfactants and finely divided solids _____the stability of the emulsion.
a) Increase
b) Decease
c) Neutralize
d) Nullify
Answer: a
Clarification: Surfactants and finely divided solids increase the stability of the emulsion. An emulsion is a heterogeneous system that consists of at least one immiscible liquid intimately dispersed in another liquid in the form of droplets, whose diameter generally exceeds 0.1 microns.

11. Organo clays are manufactured by modifying _____with quaternary amines.
a) Nitrogen
b) Bentonite
c) Potassium
d) Sodium
Answer: b
Clarification: Organo clays are manufactured by modifying bentonite with quaternary amines, a type of surfactant that contains a nitrogen ion. The nitrogen end of the quaternary amine (the hydrophilic end) is positively charged and ion exchanges onto the clay platelet for sodium or calcium.

12. What is the range of charge present on bentonite?
a) 20-30 meq/gram
b) 30-40 meq/gram
c) 40-60 meq/gram
d) 70-90 meq/gram
Answer: d
Clarification: The bentonite has a charge of 70-90 meq/gram. After it is treated with the quaternary amine, some 30-40 meq/gram remain, resulting in the organo clay also removing small amounts of the common heavy metals such as lead, copper, cadmium and nickel.

13. The design of oil/water separators is based on _________
a) Stoke’s law
b) Newton’s law
c) Boyle’s law
d) Charles’s law
Answer: a
Clarification: The design of oil/water separators is based on Stoke’s Law. The lighter oil droplets impact on the slant ribs of the media, coagulate and rise to the surface. The principle of air flotation is that oil droplets will adhere to air and gas bubbles and rise to the surface of the tank.

14. Compute the required pressure for the flotation thickener without recycle for the following information:
A/S = 0.008 mL/mg
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 302 kPa
b) 640 kPa
c) 380 kPa
d) 680 kPa
Answer: a
Clarification: The pressure is calculated by this formula A/S = 1.3 Sa (fP-1)/Sa. Then the P= p+101.35/101.35. p= 302 kPa.

15. Determine the surface area for the following information.
A/S = 0.008 mL/mg
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 34.7 m2
b) 67.7 m2
c) 38 m2
d) 72 m2
Answer: a
Clarification: Area = Flow/ surface loading rate. Area = 400 m3/d/8 L/m2/min x 1440 min/d. Therefore area = 34.7 m2.

16. Compute the required recycled rate for the flotation thickener with recycle for the following information:
A/S = 0.008 mL/mg
p = 275 atm
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 462 m3/d
b) 642 m3/d
c) 380 m3/d
d) 680 m3/d
Answer: a
Clarification: P= P+101.35/101.35. p= 3.73 atm.The pressure is calculated by this formula A/S = 1.3 Sa (fP-1)R/Sa. Q. R = 462 m3/d.

17. Determine the surface area with recycle for the following information.
A/S = 0.008 mL/mg
Temperature: 20 degree Celsius
Air solubility: 18.7 mL/L
Recycle steam pressure: 275 kPa
Fraction of saturation: 0.5
Surface Loading rate: 8 L/m2/min
Sludge flow rate: 400 m3/d
a) 40.1 m2
b) 47.7 m2
c) 38 m2
d) 72 m2
Answer: a
Clarification: P= P+101.35/101.35. p= 3.73 atm.The pressure is calculated by this formula A/S = 1.3 Sa (fP-1)R/Sa. Q. R = 462 m3/d. Area =Recycle Flow/ surface loading rate. Area = 462 m3/d x 1000 L/m3/8 L/m2/min x 1440 min/d. Therefore area = 40.1 m2.

18. What is the velocity considered for a flotation thickener to remove oil?
a) 8 L/m2/min -160 L/m2/min
b) 180 L/m2/min
c) 200 L/m2/min-250 L/m2/min
d) 2 L/m2/min-8 L/m2/min
Answer: a
Clarification: Velocity to be considered for a floatation thickener to remove oil is 8 L/m2/min-160L/m2/min. With the velocity the surface area is found out. Mostly the velocity considered is 8 L/m2/min.

250+ TOP MCQs on UASB Reactor – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “UASB Reactor – 1”.

1. UASB is an anaerobic reactor.
a) True
b) False
Answer: a
Clarification: The up-flow anaerobic sludge blanket reactor (UASB) is a single tank process in an anaerobic centralized or decentralized industrial wastewater or black water treatment system achieving high removal of organic pollutants.

2. UASB is only used for lower levels of treatment.
a) True
b) False
Answer: b
Clarification: The up-flow anaerobic sludge blanket reactors are used in industries for high removal of organic pollutants and achieve high treatment efficiency.

3. How does the wastewater enter the reactor?
a) Bottom
b) Top
c) Side
d) Top and side
Answer: a
Clarification: Wastewater enters the reactor from the bottom and flows forward. A suspended sludge blanket filters and treats the wastewater as the wastewater flows through it.

4. A sludge blanket in the reactor is ______
a) At the top
b) At the bottom
c) Absent
d) In between
Answer: d
Clarification: Wastewater enters the reactor from the bottom and flows forward. A suspended sludge blanket filters and treats wastewater as the wastewater flows through it.

5. _____ breaks down the organic matter.
a) Bacteria
b) Fungi
c) Virus
d) Algae
Answer: a
Clarification: Bacteria living in the sludge blanket breaks down organic matter by anaerobic digestion, transforming it into biogas. Solids are also retained by a filtration effect of the blanket.

6. _________ at the top prevent an outflow of the sludge blanket.
a) Rubber cock
b) Out pipe
c) Baffle
d) Stopper
Answer: c
Clarification: The up-flow regime and the motion of the gas bubbles allow mixing without mechanical assistance. Baffles at the top of the reactor allow gases to escape and prevent an outflow of the sludge blanket.

7. What is the size of the microbial granules?
a) 0.3-0.6 mm
b) 0.5-0.8 mm
c) 1-3 mm
d) 3-6 mm
Answer: c
Clarification: The sludge blanket is comprised of microbial granules i.e., small agglomeration of micro-organisms that, because of their weight, resist being washed out in the up-flow.

8. What is the product gases released at the end of the UASB process?
a) Methane
b) Methane and carbon-dioxide
c) Oxygen and carbon-dioxide
d) Carbon-dioxide
Answer: b
Clarification: The micro-organisms in the sludge layer degrade organic compounds. As a result, gases are released. The rising bubbles mix the sludge without the assistance of any mechanical parts.

9. What is the means used for the mixing of the sludge?
a) Magnetic stirrer
b) Electrically driven stirrer
c) Mechanical mixing
d) By means of bubbles
Answer: d
Clarification: The rising bubbles mix the sludge without the assistance of any mechanical parts. Upstream velocity and settling speed of the sludge is in equilibrium and forms a locally rather stable but suspended sludge blanket.

10. _______ deflects the materials downwards.
a) Baffles
b) Sloped walls
c) Deflectors
d) Separator
Answer: b
Clarification: Upstream velocity and the settling speed of the sludge is in equilibrium and forms a locally rather stable, but suspended sludge blanket. Sloped walls deflect material that reaches the top of the tank downwards.

11. A clarified effluent is extracted from the ______ of the reactor.
a) Top
b) Side
c) Bottom
d) Central pipe
Answer: a
Clarification: The clarified effluent is extracted from the top of the tank in an area above the sloped walls. A gas-liquid-solids separator separates the gas from the treated wastewater and the sludge.

12. What is the minimum pH to be maintained for the process?
a) 5.6
b) 6.3
c) 7.5
d) 8.2
Answer: b
Clarification: The pH value needs to be between 6.3 and 7.85 to allow bacteria responsible for anaerobic digestion to grow. The pH value is also important because at high pH- values, ammonia disassociates to NH3 which inhibits the growth of the methane producing bacteria.

250+ TOP MCQs on Disinfection Process – 1 and Answers

Waste Water Engineering Multiple Choice Questions on “Disinfection Process – 1”.

1. Which is the common indicator organism used in the evaluation of drinking water?
a) E. coli
b) Fecal coliform
c) Total coliform
d) Algae
Answer: c
Clarification: The most common indicator organism used in the evaluation of drinking water is Total Coliform (TC), unless there is a reason to focus on a specific pathogen.

2. Which is the common indicator organism used in the evaluation of wastewater?
a) Fecal coliform
b) Total coliform
c) E. coli
d) Algae
Answer: a
Clarification: The most common indicator organism for wastewater evaluation is fecal coliform but there has been discussion regarding the use of Escherichia coli (E. coli) or Total Coliform.

3. Surface water contains approximately 1,000 times more indicator organisms than domestic wastewater.
a) True
b) False
Answer: b
Clarification: Domestic wastewater contains approximately 1,000 times more indicator organisms than typical surface water, understanding wastewater disinfection will make it easier to understand water disinfection.

4. Which year was the Safe Drinking Water Act passed?
a) 1990
b) 1992
c) 1994
d) 1996
Answer: d
Clarification: Amendments to the Safe Drinking Water Act in 1996 mandate the development of regulations to require disinfection of groundwater as necessary.

5. The solution to pollution is _______
a) Reuse
b) Dilution
c) Detection
d) Cleaning
Answer: b
Clarification: The last 100 years have brought significant environmental advances. At the beginning of the 20th Century, water and wastewater were treated by one principle, “the solution to pollution is dilution.”

6. ________ means the deactivation or killing of pathogens.
a) Reduction
b) Disinfection
c) Oxidation
d) Pyrolysis
Answer: b
Clarification: Water disinfection means the removal, deactivation or killing of pathogenic micro-organisms. Micro-organisms are destroyed or deactivated, resulting in termination of growth and production.

7. Sterilization is a process related to disinfection
a) True
b) False
Answer: a
Clarification: Sterilization is a process related to disinfection. However, during the sterilization process, all present micro-organisms are killed, both harmful and harmless micro-organisms.

8. Which of the following is not a chemical disinfectant?
a) UV
b) Ozone
c) Chlorine
d) Bromine
Answer: a
Clarification: There are various physical, chemical and biological disinfectants used for disinfection process. Chemical disinfectants are ozone, chlorine, bromine, ammonium and hydrogen acids etc.

9. Which of the following is not a physical disinfection means?
a) Heat
b) Sound
c) Metals
d) UV
Answer: c
Clarification: For physical disinfection of water, means like heat, sound, light (UV), electronic radiations and gamma rays can be used.

10. ________ is the process where all the living micro-organisms, including bacterial spores are killed.
a) Disinfection
b) Sterilization
c) Incineration
d) Pyrolysis
Answer: b
Clarification: Sterilization can be achieved by physical, chemical and physiochemical means. Chemicals used as sterilizing agents are called chemisterilants.

11. Which of the following means cannot be used for sterilisation?
a) Physical
b) Chemical
c) Physiochemical
d) Biological
Answer: d
Clarification: There are various means used for the process of sterilization. They are physical, chemical and physiochemical. Biological means cannot be usually used for sterilization.

12. ______ is the process of elimination of most pathogenic micro-organisms.
a) Pyrolysis
b) Incineration
c) Disinfection
d) Sterilisation
Answer: c
Clarification: Chemicals used as sterilizing agents are called chemisterilants. Disinfection can be achieved by physical or chemical methods. Chemicals used in disinfection are called disinfectant.

13. What is the chemical or mechanical cleansing process called?
a) Sterilization
b) Disinfection
c) Decontamination
d) Sanitization
Answer: d
Clarification: Sanitization is the process of chemical or mechanical cleansing, applicable in public health systems. Usually used by the food industry. It reduces microbes on eating utensils to safe, acceptable levels for public health.