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# Multiple choice question for engineering

## Set 1

1. The head loss through the bed of solids of the filter can be determined by
a) Carmen-Kozney equation
b) Rose equation
c) Carmen-Kozney and Rose equation
d) Charles equation

Answer: c [Reason:] The head loss through the bed of solids of the filter can be determined by both Carmen-Kozney and Rose equation where two cases are considered, one for homogeneous mixed bed and other for stratified bed.

2. Which of the following has highest shape factor as stated by Carmen?
a) Ottawa sand
b) Pulverized coal
c) Rounded coal
d) Angular sand

Answer: a [Reason:] The shape factor as stated by Carmen is 0.95 for Ottawa sand, 0.73 for pulverized coal and angular sand and 0.82 for rounded sand.

3. Which of the following is the expression of the Carmen equation where symbols have their usual meanings?
a) hf = f* (L/(s *d)) * ((1-e)/e3) * (v2/g)
b) hf = f* (L/(s *d)) * ((1-e)/e3) * (v/g)
c) hf = f* (L/(s *d)) * ((1-e)/e) * (v2/g)
d) hf = f* (L/(s *d)) * ((1-e)/e2) * (v2/g)

Answer: a [Reason:] The expression of the Carmen equation is given by hf = f* (L/(s *d)) * ((1-e) /e3) * (v2/g) where hf =head loss, d= diameter of pipe. L= length of pipe and v=approach velocity.

4. Which of the following represents the correct relation between dimensionless friction factor f and Reynolds number?
a) f = 150* ((1-e) /R2) +1.75
b) f = 150* ((1-e) /R) +1.75
c) f = 150* ((1-e) /R3) +1.75
d) f = 150* ((1-e2) /R) +1.75

Answer: b [Reason:] The correct relation between dimensionless friction factor and Reynolds number is given by f = 150* ((1-e) /R) +1.75 Where R is the Reynolds number and the expression (1-e) represents the volume of solids.

5. Which of the following is the expression of the Rose equation where symbols have their usual meanings?
a) hf = f* (L/(s *d)) * ((1-e) /e3 ) * (v2/g)
b) hf = f* (L/(s *d)) * ((1-e) /e3 ) * (v/g)
c) hf = f* (L/(s *d)) * ((1-e) /e4 ) * (v2/g)
d) hf = f* (L/(s *d)) * ((1-e) /e2 ) * (v2/g)

Answer: a [Reason:] The expression of the Rose equation is given by hf = f* (L/(s *d)) * ((1-e4) /e3) * (v2/g) where hf =head loss, d= diameter of pipe. L= length of pipe and v=approach velocity and f= 1.067CD, Where CD is the coefficient of drag.

6. When the Reynolds number is greater than 1.9 but less than 500, the coefficient of drag CD is
a) CD = 24/R
b) CD = R/24
c) CD = 18.5/R0.6
d) CD = R0.6/18.5

Answer: c [Reason:] When the Reynolds number is greater than 1.9 but less than 500, the coefficient of drag CD is CD = 18.5/R0.6 where R is the Reynolds number.

7. The value of Reynolds number R is 1.5. The coefficient of drag is
a) 8
b) 10
c) 12
d) 16

Answer: d [Reason:] When R<1.9, the value of coefficient of drag = 24/R R=1.5, the coefficient of drag = 24/1.5 = 16.

8. State whether the following statement is true or false
Rose equation is valid for beds in which voids are clear and unobstructed.
a) True
b) False

Answer: a [Reason:] Rose equation is valid for beds in which voids are clear and unobstructed though during continues filtration, voids get clogged and head loss goes on increasing.

9. When does the particle become suspended in expanded bed?
a) When superficial velocity is greater than critical velocity
b) When superficial velocity is less than critical velocity
c) When superficial velocity is equal to critical velocity
d) When superficial velocity is constant

Answer: c [Reason:] During back washing, the bed remains fixed at low fluid velocity and as the superficial velocity increases, the lighter particles move upward and when this velocity equals the critical velocity, the particle becomes suspended.

10. The superficial velocity in a stratified bed is equal to
a) Terminal settling velocity
b) Terminal settling velocity * porosity
c) Terminal settling velocity * (porosity) 2
d) Terminal settling velocity * (porosity) 4.5

Answer: d [Reason:] A stratified bed having a non uniform sized particles are completely fluidized when for the largest particle, the superficial velocity v = vs * e4.5 where vs is the terminal settling velocity and e is the porosity.

11. Carmen-Kozney equation has been derived using which of the following equation?
a) Cole brook white equation
b) Bernoulli equation
c) Darcy Weisbach equation
d) Swamee jain equation

Answer: c [Reason:] Carmen-Kozney equation has been derived using Darcy Weisbach equation which is given by h = f*l*v2/(g*D) Where h is the head loss, D is the diameter of pipe, v is the velocity of the particle and f is the dimensionless friction factor.

## Set 2

1. __________ is an operation designed to force agitation in the fluid and induce coagulation.
a) Sedimentation
b) Flocculation
c) Disinfection
d) Aeration

Answer: b [Reason:] Flocculation is an agitating process in which destabilized particles are brought into contact to promote agglomeration.

2. The rate of change of velocity per unit distance normal to a section is called_______
a) Mean velocity
b) Average velocity

Answer: c [Reason:] Mean velocity gradient is given by G= (P/uv) 1/2 Where P= power dissipated, u= absolute viscosity, v= volume to which power is applied.

3. What is the dimension formula of mean velocity gradient?
a) 1/T
b) 1/T2
c) T
d) T2

Answer: a [Reason:] Mean velocity gradient is expressed in metre per second/m or Sec-1, so it has dimension formula of 1/T.

4. What is the desirable value of mean velocity gradient in a flocculator?
a) 20-50sec-1
b) 20-75sec-1
c) 50-100sec-1
d) 30-50sec-1

Answer: b [Reason:] The desirable value of mean velocity gradient (G) in a flocculator is 20-75sec-1 and for ‘Gt’ is 2*104 to 6*104 where ‘Gt’ is the ratio of power induced rate of flow to displace mean induced rate of flow.

5. What is the normal value of the detention period adopted in a flocculator for design purpose?
a) 30min
b) 60min
c) 90min
d) 100min

Answer: a [Reason:] The detention period in a flocculator for design purpose should be in the range of 10-40min and its normal value should be 30min.

6. The design value of the velocity of flow in a flocculator is _______
a) 0.2-0.8m/s
b) 0.3-0.5m/s
c) 0.6-0.8m/s
d) 0.1-0.5m/s

Answer: a [Reason:] The normal value of the velocity of flow in a flocculator is 0.4m/s and its range is 0.2-0.8m/s.

7. What is the detention period of a clarifier used in the treatment of water?
a) 1hour
b) 2hours
c) 3hours
d) 4hours

Answer: c [Reason:] The detention period of a clarifier is lower than in the plain sedimentation tank and its range is 2.5 to 3 hours.

8. The pulsator clarifier is a type of
a) Horizontal flow sludge tank
b) Vertical flow sludge tank
c) Circular sludge tank
d) Plain sedimentation tank

Answer: b [Reason:] The pulsator clarifier is a vertical flow sludge tank in which pulse is generated at interval of 30Sec to give rapid flow for 5-10Sec resulting in the alternative rising up of the sludge blanket.

9. Which device solved the problem of unstable hydraulic conditions and operation of sludge removal equipment?
a) Centrifugal pump
b) Pulsator clarifier
c) Tube settler
d) Flocculator

Answer: c [Reason:] Tube settler solved the problem of unstable hydraulic conditions and operation of sludge removal equipment by providing laminar flow conditions for sedimentation.

10. In which device, the primary mixing is followed by a secondary reaction zone resulting in formation of sludge blanket?
a) Centrifugal pump
b) Flocculator
c) Tube settler
d) Solid contact clarifier

Answer: d [Reason:] In Solid contact clarifier, sludge blanket is formed where straining action occurs to remove some of the finer particles. The thickness of the blanket is 1m.

11. In which type of tube settler, tubes are slightly inclined in the direction of normal flow?
a) Solid contact clarifier
b) Steeply inclined tube settler
c) Vertical tube settler
d) Horizontal tube settler

Answer: d [Reason:] In Horizontal tube settler, the tubes are slightly inclined in the direction of normal flow and the sludge settled is drained by filter backwash.They are used in small plants.

## Set 3

1. The fluoride concentration for prevention of dental caries is
a) 1mg/l
b) 2mg/l
c) 3mg/l
d) 4mg/l

Answer: a [Reason:] Dental caries occur in children. It can be prevented when the concentration of fluoride in water is between 0.7 to 1.2mg/l.

2. In which process, the fluoride content of water is raised?
a) Chlorination
b) Fluoridation
c) Defluoridation
d) Flocculation

Answer: b [Reason:] When the fluoride content is low, it will cause dental caries. So the fluoride content of water is raised and the process is called fluoridation.

3. Which of the following is not used as a fluoride compound?
a) Sodium fluoride
b) Sodium silico fluoride
c) Hydrofluosilicic acid
d) Sodium fluro carbonate

Answer: d [Reason:] Sodium fluoride, sodium silico fluoride, hydrofluosilicic acid and sodium fluro carbonate are the fluoride compounds used for fluoridation.

4. Which of the following is the pure compound?
a) Sodium fluoride
b) Sodium silico fluoride
c) Hydrofluosilicic acid
d) Sodium fluro carbonate

Answer: a [Reason:] Sodium fluoride is the most purest compound having 95-98% purity due to which, they are commonly used for fluoridation.

5. As far as safer handling is considered, which of the following is used for fluoridation?
a) Sodium fluoride
b) Sodium silico fluoride
c) Hydrofluosilicic acid
d) Sodium fluro carbonate

Answer: c [Reason:] For safer handling, hydrofluosilicic acid is preferred as even if it is spilled on the skin, it can be removed easily by washing in cold water.

6. State whether the following statement is true or false.
Fluorides in solution form are preferred over powdered form for fluoridation.
a)True
b)False

Answer: b [Reason:] Fluorides in powdered form like sodium fluoride or sodium fluosilicate are toxic and must be contained in air tight containers. Hence, they are not preferred.

7. What happens when water contains 8-20 ppm of fluoride concentration?
a) Blue baby disease
b) Crippling fluorosis
c) Dental fluorosis
d) Mottling of teeth

Answer: b [Reason:] Dental fluorosis is caused when fluoride concentration is above 3 ppm, whereas Crippling fluorosis is caused when the concentration of fluoride in water is between 8 and 20ppm.

8. The process of reducing the fluoride content from water is called _____
a) Chlorination
b) Fluoridation
c) Defluoridation
d) Flocculation

Answer: c [Reason:] When the fluoride content in water is high, it is essential to reduce the fluoride content to avoid health related problems and the process of reducing the fluoride content is called defluoridation.

9. Which of the following is not used for defluoridation?
a) Calcium phosphate
b) Copper sulfate
c) Alum
d) Bone charcoal

Answer: b [Reason:] Copper sulfate is used to remove taste, odor, color and control of algae growth and it is not used for fluoridation.

10. At which temperature, the bone is calcinated during defluoridation with calcium phosphate?
a) 100-200oC
b) 200-300oC
c) 400-600oC
d) 500-800oC

Answer: c [Reason:] During defluoridation by calcium phosphate, the bone is calcinated at 400-600oC for 10 minutes followed by mineral acid treatment.

11. One cubic metre of bone can treat how much quantity of water containing 3.5 ppm of fluoride?
a) 10m3
b) 100m3
c) 1000m3
d) 10000m3

Answer: b [Reason:] Bone is used in the filter for removal of fluorides. One cubic metre of bone can treat 100m3 water containing 3.5 ppm of fluoride.

12. Which material is used in contact filters for removal of fluorides?
a) Calcium phosphate
b) Copper sulfate
c) Synthetic tri-calcium phosphate
d) Bone charcoal

Answer: c [Reason:] Synthetic tri-calcium phosphate is used in contact filters for removal of fluorides. It is made from the milk of lime and phosphoric acid.

13. Which material is used for removing fluorides from hard water containing 3ppm of fluorides?
a) Lime
b) Copper sulfate
c) Synthetic tri-calcium phosphate
d) Bone charcoal

Answer: a [Reason:] Lime is suitable for removing fluorides from hard water containing less than 4ppm. Magnesium is also removed when this material is used for defluoridation.

14. Fluorex is a special mixture of
a) Di-calcium phosphate and carbon
b) Tri-calcium phosphate and hydroxyapatite
c) Di-calcium phosphate and phosphoric acid
d) Tri-calcium phosphate and carbon

Answer: b [Reason:] Fluorex is used for removing fluoride and it is a special mixture of tri-calcium phosphate and hydroxyapatite. It is used as a filter medium.

## Set 4

1. The normal dose of chlorine during plain chlorination is
a) 0.5-1ppm
b) 0.1-0.2ppm
c) 0.1-0.5ppm
d) 1-2ppm

Answer: a [Reason:] The normal dose of chlorine during plain chlorination is 0.5-1ppm. It should be applied when the water is clear and its turbidity does not exceed 30 ppm.

2. The normal dose of chlorine during pre chlorination is
a) 0.5-1ppm
b) 0.1-0.2ppm
c) 0.1-0.5ppm
d) 1-2ppm

Answer: c [Reason:] The normal dose of chlorine during pre chlorination is 0.5-1ppm. This results in the oxidization of organic matter and reduce the amount of coagulants required in water.

3. The normal dose of chlorine during post chlorination is
a) 0.5-1ppm
b) 0.1-0.2ppm
c) 0.1-0.5ppm
d) 1-2ppm

Answer: b [Reason:] The normal dose of chlorine during post chlorination is 0.1-0.2ppm. It is used for protection against contamination.

4. In which form of chlorination, chlorine applies to raw water supply as it enters the distribution system?
a) Plain chlorination
b) Pre chlorination
c) Super chlorination
d) Double chlorination

Answer: a [Reason:] In plain chlorination, chlorine applies to raw water supply as it enters the distribution system. It checks the growth of organic matter, algae and remove taste and odor from water.

5. _________ is the application of chlorine before filtration.
a) Plain chlorination
b) Pre chlorination
c) Super chlorination
d) Double chlorination

Answer: b [Reason:] Pre chlorination is the application of chlorine before filtration. This results in a decrease of coagulant required in water.

6. __________ is the application of chlorine at two or more points in the purification process.
a) Plain chlorination
b) Pre chlorination
c) Super chlorination
d) Double chlorination

Answer: d [Reason:] Double chlorination is the application of chlorine at two or more points in the purification process. It is applied just before when the water enters the sedimentation tank and after when it leaves the filter plant.

7. The normal dose of chlorine during break point chlorination is
a) 0.5-1ppm
b) 0.1-0.2ppm
c) 3-7ppm
d) 1-2ppm

Answer: b [Reason:] The normal dose of chlorine during post chlorination is 0.1-0.2ppm. It kills bacteria and oxidizes the organic matter present in water.

8. Which of the following regarding break point chlorination is wrong?
a) It will have adequate chlorine residual
b) It will remove manganese
c) It will remove taste and odor
d) It is determined by physical test

Answer: d [Reason:] The break point chlorination is determined by laboratory test. It is represented by instantaneous yellow color if the orthotolidine test when the orthotolidine test is applied.

9. _______ is the application of chlorine beyond the stage of the break point.
a) Plain chlorination
b) Pre chlorination
c) Super chlorination
d) Double chlorination

Answer: c [Reason:] Super chlorination is the application of chlorine beyond the stage of the break point. It is followed by a contact period of 30-60 minutes.

10. The process of removal of excess chlorine from water is
a) Plain chlorination
b) Dechlorination
c) Super chlorination
d) Double chlorination

Answer: b [Reason:] Dechlorination is the process of removal of excess chlorine from water. It can be done aeration or by the use of various chemicals like activated carbon, sodium sulfate in liquid form.

11. Which color is formed when orthotolidine is oxidized by chlorine?
a) Yellow
b) Orange
c) Green
d) Blue

Answer: a [Reason:] Orthotolidine is a colorless organic liquid that is oxidized into a yellow colored compound called Holoquinone.

## Set 5

1. The Total hardness of soft water as CaCO3 equivalent is
a) 50ppm
b) 100ppm
c) 150ppm
d) 200ppm

Answer: a [Reason:] The Total hardness of soft water as CaCO3 equivalent is 50ppm while of hard water, it is 150-300ppm.

2. The total hardness of water is 100ppm as CaCO3 equivalent. What is the type of water?
a) Soft
b) Moderately hard
c) Hard
d) Very hard

Answer: b [Reason:] When the total hardness of water is between 50 and 150ppm as CaCO3 equivalent, then the water is moderately hard.

3. The permanent hardness in water is due to the presence of
a) Sulfates, Chlorides
b) Sulfates, chlorides, nitrates
c) Carbonates and bicarbonates
d) Sulfates and carbonates

Answer: b [Reason:] The permanent hardness is due to the presence of sulfates, chlorides and nitrates of calcium and magnesium in water.

4. The temporary hardness in water is due to the presence of
a) Sulfates, Chlorides
b) Sulfates, Chlorides, nitrates
c) Carbonates and bicarbonates
d) Sulfates and carbonates

Answer: c [Reason:] The temporary hardness is due to the presence of carbonates and bicarbonates of calcium and magnesium in water.

5. Which of the following causes the temporary hardness?
a) CaSO4
b) MgSO4
c) MgCl2
d) Ca (HCO3) 2

Answer: d [Reason:] As the temporary hardness is caused due to the presence of carbonate and bicarbonates, therefore Ca (HCO3) 2 cause temporary hardness.

6. The permanent hardness is also known as
a) Pseudo hardness
b) Non carbonate hardness
c) Carbonate hardness
d) Brinell hardness

Answer: b [Reason:] The permanent hardness is also known as non carbonate hardness whereas temporary hardness is also known as carbonate hardness.

7. Which of the following is not used for removing the permanent hardness?
a) Lime soda process
b) Zeolite process
c) Demineralization