# Multiple choice question for engineering

## Set 1

1. How does the revolving field produce emf in the bars?

a) revolving field produces emf of fundamental frequency in the bars

b) revolving field produces emf of third frequency in the bars

c) revolving field produces emf of no frequency in the bars

d) revolving field produces emf of sinusoidal frequency in the bars

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2. What happens if the resistance of the end rings is negligible?

a) resistance coming in each current path is resistance of three bars

b) resistance coming in each current path is resistance of four bars

c) resistance coming in each current path is resistance of two bars

d) resistance coming in each current path is resistance of five bars

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3. What factors does the current in the bars depend on?

a) emfs, position of bars in magnetic field

b) instantaneous emfs, position of bars in magnetic field

c) emf

d) instantaneous emf

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4. The end resistance, if not negligible, will tend to distort the bar current distribution from being sinusoidal?

a) true

b) false

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5. What is the formula for the maximum current in end ring, if the current in all bars are maximum at the same time?

a) maximum current in the end ring= bars per pole * 2 * current per bar

b) maximum current in the end ring= (bars per pole / 2) * current per bar

c) maximum current in the end ring= bars per pole / 2 / current per bar

d) maximum current in the end ring= bars per pole * 2 / current per bar

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6. Given the bars per pole is 6 and the current per bar is 20 A, what is the value of the maximum current in the end rings?

a) 60 A

b) 80 A

c) 90 A

d) 70 A

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7. What is the formula for the maximum value of current through end ring, when the current is not maximum in all the bars under one pole at the same time?

a) maximum current in end ring= (2*3.14) / (bars per pole/2*no of poles) * current per bar

b) maximum current in end ring= (2/3.14) * (bars per pole/2*no of poles) / current per bar

c) maximum current in end ring= (2*3.14) * (bars per pole/2*no of poles) * current per bar

d) maximum current in end ring= (2*3.14) / (bars per pole/2*no of poles) / current per bar

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8. What is the formula for the maximum current through each bar?

a) maximum value of the current through each bar = 2 * current through each bar

b) maximum value of the current through each bar = (2 * current through each bar)1/2

c) maximum value of the current through each bar = (2 * current through each bar)2

d) maximum value of the current through each bar = (2 * current through each bar)1/3

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9. What is the formula for the rms value of the end ring current?

a) rms value of end ring current = (bars per pole * current per bar) / (3.14*no of poles)

b) rms value of end ring current = (bars per pole * current per bar) * (3.14*no of poles)

c) rms value of end ring current = (bars per pole * current per bar) / (no of poles)

d) rms value of end ring current = (bars per pole * current per bar) / (3.14+no of poles)

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10. The value of the current density is chosen for the end rings such that the desired value of rotor resistance is obtained?

a) true

b) false

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11. How is the current density of the rotor bars chosen with respect to the end rings?

a) current density of rotor bars < current density of end rings

b) current density of rotor bars > current density of end rings

c) current density of rotor bars = current density of end rings

d) current density of rotor bars <= current density of end rings

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12. What is the formula for the area of the ring?

a) area of the ring = depth of the end ring + thickness of the end ring

b) area of the ring = depth of the end ring – thickness of the end ring

c) area of the ring = depth of the end ring / thickness of the end ring

d) area of the ring = depth of the end ring * thickness of the end ring

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## Set 2

1. What is the formula for the fundamental relationship for the design of ventilation system?

a) head of air inside the machine = hydrodynamic resistance * volume of air passing2

b) head of air inside the machine = hydrodynamic resistance + volume of air passing2

c) head of air inside the machine = hydrodynamic resistance – volume of air passing2

d) head of air inside the machine = hydrodynamic resistance / volume of air passing2

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2. What is the formula for the total head produced?

a) total head produced = ∑ coefficient of hydrodynamic resistance + volume of air passing per second2

b) total head produced = ∑ coefficient of hydrodynamic resistance – volume of air passing per second2

c) total head produced = ∑ coefficient of hydrodynamic resistance * volume of air passing per second2

d) total head produced = ∑ coefficient of hydrodynamic resistance / volume of air passing per second2

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3. What are the ventilating parts in the ventilating circuits?

a) sharp or projecting inlet edges

b) inlet corners

c) variations in cross-sections of air paths

d) sharp or projecting inlet edges, inlet corners, variations in cross-sections of air paths

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4.What is the range of the coefficients of hydrodynamic resistances for the protruding edges at inlet?

a) 40-50 * 10^{-3}

b) 40-60 * 10^{-3}

c) 30-50 * 10^{-3}

d) 30-40 * 10^{-3}

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^{-3}.

5. What is the range of the coefficients of hydrodynamic resistances for the rectangular edges at inlet?

a) 10-20 * 10^{-3}

b) 30 * 10^{-3}

c) 20-30 * 10^{-3}

d) 20-25 * 10^{-3}

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^{-3}.

6. What is the range of the coefficients of hydrodynamic resistances for the rounded edges at inlet?

a) 12-20 * 10^{-3}

b) 10-20 * 10^{-3}

c) 15-20 * 10^{-3}

d) 12-30 * 10^{-3}

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^{-3}.

7. What factor/factors are required to evaluate the hydrodynamic resistance?

a) area of cross section

b) hydrodynamic coefficients

c) area of cross section or hydrodynamic coefficients

d) area of cross section and hydrodynamic coefficients

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8. How many data are required for the design of fan?

a) 3

b) 4

c) 5

d) 6

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9. How many steps are required in the design of the fan?

a) 7

b) 8

c) 9

d) 6

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10. What is the formula for the volume of air?

a) volume of air = 0.9 * losses in kW / difference of air temperature at inlet and outlet

b) volume of air = 0.9 * losses in kW * difference of air temperature at inlet and outlet

c) volume of air = 0.9 / losses in kW * difference of air temperature at inlet and outlet

d) volume of air = 1 / 0.9 * losses in kW * difference of air temperature at inlet and outlet

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11. What is the range of the difference of air temperature at inlet and outlet?

a) 11-15^{0}C

b) 10-13^{0}C

c) 12-16^{0}C

d) 14-18^{0}C

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^{0}C. The difference of air temperatures at inlet and outlet has a maximum value of 16

^{0}C.

12. What is the formula for the area of outlet opening?

a) area of outlet opening = maximum air passing per second / 0.42 * peripheral speed

b) area of outlet opening = maximum air passing per second * 0.42 * peripheral speed

c) area of outlet opening = maximum air passing per second * 0.42 / peripheral speed

d) area of outlet opening = 1/maximum air passing per second * 0.42 * peripheral speed

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13. What is the formula of the width of fan?

a) width of fan = area of outlet opening * 2.88 * outside diameter */ coefficient of utilization

b) width of fan = 1/ area of outlet opening * 2.88 * outside diameter * coefficient of utilization

c) width of fan = area of outlet opening * 2.88 * outside diameter * coefficient of utilization

d) width of fan = area of outlet opening / 2.88 * outside diameter * coefficient of utilization

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14. What is the formula for the number of blades?

a) number of blades = 3.14 * outside diameter * (1.25 – 1.5)* width of fan

b) number of blades = 3.14 / outside diameter * (1.25 – 1.5)* width of fan

c) number of blades = 3.14 * outside diameter / (1.25 – 1.5)* width of fan

d) number of blades = 3.14 * outside diameter * (1.25 – 1.5) / width of fan

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15. What is the formula for the maximum air passing per second at maximum efficiency ?

a) maximum air passing = 2 * volume of air passing per second

b) maximum air passing = volume of air passing per second

c) maximum air passing = 2 / volume of air passing per second

d) maximum air passing = volume of air passing per second / 2

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## Set 3

1. What type of coils are made use of for machines with small number of poles?

a) iron wound coils

b) wire wound coils

c) rectangular coils

d) square coils

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2. What type of strips is made use of for field coils of small alternators?

a) wood covered rectangular strips

b) bare copper strips

c) glass covered rectangular strips

d) iron strips

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3. What should be the maximum width of the edge conductors used in the large alternators?

a) 6 mm

b) 5 mm

c) 4 mm

d) 3 mm

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4. For machines with Class B insulation, how many layers of inter turn insulation is made use of and what is the distance between the layers?

a) 4, 0.18 mm

b) 3, 0.25 mm

c) 2, 018 mm

d) 2, 0.25 mm

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5. What material is the paper strips stuck with?

a) synthetic resin varnish

b) shellac

c) synthetic resin varnish and shellac

d) synthetic resin varnish or shellac

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6. What is the thickness of the flanges and what material is used in the flanges?

a) 10 mm thick, resins

b) 10 mm thick, asbestos

c) 15 mm thick, asbestos

d) 10 mm thick, bakelized asbestos

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7. Current is passed simultaneously through the conductors to raise the temperature of the field coil?

a) true

b) false

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8. During the pressing and consolidation by how much is the thickness of the interturn insulation reduced to?

a) 0.36 mm to 0.26 mm

b) 0.36 mm to 0.25 mm

c) 0.30 mm to 0.25 mm

d) 0.32 mm to 0.25 mm

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9. How many layers does the machine with Class F insulation consists of?

a) 2

b) 3

c) 4

d) 5

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10. What is the thickness of the layers of Class F insulation and what material is layers made of?

a) 0.18 mm, asbestos paper

b) 0.10 mm, asbestos paper

c) 0.18 mm, thick epoxy treated asbestos paper

d) 0.10 mm, thick epoxy treated asbestos paper

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11. What is the lamination material of the pole body and the thickness of the pole body insulation?

a) epoxy resin, 5 mm thick

b) epoxy resin. 4 mm thick

c) asbestos, 4 mm thick

d) asbestos 5 mm thick

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12. What is the range of the pressure under which the field coils are consolidated?

a) 4-10 MN per m^{2}

b) 3-10 MN per m^{2}

c) 4-12 MN per m^{2}

d) 4-15 MN per m^{2}

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^{2}. The maximum value of the pressure under which the field coils are 12 MN per m

^{2}.

13. What is the range of the exciter voltage in the field coils?

a) 50-100 V

b) 150-300 V

c) 200-400 V

d) 50-400 V

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14.The field winding should be designed for a voltage from 15-20% less than the exciter voltage?

a) true

b) false

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15. What is the formula for the voltage across each field coil?

a) voltage across each field coil = (0.8-0.85)*exciter voltage/number of poles

b) voltage across each field coil = (0.8-0.85)*exciter voltage*number of poles

c) voltage across each field coil = (0.8-0.85)/exciter voltage*number of poles

d) voltage across each field coil = (0.8-0.85)/exciter voltage/number of poles

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## Set 4

1. What does design mean?

a) creative physical relationship of theoretical concepts

b) producing the hardware

c) idea made into a drawing

d) idea laid out as thesis

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2. What are the factors which are employed in the engineering design?

a) application of science

b) application of science and technology

c) application of science and innovation

d) application of science, technology and innovation

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3. How many considerations are present in the design?

a) 6

b) 4

c) 5

d) 3

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4. What is the condition of the good design?

a) machine having operating life of 5-10 years

b) machine having operating life of 10-15 years

c) machine having operating life of 15-20 years

d) machine having operating life of 20-30 years

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5. What are the factors which are important in design when it comes to the machines used in power systems?

a) reliability

b) durability

c) cost

d) reliability and durability

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6. How many parts are used in the design of the machines?

a) 2

b) 3

c) 4

d) 5

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7. How many limitations are present in the design of the machine?

a) 7

b) 8

c) 9

d) 10

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8. How does the saturation levels limit the design?

a) saturation levels decrease the flux density

b) saturation levels increase the flux density

c) saturation levels provides no flux density

d) saturation levels provide very high flux density

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9. What happens if the insulating material is operated beyond the maximum allowable temperature?

a) the insulation gets damaged

b) the insulation is boosted

c) the insulating materials peels off

d) the lifetime is drastically reduced

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10. How many types of stresses are present in the machine?

a) 2

b) 3

c) 4

d) 5

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11. What is the relationship between efficiency and the operating costs?

a) efficiency is directly proportional to the square of the operating cost

b) efficiency is directly proportional to the operating costs

c) efficiency is indirectly proportional to the operating costs

d) efficiency is indirectly proportional to the square of the operating costs

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12. The mechanical parts design is very much important in which type of machines?

a) low speed machines

b) high speed machines

c) medium speed machines

d) low voltage machines

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13. What is the relationship between power factor, current and conductor sizes?

a) poor power factor leads to small amount of current and hence high conductor sizes should be used

b) high power factor leads to small amount of current and hence small conductor sizes should be used

c) poor power factor leads to high amount of current and hence high conductor sizes should be used

d) high factor leads to small amount of current and hence high conductor sizes should be used

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## Set 5

1. What is the formula for the mean diameter of the magnet coils?

a) mean diameter = inside diameter of coil + outer diameter of coil / 2

b) mean diameter = inside diameter of coil – outer diameter of coil / 2

c) mean diameter = inside diameter of coil * outer diameter of coil / 2

d) mean diameter = inside diameter of coil / outer diameter of coil / 2

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2. What is the formula for the outside diameter of the magnet coils?

a) outside diameter = mean diameter + 2*depth of winding

b) outside diameter = mean diameter + depth of winding

c) outside diameter = mean diameter – 2*depth of winding

d) outside diameter = mean diameter – depth of winding

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3. What is the formula for depth of winding of the magnet coils?

a) depth of winding = mean diameter of coil – inner diameter

b) depth of winding = mean diameter of coil + inner diameter

c) depth of winding = mean diameter of coil – 2* inner diameter

d) depth of winding = mean diameter of coil + 2*inner diameter

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4.What is the formula of the cross winding area of the magnet coils?

a) cross winding area = axial length of coil + depth of winding

b) cross winding area = axial length of coil – depth of winding

c) cross winding area = axial length of coil * depth of winding

d) cross winding area = axial length of coil / depth of winding

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5. What is the formula for the length of mean turn of magnet coils?

a) length of mean turns = 3.14 * (inside diameter of coil + depth of windings)

b) length of mean turns = 3.14 / (inside diameter of coil + depth of windings)

c) length of mean turns = 3.14 * (inside diameter of coil * depth of windings)

d) length of mean turns = 3.14 + (inside diameter of coil + depth of windings)

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6. What is the formula for the total heat dissipating surface of the magnet coils?

a) total heat dissipating surface = length of mean turn * depth of winding * axial length of coil

b) total heat dissipating surface = length of mean turn * depth of winding + axial length of coil

c) total heat dissipating surface = 2 * length of mean turn * (depth of winding + axial length of coil)

d) total heat dissipating surface = 2 * length of mean turn * depth of winding * axial length of coil

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7. What is the formula for the outer cylindrical heat dissipating surface of the magnet coils?

a) outer cylindrical heat dissipating surface = 3.14 * outer diameter of coil + axial length of coil

b) outer cylindrical heat dissipating surface = 3.14 + outer diameter of coil + axial length of coil

c) outer cylindrical heat dissipating surface = 3.14 / outer diameter of coil + axial length of coil

d) outer cylindrical heat dissipating surface = 3.14 * outer diameter of coil * axial length of coil

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8. What is the formula of the inner cylindrical heat dissipating surface?

a) inner cylindrical heat dissipating surface = length of mean turn * axial length of coil

b) inner cylindrical heat dissipating surface = 2 *length of mean turn * axial length of coil

c) inner cylindrical heat dissipating surface = length of mean turn / axial length of coil

d) inner cylindrical heat dissipating surface =1 / length of mean turn * axial length of coil

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9. What is the ambient temperature of the magnet coils?

a) 100C

b) 150C

c) 200C

d) 250C

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10. What is the formula for the area of the conductors of the magnet coils?

a) area of the conductors = mmf per coil * resistivity of conductor * length of mean turn * terminal voltage

b) area of the conductors = mmf per coil / resistivity of conductor * length of mean turn * terminal voltage

c) area of the conductors = mmf per coil * resistivity of conductor * length of mean turn / terminal voltage

d) area of the conductors = mmf per coil * resistivity of conductor / length of mean turn * terminal voltage

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11. What is the value of the resistivity temperature coefficient of copper?

a) 0.017 ohm per m per mm^{2}

b) 0.0173 ohm per m per mm^{2}

c) 0.01734 ohm per m per mm^{2}

d) 0.0175 ohm per m per mm^{2}

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12. What is the value of the resistance temperature coefficient of copper?

a) 0.00393 per 0C

b) 0.0040 per 0C

c) 0.00383 per 0C

d) 0.00373 per 0C

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13. What is the formula for total number of turns in the magnet coils?

a) total number of turns = mmf per coil * current

b) total number of turns = mmf per coil / current

c) total number of turns = mmf per coil – current

d) total number of turns = mmf per coil + current

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14. What is the formula for the total winding area?

a) total winding area = number of turns * area of each conductor * space factor

b) total winding area = number of turns / area of each conductor * space factor

c) total winding area = number of turns * area of each conductor / space factor

d) total winding area = 1/number of turns * area of each conductor * space factor