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Tuesday, May 24, 2016

Electrical Machines - Mixed Bag

Answer the following in brief:
1. Why do round rotor synchronous generators have small diameters and large axial length of core?
2. Why does open circuit synchronous machine have curved shape?
3. Do slot harmonics affect the torque-speed curve of a three-phase induction motor?
4. Why are brushes of a d.c. machine made of carbon?
5. Why are interpoles used in a d.c. machines?
6. What is the all-day efficiency of a transformer?
7. Why are core-type transformer used in high-voltage circuits?
8. How are iron losses in a transformer made very small?

1. Synchronous generators of round rotor types are designed mostly of 2 pole type running al 3000 rpm. They are designed for lower values of diameter in order to limit the centrifugal forces which have a profound influence on the design, to get the designed value of {D^2}L for a given output. the length of core is therefore required to be kept large.

2. The o.c.c. of a synchronous machine have a curved shaped because of saturation of magnetic core. At low values If when iron is unsaturated. the o.c.c. is almost linear but high values of If magnetic core gels saturated, hence the o.c.c. becomes curved.

3. A three phase winding carrying a sinusoidal currents produces harmonics of the order of n = 6N  \pm  1, N is an integer. The movement of the harmonics is with or against the direction of the rotation depending upon the sign. These harmonics produce torque of the same, general torque slip shape as that of fundamental but with synchronous speed corresponding to that harmonics. Thus, the torque slip characteristics will get affected.

4. The function of brushes is to collect current from commutator.

5. Inter poles are used in d.c. machines to balance or to nullify the effect of cross-magneting component of armature reaction field.

6. All  Day Efficiency = output in kWh in 24 hours/input in kWh in 24 hours

7. High voltage transformers are designed with core type construction because this provides easy maintenance and tappings can be provided more conveniently.

8. Transformers are having been made of core of superior quality and thin laminations, has lesser eddy current losses and having selecting a soft magnetic material will have low hysteresis loss and hence iron losses can be made small.

(These questions are taken from AMIE exams.)

Wednesday, May 18, 2016

Electrical Machines - Mixed Bag

Answer the following in brief:
1. Why is in a normal synchronous machine Xq smaller than Xd?
2. Resistance of dampers for synchronous generators is lower than that of dampers for synchronous motors. Why?
3. What is meant by an infinite bus?
4. Distribution transformers are designed for lower iron losses. Why?
5. What are the functions of transformer oil?
6. Do we use laminations for all iron parts of an electrical machine. If not, why?
7. Explain why in an induction motor a high value of rotor resistance is preferred at starting?
8. Why is an ordinary induction motor not started at normal voltage?
9. Why is the rotor core loss almost absent in normal operating condition of an induction motor?
10. Why is star-delta starter not suitable for high voltage induction motor?


  1. The quadrature axis reactance in a salient pole machine is smallest than the direct axis reactance, because a given current component in that axis gives rise to smaller flux, the reluctance of the magnetic path including the interpolar spaces.
  2. In synchronous generators, damper winding is provided to suppress the negative sequence field and to damp the oscillations when the machine starts hunting, hence, damper winding has lowest value. In synchronous motors the function of damper winding is to provide starting torque and to develop damping power when the machine starts hunting hence, it is designed to have high resistance.
  3. The terminal voltage and its frequency is held constant independent of load drawn from the bus. then the bus is called infinite bus.
  4. Distribution transformer are kept on for 24 hours even if load is there or not. hence continuously iron losses takes place, therefore these are designed for low value of iron losses.
  5. Functions of transformer oil are (a) To create insulation to insulated conductors and coils. (b) To provide cooling medium to dissipate or carryover heat to the tank walls.
  6. We make use of laminations for those iron pans only in which fluctuations of flux takes place. For all those parts, where, there are no fluctuations, no eddy currents are included, hence laminations arc not provided.
  7. By adding external resistance to the rotor circuit, the starting torque of the motor is increased as a result of improved power factor.
  8. An ordinary motor if. started at normal voltage, will draw very high starting current, which may be 5 to 7 times the full load current and hence motor may be damaged.
  9. In normal operating conditions of an induction motor, frequency of induced rotor emf and hence, rotor current is small. This will have very small value of iron losses because hysteresis and copper losses are both dependent on frequency.
  10. Star-delta starter is not suitable for voltage exceeding 3000 V because of the excessive number of stator turns needed for delta connection.
(Author is associated with

Saturday, May 14, 2016

Soil Mechanics - Permeability of Soil

Q.1. In terms of permeability of soil, following statement is true
(a)  clays>silts>sands
(b)  silts>sands>clays
(c)  sands>silts>clays
(d)  none of these

Q.2. Match List-I with List-II and select correct answer by using the codes given below the lists.
A.     Gravels
B.     Sand
C.     Silt
D.     Clays

1.      10-7 cm/sec.
2.      10-6 cm/sec.
3.      10-1 cm/sec.
4.      10 cm/sec.
(a)  A1,B2,C3,D4
(b)  A3,B4,C1,D2
(c)  A4,B3,C2,D1
(d)  A2,B4,C3,D1 (IES 1992)

Q.3. Constant head permeameter is used to test permeability of
(a)  silt
(b)  clay
(c)  coarse sand
(d)  fine sand

Q.4. Falling head permeameter is used to test permeability of
(a)  silt
(b)  clay
(c)  fine sand
(d)  all these

Q.5. For practically impervious type of soil, the coefficient of permeability is determined using
(a)  variable head test
(b)  constant head test
(c)  consolidation test
(d)  pumping test (IES 1993)

Q.6. Using falling head permeameter, value of permeability(k) is determined from the formula
(a)  \frac{{a.L}}{{A({t_2} - {t_1})}}.{\log _e}\frac{{{h_2}}}{{{h_1}}}
(b)  \frac{{a.L}}{{A({t_2} - {t_1})}}.{\log _e}\frac{{{h_1}}}{{{h_2}}}
(c)  \frac{{A.L}}{{a({t_2} - {t_1})}}.{\log _e}\frac{{{h_2}}}{{{h_1}}}
(d)  \frac{{a.L}}{{A({t_2} - {t_1})}}.{\log _{10}}\frac{{{h_2}}}{{{h_1}}}
Where t2 - t1 is the time taken for the head to fall from h1 to h2, A is cross-sectional area of sample, a is cross-sectional area of the standpipe and L is height of sample.

Q.7. If during a permeability test on a soil sample with a falling head permeameter, equal time  intervals are noted for drop of head from h1 to h2 and again from h2 to h3, then which one of the following relations would hold good?
(a)  h12 = h2h3
(b)  h22 = h1h3
(c)  h32 = h2h1
(d)  none of these (IES 1995)

Q.8. In an unconfined flow pumping test, acquifer is
(a)  underlain by permeable stratum
(b)  underlain by impermeable stratum
(c)  contained between permeable and impermeable stratum
(d)  none of these

Q.9. In a confined flow pumping test, acquifer is
(a)  underlain by impermeable stratum
(b)  underlain by permeable stratum
(c)  contained between impermeable and permeable stratum
(d)  contained between impermeable strata, above and below

Q.10. In an unconfined flow pumping set, pumping generates a radial flow
(a)  towards the filter well
(b)  beyond filter well
(c)  towards acquifer
(d)  towards impermeable stratum

1. c
2. c
3. c
4. d
5. d
6. b
7. b
8. b
9. d
10. a

Wednesday, April 27, 2016

Soil Mechanics - Classification of Soils

Q.1. Soils transmitted by the wind are called
(a) alluvial soil
(b) residual soil
(c) aeolin soil
(d) marine soil (AMIE Summer2000)

Q.2. A fine-grained soil transported by water and deposited at the bottom of the lake is
(a) lacustrine soil
(b) alluvial soil
(c) loess
(d) dune sand (IES 1996, AMIE Summer98)

Q.3. Match List I with List II and select the correct answer using the codes given below the lists:
List I (Type of Soil)
A. Lacustrine soil
B. Alluvial soil
C. Aeolin soil
D. Marine soil

List II (Mode of transportation and deposition)
1. Transportation by the wind
2. Transportation by running water
3. Deposited at the bottom of lakes
4. Deposited in sea water

(a) A1,B2,C3,D4
(b) A3,B2,C1,D4
(c) A3,B2,C4,D1
(d) A1,B3,C2,D4 (IES 1995)

Q.4. The description “sandy silty clay” signifies that
(a) the soil contains unequal proportions of three constituents such that sand > silt > clay
(b) the soil contains equal proportions of sand, silt and clay
(c) the soil contains unequal proportions of the three constituents such that clay > silt > sand
(d) there is no information regarding the relative proportions of the three  (GATE 1992)

Q.5. The shape of clay particle is usually
(a) angular
(b) flaky
(c) tubular
(d) rounded (GATE 1997)

Q.6. The swelling nature of black cotton soil is primarily due to the presence of
(a) Kaolinite
(b) Illite
(c) Montmorillonite
(d) Vermiculite (GATE 1993)

Q.7. The consistency of a saturated cohesive soil is affected by
(a) particle size distribution
(b) density index
(c) coefficient of permeability
(d) water content (GATE 1995)

Q.8. According to IS code, size of particles of silt will be between
(a) 0.075 mm and 4.75 mm
(b) 0.002 mm and 0.075 mm
(c) 0.001 mm and 0.06 mm
(d) 0.075 mm and 0.275 mm (AMIE Summer2000)

Q.9. Coefficient of curvature(Cc) is defined as
(a)  \frac{{{D_{30}}{D_{10}}}}{{{D_{60}}^2}}
(b)  \frac{{{D_{10}}^2}}{{{D_{60}}{D_{30}}}}
(c)  \frac{{{D_{30}}^2}}{{{D_{60}}{D_{10}}}}
(d)  \frac{{{D_{60}}^2}}{{{D_{30}}{D_{10}}}}

Q.10. Coefficient of uniformity(Cu) is
(a) D60/D10
(b) D10/D60
(c) D60/D30
(d) D60.D10 (AMIE Summer98)

1. c
2. a
3. b
4. c
5. b
6. c
7. d
8. b
9. c
10. a

Thursday, April 21, 2016

Soil Mechanics - Classification of Soils

Q.1. In soil classification, SP stands for
(a) well-graded soil
(b) poorly graded sands
(c) well graded sands
(d) inorganic clay (AMIE Summer99)

Q.2. Based on grain distribution analysis, the D10, D30 and D60 values of a given soil are 0.23 mm, 0.3 mm, and 0.41 mm respectively. As per IS code, the soil classification will be
(a) SW
(b) SP
(c) SM
(d)  SC (IES 2001)

Q.3. In soil classification, MH stands for
(a) inorganic silts of high compressibility
(b) organic silt of high compressibility
(c) well graded clay
(d) poorly graded clay (AMIE Summer99)

Q.4. Match list I(different types of soil) with List II(group symbols of I.S. classification) and select answer using codes given below the lists.
List I
A. Well graded gravel-sand mixture with little or no fines
B. Poorly graded sands or gravelly sand with little or no fines
C. Inorganic silts and very fine sands or clayey silts with low plasticity
D. Inorganic clays of high plasticity
List II
1. ML
2. CH
3. GW
4. SP
(a) A3,B1,C4,D2
(b) A3,B4,C1,D2
(c) A2,B4,C1,D3
(d) A2,B1,C4,D3 (IES 1994)

Q.5. A soil has a liquid limit of 45% and lies above the A-line when plotted on a plasticity chart. The group symbol of the soil as per IS soil classification is
(a) CH
(b) CI
(c) CL
(d) MI (IES 1997)

Q.6. Match List-I(soils) with List-II(group symbols) and select the correct answer using the codes given below the lists:

A. clayey gravel
B.  clayey sand
C. organic clay
D.  silty sand

1. SM
2. OH
3. SC
4. GC
(a) A3,B4,C2,D1
(b) A4,B3,C1,D2
(c) A4,B3,C2,D1
(d) A3,B4,C1,D2 (IES 2000)

Q.7. Consider the following statements in the context of aeolian soils:
1. The soil has low density and low compressibility.
2. The soil is deposited by the wind.
3. The soil has large permeability.
Of these statements
(a) 1,2 and 3 are correct
(b) 2 and 3 are correct
(c) 1 and 3 are correct
(d) 1 and 2 are correct (IES 1997)

Q.8. The structure of clay mineral as represented in the following figure is

(a) Illite
(b) montmorillonite
(c) silica
(d) rock

Q.9. The structure of clay mineral as represented in the following figure is of

(a) Illite
(b) montmorillonite
(c) silica
(d) rock

Q.10. When the product of rock weathering are not transported as sediments but remained in place, the soil is
(a) alluvial soil
(b) residual soil
(c) glacial soil
(d) aeolian soil

1. b
2. b
3. a
4. b
5. b
6. c
7. b
8. a
9. b
10. b

Friday, April 15, 2016

Soil Mechanics - Soil Classification

Q.1. For a soil sample, degree of shrinkage is less than 5%. The quality of the soil sample is
(a) poor
(b) very poor
(c) medium good
(d) good

Q.2. A soil having particles of nearly the same size is known as
(a) well graded
(b) uniformly graded
(c) poorly graded
(d) gap graded (GATE 1995)

Q.3. The particles size distribution curves are extremely useful for the classification of
(a) fine grained soils
(b) coarse grained soils
(c) both coarse grained and fine grained soils
(d) silts and clays (GATE 1996)

Q.4. Unified soil classification is based on
(a) grain size
(b) plasticity
(c) both(a) and (b)
(d) neither (a) nor (b)

Q.5. A soil is described by GP. the soil is
(a) silty soil
(b) inorganic soil
(c) poorly graded gravel
(d) low plasticity organic silt

Q.6. A soil is described by CI. The soil is
(a) poorly graded soil
(b) silty soil
(c) inorganic clay*
(d) none of these

Q.7. A soil is described as ML. The soil is
(a) well graded soil
(b) plastic fines soil
(c) peat
(d) low plasticity silt

Q.8. In Unified soil classification system, primary letters M, O and Pt stand for
(a) clay, sand and peat respectively
(b) organic, silt and peat
(c) silt, organic and peat
(d) none of these

Q.9. In Unified soil classification system, secondary letter M stands for
(a) well graded
(b) poorly graded
(c) Non-plastic fines
(d) low plasticity

Q.10. In Unified soil classification system, secondary letter
(a) W stands for well graded
(b) C stands for plastic fines
(c) L stands for low plasticity
(d) all of the above

1. d
2. b
3. c
4. c
5. c
6. c
7. d
8. c
9. c
10. d

Friday, April 8, 2016

Soil Mechanics - Basic Concepts

Q.1. If activity number of a soil sample is greater than 1.40, the soil would be
(a) active
(b) subjected to volume change
(c) both (a) and (b)
(d) neither (a) nor (b)

Q.2. If plasticity index is greater than 17%, the soil is
(a) plastic
(b) less plastic
(c) highly plastic
(d) stiff

Q.3. Data for a given soil sample is: Liquid limit = 60 %; Plastic limit = 30%; Flow index = 27. Toughness index of the soil will be
(a) 1.11
(b) -1.11
(c) 0.9
(d) -0.9

Q.4. Liquid index of a soil in semi-solid state would be
(a) 0
(b) 1
(c) >1
(d) negative

Q.5. Liquid index of a soil in solid state would be
(a) 0
(b) 1
(c) >1
(d) negative

Q.6. Liquid index of a soil in very stiff state, would be
(a) 0
(b) 1
(c) >1
(d) <0
Q.7. Liquidity index of a soil sample in soft state will be equal to
(a) 0
(b) 1
(c) <1
(d) >1

Q.8. Liquid index of a soil sample in liquid state(when disturbed) will be equal to
(a) 0
(b) 1
(c) <1
(d) >1

Q.9. The maximum possible value of Group Index for a soil is
(a) 20
(b) 30
(c) 40
(d) 50 (GATE 1991)

Q.10. When the natural moisture content of a soil is equal to its plastic limit, its liquidity index is equal to
(a) 0%
(b) 50%
(c) 75%
(d) 100% (AMIE Summer2000)

1. c
2. c
3. a
4. d
5. d
6. a
7. b
8. d
9. a
10. a