Conventional Question Practice Programe - IES Masteriesmaster.org/public/archive/2016/IM-1462628551.pdf · present in microprocessor 8085. ... Instruction register holds opcode of

1. (c)

2. (d)

3. (a)

4. (c)

5. (d)

6. (b)

7. (c)

8. (c)

9. (b)

10. (b)

11. (d)

12. (c)

13. (b)

14. (b)

15. (c)

16. (a)

17. (c)

18. (a)

19. (d)

20. (d)

21. (b)

22. (a)

23. (c)

24. (b)

25. (b)

26. (a)

27. (b)

28. (a)

29. (c)

30. (c)

31. (b)

32. (d)

33. (b)

34. (b)

35. (d)

36. (a)

37. (a)

38. (b)

39. (a)

40. (d)

41. (b)

42. (b)

43. (a)

44. (a)

45. (b)

46. (c)

47. (c)

48. (b)

49. (b)

50. (a)

51. (d)

52. (c)

53. (b)

54. (b)

55. (c)

56. (b)

57. (a)

58. (a)

59. (b)

60. (d)

61. (b)

62. (c)

63. (c)

64. (c)

65. (c)

66. (d)

67. (b)

68. (c)

69. (a)

70. (b)

71. (d)

72. (b)

73. (b)

74. (b)

75. (c)

76. (c)

77. (c)

78. (b)

79. (a)

80. (c)

81. (d)

82. (b)

83. (b)

84. (c)

85. (a)

86. (c)

87. (c)

88. (b)

89. (d)

90. (b)

91. (b)

92. (a)

93. (d)

94. (d)

95. (d)

96. (d)

97. (b)

98. (d)

99. (d)

100. (a)

101. (d)

102. (b)

103. (b)

104. (c)

105. (c)

106. (b)

107. (c)

108. (d)

109. (c)

110. (d)

111. (b)

112. (c)

113. (b)

114. (a)

115. (d)

116. (b)

117. (c)

118. (b)

119. (c)

120. (a)

Conventional Question Practice ProgrameDate: 26th March, 2016

ANSWERS

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(2) EE (Test-8), Objective Solutions, 26th March 2016

1. (c)PMMC type of instruments have eddy currentdamping as operating magnetic field whichproduce deflecting torque is very large.

2. (d)Let R1 and R2 be in parallel

1R =

1 2

1 1R R

1R = 1 1

10 5

R = 3.33

2R

R

= 1 22 21 2

R RR R

RR

= 1

1 1 2 2

RR R RR R R R

=3.33 3.335 1010 5

= 8.33 %

3. (a)

Voltage across meter = 3 21 10 10

= 0.1VCurrent through shunt = 5 – 0.001 = 4.999 A

Shunt resistance =0.1

4.999

= 149.9

4. (c)A rectifier type instrument measures theaverage value and it is calibrated for rms valueof sinusoidal ac.Harmonic component is

= 1/22 212.1 11.1 100

11.1

= 43.4 %

5. (d)A logarithmic scale compresses the scale, sowide range can be accumodated.

6. (b)At higher frequencies, there would be effects ofstray elements.

7. (c)

In Ferrodynamometer type instrument,

Td 1 2 cos In D´ arsonval galvanometer instrument,

Td KiIn electrodynamometer type instrument,

Td 1 2I I cos

8. (c)

Error =10 2 0.2 A100

% limiting error=0.2 1005

= 4 %

9. (b)Let the two resistance are R1 = R2 = R= 100 1%

parallel combination of R1 and R2,

Req = R2

eqR

R

= 1

2Standard deviation of the parallelcombination,

eqR = 1 2

2 2eq eq2 2

R R1 2

R RR R

where 1R and 2R are standard deviation ofR1 and R2 respectively.

eqR =2

eq 2R

R2

R

=2

212 12

=1 %2

10. (b)Temperature error and Friction error areobserved on DC measurement also.

11. (d)

In a moving iron instrument 2I

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EE (Test-8), Objective Solutions, 26th March 2016 (3)

1

2

=2

1

2

II

=22mA

1mA

= 4

2 = 14

= 200V4 = 50V

12. (c)

23A = 2 2

1 2A A ; 3 1 2A A jA

A3 = 2 212 5= 13A

13. (b)High torque to weight ratio indicates highvalue of torque or low value of weight. Boththe cases results into low friction loss.Accuracy does not depend on torque/weightratio. Ch70. (b)High torque to weight ratio indicates highvalue of torque or low value of weight. Boththe cases results into low friction loss.Accuracy does not depend on torque/weightratio. oose the correct answer

14. (b)Both Rayleigh’s current balance and lorentzmethod are absolute measurement method butRayleigh’s current balance is used for absolutemeasurement of current and Lorentz methodis used for measurement of resistance.

15. (c)

Absolute error = 2300 6V=100

So, When the voltmeter reads 222Vthe actual voltage will be

= 222 6V

= 216 to 228 VSo, option -(c) :

16. (a)Dead zone is defined as the largest change ofinput quantity for which ther is no output ofthe instrument.

And, dead time is the time before theinstrument begins to respond after themeasured quantity has been changed.

17. (c)

Im

Ish

50 A, 2000

RSince, m shI I I Now, I = 2Im

m2I = m shI I

Ish = Im

and, m mI R = shI .R

6 650 10 2000 50 10 R

R = 200018. (a)

For the instrument having weak magnetic field,air friction damping is used as it does notaffect the magnetic field of the instrument. Inall other system of damping, weak magneticfield may get distorted and gives serious error.

19. (d)

20. (d)

V

R

35V

600

1.2K

Effective resistance for voltmeter.

Reff = 600 120060 1200

= 600 1200 4001800

So, reading of voltmeter = 400 35400 R

5 = 400 35400 R

R = 2400

= 2.4K

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21. (b)Magnetic susceptibility of a paramagneticmaterial is positive and small.

22. (a)

Magnetic material

Magnetic susceptibility

DiamagneticParamagneticFerromagneticAntiferromagneticFerrimagnetic

Negative and smallPositive and smallPositive and very largePositive and smallPositive and large

23. (c)Permanent magnetic materials are difficult todemagnetize. Properties of permanent magneticmaterials :-• High retentivity• High coercivity• High permeability• Tall and wide Hysteresis loop• High saturation magnetization• High Curie temperature• High hysteresis loss because of large area of

hysteresis loop

24. (b)Magnetic susceptibility

= Relative permeability – 1= 5500 – 1 = 5499

25. (b)In diamagnetic materials, magneticsusceptibility is independent of temperaturebecause of absence of dipoles.

26. (a)Hydrogen molecule is a diamagnetic materialwhere as hydrogen atom is paramagnet.In principle any system that contains atoms,ions or molecules with unpaired spin can becalled a paramagnet. A system with unpairedspins that do not interact with each other aresystem with minimal interaction. Eachhydrogen atom has one non-interacting unpaidelectron.

27. (b)Properties of ferrites :• High resistivity

• High Permeability• High dielectric constant• Low eddy current loss• High Curie temperature• Extremely low dielectric loss

28. (a)Ferric Ion = Fe+3

26Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6

Outermost orbit: 3d6

Fe+3: 3 BMIn ferric ion, Fe+3, there are only threeunpaired electrons. Hence, magnetic momentwill be 3 Bohr magneton.

29. (c)Demagnetisation is the process in whichmagnetic field intensity (H) is negative butthe magnetic flux density (B) is positive.So, the portion bc of the curve representsdemagnetisation curve.

30. (c)Diamagnetic :

0m T

(i.e independent of temperature)Paramagnetic :

mCT

Ferromagnetic :

mC

T

( is Curie temperature)Antiferromagnetic :

mN

CT

( N is extrapolated Neel temperature)Ferrimagnetic :

mC

CT

( C is extropolated Curie temperature)31. (b)

32. (d)Ferromagnetic materials obey Curie - Weisslaw,

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m = CT

33. (b)

34. (b)Ferrimagnetic material :

35. (d)

36. (a)Permanent magnets are made from hardmagnetic materials. Hard magnetic materialshave broad hysteresis loop which have highretentivity and high coercivity.

37. (a)

38. (b)

Ferromagnetic, mC

1T

Paramagnetic, m1T

Antiferromagnetic, mC

1T

39. (a)Indexed addressing mode of addressing isvery useful for arrays. This addressing is notpresent in microprocessor 8085. The 5 types ofaddressing modes present in 8085 are.i) Immediate Addressing Mode: - An

immediate is transferred directly to theregister.Eg: - MVI A, 30H (30H is copiedinto the register A) MVI B,40H(40His copied into the register B).

ii) Register Addressing Mode: - Data iscopied from one register to anotherregister.Eg: - MOV B, A (the content of Ais copied into the register B) MOVA, C (the content of C is copied into theregister A).

iii) Direct Addressing Mode: - Data isdirectly copied from the given address tothe register.Eg: - LDA 3000H (The contentat the location 3000H is copied to theregister A).

iv) Indirect Addressing Mode: - The datais transferred from the address pointed

by the data in a register to otherregister.Eg: - MOV A, M (data istransferred from the memory locationpointed by the regiser to the accumulator).

v) Implied Addressing Mode: - This modedoesn’t require any operand. The data isspecified by opcode itself.Eg: - RAL CMP

40. (d)

8kb8 = 1kb = 1024 = 210

given 128 rows = 27

10

722

= 23 - columns

7 X address and 3 Y address are needed.

41. (b)Data bus is a bidirectional bus. The data flowsin both the direction between MPU andmemory and peripheral device. Address bus isa unidirectional bus.

42. (b)I/O address space is smaller compared tomemory mapped.I/O address space = 28

Memory mapped address space = 216

43. (a)

In memory mapped I/O MEMW or MEMRsignals are used as control signals.

44. (a)Stack pointer is a 16-bit register which indicatesthe address of stack. The stack pointer holdsthe address of the top element of data stored inthe stack. It is initialized by the programmerat the begining of a program which needs stackoperation.

45. (b)Instruction register holds opcode of presentinstruction being executed. It is a 8-bit, usernot accessible register.

46. (c)8-bit resgisters :• Accumulator• Status/Flag register• Temporary registers

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• B, C, D, E, H & L• Instruction register16-bit registers :• Program counter• Stack pointer• Increment/Decrement Address register.

47. (c)

In 8085 P , number of output pins = 27

and, number of input pins = 21

48

Note : AD0 - AD7 are taken as both input andoutput pins.

48. (b)

Mnemonics are techniques for rememberinginformation that is otherwise quite difficult torecall. It is used in microprocessor for a shortabbreviation for the operation to be performed.

eg ADD – for addition

ANI – for AND operation

49. (b)

The 8085 has six general purpose registers tostore 8-bit data. These are B, C,D,E, H, L.They can be combined as register pairs BC,DE and HL, to perform 16-bit operations.

50. (a)

Data bus is a 8-bit wide and hence 8-bitsof data can be transmitted in parallelfrom or to the microprocessor tomemory or I/O.

51. (d)

In microprocessor 8085; following are the activelow signal :

REST IN, RD, WR & INTA

52. (c)In instruction cycle, there are many machinecycles e.g. Fetch cycle, memory read cycle,memory write, I/O read, I/O write etc. And,one machine cycle consists many T-states e.g.Fetch cycle consists normally 4T-states.

53. (b)

2rmsVR

=2 22 V 1 V

R R2

P with SCR =1

2mV sin222

Vrms =

2 V V2 22

P =2 2V 1 V

2 R 4R

Ratio = 2 2 1V V

4R R 4

54. (b)The output of a Full Wave (FW) rectifier is

This waveform contains an average dccomponent and an ac component.

55. (c)

PIV = twice the voltage between one end ofsecondary winding and centre.

= 2 300= 600 V

56. (b)

The waveform of a 3 phase half-waverectification is

50 Hz

150 Hz

input

output

57. (a)

Vmean = max2V cos

cos is maximum, when = 0°cos0° = 1

so maximum occurs at = 0°

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EE (Test-8), Objective Solutions, 26th March 2016 (7)58. (a)

Peak inverse voltage occurs across the diodewhen it is not conducting and this occurs at

4t3

and is m3 E where Em is themaximum phase voltage. In terms of rmsvoltage, it becomes

= 3 2E

= 6 E

59. (b)When the load current is ripple free, then loadcurrent is constant. The supply wave form willbe square.

60. (d)In first half cycle, D, will conduct from 0 to and D2 will not. So, option (a) and (b) areeliminated after , D1 is reverse biased andD2 is forward biased and starts to conduct.Current being almost constant.

61. (b)

62. (c)Using feedback of a system,• Bandwidth increases• Distortion decreases• Gain decreases• Accuracy of the system increases• It used to stabilized the unstable system but

some cases it unstablize the stable system.

63. (c)

2G R s 1G

C(s)

2

2

G1 G R s 1G

C(s)

Transfer function =

1 2

2

G GC sR 1 Gs

64. (c)For linearity system follow the homogenity andsuperposition theorem, but system (1) having

dy ty t dt term and system (3) have 2y tterm. Hence system 1 and 3 are nonlinearsystem.

65. (c)

input

1 4

output

input output

Gain of the system is = 9 3 3 24

66. (d)

Transfer function =

C s T GR s

Sensitivity of open transfer T with respect toG is given by

TGS =

dT T dT G G. 1. 1dG G dG T G

...(i)

Transfer function =

C Gs TR 1 GHs

Sensitivity of closed loop transfer T with respectto G is given by

TGS =

dT T dT G.dG G dG T

= 21 G 1

G 1 GH1 GH1 GH

...(ii)

Hence equation (i) and (ii), clear that with anegative feedback in the closed loop controlsystem, the system sensitivity to parametervariation decreases.

67. (b)The transfer function is applicable to lineartime-invariant (LTI) system with initialconditions are zero.

68. (c)

V0(s) = iR sLI (s)R sL

0

i

V sI s =

R sLsLR 1R

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=

sL11 sLR

So, the block diagram will be,

sLI (s)i V (s)0

1R

69. (a)A tachometer acts like a differentiator, so itstransfer function is of the form KS.

70. (b)The output of a second order hold sytem toimpulse input is

h(t)

I

Tt

h u ut t t T

Taking Laplace

H(s) = sT11 es s

= sT11 es

71. (d)

The basic properties of signal flow graph arethe following :

i) The algebraic equations which are used toconstruct signal flow graph must be inthe form of cause and effect relationship.

ii) Signal flow graph is applicable to linearsystems only.

iii) A node in the signal flow graph representsthe variable or signal.

iv) A node adds the signals of all incomingbranches and transmits the sum to alloutgoing branches.

v) A mixed node which has both incomingand outgoing signals can be treated as anoutput node by adding an outgoing branchof unity transmittance

vi) A branch indicates functional dependenceof one signal on the other.

vii) The signals travel along branches only inthe marked direction and when it travelsit gets multiplied by the gain ortransmittance of the branch.

viii) The signal flow graph of system is notunique. By rearranging the systemequations different types of signal flowgraphs can be drawn for a given system.

72. (b)

Electric field due to a point charge

E = 20

Q4 r

Electric potential

VAB = B

B AA

V V E.dr

[Here rA = (0, 6, –8) and rB = (–3, 2, 6)

=B

20A

Q dr4 r

=

B

A0

Q 14 r

=B A0

1 1Qr r4

=

9 92 2 2

15 10 9 10

0 0 6 0 8 0

2 2 2

1

0 3 2 0 6 0

= 1 145

36 64 9 4 36

=1 145

10 7

= 345 1.93 V70

B 3,2,6V = A1.93 V

= 1.93 2 3.93 V

73. (b)

For parallel plate capacitor,• Charge in the plate will remain same as it

is given to the plate.

IES M

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EE (Test-8), Objective Solutions, 26th March 2016 (9)• Since, Electric field E

for a plate is 0 ,(constant) and potential difference betweenplates,

V = dE

i.e. V depends on d• Energy of the capacitor, 21E CV

2 will

change with change in ‘d’ as C and V bothare changing.

• Energy density 2E 0

1W E2

, as E is

constant, WE will be constant.So, only V and energy of capacitor will changewith change in separation.

74. (b)

Coulomb Force, F = 1 2

R2Q Q ˆK aR

= 1 2R2

0

Q Q1 a4 R

Where Ra is the unit vector along R

= 1 22

0

Q Q1 R4 R R

= 1 23

0

Q Q1 R4 R

75. (c)

• As the outer sphere (or, larger sphere) is ahollow conducting sphere, so whatever or,wherever is the charge inside, it will appearon the surface of the conducting sphere.

• So, net charge on the surface of the hollowconducting sphere is zero. Hence, there willbe no potential at any where.

76. (c)

Point B and C are at equipotential.So, ABW = ACW

77. (c)

According to gauss law = Qenclosed

= Q1 + Q2 + Q3

78. (b)

E = electric field = 2q

4 r

= 0 r

if r increases E decreases.

79. (a)Force between the two charges,

F = 2

0

(Q q)q4 d

For maximum force, dF 0dq

dFdq =

20

1 0Q q q( 1)4 d

2q = Q

q = Q2

80. (c)Electrostaic potential energy of the givensystem,

PE = 1 11 q V2 = 2

10

q1 q2 4 r

Thus, as r decreases (electron move towardone another), then potential energy (PE)increases.

81. (d)

Resistance, R = VI =

lE d

J ds

=

lE d

E ds

Capacitance, C =QV =

l

E ds

E d

So, RC =

l

l

E d E ds

E ds E d

i.e. RC =

82. (d)Total flux passing through the whole cube

= Total charge enclosed= 60 mC[According to Gauss’s Law]

The flux passing through one face of thecube

=Total flux

6

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= 60 mC6

= 10mC

83. (b)• Base width modulation (Early effect) occurs

in common-base Transistor (BJT)• MOS-capacitor works as constant voltage

source (flat band voltage) during saturationregion.

• LASER diode works on the principle ofpopulation inversion.

• Pinch-off of channel occurs in JFET whendrain current becomes zero.

84. (c)Under the influence of forward bias voltage,the free charge carriers get the energy toovercome barrier, while crossing the junction,the electrons give up the amount of energyequivalent to barrier potential. This loss ofenergy produces a voltage drops across p-njunction which is almost equal to barrierpotential, called cut-in voltage. The polaritiesof cut-in voltage are opposite of barrier potentialbut almost same in value.

85. (a)When Vin <VR, diode D is forward biased andit acts as short circuit V0 = VRWhen Vin > VR, diode D is reverse biased andacts as open circuit. V0 = Vin

86. (c)Tunnel diode current starts as voltage is applied= 0 VGermanium diode = 0.2 VSchottky diode = 0.4 VSilicon diode = 0.6 V

87. (c)Depletion region is created due to diffusion ofmajority charge carriers across junction so itdoes not contain any free charge (electron orhole). It only contains ions, negative ions on p-side and positive ions on n-side.

88. (b)Voltage across R = 40 – 15 = 25 V

Current through R = 25 V2K

= 12.5 mA

89. (d)The conductivity at any given temperature(except at absolute zero) is due to both electronsand holes, and it is given by

i = i e hn e

= e h2i

e hn e m m

wherei = conductivity of intrinsic materiale h, = relaxation time for electrons and holes

e hm ,m = mass of electron and holese h, = mobility of electron and holes

90. (b)Let x is the displacement of electron and E isapplied electric field across conductor.

F =2

e 2d xm eEdt

dxdt =

e

eE drift velocitym

As, mobility, =drift velocity

Electric field intensity

= e

eEm

E

=e

em

91. (b)To prevent thermal runaway

CEV < CCV2

92. (a)For saturated magnetic circuit, the increasein current will not lead to increase in fluxwhere as in unsaturated magnetic circuit theflux will increase with current and since speedis inversely proportional to flux so speed willbe less for unsaturated magnetic circuit.

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93. (d)

Reactance voltage didt

where i = armature current and t is thecommutation time.i2 = 2i1 and t2 = 0.5t1(Reactance voltage)2

= 2

2

it = 1

1

2i0.5t

= 1

i

4it

= 4 times

94. (d)In the method of speed variation by field fluxcontrol, we control the speed by varying thefield current. For very high speed, the currentbecomes very low. A very low field current willmake the holding electromagnet too weak toovercome the force exerted by spring and hencethe holding magnet would release the arm ofstarter and motor may stop.

95. (d)Ward Leonard method of speed control requiresminimum 3 machines. One machine acts asprime mover. One dc generator and one dcmotor is also required.

96. (d)

220 V DCM Rsh

IshIRa

Ia

Ish = sh

220 220 1AR 220

Ia =a

220 210R

=10 20 A0.5

I = Ish + Ia= 20 + 1 = 21 A

97. (b)E1 = 1 1K N

220 = 1K 750 ...(1)and E2 = 2 2K N

250 = 2K 700 ...(2)from equation (1) and (2),

250220 = 2

1

K 700K 750

1

2

= 250 750

220 700

= 1.217

98. (d)As the motor is a series motor. So, field currentis same as armature current. But as the motoris working under saturated magnetic condition,the field flux will be constant and it will notdepend on armature current.As torque in a dc motor,

aT I

Since, is constant

aT I

99. (d)

2

1

NN = 2

1

b 11 2

b 2

EE

2

1

NN = 2

1

b

b

EE

1bE = 200 – 30 × 0.5 = 185

2bE = 200 – 30Rt

300500 = t200 30R

185

Rt = 2.97

Additional resistor = t aR R == 2.97 0.5 2.47

100. (a)(b) in the direction for generator(c) Against direction of motor.

101. (d)Rapid brush wear and tear can take place dueto all the reasons stated above.

102. (b)

ERC

If

Critical resistance

Magnetising curve

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The generator voltae will not build up if thefield-circuit resitance is greater than or equalto the critical resistance.

103. (b)

Eg = NZ P60 A

Eg N

1

2

EE = 1 1

2 2

NN

E2 = 1 11

2 2

NE N

= 100 × 0.75 × 2= 150

104. (c)• Pin type insulator - As the name suggests,

the pin type insulator is mounted on a pinon the cross-arm on the pole. There is agroove on the upper end of the insulator.The conductor passes through this grooveand is tied to the insulator with annealedwire of the same material as the conductor.Pin type insulators are used for transmissionand distribution of communications, andelectric power at voltages up to 33 kV.Beyond operating voltage of 33 kV, the pintype insulators become too bulky and henceuneconomical.

• Suspension insulator - For voltages greaterthan 33 kV, it is a usual practice to usesuspension type insulators shown in Figure,consisting of a number of glass or porcelaindiscs connected in series by metal links inthe form of a string. The conductor issuspended at the bottom end of this stringwhile the top end is secured to the cross-arm of the tower. The number of disc unitsused depends on the voltage.

· Strain insulator - A dead end or anchor poleor tower is used where a straight section ofline ends, or angles off in another direction.These poles must withstand the lateral(horizontal) tension of the long straightsection of wire. In order to support thislateral load, strain insulators are used. Forlow voltage lines (less than 11 kV), shackleinsulators are used as strain insulators.However, for high voltage transmission lines,strings of cap-and-pin (suspension) insulatorsare used, attached to the crossarm in ahorizontal direction. When the tension load

in lines is exceedingly high, such as at longriver spans, two or more strings are used inparallel.

• Shackle insulator - In early days, theshackle insulators were used as straininsulators. But now a day, they arefrequently used for low voltage distributionlines. Such insulators can be used either ina horizontal position or in a vertical position.They can be directly fixed to the pole witha bolt or to the cross arm.

105. (c)106. (b)

In bundle conductor, the “self GMD” or“GMR” increases. Hence, inductancedecreases and capacitance increases.

As inductance, L = 2 × 10–7 ln GMD

GMR

C = l

02GMDnGMR

107. (c)

d d

3

1 2

d

Let 1SD = 3SD d d

2SD = 3SD d d

3SD = 3SD d d ,

where DS = GMR of conductorSo, the GMR of the bundled conductor,

GMR = 1 2 3

3 S S SD D D

= 39

SD d d

= 329

SD d

= 1 93 6

SD d/

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= 1 23 3SD d

= 1 32

SD d/

108. (d)

String efficiency = Total voltageacross string

n (max imum voltageacross any disc)

where, n = no. of discs in string

i.e. 0.80 =

max

400 / 320 V

Vmax =

400 120 0.803

=25 kV

3

Note: For 3 system, given voltage is alwaysline-to-line voltage unless or otherwise it isspecified as phase voltage.

109. (c)

12000 36.86I 30 36.86A

500 0.8Vs = Vr + IZ,whereZ = R + jXZ = 10 + j 7.5Z = 12.5 36.86

VS = 500 30 36.86 12.5 36.86

= 875 0

110. (d)

The specific resistance of ACSR conductor islarger than hard drawn copper conductor. It isaround 16 times more than that of hard drawncopper conductor.

111. (b)

Stress in the cable, g = e

Vx log D d

So, gmax = e

Vd log D d

Now, gmax will be minimum, when

ed.log D d is maximum,

i.e. eD 0d logd d

or, eDlogd

= 1

Dd = e = 2.71

112. (c)

Since, GeqHeq = 1 1 2 2G H G H

= 500 0.6 350 1.2 = 300 + 420= 720

Then, Heq = 720 7.2100

113. (b)At steady state position, Td = TCDue to balance of torques needle stop at steadystate position and underdamping is used toreduce the oscillation of pointer around steadystate position.

114. (a)To overcome the frictional torque, we need largecoil with strong magnetic field.

Td = NBAITo increase Td, large coil and strong magneticfield are used.

115. (d)Diamagnetic effect is not prsent in allmaterials. Diamagnetic is a substance whichexibit negative magnetism.

116. (b)• Iron (Fe), Cobalt (Co) and Nickel (Ni) are

ferromagnetic materials.• Magnetic properties of materials are affected

by electron spin dipole moment.Element having incompletely filled inner orbitare having non-zero value of electron spindipole moment. Both statements are correctbut ‘R’ is not correct explanation of A.

117. (c)Ferromagnetic materials have very high eddycurrent losses and poor magnetic utilization athigh frequency. So, for high frequencyapplication, ferrites are used as it has dcresistivity higher than iron.

IES M

ASTER


118. (b)Soft magnetic materials exhibit the followingfeatures by which it is used to fabricateelectromagnets:- High initial permeability- High retentivity- Low coercivity- Low hysteresis lossSoft magnetic material have also very low eddycurrent loss due to high resistivity.

119. (c)• In common base (CB) configuration in active

mode the no. of electrons reaching thecollector per unit time is proportional to theno. of electrons injected into the base, which

is function of the base-emitter voltage VBE.Hence collector current IC is proportional to

BE TV /Ve and is independent of the reversebiased base-collector voltage. Therefore thedevice appears like a constant current sourceand collector current IC is controlled by VBE.

• In common base (CB) configuration highoutput resistance and low input resistance.

120. (a)Doping materials are called impurities whichare used to increase the carrier concentrationin intrinsic (pure) semiconductor, when theyadded to intrinsic semiconductor they alteredthe crystal structure of semiconductor bymaking the covalent bonds with other atomsand release/accept free charge carriers.

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Conventional Question Practice Programe - IES Masteriesmaster.org/public/archive/2016/IM-1462628551.pdf · present in microprocessor 8085. ... Instruction register holds opcode of