vpm classes _free solved expected paper _ gate 2013 _ electrical engg
TRANSCRIPT
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7/30/2019 Vpm Classes _free Solved Expected Paper _ Gate 2013 _ Electrical Engg
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
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For IIT-JAM, JNU, GATE, NET, NIMCET and Other Entrance Exams
GATE - ELECTRICAL ENGINEERING
1-C-8, Sheela Chowdhary Road, Talwandi, Kota (Raj.) Tel No. 0744-2429714
Web Site www.vpmclasses.com [email protected]
MOCK TEST PAPER
Pattern of questions : MCQs
There are a total of 65 questions carrying 100 marks.
Questions (56-65) belongs to general aptitude (GA).
Questions (56-60) will carry 1-mark each, and
question (61-65) will carry 2-marks each
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Total marks : 100
Duration of test : 3 Hours
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Questions (1-25) will carry 1-mark each and
questions (26-55) will carry 2-marks each.
For Q.1-25 and Q.56-60 1/3 mark will be deducted
for each wrong answer.For Q.26-51 and Q. 61-65 2/3
mark will be deducted for each wrong answer.
The question pairs (Q.52, Q.53) and (Q.54, Q.55)
are linked questions.For Q.52 &54 2/3 mark will be
deducted. There is no negative marking for Q.53 &Q.55.
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Q.48-51 are common data questions.
If first question is attempted wrongly then answer of
second question will not be evaluated.
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
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1 In the matrix equation Px = q, which of the following is a necessary condition for the existence of at
least one solution for the unknown vector x?
(A) Augmented matrix [Pq] must have the same rank as matrix P
(B) Vector q must have only non - zero elements
(C) Matrix P must be singular
(D) Matrix P must be square
2. An arbitrary vector X is an eigen vector of the matrix A =
1 0 0
0 a 0
0 0 b
, if (a, b) =
(A) (0, 0)
(B) (1, 1)
(C) (0, 1)
(D) (1, 2)
3. The integration of logx.dx has the value
(A) (x log x 1)
(B) log x x
(C) x (log x 1)
(D) None of these
4. If f(x) = | x |, then the interval [1, 1], f(x) is
(A) Satisfied all the conditions of Rolles Theorem
(B) Satisfied all the conditions of Mean Value Theorem
(C) Does not satisfied the conditions of Mean Value Theorem
(D) None of these
5. Differential equation,
2
2
d x dx10 25x 0
dt dt
will have a solution of the form
(A)C1
+ C2
t)e5t
(B) C1 e2t
(C) C1
e5t + C2
e5t
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
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(D) C1 e5t + C2
e2t
Where C1
and C2
are constants.
6. For the differential equationdy
5y 0dt
with y(0) = 1, then general solution is
(A) e5t
(B) e5t
(C) 5e5t
(D) 5te
7. For | z | = 1, where C is the circle, is
(A)
(B)
(C)
(D) None of these
8. If A and B are independent and P (C)= 0, t hen A, B and C are independent
(A) True
(B) False
(C) Both (a) and (b)
(D) None of these
9. Following are the value of a function
y(x): y(1) = 5, y(0), y (1) = 8
dy
dxat x = 0 as per Newtons central difference scheme is
(A) 0
(B) 1.5
(C) 2.0
(D) 3.0
10. For any two events A and B
(A) P(B) = P(A B) + P( A B)
(B) P(A B) = P(A) +P(B) P(A B)
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
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(C) P(A/B) P(A).
(D) All of these
11. A unit step current to a network consisting of only passive elements. The voltage across the current
source observed is v(t) = (1 + e-t). The simplest possible network will consist of the elements
(A) 1 resistor and 2 capacitors
(B) 1 resistor and 2 inductors
(C) 2 resistors and 1 capacitor
(D) 2 resistor and 1 inductor
12. The energy stored in the magnetic field at a solenoid 30 cm long and 3 cm diameter wound with 1000
turns of wire carrying a current at 10 A, is
(A) 0.015 Joule
(B) 0.15 Joules
(C) 0.5 Joule
(D) 1.15 Joules
13. The current wave from in a pure resistor at 10 is shown in the given figure. Power dissipated in the
resistor is
(A) 7.29 W
(B) 52.4 W
(C) 135 W
(D) 270 W
14. A series RLC circuit consisting of R = 10 ohms, XL
= 20 ohms and XC
= 20 ohms is connected across
an ac supply of 100 V (rms). The magnitude and phase angle (with reference to supply voltage) of the
voltage across the inductive coil are respectively,
(A) 100 V, 90
(B) 100 V, - 90
(C) 200 V, 90
(D) 200V, 90
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
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15. When a unit impulse voltage is applied to an inductor of 1 H, the energy supplied by the source is
(A)
(B) 1 J
(C)1
J2
(D) 0
16. The resonant frequency of the given series circuit is
(A)1
Hz2 3
(B)1
Hz4 3
(C)1
Hz4 2
(D)1
Hz2
17. The final value of L-1
4 3 2
2s 1
s 8s 16s s
is
(A) Infinity
(B) 2
(C) 1
(D) Zero
18. If (t) denotes a unit impulse, then the laplace transform of 22
d d t
dtwill be
(A) 1
(B) s2
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
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(C) s
(D) s-2
19. The unit step response of a system is given by (1 e) u (t)
(A)e u (t)
(B) e u (t)
(C) 1
e u t
(D) e u (t)
20. If f(s) = / (s2 + 2) , then the value of dtLt
f(t)
(A) Cannot be determined(B) Is zero
(C) Is unity
(D) None of these
21. The function f(t) has a Fourier transform g(w). The Fourier transform
j tf t g t g t e dt,is
(A) 1
f
2
(B) 1
f2
(C) 2 f
(D) None of these
22. If f (t) and f (t) satisfy the Dicrichlets conditions, then f(t) can be expanded in a Fourier series
containing
(A) Only sine terms
(B) Only cosine terms
(C) Cosine terms and constant terms
(D) Sine terms and a constant term
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
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23. A single - phase starter winding when excited with ac voltage produces
(A) A single rotating field rotating as synchronous speed.
(B) Two rotating fields rotating at synchronous speed in opposite directions.
(C) Two rotating fields rotating in the same direction but at different speeds
(D) Two rotating fields rotating in opposite directions but with difference speeds.
24. The mmf of a sinusoidally distributed current sheet is
(A) Triangular shifted from the current by 90
(B) Triangular is phase with the current wave
(C) Sinusoidal shifted from the current wave by 90
(D) Sinusoidal in phase with the current wave
25. In a synchronous machine, the voltage induced by armature reaction flux acts like the voltage drop in
(A) Inductive reactance
(B) Resistance
(C) Capacitive reactance
(D) Inductive impedance
26. Synchronous reactance is
(A) The difference of armature leakage reactance and reactance equivalent of armature reaction
(B) The same as armature leakage of armature reaction
(C) The reactance equivalent of armature reaction
(D) The sum of armature leakage reactance and reactance equivalent of armature reaction
27. A synchronous motor is floating on infinite mains at no load. If its excitation is now increased, it will
draw
(A) Unity power factor current
(B) Zero power factors lagging current
(C) Zero power factor loading current
(C) No current
28. Voltage regulation of a synchronous generator calculated by synchronous impedance method is
(A) Higher than actual because of saturation of magnetic circuit
(B) Lower than actual because of saturation of magnetic circuit
(C) Nearly accurate because it takes account of magnetic circuit.
(D) Nearly accurate because the generator is normally operated with an unsaturated magnetic circuit.
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
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29. A two conductor 1 - line operates at 50 Hz. The diameter of each conductor is 4 cm and is spaced 6
m apart. What is the capacitive susceptance to neutral per km?
(A) 1.42 10-9 s/km
(B) 3.25 10-8 s/km
(C) 4.8 10-9 s/km
(D) 3.06 10-9 s/km
30. A two conductor 1 - line operates at 50 Hz. The diameter of each conductor is 20 mm and the
spacing between the conductors is 3m. The height of conductor above the ground is 6 m. The
capacitance of the line to neutral will be
(A) 9.7 pF/m
(B) 10.2 pF/m
(C) 8.7 F/km
(D) 2.4 F/m
31. A three phase line has conductor of 5 mm diameter placed at the corner of an equilateral triangle of
1.5 m side. The capacitive reactance to neutral per phase per km will be
(A) 1.25 106 /km
(B) 8.68 105/km
(C) 4.25m/ km(D) 3.66 105 /km
32. Three insulating materials with same maximum working stress and permittivities 2.5, 3.0, 4.0, are
used in a single core cable. The location of the materials with respect to the core of the cable will be
(A) 2.5, 3.0, 4.0
(B) 3.0, 2.5, 4.0
(C) 4.0, 3.0, 2.5
(D) 4.0, 2.5, 3.0
33. The incremental generating costs of two generating units are given by
IC1
= 0.1 X + 20 Rs /MWhr.
IC2
= 0.15 Y + 18 Rs/ MWhr.
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 9
Where X and Y are power (in MW) generated by the two units. For a total demand of 300 MW. The
values (in MW) of X and Y will be respectively
(A) 172 and 128
(B) 128 and 172
(C) 175 and 125
(D) 200 and 100
34. For protection of rotating machines against lighting surges a combination of
(A) Lighting conductor and capacitor are used
(B) Lighting conductor and lighting arrester are used
(C) Lighting arrester and capacitor are used
(D) Lighting arrester alone is used
35. For the system shown, the transfer function C(s) / R(s) is equal to
(A)2
10
s s 10
(B) 2 10s 11s 10
(C)2
10
s 9s 10
(D)2
10
s 2s 10
36. In the following block diagram 1 2 1
10 10G : G ;H s 3
s s 1
and H
2= 1.
The overall transfer function C/R is given by
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
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(A)2
10
11s 31s 10
(B)2
100
11s 31s 100
(C)2
100
11s 31s 10
(D)2
100
11s 31s
37. Consider a unity feedback control system with open -loop transfer function
K
G(s)s s 1
The steady state error of the system due to a unit step input is
(A) Zero
(B) K
(C) 1 / K
(D)
38. For a gain constant K, the phase - lead compositor
(A) Reduces the slope of the magnitude curve in the entire range of frequency domain
(B) Decreases the grain cross - over frequency
(C) Reduce the phase margin
(D) Reduce the resonance MP
39. A control system is as shown in the given figure. The maximum value of gain K for which the system
is stable is
(A) 3
(B) 3
(C) 4
(D) 5
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 11
40. Straight line asymptotic Bode magnitude plot for a certain system is shown in the given figure. What
will be its transfer function?
(A)4s
s s1 1
4 10
(B)2
s4 1
4
s1
10
(C)0.25
s s1 1
4 10
(D)2
0.25
s s1 1
4 10
41. A compensated wattmeter has its reading corrected for error due to the
(A) Frequency
(B) Friction
(C) Power consumed in current coil
(D) Power consumed in potential coil
42. The sensitivity of an instrument is the
(A) Smallest increment in the input that can be detected with certainty
(B) Largest input change to which the instrument fails to respond
(C) Ratio of the change in the magnitude of the output to the corresponding change in the magnitude
of the input
(D) Closeness of the output values for repeated applications of a constant input.
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 12
43. If a dc voltmeter is made from an ammeter having a full - scale deflection of 100 micro amperes, then
its sensitivity (in k - ohm. (V) Will be
(A) 1
(B) 10
(C) 100
(D) 1000
44. What of the following are advantages of instrument transformers?
(A) The readings of instruments used in conjunction with them do not depend upon their resistance,
inductance etc.
(B) The readings of instruments transformers have been standardized and the rating of instruments
used in conjunction used with them also get standardized. Therefore, there is reduction of cost
and ease in replacements.
(C) The metering circuit is electrically isolated from the power circuit thereby providing safety to
operating personnel.
(D) All of these
45. The limiting error of measurement of power consumed by and the current passing through a
resistance are 1.5 % and 1 % respectively. Then the limiting error of measurement of resistance
will be.
(A) 0.5 %
(B) 1.0 %
(C) 2.5 %
(D) 3 .5 %
46 The braking torque provided by a permanent magnet in a single phase energy meter can be changed
by
(A) Providing a magnetic shunt and changing its position
(B) Changing the distance of the permanent magnet from the center of the revolving disc
(C) Both (a) and (b)
(D) None of these
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 13
47. The power in d.c. circuit is measured with the help of ammeter and a voltmeter. The voltmeter is
connected on the load side. The power indicated by the produced to reading of two instruments (VI) is
(A) The power consumed in the load
(B) The sum of power consumed by load and the voltmeter
(C) The sum of power consumed by load and the ammeter
(D) None of these
Linked Answer Q. 48-49
The first and the second stage of a two stage RC coupled amplifier have the lower cut off frequencies
to be 100 Hz and 200 Hz respectively. Their upper cuts off frequencies are 140 kHz and 100 kHz
respectively.
48. Overall lowed cut-off frequency will be
(A) 212Hz
(B) 232 Hz
(C) 238 Hz
(D) 242 Hz
49. Overall 3 dB band-width of the amplifier will be
(A) 74.236 kHz
(B) 74.362 kHz
(C) 74.6 kHz
(D) None of these
Linked Answer Q. 50-51
The input voltage Viin the circuit shown in the given figure is a 1 kHz since-wave of 1V amplitude.
Assume ideal operational amplifiers with 15 V DC supply.
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 14
50. The peak value of V1
will be
(A) 1.06 V
(B) 1.33 V
(C) 2.33 V
(D) 3.33 V
51. The average value of Vo
will be
(A) 1.06 V
(B) 1.33 V
(C) 2.33 V
(D) 3.33 V
Common Data for Q (52-53)
A separately excited DC motor runs at 1500 rpm under no - load with 200 V applied to the armature.
The field voltage is maintained at its rated value. The speed of the motor, when it delivers a torque of
5 Nm, is 1400 rpm as shown in the figure. The rotational losses and armature reaction are neglected.
52. The armature resistance of the motor is
(A) 2
(B) 3.4
(C) 4.4
(D) 7.7
53. For the motor to deliver a torque of 2.5 Nm at 1400 rpm, the armature voltage to be applied is
(A) 125.5 V
(B) 193.3 V
(C) 200 V
(D) 241.7 V
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 15
Common Data Question 54 55
Voltage standing wave pattern in a lossless transmission line with characteristic impedance 50W and
a resistive load is shown in the figure.
l
1
4
Load
|V(z)|
54. Value of the load resistance is
(A) 50
(B) 200
(C) 12.5
(D) 0
55. Reflection coefficient is
(A) 0.6
(B) 1
(C) 0.6
(D) 0
56. No doubt, it was our own government but it was being run on borrowed ideas, using
_________solutions.
(A) Worn out
(B) Second hand
(C) Impractical
(D) Appropriate
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 16
The question below consists of pair of related words followed by four pairs of words. Select the pair that
best expresses the relation in the original pair:
57. Ratify: Approval:
(A) Mutate: change
(B) Pacify: conquest
(C) Duel: combat
(D) Appeal: authority
58. A car goes 35 km in 1 hour, next 270 km in 3 hrs. and next 80 km in1
22
hrs.Find the average speed
of the car
(A) 59.23 km/h.(B) 61.5 km/h
(C) 80 km/h
(D) None of these
59. Some critics believe that Satyajit Ray never quite came back to the great beginning he made in this
path breaking film Pather Panchali. ______have endured decades of well-traveled bad prints to
become a signpost in cinematic history.
(A) The bizarre history of its misty origins
(B) Its haunting images
(C) Its compelling munificence
(D) The breathtaking awe it inspires
Choose the most appropriate word from the options given below that is most nearly opposite in
meaning to the given word:
60. Valedictory
(A) Sad
(B) Collegiate
(C) Derivative
(D) Generosity
Each of the 11 letters A, H, I M, O,T, U V, W, X and Z appears same when looked at in mirror. They
are called symmetric letters. Other letters in the alphabet are asymmetric letters.
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 17
61. If the area of a given square ABCD is 3 find the total area of the entire figure?
(A) 452
(B) 45
(C) 48
(D) 31
62. The portion of the immunoglobulin molecule which binds the Specific antigen is formed by the amino
terminal portions of both the H and L chains.
(A) H chain(B) L chain
(C) Both (a) and (b)
(D) None of these
63. In a certain code Language
134 means good and tasty
478 means see good picture
729 means picture are faint
Which number has been used here for faint?
(A) 9
(B) 2
(C) Data are inadequate
(D) 253
64. A bag contains an equal number of one rupee, 50 paisa and 25 paisa coins. If the value of money in
the bag is Rs. 35, find the total number of coins of each type?
(A) 7
(B) 40
(C) 30
(D) 20
-
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 18
Profit to sale-Table for three companies A, B and C for 1996-97
Companies 1996 1997Total units 300000 400000
Shares A 5% 25%
B 60% 40%
C 35% 35%
Price A 10% 8%
(per unit) B 7% 14%
(in rupees) C 9% 10%
Profit A 3% 1%
(per unit) B 0.5 5%
(in rupees) C 2% 2.5
65. What is the increase in the total profits of company B in 1997?
(A) 800%
(B) 900%
(C) 750%
(D) 789%
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 19
ANSWER KEY
Que Ans Que Ans Que Ans Que Ans Que Ans
1 A 16 B 31 D 46 C 61 B
2 B 17 C 32 C 47 D 62 C
3 C 18 A 33 A 48 C 63 C
4 C 19 B 34 C 49 B 64 D
5 A 20 B 35 B 50 D 65 D6 B 21 C 36 B 51 A
7 A 22 A 37 A 52 B
8 A 23 B 38 A 53 B
9 B 24 C 39 D 54 B
10 D 25 A 40 D 55 C
11 D 26 D 41 D 56 B
12 B 27 C 42 D 57 C
13 D 28 A 43 B 58 A
14 D 29 D 44 D 59 B
15 C 30 A 45 C 60 D
Hints and Solution
1(A) According to Rouches theorem, the system is consistent if and only if the coefficient matrix and the
augmented matrix K are of the same rank, otherwise the system is inconsistent.
2.(B) Since the matrix is triangular, the eigen values are , a, b.
If (X1, X
2, X
3) is an arbitrary eigen vector, say corresponding to 1, then
-
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Page 20
1 1
2 2
3 3
1 0 0 x x
0 a 0 x 1 x
0 0 b x x
X2, X3 being not zero, we have, X1 = X1 ; a X2 = X2 which gives
a = 1
and bX = X3
which gives b = 1
(a, b) = (1, 1).
3.(C)d
logx.dx logx.x x. (logx)dxdx
= x log x 1. dx
= x log x x
= x (log x 1)
4.(C) Since f(x) = | x | is continuous is [1, 1] be it is not diff erentiable at x = 0 (1, 1)
5.(A)2
2
d x dx10 25x ]0
dt dt
(D2 + 10D + 25) X = 0
(D + 5)2 = 0
D = -5, 5
Hence solution is, (C1
+ C2t)e5t
6.(B) Given:dy
5y 0dt
dy
5 dty
Integrating, we get
loge
y = c 5t
When t = 0, y = 1.
loge1 = c 5 0
c = 0
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Page 21
logey = 5t
y = e5t
7.(A) Poles of f(z) =2
z 3
z 2z 5
are given by
z2 + 2z + 5 = 0
z =2 4i
1 2i2
Since, both poles lie outside the circle | z | = 1, therefore f(z) is analytic inside the circle
2
z 3dz 0
z 2z 5
8. (A) P(C) = 0
C =
P(A B C) = P (A B )
= P() = 0
P(A) P(B) P(C) = 0 .....Since P(C) = 0
P(A B C) = P (A) P (B) P (C)
Hence A, B, C are independent.
9. (B) 2 1
2 1at x 0
dy y y
dx x x
=y(1) y( 1) 8 5
1.51 ( 1) 2
10.(D) (a) B = (A B) ( A B)
P(B) = P(A B) + P( A B)
A B and A B are mutually exclusive.
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Page 22
P ( A B) = P(B) P (A B).
(B) P (A B) = P[A ( A B)]
= P (A) + P( A B)
= P (A) + P(B) P(A B)
(C) P (A/B) = P(A) P (B/A)
P (A) as P(B/A) 1.
11. (D) z(s) =
V s
I s
=1 1
s1s s
=s
11
s
12. (B) 6 7 42
0
10 4 10 9 10N A 4L
l 0.3
=2 59 10
0.3
Energy = 21
LI2
= 0.148 or 0.15 Joules
13. (D)
323 32 2rms
0 0
1 9 1 tI t .dt 9 27A
3 3 3 3
Power = I2R = 27 10 = 270 W
14. (D) Since the reactance are canceling each other, the circuit is purely resistive and the current is phase
with the applied voltage. The voltage across the inductor leads the current through it by 90
Since I = 100 0 / {10 + j (20 - 20) } = 10 0
LV = jL I = 20 10 90 = 200 90
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Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 23
15. (C) Current that flows is given by,
V(s) = sL1I (s)
I = s I (s) I (s) =1
s= u (t) = 1 A.
Energy supplied =1 1
1 1 J.2 2
16. (B) Leq
= L1
+ L2
+ 2M = 6H
At resonance, Leq
=1
C
or =eq
1
L .C
=1
12
f =1
Hz.4 3
17.(C)
4 3 2s s 0 s 0
s 2s 1Limf t LimsF s Lim
s 8s 16s s
3 2s 0
2s 1Lim s 8s 16s s
18.(A) Since (t) = 1
19.(B) Impulse response = td
1dt
te
20.(B) 22 2f s 0s
Lt f t Lt sF(s) 0s
s s
21.(C) Since g() = j t1
g e dt2
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Page 24
22.(A) Condition f (t) = -f(-t) implies that it has rotational symmetry.
23.(B) a single - phase starter winding when excited with ac voltage produces two rotating fields rotating at
synchronous speed in opposite directions.
24. (C) The mmf of a sinusoidally distributed current sheet is sinusoidal shifted from the current wave by 90.
25.(A) In a synchronous machine, the voltage induced by armature reaction flux acts like the voltage drop in
Inductive reactance
26. (D) Synchronous reactance is the sum of armature leakage reactance and reactance equivalent of
armature reaction
27.(C) A synchronous motor is floating on infinite mains at no load. If its excitation is now increased, it will
draw no current
28. (A) Voltage regulation of a synchronous generator calculated by synchronous impedance method is
Higher than actual because of saturation of magnetic circuit.
29.(D) Here, D = 6m, r = 2 cm = 2 10-2 M
Cn
=
0
9
2
2 16In D / r
18 10 In2 10
=9
1
318 10 In
0.01
= 9.74 10-9 F/km
Capacitive susceptance to neutral
bc
= n1
2 F.C
Xc
= 2 50 9.74 10-9 s/km
= 3.06 10-9 s / km
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Page 25
30.(A) Cn
= 0
m
s
2
D HIn In
r H
Here, r = 10 mm = 10 10-3 m,
D = 3 m, h = 6 m,
H1
= H2
= 2h = 12 m
H12
= H21
= (D2
+ 4 h2)1/2
= (32 + 4 62)1/2 = 153
Hm
= (H12
. H21
)1/2 = H12
= 153 m
Hs= (H
1. H
2)1/2 = (2h. 2h)1/2 = 2h = 12m
Cn
= 0
3
2
3 153In In10 10 12
=9
2
1
3 1218 10 In
10 153
= 9.7 10-12 F/m = 9.7 pF/m
31.(D) Here, d = 5 mm, r = 2.5 mm = 2.5 10-3 m,
D = 1.5 m
Deq = (DAB. DBC. DCA)1/3
= (1.5 1.5 1.5)1/3 = 1.5 m
Cn
= 0
eq
2
DIn
r
=9
3
1
1.518 10 In
2.5 10
=
39
1
1.5 1018 10 In
1.5
= 8.68 10-12 F/ m
= 8.68 10-9 F/km
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Page 26
= 8.68 10-3 F/km
Capacitive reactance,
c 9n1 1
X2 f C 2 50 8.68 10
= 3.66 105/km
32.(C) When all the three materials are subjected to the same maximum stress,
max
1 2 2 3 2
g2 r 2 r 2 r
or 1
r = 2
r1
= 3
r2
Since r < r1
< r2,
Therefore 1
> 2>
3
Thus, the dielectric material with highest permittivity should be placed near the conductor and other
layer in the descending order.
33.(A) For most economic load sharing
I C1
= IC2
We have X + Y = 300 MW
and 0.1 + 20 = 0.15 Y + 18
= 0.15 (300 - X) + 18
= 45 - 0.15 + 18
025 X = 43or X = 172 MW
Y = 300 - 172 = 128 MW
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 27
34.(C) The rotating machine should be protected against major and minor insulation. Major insulation is the
insulation between the winding and minor insulation means inter- turn insulation. The major isnulation
is determined by line to ground voltage across the terminal of the machine whereas minor insulation
is determined by the rate of rise of voltage. Hence to protect the rotating machine against surges
requires limiting the surge voltage magnitude at the machine terminals and sloping the wave front of
the incoming surge. To protect the major insulation, a special lighting arrester and to protect the minor
insulation a capacitor are required and connected as depicted in the give figure.
35.(B)
2
10
10s(s 1)G s
10 s 11s1 s
s s 1
2
2
2
10C s 10s 11s10R s s 11s 10
1s 11s
36.(B) Successive block diagram reduction can be
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Page 28
1 2
2 1
1 2 2
2 1
G G
C s 1 G H
G G HR s1
1 G H
= 1 2
2 1 1 2 2
G G
1 G H G G H
10 10.
C s s s 1
s 3R s 10 101 10 . .1
s 1 s s 1
=2
100
11s 31s 100
37.(A) Unity feedback control system is shown as below :
Steady state error due to unit step.
Since it is a type 1 system, hence steady as below:
Error of system due to unit step input is zero.
38.(A) Lead compensator increases the gain - crossover frequency. It also increases the phase margin. The
high frequency end of the log magnitude plot has been raised by a dB gain of 20 log 1( ).2
39.(D) The characteristics equation is
1 + GH = 0
or 1 +3 2
K0
s 3s 2s 1
or3 2
3 2
s 3s 2s K0
s 3s 2s 1
Hurwitz criterion can be applied.
s3 1 25 K
3
and (1+ K), both should be positive
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Page 29
s2 3 1+Ks1 6 1 K3
0
K should be < - 1 and less than 5
s 1 K
40.(D) Initially plot has a slope of + 20 d/B decade. So there must be zero at origin.
At = 4, slope changes to 0 dB/ decade. so pole at = 4.
Again slope at = 10 changes to - 40 dB/decade.
So there are two poles at = 10.
Now, 20 lop |K| = 0 at = 4.
Which gives, K = 0.25.
41. (D) A compensated wattmeter has its reading corrected for error due to power consumed in potential coil.
42. (D) Voltage across R1, V
1= 35 - 5 = 30 V
Current in the circuit,
I =5 5
A600 1200 400
600 1200
Rx =30 400
5
= 2.4k
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Page 30
43.(B) Sensitivity =Total of resistance of the inst
ohm./VFull scale current figure
=6
110K ohm / V
10 10
44.(D) All are correct option.
45.(C) M1
= 20 0.001 = 0.02,
Relative error at 1 A =0.02
2%1
M2
= 10 0.002 = 0.02,
Relative error =0.02
2%1
M3
= 5 0.005 = 0.025,
Relative error =0.025
2.5%1
M4
= 1 0.01 = 0.01,
Relative error =0.01
1%1
46. (C) The braking torque provided by a permanent magnet in a single phase energy meter can be changedby Providing a magnetic shunt and changing its position & Changing the distance of the permanent
magnet from the center of the revolving disc
47.(D) All of these are incorrect.
48. (C) Given,
fL1
= 100 Hz and fL2
= 200 Hz
fH1
= 140 KHz and fH2
= 100 KHz
Let fL
= Overall lower cut off frequency,
and fH
= Overall upper cut-off frequency
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Page 31
Then
1 2
2 2
H H
H H
1 1 1
2f f
1 1f f
or 2 = 1 +
1 2 1 2
42HH2 2 2 2
H H H H
f 1 1f
f . f f f
or fH
4 + 2 21 2H H
(f f ) fH
2 2 21 2H H
f . f 0
or fH
4 = (1402 + 1002) fH
2 (14000)2 = 0
fH
2 = 29600 2 829600 4 (1.96 10 )
2
= 5572
fH = 74.6 kHz
1 2
2 2
L L
L L
1 1 1
2f f1 1
f f
or 2 = 1 + 1 2 2 21 2
2 2L L
L L4 2L L
f f 1(f f ) 0
f f
or2 2 2 24 2
L L L L L L1 2 1 2f f f f (f f ) 0
or fL4 f
L2 (10000 + 40000) 1002 2002 = 0
or fL2 = 50000
2 2 250000 4 (100 200 )
2
= 57015
fL= 238 Hz
49. (B) 3 dB bandwidth = 74.6 kHz 238 Hz
= 74.362 kHz
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Page 32
50. (D) & 51. (A)
Given, Vi= sin t
= 6283 rad/sec
First op amp works as half-wave rectifier, with amplification factor
=10 5
10 5
= 3.33
V1
= 3.33 sin t, < t < 2
= 0, 0 < t <
Second op-amp network as average defector,
Vo
=3.33
= 1.06 V
Peak value of V1
= 3.33 V
Average value of Vo= 1.06 V
51.(A)
1 2 1 1x y x x x x z
0 0 1 0 1 0 10 1 1 1 1 1 1
1 0 1 1 1 1 1
1 1 0 1 0 1 1
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Page 33
Then x y z
0 0 1
0 1 1
1 0 1
1 1 1
52. (B)
Given:
No load speed
N0 = 1500 rpm, V = 200V,
T = 5 Nm, N = 1400
Emf at no load,1b
E V 200
Now b11
2 b2
EN
N E
2b2 b11
N 1400E .E 200
N 1500
= 186.67 V
But B a
m
E IT
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 34
m2 N
60
a60* 186.67* I
5
2 * 1400
Ia
= 3.926amp
2 2b a a
V E I R
22
ba
a
V ER
I
=200 186.67
3.926
= 3.395 =3.4
53. (B) Given: T = 2.5 Nm, 1400 rpm
Now
2.5 =
Ia
= 1.963 amp
V = Eb
+ Ra
Ia
= 186.66 + 1.9634 3.4
= 193.34 V
54 & 55. (B, C)
Here vmax = 4, and Vmin = 1
VSWR = 4 = S
Reflection coefficient,
S 1 4 1 3K 0.6
S 1 4 1 5
Now, R 0
R 0
Z ZK
Z Z
3ZR + 150 = 5Z R 250
ZR = 200
56.(B) No doubt, it was our own government but it was being run on borrowed ideas, using second hand
solutions.
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 35
57.(C) Ratify: Approval:: duel: combat
58.(A) Total distance covered
= 35 + 270 + 80 = 385 km.
= 1 + 3 +1
22
hrs.
= 13/2 hrs
Average speed =D 385 2
T 13
= 59.23 km/hr.
59.(B) some critics believe that Satyajit Ray never quite came back to the great beginning he made in this
path breaking film Pather Panchali. Its haunting images have endured decades of well-traveled bad
prints to become a signpost in cinematic history.
60(D) Generosity is nearly opposite to Valedictory
61.(B) Count the number of squares in the figure and multiply it by 3 i.e. 45.
62(C) The portion of the immunoglobulin molecule which binds the Specific antigen is formed by the Aminoterminal portions of both the H and L chains.
63.(C) 4 = good 7 = picture and 2 and 9 = are and faint respectively.
64.(D)X X X 7X
35 or1 2 4 4
= 35 or 7X
= 35 4 or X = 20
65(D) The increase in the total profits of company B in 1997 is 789%.
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 36
Please find below in Table the details of theCourse forGATE 2013Electrical Engineering offered by VPM CLASSES:
TABLE - GATE 2013 ENGINEERING
GATE 2013
CORRESPONDENCE COURSE
(withGATE Aptitude study material) (withoutGATE Aptitude study material)Course A Course B
Rs.7,100/- Rs.6,300/-
6 volumes of theory 6 volumes of theory
1 volume of theory covering various aspects of
GATE General Aptitude
36 Topic-wise Unit test papers(UTPs)covering the GATE syllabus
12 Topic-wise Unit test papers covering the
GATE General Aptitude syllabus
6 Volume Test Papers (VTPs) for better
practice and revision of your syllabus
36 Topic-wise Unit test papers(UTPs)
covering the GATE syllabus
12 Full length test papers (on GATE pattern)
6 Volume Test Papers (VTPs) for better
practice and revision of your syllabus
Previous 2 year solved question bank(2011-
2012)
12 Full length test papers (on GATE pattern) Hints and solutions with all test papers
Previous 2 year solved question bank(2011-
2012)
Hints and solutions with all test papers
NOTE: All prices include service tax.
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Phone: 0 7 4 4 - 2 4 2 9 7 1 4 Mobile: 9 0 0 1 2 9 7 1 1 1 , 9 8 2 9 5 6 7 1 1 4 , 9 8 2 9 5 9 7 1 1 4 , 9 0 0 1 2 9 7 2 4 2Website: www.vpmclasses.com E-Mail: [email protected] / [email protected] Address: 1- C-8 , SFS, TALWANDI , KOTA, RAJASTHAN, 32 40 05
Page 37
HOW TO APPLY
Mode of Payment of fee:Option 1 - Demand Draft
For the course you want to get enrolled with us, please prepare a Demand Draft (D.D.) of
an amount equal to the fee mentioned against your course of interest. Please prepare the
D.D. in favor of "VPM Classes" payable at any bank at Kota, Rajasthan.
Items to send:
1) Duly filled application form.
2) D.D. for the course you wish to get enrolled in
3) Xerox copy of 10th, Bachelor degree & Master Degree (if applicable) mark sheets
4) 2 additional passport sized photographs
5) Your application / roll nos. of the various exams (IIT JAM / NET / GATE / TIFR / IISc etc.)
you are appearing for (when applicable).
Please send the above items to:
VPM Classes
1-C-8, Sheela Chowdhary Road, SFS, Talwandi, Kota, Rajasthan PIN - 324005.
(Tel No.: 0744-2429714 or 09001297111 | E-mail: [email protected]
/[email protected]/[email protected]
Option 2: Direct Payment in VPM Classes Account
Please deposit the fee by cash/cheque in any of the 2 VPM Classes a/c nos. mentioned
below:
Money can be transferred from ANY bank in India to the above account numbers using the
following information:
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CSIR UGC NET, GATE (ENGINEERING), GATE (Science), IIT-JAM, UGC NET, TIFR, IISc, NIMCET, JEST etc.
Ph 0 2 2 9 M bil 9 0 0 2 9 9 8 2 9 6 9 8 2 9 9 9 0 0 2 9 2 2
VPM Classes
1-C-8, Sheela Chowdhary Road, SFS, Talwandi, Kota, Rajasthan PIN - 324005.
(Tel No.: 0744-2429714 or 09001297111 | E-mail: [email protected] /[email protected] /[email protected].).
Ideally these documents should reach us within 1 week of making the deposit in the
bank a/c.
Please visit our website www.vpmclasses.com for more information. If there are any other
questions, please feel free to send an e-mail or call us (Tel No.: 0744-2429714 or
09001297111).
Thanks & Best of Luck,
VPM Classes