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2015-1 SOLVED PAPER - 2015
GENERAL APTITUDE
QUESTION 1 TO 5 CARRY ONE MARK EACH1. Which of the following options is the closest in meaning to
the sentence below?She enjoyed herself immensely at the party.(a) She had a terrible time at the party(b) She had a horrible time at the party(c) She had a terrific time at the party(d) She had a terrifying time at the party
2. Given Set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, twonumbers are randomly selected, one from each set. What isthe probability that the sum of the two numbers equals 16?(a) 0.20 (b) 0.25(c) 0.30 (d) 0.33
3. Didn’t you buy __________ when you went shopping?(a) any paper (b) much paper(c) no paper (d) a few paper
4. Based on the given statements, select the most appropriateoption to solve the given question.If two floors in a certain building are 9 feet apart, how manysteps are there in a set of stairs that extends from the firstfloor to the second floor of the building?
Duration: 3 hrs Maximum Marks: 100
INSTRUCTIONS
1. There are a total of 65 questions carrying 100 marks.
2. This question paper consists of 2 sections, General Aptitude (GA) for 15 marks and the subject specificGATE paper for 85 marks. Both these sections are compulsory.
3. The GA section consists of 10 questions. Question numbers 1 to 5 are of 1-mark each, while question numbers 6 to10 are of 2-marks each. The subject specific GATE paper section consists of 55 questions, out of which questionnumbers 11 to 35 are of 1-mark each, while question numbers 36 to 65 are of 2-marks each.
4. Questions are of Multiple Choice Question (MCQ) or Numerical Answer type. A multiple choice question will havefour choices for the answer with only one correct choice. For numerical answer type questions, the answer is a numberand no choices will be given.
5. Questions not attempted will result in zero mark. Wrong answers for multiple choice type questions will result in
NEGATIVE marks. For all 1 mark questions, 13
mark will be deducted for each wrong answer. For all 2 marks questions,
23
mark will be deducted for each wrong answer..
6. There is NO NEGATIVE MARKING for questions of NUMERICAL ANSWER TYPE.
2015SET - 1
Statements:I. Each step is 3/4 foot high.II. Each step is 1 foot wide.(a) Statement I alone is sufficient, but statement II alone
is not sufficient.(b) Statement II alone is sufficient, but statement I alone
is not sufficient.(c) Both statements together are sufficient, but neither
statement alone is sufficient.(d) Statement I and II together are not sufficient.
5. Which one of the following combinations is incorrect?(a) Acquiescence – Submission(b) Wheedle – Roundabout(c) Flippancy – Lightness(d) Profligate – ExtravagantQUESTION 6 TO 10 CARRY TWO MARKS EACH
6. Select the alternative meaning of the underlined part of thesentence.The chain snatchers took to their heels when the policeparty arrived.(a) took shelter in a thick jungle(b) open indiscriminate fire(c) took to flight(d) unconditionally surrendered
2015 -2 SOLVED PAPER - 2015
7. The pie chart below has the breakup of the number ofstudents from different departments in an engineeringcollege for the year 2012. The proportion of male to femalestudents in each department is 5 : 4. There are 40 males inElectrical Engineering. What is the difference between thenumber of female students in the Civil department and thefemale students in the Mechanical department?
20%
40%
30%
10%
Electrical
ComputerScience
Civil
Mechanical
8. The probabilities that a student passes in Mathematics,Physics and Chemistry are m, p and c respectively. Of thesesubjects, the student has 75% chance of passing in at leastone, a 50% chance of passing in at least two and a 40%chance of passing in exactly two. Following relations aredrawn in m, p, c:I. p + m + c = 27/20II. p + m + c = 13/20III. (p) × (m) × (c) = 1/10(a) Only relation I is true(b) Only relation II is true(c) Relations II and III are true(d) Relations I and III are true
9. The number of students in a class who have answeredcorrectly, wrongly, or not attempted each question in anexam, are listed in the table below. The marks for eachquestion are also listed. There is no negative or partialmarking.
Q. No. Marks Answered Correctly
Answered Wrongly
Not attempted
1 2 21 17 62 3 15 27 23 1 11 29 44 2 23 18 35 5 31 12 1
What is the average of the marks obtained by the class inthe examination?(a) 2.290 (b) 2.970(c) 6.795 (d) 8.795
10. The given statement is followed by some courses of action.Assuming the statement to be true decide the correct option.Statement:There has been a significant drop in the water level in thelakes supplying water to the city.I The water supply authority should impose a partial
cut in supply to tackle the situation.
II The government should appeal to all the residentsthrough mass media for minimal use of water.
III The government should ban the water supply in lowerareas.
(a) Statements I and II follow.(b) Statements I and III follow.(c) Statements II and III follow.(d) All statements follow.
TECHNICAL SECTION
QUESTION 11 TO 35 CARRY ONE MARK EACH11. Consider the circuit shown in the figure. In this circuit R = 1 kW,
and C = 1mF. The input voltage is sinusoidal with a frequencyof 50 Hz, represented as a phasor with magnitude Vi andphase angle 0 radian as shown in the figure. The outputvoltage is represented as a phasor with magnitude V0 andphase angle d radian. What is the value of the output phaseangle d (in radian) relative to the phase angle of the inputvoltage?
C
v = V 0i i Ð
R
v = V < d0+–
R
0
(a) 0 (b) p(c) p/2 (d) –p/2
12. A separately excited DC generator has an armature resistanceof 0.1W and negligible armature inductance. At rated fieldcurrent and rated rotor speed, its open circuit voltage is 200V. When this generator is operated at half the rated speed,with half the rated field current, an uncharged 1000 mFcapacitor is suddenly connected across the armatureterminals. Assume that the speed remains unchanged duringthe transient. At what time (in microsecond) after thecapacitor is connected will the voltage across it reach 25 V?(a) 62.25 (b) 69.3(c) 73.25 (d) 77.3
13. In the 4 × 1 multiplexer, the output F is given by F = A Å B.Find the required input I3I2I1I0.
B
FMUX4 ×1
I0I1I2I3
A
(a) 1010 (b) 0110(c) 1000 (d) 1110
2015-3 SOLVED PAPER - 2015
14. The self inductance of the primary winding of a single phase,50 Hz transformer is 800 mH, and that of the secondarywinding is 600 mH. The mutual inductance between thesetwo windings is 480 mH. The secondary winding of thetransformer is short circuited and primary winding isconnected to a 50 Hz, single phase, sinusoidal voltagesource. The current flowing in both the windings is lessthan their respective rated currents. The resistance of bothwindings can be neglected. In this condition, what is theeffective inductance (in mH) seen by the source?(a) 416 (b) 440(c) 200 (d) 920
15. For the signal flow graph shown in the figure, which one ofthe following expressions is equal to the transfer function
( )( ) ( )12 x s 0
Y sX s
=?
X (s)1 X (s)2
Y(s)G1 G2
– 1 – 1
1
(a) ( )1
2 1
G1 G 1 G+ + (b) ( )
2
1 2
G1 G 1 G+ +
(c)1
1 2
G1 G G+ (d)
2
1 2
G1 G G+
16. Consider a HVDC link which uses thyristor based linecommutated converters as shown in the figure. For a powerflow of 750 MW from system 1 to system 2, the voltages atthe two ends, and the current, are given by: V1 = 500 kV, V2= 485 kV and I = 1.5 kA. If the direction of power flow is to bereversed (that is, from system 2 to system 1) withoutchanging the electrical connections, then which one of thefollowing combinations is feasible?
System 1 System 2I
V1 V2
+
– –
+
(a) V1 = – 500 kV, V2 = – 485 kV and I = 1.5 kA(b) V1 = – 485 kV, V2 = – 500 kV and I = 1.5 kA(c) V1 = 500 kV, V2 = 485 kV and I= –1.5 kA(d) V1 = – 500 kV, V2 = – 485 kV and I = –1.5 kA
17. The voltages developed across the 3W and 2W resistorsshow in the figure are 6 V and 2 V respectively, with thepolarity as marked. What is the power (in Watt) deliveredby the 5 V voltage source?
5 V
N22V
2W
3W
6V +–
Network
N1Network
+–
+ –
(a) 5 (b) 7(c) 10 (d) 14
18. A moving average function is given by y(t) = ( )t
t T
1 u dT
-
t tò .
If the input u is a sinusoidal signal of frequency 1 Hz
2T,
then in steady state, the output y will lag u (in degree)by__________
19. If the sum of the diagonal elements of a 2 × 2 matrix is – 6,then the maximum possible value of determinant of the matrixis ________
20. A(0 – 50 A) moving coil ammeter has a voltage drop of 0.1 Vacross its terminals at full scale deflection. The external shuntresistance (in milliohms) needed to extend its range to(0 – 500 A) is ____.
21. For the given circuit, the Thevenin equivalent is to bedetermined. The Thevenin voltage, VTh (in volt), seen fromterminal AB is
20i
2W1W i2V
1WA
B
+
–
– +
22. In the given circuit, the silicon transistor has b = 75 and acollector voltage VC = 9 V. Then the ratio of RB and RC is________
15 VRC
VC
RB
2015 -4 SOLVED PAPER - 2015
23. Consider a function 21 ˆf rr
=r
, where r is the distance from
the origin and r is the unit vector in the radial direction.The divergence of this function over a sphere of radius R,which includes the origin, is(a) 0 (b) 2p(c) 4p (d) Rp
24. A Bode magnitude plot for the transfer function G(s) of aplant is shown in the figure. Which one of the followingtransfer functions best describes the plant?
20
0
– 20
0.1 1 10 100 1k 10k 100kf(Hz)
2log|(Gj2 f)|p
(a)( )1000 s 10
s 1000+
+(b)
( )( )
10 s 10s s 1000
++
(c) ( )s 1000
10s s 10+
+ (d) ( )s 1000
10 s 10+
+
25. A random variable X has probability density function f(x)as given below:
( ) a bx for 0 x 1f x
0 otherwise+ < <ì
= íî
If the expected value E[X] = 2/3, then Pr[X < 0.5] is___________
26. In the following chopper, the duty ratio of switch S is 0.4. Ifthe inductor and capacitor are sufficiently large to ensurecontinuous inductor current and ripple free capacitorvoltage, the charging current (in Ampere) of the 5 V battery,under steady-state, is _____
+–
S L
20 V
3W
5 V+
–
27. An inductor is connected in parallel with a capacitor asshown in the figure. As the frequency of current i isincreased, the impedance (Z) of the network varies as
L
Zi
C
(a)
Z
f
Inductive
Capacitive
(b)
Z
f
Inductive
Capacitive
(c)
Z
f
InductiveCapacitive
(d)
Z
fInductive
Capacitive
28. The impulse response g(t) of a system G, is as shown in figure(a). What is the maximum value attained by the impulseresponse of two cascaded blocks of G as shown in figure (b)?
1
10
g(t)
tG G
(a) (b)
(a)23 (b)
34
(c)45 (d) 1
2015-5 SOLVED PAPER - 2015
29. A steady current I is flowing in the – x direction through
each of two infinitely long wires at y = L2
± as shown in the
figure. The permeability of the medium is m0. The Br
field at(0, L, 0) is
Current = I
y = L/2
Z
X
Y
Current = I
y = L/2
(a) 04 Iz
3 L- mp
(b) 04 Iz
3 Lmp
(c) 0 (d) 03 Iz
4 L- mp
30. Consider a one turn rectangular loop of wire placed in auniform magnetic field as shown in the figure. The plane ofthe loop is perpendicular to the field lines. The resistance ofthe loop is 0.4 W , and its inductance is negligible. Themagnetic flux density (in Tesla) is a function of time, and isgiven by B(t) = 0.25 sinwt, where w = 2p × 50 radian/second.The power absorbed (in Watt) by the loop from the magneticfield is _________.
10 cm
5 cm
31. If a continuous function f(x) does not have a root in theinterval [a, b], then which one of the following statements isTRUE?(a) f(a). f(b) = 0 (b) f(a) . f(b) < 0(c) f(a) . f(b) > 0 (d) f(a)/f(b) £ 0
32. When the Wheatstone bridge shown in the figure is usedto find the value of resistor Rx, the galvanometer G indicateszero current when R1 = 50 W , R2 = 65 W and R3 = 100 W,If R3 is known with 5%± tolerance on its nominal value of100 W. What is the range of Rx in Ohms?
R1 R2
R3 Rx
V+ –
G
(a) [123.50, 136.50] (b) [125.89, 134.12](c) [117.00, 143.00] (d) [120.25, 139.75]
33. Base load power plants areP. wind farmsQ. run-of-river plantsR. nuclear power plantsS. diesel power plants(a) P, Q and S only (b) P, R and S only(c) P, Q and R only (d) Q and R only
34. The primary mmf is least affected by the secondary terminalconditions in a(a) power transformer (b) potential transformer(c) current transformer (d) distribution transformer
35. Of the four characteristics given below, which are the majorrequirements for an instrumentation amplifier?P. high common mode rejection ratioQ. high input impedanceR. high linearityS. high output impedance(a) P, Q and R only (b) P and R only(c) P, Q and S only (d) Q, R and S only
QUESTION 36 TO 65 CARRY TWO MARKS EACH36. Two players, A and B, alternately keep rolling a fair dice.
The person to get a six first wins the game. Given that playerA starts the game, the probability that A wins the game is(a) 5/11 (b) 1/2(c) 7/13 (d) 6/11
37. The circuit shown is meant to supply a resistive load RLfrom two separate DC voltage sources. The switches S1and S2 are controlled so that only one of them is ON at anyinstant. S1 is turned on for 0.2 ms and S2 is turned on for0.3 ms in a 0.5 ms switching cycle time period. Assumingcontinuous conduction of the inductor current and negligibleripple on the capacitor voltage, the output voltage V0 (involt) across RL is _______
+ +–
S1
S2
RL V0
L
C10V 5V
+
––
2015 -6 SOLVED PAPER - 2015
38. The circuit shown in the figure has two sources connectedin series. The instantaneous voltage of the AC source (inVolt) is given by v(t) = 12sint. If the circuit is in steady state,then the rms value of the current (in Ampere) flowing in thecircuit is______
8V
V(t)
1H
1W
39. The maximum value of ‘a’ such that the matrix3 0 2
1 1 00 a 2
- -é ùê ú-ê úê ú-ë û
has three linearly independent real
eigenvectors is
(a) 23 3
(b)1
3 3
(c)1 2 3
3 3+
(d)1 33 3+
40. Determine the correctness or otherwise of the followingAssertion (A) and the Reason (R).Assertion (A): Fast decoupled load flow method gives
approximate load flow solution because ituses several assumptions.
Reason (R): Accuracy depends on the power mismatchvector tolerance.
(a) Both (A) and (R) are true and (R) is the correct reasonfor (A)
(b) Both (A) and (R) are true but (R) is not the correctreason for (A)
(c) Both (A) and (R) are false(d) (A) is false and (R) is true
41. Find the transfer function ( )( )
Y sX s
of the system given below:
G1
G2
HY(s) Y(s)
+–
+–
X(s)
+
+
(a)1 2
1 2
G G1 HG 1 HG
+- - (b)
1 2
1 2
G G1 HG 1 HG
++ +
(c) ( )1 2
1 2
G G1 H G G
++ + (d) ( )
1 2
1 2
G G1 H G G
+- +
42. Two single phase transformers T1 and T2 each rated at 500 kVAare operated in parallel. Percentage impedances of T1 and T2are (1 + j6) and (0.8 + j4.8), respectively. To share a load of1000 kVA at 0.8 lagging power factor, the contribution of T2(in kVA) is ________
43. In a linear two port network, when 10 V is applied to port 1,a current of 4 A flows through port 2 when it is short circuited.When 5 V is applied to Port 1, a current of 1.25 A flowsthrough a 1 W resistance connected across Port 2. When 3V is applied to Port 1, the current (in Ampere) through a 2 Wresistance connected across Port 2 is __________
44. A 200/400 V, 50 Hz, two winding transformer is rated at20 kVA. Its windings are connected as an auto transformerof rating 200/600 V. A resistive load of 12 W is connected tothe high voltage (600 V) side of the auto transformer. Thevalue of equivalent load resistance (in Ohm) as seen fromlow voltage side is ________
45. A parallel plate capacitor is partially filled with glass ofdielectric constant 4.0 as shown below. The dielectricstrengths of air and glass are 30 kV/cm and 300 kV/cm,respectively. The maximum voltage (in kilovolts), which canbe applied across the capacitor without any breakdown, is______
10 mm5 mm
Air, = 1.0er
Glass g = 4.0e
46. Consider the economic dispatch problem for a power plant havingtwo generating units. The fuel costs in Rs/MWh along with thegeneration limits for the two units are given below:
( ) 21 1 1 1C P 0.01P 30P 10;= + +
1100MW P 150MW£ £
( ) 22 2 2 2C P 0.05P 10P 10= + +
2100MW P 180MW£ £The incremental cost (in Rs/MWh) of the power plant whenit supplies 200 MW is __________
47. An unbalanced DC Wheatstone bridge is shown in the figure.At what value of p will the magnitude of V0 be maximum?
+ –
pR
pR R(1 + x)
E
V0+–
R
2015-7 SOLVED PAPER - 2015
(a) ( )1 x+ (b) (1 + x)
(c) ( )1
1 x+ (d) ( )1 x-
48. The op amp shown in the figure has a finite gain A = 1000and an infinite input resistance. A step voltage V1= 1 mV isapplied at the input at time t = 0 as shown. Assuming thatthe operational amplifier is not saturated, the time constant(in millisecond) of the output voltage V0 is
1 kW
R
Vi
t = 0s
1mVA = 1000
C
1 Fm
V0
+
–
–
++–
(a) 1001 (b) 101(c) 11 (d) 1
49. A solution of the ordinary differential equation2
2d y dy5 6y 0
dtdt+ + = is such that y(0) = 2 and y(1) = 3
1 3ee-
- .
The value of ( )dy 0dt is _________
50. The signum function is given by
( )x ; x 0xsgn x0; x 0
ì ¹ïíï =î
The Fourier series expansion of sgn(cos(t)) has(a) only sine terms with all harmonics.(b) only cosine terms with all harmonics(c) only sine terms with even numbered harmonics(d) only cosine terms with odd numbered harmonics.
51. The transfer function of a second order real system with aperfectly flat magnitude response of unity has a pole at(2 –j3). List all the poles and zeroes.
(a) Poles at ( )2 j3± , no zeroes
(b) Poles at ( )2 j3± - , one zero at origin
(c) Poles at ( ) ( )2 j3 , 2 j3- - + , zeroes at ( ) ( )2 j3 , 2 j3- - - + ,
(d) Poles at ( )2 j3± , zeroes at ( )2 j3- ±52. The open loop poles of a third order unity feedback system
are at 0, –1 , –2. Let the frequency corresponding to thepoint where the root locus of the system transits to unstableregion be K. Now suppose we introduce a zero in the open
loop transfer function at –3 , while keeping all the earlieropen loop poles intact. Which one of the following is TRUEabout the point where the root locus of the modified systemtransits to unstable region?(a) It corresponds to a frequency greater than K(b) It corresponds to a frequency less than K(c) It corresponds to a frequency K(d) Root locus of modified system never transits to
unstable region53. f(A, B, C, D) = PM(0, 1, 3, 4, 5, 7, 9,11, 12, 13, 14, 15) is an
maxterm representation of a Boolean function f(A, B, C, D)where A is the MSB and D is the LSB. The equivalentminimized representation of this function is
(a) ( ) ( )A C D A B D+ + + +
(b) ACD ABD+
(c) ACD ABCD ABCD+ +
(d) ( ) ( ) ( )B C D A B C D A B C D+ + + + + + + + +
54. A sustained three phase fault occurs in the power systemshown in the figure. The current and voltage phasors duringthe fault (on a common reference), after the natural transientshave died down, are also shown. Where is the fault located?
V1 V2I1
I3
I4
Q
RP
STransmission line
Transmission lineI2
V1
V2
I1I2
I3
I4
(a) Location P (b) Location Q(c) Location R (d) Location S
55. An 8-bit, unipolar Successive Approximation Register typeADC is used to convert 3.5 V to digital equivalent output.The reference voltage is +5 V. The output of the ADC, at theend of 3rd clock pulse after the start of conversion, is(a) 1010 0000 (b) 1000 0000(c) 0000 0001 (d) 0000 0011
56. A 3-phase 50 Hz square wave (6 step) VSI feeds a 3-phase,4-pole induction motor. The VSI line voltage has a dominant5th harmonic component. If the operating slip of the motorwith respect to fundamental component voltage is 0.04, theslip of the motor with respect to 5th harmonic component ofvoltage is __________
2015 -8 SOLVED PAPER - 2015
57. Consider a discrete time signal given by
[ ] ( ) [ ] ( ) [ ]n nx n 0.25 u n 0.5 u n 1= - + - -The region of convergence of its Z-transform would be(a) the region inside the circle of radius 0.5 and centered
at origin(b) the region outside the circle of radius 0.25 and centered
at origin(c) the annular region between the two circles, both
centered at origin and having radii 0.25 and 0.5(d) the entire Z plane
58. In the signal flow diagram given in the figure, u1 and u2 arepossible inputs whereas y1 and y2 are possible outputs.When would the SISO system derived from this diagram becontrollable and observable?(a) When u1 is the only input and y1 is the only output(b) When u2 is the only input and y1 is the only output(c) When u1 is the only input and y2 is the only output(d) When u2 is the only input and y2 is the only output
59. A separately excited DC motor runs at 1000 rpm on no loadwhen its armature terminals are connected to a 200 V DCsource and the rated voltage is applied to the field winding.The armature resistance of this motor is 1W . The no loadarmature current is negligible. With the motor developingits full load torque, the armature voltage is set so that therotor speed is 500 rpm. When the load torque is reduced to50% of the full load value under the same armature voltageconditions, the speed rises to 520 rpm. Neglecting therotational losses, the full load armature current (in Ampere)is _______
60. A self commutating switch SW, operated at duty cycle d isused to control the load voltage as shown in the figure
VdcVC RL
VL
SW d C
D
L
Under steady state operating conditions, the averagevoltage across the inductor and the capacitor respectively,are
(a) VL = 0 and C dc1V V
1=
- d
(b) VL = dc1 V2
and C dc1V V
1=
- d
(c) VL = 0 and C dcV V1
d=
- d
(d) VL = dcV2d
and C dcV V1
d=
- d
61. In the given circuit, the parameter k is positive, and thepower dissipated in the 2 W resistor is 12.5 W. The value ofk is _____
+–
kV0
5A
2W 5W
V010W
4V
+ –
62. A 50 Hz generating unit has H-constant of 2 MJ/MVA. Themachine is initially operating in steady state at synchronousspeed, and producing 1 pu of real power. The initial value ofthe rotor angle d is 5°, when a bolted three phase to groundshort circuit fault occurs at the terminal of the generator.Assuming the input mechanical power to remain at 1 pu, thevalue of d in degrees, 0.02 second after the fault is________
63. The figure shows a digital circuit constructed using negativeedge triggered J – K flip flops. Assume a starting state of Q2Q1 Q0 = 000. This state Q2 Q1 Q0 = 000 will repeat after _____number of cycles of the clock CLK.
Clk1 J0
K0
Q0 J1
K1
Q1 J2
K2
Q2
1 Q0 Q1 Q2
Clock Clock Clock
11
64. A DC motor has the following specifications: 10 hp, 37.5 A,230 V; flux/pole = 0.01 Wb, number of poles = 4, number ofconductors = 666, number of parallel paths = 2. Armatueresistance = 0.267W. The armature reaction is negligible androtational losses are 600 W. The motor operates from a 230V DC supply. If the moor runs at 1000 rpm, the output torqueproduced (in Nm) is _______’
65. The single-phase full-bridge Voltage Source Inverter (VSI),shown in figure, has an output frequency of 50 Hz. It usesunipolar pulse width modulation with switching frequencyof 50 kHz and modulation index of 0.7. For Vin = 100 V DC,L = 9.55 mH, C = 63.66 mF, and R = 5 W, the amplitude of thefundamental component in the output voltage V0 (in volt)under steady state is
Full-bridgeVSI C
L
VR V0R+–Vin
+
–
+
–
2015-9 SOLVED PAPER - 2015
GENERAL APTITUDE
1. (c)
2. (a) Number of possible combinations of two number
one from each set = 4 51 1 20´ =C C
Combinations having sum 16 are (2, 14), (3, 13),(4, 12) and (5, 11)
Required Probability 4 0.20
20= =
3. (a)
4. (a) I. Each step is 34
foot high.
No. of steps 4 12
39´= =
Statement I alone is sufficient to answer.5. (b) As a verb wheedle is to cajole or attempt to persuade
by flattery. As a adjective roundabout is indirectcircuitous or circumlocutionary; that doesnot do somethey in a direct way.
6. (c)7. 32 Let total number of students be x
Then, 5 20%9
´ of x = 40
5 20 409 100
x´ =
x = 360
No. of females in civil department = 4 30%9
´ of x
No. of females in Mechanical department 4 10%9
= ´ of x
Difference 4 3 49 10 10 9 5
x x xé ù= - = ´ê úë û
4 360 329 5
= ´ =
8. (a) P(atleast two) – P(exactly 2)= 0.5 – 0.4 = 0.10.75 = p + m + c + 0.1 – (0.5 + 0.11 × 2)
\ p + m + c = 0.65 + 0.7 = 1.35 = 2720
=
9. (b)21 2 15 3 11 11 1 23 2 31 5 2.970
21 15 11 23 31´ + ´ + ´ ´ + ´ + ´ =
+ + + +10. (a)
TECHNICAL SECTION
11. (d) Hint: 0 1æ ö æ ö-= + +ç ÷ ç ÷è ø è øi B
C C
R RV V VX X
.B inC
RV VR X
æ ö= ç ÷+è ø
=V0v0 ÐdÐ0–
+
C
R
C
A
B
R
v Vi i=
12. (b)2
1
2 2
1 1
b
b
E NE N
f=
f = 1 1
1 1
0.5 0.5NN
´ f´f
2bE = 0.25, 1bE = 0.25 × 200 = 50 VR = 0.1 W , C = 1000 m F = 1000 × 10–6 F = 10–3 F
tau |t| = RC = 10–3 × 0.1 = 10–4 sec
50 = 2000 410te -
- Þ t = 69.3 × 10–6
t = 69.3 m sec
13. (b)0
1
2
3
0 0 . 0
0 1 . 11 0 1.1 1 0.
A B F A B
I AB
I AB
I ABI AB
Å
0 1 2 3. . .= + + +F I AB I AB I AB I AB
I0
I1
I2
I3
F
A B
2015 -10 SOLVED PAPER - 2015
14. (a) Here, L1 = 800 mHL1 = 800 × 10–3 = 0.8 HL2 = 600 × 10–3 = 0.6 HM = 480 × 10–3 = 0.48 HNeglecting R1 and R2 ZL= 0
M
+
–
V1
R1R2
50 Hz
Zin = 1
1
VI
= (R1 + j X1 ) + 2 2
2 2 L
MR jX Z
w+ +
Zin + j X1 + 12 2
2
MjX
w = j X1 – j
2 2
2
MX
w
Zin = j 2 2
12
MXX
é ùw-ê úê úë û
Zin = j 2 2
12
MLL
é ùww -ê úwê úë û
= j2 2(2 50) 0.482 0.8 50
(2 50) 0.6
é ùp´ ´p´ ´ -ê úp´ ´ê úë û
= j [251.32 – 120.63] = j 130.37= j 130.37 = j (2p × 50)× LeffLeff = 415 mH » 416 mH
15. (b)
Y(s)
X (s) = 01 X (s)2
1 G1 G2
–1
–1
Let X1(s) = 0
Y(s)
X (s)2
G1 G2–1
–1
G1 = G2 , D1 = 1L1 = – G1 , L2 = – G1G2
1 1
2 1 2
( )( ) 1 ( )
D= =
- +GY s
X s L L2
1 1 21
G
G G G+ +
2
( )( )
Y sX s = 2
1 21 (1 )G
G G+ +16. (b)
17. (a) I1 =63 = 2 AA
I2 = 22
= 1 AA
I3 + 1 = 2 , I3 = 1 AP = 5 × 1 = 5W
18. 90u (t) = sin (wt)
w = 2pf = 2p × 1
2T Tp= ;
wT = p
cos( )( ) sin( )t t
tt Ty t d
-t
-
wté ù= wt t = ê úwtë ûò
1( )y t =p [cosw(t – T) – coswt]
=1p [coswt coswT + sinwt sinwT – coswt]
( )y t =2-
p coswt = 2p sin (90 + wt)
x(t) = sinwt f = 90°
19. 9sum of eigen values = sum of the diagonal elements= l1 + l2 = – 6Product of eigen values = determinant of matrixl1l2 = determinant of matrixThe possible eigen values are–1, – 5 ; –5, –1 ; – 8, 2–2, – 3 ; –4, –2 ; – 9, 3–3, – 1 ; –3, –3 ; – 10, 4Maximum possible values of determinant = –3 × – 3= 9
20. 0.22I1 = 50 A, I2 = 500 AI2 – I1 = 450 A450 × Rsh = 0.1
Rsh = 0.1450 = 0.22m W
21. 3.36V1 – Vth = – 20i
2015-11 SOLVED PAPER - 2015
V1 – 1i = ii – Vth = – 20 iVth = 21 i ____ (1)
1 W
2 W1 W
20 i
iA
B
V1
Ia
Vth– +
+–2V
121V-
= i + Ia
2 – i = i + Ia Þ 2 = 2 i + Ia __ (2)
2th
aV
I = ––––– (3)
From eqn (1) , (2) & (3)
2 = 221 2th thV V
´ +
2 = 4 21
42th thV V+
84 = 25 VthVth = 3.36
22. 105.13RCIE = 15 – VC –––– (1)VC – VB = RB IbVC – 0.7 = RB IbOr, RB.Ib = VC – 0.7 –––– (2)From eqn (1) , & (2)
0.715
B b C
C E C
R I VR I V
-=
-
. 0.7(1 ) 15
B b C
C b C
R I VR I V
-=
+ b -
9 0.7(1 ).15 9
B
C
RR
-æ ö= + b ç ÷-è ø
8.3766
B
C
RR
é ù= ê úë û
B
C
RR = 105.13
23. (a) F = 21
rar
Ñ.F = 2
21 1 1
( ) (sin . )sin sinrr F F
r rr q¶ ¶
+ q +¶ q ¶q q
F¶ f¶f
Ñ.F = 2
2 21 1r
rr r¶ æ ö+ç ÷¶ è ø
+ 0 + 0
Ñ.F = 024. (d) At initial cut off frequency (10 Hz).
20 = 20 log K1 = log KK = 10
G(s) = 1
1000
110
sK
s
æ ö+ç ÷è øæ ö+ç ÷è ø
=
1000101000
1010
s
s
+æ öç ÷è ø
+æ öç ÷è ø
G(s) = 10( 1000) 10 1000
1000 ( 10) 10( 10)s s
s s+ +´ =
+ +25. 0.25
( ) 1¥
-¥=ò f x dx
1
0( ) 1+ =ò a bx dx
12
0
12
bxaxé ù
+ =ê úê úë û
2ba + = 1
2a + b = 2 ––– (1)
E (x) = . ( )x f x dx¥
-¥ò
1
0
2 ( )3
x a bx dx= +ò
12
0
2 ( )3
ax bx dx= +ò
23 =
12 3
02 3
ax bxé ù+ê ú
ê úë û
23 = 2 3
a b+
23 =
3 26
a b+
4 = 3a + 2b –––– (2)From eqn (2)3a + 2b = 43a + 2 (2 – 2a) = 43a + 4a – 8a = 4a = 0 , b = 2
2015 -12 SOLVED PAPER - 2015
Pr [x < 0.5] = 0.5
0( )f x dxò =
0.5
02x dxò = 2
0.52
02
é ùê úê úë û
x
= 0.25
26. 1 V0 = DVs = 0.4 × 20 = 8 V
I0 = 0 8 5 33 3
V ER- -= = = 1 AA
27. (b) L
i
CXL = jwL
XC = 1
j Cw
Z = .+
L C
C L
X XX X =
1.
1
j Lj C
j Lj C
ww
w +w
= 21 C
j LL
w- w
= L
C
ZZ
28. (d) h(t) = g (t) × g (t) = d (t) = 1Convolution of two identical system is impulsefunction.
29. (a) Here,H = H1 + H2
= 0 0( ) ( )32 ( / 2) 22
I Iz z
LLm m
- + -p æ öpç ÷è ø
$ $
= 0 2 2( )2 3
Iz
L Lm é ù- +ê úp ë û
$
= 04( )
3I
zL
m-
p$ = – 04
3I
zL
mp
$
30. 0.196931. (c) For intermediate value Theorem
If f (a) f (b) < 0 then f (x) has atleast one root in[a, b]. Here, f (x) does not have root in the internal [a, b],it means f (a) f (b) > 0.
32. (a) Under balance condition, galvanometer G indicateszero current.So, R1 Rx = R2 R3 (Balance condition)
Rx = 2 3
1
R RR
R1 = 50 W , R2 = 65 W
R3 = 100 ± 5100100
´ = 100 ± 5
R3 (max) = 105 , R3 (min) = 95
Rx (max) = 65 105
50´
= 136.5 W
Rx (min) = 65 95
50´
= 123.5
Rx = (123.50, 136.50)33. (d)34. (c)35. (a)36. (d) Let P (A) be the probability of getting six by player A.
Let P (B) be the probability of getting six by player B.
( )A be the probability of not getting six by player A.
( )B be the probability of not getting six by player B.
P(A) = 16
5( )6
P A =
P(B) = 16
5( )6
P B =
Probability that A wins the game= ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ...P A P A P B P A P A P B P A P B P A+ + +
= 1 5 5 1 5 5 5 5 1. . . . . .6 6 6 6 6 6 6 6 6
+ +
= 2 41 5 51 ....
6 6 6
é ùæ ö æ öê + + + úç ÷ ç ÷è ø è øê úë û
= 16 2
1
516
æ ö- ç ÷è ø
= 16
. 125136
- = 1 36
6 11´ =
611
37. 7V0
10V
5V
0.2 0.5 t (msec)
V0 = 5 0.3 10 0.2
0.5´ + ´
= 1.5 2
0.5+
= 3.50.5 = 7 V
2015-13 SOLVED PAPER - 2015
38. 10 1 H
8V~
V( )t
1 W
Z(w) = R + jwLZ(w) = 1 + jw
Y(w) = 1 1( ) 1Z j
=w + w
Y(w) = 2
1
1+ w Ð –tan–1 (w)
Vin(t) = 8 + 12 sin t
i1 = 8 . 1
1 0+ Ð tan–1 0 = 8
i2 = 121 1+
sin (t – 45°) = 12 1 1sin cos2 2 2
t té ù-ê úë ûi2 = 6sin t – 6 cos titotal = i1 + i2 = 8 + 6 sin t – 6 cos t
Irms = 2 2
2 6 682 2
æ ö æ ö+ +ç ÷ ç ÷è ø è ø
= 64 16 16+ + = 10A39. (b) The characteristics equation of A is
| A – x I | = 0
3 0 21 1 00 2a
- -é ù-ê ú
ê ú-ë û –
0 00 00 0
xx
x
é ùê úê úë û
= 0
3 0 21 1 00 2
xx
a x
- - -é ù- -ê ú
ê ú- -ë û
(x + 1) (x + 2) (x + 3) + 2a = 0
a = –12
[x3 + 6x2 + 11x + 6]
a
x
dd =
12-
[3x2 + 12x + 11]
2
2d adx
= 1
2-
[6x + 12]
For Maxima or minima
a
x
dd = 0 Þ
12-
[3x2 + 12x + 11] = 0
x = – 1.42 , –2.5772
2d adx
= 1
2-
[6 × –1.42 + 12] = – 1.74
– ve sign indicate maximum2
2d adx
= 1
2-
[6 × – 2.577 + 12] = 1.731
+ ve sign indicate minimumMaximum value of a at –1.42 isa (max)
= 1
2-
[(–1.42)3 + 6 (–1.42)2 + 11(–1.42) + 6]
= 0.1924 = 1
3 340. (b)
41. (c)
G1
G2
Y(s)Y(s)+
+H
–
–
+
+
X(s)
Y(s)++
G [X(s) – H Y(s)]1
G [X(s) – H Y(s)]2
Y (s) = G1 [X(s) – H Y(s)] + G2 [X(s) –H Y (s)]Y (s) = G1 X(s) – G1 H Y(s) + G2 X(s) – G2 H Y (s)Y (s) [1 + H G1 + H G2] = [G1 + G2] X(s)
( )( )
Y sX s = 1 2
1 21G GHG HG
++ +
= 1 2
1 21 ( )G GH G G
++ +
42. 555
21
1 2T
ZS S
Z Z= ´
+Z1 = 1 + j 6 = 6.08 Ð 80.53Z2 = 0.8 + j 4.8Z1 + Z2 = 1.8 + j 10.8 = 10.94 Ð 80.53
From eqn (1); 2TS = 1000 × (6.08 80.53)(10.94 80.53)
ÐÐ
= 555.75 Ð0°» 555 kVA
43. 0.545I1 = YI1 V1 + Y12 V1I2 = Y21 V1 + Y22 V24 = Y21 × 10Y21 = 0.41.25 = 0.4 V1 + (I2 × 1W) Y221.25 = 0.4V1 + 1.25 Y22
2015 -14 SOLVED PAPER - 2015
1.25 0.4 51.25- ´
= Y22
Y22 = 1.25 2
1.25-
= 0.75
1.25-
= 3
5-
= –0.6
I2 = Y21 V1 + Y22 V2I2 = Y21 V1 + Y22 (2W × I2)I2 = Y21 V1 + Y22(2 I2)I2 [1 – 2Y22] = Y21 V1
I2 = 21 1
221 2Y V
Y- = 0.4 3
1 2 ( 0.6)´
- ´ -
I2= 1.2
1 1.2+ = 0.545 AA
44. 1.33P1 = P2V1I1 = V2 I2
2 21 2
1 2
V VR R
=
R2 = 2
21
1
VR
Væ öç ÷è ø
R2 = 2200
600æ öç ÷è ø
× 12 = 1.33 W
45. 18.75
C1= 0AdÎ
C2= 04 .Ad
Î
Ceq= 1 2
1 2
C CC C+
5 mm
5 mm
30 kV/cm
300 kV/cm
Î0
4 Î0
(1)
(2)
C1
C2
glass
air
Ceq =
0 0
0 0
4
4
A Ad d
A Ad d
Î Î
Î Î+
Ceq= 0AdÎ
. 04 Î Ad
0 (1 4)
dA
é ùê úÎ +ë û
Ceq= 45 . 0 A
dÎ
Dn= . .
= = =eqv eqvs
C V C VQpA A A
Dn= 04
5ÎA V
d A = 045
Î× ×Vd
E1 = 0
45
nDd
=Î
V
30 × 105 = 45
Vd
V = 530 5 10
4d´ ´
= 3 530 5 5 10 10
4
-´ ´ ´ ´
V = 18750 V= 18.75 kV46. 20
1
1
dCdP = 0.02P1 + 30
2
2
dCdP = 0.1P2 + 10
1
1
dCdP =
2
2
dCdP
0.02P1 + 30 = 0.1P2 + 102P1 + 3000 = 10P2 + 10002P1 – 10 P2 = –2000 –––– (1)P1 + P2 = 200 –––– (2)From Eqn (1) & (2)2P1 – 10 (200 – P1) = –20002P1 – 2000 + 10 P1 = – 2000P1 = 0 , P2 = 200
1
1
dCdP = 30
C1 = 10 [on putting P1& P2 values]Incremental cost = 30 – 10 = 20 Rs / MWh
47. (a)
V0
+
–
pR
pR R
R(1 + )x
2015-15 SOLVED PAPER - 2015
pR pR
R R(1 + ) = x Ry
+
–
–E Vo
0 1- = =
+ +RE EV
R PR P
0. . .+ æ ö
= = ç ÷+ è + øR y E yV E
Ry PR y P
0 0 0 1y E EV V V
y P p+ - é ù
= - = -ê ú+ +ë û
2 20 ( ) (1 )dV
y p yE p Edp
- -= - + + +
23 30
2 2( ) 2(1 )d V
y p yE p Edp
- -= + - +
For maxima or minima
0dVdp = 0
2( )y
y p+ = 21
(1 )p+
y = 2
1y p
pæ ö+ç ÷+è ø
y± = 1y p
p++
(1 )× + = +y p y p
+ = +y y p y p
(1 )- = -y y y p
p =(1 )
(1 )y y
y-
-
p = 1y x= +48. (d) time constant (t) = RC
t = 103×10–6 = 10–3 = 1 m sec
49. –32
2 5 6 0d y dy ydtdt
+ + =
D2 + 5D + 6 = 0D2 + 3D + 2D + 6 = 0D (D + 3) + 2 (D + 3) = 0D = –2, –3y (t) = C1e–2t + C2e–3t
y (0) = 2 = C1 + C2...... (1)
y (1) 3– (1– 3 )= e
e = C1e–2 + C2e–3
...... (2)From eqn (1) and (2)
3–1 3+ e
e= (2 – C2) e–2 + C2e–3
3–1 3+ e
e= 2 e–2 – C2e–2 + C2e–3
3–1 3+ e
e= –2 e–2 = C2e–3 + C2e–2
3–1 3 – 2+ e e
e = C2e–3 – C2e–2
3–1 e
e+
= C2e–3 – C2e–2
–1 + e = C2– C2 e–1 + e = (1– e) C2
C2 –11
ee
+=- Putting in eqn (1)
C2 = –1–1 + C1 = 2C1 = 3y(t) = 3e–2t – e–3t
( )dy tdt = –6e–2t + 3e–3t
(0)dydt = –6 + 3 = –3
50. (d) Sgn [x]
1
–1x
{1 cos 0–1 cos 0Sgn (cos ) >
<= ttt
2015 -16 SOLVED PAPER - 2015
Sgn [cos t]
It represents square wave, which is even and halfwave synmetry function. In this, sgn (cos t) containscosine terms for all odd harmonic
51. (d) Given, magnitude of transfer function is unity. Allpass filter has unity gain. Therefore graph of allpass filter can be drawn with option (d) only.
–2 + 3j
Im
2 + 3j
–2 – 3j 2 – 3 j
52. (d)53. (c) f (A, B, C, D) = pM(0, 1, 3, 4, 5, 7, 9, 11, 12, 13, 14, 15)
f (A, B, C, D) = åM(2, 6, 8, 10)
f (A, B, C, D) = ACD ABCD A BCD+ +
CD CD C
C
D
DA
CD
0
4
12
1
1
5
13
9
3
7
15
11
2
6
14
10
11
1
11
A B
A B
AB
A
A
B
BD
ABCD
54. (b)55. (a) If Vin > VDAC Þ maintain the loaded bit
Vin < VDAC Þ clear the loaded bitThe output of DAC = Resolution × Decimal equivalent
of binary Resolution = 85
2 1- » 20 mV
On 1st clock the value loaded to output register in(1000 0000)2 is (128)10then, VDAC = 128 × 20 = 2.56 VHere, 3.5 > 2.56 V Þ maintain the loaded bit so, after1st clock output is 1000 0000
On second clock pulse value loaded to outputregisteris (11000000) is (192)10 then VDAC = 192 × 20= 3.84VHere, 3.5 < 3.84 Þ Clear the loaded bitSo, after 2nd clock output is 10000000On third clock pulse value loaded to output registeris (1010 0000) is (160)10then VDAC = 160 × 20 = 3.2 VHere 3.5 > 3.2 V Þ maintain the loaded to outputSo, after 3rd clock output is 10100000
56. 0.80857. (c) x[n] = (– 0.25)n u[n] + (0.5)n u [–n – 1]
X[z] = 1 11 1
1 | 0.25 | 1 | 0.5 |z z- --- - -
| –0.25 z–1 | < 1| 0.25 z–1 | < 1| z | > 0.25| 0.5 z–1| > 1| z | < 0.5ROC : 0.25 < | z | < 0.5
58. (b)59. 100 N0 = 1000 rpm
E0 = 20 VNfull = 500 rpmE µ Nf
full
200 1000E 500
=
Efull = 100 = V – Ia la100 = 200 – Ia Ia = 100
60. (a) Under steady state condition capacitor will be opencircuited and inductor will be short circuited
61. 0.5P = I2 R
I = 12.5
2=P
R = 2.5
V0 = 2 × 2.5 = 5V2.5 + kV0 = 5kV0 = 2.5
k =2.55 =
12
= 0.5
62. 5.963. 6
First flip – flop is a mod – 2 counter, second andthird flip – flop is combined as Johnson counterhave mod – 4So, overall modulus = mod – 6 counterTherefore Q2 Q1 Q0 = 000 will repeat after 6 numberof cycles of the clock CLK.
64. 57.78
Hint : E = 60Z N P
Af
; T = outPw
65. 49.5
