Gate Electrical Solved Paper

SOLVED PAPER - 2015

2015-1

2015 SET - 1 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 specific GATE 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 to 10 are of 2-marks each. The subject specific GATE paper section consists of 55 questions, out of which question numbers 11 to 35 are of 1-mark each, while question numbers 36 to 65 are of 2-marks each. Questions are of Multiple Choice Question (MCQ) or Numerical Answer type. A multiple choice question will have four choices for the answer with only one correct choice. For numerical answer type questions, the answer is a number and no choices will be given. Questions not attempted will result in zero mark. Wrong answers for multiple choice type questions will result in 1 NEGATIVE marks. For all 1 mark questions, mark will be deducted for each wrong answer. For all 2 marks questions, 3 2 mark will be deducted for each wrong answer.. 3

4.

5.

6.

There is NO NEGATIVE MARKING for questions of NUMERICAL ANSWER TYPE.

GENERAL APTITUDE QUESTION 1 TO 5 CARRY ONE MARK EACH 1.

2.

3. 4.

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 Given Set A = {2, 3, 4, 5} and Set B = {11, 12, 13, 14, 15}, two numbers are randomly selected, one from each set. What is the probability that the sum of the two numbers equals 16? (a) 0.20 (b) 0.25 (c) 0.30 (d) 0.33 Didn’t you buy __________ when you went shopping? (a) any paper (b) much paper (c) no paper (d) a few paper Based on the given statements, select the most appropriate option to solve the given question. If two floors in a certain building are 9 feet apart, how many steps are there in a set of stairs that extends from the first floor to the second floor of the building?

5.

6.

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. Which one of the following combinations is incorrect? (a) Acquiescence – Submission (b) Wheedle – Roundabout (c) Flippancy – Lightness (d) Profligate – Extravagant QUESTION 6 TO 10 CARRY TWO MARKS EACH Select the alternative meaning of the underlined part of the sentence. The chain snatchers took to their heels when the police party arrived. (a) took shelter in a thick jungle (b) open indiscriminate fire (c) took to flight (d) unconditionally surrendered

2015 -2

7.

SOLVED PAPER - 2015

The pie chart below has the breakup of the number of students from different departments in an engineering college for the year 2012. The proportion of male to female students in each department is 5 : 4. There are 40 males in Electrical Engineering. What is the difference between the number of female students in the Civil department and the female students in the Mechanical department?

10% Civil

20% Electrical

30% Mechanical Computer Science 40% 8.

9.

The probabilities that a student passes in Mathematics, Physics and Chemistry are m, p and c respectively. Of these subjects, the student has 75% chance of passing in at least one, a 50% chance of passing in at least two and a 40% chance of passing in exactly two. Following relations are drawn in m, p, c: I. p + m + c = 27/20 II. p + m + c = 13/20 III. (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 The number of students in a class who have answered correctly, wrongly, or not attempted each question in an exam, are listed in the table below. The marks for each question are also listed. There is no negative or partial marking. Q. No.

Marks

1 2 3 4 5

2 3 1 2 5

Not Answered Answered Correctly Wrongly attempted 21 17 6 15 27 2 11 29 4 23 18 3 31 12 1

What is the average of the marks obtained by the class in the 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 the lakes 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 residents through mass media for minimal use of water. III The government should ban the water supply in lower areas. (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 EACH 11. Consider the circuit shown in the figure. In this circuit R = 1 kW, and C = 1mF. The input voltage is sinusoidal with a frequency of 50 Hz, represented as a phasor with magnitude Vi and phase angle 0 radian as shown in the figure. The output voltage is represented as a phasor with magnitude V0 and phase angle d radian. What is the value of the output phase angle d (in radian) relative to the phase angle of the input voltage? R C – +

vi = Vi Ð 0

v0= V0< d

R (a) 0 (b) p (c) p/2 (d) –p/2 12. A separately excited DC generator has an armature resistance of 0.1W and negligible armature inductance. At rated field current and rated rotor speed, its open circuit voltage is 200 V. When this generator is operated at half the rated speed, with half the rated field current, an uncharged 1000 mF capacitor is suddenly connected across the armature terminals. Assume that the speed remains unchanged during the transient. At what time (in microsecond) after the capacitor 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.

I0 I1 4 ×1 I2 MUX I3

A (a) (c)

1010 1000

F

B (b) 0110 (d) 1110

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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 secondary winding is 600 mH. The mutual inductance between these two windings is 480 mH. The secondary winding of the transformer is short circuited and primary winding is connected to a 50 Hz, single phase, sinusoidal voltage source. The current flowing in both the windings is less than their respective rated currents. The resistance of both windings can be neglected. In this condition, what is the effective 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 of the following expressions is equal to the transfer function Y (s)

X2 (s )

x1( s ) =0

? X2(s)

G1

–1

(a)

3W 2W

Network +

N1

2V

–

Network

–

N2

+ 5V

(a) (c)

5 10

(b) 7 (d) 14

1 18. A moving average function is given by y(t) = T

t

ò u ( t ) dt .

t -T

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 matrix is ________ 20. A(0 – 50 A) moving coil ammeter has a voltage drop of 0.1 V across its terminals at full scale deflection. The external shunt resistance (in milliohms) needed to extend its range to (0 – 500 A) is ____. 21. For the given circuit, the Thevenin equivalent is to be determined. The Thevenin voltage, VTh (in volt), seen from terminal AB is 20i 1W A – +

If the input u is a sinusoidal signal of frequency

X1(s)

1

– 6V +

G2

Y(s)

–1

G1 1 + G 2 (1 + G1 )

G2 (b) 1 + G (1 + G ) 1 2

G1 G2 (d) 1 + G G 1 + G1G 2 1 2 16. Consider a HVDC link which uses thyristor based line commutated converters as shown in the figure. For a power flow of 750 MW from system 1 to system 2, the voltages at the 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 be reversed (that is, from system 2 to system 1) without changing the electrical connections, then which one of the following combinations is feasible?

(c)

System 1

System 2

I + 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 resistors show in the figure are 6 V and 2 V respectively, with the polarity as marked. What is the power (in Watt) delivered by the 5 V voltage source?

2V

+

1W

–

2W

i

B 22. In the given circuit, the silicon transistor has b = 75 and a collector voltage VC = 9 V. Then the ratio of RB and RC is ________ 15 V RB

RC VC

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SOLVED PAPER - 2015

r 1 23. Consider a function f = rˆ , where r is the distance from r2

L i

the origin and ˆr 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 a plant is shown in the figure. Which one of the following transfer functions best describes the plant?

2log|(Gj2 p f)|

Z

C

Z

Inductive f

(a)

20

Capacitive

0

Z

Inductive

– 20

f

f(Hz) 0.1

(a)

(c)

1

10

(b)

100 1k 10k 100k

1000 ( s + 10 ) s + 1000 s + 1000 10s ( s + 10 )

Capacitive

10 ( s + 10 )

(b) s s + 1000 ( )

Z

s + 1000 (d) 10 ( s + 10 )

Inductive

(c)

Capacitive

25. A random variable X has probability density function f(x) as given below:

f

Z

ìa + bx for 0 < x < 1 f (x) = í otherwise î 0

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. If the inductor and capacitor are sufficiently large to ensure continuous inductor current and ripple free capacitor voltage, the charging current (in Ampere) of the 5 V battery, under steady-state, is _____

S

L

f 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 impulse response of two cascaded blocks of G as shown in figure (b)?

g(t) 3W

20 V + –

Inductive

(d)

+ –

5V

1 0

1

t

G

(a) 27. An inductor is connected in parallel with a capacitor as shown in the figure. As the frequency of current i is increased, the impedance (Z) of the network varies as

G (b)

(a)

2 3

(b)

(c)

4 5

(d) 1

3 4

2015-5

SOLVED PAPER - 2015 29. A steady current I is flowing in the – x direction through L as shown in the 2 r figure. The permeability of the medium is m0. The B field at (0, L, 0) is

each of two infinitely long wires at y = ±

G

Z

+ Y

Current = I

Current = I X

(a)

-4m 0 I zˆ 3pL

(b)

4m 0 I zˆ 3pL

-3m 0 I zˆ 4pL 30. Consider a one turn rectangular loop of wire placed in a uniform magnetic field as shown in the figure. The plane of the loop is perpendicular to the field lines. The resistance of the loop is 0.4 W , and its inductance is negligible. The magnetic flux density (in Tesla) is a function of time, and is given by B(t) = 0.25 sinwt, where w = 2p × 50 radian/second. The power absorbed (in Watt) by the loop from the magnetic field is _________. (c)

0

Rx

R3

y = L/2

y = L/2

R2

R1

(d)

V

–

(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 are P. wind farms Q. run-of-river plants R. nuclear power plants S. 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 terminal conditions in a (a) power transformer (b) potential transformer (c) current transformer (d) distribution transformer 35. Of the four characteristics given below, which are the major requirements for an instrumentation amplifier? P. high common mode rejection ratio Q. high input impedance R. high linearity S. 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 EACH

10 cm

5 cm

31. If a continuous function f(x) does not have a root in the interval [a, b], then which one of the following statements is TRUE? (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 used to find the value of resistor Rx, the galvanometer G indicates zero current when R1 = 50 W , R2 = 65 W and R3 = 100 W, If R3 is known with ±5% tolerance on its nominal value of 100 W. What is the range of Rx in Ohms?

36. Two players, A and B, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player A 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 RL from two separate DC voltage sources. The switches S1 and S2 are controlled so that only one of them is ON at any instant. S1 is turned on for 0.2 ms and S2 is turned on for 0.3 ms in a 0.5 ms switching cycle time period. Assuming continuous conduction of the inductor current and negligible ripple on the capacitor voltage, the output voltage V0 (in volt) across RL is _______ S1

L S2

10V + –

+ – 5V

C

RL

+ V0 –

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SOLVED PAPER - 2015

38. The circuit shown in the figure has two sources connected in series. The instantaneous voltage of the AC source (in Volt) is given by v(t) = 12sint. If the circuit is in steady state, then the rms value of the current (in Ampere) flowing in the circuit is______ 1W V(t)

1H

8V

39. The maximum value of ‘a’ such that the matrix é -3 0 -2 ù ê 1 -1 0 ú ê ú has three linearly independent real ëê 0 a -2 ûú eigenvectors is

(a)

2

(b)

1

3 3 3 3 1+ 2 3 1+ 3 (d) (c) 3 3 3 3 40. Determine the correctness or otherwise of the following Assertion (A) and the Reason (R). Assertion (A): Fast decoupled load flow method gives approximate load flow solution because it uses several assumptions. Reason (R): Accuracy depends on the power mismatch vector tolerance. (a) Both (A) and (R) are true and (R) is the correct reason for (A) (b) Both (A) and (R) are true but (R) is not the correct reason for (A) (c) Both (A) and (R) are false (d) (A) is false and (R) is true 41. Find the transfer function +

Y (s )

X (s )

42. Two single phase transformers T1 and T2 each rated at 500 kVA are operated in parallel. Percentage impedances of T 1 and T2 are (1 + j6) and (0.8 + j4.8), respectively. To share a load of 1000 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 flows through a 1 W resistance connected across Port 2. When 3 V is applied to Port 1, the current (in Ampere) through a 2 W resistance connected across Port 2 is __________ 44. A 200/400 V, 50 Hz, two winding transformer is rated at 20 kVA. Its windings are connected as an auto transformer of rating 200/600 V. A resistive load of 12 W is connected to the high voltage (600 V) side of the auto transformer. The value of equivalent load resistance (in Ohm) as seen from low voltage side is ________ 45. A parallel plate capacitor is partially filled with glass of dielectric constant 4.0 as shown below. The dielectric strengths of air and glass are 30 kV/cm and 300 kV/cm, respectively. The maximum voltage (in kilovolts), which can be applied across the capacitor without any breakdown, is ______

of the system given below:

G1

Air, er = 1.0

5 mm

10 mm Glass eg = 4.0

46. Consider the economic dispatch problem for a power plant having two generating units. The fuel costs in Rs/MWh along with the generation limits for the two units are given below: C1 ( P1 ) = 0.01P12 + 30P1 + 10;

100MW £ P1 £ 150MW C 2 ( P2 ) = 0.05P22 + 10P2 + 10

100MW £ P2 £ 180MW The incremental cost (in Rs/MWh) of the power plant when it 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

R(1 + x)

– X(s)

H +

–

+ Y(s)

+ V 0 –

Y(s)

+

G2

(a)

G1 G2 + 1 - HG1 1 - HG 2

G1 G2 + (b) + 1 HG1 1 + HG 2

(c)

G1 + G 2 1 + H ( G1 + G 2 )

G1 + G 2 (d) 1 - H ( G + G ) 1 2

R

pR

+

– E

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SOLVED PAPER - 2015

(1 + x )

(a)

(b) (1 + x)

1

(c)

(d)

(1 + x )

(1 - x )

48. The op amp shown in the figure has a finite gain A = 1000 and an infinite input resistance. A step voltage V1= 1 mV is applied at the input at time t = 0 as shown. Assuming that the operational amplifier is not saturated, the time constant (in millisecond) of the output voltage V0 is C 1mF R Vi 1mV

1 kW + –

– A = 1000 +

+ V0 –

t = 0s

(a) 1001 (b) 101 (c) 11 (d) 1 49. A solution of the ordinar y differential equation

loop transfer function at –3 , while keeping all the earlier open loop poles intact. Which one of the following is TRUE about the point where the root locus of the modified system transits 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 region 53. 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 equivalent minimized 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 system shown in the figure. The current and voltage phasors during the fault (on a common reference), after the natural transients have died down, are also shown. Where is the fault located?

V1

d2 y

1 - 3e dy + 5 + 6y = 0 is such that y(0) = 2 and y(1) = - 3 . 2 dt e dt

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 a perfectly 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 the point where the root locus of the system transits to unstable region be K. Now suppose we introduce a zero in the open

Transmission line

Q P

dy ( 0 ) is _________ The value of dt 50. The signum function is given by

ìx ï ;x ¹ 0 sgn ( x ) í x ï 0; x = 0 î

I3

I1 I2 R

V2

S Transmission line

I4 V2

V1 I2

I3

I1

I4 (a) Location P (b) Location Q (c) Location R (d) Location S 55. An 8-bit, unipolar Successive Approximation Register type ADC is used to convert 3.5 V to digital equivalent output. The reference voltage is +5 V. The output of the ADC, at the end 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 dominant 5th harmonic component. If the operating slip of the motor with respect to fundamental component voltage is 0.04, the slip of the motor with respect to 5th harmonic component of voltage is __________

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SOLVED PAPER - 2015

57. Consider a discrete time signal given by x [ n ] = ( -0.25 ) u [ n ] + ( 0.5 ) u [ -n - 1] n

n

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 are possible inputs whereas y1 and y2 are possible outputs. When would the SISO system derived from this diagram be controllable 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 load when its armature terminals are connected to a 200 V DC source and the rated voltage is applied to the field winding. The armature resistance of this motor is 1W . The no load armature current is negligible. With the motor developing its full load torque, the armature voltage is set so that the rotor speed is 500 rpm. When the load torque is reduced to 50% of the full load value under the same armature voltage conditions, the speed rises to 520 rpm. Neglecting the rotational losses, the full load armature current (in Ampere) is _______ 60. A self commutating switch SW, operated at duty cycle d is used to control the load voltage as shown in the figure VL

SW

d

C

VC

RL

Under steady state operating conditions, the average voltage across the inductor and the capacitor respectively, are (a)

VL = 0 and VC =

(b) VL = (c) (d)

1 Vdc 1- d

1 1 Vdc Vdc and VC = 1- d 2

VL = 0 and VC =

d Vdc 1- d

d d Vdc VL = Vdc and VC = 1- d 2

5W

2W + 4V

V0

–

+ –

10W 5A kV0

62. A 50 Hz generating unit has H-constant of 2 MJ/MVA. The machine is initially operating in steady state at synchronous speed, and producing 1 pu of real power. The initial value of the rotor angle d is 5°, when a bolted three phase to ground short circuit fault occurs at the terminal of the generator. Assuming the input mechanical power to remain at 1 pu, the value of d in degrees, 0.02 second after the fault is ________ 63. The figure shows a digital circuit constructed using negative edge triggered J – K flip flops. Assume a starting state of Q2 Q1 Q0 = 000. This state Q2 Q1 Q0 = 000 will repeat after _____ number of cycles of the clock CLK.

1 J Q0 0 Clk Clock Q0 1 K 0

D

L Vdc

61. In the given circuit, the parameter k is positive, and the power dissipated in the 2 W resistor is 12.5 W. The value of k is _____

Q1 J1 Clock Q1 K1

1

J2 1

Q2

Clock Q2 K2

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 of conductors = 666, number of parallel paths = 2. Armatue resistance = 0.267W. The armature reaction is negligible and rotational losses are 600 W. The motor operates from a 230 V DC supply. If the moor runs at 1000 rpm, the output torque produced (in Nm) is _______’ 65. The single-phase full-bridge Voltage Source Inverter (VSI), shown in figure, has an output frequency of 50 Hz. It uses unipolar pulse width modulation with switching frequency of 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 the fundamental component in the output voltage V0 (in volt) under steady state is + Vin + –

Full-bridge VSI

VR –

L

+ C

R

V0 –

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SOLVED PAPER - 2015

TECHNICAL SECTION

GENERAL APTITUDE 1.

(c)

2.

(a)

Number of possible combinations of two number

11.

æ -R ö æ R ö Vi + ç 1 + VB (d) Hint: V0 = ç ÷ X C ø÷ è XC ø è æ R VB = ç è R + XC

one from each set = 4 C1 ´ 5C1 = 20 Combinations having sum 16 are (2, 14), (3, 13), (4, 12) and (5, 11) Required Probability = 3. 4.

R

4 = 0.20 20

C

(a) (a)

9´ 4 = 12 3 Statement I alone is sufficient to answer. As a verb wheedle is to cajole or attempt to persuade by flattery. As a adjective roundabout is indirect circuitous or circumlocutionary; that doesnot do some they in a direct way.

6. 7.

(b)

(c) 32

(b)

5 ´ 20% of x = 40 9

4 ´ 30% of x 9

4 é 3x x ù 4 x Difference = ê - ú = ´ 9 ë 10 10 û 9 5 4 360 ´ = 32 9 5 P(atleast two) – P(exactly 2) = 0.5 – 0.4 = 0.1 0.75 = p + m + c + 0.1 – (0.5 + 0.11 × 2) =

9.

(b)

10.

(a)

-t

10-4 t = 69.3 m sec

4 No. of females in Mechanical department = ´ 10% of x 9

\ p + m + c = 0.65 + 0.7 = 1.35 = =

0.5 N1 ´ 0.5f1 N 2 f2 N 1 f1 = N1 ´ f1

50 = 2000 e

No. of females in civil department =

(a)

=

R

tau |t| = RC = 10–3 × 0.1 = 10–4 sec

5 20 x ´ = 40 9 100 x = 360

8.

Eb 1

+

Eb2 = 0.25, Eb1 = 0.25 × 200 = 50 V R = 0.1 W , C = 1000 m F = 1000 × 10–6 F = 10–3 F

Let total number of students be x Then,

E b2

v0=V0 Ðd

B C

12.

–

A

vi = Vi Ð0

3 I. Each step is foot high. 4

No. of steps = 5.

ö ÷ .Vin ø

27 20

21 ´ 2 + 15 ´ 3 + 11 ´ 11 ´ 1 + 23 ´ 2 + 31 ´ 5 = 2.970 21 + 15 + 11 + 23 + 31

13.

(b)

Þ t = 69.3 × 10–6

AÅ B

A B F

0 0 I0 . AB 0 0 1

I1. AB 1

1

0 I2 . AB 1 1 1 I3. AB 0

F = I 0 A B + I1. AB + I 2 . AB + I3 .AB I0 I1

F

I2 I3

A

B

2015 -10

SOLVED PAPER - 2015

14. (a) Here, L1 = 800 mH L1 = 800 × 10–3 = 0.8 H L2 = 600 × 10–3 = 0.6 H M = 480 × 10–3 = 0.48 H Neglecting R1 and R2 ZL= 0

G2 G1D1 Y (s ) = = X 2 (s ) 1 - ( L1 + L2 ) 1 + G1 + G1G2

M

R1

+

G1 = G2 , D1 = 1 L1 = – G1 , L2 = – G1G2

G2 Y (s) = X 2 (s ) 1 + G1(1 + G2 )

R2 16. (b)

V1

17. (a) I1 =

50 Hz – Zin =

2 =1A 2 I3 + 1 = 2 , I3 = 1 A P = 5 × 1 = 5W I2 =

w2 M 2 V1 = (R1 + j X1 ) + R2 + jX 2 + Z L I1 18. 90

w2 M 2 w2 M 2 Zin + j X1 + 1 = j X1 – j jX 2 X2

wT = p

é (2p´ 50)2 ´ 0.482 ù = j ê 2p´ 0.8 ´ 50 - (2p´ 50) ´ 0.6 ú ëê ûú

X1(s) = 0 G1

y (t ) =

1 [cosw(t – T) – coswt] p

1 [coswt coswT + sinwt sinwT – coswt] p

-2 2 coswt = sin (90 + wt) p p x(t) = sinwt f = 90°

19. 9 sum of eigen values = sum of the diagonal elements = l1 + l2 = – 6 Product of eigen values = determinant of matrix l 1l 2 = determinant of matrix The possible eigen values are –1, – 5 ; –5, –1 ; – 8, 2 –2, – 3 ; –4, –2 ; – 9, 3 –3, – 1 ; –3, –3 ; – 10, 4 Maximum possible values of determinant = –3 × – 3= 9

G2 Y(s)

–1 –1

Let X1(s) = 0

20. 0.22 I1 = 50 A, I2 = 500 A I2 – I1 = 450 A 450 × Rsh = 0.1

X2(s) –1

é cos(wt) ù ò sin(wt)d t = êë wt úût t -T

y (t ) =

X2(s)

1

G1

G2

Y(s)

Rsh =

0.1 = 0.22m W 450

21. 3.36 V1 – Vth = – 20i

–1

t -t

y (t ) =

=

= j [251.32 – 120.63] = j 130.37 = j 130.37 = j (2p × 50)× Leff Leff = 415 mH » 416 mH 15. (b)

1 p = ; 2T T

t

é w2 M 2 ù L w ê ú 1 j wL2 úû êë

Zin =

u (t) = sin (wt) w = 2pf = 2p ×

é w2 M 2 ù X ê ú 1 j X 2 ûú ëê

Zin =

6 =2A 3

SOLVED PAPER - 2015 V1 – 1i = i i – Vth = – 20 i Vth = 21 i

____ (1)

20 i – +

1 W V1 i

+ –

2V

24.

Vth

A

Ñ.F = 0 (d) At initial cut off frequency (10 Hz). 20 = 20 log K 1 = log K K = 10

Ia

1W

s ö æ æ 1000 + s ö K ç1 + 10 ç ÷ ÷ è 1000 ø è 1000 ø = G(s) = s ö æ æ 10 + s ö ç1 + ÷ ç ÷ è 10 ø è 10 ø

2W B

2 - V1 = i + Ia 1 2 – i = i + Ia Þ 2 = 2 i + Ia __ (2) Vth ––––– (3) 2 From eqn (1) , (2) & (3) Ia =

2 = 2´

22.

Vth Vth + 21 2

4Vth + 21 Vth 2= 42 84 = 25 Vth Vth = 3.36 105.13 RCIE = 15 – VC –––– (1) VC – VB = RB Ib VC – 0.7 = RB Ib Or, RB.Ib = VC – 0.7 –––– (2) From eqn (1) , & (2)

RB Ib VC - 0.7 = RC I E 15 - VC RB .Ib V - 0.7 = C RC (1 + b) Ib 15 - VC

23.

G(s) = 25.

10( s + 1000) 10 s + 1000 ´ = 1000 ( s + 10) 10( s + 10)

0.25 ¥

ò

f ( x ) dx = 1

-¥ 1

ò (a + bx) dx = 1 0

1

é bx 2 ù ê ax + ú =1 2 úû êë 0 b =1 2 2a + b = 2 a+

––– (1)

¥

E (x) =

ò

x. f ( x) dx

-¥

1

2 = x (a + bx) dx 3 ò 0 1

RB æ 9 - 0.7 ö = (1 + b). ç ÷ RC è 15 - 9 ø

2 = (ax + bx 2 ) dx 3 ò

RB é 8.3 ù = 76 ê ú RC ë 6 û

é ax 2 bx3 ù 2 = ê 2 + 3 ú 3 êë úû 0

RB RC = 105.13 (a) F = Ñ.F =

1 r2

ar

1 ¶ 2 1 ¶ 1 ¶F f (r Fr ) + (sin q.Fq ) + 2 ¶r sin q ¶q r sin q ¶f r

Ñ.F =

1 ¶ æ 2 1 ö çr + 2 ÷ + 0 + 0 r 2 ¶r è r ø

2015-11

0

1

2 a b = + 3 2 3 2 3a + 2b = 3 6 4 = 3a + 2b From eqn (2) 3a + 2b = 4 3a + 2 (2 – 2a) = 4 3a + 4a – 8a = 4 a=0,b=2

–––– (2)

2015 -12

SOLVED PAPER - 2015 0.5

Pr [x < 0.5] =

ò

0.5

f ( x ) dx =

0

ò

0

5 = 100 ± 5 100 R3 (max) = 105 , R3 (min) = 95

0.5

é x2 ù 2x dx = 2 ê ú êë 2 úû 0

R3 = 100 ± 100 ´

= 0.25 26. 1

Rx (max) =

V0 = DVs = 0.4 × 20 = 8 V I0 =

V0 - E 8 - 5 3 = = =1A R 3 3

L

27. (b)

C

(d) (c) (a) (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.

XL = jwL

( B) be the probability of not getting six by player B.

1 XC = jwC 1 jwC 1 j wL + j wC

j wL 2

1 - w LC

=

ZL ZC

=

I m0 (- $z ) + 2p( L / 2)

P ( A) =

5 6

P(B) =

1 6

5 6 Probability that A wins the game

= P( A) + P( A) P( B) P( A) + P( A) P( B) P( A) P( B) P( A) + ...

I m0 (- z$ ) æ 3L ö 2p ç ÷ è 2 ø

m0 I $ é 2 2 ù (- z ) ê + ú 2p ë L 3L û

4 I m0 $ 4Im0 $ (- z ) = – z = 3pL 3pL 30. 0.1969 31. (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 indicates zero current. So, R1 Rx = R2 R3 (Balance condition)

Rx =

1 6

P( B) =

28. (d) h(t) = g (t) × g (t) = d (t) = 1 Convolution of two identical system is impulse function. 29. (a) Here, H = H1 + H2 =

P(A) =

j wL.

X L. X C Z = X +X = C L

=

65 ´ 95 = 123.5 50 Rx = (123.50, 136.50)

Rx (min) = 33. 34. 35. 36.

i

65 ´105 = 136.5 W 50

R2 R3 R1

R1 = 50 W , R2 = 65 W

=

1 5 5 1 5 5 5 5 1 + . . + . . . . 6 6 6 6 6 6 6 6 6

=

2 4 ù 1é æ5ö æ5ö ê1 + ç ÷ + ç ÷ + ....ú 6ê è6ø è6ø úû ë

=

1 6

1 æ5ö 1-ç ÷ è6ø

2

=

1. 6

6 1 36 1 = = ´ 11 25 6 11 136

V0

37. 7

10V 5V

0.2 V0 =

0.5

t (msec)

5 ´ 0.3 + 10 ´ 0.2 1.5 + 2 3.5 = = =7V 0.5 0.5 0.5

2015-13

SOLVED PAPER - 2015 38.

1W

10

x = – 1.42 , –2.577

1H

d 2a

-1 [6 × –1.42 + 12] = – 1.74 2 dx – ve sign indicate maximum 2

~ V(t)

d 2a

8V

-1 [6 × – 2.577 + 12] = 1.731 2 dx + ve sign indicate minimum Maximum value of a at –1.42 is a (max) 2

Z(w) = R + jwL Z(w) = 1 + jw 1 1 = Z (w) 1 + j w

Y(w) =

1

Y(w) =

2

1+ w

=

Ð –tan–1 (w)

Vin(t) = 8 + 12 sin t

1

i1 = 8 . i2 =

1+ 0

Ð tan–1 0 = 8

12 é 1 1 ù cos t ú sin (t – 45°) = ê sin t 1 +1 2ë 2 2 û

2

æ 6 ö æ 6 ö 82 + ç ÷ +ç ÷ è 2ø è 2ø

41.

(c)

da -1 =0Þ [3x2 + 12x + 11] = 0 dx 2

G2

Y(s)

+ G2[X(s) – H Y(s)]

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) G1 + G2 G1 + G2 Y ( s) = = 1 + H (G + G ) X ( s) 1 + HG1 + HG2 1 2

42.

555 ST2 = S ´

Z1 Z1 + Z 2

Z1 = 1 + j 6 = 6.08 Ð 80.53 Z2 = 0.8 + j 4.8 Z1 + Z2 = 1.8 + j 10.8 = 10.94 Ð 80.53 From eqn (1); ST2 = 1000 ×

da -1 2 = [3x + 12x + 11] dx 2

For Maxima or minima

Y(s)

Y(s)+

+

1 a = – [x3 + 6x2 + 11x + 6] 2

dx

+

G1[X(s) – H Y(s)]

(x + 1) (x + 2) (x + 3) + 2a = 0

-1 [6x + 12] 2

G1 H

–

é -3 0 -2 ù é x 0 0ù ê 1 -1 0 ú – ê 0 x 0 ú = 0 êë 0 a -2 úû êë 0 0 x úû

=

X(s) +

64 + 16 + 16 = 10A (b) The characteristics equation of A is |A–xI|=0

2

3 3

–

2

0 -2 ù é -3 - x -1 - x 0 ú ê 1 a -2 - x ûú ëê 0

1

+

=

d 2a

-1 [(–1.42)3 + 6 (–1.42)2 + 11(–1.42) + 6] 2

(b)

i2 = 6sin t – 6 cos t itotal = i1 + i2 = 8 + 6 sin t – 6 cos t

39.

=

= 0.1924 = 40.

12

Irms =

=

43.

(6.08 Ð 80.53) (10.94 Ð 80.53)

= 555.75 Ð0° » 555 kVA 0.545 I1 = YI1 V1 + Y12 V1 I2 = Y21 V1 + Y22 V2 4 = Y21 × 10 Y21 = 0.4 1.25 = 0.4 V1 + (I2 × 1W) Y22 1.25 = 0.4V1 + 1.25 Y22

2015 -14

SOLVED PAPER - 2015

1.25 - 0.4 ´ 5 = Y22 1.25

Y22 =

1.25 - 2 -0.75 -3 = = = –0.6 1.25 1.25 5

I2 = Y21 V1 + Y22 V2 I2 = Y21 V1 + Y22 (2W × I2) I2 = Y21 V1 + Y22(2 I2) I2 [1 – 2Y22] = Y21 V1

ù d A Î0 4 Î0 A é ê ú . d d ë A Î0 (1 + 4) û

Ceq=

4 Î0 A . 5 d

Dn= ps =

Y21 V1 0.4 ´ 3 I2 = = 1 - 2 Y22 1 - 2 ´ (-0.6)

I2 =

Ceq=

1.2 = 0.545 A 1 + 1.2

Q Ceqv .V Ceqv .V = = A A A

Dn=

4 A Î0 V 4 Î0 × ×V = 5d A 5 d

E1 =

Dn 4 = V Î0 5d

30 × 105 =

44. 1.33 P1 = P2 V1I1 = V2 I2

30 ´ 5d ´105 30 ´ 5 ´ 5 ´10-3 ´105 = 4 4 V = 18750 V= 18.75 kV

V=

V12 V22 = R1 R2

46. 20 2

dC1 = 0.02P1 + 30 dP1

æ V2 ö R2 = ç ÷ R1 è V1 ø

dC2 = 0.1P2 + 10 dP2

2

æ 200 ö R2 = ç ÷ × 12 = 1.33 W è 600 ø

dC1 dC2 = dP1 dP2

45. 18.75 C1=

A Î0 d

C2=

4 Î0 .A d

Ceq=

C1 C2 C1 + C2

30 kV/cm C1 300 kV/cm C2

0.02P1 + 30 = 0.1P2 + 10 2P1 + 3000 = 10P2 + 1000 2P1 – 10 P2 = –2000 –––– (1) P1 + P2 = 200 –––– (2) From Eqn (1) & (2) 2P1 – 10 (200 – P1) = –2000 2P1 – 2000 + 10 P1 = – 2000 P1 = 0 , P2 = 200 air

dC1 = 30 dP1

(1)

Î0 glass

C1 = 10 [on putting P1& P2 values] Incremental cost = 30 – 10 = 20 Rs / MWh

5 mm 47. (a) (2)

pR

5 mm

A Î0 4 Î0 A d d = A Î0 4 Î0 A + d d

R(1 + x) +

4 Î0

– pR

Ceq

4V 5d

V0 R

SOLVED PAPER - 2015 49.

+

–3 d2y

pR

dy + 6y = 0 dt dt D2 + 5D + 6 = 0 D2 + 3D + 2D + 6 = 0 D (D + 3) + 2 (D + 3) = 0 D = –2, –3 y (t) = C1e–2t + C2e–3t y (0) = 2 = C1 + C2 ...... (1) 2

pR Vo

E –

R(1 + x) = Ry

R –

V0- =

RE E = R + PR 1 + P

V0+ =

æ y ö R. y.E =ç .E Ry + PR è y + P ø÷

+5

y (1) =

= C1e–2 + C2e–3 e3 ...... (2) From eqn (1) and (2) –1 + 3e 3

V0 = V0+

- V0-

– (1 – 3e)

e

é yE E ù =ê ú ë y + P 1+ p û

–1 + 3e e3

= (2 – C2) e–2 + C2e–3 = 2 e–2 – C2e–2 + C2e–3

dV0 = -( y + p) -2 yE + (1 + p)-2 E dp

–1 + 3e

d 2V0

–1 + 3e – 2e

dp

2

e3

= 2( y + p )-3 yE - 2(1 + p ) -3 E

e3

= C2e–3 – C2e–2 e3 –1 + e = C2– C2 e –1 + e = (1– e) C2

dV0 =0 dp

( y + p)

2

=

æ y+ pö y= ç ÷ è 1+ p ø

–1 + e Putting in eqn (1) 1- e C2 = –1 –1 + C1 = 2 C1 = 3 y(t) = 3e–2t – e–3t

1 (1 + p )

C2 =

2

2

y+ p ± y= 1+ p

dy (t ) = –6e–2t + 3e–3t dt

y × (1 + p) = y + p y + y p= y+ p

= C2e–3 – C2e–2

–1 + e

For maxima or minima

y

= –2 e–2 = C2e–3 + C2e–2

50.

dy (0) = –6 + 3 = –3 dt (d) Sgn [x]

y - y = (1 - y ) p p= p= 48.

y (1 - y ) (1 - y )

1 x

–1

y = 1+ x

(d) time constant (t) = RC t = 103×10–6 = 10–3 = 1 m sec

Sgn (cos t ) =

{

1 cos t > 0 –1 cos t < 0

2015-15

2015 -16

SOLVED PAPER - 2015 Sgn [cos t]

It represents square wave, which is even and half wave synmetry function. In this, sgn (cos t) contains cosine terms for all odd harmonic 51. (d) Given, magnitude of transfer function is unity. All pass filter has unity gain. Therefore graph of all pass filter can be drawn with option (d) only.

On second clock pulse value loaded to output register is (11000000) is (192)10 then VDAC = 192 × 20 = 3.84V Here, 3.5 < 3.84 Þ Clear the loaded bit So, after 2nd clock output is 10000000 On third clock pulse value loaded to output register is (1010 0000) is (160)10 then VDAC = 160 × 20 = 3.2 V Here 3.5 > 3.2 V Þ maintain the loaded to output So, after 3rd clock output is 10100000 56. 0.808 57. (c) x[n] = (– 0.25)n u[n] + (0.5)n u [–n – 1]

Im –2 + j 3

2+j3

–2 – j 3

2– j3

f (A, B, C, D) = A CD + ABCD + A BCD CD CD

CD

CD

CD 12

AB

0

1

3

AB

4

5

7

6

12

13

15

14

9

11

AB AB

ACD

1

1

1

1

200 1000 = E full 500

Efull = 100 = V – Ia la 100 = 200 – Ia Ia = 100 60. (a) Under steady state condition capacitor will be open circuited and inductor will be short circuited 61. 0.5 P = I2 R P 12.5 = = 2.5 2 R V0 = 2 × 2.5 = 5V 2.5 + kV0 = 5 kV0 = 2.5

I=

k=

AB D

5

» 20 mV 2 -1 On 1st clock the value loaded to output register in (1000 0000)2 is (128)10 then, VDAC = 128 × 20 = 2.56 V Here, 3.5 > 2.56 V Þ maintain the loaded bit so, after 1st clock output is 1000 0000

of binary Resolution =

8

1

58. (b) 59. 100 N0 = 1000 rpm E0 = 20 V Nfull = 500 rpm E µ Nf

10

54. (b) 55. (a) If Vin > VDAC Þ maintain the loaded bit Vin < VDAC Þ clear the loaded bit The output of DAC = Resolution × Decimal equivalent

-1

-

1- | - 0.25 z | 1- | 0.5 z -1 | | –0.25 z–1 | < 1 | 0.25 z–1 | < 1 | z | > 0.25 | 0.5 z–1| > 1 | z | < 0.5 ROC : 0.25 < | z | < 0.5

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)

AB

1

X[z] =

2.5 1 = = 0.5 5 2

62. 5.9 63. 6 First flip – flop is a mod – 2 counter, second and third flip – flop is combined as Johnson counter have mod – 4 So, overall modulus = mod – 6 counter Therefore Q2 Q1 Q0 = 000 will repeat after 6 number of cycles of the clock CLK. 64. 57.78 Hint : E = 65. 49.5

Pout fZ N P ;T= 60 A w