Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

Electrical Engineering Q. No. 1 – 25 Carry One Mark Each

1.

  t   t  , t  0 Consider g  t    , where t    t   t  , otherwise 

Here,  t  represents the largest integer less than or equal to t and  t  denotes the smallest integer greater than or equal to t. The coefficient of the second harmonic component of the Fourier series representing g(t) is _________. Key: 0 to 0

t   t   0   Exp: Given g  t        t   t  otherwise   If we plot the above signal, we get gt

1

2

3

1

Since this wave form contain hidden half wave symmetry, even harmonics does not exist. Thus coefficient of second harmonic component of Fourier series will be zero. 2.

A source is supplying a load through a 2-phase, 3-wire transmission system as shown in figure below. The instantaneous voltage and current in phase-a are Van=220sin 100t  V and

i a  10sin 100t  A, respectively. Similarly for phase-b the instantaneous voltage and current are Vbn  220cos 100t  V and i b  10cos 100t  A, respectively.

ia

a'

Van i b

b'

a

 b

Load



Source

n

 

Vbn

n'

The total instantaneous power flowing form the source to the load is (A) 2200 W

(B) 2200sin2 100t  W

(C) 440 W

(D) 2200sin 100t  cos 100t  W

Key: (A)

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

1

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

Exp: Pins tan eous  Van ia  Vbn ib

 220sin 100t 10sin 100t  t  220cos 100t 10cos 100 t   2200sin 2 100t   2200cos 2 100t   2200W 3.

A three-phase, 50Hz, star-connected cylindrical-rotor synchronous machine is running as a motor. The machine is operated from a 6.6 kV grid and draws current at unity power factor (UPF). The synchronous reactance of the motor is 30  per phase. The load angle is 30o. The power delivered to the motor in kW is _______.

Key:

835 to 842

Exp:

Vt  6.6kV

30

V I a Xs E b

Ia

V 6600 3  cos  cos30o E b  4400 volts Eb 

Ia 

E b 30o

E b cos   V

E b cos

~ cos   1

P  3.

E b .V 4400  3810.51 sin   3.  sin 30o Xs 30

P  838.31 kW

4.

For a complex number z, lim z i

(A) -2i Key: (D) Exp: lim z i

z2  1 is z3  2z  i  z 2  2 

(B) -i

(C) i

(D) 2i

z2  1 zi 2i  lim 2   2i 3 2 z  i z  2 1  2 z  2z  i  z  2 

5.

Consider an electron, a neutron and a proton initially at rest and placed along a straight line such that the neutron is exactly at the center of the line joining the electron and proton. At t=0, the particles are released but are constrained to move along the same straight line. Which of these will collide first? (A) The particles will never collide (B) All will collide together (C) Proton and Neutron (D) Electron and Neutron Key: (D) mp mn me Exp: e n p q  e q  e q0

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

2

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

The Gravitational force of alteration between any two particles shown above is made much negligible when compared to coloumbic force of alteration between electron and proton. Force of alteration F

e 2 F , accelleration q  ;q e  q p 2 4o r M

Due to this force, the electron as well as the proton will move towards each other, since me  m p , the speed and acceleration of the electron will be much greater than that of proton. This causes electron to collide with the neutron faster when compared to proton. 6.

Let z  t   x  t   y  t  , where “  ” denotes convolution. Let C be a positive real-valued constant. Choose the correct expression for z (ct). (A) c.x  ct   y  ct 

(B) x  ct   y  ct 

(C) c.x  t   y  ct 

(D) c.x  ct   y  t 

Key: (A) Exp: z  t   x  t  * y  t   z  s   x  s  .y  s  Converting into Laplace transform and applying time sealing property. 1 z  ct   z  s / c  c 1   s / c y s / c c 1 1  c  s / c y s / c  c c z  ct   c.x  ct  * y  ct 

7.

A 3-bus power system is shown in the figure below, where the diagonal elements of Y-bus matrix are Y11   j12pu, Y22   j15pu and Y33   j7pu Bus  1

Bus  2

jq

jr

jp

Bus  3

The per unit values of the line reactances p, q and r shown in the figure are (A) p  0.2, q  0.1, r  0.5 (B) p  0.2, q  0.1, r  0.5 (C) p  5, q  10, r  2

(D) p  5, q  10, r  2

Key: (B)

Exp: Y11  y10  y12  y13   j12    jq1  jr1 

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

3

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

 q1  r1  12 Y22   j15  y 20  y 21  y 23    jq1  jp1   p1  q1  15

Y33   j7  y30  y31  y32    jp1  jr1   p1  r1  7 solving P1  5,q1  10, r1  2  admit tan ces  P  0.2,q  0.1, r  0.5  reac tan ces 

The equivalent resistance between the terminals A and B is ______  .

8.

2

1

1

A 6 3

1

6

0.8

3

B

Key: 2.9 to 3.1 Exp: The Ckt will become 2 1

1

1 A

3

1 

3

6

6

1.2

B 0.8

0.8

R AB  1  1.2  0.8  3 The Boolean expression AB  AC BC simplifies to

9.

(A) BC  AC Key: (A)

(B) AB  AC  B

Exp: AB  AC  BC A 0 1

 BC  AC

10.

BC 00

01

11

(C) AB  AC

(D) AB  BC

10

1 1

1

BC

1

AC

The following measurements are obtained on a single phase load: V  220V  1%, I  5.0A  1% and W  555W  2%. If the power factor is calculated using these measurements, the worst case error in the calculated power factor in percent is ________.

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

4

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE| Key:

4 to 4

Exp:

P  VIcos  cos  

11.

GATE-2017-PAPER-I

www.gateforum.com

P 555  2% 555  2%    0.504  4% V.I  220  1%  5  1%  1100  2%

The transfer function of a system is given by,

Vo  s  Vi  s 



1 s Let the output of the system be 1 s

vo  t   Vm sin  t    for the input vi  t   Vm sin  t  . Then the minimum and maximum values of  (in radians) are respectively (A)

  and 2 2

(B)

 and 0 2

(C) 0 and

 2

(D)  and 0

Key: (D) Exp:

Vo  s  Vi  s 



1 s  H s 1 s

H    1  2Tan 1, If   0,   0 If   ,    Vo  t   Vm sin  t  2Tan 1   

12.

3 2  The matrix A   0 1  2

1 2  1 0  has three distinct eigenvalues and one of its eigenvectors is 3 0  2 Which one of the following can be another eigenvector of A? 0 (A)  0   1

0

 1 (B)  0   0 

1 (C)  0   1

1  0   1 

1 (D)  1  1 

Key: (C) 1 Exp: By the properties of Eigen values and Eigen vectors, another eigen vector of A is  0   1 The eigen vectors corresponding to distinct eigen values of a real symmetric matrix are orthogonal i.e., pair wise dot product is zero.

13.

For the power semiconductor devices IGBT, MOSFET, Diode and Thyristor, which one of the following statements is TRUE? (A) All of the four are majority carrier devices. (B) All the four are minority carrier devices (C) IGBT and MOSFET are majority carrier devices, whereas Diode and Thyristor are minority carrier devices. (D) MOSFET is majority carrier device, whereas IGBT, Diode, Thyristor are minority carrier devices.  ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India

© All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

5

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

Key: (D) Exp: MOSFET → Majority carrier device (NMOS, PMOS) Diode → both majority & minority carrier device Transister → Npn, pnp IGBT → input is MOSFET, Output is BJT 14.

Consider the unity feedback control system shown. The value of K that results in a phase margin of the system to be 30o is _______. Key: 1.01 to 1.06 Exp: PM  180  G gc

G s 

U s   

Ke  s s

Y s

Ke s s

For gc | G  s  |

K 1 

gc  K  G  s   

180  90o 

180o  90o  180o  K  60  K   1.047  3

30o  180o  K 

15.

A solid iron cylinder is placed in a region containing a uniform magnetic field such that the cylinder axis is parallel to the magnetic field direction. The magnetic field lines inside the cylinder will (A) bend closer to the cylinder axis (B) bend farther away from the axis (C) remain uniform as before (D) cease to exist inside the cylinder Key: (A) Exp: Flux always chooses less reluctance path. So flux tried to flow inside the conductor and closer to the axis of the cylinder. 16.

Let I  c   R xy2dxdy, where R is the region shown in the figure and c  6 104. The value of I equals________. y 10

R

2

1

5

x

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

6

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

Key: 0.99 to 1.01 Exp:

 xy dxdy    xy dxdy    xy dxdy 2

2

R

2

R1

R2

5



2

 

x 1 y  0

5

2

2x

 x 2   y3  5  y3  2 xy dx        x   dx   2 1  3 0 1  3  2 y 2

5

xy 2 dxdy 

2x



x 1

5 5 5  x2   1   x5  8 1 3  12      x  8x  8  dx  32  8    8    3   5 1 3 3 1  2 1  

1  24992  24992  32     32  3 5  15 2  C xy 2 dxdy   24992   104  0.99968  1 5 R OR  2x 2  R xy dxdy  x1  y0 xy dy dx   5

2

2x

5 5  y3  8   x   dx   x  x 3  dx 3  31 1  5

8  x5  8 24992      3124   3  5 1 15 15 24992 2  C  xy 2 dxdy   104  6   2.4992  0.9968  1 15 5 R

17.





Consider the system with following input-output relation y  n   1   1 x  n  n

where, x[n] is the input and y[n] is the output. The system is (A) invertible and time invariant (B) invertible and time varying (C) non-invertible and time invariant (D) non-invertible and time varying Key: (D)





Exp: Given y  n   1   1 x  n  n

For time invariance





y' n   1   1 x  n  n o   (1) n



y  n  n o   1   1

n no

 x  n  n   (2) o

Since (1) is not equal to (2) System is time variant For inverse system For each unique x  n  , there should be unique y  n  If x  n     n  1 n y  n   1   1    n  1  

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

7

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

 y 1  0 if x  n   2  n  1 y  n   1   1  2  n  1   y(1)=0 For two different inputs we have same output. Thus one to one mapping is not possible. Hence the systems is non invertible n

18.

The slope and level detector circuit in a CRO has a delay of 100 ns. The start-stop sweep generator has a response time of 50 ns. In order to display correctly, a delay line of (A) 150 ns has to be inserted into the y-channel (B) 150 ns has to be inserted into the x-channel (C) 150 ns has to be inserted into both x and y channels (D) 100 ns has to be inserted into both x and y channels

Key: (A) Exp: The delay line should be inverted in VDP (Y-channel) only. 19.

A 3-phase voltage source inverter is supplied from a 600V DC source as shown in the figure below. For a star connected resistive load of 20  per phase, the load power for 120o device conduction, in kW is __________.

20

600V



20



20

Key: 8.5 to 9.5 Exp: Vdc  600V R L  20 / Ph 120o mod e

RMS value of phase voltage  VP   0.4082Vdc  244.92V Load power 

20.

3Vph 2 3  244.922   8.99kW  9kW R 20

3 2 A closed loop system has the characteristic equation given by s  Ks   K  2  s  3  0. For

this system to be stable, which one of the following conditions should be satisfied? (A) 0 < K < 0.5 (B) 0.5 < K < 1 (C) 0 < K < 1 (D) K > 1 Key: (D) 3 2 Exp: Given CE  s  ks   k  2  s  3  0

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

8

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

For stable

k  k  2  3 k 2  2k  3  0

 k  1 k  3  0 k  1  k  3 k 1 (OR) By R-H criteria s3

1

k2

2

k

3

s

k  k  2  3

s

0

k

s0

3

k  0   k  3 k  1  0 K  0  k  1  k  3  k  1

21.

A 4 pole induction machine is working as an induction generator. The generator supply frequency is 60 Hz. The rotor current frequency is 5 Hz. The mechanical speed of the rotor in RPM is (A) 1350 (B) 1650 (C) 1950 (D) 2250

Key: (C) Exp:

Ns 

120  60  1800 rpm 4

Rotor speed should be greater than syn.speeed, to ge inductance generator mode. N  NS S r NS fr 5 1  f r  sf  s  f  60  12   

 N r  NS 1  S 1  N r  1800 1    1950 rpm  12 

22.

For the circuit shown in the figure below, assume that diodes D1, D2 and D3 are ideal. D1 R

 v1  v  t    sin 100t  V

D2

D3

R

 v2 

The DC components of voltages v1 and v2, respectively are (A) 0 V and 1 V (B) -0.5 V and 0.5 V (C) 1 V and 0.5 V

(D) 1 V and 1 V

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

9

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

Key: (B) Exp: For half wave Rectifier Vdc 

Vm 

V1  Vdc for  ve pulse  Vdc for  ve pulse  V2 

    1      1  0.5V 2    2

  0  0.5V 2

23.

A 10-bus power system consists of four generator buses indexed as G1, G2, G3, G4 and six load buses indexed as L1, L2, L3, L4, L5, L6. The generator bus G1 is considered as slack bus, and the load buses L3 and L4 are voltage controlled buses. The generator at bus G2 cannot supply the required reactive power demand, and hence it is operating at its maximum reactive power limit. The number of non-linear equations required for solving the load flow problem using Newton-Raphson method in polar form is _______. Key: 14 to 14 Exp: Total no of buses=10 Given G1=slack bus, G2=generator/PQ bus  G 3 ,G 4 are PV buses PQ buses  L1 , L 2 , L5 , L 6 (4) Voltagecontrolled PV buses  L3 , L 4 (2) Minimum no of nonlinear equations to be solved  2  10  2  4  14 R1

I

24.

The power supplied by the 25 V source in the figure shown below is ________W. Key: 248 to 252 Exp: KCL 25V I  0.4I  14

 17V  



 R2

14A

0.4I



 1.4I  14  I  10A The power supplied by 25 V = 25  10  250W 25.

In the converter circuit shown below, the switches are controlled such that the load voltage vo(t) is a 400 Hz square wave.

S3

S1 220V



LOAD



S4

 v t  0

S2

The RMS value of the fundamental component of vo(t) in volts is _______. Key : 196 to 200

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

10

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE| Exp: Vo  t  

4Vs 

GATE-2017-PAPER-I

www.gateforum.com

 4VS 1 1   sin  t  sin 3  t  sin 5  t   sin  nt     3 5   n 1,3,5 n

4 VS 2 n 4 VS Vol    0.9VS 2   0.9  220  198.069V

Vo  t 

Von 

Vs  220V



2

t

Vs  220V Q. No. 26 – 55 Carry Two Marks Each

26.

The output expression for the Karnaugh map shown below is CD AB

(A) BD  BCD Key: (D) CD Exp: 00 AB

00

01

11

10

00

0

0

0

0

01

1

0

0

1

11

1

0

1

1

10

0

0

0

0

(B) BD  AB 01

11

10

00

0

0

0

0

01

1

0

0

1

11

1

0

1

1

10

0

0

0

0

(C) BD  ABC

(D) BD  ABC

BD ABC

27.

A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4  and 0.1  respectively. The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.

Key: 9.5 to 12 Exp:

E b1  220  30  0.5   205 volts E b2  220  I a 2  0.5  R X 

Given T  N2 and   IR We know that, in series motor  T  Ia2  ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

11

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

T  Ia2  N 2

www.gateforum.com

Ia2

30A

Ia 2

N2   Ia 2  0.5. Ia1  15 Amp N1 Ia1 Eb  N 2 E b2 1   N1 E b1 2

N

0.1

0.1

0.4

220V

Rx 220V 0.4

E b1

0.5N1 220  15  0.5  R x  30   N1 205 15

E b2

 R x  10.75 

28.

The transfer function of the system Y(s)/U(s) whose state-space equations are given below is:  x 1  t   1 2   x1  t   1        ut   x 2  t   2 0  x 2  t  2 x t y  t   10  1   x 2  t 

(A)

s

s  2 2

 2s  2 

(B)

s

s  2 2

 s  4

(C)

s

s  4 2

 s  4

(D)

s

s  4 2

 s  4

Key: (D) Exp: Given  x 1  t   1 2   x1  t   1        u(t)   x 2  t   2 0   x 2  t  2 x  t   Ax  Bu

Transfer function = CSI  A B  D 1

Here D = 0 C  1 0 1 2  1  A B   2 0 2 1

s  1 2  1  T / F  1 0  s   2   2 2  1  s  2 s  1  2     1 0  2 s s4  s4   2  2s  2    1 0  2 s s4 s4 T/F 2 s s4

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

12

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE| 29.

GATE-2017-PAPER-I

www.gateforum.com

The load shown in the figure is supplied by a 400 V (line to line) 3-phase source (RYB sequence). The load is balanced and inductive, drawing 3464 VA. When the switch S is in position N, the three watt-meters W1, W2 and W3 read 577.35 W each. If the switch is moved to position Y, the readings of the watt-meters in watts will be:

(A) W1  1732and W2  W3  0

(B) W1  0, W2  1732and W3  0

(C) W1  866, W2  0, W3  866

(D) W1  W2  0and W3  1732

Key: (D) Exp: R 3 Y Supply

B

W1 3 load

W2

W3 Y S

N

N

If the switch is connected to Neutral, then each wattmeter will read 1  power. W1  W2  W3  3.Vph .I ph cos   1732.05

VR

 cos   0.51agg.    60

V4

Given that, load drawing Apparent power of 3464 VA.

3VL I L  3464

60

IR

30

VB4

3464 IL   5A 3  400

V4

VB

If the switch connected to “Y”, then W2=0

I4

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

13

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

W1  Vpc Icc cos  Vpc & Icc   VRy I R cos  VRy & I R   400  5cos  90   0W

W3  VBy IB cos  VBy & IB   400  5  cos30  1732watts 30.

Two passive two-port networks are connected in cascade as shown in figure. A voltage source is connected at port 1.

Given V1  A1V2  B1I 2 I1  C1V2  D1I 2 V2  A 2 V3  B2 I3 I 2  C 2 V3  D 2 I3 A1 ,B1 ,C1 ,D1 ,A2 ,B2 ,C2 and D2 are the generalized circuit constants. If the Thevenin equivalent

circuit at port 3 consists of a voltage source VT and impedance ZT connected in series, then V A B  B1D2 V1 A B  B1D2 (A) VT  1 , ZT  1 2 (B) VT  , ZT  1 2 A1A2 A1A2  B1C2 A1A2  B1C2 A1A2 (C) VT 

V1 A B  B1D2 , ZT  1 2 A1  A2 A1  A2

(D) VT 

V1 A B  B1D2 , ZT  1 2 A1A2  B1C2 A1A2  B1C2

Key: (D) Exp: We can write V1 ,I1 in terms of V3  I3

 V1   A1 B1   A 2 B2   V3   I   C D  C D   I  1  2 2 3   1  1  V1   A1A 2  B1C2  V3   A1B2  B1D 2  I3 I1   C1A 2  D1C 2  V3   C1B2  D1D 2  I3

To find Vth  or  Voc I3  0

V1   A1A 2  B1C2  Vth  Vth  Voc 

V1 A1A 2  B1C 2

To find ISC V3  0

V1   A1B2  B1D2  ISC  ISC  To find R th R th 

V1 A1B2  B1D2

VOC A1B2  B1D2  ISC A1A2  B1C2

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

14

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE| 31.

GATE-2017-PAPER-I

www.gateforum.com

The circuit shown in the figure uses matched transistors with a thermal voltage VT  25mV. The base currents of the transistors are negligible. The value of the resistance R in k  that is required to provide 1 A bias current for the differential amplifier block shown is ______.

Key: 170 to 174 Exp: R 

VT  IC1  ln  ; IC2  IC2 

IC1  1mA;IC2  1A R 32.

25  103 1 103  ln   172.7k 6  1 106 1 10 

The figure below shows an uncontrolled diode bridge rectifier supplied form a 220 V, 50 Hz 1-phase ac source. The load draws a constant current Io  14A. The conduction angle of the diode D1 in degrees is___________.

Key: 220 to 230 Exp: Average reduction in output voltage due to Ls

Vo  4f s Ls Io  4  50  10  103   14  28V

Vm cos   cos        for a diode,   0 V Vo  m 1  cos    220 2 28  1  cos     44.17o   conduction angle of diode  180    224.17 o Vo 

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

15

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

t

www.gateforum.com

 81

dy  5ty  sin  t  with y 1  2 . There exists a dt unique solution for this differential equation when t belongs to the interval (A) (-2,2) (B) (-10,10) (C) (-10,2) (D) (0,10) Key: (A)

33.

Consider the differential equation

sin  t  dy 5t  2 y 2 is a first order linear eq. dt t  81 t  81

Exp: D.E is

5t

I.F = e

2

 t 2 81dt

5

 e2





ln t 2 81

  t 2  81

5/2

 Solution is y  t 2  81

t y

2



5/ 2

 81 .sin tdt

5/ 2 3/ 2 sin t 2 t  81    t 2  81 sin tdt  c  t  81 2

3/2

t

2

 81

5/2



C

t

2

 81

5/2

If t  9,9 then the solution exists. Options (b), (c), (d) contain either -9 or 9 or both. So answer is option A 34.

A separately excited DC generator supplies 150 A to a 145 V DC grid. The generator is running at 800 RPM. The armature resistance of the generator is 0.1  . If the speed of the generator is increased to 1000 RPM, the current in amperes supplied by the generator to the DC grid is _______.

Key:

548 to 552



Exp:

145V





150A

Grid

Ra

0.1

E g1

0.1

E g2

800 rpm

1000 rpm E g2  Ia 2  0.1  145

E g1  150  0.1  145 E g1  160V N  E g    constant  N 2 E g2  N1 E g1 E g2 



150A Grid

Ra

145V

200  145  Ia 2  0.1 

55  Ia 2 0.1

Ia 2  550 amps

1000  160  200 volts 800

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

16

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE| 35.

GATE-2017-PAPER-I

In the circuit shown below, the maximum power transferred to the resistor R is _______ W.

3

Key: 3 to 3.1 Exp: To find Vth

5

5  6  10 21   2.1A 10 10 Vth  5  5  2.1  5.5V

5 6V

5V



Vth



To find R th



I



10V



55  2.5 55 The maximum power transferred to  R th 

R





I

36.

www.gateforum.com

Vth2 5.52   3.025W 4R th 4  2.5

5

 5

R th



Let a causal LTI system be characterized by the following differential equation, with initial rest condition dx  t  d2 y dy  7  10y  t   4x  t   5 2 dt dt dt Where x(t) and y(t) are the input and output respectively. The impulse response of the system is (u(t) is the unit step function)

(A) 2e 2t u(t)  7e 5t u  t 

(B) 2e2t u  t   7e 5t u  t 

(C) 7e2t u  t   2e 5t u  t 

(D) 7e2t u  t   2e5t u  t 

Key: (B) Exp: Given causal LTI system d2 y  t 

7dy  t 

5dx(t) dt dt dt 2  s y  s   7sy  s   10y  s   u x  s   5sx(s) 2





Y s X s



 10y  t   ux  t  

4  5s 5s  4  H s  s  7s  10  s  2  s  5  2

Inverse Laplace transform will give h  t  (impulse response).

2 7  s2 s5 h  t   2e2t u  t   7e 5t u  t  H s 

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

17

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE| 37.

GATE-2017-PAPER-I

www.gateforum.com

The approximate transfer characteristic for the circuit shown below with an ideal operational amplifier and diode will be

Key: (A) 38.

The switch in the figure below was closed for a long time. It is opened at t = 0. The current in the inductor of 2 H for t  0, is

(A) 2.5e4t Key: (A)

(B) 5e4t

Exp: at t  0





(D) 5e0.25t

8

I 6 50V

(C) 2.5e0.25t

8

IL  O 

50  5A 64 5 IL  0    2.5A 2 I

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

18

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I 8

For t  0

8 32

T

www.gateforum.com

32

2.5A

2H

L 2 1   Req 8 4

IL     0  bc3 it is a sourcefreeckt  i L  t   I L      I L  0   I L    e  t /T  2.5e 4t

39.

j20 j20    j39.9   j39.9 j20  pu The bus admittance matrix for a power system network is  j20  j20 j20  j39.9  There is a transmission line, connected between buses 1 and 3, which is represented by the circuit shown in figure.

If this transmission line is removed from service, What is the modified bus admittance matrix? j20 0    j19.9   j39.9 j20  pu (A)  j20  0 j20  j19.9 

j20 0    j39.95   j39.9 j20  pu (B)  j20  0 j20  j39.95

j20 0    j19.95   j39.9 j20  pu (C)  j20  0 j20  j19.95

j20 j20    j19.95   j39.9 j20  pu (D)  j20  j20 j20  j19.95 

Key: (C) Exp: When the line 1-3 is removed z13  0.05  z31 y13 

1   j20, 0.05

y13  y31  0

' y13  Half line shunt susceptance = j0.05 2 y' y11  new  = y11  old   y13  13   j39.9    j20   j0.05   j19.95 2 y' y33  new  = y33  old   y13  13   j39.9    j20   j0.05   j19.95 2 j20 0    j19.95  Modified Bus admittance matrix: yBus new   j20  j39.9 j20    0 j20  j19.95 pu

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

19

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE| 40.

GATE-2017-PAPER-I

www.gateforum.com

In the system whose signal flow graph is shown in the figure, U1(s) and U2 (s) are inputs. The Y(s) transfer function is U1 (s)

(A) (C)

k1 JLs  JRs  k1k 2

(B)

2

k1  U 2  R  sL 

JLs   JR  U 2 L  s  k1k 2  U 2 R 2

(D)

k1 JLs  JRs  k1k 2 2

k1  U 2  sL  R 

JLs   JR  U 2 L  s  k1k 2  U 2 R 2

Key: (A) Exp:

Y s

U1  s  U

2

s 0

By Masons gain formula Y s

U1  s 



P11 1   L1  L 2 

Here P1 

1 k1 . k 2 LJ

1  1 R1 Ls 1 1 L 2  . 2  k 2  k1 LJ s 1 k1 . 2 Y s s LJ  R 1 1 U1  s  1  .  . 1 k k 2 1 L s LJ s 2 k1 T/F 2 s LJ  sRJ  k1k 2 L1  

s  1 , a unit step input is applied at time t=0. s 1 The value of the response of the system at t=1.5 sec is __________. Key: 0.550 to 0.556

For a system having transfer function G  s  

41.

Exp:

Y s R s 



s  1 s 1

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

20

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

1 s 1 . s 1 s 1 2 Y s   s s 1 Apply Inverse L.T Y s 

y  t   u  t   2e  t u  t  y 1.5   1  2e1.5  1  0.44626 y 1.5   0.5537

42.

The magnitude of magnetic flux density (B) in micro Teslas  T  at the center of a loop of wire wound as a regular hexagon of side length 1m carrying a current (I=1A), and placed in vacuum as shown in the figure is __________.

Key: 0.65 to 0.75 Exp: For a finite length conductor B at a point P I B  o  cos 1  cos  2  1 4r

2

P

2

For a given hexagon 4 for a side 3 2 1   2  60o r

60o o

60

3 2 60o

I o I  cos 1  cos  2  4r 4  107  1  6 cos 60o  cos 60o   4  3 / 2  6.9  107

Total flux density B  6 

 0.69  107 Tesla

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

21

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE| 43.

GATE-2017-PAPER-I

www.gateforum.com

The figure shows the single line diagram of a power system with a double circuit transmission line. The expression for electrical power is 1.5 sin  , where  is the rotor angle. The system is operating at the stable equilibrium point with mechanical power equal to 1 pu. If one of the transmission line circuits is removed, the maximum value of  as the rotor swings, is 1.221 radian. If the expression for electrical power with one transmission line circuit removed is Pmax sin , the valueof Pmax, in pu is _________.

Key: 1.21 to 1.23 Exp: Given m  1.22rad  69.958

PC

P  1  sin 1  m   1.5   1   sin 1    41.81  0.729 rad  1.5  Using equal area criterion A1=A2 2

m

  Pm0  Pm1 sin  d  

1

P

max1

2

sin   Pm0  d

1.5sin 

1.5

Pm sin 

Pm =1 Pm0 =1

1 2  m



By solving above integration Pmax1 

44.

Pm0  m  1  cos 1  cos m



11.221  0.7297  1.22pu cos  41.81  cos  69.95 

A 375W, 230 V, 50 Hz capacitor start single-phase induction motor has the following constants for the main and auxiliary windings (at starting): Zm  12.50  j15.75   (main winding),

Za   24.50  j12.75  (auxiliary winding). Neglecting the magnetizing branch the value of the capacitance (in F ) to be added in series with the auxiliary winding to obtain maximum torque at starting is _______. Key: 95 to 100 Exp:

 x  xe  X  tan 1  m   tan 1  a   90  Rm   Ra   15.75  1  12.75  X c  tan 1    tan    90  12.5   24.5   12.75  X c   12.75  X c  51.562  tan 1   90   tan 1     38.43  24.5   24.5  12.75  X c  0.793  X c  32.194 24.5 1 Xc   98.87F 2  50  32.194

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

22

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

x  x 1 e , A function f(x) is defined as f (x)   where x   . Which one of the 2 1nx  ax  bx, x  1   following statements is TRUE? (A) f(x) is NOT differentiable at x=1 for any values of a and b. (B) f(x) is differentiable at x = 1 for the unique values of a and b (C) f(x) is differentiable at x = 1 for all values of a and b such that a + b = e (D) f(x) is differentiable at x = 1 for all values of a and b. Key: (A) Not matching with IIT key

45.

Exp: Lf 1 1  Lt

f  x   f 1

x 1 x 1

x 1

 Lt

x 1

ex   a  b  x 1

does not exists, for any values of a and b

 f  x  is not differentiable at x  1 , for any values of a and b.

46.

Consider a causal and stable LTI system with rational transfer function H(z). Whose 5 corresponding impulse response begins at n = 0. Furthermore, H 1  . The poles of H(z) are 4

Pk 

  2k  1   1 n exp  j  for k = 1,2,3,4. The zeros of H(z) are all at z = 0. Let g[n] = j h[n]. 4 2  

The value of g[8] equals ___________. Key: 0.06 to 0.065 Only one of the real roots of f  x   x 6  x  1 lies in the interval 1  x  2 and bisection method

47.

is used to find its value. For achieving an accuracy of 0.001, the required minimum number of iterations is ________. Key: 10 to 10 ba Exp: a  1, b  2and n  0.001 using bisection method 2

 2n  1000  n  10 is the minimum number of iterations 48.

Two parallel connected, three-phase, 50Hz, 11kV, star-connected synchronous machines A and B, are operating as synchronous condensers. They together supply 50 MVAR to a 11 kV grid. Current supplied by both the machines are equal. Synchronous reactances of machine A and machine B are 1 and 3 respectively. Assuming the magnetic circuit to be linear, the ratio of excitation current of machine A to that of machine B is ________. 11kV

Key: 2.05 to 2.13 Exp:

 syn. Condencors  current‟s supplied both the machines

50 MVAR

1

are same 2624.31  1312.159 Amps 2 As the two motors, supplying reactive power only, the phasor diagaram will be

3

 I1  I 2 

I1

I2

~

~

A

B

IL 

50 106  2624.31 3 11103

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

23

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

E  jIa Xs  Vt

I a Xs

E  V  jIa Xs

Consider magnitudes  E2   V  Ia Xs  E

 V  Ia XS 

E

Vt 2



2

90o

EA 

 6350.85  1312.159 1

EB 

 6350.85  1312.159  3

2

2

 5038.7 volts

Ia

 2414.14Volts

IfA E A 5038.7    2.086 IfB E B 2414.14 49.

The positive, negative and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu, and 0.1 pu, respectively. The generator is on open circuit with a terminal voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs at the terminals is _______ (assume fault impedance to be zero). Key: 0.1 to 0.1 3Ef Exp: If  Z0  Z1  Z2  3Zn 3 1 0.1  0.2  0.2  3Zn Zn  0.1P.U

3.75 

50.

Let the signal x  t  



  1

k 

k

k    t   be passed through an LTI system with frequency  2000 

response H   , as given in the figure below

The Fourier series representation of the output is given as (A) 4000+4000cos  2000t   4000cos  4000t  (B) 2000  2000cos  2000t   2000cos  4000t  (C) 4000cos  2000t  (D) 2000cos  2000t  Key: (C) Exp: Given x  t  is a periodic signal for which Fourier transform x   is to be calculated

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

24

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com



x    2  Dn     n n 

D n is exponential Fourier series coefficient for x(t) xt

3 2000

1 2000

3 2000

1 2000

2 2000

t

2 2000 1

1   x t   t    t    2000 

Define x  t  over one period

1 sec; o  2000 rad/sec 1000 1 1  D n  1  e  jno t o ; t o  To 2000 n D n  1000 1  e  jn   D n  1   1 1000

Where as To 

At n  0,2,4,...........Dn =0

for even values of n  0   i.e., Dn    2000 for odd values of n   x    2  D0  D1      2000   D 1    2000   D 2    4000   D 2    4000   ..................] x   D3

D1

D1

t

6000 4000 2000

Given x   is

D3

2000 4000 6000

x  

500

500



Thus the filtered output is

y    2  D1    2000    D1    2000   D1  D1  2000 y    4000        2000       2000    y  t   4000cos  2000t   ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

25

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE| 51.

GATE-2017-PAPER-I

www.gateforum.com

The logical gate implemented using the circuit shown below where. V 1 and V2 are inputs (with 0 V as digital 0 and 5 V as digital 1) and VOUT is the output is

(A) NOT Key: (B) Exp: V1 V2 0 0 0 1 1 0 1 1

(B) NOR

Q1 OFF OFF ON ON

Q2 OFF ON OFF ON

Vout 5V 0V 0V 0V

(C) NAND

(D) XOR

Logic Level 1 0 0 0

So, this logic level o/p is showing the functionality of NOR-gate. 52.

A load is supplied by a 230 V, 50 Hz source. The active power P and the reactive power Q consumed by the load are such that 1 kW  P  2kW and 1kVAR  Q  kVAR . A capacitor connected across the load for power factor correction generates 1 kVAR reactive power. The worst case power factor after power factor correction is (A) 0.447 lag (B) 0.707 lag (C) 0.894 lag (D) 1 Key: (B) P Exp: Under worst case, 2 Pmax  2kW 1 Q max  2kVAR Q 1  tan 1  45o P cos 45  0.707lag 53.

The input voltage VDC of the buck-boost converter shown below varies from 32 V to 72 V. Assume that all components are ideal, inductor current is continuous, and output voltage is ripple free. The range of duty ratio D of the converter for which the magnitude of the steady state output voltage remains constant at 48 V is 2 3 2 3 (A)  D  (B)  D  (C) 0  D  1 5 5 3 4 Key: (A)

(D)

1 2 D 3 3

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

26

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

Exp: Vdc  32V

Vdc  72V

Vo  48V

Vo  48V

Vo D  Vdc 1  D

Vo D  Vdc 1  D

48 D  32 1  D 3 D  2 1 D 3  3D  2D

2 D  3 1 D 3D  2  2D

3  5D  D 

5D  2 2 D 5

3 5

2 3 D 5 5

A three-phase, three winding  /  / Y (1.1kV/6.6kV/400 V) transformer is energized from AC mains at the 1.1 kV side. It supplies 900 kVA load at 0.8 power factor lag from the 6.6 kV winding and 300 kVA load at 0.6 power factor lag from the 400 V winding. The RMS line current in ampere drawn by the 1.1 kV winding from the mains is _______.

54.

Key:

623 to 627

Exp:

3  , 3  winding T F

//Y per phase representation



 6.6 kV

I1

1.1 kV

3

~

I2

3

900  103  78.73 Amps 3  6.6  103 I Iph  2  45.45 36.87o 3

I3

I2 

I2  KI 2 

900 kVA 0.8



300 kVA 0.6

400 3

6.6  103  45.45  I2  272.7 36  .87 o 1.1  103

300  103  433.01 Amp 3  400 I ph  I L  I3  433.01 53.13 I3 

I3 

400 3  433.01  I3  90.91 53.13o 3 1.1  10

I1  I2  I3  I1  360.87 40.91  I1  3 I1  625.05 40.91

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

27

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE| 55.

GATE-2017-PAPER-I

www.gateforum.com

Consider the line integral I   C  x 2  iy2  dz where z = x + iy. The line C is shown in the figure below.

The value of I is 1 2 (A) i (B) i 2 3 Key: (B) Exp: curve „C‟ is y  x  dy  dx



I   x2  i  x  1

0

2



 dx  idx   1  i 

(C)

3 i 4

(D)

4 i 5

1

2

 x3  2 2 x dx  2i     i 0  3 0 3 1

General Aptitude Q. No. 1 – 5 Carry One Mark Each 1.

Research in the workplace reveals that people work for many reasons_________. (A) money beside (B) beside money (C) money besides (D) besides money Key: (D) 2.

The probability that a k-digit number does NOT contain the digits 0.5, or 9 is (A) 0.3k (B) 0.6k (C) 0.7k (D) 0.9k Key: (C)

k digits Each digit can be filled in 7 ways as 0, 5 and 9 is not allowed, so each of these places can be filled by 1, 2, 3, 4, 6, 7, 8. k

 7 So required probability is   or 0.7 k.  10 

Find the smallest number y such that y  162 is a perfect cube. (A) 24 (B) 27 (C) 32 Key: (D) 3.

(D) 36

Exp: Factorization of 162 is 2  3 3 3 3

y 162 is a perfect cube y  2  3  3  3  3  Perfect cube For perfect cube 2's & 3's are two more required each.

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

28

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

4.

After Rajendra Chola returned from his voyage to Indoneisa, he _______ to visit the temple in Thanjavur. (A) was wishing (B) is wishing (C) wished (D) had wished Key: (C) 5.

Rahul, Murali, Srinivas and Arul are seated around a square table. Rahul is sitting to the left of Murali. Srinivas is sitting to the right of Arul. Which of the following pairs are seated opposite each other? (A) Rahul and Murali (B) Srinivas and Anil (C) Srinivas and Murali (D) Srinivas and Rahul Key: (C) Exp:

Srinivas Rahul

Arul Murali

Q. No. 6 – 10 Carry Two Marks Each 6.

Six people are seated around a circular table. There are at least two men and two women. There are at least three right-handed persons. Every woman has a left-handed person to her immediate right. None of the women are right-handed. The number of women at the table is (A) 2 (B) 3 (C) 4 (D) Cannot be determined Key: (A) Exp: Out of six people, 3 place definitely occupied by right handed people as atleast 2 women are there so these two will sit adjacently. Now as only one seat is left it will be occupied by a left handed man because on right side of this seat is sitting an right handed man.

R  m R  m

Lw

Lw

R  m

?

Therefore, answer should be 2 women.

7.

The expression

 x  y  | x  y |

2 (A) the maximum of x and y (C) 1 Key: (B)

is equal to (B) the minimum of x and y (D) none of the above

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

29

Downloaded From : www.EasyEngineering.net

Downloaded From : www.EasyEngineering.net

|EE|

GATE-2017-PAPER-I

www.gateforum.com

Exp: If x  y Exp 

x  y   x  y 2

 y min

If x  y Exp 

x  y  y  x 2

 The expression

8.

 x min

 x  y  x  y 2

is equal to min imum of x & y

A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25m intervals in this plot. If in a flood, the water level rises to 525m. Which of the villages P,Q,R,S,T get submerged?

(A) P, Q (C) R,S,T Key: (C)

(B) P,Q,T (D) Q,R,S

Exp: The given contour is a hill station, the peak point of this hill station is P, it is under a contour of 550. At floods, the water level is 525m. So the village of R, S and T are under a contour of 500. Therefore these villages are submerged.

9.

Arun, Gulab, Neel and Shweta must choose one shirt each from a pile of four shirts coloured red, pink, blue and white respectively. Arun dislikes the colour red and Shweta dislikes the colour white, Gulab and Neel like all the colours. In how many different ways can they choose the shirts so that no one has a shirt with a colour he or she dislikes? (A) 21 (B) 18 (C) 16 (D) 14 Key: (D) Exp: As there are 4 people A,G,N,S and 4 colours so without any restriction total ways have to be

4  4  16

Now, Arun  dislikes Red and Shweta  dislikes white So 16-2=14 ways “The hold of the nationalist imagination on our colonial past is such that anything inadequately or improperly nationalist is just not history.” Which of the following statements best reflects the author‟s opinion? (A) Nationalists are highly imaginative. (B) History is viewed through the filter of nationalism. (C) Our colonial past never happened (D) Nationalism has to be both adequately and properly imagined. Key: (B) 10.

 ICP–Intensive Classroom Program  eGATE-Live Internet Based Classes DLP  TarGATE-All India Test Series Leaders in GATE Preparation  65+ Centers across India © All rights reserved by Gateforum Educational Services Pvt. Ltd. No part of this booklet may be reproduced or utilized in any form without the written permission.

30

Downloaded From : www.EasyEngineering.net