_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India1Q. No. 1 – 5 Carry One Mark Each1. Choose the most appropriate phrase from the options given below to complete the followingsentence.India is a post-colonial country because(A) it was a former British colony(B) Indian Information Technology professionals have colonized the world(C) India does not follow any colonial practices(D) India has helped other countries gain freedomAnswer: (A)2. Who ___________ was coming to see us this evening?(A) you said (B) did you say (C) did you say that (D) had you saidAnswer: (B)3. Match the columns.Column 1 Column 2(1) eradicate (P) misrepresent(2) distort (Q) soak completely(3) saturate (R) use(4) utilize (S) destroy utterly(A) 1:S, 2:P, 3:Q, 4:R (B) 1:P, 2:Q, 3:R, 4:S(C) 1:Q, 2:R, 3:S, 4:P (D) 1:S, 2:P, 3:R, 4:QAnswer: (A)4. What is the average of all multiples of 10 from 2 to 198?(A) 90 (B) 100 (C) 110 (D) 120Answer: (B)Exp:10 190 20020 1809: (200) 9 100 1900100: 19 1990 110100__ −___ _ ___ −___5. The value of 12 12 12 .... is(A) 3.464 (B) 3.932 (C) 4.000 (D) 4.444_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India2

Answer: (C)Exp: let 12 12 12 .... y212 y y12 y y(y 4)(y 3) 0y 4, y 3_ _ _ −_ −Q.No. 6 – 10 Carry Two Marks Each6. The old city of Koenigsberg, which had a German majority population before World War 2,is now called Kaliningrad. After the events of the war, Kaliningrad is now a Russian territoryand has a predominantly Russian population. It is bordered by the Baltic Sea on the north andthe countries of Poland to the south and west and Lithuania to the east respectively. Which ofthe statements below can be inferred from this passage?(A) Kaliningrad was historically Russian in its ethnic make up(B) Kaliningrad is a part of Russia despite it not being contiguous with the rest of Russia(C) Koenigsberg was renamed Kaliningrad, as that was its original Russian name(D) Poland and Lithuania are on the route from Kaliningrad to the rest of RussiaAnswer: (B)7. The number of people diagnosed with dengue fever (contracted from the bite of a mosquito)in north India is twice the number diagnosed last year. Municipal authorities have concludedthat measures to control the mosquito population have failed in this region.Which one of the following statements, if true, does not contradict this conclusion?(A) A high proportion of the affected population has returned from neighbouring countrieswhere dengue is prevalent(B) More cases of dengue are now reported because of an increase in the Municipal Office’sadministrative efficiency(C) Many more cases of dengue are being diagnosed this year since the introduction of a newand effective diagnostic test(D) The number of people with malarial fever (also contracted from mosquito bites) hasincreased this yearAnswer: (D)8. If x is real and 2 x −2x 3 11, then possible values of 3 2 −x x −x include(A) 2, 4 (B) 2, 14 (C) 4, 52 (D) 14, 52_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India

3Answer: (D)Exp: x2 −2x 3 11_(x−4)(x2) 0_x 4,x −23 2 Values of x x xFor x 4Value 52for x 2Value 14Option D 14,52−−−9. The ratio of male to female students in a college for five years is plotted in the following linegraph. If the number of female students doubled in 2009, by what percent did the number ofmale students increase in 2009?Answer: 140%Exp:m3fm2.5fm=2.5fm'32fm' 6fm' mm3.5f% 1002.5f71.48% 140%−3.53

2.521.510.502008 2009 2010 2011 2012Ratioofmaletofemal_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India410. At what time between 6 a.m. and 7 a.m will the minute hand and hour hand of a clock makean angle closest to 60°?(A) 6: 22 a. m. (B) 6:27 a.m. (C) 6: 38 a.m. (D) 6:45 a.m.Answer: (A)Exp: Angle by minute’s hand60 min360 3601min 660o 8min48o Angle48 with number ‘6’Angle by hours hando 60 min 303022 min 226011Total Angle=48+11=59o.6.22am_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India5Q.No. 1 – 25 Carry One Mark Each1. Which one of the following statements is true for all real symmetric matrices?(A) All the eigenvalues are real. (B) All the eigenvalues are positive(C) All the eigenvalues are distinct (D) Sum of all the eigenvalues is zero.Answer: (A)

Exp: Eigen values of a real symmetric matrix are all real2. Consider a dice with the property that the probability of a face with n dots showing up isproportional to n. The probability of the face with three dots showing up is_________.Answer: 1/7Exp: Pnk.n where n 1 to 6xwe know P x 1 K 1 2 3 4 5 6 11K211required probability is P 3 3K7_ _

_3. Maximum of the real valued function 2f x x −1 3 occurs at x equal toA−B0 C1 DAnswer: (C)Exp: 1132 is negative, x 1 or x in 1 h,1f x3 x 1 is positive, x 1 or x in 1,1 h−−h is positive&smallf has local minima at x 1 and the minimum value is '0'4. All the values of the multi-valued complex function 1i, where i −1 , are(A) purely imaginary (B) real and non-negative(C) on the unit circle. (D) equal in real and imaginary partsAnswer: (B)Exp: 1cos2ki sin2kwhere k is int egeri 2ki 2ke1 eAll values are real and non negative−

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India65. Consider the differential equation222d y dyx x y 0dx dx−. Which of the following is a solutionto this differential equation for x > 0?(A) x e (B) 2 x (C) 1/x (D) ln xAnswer: (C)Exp:2222 z2Z Z 11 2 2d y dyx x y 0 is cauchy Euler equationdx dxd1 .y 0 where and z log x, x edzA.E :m 1 0 m 1,1CSolution is y C e C e C xx1is a solutionx−−−_ −−_ −6. Two identical coupled inductors are connected in series. The measured inductances for thetwo possible series connections are 380H and 240H . Their mutual inductance in H is________Answer: 35HExp: Two possible series connections are1. Aiding then L equation = L1 + L2 + 2M.2. Opposing then L equation = L1 + L2 –2ML1 + L2 + 2M = 380 ıH …(1)

L2 + L2 –2M = 240 ıH …(2)From 1 & 2, M 35H7. The switch SW shown in the circuit is kept at position ‘1’ for a long duration. At t = 0+, theswitch is moved to position ‘2’ Assuming 02 01 V V , the voltage C V t across capacitor ist /RC c 02 01 A v t V 1 e V −−−−t /RC c 02 01 B v t V 1 e V −−t /RC c 02 01 01 C v t V V 1 e V −−−−t /RC c 02 01 01 D v t V V 1 e V −−R '2'SW02 V 01 VRC C V

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India7Answer: (D)Exp: When switching is in position 1tzC eV t Initial final final value−−tRCC 01 V t V 1 e_ −−_ _ _When switch is in position 2Initial value istRCC 01 V t V 1 e_ −−_ _ _Final value is –V02t2RCC 01 02 01 V t V V V 1 e−_ −−_ _ _8. A parallel plate capacitor consisting two dielectric materials is shown in the figure. Themiddle dielectric slab is place symmetrically with respect to the plates.If the potential difference between one of the plates and the nearest surface of dielectricinterface is 2Volts, then the ratio 1 2 : is(A) 1 : 4 (B) 2 : 3 (C) 3: 2 (D) 4 : 1Answer: (C)

Exp:1 2cons tan t 2 11 22 1Q CV C VC VA VCd V1 2 1 1 22 1 1 2 1 211 2 111 2VV V 6 10V33 3 52: 3: 2_ _ _ _ 10V8V2V0V112

10Volt1 1 2 d / 2dRCc R202 VRCc 01

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India89. Consider an LTI system with transfer function1H` ss s 4If the input to the system is cos(3t) and the steady state output is Asin3t , then the valueof A is(A) 1/30 (B) 1/15 (C) 3/4 (D) 4/3Answer: (B)Exp: 1Given H ss s 421H j16000y t H j cos twhere H j00A H j31 1A3 9 16 151

_ _ 10. Consider an LTI system with impulse response h(t) = 5t e u t −. If the output of the system is2t 5t y t e u t e u t −−−then the input, x(t), is given by

3t A e u t −3t B 2e u t −5t C e u t −5t D 2e u t −Answer: (B)Exp:5t3t 5t3t1h t e u t H ss 51 1y t e e u t Y ss 3 s 5Y sH sX sY s 5 5 s 3 2X sH s 5 3 s 5 s 31s 5x t 2e u t−−−−−−_ −_ 0 cos t Hjx thty t

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India911. Assuming an ideal transformer,. The Thevenin’s equivalent voltage and impedance as seenfrom the terminals x and y for the circuit in figure areA2sint, 4B1sint, 1

C1sint, 2D2sint, 0.5Answer: AExp: xy oc Vin xyxy oc 2 sin t1 2_2xy2R 100 41 _ _ _ __ _th 2sintth R 412. A single phase, 50kVA, 1000V/100V two winding transformer is connected as anautotransformer as shown in the figure.The kVA rating of the autotransformer is _____________.Answer: 550kVAExp: Given,32A.TFr1000V50kVA,100V50 10I 500100kVA 1100 500 550kVA1sint1: 2xy1000 V100 V1100 V1100V1000V2 I 500A

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India1013. A three-phase, 4pole, self excited induction generator is feeding power to a load at a

frequency f1. If the load is partially removed, the frequency becomes f2. If the speed of thegenerator is maintained at 1500 rpm in both the cases, then(A) f1 f2 50Hz and f1 f2 (B) 1 2 f 50Hz and f 50Hz(C) 1 2 2 2 f f 50Hz and f f (D) 1 2 f 50Hz and f 50HzAnswer: (C)Exp: Initially self excited generator supply power to a load at 1 f . If load is partially removed thenslightly speed increase, also frequency 2 f2 1 f fBut both cases 1 2 f f 50Hz14. A single phase induction motor draws 12 MW power at 0.6 lagging power. A capacitor isconnected in parallel to the motor to improve the power factor of the combination of motorand capacitor to 0.8 lagging. Assuming that the real and reactive power drawn by the motorremains same as before, the reactive power delivered by the capacitor in MVAR is____________.Answer: 7MVARExp: Given, 1−Induction motor draws 12mW at 0.6pf, lagLet 11P 12mWcos 0.6pfTo improve pf, 2 cos0.81c del bycapacitor61111Q ?P 12 10cos SS 0.6S 20MVA_ _ Reactive power, 2 21 1 1 Q S −P 16MVARWhen capacitor is connected then1 226222 22 2 1

Pcos Real power drawn is sameS12 100.8SS 15MVAReactive power ,Q S P 9MVAR−_But motor should draw the same reactive power.c delby capacitorQ 16 −9 7MVAR_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India1115. A three phase star-connected load is drawing power at a voltage of 0.9 pu and 0.8 powerfactor lagging. The three phase base power and base current are 100MVA and 437.38Arespectively. The line-to line load voltage in kV is ___________.Answer: 117-120Exp: Given, 100 mVA, 437.38 AVL−L (kV) ? We know that, L L S 3 V .I6L L6LL100 10 3.V .I100 10V3 437.38V 132.001kVBut it is drawing power at a voltage of 0.9 puactualpuBaseVVVactual L L pu B V V V V0.9 132 118.8kV

−_ 16. Shunt reactors are sometimes used in high voltage transmission system to(A) limit the short circuit current through the line.(B) compensate for the series reactance of the line under heavily loaded condition.(C) limit over-voltages at the load side under lightly loaded condition.(D) compensate for the voltage drop in the line under heavily loaded condition.Answer: (C)17. The closed-loop transfer function of a system is 2 4T s .s 0.4s 4The steady state errordue to unit step input is ________.Answer: 0Exp: Steady state error for Type-1 for unit step input is 0.18. The state transition matrix for the system1 12 2x 1 0 x 1ux 1 1 x 1_ _ _ _ _ _ _ _

_ _ _ _ _ _ _ ___istt te 0Ae e_ _

_ _t2 t te 0Bt e e_ _

_ _tt te 0Cte e

_ _ _−_

t tte teD0 e_ _

_ _Answer: (C)_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India12Exp: Given1 0 1A B1 1 1_ _ _ _ _ _ _ _11 s 0 1 0SI A0 s 1 1−

−_ _ __− _ _ −_ ___ _ _ __1210s 1SI A1 1s 1 s 1−

_ _ − − _ −−_

The state transition matrixAt 1 1 e L SI A−_ −−_ _ _tAtt te 0e

te e_ _ _ _19. The saw-tooth voltage wave form shown in the figure is fed to a moving iron voltmeter. Itsreading would be close to ______________Answer: 57.73Exp:Moving iron meter reads RMS value only RMS value of saw-tooth waveform is max3Meter reads1003= 57.73 volts100 V20ms 40ms100 V20 msec 40 msect

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India1320. While measuring power of a three-phase balanced load by the two-wattmeter method, thereadings are 100W and 250 W. The power factor of the load is ______________.Answer: 0.802Exp: In two-wattmeter method,The readings are100W&250WPower factor cos1 1 21 23cos tan−

_ _ −__

_ _ __1 3 150cos tan350−

_ _ __

_ _ __0.802921. Which of the following is an invalid state in an 8-4-2-1. Binary Coded Decimal counter(A) 1 0 0 0 (B) 1 0 0 1 (C) 0 0 1 1 (D) 1 1 0 0Answer: (D)

Exp: In binary coded decimal (BCD) counter the valid states are from 0 to 9 only in binary system0000 to 1001 only. So, 1100 in decimal it is 12 which is invalid state in BCD counter.22. The transistor in the given circuit should always be in active region. Take CEsat V 0.2V.EE V 0.7V. The maximum value of RC in which can be used is __________.Answer: 22.32Exp: B5 0.7I 2.15mA2k−CCI 0.215A5 0.2R 22.320.215−RC5V100Rs 2k5V

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India1423. A sinusoidal ac source in the figure has an rms value of20V.2Considering all possiblevalues of RL, the minimum value of RS in to avoid burnout of the Zener diode is________.Answer: 300Exp: Vm 20Vzz z z zzSPP V I I 50mAV20 5R min 30050mA_

−24. A step-up chopper is used to feed a load at 400 V dc from a 250 V dc source. The inductorcurrent is continuous. If the ‘off’ time of the switch is 20s, the switching frequency of thechopper is kHz is __________.Answer: 31.25 kHzExp: 0 s off V 400v, V 250v, T 20sec, F?Given chopper in step up choppersooff66VV1 D250 250400 1 D1 D 400D 3 0.3758but T 1 D T20 10 1 3 T8T 32 sec1 1Then f 31.25HzT 32 10f 31.25kHz−−−_ −−−−~20V2S R5V1/ 4W L R

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India1525. In a constant V/f control of induction motor, the ratio V/f is maintained constant from 0 tobase frequency, where V is the voltage applied to the motor at fundamental frequency f.Which of the following statements relating to low frequency operation of the motor is TRUE?(A) At low frequency, the stator flux increases from its rated value.(B) At low frequency, the stator flux decreases from its rated value.(C) At low frequency, the motor saturates.(D) At low frequency, the stator flux remains unchanged at its rated value.Answer: (B)Exp: During constantVfcontrol, at low frequency, the voltage also applied to the induction motoris low. Hence the stator flux also decreases from its rated value.Q.No. 26 – 55 Carry Two Marks Each26. To evaluate the double integral8 y/210 y/22x ydx dy2 −_ __ _ _ __ _ _ __ _ , we make the substitution2x yu2 −_ __ __ andyv2. The integral will reduce to4 20 0

A _ _ 2 u du dv 4 10 0

B _ _ 2 u du dv

4 10 0

C _ _ u du dv 4 20 0

D _ _ u du dvAnswer: (B)Exp: 2x y y u ........ 1 and V ........ 22 2−y12 4 180y v 0 u 024 10 0y yx u 0 ; x 1 u 12 2y 0 v 0 ; y 8 v 4from 1 and 2 ,x u v ... 3 and y 2v ... 4x xu v 1 1Jacobian transformation; J 2y y 0 2u v2x ydx dy u J du dv22u du dv

_ _ _ _ __ −_ _ _ __ __ __ __ __ ___ _

_ _ _ __ __ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India1627. Let X be a random variable with probability density function

0.2, for x 1f x 0.1, for 1 x 40, otherwise_ _ ___The probability p0.5 x 5is __________Answer: 0.4Exp: 50.5

P 0.5 x 5 _ f x dx1 4 50.5 1 41 40.5 1Oppositef x dx f x dx f x dxHypotenuse0.2 x 0.1 x 00.1 0.3 0.4

_ _ _28. The minimum value of the function 3 2 f x x −3x −24x 100 in the interval [-3. 3] is(A) 20 (B) 28 (C) 16 (D) 32Answer: (B)Exp: 1 2 f x 0_x −2x −8 0x 2,4 3,3Now f 3 118 ; f 3 28and f 2 128 ; f 4 44f x is min imum at x 3 and the min imum value is f 3 28_ −−−−29. Assuming the diodes to be ideal in the figure, for the output to be clipped, the input voltage vimust be outside the range(A) −1V to −2V (B) −2V to −4V (C) 1V to −2V (D) 2V to −4VAnswer: (B)Exp: When both diodes are 0FF, i

ovv2(Not clipped).i For the clipped, v must bt ouside the range −2Vto −4V10ki V1V~ 10k2Vo V_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India1730. The voltage across the capacitor, as sown in the figure, is expressed asvt t A1 sin 1t −1 A2 sin 2t −2 The value of A1 and A2 respectively, are(A) 2.0 and 1.98 (B) 2.0 and 4.20 (C) 2.5 and 3.50 (D) 5.0 and 6.40Answer: (A)Exp: By using super position theorem,1. 1 C t −When 20 sin 10t voltage source is acting,Network function 1j c 1H j1 10j 1Rj c_1

1c1t 20 sin 10t tan 10101−−2. 2 c t −When 10 sin 5t current source is acting2 c10 0 10.2j1 0.2j−−2 c2j1 0.2j−

−c2 222t .sin 5t1 0.2−2 c 2 t 1.98 sin 5t −C 1 2 V t 2sin 10t −1.98 5t −By comparing with given expression, 12A 2.0A 1.9831. The total power dissipated in the circuit, show in the figure, is 1kW.The voltmeter, across the load, reads 200 V. The value of XL is _________.20sin10t ~11HC V t 1F 10sin5t10A 2A 1c1 X L X RLoadc2 V 200V acsource ~ X

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India18Answer: 17.34 Exp: Total power dissipated in the circuit is 1kW.P 1kW2 2 1000 I .1I .R.2 21000 2 .110 .R._ R 9.96 V 200Z 20I 10_ 2 2L Z R X_ 2 2 2L X Z −R2 2 2L X 20 −9.96_ L X 17.34 32. The magnitude of magnetic flux density B__at a point having normal distance d meters froman infinitely extended wire carrying current of l A is 0I

2nd(in SI units). An infinitelyextended wire is laid along the x-axis and is carrying current of 4 A in the +ve x direction.Another infinitely extended wire is laid along the y-axis and is carrying 2 A current in the +vey direction 0 is permeability of free space Assume Iˆ, Jˆ, Kˆ to be unit vectors along x, y andz axes respectively.Assuming right handed coordinate system, magnetic field intensity, H__at coordinate (2,1,0)will be2 A 3 kˆ weber / m2B4 ˆiA/m3C3 kˆ A/m2D0 A/mI ampsd0IB2 dy2A1z12,1,04A2

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India19Answer: (C)Exp: H Hx Hy

x x y zy y x zzI 4 2H a a a a2 2 1I 2 1

H a a a a2 2 21 1 3H 2 a2 2− _ _ −_ _ _ 33. A discrete system is represented by the difference equation1 12 2X k 1 a a 1 X kX k 1 a 1 a X k_ _ _ −_ _ _

_ _ _ _ _ _It has initial condition 1 2 X 0 1;X 0 0 . The pole location of the system for a = 1,are(A) 1 j0 (B) −1 j0 (C) 1 j0 (D) 0 j1Answer: (A)Exp: from the given difference equation,a a 1Aa 1 a_ −_ _ _The pole locations of the system for a = 1.Then1 0A2 1_ _ _ _2SI −A ._ s −1 0S 1j034. An input signal xt2 5sin100tis sampled with a sampling frequency of 400 Hz andapplied to the system whose transfer function is represented byN1Y z 1 1 z

X z N 1 z−−

−__ _ _ −_where, N represents the number of samples per cycle. The output y(n) of the system understeady state is(A) 0 (B) 1 (C) 2 (D) 5_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India20Answer: (C)Exp: xt 2 5sin100t sj j Njj j1 21j1 1r 01x nT 2 5sin 100 n.4002 5sin n , N 84Y e 1 1 eH eX e N 1 ex n x n x ndue to x ny n H e x n−−

_ _ __ _ _ _ _ _ __ −_ _ −_

1j j24 4j41 2y n 2y n H e sin n H e4H e 0y n y n y ny n 2Thus at steadystate y[n] 2 ___ ___ _35. A 10 kHz even-symmetric square wave is passed through a bandpass filter with centrefrequency at 30 kHz and 3 dB passband of 6 kHz. The filter output is(A) a highly attenuated square wave at 10kHz(B) nearly zero.(C) a nearly perfect cosine wave at 30kHz.(D) a nearly perfect sine wave at 30kHz.Answer: (C)Exp: 10 KHz even symmetric square wave have frequency component present10KHz, 30KHz, 50KHz, 70KHz[only odd harmonics due to half wave symmetry]Since bandpass filter is contered at 30KHz, 30KHz component will pass through_ filter output is nearly perfect cosine wave at 10 KHzCosine in due to reason that signal in even signal.36. A 250 V dc shunt machine has armature circuit resistance of 0.6and field circuit resistanceof125 . The machine is connected to 250 V supply mains. The motor is operated as agenerator and then as a motor separately. The line current of the machine in both the cases is50 A. The ratio of the speed as a generator to the speed as a motor is ____________._ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India21Answer: 1.27Exp: Given:

b a a E V I R250 48 0.6221.2V−−b a a E V I R250 52 0.6281.2VgmN?NWe know that g g m bgmN Eflux is constantN E281.2 N1.27221.2 N_ _37. A three-phase slip-ring induction motor, provided with a commutator winding, is shown inthe figure. The motor rotates in clockwise direction when the rotor windings are closed.If the rotor winding is open circuited and the system is made to run at rotational speed fr withthe help of prime-mover in anti-clockwise direction, then the frequency of voltage across sliprings is f1 and frequency of voltage across commutator brushes is f2. The values of f1 and f2respectively are(A) f + fr and f (B) f - fr and f (A) f - fr and f+ fr (D) f - fr and fAnswer: (A)Exp: Whenever the Rotor winding is open circuited and rotating in anti-clockwise direction then thefrequency of voltage across slip rings iss r11 rN N Pf120f f f

At the same time frequency of voltage across commutator brushes ifs2N Pf f120125250V2A 50A48AMotor Generator125250V2A 50A52A3−phase ac, fHzPr imemoverr f1 fSlip Ring Induction Motor2 f

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India2238. A 20-pole alternator is having 180 identical stator slots with 6 conductors in each slot. All thecoils of a phase are in series. If the coils are connected to realize single-phase winding, thegenerated voltage is 1 V . If the coils are reconnected to realize three-phase star-connectedwinding, the generated phase voltage is 2 V . Assuming full pitch, single-layer winding, theratio 1 V / 2 V is1A31B2C3 D2Answer: (D)Exp: Given poles, P=20Total slots = 180Total no. of conductor = 180 61080the ratio of voltage generated when the coils are connected in 1−to when the coils areconnected in 3 −, Y-connection.i.e.,

1 12 3V2V−−39. For a single phase, two winding transformer, the supply frequency and voltage are bothincreased by 10%. The percentage changes in the hysteresis loss and eddy current loss,respectively, are(A) 10 and 21 (B) -10 and 21 (C) 21 and 10 (D) -21 and 10Answer: (A)Exp: Given1−TransformerV and f are increased by 10%ne% W ?% W ?HereVfis constantn2eW fW fnnnW fas 'f ' increased by 10%W also 10%_ 2e2eeW fW 1.21fW by 21%_ 40. A synchronous generator is connected to an infinite bus with excitation voltage Ef = 1.3 pu.The generator has a synchronous reactance of 1.1 pu and is delivering real power (P) of 0.6pu to the bus. Assume the infinite bus voltage to be 1.0 pu. Neglect stator resistance. The

reactive power (Q) in pu supplied by the generator to the bus under this condition is_________._ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India23Answer: 0.109Exp: fsE 1.3 P.uX 1.1P.uP 0.Giv6puen,V 1.0puQ ?ssEVP sinX1.3 10.6 sin1We.130.5VQ Ecos Vknow t0.10hat,9X_ _ −41. There are two generators in a power system. No-load frequencies of the generators are 51.5Hz and 51Hz, respectively, and both are having droop constant of 1 Hz/MW. Total load in thesystem is 2.5 MW. Assuming that the generators are operating under their respective droop

characteristics, the frequency of the power system in Hz in the steady state is __________.Answer: 50Exp: Given, two generators in a power system has no load frequency of 51.5 & 51 Hz.drop constant=1Hz/mWTotal load=2.5 mW121 21 21 21for generator '1', f x 51.5generator '2' f x 51x 51.5 x 51x x 0.5 ...(1)Total load x x 2.5 ...(2)3By solving (1) &(2) x 1.52f 1.5 51.5 50Hz−−−−_ −_ _ −42. The horizontally placed conductors of a single phase line operating at 50 Hz are havingoutside diameter of 1.6 cm, and the spacing between centers of the conductors is 6 m. Thepermittivity of free space is 8.85410−12 . The capacitance to ground per kilometer of eachline is(A) 4.2 × 10-9F (B) 8.4 × 10-9F (C) 4.2 × 10-12F (D) 8.4 × 10-12F_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India24Answer: (B)Exp: Given, diameters of conductor =1.61mradius, r 0.8cmSpacing between conductors, d=6mPermitivity 0 8.85×10-12capacitance to ground per km ?12o 122192 2 8.85 10C 8.4 10d 6ln ln

r 0.8 10C/ km 8.4 10 F−−−− _ _ ___ __ _ _43. A three phase, 100 MVA, 25 kV generator has solidly grounded neutral. The positive,negative, and the zero sequence reactances of the generator are 0.2 pu, 0.2 pu, and 0.05 pu,respectively, at the machine base quantities. If a bolted single phase to ground fault occurs atthe terminal of the unloaded generator, the fault current in amperes immediately after thefault is _________.Answer: 15500Exp: Single line to ground fault,Fault current in f a1 I 3Ipositive sub transient circuit,aa11 2 0oEIz z z1 jj0.2 j0.2 j0.051j2.2223puj0.45−Fault current f a1 I 3 I3 j2.222 j6.666pu−−Base current =63100 102309.4 pu3 25 10

Fault circuit = pu fault circuit in pu x Base circuit in Ampf I 15396A44. A system with the open loop transfer function:2 KG ss s 2 s 2s 2is connected in a negative feedback configuration with a feedback gain of unity. For theclosed loop system to be marginally stable, the value of K is _______ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India25Answer: 5Exp: The characteristic equation 1 + G(s) = 02 k1 0s s 2 s 2s 24 3 2 _s 4s 6s 4s k 0R−H Arry:432101 6 kS4 4 0S5 k 0S20 4kS 05S k−For marginally stable, 20−4k = 020 = 4 k_k 545. For the transfer function2 5 S 4G ss s 0.25 s 4s 25

The values of the constant gain term and the highest corner frequency of the Bode plotrespectively are(A) 3.2, 5.0 (B) 16.0, 4.0 (C) 3.2, 4.0 (D) 16.0, 5.0Answer: (A)Exp: 2 5 s 4G ss s 0.25 s 4s 25If we convert it into time constants,2s5 4 14G ss 4 ss 0.25 1 25 1 .s0.25 25 5_ _ _ __ _ _ _ _ _ _ _ _ _ _ _2s3.2 14G ss 4 ss 1 1 .s0.25 25 25_ _ _ __ _ _ _ _ _ _ _Constant gain term is 3.2n 5highest corner frequency_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India2646. The second order dynamic systemdXPX Qudty RXhas the matrices P, Q and R as follows:1 1 0

P Q R 0 10 3 1−_ _ ___ −__ _ _ The system has the following controllability and observability properties:(A) Controllable and observable (B) Not controllable but observable(C) Controllable but not observable (D) Not controllable and not observableAnswer: (C)Exp: Given1 1 0P Q0 3 1_−_ _ _ _ −_ _ _For controllability:C0 1Q Q PQ1 3_ __ _ −_C Q 0controllableFor observability:T T T00 0Q R P .R1 3_ ___ ___ _ −_0 Q 0 Not observable.47. Suppose that resistors R1 and R2 are connected in parallel to give an equivalent resistor R. Ifresistors R1 and R2 have tolerance of 1% each., the equivalent resistor R for resistors R1 =300and 2 R 200will have tolerance of(A) 0.5% (B) 1% (C) 1.2% (D) 2%Answer: (B)Exp: 1 R 250 1% 1 2T1 2R RRR R2 R 300 1% T R 136.36% TRTTRE 100

RT 1 T 21 1 2 2R R R R. . 100R R R R_ _ _ _1 1 2 21 2R . R R . RR 2.5; R100 100136.36 2.5 136.36 3. .250 250 300 300_ _ _ _1%_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India2748. Two ammeters X and Y have resistances of 1.2and 1.5 respectively and they give fullscale deflection with 150 mA and 250 mA respectively. The ranges have been extended byconnecting shunts so as to give full scale deflection with 15 A. The ammeters along withshunts are connected in parallel and then placed in a circuit in which the total current flowingis 15A. The current in amperes indicated in ammeter X is __________.Answer: 10.157Exp: X and Y ammeters are connected in parallelShunt Registration of X and Y meters:shx 31.2R15 101150 __ −__ _shx R 0.01212shy 31.5R

15 101250 __ −__ _shy R 0.02542Current through X ammeter is0.02542150.01212 0.0254210.157 ampers49. An oscillator circuit using ideal op-amp and diodes is shown in the figureThe time duration for +ve part of the cycle is 1 t and for-ve part is 2 t .The value oft1 t2

e RC−

will be___________.R5V−−5V3k1k1kC o V15Ashx R1.2 mx I 150mA1.5my I 250 mAshy R

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India28Answer: 1.3Exp:1111tC max initial maxtsat sattttt1

V t V V V ewhere1 5UTP V LTP V e UTP 54 454 5 25 5 e LTP 5 12 255 155 e4 23.75 7.5 e0.5 et 0.69−−−−−−−−__ −−___ − −_ −_ __ _−−2222 2tsat satttt t20.69 1.098LTP V LTP V e5 55 5 e2 255 2.5 5 2.5 e27.5 2.5e e 3t 1.098e 5.98.−−−−−−− −_ −_ __ _−_ −_ −

50. The SOP (sum of products) form of a Boolean function is _(0,1,3,7,11), where inputs areA,B,C,D (A is MSB, and D is LSB). The equivalent minimized expression of the function isABCA CA BC DBBCA CA CC DCBCA CA CC DDBCA BA BC DAnswer: (A)Exp:The equivalent minimized expression of this function is B CA CA BC DCD00 01 11 10000111101 1AB00 1 0100 0 0 00 0 1 0C DA BA CB CEE-GATE-2014 PAPER-02| www.gateforum.com_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India2951. A JK flip flop can be implemented by T flip-flops. Identify the correct implementation.Answer: (B)Exp:n Q J K n 1 Q T0000111100110011

010101010011101000110101T QnJ QnkAnalysis:If you will observe the combinational circuit output expression which is the input for T flipflop is not matching directly, so you should go through the option. If you will solve thecombinational circuit of option (B) thenn nn n n n n n n nn nT J Q . K QJ.K JQ K.Q Q Q J.K JQ K.Q 0 Q .Q 0J.K JQ K.Q_Now, according to consensus theorem J-K will become redundant term, so it should beeliminated.Hence, n n T JQ K.Q , which in matching with our desired result and option-(B) is correctanswer.J

KClkTTflipflop−n Qn QAJKClkTTflipflop−n Qn QBJKClkTTflipflop−n Qn QCClkTTflipflop−n Qn QJKDn QJK00 01 11 100 0 110 10 1 1 0_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India3052. In an 8085 microprocessor, the following program is executedAddress location – Instruction2000H XRA A2001H MVI B,04H2003H MVI A, 03H2005H RAR2006H DCR B2007H JNZ 2005200AH HLT

At the end of program, register A contains(A) 60H (B) 30H (C) 06H (D) 03HAnswer: (A)Exp:Address location Instruction Operation2000H XRA A A00H, CY 0,Z 12001H MVI B, 04H B04H2003H MVI A, 03H A03H2005H RAR Rotate accumulator right with carry2006H DCR B Decrement content of B register byone2007H JNZ 2005H Jump on no zero to location 2005H200AH HLT0 0 0 0 0 0 1 1 0Accumulator CYInitial :value0 0 0 0 0 0 0 1RAR1B03H1 0 0 0 0 0 0 0RAR1B02H

_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India3153. A fully controlled converter bridge feeds a highly inductive load with ripple free load current.The input supply ( s v ) to the bridge is a sinusoidal source. Triggering angle of the bridgeconverter is 30O . The input power factor of the bridge is_________.Answer: 0.78Exp: For fully controlled converter bridgeThe input power factor (PF) = 0.9 cosIPF 0.9 cos30IPF 0.78_ 54. A single-phase SCR based ac regulator is feeding power to a load consisting of 5resistance and 16 mH inductance. The input supply is 230 V, 50 Hz ac. The maximum firingangle at which the voltage across the device becomes zero all throughout and the rms value ofcurrent through SCR, under this operating condition, are(A) 300 and 46 A (B) 300 and 23 A (C) 450 and 23 A (D) 450 and 32 AAnswer: (C)Exp: s V 230V, 50HzR 5 , L 16mH

The maximum firing angle at which the volt across device becomes zero is the angle atwhich device trigger i.e. minimum firing angle to converter.s i−S V ~Load1 1 0 0 0 0 0 0RAR__B__ 01H00 1 1 0 0 0 0 0RAR0Now loop will be overand HLT will execute andprogram will be overB00H_ India’s No.1 institute for GATE Training _ 1 Lakh+ Students trained till date _ 65+ Centers across India321 131XL Ltan tanR R2 50 16 10tan 45.15−−−−

_ _ _ _ _ __ _ _ _ __ _ _ _The current flowing SCR is max at their angle ie. when12 2mTrmsmTrms2Trms,1 VI sin t .d t2 2V 2 230I2z 2 5 5.042I 22.9 23A

_ _ _ ___ __ −__ ___

_55. The SCR in the circuit shown has a latching current of 40 mA. A gate pulse of 50 μs isapplied to the SCR. The maximum value of R in to ensure successful firing of the SCR is_________.Answer: 6060Exp: L I 40mtWidth of gate pulset 50secWhen SCR in ON with given pulse width of gate1 2t1 23I I IV VI 1 eR RL 200 10R 500−−−Time constant of RL circuit, 3 0.4 10−6350 103 0.4 102100 10040 10 1 e500 RR 6060−−−−

__ −__ __ _SCR100V500200mHR