B. Prosperity C. Scarcity
D. None of the above Ans. A
Sol.: Penury is the state of being very poor or extreme poverty. 2. Opposite word to CRITICIZE is:
A. Censure B. Denounce
C. Arraign D. Commend
Ans. D
Sol.: Commend → Praise formally or officially.
3. Consider the arithmetic progression (A .P.): . Then the value of (y – x) is _________.
A. 1 B. 2
C. 3 D. 4
Ans. C
Sol.: Since these are in A .P. 2b = a + c 2 × 17 = x + 3x – y – z 4x – y = 36 …(1) 2(3x – y) = 171.3x + y – 30 3x – 3y = –9 …(2) Using (1) & (2) x = 13 and y = 16 (y – x) = 3
4. Nitu is three time older than Pankaj, Niraj is half the age of Vidya, Pankaj is older than Niraj. Then choose the correct option:
A. Pankaj is older than Vidya. B. Nitu is older than Vidya.
C. Nitu may be younger than Vidya. D. None of the above.
Ans. B
Given Also P > N2 So 3P > 3N2
since N1 = 3p and
5. A file contains 2 English alphabet letters followed by three-digit numbers. Different files that can be made _________. A. 468000 B. 46800 C. 4680 D. 468 Ans. A
Sol.: Required files is:
6. If , where a1, a2, … an are
So, M = 1
7. If . Then the value of (x + y + z)
is _________.
A. 0 B. 1
C. 2a D. 4b
Ans. A Sol.:
8. Let . Then the value of is ________.
9. Consider 5 books (i.e., Hindi, English, Physics, Chemistry & Mathematics) randomly placed on a table. Hindi book is placed below Mathematics book. Physics book is placed above Chemistry book. English book is placed below Hindi book and Chemistry book is placed above Mathematics books. Which book is at bottom.
A. Hindi B. English
C. Physics D. Chemistry
Ans. B
Sol.: Let H stands for Hindi, E stands for English, P stands for Physics, C stands for Chemistry and M stands for Mathematics.
Based on data available, the order of books is:
So, English book is at bottom.
10. On the circumference of a circle there are 8 points. If ‘a’ is total number of possible quadrilaterals and ‘b’ is total number of possible triangles. Then the value of (a – b) is ________.
A. 7 B. 14
C. 21 D. 28
Ans. B
Sol.: Number of triangles = 8C3 = 56 = b Number of quadrilaterals = 8C4 = 70 = a (a – b) = 70 – 56 = 14
Using Laplace domain:
By Inverse Laplace transform, i(t) = e–20 u(t) mA
So, A = 1, B = 20 (A + B) = 21
12. If diagonal matrix entries for matrix A is a, b, c. Then the value of (a + b + c) is _______. Where matrix A is given as:
. Sol.:
(2 – λ)(2 – λ)(2 – λ) – (2 – λ) = 0 (2 – λ)[(2 – λ)2 – 1] = 0
Gives: λ = 1, 2, 3
So, diagonal matrix is [1 2 3]. So, a = 1, b = 2, c = 3
13. If Curls of vector at point A: (1, 2, -3) and point B: (2, 3, k) are orthogonal. The vector is
given as . Then the value of ‘k’ is
________. Ans.
Sol.:
Since curls are orthogonal so
15 – (k + 3) = 0 k = 12
14. In order to design 6-bit Gray to Binary converter ________be the total number of (2 × 1) MUX required to implement the converter.
A. 3 B. 6
C. 10 D. 16
Ans. C
Sol.: 6-bit number of EX-OR required = 5 One EX-OR is Produced by = 2 MUX So, Total MUX required = 5 × 2 = 10
15. If R = 1 Ω and RN (Norton equivalent resistance) seen form A-B terminal is 3 Ω. Then the value of β is _______.
Applying KCL at node V1:
Also, V1 = V – I ………(2)
Solving equation (1) & (2), we get, 2(V – I) = (1 + 1/β)I
2V = (3 + 1/β)I
Consider the following parameters as: V1 = 12 V and V2 = 2 V
For D1: Forward cut-in voltage = 0.2 V, Reverse Breakdown voltage = 13 V For D2: Forward cut-in voltage = 0.7 V, Reverse Breakdown voltage = 20 V Ans.
Sol.: For the circuit: D1 is forward biased i.e., ON, D2 is reverse biased, i.e., OFF because of 0.2 V drop across D1.
Thus, equivalent circuit looks like:
Applying KVL,
So, output voltage = 12 – 0.2 = 11.8 V
17. A complex Analytic function is defined as: Real part + Imaginary part. If real part is , Then imaginary parts is:
A. B.
C. D.
Ans. B Sol.:
So, .
18. If (199)12 + (199)11 = (N)5, Then the value of ‘N’ is __________. Ans. Sol.: (199)11 = 112 + 9 × 11 + 9 = (229)10 (199)12 + (199)11 = (490)10 So, (490)10 = (3430)5 So, N = 3430
19. Consider the waveform shown in figure below, choose the correct option about the construction of waveform:
A. DC term only B. DC and Odd sine harmonics
C. DC and Even sine harmonics D. Odd sine harmonics only Ans. B
Sol.: x(t) = Hidden odd symmetry + Half wave symmetry So, Fourier series contains DC and odd harmonics only.
20. A stone thrown vertically upward satisfies the equation s = 64t− 16t2, where s is in meters
and t is in seconds.
A. The maximum height achieved by the stone is at time t= 3 sec B. The maximum height achieved by the stone is at time t= 2 sec C. The maximum height achieved by the stone is 64 m.
D. The maximum height achieved by the stone is 56 m. Ans.
Sol.: s = 64t− 16t2
t =2sec
= − 32 < 0 ⟹ maxima smax = 64 × 2 −16 × 22 Maximum height = 64m
21. For a certain DC motor with 40 A of armature current 20 N-m. Torque is developed. Now, series field is added that increases 20% flux and observed armature current increases to 60 A. Developed torque is ________N-m.
Ans. Sol.:
22. For . The value of is ________. If its general solution is .
Ans.
Sol.: Since the solution of the differential equation is given as: .
So, roots of auxiliary equation = (–2 ± i) m = –(Sum of Roots) = –(–2 + i – 2 – i) = 4 n = Products of Roots = (–2 + i)(–2 - j) = 5 So,
23. For the BJT at 300 K temperature, early voltage measured is 100 V. Then intrinsic gain of BJT in (kV/V) is:
A. 1.414 B. 2.828
C. 3.867 D. None of the above
Ans. C
Now, VA = 100 V Given,
As T = 300 K (Given) So,
So,
24. For Lissajous Pattern, horizontal input frequency in Hz is _______; if vertical input frequency is 2 kHz provided CRO shows: Horizontal Tangencies = 4, Vertical Tangencies = 2.
Ans. Sol.:
fx = 1 kHz = 1000 Hz
25. Consider the following Binary ripple counter:
Where all flip flops are cascaded and fj shows the output frequency for j = {1, 2, …n}. If for n > Then MOD of counter is ________.
n = 12 – 1 = 11
MOD = 2n = 211 = 2048
26. Consider Fourier transform pair as , where P and Q are function of a. The value of is: A. a/2 B. a2 C. 2a D. 1 Ans. A Sol.: Given
Comparing equations (1) and (2), we get, P = 2a and Q = a2
So,
27. If K > 0, and Open loop transfer function of Control system having negative feedback system is , where a = 1, Then closed loop poles are
A. All real but repeated. B. All real but distinct. C. Two Complex and one real. D. None of the above Ans. B
Sol.: Using Root locus of given system:
So for K > 0, all roots are real and distinct.
28. Consider the following 1-phase semi converter, where load current at 10 A is observed ripple free.
_____ VAR is value of reactive power input to the converter. Ans.
Sol.:
Reactive Power Input,
29. Power rating of a certain DC machine is _________ provided: (i) Number of poles = 10
(ii) Type of wound = Lap (iii) Total conductors = 200
(iv) Per conductor current carrying capacity = 10 A (v) Per conductor average induced EMF = 10 V
A. 10 kW B. 20 kW
C. 100 kW D. 200 kW
Ans. B
Sol.: For Lap wound:
Parallel paths = No. of poles = 10 No. of conductors in each parallel path
Induced emf in 10 parallel paths = 20 × 10 = 200 V Current in each parallel path = 10 A
Total current = 10 × 10 = 100 A
30. For a three-bus system, if pre-fault voltage at all buses to be 1 p.u. and at bus 2, a 3-phase short-circuit occurs. Then during the fault bus 1, the voltage is ______p.u.
A. 0.05 B. 0.10
C. 0.15 D. 0.21
Ans. D
Sol.: For short circuit at bus-2: Fault current,
Gives: ΔV1 = j0.1103 × j7.1582 = –0.7895 p.u.
So, V1(Fault) = V1(0) + ΔV1 = 1.0 – 0.7895 = 0.2105 p.u.
31. During reverse recovery time of Thyristor junction(s) where charge carriers are swept out is/are:
A. J1 B. J2
C. J3 D. None of the above
Ans.
Sol.: During reverse recovery time free charge carriers swept out from the junction J1 and J3. 32. Suppose for a certain Control system, the Open loop transfer function is . This
control system is compensated by system having transfer function , such that Kv = Then the value of (a/b) is ________. (Where, Kv represents velocity error constant). Ans.
Sol.:
33. Using Simpson’s rule, value obtained of is _________, if function is defined as:
Ans. Sol.:
= 90
34. Generator delivering 1 p.u. power to an infinite bus and operating at 50 Hz frequency. If pre-fault maximum power delivered is 2 p.u. and during fault maximum power delivered is 0.5 p.u. The critical clearing angle in degrees if maximum power delivered after the clearance of the fault is 1.5 p.u.
Ans.
= 3.337
35. For the shown sinusoidal PWM technique, T1 and T2 are switched in complementary fashion. At 100 Hz, choose the correct option(s): If R = 8 Ω, XL = 12 Ω, modulation index, m = 0.8.
A. V01(peak) = 160 V B. V01(Peak) = 320 V C. I01(peak)= 11.04 A D. I01(Peak)= 22.18 A Ans. Sol.: mi = 0.8 < 1 So, V01(Peak) = Here, So, V01(Peak) = 0.8 × 200 = 160 V I01(Peak) =
36. For the given circuit it is observed that the dominant frequency of all high frequency is due to load capacitance. ______kHz is the value of bandwidth of the circuit.
Rin = rϖ + (1 + β)RE gm rϖ = β
Rin = (2.5 + 0.5 × 101) × 103 Rin = 53 kΩ
Now, time constant due to input side capacitance, τ = 10 nF x [0.5 × 103 + 53 kΩ ||(40 kΩ || 40 kΩ)] τ = 10–8 × 15.02 × 103
For upper cut-off frequency τ = (15 × 10–9)(2 kΩ||2 kΩ) τ = 15 × 10–9 × 103
So, B.W. = fH – fL = 10.610 – 1.059 = 9.551 kHz 37. Consider the circuit as shown in figure below:
The voltage waveform v(t) is as follows:
A. 1.16 Joules B. 2.26 Joules C. 3 .36 Joules D. None of the above Ans. B
Sol.: From the waveform:
For 0 – 5 sec:
E = E1 + E2 + E3 + E4 + E5
= 2.266 J
Ans. Sol.: At balance condition, I1 = I3 and I2 = IC + I4 Ic = I1jωc R3 Fir circuit
I1(R1 + jωL1) = I2R2 + IcR5
I1(R1 + jωL1 – jωcR3 R5) = I2 R2 …(1)
…(2) From equation (1) & (2),
and =
39. For the logic circuit shown below, having positive logic, then high logic output is achieved when A = a, B = b, C = c & D = d where a, b, c, d are either 0 or 1. Then the value of (a + b + c + d) = __________. A. 0 B. 1 C. 2 D. 3 Ans. C Sol.: Now f = 1 when x = y = z = 1 For X = 1, A = B = 1. For Y = 1, Y′ = 0 For Z = 1, Z′ = 0
Thus, C = D = 0 and X-NOR output = 0 So, A = B = 1, C = D = 0
Therefore, a = b = 1, c = d = 0 Hence, (a + b + c + d) = 2
40. Two discrete time signals x1(n) and x2(n) are convoluted, then the energy of convoluted signal is _______. Where x1(n) = {1, 2, 3} and x2(n) = {–1, 2, 2}.
If a = 4, b = 1, Then choose the correct option(s):
A. Open loop system is stable. B. Open loop system is unstable. C. Closed loop system is stable. D. Closed loop system is unstable. Ans.
Sol.: From given block diagram:
Using a = 4,
and
Using b = 1,
Poles = –3, 5
∵ Since one poles in RHS of s-plane. So, open loop system is unstable. For closed loop control system:
Using RH criteria:
42. The value of RL is ________Ω, for maximum power transfer if RA = 15 Ω, RB = 27 Ω and RC = 20 Ω. External supply voltage source value is unknown.
Ans.
Sol.: For maximum power transfer to the load resistance, RL, then RL = RTh
Using Δ to Y conversion:
or
If Primary : Secondary : Tertiary winding Ratio is n : 1 : 1. Then choose the correct option(s):
A. VA rating of transformer is . B. VA rating of transformer is . C. Load power is . D. Load power is . Ans. Sol.:
Primary waveform is:
Is(rms) = Io/n
Secondary 1: rms voltage = Vs/n, rms current = Io/√2 Tertiary 2: rms voltage = Vs/n, rms current = Io/√2 Winding volt-amp rating
Transformer VA rating
Hence, option B is correct. Load power = VoIo
Hence, option D is correct.
44. PMMC drawing 6 mA current and showing full scale deflection for 80 × 10–3 Volts potential difference. The series resistance (in Ω) to instrument, so that PMMC has to be used as voltmeter with range (0 – 300) V, is _________.
A. 15 B. 2467
C. 15000 D. 49987
Ans. D Sol.:
= 49.987 kΩ = 49987 Ω
45. For system having input-output relation. Choose the correct option(s):
where Y(s) = output, X(s) = input.
A. System is causal if converges for Re[s] = 0.75. B. System is causal if converges for Re[s] = –0.75. C. Inverse system is causal represents it is not stable. D. Inverse system is stable represents it is not causal. Ans.
Sol.:
Given, Poles are at
If system causal Re[s] > 1/2 for system to converge. So, statement A, B are false.
H–1(s) has a pole at s = 1.
When the inverse system is causal. ROC does not include jω axis, hence it is unstable. When the inverse system is stable, ROC should be Re[s] < 1, hence it is not causal. So, options C & D both are correct.
46. For the shown digital circuit, initially z = This circuit repeats z = 1 after every _______ clock cycle.
Ans. Sol.:
Initial Z = 1 (Given)
47. At t = 0, step signal of 1 V is applied and also at that time there was no change in capacitor after ‘t’ msec, the output voltage is (Take: R = 10 kΩ, RF = 1 MΩ, C = 10 nF).
A. 100(1 – e–t/10) V B. –100(1 – e–t/10) V C. 50(1 – e–t/10) V D. –50(1 – e–t/10) V Ans. B
Sol.:
Using virtual ground VA = 0 So,
IRF = 0.1 × 10–3 × 106 = 100 V As Vo = –VC VC(∞) = –100 V VC(t) = V(∞) + [V(o) – V(∞)]e–t/RC Vo(t) = –100(1 – e–t/RC) RFC = (1 MΩ) (10 πF) = 10 msec
So, Vo(t) = –100(1 – e–t/10) as ‘t’ is in msec
48. Consider a situation where moving iron voltmeter shows correct reading when 200 V supply is provided. It coil resistance is 4.5 kΩ at 20°C and coil wound with conductor having 0.003/°C temperature coefficient if temperature of coil reaches 60°C, the percentage error in reading is A%, Then the value of A is _______.
Ans. Sol.:
49. If the Open loop transfer function having negative feedback is . Obtained using following Bode plot, then the value of (A + B) is _________.
Ans.
Sol.: The initial slope = –20 dB/decade → indicates K/s
At ω = 1, slope changes from –20 dB/decade to +20 dB/decade, so this indicates term in numerator.
At ω = 3, slope changes from +20 dB/decade to –20 dB/decade, so this indicates
term .
As For
Therefore, A = 3 and B = 10 Hence, (A + B) = 3 + 10 = 13
50. For a 6-pole, 25 kW, 50 Hz, 400 V, Three-phase Induction motor. If: (i) Full load slip = 0.02
(ii) Mechanical torque loss = 30 N-m (iii) Stator losses = 1000 Watts Then choose the correct option(s):
A. Power across Air gap is 28.65 kW. B. Power across Air gap is 29.55 kW. C. Efficiency of motor is 84.32%. D. Efficiency of motor is 87.53%. Ans.
Sol.: P = 6 f = 50 Hz Vt = 400 V
Slip at full load sfl = 0.02 Stator loss = 1000 W = 1 kW
Power output at full load = 25000 = 25 kW
Synchronous speed = 1000 rpm
Shaft Torque, Tshaft =
= 243.6 N-m
Mechanical torque developed, Tmech = Tsh + TF & W = 243.6 + 30 = 273.6 N-m Power across air Gap = Pag = Tmech × ωs = 28.65 kW Power input to motor = 28.65 + 1 = 29.65 kW
= 84.32%
Then For unknown network N1. Choose the correct option(s): A. Parameter A of transmission parameter matrix is 2.5 B. Parameter B of transmission parameter matrix is 1.5 C. Parameter C of transmission parameter matrix is 1 D. Parameter D of transmission parameter matrix is 7 Ans.
Sol.: Drawing circuit using information provided,
For N2, Z-parameters are:
Using KVL at input & output side:
For cascaded network:
Hence, A = 2.5, B = -1.5, C = 1 and D = 7
52. Consider two concentric spheres, where the region between them is filled with dielectric (ϵr). If radius of larger sphere is 0.05 m, radius of smaller sphere is 0.03 m, ϵr = 5, potential of smaller sphere is +100 V and potential of larger sphere is –100 V. Then the potential at 0.04 m between spheres is
A. 0 V B. –25 V
C. –50 V D. –75 V
Ans. B
Sol.: Potential between the sphere is given as
Putting boundary conditions, we get,
Solving equation (1) & (2), we get, A = 15 and B = –400 So, potential at 0.04 m is given as
53. Six dice are thrown 729 times. _______times at least three dice shows a five or a six. Sol.:
Probability of getting at least 5 or 6:
Expected number of times in 729 throws
= 233
54. If saturated synchronous reactance of a 400 V, 6-pole, 50 Hz, Synchronous motor is 2.2 Ω, and a mechanical load disturbance causes 1 mechanical degree rotor angle slip. Then choose the correct option(s): Considering motor shaft load is 10000 Watts at 0.75 power factor (leading).
A. Synchronizing Current is 5.11 A B. Synchronizing Current is 6 .21 A C. Synchronizing Power is 4.24 kW D. Synchronizing Power is 3.13 kW Ans.
Sol.: Given load = 10000 W = 10 kW at 0.75 p.f. leading Load current,
Xs = 2.2 Ω
Now, Excitation is given as:
So, |Ef| = 260.88 V and δ = 7° electrical Mechanical Disturbance, Δδ
1 degree mechanical
= 13.66 V (Per-phase) Synchronizing Current,
= 6.21 A Synchronizing Power,
= 4.255 kW
55. The shown circular loop faces a magnetic field, which is placed in x-y plane or z = 0 plane.
If the value of current in loop is . Then ‘a’ is _________. Where, radius R is 0.1 and resistance of the loop is 5 Ω.
Ans.
Sol.: Magnetic flux is given by:
For circular loop in z = 0 (or ×-y) plane,
Voltage developed in loop is given as
Comparing with given equation, we get, a = –0.4
56. If , where C(s) = Output and R(s) = Step input. The system reaches to its 99% value of steady state at ______ sec, considering it takes 7 sec to reach 50% of its steady state value.
Ans. Sol.: Given that:
or
Its steady state value is:
At t = 7 sec,
Using these values, gives
a = 0.099
A. 5 B. 10 C. 15 D. 20 Ans. C
Sol.: Given parts = x, (20 – x) P = x3(20 – x) For maxima: 3x2(20 – x) – x3 = 0 4x3 = 60x2 x = 0, 15 So, x = 15 58. The function, f(x,y) = x2 + y2 +6x +12
A. Has minimum value at the point (-3, 0) B. Has maximum value at the point (3,0) C. Minimum value of function is 3
s = =0
At (–3, 0), rt –s2 =4 > 0 and r > 0 ∴ (-3,0) is a minimum value point Minimum value of function is f(-3,0)= 3
59. Consider a differential equation as: , provided boundary condition as at x = 0, y = Then the value of y at x = ϖ is _________.
Ans. Sol.:
Given that: y(0) = 0, therefore C = 0
60. A 6-pole, 50 Hz, 3-phase, Induction motor running on full-load develops a useful torque of 160 N-m and the rotor emf is observed to make 120 cycles/min. The Gross mechanical power developed in kW if torque loss in windage and friction is 12 N-m.
Ans.
Sol.: Synchronous speed,
or ωs = 104.7 rad/sec Rotor frequency, fr = sf
Slip,
option(s): A. X1 = 1.33 Ω B. X0 = 0.31 Ω C. X2 = 1.05 Ω D. X0 = 1 Ω Ans.
Sol.: 3-phase fault current,
X1 = 3.175 Ω L-L fault current,
X2 = 1.056 Ω L-G fault current,
62. Peak to Peak capacitor ripple voltage at 20 kHz frequency is ________Volts.
A. 0.002 B. 0.02
Ans. A
Sol.: Here, back converter is given: Vo = αVs
5 = α(12.6) α = 0.397
= 2 mV = 0.002 V
63. For a Circuit Breaker if peak restriking voltage is 100 kV and peak restriking voltage is reached in 50 × 10–6 sec. Then choose the correct option(s):
A. Average RRRV is 3.2 × 106 kV/sec. B. Average RRRV is 2 × 106 kV/sec.
C. Natural frequency of oscillations is 2.5 kHz. D. Natural frequency of oscillations is 10 kHz. Ans.
Sol.:
Average RRRV
= 10 kHz
64. For three-phase fully controlled converter, overlap angle is _____ provided: VLL = 460 V, f = 60 Hz, source inductance = 25 × 10–6 H–6, load voltage = 525 V, and load power = 500 kW.
A. 1° B. 2°
C. 3° D. 4°
Ans. C
Sol.: Load current,
Average output voltage α = 30.86° cos(α + μ) = cosα – cos(α + μ) = 0.8313 α + μ = 33.76° μ = 33.76° – 30.80° = 2.96° = 3°
65. If at generator terminals base values are 100 MVA and 15 kV, then generator voltage (in p.u.) is _______.
Ans. Sol.:
Let Ip.u. = reference phasor i.e. 0.924∠0° p.u. Generator Terminal in p.u.
= VG = Ip[Rp.u. + j(XG + XT + XL)] = 0.924[1.033 + j(1 + 0.2 + 0.2066)] = 1.6125∠53.7066
So,
