Ec Mock Gate

8/16/2019 Ec Mock Gate

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Duration: 180 MinutesDuration: 180 MinutesDuration: 180 MinutesDuration: 180 MinutesDuration: 180 Minutes Max. Marks: 100Max. Marks: 100Max. Marks: 100Max. Marks: 100Max. Marks: 100

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1. This question paper consists of 2 sections, General Aptitude (GA) for 15 marks and the subject

specific GATE paper for 85 marks. Both these sections are compulsory.

2. The GA section consists of 10 questions. Question numbers 1 to 5 are of 1-mark each, while questions

number 6 to 10 are of 2-marks each. The subject specific GATE paper section consists of 55 questions,

out of which question numbers 11 to 35 are of 1-mark each, while question numbers 36 to 65 are of

2-marks each.

3. The question paper consists multiple choice type (MCQ) questions.

4. Multiple choice type questions will have four choices against A, B, C, D out of which only one is the

correct answer. The candidate has to choose the correct answer by darkening the appropriate bubble

on Objective Response Sheet (ORS) by using Blue/Black ball point pen.

5. For all MCQ questions a wrong answer will result in deduction of 1/3 marks for a 1-mark question and

2/3 marks for a 2-marks question. All questions that are not attempted will result in zero marks.

6. Write your Name and Roll No. at the specified locations on the right half of the ORS using Blue/Black

ball point pen, darken the appropriate bubble under each digit of your registration number.

7. No charts or tables will be provided in the examination hall.

8. Choose the closest numerical answer among the choices given.

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ONMK1EC16• Delhi • Noida • Bhopal • Hyderabad • Jaipur• Lucknow • Indore • Pune • Bhubaneswar • Kolkata

Section -I (General Aptitude)

Multiple Choice Questions : Q. No. 1 - 5 carry 1 mark each

Q.1Q.1Q.1Q.1Q.1 Choose the most appropriate word from the options below to complete the following sentence.Many ancient cultures attributed disease to supernatural causes. However, modern science has largely

helped _________ such notions.

(a) impel (b) dispel

(c) propel (d) repel

Q.2Q.2Q.2Q.2Q.2 Which of the following option is the closest in meaning to the word given below?

Latitude

(a) Eligibility (b) Freedom

(c) Meticulousness (d) Coercion

Q.3Q.3Q.3Q.3Q.3 The ratio of material cost and labour cost in a small construction project is 3 : 1. If the labour cost went up

by 20% and material cost witnessed an average increase of 15%, then the new ratio of labour cost and

material cost will be

(a) 11 : 4 (b) 4 : 11

(c) 8 : 23 (d) 23 : 8

Q.4Q.4Q.4Q.4Q.4 The maximum sum of 44, 42, 40 .... is

(a) 502 (b) 504

(c) 506 (d) 508

Q.5Q.5Q.5Q.5Q.5 The average of ages of 5 children born at the interval of 3 year each is 10 years. The age of youngest child

is ________ years.

(a) 3 (b) 4(c) 5 (d) 6

Multiple Choice Questions : Q. No. 6 - 10 carry 2 marks each

Q.6Q.6Q.6Q.6Q.6 The total exports and revenues from the exports of a country are given in the two charts shown below. The

pie chart for exports shows the quantity of each item exported as a percentage of the total quantity of

exports. The pie chart for the revenues shows the percentage of the total revenue generated through

export of each item. The total quantity of exports of all the items is 500 thousand tones and the total

revenues are 250 crore rupees. Which item among the following has generated the maximum revenue

per kg?

Item 1

11%

Item 2

20%

Item 3

19%

Item 4

22%

Item 5

12%

Item 6

16%

Exports

Item 1

12%

Item 2

20%

Item 3

23% I t e m

4

6 %

Item 5

20%

Item 6

19%

Revenues

(a) Item 2 (b) Item 3

(c) Item 6 (d) Item 5


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Q.7Q.7Q.7Q.7Q.7 Excluding stoppage, the speed of bus is 54 kmph and including stoppage it is 45 km/h. For how much

time does the bus stop in an hour.

(a) 9 minutes (b) 10 minutes

(c) 12 minutes (d) 20 minutes

Q.8Q.8Q.8Q.8Q.8 Tickets numbered from 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability

that the ticket drawn has a number which is multiple of 3 or 5.

(a)

(b)

(c)

(d)

Q.9Q.9Q.9Q.9Q.9 In this question, there are 4 statements followed by 4 conclusions numbered as (a), (b), (c) and (d).

Assume the given statements to be true even if they are at variance with commonly known facts. Identify

the conclusion which follows from the given 3 statements.

S1: Some guitars are posters.

S2: All posters are doors.

S3: Some doors are tablets.

S4: All tablets are books.

Conclusions:

C1: Some doors are guitars.

C2: Some books are posters.

C3: Some tablets are guitars.

(a) only C1 follows (b) only C2 follows

(c) only C3 follows (d) only C1 and C2 follow

Q.10Q.10Q.10Q.10Q.10 A tank is filled by 3 pipes with uniform flow. The first two pipes operating simultaneously fill the tank

during which the tank is filled by the 3rd pipe alone. The second pipe fills the tank 5 hours faster than thefirst pipe and 4 hours slower than the third pipe alone. The time required by the first pipe is

(a) 15 hours (b) 3 hours

(c) 11 hours (d) 19 hours

Section -II (Technical + Engineering Mathematics)

Multiple Choice Questions : Q. No. 11 - 35 carry 1 mark each

Q.11Q.11Q.11Q.11Q.11 The cross-over distortion is found in

(a) ClassA amplifiers (b) Class B amplifier

(c) Class C amplifiers (d) Class AB amplifier

Q.12Q.12Q.12Q.12Q.12 GaAs is an example for

1. Direct Bandgap Semiconductor

2. Indirect Bandgap Semiconductor

3. Wide Bandgap Semiconductor

4. Narrow Bandgap Semiconductor

The correct statements are

(a) 1 and 3 (b) only 1

(c) 2 and 3 (d) 4 only


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Q.13Q.13Q.13Q.13Q.13 Consider the circuit given below

3 Ω1 A

2 Ω

+

–

2 Ω

2 Ω2 V

A

B

Thevenin voltage through terminalAB is

(a) 2.4 V (b) 1.4 V

(c) 3.4 V (d) 2 V

Q.14Q.14Q.14Q.14Q.14 The power spectrum of a random process X (t ) with a DC term is given by

S X (ω) = 8πδ (ω) + 1.2 πδ (ω – 2.3π) + 1.2 πδ(ω + 2.3π) + 3.6 π⋅ δ(ω – 5) + 3.6 πδ(ω + 5)

The mean and variance of X (t ) is given by

(a) 2, 9.6 (b) 2, 4.8

(c) 4, 4.8 (d) 4, 9.6

Q.15Q.15Q.15Q.15Q.15 In INTEL 8085 microprocessor, content of accumulator and registerB are 1B H and 16 H respectively. The

condition of zero and carry flag after execution of SUB B instruction will be respectively.

(a) Set, Set (b) Set, Reset

(c) Reset, Set (d) Reset, Reset

Q.16Q.16Q.16Q.16Q.16 For a certain medium the characteristic impedance is + Ω . The loss tangent of this medium is

(a)

(b)

(c) 1 (d) Data insufficient

Q.17Q.17Q.17Q.17Q.17 The auto-correlation function R x(τ) of a signal x(t ) obtain its maximum value at

(a) τ = ∞ (b) τ = 0(c) Does not depend on τ (d) Depends on the maximum values of x(t )

Q.18Q.18Q.18Q.18Q.18 In the given state model

x

x

=

− − +

x

x

y = [ ]

x

x

The transfer function is

(a)

+

+ +(b)

+

+ +

(c)

+

+(d)

+

+


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Q.19Q.19Q.19Q.19Q.19 Consider the digital circuit given below

X

Y

Z

F

The simplified expression for Boolean function f is

(a) + + (b) +

(c) ⊕ (d)

Q.20Q.20Q.20Q.20Q.20 The electric field induced inside a semiconductor with a non-uniform doping of donor atoms and having

the doping profile in one dimension is

(i) Directly proportional to temperature

(ii) Directly proportional to doping profile

(iii) Directly proportional to the derivative of doping profile

(iv ) Inversely proportional to the doping profile

Which of the following statements are correct.

(a) Only (i) and (ii) (b) Only (i) and (iii)

(c) Only (i), (iii) and (iv ) (d) Only (iii) and (iv )

Q.21Q.21Q.21Q.21Q.21 The inverse Laplace transform of a signalX (s ) is given as x(t ). Thus x(t ) is defined as

(a)

∞+

− ∞

=π ∫ x (b)

+ ∞

− ∞

=π ∫ x

(c)

σ+ ∞

σ− ∞

=π ∫ x (d)

∞

−∞

= ∫ x

Q.22Q.22Q.22Q.22Q.22 Which of the following statement INCORRECT?

(a) Each tree has (n – 1) branches where n is the number of nodes of the tree.

(b) Maximum possible rank of a tree is equal to number of nodes.

(c) Maximum number of possible tree = det([A] ⋅ [A]T

). Where [A] is the reduce incident matrix(d) The determinant of Incidence matrix of closed loop is zero.

Q.23Q.23Q.23Q.23Q.23 A square matrix is given as A =

, then the value of ( )

(a) 3 (b) 2

(c) 1 (d) 4


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Q.24Q.24Q.24Q.24Q.24 The value of

→

− x x

x

(a) –1 (b) e

(c) –log10 e (d)

Q.25Q.25Q.25Q.25Q.25 Which of the following statements is not a condition for the application of Rolle’s theorem

(a) f ( x) be continues in (a , b ) (b) f ( x) be differentiable in (a , b )

(c) f (a ) = f (b ) (d) f ( x) be continues in [a , b ]

Q.26Q.26Q.26Q.26Q.26 In the voltage regulator circuit shown below

10 kΩ

R LV Z = 9 V25 V

+

–

iz

given that iz (min)

= 0.6 mA, then the maximum value of load current is _____ mA.

(a) 5 mA (b) 3 mA

(c) 7 mA (d) 1 mA

Q.27Q.27Q.27Q.27Q.27 Consider a two port network as shown below

Network

N

+

–

V 1 V 2

I 1

I 2

+

–

V 1 = 5V 2 – 3 I 2 I

1 = 6 V2 – 2 I

2

A load resistance of 5 Ω is connected across the output port. The input impedance is(a) 0.875 Ω (b) 1.142 Ω(c) 1.875 Ω (d) 1.25 Ω

Q.28Q.28Q.28Q.28Q.28 Consider an angle modulated signal

ϕ(t ) = cos(200 πt )cos(8sin 4πt ) + sin(200 πt )sin(8sin 4πt )The maximum frequency deviation is

(a) 32 Hz (b) 16 Hz

(c) 8 Hz (d) 4 Hz

Q.29Q.29Q.29Q.29Q.29 A 2 × 1 MUX is realized using only 2 input NAND gates. The minimum number of gates required are

(a) 4 (b) 5

(c) 6 (d) 7

Q.30Q.30Q.30Q.30Q.30 The z - transform of a discrete time signal x[n ] is defined a X (z ) =

− α if x[n ] is represented as

αn – k u [n – k ] then the value of k is(a) 9 (b) 10

(c) –8 (d) –9


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Q.31Q.31Q.31Q.31Q.31 The leakage current of a transistor are I CBO = 1 µA and I CEO = 0.1 mA. If I C = 5 mA, then the value of β is(a) 100 (b) 99

(c) 101 (d) 98

Q.32Q.32Q.32Q.32Q.32 The open loop transfer function of a unity feedback system is

G (s ) =

+ +

The value of phase cross over frequency is

(a) 3.873 rad/sec (b) 2.38 rad/sec

(c) 1.23 rad/sec (d) 15 rad/sec

Q.33Q.33Q.33Q.33Q.33 The circuit shown below is of an oscillator known as Wein’s bridge oscillator. The value of

is equal to

–

+

R 1

R 2

V 0

R

C

C

R

(a) 1 (b) 2

(c) 3 (d) 4

Q.34Q.34Q.34Q.34Q.34 If = + + xi and r = , then ∇ (log r ) is equal

(a)

(b)

(c)

(d) ⋅

Q.35Q.35Q.35Q.35Q.35 The value of the integral

+ ∫ where C is the circle withz = 4 is

(a)

πi(b)

πi

(c) 4πie –2 (d)

πi

Multiple Choice Questions : Q. No. 36 - 65 carry 2 marks each

Q.36Q.36Q.36Q.36Q.36 The electric field intensity vector of a plane wave propagating in free space is given by

= ( ) ω −β x V/m. The total power passing through a circular disk of radius 4 cm on plane

z = 0 is

(a) 15 W (b) 0.15 µW(c) 1.5 mW (d) 0.15 W


8/31



Q.37Q.37Q.37Q.37Q.37 A two terminal black box contains any combination of RLC elements. The box is connected to a 220 V AC

supply. The current through the source is I . When a capacitor is inserted in series between box and

supply, the current through source is 2 I . The element can be

(a) Resistance (b) Inductor

(c) Capacitor (d) Not possible to determine

Q.38Q.38Q.38Q.38Q.38 A standard television picture requires 30 frames of 300,000 picture elements, each to be transmitted per

second. Each of these elements can assume 10 distinguishable brightness levels with equal probability.

The theoretical bandwidth of AWGN channel if the SNR at the receiver is required to be atleast 50 dB?

(a) 36 MHz (b) 9.96 MHz

(c) 5.25 MHz (d) 2.17 MHz

Q.39Q.39Q.39Q.39Q.39 In the circuit shown below, the small signal voltage gain will be (Assume β to be very large and V T = 25 mV)

12 V

V o

0.5 mA

5 kΩ

C C

V s

(a) 150 (b) 100

(c) 200 (d) 250

Q.40Q.40Q.40Q.40Q.40 A unit step input is applied to a first order system at t = 0. The steady state value of the response is

5 units and 1.57 units at t = 4 minutes respectively. If the transfer function of the system is given by

=

+ τ, then the system time constant will be

(a) 240 sec (b) 10.61 sec

(c) 636.81 sec (d) 1.507 sec

Q.41Q.41Q.41Q.41Q.41 An 8 × 1 MUX is used to implement a logical functionY as shown in the figure.

I 0

I

I

I

I

I

I

I

1

2

3

4

5

6

7

8 × 1MUX

S0

S1S2

0

D

0

D

1

D

1

D

A B C

Y

The output Y is

(a) + + (b) + +

(c) + + (d) + +


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Q.42Q.42Q.42Q.42Q.42 An antenna having directive gain of 10 dB radiates a total power of 10 kW. The electric field intensity at a

distance of 5 km is

(a) 1 V/m (b) 0.5 V/m

(c) 0.25 V/m (d) 2 V/m

Q.43Q.43Q.43Q.43Q.43 The input output relation of a continuous time signal is given as

−∞

= − τ + τ τ ∫ x

then the output of the system for an impulse input will be

(a)

h t ( )

1

t 1

(b)

h t ( )

1

t 1

(c)

h t ( )

1

t

(d)

h t ( )

1

t –1

Q.44Q.44Q.44Q.44Q.44 The Bode plot of a transfer function G (s ) is shown in the figure below:

100

40

32

20

01

–8 10

G a i n ( d B )

ω(rad/s)

The gain (20logG (s ) ) is 32 dB and –8 dB at 1 rad/s and 10 rad/s respectively. The phase is negative forall ω. ThenG (s ) is

(a)

(b)

(c)

(d)


10/31



Q.45Q.45Q.45Q.45Q.45 The Fermi function at energyE is given as

f F (E ) =

− +

which is approximated as f F (E ) =

−

when E –E F >> kT , the minimum value of (E –E

F ) above

which this approximation can be calculated within 5% of tolerance band is

(a) E – E F > kT (b) (E – E

F ) > 2kT

(c) (E – E F ) > 3kT (d) (E – E F ) > 0.5kT

Q.46Q.46Q.46Q.46Q.46 Let the transform of a signal

←→ x

with ROC R 1 and the Laplace transform of a signal

←→ x

with ROC R 2 then the Laplace transform of x1(t ) ⋅ x2(t ) is given as

(a) ∗ ∩ (b) ∗ ∪

(c)

∗ ∪π

(d)

∗ ∩π

Q.47Q.47Q.47Q.47Q.47 Consider the circuit given below

+ – 10 V R 10 Ω

10 Ω 10 Ω

10 Ω

10 Ω10 Ω

10 Ω

The maximum power that can be transferred to R is

(a) 90.9 mW (b) 28.4 mW

(c) 7.27 W (d) 13.75 mW

Q.48Q.48Q.48Q.48Q.48 Consider the synchronous digital circuit given below

J A

Q A

Clock

K A

Q A

J B

QB

K B

QB

J C

QC

K C

QC 1 11

Initially all flip-flops are cleared. The MOD count of the circuit is

(a) 4 (b) 5

(c) 6 (d) 7


11/31



Q.49Q.49Q.49Q.49Q.49 A carrier wave of frequency 1 MHz is modulated 50% by a sinusoidal frequency 5 kHz. The resulting AM

signal is transmitted through the resonant circuit shown in below figure. Which is tuned to the carrier

frequency and has quality factor of 175.

Output AMwave

C L R

The approximate percentage modulation of the modulated signal after transmitted through this circuit is

(a) 25% (b) 50%

(c) 75% (d) 100%

Q.50Q.50Q.50Q.50Q.50 1 V step function is applied to a non-inverting amplifier with a close loop gain of 10. The slew rate of

op-amp is 5 V/ µsec. The time required to reach its full scale output is

–

+

R 2

V 0

R 1

+ – 1 V

(a) 5 µsec (b) 2.5 µsec(c) 2 µsec (d) 1 µsec

Q.51Q.51Q.51Q.51Q.51 An EM wave travels in free space with electric field component

+= x

The corresponding magnetic field component (A/m) is

(a)

+ + π π (b)

+

π

(c)

+ − π π (d)

+−

π

Q.52Q.52Q.52Q.52Q.52 Consider the initially relaxed system as shown in figure below

+ – 2 ( ) Vu t

6 ( ) Vu t 1 Ω

1/2 Ω

1/4 F

1 Ω

V 2V 1

The expression for V 1(t ) is

(a) 4.8(1 – e –20/3t ) (b) 2.4(1 – e –20/3t )

(c) 3.6(1 – e –20/3t ) (d) 1.2(1 – e –20/3t )


12/31



Q.53Q.53Q.53Q.53Q.53 Which of the following equations are exact?

1.1.1.1.1. ( ) x x + x x

+ + =

2.2.2.2.2. ( )

++ + =

+

x x x

3.3.3.3.3. ( ) ( ) + + + x x x x4.4.4.4.4. xdy – yd x

(a) 1 and 2 only (b) 2 and 3 only

(c) 3 and 4 only (d) 4 and 1 only

Q.54Q.54Q.54Q.54Q.54 The density function of repairing a machine is given by

−

= x

x , where ‘ x’ is repair time in hours. The

probability that the repair time is more than 2 hours is

(a) 0.368 (b) 0.482

(c) 0.518 (d) 0.632

Q.55Q.55Q.55Q.55Q.55 The values of a and b for which the following system has no solutions are respectively

a x + y + 2z = 0

x + 2y + z = b

2 x + y + az = 0

(a) a = –1, b = 0 (b) a = 2, b = 0

(c) a = –1, b ≠ 0 (d) a = 2, b ≠ 0

Q.56Q.56Q.56Q.56Q.56 The open loop voltage gain of the op-amp shown below is equal to 500 then the close loop gain of the

op-amp is _________.

–

+

8 kΩ

V 0

2 kΩ

V in

(a) 4.95 (b) 5.95

(c) 6.95 (d) 7.95

Q.57Q.57Q.57Q.57Q.57 The close loop response curve of a second order system is shown in below figure

0.7

1.25

4ω

M ( j ω)

The rise time of the system will be

(a) 0.68 sec (b) 0.44 sec

(c) 0.216 sec (d) 0.194 sec


13/31



Q.58Q.58Q.58Q.58Q.58 For a depletion width of 2 µm the built in junction capacitance per unit area for a silicon p -n junction willbe (ε

Si = 11.7 × ε

0, ε

0 = 8.85 × 10–12 F/m)

(a) 0.5177µF/m2 (b) 51.77 µF/m2

(c) 27.36µF/m2 (d) 0.2736µF/m2

Q.59Q.59Q.59Q.59Q.59 A lossy transmission line operating at 400 MHz has the line parameters R = 3.5 Ω /m, L = 2 mH/m,G = 0 S/m and C = 120 pF/m. The phase velocity on the line is __________ × 106 m/s.

(a) 64.54 (b) 49.86

(c) 38.94 (d) 10.27

Q.60Q.60Q.60Q.60Q.60 Shown below is a digital discrete time signal with a feedback loop

Σ x[ ]n

+

+

α

DelayUnit

y n[ ]

The system is stable if α < __________.(a) 1 (b) 1.5

(c) 2 (d) 0.5

Q.61Q.61Q.61Q.61Q.61 Consider a two port network given below

V 1

10 Ω0.2 I 2

I 2

0.3 V 2

V 2

+ –+

–

4 Ω I 1

+

–

The value of h -parameter h 12 is

(a) 0.907 (b) 1.872

(c) 1.103 (d) 2.061

Q.62Q.62Q.62Q.62Q.62 In a certain telemetry system eight message signals having 1 kHz bandwidth are to be transmitted using

binary PCM with TDM. The maximum tolerable error in sampling amplitude is 0.5% of the peak signal

amplitude. The minimum transmission bandwidth required is

(assume the sampling rate must be atleast 20% above the Nyquist rate)

(a) 2.4 kHz (b) 8.4 kHz

(c) 9.6 kHz (d) 19.2 kHz


14/31



Q.63Q.63Q.63Q.63Q.63 In the figure shown below, assume that the value of g m R E >> 1, then the value of gain provided by the

circuit is

15 V

V out

C C

R 2

R 1

V s R

E = 500 Ω

R C = 2 kΩ

C C

(a) 3 (b) 5

(c) 4 (d) 6

Q.64Q.64Q.64Q.64Q.64 Match List-IList-IList-IList-IList-I (Numerical Method) with List-IIList-IIList-IIList-IIList-II (Iterative Root Equation) and select the correct answer using

codes given below the lists:List-IList-IList-IList-IList-I List-IIList-IIList-IIList-IIList-II

A.A.A.A.A. Newton-Raphson 1.1.1.1.1.

−

−−

= − x

x x x

B.B.B.B.B. Secant 2.2.2.2.2.

−

− − −− −

= − −−

x x x x

C.C.C.C.C. Regular Falsi 3.3.3.3.3.

−− −

−= − −

− x x x x

CodesCodesCodesCodesCodesAAAAA BBBBB CCCCC

(a) 1 2 3

(b) 1 3 2

(c) 3 1 2

(d) 3 2 1

Q.65Q.65Q.65Q.65Q.65 The value of

⋅ ∫ , where = + x and C is the curve y 2 = 4 x in the x y-plane from (0, 0) to

(4, 4), is

(a) 66 (b) 132

(c) 264 (d) 528


15/31


ONMK1EC16• Delhi • Noida • Bhopal • Hyderabad • Jaipur

• Lucknow • Indore • Pune • Bhubaneswar • Kolkata

1.1.1.1.1. (b)

2.2.2.2.2. (b)

3.3.3.3.3. (c)

4.4.4.4.4. (c)

5.5.5.5.5. (b)

6.6.6.6.6. (d)

7.7.7.7.7. (b)

8.8.8.8.8. (d)

9.9.9.9.9. (a)

10.10.10.10.10. (a)

11.11.11.11.11. (b)

12.12.12.12.12. (b)

13.13.13.13.13. (b)

14.14.14.14.14. (b)

15.15.15.15.15. (c)

16.16.16.16.16. (b)

17.17.17.17.17. (b)

35.35.35.35.35. (b)

36.36.36.36.36. (c)

37.37.37.37.37. (b)

38.38.38.38.38. (d)

39.39.39.39.39. (b)

40.40.40.40.40. (c)

41.41.41.41.41. (d)

42.42.42.42.42. (b)

43.43.43.43.43. (c)

44.44.44.44.44. (b)

45.45.45.45.45. (c)

46.46.46.46.46. (d)

47.47.47.47.47. (b)

48.48.48.48.48. (b)

49.49.49.49.49. (a)

50.50.50.50.50. (c)

51.51.51.51.51. (c)

18.18.18.18.18. (c)

19.19.19.19.19. (b)

20.20.20.20.20. (c)

21.21.21.21.21. (c)

22.22.22.22.22. (b)

23.23.23.23.23. (c)

24.24.24.24.24. (d)

25.25.25.25.25. (a)

26.26.26.26.26. (d)

27.27.27.27.27. (a)

28.28.28.28.28. (b)

29.29.29.29.29. (a)

30.30.30.30.30. (c)

31.31.31.31.31. (b)

32.32.32.32.32. (a)

33.33.33.33.33. (b)

34.34.34.34.34. (a)

52.52.52.52.52. (a)

53.53.53.53.53. (a)

54.54.54.54.54. (a)

55.55.55.55.55. (c)

56.56.56.56.56. (a)

57.57.57.57.57. (b)

58.58.58.58.58. (b)

59.59.59.59.59. (a)

60.60.60.60.60. (a)

61.61.61.61.61. (a)

62.62.62.62.62. (b)

63.63.63.63.63. (c)

64.64.64.64.64. (a)

65.65.65.65.65. (c)

Answer Key

MOCK TEST-1

Electronics Engineering


16/31




EXPLANATIONS

3.3.3.3.3. (c )(c )(c )(c )(c )

Initial ratio M : L

3 : 1 or 3 x : 1 xNew Ratio M : L

+ × x x :

+ × x x

3.45 x : 1.20 x

23 : 8

L : M = 8 : 23

4.4.4.4.4. (c )(c )(c )(c )(c )

Maximum sum is = [44 + 42 + 40 + .... 0]

= 2[22 + 21 .... 0]= 506

5.5.5.5.5. (b)(b)(b)(b)(b)

Let the age of youngest child is = x

then

x + ( x + 3) + ( x + 6) + ( x + 9) + ( x + 12) = 50

5 x + 30 = 50

x = 4

6.6.6.6.6. (d)(d)(d)(d)(d)

For maximum revenue per kg. Let us solve through options

A.A.A.A.A. Item 2 =

= 0.5

B.B.B.B.B. Item 3 =

= 0.6

C.C.C.C.C. Item 6 =

= 0.59

D.D.D.D.D. Item 5 =

= 0.83

Clearly item 5 is largest.

7.7.7.7.7. (b)(b)(b)(b)(b)

Due to stoppage it cover 9 km less

time for stoppage =

=

or

× = 10 minutes


17/31


18/31




15.15.15.15.15. (c )(c )(c )(c )(c )

A = 16 H 00010110

B = 1B H 00011011

2’s compliment of content of B register

1 1 1 1

0 0 0 1

1 1 1 1

1 0 1 1

1 1 1 0 0 1 0 0

–

+ 1

1 1 1 0 0 1 0 1

A – B = A + 2’s compliment of B

0 0 0 1

1 1 1 0

0 1 1 0

0 1 0 1

1 1 11 1 0 1 1

zero flag Reset

Carry generated is but since it is a subtraction operation so carry flag is Set.

16.16.16.16.16. (b)(b)(b)(b)(b)

Given η = + = 100∠30° Ω

= η∠θηloss tangent = tan 2θη

= tan60° =

17.17.17.17.17. (b)(b)(b)(b)(b)

∵ R x(τ)


19/31




20.20.20.20.20. (c )(c )(c )(c )(c )

∵ The induced electric field given as

E x

=

− ⋅

x

x x

22.22.22.22.22. (b)(b)(b)(b)(b)

Maximum possible rank of a tree is = n – 1

where n is number of nodes.

23.23.23.23.23. (c )(c )(c )(c )(c )

A = 4 – 3 = 1

thus, ( ) =

−

= 1

24.24.24.24.24. (d)(d)(d)(d)(d)

∵

→ + x

x x = e a

→

+ − x x

x = e –1 =

26.26.26.26.26. (d)(d)(d)(d)(d)

⇒ I L(max)

=

−− i = 1.6 – 0.6 = 1 mA

27.27.27.27.27. (a )(a )(a )(a )(a )

Network

N

+

–V 1 V 2

I 1 I 2

+

–5 Ω

V 1 = 5V 2 – 3 I 2

I 1 = 6V 2 – 2 I 2V 2 = – 5 I 2 at port 2

V 1 = 5(–5 I 2) – 3 I 2 = –28 I 2 I

1= 6(–5 I

2) – 2 I

2 = –32 I

2

Z in =

−= =

− I

I I = 0.875 Ω

28.28.28.28.28. (b)(b)(b)(b)(b)

The signal can be written as

ϕ(t ) = cos(200 πt – 8sin 4πt )

comparing with standard equationwe get, θ(t ) = ω

c t + φ(t )

= 200πt + (–8)sin 4πt ∴maximum frequency deviation

φ′(t ) = 8 × 4πcosπt = 32π = 100.48 rad/sec= 16 Hz


20/31




29.29.29.29.29. (a )(a )(a )(a )(a )

I 0

I 1

2 × 1 MUX

Y

S

2 × 1 MUX using NAND gate

30.30.30.30.30. (c )(c )(c )(c )(c )

X 1(z ) =

−←→ α− α

∴ X (z ) =

− − −←→ α − −− α

31.31.31.31.31. (b)(b)(b)(b)(b)

1 + β =

−

−

×= =

×

I

I

∴ β = 99

32.32.32.32.32. (a )(a )(a )(a )(a )

–180° =

− −ω ω− ° − −

90° =

−

ω ω+

ω

−

tan90° =

ω ω

+ ω

−

or

ω− = 0

or ωpc

=

= 3.873 rad/sec


21/31




33.33.33.33.33. (b)(b)(b)(b)(b)

In A wein’s bridge oscillator the condition for sustained

A =

+ = 3

∴

= 2

34.34.34.34.34. (a )(a )(a )(a )(a )

∵ ( ) ∇ =

′⋅ =

⋅ =

⋅

35.35.35.35.35. (b)(b)(b)(b)(b)

f (z ) =

+

φ=

− +now, z = –1 pole which lies insidez = 4thus couchis integral formula says

+ ∫ =

= −

πi =

πi

36.36.36.36.36. (c )(c )(c )(c )(c )

Power density =

+= =

η × π π

Power = power density × area

=

−× π × ×π

= 1.5 mW

37.37.37.37.37. (b)(b)(b)(b)(b)

Let Z is impedance of black box,

X C is reactance of capacitor and V is supply

= I ⇒

I = Z

+ = 2 I ⇒

I = Z + jX C

from here,

= Z + jX C

+ = 0

Z = –2 jX C Reactance of black box should be negative of reactance of capacitor

⇒ impedance of black box should be purely inductive


22/31




38.38.38.38.38. (d)(d)(d)(d)(d)

Information/element = log210 = 3.32 bits = 4 bits

∴ Information/picture frame = 4 × 300,000 = 12 × 105 bitsfor 30 picture per frame/sec, the channel capacity will be

= 30 × 12 × 105 = 3.6 × 107 bits/secfor white Gaussian Noise

C =

+

B =

×=

+

B = 2.167 MHz

39.39.39.39.39. (b)(b)(b)(b)(b)

I C ≈ I C = 0.5 mA

∴ g m =

−×= =

I

now, drawing the small signal model,

we get

+ –

V o

R C g V m π

V s

r π

V π

+

–

thus, V 0 = –g m V πR C [V π = –V s ]

gain =

gain = g m

R C

= 20 × 10–3 A/V × 5 × 103 = 100

40.40.40.40.40. (c )(c )(c )(c )(c )

For the first order system

c (t ) = K (1 – e–t / τ)at t = ∞

c (∞) = 5 = K Now,at t = 4 min

c (4 min) = 1.57 = 5(1 – e –4/ τ)0.686 = e –4 / τ

τ= 0.3769

or τ = 10.61 minute = 636.81 sec


23/31




41.41.41.41.41. (d)(d)(d)(d)(d)

I 0

I 1

I 2

I 3

I 4

I 5

I 6

I 7

0 2 4 6 8 10 12 14

1 3 5 7 9 11 13 15

D

D

Y =

1

AB

CD

1

11 1

1

1

1

= + +

42.42.42.42.42. (b)(b)(b)(b)(b)

10 log Gd = 10 ⇒ Gd = 10

P avg =

−× ×= = ×π π ×

P avg =

−= ×η

⇒ E = = 0.49 V/m

43.43.43.43.43. (c )(c )(c )(c )(c )

∵ y (t ) = ( )

−∞

+ − τ τ τ ∫ x

h (t ) = ( )

−∞

+ − τ δ τ τ ∫ h (t ) = (t + 1)u (t )

44.44.44.44.44. (b)(b)(b)(b)(b)

Let G (s ) =

G ( j ω) =

ω∵ Gain = 32 dB at ω = 1 rad/sec

∴

= 32

⇒ 20log K = 32⇒ K = 39.8


24/31




Now,

∵ Gain = –8 dB at ω = 10 rad/sec

∴

= –8

⇒ 10n = 100⇒ n = 2

45.45.45.45.45. (c )(c )(c )(c )(c )

− − − +

+− +

= 0.05

⇒

− − − + − = 0

⇒

− −

= 0.05

E – E F =

l ∝ 3kT

46.46.46.46.46. (d)(d)(d)(d)(d)

[ ]

⋅ x x =

∞−

− ∞ ∫ x x

now, x1(t ) =

σ+ ∞

σ− ∞π ∫

Changing variable of s → a ,we get

x1(t ) =

σ+ ∞α

σ− ∞

α απ ∫

∴ [ ] ⋅ x x =

σ+ ∞∞

α −

−∞ σ− ∞α απ ∫ ∫ x

=

σ+ ∞ ∞− α

σ− ∞ −∞

α α ⋅ απ ∫ ∫ =

σ+ ∞

σ− ∞

α − α απ ∫

=

∗π


25/31




47.47.47.47.47. (b)(b)(b)(b)(b)

+ – 10 V

A

B

10 Ω 10 Ω

10Ω

10Ω10Ω

10 Ω

I 1

R in

I 2

V Th :

R in = 10 + 10 (10 + 10 + 10) + 10R in = 27.5 Ω

I 1

=

I 2

=

V Th = V AB =

× ×

=

R Th

:

R th

10 Ω 10 Ω

10Ω

10Ω10Ω

10ΩR 1 = 10 20 =

Ω

R Th =

+ +

=

10Ω

R thR 1

10Ω

10 Ω

Maximum power can be transferred

=

= × = ×


26/31




48.48.48.48.48. (b)(b)(b)(b)(b)

Initially

1

2

3

4

Q A

0

1

0

0

1

QB

0

0

1

0

1

QC

0

0

0

1

0

⇒ MOD of the counter is 5.

49.49.49.49.49. (a )(a )(a )(a )(a )

∵ s (t ) = AC (1 + m a m (t )) cos2π f c t where, f m = 5 kHz and f c = 1 MHz

for the filter BW =

= =

∵ m (t ) lies close to the 3 dB bandwidth of the filter and therefore it is attenuated by a factor of a half.

∴ m ′(t ) =

Hence, m a

= 25%

50.50.50.50.50. (c )(c )(c )(c )(c )

= 10

∴ V 0 = 10 V

now,

= Slew rate

−− = 5 V/ µsec

−−

×= t 2

t 2 = 2 µsec

51.51.51.51.51. (c )(c )(c )(c )(c )

Propagation vector

= +

= ( ) ×µω

=

+

−+ ×

π × × × x

=

+−

− + = + π × × × ×

=

+ − π π A/m


27/31




52.52.52.52.52. (a )(a )(a )(a )(a )

Redrawing the circuit

+ –

1/2 Ω1 Ω

V s1( )

2S

4S

+ –

6S

1 Ω

− −+ =

+ = 0

+ + +

=

+

V 1(s ) =

+

V 1(t ) = ( )

−−

53.53.53.53.53. (a )(a )(a )(a )(a )

The criterion for exactness is that

∂ ∂

=∂ ∂ x where M ( x, y )d x + N ( x, y )dy = 0.

In 1,

∂∂

= ∂

=∂ x x

In 2,

∂∂

=( )

∂=

∂+ x

In 3,

∂∂

=

∂− + ≠ + =

∂ x x

x x

In 4,

∂

∂ =

∂

− ≠ = ∂ x

54.54.54.54.54. (a )(a )(a )(a )(a )

Probability =

∞

∫ x x =

∞ −

∫ x

x

= x

∞−− − =

= 0.368


28/31




55.55.55.55.55. (c )(c )(c )(c )(c )

The matrix formed by the coefficients is

Determinant = 2a 2 – 2a – 4

∴ D = 0 for a = 2 or a = –1(A) If D ≠ 0, then the system will have unique solution.

(B) If a = 2, the matrix formed by the coefficients is

The rank of matrix is 2.

Considering ‘z ’ as side unknown.

The characteristic determinant will be

The determinant of this is 0.

The system will have infinite solutions when a = 2.

(C) If a = –1, the matrix formed by the coefficients is

− −

Its rank is 2.

Considering ‘z ’ as side unknown.

The characteristic matrix is

−

The determinant of this matrix is 3b .

The system will have no solution if b ≠ 0∴ For a = –1 and b ≠ 0, the system will have no solution.

56.56.56.56.56. (a )(a )(a )(a )(a )

ACL =

+

B =

Ω=

Ω + Ω

thus substituting the values in the above expression, we get

ACL

=

=

+ × = 4.95


29/31




57.57.57.57.57. (b)(b)(b)(b)(b)

Here, ωr

= 4 rad/sec = ω − ξ ; ξ ≤ 0.707 ...(1)

m r = 1.25 =

ξ − ξ...(2)

from equation (2)

4ξ2 (1 – ξ2) = 0.64= ξ2 – ξ4 = 0.16= ξ4 + 0.16 – ξ2 = 0

ξ2 = 0.8 and 0.2ξ = 0.894, 0.447

for ξ = 0.447ω

n = 5.162 rad/sec

∴ τr

=

−π − θ π −=

ω −

=

=

58.58.58.58.58. (b)(b)(b)(b)(b)

∵ C ′ =

−

−

ε × ×=

×= 51.77 µF/m2

59.59.59.59.59. (a )(a )(a )(a )(a )

γ = + ω + ω = α + β

= ( ) ( ) − −+ π × × × × + π × × × ×= 0.034 + j 38.94 (m –1)

now, β = 38.94

⇒ V p

=

ω π × ×= = ×

β

60.60.60.60.60. (a )(a )(a )(a )(a )

∵ x[n ] + αy [n – 1] = y [n ]taking z-transform

H (z ) =

−=

− α

∴ h [n ] = αn u [n ]thus the system is convergent only if α < 1.


30/31




61.61.61.61.61. (a )(a )(a )(a )(a )

h 12 =

= I

V x

V y

V 1

10 Ω0.2 I 2

I 2

0.3 V 2

V 2

+ –+

–

4 Ω I 1

+

–

V y = V 2V

x= V 1 – 0.3 V 2

V y – V

x= 1.3 V 2 – V 1

− x = 0.2 I 2

−+ = I 2 ⇒ –V 1 + 0.425 C 2 = I 2

−= 0.2 I 2 ⇒ 0.25V 1 – 0.325 V 2 = 0.2 I 2

V 1 = –1.87 I 2 V2 = –2.06 I 2

h 12 =

= I = 0.907

62.62.62.62.62. (b)(b)(b)(b)(b)

Given q e =

−−

∆= ≤ ×

∴ L ≥ 100Thus, the number of bits required ≈ 7Since, Nyquist sampling rate = 2 f m = 2 kHz

∴ The sampling rate for each signal = f s = 1.2 × 2000 = 2400 samples/sec

The transmission bandwidth =

×= = 8.4 kHz

63.63.63.63.63. (c )(c )(c )(c )(c )

Av =

+

now g m

R E

>> 1

thus Av

≈

=

=


31/31



64.64.64.64.64. (a )(a )(a )(a )(a )

Newton Raphson method for root finding is:

xn =

−

−−

−′

x x

x

OR xn +1

=

−

′ x

x x

65.65.65.65.65. (c )(c )(c )(c )(c )

∫ =

+ ∫ x x

For curve C , y 2 = 4 xand 2y dy = 4 d x

⇒

∫ = ( )

∫ x x x + x

= ( )

+ = ∫ x x

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