Practice Test: Electronics Engineering (ECE)- 5 - Electronics and Communication Engineering (ECE)


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65 Questions MCQ Test GATE ECE (Electronics) 2024 Mock Test Series - Practice Test: Electronics Engineering (ECE)- 5

Practice Test: Electronics Engineering (ECE)- 5 for Electronics and Communication Engineering (ECE) 2023 is part of GATE ECE (Electronics) 2024 Mock Test Series preparation. The Practice Test: Electronics Engineering (ECE)- 5 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Practice Test: Electronics Engineering (ECE)- 5 MCQs are made for Electronics and Communication Engineering (ECE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Electronics Engineering (ECE)- 5 below.
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Practice Test: Electronics Engineering (ECE)- 5 - Question 1

Choose the word or phrase which is nearest in meaning to the key word:

        Pilfer

Practice Test: Electronics Engineering (ECE)- 5 - Question 2

The following question comprises two words that have a certain relationship between them followed by four pairs of words. Select the pair that has same relationship as the original pair of words

            Fox : Cunning

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 2

Fox is cunning animal similarly
Ants are Industrious.
The second word shows the quality of the first.

Practice Test: Electronics Engineering (ECE)- 5 - Question 3

Replace the phrase printed in bold to make it grammatically correct ?

    A twenty-first century economy "cannot be held" hostage by power cuts nor travel on nineteenth century roads.

Practice Test: Electronics Engineering (ECE)- 5 - Question 4

Improve the sentence with suitable options by replacing the underlined word.

 He "lay" on the grass enjoying the sunshine.

Practice Test: Electronics Engineering (ECE)- 5 - Question 5

Fill in the blanks :

 At times, we are all ______ to be mistaken.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 5

At times, we are all likely to be mistaken.

Practice Test: Electronics Engineering (ECE)- 5 - Question 6

There are two solutions of wine and water, the concentration of wine being 0.4 and 0.6 respectively. If five liters of the first solution be mixed with fifteen liters of the second, find the concentration of wine in the resultant solution.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 6

Resultant concentration of wine

Practice Test: Electronics Engineering (ECE)- 5 - Question 7

Rajiv bought some apples at the rate of 25 for Rs. 400 and an equal number at the rate of 30 for Rs. 270. He then sold all at the rate of 10 for Rs. 180. Find his profit/loss percentage

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 7

Let Rajiv bought x apples of each kind.

Total cost price of Rajiv =

Total selling price of Rajiv =

SP > CP , Hence he makes profit;

 Profit percentage =

Practice Test: Electronics Engineering (ECE)- 5 - Question 8

Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is alteast one graduate among them ?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 8

P (at least one graduate) = 1 – P (no graduate)

Practice Test: Electronics Engineering (ECE)- 5 - Question 9

2 men and 3 boys can do a piece of work in 10 days, while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work ?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 9

Let work done by 1 man in 1 day = x units

Let work done by 1 boy in 1 day = y units

2 men and 1 boy can finish work

No. of days taken by 2 men and 1 boy = = 12.5 days

Practice Test: Electronics Engineering (ECE)- 5 - Question 10

In an exam, 60% of the candidates passed in Maths and 70% candidates passed in English and 10% of the candidates failed in both the subjects, 300 candidates passed in both the subjects. Find the total number of candidates appeared in the exam, if they took test in only two subjects that is Math and English.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 10

Let total number of students appear be X.

X = 750

Practice Test: Electronics Engineering (ECE)- 5 - Question 11

In the figure, the value of resistor R is (I 5) ohms, where I is the current in amperes. The value of current I is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 11

150 – IR = 0

            150 – I (I +5) = 0

            150 – I2 - 5I = 0

            I2 +5I – 150 = 0

I2 +15I – 10I - 150 = 0

I (I +15) – 10 (I +15) = 0

(I +15) (I-10) = 0

I = 10A & I = -15A

Practice Test: Electronics Engineering (ECE)- 5 - Question 12

The velocity of propagation of electromagnetic wave in an underground cable with relative permittivity of 9 is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 12

Practice Test: Electronics Engineering (ECE)- 5 - Question 13

The function shown in the figure can be represented as

Practice Test: Electronics Engineering (ECE)- 5 - Question 14

Consider an LTI system with impulse response h(t) = e-5t u(t). Determine the output of the system for the input x(t) = 5 u(t)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 14

y(t)      = h(t) * x(t)

                    = e-5t u(t) * 5u(t)

y(t)      = (1 - e-5t) u(t)

Practice Test: Electronics Engineering (ECE)- 5 - Question 15

The value of integral   is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 15

                   = e0 ×  0

                               = 0

Practice Test: Electronics Engineering (ECE)- 5 - Question 16

A rectangular pulse x(t) is shown in the figure. Determine the value of  

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 16

According to Parswell’s theorem

Practice Test: Electronics Engineering (ECE)- 5 - Question 17

If   and f(t) is     as t → ∞ . Determine the value of K

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 17

⇒K = 2

Practice Test: Electronics Engineering (ECE)- 5 - Question 18

In the signal flow graph shown in the given figure, the value of C/R ratio is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 18

There are three path

Practice Test: Electronics Engineering (ECE)- 5 - Question 19

What is the open-loop transfer function for an unity feedback system having root locus shown in the following figure ?

Practice Test: Electronics Engineering (ECE)- 5 - Question 20

The digital circuit shown in the figure works as a  

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 20

For D flip-flop

            Qn 1 = D

            Here, D = X Qn

        So Qn 1 = X Qn

     So it becomes T flip-flop

Practice Test: Electronics Engineering (ECE)- 5 - Question 21

The simplified form of Boolean function F(A,B,C) implemented by the given 3 × 8 decoder is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 21

Practice Test: Electronics Engineering (ECE)- 5 - Question 22

The contents of the accumulator in an 8085 microprocessor is altered after the execution of the instruction

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 22

Compare operation does not affect the accumulator and ORAA does not change the content of accumulator.

Practice Test: Electronics Engineering (ECE)- 5 - Question 23

Assuming that all the diodes are ideal in figure, the current in the diode D2 is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 23

Assume both the diodes are ON

So, I1 = -1A

But current can not flow from n to p side in a diode so D1 will be off and equivalent circuit becomes

      Therefore,

Practice Test: Electronics Engineering (ECE)- 5 - Question 24

Determine the upper and lower threshold voltage of the schmitt trigger shown below :

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 24

Practice Test: Electronics Engineering (ECE)- 5 - Question 25

The transistor in the given circuit, should always be in saturation region. Take VCE (sat) = 0V and VBE = 0.7V. the minimum value of RC is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 25

Practice Test: Electronics Engineering (ECE)- 5 - Question 26

A differential amplifier has input at non inverting terminal is 1050 μV and at inverting terminal 950 μV. What is the error in the differential output if CMRR is 1000.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 26

V0 = AdVd AcmVc

Practice Test: Electronics Engineering (ECE)- 5 - Question 27

Which one of the following type of negative feedback increases the input resistance and decreases the output resistance of an amplifier ?

Practice Test: Electronics Engineering (ECE)- 5 - Question 28

The spectral density and auto correlation function of white noise are respectively.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 28

Auto corelation function and PSD are fourier transform pair

            Here, PSD = N0

Practice Test: Electronics Engineering (ECE)- 5 - Question 29

For a random variable x having the probability density function is shown in the figure, what are the variance of random variable.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 29

=  4/3

Practice Test: Electronics Engineering (ECE)- 5 - Question 30

An amplitude modulated signal is given as, S(t) = 100 Cos 2pfct 50 Cos 2pfmt.Cos 2pfct, where f fm. and fc is the carrier frequency and fm is the modulating signal frequency. The power efficiency is.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 30

S(t) = 100 Cos 2pfct 50 Cos 2pfmt.Cos 2pfct

                  = 100 [1 0.5 Cos 2pfmt] Cos 2pfct

            Here modulation index, μ = 0.5

      = 11.11%

Practice Test: Electronics Engineering (ECE)- 5 - Question 31

Signal x(t) = 4Sinc2 100t. Sin 500pt .At what sampling frequency should this signal be sampled to avoid aliasing ?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 31

So, fm = f1 f2

                       = 350 Hz

            So, fs  = 2fm

                      = 2 × 350

                      = 700 Hz

Practice Test: Electronics Engineering (ECE)- 5 - Question 32

The value of   is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 32

Practice Test: Electronics Engineering (ECE)- 5 - Question 33

The solution of differential equation dy = (1-y) dx is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 33

dy = (1 – y) dx

y = 1 ce-x

Practice Test: Electronics Engineering (ECE)- 5 - Question 34

The rank of the matrix A is 2, when K is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 34

 = 2 {3k} - 3{2k} 4{2´15 - 3´16}

                  = 0 {for any value of k}

Practice Test: Electronics Engineering (ECE)- 5 - Question 35

Given    If C is a counter-clockwise path in the z-plane such that |z - i| = 1, the value of  is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 35

Poles are at z = -1, -2

There is no any pole inside the circle, so integration will be zero

Practice Test: Electronics Engineering (ECE)- 5 - Question 36

In the circuit shown in figure, determine the maximum power absorbed by the load resistance RL

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 36

Rth  = 4 ohms

            For Vth

 

 

 

                V2 – V3 V2 – V1 12 = 0

            But V1 = 2V

So,       2 V2 – V3 – V1 12 = 0

            2 V2 – V3 – 2 12 = 0

            2 V2 – V= -10…………....2

            From equation 1 and 2

            V3 – V= 12………………..1

            -V3 2V= -10…………….2

            V2 = 2 Volt

            So V3 = 14 Volt

            So Vth = 14V

Practice Test: Electronics Engineering (ECE)- 5 - Question 37

In the circuit shown below, the switch is closed at t = 0. What is the initial value of the current through the capacitor C.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 37

At t = 0-, the equivalent circuit is shown below :

 

At t = 0 , the equivalent circuit

So, ic = 4 – 3.2 = 0.8A

Practice Test: Electronics Engineering (ECE)- 5 - Question 38

Determine the z-parameter for the network shown below.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 38

Since it is symmetrical ladder network

Practice Test: Electronics Engineering (ECE)- 5 - Question 39

At which frequency (in rad/sec) the thevenins equivalent impedance looking at terminals AB becomes purely resistive?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 39

From the circuit it is clear, that ZAB becomes real when circuit is in resonance and resonance frequency of this circuit is given as

   = 1000 rad/sec

Practice Test: Electronics Engineering (ECE)- 5 - Question 40

Consider  if  y(t) = Ae-t  for    then value of A is

Practice Test: Electronics Engineering (ECE)- 5 - Question 41

The 4-point DFT of sequence x[n] = [1,2,2,1] is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 41

Practice Test: Electronics Engineering (ECE)- 5 - Question 42

The output y(t) of a continuous-time system for the input x(t) is given by . Which one of the following is correct ?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 42

It is an integrator that generally an unstable system because integrator produces an unbounded output and for causal

            Let at t = 2

It means y(2) is depends on x(4) that means present output is depends on future input, so it is non-causal

 

Practice Test: Electronics Engineering (ECE)- 5 - Question 43

For the NMOS and PMOS shown in the figure are matched and   = 1mA/V2  and V n = -V p = 1 V. Determine V0

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 43

For the circuit, it is clear that since QN and QP are perfectly matched and are operating at |VGS| equal values of  i.e. are 2.5V, the symmetrical which dictates that V0 = 0V

Practice Test: Electronics Engineering (ECE)- 5 - Question 44

Determine the region of operation of BJT in the given circuit.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 44

Since, VEB = 0.7

            Here VB = 0, So, VE = 0.7 V

  So, VC = -10 4.6 × 1

                          = - 5.4 V

            Since, VEC = 0.7 5.4 = 6.1V

            That is greater than 0.2V So BJT is in active region.

Practice Test: Electronics Engineering (ECE)- 5 - Question 45

In the silicon BJT circuit shown below. Assume the emitter area of transistor Q2 is two time of area of Q1. The value of Io { Let b of the transistors are very large}

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 45

      = 2mA

            Since β is very large,

            So, Iref = Ic1

i.e. Ic1 = 2mA

            Also the emitter current area of transistor Q2 is times of Q1

            So, Ic2 = 2 Ic1

                       = 2 × 2 mA

                       = 4 mA

            So, I0 = +4 mA

Practice Test: Electronics Engineering (ECE)- 5 - Question 46

Determine the current io in the given circuit.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 46

Circuit can be redrawn

Practice Test: Electronics Engineering (ECE)- 5 - Question 47

Consider a bar of silicon in which a hole concentration profile describe by P(x) = P0.e-x/Lp. Find the current at x = 0, given P0 = 1016/cm3, Lp = 1 mm and cross-sectional area of the bar is 100 mm2 and Dp = 12 cm2/sec.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 47

   = 192 A/cm2

            So the current Ip,

            Ip = Jp × A

               = 192 × 10-8

               = 192 mA

Practice Test: Electronics Engineering (ECE)- 5 - Question 48

A system with the open loop transfer function   is connected in a negative feedback gain of unity. What is the value of K for which the system is unstable.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 48

            The characteristic equation

            S (S3 8S2 24S 32) K = 0

            Routh – Hurwitz array

            S4                1             24        K

            S3                8             32        O

            S2                20           K

           S' (20×32-8K) 0

            S0                K

            So for stable system,

            K > 0

            and 20×32 - 8K > 0

            K < 80

            So for unstable system    K > 80

            i.e. K = 90

Practice Test: Electronics Engineering (ECE)- 5 - Question 49

In the system shown below x(t) = e-2t u(t). Determine the output y(t) in steady state.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 49

Here, x(t) = e-2t u(t)

Practice Test: Electronics Engineering (ECE)- 5 - Question 50

The polar diagram of a conditionally stable system for open loop gain K = 1 is shown in the figure. The open loop transfer function of the system is known to be stable. The closed loop system is stable for 

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 50

System is stable in the region 0.1 to 5 on the left side of 10 as number of enrichment there is zero

            0.1K < 1 ⇒ K < 10

            5K > 1    ⇒ K > 0.2

            0.2 < K < 10

            Also 1 > 10 K

                    K < 0.1

            So for the stable system 0.2 < K < 10 and K < 0.1

Practice Test: Electronics Engineering (ECE)- 5 - Question 51

The sensitivity of transfer function  with respect to the parameter K is -0.5, Determine the value of K

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 51

8K2 10K 3 = -4K

            8K2 14K 3 = 0

            4K (2K 3) 1 (2K 3) = 0

                 (2K 3) (4K 1) = 0

            K = -1.5 and K = -0.25

Practice Test: Electronics Engineering (ECE)- 5 - Question 52

Given two binary FSK signals    i = 1, 2.The minimum frequency separation of these two signals such that they are orthogonal is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 52

For the two signals to be orthogonal

Practice Test: Electronics Engineering (ECE)- 5 - Question 53

A block code is transmitted in NRZ invent-one-one format with voltage levels

            6V → ‘1’

            -6V → ‘0’

Three code words are sent : 10010, 11110, 01001. Over these three transmitted code words the average voltage is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 53

There are total 15 digits so, In which 8 are 1’s and 7 are 0’s

So average value = = 0.4V

Practice Test: Electronics Engineering (ECE)- 5 - Question 54

A signal is given by,

            γ(t) = s(t) n(t)

            where s(t) = 20 Cos2π4000t 40 Cos2π2000t and the noise n(t) is white with power spectral density η0 = 1 W/Hz. The total received signal is put through a band pass filter with pass band between 100 to 110 Hz. The value of SNR at the output of filter is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 54

Signal power

    = 200 800

                = 1000

            and noise power, Pn = n0 × (BW)

            = 1 × 10

            = 10W

            So, SNR = 1000/10 = 100

            In dB = 10log100 = 20dB

Practice Test: Electronics Engineering (ECE)- 5 - Question 55

The base band signal m(t) is recovered from the DSB-Sc signal s(t) = m(t). Cos2πfct, by multiplying s(t) by the locally generated carrier Cos (2πfct φ). The product is passed through a LPF which rejects the double frequency signal. If the base band signal is band limited to 4KHz, what is the minimum value of fc for which m(t) can be recovered by filtering ?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 55

In order to recover m(t) from s(t) by filter method it is necessary that the lowest frequency contained in the first term of s(t) must be greater than the highest frequency contained in the second term that is

So minimum value of fc = 4 KHz

 

Practice Test: Electronics Engineering (ECE)- 5 - Question 56

A plane wave in free space  is incident normally on a large block of material  . which occupies z > 0. If the incident electric field is = 10 Cos (ωt z)     V/m, the reflected magnetic field is :

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 56

Practice Test: Electronics Engineering (ECE)- 5 - Question 57

A plane wave with    is incident normally on a thick plane conductor lying in the X-Y plane. Its conductivity is 6 × 106 s/m and surface impedance is 5 × 10-4

?.  The skin depth of conductor is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 57

Practice Test: Electronics Engineering (ECE)- 5 - Question 58

The volume of the paralleopipe whose sides are given by  is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 58

Volume of paralleopipe

2 {-1} 2 {-1 3} 0 = -2 4 = 2

Practice Test: Electronics Engineering (ECE)- 5 - Question 59

A number in 4 bit 2 5 complement representation is 1101. This number when stored using 8-bits will beA number in 4 bit 2’5 complement representation is 1101. This number when stored using 8-bits will be

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 59

The number in 4 bit 1101

So original number, 0011

            Now the original number in 8-bit,

00000011

            So, in 2s complement, 11111101

Practice Test: Electronics Engineering (ECE)- 5 - Question 60

Four variable K-map is shown in the figure. The essential prime implicants are

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 60

Practice Test: Electronics Engineering (ECE)- 5 - Question 61

An 8085 microprocessor is interfaced to a 8KB RAM as shown below. The address range of the RAM is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 61

Range of Address

 The RAM will be active when   is 0

 when, A13 A14 A15 = 011 as decoder is active low.

So the range of RAM address is from COOOH to DFFFH

Practice Test: Electronics Engineering (ECE)- 5 - Question 62

If z = x – iy and   = p iq, then     is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 62

= P iq

Z = P3 (iq)3 3P (iq) (P iq)

            x – iy = P3 – 3Pq2 i (3P2q – q3)

             X = P3 – 3Pq2

 

           

Practice Test: Electronics Engineering (ECE)- 5 - Question 63

The Newton – Raphson method is used to find the roots of the equation f(x) = x - cosπx   If the initial guess for the root is 0.5, then the value of x after the first iteration is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 63

f(x)    = X - Cosπx

f(0.5) = 0.5 – Cos 0.5π

               = 0.5

             = 0.38

Practice Test: Electronics Engineering (ECE)- 5 - Question 64

The eign value of the matrix     are,

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 64

Practice Test: Electronics Engineering (ECE)- 5 - Question 65

The value of     is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 65

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