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BDL MT Electronics Mock Test - 1 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test BDL MT Electronics Mock Test Series 2025 - BDL MT Electronics Mock Test - 1

BDL MT Electronics Mock Test - 1 for Electronics and Communication Engineering (ECE) 2025 is part of BDL MT Electronics Mock Test Series 2025 preparation. The BDL MT Electronics Mock Test - 1 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The BDL MT Electronics Mock Test - 1 MCQs are made for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BDL MT Electronics Mock Test - 1 below.
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BDL MT Electronics Mock Test - 1 - Question 1

A RL high-pass filter can be represented by the circuit where R is the internal resistance of the inductor

The frequency at which is _____Hz

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 1

From the voltage division rule

BDL MT Electronics Mock Test - 1 - Question 2

Three resistances of 1 ohm, 2 ohms and 3 ohms are connected in delta. These resistances are to replace by star connection as shown in the figure below, maintaining the same terminal conditions. The value of the highest resistance in the star will be

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 2

To convert delta to star,

BDL MT Electronics Mock Test - 1 - Question 3

______ is the maximum reverse voltage that can be applied to the pn junction ______ to the junction.

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 3

Peak inverse voltage:
The peak inverse voltage rating of a diode may be defined as the maximum value of the reverse voltage that a PN-junction (or diode) can withstand without damaging (or destruction).
When the voltage applied to a diode is more than PIV, it is likely to result in breakdown at the junction.Important PointsPIV for different rectifiers is shown below:

  • Half Wave rectifier: Vm
  • Full Wave centre tap rectifier: 2Vm
  • Full Wave Bridge rectifier: Vm
BDL MT Electronics Mock Test - 1 - Question 4

A transistor connected in a common base configuration has the following readings IE = 2 mA and IB = 20 μA. Find the current gain α.

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 4

Current amplification factor: It is defined as the ratio of the output current to the input current. In the common-base configuration, the output current is emitter current IC, whereas the input current is base current IE.
Thus, the ratio of change in collector current to the change in the emitter current is known as the current amplification factor. It is expressed by the α.

Where, IE = IC + IB
Calculation:
Given,
IE = 2 mA
IB = 20 μA = 0.02 mA
From above concept,
IC = 2 mA - 0.02 mA = 1.98 mA
Current amplification factor is given as,
α = IC/IE = 1.98/2 = 0.99

BDL MT Electronics Mock Test - 1 - Question 5

The open loop transfer function of a unity feedback system is:

In Nyquist plot, the number of encirclements of origin is:
(Take Nyquist contour in the clockwise direction)

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 5

Concept:
Number of encirclements of origin N = P – Z
Where P = open loop RHS poles
Z = open loop RHS zeros
N = no. of encirclements in counter clockwise direction

Analysis:
Since, no open loop pole lies in the RHP, P = 0
∴ N = 0 – 1 = -1
∴ 1 encirclements in clockwise direction.

BDL MT Electronics Mock Test - 1 - Question 6

In root-locus plot, the breakaway points

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 6

Concept:

1. Every branch of a root locus diagram starts at a pole (K = 0) and terminates at a zero (K = ∞) of the open-loop transfer function.
2. The root locus diagram is symmetrical with respect to the real axis.
3. A number of branches of the root locus diagram are:
N = P if P ≥ Z
= Z, if P ≤ Z
4. Number of asymptotes in a root locus diagram = |P – Z|
5. Centroid: It is the intersection of the asymptotes and always lies on the real axis. It is denoted by σ.

ΣPi is the sum of real parts of finite poles of G(s)H(s)
ΣZi is the sum of real parts of finite zeros of G(s)H(s)

6. The angle of asymptotes:
l = 0, 1, 2, … |P – Z| – 1
7. On the real axis to the right side of any section, if the sum of the total number of poles and zeros are odd, the root locus diagram exists in that section.
8. Break-in/away points: These exist when there are multiple roots on the root locus diagram.
At the breakpoints gain K is either maximum and/or minimum.
So, the roots of are the breakpoints.
Observation:
n root-locus plot, the breakaway points, must lie on the root loci.

BDL MT Electronics Mock Test - 1 - Question 7

A signal x(t) = sin [10 πt] is applied to the system whose output is y(t) = sgn [x(t)]. Then y(t) contains _____ harmonics.

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 7

Concept :

Application:

The signal y(t) satisfies:
y(t) = - y(t ± T/2) so it follows half wave odd symmetry also y(t) is odd signal.
∴ The Fourier Series of y(t) will have sine terms, odd harmonics only.

BDL MT Electronics Mock Test - 1 - Question 8

What is the octal equivalent of Hexadecimal number CD.AB?

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 8

The given hexadecimal number is: CD.AB
To convert the given Hexadecimal number, first, we need to convert it in to binary and then into octal.
Hexadecimal to binary: CD.AB = 1100 1101. 1010 1011
Binary to octal: To convert a binary number into octal, we need to make a group of three digits from right to left before the decimal pointer and left to right after the decimal pointer.
= 011 001 101. 101 010 110
= 315.526

BDL MT Electronics Mock Test - 1 - Question 9

For a n-variable Boolean logic expression, the minimum and maximum possible number of essential prime implicants are ______ and ______ respectively.

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 9

In a n-variable Boolean function,

  • The minimum number of prime implicants are 1
  • The minimum number of essential prime implicants are 0
  • The maximum number of prime implicants are 2n-1
  • The maximum number of essential prime implicants are 2n-1

Example:
Consider the following 4-variable K-map,

In the above K-map representation, all the 1’s are part of more than one group. Therefore, there are no essential prime implicants.
Consider the following 4-variable K-map,

In the above K-map representation, the maximum number of essential prime applicants = maximum number of prime applicants = 2n-1 = 24 -1 = 8

BDL MT Electronics Mock Test - 1 - Question 10

Which among the following statements are true with respect to semiconductor breakdown?

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 10
  • The avalanche breakdown is a phenomenon in which there is an increase in the number of free electrons beyond the rated capacity of the diode; This results in the flow of heavy current through the diode in reverse biased condition
  • Avalanche breakdown occurs in lightly doped diode
  • Zener breakdown mainly occurs because of a high electric field; When the high electric field is applied across the PN junction diode, then the electrons start flowing across the PN-junction; Consequently, develops little current in the reverse bias
  • The Zener breakdown occurs in heavily doped diodes

Important difference between Avalanche and Zener breakdown:

BDL MT Electronics Mock Test - 1 - Question 11
In an n-type semiconductor, as the donor concentration ND increases, the Fermi level EF:
Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 11

For a p-type semiconductor, the Fermi level lies close to the valence band and, as the p-type doping concentration increases the Fermi level shift moves more closer to the valence band indicating stronger doping of p-type impurities.

Other Related Points

Fermi level versus Temperature for constant doping:

  • At T= 0 K all donor levels are occupied. We know that at T = 0 K, Fermi level is that energy above which all the energy states are unoccupied and below which all are occupied.
  • If the Fermi level lies below Ed then that means all the donor levels are unoccupied which is not true.
  • Again, if EF lies above Ec then that means all the conduction band is occupied which is not true because at T=0 K there are no free electrons in the conduction band.
  • EF should lie between Ec and Ed.
  • At very high-temperature holes and electrons increase and the semiconductor starts to behave like an intrinsic semiconductor so Fermi-level coincides with the intrinsic level.
BDL MT Electronics Mock Test - 1 - Question 12

Assuming the OP-AMP to be ideal. the voltage gain of the amplifier shown below is

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 12

Concept:

If Op-Amp is ideal then the virtual ground concept will be applicable.

Analysis:

from the virtual ground,

VA = 0V

applying KCL at node A

VA = 0

BDL MT Electronics Mock Test - 1 - Question 13

Match List-I (Scheme of Feedback) with List-II (Performance Measure) and select the correct answer using the code given below:

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 13

There are 4 possible combinations for Voltage and Current with which we can sample at the output and mix the feedback to the input.
Sampling:

  • At the output side, we will take a sample of output since we want to check the behavior of output and we don't want to disturb the output when we take the sample.
  • That's why when the voltage is sampled, it is in parallel (as the voltage is the same in parallel) and current in series (as the current is the same in series).

Mixing:

  • In mixing end we want to affect the signal that is provided to the amplifier since that is the actual fundament of giving the feedback.
  • So, the voltage will be in series and current will be in parallel. So that they can change the input and effect the change in output.

The four basic feedback topologies are as shown:

i) Voltage-sampling voltage-mixing (series-shunt) topology.
ii) current-sampling voltage-mixing (series-series) topology.
iii) current-sampling current-mixing (shunt-series) topology.
iv) voltage-sampling current-mixing (shunt-shunt) topology.
Imortant Point

BDL MT Electronics Mock Test - 1 - Question 14

The electric potential at a point in free space due to a charge Q coulomb is Q × 1011 volts. The electric field at that point is -

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 14

CONCEPT:

  • Electric Potential (V): This is the amount of work needed to move a unit of electric charge from a reference point to a specific point in an electric field without producing an acceleration.

V = q/(4π ϵor)
q = charge , ϵo = perimitivity of free space.

  • Electric Field (E): An Electric charge produces an electric field, which is a region of space around an electrically charged particle or object in which an electric charge would feel forced.

it is given by
E = q/(4π ϵo r2)
q = charge, ϵo = permittivity of space, r is the distance from the charge

CALCULATION:

Given:
The electric potential at a point due to a charge Q is coulomb is Q × 1011.
V = Q/(4π ϵor) = Q × 1011
So r = 1/4πϵo × 1011
Now, the Electric field at that point is.
E = Q/ 4π ϵor2 = (Q (4πϵo × 1011)2)/4πϵo = 4π ϵoQ × 1022 Vm-1 ..
option 3 is correct.

BDL MT Electronics Mock Test - 1 - Question 15

For a dielectric-conductor interface where subscript 1 denotes the dielectric, the boundary condition that is NOT satisfied for static fields is

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 15

Analysis:

A conductor placed in a static field will exhibit or induce a charge on its surface so as to have zero fields inside it. i.e. for an ideal conductor, the electric & magnetic field inside it is zero.

The given situation can be visualized as shown:

A conductor dielectric interface satisfies the following boundary conditions:
H1t – H2t = Js
B1n = B2nE1t = E2t = 0
D1n – D2n = ρs
For a conductor: H2t = E2t = B2n = D2n = 0
So, H1t = Js
B1n = B2n = 0
E1t = E2t = 0
D1n = ρs

BDL MT Electronics Mock Test - 1 - Question 16

What is the value of the major cross-sectional dimension(width) of a rectangular waveguide with dominant TE10 mode propagation, if its cut off frequency is 10 GHz?

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 16

Concept:
The dominant mode in a particular waveguide is the mode having the lowest cut-off frequency.\
The cut-off frequency for a rectangular waveguide with dimension ‘a (length)’ and ‘b (width)’ is given as:

'm' and 'n' represents the possible modes.
c = speed of light = 3 × 1010 cm/s
Calculation:
TE10 mode means m = 1, n = 0
The cut - off frequency of the dominant mode TE10 of the rectangular waveguide is:
fc = c/2a
Where a is the dimension of the inner broad wall
a = 3 x 1010/2 x 10 x 109
a =  1.5cm
= 15 mm

BDL MT Electronics Mock Test - 1 - Question 17

Which modulator is used for the generation of the DSB-SC signal?

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 17

In a balanced modulator, 2 AM-modulators are connected in a way that the resultant signal does not contain the carrier spectrum, i.e. to generate a Double Side-band suppressed carrier (DSB-SC).
The circuit diagram of a balanced modulator is as shown:


sAM'(t) = Ac [1 + ka m(t)] cos 2πfct
sAM''(t) = Ac [1 - ka m(t)] cos 2πfct
The resultant DSB signal is the difference between the two, i.e.
sDSB(t) = sAM'(t) - sAM''(t)
Hence,

Important Point
1) Square law modulator
is used for the generation of conventional AM signal which includes the carrier frequency component as well.
2) An Armstrong modulator is used to generate an FM signal.
3) Envelop detector is used for AM demodulation.

BDL MT Electronics Mock Test - 1 - Question 18

If the transmission bandwidth is doubled in FM, then the figure of merit (FOM) is

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 18

Figure of merit (FOM):
A figure of merit is used to decide the quality of a communication technique
For Angle modulation:
FOM of FM is given by,

Thus, as transmission bandwidth W is doubled, FOM becomes one fourth.
Importan Point
For Amplitude modulation:
FOM = μ22 + 2
The signal to noise ratio at output Is a maximum 1/3 (for μ = 1) of the signal to noise ratio at the input.
FOM is 1 for DSB, SSB, VSB.
Phase Modulation:
FOM = 1/2β2

BDL MT Electronics Mock Test - 1 - Question 19

The following program is running on an 8085 microprocessor

After the completion of execution of the program the SP contains-

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 19

Explanation

  • LXI SP, d16 Load the immediate 16-bit data to the Stack pointer, hence SP will point to the 16-bit address given in the instruction
  • PUSH → Data of the register pair will be pushed to the stack and the stack pointer will decrement by 2
  • CALL XXXX → After the execution of CALL instruction the microprocessor will execute the subroutine at the location associated with the instruction (XXXX) and then it will transfer back control to the main program, address located by PC after encountering RET instruction in the subroutine. In this whole process the net value of SP remains unaltered.
  • POP → Top of the stack will be poped and SP will be incremented by 2.

Analysis:

Hence, The answer is 6FFE H.

BDL MT Electronics Mock Test - 1 - Question 20

General solution of the Cauchy-Euler equation  x² (d²y/dx²) - 7x (dy/dx) + 16y = 0 is

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 20

Concept:
For different roots of the auxiliary equation, the solution (complementary function) of the differential equation is as shown below.

Calculation:
Given:
x² (d²y/dx²) - 7x (dy/dx) + 16y = 0
Put x = et
⇒ t = ln x
dy/dx = dy/dt ⋅ dt/dx = dy/dt ⋅ (1/x) ⇒ x dy/dx = dy/dt = Dy
x² d²y/dx² = D(D - 1)y; x³ d³y/dx³ = D(D - 1)(D - 2)y
Now, the above differential equation becomes
D(D – 1)y – 7Dy + 16y = 0
⇒ D2y – Dy – 7Dy + 16y = 0
⇒ (D2 – 8D + 16)y = 0
Auxiliary equation:
(D2 – 8 D + 16) = 0
⇒ D = 4
The solutions for the above roots of auxiliary equations are:
y(t) = (c1 + c2 t) e4t
⇒ y(x) = (c1 + c2 ln x) x4

BDL MT Electronics Mock Test - 1 - Question 21
The diagonal of the rectangle is 17 cm and the length is 15 cm. find the area of the rectangle.
Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 21

Given:

Diagonal of rectangle = 17 cm

Length of the rectangle = 15 cm

Formula:

Diagonal2 = Length2 + Breadth2

Area of the rectangle = Length × Breadth

Calculation:

According to the formula

172 = 152 + Breadth2

⇒ Breadth2 = 289 – 225

⇒ Breadth = √64

⇒ Breadth = 8

∴ Area of the rectangle = 15 × 8 = 120 cm2

BDL MT Electronics Mock Test - 1 - Question 22

Directions: In the following question, some part of the sentence may have errors. Find out which part of the sentence has an error and select the appropriate option. If a sentence is free from error, select 'No Error'.

Neither Rohit nor his friend/(A) was doing their/(B) work properly./(C) No Error(D)
Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 22

The correct answer is B.

Explanation

  • When the subjects connected by 'or' or 'nor' are of different persons, the verb agrees with the noun that comes closer to it.
    • Ex: Neither you nor he is responsible for this. (Here the verb 'is' agrees with the third person pronoun he.)
    • Ex: Either he or you are to clean up the mess. (Here the verb 'are' agrees with the second person pronoun you.)
  • Here, the pronoun and verb come according to the closer subject His Friend.
  • We need to replace the pronoun 'their' with 'his'.
  • Hence, option 2 is the correct answer.

Thus the correct sentence is: Neither Rohit nor his friend was doing his work properly.

Other Related Points

  • When the pronouns 'Each, every, neither, either, anyone, many a, more than one(possessive adjective)' are used as the subject, the possessive case should be third person singular. They may refer to two or more than two objects or persons.
    • Each one of us is doing our duty properly. (Use 'his' in place of 'our')
    • Everyone should do one's duty. (Use 'his' in place of 'one's')
BDL MT Electronics Mock Test - 1 - Question 23

Which of the following have been mentioned as disadvantages of dams?

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 23

The correct answer is 'Displacement of people on a large scale and ecological imbalance due to restrictions on the natural flow of rivers'.Explanation

  • Let's refer to the following lines of the passage:
    • The diversion of rivers and construction of a long system of canals in a densely populated country like India will involve displacement of people on a colossal scale and the people affected are never likely to agree with such measures.
    • Preserving rivers in free-flow condition is considered ecologically necessary and the construction of large dams is now legally prohibited in Sweden and also in parts of USA.
  • Thus, from above, we can refer that the correct answer is option 4.
BDL MT Electronics Mock Test - 1 - Question 24

Which of the following statements is true?

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 24

The correct answer is 'The diversion of rivers will involve displacement of people on a colossal scale. 'Explanation

  • Let's refer to the following lines of the passage:
    • There has however been hardly any attempt at questioning the extent of damage caused or in evaluating whether the promises of food, water and prosperity for all have actually been realised.
    • The diversion of rivers and construction of a long system of canals in a densely populated country like India will involve displacement of people on a colossal scale and the people affected are never likely to agree with such measures.
  • Thus, From above, we can refer that the correct answer is option 4.
BDL MT Electronics Mock Test - 1 - Question 25

Which word is similar in meaning to displacement?

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 25

The correct answer is 'Dislocation'.Explanation

  • Let's explore the meaning of the marked word and the given word.
    • Displacement: the moving of something from its place or position.
      • Example: The recent famine in these parts has caused the displacement of tens of thousands of people.
    • Dislocation: to put out of place; put out of proper relative position; displace.
      • Example: The earthquake dislocated several buildings.
  • From above, we can refer that the correct answer is option 2.

Other Related Points

  • Inactivity: idle or inert; not active.
  • Stagnation: the state of not flowing or moving.
  • Inactive: not engaging in or involving any or much physical activity.
BDL MT Electronics Mock Test - 1 - Question 26

Direction: Choose the correct alternative which can be substituted for the below given sentence.
One who loves himself

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 26

The correct answer is 'Narcissist'Explanation

  • The word is narcissist a noun and its meaning is "someone in love with themselves"
  • So this will be the most appropriate option to replace the given sentence.
  • Thus the correct answer is option 1

Other Related Points

  • The meaning of the other given options is:
BDL MT Electronics Mock Test - 1 - Question 27

Direction: Select the segment of the sentence that contains the grammatical error. If there is no error, mark 'No error' as your answer.
Besides Ajay, (A)/ many other (B)/ were present. (C)/ No error (D)

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 27

The correct answer is B.Explanation

  • According to subject-verb agreement, a singular subject takes a singular verb and a plural subject takes a plural verb.
    • Example: Ram goes to the temple every day.
    • Example: We have decided to start a new business.
  • In the given sentence, the adjective 'many' and the verb 'were' are plural in meaning and thus, the noun should also be plural.
  • The word 'other' is not plural as it means 'the second of two people or things when the first has already been mentioned'.
  • So, 'other' should be replaced with 'others' in part B of the sentence.

Correct sentence: Besides Ajay, many others were present.
Other Related Points

  • The word 'besides' is a preposition that means 'in addition to or as well as somebody/something'.
    • Example: Besides being a banker, she was a famous writer.
BDL MT Electronics Mock Test - 1 - Question 28

Select the option that is related to the third word in the same way as the second word is related to the first word.

Boom : Collapse : : Profit : ?

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 28

The logic follows here is:

Boom is the antonyms of Collapse.

Similarly,

Profit is the antonyms of Loss

Hence, the correct answer is "loss.

BDL MT Electronics Mock Test - 1 - Question 29

In the following question below some statements are given followed by some conclusions. Taking the given statements to be true even if they seem to be at variance from commonly known facts, read all the conclusions, and then decide which of the given conclusions logically follows the given statements.
Statements:
Some A are B.
All B are C.
No A are D.
Conclusions:
I. Some A are C.
II. Some B are D.
III. Some D are A.
IV. Some C are not D.

Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 29

The least possible Venn diagram is:

Conclusions:
I. Some A are C - True (As some A are B and All B are C, Some A are C is definitely correct.)
II. Some B are D - False (There is no definite relation between B and D)
III. Some D are A - False (There is a "No" type of relation between D and A so Some type of relation is always invalid.)
IV. Some C are not D -True (As No A is D and some A are B and All B are C, Some C are not D is also true.)
Hence, Only I and IV follow.

BDL MT Electronics Mock Test - 1 - Question 30
The sum of the ages of 4 friends L, M, N and O is 60 years. What would be the total of their ages 10 years later?
Detailed Solution for BDL MT Electronics Mock Test - 1 - Question 30

The sum of the ages of 4 friends L, M, N, and O is 60 years.

L + M + N + O = 60 years

The total of 4 friends ages after 10 years will increase by 40 years, as ⇒ 10 × 4 = 40.

So, total age of 4 friends after 10 years = (60 + 40) years

100 years

Hence, the correct answer is "100".

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