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Gate Practice Test: Electrical Engineering(EE)- 10 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test - Gate Practice Test: Electrical Engineering(EE)- 10

Gate Practice Test: Electrical Engineering(EE)- 10 for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Gate Practice Test: Electrical Engineering(EE)- 10 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Gate Practice Test: Electrical Engineering(EE)- 10 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Gate Practice Test: Electrical Engineering(EE)- 10 below.
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Gate Practice Test: Electrical Engineering(EE)- 10 - Question 1

“Nobody praised her ________ nature, so she decided to mend her ways to become ________.”

The words that best fill the blanks in the above sentence are

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 1

complaisant’- willing to please or oblige others
complacent’- self-satisfied, unconcerned
Thus, the sequence in option 3 must be used to make the sentence contextually sensible.

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 2

“All the development programmes of the government have been ________ over the years by the perpetual rise in the population of the country.”

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 2

‘vitiate’- to make faulty or defective, hurt
‘ratify’- to approve and sanction formally
The sentence implies that the ever-increasing population of the country is posing a hindrance to all the developmental programmes initiated by the government.
Hence, option 1 is the correct answer.

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Gate Practice Test: Electrical Engineering(EE)- 10 - Question 3

D = ABC + BCA + CAB where A, B and C are decimal digits, then D is divisible by

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 3

D = ABC + BCA + CAB
= (100 A + 10 B + C) + (100 B + 10 C + A) + (100 C + 10 A + B)
= 111 (A + B + C)
So, D is a multiple of 111.
111 is divisible by 37 but not divisible by 29. Thus, D is divisible by 37 but not divisible by 29.

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 4

In an examination 100 questions of 1 mark each are given. After the examination, 20 questions are deleted from evaluation, leaving 80 questions with a total of 100 marks. Student A had answered 4 of the deleted questions correctly and got 40 marks, whereas student B had answered 10 of the deleted questions correctly and got 35 marks. In this situation,

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 4

In an examination 100 questions of 1 mark each are given.
After the examination, 20 questions are deleted from evaluation.
Now, total number of questions considered for evaluation = 80
Marks for each correct question = 100/80 = 1.25

Student A:
Total marks secured = 40
Number of questions answered correctly = 40/1.25 = 32
Student A had answered 4 of the deleted questions correctly
Number of questions answered correctly before the deletion of 20 questions = 32 + 4 = 36
Marks as per 100 question paper = 36

Student B:
Total marks secured = 35
Number of questions answered correctly = 35/1.25 = 28
Student A had answered 10 of the deleted questions correctly
Number of questions answered correctly before the deletion of 20 questions = 28 + 10 = 38
Marks as per 100 question paper = 38


Conclusion: B lost more than A.

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 5

Two solutions of X and Y containing ingredients A, B and C in proportions a : b : c and c : b: a, respectively are mixed. For the resultant mixture to have A, B and C in equal proportion. It is necessary that

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 5

The ratio of ingredients A, B and C = a : b : c
Let x be the factor.
The quantity of A in the solution X = ax
The quantity of B in the solution X = bx
The quantity of C in the solution X = cx

Solution - Y:
The ratio of ingredients A, B and C = c : b: a
Let x be the factor.
The quantity of A in the solution Y = cx
The quantity of B in the solution Y = bx
The quantity of C in the solution Y = ax

Mixture Solution:
The quantity of A in the mixture solution = (a + c)x
The quantity of B in the mixture solution = 2bx
The quantity of C in the mixture solution = (a + c)x
The ratio of ingredients A, B and C = (a + c)x : 2bx : (a + c)x = (a + c) : 2b : (a + c)
It is given that, the resultant mixture to have A, B and C in equal proportion.
⇒ 2b = a + c

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 6

In the given diagram of Testbook team, contents are represented in the parallelogram, technical in the rectangle and DTP in the triangle. Out of the total number of the people, the percentage of content shall be in the range of ________.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 6

Total number of people in content = total number of people in rectangle
Total number of people in rectangle = 18 + 50 + 25 = 93
Total number of people in Testbook team = 152

Hence the answer the is 61 to 75

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 7

How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12 where n is an odd integer less than 60?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 7


Since m and n are both positive and their sum is equal to
Hence the value of both there fractions must be less than
Therefore n-can take values > 48 and odd
N = {49, 51, 53, 55, 57, 59}

Now we have to check which values of ‘n’ will give integral values of ‘m’.
Equation:

For m to be on integer (n - 48) should be a multiple of 12n
The above condition is satisfied for
N = 49, 51, 57
Corresponding values of m
N = 49
M = 12 × 49
N = 51

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 8

Most successful professional football players began playing football as youth, though a significant number learned the sport later in life. No professional football player who ignored the advice of his coaches, however, has ever become successful. If the statements are true, which of the following must also be true?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 8

According to the second sentence of the passage, a necessary condition for being a successful professional football player is not ignoring the advice of coaches.

So all of the successful professional football players mentioned in the passage’s first sentence, including those who played the sport as youths, do not ignore the advice of coaches. That’s what correct answer states.

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 10 - Question 9

The diagram shows a circle with one equilateral triangle inside and one equilateral triangle outside. The ratio of areas of big triangle to the small triangle is n :1. Find the value of n.


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 9

Rotating the smaller triangle by 180°, we get the following figure.

From symmetry we can say R is the midpoint of BC.
So P and Q are also mid points of AB and AC
From the midpoint theorem:

Since two equilateral tringles are also similar triangle we know

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 10 - Question 10

The following hydrograph depicts the rainfall intensity of an area for the time period of 8 hours. Also, infiltration curve is plotted on the same graph for the given land characteristics. Hydrograph region below the infiltration curve represents infiltrated portion of rainfall. Calculate the excess of rainfall occurred.


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 10

The region above the infiltration curve represents the excess of rainfall.
∴ Excess rainfall = Area of the hydrograph above infiltration curve.
Area of hydrograph above infiltration curve (A) is given by 


∴ Excess rainfall is 11 mm.

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 11

Which of the following expression represents the given logic circuit?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 11

= A̅ B̅ C + ABC
⇒ Y = (A̅ B̅ + AB) C
⇒ Y = C(A⊙B) = C(A̅⊙B̅)

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 10 - Question 12

A Transformer is rated at 100 KVA. At full load its copper loss is 1200 W and its iron loss is 850 W. Calculate the load (in KVA) at which maximum efficiency will occur.


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 12

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 10 - Question 13

Two transformers operating in parallel having impedance of 6 % and 4 %. If the total load supplied by the combinations is 100 A, then the current supplied by 6 % impedance transformer _______ A.


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 13

This question is based on load sharing

By using current division rule

However, if the impedance is in PU then their PU value should be converted on a common base to ensure that the ohmic ratio remain unchanged.

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 14

Let x(t) = 1 – 2t. Then what is the value of  where δ(t + 1) is unit impulse at t = -1.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 14


From the property of unit pulse function,

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 15

  what is the inverse Laplace transform of the product F1(s) . F2(s)?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 15


F(s) = F1(s) . F2(s)

By using partial fractions,

By applying inverse Laplace transform,

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 16

Latching current for an SCR, inserted in between a dc voltage source of 300 V and the load, is 150 mA. Compute the minimum width of gate pulse current required to turn-on this SCR in case the load of L = 0.3 H in series with R = 30 Ω

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 16


1 – e-100t = 0.015
-e-100t = 0.985
-100t = -0.01511
t = 151.136 μsec

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 10 - Question 17

A step-up chopper shown in figure is to deliver 3A

In to the 10 Ω load. The battery voltage is 12 V, L = 20 μH, C = 100 μF and chopper frequency is 50 kHz. Determine the on-time of the chopper ________ μs


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 17

Calculation:
Vo = IoRL = 3 × 10 = 30 V
Vb = Vo (1 – D)

On time = TON = DT = 0.6 × 20 × 10-6 = 12 μs 

 

When the transistor is switched on, current ramps up in the inductor and energy is stored.

When the transistor is switched off, the inductor voltage reverses and acts together with the battery voltage to forward bias the diode, transferring energy to the capacitor.

When the transistor is switched on again, the load current is maintained by the capacitor, energy is stored in the inductor and the cycle can start again.

The value of the load voltage is increased by increasing the duty cycle or the on-time of the transistor.

During the transistor on-period, assuming an ideal inductor and transistor
Vb = VL = Ldi/dt
Vb/s = (sL) i(s)
i(s) = Vb/s2L
i(t) = (Vb/L)t = kt

Over an on-time of TON, the change of battery current will be
Δi = k TON

During the off-period
V= Vb + V= Vb + L(di/dt) = Vb + L(Δi/Δt)
= Vb + L(Vb/L)TON /Toff = Vb(1 + TON /Toff)

Let the duty cycle
D = TON T, and Toff = T – TON = T(1 - D). Now
Vo = Vb (1 – DT/T(1 – D))

which simplifies to
Vo = Vb/(1 - D)

In an ideal circuit, VbI= V0I0. Therefore, from eq.
Ib = (Vo/Vb)Ib = Io/(1 – D)
Ib = Ib + Δi/2
Ibo = Ib – Δi/2

When the steady variation of battery current has been reached, the variation will be between lo and I1, as shown in.

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 18

For the circuit, shown in the figure below, the values of i0 and i1 respectively are

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 18

By applying voltage division in the input loop,

i0 will be zero, as there is no closed path.
By applying KCL in the right-side circuit,

As 10 Ω and 5 Ω are connected in parallel, voltage across them are equal.
⇒ 10 i1 = 5 i2 ⇒ i2 = 2i1 ----(2)
From equation (1) and (2)
i1 + 2i1 = -3 ⇒ i1 = -1 A

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 19

For the two – port circuit shown in the figure, the S – domain expression for the parameter d is where 

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 19

The s-domain circuit for the given circuit is

When V2 = 0

By applying KVL in the second loop
sI2(s) + 4 (I1(s) + I2(s)) = 0
⇒ 4 I1 (s) + (s + 4) I2(s) = 0

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 20

Match List-I (Transfer function of the system) with List-II (Type and order of the system) and select the correct answer using the codes given below the Lists:

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 20

Type of the system indicates the number of poles at origin.
Order of the system indicates the total number of poles.

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 21

The feedback system shown below is stable for all values of K given by

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 21

Characteristic equation,
1 + G(s)H(s) = 0

⇒ s(s + 1)(s + 6) = 0
⇒ s3 + 7s2 + 6s + K = 0
Routh array:

For the system to be stable, there should be no sign change in the first column of Routh array.

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 22

A 60 μA meter has a resistance of 100 ohms. If the meter has to measure 100 mA, then the value of shunt resistance and the current through shunt respectively are

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 22

Given that, current through meter (Im) = 60 μA
Total required current (I) = 100 mA
Meter resistance (Rm) = 100 Ω

Current flows though shunt resistance
Ish = I - Im
= 100 × 10‑3 – 60 × 10-6
= (100 – 0.06) × 10-3
= 99.94 mA

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 23

In the measurement of power by 3 voltmeter method connected as shown below. The expression for power consumed in the load is given by

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 23

Power consumed in the load,
PL = V3 Icos θ
V1 = ILR

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 24

The output of the 4 to 1 MUX shown in the below figure is

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 24


Gate Practice Test: Electrical Engineering(EE)- 10 - Question 25

Find VDS under DC condition of the given circuit

Given rd = 15 KΩ, IDSS = 20 mA, Vp = -10V

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 25

Given:
VGS = -6V

VDS = VDD – IDRD = 40 – 3.2 × 3 = 30.4 V

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 10 - Question 26

What is the output voltage Vo (in Volt)of the given circuit?


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 26


Voltage at A point VA = 4 × 4 = 16 V
By using virtual short circuit condition VA = VB = 16 V
Voltage droop across feedback resistance V4kΩ = 4 × 4 = 16 V
Output voltage Vo = 16 + 16 = 32 V

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 10 - Question 27

A 30 kV, 50 Hz, 60 MVZ generator has the positive negative and zero sequence reactance of 0.2 pu, 0.15 pu and 0.05 pu respectively. The generator ground with a reactance of 0.05 pu. The fault current magnitude for a single line to ground fault is ________(in pu)


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 27

Given that, Z1 = 0.2 pu
Z2 = 0.15 pu
Z0 = 0.05 pu
Zn = 0.05 pu
For a G fault, fault current is given by

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 28

The distribution system shown in the figure is to be protected by over current system of protection. For proper fault discrimination, directional over current relays will be required at which locations?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 28

Directional relays are not required at supply side and at radial feeders as the power flow unidirectional but not bidirectional.
so, the directional over-current relays will be required at 2 and 3 only.

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 29

In a four-bus system the total self-admittance of bus 2 is Y22 = -j14 p.u. A transmission line having a series reactance of j 0.2 p.u. and shunt capacitance of j 0.5 p.u. is connected between buses 2 and 3. The magnitude value of y22 modified is________

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 29


y22(old) = -j 14 p.u
Z23 = j 0.2 p.u
y’23 = -j5 p.u

y22 = -j 14 – j5 + j 0.25
= -j 18.75 p.u.

Gate Practice Test: Electrical Engineering(EE)- 10 - Question 30

A fair die is rolled two times independently. Given that the outcome on the first roll is 1, the expected value of the sum of the two outcomes is

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 10 - Question 30

Let X is a random variable represents an event of rolling a die and P(X = xi) represents the probability of getting sum as xi.

While rolling a die, the probability of getting different numbers as follows.

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