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Gate Practice Test: Electrical Engineering(EE)- 14 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test - Gate Practice Test: Electrical Engineering(EE)- 14

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Gate Practice Test: Electrical Engineering(EE)- 14 - Question 1

Given 4 words labelled as (P), (Q), (R) and (S). One of the pairs given below has words which have similar meaning or have opposite meaning. Identify the correct option.

(P) instigate (Q) enquire

(R) construe (S) interpret

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 1
Construe and interpret have identical meaning (a word or action).

Thus, option C is the correct answer.

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 2

A job can be completed by 12 men in 12 days. How many extra days will be needed to complete the remaining job, if 6 men leave after working for 6 days?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 2
work done by one man in 1 day

=1 / 12×12 Work done by 12 men in 6 days = 1/2 Remaining work = 1−(1/2) = 1/2 After 6 men leave the work, time taken to complete the remaining work = Remaining Work/Remaining Men

∴ Time taken

= 12 × 12 / 6 × 2 =12days

So,

so the extra time taken =12 - 6 = 6 days

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Gate Practice Test: Electrical Engineering(EE)- 14 - Question 3

Which of the following phrases should be placed in the blank spaces so as to make a grammatically correct and meaningful sentence.

Even though the institute had lot of space.______

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 3
Answer b would be a grammatically correct sentence.
Gate Practice Test: Electrical Engineering(EE)- 14 - Question 4

In the 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 following 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.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 4
S1 is ‘particular affirmative’ type of statement whereas S2 is a universal affirmative. S1 & S2 combined lead to ‘particular affirmative’ i.e. some guitars are doors ‘and’ some doors are guitars’ i.e. C3 is not possible. C2 is also NOT possible since no conclusion is possible regarding books and posters; hence only C1 following leading to option (a).
Gate Practice Test: Electrical Engineering(EE)- 14 - Question 5

Choose one word out of the given option to replace the phrase ‘person who insists on adherence to formal rules of the literal meaning’.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 5
Pedant is a person who strictly goes by the rule of the book of the literal meaning of the written text.

Thus, option A is the correct answer.

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 6

A shopkeeper gives a discount on the marked price based on the quantity bought by a customer. Raja bought 10 pieces of a lunch box and was given a 10% discount by the shopkeeper. When Radhika bought 20 pieces of the same lunch box, she was given a 15% discount by the shopkeeper. In both the cases, the net amount of profit for the shopkeeper was identical. Which of the following is the ratio of marked price to the cost price for the lunch box?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 6

Let 'M' and 'C' be the marked price and cost price per lunch box. We can form an equation based on the given information as

i.e.

i.e. C = 8M / 10

or M:C = 10:8 or 5: 4

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 7

It would take one machine 4 hours to complete a production order and another machine 2 hours to complete the same order. If both machines work simultaneously at their respective constant rates, the time taken to complete the same order is __________ hours.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 7

Let t be the time taken by the machines when they work simultaneously.

∴ 1 / t = 1 / 4 + ½

∴ 1 / t = 3 / 4

∴ t = 4 / 3

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 8

Which of the options given below should replace the part of the sentence written in BOLD so as to get a meaningful and grammatically correct sentence?

We had not only helped them with money but also with new machinery and raw material.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 8

Answer C is corrected as not only will come after the sentence.

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 9

A fair die is thrown three times and the sum of three numbers is found to be 16. Find the probability that 5 appears on the third throw.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 9

For sum = 16, we have the following cases:

(i) 6, 6, 4

(ii) 6, 4, 6

(iii) 4, 6, 6

(iv) 5, 5, 6

(v) 5, 6, 5

(vi) 6, 5, 5

Total cases of sum (16) = 6

Favourable cases for 5 appears in 3rd throw = 2

p = 2 / 6 = 1 / 3

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

A work of classics is split up in 3 volumes-each volume having equal number of pages. It is also known that the page numbers are running across the 3 volumes. If the sum of the first page number of the 3 volumes is 1473, the number printed on the first page of volume 3 is ________.

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

Let the number of pages in each volume be ′x′. The first page of the 3 volumes will be 1,1+x and 1+2x respectively. 1473 = 1 + (1 + x) + (1 + 2x) = 3 + 3xgiving

x = 1470/3 = 490

First page of volume 3 will be having pages starting from 1 + 980 = 981

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 11

Find the value of the resistance and the inductance of the branch CD if the balance is obtained under these conditions. if the arms of an ac. maxwell bridge are given as: AB is a non-inductive resistance of 1000 Ω in parallel with a capacitor of capacitance 0.5 μF, BC is a non inductive resistance of 600Ω and CD is an inductive impedance which is unknown and DA is a non-inductive resistance of 400 Ω?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 11
The balanced condition is

R1R3 = R2R4

R3 = R2R4/R1

R3 = 600 × 400/1000 = 240Ω

L3 = CR2R4

L3 = 0.5 × 10−6 × 400 × 600

L3 = 12 × 10−2

L3 = 0.122H

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 12

A single phase semi converter is operated from 120V, 50Hz ac supply. The load current with an average value IDC is continuous and ripple free firing angle α=π/6. Determine the harmonic factor of input current?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 12
Supply rms current

= 0.91Idc

Now the rms value of supply fundamental component of input current

Harmonic factor (HF) of input current

=

=

= 0.30

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 13

A 200V dc series motor develops its rated output at 1500 rpm while taking 20 A . Armature and series field resistances are 0.6Ω and 0.4Ω respectively. To obtain rated torque at 1000 rpm, external resistance must be added is

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 13

⊤ ∝ la2 ⇒ T = kla2

For const torque armature I must be const.

Eb = 200 − 20 × (0.6 + 0.4) = 180V

la = 20A for 1000rpm. Ea = kϕw

180Ea = 1500 / 1000

Ea = 120V = 200 − 20(0.4 + 0.6 + Rext)

⇒ Rext = 200 − 120 / 20 −1 = 3Ω

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

The value of impedance of short transmission line having supply of 6.5 kV with the rating of 10 kVA is (1+j5) p.u. What is the impedance at the new base of 13 kV and 30kVA?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 14
As we know that by the formula that

So

= 0.75 + j3.75

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 15

A Signal x(t) has Fourier transform x(ω) phase and magnitude of x(ω) are shown below

Determine x(t) at t = 5sec

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 15

=

x(5) = 0.5

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 16

A 3-ɸ bridge converter is provided with line-line voltage of 400 V. A load which is resistive in nature of value 100 ohm takes 400 W of power from the converter, the input power factor will be

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 16

Load current

In A 3ϕ fully enthralled bridge inverter input rms current I or the

correct in each supply phase exist for 120∘ in every 180∘

Therefore rms value of input current

Input apparent power

= √3 × VSIS

= √3 × 400 × 1.15

= 796.72vA

∴ Input apparent power = Output power

796.72cosθ = 400

cos⁡θ = 0.5 lagging

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 17

The z-parameters of the network shown in figure are given by

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 17

z11 = v1I1∣∣t2 = 0 (Open circuit output terminal)

z11 = v1I1∣∣t2 = 0 (Open circuit output terminal)

V1 = I1R1

z11 = V1I1 = R1

z11 = v2I1∣∣I2 = 0

V2 =−αV1

V2 = −α(I1R1)

z21 = V2I1 = −αR1

z12 = v1I2∣∣I1 = 0 (Open circuit input terminal)

Since V1 = I1R1 = 0

50,Z12 = 0

z22=V2I2∣∣I1 = 0

since V1 = I1R1 = 0

Thus, α∨1 = 0

V2 = I2R2

z22 = V2I2 = R2

Alternative Method:

We have the equations for voltage V1 and V2 as

∨1 = |1R1

V2 − I2R2 + αV1=O

V2 = I2R2−αV1

From eq. (1) V2 = I2R2 − α(1R1) = (−αR1)I1 + R2I2

Comparing equation (1) and (2) to the general equation of z- parameter,

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 18

The admittance parameter matrix [y] is

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 18

(Node equation at the top left node)

(Node equation at the top right node)

Comparing equation (1) and (2) with general equations, we get z-

parameters as

[y] =

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 19

Geneticists say that they are very close to confirming the genetic roots of psychiatric illnesses such as depression and schizophrenia, and consequently, that doctors will be able to eradicate these diseases through early identification and gene therapy. On which of the following assumptions does the statement above rely?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 19
  1. A says that strategies are now available for eliminating psychiatric illnesses but it is mentioned in the very first line that the geneticists are very close but the strategy is not available till now. So, A is incorrect.

  2. The given data says that the geneticists are working on the genetic roots of psychiatric illnesses such as depression and schizophrenia, which implies that these two diseases have genetic basis. Thus, B is the correct option.

  3. C says that all human diseases can be traced back to genes and how they are expressed, but the data given in the question talks about psychiatric illnesses such as depression and schizophrenia only. So, C is also incorrect.

  4. D says that in future, genetics will be the only relevant field for identifying psychiatric illnesses, which cannot be inferred from the given data. So, D is also incorrect.

So, B is the correct option

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 20

Determine the maximum drain current in mA for the JFET in the given network, if VGS = −3V and Vp = −8V

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 20

Considering input loop, since, IG = 0A and IS = ID

VGS = −IDRS

Since,

Substituting values,

⇒ 2.5mA =

⇒ 2.5mA=IDSS(2564)

⇒ IDSS = 64/25 × 2.5mA

⇒ IDSS = 6.4mA

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 21

An ac voltmeter uses half wave rectifier and the basic meter with full scale deflection current of 1 mA and the meter resistance of 200 Ω. Calculate the multiplier resistance for a 10 V r.m.s range on the voltmeter.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 21

the meter uses half wave rectifier and input is 10 V r.m.s

Eav = 1/2 (Eav over a cycle of input)

Now Ep = √2 Erms = 14.14V

Eav = 0.6 Ep = 8.99 ≈ 9V

Therefore, Eav (output) = 1/29 = 4.5V

Edc = 0.45Erms

Rs = Edc/ldc − Rm = 0.45Erms/ldc − Rm = 0.45×10/1×10−3 − 200

Rs = 4.3KΩ

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 22

Let f(x) = x(x - 1)(x - 2) be defined in [0,0.5]. Then the value of c of the mean value theorem is

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 22

f(x) = x3 − 3x2 + 2x

f′(x) = 3x2 − 6x + 2

f′(c) = 3c2 − 6c + 2

by mean value through, we have

∴ 3c2 − 6c + 2 = 3/4

12c2 − 24c + 5=0

c = 6 ± √21/6

C = 6 − √21/6 = 0.24↔(0, 1/2)

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 23

For the flip flop configuration shown below, find the operation carried out by this configuration.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 23

Stages

It is a gray counter.

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 24

If Laplace transform of a function is given by

Find the value of f(0+) and f(∞) respectively.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 24

Initial value = 2

Final value = 6/8 = ¾

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 25

A 70 MW steam station uses coal of calorific value 5000 Kcal/kg. Thermal efficiency of the station is 40% and electrical efficiency 70 % . calculate the coal consumption per hour when the station is delivering its fuel rated output is ___x 103 kg?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 25

Given data

Overall efficiency of the power station is n overall =ηthermal ×nelectrical

= 0.4 × 0.7 = 0.28

Units generated /hr = (70×103)×1

= 70kwh

Heat produced / hr

H= electrical output in heat units /n overall

= 70 × 103 × 860/0.28 = 215 × 106Kcal(1kwh = 860Kcal)

Coal consumption 1hr=H/ calorific value

= 215 × 106/5000

= 43 × 103kg

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 26

The value of is ____

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 26

=

Putting

= -0.94

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 27

Consider two real valued signals, x(t) band-limited to [−500 Hz, 500Hz] and y(t) bandlimited to [−1kHz, 1kHz]. For z (t) = x(t). y(t), the Nyquist sampling frequency (in kHz) is __________

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 27
x(t ) is band limited to [−500Hz, 500Hz] y(t )is band limited to [−1000Hz, 1000Hz] z(t ) = x (t ).y(t )

Multiplication in the time domain results in convolution in the frequency domain.

The range of convolution in frequency domain is [−1500Hz, 1500Hz]

So, the maximum frequency present in z(t) is 1500Hz Nyquist rate is 3000Hz or 3 kHz.

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 28

Keeping the kW demand constant if the load power factor increases the kVA demand –

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 28

kW = kVA cosϕ

Kva ∝ 1/cos⁡ϕ

If the power factor increases the kVA demand will decrease.

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 29

A delayed unit step function as

Its Laplace Transform is

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 29

=

=

Gate Practice Test: Electrical Engineering(EE)- 14 - Question 30

Find the inverse Laplace transform of the function F(s) =

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 14 - Question 30

let G(s) = 2/(s + c)

⇒ g(t) = L−1{G(s)} = 2e−ct

f(t) = L−1{G(s) e−bs}

=2e− k(t − b)u(t − b)

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