Gate Practice Test: Electrical Engineering(EE)- 9 - Electrical Engineering (EE) MCQ

The sentence already mentions a trait that they are satisfied with what they have thus it becomes easy for us to choose the correct word. The meanings of the words are:
Contentment → being satisfied with what one has
Ambition →  a strong desire to do or achieve something.
Perseverance →   persistence in doing something despite difficulty or delay in achieving success.
Hunger →  a feeling of discomfort or weakness caused by lack of food, coupled with the desire to eat.
Thus, contentment is the correct answer.

The sentence mentions a person doing something as per her wishes thus the first word must be a positive reaction. 'Restive' which means 'restless' and 'jinxed' which means 'cursed' are incorrect here. Between options 1 and 3,the word 'elated' which means 'too happy' and the word 'venture' which means 'undertake a risky or daring journey or course of action' fit here correctly. The word 'cease' means 'stop' and does not convey a proper meaning. Thus option 3 is the correct answer.

The correct tense that fits here is simple present and the subject being singular, the verb should be in the correct form too. Thus 'wants' is the correct answer. The other options are incorrect as:
Is wanting ⇒ it is in present continuous tense which is used to denote an activity which is continuing
Want ⇒ subject is singular thus the verb cannot be in plural form
Was wanting ⇒ it is in past continuous tense which is used to denote an activity which was continuing in the past
Thus option 2 is the correct answer.

In this problem, cumulative frequency is given, so converting into frequency

From the table, it is clear that option b, i.e.
60% of the scholars published at least 2 papers is correct.

So -30 will be correct as per provided options.

The number affected by famine always increases with the population. Therefore, if the population increased in 2005, then the number of those affected by famine also increased.
If there was an increase in 2005, there must have been more people affected in that year than the previous year of 2004.
The available information does not allow us to draw any of the other inferences.
The number of those affected by famine could increase without a corresponding percentage of the population increase.
Neither can we draw inferences about any particular year between 2000 and 2010.
Note:
Option 2 cannot be correct because  consider a case where in 2000 the population is 100 , and people affected are 50
in 2001 the population grew to say 200, and people affected is 100
So in Both cases the percentage of people affected is 50% but the number of people affected grew.

Let, the per day tunnelling plan was ‘x’ meter and the tunnelling of 360 m was to be completed in ‘y’ day.
Therefore, xy = 360----(i)
Now, as per plan tunneling was done for four days and then rate of tunnelling was increased to
(x + 15) for (y - 4) days. One day before the planned date they have tunnelled 380 m.
∴ Number of days when the tunnelling was done at increased rate is
= (y – 4 - 1) = (y - 5) days
According to the statement given the following situations can be inferred
∴ 4x + (y - 5) (x + 15) = 380
⇒ 4x + xy + 15y – 5x – 75 = 380
⇒ 4x + 15y – 5x = 380 + 75 – xy
⇒ 15y – x = 380 + 75 – 360
⇒ 15y – x = 95     ----(ii)
Putting y value from eq (i) in eq (ii) we get

⇒ x² + 95x – 5400 = 0
On solving the above quadratic equation, we get
⇒ x = 40 m

A person can take 1^st step in any direction independently
Suppose he moves to A
Now to return to O ( initial point) in 2 steps he can move in 2 directions. Either B or f
Thus probability he will move to either B or F will be 
Suppose he moves to B
Now to return back to O
he has only 1 option
⇒ to move in BO
Probability he will move in BO direction 
Total probability he will return

Let PQR is the initial triangle and SQT is the final triangle.
ΔPQR is similar to ΔSQT
∵ ST = 0.75 PR
∴ SQ = 0.75 PQ
And QT = 0.75 QR
Initial Area = 
Final Area = 

= (0.75)² × 28 = 15.75 cm²

x² + x – 1 = 0
x² + x = 1
x(x + 1) = 1

Again squaring 

Y = AB + A(B + C) + B(B + C)
= AB + AB + AC + BB + BC
= AB + AC + B + BC
= AB + AC + B(1 + C)
= AB + AC + B
= B(A + 1) + AC
= B + AC

kVA rating of transformer = 50 kVA
Auto transformer rating = 2300 / 2530

kVA rating of auto transformer 

For the parallel operation, phase displacement angle between secondaries of both transformers must be zero. That means both the transformers must belong to same phase group.

In the given options,
Y/Zig Zag Y and Y/Y belongs to different phasor groups. Hence this pair is not suitable for parallel operation.

X₂ = 0.4375 Ω
S_starting 
R_ext = 0.3675 Ω

The Basic difinition of minimum phase system in Z-domain is it should have all poles and zerores inside the unit circle.
H(z) stable and causal:
All poles of H(z) are inside the unit circle.

All poles of  i.e. zeroes of H(z) are inside unit circle.

A system is said to be BIBO stable if it’s impulse response is absolutely summable,

Out of all the responses only  is absolutely summable.

Is not absolutely summable it is divergent.

We know that 
Ripple frequency = 2nf = 2 × 12 × 30 = 720 Hz 

Magnitude at 10 rad/sec is same as magnitude at 2 rad/sec

6 dB/octave = 20 dB/decade
y = 24.04 dB

For intersection with -180⁰ axis
∠GH = -180⁰

Given that,
Phase voltage (V_ph) = 200 V
Phase current (I_ph) = 10 A
Power factor = cos ϕ = 0.5
⇒ ϕ = cos^-1 (0.5) = 60⁰
In two wattmeter method,
W₁ = V_LI_L cos (30 + ϕ)
And
W₂ = V_LI_L cos (30 – ϕ)

Given that, I = 2 A
R_m = 500 Ω
R_sh = 20 Ω
We know that,

Voltage at A point

By using virtual short circuit condition
V_A = V_B = 8V
Apply KCL at node B

Frequency compensation is used in operational amplifiers to improve the stability of the op-amp over the input signal frequency range.

Given that, impedance = (0.01 + j 0.15) p.u.
Resistance (R) = 0.01 p.u.
Reactance (X) = 0.15 p.u.
Voltage regulation
% V.R = [R cos ϕ_r + X sin ϕ_r] × 100
= [0.01 × 0.8 + 0.15 × 0.06] × 100
= 0.098 × 100 = 9.8%

Given that, X_s = 0.2 pu
X_m = 0.05 pu
X_1eq = X_s - X_m = 0.2 – 0.05 = 0.15 pu
X_2eq = X_s - X_m = 0.2 – 0.05 = 0.15 pu
X_0eq = X_s + 2X _m = 0.2 + 2 (0.05) = 0.3 pu
X_1eq + X_0eq + X_2eq = 0.6 pu

The size of Jacobian matrix = (2n - 2 - m) × (2n - 2 - m)
Where n is number of buses
M is number of voltage controlled buses
Size of Jacobian matrix = (2 (30) - 2 - 3) × (2 (30) - 2 - 3)
= 55 × 55