GATE Mock Test: Electrical Engineering (EE)- 10 - Electrical Engineering (EE) MCQ

The correct spelling of the word is 'definitive' which means '(of a conclusion or agreement) done or reached decisively and with authority.' Thus option 2 is the correct answer.

The preposition 'about' is mandatory here thus option 1 and 4 are eliminated. Option 2 is correct as the tense present perfect continuous fits here. It conveys the meaning that the person usually talked about the place. 

Options 1 and 2 are incorrect as they change the meaning of what is mentioned. Option 3 is incorrect as with 'less' we use the adjective in positive form and not comparative form. The correct word here should be 'brave.' Option 4 is thus the correct answer.

Let number is N

N = 65 k +29

∴ Remainder = 3

The word 'here' and 'there' both can be used here but should be followed with 'waiting' as no other word can fit here. The word 'caught' is correct here as 'catch a sight' means 'to see something.' Option 2 thus has the correct combination of words. The word 'cot' means 'a small bed with high barred sides for a baby or very young child.'

For a number to be greater than 5000, d₁ should be filled with either 5 or 7

∴ Total numbers formed when the digits are repeated = 2 × 5 × 5 × 5 = 250

total cases = 250 -1 = 249 ( case of 5000 is not included)

Now, For the number to be divisible by 5, unit digit d₄ should be either 0 or 5.

∴ Total no. of ways = 2 × 5 × 5 × 2 = 100

favorable cases = 100 - 1=9 ( 5000 is not included))

∴Required Probability = 

The question asks which of the statements given in the options can weaken the argument put by the author that all bacteria from Venus must have died out.

The passage states that since there is a single strain of bacteria which exists on the Earth, they all must be belonging to the Earth or let's say a single planet. But here the author does not take into consideration (as can be argued from his theory) the fact that may be all the bacteria came from Venus and there are none which originally belong to the Earth. So this criticism as mentioned in option 3 makes the argument of the author vulnerable. 

Options 1, 2 and 4 are completely irrelevant criticisms as they do not address the main argument. The argument claims that if there were Venusian bacteria on Earth, then they must have died out by now. Whether there are bacteria originally from Earth that have also disappeared from Earth is irrelevant to the question and has no effect on the given argument.

Let a, b and c be the cost prices of the three articles A, B and C.

SP = CP + Profit (or) SP = CP – Loss

⇒ SP of A = 1.1a; SP of B = 1.2b; SP of C = 0.9c

By question,

1.1a + 0.9c = a + c ⇒ 0.1a = 0.1c ⇒ a = c

1.2b + 0.9c = 1.05(b + c) ⇒ 0.15b = 0.15c ⇒ b = c = a

Gain% = {(SP – CP)/CP} × 100

⇒ Net gain on the sale of all the articles =

∴ Net gain on the sale of all the articles = 6.66%

Option 1 is false as graph says there is decrease in students qualifying in Physics in 2015 compared to 2014.

Option 2

Let no. of students qualifying in Biology in 2013 be 100

⇒ No. of students qualifying in Biology in 2014 = 100 – 10% of 100 = 90

⇒ No. of students qualifying in Biology in 2015 = 90 + 10% of 100 = 99

∴ The number of students qualifying in Biology in 2015 is less than that in 2013

Option 3 and option 4 are incorrect since no detail is given regarding how many students qualified the subject in 2013.

Area of triangle PAB = Area of triangle RCB (By symmetry)

∴ Area of Δ PAB = ½ × PA × AB

= ½ × 2A/3 × A = A²/3

Area of ΔRCB = A²/3

Now, Area of ABCD = Area of DRQP + Area of PAB + Area of RCB + Area of PBRQ

A² = a² + A²/3 + A²/3 + Area of PBRQ

As, A/a = 3 ⇒ a = A/3

 Area of PBRQ

⇒ Area of PBRQ = 

Y(s) = L(y(t))

Apply inverse Laplace transform,

Differentiate with respect to ‘t’.

⇒ y(0) + y'(0) = 1

Given that f(z) = (x² + y²) + i 2xy

g(z) = 2xy + i (y² – x²)

To check analyticity of a function, we need to check CR equations.

u_x = v_y, u_y = - v_x

f(z) = (x² + y²) + i 2xy

u = x² + y², v = 2xy

u_x = 2_x

u_y = 2_y

v_x = 2_y

v_y = 2_x

u_x = v_y but u_y ≠ -v_x

Hence, f(z) is not analytic

g(z) = 2xy + i (y² – x²)

u = 2xy, v = y² – x²

u_x = 2y

u_y = 2x

v_x = -2x

v_y = 2y

u_x = v_y and u_y = -v_x

Hence g(z) is analytic.

A matrix is said to be singular, if determinant of that matrix is zero.

= 1 (18 – 12) - 1 (9 – 4) + 1 (3 – 2)

= 6 – 5 + 1 = 2 ≠ 0

M is non singular

(Q) A matrix can be diagonalizable when it has distinct eigen values

S is a diagonalizable matrix. Hence, has distinct eigen values.

Let S be a 3 × 3 matrix and the eigen values of s are λ₁, λ₂, λ₃

Given that, S + 5T = I

From the properties of Eigen values,

(a) If λ₁ is an eigen value of matrix A, then -λ_{1 ­}will be on eigen value of matrix -A.

(b) If λ₁ is an eigen value of matrix A, then (λ₁ + 1) will be an eigen value of matrix (A + I)

(c) If λ_1­ is an eigen value of matrix A, then  will be an eigen value of matrix  where K is a scalar.

From the above properties, eigen values of T are,

As λ₁, λ₂, λ₃ are distinct values, λ′₁,λ′₂,λ′₃ will be distinct.

Hence, matrix T is diagonalizable

So, only Q is true.

 = 0

A.E is, (D² – 3D + 2) = 0

⇒ (D – 1) (D – 2) = 0

⇒ D = 1, 2

x(t) = C₁e^t + C₂e^2t

x = 0 at t = 0

⇒ 0 = C₁ + C₂ ⇒ C₁ – C₂

x = 1 at t = 1

⇒ 1 = C₁e¹ + C₂e²

⇒ 1 = C₁e¹ + (-C₁) e²

⇒ C₁(e¹ – e²) = 1

x(2) = C₁e² + C₂e⁴

= e + e²

R.M.S value of phase voltage

In DC equivalent circuit capacitor is open

A small signal equivalent circuit is shown in fig.

Put r_d in equation (2)

The sequence is 000 to 001 to 011 to 111 to 110 to 100 and back to 001, etc. 

Multiplication in the domain become convolution in frequency domain therefore individual.

Frequency will add up

_f1 = 400 Hz, f₂ = 500 Hz

Sampling frequency f_s = 2 f_net = 2 × 900 = 1.8 KHz

Ampere law: 

for isotropic medium

Phase crossover frequency is given as 

G.M = 20 log 164 = 44.3 db

It can be observed from the Lissajous figure

y₁ = 9 units, y₂ = 10 units

V_EE = η V_BB + V_D + I_P . R_E

= 0.67 × 20 + 0.7 + 12 × 10^-6 × 1 × 10³

= 13.4 + .7 + 0.012

= 14.112

Here, 3 capacitor are connected in parallel.

∴ Total capacitance = C₁ + C₂ + C₃ = 4 + 4 + 2 = 10 μF

Now find time constant of the circuit

T = RC_eq = 20 × 5 = 100 μsec

The initial voltage across the

With closing of k, the capacitor co will start discharging 

Input resistance of voltmeter Z_L = 2000 × 10 = 20 kΩ

Output resistance of circuit Z₀ = 5 kΩ

Open circuit voltage of circuit under measurement E₀ = 6 V

Reading of voltmeter

∴ Percentage error in voltage reading

- 20%

Or 20% low

For power transformer of large rating the tapping are located in the middle portion of the winding to reduce the mechanical force affecting the winding during short-circuits.

Signal energy