GATE Mock Test Electronics Engineering (ECE)- 5 - Electronics and Communication Engineering (ECE) MCQ

Statement-2 can be used if it is known what quantity of each tank is full/empty.

Therefore, by using both statements

Let capacity of tank B is x

70/100x = 14000

= x = 20000 gallons

Solution in tank A = 80/100 × 14000

= 11200 gallons

Solution in tank A = 40/100 × 20000

= 8000 gallons

∴ Total solution = 11200 + 8000

= 19200 gallons

Base on the diagram and information available,

c = f + g …(i)

a + b +c = 36

⇒ d + e + f + g = 61-36 = 25 …(ii)

a + b + c = 36

⇒ g + 3 + g + 8 + c = 36

⇒ 2g + c = 25 = 3g + f …(iii)

∵ b + d + e + g = 32

⇒ g + 8 + d + e + g = 32

⇒ d + e + g + g = 24

⇒ d + e + f + g + g = 24 + f = 25+g

⇒ f = g +1

∵ 3g + f = 25

⇒ 4g = 24

∴ g = 6

(1) is eliminated as it is true as per line,' Inward humility is achievement, outward courtesy is virtue' of the passage.

(2) is true as per the line ,'Refining your own nature is achievement. Refining your own person is virtue'. Hence, eliminated.

(3) can also be inferred in the last lines.

(4) is the only statement that can not be inferred,

duration of tab A is opened for '90-t₁' minute.

Since the rate of inflow of water of tab A is 90 liter/minute, while the rate of inflow of water of tab B is 60 liter/minute.

So as per the given data they both fill the tank as follow

90×(90-t₁) + 60 x(t₁) =6000

8100-6000 = 30 t₁

t₁ = 2100/30 = 70 Minute

Hence Tab B is opened for 70 Minute.

It is already given that the total time of open for both tab is 90 Minute hence the duration for which Tab A is open = 90-70 = 20 Minute

Difference in time = 70-20 = 50 Minute.

Rearrange 1/3⁸ so that it can be added to 1/3⁹ . That is, try and turn 1/3⁸ into a fraction with 3⁹ in the denominator.

Multiply 1/3⁸ by 3/3 to get

Solving,

Canceling out a factor of 3 gives,

Solving this we get to know,

Vectors are linearly independent.

So, sum = ;

Here a = 3, p = 129,

So required sum =.(3¹³⁰ - 1)/2

Since now 25Ω and 10Ω are in parallel with 50V.

i₁ = 50/10 = 5A

i₂ = 50/25 = 2A

i₁ + i₂ = 7A

Differentiating both sides

d/dtx(t) = δ(t + 1) - δ(t - 4)

Taking Fourier transform on both sides

jω × (jω) = e^jω − e^−jω

Counting from MSB to LSB, the output will be 010101.

Taking laplace transfarm

sY(s) + 3Y(s) = X(s)

Y(s) = X(s)/(s+3)

Given x(t) = u(t) ⇒ X(s) = 1/s

Y(s) = 1/s(s+3) = 1/3(1/s − 1/s+3)

Y(t) = 1 - e^-3t/3

Here R(s) = (unit step input)

G(s) = 1/s(s + 2)

H(s) = 1 (Unity feed back)

∴ It has term K/S²

As the initial slope intersect at w = 5

K = w²

K = 25

New for corner frequency w_cf₁ = 2 slope changes to –20 db/d

∴ It has zero in the transfer function having time constant

T₁ = 1/ω_cf₁ = 1/2 = 0.5

For corner frequency cos⁡f₂ = 10,slope changes from –20 db/d to – 40 db/d

∴ It contain in the transfer function having time constant

T2 = 1/ω_cf₂ = 0.1

∴ Transfer function = 25(1 + 0.5S)/S²(1 + 0.1S)

For real periodic signal,

C_-k = C_k*

Thus, C_-3 = 3 - j5

fu is substituted for x and therefore duedx Separation of variables gives the next equation m(d²y/dt²) + c(dy/dt) + k(y) = c(du/dt) + k(u)

Now the function is ready to be transformed from the time domain into the frequency domain. The Laplace of d²y/dt² and dy/dt are as follows:

L(f′′) = s²L{f) − sf(0) − f′(0)

L(f′) = s.L(f) − f(0)

Assuming zero initial conditions the equations are (m(s²) + c(s) + k)⋅Y(s) = {c(s) + k)∪(s)

Let the transfer function G(s) is defined by \Upsilon(s) / U(s\) G(s) = {c(s) + k)/(m(s²) +c(s) + k)

Substituting the values in the G(s) G(s) = (8s + 6)/(s² + 8s + 6)

⇒ (8s + 6)/(s + 4)(s + 2)

From the numerator we have would give you a zero at s= - 3/4 and solving for s in the denominator gives poles at s = -4, -2

F = v x B

Since the direction of velocity v and B are perpendicular to each other as obtained from the figure shown, the resultant force will be perpendicular to both of them, i.e. the force on the moving charged particle will be in an upward direction. And as the particle is also deflected in an upward direction with the applied force, so it gives the conclusion that the particle will be positively charged.

S = f₁(J, K, Q_n) and R = f₂(J, K, Q_n)

Steps:

1. List all possible combinations of J, K and Qn (000 to 111, first three columns).

2. Fill the values of Q_n+1 using JK FF truth table (4^th column).

3. Now, we have Q_n and Q_n+1 ready, so fill the values of S and R using SR excitation table.

Using K map to simplify S and R:

S = JQ_n and R = KQ_n

So, two additional AND gates required.

There are 4 minimum state.

Or Q = P^-1 = or P = Q^-1

If zero is an eigen value of any of the matrix, then other matrix will have infinity as an eigen value.

Thus, zero is not an eigen value of either P or Q.

⇒ z = e^iθ

⇒ dz = e^iθdθ

and θ : π/2 to 3π/2

⇒ I = ie^iθ(3π/2) - e^(π/2)

∴ I = -2i

A discrete time complex exponential is periodic if ω₀/2π is a rational number. But given ω₀ = 1

ω₀/2π = 1/2π = is a irrational number, so it is non periodic.

Applying L Hospital rule because its 0/0 form

n₀ ≈ N_D

p₀ = (2.4² x 10²⁶)/(6 x 10¹⁸) = 0.96 × 10⁸/cm³ = 96 × 10⁶/cm³