GATE Instrumentation Mock Test - 4 - GATE Instrumentation MCQ

Dividing both the numerator and the denominator of the given equation 

Cross-multiplying this equation yields 
Solving yields x/y = 5.
Now, the percentage of x + 3y to the expression x - 3y makes 
Dividing both the numerator and the denominator of the expression by y yields.

Given : In the diagram ovals (contour lines) are marked at different heights (h) of a hill.

The top view of the hill is best depicts by option Q from the figure we can see AB is perpendicular to CD and the slop of the hill is divided in PQ and QR in two parts.

Slop QR is stretching linearly as it down wards and the ovals are widening from line AB towards slop QP, which we can clearly see in the top view diagram of the hill in figure Q.

Hence, the correct option is (B).

Professional is the person who does something as a part of his job. Amateur is a person who does something because he loves doing it.

Time taken by boat to go from A to B and come back to A = 6 hr
Time taken by boat to go from A to C = 4 hr
Time taken by boat to go from A to B = 2 hr
Since B is equidistant from both points A and C, time taken by boat to go from B to A = 6 - 2 = 4 hr
Time taken by boat to go from C to A = 8 hr

Emitter coupled Logic is the fastest Logic family among the given options.
Hence, the correct option is (B).

For stable,
5(9 - K) > K and K > 0
45 > 6K
0 < K < 7.5 is the range. Option (2) alone satisfies the criteria.

Let f(z) be cosπz, then f(z) is analytic within and on |z| = 3.
Now by Cauchy's integral formula,

Electric field intensity due to Conducting Sphere:

Conducting sphere charged means the charge will be uniformly distributed over the surface of the sphere. Let's consider a conducting sphere of radius R, which is charged with charge Q. As shown in the below  figure,

There are three cases to analyze the Electric field intensity of a conducting sphere,

Case 1 (r > R): Electric field intensity at a point outside the conducting sphere

The electric field intensity at a point P which is lying outside the conducting sphere at a distance of r from the center of the conducting sphere is given by 

Where
Q = charge on the surface of the conducting sphere
r = distance between the center of the sphere and the point outside the sphere.
ε₀ = absolute permittivity
From the expression, we can observe that E ∝ ( 1 / r²)
So, with the increase in distance between the center of the sphere and the point the electric field intensity decreases.
Case 2 (r = R): Electric field intensity at the surface of the conducting sphere 
We can get the value electric field intensity on the surface of the conducting sphere by taking r = R in the above case, So electric field intensity on the surface is given by

Where R is the radius of the conducting sphere

From the expression, we can observe that E ∝ ( 1 / R²)

Case 3 (r < R): Electric field inside the conducting sphere

For the conducting sphere as the charge is distributed over the surface, so there is no charge available inside the sphere. As there is no charge inside the conducting sphere there is no presence of an electric field. Therefore the electric field intensity (E) inside the conducting sphere is equal to zero.

Concept:

The boundary condition for electrostatic fields are defined as:
E_1t = E_2t (Tangential components are equal across the boundary surface)
Also, the normal component satisfies the following relation:
D_1n – D_2n = ρs
For a charge-free boundary, ρs = 0, the above expression becomes:
D_1n – D_2n = 0
D_1n = D_2n

Analysis:

Given, the electric field intensity in medium 1.
E₁ = 5a_x – 2a_y + 3a_z
Since, the medium interface lies in plane z = 0
So, we get the field components as
E_1t = 5a_x – 2a_y
And E_1n = 3a_z
Now, From the boundary condition for the electric field we have:
E_1t = E_2t
ε₁E_1n = ε₂E_2n

So, the field components in medium 2 are:
E_2t = E_1t = 5a_x – 2a_y
E_2n = 6a_z
Therefore, the net electric field intensity in medium 2 will be:
E₂ = E_2t + E_2n = 5a_x – 2a_y + 6a_z
So, the z-component of the field intensity in medium 2 is
E_2z = 6a_z

z = x+ iy

= x² + (iy)² - 2ixy


The equation 

The above eqaution represents the equation of rectangular hyperbola.
Hence the correct option is (B).

Concept:
The duty cycle for an alternating quantity describes the fraction of time that the pulse is ON in one pulse period.
Mathematically, this is defined as:
Duty Cycle = 
Also, the Time period is the sum of the ON time and OFF time, i.e.
T = T_on + T_off
Duty Cycle = T_on/T

Analysis:
Given:

Given that the duty cycle of the asymmetrical waveform is 35 %.

T_ON = 0.35 T_ON + 0.35 T_OFF
0.65 T_ON = 0.35 T_OFF

Since T_ON + T_OFF = T

T_OFF = 0.65 T
∴ T_ON = 0.35 T

During ON time, the diode is forward biased and will act as a short circuit as shown:

V₀₁ = V_s = 7 V

During OFF time, the diode is OFF and will be open-circuited as shown:

As V_s = 0 for this interval: V₀₂ = 0 V and I₀₂ = 0 Amp
Now, the average value of the output voltage will be:

Similarly, the average value of current will be:

Hence, the correct options are (3) and (4).

Taking Fourier Transform
Y(ω) = jω X(ω)
⇒ 
⇒ |H(ω)| = ω
∠H(ω) = 90°
Magnitude increases linaerly and phase is constant.

From inspection
3I₁ – I₂ – 2I₃ = V₁ ...(1)
–I₁ + 3I₂ – I₃ = V₂ ...(2)
–2I₁ – I2 + 3I₃ = –V₃ ...(3)

if Z_L  is purely resistor the for maximum power transfer 

Neper is the unit for ratios of power or currents.
N = logₑ(A₂ / A₁)

1 Np = 8.686 dB

Corresponding attenuation in dB:
A_f = 0.075 x 8.868 dB/cm/MHz = 0.65
A_L = 0.1 x 8.868 dB/cm/MHz = 0.868

Absorption loss:
Loss in dB
L_f = 0.65 x 2 x 5 = -6.5 dB
L_L = 0.868 x 2 x 5 = -8.686 dB

Not taken account:

Reflection loss

D₁, D₂, D₃ are in common cathode connection. Only D₁ can conduct D₂ and D₃ remain OFF. Let D_4, D₅ be ON.

KCL at V_x Node:
I_x = I₄ + I₅

 = V_x – 9 ⇒ V_x = 10.5 V
∵ V_x < 12 V ⇒ D₄ must be OFF

= 1.8 mA

Concept:
For a transistor in Active Mode,
⇒ i_c = β_ib
And I_e = (β + 1)i_B

Calculation:
Given, β > > 1. i.e. i_B ≈ 0.
Also i_CQ = 2mA (Given)
The DC analysis of the given amplifier is as shown:

Since, iB ≈ 0,


3R₁ = 324k - 54k
⇒ 3R₁ = 324k - 54k
⇒ 3R₁ = 270 k
R₁ = 90k

Concept:

If x is the distance moved by the movable mirror between two consecutive positions of maximum indistinctness (or distinctness), we have

λ is the average of λ₁ and λ₂

Calculation:
λ = 5893 A° = 5893 × 10-8 cm
x = 0.9884 – 0.6939
= 0.2945 mm
= 0.02945 cm

Concept:

The Fourier transform of a signal in the time domain is given as:

Analysis:
The signal is defined as:
x(t) = t, 0 ≤ t ≤ 1
x(t) = -t, -1 ≤ t ≤ 0
The Fourier transform will now be:
By ILATE rule:

f(x) is having more priority than g(x)
Order of priority is decided as
I - Inverse trigo. function
L - Logarithmic function
A - Algebric function
T - Trigonometric function
E - Exponential function

Here t is algebric function and e^-jωt is exponential function.
t = f(x) and e^-jωt is g(x).
Applying integration by parts, the above integration can be written as:

Concept:
Consider the system of m linear equations
a₁₁ x₁ + a₁₂ x₂ + … + a_1n x_n = b₁
a₂₁ x₁ + a₂₂ x₂ + … + a_2n x_n = b₂
…
a_m1 x₁ + a_m2 x₂ + … + a_mn x_n = b_m
The above equations containing the n unknowns x₁, x₂, …, x_n. To determine whether the above system of equations is consistent or not, we need to find the rank of following matrices.

A is the coefficient matrix and [A|B] is called as augmented matrix of the given system of equations.
We can find the consistency of the given system of equations as follows:
(i) If the rank of matrix A is equal to rank of an augmented matrix and it is equal to the number of unknowns, then the system is consistent and there is a unique solution.
The rank of A = Rank of augmented matrix = n
(ii) If the rank of matrix A is equal to rank of an augmented matrix and it is less than the number of unknowns, then the system is consistent and there are an infinite number of solutions.
The rank of A = Rank of augmented matrix < n
(iii) If the rank of matrix A is not equal to rank of the augmented matrix, then the system is inconsistent, and it has no solution.
The rank of A ≠ Rank of an augmented matrix
Calculation:
Given:
The given system of equations can be represented in a matrix form as shown below.

The Augmented matrix can be written by

R₃ → R₃ – R₁
R₂ → R₂ – R₁

The system to have a unique solution,

⇒ (a - 2)² – 9 ≠ 0
⇒ a – 2 ≠ ±3
⇒ a ≠ -1 or 5
The system doesn’t have a unique solution for a = -1 and 5.

Resistance is directly proportional to length.
Given that length of AB = 25 mm
By voltage division,

Concept:

Maximum Power transfer in DC circuit 
R_L = R_th
For complex load to deliver maximum power to the load
Z_L = Z^*_th
Z_L = R_th - jX_th
For Z_L = R_L + j_XL (X_L = constant)

Calculation:

We draw thevenin equivalent circuit
Applying KVL at a:

Zth (short circuiting the voltage source) = (3 + j4) ∥ (3 + j4) = (1.5 + j2) Ω
For Maximum power transfer

Equivalent circuits comes out as: