GATE EE 2022 Official Paper (Previous Year Paper)

Given:

There are two figures that represent the front and rear view of a disc.

Concept used:

Lines of symmetry are imaginary lines that pass through the center of the shape or object and divide it into identical halves.

Formula used:

Probability to does not retain same view = 1 - Probability to retain the same view

Calculation:

From the front and rear view of the disc, we conclude that the view is only symmetrical about 1-1 axis.

Therefore, the disc will retain the same front and rear views after the flipping operation after flipping about 1-1 axis out of three given axis.

∴ Probability to retain the same front and rear view of the disc = 1/3

∴ The probability that the disc DOES NOT retain the same front and rear views = 1 - 1/3 = 2/3

∴ The probability that the disc DOES NOT retain the same front and rear views after the flipping operation are 2/3.

Given:

An ant is at the bottom-left corner of a grid (point P) in the given figure.

It aims to move to the top-right corner of the grid.

Concept used:

The ant moves only along the lines marked in the grid. We have to justify each option by seeing from the ant's starting position to find the correct option.

Calculation:

The starting position of the ant is point P.

So, options 1 and 2 will be eliminated because there are no starting points at point P.

Also, according to the question, the ant moves only along the lines marked in the grid such that the current distance to the top-right corner strictly decreases.

Hence, option 4 is also incorrect.

If we justify option 3, there is the starting point P and the following line as same as marked in the grid.

Hence, option 3 is the correct answer.

The correct answer is heel/heal.

Key PointsLet's understand the meanings of these words:

• Heel: the rounded back part of the foot
• Heal: cause (a wound, injury, or person) to become sound or healthy again

The sentence conveys the meaning- An injury to your foot may take longer to get healthy again.

Thus, heel is the apt word for blank 1 and heal is the apt word for blank 2.

Complete sentence: As you grow older, an injury to your heel may take longer to heal.

The least possible Venn diagram for the given statement is:

Conclusions:

I. Some bottles are knives → True (As per Venn diagram).

II. Some knives are cups → True (As all the cups are knives then some knives will also be cups is a definite conclusion).

III. All cups are bottles → False (It is possible but not definite).

IV. All knives are cups → False (It is possible but not definite).

Hence, the correct answer is "Option 1".

Additional Information

Given:

Total distance = 500 m

Speed of P : Speed of Q = 3 : 4

Formula used:

Time = Distance/Speed

Calculation:

Let, Speed of P = 3x

Speed of Q = 4x

P already covered distance = 140 m

∴ P's required distance to cover = (500 - 140) = 360 m

Time required for P to cover remaining distance = 360/3x

P cover the distance = 500 m

Q will cover the distance in (360/3x) time = (360/3x) × 4x = (360/3) × 4 = 480 m

∴ Distance between P and Q = (500 - 480) = 20 m

∴​ The distance between P and Q (in m) when P wins the race is 20 m.

The correct answer is Humans engage in altruism due to empathy but not guilt.

Key Points

• Referring to the above lines, we understand that altruism may be attributed to empathy and has been motivated more by social bonding, familiarity, and identification of belongingness to a group.
• Thus, option 1 can be negated as it says Humans engage in altruism due to guilt but not empathy.
• Option 2 can't be the correct answer as it has been said in the passage that empathy and guilt have now been rejected.
• Option 3 is the correct answer**.** As it has been given in the last line of the passage that empathy and guilt have now been rejected, we can easily infer that Humans do not engage in altruism due to empathy but due to group identification.
• ​Altruism has been shown to be motivated more by social bonding, familiarity, and identification of belongingness to a group.
• Option 4 says that Humans engage in altruism due to empathy but not familiarity but it has been mentioned that altruism has been motivated more by social bonding, familiarity, and identification of belongingness to a group.

Thus, the CORRECT logical inference based on the information in the above passage is Humans engage in altruism due to group identification but not empathy.

Given:

The price of an item is 10% cheaper in store S compared to store M.

Delivery charge of store S = ₹ 150

There is no delivery charge of store M.

Calculation:

Let, the price of an item at store S = SP

The price of an item at store M = MP

The price of an item is 10% cheaper in store S compared to store M.

10% = 10/100 = 0.1

According to the question,

SP = (1 - 0.1)MP = 0.9MP

Delivery charge of store S = ₹ 150

There is no delivery charge of store M.

A person bought the item from store S and saved ₹ 100.

Therefore, MP - (SP + 150) = 100

⇒ MP - (0.9MP + 150) = 100

⇒ MP - 0.9MP - 150 = 100

⇒ 0.1MP = 100 + 150 = 250

⇒ MP = 250/0.1 = 2500

∴ The price of an item at store M = ₹ 2500

∴ The price of an item at store S = 0.9 × 2500 = 2250

∴ The price of the item at the online store S (in ₹) if there are no other charges than what is described above is ₹ 2250.

Given:

There is a regular convex hexagon in which the letters P, Q, R, S, T, and U are to be placed one per vertex but not necessarily in the same order.

Concept used:

We have to draw the diagram according to the question to find the correct answer from the given options.

Calculation:

According to the question by using given statements, the regular convex hexagon will be like,

From the above diagram, we can easily see that the line segment joining R and T is parallel to the line segment joining Q and S.

Hence, option 1 is correct answer. The line segment joining R and T is parallel to the line segment joining Q and S.

Given:

P bell is rung every 20 minutes.

Q bell is rung every 30 minutes.

R bell is rung every 50 minutes.

All of the three bells are rung at 12:00 PM.

Concept used:

We have to find the LCM of 20, 30, and 50 to find the time which after all three bells will ring together.

Formula used:

1 minute = 1/60 hour

Calculation:

Prime factors of 20 = 2 × 2 × 5

Prime factors of 30 = 2 × 3 × 5

Prime factors of 50 = 2 × 5 × 5

LCM of 20, 30, and 50 = 2 × 2 × 3 × 5 × 5 = 300

∴ All three bells will ring together after 300 minutes.

300 minutes = 300/60 h = 5 hours

∴ All three bells will ring together at (12:00 PM + 5 hours) = 5:00 PM

∴ If all the three bells are rung at 12:00 PM, the three bells will ring together again at 5:00 PM.

Given:

There are two identical dice with a single letter on each of the faces: Q, R, S, T, U, V

Formula used:

Probability of outcomes = Total number of favorable outcomes/Total number of possible outcomes

Calculation:

Total faces of each dice = 6

We know that in a single thrown of two dice, the number of possible outcomes = 6 × 6 = 36

Favorable outcomes can be composed of any combination of Q, U, and V are (Q, Q), (U, U), (V, V), (Q, U), (U, Q), (Q, V), (V, Q), (U, V), (V, U).

There are three favorable outcomes that come from each dice.

Total numbers of favorable outcomes can be composed of any combination of Q, U, and V = 9

∴ The probability that the outcomes on the dice can be composed of any combination of Q, U, and V = 9/36 = 1/4

∴ The probability that the outcomes on the dice were composed only of any combination of the following possible outcomes: Q, U, and V are 1/4.

Concept:

Small signal current gain is defined in common drain amplifier as

$\rm A_F = \frac{I_s}{I_g} = \frac{Source \ current}{Gate \ current}$

For FET, I_g = 0

∴ $\rm A_i = \frac{I_s}{0}$ = Infinite

Concept:

For a three-phase transmission line, sequence impedances is given as

Z₁ = Z_s - Z_m (+ve sequence impedance)

Z₂ = Z_s - Z_m (-ve sequence impedance)

Z₀ = Z_s + 2Z_m (zero sequence impedance)

Z_s = Self impedance, Z_m = Mutual impedance

From the above definition of impedances, it is clear that +ve and -ve impedance of transmission line are equal whereas zero sequence impedance is about 2-3 times of +ve sequence impedance.

From the given option, option (b) satisfies the required criteria.

Concept:

From the given bode plot, it is clear that magnitude is constant at all frequencies.

​

• Also, it is given a first-order system, hence there exists one finite pole.
• It can be an all-pass system.
• The transfer function of the all-pass system is

 

$\rm T.F = \frac{1-S}{1 + S}$

Phase angle of T.F is

ϕ = - tan^-1ω - tan^-1ω

At ω → ∞, ϕ = -180°

It is given in the figure also, at high-frequency phase is equal to -180°.

Therefore, it is an all-pass system having one pole at left and one zero at right at the same frequency.

Therefore, the correct option is (a).

Concept:

Wheatstone bridge is a device which is usually used for measurement of medium resistance whose value ranges in between 1 ohm to 0.1 Mohm.

Calculation:

Given,

R_AB = 1 kΩ ± 2.1 %

R_BC = 100 Ω ± 0.5 %

RDA = 300 Ω ± 0.4 %

Under the balanced condition, the product of opposite arms is equal.

∴ R_AB × R_CD = R_BC × R_DA

∴ $\rm R_{CD}=\frac{R_{BC};\times ;R_{DA}}{R_{AB}}$

$=\frac{(100±0.5\%)(300±0.4\%)}{(1000± 2.1\%)}$

= 30 ± 3%

∴ % Error = ± 3%

∴ R_CD = 30 ± $30 \times \frac{3}{100}$ = 30 ± 0.9 Ω

Calculation:

Given

T. F = $\rm G(S)=\frac{k}{S^2+4S-5}$

Routh-Hurwitz criteria can be used to determine the range of k for a stable system.

Characteristic equation , 1 + G(S) = 0

$\rm 1+\frac{k}{S^2+4S-5}=0$

⇒ S² + 4S - 5 + k = 0

Routh table

For a stable system, the element of the first column does not have any sign changes.

For that to happen

k - 5 > 0

or, k > 5

Therefore, option (c) is correct.

Concept:

Square matrix A is said to be skew-symmetric if a_ij = -a_ji for all i and j.

In other words, we can say that matrix A is said to be skew-symmetric if the transpose of matrix A is equal to the negative of matrix A i.e, A^T = -A.

Also, in a skew-symmetric matrix, the main diagonal elements are zero.

Explanation:

Given A = [a_ij]_{3 × 3}, a_ij = (i - j)³

To know about main diagonal elements, put i = j

∴ for i = j ⇒ a_ij = (i - i)³ = 0 ∀ i

For remaining elements, i ≠ j

∴ For i ≠ j ⇒ a_ij = (i - j)³ = (-(j - i))³

= -(j - i)³ = -a_ji

∴ Both the above conditions are satisfied.

Therefore matrix A is skew-symmetric matrix.

Concept:

In ideal opp-amp, the input gain is infinity so that the positive side of voltage is zero so the other voltage V_b is also equal to zero

Calculation:

By the figure v_out = I_out (X_c)      .........(1)

$\rm X_c=\frac{1}{\omega c_c}$

V_B = 0

Then $\rm \frac{V_{out}-V_B}{X_f}=I_{in}$

$\rm X_f=\frac{1}{\omega c_f}$

V_out - 0 = I_in (X_f)

V_out = I_in (X_f)     ........(2)

I_in (X_f) = I_out (X_c)

$\rm \frac{I_{out}}{I_{in}}=\frac{X_f}{X_c}$

$\rm \frac{I_{out}}{I_{in}}=\frac{C_c}{C_f}$

Concept:

To know the information about the type of filter, the magnitude of the transfer function at various frequencies should be known.

• A low-pass filter should pass the low-frequency component i.e., the gain should be finite at low frequency.
• A high-pass filter should pass the high-frequency component i.e., the gain should be finite at high frequency.

 

Explanation:

Let us observe the given transfer function at low and high frequency

$H(s) = \frac{As + B}{s^2 + Cs + D}$

At low frequency, s = 0

∴ $\rm H(0) = \frac{0 + B}{0 + 0 + D} = \frac{B}{D}$ i.e.  finite

∴ It can be a low pass filter.

At high frequency, s = ∞

∴ H(∞) = $\rm \frac{\frac{A}{s} + \frac{B}{s^2}}{1 + \frac{c}{s} + \frac{D}{s^2}} = 0$

Since it is offering zero gain at high frequency, it can not operate as a high pass filter.

Therefore, option (b) is correct.

Note:

For this question as per the official answer key by IIT, the answer is mentioned as B or D, there is a need for a small correction in the question to make the answer options B, that is the term positive constant should not be there in the question, because of that term we can't make the constant A, B, C, and D as zero, so integrator not possible. But we have chosen option B as the best possible answer as it is an MCQ. But in the real exam for both options marks were given.

Given:

 $\rm \vec J = J_a r^3 \hat z$

∴ $\rm I = \int_s \vec J . d \vec s ; d \vec s = r dr d \phi \hat z$

⇒ $\rm I = \int J_a r^3 \hat z . r dr d \phi \hat z$

$\rm = J_a \int_{r = 0}^r r^4 dr \int_{\phi =0}^{2π} d \phi$

$= \rm J_a \left. \frac{r^5}{5}\right|_0^r . \left. \phi \right|_0^{2π} = \frac{J_a (2 π)r^5}{5}$

As $\rm \oint \vec H. d \vec L = I_{enc} = \int_s \vec J . d \vec s$

⇒ H(2πr) = $\rm \frac{J_a (2 \pi)r^5}{5}$

⇒ $\rm H = \frac{J_a r^4}{5}$

Concept:

A single phase induction motor is not self starting, To provide starting torque, some modifications are done in the motor so that they have finite torque at starting.

Split phase induction motor: (low starting torque, low power factor)

Current in two windings are not equal therefore, the rotating field is not uniform and the starting torque is small of order of 1.5 to 2 times the rated running torque.

Its performance is noisy and power factor is poor.

→ Because of low starting torques, they are seldom used for drives requiring more than 1 KW.

Capacitor start motor: (High starting torque, low power factor)

• By choosing a capacitor of the proper rating auxiliary current made to lead by supply.
• Voltage thus increasing the starting characteristic.
• Then the capacitor is switched off. So running characteristic is poor.
• In capacitor start and run motor capacitor is present at both starting and running condition.
• Therefore its running characteristic is superior among all other 1-phase induction motor.

 

Shaded pole: low starting torque, low power factor.

Explanation:

For a diode to be forward biased, voltage at anode should be greater that the voltage at cathode.

Consider the case when V_in > 0.

Both the diodes will be ON and total source voltage will appear across output voltage

V₀ = V_in

Consider the case for V_in < 0,

Both the diode will be OFF. Hence output voltage will be zero.

The output will be similar to the o/p of half wave rectifier.

Concept:

Loss of Excitation for an Alternator:

• If excitation of alternator is lost then it will run asynchronously.
• If it happen for long duration then relative motion between stator field and rotor induces large currents in the rotor body and, therefore, there is high rate of heating of rotor surfaces.
• So, loss of excitation scheme is arranged to trip after certain time delay with the help of offset mho relay which is operated from AC current and voltage at the generator terminals.

• The relay setting is so arranged that the relay operates whenever the excitation goes below a certain value and the machine starts running asynchronously.
• Due to failure of excitation, alternator works as an induction generator, drawing reactive power from the grid and hence it operates at leading power factor.
• As a result of this, the impedance of the induction generator as seen by the relay shifts into the fourth quadrant of the R-X diagram and this impedance swings into off-set mho relay characteristic as shown in the figure given below and the relay will operate.

Concept:

Laplace Equation: It states that the Laplacian of the electric potential field is zero in a source-free region.

∇² V = 0

For spherical coordinate system:

∇² V = $\frac{1}{r^2sin\theta}[\frac {\partial}{\partial r}(\frac{r^2sin\theta}{1}.\frac {\partial V}{\partial r})]=0$

r² ${dv \over dr}$ = A, where A = constant

${dv \over dr}$ = ${A \over r^2}$

V(r) = ${-A \over r} + B$...............(i)

Case 1: at r = c, V = V₁ 

Putting values in equation (i), we get:

V₁ = ${-A \over c} + B$.....................(ii)

Case 2: at r = d, V = V₂ 

Putting values in equation (i), we get:

V₂ = ${-A \over d} + B$

V₁ - V₂ = ${-A \over c} + {A \over d}$

V1 - V2 = $A({c-d \over cd})$

A = $({V_1-V_2 \over c-d})cd$...........(iii)

Putting the value of equation (iii) in eq (ii), we will get the value of constant 'B' :

V1 = ${-A \over c} + B$

B = $V_1+{A \over c}$

B = V₁ + $({V_1-V_2 \over c-d})cd$ × ${1 \over c}$

B = V1 + $({V_1-V_2 \over c-d})d$...............(iv)

Putting values of equation (iii) and (iv) in equation (i), we get:

V(r) = $-({V_1-V_2 \over c-d})cd$ × ${1 \over r}$ + V₁ + $({V_1-V_2 \over c-d})d$

V(r) = $(V_1-V_2)cd\over(d-c)r$ + V₁ - $d({V_1-V_2 \over d-c})$

V(r) = $(V_1-V_2)cd\over(d-c)r$ + ${V_1d-V_1c-V_1d+V_2d \over d-c}$

$\rm V(r) = \frac{cd(V_1 - V_2)}{(d-c)r} + \frac{ V_2 d - V_1 c}{d - c}$

Concept:

Area of any region 'R' is given by:

A = $∬_R$dxdy

where A = area

 dx = differential length along x-axis

 by = differential length along y-axis

Area of any region 'R' is also given by " Green's Theorem " which is defined as:

∮_c Mdx + Ndy = $\iint_R$ $\left(\rm \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \rm dx dy$

Application:

Option 1: $∬_R$dxdy = Denotes the area of any region 'R' bounded by C

Option 2: From Green's Theorem

∮_c xdy = $\rm ∬_R (1 - 0) dx dy$ = $∬_R$ dxdy = Denotes the area of any region 'R' bounded by C

Option 3: From Green's Theorem

∮c ydx = $\rm ∬_R (0 - 1) dx dy$= $-∬_R$ dxdy = Does not denotes the area of any region 'R' bounded by C

Option 4: From Green's Theorem

$\rm \frac{1}{2} \oint_c (x dy - y dx)$ = $\frac{1}{2} ∬_R$ (1 - (-1) dxdy = $∬_R$ dxdy = Denotes the area of any region 'R' bounded by C

Therefore, option 3 is correct.

Concept:

Discrete-time Fourier series is given by:

$\rm x[n] = \sum_{k = 0}^{N - 1} a_k e^{j(2kn\pi/N)}$

where N = time period

In discrete-time Fourier series, coefficients are periodic with the same time period of input signal x[n].

Therefore, a_k = a_k+N

m = any positive integer

Calculation:

Given, N = 3 

a_-3 = 2 and a₄ = 1

a_k = a_k+N

a_-3 = a_-3+3

a_-3 = a₀ = 2

a₁ = a₁₊₃

a₁ = a₄ = 1

Expanding Fourier series by putting values of k = 0 and 1

$\rm x[n] = a_0 e^{j(2(0)n\pi/3)} + a_1 e^{j(2(1)n\pi/3)}$

$\rm x[n] = a_0 e^{j0} + a_1 e^{j(2n\pi/3)}$

$\rm x[n] = 2 + 1 e^{j(2n\pi/3)}$

$\rm x[n] = 1 +1 + 1 e^{j(2n\pi/3)}$

$\rm x[n] = 1 + e^{j(2n\pi/6)} e^{j(-2n\pi/6)} + 1 e^{j(2n\pi/3)}$

$\rm x[n] = 1 + e^{j(2n\pi/6)} \times (e^{j(2n\pi/6)}+e^{-j(2n\pi/6)})$

$\rm x[n] = 1 + e^{j(2n\pi/6)} \times 2{e^{j(2n\pi/6)}+e^{-j(2n\pi/6)}\over2}$

∵ $e^{j(2n\pi/6)}+e^{-j(2n\pi/6)}\over2$ = $cos \left( \frac{2 \pi}{6} n\right)$

$\rm x[n] =\rm 1 + 2e^{ \left( j \frac{2\pi}{6} n \right) }\cos \left( \frac{2 \pi}{6} n\right)$

Output impedence = $\frac{Z_{out}}{1 + A_{oL} β}$

Z_out = output impedence without feedback

A_oL = open loop gain

β = 1

⇒ Output impedence (with feedback) = $\frac{Z_{out}}{1 + A_{oL}}$

⇒ OP-AMP is behave as a low pass filter.

Low pass filter:

It passes the all small frequencies

⇒ $\left| \frac{V_0}{V_i} \right|$ will be maxed at low frequency and $\left| \frac{V_0}{V_i} \right|$ will be zero at maximum frequencies

As open loop gain decrease, the output impedence with feedback increases.

so that option (1) 

Concept:

for the given square matrix, the characteristic equation will be

|B - AI| = 0

B = Given matrix

I = Unit matrix

$\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix};$

A = Characteristic roots

Calculation:

|B - AI| = 0

$\begin{bmatrix}1&0&0\\ 0&4&-2\\ 0&1&1\end{bmatrix}$ - A $\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}$ = 0

$\rm \begin{bmatrix}1-A&0&0\\ 0&4-A&-2\\ 0&1&1-A\end{bmatrix}=0;$

Take the determinant of matrix, then 

(1 - A) [(4 - A) (1 - A) + 2] = 0

(1 - A) [4 - 4A - A + A² + 2] = 0

(1 - A) [4 - 5A + A² + 2] = 0

(1 - A) [A² - 5A + 6] = 0

A² - 5A + 6 - A³ + 5A² - 6A = 0

-A³ + 6A² - 11A + 6 = 0

A³ - 6A² + 11A = 6

A² - 6A + 11 = 6A^-1       ........(1)

Given 6A^-1 = A² + cA + dI     .........(2)

Compare 1 and 2

c = -6, d = +11

c + d = +5

r(t) = 1 + 0.1 sin(10t) → DC component ⇒ 1

$\rm G(s) = \frac{100}{s^2 + 0.1s + 100}$ ⇒ AC component ⇒ 0.1 sin(10t) 

ω ⇒ 10

r(t) = a + b sin(10t + θ)

Transfer function ⇒ |G(jω)|

$\rm G(j ω) = \frac{100}{- ω^2 + 0.1 j ω + 100}$

First take dc component and put ω = 0 in G(jω)

Then Transfer function = $\frac{100}{100} = 1$

So that value of a = 1

Second take AC component, and put ω = 10 in G(jω)

$\rm G(j \omega) = \left| \frac{100}{100 + 0.1 j \omega - \omega^2} \right|_{\omega = 10}^{x 0.1}$

Then b = 10

So that a = 1

b = 10

Concept

The number of encirclements of the critical point (-1+j0) is given by

N = P - Z

where N = no. of encirclements of critical point (-1 + j0) in anti-clockwise direction

P = No. of open-loop poles in the right half of s plane

Z = No. of closed-loop poles in the right half of s plane

For stability Z = 0

N = P 

Note: Above formula is valid only when the Nyquist contour is defined for the entire right half of the s-plane and excluding the poles at the origin.

The rotation of the contour must be clockwise and the encirclement of the critical point is in an anti-clockwise direction.

Calculation

Given, $\rm G(s) = \frac{1}{s(s + 1)}$

The open loop poles are present at s = 0,-1

Closed loop transfer function = $\frac{1}{s^2 + s + 1}$

No. of the closed pole on the right-hand side is 0.

Z = 0

But in the question, Nyquist contour is defined for the entire right half of the s-plane and includes the pole at the origin also.

Hence, the open loop pole at s = 0 is considered as the pole in the right half of the s-plane.

N = P - Z

N = 1 - 0

Z = 1

Mistake Point While applying the formula N = P - Z, check the region of the Nyquist contour whether it includes the origin or not, and then put the value of open-loop poles in the formula accordingly.

Concept:

Mason’s gain formula-  It is applied between input and output nodes only

Mason's gain formula

Transfer function ⇒ $\rm \frac{\sum_{K = 1}^n M_K Δ_K}{Δ}$

n = no. of forward path

M_K = K^th forward path gain

Δ_K = the value of Δ, which is not touching the K^th FBM_K

Δ = Determinant

Δ = 1 -(sum of the loop gain) + (sum of the gain product of two non-touching loops) - (sum of the gain product of non-touching loop) ...so on

Second-order equation of center system:

$\rm \frac{C(R)}{C(S)}$ ⇒ Transfer function ⇒ $\rm \frac{ω_n^2}{s^2 + 2 ξ ω_n + ω_n^2}$ ....(1)

Calculation:

The transfer function of given figure:

T.F. ⇒ $\rm \frac{\frac{10}{s} \times \frac{10}{s}}{1 + \frac{10}{s} + \frac{100}{s}}$

T.F ⇒ $\frac{100}{s^2 + 10s + 100}$   ...(2)

Compare equation (1) and (2)

$ω_n^2= 100$

ω_n = 10

2ξω_n = 10

ξ = 0.5

Concept -

If there exist eigenvector V corresponding to eigenvalue λ then the relation of matrix A is given by

AV = λV

The value of $\rm e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

Calculation:

$\rm e^A = 1 + A + \frac{A^2}{2!} + \frac{A^3}{3!} + ...$

multiply by 'V'

$\rm e^A.V = V+ AV + \frac{A^2V}{2!} + \frac{A^3V}{3!} + ...$

AV = λV

$\rm e^A.V = V+ λ V + \frac{λ^2V}{2!} + \frac{λ^3V}{3!} + ...$

$\rm e^A.V = V \left[ 1 + λ + \frac{λ^2}{2!} + .... \right]$

e^A.V = V.e^λ

so Both the statement are correct

Concept:

Test for Increasing and Decreasing functions:-

Let f(x) be continuous on closed internal (a, b) and differentiable on the open interval i.e a < x < b. Then if,

1. f'(x) > 0 for all x ∈ (a, b) then f(x) is Increasing on (a, b) 2. f'(x) < 0 or all x ∈ (a, b) then f(x) is Decreasing on (a, b)

Calculation:

$\rm f'(x)=\frac{d}{dx}f(x)<0$

⇒ $\rm e^x(x-1)(x-2)\frac{d}{dx}(x)-[e^0(0-1)(0-2)]\times\left[\frac{d}{dx}(0)\right]$

f'(x) = [e^x (x - 1) (x - 2)] - 0 < 0

= (x - 1) (x - 2) < 0

= x > 1 and x < 2

= x ∈ (1, 2)

Concept

Incremental cost:

The ratio of a small change in input to the corresponding small change in output is called Incremental cost

Incremental cost = $\rm \frac{dC}{dP}$

C = Fuel cost

$\rm \frac{dC}{dP}=λ$

Calculation:

Given C₁ = 350 + 6P₁ + 0.004$\rm P_1^2$

$\rm \frac{dC_1}{dP}=6+0.008P_1=λ$

Given λ = 8

6 + 0.008P₁ = 8

P₁ = 250 MW

Total Demand = 500 MW

P₁ = 250 MW

P₂ = 550 - 250 = 300 MW

(no need to calculate the value of a)

P₁ = 250 MW, P₂ = 300 MW

Concept:

GMR is defined as the effective distance over which self magnetic flux linkages occur

For a solid conductor with radius r,

GMR = r' = 0.7788r

GMR is less than the physical radius of the conductor.

In the given figure, standard conductor with four identical strands touching each other is given with equal radius r.

$\rm GMR_{a_1} = ( r' \times 2r \times 2r \times 2\sqrt 2 r)^{\frac{1}{4}}$

= 1.722 r

Since each strands are identical.

∴ GMR of conductor will be equal to GMR of strand i.e.,

GMR_cond = GMR_a1

= 1.722 r

Therefore, correct option is (d)

Concept

Ampere's law:-

It states that the line integral of magnetic field intensity around any closed path = direct current enclosed by that path

$\rm \oint_L \vec{H} \vec{dl}=\int_s \vec J.\vec {ds}$

$\rm \int_s \nabla\times \vec H.\vec {ds}\Rightarrow \int_s \vec J.\vec {ds}$

$\rm \nabla \times \vec H=\vec J$ → maxwall's equation

It is the Differential form of ampere's law

Calculation

$\rm \vec J=\nabla\times \vec H$

\(\rm |\vec J|=\begin{vmatrix}̂ i&̂ j&̂ k\\ \frac{\partial}{\partial x}&\frac{\partial}{\partial y}&\frac{\partial}{\partial z}\\ x^2& x^2y^2&x^2y^2z^2\end{vmatrix}\)

then solve the matrix

$\rm \vec J=̂ i(2x^2yz^2-0)-̂ j(2xy^2z^2-0)+̂ k(2xy^2-0)$

= 4î - 8ĵ +8k̂

$\rm |\vec j|=\sqrt{4^2+8^2+8^2}$

= 12

Current Density = 12

Explanation:

In this question, it is asked at a switching frequency of as high as 200 kHz, which device will be best suited.

Let's see all the options:-

Thyristor - It has stored charged carrier near its gate-cathode junction. Therefore, for removal of these charge carrier it takes source time. It is not suitable for fast switching.

BJT - It is having law conduction has but high switching loss. Therefore it is not suitable for high switching frequency.

MOSFET & IGBT both devices are suitable for operating at high frequency but compared to MOSFET, IGBT having less suitability at high frequency. Therefore MOSFET is best suited for above mentioned frequency.

Concept:

The transfer function for a control system is defined as:

H(s) = ${Y(s) \over X(s)}$

where H(s) = Transfer function

Y(s) = Output function

X(s) = Input function

The impulse response for a control system is also known as the transfer function of the system.

Explanation:

$\rm y(t) + \frac{1}{4} \frac{dy}{dt} = 2x(t)$

Taking Laplace Transform on both sides:

∵ $dy\over dt$ = sY(s)

$\therefore \rm Y(s) + \frac{1}{4} sY(s) = 2X(s)$

Y(s) $(1+{s\over 4})$  = 2X(s)

Y(s) = 2X(s)

${Y(s)\over X(s)}$ = ${8\over s+4}$

Taking Inverse Laplace Transform on both sides:

${ 1\over s+a}$ = e^-at u(t)

${y(t)\over x(t)}$ = 8e^-4t u(t)

Concept:

Fourier Transform of any time-domain function f(t) = ${A sin ω_ct \over π t}$ is defined by:

If f(t) is multiplied by any cosine term, then modified F(ω) is given by:

f(t) cos(ωt) = ${f(w_c-w)+f(w_c+w)\over 2}$

Calculation:

Given, f(t) = ${2sin 10π t \over π t}$

F(ω_c) is given as:

f(t) cos(40πt) = ${f(w_c-40π )+f(w_c+40π )\over 2}$

Applying shifting property in F(ω), we get f(ω_c - 40π) as:

      ---(i)

Applying shifting property in F(ω), we get f(ωc + 40π) as:

      ---(ii)

Adding equations (i) and (ii) and dividing their sum by 2, we get:

 

• input x(t) = 2 sin(10πt) + 5 cos(15πt) + 7 sin(42πt) + 4 cos(45πt) has four different frequencies 10π , 15π , 42π and 50π .
• Frequency 10π and 15π lie outside the frequency band of above shown Fourier transform and will not pass, only frequency components 42π and 50π will pass.

Therefore, output y(t) = 7 sin(42πt) + 4 cos(45πt)

Concept:

A system is said to be linear if it satisfies the homogeneity and superposition principle.

• If any input x(t) is multiplied by factor 'a' and output y(t) also gets multiplied by the same factor 'a', then the system is said to be following the homogeneity principle.
• For any input { x₁(t) + x₂(t) } to a system, if output is { y1(t) + y2(t) }, then the system is said to be following superposition principle.

A system is said to be causal if the output of the system at all instant depends only on past and present values of input.

Application:

Checking for linearity:

Condition 1: Homogeneity Principle

Condition 2: Superposition Principle

The above system equation satisfies the principle of homogeneity and superposition.

Therefore, y(t) = x(et) is linear.

Checking for causality:

y(0) = x(e⁰)

y(0) = x(1)

Since the output of the system is depending upon future values of input.

Therefore, y(t) = x(et) is non-causal.

Important Points If any operation is performed on system input x(t), then it makes the system non-linear.

In the above-given problem, operation (e^t) is being performed on the time period 't' instead of x(t), therefore the system remains to be linear in nature.

Mistake Points If exponential operation 'e' is performed on x(t) in the form of e^x(t), then the system becomes non-linear.

Concept:

The gradient of a function f(x, y, z) is given by:

$\rm ∇ f = \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j + \frac{\partial f}{\partial z} k$

where i, j, and k are the unit vectors along the x, y, and z-axis respectively.

The gradient of any function f(x, y, z) represents the direction of the greatest rate of increase of any scalar field function.

Calculation:

Given, f(x, y, z) = 4x² + 7xy + 3xz²

\(∇ f = \frac{\partial (4x^2 + 7 xy + 3xz^2)}{\partial x} ̂ i + \frac{\partial (4x^2 + 7 xy + 3xz^2)}{\partial y} ̂ j + \frac{\partial (4x^2 + 7 xy + 3xz^2)}{\partial z} ̂ k\)

∇ f = (8x + 7y + 3z²)î + (7x)ĵ + (6xz)k̂

Gradient of a function f(x, y, z) at P = (1, 0, 2) is:

∇ f = 20î + 7ĵ + 12k̂

Calculation:

In the given figure, a γ - connected balanced source is connected to γ - connected unbalanced load.

In a balanced source, voltages are equal in magnitude and phase displaced by 120°.

E_A = 100 ∠0°

E_B = 100 ∠-120°

E_C = 100 ∠120° = 100 ∠-240°

In γ - connection, phase voltage = $\rm \frac{Line\ voltage}{\sqrt3}$

Given, Z_A = 10 Ω = 10 ∠0°, ZB = 20 ∠60°

∴ $\rm I_A=\frac{E_A}{Z_A}=\frac{100 ∠ 0^{\circ}}{10∠ 0^{\circ}}=10∠ 0^\circ\$

∴ $\rm I_B=\frac{E_B}{Z_B}=\frac{100 ∠ -120^{\circ}}{20∠ 60^{\circ}}=5∠ -180^\circ\$

Since the potential difference between n and n' is zero

therefore,

$\vec{I_A}+\vec{I_B}+\vec{I_C}=0$

⇒ $\vec{I_C}=-(\vec{I_A}+\vec{I_B})\$

$=-(10∠ 0^{\circ}+5∠ -180^{\circ})$

= 5 ∠180°

∴ $\rm Z_C=\frac{E_C}{I_C}=\frac{100∠ -240^{\circ}}{5∠ 180^{\circ}}$

= 20 ∠-60° Ω

Therefore, Correct option is (b)

Concept:

Consider a vector field $\rm \vec E (x, y, z) = E_x \hat i + E_y\hat j + E_z\hat k$

Divergence of any vector field $(\rm \vec E)$ is given by:

$\rm \vec \nabla . \vec E = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z}$

Differential volume for a cartesian coordinate system (x,y,z) is defined as:

dV = dxdydz

Calculation:

$\rm \vec \nabla . \vec E = \frac{\partial (2x^2)}{\partial x} + \frac{\partial (5y)}{\partial y} + \frac{\partial (3z)}{\partial z}$

$\rm \vec \nabla . \vec E = 4x + 5 + 3$

$\rm \vec \nabla . \vec E = 4x + 8$

∭_v $(\vec \nabla . \vec E) dV$ = ∭_v (4x + 8) dxdydz

∭v $(\vec \nabla . \vec E) dV$ = $\rm \int_0^1\int_0^1\int_0^1 (4x + 8) dx dy dz$

∭v $(\vec \nabla . \vec E) dV$ = $\rm \int_0^1 (4x + 8) dx \int_0^1 dy \int_0^1 dz$

∭v $(\vec \nabla . \vec E) dV$ = $\rm \left( \frac{4x^2}{2} + 8x \right) _0^1 (y)_0^1 (z)_0^1$

∭v $(\vec \nabla . \vec E) dV$ = 10

Concept:

The circuit given in the below figure represents an integration of the input supply.

According to virtual short condition,

V⁺ = V^- = 0

Apply nodal analysis at Inverting terminal,

I₁ = I₂

$\rm \frac{0.1 - 0}{R_1} = C_1 \frac{d}{dt} (0 - V_0)$

$\Rightarrow \frac{0.1}{R_1} = - C_1 \frac{d}{dt} V_0$

Integrating on both sides, we get

$\rm V_0 = - \frac{1}{R_1C_1} \int0.1 \ dt$

$\rm V_0 = - \frac{0.1}{R_1C_1} × \ t$

or, V₀ = -k × t

The o/p will be constant at -V_EE