GATE EE 2021 Set 1

The relative pronoun who is used to refer to people when they are the subject of the verb in the relative clause. In this sentence, "who" refers to "The people" and is the subject of the verb "were".

• whose is a possessive relative pronoun.
• which is used for things or animals.
• whom is used for people as the object of a verb or preposition.

When the transparent square sheet is folded along the vertical dotted line from right to left, the pattern on the right half is mirrored onto the left half. The original pattern consists of a vertical line on the left and two diagonal lines forming a 'V' shape on the right. When folded, the 'V' shape will be reflected across the dotted line and superimposed on the left half. This results in the pattern shown in option (C).

The formula for the interior angle of a regular $n$-sided polygon is:$$\text{Interior Angle} = \frac{(n-2) \times 180^\circ}{n}$$Given $n = 10$:$$\text{Interior Angle} = \frac{(10-2) \times 180^\circ}{10} = \frac{8 \times 180^\circ}{10} = 8 \times 18^\circ = 144^\circ$$

Let $x = 11^{13}$. We are looking for an expression divisible by $(x + 1)$.
Using the algebraic properties of divisibility:Evaluating the options:
(A) $11^{26} + 1 = (11^{13})^2 + 1 = x^2 + 1$. Since $n=2$ is even, it is not divisible by $(x + 1)$.
(B) $11^{33} + 1$. Since 33 is not a multiple of 13, this is not a simple integer power of $x$.
(C) $11^{39} - 1 = (11^{13})^3 - 1 = x^3 - 1$. Since $n=3$ is odd, it is divisible by $(x - 1)$, not $(x + 1)$.
(D) $11^{52} - 1 = (11^{13})^4 - 1 = x^4 - 1$. Since $n=4$ is even, $x^4 - 1$ is divisible by $(x + 1)$ because $x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x - 1)(x + 1)(x^2 + 1)$.Thus, $11^{52} - 1$ is exactly divisible by $11^{13} + 1$.

The relationship is based on a geographical feature and its surrounding environment. An oasis is a fertile area of land found in the middle of a desert (surrounded by sand). Similarly, an island is a piece of land surrounded by water. Therefore, 'Water' is the correct logical fit.

The passage argues that sleep deprivation negatively impacts the ability to recall information and that sacrificing sleep for extra study time is counterproductive. This implies that for preparation to be effective (i.e., to ensure recall during the exam), adequate sleep is necessary. Option D directly reflects this inference. Options A, B, and C are either unsupported by the text or directly contradict the passage's logic.

The area of a square formed by joining the midpoints of the sides of a larger square is exactly half the area of that larger square.By counting the squares in the diagram, the shaded square is the $6^{\text{th}}$ square in the sequence. Thus, its area is $3.125 \text{ cm}^2$.

The probability density function (PDF) is given as $f(x) = 0.01$ for $0 \leq X \leq 100$.
This is the definition of a uniform distribution over the interval $[a, b]$, where $a=0$ and $b=100$.
For a uniform distribution, the PDF is $f(x) = \frac{1}{b-a}$.
In this case, $f(x) = \frac{1}{100-0} = \frac{1}{100} = 0.01$, which matches the given PDF.The mean (expected value) of a continuous random variable $X$ with PDF $f(x)$ is given by:
$E[X] = \int_{-\infty}^{\infty} x f(x) dx$
For a uniform distribution over $[a, b]$, the mean is also given by the formula:
$E[X] = \frac{a+b}{2}$Substituting $a=0$ and $b=100$:
$E[X] = \frac{0+100}{2} = \frac{100}{2} = 50$Thus, the mean of $X$ is $50.0$.The final answer is $\boxed{\text{50.0}}$.

Let's extract the data from the bar chart:Year 1:
Students Passed = 50
Students Failed = 10
Total Students in Year 1 = 50 + 10 = 60Year 2:
Students Passed = 60
Students Failed = 5
Total Students in Year 2 = 60 + 5 = 65Year 3:
Students Passed = 50
Students Failed = 3
Total Students in Year 3 = 50 + 3 = 53Now, let's apply the given conditions:Calculate students who took the exam for the first time in Year 2:
Total students who appeared in Year 2 = 65.
From condition 2, students who failed in Year 1 (10 students) must reappear in Year 2. From condition 3, they will pass in Year 2.
So, the 10 students who failed in Year 1 are part of the 65 students in Year 2, and they are not taking the exam for the first time.
Number of students who took the exam for the first time in Year 2 = (Total students in Year 2) - (Students who failed in Year 1 and reappeared)
= 65 - 10 = 55.Calculate students who took the exam for the first time in Year 3:
Total students who appeared in Year 3 = 53.
From condition 2, students who failed in Year 2 (5 students) must reappear in Year 3. From condition 3, they will pass in Year 3.
So, the 5 students who failed in Year 2 are part of the 53 students in Year 3, and they are not taking the exam for the first time.
Number of students who took the exam for the first time in Year 3 = (Total students in Year 3) - (Students who failed in Year 2 and reappeared)
= 53 - 5 = 48.Therefore, the number of students who took the exam for the first time in Year 2 and Year 3 respectively are 55 and 48.The final answer is $\boxed{\text{55 and 48}}$.

Let's analyze the constraints provided:

• If we use the sequence $V-S-Q$, then $R$ would be to the left of $V$ to maintain the $V-S$ adjacency, which contradicts the rule.
• Therefore, the sequence must be $Q-S-V-R$, forming the block $(QSVR)$.

4. In the block $(QSVR)$, there is exactly one car ($S$) between $Q$ and $V$. Statement (A) claims there are two cars, which is logically impossible under these constraints.

The characteristic equation of the matrix $A = \begin{bmatrix} p & q \\ q & -p \end{bmatrix}$ is given by $|A - \lambda I| = 0$.$$\begin{vmatrix} p - \lambda & q \\ q & -p - \lambda \end{vmatrix} = 0$$$$(p - \lambda)(-p - \lambda) - q^2 = 0$$$$-(p^2 - \lambda^2) - q^2 = 0$$$$\lambda^2 - (p^2 + q^2) = 0$$Given that $p^2 + q^2 = 1$, the equation becomes:$$\lambda^2 - 1 = 0 \implies \lambda = \pm 1$$Thus, the eigenvalues are 1 and -1.

Consider the polynomial $p(z) = z^3 + (1 + j)z^2 + (2 + j)z + 3$.

Given $f'(x_0) = 0$ for $x_0 \in (0, 1)$ and $f''(x) > 0$ for all $x \in (0, 1)$:Therefore, $f(x)$ has exactly one local minimum in $(0, 1)$.

To find the Thevenin voltage ($V_{th}$), we calculate the open-circuit voltage across terminals 'ab'. Since the terminals are open, the entire $5\text{A}$ current from the source flows through the $10\ \Omega$ resistor. Thus, $i_1 = 5\text{A}$. The voltage across the $10\ \Omega$ resistor is $V_{10\Omega} = 10 \times 5 = 50\text{V}$. The dependent source voltage is $3i_1 = 3 \times 5 = 15\text{V}$. Following the KVL path from 'b' to 'a', $V_{th} = V_{ab} = 10i_1 + 3i_1 = 13i_1 = 13 \times 5 = 65\text{V}$.To find the Thevenin impedance ($R_{th}$), we deactivate the independent $5\text{A}$ source (open circuit) and apply a test current $I_t$ at terminal 'a'. The current $I_t$ flows through the $2\ \Omega$ resistor, the dependent source, and the $10\ \Omega$ resistor. Here, $i_1 = -I_t$. The voltage at terminal 'a' is $V_a = 2I_t + 3i_1 + 10(-i_1) = 2I_t + 3(-I_t) + 10(I_t) = 15I_t$. Thus, $R_{th} = V_a / I_t = 15\ \Omega$. Therefore, the equivalent is 65 V in series with $15\ \Omega$.

A valid magnetic field $\vec{B}$ must satisfy the condition $\nabla \cdot \vec{B} = 0$ (Gauss's Law for Magnetism). Let's check each option:
(A) $\nabla \cdot \vec{B} = \frac{\partial}{\partial x}(10x) + \frac{\partial}{\partial y}(20y) + \frac{\partial}{\partial z}(-30z) = 10 + 20 - 30 = 0$. This is a valid magnetic field.
(B) $\nabla \cdot \vec{B} = \frac{\partial}{\partial x}(10y) + \frac{\partial}{\partial y}(20x) + \frac{\partial}{\partial z}(-10z) = 0 + 0 - 10 = -10 \neq 0$.
(C) $\nabla \cdot \vec{B} = \frac{\partial}{\partial x}(10z) + \frac{\partial}{\partial y}(20y) + \frac{\partial}{\partial z}(-30x) = 0 + 20 + 0 = 20 \neq 0$.
(D) $\nabla \cdot \vec{B} = \frac{\partial}{\partial x}(10x) + \frac{\partial}{\partial y}(-30z) + \frac{\partial}{\partial z}(20y) = 10 + 0 + 0 = 10 \neq 0$.

Therefore, the system is non-linear and time-invariant.

The impulse response of two cascaded LTI systems is the convolution of their individual impulse responses.
$h[n] = h_1[n] * h_2[n]$
$h[n] = (\delta[n-1] + \delta[n+1]) * (\delta[n] + \delta[n-1])$
Using the property $\delta[n-k] * \delta[n-m] = \delta[n-(k+m)]$:
$h[n] = \delta[n-1]*\delta[n] + \delta[n-1]*\delta[n-1] + \delta[n+1]*\delta[n] + \delta[n+1]*\delta[n-1]$
$h[n] = \delta[n-1] + \delta[n-2] + \delta[n+1] + \delta[n]$
Rearranging terms: $h[n] = \delta[n-2] + \delta[n-1] + \delta[n] + \delta[n+1]$.
This matches option (C).

• Damper bars (P) are used in Synchronous machines (U) to provide damping torque and mitigate Hunting (X).
• Skewed rotor slots (Q) are used in Induction motors (S) to reduce the effect of space harmonics, which helps in mitigating Magnetic locking (Y) (cogging).
• Compensating windings (R) are used in DC machines (V) to neutralize the cross-magnetizing effect of Armature reaction (Z) in the interpolar region.

Thus, the correct combination is P-U-X, Q-S-Y, R-V-Z.

In the Newton-Raphson load flow method (polar form), the Jacobian matrix $J$ is:$$\begin{bmatrix} \Delta P \\ \Delta Q \end{bmatrix} = \begin{bmatrix} H & S \\ M & R \end{bmatrix} \begin{bmatrix} \Delta \delta \\ \Delta V \end{bmatrix}$$Where:

• $\Delta P$ is the vector of active power mismatches for all buses except the slack bus. Dimension: $(N-1) \times 1$.
• $\Delta Q$ is the vector of reactive power mismatches for PQ buses only. Dimension: $N_L \times 1$.
• $\Delta \delta$ is the vector of voltage angle corrections for all buses except the slack bus. Dimension: $(N-1) \times 1$.
• $\Delta V$ is the vector of voltage magnitude corrections for PQ buses only. Dimension: $N_L \times 1$.

The sub-matrix $M$ relates $\Delta Q$ to $\Delta \delta$:
$M = \frac{\partial Q}{\partial \delta}$
Its dimension is (number of $Q$ equations) $\times$ (number of $\delta$ variables) = $N_L \times (N-1)$.

For economic operation when network losses are ignored, the incremental fuel costs of the generators must be equal:$$\frac{dF_1}{dP_1} = \frac{dF_2}{dP_2}$$$$40 + 0.2P_1 = 32 + 0.4P_2$$$$0.2P_1 - 0.4P_2 = -8$$$$P_1 - 2P_2 = -40 \quad \text{--- (1)}$$The total load requirement is:$$P_1 + P_2 = 260 \quad \text{--- (2)}$$Subtracting equation (1) from (2):$$(P_1 + P_2) - (P_1 - 2P_2) = 260 - (-40)$$$$3P_2 = 300 \Rightarrow P_2 = 100 \text{ MW}$$Substituting $P_2 = 100$ into equation (2):$$P_1 + 100 = 260 \Rightarrow P_1 = 160 \text{ MW}$$Thus, the generations are $P_1 = 160$ MW and $P_2 = 100$ MW. Correct option is (B).

From the block diagram, the error signal $E(s)$ is the difference between the reference input $R(s)$ and the feedback signal $B(s)$:$$E(s) = R(s) - B(s)$$The feedback signal $B(s)$ is the output $C(s)$ multiplied by the feedback gain $H(s)$:$$B(s) = C(s)H(s)$$The output $C(s)$ is the error signal $E(s)$ multiplied by the forward gain $G(s)$:$$C(s) = E(s)G(s)$$Substituting $C(s)$ into the expression for $B(s)$:$$B(s) = E(s)G(s)H(s)$$Now substitute $B(s)$ back into the error equation:$$E(s) = R(s) - E(s)G(s)H(s)$$$$E(s) + E(s)G(s)H(s) = R(s)$$$$E(s)[1 + G(s)H(s)] = R(s)$$$$\frac{E(s)}{R(s)} = \frac{1}{1 + GH}$$Correct option is (C).

• Maxwell bridge is specifically used for the measurement of medium inductance (coils with $1 < Q < 10$).
• Schering bridge is used for measuring capacitance and dissipation factor.
• Kelvin bridge is used for measuring very low resistances.
• Wien bridge is primarily used for measuring frequency and can also measure capacitance.

Therefore, inductance is measured by Maxwell bridge. Correct option is (B).

For two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to intersect orthogonally, the condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
Circle 1: $x^2 + y^2 - 1 = 0 \Rightarrow g_1 = 0, f_1 = 0, c_1 = -1$.
Circle 2: $(x - 1)^2 + (y - 1)^2 - r^2 = 0 \Rightarrow x^2 + y^2 - 2x - 2y + 2 - r^2 = 0 \Rightarrow g_2 = -1, f_2 = -1, c_2 = 2 - r^2$.
Applying the condition: $2(0)(-1) + 2(0)(-1) = -1 + 2 - r^2 \Rightarrow 0 = 1 - r^2 \Rightarrow r^2 = 1$.
Since $(u, v)$ is the point of intersection, it lies on both circles:
1) $u^2 + v^2 = 1$
2) $(u - 1)^2 + (v - 1)^2 = r^2 = 1 \Rightarrow u^2 - 2u + 1 + v^2 - 2v + 1 = 1$
Substitute $u^2 + v^2 = 1$ into the second equation:
$1 - 2u - 2v + 2 = 1 \Rightarrow 3 - 2(u + v) = 1 \Rightarrow 2(u + v) = 2 \Rightarrow u + v = 1$.