GATE EE 2014 Set 2

Post-colonialism refers to the period following the end of colonial rule. India is described as a post-colonial country because it was previously a colony under British rule until its independence in 1947.

The sentence requires an interrogative phrase to complete the question. 'Who did you say was coming...' is the grammatically correct structure for asking about someone's previous statement regarding an arrival.

Matching the words with their meanings:Therefore, the correct matching is 1:S, 2:P, 3:Q, 4:R.

The multiples of 10 between 2 and 198 are 10, 20, 30, ..., 190.
This is an arithmetic progression (AP) where:
First term ($a$) = 10
Last term ($l$) = 190
Common difference ($d$) = 10To find the number of terms ($n$):
$l = a + (n - 1)d$
$190 = 10 + (n - 1)10$
$180 = (n - 1)10$
$18 = n - 1$
$n = 19$The average of an AP is given by $\frac{a + l}{2}$:
Average = $\frac{10 + 190}{2} = \frac{200}{2} = 100$.

Let $x = \sqrt{12 + \sqrt{12 + \sqrt{12 + \dots}}}$
Squaring both sides, we get:
$x^2 = 12 + \sqrt{12 + \sqrt{12 + \dots}}$
$x^2 = 12 + x$
$x^2 - x - 12 = 0$
$(x - 4)(x + 3) = 0$This gives $x = 4$ or $x = -3$.
Since the square root of a positive sum must be positive, $x = 4$.

The passage states that Kaliningrad is a Russian territory but is bordered by the Baltic Sea, Poland, and Lithuania. This geographical description implies that it is an exclave, separated from the main body of Russia by other countries and the sea. Thus, it is not contiguous with the rest of Russia. (A) is incorrect because the passage says it had a German majority before the war.
(C) is incorrect because the passage doesn't state Kaliningrad was the original Russian name.
(D) is not necessarily true based solely on the text provided, as there could be sea or air routes.

The conclusion is that mosquito control measures failed. We need a statement that does not contradict this.
(A) Suggests cases are imported, which would contradict the failure of local control measures.
(B) Suggests the increase is due to better reporting, not necessarily more mosquitoes.
(C) Suggests the increase is due to better testing, not necessarily more mosquitoes.
(D) Malaria is also a mosquito-borne disease. An increase in malaria cases supports the idea that the mosquito population has increased generally, which aligns with the conclusion that control measures failed. Thus, it does not contradict the conclusion.

Given $|x^2 - 2x + 3| = 11$.
Case 1: $x^2 - 2x + 3 = 11 \Rightarrow x^2 - 2x - 8 = 0 \Rightarrow (x-4)(x+2) = 0 \Rightarrow x = 4$ or $x = -2$.
Case 2: $x^2 - 2x + 3 = -11 \Rightarrow x^2 - 2x + 14 = 0$. The discriminant $D = (-2)^2 - 4(1)(14) = 4 - 56 = -52 < 0$. No real roots for $x$.
Now evaluate $|-x^3 + x^2 - x|$ for the real values of $x$:
For $x = 4$: $|-(4)^3 + (4)^2 - 4| = |-64 + 16 - 4| = |-52| = 52$.
For $x = -2$: $|-(-2)^3 + (-2)^2 - (-2)| = |8 + 4 + 2| = |14| = 14$.
Thus, the possible values include 14 and 52.

From the graph, the ratio of male ($M$) to female ($F$) students is:
In 2008: $R_1 = \frac{M_{2008}}{F_{2008}} = 2.5 \Rightarrow M_{2008} = 2.5 F_{2008}$
In 2009: $R_2 = \frac{M_{2009}}{F_{2009}} = 3.0 \Rightarrow M_{2009} = 3.0 F_{2009}$
Given that the number of female students doubled in 2009: $F_{2009} = 2 F_{2008}$
Substitute $F_{2009}$ into the equation for $M_{2009}$:
$M_{2009} = 3.0 \times (2 F_{2008}) = 6.0 F_{2008}$
Percentage increase in male students = $\frac{M_{2009} - M_{2008}}{M_{2008}} \times 100$
$= \frac{6.0 F_{2008} - 2.5 F_{2008}}{2.5 F_{2008}} \times 100 = \frac{3.5}{2.5} \times 100 = 1.4 \times 100 = 140\%$.

The angle $\theta$ between the hour hand and the minute hand of a clock at time $h:m$ is given by the formula:$$\theta = |30h - 5.5m|$$For the time between $6\text{ a.m.}$ and $7\text{ a.m.}$, we have $h = 6$. We want to find the time $m$ such that $\theta$ is closest to $60^\circ$.Evaluating the angle for each given option:

• Option (A): $6:22\text{ a.m.}$

$m = 22$
$\theta = |30(6) - 5.5(22)| = |180 - 121| = 59^\circ$
Difference from $60^\circ = |60 - 59| = 1^\circ$

• Option (B): $6:27\text{ a.m.}$

$m = 27$
$\theta = |30(6) - 5.5(27)| = |180 - 148.5| = 31.5^\circ$
Difference from $60^\circ = |60 - 31.5| = 28.5^\circ$

• Option (C): $6:38\text{ a.m.}$

$m = 38$
$\theta = |30(6) - 5.5(38)| = |180 - 209| = |-29| = 29^\circ$
Difference from $60^\circ = |60 - 29| = 31^\circ$

• Option (D): $6:45\text{ a.m.}$

$m = 45$
$\theta = |30(6) - 5.5(45)| = |180 - 247.5| = |-67.5| = 67.5^\circ$
Difference from $60^\circ = |60 - 67.5| = 7.5^\circ$Comparing the differences, $1^\circ$ is the smallest. Therefore, the angle is closest to $60^\circ$ at $6:22\text{ a.m.}$

To determine which statement does not contradict the municipal authorities' conclusion, let us first analyze the core argument and the conclusion.

Analysis of the Argument

• Observation: The number of diagnosed dengue cases in north India has doubled compared to last year.
• Conclusion: Measures to control the mosquito population have failed in this region.

The conclusion relies on the assumption that the increase in diagnosed cases is directly caused by an increase in the mosquito population (due to failed control measures). To contradict (weaken) this conclusion, a statement must provide an alternative explanation for the rise in diagnosed cases that does not involve a failure in local mosquito control.
To not contradict (support or remain neutral/consistent with) the conclusion, a statement must align with the idea that the local mosquito population has indeed increased or that the control measures have failed.---

Evaluating the Options

• Option A: "A high proportion of the affected population has returned from neighbouring countries where dengue is prevalent"
• Impact: This suggests that the increase in cases is due to infections contracted abroad, not because of a failure in local mosquito control. This contradicts the authorities' conclusion.
• Option B: "More cases of dengue are now reported because of an increase in the Municipal Office’s administrative efficiency"
• Impact: This implies that the actual number of infections might not have doubled; rather, the reporting system has simply become better at documenting them. This provides an alternative administrative reason for the rise, contradicting the conclusion that mosquito control failed.
• Option C: "Many more cases of dengue are being diagnosed this year since the introduction of a new and effective diagnostic test"
• Impact: This suggests the rise in diagnosed cases is due to improved medical detection capabilities rather than an actual increase in transmission by local mosquitoes. This contradicts the authorities' conclusion.
• Option D: "The number of people with malarial fever (also contracted from mosquito bites) has increased this year"
• Impact: Malaria, like dengue, is a vector-borne disease transmitted by mosquito bites. An increase in malaria cases strongly indicates a general rise in the mosquito population or increased mosquito activity in the region. This directly supports and does not contradict the conclusion that measures to control the mosquito population have failed.

---

Conclusion

Only Option D is consistent with and does not contradict the municipal authorities' conclusion.

A fundamental property of real symmetric matrices is that all their eigenvalues are real. They are not necessarily positive (could be zero or negative), not necessarily distinct (can have repeated eigenvalues), and their sum is the trace of the matrix, which is not necessarily zero.

Let the probability of face $n$ showing up be $P(n)$.
Given that $P(n) \propto n$, we can write $P(n) = k \cdot n$ for some constant $k$.
For a standard six-sided dice, the sum of probabilities of all possible outcomes must be 1:$$\sum_{n=1}^{6} P(n) = 1$$$$k(1 + 2 + 3 + 4 + 5 + 6) = 1$$$$k(21) = 1 \implies k = \frac{1}{21}$$The probability of the face with three dots showing up is:$$P(3) = k \cdot 3 = \frac{3}{21} = \frac{1}{7}$$Calculating the decimal value:$$\frac{1}{7} \approx 0.1428$$The answer lies within the range $[0.13, 0.15]$. Official answer key specifies the range 0.13 to 0.15.

The function $f(x) = (x - 1)^{2/3} = \sqrt[3]{(x - 1)^2}$.
Since $(x - 1)^2 \geq 0$ for all real $x$, the minimum value of the function is 0.
This minimum occurs when $(x - 1)^2 = 0$, which implies $x = 1$.

The multi-valued complex function $1^i$ can be expressed as:
$1^i = (e^{i2n\pi})^i = e^{i^2 2n\pi} = e^{-2n\pi}$ for $n = 0, \pm 1, \pm 2, \dots$
Since $e^x$ is always real and positive for any real $x$, all values of $1^i$ are real and non-negative.

This is a Cauchy-Euler differential equation. Let $y = x^m$. Substituting into the equation gives the characteristic equation:
$m(m-1) + m - 1 = 0$
$m^2 - m + m - 1 = 0$
$m^2 - 1 = 0 \Rightarrow m = \pm 1$
The general solution is $y = c_1 x + c_2 x^{-1}$.
Among the given options, $1/x$ (which corresponds to $m = -1$) is a solution.

Let the self-inductances of the two identical inductors be $L_1 = L_2 = L$ and their mutual inductance be $M$.
The equivalent inductance for series aiding connection is $L_{eq1} = L_1 + L_2 + 2M = 2L + 2M = 380\ \mu\text{H}$.
The equivalent inductance for series opposing connection is $L_{eq2} = L_1 + L_2 - 2M = 2L - 2M = 240\ \mu\text{H}$.
Subtracting the two equations:
$(2L + 2M) - (2L - 2M) = 380 - 240$
$4M = 140$
$M = 35\ \mu\text{H}$.

To find the ratio of the permittivities $\epsilon_1 : \epsilon_2$, we can model the system as three capacitors connected in series.

1. System Geometry and Parameters

Let the area of the plates be $A$. The total distance between the plates is $d$.
The middle dielectric slab of permittivity $\epsilon_2$ has a thickness of $d/2$ and is placed symmetrically. Thus, the remaining space of thickness $d - d/2 = d/2$ is shared equally by the two outer regions of permittivity $\epsilon_1$.This gives the thicknesses of the three regions as:

• First region (permittivity $\epsilon_1$): $d_1 = \frac{d}{4}$
• Middle region (permittivity $\epsilon_2$): $d_2 = \frac{d}{2}$
• Third region (permittivity $\epsilon_1$): $d_3 = \frac{d}{4}$

2. Capacitance of Each Region

The capacitance of each of the three regions can be written as:$$C_1 = \frac{\epsilon_1 A}{d_1} = \frac{\epsilon_1 A}{d/4} = \frac{4 \epsilon_1 A}{d}$$$$C_2 = \frac{\epsilon_2 A}{d_2} = \frac{\epsilon_2 A}{d/2} = \frac{2 \epsilon_2 A}{d}$$$$C_3 = \frac{\epsilon_1 A}{d_3} = \frac{\epsilon_1 A}{d/4} = \frac{4 \epsilon_1 A}{d}$$Since the regions are in series, the same electric displacement field $D$ exists across all regions, meaning the charge $Q$ on each equivalent series capacitor is the same:$$Q = C_1 V_1 = C_2 V_2 = C_3 V_3$$where $V_1$, $V_2$, and $V_3$ are the potential drops across the respective regions.

3. Using the Potential Distribution

The total potential difference across the capacitor is:$$V_{\text{total}} = 10\text{ V} - 0\text{ V} = 10\text{ V}$$The potential drop across the first region (between one of the plates and the nearest dielectric interface) is given as:$$V_1 = 2\text{ V}$$By symmetry, the potential drop across the third region is also:$$V_3 = V_1 = 2\text{ V}$$The potential drop across the middle region $V_2$ is:$$V_2 = V_{\text{total}} - (V_1 + V_3) = 10 - (2 + 2) = 6\text{ V}$$

4. Finding the Ratio $\epsilon_1 : \epsilon_2$

Using the charge equality relation $C_1 V_1 = C_2 V_2$:$$\left(\frac{4 \epsilon_1 A}{d}\right) V_1 = \left(\frac{2 \epsilon_2 A}{d}\right) V_2$$Substitute the values of $V_1 = 2\text{ V}$ and $V_2 = 6\text{ V}$:$$4 \epsilon_1 (2) = 2 \epsilon_2 (6)$$$$8 \epsilon_1 = 12 \epsilon_2$$Solving for the ratio $\epsilon_1 : \epsilon_2$:$$\frac{\epsilon_1}{\epsilon_2} = \frac{12}{8} = \frac{3}{2}$$Thus, the ratio $\epsilon_1 : \epsilon_2$ is $3:2$.Correct Option: C

To find the steady-state response of the linear time-invariant (LTI) system to a sinusoidal input, we use the frequency response of the system.The transfer function of the system is given by:$$H(s) = \frac{1}{s(s + 4)}$$

Step 1: Find the frequency response $H(j\omega)$

Substitute $s = j\omega$ into the transfer function:$$H(j\omega) = \frac{1}{j\omega(j\omega + 4)}$$

Step 2: Evaluate $H(j\omega)$ at the input frequency

The input to the system is $x(t) = \cos(3t)$, which has an angular frequency of $\omega = 3\text{ rad/s}$.Substituting $\omega = 3$ into $H(j\omega)$:$$H(j3) = \frac{1}{j3(j3 + 4)} = \frac{1}{j3(4 + j3)}$$

Step 3: Calculate the magnitude $|H(j3)|$

The magnitude of the frequency response at $\omega = 3$ is:$$|H(j3)| = \frac{1}{|j3| \cdot |4 + j3|}$$Calculate the individual magnitudes:

• $|j3| = 3$
• $|4 + j3| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

Thus, the magnitude is:$$|H(j3)| = \frac{1}{3 \times 5} = \frac{1}{15}$$

Step 4: Determine the amplitude $A$ of the steady-state output

For a stable LTI system, if the input is $x(t) = X_m \cos(\omega t + \theta)$, the steady-state output is:$$y(t) = X_m |H(j\omega)| \cos(\omega t + \theta + \angle H(j\omega))$$Here, the input amplitude is $X_m = 1$. The steady-state output is given as $y(t) = A \sin(3t + \alpha)$, which can be written in terms of a phase-shifted cosine wave. The amplitude of the output waveform is determined solely by the product of the input amplitude and the magnitude of the transfer function at the operating frequency:$$A = X_m |H(j3)| = 1 \times \frac{1}{15} = \frac{1}{15}$$Therefore, the value of $A$ is $\frac{1}{15}$.Correct Option: B

The transfer function of the system is the Laplace transform of the impulse response:
$H(s) = \mathcal{L}\{h(t)\} = \frac{1}{s+5}$The Laplace transform of the output is:
$Y(s) = \mathcal{L}\{y(t)\} = \mathcal{L}\{e^{-3t} u(t) - e^{-5t} u(t)\} = \frac{1}{s+3} - \frac{1}{s+5} = \frac{(s+5) - (s+3)}{(s+3)(s+5)} = \frac{2}{(s+3)(s+5)}$The input in the s-domain is given by:
$X(s) = \frac{Y(s)}{H(s)} = \frac{2}{(s+3)(s+5)} \cdot (s+5) = \frac{2}{s+3}$Taking the inverse Laplace transform to find $x(t)$:
$x(t) = \mathcal{L}^{-1}\{X(s)\} = 2e^{-3t} u(t)$

For an ideal transformer with turns ratio $n = \frac{N_2}{N_1} = \frac{2}{1} = 2$:$V_{th} = n \cdot V_{in} = 2 \cdot \sin(\omega t)$$Z_{th} = n^2 \cdot Z_{primary} = 2^2 \cdot 1 \Omega = 4 \Omega$Thus, the equivalent is $2 \sin(\omega t)$ and $4 \Omega$.

Given a two-winding transformer with rating $50\text{ kVA}$, $1000\text{ V}/100\text{ V}$.

• Rated current of the $1000\text{ V}$ winding: $I_{1000} = \frac{50 \times 10^3}{1000} = 50\text{ A}$
• Rated current of the $100\text{ V}$ winding: $I_{100} = \frac{50 \times 10^3}{100} = 500\text{ A}$

2. Analyze the autotransformer connection:

• From the figure, the $1000\text{ V}$ winding is the common winding and the $100\text{ V}$ winding is the series winding.
• The input voltage is $V_L = 1000\text{ V}$.
• The output voltage is $V_H = 1000\text{ V} + 100\text{ V} = 1100\text{ V}$.
• The series winding ($100\text{ V}$) carries the output current. Its maximum rated current is $500\text{ A}$.

3. Calculate the kVA rating:

• $S_{auto} = V_H \times I_{series} = 1100\text{ V} \times 500\text{ A} = 550,000\text{ VA} = 550\text{ kVA}$.

Alternatively, using the transformation ratio $a = \frac{V_H}{V_L} = \frac{1100}{1000} = 1.1$:$$S_{auto} = \left( \frac{a}{a-1} \right) S_{2w} = \left( \frac{1.1}{1.1-1} \right) \times 50 = 11 \times 50 = 550\text{ kVA}$$The calculated value is $550$, which falls within the specified range of $545$ to $555$.

For an induction machine, the synchronous speed is given by $N_s = \frac{120f}{P}$. For a 4-pole machine at $50 \text{ Hz}$, $N_s = 1500 \text{ rpm}$.In generator mode, the rotor speed $N$ must be greater than the synchronous speed $N_s$ ($N > N_s$). Given that the generator speed is maintained at $N = 1500 \text{ rpm}$, we must have $N_s < 1500 \text{ rpm}$.
$N_s = \frac{120f}{4} = 30f < 1500 \implies f < 50 \text{ Hz}$.
Thus, both $f_1$ and $f_2$ must be less than $50 \text{ Hz}$.When the load is partially removed, the electrical power output decreases. For an induction generator, this corresponds to a decrease in the magnitude of the negative slip $|s|$.
$s = \frac{N_s - N}{N_s}$. Since $N > N_s$, $s$ is negative.
As $|s|$ decreases, $N_s$ must increase towards $N$. Since $N_s = 30f$, an increase in $N_s$ implies an increase in frequency $f$. Therefore, $f_2 > f_1$.Combining these, $f_1, f_2 < 50 \text{ Hz}$ and $f_2 > f_1$.

Given:
Real power $P = 12 \text{ MW}$
Initial power factor $\cos \phi_1 = 0.6 \text{ lagging} \implies \tan \phi_1 = \frac{\sqrt{1 - 0.6^2}}{0.6} = \frac{0.8}{0.6} = 1.333$
Initial reactive power $Q_1 = P \tan \phi_1 = 12 \times 1.333 = 16 \text{ MVAR}$Target power factor $\cos \phi_2 = 0.8 \text{ lagging} \implies \tan \phi_2 = \frac{\sqrt{1 - 0.8^2}}{0.8} = \frac{0.6}{0.8} = 0.75$
Target reactive power $Q_2 = P \tan \phi_2 = 12 \times 0.75 = 9 \text{ MVAR}$The reactive power delivered by the capacitor is the difference between the initial and final reactive power:
$Q_c = Q_1 - Q_2 = 16 - 9 = 7 \text{ MVAR}$.

Given:
$S_{base} = 100 \text{ MVA}$
$I_{base} = 437.38 \text{ A}$For a three-phase system, the relationship between base power, base voltage (line-to-line), and base current is:
$S_{base} = \sqrt{3} \times V_{base} \times I_{base}$
$V_{base} = \frac{S_{base}}{\sqrt{3} \times I_{base}} = \frac{100 \times 10^6}{\sqrt{3} \times 437.38} \approx 132003.5 \text{ V} \approx 132 \text{ kV}$The actual line-to-line load voltage is:
$V_{actual} = V_{pu} \times V_{base} = 0.9 \times 132 \text{ kV} = 118.8 \text{ kV}$.

Shunt reactors are inductive elements connected in parallel to the transmission line. They are primarily used to absorb the reactive power generated by the line capacitance under light load or no-load conditions (known as the Ferranti effect), thereby limiting the over-voltages at the receiving end.

The steady state error $e_{ss}$ for a unit step input $R(s) = \frac{1}{s}$ is given by:$$e_{ss} = \lim_{s \to 0} s E(s) = \lim_{s \to 0} s R(s) [1 - T(s)]$$Substituting $R(s) = \frac{1}{s}$ and the given $T(s)$:$$e_{ss} = \lim_{s \to 0} s \cdot \frac{1}{s} \left[ 1 - \frac{4}{s^2 + 0.4s + 4} \right] = 1 - T(0)$$$$T(0) = \frac{4}{0^2 + 0.4(0) + 4} = \frac{4}{4} = 1$$$$e_{ss} = 1 - 1 = 0$$