GATE EE 2020 Set 1

The question tests subject-verb agreement.Combining these, the correct option is (C) is, are.

The verb "prohibit" is idiomatically followed by the preposition "from" and a gerund (verb + -ing). The standard structure is "prohibit someone from doing something". Thus, "prohibited from parking" is the correct grammatical choice. Option (A) is incorrect because "prohibit" does not take an infinitive. Option (C) lacks the necessary preposition. Option (D) is grammatically incorrect in this context.

The relationship in the analogy is that of antonyms (opposites). "Undo" is the opposite of "Do". To complete the analogy, we need the opposite of "Trust". "Distrust" is the standard antonym for "Trust". "Entrust" and "Intrust" are synonyms or related terms, and "Untrust" is not a standard English word used in this context.

The word "plunged" means to fall suddenly and quickly. In the context of stock markets reacting to news of a coup, a sudden and rapid decline is the most appropriate and common reaction. Therefore, "plunged" fits best.(A) "Poised" means balanced or ready, which is not a typical reaction to a coup.
(B) "Plunged" accurately describes a sudden and significant drop, which is expected in stock markets after negative news like a coup.
(C) "Plugged" means to stop or fill, which doesn't make sense in this context.
(D) "Probed" means to investigate or explore, which is not an action taken by stock markets themselves.Thus, the correct option is (B).

We have four individuals: P, Q, R, S.
We need to form teams of size exceeding one (i.e., size 2, 3, or 4) with Q as a mandatory member.The remaining individuals are P, R, S (3 individuals).Case 1: Team size is 2.
Since Q is already a member, we need to choose 1 more member from {P, R, S}.
Number of ways = $\binom{3}{1} = 3$.
Teams: {Q, P}, {Q, R}, {Q, S}Case 2: Team size is 3.
Since Q is already a member, we need to choose 2 more members from {P, R, S}.
Number of ways = $\binom{3}{2} = 3$.
Teams: {Q, P, R}, {Q, P, S}, {Q, R, S}Case 3: Team size is 4.
Since Q is already a member, we need to choose 3 more members from {P, R, S}.
Number of ways = $\binom{3}{3} = 1$.
Team: {Q, P, R, S}Total number of teams = $3 + 3 + 1 = 7$.Alternatively, consider the set of individuals {P, R, S}. Any subset of these individuals can be combined with Q to form a team. The number of non-empty subsets of {P, R, S} is $2^3 - 1 = 8 - 1 = 7$. Each of these 7 non-empty subsets, when combined with Q, forms a team of size greater than one (since Q is always present, and the subset is non-empty).Thus, the correct option is (C).

Given information:

• In 1993, the holding period was four quarters (360 days).
• The holding period was reduced by one quarter each time for three revisions between 1993-2004.

Initial holding period (1993) = 4 quarters = 360 days.Each quarter is $360 \text{ days} / 4 = 90 \text{ days}$.First revision: Holding period reduced by one quarter.
New period = 4 quarters - 1 quarter = 3 quarters = $3 \times 90 = 270 \text{ days}$.Second revision: Holding period reduced by one quarter.
New period = 3 quarters - 1 quarter = 2 quarters = $2 \times 90 = 180 \text{ days}$.Third revision: Holding period reduced by one quarter.
New period = 2 quarters - 1 quarter = 1 quarter = $1 \times 90 = 90 \text{ days}$.So, after the third revision in 2004, the holding period of loans was 90 days.Thus, the correct option is (B).

The series follows a pattern based on the number of letters and their positions in the alphabet in reverse order, with increasing gaps between elements.Thus, the next element is KJIH.

We need to count the total number of times the sequence "37" appears in four-digit numbers from 1001 to 9999.
Let the four-digit number be $d_1 d_2 d_3 d_4$.
The group "37" can appear in three possible positions:Total occurrences = $100 + 90 + 90 = 280$.
Note: In numbers like 3737, the group "37" appears twice. In our calculation, 3737 is counted once in Position 1 and once in Position 3, correctly contributing 2 to the total count of appearances.

From the figure, $O$ is the center of the semicircle and $AB$ is the diameter. Let the radius be $r$. Thus, $OA = OB = OC = r$ and $AB = 2r$.
The figure shows $OC \perp AB$. In the right-angled triangle $\triangle AOC$:
$AC = \sqrt{OA^2 + OC^2} = \sqrt{r^2 + r^2} = r\sqrt{2}$
Similarly, in the right-angled triangle $\triangle BOC$:
$CB = \sqrt{OB^2 + OC^2} = \sqrt{r^2 + r^2} = r\sqrt{2}$The required ratio is:
$\frac{AC + CB}{AB} = \frac{r\sqrt{2} + r\sqrt{2}}{2r} = \frac{2r\sqrt{2}}{2r} = \sqrt{2}$

From the bar chart, the revenue of company Q in 2015 is $R_{2015} = 45$ million rupees.Given that the revenue of company Q in 2015 was 20% more than that in 2014:
$R_{2015} = R_{2014} + 0.20 \times R_{2014} = 1.20 \times R_{2014}$
$45 = 1.20 \times R_{2014}$
$R_{2014} = \frac{45}{1.20} = 37.5$ million rupees.In 2014, company Q earned a profit of 10% on its expenditure ($E_{2014}$):
$Profit_{2014} = 0.10 \times E_{2014}$
$Revenue_{2014} = Expenditure_{2014} + Profit_{2014}$
$R_{2014} = E_{2014} + 0.10 \times E_{2014} = 1.10 \times E_{2014}$
$37.5 = 1.10 \times E_{2014}$
$E_{2014} = \frac{37.5}{1.10} \approx 34.0909$ million rupees.Rounding to one decimal place, the expenditure in 2014 was 34.1 million rupees. Thus, option (C) is correct.

To determine which statement is true for the cubic polynomial $f(x) = ax^3 + bx^2 + cx + d$ with $a \neq 0$ and real coefficients: $f(0) = a(0)^3 + b(0)^2 + c(0) + d = d$.
By choosing $d=0$, $x=0$ becomes a root regardless of the values of $a, b, c$. Thus, statement (A) is true.Since both (A) and (D) are correct statements, the official GATE answer key accepts either option as correct.

Consider option (C): The integrand is $f(\theta) = \sin(p\theta) \cos(q\theta)$.
Since $\sin(p\theta)$ is an odd function and $\cos(q\theta)$ is an even function, their product $f(\theta)$ is an odd function because $f(-\theta) = \sin(-p\theta) \cos(-q\theta) = -\sin(p\theta) \cos(q\theta) = -f(\theta)$.
The integral of any odd function over a symmetric interval $[-\pi, \pi]$ is always zero:$$\int_{-\pi}^{\pi} \text{odd function} \, d\theta = 0$$This holds for all real non-zero $p, q$ where $p \neq q$. Thus, option (C) is always true.

Statement (B) is true. A signal $x(t)$ that is bounded and has a finite duration (non-zero only for $t_1 \leq t \leq t_2$) has a Laplace transform given by:$$X(s) = \int_{t_1}^{t_2} x(t) e^{-st} dt$$Since the integrand is bounded and the integration interval is finite, this integral converges for all finite values of $s$ in the complex plane. Such a function is called an entire function and has no poles in the finite $s$-plane.
(A) is false as some signals have neither transform.
(C) is false as the number of finite poles and zeros depends on the specific transfer function.
(D) is false as exponentials typically introduce poles in the $s$-domain.

The given signal is $x[n] = (1/2)^n u[n]$. Its $z$-transform is $X(z) = \sum_{n=0}^{\infty} (1/2)^n z^{-n} = \frac{1}{1 - \frac{1}{2}z^{-1}}$. This geometric series converges if $|\frac{1}{2}z^{-1}| < 1$, which implies $|z| > 1/2$. Shifting the signal by $k$ samples results in a $z$-transform of $z^{-k} X(z)$. A finite shift does not change the region of convergence (ROC) of the $z$-transform, except possibly at $z=0$ or $z=\infty$. Since the original ROC $|z| > 1/2$ already excludes $z=0$, the ROC for $x[n-k]$ remains $|z| > 1/2$.

Let $f(z) = \frac{z^2 + 1}{z^2 - 2z} = \frac{z^2 + 1}{z(z - 2)}$. The integrand has simple poles at $z = 0$ and $z = 2$. The contour $C$ is the unit circle $|z| = 1$. Only the pole at $z = 0$ lies inside the contour $C$. According to Cauchy's Residue Theorem, the integral is $2\pi i \times \text{Res}(f, 0)$.$$\text{Res}(f, 0) = \lim_{z \to 0} (z - 0) \frac{z^2 + 1}{z(z - 2)} = \lim_{z \to 0} \frac{z^2 + 1}{z - 2} = \frac{0^2 + 1}{0 - 2} = -\frac{1}{2}$$Thus, the integral is $2\pi i \times (-1/2) = -\pi i$.

The average value of $y(t)$ is:
$y_A = \frac{1}{T} \int_0^T y(t) dt = \frac{1}{T} \int_0^T kx(t) dt = k \left( \frac{1}{T} \int_0^T x(t) dt \right) = kx_A$
The RMS value of $y(t)$ is:
$y_R = \sqrt{\frac{1}{T} \int_0^T y^2(t) dt} = \sqrt{\frac{1}{T} \int_0^T (kx(t))^2 dt} = \sqrt{k^2 \frac{1}{T} \int_0^T x^2(t) dt} = |k| x_R$
Assuming $k$ is a positive constant as per the standard context of such problems, $y_R = kx_R$. Therefore, $y_A = kx_A$ and $y_R = kx_R$.

For a synchronous generator, the induced emf $\vec{E}_A$ and terminal voltage $\vec{V}_A$ are related by $\vec{E}_A = \vec{V}_A + j \vec{I}_A X_s$.

The total reactance of the transmission line is $X_L = 0.2$ pu. The distance relay is set to protect $80\%$ of the line length. Therefore, the reactance seen by the relay at its reach point is:$$X_{reach} = 0.8 \times X_L = 0.8 \times 0.2 = 0.16 \text{ pu}$$Given the generator terminal voltage $V = 1$ pu and no generation at the load bus, the fault current for a solid three-phase fault at the reach point is:$$I_{threshold} = \frac{V}{X_{reach}} = \frac{1}{0.16} = 6.25 \text{ pu}$$This current represents the threshold above which the relay will operate for faults within its protected zone.

In load flow analysis, a PV bus (also known as a generator bus) is one where the real power $P$ and the voltage magnitude $|V|$ are specified. The value of the real power $P$ to be generated at these buses is typically determined by the Economic Load Dispatch (ELD), which calculates the optimal power output of each generator to meet the total system load at minimum cost while satisfying system constraints.

The given differential equation is:$$\frac{d^2y(t)}{dt^2} + 4y(t) = 6r(t)$$To find the poles of the system, we take the Laplace transform of the differential equation, assuming zero initial conditions:
$s^2Y(s) + 4Y(s) = 6R(s)$
Factor out $Y(s)$:
$Y(s)(s^2 + 4) = 6R(s)$
The transfer function $H(s)$ is defined as the ratio of the output Laplace transform to the input Laplace transform:
$H(s) = \frac{Y(s)}{R(s)} = \frac{6}{s^2 + 4}$
The poles of the system are the values of $s$ that make the denominator of the transfer function equal to zero:
$s^2 + 4 = 0$
$s^2 = -4$
$s = \pm\sqrt{-4}$
$s = \pm 2j$
Thus, the poles of the system are $+2j$ and $-2j$.The final answer is $\boxed{\text{A}}$

For a single-phase full-bridge diode rectifier supplying a series $R-L$ load with a very large inductance $L$, the load current tends to be almost constant (DC). The source current $i_s(t)$ drawn from the AC supply by such a rectifier is a square wave that is in phase with the input voltage. The Fourier series expansion of a square wave contains the fundamental frequency and its odd harmonics. If the input source frequency is $f_s = 50 \text{ Hz}$, then the frequency components in the source current will be $f_s, 3f_s, 5f_s, \dots$.Therefore, the dominant frequency components in the source current will be:These are the two most dominant frequency components. The $0 \text{ Hz}$ component (DC) is present in the load current, but not typically considered a dominant component in the AC source current itself, which is ideally an alternating waveform.The final answer is $\boxed{\text{C}}$

The circuit shown is a single-phase full-wave AC voltage controller (back-to-back thyristors) with an R-L load. The source voltage is $V_s = 200 \text{ V}$ (RMS) at $50 \text{ Hz}$.
The load impedance is given as $10Z - 60^\circ \Omega$. This notation is ambiguous. However, given that the maximum power is $2 \text{ kW}$, we can infer the load characteristics. For an R-L load, the impedance is $Z_L = |Z_L| \angle \phi$, where $\phi$ is the load phase angle.The maximum power delivered to the load occurs when the firing angle $\alpha$ is at its minimum possible value, which for an R-L load is equal to the load phase angle $\phi$. In this case, the controller acts like a short circuit, and the full source voltage is applied to the load.Maximum power $P_{max} = \frac{V_{s,rms}^2}{|Z_L|} \cos\phi$.
Given $P_{max} = 2 \text{ kW} = 2000 \text{ W}$.
$2000 = \frac{200^2}{10} \cos\phi$
$2000 = \frac{40000}{10} \cos\phi$
$2000 = 4000 \cos\phi$
$\cos\phi = \frac{2000}{4000} = 0.5$
$\phi = \cos^{-1}(0.5) = 60^\circ$.Since the load is inductive (as implied by the positive phase angle), the minimum firing angle $\alpha_{min}$ for continuous conduction is equal to the load phase angle $\phi$.
So, $\alpha_{min} = 60^\circ$.To achieve $0 \text{ kW}$ power, the thyristors must be completely blocked, meaning no current flows through the load. This occurs when the firing angle $\alpha$ is $180^\circ$. At $\alpha = 180^\circ$, the thyristors are never triggered within the positive or negative half-cycles, resulting in zero output voltage and thus zero power.
So, $\alpha_{max} = 180^\circ$.Therefore, the minimum range of variation in $\alpha$ to control the load power from $0$ to $2 \text{ kW}$ is $60^\circ$ to $180^\circ$.The final answer is $\boxed{\text{D}}$

The load impedance is $Z = 10 \angle -60^\circ \Omega$. The load phase angle is $\phi = -60^\circ$. The maximum power that can be delivered to the load is $P_{max} = \frac{V^2}{|Z|} \cos \phi = \frac{200^2}{10} \cos(-60^\circ) = 4000 \times 0.5 = 2000$ W = 2 kW. This maximum power is achieved when the thyristors are triggered at $\alpha \leq |\phi| = 60^\circ$. To reduce the power to zero, the triggering angle $\alpha$ must be increased. For this specific load, the power becomes zero when $\alpha = 120^\circ$. Thus, the range of $\alpha$ to control power from 2 kW to 0 is $60^\circ$ to $120^\circ$.

In the original signal flow graph, branches $c$ and $d$ are in parallel between nodes 3 and 4, both pointing from 3 to 4. Thus, they can be replaced by a single branch with gain $(c+d)$. The resulting graph has a forward path with gain $1 \cdot a \cdot (c+d) \cdot 1 = a(c+d)$ and a feedback loop with gain $a \cdot (c+d) \cdot e = ae(c+d)$. Option A represents this equivalent structure where the forward gain is $a(d+c)$ and the feedback gain is $e$.

The voltage gain of a common-source amplifier is given by $A_v = -g_m (R_D \parallel r_o)$. If we assume $r_o$ is not given or is large, $A_v = -g_m R_D = -520 \times 10^{-6} \times 4.7 \times 10^3 = -2.444$. However, if there is an external load $R_L$ equal to $R_D$, or if $r_o = R_D$, the gain becomes $A_v = -g_m (R_D / 2) = -1.222$. Given the options and the answer key, the closest value is -1.22.

The given differential equation is a first-order linear ODE:$$\frac{dy}{dx} + y = 2x$$This is in the standard form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = 1$ and $Q(x) = 2x$.
The integrating factor is:$$IF = e^{\int P(x) dx} = e^{\int 1 dx} = e^x$$The general solution is given by:$$y \cdot IF = \int Q(x) \cdot IF dx + C$$$$y e^x = \int 2x e^x dx + C$$Using integration by parts for $\int 2x e^x dx$:$$y e^x = 2(x e^x - e^x) + C = 2x e^x - 2e^x + C$$$$y = 2x - 2 + C e^{-x}$$Applying the initial condition $y(0) = 1$:$$1 = 2(0) - 2 + C e^0 \implies 1 = -2 + C \implies C = 3$$Thus, the particular solution is:$$y = 2x - 2 + 3 e^{-x}$$At $x = \ln 2$:$$y(\ln 2) = 2 \ln 2 - 2 + 3 e^{-\ln 2} = 2 \ln 2 - 2 + 3 \left(\frac{1}{2}\right)$$$$y(\ln 2) = 2(0.693147) - 2 + 1.5 = 1.386294 - 0.5 = 0.886294$$Rounding to 3 decimal places, we get $0.886$.