GATE EC 2019 Set 1

The sentence requires subject-verb agreement for two different subjects:Combining these, the correct pair is uses, include, which corresponds to option (B).

The preposition 'at' is used for specific points in time, such as 'at dawn', 'at noon', 'at midnight', 'at 3 PM'.Let's analyze the options:
(A) 'in' is used for longer periods (e.g., in the morning, in July, in 2019).
(B) 'at' is correct for specific times like 'dawn'.
(C) 'on' is used for specific days or dates (e.g., on Monday, on July 4th).
(D) 'under' indicates position, not time.Therefore, the correct preposition is 'at'.

This is a work-rate problem. We need to find the combined rate of work for both machines.Let the total work be 1 unit (one production order).Machine 1 completes the order in 4 hours.
So, the rate of Machine 1 ($R_1$) = $\frac{1 \text{ order}}{4 \text{ hours}} = \frac{1}{4}$ order/hour.Machine 2 completes the order in 2 hours.
So, the rate of Machine 2 ($R_2$) = $\frac{1 \text{ order}}{2 \text{ hours}} = \frac{1}{2}$ order/hour.When both machines work simultaneously, their rates add up.
Combined rate ($R_{combined}$) = $R_1 + R_2 = \frac{1}{4} + \frac{1}{2}$
To add these fractions, find a common denominator, which is 4.
$R_{combined} = \frac{1}{4} + \frac{2}{4} = \frac{1+2}{4} = \frac{3}{4}$ order/hour.The time taken to complete the same order (1 unit of work) when working together is the reciprocal of the combined rate.
Time ($T$) = $\frac{1}{R_{combined}} = \frac{1}{3/4} = \frac{4}{3}$ hours.Therefore, the time taken to complete the same order is $4/3$ hours.

The total number of books is 5: P, Q, R, S, T.
The problem states that books R and S are to be arranged first and second, respectively, from the right side of the shelf.
Let's visualize the shelf with 5 positions:
_
From the right side, the first position is R and the second is S:
_ S R
Now, the remaining books are P, Q, and T. There are 3 remaining positions for these 3 books.
The number of different ways to arrange 3 distinct items in 3 distinct positions is given by $3!$ (3 factorial).
$3! = 3 \times 2 \times 1 = 6$.
Therefore, P, Q, and T can be arranged in 6 different orders.The final answer is $\boxed{6}$

The sentence describes a situation where someone is imagining a grim scenario because another person has not returned home. Let's analyze the options:

• (A) concluded: To conclude means to arrive at a judgment or opinion by reasoning. While she might reason, the word 'pictured' better conveys the formation of a mental image.
• (B) looked: To look implies a physical act of seeing, which is not possible in this context as the person is not home.
• (C) notice: To notice means to become aware of something, usually through observation. This doesn't fit the context of imagining a future event.
• (D) pictured: To picture means to form a mental image of something; to imagine. This word perfectly fits the context of someone imagining a terrible outcome.

Therefore, the most appropriate word to complete the sentence is "pictured".

Let the four positions in the line be P1, P2, P3, P4 from left to right. The people are Rahul (R), Mathew (M), Seema (S), and Lohit (L). The professions are Engineer (E), Doctor (D), Teacher (T), and Dancer (Da).Let's analyze the clues:Let's combine clues:
From Clue 2, we have two main cases for the ends:
Case 1: Lohit is at P1, and the Engineer is at P4.
Line: L ENow, consider the (Doctor) Seema (D S) block from Clue 5. The Doctor is a person, and Seema is a person. So, the D S block refers to the positions of the Doctor and Seema.
Possible positions for (D S) block:

• If (D S) is (P1, P2): This would mean Lohit is the Doctor, and Seema is at P2. So, L(D) S _ E. The remaining people are R and M. The remaining professions are T and Da. So, R and M must be T and Da. The line is L(D) S R M(E) (R and M fill the remaining spots).

Let's check this arrangement: L(D) S R M(E)

• Clue 1: M is not next to S. (M is at P4, S is at P2. They are not adjacent. Consistent.)
• Clue 2: Two people between L and E. (L is at P1, M is E at P4. S and R are between them. Consistent.)
• Clue 3: R is not D. (R is at P3, L is D. Consistent.)
• Clue 4: T and Da are next to each other. (S and R are T and Da, and they are adjacent. Consistent.)
• Clue 5: (Doctor) Seema. (L is the Doctor at P1, S is at P2. L is to the left of S. Consistent.)

This arrangement is consistent with all clues.
In this arrangement, Mathew (M) is the Engineer.Let's quickly check other possibilities for completeness, though we found a consistent solution.

• If (D S) is (P2, P3): L D S E. This means the person at P2 is the Doctor, and Seema is at P3. So, L (Doctor) S E. The remaining people are R and M. The remaining professions are T and Da. So, R and M are T and Da. The line is L R(D) S M(E) or L M(D) S R(E).
• If L R(D) S M(E): R is the Doctor. This contradicts Clue 3 (Rahul is not a doctor). So, this is not possible.
• If L M(D) S R(E): M is the Doctor. S is at P3. M is at P2. This violates Clue 1 (Mathew is not next to Seema). So, this is not possible.
• If (D S) is (P3, P4): L _ D S. This means the person at P3 is the Doctor, and Seema is at P4. But P4 is the Engineer. So, Seema is the Engineer. The line is L M R(D) S(E) or L R M(D) S(E).
• If L M R(D) S(E): S is at P4. M is at P2. This violates Clue 1 (Mathew is not next to Seema). So, this is not possible.
• If L R M(D) S(E): S is at P4. R is at P2. M is at P3. This violates Clue 1 (Mathew is not next to Seema). So, this is not possible.

Case 2: Engineer is at P1, and Lohit is at P4.
Line: E L

• If (D S) is (P1, P2): This means the Engineer is the Doctor. This is impossible as they are distinct professions.
• If (D S) is (P2, P3): E D S L. This means the person at P2 is the Doctor, and Seema is at P3. The remaining people are R and M. The remaining professions are T and Da. So, R and M are T and Da. The line is M(E) R(D) S L or R(E) M(D) S L.
• If M(E) R(D) S L: R is the Doctor. This contradicts Clue 3 (Rahul is not a doctor). So, this is not possible.
• If R(E) M(D) S L: M is the Doctor. S is at P3. M is at P2. This violates Clue 1 (Mathew is not next to Seema). So, this is not possible.
• If (D S) is (P3, P4): E _ D S. This means the person at P3 is the Doctor, and Seema is at P4. But P4 is Lohit. So, Seema is Lohit. This is impossible as they are distinct people.

The only consistent arrangement is from Case 1, where Lohit is at P1 (Doctor), Seema is at P2 (Teacher/Dancer), Rahul is at P3 (Teacher/Dancer), and Mathew is at P4 (Engineer).Therefore, Mathew is the Engineer.The final answer is $\boxed{\text{Mathew}}$

Let the total population be $P$.In 2001:

• Female population = $40\%$ of $P = 0.4P$ (from Panel (b))
• Male population = $60\%$ of $P = 0.6P$ (from Panel (b))
• Female illiterates = $60\%$ of $0.4P = 0.24P$ (from Panel (a))
• Female literates = $0.4P - 0.24P = 0.16P$
• Male illiterates = $50\%$ of $0.6P = 0.30P$ (from Panel (a))
• Male literates = $0.6P - 0.30P = 0.30P$
• Total literates in 2001 ($L_{2001}$) = $0.16P + 0.30P = 0.46P$

In 2011:

• Female population = $50\%$ of $P = 0.5P$ (from Panel (c))
• Male population = $50\%$ of $P = 0.5P$ (from Panel (c))
• Female illiterates = $40\%$ of $0.5P = 0.20P$ (from Panel (a))
• Female literates = $0.5P - 0.20P = 0.30P$
• Male illiterates = $40\%$ of $0.5P = 0.20P$ (from Panel (a))
• Male literates = $0.5P - 0.20P = 0.30P$
• Total literates in 2011 ($L_{2011}$) = $0.30P + 0.30P = 0.60P$

Percentage Increase:

• Increase in literates = $L_{2011} - L_{2001} = 0.60P - 0.46P = 0.14P$
• Percentage increase = $\frac{0.14P}{0.46P} \times 100 = \frac{14}{46} \times 100 \approx 30.43\%$

Thus, the correct option is (A).

The passage explicitly states that the history written by British historians was "extremely well documented and researched". It also mentions that it was "not always impartial", which implies that it was biased in certain instances or "sometimes biased". Option (C) is the only one that correctly captures both these points from the text.

The hour hand of a clock moves $360^\circ$ in $12$ hours. Therefore, the rate of movement is $\frac{360^\circ}{12} = 30^\circ$ per hour.For consultant $P$:
Total time duration = $\frac{210^\circ}{30^\circ/\text{hr}} = 7$ hours.
Unpaid breaks = $2 \times 15$ minutes = $30$ minutes = $0.5$ hours.
Billable hours = $7 - 0.5 = 6.5$ hours.For consultant $Q$:
Total time duration = $\frac{240^\circ}{30^\circ/\text{hr}} = 8$ hours.
Unpaid breaks = $20$ minutes = $\frac{20}{60}$ hours = $\frac{1}{3}$ hours.
Billable hours = $8 - \frac{1}{3} = \frac{23}{3}$ hours.Total Calculation:
Total billable hours = $6.5 + \frac{23}{3} = \frac{13}{2} + \frac{23}{3} = \frac{39 + 46}{6} = \frac{85}{6}$ hours.
Total payment to consultants = $\text{Rate} \times \text{Total Hours} = 200 \times \frac{85}{6} = \frac{100 \times 85}{3} = \frac{8500}{3} \approx 2833.33$ USD.Remaining budget = $\text{Total Budget} - \text{Total Payment} = 3000 - 2833.33 = 166.67$ USD.Thus, the correct option is (B).

Based on the conditions provided in the problem:Therefore, option (C) correctly represents the current locations.

A function is analytic over the entire complex plane if it is an entire function.

• (A) $\ln(z)$ has a branch point at $z=0$ and is not analytic on a branch cut.
• (B) $e^{1/z}$ has an essential singularity at $z=0$.
• (C) $\frac{1}{1-z}$ has a simple pole at $z=1$.
• (D) $\cos(z) = \frac{e^{iz} + e^{-iz}}{2}$ is a sum of two entire functions ($e^{iz}$ and $e^{-iz}$), hence it is analytic everywhere in the complex plane.

Case 1: $n = -1$
$$\frac{dy}{dx} = -\left(\frac{x}{y}\right)^{-1} = -\frac{y}{x}$$$$\frac{dy}{y} = -\frac{dx}{x}$$Integrating both sides: $\ln y = -\ln x + C \Rightarrow \ln(xy) = C \Rightarrow xy = K$. This represents a family of rectangular hyperbolas.Case 2: $n = +1$
$$\frac{dy}{dx} = -\frac{x}{y}$$$$y dy = -x dx$$Integrating both sides: $\frac{y^2}{2} = -\frac{x^2}{2} + C \Rightarrow x^2 + y^2 = 2C = R^2$. This represents a family of circles centered at the origin.

Given $P(z) = H(z)H\left(\frac{1}{z}\right)$ and $h[n]$ is a real-valued signal.Given that $z_1 = \frac{1}{2} + \frac{1}{2}j$ is a zero of $P(z)$. This means $z_1$ is a zero of either $H(z)$ or $H\left(\frac{1}{z}\right)$.Case 1: $z_1 = \frac{1}{2} + \frac{1}{2}j$ is a zero of $H(z)$

• By conjugate symmetry, $z_1^* = \frac{1}{2} - \frac{1}{2}j$ is also a zero of $H(z)$.
• The corresponding zeros of $H\left(\frac{1}{z}\right)$ are:
• $\frac{1}{z_1} = \frac{1}{0.5 + 0.5j} = \frac{2}{1 + j} = \frac{2(1 - j)}{2} = 1 - j$
• $\frac{1}{z_1^*} = \frac{1}{0.5 - 0.5j} = \frac{2}{1 - j} = \frac{2(1 + j)}{2} = 1 + j$

Case 2: $z_1 = \frac{1}{2} + \frac{1}{2}j$ is a zero of $H\left(\frac{1}{z}\right)$

• Then $\frac{1}{z_1} = 1 - j$ is a zero of $H(z)$.
• By conjugate symmetry, $(1 - j)^* = 1 + j$ is also a zero of $H(z)$.
• The corresponding zeros of $H\left(\frac{1}{z}\right)$ are $\frac{1}{1 - j} = 0.5 + 0.5j$ and $\frac{1}{1 + j} = 0.5 - 0.5j$.

In both cases, the complete set of four zeros for $P(z)$ is $\{0.5 \pm 0.5j, 1 \pm j\}$.Plot (D) correctly represents these four zeros in the complex z-plane: two zeros inside the unit circle at $0.5 \pm 0.5j$ (magnitude $1/\sqrt{2}$) and two zeros outside the unit circle at $1 \pm j$ (magnitude $\sqrt{2}$).

According to the Reciprocity Theorem, for a linear, passive, and bilateral network, the ratio of excitation to response remains constant even if the positions of excitation and response are interchanged. In case (a):
Excitation $V_1 = 5\text{ V}$ at Port 1
Response $I_2 = 1\text{ A}$ at Port 2 (short-circuited)
Transfer conductance $G_{21} = \frac{I_2}{V_1} = \frac{1}{5} = 0.2\text{ S}$In case (b):
Excitation $V_2 = 5\text{ V}$ at Port 2
Response $I_1$ at Port 1 (short-circuited)
By reciprocity, $G_{12} = G_{21} = 0.2\text{ S}$
$I_1 = G_{12} \times V_2 = 0.2 \times 5 = 1\text{ A}$.

Partial fractions.$$Y(s) = \frac{3-s}{s(s+1)(s+3)} = \frac{A_0}{s} + \frac{A_1}{s+1} + \frac{A_3}{s+3}$$

• $A_0 = \left.\dfrac{3-s}{(s+1)(s+3)}\right|_{s=0} = \dfrac{3}{3} = 1$
• $A_1 = \left.\dfrac{3-s}{s(s+3)}\right|_{s=-1} = \dfrac{4}{(-1)(2)} = -2$
• $A_3 = \left.\dfrac{3-s}{s(s+1)}\right|_{s=-3} = \dfrac{6}{(-3)(-2)} = 1$

So the complete step response is$$y(t) = u(t) - 2e^{-t}u(t) + e^{-3t}u(t).$$Forced vs natural response. The forced response is the part contributed by the pole of the input ($1/s$ at $s=0$), i.e. $u(t)$ (equivalently $G(0)\,u(t) = 1\cdot u(t)$). The terms $-2e^{-t}u(t) + e^{-3t}u(t)$ come from the system poles and form the natural (transient) response.

• (A) is the complete response (forced + natural); the official key also accepts it.
• (B) has the wrong DC level ($G(0)=1$, not 2).
• (C) $2u(t)$ has the wrong amplitude.
• (D) $u(t)$ is the forced response.

Answer: (D) $u(t)$ (officially, (D) or (A) is accepted)

The number of poles and zeros can be determined by the change in slope at each corner frequency:Total number of poles $N_p = 1 + 2 + 2 + 1 = 6$.
Total number of zeros $N_z = 1 + 2 = 3$.

Let the 4-bit message be $m = [m_3 m_2 m_1 m_0]$ and the 7-bit codeword be $c = [c_6 c_5 c_4 c_3 c_2 c_1 c_0]$.
Since the encoder mapping is linear, it can be represented by a generator matrix $G$ such that $c = mG$.Given mappings:We need to find the codeword for message $m_3 = [0010]$.Notice that $m_3 = m_2 - m_1$ (in binary arithmetic, which is XOR for bits).
$m_2 = [0011]$
$m_1 = [0001]$
$m_2 \oplus m_1 = [0011] \oplus [0001] = [0010]$.Due to linearity, if $c = mG$, then $(m_a \oplus m_b)G = m_a G \oplus m_b G = c_a \oplus c_b$.
Therefore, the codeword for $m_3$ will be $c_3 = c_2 \oplus c_1$.$c_2 = [1100110]$
$c_1 = [0000111]$
$c_3 = c_2 \oplus c_1 = [1100110] \oplus [0000111]$Performing bitwise XOR:
$1 \oplus 0 = 1$
$1 \oplus 0 = 1$
$0 \oplus 0 = 0$
$0 \oplus 0 = 0$
$1 \oplus 1 = 0$
$1 \oplus 1 = 0$
$0 \oplus 1 = 1$So, $c_3 = [1100001]$.The final answer is $\boxed{\text{1100001}}$

In thermal equilibrium, the Fermi level ($E_F$) must be constant (flat) throughout the entire semiconductor device. This immediately eliminates option (B), where the Fermi level is not flat.The device has a $p-n-n^+-p^{++}$ configuration. The position of the Fermi level relative to the band edges ($E_C$ and $E_V$) depends on the doping type and concentration:Evaluating the remaining options:

• Option (A): Shows a flat $E_F$. In the $p$ region, $E_F$ is near $E_V$. In the $n$ region, $E_F$ is near $E_C$. In the $n^+$ region, the conduction band $E_C$ moves even closer to $E_F$. In the $p^{++}$ region, the valence band $E_V$ moves very close to $E_F$. This perfectly matches the doping profile.
• Option (C): Shows a flat $E_F$, but it fails to distinguish between the $n$ and $n^+$ regions or the $p$ and $p^{++}$ regions correctly in terms of band bending relative to the Fermi level.
• Option (D): Shows $E_F$ in the middle of the bandgap for the $p$ region, which would represent an intrinsic semiconductor, not a $p$-doped one.

Therefore, option (A) is the correct equilibrium band diagram.

Identify the regions of the planar structure:

• Emitter (E): the $n^{++}$ well contacted by E.
• Base (B): the $p^+$ region surrounding the emitter.
• Collector (C): the lightly doped $n$ region (epi layer). It is reached through the $n^{++}$ sinker and the $n^+$ buried layer.

So the transistor is $n^{++}$-$p^+$-$n$, an npn BJT. Its symbol has the emitter arrow pointing outward.The base metal contact is drawn overlapping the edge of the $p^+$ region onto the lightly doped $n$ collector. A metal contact on lightly doped $n$-silicon forms a Schottky diode, with the metal (base terminal) as anode and the $n$ collector as cathode. This is the Schottky-clamped transistor, where a diode from base (anode) to collector (cathode) prevents deep saturation.Check the options:

• (A) npn transistor with a diode whose anode is at B and cathode at C. Correct.
• (B) npn, but the diode points from C to B. Wrong polarity.
• (C) diode B to C, but the transistor symbol is pnp (arrow into the emitter). Wrong.
• (D) pnp transistor and reversed diode. Wrong.

Answer: (A)

With $\phi_{ms} = 0$ and no oxide charge, $V_{FB} = \phi_{ms} - Q_{ox}/C_{ox} = 0$. Capacitance falls as $V_G$ becomes positive, so the substrate is p-type.

• Accumulation: a large negative $V_G$ draws holes to the surface, and $C = C_{ox}$ (maximum). This is P.
• Flat band: $V_G = V_{FB} = 0$. This is Q.
• Inversion: a large positive $V_G$ forms an electron inversion layer. At high frequency the capacitance stays at its minimum. This is R.

So flat-band, inversion and accumulation are Q, R, P.Answer: (B)

Treat $Q$ as the line-charge density (C/m). The field of an infinite line charge is radial: $\vec E = \dfrac{Q}{2\pi\epsilon_0 r}\hat r$.

• Curved face (a quarter of the full cylinder, $\phi$ from 0 to $\pi/2$, at radius $r$):

$$\Phi = \int_0^{\pi/2}\int_0^H \frac{Q}{2\pi\epsilon_0 r}\, r\,dz\,d\phi = \frac{Q}{2\pi\epsilon_0}\cdot\frac{\pi}{2}\cdot H = \frac{HQ}{4\epsilon_0}$$

• Flat rectangular faces: they contain the axis, so $\vec E$ lies in their plane and gives zero flux.
• Top and bottom faces: $\vec E \perp \hat z$, so zero flux.

Equivalently, a full cylinder of height $H$ carries flux $QH/\epsilon_0$ (Gauss's law), and by symmetry a quarter of it passes through the quarter-cylinder.(A) is the full-cylinder flux. (C) and (D) are dimensionally wrong.Answer: (B) $\dfrac{HQ}{4\epsilon_0}$

This question requires matching the terms from the differential form of Maxwell's equations.Summarizing the matches:

• 1 $\rightarrow$ Q
• 2 $\rightarrow$ R
• 3 $\rightarrow$ P
• 4 $\rightarrow$ S

This corresponds to the option (B) 1 - Q, 2 - R, 3 - P, 4 - S.

With $\beta_n = \beta_p$ the inverter is symmetric and the switching threshold is $V_M \approx V_{DD}/2$.Increasing the pMOS width increases $\beta_p$, so $\beta_p > \beta_n$. The switching threshold$$V_M = \frac{V_{Tn} + \sqrt{\beta_p/\beta_n}\,(V_{DD} - |V_{Tp}|)}{1 + \sqrt{\beta_p/\beta_n}}$$increases, i.e. the whole transition region of the VTC shifts to the right. Both $V_{IL}$ and $V_{IH}$ (the unity-gain points) move to higher input voltages.With $V_{OL} = 0$ and $V_{OH} = V_{DD}$:

• $NM_L = V_{IL} - V_{OL}$ increases, because $V_{IL}$ increases.
• $NM_H = V_{OH} - V_{IH}$ decreases, because $V_{IH}$ increases.

A square-law simulation (e.g. $V_{DD}=2.5$ V, $V_T = 0.5$ V, $\beta_p/\beta_n$ from 1 to 2) confirms that $V_{IL}$ rises and $V_{DD}-V_{IH}$ falls.

• (B) is the effect of making the nMOS stronger, not the pMOS.
• (C) and (D) are wrong because the shift helps one margin at the expense of the other.

Answer: (A) NM$_L$ increases and NM$_H$ decreases.