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, $G_{12} = G_{21} = 0.2\text{ S}$
$I_1 = G_{12} \times V_2 = 0.2 \times 5 = 1\text{ A}$.

The circuit shown is a 6-transistor (6T) Static Random Access Memory (SRAM) cell. It consists of two cross-coupled inverters, which form a bistable element capable of storing one bit of information. This fundamental memory-holding functionality is characteristic of a latch. The additional transistors (connected to inputs A and B) are used for controlling read and write operations to this memory cell.

• (A) Latch: A latch is a basic memory element that can store one bit. The core of an SRAM cell is indeed a latch.
• (B) XNOR: XNOR is a logic gate, not a memory element.
• (C) SRAM Cell: This is the most specific and accurate description of the circuit shown, as it is a standard 6T SRAM cell.
• (D) XOR: XOR is a logic gate, not a memory element.

Given the instruction to select 'A' as the correct option for this MCQ, 'Latch' is a correct, albeit more general, description of the circuit's functionality, as an SRAM cell is a type of latch.

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.

The given structure shows a Bipolar Junction Transistor (BJT).
Let's identify the regions and their doping types:

• The top layer is labeled 'B' and is $p^+$. This is the Base.
• The region below 'B' is 'n'. This is the Collector.
• The region on the right, labeled 'C', is $n^{++}$. This is the Emitter.
• The region on the left, labeled 'E', is $n^{++}$. This is also the Emitter.

Wait, let's re-examine the structure carefully. It's a planar BJT structure.The structure is $n^{++}$ (Emitter) - $p^+$ (Base) - $n$ (Collector). This indicates an NPN transistor.Now let's look at the circuit symbols for an NPN transistor:

• The arrow on the emitter points outwards from the base.
• The base is connected to the $p^+$ region.
• The collector is connected to the $n$ region.

Let's check the options:

• Option (A): Shows an NPN transistor. The base is connected to 'B', the collector to 'C', and the emitter to 'E'. This matches the NPN structure identified.
• Option (B): Shows a PNP transistor (arrow points inwards). This is incorrect.
• Option (C): Shows an NPN transistor, but the connections are swapped for collector and emitter compared to the labels in the diagram. The 'C' terminal is connected to the emitter and 'E' to the collector. This is incorrect based on the physical layout and typical BJT labeling.
• Option (D): Shows a PNP transistor. This is incorrect.

Therefore, option (A) correctly represents the NPN BJT structure shown.The final answer is $\boxed{A}$

The C-V curve of a MOS capacitor shows the capacitance variation with the applied gate voltage ($V_G$).The question asks for the points representing flat-band, inversion, and accumulation conditions, respectively. This corresponds to the sequence: Q, R, P.Therefore, option (B) is the correct answer.

The problem asks for the electric flux through the surface of a quarter-cylinder due to an infinite line charge along its axis.Let the linear charge density of the infinite line charge be $\lambda$. The problem states the charge density is Q. Based on the units in the options, Q must represent the linear charge density, so $\lambda = Q$ (in Coulombs per meter).According to Gauss's Law, the total electric flux $\Phi_{total}$ emanating from a charge $Q_{enc}$ enclosed within a closed surface is given by:$$\Phi_{total} = \oint \vec{E} \cdot d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$$Consider a Gaussian surface in the form of a full cylinder of height H and radius R, coaxial with the line charge. The charge enclosed by this full cylinder is $Q_{enc} = \lambda H = QH$.The total flux emanating from this segment of the line charge is:$$\Phi_{total} = \frac{QH}{\epsilon_0}$$Due to the cylindrical symmetry of the infinite line charge, the electric field $\vec{E}$ is directed radially outward from the line charge. That is, $\vec{E} = E(r) \hat{r}$.This total flux passes through the curved surface of the full cylinder. The flux through the top and bottom flat circular surfaces is zero because the electric field vector $\vec{E}$ is parallel to these surfaces (i.e., perpendicular to their area vectors).Now, consider the quarter-cylinder of height H. It subtends an angle of $\frac{\pi}{2}$ radians (or 90 degrees), which is $\frac{1}{4}$ of the full circle ($2\pi$ radians).By symmetry, the flux is distributed uniformly in the azimuthal direction. Therefore, the flux passing through the curved surface of the quarter-cylinder is exactly $\frac{1}{4}$ of the total flux.Flux through curved surface of quarter-cylinder = $\frac{1}{4} \Phi_{total} = \frac{1}{4} \left( \frac{QH}{\epsilon_0} \right) = \frac{QH}{4\epsilon_0}$.The question asks for the flux $\int \vec{E} \cdot d\vec{a}$ through the quarter-cylinder. This refers to the flux through the surface area of the quarter-cylinder. The surface consists of the curved part, two flat rectangular sides, and two flat quarter-circle ends (top and bottom).

• Flux through curved surface: As calculated above, it is $\frac{QH}{4\epsilon_0}$.
• Flux through flat rectangular sides: The area vectors of these sides are in the azimuthal direction ($\hat{\phi}$), while the electric field is in the radial direction ($\hat{r}$). Since $\hat{r} \cdot \hat{\phi} = 0$, the flux through these two sides is zero.
• Flux through top and bottom quarter-circle ends: The area vectors of these ends are in the axial direction ($\hat{z}$), while the electric field is in the radial direction ($\hat{r}$). Since $\hat{r} \cdot \hat{z} = 0$, the flux through these ends is also zero.

Therefore, the total flux through the entire surface of the quarter-cylinder is just the flux through its curved part, which is $\frac{QH}{4\epsilon_0}$.This matches option (B).

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.

When the width of the pMOS transistor ($W_p$) is increased, its transconductance parameter $\beta_p = \mu_p C_{ox} (W_p/L_p)$ increases. This makes the pMOS transistor stronger.For a CMOS inverter, the switching threshold voltage ($V_M$) is given by:$$V_M = \frac{V_{Tn} + \sqrt{\frac{\beta_p}{\beta_n}}(V_{DD} - |V_{Tp}|)}{1 + \sqrt{\frac{\beta_p}{\beta_n}}}$$
where $V_{Tn}$ is the nMOS threshold voltage and $V_{Tp}$ is the pMOS threshold voltage.If $\beta_p$ increases, the ratio $\sqrt{\frac{\beta_p}{\beta_n}}$ increases. This will cause $V_M$ to shift towards $V_{DD}$.Noise Margins are defined as:
NM$_L$ (Low Noise Margin) = $V_{IL} - V_{OL}$
NM$_H$ (High Noise Margin) = $V_{OH} - V_{IH}$For an ideal CMOS inverter, $V_{OL} = 0$ and $V_{OH} = V_{DD}$.
$V_{IL}$ is the input voltage at which the slope of the transfer characteristic is -1 in the low-to-high transition region.
$V_{IH}$ is the input voltage at which the slope of the transfer characteristic is -1 in the high-to-low transition region.When the pMOS becomes stronger (larger $\beta_p$), the inverter's switching point ($V_M$) shifts towards $V_{DD}$. This means the output will transition from high to low at a higher input voltage. Consequently, $V_{IH}$ will increase, and $V_{IL}$ will decrease.Since $V_{OH}$ and $V_{OL}$ remain $V_{DD}$ and $0$ respectively (assuming ideal output swings):
NM$_L = V_{IL} - V_{OL} = V_{IL} - 0 = V_{IL}$. If $V_{IL}$ decreases, NM$_L$ decreases.
NM$_H = V_{OH} - V_{IH} = V_{DD} - V_{IH}$. If $V_{IH}$ increases, NM$_H$ decreases.However, the question asks about the effect on noise margins. Let's re-evaluate the effect of $V_M$ shift.If $V_M$ shifts towards $V_{DD}$, it means the inverter is more easily pulled down by the nMOS for a given input, or it takes a higher input to pull the output low. This makes the inverter more sensitive to low input voltages and less sensitive to high input voltages.More precisely, increasing $W_p$ (and thus $\beta_p$) makes the pMOS stronger relative to the nMOS. This shifts the VTC (Voltage Transfer Characteristic) to the right (i.e., $V_M$ increases). A higher $V_M$ means the inverter switches at a higher input voltage.When $V_M$ increases:
$V_{IL}$ (the point where the slope is -1 on the rising edge of the VTC) will decrease.
$V_{IH}$ (the point where the slope is -1 on the falling edge of the VTC) will increase.NM$_L = V_{IL} - V_{OL}$. Since $V_{OL}$ is typically 0V, NM$_L = V_{IL}$. If $V_{IL}$ decreases, NM$_L$ decreases.
NM$_H = V_{OH} - V_{IH}$. Since $V_{OH}$ is typically $V_{DD}$, NM$_H = V_{DD} - V_{IH}$. If $V_{IH}$ increases, NM$_H$ decreases.This implies both NM$_L$ and NM$_H$ decrease. This contradicts option B.Let's reconsider the definition of equal rise and fall times. Equal rise and fall times imply $\beta_n = \beta_p$ (or $\mu_n W_n = \mu_p W_p$ for equal $L$). If $W_p$ is increased, then $\beta_p$ becomes greater than $\beta_n$. This makes the pMOS stronger than the nMOS.When pMOS is stronger, the VTC shifts to the right (higher $V_M$).For a CMOS inverter, the noise margins are:
NM$_L = V_{IL} - V_{OL}$
NM$_H = V_{OH} - V_{IH}$If the VTC shifts to the right (higher $V_M$):

• The output will stay high for a wider range of low input voltages, meaning $V_{IL}$ decreases. Thus, NM$_L$ decreases.
• The output will transition to low at a higher input voltage, meaning $V_{IH}$ increases. Thus, NM$_H$ decreases.

This result (both decrease) is not among the options. Let's check common interpretations or specific conditions.In some contexts, $V_{IL}$ and $V_{IH}$ are defined relative to $V_M$. If $V_M$ shifts, the symmetry of the noise margins changes.Let's re-examine the effect of a stronger pMOS on the VTC and noise margins. A stronger pMOS means the pull-up is stronger. This shifts the switching threshold $V_M$ towards $V_{DD}$.If $V_M$ increases (shifts right):

• The region where the output is high ($V_{OH}$) extends to higher input voltages. This means $V_{IL}$ (the point where the output starts to drop significantly from $V_{OH}$) will increase. So, NM$_L$ increases.
• The region where the output is low ($V_{OL}$) starts at a higher input voltage. This means $V_{IH}$ (the point where the output has dropped significantly to $V_{OL}$) will also increase. So, NM$_H$ decreases (since $V_{DD} - V_{IH}$).

Let's verify this with a typical VTC. If the VTC shifts right, the point where the slope is -1 on the left side ($V_{IL}$) moves right (increases). The point where the slope is -1 on the right side ($V_{IH}$) also moves right (increases).So, if $V_{IL}$ increases, NM$_L$ increases.
If $V_{IH}$ increases, NM$_H$ decreases.This matches option (B).Final check:

• Increasing $W_p$ makes pMOS stronger.
• Stronger pMOS shifts $V_M$ towards $V_{DD}$ (i.e., $V_M$ increases).
• A higher $V_M$ means the inverter is more robust against low input noise (it takes a higher input to make the output drop from $V_{OH}$). So, $V_{IL}$ increases, leading to an increase in NM$_L$.
• A higher $V_M$ also means the inverter is less robust against high input noise (it switches to low at a higher input, but the $V_{IH}$ point also shifts right, reducing the margin from $V_{DD}$). So, $V_{IH}$ increases, leading to a decrease in NM$_H$.

Therefore, NM$_L$ increases and NM$_H$ decreases.The final answer is $\boxed{B}$