GATE EC 2016 Set 2

When comparing the vertical height of human beings, the correct adjective to use is taller.

• Longer is typically used for horizontal measurements or duration.
• Elongated refers to something that has been stretched out or is unusually long in shape.
• Lengthier is usually used for duration (e.g., a lengthier discussion) or physical length of objects.

The correct word is felicitated, which means to congratulate or honor someone for an achievement or service.

• Facilitated means to make a process or action easier.
• Fantasized means to imagine or daydream about something.
• Facillitated is an incorrect spelling of facilitated.

The idiom "the rest is history" is used to indicate that the events that followed a specific point in time are very well known or famous and do not need to be explained in detail. Therefore, "rest is well known" is the correct interpretation.

The given statement is $(9 \text{ inches})^{1/2} = (0.25 \text{ yards})^{1/2}$.
To remove the square root, we can square both sides of the equation:
$( (9 \text{ inches})^{1/2} )^2 = ( (0.25 \text{ yards})^{1/2} )^2$
$9 \text{ inches} = 0.25 \text{ yards}$Now, let's check the given options:
(A) $3 \text{ inches} = 0.5 \text{ yards}$
If $9 \text{ inches} = 0.25 \text{ yards}$, then $3 \text{ inches} = \frac{0.25}{3} \text{ yards} \approx 0.0833 \text{ yards}$. This option is incorrect.(B) $9 \text{ inches} = 1.5 \text{ yards}$
This contradicts our derived statement $9 \text{ inches} = 0.25 \text{ yards}$. This option is incorrect.(C) $9 \text{ inches} = 0.25 \text{ yards}$
This statement is exactly what we derived from the given information. This option is correct.(D) $81 \text{ inches} = 0.0625 \text{ yards}$
If $9 \text{ inches} = 0.25 \text{ yards}$, then $81 \text{ inches} = 9 \times (9 \text{ inches}) = 9 \times (0.25 \text{ yards}) = 2.25 \text{ yards}$. This option is incorrect.Thus, the only true statement is (C).

Let the efficiency of S, E, and F be $x$ units/hour.
According to the problem statement, M works with twice the efficiency of others.
So, Efficiency of M ($E_M$) = $2x$ units/hour.
Efficiency of E ($E_E$) = $x$ units/hour.Let the number of days E worked be $D_E$.
According to the problem statement, M worked for half as many days as E worked.
So, Number of days M worked ($D_M$) = $D_E / 2$.Shift hours per day:
S and M have 6 hour shifts in a day.
E and F have 12 hour shifts in a day.Total hours worked by M = $D_M \times 6 \text{ hours/day} = (D_E / 2) \times 6 = 3D_E$ hours.
Total hours worked by E = $D_E \times 12 \text{ hours/day} = 12D_E$ hours.Contribution is calculated as Efficiency $\times$ Total hours worked.Contribution of M ($C_M$) = $E_M \times (\text{Total hours by M})$
$C_M = (2x) \times (3D_E) = 6xD_E$ units.Contribution of E ($C_E$) = $E_E \times (\text{Total hours by E})$
$C_E = (x) \times (12D_E) = 12xD_E$ units.The ratio of contribution of M to contribution of E is:
$\frac{C_M}{C_E} = \frac{6xD_E}{12xD_E} = \frac{6}{12} = \frac{1}{2}$So, the ratio is 1:2.

The question asks for the number of students who like to read books OR play sports. This means we need to find the total number of students in the union of the 'Read books' set and the 'Play sports' set, which includes students who like only reading books, only playing sports, or both.From the Venn diagram:Students who like 'Read books' only: 13
Students who like 'Read books' and 'Watch TV' only: 12
Students who like 'Read books' and 'Play sports' only: 44
Students who like 'Read books', 'Watch TV', and 'Play sports': 7Students who like 'Play sports' only: 15
Students who like 'Play sports' and 'Watch TV' only: 17To find the number of students who like to read books OR play sports, we sum all unique numbers within the 'Read books' circle and 'Play sports' circle:Number of students = (Read books only) + (Read books and Watch TV only) + (Read books and Play sports only) + (Read books, Watch TV, and Play sports) + (Play sports only) + (Play sports and Watch TV only)Number of students = $13 + 12 + 44 + 7 + 15 + 17$
Number of students = $25 + 44 + 7 + 15 + 17$
Number of students = $69 + 7 + 15 + 17$
Number of students = $76 + 15 + 17$
Number of students = $91 + 17$
Number of students = $108$Alternatively, using the Principle of Inclusion-Exclusion:
$|A \cup B| = |A| + |B| - |A \cap B|$
Where A = 'Read books' set and B = 'Play sports' set.$|\text{Read books}| = 13 + 12 + 7 + 44 = 76$
$|\text{Play sports}| = 15 + 17 + 7 + 44 = 83$
$|\text{Read books} \cap \text{Play sports}| = 44 + 7 = 51$$|\text{Read books} \cup \text{Play sports}| = 76 + 83 - 51 = 159 - 51 = 108$Therefore, the number of students who like to read books or play sports is 108.

Let's analyze each statement based on the provided text: The text states: "conventional ways of knowledge production have become obsolete." This refers to the ways of knowledge production, not the social science disciplines themselves. Therefore, this statement cannot be directly inferred. The text says: "Social science disciplines were in existence in an amorphous form until the colonial period when they were institutionalized." The phrase "until the colonial period" implies that they existed before the colonial period, even if not in an institutionalized form. Therefore, this statement can be logically inferred. The text states: "In varying degrees, they were intended to further the colonial interest." This indicates that during the colonial period, they served colonial interests to some extent. However, it does not imply that they always promote colonialism, especially considering the context of "economic rise of postcolonial countries" and "conventional ways of knowledge production have become obsolete." Therefore, this statement cannot be logically inferred. The text discusses the historical development and current obsolescence of certain knowledge production methods but does not provide any information or argument about the necessity for social science to maintain disciplinary boundaries. Therefore, this statement cannot be logically inferred.Based on the analysis, only statement (ii) can be logically inferred from the given text.The final answer is $\boxed{\text{A}}$

To find the actual time from a mirror image, we subtract the mirror time from 11:60 (which is equivalent to 12:00).Mirror time shown = 1:30Actual time (two and a quarter hours back) = 11:60 - 1:30 = 10:30The problem states that this time (10:30) was "two and a quarter hours back".
Two and a quarter hours = 2 hours and 15 minutes.To find the current time, we add 2 hours and 15 minutes to 10:30.Current Time = 10:30 + 2 hours 15 minutes
Current Time = (10 + 2) hours : (30 + 15) minutes
Current Time = 12 hours : 45 minutes
Current Time = 12:45The final answer is $\boxed{\text{D}}$

Let the starting location be the origin $(0,0)$ on a Cartesian coordinate system.For person M: Change in x-coordinate = $10 \cos(45^\circ) = 10 \times \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} \approx 7.071$ km.
Change in y-coordinate = $10 \sin(45^\circ) = 10 \times \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} \approx 7.071$ km.
Final position of M ($P_M$) = $(10 + 7.071, 0 + 7.071) = (17.071, 7.071)$.For person N: Change in x-coordinate = $4 \cos(-45^\circ) = 4 \times \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} \approx 2.828$ km.
Change in y-coordinate = $4 \sin(-45^\circ) = 4 \times (-\frac{1}{\sqrt{2}}) = -\frac{4}{\sqrt{2}} \approx -2.828$ km.
Final position of N ($P_N$) = $(0 + 2.828, -5 - 2.828) = (2.828, -7.828)$.Shortest distance between M and N:
Using the distance formula $D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:
$D = \sqrt{(17.071 - 2.828)^2 + (7.071 - (-7.828))^2}$
$D = \sqrt{(14.243)^2 + (7.071 + 7.828)^2}$
$D = \sqrt{(14.243)^2 + (14.899)^2}$
$D = \sqrt{202.863 + 221.980}$
$D = \sqrt{424.843}$
$D \approx 20.6117$ km.Rounding to two decimal places, the shortest distance is 20.61 km.The final answer is $\boxed{\text{C}}$

Let the length of the wire be $L = 340$ mm. Suppose the wire is cut into two pieces of lengths $x$ and $L - x$.$$A'(x) = \frac{2x}{16} + \frac{2(340 - x)(-1)}{18} = 0$$$$\frac{x}{8} - \frac{340 - x}{9} = 0$$$$9x = 8(340 - x)$$$$9x = 2720 - 8x$$$$17x = 2720 \implies x = 160 \text{ mm}$$

To determine which statement is true, we start with the given equation:$$(9 \text{ inches})^{1/2} = (0.25 \text{ yards})^{1/2}$$To eliminate the fractional exponent (the square root), we square both sides of the equation:$$\left[ (9 \text{ inches})^{1/2} \right]^2 = \left[ (0.25 \text{ yards})^{1/2} \right]^2$$Using the algebraic identity $(x^a)^b = x^{a \cdot b}$, we have:$$(9 \text{ inches})^{1/2 \times 2} = (0.25 \text{ yards})^{1/2 \times 2}$$This simplifies directly to:$$9 \text{ inches} = 0.25 \text{ yards}$$Thus, the statement in Option C is mathematically equivalent to the given equation and is therefore TRUE.Correct Answer: Option C

A matrix has zero as an eigenvalue if and only if its determinant is zero.
Given the matrix $A = \begin{bmatrix} 3 & 2 & 4 \\ 9 & 7 & 13 \\ -6 & -4 & -9 + x \end{bmatrix}$, we calculate its determinant $\det(A)$ and set it to zero.$$\det(A) = 3 \begin{vmatrix} 7 & 13 \\ -4 & -9+x \end{vmatrix} - 2 \begin{vmatrix} 9 & 13 \\ -6 & -9+x \end{vmatrix} + 4 \begin{vmatrix} 9 & 7 \\ -6 & -4 \end{vmatrix}$$$$\det(A) = 3 [7(-9+x) - (-4)(13)] - 2 [9(-9+x) - (-6)(13)] + 4 [9(-4) - (-6)(7)]$$$$\det(A) = 3 [-63 + 7x + 52] - 2 [-81 + 9x + 78] + 4 [-36 + 42]$$$$\det(A) = 3 [7x - 11] - 2 [9x - 3] + 4 [6]$$$$\det(A) = 21x - 33 - 18x + 6 + 24$$$$\det(A) = 3x - 3$$For zero to be an eigenvalue, $\det(A) = 0$:$$3x - 3 = 0 \implies 3x = 3 \implies x = 1$$Thus, the value of $x$ is 1.

A function $f(z)$ is analytic if it satisfies the Cauchy-Riemann equations. The term $2z^3$ is a polynomial in $z$ and is analytic everywhere. For $f(z)$ to be analytic, the term $b|z|^3$ must also be analytic.
$|z|^3 = (z\bar{z})^{3/2}$.
Since $|z|^3$ depends on $\bar{z}$, it is not analytic anywhere except possibly at the origin. For the entire function $f(z)$ to be analytic in a domain, the coefficient $b$ must be zero.

To find the behavior of the function, we examine its derivative:
$f'(x) = 3x^2 - 6x = 3x(x - 2)$.
The critical points are $x = 0$ and $x = 2$.

• For $x < 0$, $f'(x) > 0$ (function is increasing).
• For $0 < x < 2$, $f'(x) < 0$ (function is decreasing).
• For $x > 2$, $f'(x) > 0$ (function is increasing).

In the interval $[-1, 3]$, the function increases from $x = -1$ to $x = 0$, decreases from $x = 0$ to $x = 2$, and increases again from $x = 2$ to $x = 3$. Thus, the behavior is: increases, then decreases and increases again.

To find the angle $\theta$ that the vector $\mathbf{I}$ makes with the $x$-axis when its magnitude is minimized, we analyze the components of the vector:$$\mathbf{I}(t) = I_x(t)\hat{\mathbf{x}} + I_y(t)\hat{\mathbf{y}}$$where:$$I_x(t) = 15 \cos(\omega t)$$$$I_y(t) = 5 \sin(\omega t)$$

Step 1: Express the squared magnitude of the vector

The squared magnitude of the vector $\mathbf{I}$ is given by:$$|\mathbf{I}|^2 = I_x^2 + I_y^2 = 225 \cos^2(\omega t) + 25 \sin^2(\omega t)$$Using the trigonometric identity $\cos^2(\omega t) = 1 - \sin^2(\omega t)$, we can rewrite $|\mathbf{I}|^2$ in terms of $\sin^2(\omega t)$:$$|\mathbf{I}|^2 = 225 (1 - \sin^2(\omega t)) + 25 \sin^2(\omega t)$$$$|\mathbf{I}|^2 = 225 - 200 \sin^2(\omega t)$$

Step 2: Minimize the magnitude

To minimize $|\mathbf{I}|^2$ (and thus minimize $|\mathbf{I}|$), we must maximize the term being subtracted, which is $200 \sin^2(\omega t)$. Since the maximum value of $\sin^2(\omega t)$ is $1$, the minimum value of $|\mathbf{I}|^2$ occurs when:$$\sin^2(\omega t) = 1$$This condition implies:$$\cos^2(\omega t) = 0 \implies \cos(\omega t) = 0$$At this minimum point, the components of the vector $\mathbf{I}$ are:$$I_x = 15 \cos(\omega t) = 0$$$$I_y = 5 \sin(\omega t) = \pm 5$$Thus, the vector at its minimum magnitude is:$$\mathbf{I}_{\text{min}} = \pm 5 \hat{\mathbf{y}}$$

Step 3: Calculate the angle $\theta$ with the $x$-axis

The angle $\theta$ that the vector makes with the positive $x$-axis is given by:$$\tan\theta = \frac{I_y}{I_x}$$Since $I_x = 0$ and $I_y \neq 0$, the vector lies entirely along the $y$-axis. For $0 \leq \theta \leq 180^\circ$:

• If $I_y = 5$, the vector points along the positive $y$-axis, corresponding to $\theta = 90^\circ$.
• If $I_y = -5$, the vector points along the negative $y$-axis, corresponding to $\theta = 270^\circ$ (or $-90^\circ$), which lies outside the specified range $[0, 180^\circ]$.

Thus, the unique angle within the interval $[0, 180]$ is:$$\theta = 90^\circ$$

To find the value of the constant current load $I_L$ that maximizes the power absorbed by it, we analyze the given circuit.

Step 1: Express the Voltage across the Current Load

The circuit consists of a constant voltage source $V_S$ in series with a resistor $R$ and a constant current load $I_L$. By applying Kirchhoff's Voltage Law (KVL) around the loop, the voltage $V_L$ across the constant current load is given by:$$V_L = V_S - I_L R$$

Step 2: Formulate the Power Absorbed by the Load

The power $P_L$ absorbed by the constant current load is the product of the voltage across it and the current flowing through it:$$P_L = V_L \cdot I_L$$Substituting the expression for $V_L$ into the power equation:$$P_L = (V_S - I_L R) I_L$$$$P_L = V_S I_L - I_L^2 R$$

Step 3: Maximize the Power with respect to $I_L$

To find the value of $I_L$ that maximizes $P_L$, we differentiate $P_L$ with respect to $I_L$ and set the derivative to zero:$$\frac{dP_L}{dI_L} = 0$$Differentiating the expression:$$\frac{d}{dI_L} \left( V_S I_L - I_L^2 R \right) = V_S - 2 I_L R = 0$$Solving for $I_L$:$$2 I_L R = V_S$$$$I_L = \frac{V_S}{2R}$$

Step 4: Verify the Maximum Condition

We check the second derivative of $P_L$ with respect to $I_L$:$$\frac{d^2P_L}{dI_L^2} = -2R$$Since $R > 0$, the second derivative is negative ($-2R < 0$), confirming that $I_L = \frac{V_S}{2R}$ maximizes the power absorbed by the load.Correct Option: B

Thus, the correct option is (D).

In a parallel RLC circuit, the resonant frequency is given by:$$\omega_0 = \frac{1}{\sqrt{LC}}$$Given:
$R = 10 \Omega$
$L = 10 \text{ mH} = 10 \times 10^{-3} \text{ H} = 10^{-2} \text{ H}$
$C = 10 \mu\text{F} = 10 \times 10^{-6} \text{ F} = 10^{-5} \text{ F}$Calculating $\omega_0$:$$\omega_0 = \frac{1}{\sqrt{10^{-2} \times 10^{-5}}} = \frac{1}{\sqrt{10^{-7}}} = \sqrt{10^7} = 10^3 \sqrt{10} \approx 3162.28 \text{ rad/s}$$At resonance, the impedance of the parallel combination is purely resistive and equal to $R$. The voltage across the parallel branches is:$$V = I_R \cdot R$$The magnitude of the current through the inductor is:$$|\mathbf{I}_L| = \frac{V}{X_L} = \frac{V}{\omega_0 L} = \frac{I_R \cdot R}{\omega_0 L}$$The required ratio is:$$\frac{|\mathbf{I}_L|}{|\mathbf{I}_R|} = \frac{R}{\omega_0 L} = \frac{10}{3162.28 \times 10^{-2}} = \frac{10}{31.6228} \approx 0.3162$$The value is approximately 0.316, which falls within the specified range of 0.30 to 0.34.

For a T-network with series arm impedances $Z_a, Z_b$ and shunt arm impedance $Z_c$, the z-parameters are defined as:
$Z_{11} = Z_a + Z_c$
$Z_{12} = Z_{21} = Z_c$
$Z_{22} = Z_b + Z_c$Given the z-parameter matrix:$$\mathbf{Z} = \begin{bmatrix} 2j\omega & j\omega \\ j\omega & 3 + 2j\omega \end{bmatrix}$$By comparing the elements:Substitute $Z_c = j\omega$ into the equation for $Z_{22}$:$$Z_b + j\omega = 3 + 2j\omega$$$$Z_b = 3 + j\omega$$It is given that $Z_b(j\omega) = R_b + j\omega$. Comparing the real parts:$$R_b = 3$$The value of $R_b$ is 3, which is within the specified range of 2.98 to 3.02.

The energy E of a continuous-time signal $x(t)$ is given by the integral of its squared magnitude over all time:$$E = \int_{-\infty}^{\infty} |x(t)|^2 dt$$For the given signal $x(t) = \frac{\sin(4\pi t)}{4\pi t}$, the energy is:$$E = \int_{-\infty}^{\infty} \left| \frac{\sin(4\pi t)}{4\pi t} \right|^2 dt$$We can solve this using Parseval's theorem, which states that the energy of a signal is equal to the energy of its Fourier transform:$$E = \int_{-\infty}^{\infty} |x(t)|^2 dt = \int_{-\infty}^{\infty} |X(f)|^2 df$$where $X(f)$ is the Fourier Transform of $x(t)$.The signal $x(t)$ is a form of the sinc function. The Fourier transform of a rectangular pulse is a sinc function, and by duality, the Fourier transform of a sinc function is a rectangular pulse. The relevant Fourier transform pair is:$$A \cdot \text{sinc}(2Wt) \iff \frac{A}{2W} \text{rect}\left(\frac{f}{2W}\right)$$where the normalized sinc function is $\text{sinc}(u) = \frac{\sin(\pi u)}{\pi u}$ and the rectangular function is $\text{rect}(v) = 1$ for $|v| < 1/2$ and 0 otherwise.Let's express the given signal $x(t)$ in terms of the normalized sinc function:$$x(t) = \frac{\sin(4\pi t)}{4\pi t} = \frac{\sin(\pi(4t))}{\pi(4t)} = \text{sinc}(4t)$$Comparing this with $A \cdot \text{sinc}(2Wt)$, we have $A=1$ and $2W=4$, which means $W=2$.Now, we can find the Fourier transform $X(f)$:$$X(f) = \frac{A}{2W} \text{rect}\left(\frac{f}{2W}\right) = \frac{1}{4} \text{rect}\left(\frac{f}{4}\right)$$The function $\text{rect}(f/4)$ is 1 for $|f/4| < 1/2$, i.e., $|f| < 2$, and 0 otherwise. So, the Fourier transform is:$$X(f) = \begin{cases} 1/4 & \text{for } -2 < f < 2 \\ 0 & \text{otherwise} \end{cases}$$Now, we calculate the energy using the integral in the frequency domain:$$E = \int_{-\infty}^{\infty} |X(f)|^2 df = \int_{-2}^{2} \left(\frac{1}{4}\right)^2 df$$$$E = \int_{-2}^{2} \frac{1}{16} df = \frac{1}{16} [f]_{-2}^{2}$$$$E = \frac{1}{16} (2 - (-2)) = \frac{1}{16} (4) = \frac{4}{16} = \frac{1}{4} = 0.25$$Alternatively, we can use the standard integral formula:$$\int_{-\infty}^{\infty} \left(\frac{\sin(at)}{t}\right)^2 dt = a\pi$$Our energy integral is:$$E = \int_{-\infty}^{\infty} \left(\frac{\sin(4\pi t)}{4\pi t}\right)^2 dt = \frac{1}{(4\pi)^2} \int_{-\infty}^{\infty} \left(\frac{\sin(4\pi t)}{t}\right)^2 dt$$Here, $a = 4\pi$. Applying the formula:$$E = \frac{1}{16\pi^2} \cdot (4\pi \cdot \pi) = \frac{4\pi^2}{16\pi^2} = \frac{1}{4} = 0.25$$Both methods give the energy as 0.25.

The Ebers-Moll model is a large-signal mathematical model for the bipolar junction transistor (BJT). It is derived from the diffusion equations and is valid for all regions of operation, including forward-active, reverse-active, saturation, and cut-off. It treats the BJT as two back-to-back diodes with current-controlled current sources.

The question asks to identify the incorrect statement regarding the resistance of a long-channel NMOS transistor biased in the linear region.The drain current ($I_D$) for an NMOS transistor in the linear (or triode) region is given by the equation:$$I_D = \mu_n C_{ox} \frac{W}{L} \left[ (V_{GS} - V_{th})V_{DS} - \frac{V_{DS}^2}{2} \right]$$where:

• $\mu_n$ is the electron mobility
• $C_{ox}$ is the gate oxide capacitance per unit area
• $W$ is the channel width
• $L$ is the channel length
• $V_{GS}$ is the gate-to-source voltage
• $V_{th}$ is the threshold voltage
• $V_{DS}$ is the drain-to-source voltage

The transistor is used as a resistance, which is defined as $R_{DS} = \frac{V_{DS}}{I_D}$.
Given that $V_{DS} = 50$ mV, which is a small value, the condition $V_{DS} \ll 2(V_{GS} - V_{th})$ for linear region operation is satisfied, and we can approximate the current by neglecting the $V_{DS}^2$ term.$$I_D \approx \mu_n C_{ox} \frac{W}{L} (V_{GS} - V_{th})V_{DS}$$The resistance $R_{DS}$ is then:$$R_{DS} = \frac{V_{DS}}{I_D} \approx \frac{1}{\mu_n C_{ox} \frac{W}{L} (V_{GS} - V_{th})}$$Now, let's analyze each statement based on this expression for $R_{DS}$:(A) If the device width $W$ is increased, the denominator increases, so the resistance $R_{DS}$ decreases. This statement is correct.(B) If the threshold voltage $V_{th}$ is reduced, the term $(V_{GS} - V_{th})$ increases. This increases the denominator, so the resistance $R_{DS}$ decreases. This statement is correct.(C) If the device length $L$ is increased, the denominator decreases (since $L$ is in the denominator of the $\frac{W}{L}$ term), so the resistance $R_{DS}$ increases. This statement is correct.(D) If $V_{GS}$ is increased, the term $(V_{GS} - V_{th})$ increases. This increases the denominator, so the resistance $R_{DS}$ decreases. The statement says the resistance increases, which is contrary to our finding. This statement is NOT correct.The question asks for the statement that is NOT correct. Therefore, option (D) is the correct answer.