GATE EE 2016 Set 2

The phrasal verb "bear with" means to be patient or tolerant with someone. "Bare" as a verb means to uncover, and "bore" is the past tense of bear or means to make a hole. Therefore, "bear with" is the grammatically correct choice in this context.

The correct spelling is "Manageable". It is formed by adding the suffix "-able" to the root word "manage". In English, when adding "-able" to a word ending in a soft "g" (like manage), the silent "e" is retained to preserve the soft sound.

In the given sequence (13, 23, 33, 43, 53):

• 13, 23, 43, and 53 are prime numbers (they have no factors other than 1 and themselves).
• 33 is a composite number because it is divisible by 3 and 11 ($3 \times 11 = 33$).

Thus, 33 is the odd one out.

From the given statements:Combining (2) and (3), we can conclude that R2D2 is the only robot capable of repairing aeroplanes. This matches option (B). Options (A) and (C) are incorrect because the statements don't imply that R2D2 only repairs aeroplanes; it might perform other tasks as well. Option (D) is incorrect as the statements don't claim R2D2 is the only robot in existence.

Given equation: $|9y - 6| = 3$
This implies two cases:
Case 1: $9y - 6 = 3$
$9y = 9 \implies y = 1$
Then, $y^2 - \frac{4y}{3} = 1^2 - \frac{4(1)}{3} = 1 - \frac{4}{3} = -\frac{1}{3}$Case 2: $9y - 6 = -3$
$9y = 3 \implies y = \frac{1}{3}$
Then, $y^2 - \frac{4y}{3} = (\frac{1}{3})^2 - \frac{4(\frac{1}{3})}{3} = \frac{1}{9} - \frac{4}{9} = -\frac{3}{9} = -\frac{1}{3}$In both cases, the value is $-\frac{1}{3}$.

Step 1: Identify Large and Small plants based on Installed Capacity (IC).
Large plants (IC $\geq$ 200): Plants 1, 4, 8, 9.
Small plants (IC < 200): Plants 2, 3, 5, 6, 7.Step 2: Calculate total Actual Production (AP) for Large plants.
Total AP (Large) = AP(1) + AP(4) + AP(8) + AP(9)
Total AP (Large) = 160 + 190 + 230 + 190 = 770 tonnes.Step 3: Calculate total Actual Production (AP) for Small plants.
Total AP (Small) = AP(2) + AP(3) + AP(5) + AP(6) + AP(7)
Total AP (Small) = 150 + 160 + 120 + 100 + 120 = 650 tonnes.Step 4: Find the difference.
Difference = 770 - 650 = 120 tonnes.

The paragraph presents two pieces of information:From these two statements, we can logically infer that despite the perception among students, mechanical engineering is indeed a profession well-suited for women, especially those with higher qualifications, given their high success rate. Therefore, option (C) is the most direct and logical inference.Let's analyze other options:
(A) "Many students have misconceptions regarding various engineering disciplines." While the first statement suggests a misconception about mechanical engineering, the paragraph does not provide information to generalize this to "various engineering disciplines."
(B) "Men with advanced degrees in mechanical engineering believe women are well suited to be mechanical engineers." The paragraph does not provide any information about the beliefs of men with advanced degrees.
(D) "The number of women pursuing higher degrees in mechanical engineering is small." The paragraph states that 99% of such women (women with masters or higher degrees in mechanical engineering) were successful, but it does not give any indication of the total number or proportion of women pursuing these degrees.

Let's analyze each statement based on the given sentence:Sentence: "Sourya committee had proposed the establishment of Sourya Institutes of Technology (SITs) in line with Indian Institutes of Technology (IITs) to cater to the technological and industrial needs of a developing country."(i) In the initial years, SIT students will get degrees from IIT.
This cannot be inferred. "In line with IITs" means similar in structure, quality, or objective, not that they would issue IIT degrees.(ii) SITs will have a distinct national objective.
This can be inferred. The phrase "to cater to the technological and industrial needs of a developing country" clearly indicates a specific national objective for the SITs.(iii) SIT like institutions can only be established in consultation with IIT.
This cannot be inferred. The sentence states SITs are "in line with IITs," which implies a similar model or standard, but it does not state that consultation with IITs is a prerequisite or the only way to establish such institutions.(iv) SITs will serve technological needs of a developing country.
This can be directly inferred. The sentence explicitly states that SITs are established "to cater to the technological and industrial needs of a developing country."Therefore, statements (ii) and (iv) can be logically inferred from the given sentence.Comparing with the options:
(A) (iii) and (iv) only. (iii) is incorrect.
(B) (i) and (iv) only. (i) is incorrect.
(C) (ii) and (iv) only. Both (ii) and (iv) are correct inferences.
(D) (ii) and (iii) only. (iii) is incorrect.Thus, option (C) is the correct answer.

This is a binomial probability problem. We are given:

• Probability of success (making a free throw), $p = 60\% = 0.6$
• Number of attempts (trials), $n = 10$
• Number of successful free throws desired, $k = 6$

The probability of failure (missing a free throw) is $q = 1 - p = 1 - 0.6 = 0.4$.The binomial probability formula is given by:
$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$Substitute the given values:
$P(X=6) = \binom{10}{6} (0.6)^6 (0.4)^{10-6}$
$P(X=6) = \binom{10}{6} (0.6)^6 (0.4)^4$First, calculate the binomial coefficient $\binom{10}{6}$:
$\binom{10}{6} = \frac{10!}{6!(10-6)!} = \frac{10!}{6!4!} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6! \times 4 \times 3 \times 2 \times 1} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$Next, calculate the powers:
$(0.6)^6 = 0.046656$
$(0.4)^4 = 0.0256$Now, multiply these values:
$P(X=6) = 210 \times 0.046656 \times 0.0256$
$P(X=6) = 210 \times 0.0011943936$
$P(X=6) = 0.250822656$Rounding to four decimal places, we get $0.2508$.Comparing this with the given options:
(A) 0.2508
(B) 0.2816
(C) 0.2934
(D) 0.6000The calculated probability matches option (A).The final answer is $\boxed{\text{0.2508}}$

To find the units digit of the expression $211^{870} + 146^{127} \times 3^{424}$, we determine the units digit of each term individually:Now, combine these results according to the expression:
Units digit = (Units digit of $211^{870}$) + (Units digit of $146^{127} \times$ Units digit of $3^{424}$)
Units digit = $1 + (6 \times 1) = 1 + 6 = 7$.Therefore, the numeral in the units position is 7.

The Karnaugh map is given as:We can identify two groups of 1s:Combining these, the simplified Boolean expression is $A + \bar{C}$.

The given circuit is an inverting amplifier configuration. The transfer function is $H(s) = -\frac{Z_f}{Z_{in}}$.

• $Z_{in} = R_1$
• $Z_f = R_2 \parallel \frac{1}{sC} = \frac{R_2}{1 + sR_2C}$

Substituting these into the transfer function:
$H(s) = -\frac{R_2/R_1}{1 + sR_2C}$This is the standard form of a first-order low-pass filter transfer function, where the DC gain is $-R_2/R_1$ and the cutoff frequency is $\omega_c = \frac{1}{R_2C}$.

Given:
Primary voltage, $V_1 = 100\text{ V}$ (rms)
Turns ratio, $a = \frac{N_1}{N_2} = \frac{1}{100}$
Primary reactance, $X_L = 10\ \Omega$
Secondary resistance, $R = 80\ \text{k}\Omega = 80 \times 10^3\ \Omega$
Secondary capacitive reactance, $X_C = 40\ \text{k}\Omega = 40 \times 10^3\ \Omega$The secondary impedance is $Z_2 = R - jX_C = (80 - j40) \times 10^3\ \Omega$.Reflecting the secondary impedance to the primary side:
$Z_2' = a^2 Z_2 = \left(\frac{1}{100}\right)^2 \times (80 - j40) \times 10^3 = 10^{-4} \times (80 - j40) \times 10^3 = 8 - j4\ \Omega$.The total impedance seen by the primary source is:
$Z_{total} = jX_L + Z_2' = j10 + (8 - j4) = 8 + j6\ \Omega$.The magnitude of the total impedance is:
$|Z_{total}| = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\ \Omega$.The rms value of the primary current $I$ is:
$I = \frac{V_1}{|Z_{total}|} = \frac{100}{10} = 10\text{ A}$.

$sY(s) + \frac{1}{6} Y(s) = 3X(s) \Rightarrow H(s) = \frac{Y(s)}{X(s)} = \frac{3}{s + 1/6}$.$A = \frac{9}{-1/6 + 1/3} = \frac{9}{1/6} = 54$.
$B = \frac{9}{-1/3 + 1/6} = \frac{9}{-1/6} = -54$.

• $\frac{\partial u}{\partial x} = 2$
• $\frac{\partial v}{\partial y} = 0$

Since $2 \neq 0$, the first C-R equation is not satisfied. Thus, $f(z)$ is not analytic anywhere.

Given the matrix equation $P^3 = P$. The eigenvalues $\lambda$ of matrix $P$ must satisfy the same polynomial equation as the matrix itself:$$\lambda^3 = \lambda$$$$\lambda^3 - \lambda = 0$$$$\lambda(\lambda^2 - 1) = 0$$$$\lambda(\lambda - 1)(\lambda + 1) = 0$$The possible eigenvalues are $\lambda \in \{0, 1, -1\}$.For a $3 \times 3$ matrix, any combination of three values from this set can be its eigenvalues.

• Option (A) $\{1, 1, -1\}$ consists of values from the set $\{0, 1, -1\}$.
• Option (D) $\{0, 1, -1\}$ consists of values from the set $\{0, 1, -1\}$.

Options (B) and (C) contain complex numbers that do not satisfy the equation $\lambda^3 = \lambda$. Therefore, both (A) and (D) are correct sets of eigenvalues.

The given differential equation is a second-order linear homogeneous equation with constant coefficients: $y''(t) + 2y'(t) + y(t) = 0$. The characteristic equation is $r^2 + 2r + 1 = 0$, which has a repeated root $r = -1$. The general solution is $y(t) = (C_1 + C_2 t) e^{-t}$. Applying the initial conditions:Thus, $y(t) = t e^{-t}$. Since the solution is for $t > 0$, we multiply by the unit step function $u(t)$, giving $y(t) = t e^{-t} u(t)$.

To evaluate the line integral $\int_C (2xy^2 dx + 2x^2y dy + dz)$, we first check if the vector field $\mathbf{F} = (2xy^2, 2x^2y, 1)$ is conservative. The curl of $\mathbf{F}$ is:$$\nabla \times \mathbf{F} = \left(\frac{\partial 1}{\partial y} - \frac{\partial (2x^2y)}{\partial z}, \frac{\partial (2xy^2)}{\partial z} - \frac{\partial 1}{\partial x}, \frac{\partial (2x^2y)}{\partial x} - \frac{\partial (2xy^2)}{\partial y}\right) = (0-0, 0-0, 4xy-4xy) = (0, 0, 0)$$Since the curl is zero, the field is conservative. We find the potential function $\phi(x, y, z)$ such that $\nabla \phi = \mathbf{F}$.
Integrating $F_x = 2xy^2$ with respect to $x$ gives $\phi = x^2y^2 + g(y, z)$.
Differentiating with respect to $y$ gives $2x^2y + \frac{\partial g}{\partial y} = 2x^2y$, so $g$ is a function of $z$ only.
Differentiating with respect to $z$ gives $g'(z) = 1$, so $g(z) = z + C$.
Thus, $\phi(x, y, z) = x^2y^2 + z$.
The line integral is $\phi(1, 1, 1) - \phi(0, 0, 0) = (1^2 \cdot 1^2 + 1) - (0^2 \cdot 0^2 + 0) = 2$.

A function $f(x)$ is defined as odd if $f(-x) = -f(x)$. For an odd periodic function, the Fourier series coefficients are calculated as follows:Therefore, the general form of the Fourier series for an odd function contains only sine terms: $f(x) = \sum_{k=1}^{\infty} b_k \sin(kx)$.

In a series circuit with a resistor $R$ and a coil (having resistance $r$ and inductance $L$), the phasor relationship for the source voltage $V_s$ is:
$\vec{V}_s = \vec{V}_R + \vec{V}_C$
Where $\vec{V}_R = I R \angle 0^\circ$ and $\vec{V}_C = I(r + jX_L) = V_C \angle \theta$, with $0^\circ \leq \theta \leq 90^\circ$ for an inductive coil.
The magnitude of the source voltage is given by:
$V_s^2 = V_R^2 + V_C^2 + 2 V_R V_C \cos \theta$
Since $0 \leq \cos \theta \leq 1$, the plausible values must satisfy $V_R^2 + V_C^2 \leq V_s^2 \leq (V_R + V_C)^2$.
Given $V_s = 100\text{ V}$, then $V_s^2 = 10000$.
Checking the options:
(A) $65^2 + 35^2 = 4225 + 1225 = 5450 \leq 10000$. (Possible, but implies $\theta = 0^\circ$, a purely resistive coil)
(B) $50^2 + 50^2 = 2500 + 2500 = 5000 \leq 10000$. (Possible, but implies $\theta = 0^\circ$, a purely resistive coil)
(C) $60^2 + 90^2 = 3600 + 8100 = 11700 > 10000$. (Impossible for a passive RL circuit)
(D) $60^2 + 80^2 = 3600 + 6400 = 10000$. (Possible, implies $\theta = 90^\circ$, an ideal inductive coil)Option (D) is the most plausible as it represents a standard ideal inductive coil and fits the Pythagorean triple $60-80-100$.

Given the instantaneous voltage and current:
$v(t) = 100 \sin(\omega t)$ V
$i(t) = 10 \sin(\omega t - 60^\circ) + 2 \sin(3\omega t) + 5 \sin(5\omega t)$ AThe average power $P_{avg}$ is defined as:
$P_{avg} = \frac{1}{T} \int_{0}^{T} v(t) i(t) dt$In a circuit with harmonics, the total average power is the sum of the average powers produced by each harmonic component of the same frequency:
$P_{avg} = \sum_{n=1}^{\infty} V_{n,rms} I_{n,rms} \cos(\phi_n)$
where $\phi_n$ is the phase difference between the $n^{th}$ harmonic voltage and current.From the given expressions: $V_{1,max} = 100$ V $\implies V_{1,rms} = \frac{100}{\sqrt{2}}$ V
$I_{1,max} = 10$ A $\implies I_{1,rms} = \frac{10}{\sqrt{2}}$ A
Phase of voltage $\theta_{v1} = 0^\circ$
Phase of current $\theta_{i1} = -60^\circ$
Phase difference $\phi_1 = \theta_{v1} - \theta_{i1} = 0^\circ - (-60^\circ) = 60^\circ$
$P_1 = V_{1,rms} I_{1,rms} \cos(\phi_1) = \left( \frac{100}{\sqrt{2}} \right) \left( \frac{10}{\sqrt{2}} \right) \cos(60^\circ) = \frac{1000}{2} \times 0.5 = 250$ W The voltage $v(t)$ has no components at $3\omega$ or $5\omega$ (i.e., $V_3 = 0, V_5 = 0$). Therefore, these current harmonics do not contribute to the average power.
$P_3 = 0 \times I_{3,rms} \cos(\phi_3) = 0$
$P_5 = 0 \times I_{5,rms} \cos(\phi_5) = 0$Total average power:
$P_{avg} = P_1 + P_3 + P_5 = 250 + 0 + 0 = 250$ W.

For a fault on the transmission line between the middle bus and the right bus, the primary relays are $R_4$ and $R_5$. If $R_4$ fails to operate, the fault must be cleared by the next upstream relay in that direction, which is $R_1$. Similarly, if $R_5$ fails, the backup is $R_6$. Therefore, the remote backup relays for the protection zone involving $R_4$ are $R_1$ and $R_6$.

In the Newton-Raphson load flow method using polar coordinates, the size of the Jacobian matrix is $(2n - m - 2) \times (2n - m - 2)$, where $n$ is the total number of buses and $m$ is the number of PV (generator) buses (excluding the slack bus).
Given $n = 100$ and total generator buses = 10. Assuming one generator bus is the slack bus, $m = 10 - 1 = 9$.
Size $= (2 \times 100 - 9 - 2) = 200 - 11 = 189$.
Thus, the size is $189 \times 189$.