GATE CE 2014 Set 1

The word "comprehension" refers to the action or capability of understanding something.

• (A) understanding: This is the closest synonym as it directly relates to the ability to grasp information.
• (B) meaning: Refers to the significance of something, not the act of grasping it.
• (C) concentration: Refers to the focus of attention.
• (D) stability: Refers to the state of being stable.

Therefore, the correct option is (A).

The sentence describes a positive character trait ("ability to forgive").

• (A) vice: Refers to a moral flaw or bad habit.
• (B) virtues: Refers to behavior showing high moral standards. Forgiveness is a classic moral virtue.
• (C) choices: Refers to acts of selection, which doesn't fit the context of a character trait.
• (D) strength: While forgiveness can be a strength, "virtues" is the more precise term for moral qualities in this context.

Therefore, the correct option is (B).

The passage states that Rajan was unhappy because Sajan decided to work alone. This reaction of unhappiness logically implies that Rajan had a prior expectation or belief that they would be collaborating on the project.

• (A) is an overgeneralization; we only know about this specific project.
• (B) is not supported by the text; we don't know how the group was intended to be formed.
• (C) is contradicted by the text, as Sajan chose to work independently.
• (D) is the most direct and valid inference based on Rajan's expectation of partnership.

Therefore, the correct option is (D).

Given the equation of the curve: $y = 5x^2 + 3$
To find the slope of the tangent, we differentiate $y$ with respect to $x$:$$\frac{dy}{dx} = \frac{d}{dx}(5x^2 + 3) = 10x$$At the point $(0, 3)$, the slope $m$ is:$$m = \left. \frac{dy}{dx} \right|_{x=0} = 10(0) = 0$$A line with a slope of $0$ is a horizontal line, which is parallel to the $x$-axis.
Therefore, the tangent at $x = 0, y = 3$ is parallel to the $x$-axis.

Fixed daily cost = Rs 50,000
Variable cost = Rs 800Q, where $Q$ is production in tonnes.
Total daily cost $C = 50,000 + 800Q$
For a daily production of $Q = 100$ tonnes:
Total cost $C = 50,000 + 800(100) = 50,000 + 80,000 = \text{Rs } 1,30,000$
Cost of production per tonne = $\frac{\text{Total Cost}}{\text{Total Production}} = \frac{1,30,000}{100} = \text{Rs } 1300$.

Let's examine the numerical positions of the letters in the alphabet for each group:(A) ALRVX: A(1), L(12), R(18), V(22), X(24)
Differences: $12-1=11$, $18-12=6$, $22-18=4$, $24-22=2$. Pattern: $+11, +6, +4, +2$.(B) EPVZB: E(5), P(16), V(22), Z(26), B(2 or 28)
Differences: $16-5=11$, $22-16=6$, $26-22=4$, $28-26=2$. Pattern: $+11, +6, +4, +2$.(C) ITZDF: I(9), T(20), Z(26), D(4 or 30), F(6 or 32)
Differences: $20-9=11$, $26-20=6$, $30-26=4$, $32-30=2$. Pattern: $+11, +6, +4, +2$.(D) OYEIK: O(15), Y(25), E(5 or 31), I(9 or 35), K(11 or 37)
Differences: $25-15=10$, $31-25=6$, $35-31=4$, $37-35=2$. Pattern: $+10, +6, +4, +2$.Groups A, B, and C follow the same pattern of differences ($+11, +6, +4, +2$). Group D is the odd one because its first difference is $+10$.

$3x + 4x + 5x + 6x = 18x = 360^\circ \Rightarrow x = 20^\circ$.
Smallest angle of quadrilateral $= 3 \times 20^\circ = 60^\circ$.
Largest angle of quadrilateral $= 6 \times 20^\circ = 120^\circ$. The angles of the triangle are $40^\circ, 60^\circ, 80^\circ$. The second largest is $60^\circ$.

Let the population of country Y be $N$. Then the population of country X is $3N$.
Total population $= N + 3N = 4N$.
Number of people taller than 6 ft in X $= 1\% \text{ of } 3N = 0.01 \times 3N = 0.03N$.
Number of people taller than 6 ft in Y $= 2\% \text{ of } N = 0.02 \times N = 0.02N$.
Total number of people taller than 6 ft $= 0.03N + 0.02N = 0.05N$.
Overall percentage $= \frac{0.05N}{4N} \times 100 = \frac{5}{4} = 1.25\%$.

Step-by-step analysis of the statements based on the provided rainfall chart:Since statements (i) and (iii) are true, the correct option is (B).

To evaluate the limit L = \lim_{x \to $\infty$} \left( \frac{x + \sin x}{x} \right)we can simplify the expression inside the limit by dividing each term in the numerator by the denominator:$\frac{x + \sin x}{x} = \frac{x}{x} + \frac{\sin x}{x} = 1 + \frac{\sin x}{x}$Now, we apply the limit to each term:L = \lim_{x \to $\infty$} \left( 1 + \frac{\sin x}{x} \right) = \lim_{x \to $\infty$} (1) + \lim_{x \to $\infty$} \left( \frac{\sin x}{x} \right)To evaluate the second limit, we use the Squeeze Theorem: $$-1 \le \sin x \le 1$$ $ -\frac{1}{x} \le \frac{\sin x}{x} \le \frac{1}{x}$ \lim_{x \to $\infty$} \left( -\frac{1}{x} \right) = 0 \quad \text{and} \quad \lim_{x \to $\infty$} \left( \frac{1}{x} \right) = 0By the Squeeze Theorem, it follows that:\lim_{x \to $\infty$} \left( \frac{\sin x}{x} \right) = 0Substituting this back into the original limit expression:$$L = 1 + 0 = 1$$Correct Option: C

To find the value of the product $K^T J K$, we perform the matrix multiplications step by step.The given matrices are:$$J = \begin{bmatrix} 3 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 6 \end{bmatrix} \quad \text{and} \quad K = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$$First, we find the transpose of matrix $K$:$$K^T = \begin{bmatrix} 1 & 2 & -1 \end{bmatrix}$$

Step 1: Compute the product $J K$

Multiply the $3 \times 3$ matrix $J$ by the $3 \times 1$ column vector $K$:$$J K = \begin{bmatrix} 3 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 6 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$$Calculating each element of the resulting $3 \times 1$ vector:

• First row:

$$(3 \times 1) + (2 \times 2) + (1 \times -1) = 3 + 4 - 1 = 6$$

• Second row:

$$(2 \times 1) + (4 \times 2) + (2 \times -1) = 2 + 8 - 2 = 8$$

• Third row:

$$(1 \times 1) + (2 \times 2) + (6 \times -1) = 1 + 4 - 6 = -1$$Thus, $$J K = \begin{bmatrix} 6 \\ 8 \\ -1 \end{bmatrix}$$---

Step 2: Compute the product $K^T (J K)$

Now, multiply the $1 \times 3$ row vector $K^T$ by the $3 \times 1$ column vector $J K$:$$K^T J K = \begin{bmatrix} 1 & 2 & -1 \end{bmatrix} \begin{bmatrix} 6 \\ 8 \\ -1 \end{bmatrix}$$$$K^T J K = (1 \times 6) + (2 \times 8) + (-1 \times -1)$$$$K^T J K = 6 + 16 + 1$$$$K^T J K = 23$$The value of the product $K^T J K$ is 23.

To find the probability that the daily evaporation $E$ lies between $2\text{ mm/day}$ and $4\text{ mm/day}$, we integrate the probability density function $f(E)$ over the interval $[2, 4]$.The probability density function is given by:$$f(E) = \begin{cases} \frac{1}{5} & 0 \le E \le 5 \\ 0 & \text{otherwise} \end{cases}$$The required probability $P(2 \le E \le 4)$ is:$$P(2 \le E \le 4) = \int_{2}^{4} f(E) \, dE$$Substitute the value of $f(E) = \frac{1}{5}$ for the interval $[2, 4]$:$$P(2 \le E \le 4) = \int_{2}^{4} \frac{1}{5} \, dE$$Evaluate the integral:$$P(2 \le E \le 4) = \left[ \frac{E}{5} \right]_{2}^{4}$$$$P(2 \le E \le 4) = \frac{4}{5} - \frac{2}{5} = \frac{2}{5}$$$$P(2 \le E \le 4) = 0.4$$The probability that $E$ lies between $2$ and $4\text{ mm/day}$ is 0.4.

The sum of the eigenvalues of a square matrix is equal to the trace of the matrix, which is the sum of its diagonal elements.Given matrix $[M] = \begin{bmatrix} 215 & 650 & 795 \\ 655 & 150 & 835 \\ 485 & 355 & 550 \end{bmatrix}$Sum of eigenvalues = Trace($M$) = $215 + 150 + 550 = 915$.Therefore, the correct option is (A).

The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by the formula:$$\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$$Substituting the given coordinates $(1, 0), (2, 2),$ and $(4, 3)$:$$\text{Area} = \frac{1}{2} |1(2 - 3) + 2(3 - 0) + 4(0 - 2)|$$$$\text{Area} = \frac{1}{2} |1(-1) + 2(3) + 4(-2)|$$$$\text{Area} = \frac{1}{2} |-1 + 6 - 8|$$$$\text{Area} = \frac{1}{2} |-3| = 1.5 = \frac{3}{2}$$Therefore, the correct option is (A).

Matching the concepts from structural design and plastic analysis:

• P (Factor to decrease ultimate strength to design strength): In limit state design, the design strength is obtained by dividing the characteristic (ultimate) strength by a Material partial safety factor (3). Thus, P matches with 3.
• Q (Factor to increase working load to ultimate load for design): The Load factor (4) is the ratio of the ultimate load to the working load. It is used to increase the working load to the design ultimate load. Thus, Q matches with 4.
• R (Statical method of ultimate load analysis): According to the lower bound theorem of plastic analysis, the statical method provides a Lower bound on ultimate load (2). Thus, R matches with 2.
• S (Kinematical mechanism method of ultimate load analysis): According to the upper bound theorem of plastic analysis, the kinematical (mechanism) method provides an Upper bound on ultimate load (1). Thus, S matches with 1.

The correct combination is P-3, Q-4, R-2, S-1, which corresponds to option (C).

The shear center of a section is the point through which if a load is applied, the section will undergo only bending and no twisting. For a mono-symmetric section like a channel, the shear center lies on the axis of symmetry but outside the web. In the given figure, point P is the only point located on the axis of symmetry and outside the section, which is the characteristic location for a channel's shear center.

Using the Principle of Virtual Work for the collapse mechanism:

According to IS 456:2000, Clause 34.4, the design bearing strength of concrete in the limit state method is $0.45 f_{ck}$.
Given the concrete grade is M20, the characteristic strength $f_{ck} = 20$ N/mm$^2$.
Bearing strength $= 0.45 \times 20 = 9$ N/mm$^2$.

According to IS 800:2007, Clause 9.2.2, a section is considered to be subjected to 'high shear' when the design shear force $V$ exceeds $60\%$ of the design shear strength $V_d$ (or $V_s$). In such cases, the reduction in moment capacity due to the presence of high shear must be accounted for.

The degree of static indeterminacy $D_s$ for a 2D frame is given by:
$D_s = (3m + r) - 3j - r_r$
Where:

• $m$ = number of members = 3 (PQ, QR, and cable RS)
• $r$ = number of external reactions = 1 (at P) + 3 (at S) = 4
• $j$ = number of joints = 4 (P, Q, R, S)
• $r_r$ = number of internal releases. Since the cable RS is a link, it is pinned at R and S. This provides 2 releases (one at each end of the cable).

$D_s = (3 \times 3 + 4) - (3 \times 4) - 1$ (assuming only one pin at R for the cable connection to the rigid frame)
$D_s = 13 - 12 - 1 = 0$Alternatively, considering the frame PQR as a single rigid body (since Q is rigid), it has 3 degrees of freedom. The roller at P removes 1 DOF. The cable RS (acting as a link) removes 1 DOF at R. This leaves 1 DOF, but the geometry prevents motion, making it statically determinate ($D_s = 0$).

For a section at distance $a$ from the left support in a simply supported beam of length $L$, the Influence Line Diagram (ILD) for shear force is defined as:

• For $0 \leq x < a$: $ILD = -\frac{x}{L}$
• For $a < x \leq L$: $ILD = 1 - \frac{x}{L}$

Given $a = \frac{L}{4}$:

• Just to the left of the section ($x \to \frac{L}{4}^-$): $ILD = -\frac{L/4}{L} = -0.25$
• Just to the right of the section ($x \to \frac{L}{4}^+$): $ILD = 1 - \frac{L/4}{L} = 0.75$

The total jump at the section is $0.75 - (-0.25) = 1.0$. Diagram P correctly represents these values.

The area ratio ($A_r$) is a parameter used to evaluate the degree of disturbance of a soil sample. It is defined as the ratio of the cross-sectional area of the sampler wall to the cross-sectional area of the soil sample (which is equal to the inner area of the sampler).$$A_r = \frac{A_{outer} - A_{inner}}{A_{inner}} \times 100$$Substituting the areas in terms of diameters:$$A_r = \frac{\frac{\pi}{4}D_o^2 - \frac{\pi}{4}D_i^2}{\frac{\pi}{4}D_i^2} = \frac{D_o^2 - D_i^2}{D_i^2}$$For a sample to be considered undisturbed, the area ratio should ideally be less than 10%.

In an Unconsolidated Undrained (UU) test on a saturated cohesive soil, the total stress friction angle ($\phi_u$) is zero. This occurs because the increase in confining pressure is entirely balanced by an equal increase in pore water pressure, meaning the effective stress remains constant. Since the effective stress does not change, the shear strength remains constant regardless of the cell pressure, resulting in a horizontal Mohr-Coulomb failure envelope ($\phi = 0$). While the Unconfined Compression (UC) test also yields $\phi = 0$, it is a specific case of the UU test where the confining pressure is zero, whereas the question refers to the general triaxial test category.

Negative skin friction is a downward drag force exerted on a pile by the surrounding soil when the soil settles more than the pile itself. This typically happens in soft clays or recently placed fills. Because this force acts in the same direction as the structural load (downwards), it adds to the total load the pile must support. This reduces the net allowable load that the pile can safely carry from the superstructure.

Factor of Safety ($FS$) for an infinite slope with seepage parallel to the slope is given by:$$FS = \frac{\gamma'}{\gamma_{sat}} \cdot \frac{\tan \phi'}{\tan \beta}$$where $\gamma' = \gamma_{sat} - \gamma_w = 18 - 10 = 8 \text{ kN/m}^3$. Given $FS = 1.5$, $\phi' = 34^\circ$, and $\gamma_{sat} = 18 \text{ kN/m}^3$:$$1.5 = \frac{8}{18} \cdot \frac{\tan 34^\circ}{\tan \beta}$$$$\tan \beta = \frac{8 \cdot \tan 34^\circ}{18 \cdot 1.5} = \frac{8 \cdot 0.6745}{27} \approx 0.19985$$$$\beta = \tan^{-1}(0.19985) \approx 11.3^\circ$$

Given:
Diameter of the large section ($D_1$) = $15\text{ cm}$
Velocity at the large section ($V_1$) = $2.5\text{ m/s}$
Diameter of the small section ($D_2$) = $5\text{ cm}$For an incompressible homogeneous fluid flowing steadily, the continuity equation is applied:$$A_1 V_1 = A_2 V_2$$$$\frac{\pi}{4} D_1^2 V_1 = \frac{\pi}{4} D_2^2 V_2$$$$D_1^2 V_1 = D_2^2 V_2$$Substituting the given values into the equation:$$(15)^2 \times 2.5 = (5)^2 \times V_2$$$$225 \times 2.5 = 25 \times V_2$$$$V_2 = \frac{225 \times 2.5}{25}$$$$V_2 = 9 \times 2.5$$$$V_2 = 22.5\text{ m/s}$$The velocity of the fluid at the $5\text{ cm}$ portion is $22.5\text{ m/s}$, which falls within the range of 22 to 23.

A flow duration curve (FDC) is a cumulative frequency curve that shows the percentage of time that specified discharges were equaled or exceeded during a given period. It is a plot of flow versus the percentage of time that flow is exceeded.

In reservoir routing through an uncontrolled spillway, the peak of the outflow hydrograph occurs at the point where the inflow rate equals the outflow rate ($I = Q$). This intersection point always lies on the recession limb of the inflow hydrograph because the reservoir storage (and thus outflow) continues to increase as long as the inflow rate is greater than the outflow rate.

Kinematic viscosity ($\nu$) is defined as the ratio of dynamic viscosity ($\mu$) to density ($\rho$).$$\nu = \frac{\mu}{\rho}$$The SI unit for kinematic viscosity is $m^2/s$. Therefore, its dimensional formula is $[L^2 T^{-1}]$, which can be written as $\frac{L^2}{T}$.

Toxic metals commonly found in water include Arsenic (As), Lead (Pb), Mercury (Hg), Cadmium (Cd), and Chromium (Cr). Iron (Fe), Manganese (Mn), and Magnesium (Mg) are common nontoxic metals found in natural water. While they may cause aesthetic issues (like staining or hardness), they are not considered toxic at typical concentrations found in natural sources.

The complete oxidation of methane ($CH_4$) is represented by the following balanced chemical equation:$$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$1. Molecular Weights:

• Molecular weight of $CH_4 = 12 + (4 \times 1) = 16$ g/mol
• Molecular weight of $CO_2 = 12 + (2 \times 16) = 44$ g/mol

2. Stoichiometric Calculation:
According to the balanced equation, 1 mole of $CH_4$ (16 g) reacts to produce 1 mole of $CO_2$ (44 g).
Therefore, 16 kg of $CH_4$ will produce 44 kg of $CO_2$.3. Mass of $CO_2$ for 1 kg of $CH_4$:$$\text{Mass of } CO_2 = \frac{44}{16} \times 1 \text{ kg} = 2.75 \text{ kg}$$The amount of $CO_2$ generated is 2.75 kg, which falls within the accepted range of 2.7 to 2.8.

The Peak Hour Factor (PHF) for a 15-minute interval is defined as:$$PHF_{15} = \frac{V}{4 \times V_{15}}$$where $V$ is the total hourly volume and $V_{15}$ is the maximum volume during any 15-minute interval within that hour.The minimum value of PHF occurs when all the traffic for the entire hour occurs within a single 15-minute interval. In this extreme case, $V = V_{15}$.$$PHF_{\text{min}} = \frac{V}{4 \times V} = 0.25$$

Analysis of statements:

• P is TRUE: If a slab is weightless and has free edges, there is no restraint to thermal expansion, contraction, or curling. Without restraint, no internal stresses are developed.
• Q is TRUE: During the night, the top surface is cooler than the bottom, causing the slab to curl upwards (concave). The self-weight of the slab acts as a restraint against this upward curling, which induces compressive stress at the bottom of the slab.
• R is FALSE: During the day, the top surface is hotter than the bottom, causing the slab to curl downwards (convex). The self-weight resists this downward curling, which induces tensile stress at the bottom of the slab.

Thus, statements P and Q are true.

Given:
$RL_P = 49.600$ m
$RL_Q = 51.870$ m
Distance $PQ = 20$ m
Target contour $RL = 51.000$ mDifference in RL between P and Q = $51.870 - 49.600 = 2.270$ m
Difference in RL between P and the target contour = $51.000 - 49.600 = 1.400$ mUsing linear interpolation, the distance from P is:
Distance $= \frac{\text{Difference in RL from P to target}}{\text{Total difference in RL}} \times \text{Total distance}$
Distance $= \frac{1.400}{2.270} \times 20 \approx 12.3348$ mThus, the distance is approximately 12.33 m.