GATE CE 2016 Set 1

The word 'until' is a negative conjunction, meaning 'up to the time that'. Using 'does not' or 'doesn't' after 'until' creates a double negative, which is grammatically incorrect. Therefore, options (A) and (B) are wrong. In option (C), 'minister' is a singular subject, so the verb should be 'meets' (singular) instead of 'meet' (plural). Thus, option (D) is the most suitable sentence.

A paraphrase is a restatement of the meaning of a text or passage using other words. A paradox is a seemingly absurd or self-contradictory statement. A paradigm is a typical example or pattern of something. Paraffin is a flammable, whitish, translucent, waxy solid.

The statement is figurative because it uses metaphorical language (moving the world with a lever) to express a concept (the power of leverage) rather than describing a literal physical possibility. 'Literal' would mean he actually intended to move the planet. 'Collateral' and 'figurine' are irrelevant in this context.

From the given words:

• 'relftaga' = carefree
• 'otaga' = careful
• 'fertaga' = careless

We can observe that 'taga' corresponds to 'care'. In the English words 'carefree', 'careful', and 'careless', 'care' is the first part. In the artificial language, 'taga' is the second part. This indicates a reversal of the order of morphemes.

• 'relf' = free
• 'o' = ful
• 'fer' = less

Now, for 'aftercare':
In English, 'care' is the second part (after + care). Following the reversal rule, 'taga' (care) should be the first part in the artificial language. Among the options, only tagazen starts with 'taga'. Therefore, 'zen' would mean 'after'.

To find which product contributes the greatest fraction to the total revenue, we calculate the total revenue for each product over the four quarters.Total Quantity Sold:

• Elegance: $27300 + 25222 + 28976 + 21012 = 102510$
• Smooth: $20009 + 19392 + 22429 + 18229 = 80059$
• Soft: $17602 + 18445 + 19544 + 16595 = 72186$
• Executive: $9999 + 8942 + 10234 + 10109 = 39284$

Total Revenue (Price $\times$ Total Quantity):

• Elegance: $48 \times 102510 = 4,920,480$
• Smooth: $63 \times 80059 = 5,043,717$
• Soft: $78 \times 72186 = 5,630,508$
• Executive: $173 \times 39284 = 6,796,132$

Comparing the revenues, the Executive razor generates the highest revenue, and thus contributes the greatest fraction to the total revenue of the company.

The statement "If this is not an indication... nothing else is" is a rhetorical device used to emphasize that the subject (linguistic pluralism on currency) is a definitive and strong indicator of the nation's diversity. Option (D) captures this. Option (A) is incorrect because the text says "at least seventeen", not exactly. Option (B) is incorrect because the rhetorical emphasis does not literally mean no other indicators exist. Option (C) is an observation not supported by the logical structure of the argument.

Step 1: Analyze the 'always' relationships given in the statements.

• Statement I: P always beats Q ($P > Q$)
• Statement IV: R always loses to Q, which means Q always beats R ($Q > R$)
• Statement II: R always beats S ($R > S$)

Combining these, we get a chain of 'always' victories: $P > Q > R > S$.Step 2: Analyze Statement III.

• Statement III: S loses to P only sometimes. This implies that S beats P (or draws) most of the time. This creates a non-transitive relationship where $P$ beats $Q$, $Q$ beats $R$, $R$ beats $S$, but $S$ beats $P$ frequently.

Step 3: Evaluate the inferences.

• Inference (i): '**P** is likely to beat all the three other players'. Since P loses to S most of the time, we cannot logically infer that P is likely to beat all three players without more specific frequency data, especially since one player ($S$) has a winning record against $P$.
• Inference (ii): '**S** is the absolute worst player in the set'. Since S beats the 'best' player P most of the time, S cannot be considered the absolute worst player in the set.

Conclusion: Neither inference (i) nor (ii) logically follows from the given statements. Thus, the correct option is (D).

According to the factor theorem, $(x - a)$ is a factor of a polynomial $f(x)$ if and only if $f(a) = 0$.
Given $f(x) = 2x^7 + 3x - 5$.
Let's test option (B) where $x - 1 = 0 \implies x = 1$:
$f(1) = 2(1)^7 + 3(1) - 5 = 2 + 3 - 5 = 0$.
Since $f(1) = 0$, $(x - 1)$ is a factor of $f(x)$.

Let $N$ be the number of cycles to failure and $L$ be the load. The problem states that $N$ decreases exponentially with an increase in $L$, which can be modeled as:
$N = A e^{-kL}$
Taking the natural logarithm on both sides:
$\ln N = \ln A - kL$Given data:$\ln 100 = \ln A - 80k$ --- (Eq. 1)$\ln 10000 = \ln A - 40k$ --- (Eq. 2)Subtracting Eq. 1 from Eq. 2:
$\ln 10000 - \ln 100 = (-40k) - (-80k)$
$\ln(10000/100) = 40k$
$\ln 100 = 40k$
$k = \frac{\ln 100}{40} = \frac{2 \ln 10}{40} = \frac{\ln 10}{20}$$\ln 5000 = \ln A - kL_3$ --- (Eq. 3)From Eq. 2, we have $\ln A = \ln 10000 + 40k$. Substituting this into Eq. 3:
$\ln 5000 = (\ln 10000 + 40k) - kL_3$
$\ln 5000 - \ln 10000 = k(40 - L_3)$
$\ln(5000/10000) = k(40 - L_3)$
$\ln(0.5) = k(40 - L_3)$
$-\ln 2 = \frac{\ln 10}{20} (40 - L_3)$
$40 - L_3 = -20 \frac{\ln 2}{\ln 10} = -20 \log_{10} 2$
$L_3 = 40 + 20 \log_{10} 2$
Using $\log_{10} 2 \approx 0.3010$:
$L_3 = 40 + 20(0.3010) = 40 + 6.02 = 46.02$Therefore, the load is 46.02 units. The correct option is (B).

Given the function $f(x) = 3x - e^x + \sin x = 0$.
Its derivative is $f'(x) = 3 - e^x + \cos x$.
The Newton-Raphson iteration formula is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$.
Given $x_0 = 0.333$:$f(0.333) = 3(0.333) - e^{0.333} + \sin(0.333)$
$f(0.333) \approx 0.999 - 1.395147 + 0.32688 \approx -0.069267$$f'(0.333) = 3 - e^{0.333} + \cos(0.333)$
$f'(0.333) \approx 3 - 1.395147 + 0.94506 \approx 2.549913$$x_1 = 0.333 - \frac{-0.069267}{2.549913} \approx 0.333 + 0.027164 \approx 0.360164$
Rounding to three decimal places, the next approximation is 0.360.

A second-order partial differential equation of the form $A\frac{\partial^2 P}{\partial x^2} + B\frac{\partial^2 P}{\partial x \partial y} + C\frac{\partial^2 P}{\partial y^2} + \dots = 0$ is classified by the discriminant $D = B^2 - 4AC$.
From the given equation:
$A = 1$ (coefficient of $\frac{\partial^2 P}{\partial x^2}$)
$B = 3$ (coefficient of $\frac{\partial^2 P}{\partial x \partial y}$)
$C = 1$ (coefficient of $\frac{\partial^2 P}{\partial y^2}$)
$D = 3^2 - 4(1)(1) = 9 - 4 = 5$.
Since $D > 0$, the equation is hyperbolic.

If the sum of elements in each column of a square matrix $M$ is a constant $k$, then $k$ is an eigenvalue of $M$. This is because the sum of elements in each row of the transpose matrix $M^T$ is $k$, which means $M^T \mathbf{1} = k \mathbf{1}$ where $\mathbf{1}$ is a vector of ones. Thus $k$ is an eigenvalue of $M^T$. Since $M$ and $M^T$ have the same eigenvalues, $k$ is an eigenvalue of $M$. Here $k=1$, so 1 is an eigenvalue.

A Type II error occurs when one fails to reject a null hypothesis that is actually false. In other words, it is the acceptance of a false null hypothesis.Option (A) states: "acceptance of the null hypothesis when it is false and should be rejected." This perfectly matches the definition of a Type II error.Option (B) describes a Type I error (rejecting a true null hypothesis).
Option (C) describes a correct decision (rejecting a false null hypothesis).
Option (D) is the same as (A).Therefore, the correct definition is given in option (A).

The given partial differential equation is the one-dimensional heat equation:
$\frac{\partial u}{\partial t} = \alpha \frac{\partial^2 u}{\partial x^2}$We can solve this using the method of separation of variables. Let $u(x,t) = X(x)T(t)$.
Substituting this into the PDE:
$X(x)T'(t) = \alpha X''(x)T(t)$Dividing by $\alpha X(x)T(t)$:
$\frac{T'(t)}{\alpha T(t)} = \frac{X''(x)}{X(x)} = -k^2$ (where $-k^2$ is the separation constant)This gives two ordinary differential equations:So, $u(x,t) = C_T e^{-\alpha k^2 t} (C_1 \cos(kx) + C_2 \sin(kx))$. This form is for oscillatory solutions in space and decaying in time, typically for boundary conditions like $u(0,t)=u(L,t)=0$.However, the options provided suggest a different form for $X(x)$ and $T(t)$. Let's re-evaluate the separation constant. If we choose the separation constant as $k^2$ (positive) instead of $-k^2$:
$\frac{T'(t)}{\alpha T(t)} = \frac{X''(x)}{X(x)} = k^2$This gives:Combining these, the general solution would be $u(x,t) = C_T e^{\alpha k^2 t} (C_1 e^{kx} + C_2 e^{-kx})$.Let's compare this with the given options. Let $k_{new} = \sqrt{\alpha} k$. Then the time part is $e^{k_{new}^2 t}$.
Or, if we use $k$ as the constant in the options, then the time part is $e^{kt}$ and the spatial part involves $\sqrt{k/\alpha}$.Let's assume the separation constant is $k$ (as used in the options for the time component).
So, $\frac{T'(t)}{\alpha T(t)} = \frac{X''(x)}{X(x)} = k$.Let's assume the separation constant is such that $T(t)$ is $e^{kt}$. This means $\frac{T'(t)}{T(t)} = k$. So, $\frac{X''(x)}{X(x)} = \frac{k}{\alpha}$.Then, $X''(x) = \frac{k}{\alpha} X(x)$. The characteristic equation is $r^2 - \frac{k}{\alpha} = 0$, so $r = \pm \sqrt{\frac{k}{\alpha}}$.
Thus, $X(x) = C_1 e^{\sqrt{k/\alpha} x} + C_2 e^{-\sqrt{k/\alpha} x}$.Combining $T(t) = C_T e^{kt}$ and $X(x) = C_1 e^{\sqrt{k/\alpha} x} + C_2 e^{-\sqrt{k/\alpha} x}$, the solution is:
$u(x,t) = C_T e^{kt} (C_1 e^{\sqrt{k/\alpha} x} + C_2 e^{-\sqrt{k/\alpha} x})$.This matches option (B) if we absorb $C_T$ into the constants $C_1$ and $C_2$, or consider $C_T$ as the $C$ outside the bracket.Option (A) and (D) have $\cos(kt)$ or $\sin(kt)$ for the time component, which would arise if the separation constant for $T(t)$ was negative, leading to a decaying exponential for $X(x)$ (not hyperbolic sines/cosines or real exponentials).
Option (C) has $e^{kt}$ for time, but $\cos$ and $\sin$ for space, which would arise if the separation constant for $X(x)$ was negative, i.e., $X''(x) = -\frac{k}{\alpha} X(x)$. This would lead to $X(x) = C_1 \cos(\sqrt{k/\alpha} x) + C_2 \sin(\sqrt{k/\alpha} x)$. This is also a valid form depending on boundary conditions, but option (B) is also a valid general form.Given the options, option (B) is a standard form for solutions to the heat equation, particularly for problems with infinite domains or specific boundary conditions where exponential growth/decay in space is possible.

Step 1: Identify the support reactions. Joint B is a pin support, which has two reaction components: $H_B$ (horizontal) and $V_B$ (vertical). Joint C is a roller support, which has one vertical reaction component: $V_C$.Step 2: Determine the geometry. The truss is composed of equilateral triangles with side length $L$.

• The horizontal distance from B to D is $L$, and from D to C is $L$. Thus, the total span $BC = 2L$.
• The horizontal distance from B to G can be calculated by summing the horizontal projections of the members: $x_G = L \cos 60^\circ + L + L \cos 60^\circ = 0.5L + L + 0.5L = 2L$.
• Since $x_G = 2L$ and $x_C = 2L$, the load $P$ at joint G acts directly above the support at C.

Step 3: Apply static equilibrium equations for the entire truss.

• Sum of horizontal forces: $\sum F_x = 0 \implies H_B = 0$ (as there are no other horizontal forces).
• Sum of moments about joint B: $\sum M_B = 0 \implies P \times (2L) - V_C \times (2L) = 0 \implies V_C = P$.
• Sum of vertical forces: $\sum F_y = 0 \implies V_B + V_C - P = 0 \implies V_B + P - P = 0 \implies V_B = 0$.

Step 4: Conclusion. The reactions are $V_B = 0$, $H_B = 0$, and $V_C = P$. This corresponds to option (A).

According to IS 456:2000, the nominal shear stress $\tau_v$ in beams shall not exceed $\tau_{c \text{ max}}$. This check is provided to prevent the possibility of crushing of concrete by diagonal compression in the web before the shear reinforcement can yield.

For a semi-compact section, the design flexural capacity ($M_d$) is determined by the elastic section modulus ($Z_e$) and the design bending compressive stress ($f_{bd}$).Given data:

• Elastic section modulus, $Z_e = 500\text{ cm}^3 = 500 \times 10^3\text{ mm}^3$
• Plastic section modulus, $Z_p = 650\text{ cm}^3$ (Note: For semi-compact sections, the capacity is governed by the elastic section modulus $Z_e$ rather than the plastic section modulus $Z_p$)
• Design bending compressive stress, $f_{bd} = 200\text{ MPa} = 200\text{ N/mm}^2$

Calculation:
The design flexural capacity is given by the formula:$$M_d = Z_e \cdot f_{bd}$$Substituting the values:$$M_d = (500 \times 10^3\text{ mm}^3) \times (200\text{ N/mm}^2)$$$$M_d = 100,000,000\text{ N-mm}$$To express the capacity in kNm:$$M_d = \frac{100,000,000}{10^6}\text{ kNm} = 100\text{ kNm}$$Thus, the design flexural capacity of the section is $100\text{ kNm}$.

Bull's trench kiln is a continuous kiln used for burning bricks. It is an elliptical or circular trench dug in the ground where bricks are stacked and fired in a continuous process.

Tricalcium silicate ($C_3S$) is responsible for the early strength of concrete (within 7 days). Tricalcium aluminate ($C_3A$) is responsible for the initial setting (flash set). Since the question asks for both, and $C_3S$ is the primary strength contributor, (B) is the most appropriate choice.

In a Consolidated Undrained (CU) test:Thus, pore pressure is zero only at the end of the consolidation stage.

Given: Plastic range is 26-48%. This means Plastic Limit $w_P = 26\%$ and Liquid Limit $w_L = 48\%$.
Plasticity Index $I_P = w_L - w_P = 48 - 26 = 22\%$.
According to IS classification:Since $I_P (22\%) > 20.44\%$, the soil is above the A-line, hence it is Clay (C).Therefore, the soil is classified as CI.

The depth of tension crack in a cohesive soil is given by $z_c = \frac{2c}{\gamma \sqrt{K_a}}$. The maximum depth of unsupported excavation is twice the depth of the tension crack, i.e., $H_c = 2 z_c = \frac{4c}{\gamma \sqrt{K_a}}$.

The volume of direct runoff is equal to the area under the direct runoff hydrograph (DRH).Given:

• Rainfall excess depth ($R$) = $5\text{ cm} = 0.05\text{ m}$
• The DRH is a triangle with base ($B$) = $6\text{ hours}$ and peak flow ($Q_p$) = $1.0\text{ m}^3/\text{s}$.

Step 1: Calculate the volume of runoff ($V$):$$V = \text{Area under DRH} = \frac{1}{2} \times \text{Base} \times \text{Peak Flow}$$$$V = \frac{1}{2} \times (6 \times 3600\text{ s}) \times 1.0\text{ m}^3/\text{s} = 10800\text{ m}^3$$Step 2: Calculate the catchment area ($A$):$$A = \frac{V}{R} = \frac{10800\text{ m}^3}{0.05\text{ m}} = 216000\text{ m}^2$$Step 3: Convert the area to hectares ($1\text{ hectare} = 10000\text{ m}^2$):$$A = \frac{216000}{10000} = 21.6\text{ hectares}$$Thus, the area of the catchment is $21.6\text{ hectares}$.

Hydrologic flood routing (lumped routing) uses only the continuity equation. Hydraulic flood routing (distributed routing) uses both the continuity and momentum equations (Saint-Venant equations). The energy equation is not typically used for flood routing.

For a hydraulic jump in a horizontal rectangular channel, the ratio of sequent depths $y_2$ (post-jump) and $y_1$ (pre-jump) is given by the Belanger equation:$$\frac{y_2}{y_1} = \frac{1}{2} \left( \sqrt{1 + 8 Fr_1^2} - 1 \right)$$Given the pre-jump Froude Number $Fr_1 = 10$:$$\frac{y_2}{y_1} = \frac{1}{2} \left( \sqrt{1 + 8(10)^2} - 1 \right)$$$$\frac{y_2}{y_1} = \frac{1}{2} \left( \sqrt{1 + 800} - 1 \right)$$$$\frac{y_2}{y_1} = \frac{1}{2} \left( \sqrt{801} - 1 \right)$$$$\frac{y_2}{y_1} \approx \frac{1}{2} (28.3019 - 1)$$$$\frac{y_2}{y_1} = \frac{27.3019}{2} = 13.65095$$Rounding to two decimal places, the ratio of sequent depths is approximately 13.65, which falls within the specified range of 13.6 to 13.7.

Photochemical oxidants, such as ozone ($O_3$), are formed in the atmosphere through complex chemical reactions involving nitrogen oxides ($NO_X$) and volatile organic compounds (VOCs) in the presence of sunlight. These are known as precursors to photochemical smog.

Crown corrosion in sewers occurs when hydrogen sulfide ($H_2S$) gas, produced by anaerobic decomposition of organic matter in sewage, escapes into the sewer atmosphere. It condenses on the moist upper surface (crown) of the sewer pipe, where aerobic bacteria (Thiobacillus) oxidize it to sulfuric acid ($H_2SO_4$). The acid then reacts with the calcium hydroxide in the concrete, leading to structural degradation.

$A_s = \pi \cdot d \cdot L = \pi \cdot 0.45 \text{ m} \cdot 7.5 \text{ m} \approx 10.6029 \text{ m}^2$ $v = 2.0 \text{ m/min} = \frac{2.0}{60} \text{ m/s} = \frac{1}{30} \text{ m/s}$ $A_{total} = \frac{Q}{v} = \frac{10 \text{ m}^3/\text{s}}{1/30 \text{ m/s}} = 300 \text{ m}^2$ $n = \frac{A_{total}}{A_s} = \frac{300}{10.6029} \approx 28.29$ $n = 29$ bags.

• P. Activated sludge process: Primarily utilizes heterotrophic bacteria (3) to decompose organic matter.
• Q. Rising of sludge: This phenomenon is typically caused by denitrification (4) in the secondary clarifier, where nitrogen gas bubbles attach to sludge flocs and cause them to float.
• R. Conventional nitrification: This process is carried out by autotrophic bacteria (2) such as Nitrosomonas and Nitrobacter.
• S. Biological nitrogen removal: This comprehensive process involves both nitrifiers and denitrifiers (1) to convert ammonia to nitrogen gas.

Therefore, the correct match is P-3, Q-4, R-2, S-1.

In the Marshall method of mix design, the following properties are plotted against bitumen content to determine the optimum binder content:Since graphs $P, Q,$ and $S$ are theoretically plausible, the correct option is (B).