GATE CE 2026 Set 2

The sentence describes a situation where the team achieved a good score (more than 300 runs) but still had players who needed improvement. The word 'scored' fits the past tense action of achieving runs. The conjunction 'despite' or 'in spite of' is needed to introduce a contrasting idea (the rain) that did not prevent the team from scoring well, or to introduce the contrasting idea of players needing improvement despite the good score.Let's analyze the options:
(A) "score; despite": "score" is a noun here, not a verb. "The team score" is grammatically incorrect in this context.
(B) "scoring; instead of": "scoring" is a gerund, which could work, but "instead of rains" implies that scoring happened instead of rain, which doesn't make sense in the context of the second part of the sentence.
(C) "scored; despite": "scored" is the correct past tense verb. "despite rains" correctly introduces the contrasting element of rain. The overall sentence structure and meaning are coherent: 'The team scored more than 300 runs in 20 overs despite rains. However, some players needed to improve their batting skills.' This makes perfect sense.
(D) "scoring; in spite of": Similar to (B), "scoring" is a gerund. While "in spite of rains" is grammatically correct, "scored" is a more direct and appropriate verb for the first blank.Therefore, option (C) provides the most grammatically correct and contextually appropriate words.

Using the base change formula $\log_a b = \frac{\log_c b}{\log_c a}$, we can rewrite the equation as:$$\log_2 x + \frac{\log_2 x}{\log_2 \sqrt{2}} = 48$$Since $\sqrt{2} = 2^{1/2}$, we have $\log_2 \sqrt{2} = 1/2$. Substituting this back:$$\log_2 x + \frac{\log_2 x}{1/2} = 48$$$$\log_2 x + 2 \log_2 x = 48$$$$3 \log_2 x = 48$$$$\log_2 x = 16$$$$x = 2^{16}$$Thus, the correct option is (A).

To determine the next figure in the sequence, we analyze the movement of each element within the $3 \times 3$ grid:

• Figure 1: Row 2, Column 1
• Figure 2: Row 3, Column 1
• Figure 3: Row 1, Column 1 (overlapping with the arrow)
• Pattern: The circle moves down by one cell in the first column and cycles back to the top. Following this, the next position must be Row 2, Column 1.

3. Triangle (▲):

• Figure 1: Row 1, Column 3
• Figure 2: Row 2, Column 3
• Figure 3: Row 3, Column 3
• Pattern: The triangle moves down by one cell in the third column.

Evaluating the Options:

• Option (A): Places the circle at (2, 1) and the triangle at (3, 3). This matches the established movement of the circle.
• Option (B): Places the triangle at (2, 1) and the circle at (3, 3). Incorrect.
• Option (C): Places the circle at (2, 2). Incorrect.
• Option (D): Places the circle at (3, 1). Incorrect.

Since only Option (A) correctly identifies the next position of the circle based on its cyclic movement in the first column, it is the correct answer.

The original statement is: "All the mangoes in the basket are good."If this statement is false, it means that it is NOT the case that all mangoes are good. The negation of "All A are B" is "Some A are not B" or "There exists at least one A that is not B."Let's apply this to the given statement:
If "All the mangoes in the basket are good" is false, then it must be true that "There exists at least one mango in the basket that is not good."Let's evaluate the options:
(A) "All the mangoes in the basket are not good." This is a stronger statement than the negation. If the original statement is false, it doesn't necessarily mean all mangoes are bad; it just means not all are good. Some could still be good.
(B) "No mango in the basket is good." This is equivalent to "All mangoes are not good," which is the same as option (A) and is not necessarily true.
(C) "In the basket, some of the mangoes are good and some are not good." This is one possibility if the original statement is false, but it's not necessarily true. It's possible that only one mango is not good, and all others are good. Or it's possible that all mangoes are not good. This statement implies a mix, which isn't guaranteed.
(D) "There exists at least one mango in the basket that is not good." This is the direct and necessary negation of the original statement. If it's false that all mangoes are good, then it must be true that at least one mango is not good.Therefore, option (D) is the necessarily true statement.

Let the four numbers be $a, b, c, d$.
(S1) $\implies a + b + c + d = 100$
(S2) $\implies a, b, c, d \leq 40$
(S3) $\implies a, b, c, d \geq 20$Evaluating the options:
(A) If (S1) and (S2) are true, can we violate (S3)? Yes. If $b=c=d=40$, then $a = 100 - 120 = -20$, which is not $\geq 20$.
(B) If (S2) and (S3) are true, the average can be any value between 20 and 40. It is not necessarily 25.
(C) If (S1) and (S3) are true, then for any number, say $a$:
$a = 100 - (b + c + d)$. Since $b, c, d \geq 20$, their sum is at least 60. Thus, $a \leq 100 - 60 = 40$. This applies to all four numbers, so (S2) is necessarily true.
(D) (S1) alone does not restrict the numbers to be $\geq 20$.Therefore, (S1) and (S3) together imply (S2).

The sentence requires appropriate articles and conjunctions/prepositions:The correct sequence is: a; an; and; at. This matches option (B).

Let the quantity of product T sold be $x$.
From the given observations:

• Quantity of S sold = $2x$
• Quantity of R sold = $3x$
• Quantity of Q sold = $4x$

Let the quantity of product P sold be $y$.
Total items sold = $y + 4x + 3x + 2x + x = y + 10x = 250$ ---(1)Total revenue = $(100 \times y) + (50 \times 4x) + (40 \times 3x) + (60 \times 2x) + (60 \times x)$
Total revenue = $100y + 200x + 120x + 120x + 60x = 100y + 500x$Average selling price = $\frac{\text{Total Revenue}}{\text{Total Quantity}} = \frac{100y + 500x}{250} = 60$
$100y + 500x = 15000 \Rightarrow y + 5x = 150$ ---(2)Subtracting equation (2) from equation (1):
$(y + 10x) - (y + 5x) = 250 - 150$
$5x = 100 \Rightarrow x = 20$Substituting $x = 20$ into equation (2):
$y + 5(20) = 150 \Rightarrow y + 100 = 150 \Rightarrow y = 50$Thus, the quantity of product P sold is 50.

Initial state (Figure i):
Length of string P = $l$.
String K is a semicircular arc with diameter $l$. Length of K = $\frac{1}{2} \pi l$.Final state (Figure ii):
Both strings are shortened by $x$.
Length of shortened string P = $l - x$.
Length of shortened string K = $\frac{\pi l}{2} - x$.Shortened string K forms a full circle with diameter equal to the length of shortened string P ($l - x$).
Circumference of circle = $\pi \times \text{diameter} = \pi(l - x)$.Equating the length of shortened string K to the circumference:
$\frac{\pi l}{2} - x = \pi(l - x)$
$\frac{\pi l}{2} - x = \pi l - \pi x$
$\pi x - x = \pi l - \frac{\pi l}{2}$
$x(\pi - 1) = \frac{\pi l}{2}$
$\frac{x}{l} = \frac{\pi}{2(\pi - 1)}$

Let the people be Meritorius (M), Brother (B), Son (S), and Daughter (D). Siblings are (M, B) and (S, D).From (i), D and B are in the same column. Let's say Column 1. Then M and S must be in Column 2.
From (iii), the Best and Worst orators are in the same row.Let's test the case where Best = Brother (B). Then Worst must be Meritorius (M) to be in the same row (since B is in Col 1 and M is in Col 2).
Suppose B is at Row 2, Col 1 and M is at Row 2, Col 2.
Then D must be at Row 1, Col 1 (same column as B).
S must be at Row 1, Col 2 (remaining spot).Check condition (ii): Son (S) is seated diagonally across the sibling of the worst orator.
Worst orator is M. Sibling of M is B.
S is at (1, 2) and B is at (2, 1). They are indeed diagonally across from each other.This configuration satisfies all conditions:

• D and B in Col 1 (i)
• S (1,2) diagonal to B (2,1), where B is sibling of Worst M (ii)
• Best B and Worst M in Row 2 (iii)

Therefore, the best orator is Meritorius’ brother.

To determine which pattern (P, Q, R, or S) generates the main figure, we need to identify the fundamental repeating unit (tile) of the tessellation.

• Pattern P: Shows a specific 2x2 arrangement, but the relative positions of the blocks do not match the continuous tiling of the main figure when repeated.
• Pattern Q: Shows a 2x2 arrangement where the blocks are positioned such that repeating this entire block horizontally and vertically perfectly reconstructs the main figure's interlocking structure.
• Pattern R and S: These show different orientations or arrangements that do not align with the specific shift seen in the main figure.

3. Conclusion: Pattern Q is the correct fundamental unit that generates the larger figure through simple translation. Therefore, option (B) is correct.

For any matrix $A$, if its eigenvalues are $\lambda_1, \lambda_2, \dots, \lambda_n$, then the eigenvalues of $A^k$ are $\lambda_1^k, \lambda_2^k, \dots, \lambda_n^k$. The trace of a matrix is the sum of its eigenvalues. Given eigenvalues of $A$ are 1, 2, and 3, the eigenvalues of $A^2$ are $1^2=1$, $2^2=4$, and $3^2=9$. Therefore, Trace($A^2$) = $1 + 4 + 9 = 14$.

Let $f(x) = a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$ be the fifth-degree polynomial where $a_i > 0$. For $x > 0$, the second derivative $f''(x) = 20a_5x^3 + 12a_4x^2 + 6a_3x + 2a_2$ is always positive. This means the function is concave up.Therefore, both statements are FALSE.

The given function $f(x)$ is an odd function because $f(-x) = -f(x)$ for the interval $(-\pi, \pi)$. For an odd function, the Fourier coefficients $a_0$ and $a_n$ are zero.
The coefficient $b_n$ is calculated as:
$b_n = \frac{2}{\pi} \int_{0}^{\pi} f(x) \sin(nx) dx = \frac{2}{\pi} \int_{0}^{\pi} (1) \sin(nx) dx$
$b_n = \frac{2}{\pi} \left[ -\frac{\cos(nx)}{n} \right]_{0}^{\pi} = \frac{2}{n\pi} (1 - \cos(n\pi))$
For even $n$, $b_n = 0$. For odd $n$, $b_n = \frac{4}{n\pi}$.
Substituting these into the Fourier series $f(x) = \sum_{n=1}^{\infty} b_n \sin(nx)$:
$f(x) = \frac{4}{\pi} \sin x + \frac{4}{3\pi} \sin 3x + \frac{4}{5\pi} \sin 5x + \dots$
$f(x) = \frac{4}{\pi} \left[ \sin x + \frac{\sin 3x}{3} + \frac{\sin 5x}{5} + \dots \right]$

Given:
Effective depth, $d = 550$ mm
Actual depth of neutral axis, $x_u = 302$ mm
Limiting depth ratio, $\frac{x_{u,max}}{d} = 0.456$
Calculating limiting depth of neutral axis:
$x_{u,max} = 0.456 \times 550 = 250.8$ mm
Comparing $x_u$ and $x_{u,max}$:
Since $x_u (302 \text{ mm}) > x_{u,max} (250.8 \text{ mm})$, the section is over-reinforced.
In an over-reinforced section, the concrete reaches its ultimate strain before the steel yields, leading to a sudden and brittle compression failure.

According to IS 800:2007 Clause 8.2.1.2, for sections with high shape factors (like solid circular or rectangular sections), the plastic moment capacity can be significantly higher than the yield moment. To prevent excessive plastic deformation at serviceability (working) loads, the design moment capacity $M_d$ is limited to $1.2 Z_e f_y / \gamma_{m0}$ for simply supported beams and $1.5 Z_e f_y / \gamma_{m0}$ for cantilever beams.

In a plane strain problem in the $X-Y$ plane, the strain in the $Z$ direction and the shear strains in the $XZ$ and $YZ$ planes are zero. Thus, $\epsilon_{zz} = \epsilon_{xz} = \epsilon_{yz} = 0$. Option (A) is the condition for plane stress.

A state of pure shear stress is characterized by a stress condition where the principal stresses are equal in magnitude but opposite in sign, i.e., $\sigma_1 = -\sigma_2$. This results in a Mohr's circle centered at the origin of the $\sigma$-$\tau$ plane.In option (A), the element is subjected to a normal tensile stress of 40 units in the x-direction ($\sigma_x = 40$) and a normal compressive stress of 40 units in the y-direction ($\sigma_y = -40$), with no shear stress on these planes ($\tau_{xy} = 0$). The principal stresses are thus $\sigma_1 = 40$ and $\sigma_2 = -40$. Since $\sigma_1 = -\sigma_2$, this represents a state of pure shear stress.Other options represent different states:

• (B) represents hydrostatic tension ($\sigma_x = \sigma_y = 40$).
• (C) represents a general state of stress with both normal and shear components.
• (D) represents a state of hydrostatic tension (or compression depending on arrow direction).

To find the upward vertical reaction at support $L$, we can analyze the beam by segments separated by the internal hinges $G$ and $J$. Segment $GJ$ is a simply supported beam between hinges $G$ and $J$ with a length of $4\text{ m}$ ($2\text{ m} + 2\text{ m}$). A point load of $30\text{ kN}$ acts at its center $H$.
The vertical reactions at $G$ and $J$ due to this load are:
$$V_{G,GJ} = V_{J,GJ} = \frac{30}{2} = 15\text{ kN}$$ This segment is fixed at $L$ and has a hinge at $J$. The forces acting on it are:

• The downward reaction from segment $GJ$ at hinge $J$, which is $15\text{ kN}$.
• A uniformly distributed load (UDL) of $4\text{ kN/m}$ over the segment $KL$ (length $= 2\text{ m}$). Total UDL $= 4 \times 2 = 8\text{ kN}$.

3. Calculate the vertical reaction at $L$ ($R_L$):
Apply vertical equilibrium ($\sum F_y = 0$) to segment $JL$:
$$R_L - 15 - 8 = 0$$ $$R_L = 23\text{ kN}$$Therefore, the upward vertical reaction at support $L$ is $23\text{ kN}$.

Given:

• Culturable Command Area ($CCA$) = 10000 hectares
• Base period ($B$) = 140 days
• Watering depth (Delta, $\Delta$) = 40 cm = 0.4 m

Step 1: Calculate the Duty ($D$) of the water.
The relationship between Duty, Base period, and Delta is given by the formula:$$D = \frac{8.64 \times B}{\Delta}$$Where $D$ is in hectares/cumec, $B$ is in days, and $\Delta$ is in meters.$$D = \frac{8.64 \times 140}{0.4} = \frac{1209.6}{0.4} = 3024 \text{ hectares/cumec}$$Step 2: Calculate the outlet discharge ($Q$).
The discharge required is the ratio of the area to the duty:$$Q = \frac{CCA}{D}$$$$Q = \frac{10000}{3024} \approx 3.307 \text{ m}^3/\text{s}$$The calculated discharge of approximately 3.307 $m^3/s$ lies between 3 and 4. Thus, option (A) is correct.

A mass curve of rainfall is a plot of cumulative rainfall against time. The slope of the mass curve at any point represents the intensity of rainfall at that instant ($i = dP/dt$).
(I) When it is raining, the cumulative rainfall increases, so the slope is positive. Thus, statement (I) is TRUE.
(II) During a dry spell (no rain), the cumulative rainfall remains constant, so the slope is zero. Thus, statement (II) is TRUE.

According to Casagrande's plasticity chart, the boundaries for inorganic clays are defined by the A-line and the U-line: $$I_p = 0.73(w_L - 20)$$ For a liquid limit ($w_L$) of $40 \%$:
$$I_p = 0.73(40 - 20) = 0.73 \times 20 = 14.6 \%$$ $$I_p = 0.9(w_L - 8)$$ For a liquid limit ($w_L$) of $40 \%$:
$$I_p = 0.9(40 - 8) = 0.9 \times 32 = 28.8 \%$$Inorganic clays lie between the A-line and the U-line on the plasticity chart. Therefore, for a liquid limit of $40 \%$, the plasticity index ($I_p$) ranges between $14.6 \%$ and $28.8 \%$.Hence, the correct option is (B).

The correct matches are:
(I) Converse-Labarre formula is used to calculate Pile group efficiency (Q).
(II) Hiley formula is a well-known dynamic formula for determining Pile capacity (P).
(III) Newmark’s influence chart is used for calculating the increase in vertical stress below a footing (S) of any shape.
(IV) Skempton’s equation provides the bearing capacity factor (R) for foundations in saturated clay.Therefore, the correct sequence is (I)-(Q), (II)-(P), (III)-(S), (IV)-(R), which corresponds to option (A).

On a dual carriageway (one-way traffic), the desired sight distance is typically the Intermediate Sight Distance (ISD) to allow for safe maneuvers.
Under normal conditions: $\text{Desired Sight Distance} = \text{ISD} = 2 \times \text{SSD}$.
Given $\text{ISD} = 360$ m, we find $\text{SSD} = \frac{360}{2} = 180$ m.
When traffic is diverted to the other side, the road section operates with two-way traffic. The minimum sight distance required for safe operation is the Stopping Sight Distance (SSD). Since SSD is a property of the design speed and reaction time, it remains $180$ m.

In railway engineering:Thus, statement (A) is true.

A wind rose diagram is a graphical tool used by meteorologists and airport engineers to give a succinct view of how wind speed and direction are typically distributed at a particular location. In airport design, it is primarily used to determine the orientation of a runway to ensure that the crosswind component is within acceptable limits for the majority of the time (usually 95% coverage).

In surveying:

• Statement (A) is TRUE: A level surface is a curved surface which is at every point perpendicular (normal) to the direction of gravity (plumb line) at that point.
• Statement (B) is FALSE: Level surfaces are equipotential surfaces and they are parallel to each other; they cannot cross.
• Statement (C) is FALSE: An ellipsoid is a mathematically defined regular surface used as a reference for coordinates.
• Statement (D) is FALSE: The geoid is an irregular equipotential surface of the Earth's gravity field that best fits mean sea level.

The correct matching of air pollutants to their treatment technologies is:Thus, the correct sequence is 1-P, 2-S, 3-Q, 4-R, which corresponds to option (A).

The average specific gravity of solids ($G_s$) is calculated using the mass fractions of its components:$$\frac{1}{G_s} = \frac{w_{fs}}{G_{fs}} + \frac{w_{vs}}{G_{vs}}$$Where:

• $w_{fs} = \frac{1}{3}$ (mass fraction of fixed solids)
• $G_{fs} = 2.5$ (specific gravity of fixed solids)
• $w_{vs} = \frac{2}{3}$ (mass fraction of volatile solids)
• $G_{vs} = 1.0$ (specific gravity of volatile solids)

Substituting the values:$$\frac{1}{G_s} = \frac{1/3}{2.5} + \frac{2/3}{1.0} = \frac{1}{7.5} + \frac{2}{3} = \frac{1 + 5}{7.5} = \frac{6}{7.5} = 0.8$$$$G_s = \frac{1}{0.8} = 1.25$$The value $1.25$ lies between $1.2$ and $1.3$.

Oxidation ponds (also known as stabilization ponds) rely on a symbiotic relationship between aerobic bacteria and algae. Algae produce oxygen through photosynthesis, which is utilized by aerobic bacteria to decompose organic matter. In turn, the bacteria release carbon dioxide and nutrients that support algal growth.

The given equation is the one-dimensional wave equation:$$\frac{\partial^2 u}{\partial x^2} - \frac{\partial^2 u}{\partial y^2} = 0$$The general solution is of the form $u(x, y) = f(x + y) + g(x - y)$.

• Option (A): $u = (x + y)^5$. Here $u_{xx} = 20(x+y)^3$ and $u_{yy} = 20(x+y)^3$. Thus $u_{xx} - u_{yy} = 0$. Correct.
• Option (B): $u = (x - 2y)^3$. Here $u_{xx} = 6(x-2y)$ and $u_{yy} = 24(x-2y)$. Thus $u_{xx} - u_{yy} \neq 0$. Incorrect.
• Option (C): $u = \cos(x + y)$. Here $u_{xx} = -\cos(x+y)$ and $u_{yy} = -\cos(x+y)$. Thus $u_{xx} - u_{yy} = 0$. Correct.
• Option (D): $u = \sin(x - 2y)$. Here $u_{xx} = -\sin(x-2y)$ and $u_{yy} = -4\sin(x-2y)$. Thus $u_{xx} - u_{yy} \neq 0$. Incorrect.

Therefore, (A) and (C) are valid solutions.

The determinant of a square matrix is equal to the product of its eigenvalues.
Given eigenvalues are:
$\lambda_1 = -1.547$
$\lambda_2 = 12.330$
$\lambda_3 = 4.711$The determinant $|A|$ is calculated as:
$|A| = \lambda_1 \times \lambda_2 \times \lambda_3$
$|A| = (-1.547) \times (12.330) \times (4.711) \approx -89.853$The absolute value of the determinant is:
$||A|| = |-89.853| = 89.853$Rounding off to two decimal places, we get $89.85$.

Given:
Return period ($T$) = 15 years
Time period ($n$) = 10 yearsThe probability of occurrence in any single year is $p = \frac{1}{T} = \frac{1}{15}$.
The probability of non-occurrence in a single year is $q = 1 - p = 1 - \frac{1}{15} = \frac{14}{15}$.The probability that the storm occurs at least once in $n$ years (Risk) is given by:
$P = 1 - q^n$
$P = 1 - \left(\frac{14}{15}\right)^{10}$
$P = 1 - 0.497445 \approx 0.502555$In percentage:
$P(\%) = 0.502555 \times 100 = 50.26\%$Rounding off to two decimal places, the answer is $50.26$.