GATE CE 2025 Set 1

The preposition 'on' is used when referring to programs or shows being broadcast via television. Standard English usage is 'on television'.

'Summarily' means in a summary manner; immediately and without the customary formalities. In the context of disciplinary action like dismissal from service due to proven guilt, it is the most appropriate term.

The Taylor series expansion for $e^x$ is given by:$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$$Setting $x = 1$:$$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$$Subtracting 1 from both sides:$$e - 1 = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$$This matches the given series.

A cube has 8 vertices, and each vertex has a degree of 3 (3 edges meet at each vertex). In graph theory, a graph can be traversed in a single continuous path (Eulerian path) if it has at most 2 vertices of odd degree. Here, all 8 vertices have odd degrees. To cover all edges with the minimum number of separate paths, we need $\frac{\text{number of odd-degree vertices}}{2} = \frac{8}{2} = 4$ paths. To obtain 4 separate pieces from a single 12m wire, we need 3 cuts.

The sequence of figures shows a pattern in both the shaded area and the orientation of the shading lines:The natural progression for the shaded area is cumulative (adding one quadrant clockwise in each step), and the shading lines alternate between vertical and horizontal. Following this, Figure IV should have all four quadrants shaded with horizontal lines. However, none of the options show a fully shaded circle.In such visual reasoning problems, if the direct progression is not available, we look for an oscillating or symmetric pattern: I $\rightarrow$ II $\rightarrow$ III $\rightarrow$ II $\rightarrow$ I. In this case, the figure at position IV would be identical to Figure II.Figure II consists of the top half (top-left and top-right quadrants) shaded with horizontal lines. Option (B) matches this exactly. Therefore, (B) is the correct option.

The passage explicitly states, "Every day the wealthier nations are pulling down one or another in their capitals and their great towns...". This directly supports statement (B).

• Statement (A) is incorrect because the author takes delight in "crooked streets", not in the act of "wondering".
• Statement (C) is incorrect as it misinterprets the metaphorical phrase "islands of the past".
• Statement (D) is incorrect because the author suggests that broad ways are needed for commerce but they destroy history, rather than protecting it.

The minute hand gains $55$ minutes on the hour hand in $60$ minutes.
So, the hands coincide every $60 + \frac{60}{55} = 60 + \frac{12}{11} = \frac{660+12}{11} = \frac{720}{11}$ minutes.Alternatively, the relative speed of the minute hand with respect to the hour hand is $5.5$ degrees per minute ($6$ degrees/min for minute hand, $0.5$ degrees/min for hour hand).The hands coincide when the angle between them is $0$ degrees. They coincide $11$ times in $12$ hours (or $12$ times if we count the coincidence at 12:00 twice for a 12-hour cycle, but typically it's $11$ distinct coincidences). The time between two consecutive coincidences is $\frac{360}{5.5 \times 11} = \frac{720}{11}$ minutes.Let's verify the time between coincidences. The hands coincide at 12:00. The next coincidence is between 1 and 2. At 1:00, the minute hand is at 12 and the hour hand is at 1. The minute hand needs to cover $30$ degrees to catch up. Time taken = $\frac{30}{5.5} = \frac{60}{11}$ minutes. So, the first coincidence after 12:00 is at $1$ hour and $\frac{60}{11}$ minutes, i.e., $1:5\frac{5}{11}$.The time interval between two consecutive coincidences of the hour and minute hands is $\frac{720}{11}$ minutes.Rohit enters the restaurant at about 1 PM and the hands are coinciding. This would be at $1$ hour and $\frac{60}{11}$ minutes past 12, i.e., $1:05\frac{5}{11}$ PM.He leaves after about an hour, and the hands are again coinciding. The next coincidence after $1:05\frac{5}{11}$ PM would be $\frac{720}{11}$ minutes later.Time spent at the restaurant = Time between two consecutive coincidences = $\frac{720}{11}$ minutes.$\frac{720}{11} = 65\frac{5}{11}$ minutes.The final answer is $\boxed{C}$

The provided color model defines the following relationships:Evaluating the options:

• In Option (A), the labels correctly match these intersections: the overlap of Y and M is R, M and Cy is Bl, Cy and Y is G, and the center is K. This is perfectly consistent with the model.
• Other options misplace these labels (e.g., swapping Bl and R or placing K in a dual intersection).

The equations of the two circles are:Subtracting equation (1) from equation (2):
$(x^2 + y^2 - 2x - 2y + 1) - (x^2 + y^2 - x) = 0$
$-x - 2y + 1 = 0 \Rightarrow x = 1 - 2y$Substitute $x = 1 - 2y$ into equation (1):
$(1 - 2y)^2 + y^2 - (1 - 2y) = 0$
$1 - 4y + 4y^2 + y^2 - 1 + 2y = 0$
$5y^2 - 2y = 0 \Rightarrow y(5y - 2) = 0$This gives $y = 0$ or $y = 0.4$.

• If $y = 0$, $x = 1 - 2(0) = 1$. The point is $(1, 0)$.
• If $y = 0.4$, $x = 1 - 2(0.4) = 0.2$. The point is $(0.2, 0.4)$.

Comparing with the options, $(0.2, 0.4)$ is option (B).

4-fold rotational symmetry requires the object to look identical after a rotation of $\frac{2\pi}{4} = 90^\circ$.In Option (B), the dots are placed as follows:

• Top arm: dot on the right side.
• Right arm: dot on the bottom side.
• Bottom arm: dot on the left side.
• Left arm: dot on the top side.

Rotating $90^\circ$ clockwise:

• The top arm moves to the right arm position, and its dot (right side) moves to the bottom side of the new position. This matches the original right arm.
• The right arm moves to the bottom arm position, and its dot (bottom side) moves to the left side. This matches the original bottom arm.
• The pattern continues for all arms, making the object identical to the original. Thus, (B) has 4-fold symmetry.

In other options, the relative positions of the dots do not map onto each other under a $90^\circ$ rotation.

Given that $\lambda$ is an eigenvalue of $A$ with eigenvector $x$, we have $Ax = \lambda x$.
For the matrix $B = A - 2I$, we can find its eigenvalue corresponding to $x$ as follows:$$Bx = (A - 2I)x = Ax - 2Ix = \lambda x - 2x = (\lambda - 2)x$$Thus, the eigenvalue of $B$ corresponding to the eigenvector $x$ is $\lambda - 2$.

For the system $Ax = b$ to be solvable, the vector $b$ must lie in the column space of $A$. This is equivalent to saying that the augmented matrix $[A|b]$ must have the same rank as $A$.
Performing row operations on the augmented matrix:$$\begin{bmatrix} 1 & 1 & b_1 \\ 1 & 3 & b_2 \\ -2 & -3 & b_3 \end{bmatrix} \xrightarrow{R_2 \to R_2 - R_1, R_3 \to R_3 + 2R_1} \begin{bmatrix} 1 & 1 & b_1 \\ 0 & 2 & b_2 - b_1 \\ 0 & -1 & b_3 + 2b_1 \end{bmatrix}$$$$\xrightarrow{R_3 \to R_3 + 0.5R_2} \begin{bmatrix} 1 & 1 & b_1 \\ 0 & 2 & b_2 - b_1 \\ 0 & 0 & b_3 + 2b_1 + 0.5(b_2 - b_1) \end{bmatrix}$$For the system to be consistent, the last row must be zero:$$b_3 + 2b_1 + 0.5b_2 - 0.5b_1 = 0$$$$1.5b_1 + 0.5b_2 + b_3 = 0$$Multiplying by 2:$$3b_1 + b_2 + 2b_3 = 0$$

The function $f(x)$ is an even function over the interval $(-2, 2)$, so its Fourier series will only contain cosine terms ($b_n = 0$). The period is $T = 4$, so $L = T/2 = 2$.$$a_0 = \frac{1}{L} \int_{-L}^{L} f(x) dx = \frac{1}{2} \int_{-1}^{1} 2k dx = \frac{1}{2} [2kx]_{-1}^{1} = \frac{1}{2} (2k - (-2k)) = 2k$$So, the constant term is $a_0/2 = k$.$$a_n = \frac{1}{L} \int_{-L}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx = \frac{1}{2} \int_{-1}^{1} 2k \cos\left(\frac{n\pi x}{2}\right) dx = k \left[ \frac{2}{n\pi} \sin\left(\frac{n\pi x}{2}\right) \right]_{-1}^{1}$$$$a_n = \frac{2k}{n\pi} \left( \sin\frac{n\pi}{2} - \sin\left(-\frac{n\pi}{2}\right) \right) = \frac{4k}{n\pi} \sin\frac{n\pi}{2}$$For $n=1, a_1 = \frac{4k}{\pi}$; for $n=2, a_2 = 0$; for $n=3, a_3 = -\frac{4k}{3\pi}$; for $n=5, a_5 = \frac{4k}{5\pi}$.
The series is:$$f(x) = k + \frac{4k}{\pi} \cos\frac{\pi x}{2} - \frac{4k}{3\pi} \cos\frac{3\pi x}{2} + \frac{4k}{5\pi} \cos\frac{5\pi x}{2} - \dots$$This matches option (C).

First, calculate the expected value $E[X]$:$$E[X] = \sum x_i P(X=x_i) = 0(0.4) + 1(0.3) + 7(0.1) + 11(0.1) + 12(0.1)$$$$E[X] = 0 + 0.3 + 0.7 + 1.1 + 1.2 = 3.3$$Next, calculate $E[X^2]$:$$E[X^2] = \sum x_i^2 P(X=x_i) = 0^2(0.4) + 1^2(0.3) + 7^2(0.1) + 11^2(0.1) + 12^2(0.1)$$$$E[X^2] = 0 + 0.3 + 4.9 + 12.1 + 14.4 = 31.7$$The variance is given by:$$Var(X) = E[X^2] - (E[X])^2 = 31.7 - (3.3)^2 = 31.7 - 10.89 = 20.81$$

According to IS 456:2000, Annex D-1.8, for two-way slabs with continuous edges, the reinforcement in the edge strip shall be at least equal to the minimum reinforcement specified in clause 26.5.2.1. Clause 26.5.2.1 defines the minimum cross-sectional area of reinforcement for slabs (0.15% for mild steel and 0.12% for HYSD bars). Thus, the reinforcement in the edge strip is based on the prescribed minimum cross-sectional area.

The compressive strength of concrete is significantly influenced by its moisture state. Saturated concrete specimens typically exhibit lower compressive strength (often by 10-20%) compared to air-dried specimens of the same concrete. This is primarily due to the development of internal pore water pressure during loading in saturated specimens, which reduces the effective stress and promotes crack propagation. Therefore, statement (B) is false.

For a normally consolidated clay, the effective cohesion $c'$ is zero. The failure envelope passes through the origin. The relationship between the principal stresses at failure and the angle of internal friction $\phi'$ is given by:$$\sin \phi' = \frac{\sigma'_1 - \sigma'_3}{\sigma'_1 + \sigma'_3}$$From the provided Mohr circle:
$\sigma'_3 = 50$ kPa
$\sigma'_1 = 100$ kPa
Substituting these values:$$\sin \phi' = \frac{100 - 50}{100 + 50} = \frac{50}{150} = \frac{1}{3}$$$$\phi' = \sin^{-1} \left( \frac{1}{3} \right)$$Thus, option (B) is correct.

In the Indian Standard Soil Classification System (ISSCS), fine-grained soils are classified using the plasticity chart which plots the Plasticity Index ($I_p$) against the Liquid Limit ($w_L$).

• $w_L < 35\%$: Low plasticity ($L$)
• $35\% < w_L < 50\%$: Intermediate/Medium plasticity ($I$)
• $w_L > 50\%$: High plasticity ($H$)

From the provided figure, region X is located:

• Above the A-line, which indicates it is an inorganic clay ($C$).
• To the right of the vertical line at $w_L = 50\%$, which indicates high plasticity ($H$).

Therefore, X represents inorganic clay with high plasticity ($CH$).

To find the dimensionless combination, we check the dimensions of each variable:

• Velocity ($V$): $[LT^{-1}]$
• Length scale ($l$): $[L]$
• Acceleration due to gravity ($g$): $[LT^{-2}]$

Checking option (A):$$\frac{V^2}{gl} = \frac{(LT^{-1})^2}{(LT^{-2})(L)} = \frac{L^2T^{-2}}{L^2T^{-2}} = [M^0L^0T^0]$$This is a dimensionless quantity. It represents the square of the Froude number ($Fr^2$).

The standard procedure to derive a total flood hydrograph is:Thus, the correct sequence is P-R-S-Q.

The 'Log Kill' or log reduction is defined as the base-10 logarithm of the ratio of the initial concentration to the final concentration.$$\text{Log Kill} = \log_{10} \left( \frac{N_0}{N_t} \right)$$Given:

• Initial concentration ($N_0$) = $10780 \text{ cfu/100 ml}$
• Final concentration ($N_t$) = $23 \text{ cfu/100 ml}$

$$\text{Log Kill} = \log_{10} \left( \frac{10780}{23} \right)$$$$\text{Log Kill} = \log_{10} (468.6956...)$$$$\text{Log Kill} \approx 2.6709$$The value is closest to 2.67.

The combustion reaction for a hydrocarbon $C_n H_m$ in air is:$$C_n H_m + \left(n + \frac{m}{4}\right) (O_2 + 3.78 N_2) \rightarrow n CO_2 + \frac{m}{2} H_2 O + \left(n + \frac{m}{4}\right) \times 3.78 N_2$$ $$M_{fuel} = 12n + m$$ Moles of air per mole of $O_2$ = $1 + 3.78 = 4.78$
$$M_{air} = \left(n + \frac{m}{4}\right) \times 4.78 \times 28.8 = \frac{4n + m}{4} \times 137.664 = (4n + m) \times 34.416$$ $$\text{Ratio} = \frac{M_{fuel}}{M_{air}} = \frac{12n + m}{34.416(4n + m)} = \frac{1}{34.416} \times \frac{12n + m}{4n + m} \approx 0.029056 \times \frac{12n + m}{4n + m} \approx 0.0291 \frac{12n + m}{4n + m}$$This matches option (D).

The Peak Hour Factor ($PHF$) for a 15-minute interval is defined as:$$PHF_{15} = \frac{V}{4 \times V_{15,max}}$$where $V$ is the total hourly volume and $V_{15,max}$ is the maximum volume in any 15-minute interval within that hour.Given that all vehicles arrive within a 10-minute period, this entire volume $V$ must be contained within a single 15-minute interval. Therefore, $V_{15,max} = V$.Substituting this into the formula:$$PHF_{15} = \frac{V}{4 \times V} = \frac{1}{4} = 0.25$$Thus, the correct option is (A).

Let's evaluate each statement:
(A) True: Penetration is measured as the depth (in units of $0.1$ mm) to which a standard needle penetrates a bitumen sample under specified conditions.
(B) True: The softening point of bitumen is determined using the Ring and Ball apparatus.
(C) False: The softening point is the temperature ($^\circ C$) at which the bitumen attains a particular degree of softening. It is not measured in units of time.
(D) True: Ductility is measured as the distance (length in cm) to which a standard briquette of bitumen can be stretched before breaking.Since statement (C) is false, it is the correct answer.

The maximum degree of curvature for a railway track is dependent on the gauge. For Indian Railways, the standard maximum degree of curvature for Broad Gauge (BG) is $10^\circ$. However, for Narrow Gauge (NG), which is typically used in mountainous regions to negotiate sharp curves, the maximum degree of curvature is $40^\circ$. Therefore, in the context of mountainous regions, $40^\circ$ is the standard maximum value.

The error due to the curvature of the earth ($C_c$) is given by the formula:$$C_c = \frac{d^2}{2R}$$where $d$ is the horizontal distance and $R$ is the radius of the earth. From this expression, it is clear that the error is directly proportional to the square of the distance, i.e., $C_c \propto d^2$.

The quadrantal bearing (QB) is given as N$30^\circ$ W. This indicates that the line is in the 4th quadrant (North-West). The Whole Circle Bearing (WCB) is measured clockwise from the North direction ($0^\circ$). For the 4th quadrant:$$\text{WCB} = 360^\circ - \text{QB}$$$$\text{WCB} = 360^\circ - 30^\circ = 330^\circ$$

A partial differential equation is second-order if the highest derivative is of order 2. It is linear if the dependent variable and its derivatives appear only to the first power and are not multiplied together. It is homogeneous if every term contains the dependent variable or its derivatives.(A) $$\frac{\partial^2 u}{\partial t^2} = c^2 \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right) + xy$$: Second-order, linear, but non-homogeneous due to the term $xy$.
(B) $$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = 0$$: Second-order, linear, and homogeneous (Laplace equation).
(C) \frac{\partial u}{\partial t} = c \frac{\partial u \partial x}: First-order.
(D) $$\left( \frac{\partial^2 u}{\partial t^2} \right)^2 = c^2 \frac{\partial^2 u}{\partial x^2}$$: Second-order, but non-linear because the second derivative is squared.

The frame and the loading are symmetric about the vertical axis passing through member CD.

• Due to symmetry, the rotation at joint C must be zero ($\theta_C = 0$). Thus, statement (C) is true.
• For a symmetric structure under symmetric loading, there is no horizontal sway ($\Delta_C = 0$).
• Member CD is fixed at D and connected to joint C. Since $\theta_C = 0$, $\Delta_C = 0$, $\theta_D = 0$, and $\Delta_D = 0$, the member CD undergoes no flexural deformation. Therefore, the bending moments $M_{CD} = M_{DC} = 0$ and the shear force $V_{CD} = 0$. Thus, statement (B) is true.

2. Quantitative Analysis (Slope-Deflection Method):

• At joint B, the equilibrium equation is $M_{BA} + M_{BC} = 100\text{ kN.m}$.
• $M_{BA} = \frac{4(2EI)}{6} \theta_B = \frac{4}{3} EI \theta_B$
• $M_{BC} = \frac{2(4EI)}{8} (2\theta_B + \theta_C) = 2EI \theta_B$ (since $\theta_C = 0$)
• Substituting into the equilibrium equation: $(\frac{4}{3} + 2) EI \theta_B = 100 \Rightarrow \frac{10}{3} EI \theta_B = 100 \Rightarrow EI \theta_B = 30$.
• The bending moment in BC at end C is $M_{CB} = \frac{2(4EI)}{8} (2\theta_C + \theta_B) = EI \theta_B = 30\text{ kN.m}$. Since $30 < 50$, statement (D) is false.
• The shear force in member BC is $V_{BC} = \frac{M_{BC} + M_{CB}}{L_{BC}} = \frac{2(30) + 30}{8} = \frac{90}{8} = 11.25\text{ kN}$.
• By symmetry, the shear force in member CG is also $11.25\text{ kN}$. Both act downwards on joint C. The axial force in member CD is the sum of these shears: $P_{CD} = 11.25 + 11.25 = 22.5\text{ kN} \neq 0$. Thus, statement (A) is false.