GATE Civil Engineering (CE) Official Paper (Held On: 04 Feb, 2024 Shift 2) (Previous Year Paper)

Explanation:

• Drizzle means intermittant rain

• Downpour means heavy rain

• Quarrel means an angry argument

• Feud means an angry and bitter argument

• Bicker means to argue about things that are not important.

Now its very clear that

drizzle → rain → downpour is analogous to bicker → quarrel → feud

Explanation:

Explanation:

• stands out
• standing down
• by her writtings
• for the freedom of speech is the best filling

so option (c) is correct

Explanation:

Actual result = pq

wrong result  = p/q

\(\frac{\text { Wrong result }}{\text { Actual result }}=\) = 0.2 (result reduced by 80%)

\(\frac{p / q}{p q} =0.2=\frac{2}{10}=\frac{1}{5}\)

\(\frac{1}{q^2} =\frac{1}{5} \Rightarrow q=\sqrt{5}\)

Explanation:

From above Venn diagram its clear that (ii) and (iii) are valid deductions.

Explanation:

As per given data:

Let the radius of cylinder = R

and the radius of chalk = r

2r + 2r + 2r = 2R

r = (R/3)

Volume of cylinder = πR²h

Volume of chalk \(=π\left(\frac{\mathrm{R}}{3}\right)^2 \mathrm{~h}\)

\(\frac{\text { Volume of occupied space }}{\text { Volume of empty space }} =\frac{7 π \frac{\mathrm{R}^2}{9} h}{π R^2 h-\frac{7 π R^2}{9}} \)

\( =\frac{7 / 9}{2 / 9}=\left(\frac{7}{2}\right)\)

Explanation:

For any 5000 increment in steps/day the largest risk reduction occurs on going from 0 to 5000.

• Observing the graph, the curve shows a steep decline from 0 to 5000 steps/day, indicating a significant reduction in the mortality risk of cardiovascular disease.
• As the number of steps increases beyond 5000, the curve flattens, indicating smaller reductions in risk for each additional 5000 steps/day increment.

Therefore, the largest reduction in risk happens when increasing steps from 0 to 5000 steps/day compared to any other 5000-step increment.

Explanation:

When the cube is rotated that is held with one of the four body diagonals aligned to the vertical axis

Explanation:

1 + 3 = 4 = 2²

1 + 3 + 5 = 9 = 3²

1 + 3 + 5 + 7 = 4²

(1 + 3 + 5 + 4_________ 2 times) = 20²

\(\frac{20^2}{20^2}=1\)

Explanation:

Let no. of girls in 8^th = no. of girls in 9^th

= x

\(\frac{\text { No. of girls in } 8^{\text {th }} \text { class }}{\text { No. of girls in } 9^{\text {th }} \text { class }}=\frac{\text { No. of boys in } 9^{\text {th }} \text { class }}{\text { No. of girls in } 9^{\text {th }} \text { class }}\)

\(\frac{x}{450-x} =\frac{360-x}{x}\)

x² = (450 - x) (360 - x)

x² = 16200 - 450x - 360x + x²

\(x =\left(\frac{16200}{810}\right)\)

= 200

Explanation:

Anaerobic decomposition of glucose occurs in the absence of oxygen, leading to the production of:

• Carbon dioxide (CO₂)
• Methane (CH₄)
• Water (H₂O) (as a byproduct in some cases)

However, Hydrogen sulfide (H₂S) is not a direct product of anaerobic glucose decomposition. H₂S is typically produced by sulfate-reducing bacteria when they break down sulfur-containing compounds, which is a different biochemical process.

Explanation:

• Various types of newtonian & non-newtonian fluids are shown in the figure.
• Fluids which obeys Newton's law of viscosity \(\left(\tau=\mu \frac{d u}{d y}\right)\) are called Newtonian Fluids and those fluids which do not obey this rule are called Non-Newtonian Fluids.
• General relationship between shear stress and velocity gradient is given by \(\tau=A\left(\frac{d u}{d y}\right)^n+B\)
• In the figures shown above, slope of the curve is called apparent viscosity.
• Fluid for which apparent viscosity increases with du/dy are called Dilatant.
• Dilatant fluids are also called shear thickening fluids. Examples of dilatant fluids are solution with suspended starch or sand, sugar in water.
• Fluids for which apparent viscosity decreases with du/dy are called Pseudo Plastic.
• Pseudo plastic fluid are also called shear thinning fluid. Examples are paints, polymer solutions, blood, paper pulp, syrup, molasses, milk, gelatine.
• Bingham Plastic (ideal plastic) fluids require a certain minimum shear stress t_y (yield stress) before they start flowing. Examples: tooth paste, sewage sludge, drilling mud have time dependent Newtonian Behaviour.

Explanation:

The fourth-order Central Divided Difference formula for the second derivative of a function \( f(x) \) is given by:

\( f''(x) = \frac{1}{12h^2} \left[ -f_{i+2} + 16f_{i+1} - 30f_i + 16f_{i-1} - f_{i-2} \right] \)

Now, comparing this with the given options:

• \( (a) \quad \frac{1}{12h^2} \left[ f_{i+2} + 16f_{i+1} - 30f_i + 16f_{i-1} - f_{i-2} \right] \) (Incorrect sign for \( f_{i+2} \))
• \( (b) \quad \frac{1}{12h^2} \left[ -f_{i+2} - 16f_{i+1} + 30f_i - 16f_{i-1} - f_{i-2} \right] \) (Incorrect sign for several terms)
• \( (c) \quad \frac{1}{12h^2} \left[ -f_{i+2} + 16f_{i+1} - 30f_i + 16f_{i-1} + f_{i-2} \right] \) (Incorrect sign for \( f_{i-2} \))
• \( (d) \quad \frac{1}{12h^2} \left[ -f_{i+2} + 16f_{i+1} - 30f_i + 16f_{i-1} - f_{i-2} \right] \) (Correct expression)

Thus, the correct answer is option (d).

Explanation:

The effective gradient is given by

\(=\frac{103.6-98}{2000} \times 100=0.28 %\)

Explanation:

Given data:

Elastic section modulus = 150.92 cm3

Shape factor \(=\frac{Z_p}{Z_e}\)

Plastic Section Modulus is given by:

\(Z_p =\frac{A}{2}\left(\bar{y}_1+\bar{y}_2\right)\)

\(=\frac{2 \times 12 \times 3}{2}(6+1.5)=270\)

The shape Factor is given by:

\(S F =\frac{270}{150.92}=1.79\)

Explanation:

The most probable value with equal weightage is given by:

= \(30°30'00''+30°29'40''+30°30'20''+30°31'20'' \over 4\) = 30°30'20''

Explanation:

As per given data:

\(P: \frac{d y}{d x}=\frac{x^4+3 x^2 y^2+2 y^4}{x^3 y} \)

\(\frac{d y}{d x}=\left(\frac{x^4}{x^3 y}\right)+\left(\frac{3 x^2 y^2}{x^3 y}\right)+\left(\frac{2 y^4}{x^3 y}\right) \)

\(\frac{d y}{d x}=\left(\frac{x}{y}\right)+3\left(\frac{y}{x}\right)+2\left(\frac{y}{x}\right)^3 \)

\(\frac{d y}{d x}=\left(\frac{1}{y / x}\right)+3\left(\frac{y}{x}\right)+2\left(\frac{y}{x}\right)^3 \)

\(\frac{d y}{d x}=f\left(\frac{y}{x}\right)\)

So P is a homogenous ODE

\(Q: \frac{d y}{d x}=\frac{-y^2}{x^2} \)

x²dy = -y²dx

y²dx + x²dy = 0

\(M=y^2 \quad \frac{d M}{\partial y}=2 y\)

\(N=x^2 \quad \frac{\partial N}{\partial x}=2x\)

So Q is non exact ODE 

∴ The correct answer is option (3).

Explanation:

1. Flexible footing over clayey soil: 

In flexible footing, the contact pressure at the interface between footing and soil is uniformly distributed producing dish-shape pattern in clayey soil.

2. Flexible footing over Granular soil:

In granular soil, modulus of elasticty (E_s) varies across the width being maximum at the centre and minimum at edge. As E is maximum at centre, defflection is less at centre. As E is less at edge deflection is more at edge.

3. Rigid footing on Clayey soil:

In case of flexible footing, deflection is more at centre. Hence pressure developed at centre is less. Deflection is less in flexible footing at edge, hence in rigid footing pressure developed is more at edge.

4. Rigid footing on Granular soil

E×planation:

Given Information:

• The pavement is jointed plain concrete with a slab size of 3.5m × 4.5m.
• Different stresses acting on the slab:
• Wheel load stress at different locations:
• Interior: SiwlSwli S^i_{wl}

​

• Edge: SewlSwle S^e_{wl}

​

• **Corner: SCwlSwlC S^C_{wl}

​**

• Warping stress at different locations:
• Interior: SitSti S^i_t

​

• Edge: SetSte S^e_t

​

• **Corner: SCtStC S^C_t

​**

• Frictional stress: SfSf S_f

​

Understanding Critical Stress Condition at Summer Midnight

• At midnight during summer, the pavement slab cools down, leading to negative temperature gradients:
• Warping stress at the corner (SCtStC S^C_t

) is negative (due to contraction).

• Wheel load stress at the corner (SCwlSwlC S^C_{wl}

) is always positive.

• **Frictional stress (SfSf S_f

) does not significantly affect corner stresses**.

Since the warping stress is negative, the critical stress at the corner is simply the sum of the wheel load stress and the warping stress:

\(\mathrm{S}_{\mathrm{wl} }^{\mathrm{C}}+\mathrm{S}_{\mathrm{t}}^{\mathrm{C}}\)​

Explanation:

ΣM_Q↓= 0 ⇒ R_P × 3 - 3 × 2 = 0 

RP = 2kN & R_Q = 1 kN

As beam PQ is rigid & for no rotation settlement at both P & Q should be same.

ΔP = ΔQ

\(\frac{\mathrm{R}_{\mathrm{P}}}{\mathrm{~K}_{\mathrm{P}}} =\frac{\mathrm{R}_{\mathrm{Q}}}{\mathrm{~K}_{\mathrm{Q}}} \)

\(⇒ \mathrm{~K}_{\mathrm{Q}} =\frac{\mathrm{R}_{\mathrm{Q}}}{\mathrm{R}_{\mathrm{P}}} × \mathrm{K}_{\mathrm{P}}=\frac{1}{2} × 100=50 \mathrm{kN} / \mathrm{m}\)

The value of the spring constant KQ (in kN/m) for the spring at the end Q is 50 kN/m.

Explanation:

As peer given data:

f(x) = x3 - 27x + 4

​f'(x) = 3x2 - 27

​f'(x) = 0

3x2 - 27 = 0

x = ±3

x = 3 ∈ (1, 6)

So at x = 3 function has point at local minima.

Explanation:

The Working Stress Method (WSM) is a traditional structural design approach that uses elastic theory and ensures that stresses in materials remain within their permissible limits under working loads. It does not incorporate safety factors on loads, but instead applies safety factors to material strength.

Additional Information

• Ultimate Load Method (ULM) applies a factor of safety on loads to ensure failure does not occur at the ultimate stage.
• Load Factor Method is another term for ULM, incorporating load factors for safety.
• Limit State Method (LSM) considers safety factors on both loads and material strengths, making it the most advanced and widely used approach.

Explanation:

The given partial differential equation is the Laplace equation:

\(\frac{\partial^2 T}{\partial x^2}+\frac{\partial^2 T}{\partial y^2}=0\)

This equation describes steady-state heat conduction in a 2D square domain.

Given Boundary Conditions:

• The square domain is 2m × 2m.
• Three boundaries are kept at T = 50.
• The fourth boundary is maintained at T = 100.
• We need to determine the temperature at the center of the domain.

Concept of Temperature Distribution:

• The Laplace equation ensures that the temperature field follows a smooth variation.
• In a square plate with three sides at a lower temperature and one side at a higher temperature, the center temperature will be an average of the boundary conditions due to symmetry.

Calculation:

Using approximate averaging techniques for steady-state conduction in such problems, the center temperature TcenterT_{$\text{center}$}Tcenter​ is roughly estimated as:

T_center​ ≈ [3(50)+100]/4 ​= [150 + 100]/4 ​= 250/4 ​= 62.5

Explanation:

The shear center is the point through which the applied transverse shear force must pass to avoid any twisting of the section.

Key Observations:

1. The section consists of a vertical web and two unequal flanges.
2. The top flange is asymmetric, extending further to the right than the bottom flange.
3. The shear center shifts towards the more extended flange to balance the torque effects.

Why is the Shear Center at Point R?

• For thin-walled open sections like C-sections or asymmetric I-sections, the shear center is generally located towards the larger flange.
• Since the top flange extends more to the right, the shear center must shift rightward along the X-X axis.
• Point R is located towards the extended top flange, making it the correct location for the shear center.

Explanation:

Given data:

Length of pile (L) = 10 m

Diameter of pile (D) = 0.7 m

Cohesion of clay (C₄) = 50 kPa

Adhesion factor (α) = 0.5 

For clay, (P) skin friction resistance = f_s A_s

P = α C_uA_s +

Taking unit wt of concrete = 25 kN/m³

Uplift pull out capacity = skin friction resistance + wt. of pile

\(P =0.5 \times 50 \times \pi \times 0.7 \times 10+\frac{\pi(0.7)^2}{4} \times \gamma_c \)

\(P =0.5 \times 50 \times \pi \times 0.7 \times 10+\frac{\pi(0.7)^2}{4} \times 25\)

P = 550 kN

Explanation:

The concept of raising the outer edge above the inner edge in horizontal curves is known as super-elevation (cant or banking) and is primarily used in:

1. Highways – To counteract centrifugal force on vehicles, reducing the risk of skidding and overturning.
2. Railways – To provide a smooth transition for trains on curves, reducing lateral forces and rail wear.

However, taxiways generally do not require super-elevation because aircraft move at lower speeds on taxiways, and the main concern is ensuring smooth, level surfaces for ground maneuvering rather than counteracting lateral forces.

Thus, the correct choice is Highways and Railways only

Explanation:

• P. NO₂ → iv. NH₃ injection

(Selective catalytic reduction (SCR) or selective non-catalytic reduction (SNCR) involves NH₃ injection to reduce NO₂ emissions.)

• Q. SO₂ → iii. Lime scrubbing

(Lime scrubbing is used in flue gas desulfurization (FGD) to remove SO₂ from emissions.)

• R. CO → i. Flaring

(Flaring is used to burn off excess CO and other hydrocarbons in industrial processes.)

• S. Particles → ii. Cyclonic separator

(Cyclonic separators help in removing particulate matter from air streams.)

Explanation:

According to the properties of a matrix

(i) (A + B)T = AT + BT

The sum of the transpose of matrices is equal to the transpose of the sum of two matrices.

(ii) (AB)T = AT BT

The product of the transpose of two matrices in reverse order is equal to the transpose of the product of them.

Explanation:

• P. Bubble tube → iv. Levelling

(The bubble tube is used in leveling instruments to check and adjust the horizontal level.)

• Q. Plumb bob → iii. Centering

(A plumb bob helps in centering the instrument precisely over a survey point.)

• R. Tangent screw → ii. Minor movements

(Tangent screws allow fine adjustments and minor movements for precise alignment.)

• S. Stadia cross-wire → i. Tacheometry

(Stadia cross-wires are used in tacheometry for measuring distances using a theodolite or level.)

Explanation:

Statement P: "In a Pure project organization, the project manager maintains complete authority and has maximum control over the project."

True

• A pure project organization (also called a projectized organization) gives the project manager full authority over project execution, decision-making, and resources.
• The project team reports directly to the project manager, as confirmed by standard project management literature.

Statement Q: "A matrix organization structure facilitates quick response to changes, conflicts, and project needs."

True

• As per the matrix organization structure description in the images, one of its advantages is:
• "The structure facilitates quick response to changes, conflicts, and project needs."
• The matrix structure combines functional and project-based management, ensuring efficient adaptability to project demands and conflicts.

Explanation:

Discharge (Q) = 30 m³/s

Width of channel (B) = 2m

Bed slope = 1 in 300

Energy correction factor (α) = 1.1

g = 10 m/s²

Discharge per unit width \((q)=\frac{Q}{B}\)

\(=\frac{30}{2}=15 \mathrm{~m}^3 / \mathrm{sec} / \mathrm{m}\)

Critical depth does not depend on slope or the roughness of channel.

Critical depth is given by

 \(\left(y_c\right)=\left(\frac{q^2 \cdot \alpha}{g}\right)^{1 / 3}\)

\(=\left(\frac{1.1 \times 15^2}{10}\right)^{1 / 3}=2.914 \mathrm{~m} \simeq 2.91 \mathrm{~m}\)

Additional InformationTotal energy \((E)=y+\frac{\alpha V^2}{2 g}\)

\(=y+\frac{\alpha q^2}{y^2 \times 2 g} \)

\(\frac{d E}{d y} =1+\frac{\alpha q^2}{2 g} \frac{(-2)}{y^3}\)

For critical low, putting \(\frac{d E}{d y}=0\)

\(\frac{\alpha q^2}{g y_c^3}=1\)

\(\therefore \quad y_c=\left(\frac{\alpha q^2}{g}\right)^{1 / 3}\)

Explanation:

For normally consolidated clays A = 0.5 - 1

• For OC clays, A = f (OCR), for heavily over consolidated clays, A < 0.

Additional Information

Pore Pressure Parameters:

(i) Sometimes it is not possible to determine pore pressure practically, then the theoretical approach given by skempton can be adopted.

(ii) Pore pressure parameters A and B are empirical coefficients that are used to express the response of pore pressure to changes in vertical pressure and lateral pressure under "undrained condition".

Parameter B:

(i) This parameter is defined under cell pressure stage and represents the ratio of change in pore pressure to the change in cell pressure.

B = ΔUc/Δσc

(ii) B varies from o to 1 depending on the degree of saturation. "B" is zero for dry soil and is equals unity for fully saturated soil.

Parameter A:

(i) This parameter is valid in deviator stage and is defined in terms of another parameter A̅, such that

A̅ = A.B

(ii) The parameter A̅ represents the ratio of change in pore pressure to change in deviator stress during shear stage.

\(\bar A = \dfrac{{\Delta {U_d}}}{{\Delta {\sigma _d}}} = \dfrac{{\Delta {U_d}}}{{\Delta ({\sigma _1} - {\sigma _3})}} = \dfrac{{\Delta {U_d}}}{{\Delta {\sigma _1} - \Delta {\sigma _3}}}\)

(iii) The parameter depends upon strain in soil, degree of saturation, over consolidated ratio, stratification of soil, sample disturbance, etc. Its value may be as low as -0.5 for over consolidated soil with high O.C.R. to as high as 3 or loose saturated sand.

Explanation:

Given Data:

• Required scale of aerial photograph = 1 : 13,000
• Permissible flying height above datum = 3000 m
• Average terrain elevation above datum = 1050 m

Now,

Scale  \(=\frac{1}{13000}\)

Hence, 

\(\mathrm{S} =\frac{\mathrm{f}}{\mathrm{H}-\mathrm{h}} \)

\(\frac{1}{13000} =\frac{\mathrm{f}}{3000-1050} \)

∴   f = 0.15 m = 150 mm

Explanation:

As per given data:

• Budgeted Cost of Work Scheduled (BCWS) = Rs. 200
• Budgeted Cost of Work Performed (BCWP) = Rs. 150
• Actual Cost of Work Performed (ACWP) = Rs. 190

Schedule Variance is given by

= BCWP - BCWS 

= 150 - 200

= -50

Explanation:

Grashoff method is used to determine the bending moment in two way RCC slab. It is used for load distribution when the slab is simply supported on all four edges and the corners are not held down.

As per Rankine Grashoff method

\(M_{u x}=\frac{r^4}{8\left(1+r^4\right)} w \epsilon_x^2\)

\(M_{u y}=\frac{r^2}{8\left(1+r^4\right)} w \epsilon_x^2\)

Here \(r=\frac{\ell_y}{\ell_x} \)

\(\Rightarrow \quad \frac{\mathrm{M}_{\mathrm{ux}}}{\mathrm{M}_{\mathrm{uy}}}=\mathrm{r}^2=\left(\frac{\ell_{\mathrm{y}}}{\ell_{\mathrm{x}}}\right)^2\)

Explanation:

The effective interest rate (in percentage) is

\(i_{\text {eff }} =\lim _{m \rightarrow \infty}\left(1+\frac{0.05}{m}\right)^m-1 \)

\(=e^{\lim _{m \rightarrow \infty}\left(1+\frac{0.05}{m}\right) \times m-1}\)

\(=e^{\lim _{m \rightarrow \infty}\left(\frac{0.05}{m} \times m\right)-1}\)

= e^0.05 - 1 = 0.05127 or (5.127%)

Explanation:

Given Data:

• Total patients = 100
• Probability of a heart attack without medicine X = 80% = 0.8
• Medicine X reduces the probability of a heart attack by 50%.
• Number of patients taking medicine X = 50
• Number of patients not taking medicine X = 50

Step 1: Probability of a heart attack for patients taking medicine X

• Without medicine X, probability of a heart attack = 0.8
• With medicine X, probability of a heart attack = 0.8 × 0.5 = 0.4

Step 2: Probability of selecting a patient who takes medicine X and has a heart attack

• Probability that a randomly selected patient takes medicine X = 50/100 = 0.5
• Probability that a patient who takes medicine X has a heart attack = 0.4

Thus, the required probability:

P(Takes medicine X and has a heart attack)=P(Takes medicine X)×P(Heart attack | Takes medicine X)

= 0.5 × 0.4 = 0.2 (or 20%)

Explanation:

As per given data:

• Tank dimensions: 2 m × 1.5 m
• Tank height: 6 m
• Orifice diameter: 100 mm = 0.1 m
• Initial water level: 5 m
• Final water level: 3.5 m
• Discharge coefficient (Cₓ): 0.6
• Acceleration due to gravity (g): 10 m/s²

Let at any instant depth of liquid in the tank is 'h' m and in time (dt), the depth falls by (-dh)

Q.dt = A₀(-dh) 

\(\Rightarrow\left(C_d \cdot\left(\frac{\pi d^2}{4}\right) \sqrt{2 g h}\right) d t=A_0(-d h)\)

\(t=\int d t=\frac{A_0}{C_d \cdot \frac{\pi d^2}{4} \sqrt{2 g}} \int_{H_1}^{H_2}\left(\frac{-d H}{\sqrt{h}}\right) \)

\(t=\frac{A_0}{C_d \cdot \frac{\pi d^2}{4} \sqrt{2 g}}\left[2\left(\sqrt{H_1}-\sqrt{H_2}\right)\right]\)

\(t=\left[\frac{2 \times 1.5 m^2}{0.6 \times \frac{\pi}{4}(0.1)^2 m^2 \sqrt{2 \times 10 \frac{m}{s^2}}} \times 2(\sqrt{5}-\sqrt{3.5}) \mathrm{m}^{1 / 2}\right] \mathrm{sec} \)

t = 103.985 sec

Explanation:

cotθ  = 1.25

θ = 0.675 rad

n = 0.013

\(\mathrm{Q} =25 \mathrm{~m}^3 / \mathrm{s}, \mathrm{~S}=\frac{1}{1500}\)

Using manning's equation: 

Area = y² (θ + cot θ) \

A = 1.925 y²

\(R =\frac{y}{2} \)

\(Q =\frac{A}{n} R^{2 / 3}(S)^{1 / 2}\)

\(25 =\frac{1.925 y^2}{0.013}\left(\frac{y}{2}\right)^{2 / 3}\left(\frac{1}{1500}\right)^{1 / 2} \)

⇒ y = 2.40 m

The flow depth (in meters) is 2.40.

Explanation:

Step 1: Calculate HI at Point P

Given:

• Benchmark (BM) at P = 200.000 m
• Back Sight (BS) at P = (-2.050 m)

HI=RL of P+BS

HI=200.000+(−2.050)=197.950 m

Step 2: Calculate RL of Q

• Fore Sight (FS) at Q = 0.950 m

RL of Q=HI−FS

RL of Q=197.950−0.950=197.000 m

Step 3: Calculate New HI at Q

• Back Sight (BS) at Q = 1.050 m

HI at Q=RL of Q+BS

HI at Q=197.000+1.050=198.050 m

Step 4: Calculate RL of R

• Fore Sight (FS) at R = (-1.655 m)

RL of R=HI at Q−FS

RL of R=198.050−(−1.655)

RL of R=198.050+1.655=199.705 m

Explanation:

As per given data:

The beam is determinate.

ILD for BM at 'P' is given by

Hence when the point load is at 'Q', the point of max ordinate for ILD for BM at P, BM at P will be max.

⇒ M_Pmax = 10 × 15 = 150 kN-m

Explanation:

As per given data:

width of the channel (B) = 4 m

discharge (Q) = 2m³/s

depth of flow (y) = 0.4 m

Specific energy at section (1) - (1) (E)

\(=\mathrm{y}_1+\frac{\mathrm{V}_1^2}{2 \mathrm{~g}}\)

\(=0.4+\frac{\mathrm{Q}^2}{\mathrm{~B}_1^2 \times 2 \mathrm{~g} \times \mathrm{y}_1^2} \)

\(=0.4+\frac{2^2}{4^2 \times 2 \times 9.81 \times 0.4^2}\)

= 0.4796 m

When the channel section is contracted to a minimum width and for constant discharge Q. The flow over the contracted section will be critical flow.

\(\mathrm{E} =\mathrm{E}_{\mathrm{c}}=\frac{3}{2} \cdot \mathrm{y}_{\mathrm{C}}\)

\(0.4796 =\frac{3}{2} \times \mathrm{y}_{\mathrm{C}}\)

\({y}_{\mathrm{c}}=0.4796 \times \frac{2}{3}=0.3197 \mathrm{~m}\)

For critical flow conditions,

\(\frac{\mathrm{Q}^2 \mathrm{~T}}{\mathrm{gA}^3}=1\)

\(\frac{\mathrm{Q}^2 \mathrm{~B}_{\min }}{\mathrm{gB} \mathrm{~B}_{\min }^3 y_C^3}=1\)

\({B}_{\min }^2 =\frac{\mathrm{Q}^2}{9 \mathrm{y}_{\mathrm{c}}^3}=1\)

\(\mathrm{~B}_{\min } =\left(\frac{\mathrm{Q}^2}{g \mathrm{y}_{\mathrm{c}}^3}\right)^{1 / 3}\)

\(=\left(\frac{2^2}{9.81 \times 0.3197^3}\right)^{1 / 2} \)

= 3.5325 m ≃ 3.53 m

Explanation:

As per given data:

Hence 85^th percentile speed or safe speed = 55 km/hr

Explanation:

As per given data:

Let us consider the stiffness of the beam as k_b.

Here both the stiffness elements are in parallel.

k_eq = k₁ + k₂

\( =\mathrm{k}+\frac{48 \mathrm{EI}}{(2 \ell)^3}=\mathrm{k}+\frac{6 \mathrm{EI}}{\ell^3}\)

Natural frequency is given by

\(\omega_{\mathrm{n}}=\sqrt{\frac{\mathrm{k}_{\mathrm{eq}}}{\mathrm{~m}}}\)

\(=\sqrt{\frac{\left(\mathrm{k}+\frac{6 \mathrm{EI}}{\ell^3}\right)}{\mathrm{m}}}=\sqrt{\frac{\mathrm{k} \ell^3+6 \mathrm{EI}}{\mathrm{~m} \ell^3}}\)

Explanation:

Statement P:

"When a rough retaining wall moves toward the backfill, the wall friction force/resistance mobilizes in the upward direction along the wall."

True

• When a rough retaining wall moves toward the backfill, it activates passive earth pressure conditions.
• In this scenario, wall friction mobilizes in an upward direction due to the interaction between the soil and the wall.

Statement Q:

"Most of the earth pressure theories calculate the earth pressure due to surcharge by neglecting the actual distribution of stresses due to surcharge."

True

• Earth pressure theories, such as Rankine’s and Coulomb’s theories, generally assume uniform stress distribution due to surcharge load.
• The actual variation of stresses due to surcharge is neglected for simplicity in calculations.

Explanation:

Given data:

Discharge, Q = 10 MLD

Detention period, t_d = 2 hr

Overflow rate, V₀ = 40,000 litres/m²/day 

The surface area of the tank is given by

\(=\frac{\text { Discharge }}{\text { Over flow rate }}=\frac{Q}{V_0}\)

\(\text { S.A. }=\frac{10 × 10^6 \mathrm{~L} / \text { day }}{40,000 \mathrm{~L} / \mathrm{m}^2 / \text { day }}\)

S.A = 250 m²

The volume of tank is given by

(V) = Q × t_d

\(=\frac{10 × 10^6 × 10^{-3}}{24} × 2 \mathrm{~m}^3\)

V = 833.33 m³

Hence, Height of water in the tank is

H \(=\frac{V}{S . A}\)

Height of setting zone is

\(H=\frac{833.33 m^3}{250 m^2} \)

H = 3.33 m

Additional Information

In reality, height of water in tank OR height of settling zone is estimated by above approach but if the bottom of sedimentation tank is assumed to be sloping or hoppered (i.e sludge zone is also considered) then height of water in tank is determined as follows:

Using,

Volume of tank \((\mathrm{V})=\mathrm{D}^2\left(\frac{\pi}{4} \mathrm{H}+0.011 \mathrm{D}\right)\)    ..........(1)

As,

\(S . A . =\frac{\pi}{4} D^2=250 m^2\)

D =17.835 m

So, \(833.33=(17.835)^2\left(\frac{\pi}{4} \mathrm{H}+0.011(17.835)\right)\)

So, H = 3.085 m

Explanation:

\(A =\left[\begin{array}{lll} 2 & 1 & 4 \ 1 & 0 & 3 \end{array}\right] \)

\(B =\left[\begin{array}{cc} -1 & 0 \ 2 & 3 \ 1 & 4 \end{array}\right]\)

\(A B =\left[\begin{array}{ll} -2+2+4 & 0+3+16 \ -1+0+3 & 0+0+12 \end{array}\right] \)

\(=\left[\begin{array}{ll} 4 & 19 \ 2 & 12 \end{array}\right] \)

The determinant of the matrix AB is

= 48 - 38 = 10

Explanation:

The minimum compressive strain in the column is given by

\(ε_{\mathrm{c}_1 \max }=0.0035-0.75 ε_{\mathrm{c}_1 \min }\)

\(ε_{\mathrm{c}_1 \min }=\) strain in the least compressed fibre 

Calculation:

In the strain diagram,

From triangle ΔOAD and ΔOBC

\(\Rightarrow \frac{\varepsilon_{c_1 \max }}{1.1 \mathrm{D}}=\frac{\varepsilon_{\mathrm{c}_1 \min }}{0.1 \mathrm{D}} \Rightarrow \varepsilon_{\mathrm{c}_1 \min }=\frac{1}{11} \varepsilon_{\mathrm{c}_1 \max }\)

\(\varepsilon_{\mathrm{c}_1 \max }=0.0035-0.75 \times \frac{1}{11} \varepsilon_{\mathrm{c}_1 \max }\)

\(\Rightarrow \varepsilon_{\mathrm{c}_1 \max }=3.276 \times 10^{-3}=3.28 \times 10^{-3}\)

Hence, The maximum compressive strain (εc) at highly compressive extreme fiber in concrete, where there is no tension in the section, is 3.28 × 10-3

Explanation:

The Sight distance is given by:

(S) \(=V \cdot t_a+\frac{V^2}{2 g \mu} \)

\(=\left(\frac{5}{18} \times 80 \times 2\right)+\frac{\left(\frac{5}{18} \times 80\right)^2}{2 \times 9.8 \times 0.36} \)

= 44.44 + 69.92 = 114.36 m

Now, Let assume L > S 

The design length of the summit curve (in meters) to accommodate the stopping sight distance is

\(L =\frac{N S^2}{2\left(\sqrt{h_1}+\sqrt{h_2}\right)^2} \)

L \(=\frac{0.08 \times 1140.36^2}{2(\sqrt{2.3}+\sqrt{0.35})^2}=117.7\)

Explanation:

Organic waste = C240 O200 H180 N5S

Weight of organic waste = 1 kg

Burning of organic waste converts

sulphur into sulphur dioxide as follows:

C240 O200 H180 N5S + O₂  → xCO₂ + yH₂O + zH₂O + SO₂

(6362 g)                                                                      (64 g)

Now, 6362g of organic waste produces 64 g of sulphur dioxide.

So,

1 kg of organic waste produces \(\frac{64}{6362} \times 1000 \mathrm{gm}\) of SO₂

Hence,

Amount of SO₂ produced=10 grams/kg

Explanation:

The correct statements regarding pavement material characterization are:

1. "In compacted bituminous mix, voids in the mineral aggregate (VMA) is equal to the sum of total volume of air voids (Vᵥ) and total volume of bitumen (Vᵦ)."

Correct - VMA represents the total void space within the aggregate structure, including both air voids and bitumen. 2. "The toughness and hardness of road aggregates are determined by Los Angeles abrasion test and aggregate impact test, respectively."

Incorrect -

• The Los Angeles Abrasion Test measures the abrasion resistance (not toughness).
• The Aggregate Impact Test measures toughness (resistance to sudden impact), whereas hardness is generally evaluated using crushing strength tests.

3. "The load penetration curve of the CBR test may need origin correction due to the non-vertical penetrating plunger of the loading machine."

Correct - A non-vertical plunger can cause errors in the initial part of the CBR curve, requiring origin correction. 4. "Grading of normal (unmodified) bitumen binders is done based on viscosity test results."

Correct - Bitumen binders are graded based on viscosity tests, which determine their performance in different temperature conditions.

Explanation:

Let the settling velocity of the anthracite pacticle at top, silica pacticle at middle & ilmenite pacticle at bottom, after back washing be (V_S)_T, (VS), (VS)_B respectively.

For Middle silica sand

\(v_S =\frac{\left(G_{s_1}-1\right) \gamma_w D_1^2}{18 \mu} \)

vS = K(2.6 - 1)(0.2)2 

v_S = K(1.6)(0.2)² where \(\mathrm{K}=\frac{\gamma_{\mathrm{w}}}{18 \mu}\)

For Top Anthracite

(V_S)_T = K\( \left(\mathrm{G}_{\mathrm{s}_2}-1\right)\) (D₂)²

(VS)T = K(0.5) (D2)2

As anthracite lies above the silica layer

So, anthracite particles should have a settling velocity less than silica particles

(V_S)_T < V_S

So, K(0.5) (D2)2 < K(1.6) (0.2)²

D₂ or \(D_{\text {Top }}<\sqrt{\frac{1.6 \times(0.2)^2}{0.5}}\)

D₂ or D_Top < 0.357 mm

For bottom ilmenite sand

(V_s)_B = K\(\left(G_{s_3}-1\right)\left(D_3^2\right)\)

(V_S)_B = K(4.2-1) \(\left(D_3^2\right)\)

As, ilmenite layer lies below silica layers

So, ilemite particles should have a settling velocity greater than silica

i.e., (V_S)_B > V_S

\(K(3.2)\left(D_3^2\right)>K(1.6)(0.2)^2\)

\(D_3 \text { or } D_{\text {bottom }}>\sqrt{\frac{1.6 \times(0.2)^2}{3.2}}\)

D₃ or D_bottom > 0.141 mm

Hence, diameter of anthracite particle is slightly greater than 0.35 mm & diameter of ilmenite particle is slightly greater than 0.141 mm

So (d) is the correct option.

Explanation:

Assume ϕ ≤  i_lowest 

\(ϕ =\left(\frac{\mathrm{P}-\mathrm{R}}{\mathrm{t}}\right)_{\text {eff. duration }} \)

\(=\frac{11-6}{8}=0.625\)

Assume 0.625 < ϕ ≤ 0.8

\(ϕ =\left(\frac{\mathrm{P}-\mathrm{R}}{\mathrm{t}}\right)_{\text {eff. }}\)

\(=\frac{10.5-6}{7}=0.643\)

Explanation:

As per given data:

(ΔH) total head loss = 15 m

(n_d) number of equipotential drops = 10

(I) length of flow field = 3 m

(G) specific gravity = 2.65

(FOS) factor of safety = 2

(e) void ratio = ?

(Δh) an equipotential drop of head

\(=\frac{\text { Total head loss }}{\text { no. of equipotential drop }} \)

\(Δ \mathrm{h} =\frac{Δ \mathrm{H}}{\mathrm{~N}_{\mathrm{d}}}=\frac{15}{10}=1.5 \mathrm{~m}\)

FOS against the quicksand condition

\(=\frac{\mathrm{i}_{\mathrm{cr}}}{\mathrm{i}}\)

(i) Hydraulic gradient

\(=\frac{\Delta h}{l}=\frac{1.5}{3}=0.5\)

Now,

\(2=\frac{i_{\mathrm{cr}}}{0.5}\)

The critical hydraulic gradient = 1

The critical hydraulic gradient is given by

\(\left(\mathrm{i}_{\mathrm{cr}}\right)=\frac{\mathrm{G}-1}{1+\mathrm{e}}\rightarrow1=\frac{2.65-1}{1+\mathrm{e}}\)

(e)void ratio = 0.65

Explanation:

(a) \(\vec{p} \times(\vec{q} \times \vec{r})=(\vec{p} \cdot \vec{r}) \vec{q}-(\vec{p} \cdot \vec{q}) \vec{r}\)

(This is always true for any three given vectors)

(b) We know that \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\) is always true but \(\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\) because \(\vec{a} \times \vec{b}=-\vec{b} \times \vec{a}\)

This can be true only when \(\vec{a} \times \vec{b}=0\)

So, \(\quad \vec{r} \cdot(\vec{p} \times \vec{q})=\vec{r} .(\vec{q} \times \vec{p})\)

\(\vec{r} \cdot(\vec{p} \times \vec{q})=-\vec{r} \cdot(\vec{p} \times \vec{q})\)

This can be true if

\(\vec{r} \cdot(\vec{p} \times q) =0\)

\(\vec{p} \times \vec{q} =\hat{i}-2 \hat{j}+j\)

\((\vec{r} \cdot \vec{p} \times \vec{q})=0\) this is true in this case

(c) \(\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{r}})\) can't be equal to \((\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}) \times \overrightarrow{\mathrm{r}}\) because \(\vec{p} \times(\vec{q} \times \vec{r}) \perp \vec{p}\) and \((\vec{p} \times \vec{q}) \times \vec{r}) \perp \vec{r}\)

So, \((\vec{p} \times(\vec{q} \times \vec{r}) \neq(\vec{p} \times \vec{q})) \times \vec{r}\)

(d) \(\overrightarrow{\mathrm{p}} \times(\overrightarrow{\mathrm{q}} \times \overrightarrow{\mathrm{r}})+\overrightarrow{\mathrm{q}} \times(\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}})+\overrightarrow{\mathrm{r}} \times(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})=\overrightarrow{0}\)

\((\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{r}}) \overrightarrow{\mathrm{q}}-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{r}}+(\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{p}}) \overrightarrow{\mathrm{p}}-(\overrightarrow{\mathrm{q}} \cdot \overrightarrow{\mathrm{r}}) \overrightarrow{\mathrm{p}}+(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{p}}-(\overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{q}}) \overrightarrow{\mathrm{q}} \)

0 = 0

\(\because \quad \vec{p} \cdot \vec{r} =\vec{r} \cdot \vec{p} \)

\(\vec{p} \cdot \vec{q} =\vec{q} \cdot \vec{p} \)

\(\vec{q} \cdot \vec{r} =\vec{r} \cdot \vec{q}\)

(Hence this is proved)

Explanation:

Given Data:

• Normal duration: 15 days
• Normal cost: Rs. 30,000
• Crashed duration: 12 days
• Crashed cost: Rs. 54,000
• Indirect cost: Rs. 8,500 per day (for P) and Rs. 10,000 per day (for Q)

Cost slope (Crashing cost per day) is given by:

\(\mathrm{C} / \mathrm{s}=\frac{54000-30000}{15-12}=8000 \mathrm{Rs} / \text { day }\)

Crashing is economically beneficial if the savings in indirect cost per day exceed the direct cost of crashing per day.

• If the indirect cost per day is Rs. 8,500 (Statement P):

Since Rs. 8,500 > Rs. 8,000, it is economically advisable to crash. So, P is true.

• If the indirect cost per day is Rs. 10,000 (Statement Q):

Since Rs. 10,000 > Rs. 8,000, it is also economically advisable to crash. So, Q is true.

Explanation:

For the closed traverse,

ΣL = 0, 4cos30° + 4cos90° + Lcos240° = 0 

L = 6.928 m ≈ 7 m

The straight-line path length RP (in meters) is 7 m.

Explanation:

As per given data:

σ_temp = EαΔT

\(σ_{\mathrm{cr}} =\frac{\pi^2 \mathrm{E}}{\lambda_{\text {eff }}^2}=\frac{\pi^2 \mathrm{E}}{\left(\ell_{\text {eff }} / r\right)^2}\)

For given support \(\rightarrow \ell_{\text {eff }}=\frac{1}{2}\)

Hence,

\(\mathrm{E} α Δ \mathrm{~T} =\frac{\pi^2 \mathrm{E}}{\left(\frac{I}{2 r}\right)^2} \)

\(Δ \mathrm{~T} =\frac{\pi^2 \times 4}{α 5\left(\frac{I}{2}\right)^2}=\frac{\pi^2 \times 4}{12 \times 10^{-6} \times 105^2} \)

The differential increase in temperature (rounded off to the nearest integer) at which the bar buckles is

ΔT = 298°C

Explanation:

As per given data:

To replace the pipe, the head loss has to be the same

h_fp = h_fQ

_{\(\Rightarrow \frac{f_{f p} L Q^2}{12.1 D_P^5}=\frac{f_{f Q} L Q^2}{12.1 D_Q^5}\)}

\(\Rightarrow \quad \frac{0.04 \times 500 \times Q^2}{12.1 \times 1^5}=\frac{0.01 \times 500 \times Q^2}{12.1 \times D_Q^5} \)

\(\Rightarrow \quad D_Q^5=\frac{0.01}{0.04}\)

\(\Rightarrow \quad D_Q=\left(\frac{0.01}{0.04}\right)^{1 / 5}=0.7579 \mathrm{~m} \)

The diameter of the pipeline Q (in meters) is ≃ 0.76 m

Explanation:

The equilibrium cant is given by

\(=\frac{\mathrm{GV}_{\mathrm{eq}}^2}{127 \mathrm{R}}=\frac{1.75 \times 70^2}{127 \times 1080}=0.062518 \mathrm{~m}=62.52 \mathrm{~mm}\)

With the allowable cant deficiency of 100 mm, the Maximum speed of train can run with theoretical cant = 62.52 + 100 = 162.52 mm

The maximum restricted speed (in kmph) of the express trains running on this track is

\(∴ 162.52 \times 10^{-3} =\frac{1.75 \times \mathrm{V}_{\mathrm{m}}^2}{127 \times 1080} \)

∴  V_m = 112.86 kmph ≃ 113 kmph

Explanation:

In stress-path triaxial testing, we can control both radial and axial stress.

Failure situation

\(\sigma_1=\sigma_3 \tan ^2\left(45^{\circ}+\frac{ϕ}{2}\right)+2 C \tan \left(45^{\circ}+\frac{ϕ}{2}\right) \)

\(300=100 \tan ^2\left(45^{\circ}+\frac{ϕ}{2}\right)+0\)

[Since for drained test C = 0]

⇒ ϕ = 30°

The angle between the normal to major plane and the failure plane is \(45+\frac{ϕ}{2}\)

The angle between the normal to minor plane and the failure plane is \(45^{\circ}-\frac{ϕ}{2}\)

The angle between minor plane and failure plane is

\(=45-\frac{ϕ}{2}=45^{\circ}-\frac{30^{\circ}}{2}=30^{\circ}\)

Explanation:

Statement P:

"Compacted fine-grained soils with flocculated structure have isotropic permeability."

True - Fine-grained soils compacted on the dry side of optimum moisture content tend to have a flocculated structure (random arrangement). This structure provides approximately equal permeability in all directions (isotropic permeability), as confirmed by the provided explanation.

Statement Q:

"Phreatic surface/line is the line along which the pore water pressure is always maximum."

False - The phreatic line represents the topmost flow line below which seepage occurs in a dam body. However, pore water pressure is not maximum along the phreatic line; rather, it is atmospheric at this level. The maximum pore water pressure occurs below the phreatic line, not along it.

Statement R:

"The piping phenomenon occurring below the dam foundation is typically known as blowout piping."

False -

• Piping phenomenon refers to the erosion of soil due to excessive seepage forces.
• The correct term for piping failure below the dam foundation is "heave" or "boiling", not blowout piping. Blowout typically refers to failure due to excessive water pressure in confined aquifers rather than dam foundation piping.

Explanation:

The critical path is given by:

A-K-L-H

Project duration = 30 days

Hence, The total duration (in days) of the project is 30 days.

Explanation:

Given data:

maximum density (γ_d)_max = 1.8 g/cm³

maximum density (γd)max = 1.3 g/cm3

specific gravity (G) = 2.7

(I_D) relative density is given by

\(=\frac{γ_{d \max }\left(γ_d-γ_{d \text { min }}\right)}{γ_d\left(γ_{d \text { max }}-γ_{d \text { min }}\right)}\)

(n) percentage of voids

\(=\frac{\left(V_v\right) \text { volume of voids } \times 100}{\left(V_T\right) \text { Total volume }}=50 %\)

Porosity (n) = 0.5 

void ratio \((e)=\frac{n}{1-n}=\frac{0.5}{1-0.5}=1\)

\(\left(\gamma_d\right)_{\text {insitu }}=\frac{G \cdot \gamma_w}{1+e}=\frac{2.7 \times 1}{1+1}=1.35 \mathrm{~g} / \mathrm{cm}^3\)

\(\left(I_D\right)=\frac{1.8(1.35-1.3)}{1.35(1.8-1.3)}=0.13\)

Relative density (I_D) = 0.13

Explanation:

As per given data:

The magnitude of the vertical displacement (in mm) at point Q is

\(\Delta_Q=\frac{P L_1^3}{3 \mathrm{EI}}+\frac{\mathrm{M} L_2^2}{2 \mathrm{EI}}\)

\(Δ_{\mathrm{Q}}=\frac{(15 \mathrm{kN})(3 \mathrm{~m})^3}{3 × 2 × 10^5 \mathrm{kNm}^2}+\frac{(22.5 \mathrm{kNm})(3m)^2}{2 × 2 × 10^5}\)

Δ_Q = (6.75 × 10^-4 + 5.0625 × 10^-4) m

ΔQ = 1.181 mm