GATE Civil Engineering (CE) Official Paper (Held On: 04 Feb, 2024 Shift 1) (Previous Year Paper)

Explanation:

The analogy works based on the increasing intensity of the actions described by the words. Here's the breakdown:

1. Simmer → Seethe → Smolder:

• Simmer suggests a mild or low-level intensity.
• Seethe indicates a higher intensity, where things are boiling or agitated.
• Smolder refers to something burning slowly with little flame, implying an intense state of heat, though not yet fully destroyed or burnt.

Now, for the analogy of [break → raze → _______]:

• Break suggests a low-level action, where something is damaged or disrupted.
• Raze involves completely destroying or leveling something, a much stronger action.
• The final word must indicate a complete or intense form of destruction.

The appropriate word to complete the analogy is obliterate, which means to completely destroy or wipe out, just as smolder represents a highly intense state of burning.

So, the solution is: break → raze → obliterate

Explanation:

Minute hand covers 6° angle in 1 min.

Second hand covers 360° angle in 1 min.

Means 354° angle gained by second hand over min. hand in 1 min. of time

So, 360° angle gained in $\frac{360}{354}$ min.

(which is required time for one crossing)

But first crossing taking $\left(\frac{30}{354}\right)$ min. only

So, total time taken in 50

Crossings = $1 \times \frac{30}{354}+49 \times \frac{360}{354}=49.91$ min.

So, no. of crossings = 50

Explanation:

Let no. of houses on each side = n

The sum of odd-numbered houses

= (301 + 303 + 305, …, n)

= $\frac{n}{2}[2 \times 301+(n-1) 2]$

= $\frac{n}{2}[2 n+600]=n(n+300)$

The sum of even-numbered houses

= 302 + 304 + 306, …, n

= $\frac{n}{2}[2 \times 302+(n-1) 2]$

= $\frac{n}{2}[2 n+602]$

= n[n + 301]

According to question

n[n + 301] – n[n + 300] = 27

n[n + 30] – n – 300] = 27

n = 27

n[n + 301] – n[n + 300] = 27

n[n + 30] – n – 300] = 27

n = 27

Alternate method:

If one house each side difference = 302 – 301 = 1

If two house each side difference = (302 + 304) – (301 + 303) = 2

If three houses each side difference = (302 + 304 + 306) – (301 + 303 + 39) = 3

So, to make difference equal to 27 number of houses on each side must be 27.

Explanation:

7 – 3 = 4 = 4 × 1

15 – 7 = 8 = 4 × 2

x – 15 = 16 = 8 × 2

63 – x = 32 = 16 × 2

127 – 63 = 64 = 32 × 2

255 – 127 = 128 = 64 × 2

x – 15 = 16 ⇒ x = 31

Explanation:

$\rm \left(\frac{p}{q}\right)^{p / q}=p^{\left(\frac{p}{q}-1\right)}$

$\rm \frac{p^{p / q}}{q^{p / q}}=p^{p / q} \times p^{-1}$

$\rm \left(\frac{p^{p / q}}{q^{p / q}}-\frac{p^{p / q}}{p}\right)=0$

$\rm p^{p / q}\left(\frac{1}{q^{p / q}}-\frac{1}{p}\right)=0$

$\rm \left(\frac{1}{q^{p / q}}-\frac{1}{p}\right)=0$

(∵ pp/q ≠ 0)

$\rm \frac{p-q^{p / q}}{p \cdot q^{p / q}}=0$

⇒ p = q^p/q

p = (q^1/q)^p

p^1/p = q^1/q

q^p = p^q

Explanation:

Capacity factor of T1 = $\left(\frac{10000}{20}\right) \times \frac{1}{1000}=0.5$

Capacity factor of T2 = $\left(\frac{9000}{30}\right) \times \frac{1}{1000}=0.3$

Capacity factor of T3 = $\left(\frac{7000}{15}\right) \times \frac{1}{1000}=0.47$

Capacity factor of T4 = $\left(\frac{12000}{40}\right) \times \frac{1}{1000}=0.3$

Hence, the capacity of power generation technology T1 (0.5) is highest.

Explanation:

Step-by-Step Calculation:

1. Treat each twin pair as a unit → We have 3 units to arrange.
2. Circular arrangement of 3 units →(3−1)! = 2! = 2.
3. Arrangement of distinct twin pairs →3! ways.
4. Twins within each pair are indistinguishable → No extra arrangement needed.
5. Empty seats are fixed automatically.

Final Answer:

(3−1)!×3! = 2! × 6 = 12

Explanation:

Step-by-Step Analysis

• Cell (1,4): The number 2 at (1,3) means it already has 2 Xs in its vicinity, so (1,4) cannot be X.
• Cell (2,4): The number 3 at (2,3) already has X at (2,2) and (3,2), so (2,4) can be X.
• Cell (3,4): The number 4 at (3,3) means it needs 4 Xs in its vicinity. Since it has Xs at (3,2) and (2,2), placing an X at (3,4) would fulfill the condition.
• Cell (4,4): The X at (4,3) satisfies all conditions, meaning (4,4) cannot be X.

Hence, the maximum number of Xs in the last column is 2.

Explanation:

The distance from the earth to the moon

= 240000 mile (estimated)

$\rm \cos 89.85=\frac{240000}{S E}$

SE = 91673351.94 mile

$\rm \frac{E S}{E M}=\left(\frac{91673351.94}{240000}\right)=381.97=382$

Explanation:

According to subject-verb agreement,

singular noun takes singular verb and plural noun takes plural verb.

“athletics” is placed as a game (singular common noun) so singular verb form is used.

‘crowd” is a singular noun, so singular verb form is used.

“cross-steps” is a plural form of noun, so it will be followed with a plural form of verb.

“his body” is a singular noun so singular form of verb is used.

Hence, option a is correct.

Explanation:

The sum of the interior angle is given by

= (2n – 4) × 90

For 3 sides, the sum of the interior angle is

= (2×3 – 4) × 90 = 180°

For 4 sides, the sum of the interior angle is

= (2×4 – 4) × 90 = 360°

Hence, the sum of the interior angles increases by 

= 360° - 180° = 180°

Explanation:

As per given equation

$\rm \frac{\partial^{2} u}{\partial x^{2}}=2$ …(i)

$\rm \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)=2$

Integrating both sides with respect to x

$\rm \frac{\partial u}{\partial x}=2 x+f(y)$

Again integrating with respect to x

u = x² + xf(y) + f(y)

u = x² + xf(y) + h(y)

Explanation:

The degrees of freedom (DOF) in an open channel flow system refer to the number of independent variables required to describe the flow conditions.

For a natural open channel flow with a mobile bed, the key parameters influencing the system are:

1. Flow depth (y)
2. Velocity (V)
3. Slope of the bed (S)
4. Sediment concentration (C)

Since these four variables are independently adjustable in a mobile bed channel, the number of degrees of freedom is 4.

Explanation:

Ductility Test (True)

• The ductility test measures the ability of bitumen to stretch under tensile stress, assessing adhesiveness and flexibility.

Flash Point (True)

• The flash point is the minimum temperature at which bitumen vapors ignite momentarily when exposed to a flame.

Kinematic Viscosity (True)

• Kinematic viscosity represents the resistance to flow under gravity, crucial for evaluating bitumen’s workability at different temperatures.

Softening Point of Bitumen (False)

• Softer grade bitumen has a lower softening point, meaning it softens at a lower temperature compared to harder grades.
• Harder bitumen has a higher softening point, making it more resistant to deformation at high temperatures.
• Thus, this statement is incorrect.

Explanation:

Carbon (C) (Increases Strength)

• Strengthens steel by increasing hardness and tensile strength.
• However, excessive carbon reduces ductility and weldability.

Manganese (Mn) (Increases Strength)

• Improves strength, toughness, and wear resistance.
• Helps in deoxidation and grain refinement.

Chlorine (Cl) (Does Not Increase Strength)

• Harmful impurity in steel.
• Causes corrosion and brittleness, reducing strength.

Sulphur (S) (Does Not Increase Strength)

• Weakens steel by making it brittle and reducing ductility.
• Increases machinability but is considered an undesirable impurity in structural steel.

Explanation:

Zenith angle (Z) is measured from the vertical (0° at zenith downward to 180° at nadir).

Vertical angle (V) is measured from the horizontal (0° at the horizon, +ve above, -ve below).

Conversion Formula: V=90^∘−Z

Given Data:

Zenith Angle (Z) = 93°00'00"

Using the formula: 

V=90^∘−93^∘=−3^∘00′00"

Explanation:

• Speed of the car: 60 km/hr
• Downward gradient of the highway: 2%
• Distance to the stopped vehicle: 130 m
• Brake efficiency: 60%
• Coefficient of friction: 0.7
• Driver's reaction time: 2.5 seconds
• Acceleration due to gravity: 9.81 m/s²

The distance (in meters) required by the driver to bring the car to a safe stop is

$\mathrm{SSD} =\mathrm{vt}_{\mathrm{r}}+\frac{\mathrm{v}^{2}}{2 \mathrm{~g}(\mu-0.01 \mathrm{n})}$

SSD = $\left(\frac{5}{18} \times 60\right) \times 2.5+\frac{\left(\frac{5}{18} \times 60\right)^{2}}{2 \times 9.81(0.6 \times 0.7-0.01 \times 2)}$

SSD = 41.67 + 35.4 = 77.07 m

Explanation:

The given equation is

x5 – 5x4 – 10x3 + 50x2 + 9x – 45 = 0

Now, 

f(0) = 0 – 45 = Negative

f(2) = 2⁵ – 5 × 2⁴ – 10 × 2³ + 50 × 2² + 9 × 2 – 45

= 32 – 65 – 80 + 200 + 18 – 45

= 60 = Positive

Hence, If f(x) is continuous on [a, b] and f(a) and f(b) are of opposite sign then there exist atleast one point c on [a, b] such that f(c) = 0 so atleast one root lying between (0, 2).

Explanation:

The most general case of second-order partial differential equation

$\rm A \frac{\partial^{2} u}{\partial x}+B \frac{\partial^{2} u}{\partial x \partial y}+C \frac{\partial^{2} u}{\partial y^{2}}+D \frac{\partial u}{\partial x}+E \frac{\partial u}{\partial y}+f u=G$ …(i)

Given that the partial differential equation is

$\rm x \frac{\partial^{2} f}{\partial x^{2}}+y \frac{\partial^{2} f}{\partial y^{2}}=\frac{x^{2}+y^{2}}{2}$ …(ii)

Now, Comparing (i) and (ii)

A = x and B = 0, C = y

Discrimanant = B² – 4AC = 0 – 4 × x × y = –4xy

–4yx > 0 if x < 0 and y > 0

So, the given equation is hyperbolic if x < 0 and y > 0.

Similarly, we can prove if x > 0, y > 0 given partial differential equation is elliptic also.

Explanation:

While removing the props, we need to ensure that the beam type does not change i.e. if it is a cantilever beam then the props shall be removed in such a manner that the beam at the intermediate stage also shows cantilever behaviour.

If we remove prop ‘S1’ first then tensile stress will be generated at the bottom and no structural reinforcement is there to take care of this tension.

Hence, The correct sequence of removing the Shores/Props is

S5 → S4 → S3 → S2 → S1

Explanation:

Bulky particles are formed mostly by mechanical weathering of rocks and minerals. Geologists use such terms as angular, subangular, subrounded and rounded to describe the shape of bulky particles. Small sand particles located close to their origin are generally very angular sand particles carried by wind and water for a long distance can be subangular to rounded in shape.

Activity Number:

It is used to study the swelling behaviour.

Activity number (A_C) is given by

= $\frac{\text { Plasticity index }}{\text {\% of clay sized particles in soil }}$

Different state of activity of soil

A_C < 0.75 → inactive

0.75 < AC < 1.25 → normal active

AC > 1.25 → active

Active means more prone to volume change.

Explanation:

Statement 1: "The Vernier of the vertical circle must read zero when the line of collimation is vertical."

• Incorrect: The Vernier reading depends on the initial instrument setup and may not always read zero when the collimation is vertical.

Statement 2: "The axis of the plate level must lie in a plane parallel to the vertical axis."

• Incorrect: As per the fundamental relations, the plate level axis should be perpendicular to the vertical axis, not parallel.

Statement 3: "The line of collimation must be perpendicular to the horizontal axis at its intersection with the vertical axis."

• Correct: As per the fundamental relations (point 5 in the reference image), the line of sight (collimation) must be perpendicular to the horizontal axis at its intersection with the vertical axis.

Statement 4: "The axis of altitude level must be perpendicular to the line of collimation."

• Incorrect: The axis of the telescope (altitude level) must be parallel to the line of sight (collimation), not perpendicular.

Explanation:

Statement P:

"Client’s preliminary estimate is used for budgeting costs toward the end of the planning and design phase."

• A preliminary estimate (also known as an approximate or budget estimate) is prepared in the early planning phase.
• It provides a rough cost assessment based on historical data, unit costs, or area-based estimates.
• This estimate is refined towards the end of the planning and design phase to align with the budget before execution.
• Hence, Statement P is correct.

Statement Q:

"Client’s detailed estimate is used for controlling costs during the execution of the project."

• A detailed estimate is prepared after the design phase and before project execution.
• It includes:
• Material, labor, and equipment costs
• Overheads, contingencies, and profit margins
• Bill of Quantities (BOQ) for contracts and tendering
• This estimate helps in cost control during execution by ensuring the project stays within budget.
• Thus, Statement Q is correct

Explanation:

$\sigma=\left[\begin{array}{ccc} 10 & 0 & 0 \ 0 & 40 & 0 \ 0 & 0 & 0 \end{array}\right]$ = $\left[\begin{array}{ccc} \sigma_{\mathrm{xx}} & \tau_{\mathrm{xy}} & \tau_{\mathrm{xz}} \ \tau_{\mathrm{yx}} & \sigma_{\mathrm{yy}} & \tau_{\mathrm{yz}} \ \tau_{\mathrm{zx}} & \tau_{\mathrm{zy}} & \sigma_{\mathrm{zz}} \end{array}\right]$

σ1 = 40, σ2 = 10, σ3 = 0 

The maximum shear stress at the point is given by

\(\sigma_{\text {max }, \text { abs }}=\max \left{\frac{\left|\sigma_{1}-\sigma_{2}\right|}{2}, \frac{\left|\sigma_{2}-\sigma_{3}\right|}{2}, \frac{\left|\sigma_{3}-\sigma_{1}\right|}{2}\right} \)

\(\sigma_{\text {max }, \text { abs }}=\max \left{\left|\frac{40-10}{2}\right|,\left|\frac{\mid 0-0}{2}\right|,\left|\frac{0-40}{2}\right|\right}\)

τmax, abs = 20 MPa

Explanation:

P(A ∪ B ∪ C) → Probability that the student passes is atleast one subject. 

$\rm P(A \cup B \cup C)=\frac{1}{3}+\frac{1}{6}+\frac{4}{9}=\frac{6+3+8}{18}=\left(\frac{17}{18}\right)$

Explanation:

Give data:

For pure clay, ϕ = 0

Nc = 5.7, Nq = 1, Nγ = 0

The ultimate bearing capacity (q_u) is given by

= 5.7C + q

The net ultimate bearing capacity (q_nu) is given by

= 5.7C + q – q = 5.7C

Since cohesion is given to be constant, net ultimate bearing capacity will not change with water table fluctuations.

Explanation:

Primary pollutant: Pollutants which are emitted directly from the identifiable sources, either from the natural hazardous events like dust storms, volcanoes, etc or from human activities like burning of wood, coal, oil in homes or industries or automobiles, etc.

The important primary air pollutants are:

1. Oxides of sulphur, particularly the sulphur dioxide (SO2);
2. Oxides of carbon like carbon monoxide (CO) and carbon dioxide (CO2), particularly the carbon monoxide (CO); 3. Oxides of nitrogen, like NO, NO2, NO3 (expressed as NOx);
3. Volatile organic compounds, mostly hydrocarbons; and
4. Suspended particulate matter (SPM).

These primary pollutants often react with one another or with water vapour, aided and abetted by the sunlight, to form entirely a new set of pollutants, called the secondary pollutants.

These are the chemical substances, which are produced from the chemical reactions of natural or anthropogenic pollutants or due to their oxidation, etc., caused by the energy of the sun.

The important secondary pollutants are:

(i) Sulphuric acid (H2SO4);

(ii) Ozone (O3);

(iii) Formaldehydes; and

(iv) Peroxy-acyl-nitrate (PAN); etc.

Primary Pollutants: Co, CO2, SOX, NOX, Aerosols, Bacteria, Pollens, Suspended particulates, Fly ash, Mist, Dew, Mist, Soot, Volcanoes, etc.

Secondary Pollutants: Chlorofluorocarbon, Ozone, Smog, Acid rain (H2SO4, HNO3, H2CO3), PAN (Peroxyacetyl Nitrate), PBN (Peroxybutyl Nitrate), PPN (Peroxypropyl Nitrate)

Hence,

Lead is a particulate matter (PM) and its main source in atmosphere is automobiles.

SO₂ is first abundant atmospheric contaminant in many cities. It is produced by chemical interaction between sulphur and oxygen.

Explanation:

Basic Runway length:

It is the length of runway under the following assumed conditions at the aircraft

(i) Airport altitude at sea level.

(ii) Temperature at the airport is standard (15° C)

(iii) Runway is leveled in the longitudinal direction.

(iv) No wind is blowing on runway.

(v) Aircraft is loaded to its full loading capacity.

(vi) There is no wind blowing enroute to the destination.

(vii) Enroute temperature is standard.

Correction for Elevation, Temperature and Gradient:

(i) Correction for Elevation: Basic runway length is increased at the rate of 7% per 300 m rise in elevation above the mean sea level.

(ii) Correction for Temperature: 

Airport reference temperature = ${T_a} + \frac{{{T_m} - {T_a}}}{3}$

Where, Ta = Monthly mean of average daily temperature

Tm = Monthly mean of the maximum daily temmperature for the same month of the year

Total correction for elevation plus temperature ⇒ 35% of basic runway length

(c) Correction for Gradient:

(i) Steeper gradient results in greater consumption of energy and as such longer length of runway is required to attain the desired ground speed.

(ii) After having been corrected for elevation and temperature should be further increased at the rate of 20% for every 1% of effective gradient.

Explanation:

Ferrous metals (P) → Magnetic separator (iv)

• Ferrous metals (like iron and steel) are separated using a magnetic separator.

Aluminum and copper (Q) → Eddy current separator (iii)

• Non-ferrous metals (like aluminum and copper) are separated using eddy current separators.

Food waste (R) → Rapid composting (ii)

• Organic waste like food is treated using rapid composting.

Cardboard (S) → Incineration (i)

• Cardboard can be incinerated as part of waste-to-energy processes.

Explanation:

Given data:

Characteristic strength = 30 MPa

No. of cubes = 40

Hence, Characteristic strength is the strength below which not more than 5% of the test results are expected to fall.

95% of the test results are having strength ≥ characteristic strength.

0.95 × 40 = 38

38 = 38

The minimum number of specimens having at least 30 MPa strength is = 38

Explanation:

1. Understanding Under-Reinforced Sections (Reference: Limit State Design of Reinforced Concrete by Varghese & IS 456:2000)

• A balanced section is when both concrete in compression and steel in tension reach their limit state at the same time.
• A section becomes under-reinforced when steel in tension yields before the concrete in compression fails. This makes the failure ductile, which is desirable for safety.

2. Effect of Changes in Reinforcement Area

• Decreasing the area of tension reinforcement
  

Correct (Under-Reinforced) → Less steel means it reaches yield stress earlier, leading to ductile failure.

• Decreasing the area of compression reinforcement
  

Incorrect (Not necessarily Under-Reinforced) → This affects compression strength, but the section may still remain balanced or shift towards an over-reinforced state.

• Increasing the area of tension reinforcement
  

Incorrect (Over-Reinforced) → More steel means higher resistance, making the section over-reinforced.

• Increasing the area of compression reinforcement
  

Correct (Under-Reinforced) → Adding compression reinforcement improves the compression zone, making the steel in tension yield earlier, shifting failure towards a ductile mechanism.

Explanation:

Step 1: Understanding the Given Stress States

• P: Pure shear stress (equal and opposite shear forces) → Mohr’s circle should be symmetric about the τ-axis.
• Q: Equal biaxial tension (same normal stress in both directions) → Mohr’s circle should be centered at the same positive stress value.
• R: Uniaxial tensile stress in one direction → Mohr’s circle should have one normal stress component.
• S: Equal biaxial compression (same normal stress in both directions but negative) → Mohr’s circle should be a point on the σ-axis.

Step 2: Matching with Mohr’s Circles

• P (Pure shear) → III (Circle symmetric about τ-axis).
• Q (Equal biaxial tension) → IV (Circle centered at positive σ with equal principal stresses).
• R (Uniaxial tension) → I (Circle with normal stress +10).
• S (Equal biaxial compression) → II (Single point on σ-axis at -10).

Explanation:

As per given data:

\(\rm \frac{T_{P}}{T_{R}}=4\ &\ \frac{J_{1}}{J_{2}}=2\)

As P & R are fixed supports,

Hence, 

|ϕ_QP| = |ϕ_RQ|

\(\rm \frac{T_{\mathrm{P}} \times \mathrm{L}{1}}{J{1} \times G_{1}}=\frac{T_{R} \times L_{2}}{G_{2} \times J_{2}}\)

$\rm \Rightarrow \frac{L_{1}}{L_{2}}=\frac{T_{R}}{T_{P}} \times \frac{J_{1}}{J_{2}} \times\left(\frac{G_{P}}{G_{2}}\right)^{1}$

$\rm \frac{L_{1}}{L_{2}}=\frac{1}{4} \times 2=\frac{1}{2}=0.50$

CONCEPT:

The relationship between speed ( u )  and density ( k ) is given by the

\({\bf{u}} = {{\bf{u}}{\bf{f}}} - \left( {\frac{{{{\bf{v}}{\bf{f}}}}}{{{{\bf{k}}_{\bf{j}}}}}} \right) \times {\bf{k}}\)

u - Mean speed at density k ( m/s )

k- Density of stream ( veh/km )

uf - Free mean speed

kj – Jam density

Relation between flow, speed, and density is given by,

q = k × u

q – Flow given in veh/hr

Space headway is defined as the distance between corresponding points of two successive vehicles at any given time. It can also be taken as the average space occupied by each vehicle.

${\bf{k}} = \frac{{1000}}{{{\bf{Space}};{\bf{headway}}}}$

At capacity or maximum flow,

\({{\rm{k}}0} = \frac{{{{\rm{k}}{\rm{j}}}}}{2}\)

${u_0} = \frac{{{{\rm{u}}_{\rm{f}}}}}{2}$

\({{\bf{q}}_{{\bf{max}}}} = {{\bf{k}}0} \times {{\bf{u}}0} = \frac{{{{\bf{k}}{\bf{j}}} \times {{\bf{u}}{\bf{f}}}}}{4}\)

qmax – Maximum flow or flow at capacity

k0, u0 – Density and speed at capacity

Calculation:

$\rm q_{\max }=\frac{V_{f} K_{j}}{4}=\frac{60 \times \frac{1000}{8}}{4}=1875 \mathrm{veh} / \mathrm{hr}$

Explanation:

As per Given data:

$\rm \vec{v}=3 \hat{i}+(5 x) \hat{j}$

Here, u = 3, v = (5x)

Check for incompressibility 

$\rm \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=\frac{\partial(3)}{\partial x}+\frac{\partial(5 x)}{\partial y}=0$

Since, $\rm \frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

Fluid is incompressible.

Check for rotationality

$\rm \omega_{z}=\frac{1}{2}\left[\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y}\right]$

$\rm =\frac{1}{2}\left[\frac{\partial(5 x)}{\partial x}-\frac{\partial(3)}{\partial y}\right]$

$=\frac{1}{2}[5-0]=2.5 \neq 0$

So, flow is rotational.

Solution:

To verify the given statements about vector fields and scalar fields:

Given Data:

Scalar field $r$ and vector field $\mathbf{p}$ are defined as follows:

$r = 6x^2 + 4y^2 - z^2 - 9yz - 2xy + 3xz - yz$

$\mathbf{p} = (2x^2 - 3xy + z^2) \mathbf{i} + (2y^2 - 3yz + x^2) \mathbf{j} + (2z^2 - 3xz + x^2) \mathbf{k}$

Step 1: Analyze Statement P (Curl of the gradient of the scalar field $r$ is a null vector)

Compute the gradient of scalar field $r$:

$\nabla r = (12x - 2y + 3z) \mathbf{i} + (8y - 10z - x) \mathbf{j} + (-2z - 9y + 3x) \mathbf{k}$

Compute the curl of this gradient vector:

$\nabla \times (\nabla r) = (0, 0, 0)$ (Since the curl of any gradient is always zero.)

Statement P is confirmed true.

Step 2: Analyze Statement Q (Divergence of curl of the vector field $\mathbf{p}$ is zero)

Compute the curl of vector field $\mathbf{p}$:

$\nabla \times \mathbf{p} = (3y, 5z, 4x)$

Compute the divergence of this curl vector:

$\nabla \cdot (\nabla \times \mathbf{p}) = 0$ (As the divergence of any curl is always zero.)

Statement Q is confirmed true.

Conclusion: Both statements P and Q are true based on vector calculus identities.

Correct Option: Both P and Q are true.

Explanation:

Given data

Emission standard (based on 11% oxygen)

For HCl = 50 mg/Nm³

The measured concentration of HCl and oxygen in flue gas are 42 mg/Nm³ and 13% respectively.

As per CPCB,

The calculated emission concentration at the standard percentage oxygen concentration (ES)

$\mathrm{ES}=\frac{21-\mathrm{OS}}{21-\mathrm{OM}} \times \mathrm{EM}$

EM = Measured emission concentration at the standard percentage oxygen concentration

OS = Standard oxygen concentration

OM = Measured oxygen concentration

So, 

$\mathrm{ES}=\frac{21-11}{21-13} \times 42=52.5 \mathrm{mg} / \mathrm{Nm}^{3}>50 \mathrm{mg} / \mathrm{Nm}^{3}$

Here, no compliance as the corrected HCl emission is greater than the emission standard.

Solution:

To calculate the eigenvalues of the given matrix:

Given Data:

$A = \begin{bmatrix} 2 & 1 & 1 \ 1 & 4 & 1 \ 1 & 1 & 2 \end{bmatrix}$

Step 1: Write the characteristic equation

$\text{det}(A - \lambda I) = 0$

Subtract $\lambda$ along the diagonal elements:

$\text{det} \begin{bmatrix} 2-\lambda & 1 & 1 \ 1 & 4-\lambda & 1 \ 1 & 1 & 2-\lambda \end{bmatrix} = 0$

Step 2: Expand the determinant

Expand using the first row:

$(2-\lambda) \cdot \text{det} \begin{bmatrix} 4-\lambda & 1 \ 1 & 2-\lambda \end{bmatrix} - 1 \cdot \text{det} \begin{bmatrix} 1 & 1 \ 1 & 2-\lambda \end{bmatrix} + 1 \cdot \text{det} \begin{bmatrix} 1 & 4-\lambda \ 1 & 1 \end{bmatrix} = 0$

Simplify the determinant step-by-step to get the cubic equation:

$-\lambda^3 + 9\lambda^2 - 26\lambda + 30 = 0$

Step 3: Solve the cubic equation

Using numerical methods or a solver, the roots of the equation are:

$\lambda_1 = 5, , \lambda_2 = 1, , \lambda_3 = 2$

The eigenvalues of the matrix are 5, 1, and 2.

Explanation:

• Swell potential of soil decreases with an increase in the shrinkage limit.
• Correct: The shrinkage limit is the maximum water content at which further reduction in water content does not lead to a reduction in the volume of the soil. A higher shrinkage limit indicates that the soil can retain more water without additional volume change, thus leading to a decrease in swell potential as the soil's capacity to absorb water decreases.
• In electrical resistivity tomography, the depth of current penetration is half of the spacing between the electrodes.
• Incorrect: Typically, the depth of current penetration in electrical resistivity tomography is approximately equal to the spacing between the electrodes, not half.
• Both loose and dense sands with different initial void ratios can attain similar void ratio at large strain during shearing.
• Correct: Both initially loose and initially dense sands can approach a constant void ratio, known as the critical void ratio, when subjected to large shearing strains. This behavior is a characteristic of granular materials like sand under shearing stress.
• Among the several corrections to be applied to the SPT-N value, the dilatancy correction is applied before all other corrections.
• Incorrect: Dilatancy correction is generally applied after corrections for overburden pressure have been made to the SPT-N values.

Explanation:

As per given data:

$\text { Scale } \Rightarrow \frac{1}{1000}$

Horizontal distance on map = 10 mm

Now, Horizontal distance between two contour on ground

$=\frac{10}{\text { Scale }}=10 \times \frac{1000}{1} \mathrm{~mm}=10 \mathrm{~m}$

Hence,$\text { Slope }=\frac{\text { Vertical }}{\text { Horizontal }} \times 100=\frac{1 \mathrm{~m}}{10 \mathrm{~m}} \times 100=10 \%$

Explanation:

As per given data:

(ϕ) friction angle = 30°

For active state

Failure plane inclination with major principal plane (horizontal)

$\Rightarrow \theta_{c}=45^{\circ}+\frac{\phi}{2}=45^{\circ}+\frac{30^{\circ}}{2}$

θc = 60° 

For passive state

Failure plane inclination with major principal plane

$\left(\theta_{\mathrm{C}}\right)=45^{\circ}+\frac{\phi}{2}$

$\theta_{\mathrm{c}}=45^{\circ}+\frac{30^{\circ}}{2}=60^{\circ}$

Hence inclination of failure planes with major principal plane for both active & passive conditions is 60°

Explanation:

As per the given details:

Bending Moment at hinge = 0 

$\Rightarrow \mathrm{R}_{\mathrm{A}} \ell-\frac{\mathrm{q} \ell^{2}}{2}=0$

$\mathrm{R}_{\mathrm{A}}=\frac{\mathrm{q} \ell}{2}$ upward i.e. (+)ve

Now, equilibrium in the vertical direction

∑F_V = 0 ⇒ R_A + R_B + R_C = 2qℓ

⇒ R_B + R_C = 2qℓ - R_A

$=2 \mathrm{q} \ell-\frac{\mathrm{q} \ell}{2}=1.5 \mathrm{q} \ell$ …(A)

Moment equilibrium at B

∑M_B = 0

⇒ RA(2ℓ) - q(2ℓ)ℓ - Rcℓ = 0

$\frac{\mathrm{q} \ell}{2}(2 \ell)-2 \mathrm{q} \ell^{2}-\mathrm{R}_{\mathrm{C}} \ell=0$

-qℓ² - R_Cℓ = 0

R_C = qℓ

⇒ RC = qℓ downward

⇒ From (A), R_B = 1.5qℓ + qℓ = 2.5qℓ

⇒ R_B = 2.5 qℓ upward i.e. (+)ve