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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:

Step 1

x₀ = 1, y₀ = 0 

\(\rm \rightarrow \frac{y_{1}-y_{0}}{x_{1}-x_{0}}= 0.3010\)

\(\rm \frac{y_{(1,2)}-y_{(0,1)}}{x_{2}-x_{1}}=-0.1249\)

Step 2

x₁ = 2, y₁ = 0.3010

\(\rm \rightarrow \frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}=0.1761\)

Step 3

​The value of f(1.5) estimated using second-order Newton’s interpolation formula is

f(x) = y₀ + y_{(0, 1)}(x – x₀) + y_{(0, 12)}(x – x₀)(x – x₁) 

f(x) = 0 + 0.3010(x – 1) + (–0.1249) × (x – 1)(x – 2) 

f(x) = 0.3010(x – 1) – 0.1249(x – 1)(x – 2)

f(1.5) = 0.3010 × (1.5 – 1) – 0.1249(1.5 – 1)(1.5 – 2)

f(1.5) = 0.1505 + 0.03125 = 0.1817 ≃ 0.18

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:

Given data:

Degree of saturation (S) = 75%

Specific gravity of soil (G) = 2.68

Moisture content (W) = 17%

Unit weight of water (γ_w) = 10 kN/m³

As per the relation between S, e, W, and G

Se = WG

0.75 × e = 0.17×2.68

(e) void ratio = 0.607

Hence, The dry unit weight is given by

\(\rm \gamma_{d}=\frac{G \cdot \gamma_{w}}{1+e}=\frac{2.68 \times 10}{1+0.607}\)

γd = 16.68 kN/m³

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.