• Types of Shallow Foundations:

41. What is the purpose of a foundation in building construction, and what are
the common types of foundations used? a) Spread Footing (Isolated Footing):
Purpose of a Foundation in Building Construction: o Supports individual columns and spreads the load over a wide
A foundation is the lowest part of a building structure that transfers loads safely to area.
the ground. It plays a vital role in ensuring the stability, strength, and durability of o Common in reinforced concrete structures.
the building. b) Strip Footing:

The key purposes of a foundation are: o Continuous footing under load-bearing walls.
o Used in residential and commercial buildings.
1. Load Distribution:
c) Raft or Mat Foundation:
o The foundation spreads the weight of the structure over a large area
o A thick concrete slab that covers the entire area of the building.
to prevent excessive stress on the soil.
o Used where soil has low bearing capacity to distribute loads over a
o This prevents overloading and settlement of the building.
large surface.
2. Structural Stability:
2. Deep Foundations:
o It provides firm anchorage to the structure, making it resistant to
wind, earthquakes, and vibrations. Used when the surface soil has low strength, and the load needs to be transferred to
o Prevents overturning and lateral movement due to external forces. deeper, stronger strata.
3. Prevention of Differential Settlement:
• Suitable for: Skyscrapers, bridges, heavy industrial structures.
o Differential settlement occurs when different parts of a structure
sink unevenly, leading to cracks and structural failure. • Types of Deep Foundations:
o The foundation ensures uniform load transfer, reducing risks. a) Pile Foundation:
4. Moisture Protection:
o Long cylindrical columns made of concrete, steel, or timber driven
o Prevents groundwater infiltration into the structure, which can deep into the ground.
cause dampness, corrosion, and weakening of materials. o Used in soft clay, loose sand, or waterlogged areas.
5. Protection from Frost Action: b) Pier Foundation:
o In cold regions, freezing and thawing of soil can lead to frost o Large diameter vertical columns drilled deep into the ground.
heaving, causing foundation movement.
o Used in massive structures like towers and bridges.
o A well-designed foundation prevents such damage.
c) Caisson Foundation:
Common Types of Foundations Used in Construction:
o Used for underwater construction (e.g., bridges, piers, ports).
Foundations are classified into Shallow Foundations and Deep Foundations, o A large watertight chamber is placed underwater to support the
depending on the depth and load-bearing capacity of the soil. structure.

1. Shallow Foundations:

Used when good soil is available near the surface (up to 3m depth).

• Suitable for: Light to medium structures (e.g., houses, small buildings).

42. What is the significance of "Sustainable Construction Materials" in modern • Recycled Steel & Aluminum (low-energy alternatives).
building practices?
43. Describe the key characteristics and advantages of steel as a construction
What are Sustainable Construction Materials? material.
Sustainable construction materials are those that reduce environmental impact while Key Characteristics of Steel:
maintaining high performance in buildings. They are designed to: 1. High Strength-to-Weight Ratio:
• Minimize resource depletion (e.g., recycled materials). o Steel is stronger than most materials while remaining lightweight.
• Improve energy efficiency (e.g., insulated materials). o Used in high-rise buildings, bridges, and industrial structures.
• Reduce carbon footprint (e.g., low-emission production methods). 2. Ductility and Toughness:
Significance of Sustainable Construction Materials: o Can bend or stretch without breaking, providing earthquake
1. Energy Efficiency: resistance.
o Absorbs energy, preventing brittle failure.
o Green materials like aerated concrete blocks improve insulation,
3. Durability and Corrosion Resistance:
reducing heating and cooling costs.
o Cool roofing materials reflect sunlight, reducing urban heat effects. o Stainless steel contains chromium, preventing rusting.
2. Reduced Carbon Emissions: o Can last decades with proper maintenance.
4. Weldability and Fabrication:
o Traditional cement production contributes 8% of global CO₂
emissions. o Easily cut, shaped, and welded into complex structures.
o Using alternatives like fly ash-based cement or geopolymer o Prefabrication allows fast construction.
concrete significantly reduces CO₂ output.
5. Recyclability and Sustainability:
3. Resource Conservation:
o 100% recyclable without losing strength.
o Bamboo, recycled steel, and reclaimed wood reduce deforestation
o Reduces the environmental impact of construction.
and mining.
o Hempcrete (hemp-based concrete) provides carbon Advantages of Steel in Construction:
sequestration benefits.
4. Health Benefits: ✔ 1. Structural Efficiency:

o Low-VOC paints and adhesives improve indoor air quality. • High load-bearing capacity.
o Non-toxic insulation materials prevent health hazards. • Allows for large spans and open spaces (e.g., airports, stadiums).
5. Waste Reduction & Circular Economy:
✔ 2. Speed of Construction:
o Recycled plastic bricks and reused demolition debris reduce
landfill waste. • Prefabricated steel components allow rapid assembly.
o Encourages circular construction (materials can be reused in future • Reduces labor costs and project timelines.
projects).
✔ 3. Design Flexibility:
Examples of Sustainable Construction Materials:
• Can be molded into various shapes, allowing modern architectural designs.
• Green Concrete (uses fly ash, slag, or rice husk ash).
• Used in curved structures, cantilevers, and lightweight bridges.
• Bamboo & Engineered Wood (rapidly renewable).
• Solar Panels & Smart Glass (energy efficiency). ✔ 4. Seismic and Fire Resistance:
 • Ductility makes it suitable for earthquake-prone zones. Where:
• Fire-resistant steel can withstand high temperatures.
• ε (Strain) = Change in length/original length.
✔ 5. Sustainable and Cost-Effective: • ΔL (Change in length) = Deformation due to force.
• L (Original length) = Initial length of material.
• Reduces material waste.
• Long lifespan with minimal maintenance. Types of Strain:
✔ Tensile Strain – Increase in length under tensile stress.
Applications of Steel in Construction: ✔ Compressive Strain – Decrease in length under compressive stress.
✔ Skyscrapers (Burj Khalifa, Empire State Building). ✔ Shear Strain – Angular distortion under shear stress.
✔ Bridges (Golden Gate Bridge, Sydney Harbour Bridge). 3. Stress-Strain Curve (Material Behavior)
✔ Industrial Buildings (Warehouses, Factories).
• Elastic Region: Material returns to original shape after removing force.
44. Explain the concept of stress and strain in materials. • Plastic Region: Permanent deformation occurs.
• Ultimate Stress Point: Maximum stress the material can withstand.
Definition of Stress and Strain
• Fracture Point: Material breaks under extreme stress.
Stress and strain are fundamental concepts in Strength of Materials and structural
engineering. They describe how materials behave under applied forces. Applications of Stress & Strain in Engineering:

1. Stress (σ): ✔ Used in material selection (e.g., steel vs. concrete in bridges).
✔ Determines load-bearing capacity in buildings.
Stress is the internal resistance developed within a material when subjected to an ✔ Essential in designing earthquake-resistant structures.
external force. It is expressed as force per unit area:
45. How does the "method of joints" contribute to analyzing truss structures?
𝐹
𝜎= What is the "Method of Joints"?
𝐴
Where: The method of joints is a structural analysis technique used to determine forces in
individual truss members by solving equilibrium equations at each joint.
• σ (Stress) = Force per unit area (N/m² or Pascal).
• F (Force applied) = External force acting on the material (N). Key Assumptions:
• A (Cross-sectional area) = Area over which the force is applied (m²). ✔ Truss members are pin-jointed (ideal assumption).
Types of Stress: ✔ Loads are applied only at joints.
✔ Tensile Stress (Pulling force) – Increases length. ✔ Weight of members is negligible compared to external loads.
✔ Compressive Stress (Pushing force) – Reduces length.
Steps in the "Method of Joints" Analysis:
✔ Shear Stress (Sliding force) – Causes deformation along the plane.
Step 1: Identify a Joint with at Most Two Unknown Forces
2. Strain (ε):
• Start from a joint where only two unknown forces exist.
Strain is the measure of deformation (change in shape or size) due to applied stress.
It is a dimensionless ratio given by: • Use Newton’s Second Law (ΣF = 0).

𝛥𝐿 Step 2: Apply Equilibrium Equations

𝜀= For each joint:
𝐿
• ΣFx = 0 (Sum of horizontal forces = 0).
 • ΣFy = 0 (Sum of vertical forces = 0). o Strong soils (rock, gravel) → Shallow foundations (spread
footing).
Step 3: Solve for Unknown Member Forces o Weak soils (clay, silt) → Deep foundations (pile foundation).
• If force direction pulls the joint, it's tension (+ve). 2. Load & Structural Type:
• If force direction pushes the joint, it's compression (-ve). o Light buildings → Strip or raft foundations.
Step 4: Repeat for Other Joints o Heavy structures → Pile or caisson foundations.
Continue the process for remaining joints until all forces are found. 3. Water Table Level:

Importance of the "Method of Joints" in Structural Analysis: o High water table → Deep foundation required.
o Prevents foundation settlement & flooding issues.
✔ Determines internal forces in trusses (e.g., bridges, towers). 4. Seismic & Wind Considerations:
✔ Helps in optimizing truss design for strength & weight efficiency.
✔ Used in aerospace, mechanical, and civil engineering projects. o Earthquake-prone areas → Deep foundations (pile, raft) for stability.
o High-wind zones → Strong anchoring required.
46. What is the purpose of a truss in structural engineering, and how does it 5. Cost & Availability of Materials:
differ from a beam?
o Shallow foundations are cheaper than deep foundations.
1. Purpose of a Truss in Structural Engineering:
o Budget constraints affect foundation choice.
A truss is a structural framework composed of triangular units designed to: 6. Site Constraints & Accessibility:
✔ Distribute loads efficiently with minimal material use.
✔ Provide stability & resistance against bending & shear forces. o Urban sites → Shallow foundation due to space limits.
✔ Support long spans (e.g., bridges, roofs, transmission towers). o Remote locations → Prefabricated foundations may be suitable.
Example of Foundation Selection:
2. Key Differences Between a Truss and a Beam:
Feature Truss Beam ✔ Skyscrapers → Pile Foundations (tall buildings).
Structural Made of interconnected triangular A solid horizontal ✔ Residential Houses → Strip Footings (low-rise buildings).
Form members. member.
48. What is the purpose of "soil testing" in foundation engineering, and what
Load Transfers loads through axial forces Transfers loads through types of tests are commonly performed?
Transfer (tension/compression). bending & shear forces.
Purpose of Soil Testing in Foundation Engineering
Weight Lighter, material-efficient. Heavier, uses more
material. Soil testing is a critical step in foundation engineering that determines the suitability
Applications Roofs, bridges, transmission towers. Floors, ceilings, bridges. of soil for supporting a structure. The primary purposes include:

Example: ✔ Assessing Soil Bearing Capacity – Determines the load-carrying capacity.

✔ Trusses are used in suspension bridges (Golden Gate Bridge). ✔ Understanding Soil Composition & Type – Identifies sand, clay, silt, or gravel
✔ Beams are used in floor slabs of buildings. content.
✔ Evaluating Moisture Content & Drainage Properties – Prevents issues like
47. What factors should be considered when selecting a suitable foundation type settlement & water logging.
for a building? ✔ Predicting Soil Behavior Under Load – Helps in selecting the right foundation
type (shallow or deep).
Key Factors Influencing Foundation Selection:
✔ Checking for Contaminants – Ensures soil is free from harmful chemicals
1. Soil Type & Bearing Capacity: affecting stability.
Common Soil Tests in Foundation Engineering 4. Seismic & Wind Considerations
1. Field Tests • Seismic-prone areas → Raft foundations to distribute loads.
• Cyclone-prone zones → Strong anchoring required.
✔ Standard Penetration Test (SPT): Measures soil resistance using a split-barrel
sampler. 5. Cost & Material Availability
✔ Cone Penetration Test (CPT): Determines soil strength by pushing a cone-shaped • Shallow foundations are cheaper but less effective for poor soil.
device into the ground.
• Deep foundations are expensive but required for high loads.
✔ Plate Load Test: Estimates soil bearing capacity by applying a load on a steel
plate. ✔ Example: Skyscrapers require pile foundations, while houses use strip footings.
2. Laboratory Tests 50. Explain the fracture in cast iron during compression test.
✔ Moisture Content Test: Measures the percentage of water in soil. Fracture Behavior of Cast Iron Under Compression
✔ Atterberg Limits Test: Determines plasticity index of fine-grained soils. • Cast iron is brittle, so it fails suddenly under load.
✔ Compaction Test (Proctor Test): Establishes optimum moisture content for • In compression, cast iron undergoes crushing failure.
maximum soil density.
• Fracture pattern: Exhibits conical fracture at 45° angles.
✔ Permeability Test: Assesses how well water flows through soil (important for
drainage). Reasons for Fracture in Cast Iron
3. Chemical Tests ✔ High Carbon Content (~2-4%) makes it brittle.
✔ Lack of Ductility – Cannot deform before breaking.
✔ pH Test: Checks soil acidity or alkalinity.
✔ Weak in Tension & Shear – Cracks propagate easily.
✔ Sulphate & Chloride Content Test: Ensures soil won’t corrode foundations.
Failure Types in Cast Iron
Importance of Soil Testing
• Prevents foundation failure & settlement. ✔ Brittle Fracture – Sudden failure without warning.
• Ensures cost-effective foundation design. ✔ Shear Failure – Due to diagonal shear forces.
• Essential for designing safe & durable structures.
✔ Example: Cast iron columns fail under heavy axial loads.
49. What are the key factors that influence the choice of foundation type for a
building? 51. Explain the phenomenon of bulking of sand.

1. Soil Conditions & Bearing Capacity What is Bulking of Sand?

• Strong soil (rock, gravel) → Shallow foundations (strip footings, raft Bulking is the increase in sand volume due to moisture presence.
foundation).
• Weak soil (clay, silt) → Deep foundations (pile, caisson). ✔ Occurs in fine sand with moisture content below 8%.
✔ Air voids form between sand particles, increasing volume.
2. Building Load & Structure Type
✔ Affects concrete mix design – leads to incorrect proportioning.
• Light structures (single-story houses) → Shallow foundation.
• Heavy structures (high-rise buildings, bridges) → Deep foundation. Bulking Effect on Construction

3. Groundwater Table ✔ Underestimation of Sand Volume – Leads to poor mix.

• High water table → Requires pile foundation to prevent water seepage. ✔ Incorrect Water-Cement Ratio – Reduces strength.
• Low water table → Allows for shallow foundation. ✔ Solution: Moisture correction is applied before using sand.
52. Write a short note on a parabolic arch subjected to pure compression due to ✔ Example: Skyscrapers, bridges, offshore structures.
UDL.
2. Caisson Foundation
Parabolic Arch & Uniformly Distributed Load (UDL)
✔ Large watertight chamber sunk into the ground.
✔ Parabolic arches are structurally efficient for UDL. ✔ Used for bridges, piers, and marine structures.
✔ Under pure compression, no bending moment develops. ✔ Types:
✔ All forces are axial, reducing stress concentration.
• Open caisson – Hollow structure filled with concrete.
✔ Example: Used in bridges, viaducts, and ancient masonry structures. • Pneumatic caisson – Used in underwater construction.
53. Explain the characteristic strength of concrete in an actual structure.
✔ Example: Bridge piers in deep water.
Definition of Characteristic Strength (f_ck)
3. Well Foundation
✔ Defined as the strength below which only 5% of test results fall.
✔ Deep well-like structures for bridge piers.
✔ Measured in MPa (N/mm²) – Standard 28-day cube strength.
✔ Used in riverbeds and marine works.
Factors Affecting Concrete Strength in Structures
✔ Example: Foundation of large suspension bridges.
✔ Mix Design – Water-cement ratio affects strength.
Comparison of Deep Foundations
✔ Curing Conditions – Insufficient curing reduces strength.
✔ Compaction – Poor compaction leads to weak concrete. Foundation Type Use Case Example
✔ Age of Concrete – Strength increases over time. Pile Foundation Weak soil at shallow depth Skyscrapers
Caisson Foundation Underwater construction Bridge piers
✔ Example: M25 grade concrete has a characteristic strength of 25 MPa.
Well Foundation Riverbed construction Large bridges
54. Explain permanent dimension changes due to loading of concrete.
Causes of Permanent Dimension Changes 56. Determine the bearing capacity of a square footing with dimensions 2m × 2m,
located at a depth of 1.5 meters in cohesive soil with a cohesion of 20 kPa and an
✔ Shrinkage – Reduction in volume due to drying.
angle of internal friction of 30 degrees. Use Terzaghi’s bearing capacity equation.
✔ Creep – Long-term deformation under sustained load.
✔ Thermal Expansion – Changes due to temperature fluctuations. Solution:

✔ Example: Concrete columns undergo creep in bridges over time. Terzaghi’s bearing capacity equation for a square footing is given by:

55. Explain the various types of deep foundation. 𝑞𝑢 = 𝑐𝑁𝑐 + 𝛾𝐷𝑓 𝑁𝑞 + 0.5𝛾𝐵𝑁𝛾

1. Pile Foundation Where:

✔ Used when soil is weak at shallow depth. • 𝑞𝑢 = Ultimate bearing capacity

✔ Transfers load to stronger soil or rock below. • 𝑐 = Cohesion of soil = 20 kPa
✔ Types: • 𝛾 = Unit weight of soil (assume 18 kN/m³ if not given)
• 𝐷𝑓 = Depth of footing = 1.5 m
• End-bearing pile – Load transferred to hard strata.
• 𝐵 = Width of footing = 2 m
• Friction pile – Load resisted by skin friction.
• 𝜙 = Angle of internal friction = 30°
 • 𝑁𝑐 , 𝑁𝑞 , 𝑁𝛾 = Bearing capacity factors for 𝜙 = 30∘ 𝑀𝑐
𝜎=
𝐼
From Terzaghi’s bearing capacity factor tables:
Where:
• 𝑁𝑐 = 30.14
• 𝑀 = Bending moment = 60 kNm = 60 × 10^3 Nm
• 𝑁𝑞 = 18.4
• 𝑏 = Width of beam = 300 mm = 0.3 m
• 𝑁𝛾 = 22.4
• 𝑑 = Depth of beam = 150 mm = 0.15 m
Substituting values: • 𝐼 = Moment of Inertia for a rectangular section:

𝑞𝑢 = (20 × 30.14) + (18 × 1.5 × 18.4) + (0.5 × 18 × 2 × 22.4) 𝑏𝑑 3 0.3(0.15)3

𝐼= = = 8.44 × 10−5 m4
12 12
𝑞𝑢 = 602.8 + 497.2 + 403.2
𝑑 0.15
• 𝑐 = Distance from neutral axis = = = 0.075 m
𝑞𝑢 = 1503.2 kPa 2 2

Ultimate Bearing Capacity: 1503.2 kPa Substituting values:

Safe Bearing Capacity (assuming factor of safety = 3): (60 × 103 )(0.075)
𝜎=
8.44 × 10−5
𝑞𝑢 1503.2
𝑞𝑠𝑎𝑓𝑒 = = = 501.07 kPa 4500
3 3 𝜎=
8.44 × 10−5
57. Calculate the water-cement ratio for a concrete mix with a target compressive
strength of 30 MPa, using Portland cement with a water requirement of 0.4 for a 𝜎 = 53.34 MPa
specific mix design.
Final Answer: Maximum bending stress = 53.34 MPa
Solution:
59. The volume of the cement required for 10m³ of brickwork in 1:6 cement
The water-cement ratio (w/c) is given by: mortar is approximately equal to?

𝑊𝑎𝑡𝑒𝑟 𝐶𝑜𝑛𝑡𝑒𝑛𝑡 Solution:

𝑤/𝑐 =
𝐶𝑒𝑚𝑒𝑛𝑡 𝐶𝑜𝑛𝑡𝑒𝑛𝑡
In a 1:6 mix, cement : sand = 1:6.
From empirical relationships, the w/c ratio for a 30 MPa strength concrete using
ordinary Portland cement (OPC) is approximately 0.45. ✔ Total mix proportion = 1 + 6 = 7

If the mix design specifies a water requirement of 0.4: ✔ Volume of cement in mortar = Total volume × (Cement ratio / Total
proportion)
w/c ratio = 0.4
Considering 30% mortar voids in brickwork, the actual mortar required is:
Final Answer: Water-cement ratio = 0.4
𝑉mortar = 10 × 0.3 = 3 m3
58. A steel beam with a length of 6 meters and a cross-sectional area of 300 mm ×
150 mm is subjected to a bending moment of 60 kNm. Calculate maximum Cement volume:
bending stress in the beam in MPa.
1
Solution: 𝑉cement = 3 × = 0.428 m3
7
The maximum bending stress is given by Bending Stress Formula: Since 1 bag of cement = 0.035 m³,
 0.428 Significance of Specific Energy
No. of bags = = 12.2 ≈ 12 bags
0.035 1. Critical Flow Condition: At minimum specific energy, the flow is in
critical state.
Final Answer: 12 Bags of Cement
2. Subcritical and Supercritical Flow:
60. A 300 mm square bearing plate settles by 15 mm in a plate load test on a o If E > E_{critical}, flow is subcritical (slow, deep).
cohesive soil when the intensity of loading is 0.2 N/mm². Calculate the settlement o If E < E_{critical}, flow is supercritical (fast, shallow).
of a prototype shallow footing 1m square under the same intensity of loading. 3. Hydraulic Jump Analysis: Used to determine changes in flow depth.
Solution: 4. Efficient Channel Design: Helps in designing energy-efficient channels.

Settlement in a plate load test follows Terzaghi’s Scale Factor Equation: 42. What is the significance of the Darcy-Weisbach equation in pipe flow
analysis?
𝐵𝑓 Darcy-Weisbach Equation
𝑆𝑓 = 𝑆𝑝 × ( )
𝐵𝑝
𝐿 𝑣2
Where: ℎ𝑓 = 𝑓
𝐷 2𝑔
• 𝑆𝑓 = Settlement of footing (to be calculated) Where:
• 𝑆𝑝 = Settlement of plate = 15 mm
• ℎ𝑓 = Head loss due to friction (m)
• 𝐵𝑓 = Width of footing = 1000 mm
• 𝑓 = Darcy-Weisbach friction factor (dimensionless)
• 𝐵𝑝 = Width of plate = 300 mm
• 𝐿 = Length of pipe (m)
1000 • 𝐷 = Diameter of pipe (m)
𝑆𝑓 = 15 × ( )
300 • 𝑣 = Velocity of fluid (m/s)
• 𝑔 = Acceleration due to gravity (9.81 m/s²)
𝑆𝑓 = 15 × 3.33 = 50 mm
Significance
Final Answer: Settlement of footing = 50 mm 1. Determining Frictional Losses: Used for calculating energy loss due to
friction.
41. Explain the concept of specific energy in open-channel flow.
2. Pipe Flow Design: Helps in selecting proper pipe diameters.
Concept of Specific Energy
3. Pump Power Calculation: Used in designing pumping systems.
Specific energy (𝐸) in open-channel flow is defined as the total energy per unit weight 4. Efficient Water Supply Systems: Helps in minimizing energy loss.
of water with respect to the channel bed. It is given by:
43. Explain the purpose of a hydraulic jump in open-channel flow and its
𝑣 2 applications.
𝐸 =𝑦+ Hydraulic Jump
2𝑔

Where: A hydraulic jump is a sudden transition from supercritical to subcritical flow,

causing an increase in depth and energy dissipation.
• 𝐸 = Specific energy (m)
Purpose of Hydraulic Jump
• 𝑦 = Depth of flow (m)
1. Energy Dissipation: Reduces kinetic energy in spillways.
• 𝑣 = Velocity of flow (m/s)
2. Erosion Control: Protects downstream channels from scouring.
• 𝑔 = Acceleration due to gravity (9.81 m/s²)
3. Mixing of Fluids: Helps in aeration of water in sewage treatment plants.
4. Flow Control: Regulates flow depth in irrigation canals.
44. Define the concept of "consumptive use" in crop irrigation. 3. Weed Control: Different crops prevent weed growth.
Consumptive Use (Evapotranspiration) 4. Improved Water Use: Reduces excessive water consumption.

It is the total amount of water used by plants for transpiration and evaporation 47. Explain the term "infiltration" in irrigation. What factors affect the rate of
from soil. infiltration?
Definition
𝐶𝑈 = 𝐸𝑇𝑐 = 𝐸𝑇𝑜 × 𝐾𝑐
Infiltration is the process by which water enters the soil surface and percolates
Where: downward.
• 𝐸𝑇𝑐 = Crop evapotranspiration Factors Affecting Infiltration
• 𝐸𝑇𝑜 = Reference evapotranspiration 1. Soil Texture: Sandy soils have high infiltration, clay soils have low.
• 𝐾𝑐 = Crop coefficient 2. Vegetation Cover: Roots create pathways for water movement.
Factors Affecting Consumptive Use 3. Soil Moisture Content: Dry soil absorbs more water initially.
1. Climatic Conditions (temperature, humidity). 4. Rainfall Intensity: Heavy rain causes runoff, reducing infiltration.
2. Type of Crop (water-intensive crops require more). 48. Explain the purpose and functioning of a wastewater treatment plant's
3. Soil Properties (porous soil absorbs more water). primary treatment stage.
4. Irrigation Methods (drip irrigation reduces water loss).
Purpose
45. Explain the term "duty" in irrigation. How is it calculated?
The primary treatment stage removes solid waste, grease, and sediments from
Definition of Duty wastewater.
"Duty" refers to the area of land (in hectares) that can be irrigated with a unit Functioning
discharge (in cubic meters per second) throughout the growing season. 1. Screening: Removes large debris.
𝐴 2. Grit Removal: Separates sand and gravel.
𝐷= 3. Sedimentation: Suspended solids settle down.
𝑄

Where: 49. What are the different types of cross-drainage works that are required when
a canal crosses a natural drainage?
• 𝐷 = Duty (hectares/cumec) 1. Aqueduct: Canal passes over a river.
• 𝐴 = Area irrigated (hectares) 2. Super Passage: River passes over a canal.
• 𝑄 = Discharge (cumec) 3. Syphon Aqueduct: Canal over a river with siphoning action.
Factors Affecting Duty 4. Level Crossing: Canal and river meet at the same level.
1. Soil Type: Sandy soil requires more water. 50. Explain energy possessed by liquid in flow.
2. Climate: High temperatures increase water demand. 𝑣2
3. Crop Type: Rice requires more water than wheat. 1. Kinetic Energy ( ): Due to velocity.
2𝑔
4. Irrigation Efficiency: Lining of canals improves duty. 2. Potential Energy (𝑧): Due to height.
𝑃
46. What is the primary purpose of a "crop rotation" strategy in irrigation 3. Pressure Energy ( ): Due to pressure forces.
𝛾
agriculture?
51. Explain classification of triangulation.
Purpose of Crop Rotation
1. Soil Fertility Maintenance: Different crops use different nutrients. 1. Primary Triangulation: Large-area surveys.
2. Pest and Disease Control: Reduces soil-borne pests. 2. Secondary Triangulation: Smaller accuracy than primary.
 3. Tertiary Triangulation: Used in local surveys. Final Answer
52. Write the disadvantages of direct shear test. The ruling design radius of the curve is 357.8 m.
1. No Control Over Drainage: Cannot measure pore water pressure.
57. What is the true area (in acres), if the area calculated by the chain which is
2. Non-Uniform Stress Distribution: Causes errors. found to be 0.8 link too long is 100 acres?
3. Shear Plane is Predefined: Weakens accuracy.
Formula for Corrected Area
53. What are the various causes of failure of weirs?
𝐿𝑐 2
1. Piping Failure: Due to seepage. 𝐴𝑐 = 𝐴𝑚 × ( )
𝐿𝑚
2. Scour Failure: Erosion of foundation.
3. Sliding Failure: Due to water pressure. Where:
54. What is the effect of temperature on viscosity of fluid? • 𝐴𝑐 = True area
1. In Liquids: Viscosity decreases with temperature. • 𝐴𝑚 = Measured area = 100 acres
2. In Gases: Viscosity increases with temperature. • 𝐿𝑐 = Correct chain length = standard length (1 link = 20 cm)
55. What are the properties of flow net? • 𝐿𝑚 = Measured chain length = (1 + 0.8/100) = 1.008 links
1. Flow and Equipotential Lines are Perpendicular. Substituting Values
2. Equal Energy Drop in Each Flow Channel. 1 2
3. Curvilinear Squares are Formed. 𝐴𝑐 = 100 × ( )
1.008
56. The ruling design speed on a curve is 100 km/h and the superelevation on the
𝐴𝑐 = 100 × (0.992)2
curve is 7%. Calculate the ruling design radius (m) of the curve. Take the
coefficient of lateral friction as 0.15. 𝐴𝑐 = 100 × 0.984 = 98.4 acres
Formula for Radius of a Horizontal Curve
Final Answer
𝑉2
𝑅= The true area is 98.4 acres.
127(𝑒 + 𝑓)
58. Calculate the difference in pressure head, measured by a mercury-water
Where:
differential manometer for a 20 cm difference of mercury head.
• 𝑅 = Radius of curve (m) Formula for Pressure Head Difference
• 𝑉 = Speed (km/h) = 100 km/h Difference in mercury head × Specific gravity of mercury
• 𝑒 = Superelevation = 7% = 0.07 ℎ=
Specific gravity of water
• 𝑓 = Coefficient of lateral friction = 0.15
Substituting Values Given:

1002 • Difference in mercury head = 20 cm = 0.2 m

𝑅= • Specific gravity of mercury = 13.6
127(0.07 + 0.15)
10000 ℎ = 0.2 × 13.6
𝑅=
127 × 0.22 ℎ = 2.72 m
10000
𝑅= = 357.8 m
27.94
Final Answer 𝑉 = 0.549 × 1.113 = 0.611 m/s

The pressure head difference is 2.72 m of water. 2. Hydraulic Mean Depth

59. Determine the Biochemical Oxygen Demand (BOD) removal efficiency of a 1/3
𝑄2
wastewater treatment plant. If the influent BOD is 200 mg/L and the effluent 𝑅 = 0.47 × ( )
𝑓
BOD is 20 mg/L, what is the percentage removal?
Formula for BOD Removal Efficiency 1/3
52
𝑅 = 0.47 × ( )
Influent BOD − Effluent BOD 1.24
BOD Removal Efficiency = ( ) × 100
Influent BOD
𝑅 = 0.47 × (20.16)1/3
Given:
𝑅 = 0.47 × 2.73 = 1.28 m
• Influent BOD = 200 mg/L
3. Wetted Perimeter
• Effluent BOD = 20 mg/L
200 − 20 𝑃 = 4.75√𝑄
BOD Removal Efficiency = ( ) × 100
200
𝑃 = 4.75 × √5
180
=( ) × 100 𝑃 = 4.75 × 2.236 = 10.63 m
200
= 90% 4. Channel Width

Final Answer 𝐵 = 5.4𝑄1/2

The BOD removal efficiency is 90%. 𝐵 = 5.4 × √5

60. Design a Lacey channel to carry 5m³/sec through 0.5 mm sand. 𝐵 = 5.4 × 2.236 = 12.07 m
Lacey's Silt Factor Formula Final Design Parameters
𝑓 = 1.76√𝑑 • Velocity = 0.611 m/s
• Hydraulic Mean Depth = 1.28 m
Where: • Wetted Perimeter = 10.63 m
• 𝑑 = Mean diameter of silt in mm = 0.5 mm • Channel Width = 12.07 m
1. What is concrete? Define the initial setting time of cement.
𝑓 = 1.76√0.5
Concrete is a mixture of cement, sand, aggregates, and water that hardens over time.
𝑓 = 1.76 × 0.707 = 1.24 The initial setting time of cement is the time taken for it to start hardening and lose
Lacey’s Regime Relations plasticity, typically 30 minutes for Ordinary Portland Cement (OPC).
1. Mean Velocity

𝑉 = 0.549√𝑓

𝑉 = 0.549 × √1.24
2. Describe the term ‘Beam of Uniform Strength’. What is the common method 8. Define Bond Length of Steel Bar. Why is it important?
of obtaining it?
Bond length is the length of a steel bar required for effective stress transfer between
A Beam of Uniform Strength is designed so that its bending stress remains constant steel and concrete. It is important for avoiding slippage, ensuring structural
throughout its length. This is achieved by varying the cross-section or stability, and preventing cracks.
reinforcement, commonly done by increasing depth where bending moment is
higher. 9. What do you understand by Detailed Estimate? What is the importance of
preparing it?
3. Define Stress and Strain.
A Detailed Estimate includes quantities, costs, and specifications of all materials
• Stress is the internal resistance per unit area to an applied external force, and labor required. It is important for budget planning, cost control, resource
measured in N/m² (Pascal). allocation, and project execution.
• Strain is the deformation per unit length due to stress, having no unit.
10. How is the area required for Isolated Column Footing assessed? What are the
4. Describe Active Earth Pressure and Passive Earth Pressure. considerations for ascertaining its depth?
• Active Earth Pressure occurs when the soil mass moves away from a
retaining structure, reducing pressure. Area is determined by dividing the column load by the soil’s safe bearing capacity.
Depth depends on:
• Passive Earth Pressure occurs when the soil mass moves toward the
structure, increasing pressure. • Load intensity
5. What is a Foundation? Differentiate between Shallow and Deep Foundations. • Soil type
• Frost depth
A foundation transfers the structure’s load to the ground safely. • Settlement criteria
• Shallow Foundation: Depth ≤ Width, e.g., spread footing, raft. 11. Define Construction Planning. What are the methods of Construction
• Deep Foundation: Depth > Width, e.g., pile, pier. Planning and Scheduling?
6. What do you understand by Workability of Concrete? What are the common Construction Planning is organizing resources, materials, and timelines for efficient
factors affecting it? project execution.
Methods:
Workability is the ease with which concrete can be mixed, placed, and compacted
without segregation. • Bar charts
Factors affecting workability: • Critical Path Method (CPM)
• Water-cement ratio • PERT (Program Evaluation Review Technique)
• Aggregate size & shape 12. Define Safe Bearing Capacity of Soil. What is its importance in designing the
• Admixtures foundation of a structure?
• Mixing time
Safe Bearing Capacity (SBC) is the maximum load soil can carry without failure. It
7. What is a Singly Reinforced Beam? Why is Steel generally used in RCC work? is crucial for foundation design to prevent settlement, tilting, or structural failure.

A Singly Reinforced Beam has reinforcement only in the tension zone. 13. Define Pile Foundation. How does the pile transfer load to the soil?
Steel is used in RCC because it has high tensile strength, good bond with concrete,
and similar thermal expansion properties. A Pile Foundation is a deep foundation that transfers loads to deeper, stronger soil
layers. Load transfer occurs through:

• Friction (skin friction piles)

• End-bearing (load transferred to rock or hard soil)
 𝑊𝐿
• At Support B: 𝑉𝐵 = −
2

14. Why is Damp Proofing done in buildings? What are the general causes of Maximum Bending Moment:
dampness in a building?
The bending moment is maximum at the center of the beam:
Damp proofing prevents moisture penetration into buildings, protecting walls and
floors. 𝑊𝐿2
𝑀max =
Causes of dampness: 8

• Capillary action from groundwater Shear Force and Bending Moment Diagram:
• Poor drainage • Shear Force Diagram (SFD): A triangular variation from +WL/2 at A to -
WL/2 at B.
• Defective roofing
• Bending Moment Diagram (BMD): A parabolic curve with maximum
• Leakage in pipes
moment at midspan.
• Condensation due to poor ventilation
2. A hollow rectangular section is having outer width of 20 mm and depth of 12
15. Differentiate between Retaining Wall and Breast Wall. mm and the inner width of 16 mm and depth of 9 mm. Calculate the moment of
• Retaining Wall: Holds back soil or water at different levels, preventing inertia both along ‘X-axis’ and ‘Y-axis’ passing through its center.
landslides. Solution:
• Breast Wall: Supports cut slopes on roads, preventing erosion.
• Retaining walls bear active earth pressure, while breast walls face passive Moment of inertia for a rectangular section is given by:
pressure. 𝑏𝐷3 − 𝑏𝑑 3 𝐷𝑏 3 − 𝑑𝑏 3
𝐼𝑥 = , 𝐼𝑦 =
12 12
1. A simply supported beam of length of L mm is loaded with a uniformly Given Data:
distributed load of W N/mm over its entire length. Calculate the reactions,
• Outer dimensions: Width 𝑏 = 20 mm, Depth 𝐷 = 12 mm
maximum bending moment, and maximum shear force. Draw the bending
moment and shear force diagram. • Inner dimensions: Width 𝑑 = 16 mm, Depth 𝑑 = 9 mm
Solution: Moment of Inertia About X-axis:

For a simply supported beam subjected to a uniformly distributed load (UDL) W 20 × 123 − 16 × 93
𝐼𝑥 =
N/mm over its entire length L, we calculate: 12

Reactions at the Supports: 20 × 1728 − 16 × 729

𝐼𝑥 =
12
By symmetry, the reactions at both supports are:
34560 − 11664 22896
𝑊𝐿 𝐼𝑥 = = = 1908 mm4
12 12
𝑅𝐴 = 𝑅𝐵 =
2
Moment of Inertia About Y-axis:
Maximum Shear Force:
12 × 203 − 9 × 163
Shear force varies linearly. 𝐼𝑦 =
12
𝑊𝐿 12 × 8000 − 9 × 4096
• At Support A: 𝑉𝐴 = + 𝐼𝑦 =
2
12
• At Mid-span: 𝑉 = 0
 96000 − 36864 59136 4. A steel plate is bent into a circular arc of radius 10 meters. If the plate section
𝐼𝑦 = = = 4928 mm4 is 120 mm wide and 20 mm thick, find the maximum stress induced and the
12 12
bending moment which can produce this stress. Take E = 2 × 10⁵ N/mm².
3. A hollow circular mild steel column of external diameter 300 mm and internal
diameter 250 mm carries an axial load of 1500 KN. Determine the compressive Solution:
stress in the column. If the initial length of the column is 3.75m, find the decrease Bending Stress Formula:
in length of the column. Take E = 2×10⁵ N/mm².
Solution: 𝐸
𝜎= ×𝑦
𝑅
Compressive Stress Formula:
ℎ
where 𝑦 = is the distance from the neutral axis.
𝑃 2
𝜎=
𝐴 Given Data:
where A is the cross-sectional area of the hollow circular column: • Radius 𝑅 = 10 m = 10000 mm
𝜋 2 • Width 𝑏 = 120 mm
𝐴= (𝐷 − 𝑑 2 )
4 • Thickness ℎ = 20 mm
Given Data: • E = 2 × 10⁵ N/mm²
ℎ 20
• External Diameter 𝐷 = 300 mm 𝑦= = = 10 mm
2 2
• Internal Diameter 𝑑 = 250 mm
• Load 𝑃 = 1500 kN = 1500 × 103 N 2 × 105
𝜎= × 10 = 200 N/mm2
𝜋 10000
𝐴= (3002 − 2502 )
4 Bending Moment:
𝜋 𝐸𝐼
𝐴= (90000 − 62500)
4 𝑀=
𝑅
𝜋
𝐴= (27500) = 21600.47 mm2 Moment of Inertia for rectangular section:
4
Compressive Stress: 𝑏ℎ3 120 × 203
𝐼= = = 80000 mm4
12 12
1500 × 103
𝜎= = 69.44 N/mm2 2 × 105 × 80000
21600.47 𝑀= = 1600 kNmm
10000
Decrease in Length:
5. A hollow steel cantilever pipe of 2.4 metre length and 100 mm outer diameter
𝜎𝐿 is to support two concentrated loads of 2.15 KN each placed at center and free
𝛥𝐿 = end of the cantilever pipe. Determine the minimum thickness of the pipe if the
𝐸
allowable bending stress is 160 N/mm².
69.44 × 3750
𝛥𝐿 = Solution:
2 × 105
Maximum bending moment at fixed end:
𝛥𝐿 = 1.30 mm
𝑀 =𝑃×𝐿
Total load = 2.15 + 2.15 = 4.3 kN 5. Define Unit Hydrograph.

𝑀 = 4.3 × 2400 = 10320 Nmm A Unit Hydrograph is a hydrograph representing the runoff resulting from one unit
(1 cm) of effective rainfall occurring uniformly over a drainage basin for a specific
Section modulus: duration. It is used for flood forecasting and watershed analysis.
𝜋
𝑍= (𝐷4 − 𝑑 4 )/𝐷 6. What is the yield of a well? What tests are performed to find the yield?
32 Explain any one test.
Using 𝜎 = 𝑀/𝑍 and solving for t, • Yield of a well is the amount of water that can be extracted per unit time.
• Tests include: Pumping test, Recuperation test, and Step-drawdown test.
𝑡 ≈ 4.5 mm
• Recuperation Test: Measures the rise in water level after pumping stops,
helping estimate the recharge capacity.

1. What is an ideal fluid and its characteristics? 7. How does a flood control reservoir manage floods?

An ideal fluid is incompressible and has zero viscosity. It offers no internal resistance A flood control reservoir stores excess rainfall and releases it gradually, reducing
to flow and does not exhibit turbulence. It follows Bernoulli’s principle and is a peak flood discharge. It has spillways and gates to regulate flow, preventing
theoretical concept since all real fluids have some viscosity. downstream flooding and damage.

2. If a fluid flows through a pipe, what are the various energy losses in pipes? 8. Name some important primary air pollutants and secondary air pollutants.
• Primary Pollutants: Emitted directly from sources (e.g., CO, SO₂, NOx,
Energy losses in pipe flow include: Particulate Matter, Lead).
1. Major losses due to pipe friction (Darcy-Weisbach equation). • Secondary Pollutants: Formed by chemical reactions in the atmosphere
2. Minor losses due to bends, expansions, contractions, valves, and fittings. (e.g., Ozone, Smog, Peroxyacetyl Nitrate, Sulfuric acid).
These losses reduce fluid velocity and pressure. 9. Describe the preventive measures that can be adopted to tackle water
3. Write a short note on the number of methods employed to ascertain the pollution.
average velocity of flow in an open channel. • Industrial wastewater treatment
• Proper sewage disposal
Methods to determine velocity in an open channel include:
• Use of eco-friendly fertilizers and pesticides
• Surface Float Method • Rainwater harvesting
• Pitot Tube Method • Public awareness and regulation enforcement
• Velocity-area Method
10. Write a short note on sanitary landfill.
• Current Meter Method
• Acoustic Doppler Method A sanitary landfill is a controlled waste disposal method where garbage is compacted
Each method measures velocity at different depths to estimate the average in layers and covered with soil. It prevents groundwater contamination, controls odors,
velocity. and reduces pests. Leachate and gas management systems are used for
environmental safety.
4. What is evapotranspiration and effective rainfall?
11. Write important reasons for providing road drainage.
• Evapotranspiration: The combined loss of water from soil and plants due to
evaporation and transpiration. • Prevents water accumulation, reducing road damage.
• Effective Rainfall: The portion of total rainfall available for plant use after • Avoids pothole formation and structural failures.
losses due to surface runoff and deep percolation. • Enhances traffic safety by preventing skidding.
• Protects pavement strength and subgrade stability.
12. What are the prime causes of failures in sub-grade and sub-base of flexible • Discharge, 𝑄 =?
pavement? Solution:
• Poor drainage leading to water accumulation.
Using Darcy-Weisbach equation,
• Weak soil compaction causing settlement.
• Excessive traffic loads exceeding design limits. 4𝑓𝐿𝑉 2
• Frost action causing pavement cracks. ℎ𝑓 =
2𝑔𝑑
• Inferior materials in base/sub-base layers.
where,
13. What is the Uni-gauge policy of Indian Railways? 𝑔 = 9.81𝑚/𝑠 2 (acceleration due to gravity).
The Uni-gauge policy aims to standardize broad gauge (1676 mm) across India, Velocity,
replacing meter and narrow gauges. It improves interoperability, reduces
maintenance costs, and enhances speed and safety. ℎ𝑓 ⋅ 2𝑔 ⋅ 𝑑
𝑉=√
14. Write a list of instruments and materials used in plane table surveying. 4𝑓𝐿
• Instruments: Plane table, Alidade, Spirit level, Compass, U-fork with plumb
Substituting the given values:
bob, Tripod.
• Materials: Drawing sheets, Pencils, Pins, Clips, Erasers.
2.0 × 2 × 9.81 × 0.2
15. What are the methods for finding the reduced level of a point? Write in detail 𝑉=√
4 × 0.01 × 500
about any one method.

Methods: 7.848
𝑉=√ = √0.3924 = 0.626𝑚/𝑠
20
• Height of Instrument (HI) Method
• Rise and Fall Method Discharge,
Height of Instrument (HI) Method: 𝜋𝑑 2
𝑄 =𝐴⋅𝑉 = ×𝑉
• The instrument height is determined using benchmark elevation. 4
• Staff readings at various points are taken. 𝜋(0.2)2
𝑄= × 0.626
• The reduced level (RL) is found using: 4
𝑅𝐿 = 𝐻𝐼 − 𝑆𝑡𝑎𝑓𝑓𝑅𝑒𝑎𝑑𝑖𝑛𝑔 3.1416 × 0.04
𝑄= × 0.626
This method is simple, quick, and widely used in leveling. 4

1. The difference of head observed between two ends of a 500m long pipe with 𝑄 = (0.03142) × 0.626
200mm diameter is 2.0 metre. Neglecting minor losses, calculate the discharge
𝑄 = 0.0197𝑚3 /𝑠 = 19.7𝐿/𝑠
flowing through the pipe, considering Darcy’s coefficient of friction as 0.01.
Given Data: Final Answer: The discharge flowing through the pipe is 19.7 L/s.
• Pipe length, 𝐿 = 500𝑚
• Diameter, 𝑑 = 200𝑚𝑚 = 0.2𝑚
• Head difference, ℎ𝑓 = 2.0𝑚
• Darcy’s friction factor, 𝑓 = 0.01
2. The culturable command area of a water course is 1500 ha. Intensities of 𝑦0 = 0, 𝑦1 = 2.50, 𝑦2 = 3.50, 𝑦3 = 5.00, 𝑦4 = 4.60, 𝑦5 = 3.20, 𝑦6 = 0
irrigation for sugarcane and wheat crops are 30% and 40% respectively. The
duties of the crop at the head of the water course are 700 ha/cumec and 1600 Interval, ℎ = 10𝑚
ha/cumec respectively. Calculate the discharge required at the head of water
course. Solution Using Simpson’s Rule:

Given Data: ℎ
𝐴= [𝑦 + 4(𝑦1 + 𝑦3 + 𝑦5 ) + 2(𝑦2 + 𝑦4 ) + 𝑦6 ]
• Culturable command area (CCA) = 1500 ha 3 0
• Irrigation intensity: Substituting values:
o Sugarcane = 30%
o Wheat = 40% 10
𝐴= [0 + 4(2.50 + 5.00 + 3.20) + 2(3.50 + 4.60) + 0]
• Duties at the head of watercourse: 3
o Sugarcane = 700 ha/cumec 10
𝐴= [0 + 4(10.70) + 2(8.10) + 0]
o Wheat = 1600 ha/cumec 3
Solution: 10
𝐴= [42.8 + 16.2]
The discharge required is given by: 3

𝐶𝐶𝐴 × 𝐼𝑛𝑡𝑒𝑛𝑠𝑖𝑡𝑦 10
𝐷= 𝐴= × 59 = 196.67𝑚2
𝐷𝑢𝑡𝑦 3
Final Answer: The area is 196.67 m².
For sugarcane,
4. Estimate the theoretical maximum capacity of a traffic lane with one-way
1500 × 30
𝐷sugarcane = traffic flow at 50 kmph. Assume 𝑆𝑔 = 0.278𝑉𝑡, 𝑡 = 0.8𝑠, and vehicle length 𝑙 =
100 × 700 4.0𝑚.
450 Given Data:
𝐷sugarcane = = 0.6429 cumecs
700 • Speed of traffic, 𝑉 = 50𝑘𝑚𝑝ℎ = 13.89𝑚/𝑠
For wheat, • Reaction time, 𝑡 = 0.8𝑠
• Vehicle length, 𝑙 = 4.0𝑚
1500 × 40
𝐷wheat = Solution:
100 × 1600
600 Space gap,
𝐷wheat = = 0.375 cumecs
1600 𝑆𝑔 = 0.278 × 13.89 × 0.8
Total discharge required:
𝑆𝑔 = 3.08𝑚
𝐷total = 0.6429 + 0.375 = 1.0179 ≈ 1.02 cumecs
Total space occupied per vehicle:
Final Answer: The total discharge required is 1.02 cumecs.
𝑆total = 𝑆𝑔 + 𝑙 = 3.08 + 4.0 = 7.08𝑚
3. Compute the area between the chain line and the irregular boundary using
Simpson’s Rule. Traffic capacity:

Given Offsets: 3600 3600

𝑞= = = 508.47 ≈ 508 vehicles/hour
𝑆total 7.08
Final Answer: The maximum capacity is 508 vehicles/hour.

5. Determine the Plume height (Δh) of a stack using Holland’s equation.

Given Data:

• Stack height, 𝐻 = 200𝑚

• Stack gas velocity, 𝑉𝑠 = 12𝑚/𝑠
• Stack gas temperature, 𝑇𝑠 = 150∘ 𝐶
• Ambient temperature, 𝑇𝑎 = 30∘ 𝐶
• Wind velocity, 𝑈 = 3𝑚/𝑠
Solution Using Holland’s Formula:
1.6𝑉𝑠 𝐻
𝛥ℎ =
𝑈√𝑇𝑠 − 𝑇𝑎

1.6 × 12 × 200
𝛥ℎ =
3 × √150 − 30
3840
𝛥ℎ =
3 × 10.95
3840
𝛥ℎ = = 117.0𝑚
32.85
Final Answer: Plume height is 117m, so total stack height = 317m.