Junior Engineer • 2025 • PAPER-IV B (Technical Subject)
What is Poisson’s ratio?
A)
Ratio of lateral strain to axial strain
B)
Ratio of shear stress to shear strain
C)
Ratio of axial strain to lateral strain
D)
Ratio of stress to strain
Poisson's ratio (denoted by the Greek letter ν - nu) is a fundamental material property in the study of Strength of Materials. When a material is stretched in one direction (axial direction), it tends to contract in the other two directions (lateral directions), and vice versa.
1. **Definition**: Poisson's ratio is defined as the ratio of the lateral strain (strain perpendicular to the direction of loading) to the longitudinal or axial strain (strain in the direction of loading).
2. **Formula**: ν = -(Lateral Strain / Axial Strain). The negative sign is typically used in the formula because axial extension usually results in lateral contraction.
Evaluating the options:
- **Option a**: Correct. This matches the definition of Poisson's ratio.
- **Option b**: Incorrect. The ratio of shear stress to shear strain is defined as the Modulus of Rigidity (G).
- **Option c**: Incorrect. This is the reciprocal of Poisson's ratio.
- **Option d**: Incorrect. The ratio of stress to strain (within the elastic limit) is defined as Young's Modulus of Elasticity (E).
Junior Engineer • 2025 • PAPER-IV B (Technical Subject)
Why are under-sluices provided in weirs and barrages?
A)
To divert excess water during floods
B)
To control the movement of fish in the river
C)
To remove sediment from the riverbed and prevent silting
D)
To allow boats and ships to pass through
Under-sluices are gated openings provided at the lower level of a weir or barrage, typically located adjacent to the canal head regulator. Their primary engineering purpose is to create a 'scouring' effect. By opening these gates, water flows at a high velocity along the riverbed, which flushes out accumulated silt and sediment (sediment exclusion). This prevents the riverbed from rising (aggradation) and ensures that the water entering the canal is relatively free of sediment.
Analysis of other options:
- Option (a) is incorrect because excess floodwater is typically handled by the main spillway or crest gates of the barrage, not specifically by the under-sluices.
- Option (b) is incorrect because structures designed for fish movement are called 'fish ladders' or 'fish ways'.
- Option (d) is incorrect because structures allowing boats to pass are called 'locks'.
Junior Engineer • 2025 • PAPER-IV B (Technical Subject)
Why is a detailed estimate preferred over a preliminary estimate?
A)
Because it provides only an approximate cost
B)
Because it includes labor cost only
C)
Because it provides accurate cost analysis for execution
D)
Because it is easier to prepare than a preliminary estimate
In civil engineering estimation and costing, there are two primary stages of estimation: Preliminary (Approximate) and Detailed.
1. **Preliminary Estimate:** This is prepared in the initial stages of a project to obtain administrative approval. It is based on rough data, similar previous projects, or plinth area rates. Its purpose is to get a rough idea of the budget, not for actual construction execution.
2. **Detailed Estimate:** This is prepared after the detailed design and drawings are finalized. It involves the preparation of a 'Quantity Take-off' (Measurement Sheet) and a 'Detailed Abstract of Cost'. It accounts for every single item of work, including materials, labor, and overheads, based on the current Schedule of Rates (SOR).
**Analysis of Options:**
- **Option (a)** is incorrect because providing an approximate cost is the characteristic of a preliminary estimate, not a detailed one.
- **Option (b)** is incorrect because a detailed estimate includes all costs (materials, labor, plant, machinery, and contingencies), not just labor.
- **Option (c)** is correct because the primary purpose of a detailed estimate is to provide a precise and accurate cost analysis, which is essential for tendering, budgeting, and the actual execution of the work on site.
- **Option (d)** is incorrect because a detailed estimate is significantly more time-consuming and complex to prepare than a preliminary estimate.
Junior Engineer • 2025 • PAPER-IV B (Technical Subject)
Why is the moment method used in calculating earthwork?
A)
To reduce the cost of excavation
B)
To measure soil hardness
C)
To provide an efficient way to estimate volume changes in excavation
D)
To estimate the number of laborers needed
In civil engineering estimation and costing, the 'Moment Method' (also known as the Mid-Sectional Area Method or a variation of the Prismoidal/Trapezoidal rule application) is used to calculate the volume of earthwork for embankments, cuttings, and roads.
1. **Purpose**: Earthwork involves calculating the volume of soil to be excavated (cutting) or filled (filling). Because the cross-sectional area of a road or canal changes gradually along its length, a simple 'length x area' calculation is inaccurate.
2. **Mechanism**: The moment method involves taking cross-sections at regular intervals and calculating the 'moments' of these areas about a specific point or using the average of sectional areas to find the total volume. This provides a mathematically efficient and accurate way to estimate the total volume of material moved.
3. **Analysis of Options**:
- **Option a**: The method is for *calculating* the volume, not for *reducing* the actual cost of physical excavation.
- **Option b**: Soil hardness is measured by geotechnical tests (like SPT or CBR), not by a volumetric calculation method.
- **Option c**: This is correct as it describes the primary function of the method: estimating the volume of earthwork (excavation/filling).
- **Option d**: Labor estimation is a separate step in costing based on the total volume calculated, not the method used to find that volume.
Junior Engineer • 2025 • PAPER-IV B (Technical Subject)
Why is seepage control essential in earthen dams?
A)
To maintain a stable water level in the reservoir
B)
To prevent water loss and structural failure due to internal erosion
C)
To increase the storage capacity of the dam
D)
To improve the water quality in the reservoir
In earthen dams, seepage is inevitable because soil is porous. However, uncontrolled seepage can lead to a phenomenon known as 'piping' or internal erosion. Piping occurs when water flowing through the dam carries soil particles with it, creating hollow tubes or 'pipes' within the embankment. If left unchecked, these pipes enlarge, leading to a rapid loss of structural integrity and eventual collapse of the dam. Therefore, seepage control measures (such as clay cores, grout curtains, or chimney drains) are essential to ensure the stability of the structure and minimize water loss.
Analysis of other options:
- Option (a) is incorrect because water level is maintained by the dam's height and spillway management, not by seepage control.
- Option (c) is incorrect because storage capacity is determined by the reservoir's geometry and the dam's height, not by the control of seepage through the embankment.
- Option (d) is incorrect because seepage control relates to structural stability and water quantity, not the chemical or biological quality of the reservoir water.
Junior Engineer • 2025 • PAPER-IV B (Technical Subject)
Which type of cement is best suited for marine construction?
A)
Ordinary Portland Cement (OPC)
B)
Rapid Hardening Cement
C)
Sulphate Resistant Cement
D)
White Cement
Marine environments are characterized by high concentrations of chlorides and sulphates (specifically magnesium and sodium sulphates) present in seawater. When these sulphates react with the tricalcium aluminate (C3A) in standard cement, they form ettringite, which causes expansion, cracking, and eventual disintegration of the concrete (sulphate attack). Sulphate Resistant Cement (SRC) is specifically formulated with a low C3A content (typically below 5%) to minimize this reaction, making it the most durable choice for marine construction.
Analysis of other options:
- Ordinary Portland Cement (OPC): Susceptible to sulphate attack and chloride penetration, leading to reinforcement corrosion and concrete spalling in marine environments.
- Rapid Hardening Cement: Designed for high early strength (e.g., road repairs), not for chemical resistance against seawater.
- White Cement: Primarily used for architectural and decorative purposes due to its color; it does not possess specific chemical resistance to sulphates.
Junior Engineer • 2025 • PAPER-IV B (Technical Subject)
A road has a width of 6m and an average thickness of 20 cm. The length of the road is 1 km. What is the total volume of material required?
A)
1000 m³
B)
1200 m³
C)
1500 m³
D)
1800 m³
To find the total volume of material required for the road, we use the formula for the volume of a rectangular prism: Volume = Length × Width × Thickness.
1. **Identify and Convert Units to Meters (m):**
- Length (L) = 1 km = 1,000 m
- Width (W) = 6 m
- Thickness (T) = 20 cm = 20 / 100 = 0.2 m
2. **Calculate the Volume:**
- Volume = 1,000 m × 6 m × 0.2 m
- Volume = 6,000 m² × 0.2 m
- Volume = 1,200 m³
Comparing this result to the given options:
- Option (a) 1000 m³ is incorrect.
- Option (b) 1200 m³ matches our calculation.
- Option (c) 1500 m³ is incorrect.
- Option (d) 1800 m³ is incorrect.
Junior Engineer • 2025 • PAPER-IV B (Technical Subject)
Which law governs the permeability of soil?
A)
Bernoulli’s Law
B)
Stokes’ Law
C)
Darcy’s Law
D)
Newton’s Law
Permeability refers to the property of a soil mass that allows water to flow through its interconnected voids. This process is governed by Darcy's Law, which states that the velocity of flow (v) through a porous medium is directly proportional to the hydraulic gradient (i), expressed as v = ki, where 'k' is the coefficient of permeability.
Why other options are incorrect:
- Bernoulli’s Law: Relates to the conservation of energy in a flowing fluid (pressure, velocity, and elevation).
- Stokes’ Law: Governs the settling velocity of spherical particles in a viscous fluid, used primarily in hydrometer analysis for soil grain size distribution.
- Newton’s Law: Generally refers to laws of motion or cooling, which are not the governing principles for soil permeability.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
In a simply supported beam carrying a uniformly distributed load, where is the maximum bending moment?
A)
At the ends of the beam
B)
At the midpoint of the beam
C)
At one-quarter of the beam length
D)
At any random point
Option B is correct: At the midpoint of the beam.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
Why are cutoff walls provided in hydraulic structures like dams and weirs?
A)
To increase the storage capacity of the reservoir
B)
To allow fish to migrate upstream
C)
To improve the aesthetic appearance of the structure
D)
To prevent seepage and reduce piping failure
Option D is correct: To prevent seepage and reduce piping failure.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
What is the main difference between a gravity dam and an arch dam?
A)
Gravity dams rely on their weight, while arch dams transfer forces to the valley walls
B)
Gravity dams require strong abutments, while arch dams do not
C)
Gravity dams are more suitable for narrow valleys than arch dams
D)
Gravity dams require less construction material than arch dams
Option A is correct: Gravity dams rely on their weight, while arch dams transfer forces to the valley walls.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
How does a siphon aqueduct differ from a normal aqueduct?
A)
A siphon aqueduct allows the river to flow above the canal
B)
A siphon aqueduct allows the canal to pass underneath a river by siphon action
C)
A siphon aqueduct requires more energy to operate than a normal aqueduct
D)
A siphon aqueduct is only used in flat terrains
Option B is correct: A siphon aqueduct allows the canal to pass underneath a river by siphon action.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
A concrete beam is designed for a compressive strength of 30 MPa. What is its approximate tensile strength?
A)
1.5 MPa
B)
3.8 MPa
C)
4.5 MPa
D)
6.2 MPa
Option B is correct: 3.8 MPa.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
Why are under-sluices provided in weirs and barrages?
A)
To divert excess water during floods
B)
To control the movement of fish in the river
C)
To remove sediment from the riverbed and prevent silting
D)
To allow boats and ships to pass through
Option C is correct: To remove sediment from the riverbed and prevent silting.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
How much fine aggregate is required if a mix contains 1:2:4 (Cement:Sand:Coarse Aggregate) and cement weighs 300 kg?
A)
400 kg
B)
500 kg
C)
600 kg
D)
700 kg
Option C is correct: 600 kg.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
Why are energy dissipaters necessary at the downstream of spillways?
A)
To increase the velocity of water flow
B)
To prevent erosion and structural damage caused by high-energy water
C)
To store additional water for irrigation purposes
D)
To divert water into different canals
Option B is correct: To prevent erosion and structural damage caused by high-energy water.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
What is the function of coarse aggregates in concrete?
A)
Reduce shrinkage
B)
Increase workability
C)
Increase strength and durability
D)
Act as a binding agent
Option A is correct: Reduce shrinkage.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
What is the primary function of a head regulator in a canal system?
A)
To allow excess water to escape from the canal
B)
To regulate and control the entry of water into the canal from the river
C)
To store water for future use
D)
To measure the discharge of water entering the canal
Option B is correct: To regulate and control the entry of water into the canal from the river.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
A dam is 80 m high, and the density of water is 1000 kg/m³. What is the hydrostatic pressure at the base of the dam? (g = 9.81 m/s²)
A)
500 kPa
B)
685 kPa
C)
785 kPa
D)
980 kPa
Option D is correct: 980 kPa.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
Mohr’s Circle is used for which analysis?
A)
Beam bending
B)
Column buckling
C)
Principal stresses and strains
D)
Torsional analysis
Option C is correct: Principal stresses and strains.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
Why do earthen dams require more maintenance compared to concrete dams?
A)
Because they are expensive to construct
B)
Because they are susceptible to seepage and erosion
C)
Because they do not store water efficiently
D)
Because they cannot withstand high pressure
Option C is correct: Because they do not store water efficiently.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
A river carries a discharge of 800 cumecs. If the average depth of flow is 4 m and the width is 50 m, what is the average velocity of the river?
A)
3m/s
B)
4m/s
C)
5m/s
D)
6m/s
Option C is correct: 5m/s.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
Why is river training necessary in hydraulic engineering?
A)
To increase the river’s flow velocity
B)
To control river meandering and prevent erosion
C)
To reduce the depth of the river channel
D)
To increase sediment deposition in floodplains
Option B is correct: To control river meandering and prevent erosion.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
A cantilever beam of length 3 m is subjected to a uniformly distributed load of 5 kKN/m. Using Macaulay’s method, what is the slope at the free end? (EI = 5,000 kNm²)
A)
0.01 rad
B)
0.02 rad
C)
0.03 rad
D)
0.04 rad
Step-by-step solution using Macaulay's Method (or standard beam theory for a cantilever):
1. **Given Data**:
- Length of beam (L) = 3 m
- Uniformly Distributed Load (w) = 5 kN/m
- Flexural Rigidity (EI) = 5,000 kNm²
2. **Bending Moment Equation**:
Let the origin be at the fixed support (x = 0). For a cantilever with a UDL, the bending moment at any section x from the fixed end is:
M(x) = -w(L - x)² / 2
Alternatively, using Macaulay's notation starting from the fixed end, the moment is M(x) = -wLx + (wx²/2) + M_fixed. However, the simplest approach for a cantilever is to integrate the load:
EI(d²y/dx²) = M(x) = -w(L - x)² / 2
3. **Integration for Slope (θ)**:
EI(dy/dx) = ∫ [-w(L - x)² / 2] dx
EI(dy/dx) = -w/2 * [-(L - x)³ / 3] + C₁
EI(dy/dx) = w(L - x)³ / 6 + C₁
4. **Boundary Condition for Slope**:
At the fixed support (x = 0), the slope (dy/dx) = 0.
0 = w(L)³ / 6 + C₁ => C₁ = -wL³ / 6
5. **Slope at the Free End (x = L)**:
EI(θ_free) = w(L - L)³ / 6 + C₁
EI(θ_free) = 0 + (-wL³ / 6)
θ_free = -wL³ / (6EI)
6. **Calculation**:
|θ_free| = (5 kN/m * (3 m)³) / (6 * 5,000 kNm²)
|θ_free| = (5 * 27) / 30,000
|θ_free| = 135 / 30,000
|θ_free| = 0.0045 rad
Wait, let me re-evaluate the standard formula for cantilever UDL slope: θ = wL³/3EI.
Let's check:
θ = (5 * 3³) / (3 * 5000) = 135 / 15000 = 0.009 rad.
Let's re-read the options. The options are 0.01, 0.02, 0.03, 0.04.
If we use the formula θ = wL³/3EI, we get 0.009, which rounds to 0.01 (Option a).
However, let's check if the load was meant to be different or if there's a calculation error.
If w = 15 kN/m, θ = (15*27)/(3*5000) = 0.027 ≈ 0.03.
If L = 4m, θ = (5*64)/(3*5000) = 0.021.
Let's re-calculate wL³/3EI with the given values:
(5 * 27) / (3 * 5000) = 135 / 15000 = 0.009 rad.
Looking at the options provided in typical competitive exams for this specific problem set, there is often a typo in the provided EI or Load. If we assume the standard formula θ = wL³/3EI and the result is 0.009, the closest option is 0.01.
But let's check the integration again.
M = -wx(L - x/2) if x is from free end.
θ = ∫ (wx(L-x/2)/EI) dx = (wLx²/2EI - wx³/6EI).
At x=L: θ = wL³/2EI - wL³/6EI = wL³/3EI.
Calculation: (5 * 27) / (3 * 5000) = 0.009 rad.
Given the options, 0.009 is closest to 0.01. However, if the question intended EI = 1500 or w = 16.6, the answer would change. Based on the provided values, the mathematical result is 0.009 rad, which rounds to 0.01 rad.
Junior Engineer • 2025 • Paper-IV A (Technical Subject)
In a stress-strain curve, what does the yield point represent?
A)
The point where the material fractures
B)
The point where plastic deformation begins
C)
The maximum stress the material can withstand
D)
The point where the material returns to its original shape
Option B is correct: The point where plastic deformation begins.