Which of the following set of terms does not relate to operation of a theodolite? A) Transiting and inverting B) Face left and face right C) Right swing and left swing D) Gauging and sounding
💡Correct Answer: Option D (Gauging and sounding)
Transiting (or plunging/inverting), face left/face right observations, and swinging the telescope (right or left swing) are all fundamental operations and terms associated with a transit theodolite. In contrast, 'gauging' (measuring stream flow or water stage) and 'sounding' (measuring the depth of a water body below its surface) are terms used in hydrographic surveying and river gauging, not in the operation of a theodolite.
If the whole circle bearing is 315°20', its quadrantal bearing would be A) S 36°30' W B) N 44°40' W C) N 57°24' W D) S 60°40' W
💡Correct Answer: Option B (N 44°40' W)
A Whole Circle Bearing (WCB) of 315°20' lies in the fourth quadrant (North-West quadrant, between 270° and 360°). The Quadrantal Bearing (QB) or Reduced Bearing (RB) is measured relative to the North meridian: QB = N (360° - WCB) W = N (360°00' - 315°20') W = N 44°40' W.
Q3
Civil EngineeringSurveying - Levelling
Error due to inclination of line of collimation in levelling across a river can be eliminated by A) Reversion B) Reciprocal ranging C) Reciprocal levelling D) Keeping level in middle
💡Correct Answer: Option C (Reciprocal levelling)
When levelling across obstacles such as wide rivers, ravines, or ponds where setting up the level midway is impossible, reciprocal levelling is employed. By taking staff readings on both banks from instrument setups on opposite banks and averaging the true differences, reciprocal levelling completely eliminates collimation error, earth's curvature error, and uniform atmospheric refraction error.
Q4
Civil EngineeringSurveying - Chain Corrections
The length of a survey line when measured with a chain of 20 m nominal length was found to be 841.5 m. If the chain used is 0.1 m too long, the correct length of the measured line is A) 845.7 m B) 837.3 m C) 841.6 m D) 839.4 m
💡Correct Answer: Option A (845.7 m)
True distance = Measured distance (Actual chain length / Nominal chain length). Nominal chain length L = 20 m. Actual chain length L' = 20 + 0.1 = 20.1 m. Measured distance = 841.5 m. Correct length = 841.5 (20.1 / 20) = 841.5 * 1.005 = 845.7075 m ≈ 845.7 m.
In any closed traverse, if the survey work is error free, then I. The algebraic sum of all the latitudes should be equal to zero II. The algebraic sum of all the departures should be equal to zero III. The sum of the northings should be equal to the sum of the southings Which of the above statements are correct? A) I and II B) I and III C) II and III D) I, II and III
💡Correct Answer: Option D (I, II and III)
For an error-free closed traverse: (I) The algebraic sum of all latitudes is zero (ΣL = 0). (II) The algebraic sum of all departures is zero (ΣD = 0). (III) Since latitude = Northing (+) - Southing (-), ΣL = 0 directly implies that the total Northings equal the total Southings (Σ Northings = Σ Southings). Therefore, statements I, II, and III are all correct.
Q6
Civil EngineeringSurveying - Levelling Errors
Consider the following statements: Reciprocal leveling eliminates the effect of I. Errors due to earth's curvature II. Errors due to atmospheric refraction III. Mistakes in taking levelling staff readings IV. Error due to line of collimation Which of these statements are correct? A) I, II and III B) I, III and IV C) II, III and IV D) I, II and IV
💡Correct Answer: Option D (I, II and IV)
Reciprocal levelling eliminates systematic errors: (I) Error due to earth's curvature, (II) Error due to uniform atmospheric refraction, and (IV) Error due to inclination of the line of collimation. However, it cannot eliminate random personal mistakes or blunders in staff reading (statement III). Thus, statements I, II, and IV are correct.
Consider the following statements about the characteristics of contours: I. Closed contour lines with higher values inside show a lake II. Contour is an imaginary line joining points of equal elevations III. Closely spaced contours indicate steep slope IV. Contour lines can cross each other in case of an overhanging cliff Which of these statements are correct? A) II, III and IV B) I and II C) I and IV D) I, II and III
💡Correct Answer: Option A (II, III and IV)
Statement I is incorrect because closed contours with higher values inside represent a hill or ridge (lower values inside represent a pond, lake, or depression). Statement II is correct by definition. Statement III is correct as close contours indicate a steep slope. Statement IV is correct because contour lines cross each other only in the exceptional case of an overhanging cliff. Thus, statements II, III, and IV are correct.
Q8
Civil EngineeringSurveying - Classification of Surveys
Match List - I with List-II and select the correct answer using the codes given below the lists: List - I (Type of survey) i. Topographical survey ii. Reconnaissance survey iii. Cadastral survey iv. Archaeological survey List - II (Purpose) 1. To determine boundaries of fields, houses, etc. 2. To find relics of antiquity 3. To determine natural features of a country 4. To determine possibility and rough cost of the surveying system to be adopted Choose the correct option: A) (i) - 3, (ii) - 4, (iii) - 1, (iv) - 2 B) (i) - 3, (ii) - 1, (iii) - 4, (iv) - 2 C) (i) - 2, (ii) - 4, (iii) - 1, (iv) - 3 D) (i) - 2, (ii) - 1, (iii) - 4, (iv) - 3
💡Correct Answer: Option A ((i) - 3, (ii) - 4, (iii) - 1, (iv) - 2)
Matching of survey types with their purposes: (i) Topographical survey: To determine natural features of a country (rivers, hills, lakes) -> 3. (ii) Reconnaissance survey: To determine the possibility and rough cost of the project/survey scheme -> 4. (iii) Cadastral survey: To fix and determine property boundaries of fields, plots, and houses -> 1. (iv) Archaeological survey: To unearth relics of antiquity -> 2. Hence, the correct matching is (i)-3, (ii)-4, (iii)-1, (iv)-2.
Match List - I with List-II and select the correct answer using the codes given below the lists: List - I (Statement) i. Accurate centering in plane table surveying is necessary for ii. Exact orientation is more important than accurate centering for iii. The intersection method of plane table surveying is particularly employed for iv. Plane table survey is useful for List - II (Situation) 1. Inaccessible points 2. Open country with good intervisibility 3. Large scale maps 4. Small scale maps Choose the correct option: A) (i) - 3, (ii) - 4, (iii) - 1, (iv) - 2 B) (i) - 4, (ii) - 3, (iii) - 2, (iv) - 1 C) (i) - 1, (ii) - 2, (iii) - 4, (iv) - 3 D) (i) - 3, (ii) - 2, (iii) - 1, (iv) - 4
💡Correct Answer: Option A ((i) - 3, (ii) - 4, (iii) - 1, (iv) - 2)
Plane table surveying principles: (i) Accurate centering is critical for large scale maps (3) because small centering discrepancies are plottable at large scales. (ii) Exact orientation is more important than accurate centering for small scale maps (4) where centering errors become negligible on paper. (iii) The intersection method is particularly employed for locating inaccessible points (1) like distant hills, river points, etc. (iv) Plane table surveying is most useful for open country with good intervisibility (2). Therefore, the correct code is (i)-3, (ii)-4, (iii)-1, (iv)-2.
Q10
Civil EngineeringSurveying - Linear Measurements & Taping
The following steps are necessary to obtain sufficient accuracy with the tape: I. Keeping uniform tension on tape for each measurement II. "Breaking" tape on slopes are necessary to keep the tape level III. Keeping accurate count of the stations IV. Keeping the tape on the line being measured The correct sequence of these steps is: A) IV, II, I, III B) II, III, IV, I C) IV, I, II, III D) III, II, I, IV
💡Correct Answer: Option C (IV, I, II, III)
In standard field chaining and taping procedures: First, the tape must be aligned on the survey line being measured (IV: alignment). Next, uniform pull/tension must be applied to the tape (I: standard tension). Then, when encountering sloping ground, the tape is 'stepped' or 'broken' into horizontal segments to measure horizontal distance (II: leveling/breaking tape). Finally, an accurate count of tape lengths/stations is kept by the leader and follower (III: tallying stations). Thus, the correct operational sequence is IV, I, II, III.
Q11
Civil EngineeringMechanics of Materials - Stress and Strain
Robert Hooke discovered experimentally that within elastic limit A) Stress = strain B) Stress * strain = 1 C) Stress / Strain = a constant D) None of the above
💡Correct Answer: Option C (Stress / Strain = a constant)
Hooke's Law states that within the elastic limit of a material, stress is directly proportional to strain: σ ∝ ε, which means σ / ε = E (Young's Modulus, a constant). Thus, Stress / Strain = a constant.
Q12
Civil EngineeringMechanics of Materials - Principal Stresses & Shear
A circular bar is subjected to an axial pull of 100 kN. If the maximum intensity of shear stress on any oblique plane is not to exceed 60 MN/m², determine the diameter of the bar: A) 21.6 mm B) 32.6 mm C) 45.6 mm D) 48.6 mm
💡Correct Answer: Option B (32.6 mm)
For a member under pure uniaxial tension P, the maximum shear stress on an oblique plane occurs at 45° and is given by: τ_max = σ / 2 = P / (2 A). Given P = 100 kN = 100 10³ N, and τ_max ≤ 60 MN/m² = 60 N/mm². Therefore, P / (2 A) ≤ 60 => A ≥ P / (2 60) = 100000 / 120 = 833.33 mm². Since A = (π/4) d², d = √((4 833.33) / π) = √(1061.03) = 32.57 mm ≈ 32.6 mm.
Q13
Civil EngineeringMechanics of Materials - Shear Stress in Beams
In case of a circular section, the maximum shear stress is _______ percent more than the mean shear stress. A) 10 B) 20 C) 33.33 D) 66.66
💡Correct Answer: Option C (33.33)
For a solid circular cross-section subjected to transverse shear force, the maximum shear stress occurs at the neutral axis and is given by: τ_max = (4/3) * τ_avg. Thus, τ_max is (4/3 - 1) = 1/3 = 33.33% greater than the mean shear stress.
Q14
Civil EngineeringMechanics of Materials - Combined Direct & Bending Stress
In a rectangular section, the stress will be of the same sign throughout the section if the load lies within the A) Middle third of the section B) Middle half C) Middle fourth D) None of the above
💡Correct Answer: Option A (Middle third of the section)
To avoid tension in a masonry or unreinforced rectangular section under eccentric axial load (i.e. to keep compressive stress throughout the section), the eccentricity must satisfy e ≤ d / 6. The load must lie within the central kernel or core of width b/3 and depth d/3, which is the famous 'middle-third rule'.
Q15
Civil EngineeringMechanics of Materials - Deflection of Beams
A cantilever 1.5 m long carries a uniformly distributed load over the entire length. Find the deflection at the free end if the slope at the free end is 1.5°. A) 21.88 mm B) 29.45 mm C) 36.48 mm D) 41.54 mm
💡Correct Answer: Option B (29.45 mm)
For a cantilever beam with UDL over its entire length L: Slope at free end θ = w L³ / (6 E I). Deflection at free end δ = w L⁴ / (8 E I). The ratio gives: δ = (3/4) L θ. Here, L = 1.5 m = 1500 mm, and θ = 1.5° = 1.5 (π / 180) = 0.0261799 rad. Therefore, δ = 0.75 1500 * 0.0261799 = 29.452 mm ≈ 29.45 mm.
Q16
Civil EngineeringMechanics of Materials - Methods of Slope & Deflection
The slope and deflection at a section in a loaded beam can be found out by which of the following methods? A) Double integration method B) Moment area method C) Macaulay's method D) Any of the above
💡Correct Answer: Option D (Any of the above)
Slope and deflection in beams can be determined by several established structural analysis methods, including the Double Integration method, Mohr's Moment-Area theorems, Macaulay's method, the Conjugate Beam method, and Castigliano's strain energy method. Therefore, any of the above can be used.
Q17
Civil EngineeringMechanics of Materials - Material Properties
The material that exhibits the same elastic properties in all directions at a point is said to be A) homogeneous B) orthotropic C) viscoelastic D) isotropic
💡Correct Answer: Option D (isotropic)
An isotropic material has identical elastic properties (such as modulus of elasticity, Poisson's ratio) in all directions at a given point. In contrast, a homogeneous material has identical physical properties at all points throughout its volume, and an orthotropic material has different properties along mutually orthogonal directions.
Q18
Civil EngineeringMechanics of Materials - Principal Stresses
If principal stresses in a two-dimensional case are -10 MPa and 20 MPa respectively, then maximum shear stress at the point is A) 10 MPa B) 15 MPa C) 20 MPa D) 30 MPa
💡Correct Answer: Option B (15 MPa)
In 2D plane stress, maximum in-plane shear stress is given by: τ_max = (σ₁ - σ₂) / 2. Here, σ₁ = 20 MPa and σ₂ = -10 MPa. Therefore, τ_max = [20 - (-10)] / 2 = 30 / 2 = 15 MPa.
Q19
Civil EngineeringStructural Analysis - Truss Member Forces
All the members of the planar truss (see figure), have the same properties in terms of area of cross-section (A) and modulus of elasticity (E). For the loads shown on the truss, the statements that correctly represents the nature of forces in the members of the truss is: A) There are 3 members in tension, and 2 members in compression B) There are 2 members in tension, and 2 members in compression, 1 zero-force member C) There are 2 members in tension, 1 member in compression, and 2 zero-force members D) There are 2 members in tension, and 3 zero-force members
💡Correct Answer: Option B (There are 2 members in tension, and 2 members in compression, 1 zero-force member)
Analysis of the planar 5-member square truss: Under the equal and opposite external couple loads P acting horizontally at diagonal joints, joint equilibrium reveals that 2 members are in tension, 2 members are in compression, and 1 member has zero force. Thus, statement B correctly describes the force distribution.
Q20
Civil EngineeringMechanics of Materials - Bending Relationships
The following statements are related to bending of beams: I. The slope of the bending moment diagram is equal to the shear force II. The slope of the shear force diagram is equal to the load intensity III. The slope of the curvature is equal to the flexural rotation IV. The second derivative of the deflection is equal to the curvature The only FALSE statement is: A) I B) II C) III D) IV
💡Correct Answer: Option C (III)
From beam bending relationships: I. dM/dx = V (Slope of BMD is shear force) -> TRUE. II. dV/dx = -w (Slope of SFD is load intensity) -> TRUE. IV. d²y/dx² = κ (Second derivative of deflection is curvature) -> TRUE. III. The flexural rotation/slope is θ = dy/dx, while curvature is κ = dθ/dx = d²y/dx². The slope of curvature (dκ/dx) is related to shear force divided by EI, NOT equal to flexural rotation. Hence, statement III is the only FALSE statement.
Q21
Civil EngineeringMechanics of Materials - Torsion of Shafts
A circular solid shaft of span L = 5 m is fixed at one end and free at the other end. A twisting moment T = 100 kN-m is applied at the free end. The torsional rigidity GJ is 50000 kN-m²/rad. Following statements are made for this shaft: 1. The maximum rotation is 0.01 rad 2. The torsional strain energy is 1 kN-m With reference to the above statements, which of the following applies? A) Both statements are true B) Statement 1 is true but 2 is false C) Statement 2 is true but 1 is false D) Both statements are false
💡Correct Answer: Option B (Statement 1 is true but 2 is false)
Calculations for torsional deformation and strain energy: 1. Maximum angle of twist θ = T L / (GJ) = (100 5) / 50000 = 500 / 50000 = 0.01 rad. Statement 1 is TRUE. 2. Torsional strain energy U = (1/2) T θ = (1/2) 100 0.01 = 0.5 kN-m. Since the statement claims 1 kN-m, Statement 2 is FALSE. Therefore, Statement 1 is true but 2 is false.
Q22
Civil EngineeringMechanics of Materials - Principal Planes & Stress Invariants
Consider the following statements: I. On a principal plane, only normal stress acts II. On a principal plane, both normal and shear stresses act III. On a principal plane, only shear stress acts IV. Isotropic state of stress is independent of frame of reference Which of these statements is/are correct? A) I and IV B) II only C) II and IV D) II and III
💡Correct Answer: Option A (I and IV)
By definition, a principal plane is a plane on which shear stress is identically zero, meaning only normal stress acts (Statement I is TRUE, Statements II and III are FALSE). In an isotropic (hydrostatic) state of stress, normal stress is identical in all directions and shear stress is zero on all planes; hence, it is invariant and independent of the coordinate frame of reference (Statement IV is TRUE). Thus, statements I and IV are correct.
Q23
Civil EngineeringMechanics of Materials - Pure Bending Theory
The "Plane section remains plane" assumption in bending theory implies A) Stress profile is linear B) Strain profile is linear C) Both profiles are linear D) Shear deformation is neglected
💡Correct Answer: Option B (Strain profile is linear)
The Bernoulli-Euler assumption that 'plane sections normal to the neutral axis before bending remain plane and normal to the neutral axis after bending' is purely a kinematic assumption. It directly implies that longitudinal strain varies linearly with distance from the neutral axis (linear strain profile). The stress profile is linear only if Hooke's law additionally holds.
Q24
Civil EngineeringMechanics of Materials - Shear Force & Bending Moment Diagrams
List-I shows different loads acting on a beam and List-II shows different bending moment distributions. Match the load with the corresponding bending moment diagram. A) A - 4, B - 2, C - 1, D - 3 B) A - 4, B - 3, C - 1, D - 2 C) A - 2, B - 3, C - 4, D - 1 D) None of the above
💡Correct Answer: Option D (None of the above)
Matching loads with bending moment diagrams: A (Concentrated couple at center) -> BMD with vertical step at center (Diagram 2). B (Central concentrated point load) -> Triangular BMD with apex at center (Diagram 4). C (Uniformly distributed load) -> Parabolic curved BMD (Diagram 1). D (Uniformly varying triangular load) -> Cubic parabolic BMD (Diagram 3). The correct combination is A-2, B-4, C-1, D-3. Since this exact permutation is not present among options A, B, or C, the correct choice is D (None of the above).
Q25
Civil EngineeringMechanics of Materials - Columns & Buckling Load
Four columns of the same material and having identical geometric properties are supported in different ways as shown below: (I) Pinned-Pinned (II) Fixed-Free (III) Fixed-Fixed (IV) Fixed-Pinned It is required to arrange these four columns in the increasing order of their respective first buckling loads. The correct order is given by A) I, II, III, IV B) III, IV, I, II C) II, I, IV, III D) I, II, IV, III
💡Correct Answer: Option C (II, I, IV, III)
Euler critical buckling load is P_cr = π²EI / L_e²: (II) One end fixed, other free: L_e = 2L => P_cr = 0.25 (π²EI / L²).
(I) Both ends pinned: L_e = L => P_cr = 1.0 (π²EI / L²). (IV) One end fixed, other pinned: L_e = L / √2 => P_cr = 2.0 (π²EI / L²).
(III) Both ends fixed: L_e = 0.5L => P_cr = 4.0 (π²EI / L²). Arranging in increasing order: II < I < IV < III. Thus, the correct sequence is II, I, IV, III.
Q26
Civil EngineeringIrrigation Engineering - Water Requirements of Crops
Find the delta for a crop when its duty is 864 hectares/cumec on the field, the base period of this crop is 120 days. A) 104 cm B) 120 cm C) 138 cm D) 720 cm
💡Correct Answer: Option B (120 cm)
The relationship between delta (Δ), duty (D), and base period (B) is: Δ = (8.64 B) / D (in meters). Here, B = 120 days and D = 864 hectares/cumec. Δ = (8.64 120) / 864 = 1036.8 / 864 = 1.20 meters = 120 cm.
Q27
Civil EngineeringIrrigation Engineering - Efficiencies of Irrigation
The depth of penetration along the length of a boarder strip at points 30 meters apart were probed. Their observed values are 2.0, 1.9, 1.8, 1.6 and 1.5 meters. Compute the water distribution efficiency. A) 0.168 B) 0.905 C) 1.760 D) 2.105
💡Correct Answer: Option B (0.905)
Water distribution efficiency is defined by Christiansen's equation: η_d = (1 - y/d), where d is the average depth of water penetration and y is the average numerical deviation from the mean depth. Mean depth d = (2.0 + 1.9 + 1.8 + 1.6 + 1.5) / 5 = 8.8 / 5 = 1.76 m. Absolute deviations |d_i - d|: |2.0 - 1.76| = 0.24 |1.9 - 1.76| = 0.14 |1.8 - 1.76| = 0.04 |1.6 - 1.76| = 0.16 |1.5 - 1.76| = 0.26 Mean deviation y = (0.24 + 0.14 + 0.04 + 0.16 + 0.26) / 5 = 0.84 / 5 = 0.168 m. η_d = 1 - (0.168 / 1.76) = 1 - 0.09545 = 0.90455 ≈ 0.905 (or 90.5%).
Q28
Civil EngineeringIrrigation Engineering - Land Drainage
A tile drainage system draining 12 hectares, flows at a design capacity for two days, following a storm. If the system is designed using a drainage coefficient of 1.25 cm, how many cubic meters of water will be removed during this period? A) 1000 m³ B) 2000 m³ C) 3000 m³ D) 4000 m³
💡Correct Answer: Option C (3000 m³)
Drainage coefficient is the depth of water removed from the drainage area in 24 hours (1 day). Drainage coefficient = 1.25 cm/day = 0.0125 m/day. Drainage area = 12 hectares = 12 10⁴ m² = 120,000 m². In 2 days, total depth removed = 2 0.0125 m = 0.025 m. Total volume of water removed = Area Depth = 120,000 m² 0.025 m = 3000 m³.
The recharging rate of groundwater depends upon _______ and on the depth of water stored, and is generally less. A) the permeability of the spread area B) Renney type wells C) Infiltration tube springs D) None of the above
💡Correct Answer: Option A (the permeability of the spread area)
In groundwater artificial recharge techniques (such as spreading basins or recharge ponds), the infiltration and recharging rate depends directly on the hydraulic conductivity and permeability of the soil in the spreading area as well as the hydraulic head (depth of water stored).
Q30
Civil EngineeringIrrigation & Hydrology - Dams and Reservoirs
The lake of water which is formed upstream is often called a A) creep B) reservoir C) side flanks D) swift frinity
💡Correct Answer: Option B (reservoir)
A large artificial or natural body of water formed upstream of a dam, weir, or barrage constructed across a river or stream is called a reservoir (or storage reservoir).
Q31
Civil EngineeringIrrigation Engineering - Cross Drainage Works
The following statements are related to cross drainage works and their types: Statement - 1: The culvert length or width of aqueduct is maximum in Type I and minimum in Type III. An intermediate value exists in Type II. Statement - 2: In Type III, earthen section of the canal is discontinued and the canal water is carried in a masonry or a flumed trough in this case. Choose the correct option: A) Statement 1 is correct and Statement 2 is incorrect B) Statement 2 is correct and Statement 1 is incorrect C) Statement 1 and 2 are correct D) Statement 1 and 2 are incorrect
💡Correct Answer: Option C (Statement 1 and 2 are correct)
Classification of aqueducts based on fluming: Statement 1: In Type I aqueducts, the canal section is not flumed (earthen banks continue), making the culvert/barrel length maximum. In Type III, maximum fluming is adopted, making the culvert length minimum, while Type II is intermediate. Hence Statement 1 is correct. Statement 2: In Type III aqueducts, the canal's earthen banks are completely discontinued and canal flow is carried across in a concrete or masonry trough (flumed section). Hence Statement 2 is also correct. Therefore, both Statement 1 and Statement 2 are correct.
Q32
Civil EngineeringIrrigation Engineering - River Training Works
Consider the following Statements: Statement - 1: The length of the guide bank on the downstream side should be between 0.1 L to 0.2 L. L indicates the length of structure between the abutments. Statement - 2: The top level of guide banks is governed by HFL, afflux, velocity head, and freeboard. It can be obtained by adding all these four values. Choose the correct option: A) Statement 1 is correct and Statement 2 is incorrect B) Statement 2 is correct and Statement 1 is incorrect C) Statement 1 and 2 are correct D) Statement 1 and 2 are incorrect
💡Correct Answer: Option C (Statement 1 and 2 are correct)
River training and guide bank design principles: Statement 1: According to Spring's classic guidelines for guide banks, the downstream length is kept between 0.1 L to 0.2 L (where L is the waterway length between bridge/barrage abutments). Statement 1 is correct. Statement 2: The top level of a guide bank is fixed by taking the High Flood Level (HFL) and adding afflux, velocity head, and suitable freeboard (normally 1.5 m to 2.0 m). Statement 2 is also correct. Therefore, both Statement 1 and 2 are correct.
Q33
Civil EngineeringIrrigation Engineering - Cross Drainage Works
The drainage water intercepting the canal can be disposed in different ways. And the type of cross-drainage works with their function is given below. Match List - I with List-II and select the correct answer using the codes given below the lists: List - I (Function/Places) 1. Bypassing the canal over the drainage 2. Bypassing the canal below the drainage 3. Bypassing the drain through a canal List - II (Accomplished) i. Either through a level crossing or through inlets and outlets ii. Either through a super-passage or through a canal syphon iii. Either through an aqueduct or through a syphon-aqueduct Choose the correct option: A) (1) - i, (2) - ii, (3) - iii B) (1) - ii, (2) - iii, (3) - i C) (1) - iii, (2) - ii, (3) - i D) (1) - iii, (2) - i, (3) - ii
💡Correct Answer: Option C ((1) - iii, (2) - ii, (3) - i)
Classification of cross-drainage works based on relative bed levels: 1. Canal over drainage: Aqueduct or Syphon Aqueduct -> (iii) 2. Canal below drainage: Super passage or Canal Syphon -> (ii) 3. Canal and drainage at same level: Level crossing or Inlets and Outlets -> (i) Thus, the correct matching is (1)-iii, (2)-ii, (3)-i.
Q34
Civil EngineeringIrrigation Engineering - River Training Types
Match List - I with List-II and select the correct answer using the codes given below the lists: List - I (Classification of River Training) 1. High water training 2. Low water training 3. Mean water training List - II (Purpose) i. Sediment (efficient disposal of suspended load and bed load) ii. Discharge (flood control) iii. Depth (provide sufficient water depth) Choose the correct option: A) (1) - i, (2) - ii, (3) - iii B) (1) - ii, (2) - iii, (3) - i C) (1) - iii, (2) - i, (3) - ii D) (1) - iii, (2) - ii, (3) - i
💡Correct Answer: Option B ((1) - ii, (2) - iii, (3) - i)
Classification of river training works by primary objective: 1. High water training (Training for Discharge): Aimed at flood disposal and flood control without causing overtopping -> (ii) 2. Low water training (Training for Depth): Aimed at maintaining navigable water depth during low-flow periods -> (iii) 3. Mean water training (Training for Sediment): Aimed at efficient movement and disposal of sediment/bed load to maintain channel stability -> (i) Therefore, the correct match is (1)-ii, (2)-iii, (3)-i.
Q35
Civil EngineeringHydrology - Hydrological Cycle
_______ evaporates water, and (ii) _______ by causing and controlling winds, (iii) _______ circulates the evaporated water vapour, and thus, helping in its precipitation at different places. Choose the correct option to fill out i, ii, iii in order. A) Sun, hydro cycle, filtration B) Sun, Coriolis force, filtration C) Sun, cycle, precipitation D) Sun, Coriolis force, reprecipitation
💡Correct Answer: Option C (Sun, cycle, precipitation)
In the hydrologic cycle: The (i) Sun radiates thermal energy that evaporates water from oceans and lakes; atmospheric circulation and wind patterns drive the (ii) cycle (or wind currents) that transport evaporated water vapour; leading to cloud condensation and (iii) precipitation over land and water bodies. Thus, the sequence 'Sun, cycle, precipitation' completes the sentence.
In a domestic wastewater sample, COD and BOD were measured. Generally, which of the following statement is true for their relative magnitude? A) COD = BOD B) COD > BOD C) COD < BOD D) None of the above
💡Correct Answer: Option B (COD > BOD)
Chemical Oxygen Demand (COD) measures the oxygen required to oxidize both biodegradable and non-biodegradable organic matter using a strong chemical oxidant (potassium dichromate). Biochemical Oxygen Demand (BOD) measures only the biologically degradable organic matter. Hence, COD is always greater than BOD for wastewater (COD > BOD).
Q37
Civil EngineeringEnvironmental Engineering - Water Treatment Sequence
The potable water is prepared from turbid surface water by adopting the following treatment sequence: A) Turbid surface water - Coagulation - Flocculation - Sedimentation - Filtration - Disinfection - Storage and Supply B) Turbid surface water - Sedimentation - Coagulation - Flocculation - Filtration - Disinfection - Storage and Supply C) Turbid surface water - Flocculation - Coagulation - Filtration - Disinfection - Sedimentation - Storage and Supply D) Turbid surface water - Coagulation - Flocculation - Filtration - Disinfection - Sedimentation - Storage and Supply
💡Correct Answer: Option A (Turbid surface water - Coagulation - Flocculation - Sedimentation - Filtration - Disinfection - Storage and Supply)
The standard conventional water treatment plant flow sequence for turbid surface raw water is: Raw Water -> Coagulation (rapid mixing of coagulant) -> Flocculation (slow mixing to form flocs) -> Sedimentation (settling of flocs in clarifier) -> Filtration (sand/media filters) -> Disinfection (chlorination) -> Storage and Distribution.
Q38
Civil EngineeringEnvironmental Engineering - BOD Calculation
A 1 % solution of sewage sample is incubated for 5 days at 20°C. The depletion of oxygen was found to be 3 ppm. Determine the BOD of raw sewage. A) 60 ppm B) 150 ppm C) 300 ppm D) 600 ppm
💡Correct Answer: Option C (300 ppm)
BOD₅ is calculated as: BOD₅ = (DO depletion) Dilution Factor. A 1% solution means 1 volume of sewage in 100 volumes of total mixture, giving a dilution factor (DF) = 100 / 1 = 100. Given DO depletion = 3 ppm (or mg/L). BOD₅ = 3 ppm 100 = 300 ppm.
As per Noise Pollution (Regulation and Control) Rules 2000 of India, the day time noise limit for a residential zone, expressed in dB(A) L_eq, is A) 45 B) 55 C) 65 D) 75
💡Correct Answer: Option B (55)
According to the Noise Pollution (Regulation and Control) Rules, 2000 of India: - Industrial Area: Day = 75 dB(A), Night = 70 dB(A) - Commercial Area: Day = 65 dB(A), Night = 55 dB(A) - Residential Area: Day = 55 dB(A), Night = 45 dB(A) - Silence Zone: Day = 50 dB(A), Night = 40 dB(A) Therefore, for residential zones during daytime, the limit is 55 dB(A).
Q40
Civil EngineeringEnvironmental Engineering & Construction - Trench Excavation
When the depth of trench exceeds 1.5 to 2 m, and when excavation is made with sides vertical, it becomes necessary to support the side by sheeting and bracing. This operation is known as A) shafts of trench B) electro-osmosis C) boning of trench D) timbering of trench
💡Correct Answer: Option D (timbering of trench)
Supporting the vertical sides of deep excavated trenches (depths > 1.5 m) using timber or steel sheeting, walings, struts, and bracing to prevent soil collapse is termed 'timbering of trenches' (also known as shoring or trench sheeting).
The following two statements are relevant to method of sewage disposal: Statement - 1: The three principal processes of land treatment of wastewater are (i) Broad irrigation or sewage farming, (ii) Rapid infiltration, and (iii) Overland runoff. Statement - 2: The ratio of quantity of receiving water to that of wastewater or effluent discharge is called the dilution factor. Choose the correct option: A) Statement 1 is correct and Statement 2 is incorrect B) Statement 2 is correct and Statement 1 is incorrect C) Statement 1 and 2 are correct D) Statement 1 and 2 are incorrect
💡Correct Answer: Option C (Statement 1 and 2 are correct)
Methods of sewage disposal: Statement 1: The three primary recognized methods of land application/treatment of wastewater are slow-rate infiltration (broad irrigation/sewage farming), rapid infiltration, and overland flow/runoff. Thus, Statement 1 is correct. Statement 2: When disposing sewage by dilution into a natural water body, the dilution factor is defined as the ratio of the discharge/volume of receiving clean water to that of the incoming wastewater or effluent. Thus, Statement 2 is correct. Therefore, both Statement 1 and 2 are correct.
Q42
Civil EngineeringEnvironmental Engineering - Water Quality Testing
Match List - I with List-II and select the correct answer using the codes given below the lists: List - I (Water properties) i. Alkalinity ii. Hardness iii. Chlorine iv. Dissolved Oxygen List - II (Titrants) 1. N/35.5 AgNO₃ 2. N/40 Na₂S₂O₃ 3. N/50 H₂SO₄ 4. N/50 EDTA Choose the correct option: A) (i) - 2, (ii) - 4, (iii) - 3, (iv) - 1 B) (i) - 3, (ii) - 4, (iii) - 1, (iv) - 2 C) (i) - 1, (ii) - 2, (iii) - 4, (iv) - 3 D) (i) - 4, (ii) - 3, (iii) - 2, (iv) - 1
💡Correct Answer: Option B ((i) - 3, (ii) - 4, (iii) - 1, (iv) - 2)
Standard environmental engineering titrimetric methods: (i) Alkalinity is determined by titration against N/50 H₂SO₄ (with phenolphthalein and methyl orange indicators) -> 3. (ii) Hardness is determined by complexometric titration against N/50 EDTA (with EBT indicator) -> 4. (iii) Chloride / Chlorine test (Mohr's method) uses standard silver nitrate N/35.5 AgNO₃ -> 1. (iv) Dissolved Oxygen (Winkler iodometric method) uses standard sodium thiosulfate N/40 Na₂S₂O₃ -> 2. Thus, the correct matching is (i)-3, (ii)-4, (iii)-1, (iv)-2.
Q43
Civil EngineeringEnvironmental Engineering - Air Pollution Effects
Match List - I with List-II and select the correct answer using the codes given below the lists: List - I (Materials and Air pollutants) 1. Materials: Metals; and Pollutants: SO₂ 2. Materials: Paper; and Pollutants: SO₂ 3. Materials: Textiles; and Pollutants: SO₂ List - II (Effects) i. Embrittlement ii. Reduction in tensile strength iii. Tarnishing of surfaces Choose the correct option: A) (1) - i, (2) - ii, (3) - iii B) (1) - ii, (2) - iii, (3) - i C) (1) - iii, (2) - i, (3) - ii D) (1) - iii, (2) - ii, (3) - i
💡Correct Answer: Option C ((1) - iii, (2) - i, (3) - ii)
Air pollution effects of sulfur dioxide (SO₂) on engineering and everyday materials: 1. Metals exposed to SO₂ and moisture undergo corrosion, pitting, and tarnishing of surfaces -> (iii). 2. Paper absorbs SO₂ which oxidizes to sulfuric acid, causing acid degradation, yellowing, and embrittlement -> (i). 3. Textiles (fibers like cotton, nylon) undergo acid hydrolysis resulting in significant reduction in tensile strength -> (ii). Therefore, the correct match is (1)-iii, (2)-i, (3)-ii.
Q44
Civil EngineeringEnvironmental Engineering - Water Treatment Methods
Match List - I with List-II and select the correct answer using the codes given below the lists: List - I (Type of water impurity) i. Hardness ii. Brackish water from sea iii. Residual MPN from filters iv. Turbidity List - II (Method of treatment) 1. Reverse Osmosis 2. Chlorination 3. Zeolite treatment 4. Coagulation, Flocculation and Filtration Choose the correct option: A) (i) - 1, (ii) - 4, (iii) - 3, (iv) - 2 B) (i) - 4, (ii) - 3, (iii) - 1, (iv) - 2 C) (i) - 2, (ii) - 1, (iii) - 4, (iv) - 3 D) (i) - 3, (ii) - 1, (iii) - 2, (iv) - 4
💡Correct Answer: Option D ((i) - 3, (ii) - 1, (iii) - 2, (iv) - 4)
Water treatment matching: (i) Hardness: Removed by ion-exchange / Zeolite treatment -> 3. (ii) Brackish water from sea: Desalinated using Reverse Osmosis (RO) -> 1. (iii) Residual MPN (coliform bacteria) from filters: Disinfected using Chlorination -> 2. (iv) Turbidity: Removed by Coagulation, Flocculation, and Filtration -> 4. Thus, the correct match is (i)-3, (ii)-1, (iii)-2, (iv)-4.
Q45
Civil EngineeringEnvironmental Engineering - Sedimentation Theory
Particles may settle out of a suspension in four ways, depending upon the concentration of the suspension and the flocculating properties of the particles. Arrange the four types in order with respect to decreasing of settling depth vs increasing of settling time. Choose the correct order. A) Discrete settling, Flocculant settling, Zone settling, Compression settling B) Compression settling, Flocculant settling, Discrete settling, Zone settling C) Zone settling, Flocculant settling, Discrete settling, Compression settling D) Discrete settling, Zone settling, Flocculant settling, Compression settling
💡Correct Answer: Option A (Discrete settling, Flocculant settling, Zone settling, Compression settling)
Settling phenomena are categorized into four types with increasing solid concentration and flocculation interaction: Type 1: Discrete settling (individual unhindered particles, settles fastest, minimum time) Type 2: Flocculant settling (dilute suspension with agglomerating particles) Type 3: Zone / Hindered settling (intermediate concentration, particles settle as a blanket) Type 4: Compression settling (highest solids concentration, lower blanket layers compressed under weight, longest settling time). Arranged in this sequential order from dilute to concentrated: Discrete settling, Flocculant settling, Zone settling, Compression settling.
Gap grading of aggregate is one, in which A) At least one intermediate aggregate fraction is absent B) % passing fall within a narrow limit of size fractions C) Combines different fractions of fine and coarse aggregates D) All the aggregate are of same size
💡Correct Answer: Option A (At least one intermediate aggregate fraction is absent)
Gap-graded aggregate refers to an aggregate gradation where one or more intermediate particle size fractions are deliberately omitted or absent from the continuous grading curve (indicated by a horizontal plateau on the gradation chart).
Q47
Civil EngineeringRCC Design - Limit State Assumptions (IS 456)
As per limit state design method the ultimate strain in the outermost compression fiber of concrete in bending is taken as A) 0.0020 B) 0.0120 C) 0.0035 D) 0.0520
💡Correct Answer: Option C (0.0035)
According to Clause 38.1(b) of IS 456:2000, in the limit state of collapse in flexure, the maximum strain in concrete at the outermost compression fiber is taken as 0.0035.
Q48
Civil EngineeringStructural Analysis - Static Determinacy of Trusses
A frame having (M) number of member and joints (J) is said to be perfect frame if it follows the equation as given below A) J = 2M - 3 B) M = 2J - 3 C) M = 3J - 2 D) J = M + 3
💡Correct Answer: Option B (M = 2J - 3)
A pin-jointed plane frame or truss having M members and J joints is statically determinate and stable (called a perfect frame) when it satisfies the condition M = 2J - 3. If M < 2J - 3 it is deficient, and if M > 2J - 3 it is redundant.
Q49
Civil EngineeringMechanics of Materials - Cantilever Formulas
Consider the following pairs, for a cantilever beam of length (l), loaded by point load (w) at free end and flexural rigidity (EI): i. Maximum Bending Moment - 1. wl ii. Strain Energy - 2. wl² / (2EI) iii. Maximum slope - 3. wl³ / (3EI) iv. Maximum deflection - 4. w²l³ / (6EI) Which of the pair given below is correctly matched? A) (i) - (1), (ii) - (4), (iii) - (3), (iv) - (2) B) (i) - (1), (ii) - (4), (iii) - (2), (iv) - (3) C) (i) - (4), (ii) - (2), (iii) - (1), (iv) - (3) D) (i) - (4), (ii) - (3), (iii) - (1), (iv) - (2)
💡Correct Answer: Option B ((i) - (1), (ii) - (4), (iii) - (2), (iv) - (3))
For a cantilever beam of length l carrying a point load w at the free end: (i) Maximum Bending Moment at fixed end M_max = w l -> 1
(ii) Total Strain Energy U = ∫ (M² dx) / (2EI) = w²l³ / (6EI) -> 4
(iii) Maximum slope at free end θ_max = w l² / (2EI) -> 2 (iv) Maximum deflection at free end δ_max = w * l³ / (3EI) -> 3 Therefore, the correctly matched pair is (i)-(1), (ii)-(4), (iii)-(2), (iv)-(3).
Q50
Civil EngineeringConcrete Technology - Chemistry of Cement Hydration
The constituent compounds of cement in decreasing order of rate of hydration are A) C₂S, C₃S and C₃A B) C₃S, C₃A and C₂S C) C₃A, C₂S and C₃S D) C₃A, C₃S and C₂S
💡Correct Answer: Option D (C₃A, C₃S and C₂S)
The rate of hydration of Bogue's compounds in Portland cement follows the order: C₄AF > C₃A > C₃S > C₂S. Among the three given compounds (C₃A, C₃S, C₂S), the decreasing order of rate of hydration is C₃A > C₃S > C₂S.
Q51
Civil EngineeringStructural Design - Indian Standard Codes
Which of the following Indian Standard (IS) code used for wind load analysis for designing building structure A) IS 456 B) IS 800 C) IS 875 D) IS 1893
💡Correct Answer: Option C (IS 875)
IS 875 (specifically Part 3) is the Indian Standard Code of Practice for Design Loads (other than earthquake) for Buildings and Structures - Part 3: Wind Loads. IS 456 is for plain and reinforced concrete, IS 800 is for general construction in steel, and IS 1893 is for earthquake resistant design.
Q52
Civil EngineeringRCC Design - Cantilever Beams
In a cantilever RCC beam design, tensile reinforcement is provided A) On the top of the beam B) On the bottom of the beam C) In the middle of the beam D) On the top and bottom of the beam
💡Correct Answer: Option A (On the top of the beam)
A cantilever beam subjects its top fibers to tensile stresses and its bottom fibers to compressive stresses due to negative (hogging) bending moments throughout its span. Because concrete is weak in tension, main tensile reinforcement must be provided at the top of the beam.
Q53
Civil EngineeringBuilding Materials - Lime
For construction of structure under water the type of lime used is A) Hydraulic lime B) Fat lime C) Quick lime D) Pure lime
💡Correct Answer: Option A (Hydraulic lime)
Hydraulic lime contains clay (silica and alumina, usually 5% to 30%) which enables it to set and harden through chemical hydration even under water and in thick damp masonry where air is absent. Fat lime or pure lime hardens only by absorbing carbon dioxide from air and cannot set under water.
Q54
Civil EngineeringBuilding Materials - Bricks
The standard size of the brick as per IS standard is A) 10 x 9 x 9 cm B) 18 x 9 x 9 cm C) 19 x 9 x 9 cm D) 23 x 11.5 x 7.5 cm
💡Correct Answer: Option C (19 x 9 x 9 cm)
According to IS 1077, the standard modular size of a common burnt clay building brick is 19 cm x 9 cm x 9 cm (or 190 mm x 90 mm x 90 mm). With 10 mm mortar thickness, the nominal modular size is 20 cm x 10 cm x 10 cm.
Q55
Civil EngineeringConcrete Technology - Compaction and Defects
Inadequate compaction during concrete casting results in A) Segregation B) Rutting C) Bleeding D) Honey combing
💡Correct Answer: Option D (Honey combing)
Insufficient or inadequate compaction leaves large entrapped air voids and pockets between coarse aggregate particles where cement paste fails to penetrate, creating a porous, hollow appearance known as 'honey combing' on the concrete surface.
Side face reinforcement is required when the depth of the web of a beam is A) Greater than 750 mm B) Greater than 900 mm C) Greater than 950 mm D) Greater than 1000 mm
💡Correct Answer: Option A (Greater than 750 mm)
According to Clause 26.5.1.3 of IS 456:2000, where the depth of the web in a beam exceeds 750 mm, side face reinforcement of not less than 0.1% of the web area must be provided, distributed equally on both faces at a spacing not exceeding 300 mm or web thickness, whichever is less (reduced to 450 mm if subjected to torsion).
Q57
Civil EngineeringMechanics of Materials - Shear Stress Distribution
Distribution of shear stress intensity over a rectangular section of a loaded beam follows diagrammatically A) A circular curve B) A straight line C) A parabolic curve D) An elliptical curve
💡Correct Answer: Option C (A parabolic curve)
Transverse shear stress across depth y of a beam is given by τ = (V A y_bar) / (I b). For a rectangular cross-section of width b and depth d, τ = [V / (2I)] [(d/2)² - y²], which describes a parabolic variation with maximum intensity at the neutral axis (y = 0) and zero at top and bottom fibers.
The compressive strength of 100 mm cube as compared to 150 mm cube is always A) More B) Less C) Same D) Unpredictable
💡Correct Answer: Option A (More)
Due to specimen size effects and platen restraint friction extending over a larger percentage of height in smaller cubes, the apparent compressive strength of a smaller 100 mm concrete cube is approximately 5% to 10% higher (factor ~1.05 to 1.10) than that of a standard 150 mm cube.
Which of the following is a tensile test of a cylindrical concrete sample, A) Compaction factor test B) Le Chatelier's test C) Splitting test D) Autoclave test
💡Correct Answer: Option C (Splitting test)
The split cylinder tensile test (IS 5816 / ASTM C496) is an indirect tension test where a cylindrical concrete specimen (150 mm diameter x 300 mm long) is subjected to a compressive line load along two diametrically opposite lines, inducing uniform tensile stresses perpendicular to the diameter and causing splitting along the vertical diameter (f_ct = 2P / (π D L)).
The main reinforcement of a RC slab consists of 6 mm bars at 10 cm spacing. If it is desired to replace 6 mm bars by 12 mm bars, then spacing of 12 mm bars should be A) 10 cm B) 40 cm C) 60 cm D) 120 cm
💡Correct Answer: Option B (40 cm)
For equal total steel reinforcement area per unit width: A_st = (a_s 1000) / S = constant => S ∝ a_s ∝ d². Ratio of cross-sectional area of 12 mm bar to 6 mm bar: a_s2 / a_s1 = (12 / 6)² = 2² = 4. Therefore, the new spacing must be: S₂ = S₁ 4 = 10 cm * 4 = 40 cm.
The void ratio of a fully saturated soil sample having a porosity of 0.3 is A) 0.26 B) 0.66 C) 0.43 D) 1.0
💡Correct Answer: Option C (0.43)
The relationship between void ratio (e) and porosity (n) is: e = n / (1 - n). Given porosity n = 0.3: e = 0.3 / (1 - 0.3) = 0.3 / 0.7 = 3 / 7 = 0.42857 ≈ 0.43.
Q62
Civil EngineeringGeotechnical Engineering - Density Relationships
The soil has a bulk density of 55 kN/m³ and water content 10%. The dry density of soil is A) 5.5 kN/m³ B) 10 kN/m³ C) 50 kN/m³ D) 55 kN/m³
💡Correct Answer: Option C (50 kN/m³)
Dry density γ_d is related to bulk density γ and moisture content w by: γ_d = γ / (1 + w). Here γ = 55 kN/m³ and w = 10% = 0.10. Therefore, γ_d = 55 / (1 + 0.10) = 55 / 1.10 = 50 kN/m³.
Lime stabilization is very effective in treating A) Sandy soil B) Silty soil C) Non-plastic soil D) Plastic clayey soil
💡Correct Answer: Option D (Plastic clayey soil)
Lime stabilization relies on cation exchange and pozzolanic reactions between lime (calcium hydroxide) and the silica/alumina in clay minerals. It causes flocculation, sharply reduces the plasticity index, controls swell-shrink behavior, and increases bearing capacity. Thus, it is most effective in highly plastic clayey soils.
Soil which contains finest grain particles A) Fine sand B) Sand C) Clay D) Silt
💡Correct Answer: Option C (Clay)
As per the Indian Standard Soil Classification System (IS 1498): Gravel (> 4.75 mm), Sand (0.075 mm to 4.75 mm), Silt (0.002 mm to 0.075 mm), and Clay (< 0.002 mm or < 2 μm). Clay has the finest particle size.
The minimum water content at which the soil just begins to crumble when rolled into threads of 3 mm in diameter, is known as A) Liquid limit B) Plastic limit C) Shrinkage limit D) Permeability limit
💡Correct Answer: Option B (Plastic limit)
The Plastic Limit (PL or w_p) of a soil is defined as the minimum moisture content at which the soil can be rolled into threads of 3 mm diameter without crumbling, transitioning from the plastic state to the semi-solid state.
Q66
Civil EngineeringGeotechnical Engineering - Consistency Indices
Consider soil as a three-phase system, the ratio of (Liquid Limit - Natural water Content) to Plasticity Index for the soil mass is called A) Liquidity index B) Shrinkage ratio C) Consistency index D) Toughness index
💡Correct Answer: Option C (Consistency index)
The Consistency Index (or Relative Consistency, I_c) is defined as: I_c = (w_L - w) / I_p = (Liquid Limit - Natural water Content) / Plasticity Index. (In contrast, Liquidity Index is I_l = (w - w_P) / I_p).
Q67
Civil EngineeringGeotechnical Engineering - Grain Size Distribution
Which of the following is a measure of particle size range? A) Effective size B) Uniformity coefficient C) Effective diameter D) Elongation index
💡Correct Answer: Option B (Uniformity coefficient)
The Uniformity Coefficient (C_u = D₆₀ / D₁₀) is a numerical measure of the particle size range or width of grain sizes present in a soil mass. A higher C_u indicates a wide range of particle sizes (well-graded soil), while C_u close to 1 indicates a uniform soil of narrow particle size range.
Q68
Civil EngineeringGeotechnical Engineering - Proctor Compaction Test
From a laboratory proctor compaction test data, the mass of the soil and water content is found to be 1100 g and 10% respectively. The volume of the proctor mould used is 1000 cm³, then the dry density (γ_d) of the soil is A) 1 g/cm³ B) 1.10 g/cm³ C) 1.50 g/cm³ D) 1.15 g/cm³
💡Correct Answer: Option A (1 g/cm³)
Bulk density ρ = Total Mass / Mould Volume = 1100 g / 1000 cm³ = 1.10 g/cm³. Moisture content w = 10% = 0.10. Dry density ρ_d = ρ / (1 + w) = 1.10 / (1 + 0.10) = 1.10 / 1.10 = 1.0 g/cm³.
Q69
Civil EngineeringGeotechnical Engineering - Shear Strength of Soil
In an unconfined compression test, a saturated clay sample, fails under a load of 150 N. Consider final cross-section area (A_f) = 2250 mm². Then the shear resistance (cohesion) of the sample is A) 66.21 kPa B) 34.95 kPa C) 33.33 kPa D) 15 kPa
💡Correct Answer: Option C (33.33 kPa)
In an unconfined compression test on saturated clay (φ = 0): Unconfined compressive strength q_u = Failure Load / A_f = 150 N / (2250 10⁻⁶ m²) = (150 / 2250) 10⁶ Pa = 66,666.67 Pa = 66.67 kPa. Undrained shear strength (cohesion) c_u = q_u / 2 = 66.67 kPa / 2 = 33.33 kPa.
Consider the following pairs: i. Standard Proctor Test - 1. [C, φ], apparent cohesion and angle of internal friction ii. Modified Proctor Test - 2. 03 layers, 310 mm height of drop iii. Direct Shear Test - 3. 05 layers, 450 mm height of drop iv. Triaxial Test - 4. [C', φ'], strength parameter Which of the pair given below is correctly matched? A) (i) - (1), (ii) - (2), (iii) - (4), (iv) - (3) B) (i) - (2), (ii) - (3), (iii) - (1), (iv) - (4) C) (i) - (3), (ii) - (2), (iii) - (1), (iv) - (4) D) (i) - (4), (ii) - (3), (iii) - (1), (iv) - (2)
💡Correct Answer: Option B ((i) - (2), (ii) - (3), (iii) - (1), (iv) - (4))
Matching geotechnical tests with their parameters and specifications: (i) Standard Proctor Test: Compaction in 3 layers with 310 mm hammer drop -> 2 (ii) Modified Proctor Test: Heavy compaction in 5 layers with 450 mm hammer drop -> 3 (iii) Direct Shear Test: Yields total/apparent shear parameters [c, φ] -> 1 (iv) Triaxial Test: Measures pore water pressure to evaluate effective shear parameters [c', φ'] -> 4 Therefore, the correct code is (i)-2, (ii)-3, (iii)-1, (iv)-4.
Q71
Civil EngineeringFluid Mechanics - Fluid Properties & Newton's Law
A fluid is said to be Newtonian fluid when its shear stress is A) Inversely proportional to the velocity gradient B) Proportional to the velocity gradient C) Independent to the velocity gradient D) Inversely proportional to the shear stress rate
💡Correct Answer: Option B (Proportional to the velocity gradient)
According to Newton's law of viscosity, a fluid is Newtonian when the shear stress τ is directly proportional to the rate of shear strain or velocity gradient: τ = μ * (du/dy).
Q72
Civil EngineeringFluid Mechanics - Fluid Units & Dimensions
Pascal-second (Pa.s) is the unit of A) Pressure B) Specific gravity C) Dynamic viscosity D) Compressibility
💡Correct Answer: Option C (Dynamic viscosity)
Pascal-second (Pa·s = N·s/m² = kg/(m·s)) is the SI unit of dynamic (or absolute) viscosity μ. (Kinematic viscosity ν has the SI unit m²/s, pressure has Pa or N/m², and specific gravity is dimensionless).
If specific gravity of a fluid is 0.5, and dynamic viscosity is 0.5 poise, then the kinematic viscosity of that fluid is A) 0.25 stokes B) 0.50 stokes C) 1.0 stokes D) 25 stokes
💡Correct Answer: Option C (1.0 stokes)
In CGS units: 1 poise = 1 g/(cm·s). Since water has density 1 g/cm³, a fluid of specific gravity S = 0.5 has density ρ = 0.5 g/cm³. Kinematic viscosity ν = μ / ρ = 0.5 poise / 0.5 g/cm³ = 1.0 cm²/s = 1.0 stokes.
Q74
Civil EngineeringFluid Mechanics - Hydrostatic Pressure & Force
A uniformly tapering vessel is filled with liquid of density 900 kg/m³. Then the force that act on the base of the vessel due to the liquid (g = 10 m/sec²) is: (Given: Height h = 0.4 m, Base Area = 2 x 10⁻³ m², Top Area = 10⁻³ m²) A) 3.6 N B) 7.2 N C) 9.8 N D) 15.4 N
💡Correct Answer: Option B (7.2 N)
Hydrostatic pressure at the flat bottom base of the vessel depends only on the depth of liquid h and liquid density ρ (hydrostatic paradox): P = ρ g h = 900 kg/m³ 10 m/s² 0.4 m = 3600 N/m². Total force on the horizontal base area = P A_base = 3600 N/m² (2 * 10⁻³ m²) = 7.2 N.
The velocity of the upper layer of water in a river is 36 km h⁻¹. Shearing stress between horizontal layers of water is 10⁻³ N m⁻². Consider co-efficient of viscosity of water is 10⁻² Pa-s. Then depth of the river is A) 100 m B) 200 m C) 360 m D) 3600 m
💡Correct Answer: Option A (100 m)
Velocity of surface layer v = 36 km/h = 36 (5/18) = 10 m/s. Assuming river bed velocity is 0 and velocity varies linearly with depth y: Velocity gradient dv/dy = v / y = 10 / y. From Newton's law: τ = μ (dv/dy) => 10⁻³ N/m² = 10⁻² Pa·s * (10 / y) => 10⁻³ = 10⁻¹ / y => y = 10⁻¹ / 10⁻³ = 100 m.
Q76
Civil EngineeringFluid Mechanics - Pressure Intensity in Layered Fluids
Open tank contains 1 m deep water with 50 cm depth of oil of specific gravity 0.6 above it, then the intensity of pressure at the bottom of tank is (taken, g = 1000 cm s⁻²) A) 3 kN/m² B) 13 kN/m² C) 30 kN/m² D) 40 kN/m²
💡Correct Answer: Option B (13 kN/m²)
Pressure intensity at tank bottom: P = P_oil + P_water = (ρ_oil g h_oil) + (ρ_water g h_water). Given g = 1000 cm/s² = 10 m/s², h_oil = 50 cm = 0.5 m, ρ_oil = 0.6 1000 = 600 kg/m³, h_water = 1.0 m, ρ_water = 1000 kg/m³. P_oil = 600 10 0.5 = 3000 N/m² = 3 kN/m². P_water = 1000 10 * 1.0 = 10,000 N/m² = 10 kN/m². Total pressure P = 3 + 10 = 13 kN/m².
The following figure shows a venturimeter, through which water is flowing. The speed of water at "X" is 2 cm s⁻¹ and height difference between "X" and "Y" is 5.1 mm. Then the speed of water at "Y" is (taken, g = 1000 cm s⁻²) A) 23 cm s⁻¹ B) 32 cm s⁻¹ C) 101 cm s⁻¹ D) 1024 cm s⁻¹
💡Correct Answer: Option B (32 cm s⁻¹)
Applying Bernoulli's equation between inlet section X and throat section Y: P_X / (ρ g) + v_X² / (2g) = P_Y / (ρ g) + v_Y² / (2g) => (P_X - P_Y) / (ρ g) = (v_Y² - v_X²) / (2g). The piezometric head difference h = (P_X - P_Y) / (ρ g) = 5.1 mm = 0.51 cm. Given g = 1000 cm/s² and v_X = 2 cm/s: v_Y² - v_X² = 2 g h = 2 1000 0.51 = 1020 cm²/s². Therefore, v_Y² = 1020 + v_X² = 1020 + 2² = 1020 + 4 = 1024 cm²/s². Speed v_Y = √1024 = 32 cm s⁻¹.
Q78
Civil EngineeringFluid Mechanics - Torricelli's Law & Orifice Flow
An open tank filled with water (density = ρ and g = 9.81 m/sec²) has a narrow hole at a depth of "h" below the water surface. Then the velocity of water flowing out is A) √(2gh) B) ρgh C) 2gh D) gh
💡Correct Answer: Option A (√(2gh))
According to Torricelli's Law of efflux, the theoretical velocity of liquid flowing out of a small orifice or narrow hole at depth h below the free liquid surface under gravity is given by v = √(2gh).
Q79
Civil EngineeringFluid Mechanics - Reynolds Number & Pipe Flow
The Reynold's number for fluid flow in a pipe is independent of A) The viscosity of the fluid B) The velocity of the fluid C) The length of the pipe D) The diameter of the pipe
💡Correct Answer: Option C (The length of the pipe)
Reynolds number for internal flow in a circular pipe is defined as Re = (ρ v D) / μ = (v * D) / ν, where ρ is fluid density, v is mean flow velocity, D is internal pipe diameter, and μ is dynamic viscosity. It is completely independent of the length of the pipe.
Q80
Civil EngineeringFluid Mechanics - Hydrostatic Force & Centre of Pressure
The point in the immersed body through which the resultant of the total pressure of the liquid be taken to act is known as A) Metacenter B) Centre of gravity C) Centre of buoyancy D) Centre of pressure
💡Correct Answer: Option D (Centre of pressure)
The Center of Pressure is defined as the point on an immersed surface through which the resultant total hydrostatic pressure force acts. (Centre of buoyancy is the centroid of the displaced volume of liquid).
Bernoulli's principle is derived from the law of conservation of A) Energy B) Mass C) Linear momentum D) Angular momentum
💡Correct Answer: Option A (Energy)
Bernoulli's principle is derived by integrating Euler's equation of motion along a streamline for steady, incompressible, frictionless flow, and represents the law of conservation of energy (pressure energy + kinetic energy + potential energy = constant).
Q82
Civil EngineeringFluid Mechanics - Surface Tension & Surface Energy
If a mercury drop is divided into 8 equal parts, then its total surface energy A) Remains same B) Becomes twice C) Decreases by a factor of 8 D) Increases by a factor of 8
💡Correct Answer: Option B (Becomes twice)
Let initial drop radius be R. When split into n = 8 identical drops of radius r: Conservation of volume gives (4/3)π R³ = 8 (4/3)π r³ => R³ = 8 r³ => r = R / 2. Initial surface area A₁ = 4π R²; initial surface energy U₁ = T 4π R². Total surface area of 8 smaller drops A₂ = 8 (4π r²) = 8 4π (R/2)² = 8 4π (R²/4) = 2 (4π R²) = 2 A₁. Thus, the total surface energy becomes twice (U₂ = 2 U₁).
The continuity equation is based on which of the following principle A) Conservation of mass B) Conservation of static energy C) Conservation of kinetic energy D) Conservation of momentum
💡Correct Answer: Option A (Conservation of mass)
The continuity equation (A₁V₁ = A₂V₂ for incompressible flow, or ∂ρ/∂t + ∇·(ρV) = 0) is directly derived from the fundamental physical principle of Conservation of Mass.
Q84
Civil EngineeringFluid Mechanics - Pipe Flow Energy Losses
The major causes for the loss of energy in long pipe is due to A) Sudden contraction B) Sudden enlargement C) Loss at the exit of the pipe D) Friction in the pipe
💡Correct Answer: Option D (Friction in the pipe)
In long pipelines, pipe wall friction (given by Darcy-Weisbach equation: h_f = 4fLV² / (2gD)) is the major loss of head/energy, whereas losses due to sudden enlargement, contraction, entrance, exit, and fittings are minor losses and are negligible compared to friction over long lengths.
Consider the following pairs: i. Pitot tube - 1. Flow and viscous property of fluid ii. Rheometer - 2. Flowing fluid velocity at any point iii. Hydrometer - 3. Pressure of the pipeline iv. Manometer - 4. Specific gravity or density of a liquid Which of the pair given below is correctly matched? A) (i) - (1), (ii) - (2), (iii) - (3), (iv) - (4) B) (i) - (1), (ii) - (4), (iii) - (3), (iv) - (2) C) (i) - (2), (ii) - (1), (iii) - (4), (iv) - (3) D) (i) - (4), (ii) - (3), (iii) - (2), (iv) - (1)
💡Correct Answer: Option C ((i) - (2), (ii) - (1), (iii) - (4), (iv) - (3))
Matching measuring instruments with their functions: (i) Pitot tube: Measures flowing fluid velocity at a point -> 2 (ii) Rheometer: Measures flow and rheological/viscous properties of fluids -> 1 (iii) Hydrometer: Measures specific gravity or density of a liquid -> 4 (iv) Manometer: Measures pressure of fluid in a pipeline -> 3 Therefore, the correct matched combination is (i)-(2), (ii)-(1), (iii)-(4), (iv)-(3).
Q86
Civil EngineeringHighway Engineering - CBR Test & Compaction
If N = Number of layers, W = Weight of the hammer, H = Height of fall, then which of the following option is correct for California Bearing Ratio (CBR) Test using heavy compaction? A) N = 3, W = 4.89 kg and H = 31 cm B) N = 3, W = 4.89 kg and H = 45 cm C) N = 5, W = 4.89 kg and H = 45 cm D) N = 5, W = 4.89 kg and H = 31 cm
💡Correct Answer: Option C (N = 5, W = 4.89 kg and H = 45 cm)
According to IS 2720 (Part 8 and Part 16) for heavy compaction in a CBR mould (modified compaction): The soil is compacted in N = 5 equal layers, using a hammer of weight W = 4.89 kg (or 4.90 kg) with a free drop height H = 45 cm (450 mm), with 56 blows per layer. Thus, N = 5, W = 4.89 kg, and H = 45 cm.
Q87
Civil EngineeringGeotechnical & Highway Engineering - Particle Size Distribution
The coefficient of uniformity is defined as A) Cu = D₃₀ / D₁₀ B) Cu = D₆₀ / D₁₀ C) Cu = [D₃₀]² / D₁₀ D) Cu = D₃₀ / (D₁₀ * D₆₀)
💡Correct Answer: Option B (Cu = D₆₀ / D₁₀)
The Coefficient of Uniformity (C_u) is defined as the ratio of D₆₀ (particle size such that 60% of soil particles are finer) to D₁₀ (effective size such that 10% are finer): C_u = D₆₀ / D₁₀. (Coefficient of curvature is C_c = [D₃₀]² / (D₆₀ * D₁₀)).
For a vehicle negotiating a curve, centrifugal force is i. Directly proportional to the weight of vehicle ii. Inversely proportional to square of velocity iii. Inversely proportional to radius of curve Which of the above statements are true? A) (i) and (ii) B) (ii) and (iii) C) (i) and (iii) D) (i), (ii) and (iii)
💡Correct Answer: Option C ((i) and (iii))
Centrifugal force acting on a vehicle traversing a horizontal curve of radius R at speed v is given by: F = m v² / R = (W v²) / (g * R). Hence, centrifugal force is: (i) Directly proportional to the weight W of the vehicle (Statement i is TRUE) (ii) Directly proportional to the square of velocity v² (Statement ii is FALSE as it claims inversely proportional) (iii) Inversely proportional to the radius of curve R (Statement iii is TRUE). Therefore, statements (i) and (iii) are true.
The penetration test of bitumen is conducted at X, measuring the penetration of needle loaded with Y load for Z time. The correct values are A) X = 20°C; Y = 50 g; Z = 10 sec B) X = 25°C; Y = 100 g; Z = 5 sec C) X = 27°C; Y = 100 g; Z = 60 sec D) X = 30°C; Y = 150 g; Z = 10 sec
💡Correct Answer: Option B (X = 25°C; Y = 100 g; Z = 5 sec)
According to IS 1203, the standard bitumen penetration test is conducted at a constant water bath temperature X = 25°C, with a total needle assembly load Y = 100 g, releasing the needle for a duration Z = 5 seconds. The depth of penetration is measured in tenths of a millimeter (1/10 mm).
For carrying out bituminous patch work during rainy season, the most suitable binder is A) Bitumen B) Bituminous emulsion C) Cutback bitumen D) Road tar
💡Correct Answer: Option B (Bituminous emulsion)
Bituminous emulsion contains microscopic droplets of bitumen suspended in water with an emulsifying agent. Because water is already its continuous carrier medium, it does not require heating and coats damp/wet aggregate surfaces exceptionally well, making it ideal for road maintenance and cold patch repair during wet and rainy weather.
Q91
Civil EngineeringHighway Engineering - Pavement Construction Sequence
Pick the right sequence of the different layers of a flexible pavement construction: i. Base course construction ii. Subbase course construction iii. Surface course and seal coat application iv. Natural subgrade and compacted subgrade construction v. Prime coat, binder course and tack coat application Choose the correct answer from the options given below: A) (i), (ii), (iii), (iv), (v) B) (i), (iii), (v), (ii), (iv) C) (iv), (ii), (i), (v), (iii) D) (iv), (v), (i), (ii), (iii)
💡Correct Answer: Option C ((iv), (ii), (i), (v), (iii))
The bottom-up sequence of constructing a flexible pavement is: 1. Subgrade preparation: Natural and compacted subgrade -> (iv) 2. Subbase course: Granular subbase (GSB) -> (ii) 3. Base course: Water Bound Macadam (WBM) or Wet Mix Macadam (WMM) -> (i) 4. Intermediate coats and binder course: Prime coat, Dense Bituminous Macadam (DBM) binder course, and tack coat -> (v) 5. Surfacing: Bituminous concrete surface course and seal coat -> (iii) Therefore, the correct sequence is (iv), (ii), (i), (v), (iii).
Alligator or map cracking is the most common type of failure in A) Bituminous surfacing B) Water Bound Macadam (WBM) surfacing C) Wet Mix Macadam (WMM) surfacing D) Telford and Macadam surfacing
💡Correct Answer: Option A (Bituminous surfacing)
Alligator cracking (or crocodile/map cracking) is a fatigue failure characterized by interconnected cracks resembling the skin of an alligator. It is the most common distress in flexible pavement bituminous surfacing, caused by repeated tensile strain under traffic wheel loads due to inadequate structural support or fatigue of the asphalt layer.
A vehicle was stopped in 2 second by fully jamming the break and skid marks measured 20 m then average skid resistance is (consider g = 10 m/sec²) A) 0.35 B) 0.50 C) 1.0 D) 2.2
💡Correct Answer: Option C (1.0)
Under uniform deceleration 'a' to rest in time t = 2 s over skid distance s = 20 m: The average velocity during braking is v_avg = v / 2 = s / t => v / 2 = 20 / 2 = 10 m/s => initial velocity v = 20 m/s. Deceleration a = v / t = 20 / 2 = 10 m/s² (or s = (1/2) a t² => 20 = (1/2) a 2² => 20 = 2a => a = 10 m/s²). The skid resistance (coefficient of friction f) is given by: f = a / g = 10 / 10 = 1.0.
Q94
Civil EngineeringHighway Engineering - Hill Road Structures
The wall which are necessary on the hill side of roadway where natural earth has to be retained from sliding is known as A) Retaining wall B) Breast wall C) Parapet wall D) Cavity wall
💡Correct Answer: Option B (Breast wall)
On hill roads: A Breast Wall is constructed on the cut/hill side of the roadway to protect and retain the natural cut hill slope from sliding and erosion. (A Retaining Wall is constructed on the valley side to retain the roadway embankment fill).
Consider the following pairs: i. Los Angeles abrasion - 1. Gradation ii. Size - 2. Toughness iii. Durability - 3. Hardness iv. Impact - 4. Soundness Which of the pair given below is correctly matched? A) (i)-(2), (ii)-(1), (iii)-(3), (iv)-(4) B) (i)-(2), (ii)-(4), (iii)-(3), (iv)-(1) C) (i)-(3), (ii)-(1), (iii)-(4), (iv)-(2) D) (i)-(3), (ii)-(4), (iii)-(1), (iv)-(2)
💡Correct Answer: Option C ((i)-(3), (ii)-(1), (iii)-(4), (iv)-(2))
Road aggregate test parameters: (i) Los Angeles abrasion test: Measures resistance to wear/abrasion -> Hardness (3) (ii) Sieve analysis / Size: Measures particle size distribution -> Gradation (1) (iii) Soundness test: Measures resistance to weathering/durability -> Soundness (4) (iv) Aggregate Impact test: Measures resistance to impact/shock -> Toughness (2) Hence, the correct matching is (i)-(3), (ii)-(1), (iii)-(4), (iv)-(2).
The fundamental relationship between space-mean speed (u), traffic density (k) and traffic flow (q) is A) q = u k
B) q = k / u
C) q = u / k
D) q = 1 / (u k)
💡Correct Answer: Option A (q = u * k)
The fundamental equation of traffic flow theory is: q = u * k, where q is traffic flow or volume (vehicles/hour), u is space-mean speed (km/hour), and k is traffic density (vehicles/km).
Q97
Civil EngineeringRailway Engineering - Permanent Way & Rail Types
Which of the following rails is most preferred by Indian railway A) Flat footed rails B) Bull head rails C) Double headed rails D) Thin headed rails
💡Correct Answer: Option A (Flat footed rails)
Flat-footed rails (Vignoles rails) are universally adopted by Indian Railways (accounting for over 90% of tracks) because of their broad base, high lateral stiffness, easy fixing directly to sleepers with simple fastenings, and economy without requiring separate chairs.
Gauge is the horizontal distance measured between A) Centre to center of two rails B) Inner or running faces of two rails C) Outer to inner face of two rails D) Inner to outer face of two rails
💡Correct Answer: Option B (Inner or running faces of two rails)
Track gauge is defined as the clear minimum horizontal distance between the inner (running) faces of the two rails of a railway track, measured at a specified depth below the top surface of the rail head.
Two major constituents in the composition of steel used in rail are A) Carbon and Silicon B) Carbon and Sulphur C) Carbon and Manganese D) Carbon and Lithium
💡Correct Answer: Option C (Carbon and Manganese)
Standard rail steel (such as 880 grade or medium manganese rail steel used in Indian Railways) primarily consists of Carbon (approx. 0.60% to 0.80% for high strength and hardness) and Manganese (approx. 0.80% to 1.30% for toughness and wear resistance).
The rising gradient on which a moving train takes the advantage of the preceding falling gradient, usually referred as A) Pusher gradient B) Neutral gradient C) Ruling gradient D) Momentum gradient
💡Correct Answer: Option D (Momentum gradient)
A gradient that is steeper than the ruling gradient but can be negotiated by a train because it is preceded by a falling gradient, allowing the train to build kinetic energy/momentum to overcome the steep incline, is called a momentum gradient.
Q101
Civil EngineeringRailway Engineering - Creep of Rails
The longitudinal movement of rails in a track is technically referred as, A) Hogging B) Buckling C) Creeping D) Tilting
💡Correct Answer: Option C (Creeping)
Creep of rails is the gradual longitudinal movement of rails in the track in the direction of traffic motion, caused by dynamic wave action of wheel loads, starting/braking acceleration, and temperature variations.
Q102
Civil EngineeringRailway Engineering - Defects in Rails
The slipping of driving wheels of locomotives on the rail surface causes, A) Wheel burns B) Hogging of rail C) Corrugation of rail D) Buckling of rail
💡Correct Answer: Option A (Wheel burns)
When locomotive driving wheels slip or spin in place upon starting on a steep grade or slippery rail head, extreme friction heat melts and tears the top surface of the rail steel, creating severe localized indentations known as 'wheel burns' (or engine burns).
Q103
Civil EngineeringRailway Engineering - Rail Wear
The wear of the rail is maximum at A) Tunnel B) Sharp curve C) Coastal area D) Square crossing
💡Correct Answer: Option B (Sharp curve)
Rail wear is greatest on sharp curves due to heavy lateral wheel flange thrust against the inner side of the outer rail head (side wear), slipping of wheel treads during traversing curves, and severe centrifugal forces.
Which among the following organization is the Research and Development (R&D) wing of Indian Railways A) IRCTC B) DRDO C) RDSO D) CRIS
💡Correct Answer: Option C (RDSO)
Research Designs and Standards Organisation (RDSO), located in Lucknow, functions as the sole Research and Development (R&D) organization and technical advisor to the Ministry of Railways, Indian Railways.
Consider the following pairs: i. Ballast - 1. Rail fastenings with concrete sleeper ii. Sleeper - 2. To hold the rail in proper gauge iii. Jim crow - 3. To bend the rail iv. Pandrol clip - 4. To provide cushion effect to the track Which of the pair given below is correctly matched? A) (i)-(2), (ii)-(1), (iii)-(3), (iv)-(4) B) (i)-(2), (ii)-(4), (iii)-(3), (iv)-(1) C) (i)-(3), (ii)-(1), (iii)-(4), (iv)-(2) D) (i)-(4), (ii)-(2), (iii)-(3), (iv)-(1)
💡Correct Answer: Option D ((i)-(4), (ii)-(2), (iii)-(3), (iv)-(1))
Matching railway components and tools with their functions: (i) Ballast: Provides elasticity and cushion effect to the track -> 4 (ii) Sleeper: Holds the rails to proper gauge and level -> 2 (iii) Jim crow: Mechanical appliance used to bend or straighten rails -> 3 (iv) Pandrol clip: Elastic rail fastening used with concrete sleepers -> 1 Therefore, the correct code is (i)-(4), (ii)-(2), (iii)-(3), (iv)-(1).
Q106
Civil EngineeringBuilding Construction - Brick Masonry Terms
_______ is the portion of a brick obtained by cutting the brick longitudinally into two equal parts? A) Queen closer B) King closer C) Half bat D) Bevelled closer
💡Correct Answer: Option A (Queen closer)
A Queen Closer is the piece of brick obtained by cutting a full brick lengthwise (longitudinally) into two equal halves. (A King closer is obtained by cutting off a corner triangular piece between middle of length and middle of width, while a half bat is obtained by cutting across the width).
Q107
Civil EngineeringEstimating and Costing - Methods of Building Estimation
In long and short wall method of estimation, the length of long wall is the center to center distance between the wall and A) breadth of the wall B) half breadth of the wall on each side C) one forth breadth of the wall on each side D) one fifth breadth of the wall on each side
💡Correct Answer: Option B (half breadth of the wall on each side)
In the long and short wall method of estimation, the out-to-out length of a long wall is obtained by adding half breadth of the wall on each side to the center-to-center distance (i.e. Length of long wall = c/c distance + 2 * (b/2) = c/c distance + half breadth of the wall on each side).
Q108
Civil EngineeringEstimating and Costing - Units of Measurement (IS 1200)
The Damp Proof Course (D.P.C) is measured in A) Cubic meter B) Square meter C) Meters D) Rupees per meter
💡Correct Answer: Option B (Square meter)
According to IS 1200 (Method of Measurement of Building and Civil Engineering Works), Damp Proof Course (DPC) has a specified thin thickness (normally 2.5 cm to 4 cm) and is measured in area units of Square meters (sq m or m²).
Q109
Civil EngineeringEstimating and Costing - Contracts and Tenders
While submitting a tender the contractor is required to deposit some amount with the department, as guarantee of the tender, known as A) Bank guarantee B) Caution money C) Security deposit D) Earnest money
💡Correct Answer: Option D (Earnest money)
Earnest Money Deposit (EMD, usually 1% to 2% of the estimated cost) is the monetary deposit required from contractors when submitting tenders to guarantee that the bidder is serious and will not back out of the bid if awarded. (Security deposit is deducted later from bills as performance security).
Estimated value of a built up property at the end of its useful life without being dismantled is known as, A) Scrap value B) Salvage Value C) Market value D) Book value
💡Correct Answer: Option B (Salvage Value)
Salvage Value is the estimated value of a property/utility at the end of its useful life period without being dismantled into pieces. In contrast, Scrap Value (junk value) is the value realized when the structure is completely demolished/dismantled and sold as scrap materials.
Q111
Civil EngineeringConstruction Management - CPM Network Elements
In a network diagram, an activity having a dotted arrow between two activity is referred as, A) Important activity B) Risky activity C) Dummy activity D) Delayed activity
💡Correct Answer: Option C (Dummy activity)
In CPM/PERT arrow network diagrams, a dummy activity is an artificial activity represented by a dashed or dotted arrow that consumes zero time and zero resources, used solely to establish logical precedence relationships and maintain unique node identification.
In the following network, choose the Critical Path from given options, (the number on the arrow shows duration of activity) Network nodes and durations: A -> B = 2 B -> C = 3 B -> D = 3 C -> D = 2 C -> E = 4 D -> E = 3 E -> F = 2 The correct option is: A) A-B-C-D-E-F B) A-B-D-E-F C) A-B-C-E-F D) A-B-D-C-E-F
💡Correct Answer: Option A (A-B-C-D-E-F)
Calculating the total duration for all paths through the network: Path 1: A-B-C-D-E-F = 2 + 3 + 2 + 3 + 2 = 12 Path 2: A-B-D-E-F = 2 + 3 + 3 + 2 = 10 Path 3: A-B-C-E-F = 2 + 3 + 4 + 2 = 11 The critical path is the path with the longest total duration, which is A-B-C-D-E-F with duration 12.
Q113
Civil EngineeringConstruction Management - PERT Expected Time & Variance
If the optimistic time (t_o), most likely or probable time (t_m), and pessimistic time (t_p), for activity (A) are 2, 5 and 14 days respectively, then expected duration and variance of the activity are A) 2 and 5 days B) 6 and 4 days C) 14 and 7 days D) 16 and 5 days
As per Critical Path Methodology (CPM) the earliest start time for an event (I) is 10 weeks. Activity (I-J) takes 4 weeks for completion. Event (J) starts after 20 weeks. Then the total float for activity (I-J) is, A) 2 weeks B) 6 weeks C) 12 weeks D) 16 weeks
💡Correct Answer: Option B (6 weeks)
Total float is given by TF = T_L^J - T_E^I - t_ij. Here, Earliest event time for I is T_E^I = 10 weeks, activity duration t_ij = 4 weeks, and completion/latest occurrence time for event J is T_L^J = 20 weeks. Total float = 20 - (10 + 4) = 20 - 14 = 6 weeks.
The difference between Late Start and Early Start Time of an activity is referred as, A) Float B) Progress of an activity C) Task of a work D) PERT task
💡Correct Answer: Option A (Float)
Total Float is defined as the maximum time by which an activity can be delayed without affecting the overall project completion time: Total Float = Late Start Time (LST) - Early Start Time (EST) = Late Finish Time (LFT) - Early Finish Time (EFT).
A scheduled activity may begin 10 days before the predecessor activity finishes. Then it's referred as an example of, A) Finish to start B) Finish to finish C) Start to finish D) Start to start
💡Correct Answer: Option A (Finish to start)
In project scheduling, the basic predecessor-successor logical relationship is Finish-to-Start (FS), where the successor normally starts after the predecessor finishes. When the successor is scheduled to begin 10 days prior to the predecessor's finish, it represents a Finish-to-Start relationship with a 10-day lead (FS - 10 days lead).
Following is a Rate list for an earth excavation work: Item 1: Manual excavation | Fixed Rate: Rs. 10 /cum | Variable rate: Nil Item 2: Machine excavation | Fixed Rate: Rs. 4000 fixed | Variable rate: Rs. 2 /cum As per above rate list, estimated quantity of earth, for which the cost of excavation by machine will be equal to the cost of manual excavation, considering variable rate is A) 500 cum B) 1000 cum C) 2500 cum D) 3000 cum
💡Correct Answer: Option A (500 cum)
Let Q be the quantity of earth in cubic meters (cum) at the break-even point where costs are equal: Cost of manual excavation = 10 Q
Cost of machine excavation = 4000 (fixed) + 2 Q Equating both costs: 10 Q = 4000 + 2 Q => 8 * Q = 4000 => Q = 4000 / 8 = 500 cum.
The cost slope is defined as A) (Crash cost - Normal cost) / Normal time B) (Crash cost - Normal cost) / Crash time C) Crash cost / (Normal time - Crash time) D) (Crash cost - Normal cost) / (Normal time - Crash time)
💡Correct Answer: Option D ((Crash cost - Normal cost) / (Normal time - Crash time))
In project crashing, cost slope is the direct cost increase per unit time saved by crashing an activity, defined mathematically as: Cost Slope = (Crash Cost - Normal Cost) / (Normal Time - Crash Time) = ΔC / Δt.
Q119
Civil EngineeringConstruction Management - Statistical Quality Control (IS 456)
Which among the following, means statistical quality control of RCC work at civil construction site? A) Minimizing cost for removal of defective work B) Applying theory of probability to sample testing or inspection C) Minimizing in wastage of inspection cost D) Calculating risk probably to minimize utilization
💡Correct Answer: Option B (Applying theory of probability to sample testing or inspection)
Statistical Quality Control (SQC) in civil construction involves applying the mathematical principles of probability theory and sampling distributions (e.g., standard deviation, characteristic strength at 95% confidence limit) to sample inspection and destructive cube testing to assess and assure concrete mix quality for the entire batch.
Consider the following pairs: i. Networking - 1. actual performance of the task and consume time or resources, represented by an arrow ii. Activity - 2. an arrow diagram obtained by connecting all the activities of a project in a logical sequence iii. Event - 3. sequential relationship between other activities but does not consume resources, and represented by a circle iv. Dummy activity - 4. start or completion of tasks and does not consume time or resources, and represent precedence relationship between real activities Which of the pair given below is correctly matched? A) (i)-(2), (ii)-(1), (iii)-(3), (iv)-(4) B) (i)-(2), (ii)-(4), (iii)-(3), (iv)-(1) C) (i)-(3), (ii)-(1), (iii)-(4), (iv)-(2) D) (i)-(3), (ii)-(4), (iii)-(1), (iv)-(2)
💡Correct Answer: Option A ((i)-(2), (ii)-(1), (iii)-(3), (iv)-(4))
Matching CPM network concepts with their definitions: (i) Networking: An arrow diagram obtained by connecting all the activities of a project in a logical sequence -> 2 (ii) Activity: Actual performance of a task that consumes time or resources, represented by an arrow -> 1 (iii) Event: Instant of start or completion of a task, consumes no time or resources, represented by a circle/node -> 3 (or 4 in standard text definitions, here paired logically as (i)-2, (ii)-1) (iv) Dummy activity: Artificial activity establishing sequential/precedence relationships without consuming resources -> 4 (or 3) Evaluating the provided multiple-choice option combinations, option A: (i)-(2), (ii)-(1), (iii)-(3), (iv)-(4) is the intended matching pair where Networking is 2 and Activity is 1.
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About JKSSB Junior Engineer (Civil) 2025 Previous Year Paper
This page provides the full solved question paper for the JKSSB Junior Engineer (Civil) examination conducted in 2025. Every MCQ is presented with verified answer keys and detailed bilingual explanations to support concept building.
Key subjects covered in this paper include Civil Engineering. Practicing authentic previous year questions is the proven way to master question patterns and improve accuracy for upcoming exams across Jammu & Kashmir.
