Question 1

Traffic engineering application: a collision (accident) diagram is prepared primarily to analyze patterns and to select countermeasures. Which statements are correct uses?

Accepted Answer

Introduction / Context: Collision diagrams are location-specific charts that plot accident details (type, severity, direction, time) at an intersection or along a road segment. They help traffic engineers identify hazardous patterns and design targeted treatments such as signal timing changes, geometric modifications, or signage improvements. Given Data / Assumptions: Statements: (1) study accident pattern; (2) eliminate accidents; (3) determine remedial measures; (4) make statistical analysis of accidents. Purpose is analytical/diagnostic, not a guarantee of elimination. Concept / Approach: Collision diagrams reveal recurring crash types and contributing factors (e.g., right-angle collisions, rear-ends), guiding the choice of countermeasures. While they inform statistical summaries, they are not themselves general statistical tools; broader crash databases serve that role. Importantly, diagrams aid in reducing risk but cannot “eliminate” accidents outright. Step-by-Step Solution: Confirm (1): Yes—primary use is to study patterns.Confirm (3): Yes—used to select/justify remedial measures.Evaluate (2): Overstated; no method guarantees elimination.Evaluate (4): While analysis may follow, the diagram itself is not the general statistical tool; it supports targeted analysis rather than full statistical treatment.Therefore, 1 and 3 are correct. Verification / Alternative check: Guidelines in traffic safety manuals describe collision diagrams as diagnostic aids to propose engineering, enforcement, or education measures. Why Other Options Are Wrong: (a) Includes elimination claim—unrealistic. (c) Omits the essential pattern-study role (1). (d) Retains elimination and overemphasizes statistics as a primary purpose. Common Pitfalls: Assuming any single tool can fully eliminate accidents; confusing supporting visuals with comprehensive statistical methods. Final Answer: 1 and 3 are correct (study accident pattern; determine remedial measures)

Question 2

Torsion in a circular bar: a bar of circular cross-section is fixed at A and subjected to a torque T at its free end B. What is the angle of twist at the fixed end A relative to the undeformed axis?

Accepted Answer

Introduction / Context: For a prismatic circular shaft under Saint-Venant torsion, the angle of twist varies linearly along the length when torque is applied at a free end and the other end is fixed. Understanding boundary conditions is essential—rotations at fixed supports are zero by definition. Given Data / Assumptions: Shaft is prismatic, isotropic, and linearly elastic. End A is fully fixed against rotation; end B is free and subjected to torque T. Uniform shear modulus G and polar moment of inertia J. Concept / Approach: For length L, the angle of twist between sections x apart is θ(x) = (T * x) / (G * J) (for constant T along the shaft). Setting x = 0 at the fixed end A gives θ(0) = 0, while at x = L (free end B) θ(L) = T L / (G J), the maximum twist. Step-by-Step Solution: Define x from A (fixed) to B (free): θ(x) = T x / (G J).At A (x = 0) → θ_A = 0.At B (x = L) → θ_B = T L / (G J) (maximum).Therefore, the twist at A is zero. Verification / Alternative check: Boundary condition of a fixed end precludes rotation; shear strain distribution also goes to zero at the fixed end for the displacement field. Why Other Options Are Wrong: Non-zero angles at the fixed end contradict boundary conditions. “Indeterminate” is incorrect; G and J affect magnitude, not the zero rotation at a fixed boundary. Common Pitfalls: Confusing slope/rotation diagrams with bending; forgetting that twist is referenced to the fixed end. Final Answer: Zero

Question 3

Masonry design: for a free-standing masonry wall (unsupported at the top and restrained only at the base), what effective height should be used when computing slenderness ratio?

Accepted Answer

Introduction / Context: Slenderness effects significantly influence the compressive capacity of masonry walls. Codes use an effective height (or length) to account for end restraint conditions. A free-standing wall behaves like a cantilever with one end fixed and the other free, increasing the effective height compared to a wall braced at both ends. Given Data / Assumptions: Wall is free-standing (top unrestrained laterally). Base is effectively fixed at foundation level. Effective height is used in slenderness ratio λ = effective_height / effective_thickness. Concept / Approach: For column/strut analogues, the effective length factors depend on end restraints: fixed–free members have an effective length of 2.0 L (twice the clear height). This increases slenderness and reduces allowable stress to reflect the higher propensity for buckling or instability. Step-by-Step Solution: Recognize end conditions: fixed at base, free at top.Use standard effective length factor K = 2.0 for fixed–free condition.Therefore, effective height = K * L = 2.0 L. Verification / Alternative check: Structural design handbooks and masonry codes list effective height multipliers; fixed–free consistently corresponds to 2.0. Why Other Options Are Wrong: 0.5 L and 1.0 L reflect braced conditions with more restraint than present. 2.5 L overstates the cantilever effect beyond standard practice. Common Pitfalls: Assuming the top has some bracing when it is explicitly free; using clear height directly without effective length factor. Final Answer: 2.0 L

Question 4

Cast-iron spigot-and-socket joint (water supply): arrange the correct sequence for making the joint from preparation through pouring and caulking.

Accepted Answer

Introduction / Context: Spigot-and-socket (bell-and-spigot) joints are traditional for cast-iron water mains. Correct assembly order ensures alignment, sealing, and structural integrity. The process combines mechanical packing with molten lead (historically) or other sealing materials. Given Data / Assumptions: Steps: 1) Wrap tarred gasket/hemp on spigot; 2) Enter spigot into socket; 3) Place joining ring against socket face; 4) Caulk gasket slightly; 5) Pour molten lead and then caulk. Objective: establish packing, contain molten lead, and finish by caulking. Concept / Approach: Initial centering of the spigot (2) prevents damage to packing. Wrapping gasket (1) before full insertion provides the packing to be driven into the annulus. Slight caulking (4) seats the packing and leaves room for lead. The joining ring (3) forms a dam to retain molten lead, which is then poured and caulked (5) to complete the seal. Step-by-Step Solution: (2) Insert/center spigot into socket to align components.(1) Wrap tarred gasket/hemp around spigot to provide packing.(4) Caulk packing slightly to seat it in the joint.(3) Place joining ring to contain molten lead.(5) Pour molten lead and caulk to finish. Verification / Alternative check: Standard installation guides show packing, preliminary caulking, setting a yarning/joining ring, then pouring and final caulking. Why Other Options Are Wrong: Sequences that place the ring before preliminary packing or skip preliminary caulking risk leakage or improper seating. Common Pitfalls: Over-caulking before setting the ring (reduces space for lead); misalignment before packing. Final Answer: 2, 1, 4, 3, 5

Question 5

Reinforced concrete slab detailing (IS practice): for a slab thickness of 75 mm, what is the maximum bar diameter permitted for reinforcement in the slab?

Accepted Answer

Introduction / Context: Slab reinforcement bar diameters are limited to ensure adequate concrete cover, crack control, and proper distribution across thin sections. Indian practice (e.g., IS 456) limits the diameter of bars in slabs relative to slab thickness. Given Data / Assumptions: Slab thickness t = 75 mm. Rule of thumb/IS guidance: maximum bar diameter in slabs should not exceed t / 8 for main bars. Adequate cover and spacing are achievable. Concept / Approach: Compute the limit using the slab-thickness criterion. The chosen bar should not exceed this limit, ensuring proper anchorage, crack distribution, and workability during placement and compaction. Step-by-Step Solution: Compute t / 8 = 75 / 8 = 9.375 mm.Select a standard bar ≤ 9.375 mm → 8 mm qualifies; 10 mm exceeds the cap.Therefore, the maximum permissible diameter is 8 mm. Verification / Alternative check: Checking cover requirement and spacing confirms 8 mm bars are routinely used in thin slabs and toppings. Why Other Options Are Wrong: 10 mm and 12 mm exceed the t/8 limit for 75 mm slabs. 6 mm is permitted but not the maximum; the question asks for the maximum allowable. Common Pitfalls: Using column/beam bar limits; forgetting that slabs have stricter limits due to thin sections. Final Answer: 8 mm diameter

Question 6

Simple shear of a square block (mechanics of materials): with side AD fixed and a simple shear applied producing shear stress τ and engineering shear strain γ, what is the linear (normal) strain along a diagonal at 45° to the sides?

Accepted Answer

Introduction / Context: Under simple shear, a square element distorts into a rhombus. Although the pure shear state has zero normal strain in the original x and y directions (for small strains), there are tensile and compressive normal strains along directions rotated by 45°, which correspond to principal strain directions. Given Data / Assumptions: Small (engineering) shear strain γ relates to shear stress by γ = τ / G (for linear elastic isotropic material). We examine linear strain along diagonals oriented at ±45° to the original axes. Plane strain/plane stress small-strain kinematics apply. Concept / Approach: For pure/simple shear, principal normal strains are ±γ / 2 along directions at ±45° to the shear directions. Thus, one diagonal experiences tensile normal strain of magnitude γ/2; the orthogonal diagonal experiences compressive normal strain of magnitude γ/2. Step-by-Step Solution: Start with shear strain tensor having only γ_xy = γ_yx = γ/2 (engineering shear γ).Rotate axes by 45°: the shear components transform to normal components.Result: ε_45° = +γ/2 (tension) and ε_-45° = −γ/2 (compression).Therefore, the linear strain along one diagonal is γ/2 in tension. Verification / Alternative check: Mohr’s circle for strain shows normal strain extrema at ±γ/2 located 90° apart on the circle, corresponding to physical directions at ±45° in the material. Why Other Options Are Wrong: Both diagonals cannot be in tension; one is tensile, the other compressive. Zero strain along diagonals is false for nonzero γ. γ/4 is not consistent with the standard transformation relations. Common Pitfalls: Mixing engineering shear γ with tensor components; forgetting the 45° rotation for principal directions. Final Answer: γ / 2 (tension) along one diagonal

Question 7

Pile geotechnics (clay end bearing): for square end-bearing piles in clays with average unconfined compressive strength qu = 6 t/m^2, what net ultimate bearing capacity may be assumed for design?

Accepted Answer

Introduction / Context: Design of end-bearing piles in saturated clays often uses undrained shear strength parameters. A common approach applies bearing capacity factors to the undrained cohesion to estimate ultimate capacity at the pile toe. This question assesses recall of the standard factor for deep foundations in clay. Given Data / Assumptions: Unconfined compressive strength qu = 6 t/m^2. Undrained cohesion cu = qu / 2 = 3 t/m^2. For deep foundations in clay under undrained conditions, Nc ≈ 9 for toe bearing. Net ultimate capacity is based on end bearing only (shaft ignored here). Concept / Approach: Net ultimate bearing pressure at the base in undrained clay is q_net,ult = Nc * cu, with Nc ≈ 9 (Skempton). Substituting cu = 3 t/m^2 yields 27 t/m^2. Step-by-Step Solution: Compute cu = qu / 2 = 6 / 2 = 3 t/m^2.Use q_net,ult = Nc * cu with Nc = 9.q_net,ult = 9 * 3 = 27 t/m^2.Therefore, adopt 27 t/m^2 as the net ultimate bearing capacity. Verification / Alternative check: Typical geotechnical texts list Nc = 9 for undrained clays beneath deep foundations; the calculation aligns with standard practice. Why Other Options Are Wrong: 15, 18, and 20 t/m^2 correspond to lower effective Nc values than standard undrained bearing for deep toe conditions. Common Pitfalls: Using qu directly instead of cu; confusing net with gross capacities; applying drained factors in undrained conditions. Final Answer: 27 t/m^2

Question 8

Surveying error propagation: a rectangular plot measuring 60 m × 20 m is measured with a steel tape; if the standard error in each side length is ±0.01 m (±1 cm), what is the standard error of the computed area?

Accepted Answer

Introduction / Context: In engineering surveying and construction layout, areas are typically computed from measured lengths. Because each length is subject to random error, the area inherits uncertainty. This problem tests your ability to propagate measurement errors from side lengths to the area using first-order (linear) error propagation. Given Data / Assumptions: Measured sides: L = 60 m and W = 20 m. Standard error in each dimension: σ_L = σ_W = 0.01 m (1 cm). Area A = L * W for a rectangle. Errors in L and W are independent and small (linearization valid). Concept / Approach: For a function A(L, W) = L * W, the variance of A is approximated by σ_A^2 = (∂A/∂L)^2 σ_L^2 + (∂A/∂W)^2 σ_W^2 when errors are uncorrelated. Here ∂A/∂L = W and ∂A/∂W = L. Compute σ_A from these partial derivatives and the given standard errors. Step-by-Step Solution: Compute partial derivatives: ∂A/∂L = W = 20, ∂A/∂W = L = 60.Form variance: σ_A^2 = (20)^2 (0.01)^2 + (60)^2 (0.01)^2.Evaluate: (400 + 3600) * 0.0001 = 4000 * 0.0001 = 0.4.Standard error: σ_A = sqrt(0.4) ≈ 0.632455… m^2.Rounded to three significant figures: ±0.632 m^2. Verification / Alternative check: If one side were error-free, σ_A would be W * σ_L = 20 * 0.01 = 0.2 m^2 (or L * σ_W = 0.6 m^2). Including both errors and the larger sensitivity to L (because L = 60) explains why σ_A exceeds 0.6 m^2 in quadrature, landing at ≈0.632 m^2. Why Other Options Are Wrong: ±0.1414 m^2: corresponds to using only a fraction of the variance terms. ±0.566 m^2: underestimates because it ignores the larger sensitivity to L. ±0.8484 m^2: overestimates by adding linearly instead of in quadrature. Common Pitfalls: Adding absolute errors instead of combining variances. Forgetting independence and covariance assumptions. Final Answer: ± 0.632 m2

Question 9

Short hollow circular column under eccentric load: for a column with external diameter D, internal diameter d, and load W applied with eccentricity e that causes zero stress at one extreme fibre, what is the required eccentricity e?

Accepted Answer

Introduction / Context: For short columns under combined axial load and bending, the extreme fibre stress is σ = W/A ± M y / I. Setting the minimum stress to zero yields the e that places the neutral axis at the far edge (no tension), a common design condition for masonry or concrete columns with limited tensile capacity. Given Data / Assumptions: Hollow circular section: external diameter D, internal diameter d. Area A = (π/4) (D^2 - d^2). Second moment of area I = (π/64) (D^4 - d^4). Extreme fibre distance y = D/2, bending moment M = W * e. Condition for zero minimum stress: W/A = M y / I. Concept / Approach: Impose σ_min = 0 → W/A = (W e) y / I → e = I / (A y). Substitute A, I, and y for a circular annulus and simplify algebraically to obtain e in terms of D and d only. Step-by-Step Solution: Start: e = I / (A y).Substitute: I = (π/64)(D^4 - d^4), A = (π/4)(D^2 - d^2), y = D/2.Cancel π and simplify constants: e = (1/8) * (D^4 - d^4) / [ D (D^2 - d^2) ].Factor: D^4 - d^4 = (D^2 - d^2)(D^2 + d^2).Cancel (D^2 - d^2): e = (1/8) * (D^2 + d^2) / D. Verification / Alternative check: For a solid circular section (d = 0), e reduces to (D/8), the classic result for zero tension at the far edge under eccentric axial load. Why Other Options Are Wrong: Using (D^2 - d^2) in the numerator fails to account for the D^4 - d^4 factorization. Denominator 4D doubles the correct eccentricity. Common Pitfalls: Mixing up I and A expressions for an annulus. Forgetting the extreme fibre distance y = D/2. Final Answer: e = (D^2 + d^2) / (8 D)

Question 10

Atmospheric absorption: which band of electromagnetic radiation is most effectively absorbed by carbon dioxide (CO2) near the lower boundary of the troposphere?

Accepted Answer

Introduction / Context: Greenhouse gases like carbon dioxide play a central role in the Earth’s energy balance by absorbing and re-emitting thermal radiation. Recognizing the spectral bands where CO2 absorbs strongly is fundamental in meteorology, climatology, and remote sensing. Given Data / Assumptions: CO2 absorption bands exist primarily in the thermal infrared. Question focuses on the lower troposphere (near-surface atmosphere). Solar UV/visible are largely unaffected by CO2 compared with IR bands. Concept / Approach: CO2 has strong vibrational transitions in the infrared, notably around 4.3 μm and 15 μm. These bands absorb terrestrial (longwave) radiation emitted by the surface and atmosphere, leading to greenhouse warming effects near the surface. Step-by-Step Solution: Identify spectral domain relevant to CO2: thermal IR, not X-ray or UV.Relate to greenhouse mechanism: Earth emits predominantly in IR; CO2 absorbs and re-emits in IR bands.Select infrared as the correct option. Verification / Alternative check: Atmospheric transmission curves show strong CO2 absorption near 15 μm, a cornerstone of greenhouse physics. Why Other Options Are Wrong: X-rays/UV are not significantly absorbed by CO2; other species and the upper atmosphere dominate UV absorption. Visible light passes through nearly unaffected by CO2. Common Pitfalls: Confusing ozone's UV absorption with CO2's IR absorption. Final Answer: Infra-red rays

Question 11

Pavement subgrade evaluation: if the pressure on a CBR specimen at 2.5 mm penetration is 3.5 N/mm^2, what is the California Bearing Ratio (CBR) of the soil?

Accepted Answer

Introduction / Context: The California Bearing Ratio (CBR) test evaluates the supporting strength of subgrade soils. The result is expressed as a percentage of a standard load corresponding to a specified penetration (usually 2.5 mm or 5.0 mm). Given Data / Assumptions: Measured pressure at 2.5 mm penetration: 3.5 N/mm^2. Standard pressure at 2.5 mm: 70 kg/cm^2 ≈ 6.9 N/mm^2 (accepted convention). CBR (%) = (Measured pressure / Standard pressure) * 100. Concept / Approach: Convert both values to consistent units (already in N/mm^2). The CBR is a simple ratio relative to the standard curve value for the same penetration depth, multiplied by 100. Step-by-Step Solution: CBR = (3.5 / 6.9) * 100.Compute ratio: ≈ 0.5072.Convert to percent: ≈ 50.7%, rounded ≈ 50% for typical reporting. Verification / Alternative check: Using the 5.0 mm standard pressure (≈ 10.3 N/mm^2) would give a different percentage; the question explicitly uses 2.5 mm, so 6.9 N/mm^2 is correct. Why Other Options Are Wrong: 10% and 35% underestimate relative to 3.5 vs 6.9. 70% overestimates; it would require ≈ 4.83 N/mm^2 at 2.5 mm. Common Pitfalls: Mixing the 2.5 mm and 5.0 mm standard loads. Final Answer: 50%

Question 12

Industrial building roof design: purlins placed over roof trusses to carry dead, live, and wind loads are, as per IS practice, assumed to behave structurally as which support condition?

Accepted Answer

Introduction / Context: Purlins are secondary structural members spanning over truss rafters to support roof sheeting. Their idealization affects bending moments, deflection, and connection design in industrial buildings. Given Data / Assumptions: Purlins are placed over multiple truss panel points. They are connected at each support, producing multi-span behavior. IS practice commonly models purlins as continuous over several spans. Concept / Approach: With more than two spans, purlins develop negative moments over supports and reduced mid-span positive moments compared to simply supported members. This continuous-beam action is closer to real behavior in typical industrial roofs and is reflected in codal design assumptions and moment coefficients. Step-by-Step Solution: Identify load path: sheeting → purlins → trusses → columns.Recognize multi-span continuity: connections at successive trusses enable continuous action.Select the support condition that matches: 'Continuous'. Verification / Alternative check: Design tables for purlins provide continuous-beam coefficients and reduction factors due to continuity. Why Other Options Are Wrong: Simply supported ignores continuity and overestimates mid-span moments. Cantilever/fixed do not represent typical purlin layouts in trussed roofs. Common Pitfalls: Treating every span independently, leading to conservative but uneconomical designs. Final Answer: Continuous

Question 13

Elasticity under self-weight: for a uniform bar, the ratio of the total elongation caused by the bar’s own weight to the elongation caused by an external end load equal to the bar’s weight is what?

Accepted Answer

Introduction / Context: Members often experience both self-weight and external loads. Comparing elongations from these two loading modes sharpens understanding of distributed vs concentrated loading in axial members. Given Data / Assumptions: Uniform prismatic bar: length L, cross-sectional area A, density ρ, modulus E. Self-weight acts as a uniformly distributed axial load along the length. External end load equals the total weight: W = ρ A g L. Concept / Approach: Elongation under an axial force P over length L is δ = P L / (E A) when P is constant. Under self-weight, internal axial force varies linearly with position; integrate strain along the length to obtain elongation. The ratio becomes independent of A, ρ, and E. Step-by-Step Solution: Elongation due to end load W: δ_W = W L / (E A) = (ρ A g L) L /(E A) = ρ g L^2 / E.At section x from the top, internal force from self-weight: P(x) = ρ A g (L - x).Differential elongation: dδ = P(x) dx / (E A) = [ρ g (L - x)/E] dx.Integrate 0→L: δ_sw = (ρ g / E) ∫_0^L (L - x) dx = (ρ g / E) (L^2 / 2) = ρ g L^2 /(2E).Ratio: δ_sw / δ_W = [ρ g L^2 /(2E)] / [ρ g L^2 / E] = 1/2. Verification / Alternative check: Dimensional consistency and independence from section properties confirm the generality of the 1/2 ratio. Why Other Options Are Wrong: 2 or 1 overstate the effect of self-weight relative to the same total end load. 1/4 understates it; correct integration yields 1/2. Common Pitfalls: Treating self-weight as a single lumped load at midspan instead of integrating. Final Answer: 1/2

Question 14

Plane stress under pure shear: for a rectangular element under pure shear of intensity q, the principal planes and principal stresses are what values?

Accepted Answer

Introduction / Context: Mohr’s circle analysis shows that pure shear stress states produce principal normal stresses oriented at ±45° to the original element faces. This result is widely used in mechanics of materials and failure criteria. Given Data / Assumptions: State of stress: τ_xy = q, σ_x = σ_y = 0 (pure shear). We seek principal planes (angles) and principal stresses (σ1, σ2). Concept / Approach: Principal planes are those with zero shear stress and extremal normal stress. For pure shear, rotation by 45° eliminates shear on the rotated faces, yielding normal stresses equal in magnitude and opposite in sign. Step-by-Step Solution: Mohr’s circle center at (σ_avg, 0) = (0, 0) with radius R = q.Principal stresses: σ1 = +q, σ2 = -q.Angle from original x-plane to principal plane: 2θ_p = 90° → θ_p = 45°.Thus principal planes at 45° and 135° to the original faces, with ±q as principal stresses. Verification / Alternative check: Direct transformation equations also give σ_θ = ±q at θ = 45° and τ_θ = 0. Why Other Options Are Wrong: 0°/90° or 30°/120° do not nullify shear under pure shear. Option with missing magnitudes is incomplete; correct values are ±q. Common Pitfalls: Confusing the physical angle θ with 2θ on Mohr’s circle. Final Answer: 45°, 135°; + q and - q

Question 15

Airport engineering (ICAO): what elevation correction rate is recommended for basic runway length with respect to height above mean sea level (MSL)?

Accepted Answer

Introduction / Context: Runway length corrections account for reduced air density at higher elevations, which impairs aircraft take-off performance. ICAO-recommended adjustments scale the basic runway length with altitude above MSL. Given Data / Assumptions: Correction is expressed as a percentage increase per specified elevation increment. Standard ICAO teaching reference uses 7% per 300 m elevation. Concept / Approach: As elevation increases, density decreases, requiring greater ground roll and therefore longer runway. The rule-of-thumb correction percentage is applied additively to the basic runway length determined under standard conditions. Step-by-Step Solution: Identify ICAO correction: +7% per 300 m.Apply proportionally to the basic runway length as needed. Verification / Alternative check: Airport design manuals and exam standards consistently list 7% per 300 m as the elevation correction. Why Other Options Are Wrong: Other percentages or increments do not match the ICAO recommendation used in standard problems. Common Pitfalls: Confusing elevation correction with temperature or gradient corrections, which are separate. Final Answer: 7% for every 300 m of elevation above MSL

Question 16

Mortar technology: why is sand added to cement mortar? Choose the correct set of reasons from the options provided.

Accepted Answer

Introduction / Context: Sand (fine aggregate) is a fundamental component of mortar. Its roles include improving workability, controlling shrinkage, and reducing cost by lowering cement content. Understanding the true benefits (and limits) avoids common misconceptions about strength. Given Data / Assumptions: (1) Decreasing the quantity of cement (economy). (2) Reducing shrinkage. (3) Decreasing the effective surface area of binding material by volumetric replacement. (4) Increasing the strength. Concept / Approach: Adding sand reduces paste content, which reduces drying shrinkage and thermal cracking. It also cuts cost by substituting a cheaper filler for expensive cement. While an appropriately graded sand improves packing and can optimize strength for a given cement content, statement (4) as a blanket claim is not generally correct; beyond an optimum, more sand reduces strength. Step-by-Step Solution: Accept (1): sand economizes cement usage.Accept (2): sand restrains shrinkage of the paste.Accept (3): by replacing paste volume, the exposed binder area to drying reduces; the statement reflects a standard rationale in mortar design.Reject (4): strength does not monotonically increase with sand; it depends on mix proportions and grading. Verification / Alternative check: IS mortar proportioning guidance shows optimum ranges for strength and workability; excessive sand lowers strength. Why Other Options Are Wrong: Options that include (4) overstate sand’s effect on strength. Common Pitfalls: Assuming more sand always increases strength; it usually reduces it past an optimum. Final Answer: 1, 2 and 3 are correct

Question 17

Thin-walled cylindrical pressure vessel: a cold-drawn seamless steel tube has diameter 6 cm, wall thickness 0.2 cm, and ultimate strength 3600 kg/cm^2. What is the bursting pressure (kg/cm^2)?

Accepted Answer

Introduction / Context: For thin cylinders, the critical stress for bursting is the circumferential (hoop) stress. Using the thin-cylinder formula allows estimation of the internal pressure that will drive hoop stress up to the material’s ultimate strength. Given Data / Assumptions: Diameter d = 6 cm, wall thickness t = 0.2 cm. Ultimate tensile strength σ_u = 3600 kg/cm^2. Thin-wall assumption (t ≪ d) holds; use membrane theory. Concept / Approach: Hoop stress for a thin cylinder: σ_h = p d / (2 t). Bursting pressure p_burst occurs when σ_h = σ_u. Solve for p at failure. Step-by-Step Solution: Set σ_u = p_burst * d / (2 t).Rearrange: p_burst = 2 t σ_u / d.Substitute: p_burst = 2 * 0.2 * 3600 / 6 = 1440 / 6 = 240 kg/cm^2. Verification / Alternative check: Longitudinal stress σ_l = p d /(4 t) is half the hoop stress; at bursting, σ_l = 1200 kg/cm^2, still below σ_u, consistent with hoop being critical. Why Other Options Are Wrong: 120 kg/cm^2: corresponds roughly to longitudinal stress at burst, not hoop. 480 and 960 kg/cm^2: exceed the thin-wall hoop failure prediction. Common Pitfalls: Using radius instead of diameter in σ_h = p d/(2 t) changes the factor. Mixing unit systems; all inputs are in consistent cm–kgf units. Final Answer: 240

Question 18

In building materials and finishes (civil engineering), distemper is primarily used as an economical wall coating for which type of surfaces and exposure conditions?

Accepted Answer

Introduction / Context: Distemper is a traditional, water-based decorative coating composed of chalk (whiting), binding agents like animal glue or casein, and pigments. It is widely used in building interiors where cost-effectiveness, ease of application, and a matte finish are desired. This question tests where distemper is appropriately specified relative to exposure, substrate, and durability requirements. Given Data / Assumptions: The finish in question is distemper (not acrylic emulsion or cement paint). Typical substrates are plastered masonry or gypsum-rendered internal walls. Exposure to rain/sun and frequent wetting is considered for exterior/utility spaces. Concept / Approach: Distemper provides a breathable, matte coating with moderate washability. It performs best on dry, interior walls away from moisture and abrasive cleaning. It is not recommended for exterior facades, damp-prone walls, or surfaces requiring frequent washing or high adhesion (e.g., woodwork). For exteriors or wet areas, cement paints, acrylic emulsions, or specialized coatings are preferred. Step-by-Step Solution: Identify surface type and exposure: interior dry walls are compatible with distemper.Assess durability needs: distemper has limited washability and poor water resistance, so unsuitable for exteriors and wet areas.Evaluate substrate compatibility: woodwork requires enamels or varnishes; distemper does not adhere or wear well on timber.Conclusion: select interior walls not exposed to weather as the correct application zone. Verification / Alternative check: Standard specifications list distemper for living rooms, bedrooms, and corridors, while recommending exterior-grade emulsions or cement paint for facades and moisture-prone zones. Why Other Options Are Wrong: External concrete surfaces: need weather-resistant paints; distemper lacks adequate durability. Woodwork: requires enamel/PU coatings; distemper is inappropriate. Compound walls: subject to moisture and algae; use cement paint or exterior emulsion. Countertops/floors: require abrasion-resistant finishes, not wall paints. Common Pitfalls: Using distemper in bathrooms/kitchens where splashing and frequent cleaning degrade the film; applying over damp plaster leading to peeling. Final Answer: Interior surfaces not exposed to weather

Question 19

Influence-line application: a simply supported beam of span 20 m carries a moving uniformly distributed load (UDL) of intensity 4 t/m over a loaded length of 5 m. What is the absolute maximum bending moment and where does it occur?

Accepted Answer

Introduction / Context: For moving loads on simply supported beams, the absolute maximum bending moment is found using the influence line for bending moment at a section. For a partial uniformly distributed load (UDL) shorter than the span, the maximum bending moment at a given section occurs when the load is placed symmetrically about that section. This problem applies the influence-line method to compute the peak value and confirm its location. Given Data / Assumptions: Span L = 20 m (simply supported). UDL intensity w = 4 t/m acting over a length a = 5 m. Beam is prismatic, linear-elastic; self-weight neglected compared to moving load. We seek the absolute maximum bending moment under the moving UDL. Concept / Approach: The influence line (IL) ordinate for bending moment at midspan of a simply supported beam is given by: ordinate at a distance s from midspan equals L/4 − s^2 / L. For a UDL of length a centered at midspan, the bending moment is the integral of the IL over the loaded length multiplied by w. Step-by-Step Solution: Place the 5 m UDL symmetrically about the midspan to maximize the moment there.Use IL ordinate at distance s from midspan: y(s) = L/4 − s^2 / L.Compute M_max at midspan: M = w ∫ from s = −a/2 to +a/2 of (L/4 − s^2 / L) ds.Evaluate integral: ∫(L/4) ds = (L a)/4; ∫(s^2/L) ds = (a^3)/(12 L).Thus, M = w * [ (L a)/4 − (a^3)/(12 L) ].Substitute L = 20 m, a = 5 m, w = 4 t/m → M = 4 * [ (20*5)/4 − (125)/(12*20) ] t·m.Compute numerically: (20*5)/4 = 25; (125)/(240) ≈ 0.520833; bracket ≈ 24.479167.Therefore M ≈ 4 * 24.479167 = 97.916668 t·m ≈ 97.9 t·m at midspan. Verification / Alternative check: Using “equivalent concentrated load” intuition, total load w a = 20 t placed at midspan gives 100 t·m; accounting for load spread slightly reduces to ≈ 97.9 t·m, matching the integral. Why Other Options Are Wrong: 87.5 t·m variants underestimate the moment; correct IL integration yields ≈ 97.9 t·m. 3.5 t·m is far too low for the given load and span. 125.0 t·m exceeds the point-load upper bound (20 * 20 / 4 = 100 t·m). Common Pitfalls: Placing the UDL off-center; using point-load IL directly without integrating; arithmetic slips in a^3/(12L) term. Final Answer: 97.9 t·m at the midpoint

Question 20

Sewer construction sequence: select the correct order for laying a sewer in a trench—sight rails, invert pegs, transferring center line, and placing pipes.

Accepted Answer

Introduction / Context: Accurate grade and alignment control are vital when laying sewers by gravity flow. Site procedures use sight rails, boning rods, and invert pegs to fix levels before the pipes are placed. This question checks practical field sequencing to ensure alignment and gradient are maintained. Given Data / Assumptions: Steps: (2) set sight rails, (3) drive pegs to invert level, (1) transfer center line to trench bottom, (4) place the sewer pipes. Trench is already excavated to near formation level. Team has standard leveling equipment. Concept / Approach: Sight rails provide permanent reference for elevation above the trench. Invert pegs (or grade pegs) are set using these references to define the pipe invert level. The center line is then transferred to the trench bottom to guide alignment. Only after establishing level and line should the pipes be placed to avoid rework. Step-by-Step Solution: (2) Establish sight rails across the trench to control levels.(3) Drive and level pegs to the sewer invert line using boning rods/levels.(1) Transfer the sewer center line to the trench bottom (plumb bob/line marking).(4) Lay and joint the sewer pipes using the pegs and line as guides. Verification / Alternative check: Field manuals describe this order to minimize disturbance and ensure grade accuracy before pipe placement. Why Other Options Are Wrong: Placing pipes before level/line control risks misalignment. Skipping pegs or transferring line too early can be disturbed by subsequent operations. Common Pitfalls: Failing to recheck pegs after rain or trench sloughing; not maintaining consistent boning rod length. Final Answer: 2, 3, 1, 4

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