Q: For a reinforced concrete (R.C.) water tank designed with M 25 concrete, what is the permissible bending tensile stress in concrete for the vertical wall to control cracking under service conditions?

Introduction / Context: Water-retaining structures are designed for crack control under service loads because leakage and durability are critical. Codes specify permissible tensile stresses in concrete (in bending and direct tension) that are lower than ordinary building limits to minimize crack widths, especially on the liquid face of walls and slabs. Given Data / Assumptions: Concrete grade: M 25 (characteristic compressive strength fck = 25 N/mm2). Member: vertical wall of an R.C. water tank. We seek permissible bending tensile stress for service design of liquid-retaining structures. Concept / Approach: For liquid-retaining structures (e.g., per IS guidelines for water tanks), permissible tensile stresses in concrete are intentionally restricted to limit crack initiation/width. Typical values for M 25 are about 1.8 N/mm2 in bending tension (on the liquid face) and even lower in direct tension. Higher values like 8.5 N/mm2 would be incompatible with crack control requirements and are not used for serviceability checks in water tanks. Step-by-Step Solution: Identify structure type → water-retaining R.C. tank requires stricter crack control.Recall permissible bending tension for M 25 → approximately 1.8 N/mm2.Select the matching option: 1.8 N/mm2. Verification / Alternative check: Design examples for liquid-retaining walls consistently adopt about 1.8 N/mm2 bending tensile limit for M 25 to satisfy leakage and durability criteria under sustained hydrostatic pressure. Why Other Options Are Wrong: 8.5 N/mm2, 2.5 N/mm2: Too high for liquid face service conditions; risk of cracking. 6.0 N/mm3: Wrong units; tensile stress should be N/mm2. 0.8 N/mm2: Overly conservative for M 25 bending tension. Common Pitfalls: Using general building code tensile limits instead of water-retaining criteria. Ignoring differential temperature/shrinkage which also influences cracking. Final Answer: 1.8 N/mm2

Q: For community water supply, what is the most appropriate single treatment for water drawn from a deep tube well (typically clear and low in turbidity)?

Introduction / Context: Groundwater from deep tube wells is usually low in turbidity and free of pathogenic contamination due to natural filtration through geological strata. However, it may still contain microorganisms or viruses and must meet drinking water safety standards. Choosing the minimal, effective treatment avoids unnecessary costs while ensuring public health. Given Data / Assumptions: Source: deep tube well (confined aquifer), generally clear with low suspended solids. No significant iron, manganese, or hardness issues indicated in the question. Objective: select the single essential treatment step for safe supply. Concept / Approach: Because turbidity is already low, coagulation, flocculation, and sedimentation are typically unnecessary. Filtration may not be required when particle load is negligible and well construction is proper. The indispensable step is disinfection (e.g., chlorination, UV), providing residual protection in the distribution system and inactivating potential pathogens that may enter during pumping or storage. Step-by-Step Solution: Assess raw water: deep well → low turbidity, low organics.Eliminate treatments aimed at solids removal (pre-settling, coagulation) absent turbidity.Retain the critical barrier: disinfection to assure microbiological safety.Choose “disinfection only.” Verification / Alternative check: Design manuals specify chlorination or UV as the required step for clear groundwater; additional treatments are added only if specific contaminants (iron, fluoride, arsenic) exceed limits. Why Other Options Are Wrong: Pre-settling/coagulation/filtration: Target turbidity and colloids; often unnecessary for deep groundwater. Aeration: Used for iron/manganese/CO2 removal—only when such issues exist. Common Pitfalls: Skipping residual disinfectant, leaving the system vulnerable to recontamination. Over-treating without a water quality basis, raising O&M costs. Final Answer: disinfection only

Question 1

In reinforced concrete design as per civil engineering practice, a pedestal is a compression member whose effective length does not exceed its least lateral dimension by more than which multiple?

Accepted Answer

Introduction / Context: In reinforced concrete (R.C.) structures, compression members are classified based on their geometry and effective length because buckling risk and design assumptions depend on slenderness. Very short compression members that primarily carry load in direct compression are termed pedestals, while longer members are designed and detailed as columns with slenderness considerations. Given Data / Assumptions: Member type: reinforced concrete compression member. Classification depends on effective length relative to the least lateral dimension. We seek the limiting multiple that defines a pedestal. Concept / Approach: Standards and textbooks define a pedestal as a very short compression member where slenderness effects are negligible. The commonly adopted criterion is that the effective length (distance between points of contraflexure for buckling about the critical axis) does not exceed three times the least lateral dimension. Within this limit, design may treat the member as a block in direct compression with appropriate detailing for load transfer and bearing. Step-by-Step Solution: Identify classification rule → pedestal vs. column depends on slenderness.Use the accepted limit: effective length ≤ 3 * (least lateral dimension).Select the numerical multiple that matches this rule → 3 times. Verification / Alternative check: Design manuals distinguish pedestals (stubby supports under base plates, beams, or walls) from columns. The 3-times rule is widely cited in R.C. design guides to separate direct-compression behavior from potential buckling behavior. Why Other Options Are Wrong: 12, 16 times: These are slenderness-type limits suitable for columns, not pedestals. 8 or 6 times: Still too large for the pedestal definition; buckling effects can be non-negligible. Common Pitfalls: Confusing effective length with clear height; boundary conditions affect effective length. Ignoring load transfer and bearing checks even for short pedestals (e.g., bearing stress under base plates). Final Answer: 3 times

Question 2

In geotechnical site investigations, a vane shear tester is normally used to determine the in-situ undrained shear strength of which soil type?

Accepted Answer

Introduction / Context: The field vane shear test is a rapid, minimally disturbing method for assessing undrained shear strength, especially where sampling quality is difficult to maintain. Choosing the correct soil type for which the vane test is appropriate is crucial for obtaining reliable design parameters for foundations, embankments, and excavations. Given Data / Assumptions: Tool: field vane shear tester (four-bladed vane). Measurement: undrained shear strength (su) in-situ. Environment: saturated fine-grained soils vs. coarse or granular soils. Concept / Approach: The vane test measures the torque required to cause shear failure along a cylindrical surface in the soil around the vane. It relies on the soil’s cohesion and is valid when drainage is negligible during the short test duration. This aligns with soft, saturated clays where undrained conditions prevail. In sands or gravels, disturbance, drainage, and particle interlock invalidate the test; in stiff clays, remolding and rate effects can lead to scatter, and alternative tools (e.g., CPTU correlations or high-quality sampling with laboratory triaxial) may be preferable. Step-by-Step Solution: Identify soil requiring undrained, cohesive behavior → soft, saturated clays.Match test principle (torque-induced undrained failure) to soil response.Select the most appropriate option → soft clays. Verification / Alternative check: Standard practice recommends the vane test primarily for very soft to soft clays and organic clays/silts; reported sensitivity can also be obtained by comparing peak and remolded strengths. Why Other Options Are Wrong: Sand/gravel: Drained, non-cohesive; vane readings are unreliable. Stiff clays: Possible but less reliable; sample disturbance and strain-rate dependency complicate interpretation. Silty sand: Predominantly frictional; vane test is unsuitable. Common Pitfalls: Testing in layered soils where sand seams cause premature failure planes. Not correcting for plasticity or sensitivity when interpreting results. Final Answer: soft clays

Question 3

A smooth nozzle discharges water under head H with outlet area a and discharge coefficient cd = 1.0. A vertical plate experiences force Fj when held across the free jet, and force Fn when pressed against the nozzle to stop the flow. What is the rati…

Accepted Answer

Introduction / Context: Fluid jet forces on surfaces are classic momentum problems in hydraulics. Comparing the force on a plate in a free jet versus the force required to arrest the jet at the nozzle exit highlights the difference between momentum flux and stagnation pressure action. Understanding this ratio is useful for designing gates, plates, and nozzle end-stops. Given Data / Assumptions: Nozzle outlet area = a, discharge coefficient cd = 1.0 (ideal discharge). Head H gives jet speed v = sqrt(2 g H); thus discharge Q = a * v. Case 1: Plate in the free jet (jet is brought to rest at the plate). Case 2: Plate held against the nozzle to stop flow right at the exit. Concept / Approach: For a free jet arrested by a plate, the force equals the rate of change of momentum: F = rho * Q * v = rho * a * v^2. For a plate closing the nozzle, the jet is not fully developed; the fluid is brought to rest at the exit with stagnation (dynamic) pressure p = 0.5 * rho * v^2 acting over area a, so F = p * a = 0.5 * rho * a * v^2. Therefore, the ratio Fj/Fn equals 2. Step-by-Step Solution: Compute v from head: v = sqrt(2 g H).Momentum force on plate in free jet: Fj = rho * a * v^2.Nozzle-blocking force: Fn = (1/2) * rho * v^2 * a.Take ratio: Fj/Fn = (rho a v^2) / (0.5 rho a v^2) = 2. Verification / Alternative check: The result is independent of H and a because both cases scale with rho * a * v^2. Real nozzles with cd ≠ 1.0 only change v slightly; the ratio remains approximately 2 for ideal assumptions. Why Other Options Are Wrong: 1 or 1/2 or 3/2 or 1/4: These contradict the momentum versus stagnation pressure comparison which yields exactly 2 under ideal conditions. Common Pitfalls: Applying pressure times area (0.5 rho v^2 a) to the free jet case; that underestimates the momentum force. Using gauge vs. absolute pressure incorrectly—forces depend on differences. Final Answer: 2

Question 4

The wall shown in the below figure has failed. The cause of failure or the error made in the design of the failded wall is :

Accepted Answer

overturning

Question 5

In one-dimensional consolidation testing (oedometer), the slope of the e–log p curve for a soil mass is used to determine which compressibility parameter?

Accepted Answer

Introduction / Context: Compressibility parameters from the oedometer (consolidation) test are essential for predicting settlement of clay layers. The void ratio (e) versus logarithm of vertical effective stress (log p) plot captures how soil structure collapses under load and is the basis for estimating primary consolidation settlement in fine-grained soils. Given Data / Assumptions: Test: one-dimensional consolidation (oedometer) under incrementally applied loads. Plot: e versus log p (p is vertical effective stress). We seek the parameter represented by the slope of the virgin compression line. Concept / Approach: The compression index Cc quantifies the virgin (normally consolidated) portion of the e–log p curve: Cc = (e1 − e2) / log10(p2/p1). It is used in settlement formulae to compute change in void ratio and corresponding strain for stress increases beyond the preconsolidation pressure. Other parameters relate to different aspects: Cv governs the rate (time) of consolidation; mv is the coefficient relating strain to stress in linear terms over small increments; k is hydraulic conductivity, not a direct slope on the e–log p plot. Step-by-Step Solution: Identify the e–log p plot from consolidation test.Recall definition: Cc is slope of the virgin compression line on this plot.Select the corresponding option: compression index, Cc. Verification / Alternative check: Recompression index Cr is the slope of the unloading–reloading (overconsolidated) portion and is smaller than Cc. This distinction confirms that the slope in the normally consolidated region reflects Cc. Why Other Options Are Wrong: k: Derived from permeability tests, not the e–log p slope. Cv: Obtained from time–settlement curves using Terzaghi’s theory. mv: Incremental parameter; not the e–log p virgin-line slope. Common Pitfalls: Confusing Cc and Cr or using natural logarithms without converting constants. Reading slopes over mixed regions (preconsolidation break not identified). Final Answer: compression index, Cc

Question 6

For turbulent flow in circular pipes, which statement about Prandtl’s mixing length is correct with respect to its variation near the wall and across the radius?

Accepted Answer

Introduction / Context: Prandtl’s mixing length hypothesis models turbulent momentum exchange by analogy with molecular diffusion, introducing a characteristic length scale l that represents the average distance over which eddies transport momentum before losing identity. Understanding how l varies in wall-bounded flows like pipes is essential for velocity profile modeling and shear stress estimation. Given Data / Assumptions: Flow: steady, fully developed turbulent flow in a smooth circular pipe. We consider the qualitative behavior of mixing length l across the radius. Wall effects dominate near the boundary; core flow differs. Concept / Approach: In wall-bounded turbulence, mixing length increases with distance from the wall: l ≈ κ * y in the log-law region, where y is the wall-normal distance and κ is the von Kármán constant (~0.4). At the wall itself, fluctuations are suppressed and l → 0. Therefore, l is not a universal constant and certainly not independent of radius. While shear stress informs velocity gradients, the canonical variation of l is tied to position, not fixed by a single shear value. Step-by-Step Solution: Recognize boundary-layer behavior → l grows with y away from the wall.At y = 0 (the pipe wall), l must vanish due to the no-slip boundary condition.Conclude the correct qualitative statement: l is zero at the pipe wall. Verification / Alternative check: Empirical velocity profiles (log-law) and eddy viscosity models (νt = l^2 |du/dy|) both require l = 0 at the wall to avoid non-physical finite turbulent shear right at the boundary. Why Other Options Are Wrong: Universal constant/independent of radius: Contradicts position dependence (l ∝ y in near-wall region). Independent of shear stress: Eddy viscosity depends on gradients; l is coupled to flow structure. Maximum at wall: Opposite of physical reality; turbulence production is suppressed at the wall. Common Pitfalls: Assuming l is constant across the section; that leads to unrealistic profiles. Ignoring viscous sublayer where turbulence is damped and l remains small. Final Answer: zero at the pipe wall

Question 7

A steady discharge of 1.0 m^3/s flows uniformly in a rectangular channel of width 1.0 m and flow depth 0.25 m. Based on this uniform flow regime, the channel bed slope is best classified as which type?

Accepted Answer

Introduction / Context: In open-channel hydraulics, flow regime (subcritical, critical, supercritical) relates to the Froude number. For a given discharge and geometry, uniform flow at a particular depth occurs only when the bed slope equals the normal slope for that regime. Recognizing whether a slope is mild, steep, critical, or adverse follows from checking the Froude number at the uniform depth. Given Data / Assumptions: Discharge Q = 1.0 m^3/s. Rectangular channel width b = 1.0 m. Uniform depth y = 0.25 m. Gravity g ≈ 9.81 m/s^2; friction and roughness chosen so this depth is the normal depth. Concept / Approach: Compute mean velocity v = Q / (b * y) = 1 / (1 * 0.25) = 4 m/s. The critical speed scale is sqrt(g * y) = sqrt(9.81 * 0.25) ≈ 1.566 m/s. The Froude number Fr = v / sqrt(g y) = 4 / 1.566 ≈ 2.56 > 1, indicating supercritical flow. For uniform supercritical flow to exist, the bed slope must be steeper than the critical slope; such channels are termed “steep” slopes. Mild slopes correspond to subcritical normal depths, while adverse slopes rise in the flow direction and cannot sustain the given uniform regime at this depth. Step-by-Step Solution: Compute velocity: v = 4 m/s.Compute wave celerity scale: sqrt(g y) ≈ 1.566 m/s.Froude number: Fr ≈ 2.56 → supercritical.Uniform supercritical regime ⇒ bed slope classification = steep. Verification / Alternative check: If the slope were mild, the normal depth would be subcritical (Fr < 1). Because the given uniform depth produces Fr > 1, only a steep slope is consistent with uniform flow at that depth. Why Other Options Are Wrong: Mild: Would yield subcritical normal depth (Fr < 1). Critical: Corresponds to Fr = 1, not 2.56. Adverse/horizontal: Not compatible with the stated uniform, supercritical regime. Common Pitfalls: Mixing up regime names with bed slope labels (mild vs. steep). Forgetting that regime is evaluated at normal (uniform) conditions. Final Answer: steep

Question 8

According to the effective stress principle in soil mechanics, consolidation of saturated soils primarily depends on which stress quantity?

Accepted Answer

Introduction / Context: Consolidation is the time-dependent compression of saturated fine-grained soils due to expulsion of pore water under sustained loading. Terzaghi’s effective stress principle is foundational: soil skeleton deformation—and thus long-term settlement—depends on effective stress, not total stress alone. Given Data / Assumptions: Saturated soil under load; drainage allowed over time. Total stress σ, pore water pressure u, effective stress σ′ where σ′ = σ − u. Primary interest: what stress governs consolidation magnitude. Concept / Approach: When load is applied, total stress rises instantly, but pore pressure initially carries part or all of the increment (undrained). With time, water drains, pore pressure dissipates, and effective stress increases. Settlement occurs as the soil skeleton takes more load. Therefore, final consolidation settlement correlates with change in effective stress and compressibility (Cc, mv), not directly with total stress or pore pressure alone. Step-by-Step Solution: State effective stress: σ′ = σ − u.Recognize consolidation mechanism: dissipation of u → increase in σ′.Conclude governing quantity: consolidation is a function of effective stress. Verification / Alternative check: Laboratory oedometer tests interpret settlement using effective stress changes and the e–log p relationship, confirming σ′ control. Different drainage boundaries change the rate (Cv), not the governing stress measure for magnitude. Why Other Options Are Wrong: “Does not depend on the present stress”: False; stress level and history (preconsolidation) are critical. “Function of pore water pressure” alone: Only transiently; long-term settlement ties to σ′. Total stress alone ignores pore pressure. Independence from drainage is incorrect; drainage controls time rate (but not the governing stress definition). Common Pitfalls: Confusing immediate (elastic) settlement with consolidation. Ignoring preconsolidation pressure which alters compressibility. Final Answer: is a function of the effective stress

Question 9

For a reinforced concrete (R.C.) water tank designed with M 25 concrete, what is the permissible bending tensile stress in concrete for the vertical wall to control cracking under service conditions?

Accepted Answer

Introduction / Context: Water-retaining structures are designed for crack control under service loads because leakage and durability are critical. Codes specify permissible tensile stresses in concrete (in bending and direct tension) that are lower than ordinary building limits to minimize crack widths, especially on the liquid face of walls and slabs. Given Data / Assumptions: Concrete grade: M 25 (characteristic compressive strength fck = 25 N/mm2). Member: vertical wall of an R.C. water tank. We seek permissible bending tensile stress for service design of liquid-retaining structures. Concept / Approach: For liquid-retaining structures (e.g., per IS guidelines for water tanks), permissible tensile stresses in concrete are intentionally restricted to limit crack initiation/width. Typical values for M 25 are about 1.8 N/mm2 in bending tension (on the liquid face) and even lower in direct tension. Higher values like 8.5 N/mm2 would be incompatible with crack control requirements and are not used for serviceability checks in water tanks. Step-by-Step Solution: Identify structure type → water-retaining R.C. tank requires stricter crack control.Recall permissible bending tension for M 25 → approximately 1.8 N/mm2.Select the matching option: 1.8 N/mm2. Verification / Alternative check: Design examples for liquid-retaining walls consistently adopt about 1.8 N/mm2 bending tensile limit for M 25 to satisfy leakage and durability criteria under sustained hydrostatic pressure. Why Other Options Are Wrong: 8.5 N/mm2, 2.5 N/mm2: Too high for liquid face service conditions; risk of cracking. 6.0 N/mm3: Wrong units; tensile stress should be N/mm2. 0.8 N/mm2: Overly conservative for M 25 bending tension. Common Pitfalls: Using general building code tensile limits instead of water-retaining criteria. Ignoring differential temperature/shrinkage which also influences cracking. Final Answer: 1.8 N/mm2

Question 10

For community water supply, what is the most appropriate single treatment for water drawn from a deep tube well (typically clear and low in turbidity)?

Accepted Answer

Introduction / Context: Groundwater from deep tube wells is usually low in turbidity and free of pathogenic contamination due to natural filtration through geological strata. However, it may still contain microorganisms or viruses and must meet drinking water safety standards. Choosing the minimal, effective treatment avoids unnecessary costs while ensuring public health. Given Data / Assumptions: Source: deep tube well (confined aquifer), generally clear with low suspended solids. No significant iron, manganese, or hardness issues indicated in the question. Objective: select the single essential treatment step for safe supply. Concept / Approach: Because turbidity is already low, coagulation, flocculation, and sedimentation are typically unnecessary. Filtration may not be required when particle load is negligible and well construction is proper. The indispensable step is disinfection (e.g., chlorination, UV), providing residual protection in the distribution system and inactivating potential pathogens that may enter during pumping or storage. Step-by-Step Solution: Assess raw water: deep well → low turbidity, low organics.Eliminate treatments aimed at solids removal (pre-settling, coagulation) absent turbidity.Retain the critical barrier: disinfection to assure microbiological safety.Choose “disinfection only.” Verification / Alternative check: Design manuals specify chlorination or UV as the required step for clear groundwater; additional treatments are added only if specific contaminants (iron, fluoride, arsenic) exceed limits. Why Other Options Are Wrong: Pre-settling/coagulation/filtration: Target turbidity and colloids; often unnecessary for deep groundwater. Aeration: Used for iron/manganese/CO2 removal—only when such issues exist. Common Pitfalls: Skipping residual disinfectant, leaving the system vulnerable to recontamination. Over-treating without a water quality basis, raising O&M costs. Final Answer: disinfection only

Question 11

In fluid mechanics, when will shear stress develop on a fluid element? Choose the condition that actually produces shear (i.e., a velocity gradient) within the fluid.

Accepted Answer

Introduction / Context: Understanding when shear stress develops inside a fluid is fundamental for civil, mechanical, and environmental engineering applications. Shear arises from relative motion between adjacent fluid layers, which requires both viscosity and a velocity gradient. This concept underpins pipe-flow headloss, boundary layers on structures, and open-channel resistance. Given Data / Assumptions: Shear stress in a Newtonian fluid is proportional to viscosity and the rate of deformation (velocity gradient). Static fluids or fluids under uniform linear acceleration experience only pressure fields, not internal shear. Inviscid models assume zero viscosity and therefore no shear resistance. Concept / Approach: For Newtonian fluids, the constitutive relation is: shear_stress = mu * (du/dy). If mu = 0 (inviscid) or du/dy = 0 (no velocity gradient), shear_stress = 0. Only when mu > 0 and the velocity varies spatially does shear arise. Hence, viscous, non-uniform flow is the necessary condition for shear inside the fluid continuum. Step-by-Step Solution: Identify the requirement for shear: both viscosity (mu > 0) and a velocity gradient (du/dy ≠ 0).Check uniform linear acceleration: fluid moves as a rigid body; du/dy = 0 → no shear.Check inviscid assumption: mu = 0 → no shear regardless of velocity field.Conclude that viscous, non-uniform flow produces shear due to finite mu and nonzero du/dy. Verification / Alternative check: Classic examples—Couette flow and Poiseuille flow—show shear proportional to velocity gradient; remove the gradient or viscosity and shear vanishes. Why Other Options Are Wrong: Uniform linear acceleration: induces hydrostatic-like pressure variation only.Inviscid fluid: by definition cannot support shear stress.Fluid at rest: no motion, no gradient, hence no shear. Common Pitfalls: Confusing body accelerations with internal deformation; only deformation rates create shear in viscous fluids. Final Answer: The fluid is viscous and the flow is non-uniform, creating a velocity gradient

Question 12

In a plate bearing test, the modulus of subgrade reaction (k) is taken from the load–deflection curve at a standard settlement. Which settlement value (in centimetres) should be used for computing k?

Accepted Answer

Introduction / Context: The modulus of subgrade reaction, k, is a key input for Westergaard plate theory and rigid pavement design. In practice, k is derived from the load–deflection curve of a plate load test at a standard, small deflection so that the response approximates the linear range and represents near-surface stiffness relevant to slabs-on-grade. Given Data / Assumptions: Field plate load test provides pressure vs. settlement (load–deflection curve). We must select a standard settlement to compute k = pressure / settlement. Rigid pavement design practices typically reference a small elastic deflection. Concept / Approach: To allow comparison across sites and tests, standards adopt a reference settlement of 1.25 mm, which equals 0.125 cm. Using this point, the secant stiffness to the origin approximates the subgrade spring constant per unit area. Larger settlements would capture nonlinear or bearing failure effects, not the desired near-elastic response for slab design. Step-by-Step Solution: Read the load–deflection curve from the plate test.Locate settlement = 1.25 mm = 0.125 cm.Compute k = q / s at this point (units consistent, e.g., kN/m^3). Verification / Alternative check: Back-calculate slab corners/edges deflection using computed k and compare with observed slab behavior for reasonableness. Why Other Options Are Wrong: 0.175–0.375 cm: larger settlements enter nonlinear ranges and vary with plate size and soil yielding.0.250 cm: too large for the standard definition of k used in rigid pavement design. Common Pitfalls: Mixing units (mm vs cm) when computing k; not correcting for plate diameter when correlating to larger slabs. Final Answer: 0.125 cm

Question 13

A rapid-mix unit treats 28,800 m^3/day of raw water (density 1000 kg/m^3, kinematic viscosity 1×10^-6 m^2/s). Alum at 35 mg/L is blended with a velocity gradient G = 900 s^-1 and a detention time of 2 minutes. What power input (in watts) is required…

Accepted Answer

Introduction / Context: Rapid mixing in water treatment provides intense, short-duration energy to disperse coagulants uniformly. The design power depends on the target velocity gradient G, the fluid viscosity, and the tank volume. This problem applies the standard relationship between G and power dissipation to size the mixer drive. Given Data / Assumptions: Flow, Q = 28,800 m^3/day. Detention time, t = 2 minutes = 120 s. Velocity gradient, G = 900 s^-1. Water density = 1000 kg/m^3; kinematic viscosity ν = 1×10^-6 m^2/s ⇒ dynamic viscosity μ = ρ * ν = 0.001 N·s/m^2. Concept / Approach: For rapid mixing, power P relates to velocity gradient by P = μ * G^2 * V, where V is the tank volume. Volume is determined from V = Q * t. Ensure all units are SI for a result in watts (N·m/s). Step-by-Step Solution: Convert flow: Q = 28,800 / 86,400 = 0.3333 m^3/s.Compute volume: V = Q * t = 0.3333 * 120 = 40 m^3.Use μ = 0.001 N·s/m^2 and G = 900 s^-1.Compute power: P = μ * G^2 * V = 0.001 * (900)^2 * 40 = 0.001 * 810,000 * 40 = 32,400 W. Verification / Alternative check: Express as kW: 32,400 W ≈ 32.4 kW. Compare with typical rapid-mix power densities (≈ 1–10 kW per 100 m^3 at high G, scaling reasonably for 40 m^3). Why Other Options Are Wrong: 32.4 or 36 W: off by three orders of magnitude due to unit mishandling.324 W: still too low; ignores G^2 scaling.3.24×10^6 W: unrealistically high for a 40 m^3 tank. Common Pitfalls: Forgetting to convert day to seconds for Q; using ν instead of μ; omitting the square on G. Final Answer: 32400

Question 14

In concrete mix design and quality control, the characteristic strength fck is defined as the compressive strength below which not more than what percentage of test results are expected to fall?

Accepted Answer

Introduction / Context: Characteristic strength, denoted fck, is the statistical backbone of design in reinforced concrete. It links laboratory test variability to field safety by defining a strength level that most test results will exceed. Codes use this concept to choose mix proportions and set acceptance criteria for site cubes or cylinders. Given Data / Assumptions: Concrete strength tests show normal-like variability. Design employs a lower fractile (a percentile) to account for scatter. Quality control aims so that only a small fraction of results fall below fck. Concept / Approach: By definition, characteristic strength is the value below which not more than 5% of test results are expected to fall. This corresponds to the 5th percentile of the population distribution. Mixes are proportioned so that the mean exceeds fck by a margin related to standard deviation, ensuring compliance despite fluctuations in materials and workmanship. Step-by-Step Solution: Recognize fck as a fractile, not the mean.Use acceptance criteria that relate target mean = fck + t * s, where s is standard deviation and t depends on selected risk level.Confirm that the adopted percentile is 5%, hence Option “5% of results fall.” Verification / Alternative check: Check cube test histograms; approximately 95% of compliant results should exceed fck if the process is centered at the target mean. Why Other Options Are Wrong: 10% or 2%: not the conventional percentile used for fck in standard practice.“None of the above”: incorrect because 5% is the standard definition. Common Pitfalls: Treating fck as an average; ignoring required margin above fck in mix design. Final Answer: 5% of results fall

Question 15

An iceberg floats in seawater with 15% of its volume above the surface. If the specific weight of seawater is 10.5 kN/m^3, what is the specific weight of the iceberg (in kN/m^3)?

Accepted Answer

Introduction / Context: Floating bodies obey Archimedes’ principle: the weight of displaced fluid equals the weight of the body. The fraction of volume submerged links the body’s specific weight to that of the fluid. This problem applies buoyancy to estimate the density (specific weight) of ice from its freeboard (portion above water). Given Data / Assumptions: Fraction above water = 15% ⇒ fraction submerged = 85% = 0.85. Specific weight of seawater, gamma_w = 10.5 kN/m^3. Hydrostatic equilibrium (no vertical acceleration). Concept / Approach: At floatation, weight of iceberg per unit volume equals weight of displaced seawater per unit volume times the submerged fraction. Therefore gamma_ice = submerged_fraction * gamma_w. The above-water percentage simply complements to 1 (or 100%). Step-by-Step Solution: Compute submerged fraction: 1 − 0.15 = 0.85.Apply equilibrium: gamma_ice = 0.85 * 10.5 kN/m^3.Calculate: gamma_ice = 8.925 kN/m^3 ≈ 8.93 kN/m^3. Verification / Alternative check: Compare with typical freshwater ice values (≈ 9.0 kN/m^3), consistent with the result given slightly denser seawater. Why Other Options Are Wrong: 12.52: exceeds seawater; object would sink.9.81: close to pure water density; would give smaller freeboard than 15% in seawater.7.83 or 6.30: too low given the observed freeboard. Common Pitfalls: Using 15% instead of 85% for the submerged fraction; confusing specific weight with specific gravity. Final Answer: 8.93

Question 16

Highway design (IRC): For a national highway on plain terrain with ruling design speed 100 km/h, maximum superelevation e = 0.07 and lateral friction f = 0.15, the ruling minimum radius of the horizontal curve is closest to what value?

Accepted Answer

Introduction / Context: Horizontal curve radius ensures that lateral acceleration is balanced by superelevation and side friction at the chosen design speed. IRC practice provides a simple formula relating speed, superelevation, and friction to the minimum safe radius. Given Data / Assumptions: Design speed V = 100 km/h. Maximum superelevation e = 0.07. Coefficient of lateral friction f = 0.15. Use standard IRC formula for R (with V in km/h). Concept / Approach: The IRC relation is R = V^2 / [225 * (e + f)], where R is in metres. This assumes small-angle approximations and combines gravitational component on the bank with frictional resistance against centripetal demand. Step-by-Step Solution: Compute e + f = 0.07 + 0.15 = 0.22.Compute denominator: 225 * 0.22 = 49.5.Compute R: 100^2 / 49.5 = 10,000 / 49.5 ≈ 202 m.Select the closest option provided: 250 m (closest to ≈ 200 m among choices). Verification / Alternative check: A sensitivity check with modest rounding of f or e (field practicability) still yields a radius near 200–230 m, supporting selection of 250 m as the nearest listed value. Why Other Options Are Wrong: 360 m and 300 m: materially larger than the computed ruling minimum.36 m: orders of magnitude too small for 100 km/h. Common Pitfalls: Using V in m/s without converting the formula constant; forgetting to include both e and f. Final Answer: 250 m

Question 17

In saturated soft clay under total stress conditions (φu = 0), what is the theoretical depth of a tension crack expressed in terms of undrained cohesion c_u and unit weight γ?

Accepted Answer

Introduction / Context: Tension cracks can develop near the crest of slopes or excavations in purely cohesive soils (φu = 0) because the soil has finite tensile capacity related to its undrained shear strength. Estimating the crack depth informs safe excavation and stability checks, including the possibility of block toppling or piping paths for water. Given Data / Assumptions: Undrained condition: φu = 0 (total stress analysis). Soil has undrained cohesion c_u and total unit weight γ. No surcharge and a vertical face idealization near the crest. Concept / Approach: For φu = 0 soils, the maximum tensile stress that can be mobilized at the ground surface is related to c_u. Equilibrium of a thin vertical slice leads to a limiting crack depth proportional to c_u/γ. Classical analysis gives z_tc = 2 c_u / γ. Beyond this depth, vertical total stress exceeds the tensile capacity and crack propagation ceases. Step-by-Step Solution: Assume a potential vertical crack of depth z at the crest.Relate vertical total stress at depth z: σ_v = γ * z.Equate available tensile resistance ≈ 2 * c_u (cohesion acting on both crack faces) to σ_v at the limit.Solve: γ * z_tc = 2 * c_u ⇒ z_tc = 2 c_u / γ. Verification / Alternative check: Check dimensions: c_u (kPa) / γ (kN/m^3) yields metres; result is dimensionally consistent. Why Other Options Are Wrong: c_u/γ: underestimates depth by factor of 2; 4 c_u/γ overestimates.(c_u * γ)/2 and √(c_u/γ): dimensionally incorrect for a length. Common Pitfalls: Mixing drained (φ′) and undrained (φu) parameters; forgetting that the derivation uses total stresses and undrained cohesion. Final Answer: z_tc = 2 c_u / γ

Question 18

Routine monitoring of the microbial quality of treated piped drinking water supplies is primarily carried out using which test?

Accepted Answer

Introduction / Context: Public health surveillance of water supplies relies on indicator organisms that signal fecal contamination and treatment efficacy. Direct routine testing for every pathogen is impractical; instead, coliform bacteria (particularly E. coli) serve as reliable indicators of sanitary quality in distribution systems. Given Data / Assumptions: Treated, piped drinking water supply under routine monitoring. Indicator approach preferred over exhaustive pathogen identification. Laboratory capacity typical of municipal water quality programs. Concept / Approach: The Coliform MPN test estimates the most probable number of coliform organisms per 100 mL using serial dilution and fermentation tubes or membrane filtration counts. The presence/absence and counts of coliforms correlate with the likelihood of fecal contamination and disinfection failures. Heterotrophic plate counts are supplementary process indicators but do not directly indicate fecal pollution. Step-by-Step Solution: Collect representative distribution system samples aseptically.Perform MPN or membrane filtration for total coliforms and E. coli as per standards.Interpret results against regulatory criteria (e.g., zero E. coli in 100 mL). Verification / Alternative check: Cross-check with chlorine residuals and turbidity; spikes in HPC may prompt operational reviews but are not fecal indicators per se. Why Other Options Are Wrong: Microscopy and heterotrophic plate counts lack specificity to fecal contamination.Identifying all pathogens or universal viral assays are infeasible for routine monitoring. Common Pitfalls: Relying solely on HPC; ignoring sample preservation and holding times that can alter bacterial counts. Final Answer: Coliform Most Probable Number (MPN) test

Question 19

Elasticity under a surface point load: directly beneath a point load P at depth z in a homogeneous, isotropic, elastic half-space, the vertical stress σz can be expressed as σz = k * P / z^2. Which exact expression is correct?

Accepted Answer

Introduction / Context: Stress distribution in soils from concentrated surface loads is often estimated using Boussinesq's elastic solution for a semi-infinite half-space. Directly under a point load, the formula simplifies to a closed form that decays with the square of depth, a key tool in foundation settlement and stress bulb calculations. Given Data / Assumptions: Homogeneous, isotropic, linear elastic half-space. Surface point load P applied at r = 0. Interest in vertical stress σz along the load axis at depth z. Concept / Approach: The general Boussinesq expression for σz at radial distance r is σz = (3 P / 2 π) * (z^3) / (r^2 + z^2)^(5/2). Setting r = 0 (directly below the load) reduces the expression to σz = (3 P) / (2 π z^2). This highlights the 1/z^2 decay with depth and the proportionality constant k = 3/(2π). Step-by-Step Solution: Start with Boussinesq's axisymmetric solution for a point load.Substitute r = 0 to obtain the on-axis stress.Simplify to σz = (3 P) / (2 π z^2), identifying k = 3/(2π) ≈ 0.4775. Verification / Alternative check: Check dimensionality: P (N) / z^2 (m^2) yields N/m^2, consistent for stress. Why Other Options Are Wrong: Other constants (1/π, 1/2π, 3/π) do not follow from the derivation and would over- or under-estimate stress significantly.Option with P z / (2π) is dimensionally incorrect for stress. Common Pitfalls: Applying the on-axis formula to off-axis points; ignoring influence factors or layered soil conditions in practical designs. Final Answer: σz = (3 P) / (2 π z^2)

Question 20

On sag (valley) vertical curves for highways, the available sight distance is primarily determined on what basis?

Accepted Answer

Introduction / Context: Vertical curve design ensures drivers can see far enough ahead to stop safely. For summit (crest) curves, the limiting condition is daytime line-of-sight over the crest. For sag (valley) curves, the constraint is the reach of vehicle headlights at night, because the road surface falls away and objects are illuminated only within the headlight beam geometry. Given Data / Assumptions: Standard headlight mounting height and upward beam angle per highway design practice. Stopping sight distance (SSD) criteria apply for safety. Driver eye height and obstacle height are accounted for differently on crest vs. sag curves. Concept / Approach: On sag curves at night, available sight distance equals the illuminated distance along the roadway, which depends on headlight height and beam spread. The design uses headlight sight distance as the control value, ensuring the provided sight distance equals or exceeds SSD at the design speed. Therefore, night-time conditions govern sag curve length and K-value selection. Step-by-Step Solution: Identify curve type: sag (valley), not crest.Set the control: headlight sight distance at night.Choose curve length/K to provide SSD ≥ required at design speed under the headlight constraint. Verification / Alternative check: Confirm by plotting headlight beam and roadway profile; the tangent point of the beam controls visible distance. Why Other Options Are Wrong: Design speed, obstacle height, or driver eye height alone are insufficient without considering night-time illumination on sag curves. Common Pitfalls: Applying crest-curve formulas to sag curves; ignoring headlight aim and mounting height in calculations. Final Answer: Night-time driving conditions governed by headlight sight distance

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