Question 1

Geometric Design – Ideal Summit (Crest) Curve Form For vertical alignment at a summit (crest), which ideal mathematical curve is adopted to provide a constant rate of change of grade and comfortable sight distance transitions?

Accepted Answer

Introduction / Context: Vertical curves connect two grades in highway design. At a summit (crest), the curve must ensure comfort, safety, adequate sight distance, and smooth appearance. The chosen curve should deliver a uniform rate of change of slope to minimize jerk on vehicles and facilitate straightforward setting-out in the field. Given Data / Assumptions: Location: summit (crest) vertical curve between two grades. Design needs: constant rate of change of grade and adequate stopping/overtaking sight distance. Constructability: simple field layout and easy computation of offsets. Concept / Approach: A simple parabola y = ax^2 + bx + c provides a constant rate of change of grade because the second derivative is constant. This property directly meets comfort (jerk control) and visibility requirements. Additionally, parabolic curves are straightforward to stake out using equal chord offsets and allow direct sight-distance checks at the crest. Step-by-Step Solution: 1) Select a curve with constant second derivative (constant change of grade) → parabola.2) Ensure sight distance criteria at the summit are satisfied by adequate length L of the parabola.3) Use standard equations for mid-ordinate and offsets to set out the curve in the field. Verification / Alternative check: Design manuals universally adopt parabolic vertical curves for both summits and sags due to the constant rate-of-grade property and computational convenience. Why Other Options Are Wrong: Spiral and lemniscate are used in horizontal alignment transitions, not vertical crest curves. Circle gives constant radius but not constant rate of change of grade in vertical alignment, leading to non-uniform comfort. Common Pitfalls: Mixing horizontal transition concepts with vertical design; using inadequate length causing restricted sight distance at the crest. Final Answer: Parabola

Question 2

Water Quality – Interpreting Excess Nitrates in River Water What does a high concentration of nitrates (NO3−) in river water most commonly indicate regarding sewage pollution timing and oxidation stage?

Accepted Answer

Introduction / Context: Assessing organic pollution in surface waters relies on indicator species and nitrogen species transformation. Sewage introduces reduced nitrogen (ammonia), which is oxidized to nitrites and then nitrates by nitrifying bacteria. Understanding which nitrogen form dominates provides insight into how long ago the pollution occurred and how far oxidation has progressed. Given Data / Assumptions: Observed: elevated nitrate concentration (NO3−) in river water. Oxygenated conditions allow nitrification over time. No unusual industrial nitrate sources are assumed. Concept / Approach: Organic pollution timeline: fresh sewage → ammonia (NH3/NH4+) peak; as microbial oxidation proceeds, nitrites (NO2−) appear; with further oxidation, nitrates (NO3−) accumulate. Therefore, high nitrates typically indicate that pollution is not fresh but has undergone complete nitrification, pointing to past sewage contamination rather than a recent discharge. Step-by-Step Solution: 1) Identify nitrogen species sequence: NH3/NH4+ → NO2− → NO3−.2) Excess NO3− implies nitrification is largely complete.3) Conclusion: the water was polluted in the past; current conditions show oxidized end products. Verification / Alternative check: River self-purification models and standard water-quality texts align nitrate prevalence with downstream recovery zones following an upstream sewage input, after DO has been restored and nitrification proceeds. Why Other Options Are Wrong: Recent/immediate pollution: would show elevated ammonia and possibly nitrite with low DO. No pollution: contradicted by elevated nitrate in the absence of non-sewage sources. Common Pitfalls: Ignoring agricultural runoff as an alternate nitrate source; not corroborating with BOD, DO, or coliform counts; misreading nitrite as nitrate. Final Answer: Past pollution of water with sewage

Question 3

On-Site Sanitation – What Is a Septic Tank? Choose the correct characterization of a household/community septic tank process for primary on-site sewage treatment.

Accepted Answer

Introduction / Context: Septic tanks are widely used in decentralized sanitation for households and small communities. They provide primary treatment of domestic sewage before effluent dispersal to a soak pit, leach field, or further polishing units. Understanding the biological process inside a septic tank is essential for design, operation, and health protection. Given Data / Assumptions: Enclosed tank with limited oxygen ingress. Domestic wastewater with settleable and floatable solids. Effluent directed to subsurface infiltration for further attenuation. Concept / Approach: Septic tanks operate primarily under anaerobic conditions. Settleable solids form sludge, floatables form scum, and the supernatant undergoes anaerobic digestion by microbes, reducing BOD and volatile solids. Aerobic processes require free oxygen and are characteristic of trickling filters or activated-sludge systems, not sealed septic tanks. Step-by-Step Solution: 1) Identify environmental condition inside tank: oxygen-limited ⇒ anaerobic.2) Mechanisms: sedimentation + anaerobic digestion of organics.3) Outcome: partial BOD reduction, solids stabilization; effluent still needs soil infiltration/polishing. Verification / Alternative check: Field performance data show typical effluent BOD reductions of the order of 30–50% within septic tanks, consistent with anaerobic degradation and long detention times. Why Other Options Are Wrong: Aerobic method: contradicts sealed, oxygen-limited environment. Physical only / physicochemical: while settling is physical, the core treatment is biological (anaerobic), not chemical addition or aeration. Common Pitfalls: Expecting septic effluent to be fully treated; neglecting desludging schedules; ventilating the tank in ways that disrupt anaerobic conditions or cause odor issues near habitations. Final Answer: An anaerobic method of on-site treatment

Question 4

Reinforced Concrete – Beam–Column Interaction (P–M curve) In a reinforced concrete beam–column, when do flexural (moment) capacity and axial (compressive) capacity increase together as axial load rises?

Accepted Answer

Introduction / Context: Beam–column elements resist combined axial load and bending moment. Their strength interaction is described by the axial load–bending moment (P–M) curve. This question asks where on that curve both axial strength and flexural strength rise together as axial load increases. Given Data / Assumptions: Reinforced concrete rectangular section under combined compression and uniaxial bending. Usual strain compatibility and stress block assumptions for design. Definitions: “tension-controlled” when steel yields first; “compression-controlled” when concrete crushes first; “balanced” point separates these regimes. Concept / Approach: Starting from pure bending (P ≈ 0), adding modest axial compression generally increases moment capacity because additional compression shifts the neutral axis and enhances the compressive block without sacrificing tension steel effectiveness. This continues up to the balanced point. In that regime, the governing failure is steel yielding in tension (tension-controlled). Beyond the balanced point, further increases in axial load reduce available lever arm and the governing failure transitions to concrete crushing; moment capacity then begins to decrease. Step-by-Step Solution: 1) Begin at P ≈ 0 (pure bending) → section is tension-controlled; steel yields at failure.2) Increment P slightly → compression block enlarges; lever arm remains adequate; moment capacity increases.3) Reach the balanced point → steel strain at yield and concrete at ultimate concurrently.4) Move beyond balanced into compression-controlled → concrete crushing governs; moment capacity decreases with further P. Verification / Alternative check: Design charts and software-generated P–M interaction diagrams show the rising limb (tension-controlled) where both P and M increase together, peaking near the balanced region, followed by a descending limb as concrete crushing dominates. Why Other Options Are Wrong: Option A: The phenomenon is not tied to “elastic limit” wording; design is based on ultimate limit states and strain compatibility. Option C: In compression-controlled regions, moment capacity trends down with increasing P. Option D: The P–M curve clearly shows a region where both increase together. Common Pitfalls: Confusing the balanced point with maximum moment under zero axial load, or assuming axial compression always decreases moment capacity. Final Answer: When yielding of the tension reinforcement governs the section (tension-controlled region)

Question 5

Hydrometry – Cup/propeller current meter calibration and use Given two calibration points (50 s counts): 12 rev → 0.25 m/s and 30 rev → 0.46 m/s. What velocity will 50 revolutions in 1 minute indicate?

Accepted Answer

Introduction / Context: Current meters are calibrated by relating flow velocity v to rotation rate N (revolutions per second). A linear relation v = a + b * N is commonly used over practical ranges. With two calibration points, we can determine a and b and then compute velocity for any observed count rate. Given Data / Assumptions: Calibration point 1: 12 revolutions in 50 s → N1 = 12/50 rps, v1 = 0.25 m/s. Calibration point 2: 30 revolutions in 50 s → N2 = 30/50 rps, v2 = 0.46 m/s. Observation: 50 revolutions in 60 s → N = 50/60 rps. Assume linear v–N relation: v = a + b * N within this range. Concept / Approach: Use the two given calibration pairs to solve for a and b. Then substitute the measured N to get the indicated velocity. This is the standard field practice for converting meter counts to flow velocity in open-channel gauging. Step-by-Step Solution: 1) Compute N1 = 12/50 = 0.24 rps; N2 = 30/50 = 0.60 rps.2) Solve a + b * 0.24 = 0.25 and a + b * 0.60 = 0.46 → subtracting gives b * (0.60 − 0.24) = 0.21 → b = 0.21 / 0.36 = 0.58333.3) Back-substitute for a: a = 0.25 − 0.58333 * 0.24 = 0.11.4) Observation N = 50/60 = 0.83333 rps → v = a + b * N = 0.11 + 0.58333 * 0.83333 ≈ 0.596 ≈ 0.60 m/s. Verification / Alternative check: Check reasonableness: 0.60 m/s lies above 0.46 m/s, consistent with the higher rotation rate (50 in 60 s is faster than 30 in 50 s). Linear interpolation/extrapolation is mild and within calibration range. Why Other Options Are Wrong: 0.42 and 0.50 are too low for the higher rotation rate; 0.73 is too high relative to the linear calibration. Common Pitfalls: Using revolutions per minute directly without converting to rps in the calibrated formula, or mixing up the 50 s calibration period with the 60 s observation period. Final Answer: 0.60

Question 6

Riveted Joint – Net section in a zig-zag pattern Two 150 mm × 10 mm plates joined by three 20 mm rivets in zig-zag; pitch = 75 mm, gauge = 60 mm, allowable tensile stress = 150 MPa. Maximum allowable tensile force?

Accepted Answer

Introduction / Context: In tension, a riveted (or bolted) plate connection fails on the weakest net section. For zig-zag riveting, the net width is increased by a stagger correction. This problem checks the correct evaluation of net section and allowable tensile capacity using working stress design. Given Data / Assumptions: Plate width b = 150 mm, thickness t = 10 mm. Three rivets of nominal diameter 20 mm arranged in zig-zag. Pitch p = 75 mm, gauge g = 60 mm. Allowable tensile stress σ_allow = 150 MPa (N/mm2). Hole diameter d0 ≈ d + 1.5 mm → 21.5 mm (typical practice). Concept / Approach: For a zig-zag path cutting two holes on alternate rows, net width = b − n * d0 + Σ(p^2 / (4 * g)) for each staggered pair crossed. Here, n = 2 holes and there is one stagger term because the critical section passes diagonally through adjacent rows. Step-by-Step Solution: 1) Net width = 150 − 2 * 21.5 + (75^2) / (4 * 60).2) Compute stagger term: 75^2 / (240) = 5625 / 240 = 23.4375 mm.3) Net width = 150 − 43 + 23.4375 = 130.4375 mm.4) Net area A_n = 130.4375 * 10 = 1304.375 mm2.5) Allowable tension = σ_allow * A_n = 150 * 1304.375 = 195,656 N ≈ 195.66 kN. Verification / Alternative check: Any section cutting three holes (without adequate stagger terms) gives a lower net width due to excessive deductions and is typically not the governing path here with the given layout. Why Other Options Are Wrong: 195.00 and 192.75 kN round off or under-estimate due to ignoring the stagger correction or rounding area too early; 225.00 kN exceeds the allowable capacity for the given dimensions and stress. Common Pitfalls: Using nominal rivet diameter instead of hole diameter for deductions, or forgetting the stagger addition p^2/(4g). Final Answer: 195.66 kN

Question 7

Soil Consistency – Consistency Index (Ic) for a clayey soil Choose the correct expression in terms of LL (liquid limit), PL (plastic limit), PI (plasticity index), and w (natural water content).

Accepted Answer

Introduction / Context: Consistency Index (Ic) indicates the relative consistency (firmness) of cohesive soils at their in-situ water content. It places the natural water content w within the Atterberg Limit range and is widely used to judge stiffness and likely behavior during excavation and foundation work. Given Data / Assumptions: LL = Liquid limit, PL = Plastic limit, PI = LL − PL. w = natural (field) water content of the clay. Units are consistent (percent for all limits and w). Concept / Approach: Ic measures how far below the liquid limit the current water content lies, normalized by the plastic range. Thus Ic = (LL − w) / (LL − PL) = (LL − w) / PI. Values near 1 indicate dry, stiff conditions close to PL; values near 0 indicate very soft conditions near LL; negative values suggest w > LL. Step-by-Step Solution: 1) Compute PI = LL − PL.2) Subtract: LL − w.3) Form ratio: Ic = (LL − w) / PI.4) Interpret Ic: the larger the Ic, the stiffer (at same soil type). Verification / Alternative check: If w = PL → Ic = (LL − PL)/PI = 1 (stiff). If w = LL → Ic = 0 (very soft). These boundary checks confirm the formula's logic. Why Other Options Are Wrong: (b) is liquidity index (not Ic). (c) LL − PL is PI, not Ic. (d) 0.5 w is unrelated to Atterberg concepts. Common Pitfalls: Confusing consistency index with liquidity index, or mixing absolute and percentage units mid-calculation. Final Answer: Ic = (LL − w) / PI

Question 8

Environmental Engineering – Minimum dissolved oxygen for aquatic life in rivers What is the minimum DO (ppm) generally required to sustain aquatic life?

Accepted Answer

Introduction / Context: Dissolved oxygen (DO) is a primary indicator of river health. Fish and aerobic microorganisms require oxygen; when DO falls too low, stress and mortality occur, and septic conditions may develop. This question targets the commonly cited minimum DO for sustaining aquatic life in flowing waters. Given Data / Assumptions: Riverine environment with typical temperature ranges. Focus on minimum threshold for survival, not optimal levels. DO expressed as mg/L (ppm). Concept / Approach: While desirable DO levels for healthy ecosystems are higher (often 5–8 mg/L or more), a widely used minimum survival threshold is about 4 mg/L. Below this, many fish species experience stress; prolonged exposure risks fish kills and anaerobic conditions. Step-by-Step Solution: 1) Identify threshold for survival → DO ≈ 4 mg/L.2) Recognize that design/operational targets (e.g., effluent standards, aeration practice) usually aim at higher DO to maintain a safety margin.3) Note temperature dependence: warmer water holds less oxygen, so maintaining ≥ 4 mg/L can be more challenging in summer. Verification / Alternative check: Field monitoring programs and fisheries guidance often flag 4 mg/L as a critical minimum, with advisories to maintain 5–6 mg/L where feasible. Why Other Options Are Wrong: 0 mg/L is anoxic and fatal; 2 mg/L is hypoxic for most fish; 8 mg/L is a healthy target but not the minimum survival threshold. Common Pitfalls: Ignoring diurnal DO swings due to photosynthesis/respiration, or not accounting for higher oxygen demand downstream of wastewater discharges. Final Answer: 4

Question 9

Soil Gradation Terminology – Name for soils with particles of nearly the same size Select the correct gradation description.

Accepted Answer

Introduction / Context: Particle-size distribution influences density, permeability, and strength. When most particles have nearly the same size, the gradation term indicates a narrow distribution, which bears on compaction and drainage behavior. Given Data / Assumptions: Soil with narrow band of particle sizes (little spread across sieves). Standard geotechnical terminology. Concept / Approach: “Uniformly graded” describes soils whose grains are nearly the same size. Such soils are also “poorly graded,” a broader class that includes uniform and gap-graded distributions. However, the most precise term for nearly same size is “uniformly graded.” Step-by-Step Solution: 1) Assess distribution → narrow (sharp peak on gradation curve).2) Assign term → uniformly graded.3) Recognize implications → higher void ratio and drainage; lower stability without fines. Verification / Alternative check: Uniform soils show low coefficients of uniformity (Cu) and curvature (Cc) values indicating lack of wide size spread. Why Other Options Are Wrong: Well graded implies a wide, continuous distribution. Poorly graded is technically correct but less specific; gap graded indicates missing intermediate sizes. Common Pitfalls: Equating “poorly graded” exclusively with uniform; while related, “uniform” is the exact descriptor for near single-size soils. Final Answer: Uniformly graded

Question 10

Torsion of Circular Shafts – Effect of scaling torque and diameter For a solid circular shaft (diameter d) with maximum shear stress τ under torque T, if torque becomes 4T and diameter becomes 2d, what is the new maximum shear stress?

Accepted Answer

Introduction / Context: The torsional shear stress in a solid circular shaft depends on applied torque and the polar section modulus. Understanding how stress scales with changes in torque and size is key to quick design checks and optimization. Given Data / Assumptions: Original shaft: solid, diameter d, torque T → maximum shear τ. New condition: torque 4T, diameter 2d. Elastic torsion theory applies (no yielding). Concept / Approach: For a solid circular shaft: τ_max = 16 * T / (π * d^3). The stress is directly proportional to T and inversely proportional to d^3. Scale T and d accordingly to find the new stress τ′. Step-by-Step Solution: 1) Original: τ = 16 * T / (π * d^3).2) New: τ′ = 16 * (4T) / [π * (2d)^3] = 64T / (π * 8 d^3) = 8T / (π * d^3).3) Ratio: τ′ / τ = [8T / (π d^3)] / [16T / (π d^3)] = 1/2 → τ′ = τ / 2. Verification / Alternative check: Dimensional check confirms stress halves because diameter cubed increases by 8 while torque increases by 4 (net factor 4/8 = 1/2). Why Other Options Are Wrong: 2τ and τ overestimate stress by ignoring the strong d^3 effect; τ/4 would imply torque unchanged or diameter growth even larger. Common Pitfalls: Forgetting the cubic dependence on diameter or confusing polar moment with second moment of area. Final Answer: τ/2

Question 11

Plasticity Chart – Local “family” of soils in a restricted area For soils with common geological origin from one locality, their LL–PI values usually plot as:

Accepted Answer

Introduction / Context: On the plasticity chart (PI vs LL), soils from a single geological formation often share mineralogy and particle characteristics. Consequently, their Atterberg data tend to align along a “local line.” Given Data / Assumptions: Soils sampled within a restricted geographic area and common origin. Atterberg limits tested consistently. Concept / Approach: Casagrande's plasticity chart distinguishes clays and silts using the A-line. Local families of soils often fall along a straight trend roughly parallel to the A-line, reflecting consistent clay mineral proportions and activity. Step-by-Step Solution: 1) Plot LL on x-axis and PI on y-axis for the soil set.2) Observe clustering → trend approximates a straight line.3) Slope similar to A-line indicates mineralogical similarity within the family. Verification / Alternative check: Many site datasets show tight linear bands; departures often indicate contamination, different depositional histories, or testing inconsistencies. Why Other Options Are Wrong: (a) Describes a region, not a line; (c) perpendicular trend lacks basis; (d) ignores the observed geological consistency. Common Pitfalls: Mixing samples from unrelated formations or misclassifying by not checking the A-line offset (clay vs silt). Final Answer: A straight line roughly parallel to the A-line

Question 12

Wastewater Treatment – Purpose of a trickling filter Identify the primary function of a trickling filter in sewage treatment.

Accepted Answer

Introduction / Context: Trickling filters are attached-growth biological reactors used in secondary treatment of wastewater. The process reduces biochemical oxygen demand (BOD) by supporting aerobic biofilms on media over which settled sewage is distributed and aerated. Given Data / Assumptions: Influent is primary-treated sewage with most settleable solids removed. Filter medium provides surface for microbial growth. Air flows through voids to maintain aerobic conditions. Concept / Approach: The rotating distributor spreads wastewater as a thin film over media. Biofilms metabolize dissolved and colloidal organics, converting them to biomass, carbon dioxide, and water. Sloughed biofilm and remaining suspended solids are captured downstream in secondary clarifiers. Step-by-Step Solution: 1) Distribute settled sewage over media → thin film forms.2) Provide ventilation → aerobic microorganisms oxidize organics.3) Collect sloughed solids → secondary clarifier removes biomass.4) Effluent has significantly reduced BOD and suspended solids. Verification / Alternative check: Typical BOD removal across trickling filters ranges from moderate to high depending on loading (standard vs high-rate designs), confirming its role as a biological oxidation unit. Why Other Options Are Wrong: (a), (c), and (d) describe physical filtration in water treatment, not the biological action in wastewater secondary treatment. Common Pitfalls: Confusing trickling filters with rapid sand filters, or ignoring the need for secondary clarification after the filter. Final Answer: Biological oxidation process to remove BOD from sewage

Question 13

Activated sludge process — identify the critical design/operating factors (Consider flow peaking and organic loading while sizing aeration and clarifier units.)

Accepted Answer

Introduction / Context: The activated sludge process is sensitive to both hydraulic surges and organic loading. Aeration basin sizing, oxygen transfer, and secondary clarifier capacity must accommodate peak flows and the highest expected substrate (BOD/COD) loads. This question asks which factors are truly critical for reliable design and operation. Given Data / Assumptions: Conventional activated sludge system with primary treatment upstream. Performance depends on mixed liquor (MLVSS), F/M ratio, sludge age (SRT), and clarifier solids flux. Design must withstand short-term peaks without washout or loss of effluent quality. Concept / Approach: Critical design conditions are tied to the worst credible scenarios. Hydraulic peaking (maximum hourly flow) governs reactor detention time reduction and clarifier surface overflow rate. Maximum daily organic load governs oxygen demand, F/M ratio, and risk of filamentous growth or bulking if under-aerated. Minimum flows are generally less critical, as control strategies (airflow, recycle) can be turned down safely without overloading units. Step-by-Step Solution: 1) Identify hydraulic driver → peak hour flow dictates minimum effective HRT and clarifier SOR.2) Identify organic driver → max daily BOD/COD sets oxygen transfer rate and diffuser/motor sizing.3) Confirm sludge handling → return activated sludge (RAS) and waste (WAS) rates must remain feasible at peaks.4) Therefore pair the two maxima: peak hour flow + peak daily organic load. Verification / Alternative check: Check solids flux at the secondary clarifier under Q_peak with elevated MLSS; confirm that the clarifier limiting flux is not exceeded and that aeration power covers the highest OUR (oxygen uptake rate) day. Why Other Options Are Wrong: Only maximum flow (a) ignores oxygen demand; (b) adds minimum flow which is less critical; (d) focuses on minima which rarely govern design. Common Pitfalls: Using average-day values; ignoring I/I-driven peaks; not coordinating aeration power turndown and RAS/WAS strategies. Final Answer: maximum hourly flow rate and maximum daily organic load

Question 14

Limit state strain profile — compression steel strain in a rectangular RC section (Given effective cover to compression steel = d' and neutral axis depth from compression face = x_u.)

Accepted Answer

Introduction / Context: In limit state design of reinforced concrete, plane sections remain plane and the extreme compression-fiber strain is taken as 0.0035 at the ultimate limit state. Knowing the neutral axis depth allows computing strains at any fiber, including the compression steel at an effective cover d' from the compression face. Given Data / Assumptions: Rectangular RC section in flexure at ULS. Extreme compression-fiber strain = 0.0035. Neutral axis depth from compression face = x_u. Compression steel centerline located at distance d' from compression face. Concept / Approach: With the plane-sections assumption, the strain distribution is linear from 0.0035 at the compression face to 0 at the neutral axis. Therefore, the strain at any fiber a distance y from the compression face is epsilon(y) = 0.0035 * (1 - y/x_u). Substituting y = d' yields the strain at the compression steel level. Step-by-Step Solution: 1) Adopt linear strain: epsilon(y) = 0.0035 * (1 - y/x_u).2) Set y = d' for the compression steel layer.3) Compute epsilon_s' = 0.0035 * (1 - d'/x_u).4) Choose the matching option. Verification / Alternative check: At y = 0 (compression face), epsilon = 0.0035; at y = x_u, epsilon = 0, confirming linearity and the adopted formula. Why Other Options Are Wrong: Options with 0.002 apply to serviceability strains, not ULS; options using x_u/d' invert the ratio, giving nonphysical results (negative or excessive strains). Common Pitfalls: Confusing effective cover d' with overall cover; mixing service and ultimate strain limits; assuming nonlinear strain without justification. Final Answer: 0.0035 (1 - d'/x_u)

Question 15

Water quality testing — what does the MPN index measure? (Interpretation of Most Probable Number counts in sanitary analysis.)

Accepted Answer

Introduction / Context: The Most Probable Number (MPN) method is a statistical, culture-based approach widely used in sanitary microbiology. Regulators apply MPN to estimate microbial contamination in drinking water, recreational waters, and treated effluents. Given Data / Assumptions: Multiple-tube fermentation or enzyme-substrate methods are employed. Incubation targets coliform growth characteristics (gas, acid, fluorescence). Result reported as MPN per 100 mL. Concept / Approach: Indicator organisms such as total coliforms and Escherichia coli serve as surrogates for fecal contamination risk. The MPN index estimates the probable concentration of these bacteria by assessing growth patterns across serial dilutions and consulting statistical MPN tables. Step-by-Step Solution: 1) Inoculate serial dilutions into selective media.2) After incubation, record positive tubes (gas/acid/fluorescence).3) Use the pattern of positives to read an MPN value from standard tables.4) Conclude that MPN quantifies coliform bacteria presence. Verification / Alternative check: Compare with membrane filtration counts (CFU/100 mL) for coliforms; both target the same indicator group. Why Other Options Are Wrong: BOD5 measures biochemical oxygen demand; DO is oxygen concentration; hardness reflects multivalent cations (Ca2+, Mg2+). None of these are quantified by MPN. Common Pitfalls: Confusing total coliforms with fecal coliforms/E. coli thresholds; misinterpreting MPN as exact rather than probabilistic. Final Answer: Coliform bacteria

Question 16

Drinking water standards — health effect of excess fluoride (If fluoride concentration exceeds 1.5 mg/L, what condition is typically caused?)

Accepted Answer

Introduction / Context: Fluoride in drinking water has a narrow beneficial-to-adverse window. While low levels can reduce dental caries, concentrations above recommended limits lead to characteristic health issues, most notably dental fluorosis in children. Given Data / Assumptions: Permissible limit considered: about 1.5 mg/L. Exposure is chronic via drinking water. Population includes children whose teeth are developing. Concept / Approach: Fluoride integrates into enamel during tooth formation. Excess interferes with ameloblast activity, producing hypomineralization and mottling of enamel (dental fluorosis). At still higher long-term intakes, skeletal fluorosis can occur, but the earliest, most typical manifestation above 1.5 mg/L is dental fluorosis. Step-by-Step Solution: 1) Identify the contaminant: fluoride.2) Compare with permissible limit: > 1.5 mg/L.3) Associate the well-known health effect at these levels: dental fluorosis.4) Select option matching this outcome. Verification / Alternative check: Public health guidelines link higher-than-permissible fluoride to brown mottling, pitting, and chalky white patches on teeth. Why Other Options Are Wrong: Cardiovascular disease, methemoglobinemia (nitrate-related), and hepatitis are not characteristic outcomes of fluoride at this range. Common Pitfalls: Confusing fluoride with nitrite/nitrate effects; assuming any dental discoloration is due to poor hygiene rather than fluoride exposure history. Final Answer: dental fluorosis

Question 17

Dimensional analysis — units of pressure gradient in fluid flow (Determine the dimensions of dp/dx for a continuum fluid.)

Accepted Answer

Introduction / Context: Dimensional consistency is a cornerstone in fluid mechanics. Pressure gradients drive flow and appear in the Navier–Stokes momentum equation alongside inertial and viscous terms; getting their dimensions right helps verify derivations and unit conversions. Given Data / Assumptions: Pressure p = force/area. Force has dimensions M L T^-2. Gradient dp/dx divides pressure by length. Concept / Approach: Start from pressure dimensions: p = (M L T^-2) / L^2 = M L^-1 T^-2. Taking a spatial derivative divides by L, so dp/dx = (M L^-1 T^-2) / L = M L^-2 T^-2. This matches the body-force-per-volume units in momentum balance (e.g., -dp/dx as a driving term). Step-by-Step Solution: 1) Pressure: p = Force / Area = (M L T^-2) / L^2.2) Simplify → p = M L^-1 T^-2.3) Apply gradient: divide by L → dp/dx = M L^-2 T^-2.4) Choose the option with M L^-2 T^-2. Verification / Alternative check: Compare with viscous term mu * d^2u/dx^2, where mu has M L^-1 T^-1 and second derivative yields L^-2 T^-1; the product gives M L^-3 T^-2 times dimension of velocity L T^-1—consistent at equation level. Why Other Options Are Wrong: (a) sign/exponents incorrect; (b) corresponds to pressure/volume, not gradient; (d) has inverted mass dimension. Common Pitfalls: Forgetting that taking a derivative with respect to length introduces an extra L^-1; mixing up dynamic and kinematic quantities. Final Answer: M L^-2 T^-2

Question 18

Pile foundations — conditions leading to negative skin friction (Identify when downward drag develops along the pile shaft.)

Accepted Answer

Introduction / Context: Negative skin friction (downdrag) is a geotechnical condition where the soil moves downward relative to the pile, imposing additional compressive drag on the shaft. Designers must account for this extra load in axial capacity checks and settlement analysis. Given Data / Assumptions: Single or group piles embedded in stratified soils. Possible consolidation or fill placement above the pile's neutral plane. Pile tip may be in a firm bearing stratum. Concept / Approach: Downdrag occurs when the soil mass experiences settlement (e.g., consolidation of soft clay under new surcharge or lowering of groundwater) that exceeds the settlement of the pile. Relative downward movement mobilizes shear along the pile in the downward direction, adding to structural load and reducing the available safety margin. Step-by-Step Solution: 1) Identify mechanisms that cause soil settlement (surcharge, consolidation, liquefaction effects).2) Compare movements: if soil settles more than the pile shaft, drag develops downward.3) Include downdrag as an additional permanent action in design load combinations.4) Mitigate via sleeves, bitumen coating, or placing pile tips deeper into non-settling strata. Verification / Alternative check: Load-transfer (t–z) curves show negative skin friction where relative displacement is downward; neutral plane lies where shaft friction changes sign. Why Other Options Are Wrong: (a) Upward drag is the opposite of downdrag; (c) firm continuous soil without settlement does not create negative friction; (d) driving onset does not by itself cause downdrag. Common Pitfalls: Ignoring downdrag in serviceability; double-counting drag as both load and resistance; not checking group effects. Final Answer: the surrounding soil settles more than the pile

Question 19

Flexure theory — consequence of “plane sections remain plane” (What strain/stress distribution follows from the Bernoulli–Euler assumption?)

Accepted Answer

Introduction / Context: The classical bending theory assumes that plane sections normal to the neutral axis before bending remain plane and normal after bending. This foundational hypothesis underpins linear strain distributions used in both steel and concrete flexure design. Given Data / Assumptions: Small deflections and linear geometry changes. Bernoulli–Euler kinematics (no warping of cross sections). Material behavior can later be linked via constitutive laws (e.g., Hooke’s law). Concept / Approach: If plane cross sections remain plane, longitudinal fibers separated by a distance y from the neutral axis must change length proportionally to y when the member bends to a curvature kappa. Therefore, axial strain varies linearly with y: epsilon(y) = kappa * y. Stress distribution is then obtained by applying the material law (elastic or nonlinear), but linear strain is the immediate kinematic consequence. Step-by-Step Solution: 1) Assume planar cross section after bending → no shear distortion across depth.2) Impose compatibility with curvature → fiber extension/compression ∝ distance from NA.3) Conclude epsilon(y) varies linearly across the depth.4) Apply stress–strain law separately to get stress distribution (elastic case → linear stress). Verification / Alternative check: Beam bending formula M/I = sigma/y = E/R derives from linear strain and Hooke’s law; experiments confirm linear strain in elastic range. Why Other Options Are Wrong: (a) Uniform strain contradicts curvature; (b) uniform stress would imply zero bending; (d) stress proportional to strain is a material law, not a direct result of the plane-sections assumption. Common Pitfalls: Conflating kinematic assumption with constitutive law; assuming linear stress in plasticity where material nonlinearity alters stress despite linear strain. Final Answer: linearly varying strain over the cross section

Question 20

Centrifugal pump — compute specific speed (Ns) (Given Q = 44 m^3/s, H = 36 m, N = 1450 rpm; use Ns = N * sqrt(Q) / H^(3/4).)

Accepted Answer

Introduction / Context: Specific speed (Ns) classifies pump impellers by geometry and performance similarity. It guides selection between radial-, mixed-, and axial-flow designs and helps anticipate efficiency ranges and NPSH behavior. Given Data / Assumptions: Discharge Q = 44 m^3/s (Cumecs). Total head H = 36 m. Rotational speed N = 1450 rpm. Formula: Ns = N * sqrt(Q) / H^(3/4) (SI-based convention typical in many texts). Concept / Approach: Compute sqrt(Q) and H^(3/4), then evaluate Ns. Larger Q and lower H increase Ns (tending toward axial-flow), while smaller Q and higher H reduce Ns (radial-flow). Step-by-Step Solution: 1) Compute sqrt(Q) = sqrt(44) ≈ 6.633.2) Compute H^(3/4): 36^(0.75) = (36^(1/2)) * (36^(1/4)) ≈ 6 * 2.449 = 14.694.3) Evaluate Ns = 1450 * 6.633 / 14.694 ≈ 1450 * 0.4513 ≈ 654.4) Select option 654. Verification / Alternative check: Order-of-magnitude: With large Q and moderate H, Ns in the hundreds is expected, consistent with mixed/axial tendencies. Why Other Options Are Wrong: 509 and 700 deviate from the precise calculation; 90 is far too low given the high discharge and moderate head. Common Pitfalls: Mistaking unit systems or using US customary definitions; misapplying exponent 3/4; rounding too early in intermediate steps. Final Answer: 654

GATE Exam Questions Questions