Question 1

An open well in coarse sand is expected to yield 0.0059 m^3/s at a working depression head of 3 m. Estimate the minimum well diameter that should be adopted (choose the closest).

Accepted Answer

Introduction / Context:Sizing open (dug) wells in granular formations is an application of unconfined well hydraulics. Designers select a diameter that can safely deliver the target discharge under the expected working drawdown, with reasonable entrance velocities and head losses. Given Data / Assumptions: Formation: coarse sand (high permeability). Required discharge Q ≈ 0.0059 m^3/s. Working depression head (drawdown at the well) h ≈ 3 m. Adopt conventional empirical sizing used in practice for open wells in coarse sands. Concept / Approach:For preliminary sizing of open wells in coarse granular materials, empirical design relations relate Q to well diameter D and working drawdown h, while implicitly accounting for permeability and inflow conditions. Designers select the nearest standard diameter that provides the required yield with acceptable velocities at the well face and manageable head losses. Step-by-Step Solution: Use standard empirical relations and tables for open wells in coarse sands at h ≈ 3 m.Find the diameter bracket that corresponds to Q ≈ 0.0059 m^3/s.Typical preliminary sizing points toward D around the mid-2 m range for such yields in coarse sand under ~3 m head.Select 2.50 m as the minimum practical standard diameter meeting the discharge safely. Verification / Alternative check:Check entrance velocity (Q / inflow area) and compare with recommended limits for sandy formations; a D of 2.5 m provides conservative velocities with 3 m drawdown in coarse sand for the stated Q. Why Other Options Are Wrong: 2.00–2.25 m: Tends to give higher approach velocities and head losses for the same Q and h. 2.75–3.00 m: Oversized for the stated yield, increasing cost without proportional benefit. Common Pitfalls: Ignoring entrance losses and sand movement risk when choosing too small a diameter. Relying on a single sizing equation without cross-checking against practical standard sizes. Final Answer:2.50 m

Question 2

Urban and rural water-supply engineering typically includes which of the following integrated components?

Accepted Answer

Introduction / Context:Water-supply systems are end-to-end solutions that start at the source and terminate at the consumer, while also meeting firefighting and public safety needs. This question checks recognition of the complete scope of a standard water-supply scheme. Given Data / Assumptions: Typical municipal or institutional water-supply scheme. Service levels include domestic, commercial, and fire-fighting demand. Concept / Approach:A water-supply project normally encompasses: raw-water collection, transmission, treatment (clarification, filtration, disinfection), storage, and distribution networks (mains, sub-mains, service connections). Fire-fighting capability via hydrants is an integral design objective influencing pipe sizing, pressure zones, and storage volumes. Step-by-Step Solution: Identify core components: source to tap requires collection, conveyance, treatment.Distribution network design includes mains, sub-mains, and branches sized for peak and fire flows.Safety component: hydrants provide fire-fighting flows and pressures.Therefore, the complete answer is 'All of the above'. Verification / Alternative check:Standard design manuals specify hydrant spacing, minimum residual pressures, and storage for fire demand—integral to the system, not an addon. Why Other Options Are Wrong: Any subset omits required elements for a fully functional and code-compliant system. Common Pitfalls: Underestimating fire-flow requirements when sizing mains/reservoirs. Ignoring water quality management from source through distribution. Final Answer:All of the above

Question 3

Compare production rates: the yield (filtration rate) of a rapid gravity (rapid sand) filter is typically how many times that of a slow sand filter?

Accepted Answer

Introduction / Context:Water-treatment plants use either slow sand filters (very low rates, high biological action) or rapid gravity filters (much higher rates with prior coagulation–flocculation). Understanding their typical filtration-rate ratio guides footprint and process selection. Given Data / Assumptions: Slow sand filters: ~2–5 m^3/m^2/day (typical design band). Rapid gravity filters: ~80–120 m^3/m^2/day (post-coagulation typical band). We are asked for a representative multiple (order-of-magnitude comparison). Concept / Approach:Taking representative mid-range values (e.g., SSF ≈ 3–4 m^3/m^2/day and RGF ≈ 90–100 m^3/m^2/day) gives a ratio on the order of 20–30. A conservative standard teaching value often cited is about 20 times, which safely represents the magnitude difference for many exam contexts. Step-by-Step Solution: Choose SSF typical ≈ 4 m^3/m^2/day; choose RGF typical ≈ 80–100 m^3/m^2/day.Compute ratio ≈ 80/4 to 100/4 = 20–25.Select the closest standard option → 20 times. Verification / Alternative check:Design manuals often present RGF throughput an order of magnitude greater than SSF; the 20× figure sits well within the usual range. Why Other Options Are Wrong: 10× or 15×: Understate the typical difference. 30× or 35×: Possible for some plants, but higher than a standard benchmark used in teaching questions. Common Pitfalls: Comparing instantaneous backwash/production cycles rather than net filtration rate. Ignoring pretreatment needs that enable RGF's higher rates. Final Answer:20 times

Question 4

Classify water using pH: identify the correct statements about neutrality, acidity, and alkalinity with respect to pH values.

Accepted Answer

Introduction / Context:pH is a master variable in water quality affecting corrosion, coagulation, disinfectant efficacy, and taste/odor. Correct classification of water based on pH is essential in environmental and municipal engineering. Given Data / Assumptions: pH scale ranges nominally from 0 (strongly acidic) to 14 (strongly alkaline). Pure water at 25°C has pH ≈ 7 (neutral). Concept / Approach:By definition, pH = −log10[H+]. At 25°C, [H+] = 10^−7 mol/L corresponds to neutral pH 7. Values below 7 indicate higher proton activity (acidic), and values above 7 indicate lower proton activity (alkaline/basic). Step-by-Step Solution: Confirm neutrality at pH 7 (true).Recognize acidity for pH < 7 (true).Recognize alkalinity for pH > 7 (true).Therefore, the inclusive correct choice is 'All the above'. Verification / Alternative check:Any standard water chemistry text or drinking-water standards describe these pH classifications. Why Other Options Are Wrong: Options listing subsets omit one or more correct statements; the complete statement is required. Common Pitfalls: Confusing alkalinity (as titratable carbonate/bicarbonate capacity) with simply having pH > 7; while related, they are distinct concepts. Final Answer:All the above

Question 5

For estimating groundwater flow velocity in porous media, which relation is the most commonly used non-empirical (physics-based) formula?

Accepted Answer

Introduction / Context:Groundwater hydraulics relies on constitutive laws that link discharge to hydraulic gradient and medium properties. Among various relations, one stands out as a first-principles, experimentally verified law valid for laminar flow in porous media. Given Data / Assumptions: Laminar flow through saturated porous media. Homogeneous, isotropic aquifer assumption for simplicity. Concept / Approach:Darcy’s law states that discharge Q is proportional to hydraulic gradient i and cross-sectional area A with proportionality constant K (hydraulic conductivity):Q = K * A * iThus, the seepage (Darcy) velocity v = Q/A = K * i. This relation is physics-based (non-empirical in the sense of a governing law within its validity range) and underpins aquifer analysis, pumping-test interpretation, and numerical groundwater models. Step-by-Step Solution: Identify the foundational law for groundwater flow → Darcy’s law.Recognize that alternatives (Slichter/Hazen) are empirical correlations for specific conditions.Select 'Darcy’s formula' as the correct answer. Verification / Alternative check:Darcy’s law is universally taught as the governing relation in hydrogeology for laminar flows, forming the basis of much of groundwater engineering. Why Other Options Are Wrong: Slichter/Hazen: Empirical relations tailored to certain grain sizes/temperatures. Lacey: Regime channel design; not a groundwater flow law. Common Pitfalls: Using Darcy’s velocity as pore velocity without dividing by effective porosity. Final Answer:Darcy’s formula

Question 6

What is a commonly adopted upper limit for the depth of sedimentation (clarifier) tanks in water-treatment plants?

Accepted Answer

Introduction / Context:Primary sedimentation (clarification) removes settleable solids before filtration. Tank depth influences settling behavior, sludge storage, scraper design, and structural cost. Designers typically observe both minimum and maximum depth limits. Given Data / Assumptions: Conventional horizontal-flow or circular sedimentation tanks for potable water treatment. Standard design practice bounds depth to aid operation and economy. Concept / Approach:While effective water depth is often in the 3–4 m range for potable-water clarifiers, an upper limit is generally placed to control construction cost, mixing characteristics, sludge removal mechanisms, and headloss constraints. Many references indicate that depths should not exceed about 5 m for conventional clarifiers in municipal practice. Step-by-Step Solution: Recall typical design depth band ~3–4 m.Recognize the practical maximum limit commonly cited ~5 m.Select the option 5 m as the adopted upper limit. Verification / Alternative check:Design manuals specify similar ranges; greater depths complicate sludge removal and offer little settling benefit compared with increased surface area. Why Other Options Are Wrong: 2–4 m: Within typical design depths, but not an 'upper limit'. 6 m: Exceeds commonly adopted upper bounds for conventional plants. Common Pitfalls: Confusing clarifier depth with detention time or surface overflow rate, which are controlled predominantly by plan area and flow. Final Answer:5 m

Question 7

As per IS 1172:1963 provisions for institutional accommodations, a minimum of 135 litres per capita per day (lpcd) should be provided for which facilities?

Accepted Answer

Introduction / Context:Indian Standard IS 1172 recommends per-capita water requirements for various occupancies to ensure adequate domestic use, hygiene, and service demands. Knowledge of these norms is essential for water-supply planning and plumbing design. Given Data / Assumptions: IS 1172:1963 (and subsequent revisions) provides benchmark lpcd values. Question references 135 lpcd for specified institutional accommodations. Concept / Approach:Boarding schools, nurses’ homes/medical quarters, and hostels are all residential-type occupancies with shared facilities and continuous occupancy patterns. IS 1172 prescribes minimum lpcd figures, with 135 lpcd a commonly quoted baseline for residential and similar institutional uses under normal supply conditions. Step-by-Step Solution: Identify each listed occupancy as qualifying for the 135 lpcd provision.Recognize that all three fall under institutional/residential categories with similar usage profiles.Select the inclusive option 'All the above'. Verification / Alternative check:Design handbooks citing IS 1172 list these occupancies at or around 135 lpcd (subject to local by-laws and service levels). Why Other Options Are Wrong: Any subset fails to include all qualifying institutional categories enumerated in the question. Common Pitfalls: Confusing peak/instantaneous demands (fixture units) with daily per-capita allocation. Final Answer:All the above

Question 8

In municipal water treatment, for which filter configuration is a high backwash (wash-water) velocity required to effectively expand and scour the media bed?

Accepted Answer

Introduction: Backwashing is essential for restoring headloss and removing trapped solids from granular media filters. The hydraulics of backwash depend on the underdrain system. When standard strainers/nozzles are absent, higher wash-water velocities are generally needed to achieve uniform expansion and adequate scouring of the bed. Given Data / Assumptions: Compare rapid gravity filters with and without strainers. Slow sand filters are normally cleaned by scraping, not by backwashing. Objective: identify the case requiring high backwash velocity. Concept / Approach: In a rapid gravity filter, the underdrain distributes wash water. With nozzles/strainers, distribution is uniform, so design backwash rates (e.g., 40–60 cm/min at 20 °C, adjusted for temperature and media) suffice to fluidize the bed. Without dedicated strainers, distribution becomes less uniform; to achieve the same degree of bed expansion and lift heavier agglomerates, the gross velocity must be increased to compensate for maldistribution and local short-circuiting. Step-by-Step Solution: Identify which systems are routinely backwashed: rapid gravity filters yes; slow sand filters no (scraping is used). Recognize the underdrain role: strainers help uniform flow and lower the required gross velocity. Conclude that absence of strainers requires higher wash-water velocity to achieve target expansion. Verification / Alternative check: Design texts recommend adjusting backwash rates for distribution quality; poorer distribution (no nozzles) implies increasing the applied rate to reach similar bed expansion percentage. Why Other Options Are Wrong: Rapid gravity filter with strainers: standard, requires lower design velocities due to uniform distribution. Slow sand with/without strainers: slow sand filters are not backwashed; they are refurbished by scraping the schmutzdecke and top sand. None of these: incorrect because the no-strainer rapid filter clearly demands higher velocity. Common Pitfalls: Assuming all filters are cleaned identically; slow sand and rapid sand have different O&M methods. Ignoring distribution uniformity when setting backwash rates. Final Answer: Rapid gravity filter without strainers in the underdrain.

Question 9

In water treatment, aeration is primarily carried out to remove which of the following from raw water?

Accepted Answer

Introduction: Aeration is a unit operation where water is brought into intimate contact with air to strip volatile gases and oxidize certain dissolved constituents. It improves aesthetic quality and prepares water for downstream processes. Given Data / Assumptions: Options list odour, colour, bacteria, hardness, turbidity. We seek the principal target of aeration in conventional treatment. Concept / Approach: Aeration removes or reduces volatile compounds (like H2S) and dissolved gases (CO2), thereby eliminating odours and correcting corrosivity (via CO2 stripping). It can also oxidize ferrous iron and manganous ions to forms more amenable to subsequent removal. It is not a disinfectant, softener, or clarifier by itself. Step-by-Step Solution: List typical aeration objectives: remove odorous gases (H2S), strip CO2, raise DO, oxidize Fe/Mn. Map each option: odour is directly addressed by stripping; colour from humic substances needs coagulation/adsorption; bacteria require disinfection; hardness requires softening; turbidity needs coagulation–flocculation–sedimentation/filtration. Select odour as the best single answer. Verification / Alternative check: Plant practice shows marked odour reduction after cascade or diffused aeration, even before coagulation. Why Other Options Are Wrong: Colour: Only marginally affected; primary removal is via coagulation/activated carbon. Bacteria: Aeration is not a disinfecting step. Hardness: Requires precipitation/ion exchange, not aeration. Turbidity: Needs clarification and filtration. Common Pitfalls: Confusing increased dissolved oxygen with microbial kill; DO does not equal disinfection. Expecting aeration to remove all tastes/colours regardless of cause. Final Answer: Odour.

Question 10

What is the maximum permissible concentration of nitrites (as NO2-) in public water supplies?

Accepted Answer

Introduction: Nitrite (NO2-) is an intermediate in the nitrogen cycle and a health-significant parameter. Elevated nitrite indicates recent contamination or nitrification within the distribution system and can pose risks such as methemoglobinemia. Given Data / Assumptions: Public supply standard is being referenced. Answer choices: 0.5, 1.0, 1.5, and 200 ppm. Concept / Approach: Potable water standards limit nitrite stringently—values are typically in sub-ppm range. A value such as 200 ppm is grossly excessive for any health-based standard; choices near or below 1 ppm are plausible. The commonly cited permissible value for nitrite as NO2- in basic MCQ references is 0.5 ppm for supplies. Step-by-Step Solution: Recognize that nitrite standards are far more restrictive than nitrate standards. From the given options, 0.5 ppm aligns with conservative potable limits. Select 0.5 ppm as the maximum permissible value among choices. Verification / Alternative check: Cross-compare with typical textbook tables where nitrite limits are ≤ 1 ppm; 0.5 ppm is widely referenced in exam-oriented materials. Why Other Options Are Wrong: 1.0 ppm and 1.5 ppm: Less conservative than commonly prescribed limits. 200 ppm: Orders of magnitude too high for drinking water; unrealistic. Common Pitfalls: Confusing nitrate limits (~45 ppm as NO3-) with nitrite limits (much lower). Ignoring the difference between “as NO2-” versus “as N.” Final Answer: 0.5 ppm.

Question 11

According to Kuichling’s empirical relation, the required fire-demand Q (in L/min) for a town of population P (in thousands) is given by which expression?

Accepted Answer

Introduction: Fire-demand formulas provide a quick estimate of the rate at which water must be supplied for firefighting. Kuichling’s is a classic empirical relation used in many water-supply design problems and competitive examinations. Given Data / Assumptions: P denotes population in thousands (dimensionless in the formula). Q is desired in litres per minute (L/min). Concept / Approach: Kuichling derived a square-root dependence of fire demand on town size, reflecting that firefighting requirements scale sublinearly with population. The metric version widely quoted is Q = 3182 * sqrt(P), with Q in L/min and P in thousands. Step-by-Step Solution: Identify the correct dependence: Q ∝ sqrt(P), not linear in P. Select the metric coefficient 3182 for L/min units. Form the expression Q = 3182 * sqrt(P). Verification / Alternative check: Imperial versions expressed in gal/min convert to the metric coefficient near 3182 when multiplied by 3.785; the square-root scaling remains the hallmark of the formula. Why Other Options Are Wrong: 5663 * sqrt(P), 2050 * sqrt(P): Incorrect coefficients for the stated units. 3182 * P and 1136 * P: Wrong dependence (linear in P instead of square-root). Common Pitfalls: Using P as absolute population rather than in thousands. Mixing imperial and metric constants without conversion. Final Answer: Q = 3182 * sqrt(P).

Question 12

Fire hydrant design: which statement below is incorrect with respect to spacing, pressure, and availability?

Accepted Answer

Introduction: Fire hydrant systems must meet criteria for spacing, pressure, and duration during emergencies. However, hydrants are not meant for ordinary continuous consumption; they are emergency outlets designed for intermittent, high-rate withdrawal during fires. Given Data / Assumptions: Typical urban hydrant spacing is 100–150 m for accessibility. Minimum residual pressure at a hydrant commonly targeted ≈ 1.5 kg/cm^2. The contentious statement suggests prolonged constant-use supply through hydrants. Concept / Approach: Hydrants are emergency fixtures, not consumer delivery points. Design guidance sets residual pressures and flow durations for firefighting scenarios, but “constant use” is not an objective for hydrants. They are isolated via valves and used only during incidents or testing. Step-by-Step Solution: Affirm spacing convention (100–150 m) as reasonable for urban grids. Affirm residual pressure ≈ 1.5 kg/cm^2 as a planning minimum. Identify the incorrect statement: long-duration constant use via hydrants contradicts the emergency-only intent. Verification / Alternative check: Municipal manuals specify hydrant operating pressures/flows for firefighting duration (e.g., a few hours) but explicitly restrict routine domestic/industrial draw from hydrants. Why Other Options Are Wrong: 100–150 m spacing: Standard practice in many jurisdictions. Minimum pressure ≈ 1.5 kg/cm^2: Common planning benchmark. None of these: Incorrect because one statement is demonstrably wrong. Common Pitfalls: Conflating hydrant design duration (for firefighting) with “constant use.” Ignoring that hydrant outlets are normally locked/valved off. Final Answer: Water at 1.0–1.5 kg/cm^2 is made available for 4–5 hours for constant use.

Question 13

In dam economics, the “economic height” is the height at which which criterion is satisfied?

Accepted Answer

Introduction: Selecting dam height balances incremental construction cost against incremental storage benefit. The “economic height” is the result of optimizing this balance for minimum cost per unit of useful storage. Given Data / Assumptions: Cost rises with height due to larger structures and appurtenances. Storage also rises with height, but with diminishing returns at higher elevations. Concept / Approach: Define unit storage cost C_u = Total cost / Live storage. The economic height occurs at the minimum of C_u with respect to height, where marginal cost per extra unit storage equals the average cost per unit storage. Step-by-Step Solution: Relate cost curve (increasing, convex) and storage curve (increasing, concave). Compute/visualize C_u vs. height; identify its minimum. Interpret this minimum as the economic height because it yields the cheapest storage per unit volume. Verification / Alternative check: Benefit–cost analysis similarly points to the height where additional storage no longer justifies its extra cost, aligning with the minimum of C_u. Why Other Options Are Wrong: Least silting/freeboard: Operational/safety constraints, not the economic optimum definition. Maximum unit cost: Opposite of optimization goal. None of these: Incorrect as the definition is standard. Common Pitfalls: Confusing economic height with safe hydraulic height (flood routing and freeboard). Optimizing total cost rather than unit storage cost. Final Answer: Cost of the dam per unit of storage is minimum.

Question 14

Sedimentation design: which of the following statements are correct for conventional settling tanks?

Accepted Answer

Introduction: Sedimentation removes suspended solids by gravity prior to filtration. Design ranges for detention time and permissible horizontal velocities promote quiescent conditions without short-circuiting or re-suspension. Given Data / Assumptions: Plain sedimentation (no coagulant) vs. coagulation–sedimentation. Typical design ranges are being tested. Concept / Approach: Plain sedimentation requires longer detention for natural settleability (4–8 h). With coagulation, floc forms and settles faster (2–4 h). Horizontal velocities are limited (~0.3 m/min) to maintain laminar/near-laminar flow and avoid scour. Step-by-Step Solution: Validate detention times for both cases against standard design practice. Check horizontal velocity limit typical of rectangular basins. All statements match design norms; choose “All the above”. Verification / Alternative check: Design manuals corroborate similar ranges; exact values vary with temperature, particle size, and floc properties. Why Other Options Are Wrong: Single statements omit the rest of the correct design guidelines; the comprehensive option is right. Common Pitfalls: Applying plain-sedimentation detention times to coagulated water (overdesign). Allowing excessive horizontal velocities that resuspend floc. Final Answer: All the above.

Question 15

In natural surface waters, which effect of sunlight is most directly beneficial for water quality?

Accepted Answer

Introduction: Sunlight influences aquatic ecosystems through heating and by enabling photosynthesis. The water-quality parameter most directly and consistently improved by sunlight during daylight is dissolved oxygen (DO), generated by photosynthetic organisms. Given Data / Assumptions: Options include effects on bacteria, algae, DO, and turbidity. We focus on the net, direct effect commonly leveraged in self-purification. Concept / Approach: Photosynthesis converts CO2 and water into organic matter and oxygen in the presence of light. Daytime algal and macrophyte activity raises DO, supporting aerobic decomposition and aquatic life. Sunlight does not reliably reduce turbidity; algal blooms may even increase it. UV has limited disinfection in natural waters due to shading and short exposure. Step-by-Step Solution: Identify the process most directly tied to sunlight: photosynthesis. Link photosynthesis to increased DO concentrations during the photoperiod. Select “Increases dissolved oxygen content”. Verification / Alternative check: Diurnal DO curves show daytime peaks and nighttime declines, reflecting photosynthesis and respiration cycles. Why Other Options Are Wrong: Helps growth of bacteria: not a typical, direct benefit; many bacteria prefer low-light or are light-insensitive. Impedes growth of algae: the opposite—sunlight promotes algal growth. Reduces turbidity: not a direct effect; can be neutral or negative during blooms. Common Pitfalls: Assuming sunlight disinfects bulk water; natural UV exposure is insufficient for consistent pathogen removal. Final Answer: Increases dissolved oxygen content.

Question 16

In storm drainage design, the time of concentration is best defined as which of the following?

Accepted Answer

Introduction: Time of concentration (Tc) controls design rainfall intensity in rational and unit-hydrograph methods. It represents the longest travel path governing the response time of a catchment. Given Data / Assumptions: We need the operational definition used in hydrologic design. Remote point refers to the point with the maximum travel time to the outlet. Concept / Approach: Tc is the time for a drop of water to travel from the hydraulically most remote point in the watershed to the outlet. For storms with duration ≥ Tc, the entire catchment contributes to peak flow; for shorter storms, only parts near the outlet contribute at the same instant. Step-by-Step Solution: Eliminate definitions about storm onset or generic duration—they do not define Tc. Select the path-based travel-time definition tied to hydraulics and topography. Choose the statement referencing the most remote point to the drain/outlet. Verification / Alternative check: Empirical Tc formulas (Kirpich, NRCS) compute travel times from flow paths and slopes, consistent with this definition. Why Other Options Are Wrong: Storm start/duration: meteorological, not hydraulic response definitions. All runoff reaching after the storm: not the standard definition of Tc. Common Pitfalls: Confusing basin lag time with Tc; they are related but not identical. Final Answer: The travel time for storm water from the hydraulically most remote point of the catchment to the inlet/outlet.

Question 17

Logistic growth (Pearl–Reed) forecasting: given populations P0 at time t0, P1 at time t1, and P2 at time t2 = 2 t1, what is the saturation (carrying capacity) Ps of the city?

Accepted Answer

Introduction: Logistic (Pearl–Reed) population forecasting assumes growth slows as population approaches a saturation (carrying capacity) Ps. Using three known populations at equally spaced times, Ps can be estimated in closed form. Given Data / Assumptions: Known: P0 at t0, P1 at t1, P2 at t2 with t2 = 2 t1 (equal time intervals). Model: Logistic growth toward Ps. Concept / Approach: Under the logistic law, 1/P varies linearly with time when expressed appropriately. Using three consecutive observations at equal time spacing, eliminating the intrinsic growth rate yields Ps in terms of P0, P1, P2. The standard result used in demographic engineering texts is Ps = (P1^2 - P0 * P2) / (2 * P1 - P0 - P2). Step-by-Step Solution: Start with the logistic form: dP/dt = r * P * (1 - P/Ps). Solve to get: P(t) = Ps / (1 + A * e^{-r t}). Write equations at t0, t1, t2 and eliminate A and r using equal spacing. Rearrange to obtain Ps = (P1^2 - P0 * P2) / (2 * P1 - P0 - P2). Verification / Alternative check: Plugging trial values (monotonically increasing P0 < P1 < P2) yields a finite Ps slightly above P2, consistent with saturation behavior. Why Other Options Are Wrong: Other algebraic forms shown do not reduce to the logistic Ps when tested; signs/coefficients are inconsistent with the elimination step. “None of these” is incorrect because a standard closed-form expression exists and is listed. Common Pitfalls: Applying the formula when observation intervals are unequal—this derivation assumes equal spacing. Confusing geometric growth with logistic growth. Final Answer: Ps = (P1^2 - P0 * P2) / (2 * P1 - P0 - P2).

Question 18

In water treatment using slow sand filters, what is the maximum raw-water turbidity (in mg/L) that can generally be handled effectively without prior coagulation or pre-treatment?

Accepted Answer

Introduction / Context:Slow sand filtration is a biological-physical process used for treating relatively clear surface sources. It relies on a biologically active layer (schmutzdecke) and fine sand to remove suspended and colloidal matter along with pathogens. This question tests practical design knowledge about the turbidity range that slow sand filters can accept without pre-treatment. Given Data / Assumptions: Standard slow sand filters with effective size sand and typical filtration rates of 0.1 to 0.3 m/h. Typical design guidance that slow sand filters treat low-turbidity water directly. No coagulation–flocculation or other pre-treatment assumed. Concept / Approach:Slow sand filters are best suited for low and fairly stable turbidity. When turbidity is too high, the schmutzdecke clogs rapidly, head loss rises quickly, and filter runs become uneconomical. Many practice-oriented texts indicate that raw-water turbidity should generally be below about 50 mg/L; a conservative upper bound commonly used in exam problems is about 60 mg/L without pre-treatment. Step-by-Step Solution: Recognize that slow sand filters handle low-turbidity waters most effectively.Adopt a conservative practical limit close to 50–60 mg/L when no pre-treatment is provided.Among the options, 60 mg/L is the closest valid ceiling without coagulation. Verification / Alternative check:Where turbidity is frequently higher, plants add pre-treatment (coagulation–flocculation–sedimentation) before slow sand filtration. This aligns with operational experience worldwide. Why Other Options Are Wrong: 75 mg/L and 150 mg/L: Too high for direct slow sand filtration; would cause rapid clogging. 100 g/L: Unit error and unrealistically large turbidity. None of these: Incorrect because 60 mg/L is a standard practical upper bound used in design questions. Common Pitfalls: Confusing slow sand (biologically driven) with rapid gravity filters that follow coagulation. Ignoring head-loss and short filter runs at higher turbidity. Final Answer:60 mg/L

Question 19

For an unconfined (water-table) tube well discharging Q with drawdown s through a screen (strainer) of length L and well radius r_w, with radius of influence R, which expression gives the required screen length in terms of aquifer hydraulic conducti…

Accepted Answer

Introduction / Context:Design of unconfined tube wells requires selecting an adequate screen length to safely pass the target discharge at the allowable drawdown. This uses steady radial-flow relations (Thiem equation) adapted to water-table conditions and distributed inflow along the screen length. Given Data / Assumptions: Unconfined aquifer with hydraulic conductivity K. Well radius r_w and radius of influence R (R >> r_w). Steady discharge Q with drawdown s at the well under design pumping. Uniform inflow along screen length L (first-approximation design). Concept / Approach:For unconfined flow, the Thiem relation connects discharge to head difference and geometry. Spreading the inflow over the screen length gives an average entrance flux. Rearrangement provides the screen length L necessary to limit entrance flux so that the overall drawdown is s for discharge Q. The commonly used expression is:L = (Q / (2 * pi * K * s)) * ln(R / r_w)with natural logarithm ln(). Step-by-Step Solution: Start with steady radial flow for unconfined conditions and distribute inflow over L.Relate Q, K, s, and log term ln(R / r_w) from Thiem.Solve for L to obtain: L = (Q / (2 * pi * K * s)) * ln(R / r_w). Verification / Alternative check:Dimensional check: Q has L^3/T; denominator 2 * pi * K * s is L/T; multiplying by the dimensionless ln term yields L (length), as required. Why Other Options Are Wrong: Options with inverted fraction or missing 2 * pi lead to incorrect scaling. Replacing ln(R / r_w) with (R / r_w) is incorrect; radial flow solution yields the logarithmic term. Common Pitfalls: Using confined-aquifer formulas directly without adapting to water-table conditions. Forgetting the logarithmic dependence on R / r_w. Final Answer:L = (Q / (2 * pi * K * s)) * ln(R / r_w)

Question 20

According to IS 1172:1963 (water supply standards in India), the recommended per capita water requirement for average domestic purposes is _____ litres per person per day (lpcd).

Accepted Answer

Introduction / Context:Per capita water demand standards are used by planners to size sources, treatment plants, and distribution systems. IS 1172 is the primary Indian standard prescribing typical residential and institutional demands under normal conditions. Given Data / Assumptions: Average domestic use under normal (non-scarcity) conditions. Standard Indian practice as per IS 1172:1963 (and widely cited figures thereafter). Concept / Approach:IS 1172 recommends a baseline of 135 lpcd for domestic purposes in urban settings, considering drinking, cooking, bathing, washing, flushing, etc. This benchmark serves as a planning norm (subject to local by-laws and service levels). Step-by-Step Solution: Identify the relevant standard: IS 1172.Recall its commonly applied residential benchmark.Select 135 litres per person per day as the recommended value. Verification / Alternative check:Many state manuals and CPHEEO guidelines echo the 135 lpcd figure for residential planning. Why Other Options Are Wrong: 50–105 lpcd are lower than the commonly cited IS 1172 norm for average urban domestic use. Common Pitfalls: Confusing domestic lpcd with institutional or intermittent-supply adjustments. Final Answer:135 litres

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