Question 1

Rain sampling containers used in field hydrology Polythene (plastic) bottles commonly used for collecting rainwater samples in precipitation chemistry and quality monitoring typically have which of the following capacities?

Accepted Answer

Introduction / Context: Rainwater sampling for chemical analysis, acid rain studies, or general precipitation monitoring often uses inert, non-reactive containers to avoid contamination. High-density polythene (HDPE) bottles are standard due to durability, low cost, and chemical compatibility. Given Data / Assumptions: Sampling programs range from event-based to composite (weekly/monthly) collection. Required volume depends on the number of analytes, detection limits, and laboratory protocols. Polythene (HDPE) containers are acceptable for most major ions and nutrients when properly cleaned and preserved. Concept / Approach: Field kits include bottles across a range of sizes to accommodate different sampling durations, rainfall intensities, and analytical suites. Smaller 2 L bottles suit grab samples, 4 L for daily/event composites, and 10 L for high-rainfall events or multiple replicates. Choice is operational rather than theoretical, so multiple capacities are common and correct. Step-by-Step Solution: Identify commonly supplied HDPE bottle sizes: ~2 L (grab), ~4 L (composite), ~10 L (high-volume).Confirm that any of these volumes may be used depending on program needs and rainfall amounts.Therefore, the inclusive option “All the above” best reflects field practice. Verification / Alternative check: Vendor catalogs and precipitation networks (e.g., academic or environmental agencies) list bottles in these standard capacities, often interchangeable with appropriate rain collectors and funnels. Why Other Options Are Wrong: Picking only one size ignores variability in program requirements. “None of these” is incorrect because these sizes are widely used and available. Common Pitfalls: Using reactive containers (e.g., some metals or glass for specific analytes) without compatibility checks. Neglecting preservation, headspace, or chain-of-custody that may affect lab results. Final Answer: All the above.

Question 2

Thiem equation for a confined aquifer (steady radial flow) In a pumping test on a confined aquifer of saturated thickness b, a steady discharge Q is pumped from a well. Two observation wells at radii r1 and r2 from the pumped well show piezometric h…

Accepted Answer

Introduction / Context: The Thiem equation is a classical steady-state solution for radial flow toward a pumping well. It allows hydrogeologists to estimate the hydraulic conductivity (permeability K) of an aquifer from discharge and drawdown observations in two observation wells. Given Data / Assumptions: Confined aquifer of uniform thickness b and homogeneous/isotropic properties. Steady pumping at constant discharge Q. Two observation wells at radii r1 and r2 with observed piezometric heads h1 and h2. Dupuit–Forchheimer assumptions: predominantly horizontal flow, negligible vertical gradients near observation points. Concept / Approach: For a confined aquifer under steady radial flow, the head distribution satisfies a logarithmic decline with radius. Integrating Darcy’s law in cylindrical coordinates between r1 and r2 gives a direct relationship among Q, K, b, and the measured heads. Step-by-Step Solution: Start with Darcy’s law for radial flow: Q = 2 * π * r * b * K * dh/dr.Separate and integrate: ∫(h1→h2) dh = (Q / (2 * π * b * K)) ∫(r1→r2) (dr / r).This yields h2 - h1 = (Q / (2 * π * b * K)) * ln(r2 / r1).Solve for K: K = (Q / (2 * π * b)) * ln(r2 / r1) / (h2 - h1). Verification / Alternative check: Dimension check: Q has L^3/T; denominator 2 * π * b has L; the remaining ratio ln(r2/r1)/(h2 - h1) has 1/L; overall gives L/T as expected for K. Why Other Options Are Wrong: Option B is for unconfined Thiem (uses h^2), not confined. Option C is the algebraic inverse (K in denominator of Q); it gives 1/K form. Option D omits thickness b and misplaces terms. “All the above” is false because only one equation is correct for the confined case posed. Common Pitfalls: Mixing the confined and unconfined forms (h vs h^2), using base-10 log instead of natural log without adjusting constants, and selecting observation wells too close to the pumped well (well losses). Final Answer: K = (Q / (2 * π * b)) * ln(r2 / r1) / (h2 - h1)

Question 3

Dupuit method for pumping-test analysis The classical Dupuit (Dupuit–Thiem) analysis for estimating aquifer parameters is based on observations from how many observation wells?

Accepted Answer

Introduction / Context: The Dupuit–Thiem method is a steady-state approach to analyze radial flow toward a pumping well. It relies on piezometric head measurements at known radii to compute hydraulic conductivity. Given Data / Assumptions: Horizontal, radial flow under Dupuit assumptions (small vertical gradients). Homogeneous, isotropic aquifer with steady pumping. Two observation wells located at known distances from the pumped well. Concept / Approach: With two observation wells, the logarithmic head–radius relationship can be written between radii r1 and r2, eliminating the unknown integration constant and directly solving for K using the observed heads. Step-by-Step Solution: Measure heads h1 and h2 at r1 and r2.Use the Thiem equation appropriate to the aquifer type.Solve for K or transmissivity as required. Verification / Alternative check: Using more than two wells provides redundancy and enables least-squares fitting to the theoretical straight line of head vs ln(r). Why Other Options Are Wrong: One well cannot define the slope of the ln(r) relation. Three wells are useful but not the basis of the original formula. No observation well is impossible for parameter estimation. Common Pitfalls: Placing observation wells too close to the pumping well (well losses dominate) or near boundaries which distort radial symmetry. Final Answer: Two observation wells

Question 4

Subsurface hydrology terminology Pick the correct set of statements about groundwater zones and pressures.

Accepted Answer

Introduction / Context: Hydrogeology distinguishes layers by their moisture content and pressure conditions. Correct terminology helps interpret well logs, pumping tests, and infiltration processes. Given Data / Assumptions: Unconfined conditions for the water table concept. Standard definitions used in groundwater hydrology. Concept / Approach: The subsurface is divided broadly into a zone of aeration (vadose) above the water table and a zone of saturation below. Below the water table, pores are fully saturated and the pore water pressure is positive (hydrostatic at rest). Step-by-Step Solution: Identify zone nomenclature: above WT ⇒ aeration; below WT ⇒ saturation.Define “groundwater”: water stored and flowing in the saturated zone.State pressure condition: saturated zone water is under hydrostatic pressure at rest. Verification / Alternative check: Textbook phreatic surface definition: the water table is where pore pressure equals atmospheric (gauge = 0). Below it, pore pressure is positive and increases with depth according to hydrostatics. Why Other Options Are Wrong: Not applicable; options a–d are all correct, making option e the best choice. Common Pitfalls: Confusing capillary fringe (often saturated but under suction) as part of the vadose zone above the water table. Final Answer: All of the above

Question 5

Definition of a unit hydrograph (runoff depth basis) A unit hydrograph is the direct-runoff hydrograph produced by a rainstorm of a specified duration that generates a runoff depth of:

Accepted Answer

Introduction / Context: The unit hydrograph (UH) concept is a cornerstone of surface-water hydrology. It represents the time distribution of direct runoff resulting from a unit depth of effective rainfall of a given duration over a catchment. Given Data / Assumptions: Catchment is linear and time-invariant during the event (superposition and proportionality hold approximately). “Unit” depth conventionally means 1 cm (10 mm) of effective rainfall excess. Specified duration refers to the rainfall excess duration (e.g., 1 hour UH, 4 hour UH). Concept / Approach: A UH is defined so that its ordinates scale linearly with rainfall excess. If a storm produces D cm of effective rainfall of the same duration, the resulting direct-runoff hydrograph is D times the UH ordinates. Step-by-Step Solution: State the definition: “unit” runoff depth is 1 cm = 10 mm.Specify that duration is fixed (e.g., T-hour UH).Thus, the correct option is the depth equal to 10 mm. Verification / Alternative check: Hydrology handbooks and standard texts uniformly adopt 1 cm as the unit runoff depth in metric practice; some analyses may use 1 inch in US customary units. Why Other Options Are Wrong: 5, 20, 25, and 30 mm are not the conventional “unit” depth; they would define non-standard hydrographs. Common Pitfalls: Confusing rainfall depth (gross) with effective rainfall (excess). The UH is tied to runoff excess after abstraction losses. Final Answer: 10 mm (i.e., 1 cm)

Question 6

Two-point method for mean stream velocity For a stream of depth h, the average mean velocity is often estimated as the average of current-meter readings taken at depths:

Accepted Answer

Introduction / Context: Accurate discharge measurement in open channels often uses vertical velocity profiles. The two-point method is a practical compromise between accuracy and effort. Given Data / Assumptions: Relatively uniform vertical velocity profile (no extreme stratification). Current meter provides point velocities at specified depths along a vertical. Depth h at the vertical is known. Concept / Approach: Empirical studies show that averaging velocities at 0.2 h and 0.8 h approximates the depth-averaged velocity well. When only one point is feasible, the 0.6 h method is common; with three points, 0.2 h, 0.6 h, and 0.8 h may be used with weights. Step-by-Step Solution: Measure v(0.2 h) and v(0.8 h).Compute mean velocity v̄ ≈ [v(0.2 h) + v(0.8 h)] / 2.Use Q = Σ(v̄ * area slice) across the cross-section to get discharge. Verification / Alternative check: The one-point 0.6 h method yields v̄ ≈ v(0.6 h) and is slightly less accurate than the two-point average for many natural streams. Why Other Options Are Wrong: 0.1/0.9 or 0.3/0.7 and 0.4/0.6 pairs are not the standard two-point recommendation and can bias v̄. Surface-only readings neglect lower-depth shear and are unreliable. Common Pitfalls: Ignoring secondary currents near banks or measuring too close to bed/surface where meter errors increase. Final Answer: 0.2 h and 0.8 h

Question 7

Physical states of water in the hydrologic cycle In nature, water may occur as which of the following states?

Accepted Answer

Introduction / Context: The hydrologic cycle continuously transforms water among phases in the atmosphere, on land, and in the oceans. Recognizing the three physical states helps explain processes like precipitation, runoff, and evapotranspiration. Given Data / Assumptions: Standard atmospheric and terrestrial conditions. Phase changes governed by temperature and pressure variations. Concept / Approach: Water exists naturally as liquid (rivers, lakes), solid (glaciers, snowpack), and vapour (atmospheric humidity). Phase transitions—melting, freezing, evaporation, condensation, sublimation—drive energy and mass exchanges. Step-by-Step Solution: Identify examples of each phase: rain/streams (liquid), snow/ice (solid), humidity/clouds (vapour/condensed droplets).Acknowledge transformations across these states via weather and climate processes. Verification / Alternative check: Observational hydrology and meteorology document all three states in daily measurements (streamflow, snowfall, humidity). Why Other Options Are Wrong: Options a–c list individual states; the comprehensive correct choice is “All of the above”. Common Pitfalls: Equating “vapour” with “cloud droplets”; clouds are liquid micro-drops or ice crystals suspended in air, while water vapour is invisible gas. Final Answer: All of the above

Question 8

Dickens formula constant for high-flood estimation In Dickens’ regional flood formula Q = C * A^(3/4), an average value of the constant C commonly adopted for high-flood estimates is approximately:

Accepted Answer

Introduction / Context: Empirical regional flood formulas like Dickens’ are used for preliminary peak-discharge estimation where gauge data are sparse. The coefficient C captures regional rainfall and catchment response. Given Data / Assumptions: Dickens’ formula: Q = C * A^(3/4), with Q in m^3/s and A in km^2. “Average” value of C for high-flood estimation in many Indian regions. Concept / Approach: While C varies geographically (mountainous vs coastal vs inland basins), guidance texts often suggest a representative average near 11 to 12 for high floods where local calibration is unavailable. Step-by-Step Solution: Recognize that among the given choices, 11.5 best represents the widely cited average for high floods. Verification / Alternative check: Regional hydrology manuals list ranges (e.g., 6–30) with typical values around 11 for many basins; selection depends on local context. Why Other Options Are Wrong: 6.5–10.5 are on the lower side and may underpredict high floods in many regions. Values significantly different should be justified by local calibration rather than treated as average. Common Pitfalls: Using a single C universally without checking basin characteristics; empirical formulas are rough tools. Final Answer: 11.5

Question 9

Basic rainfall descriptors Pick the correct set of definitions relating to rainfall characteristics.

Accepted Answer

Introduction / Context: Hydrologic design uses three basic rainfall descriptors: intensity, duration, and frequency. Understanding these terms underpins IDF (Intensity–Duration–Frequency) analysis and stormwater design. Given Data / Assumptions: Standard meteorological definitions. Frequency often expressed via return period (e.g., 10-year event). Concept / Approach: Intensity is a rate (e.g., mm/h), duration is the time span over which that intensity is considered, and frequency/return period describes how often events of that magnitude are expected statistically. Step-by-Step Solution: State intensity definition clearly (depth/time).State duration as time length of rainfall occurrence.State frequency as occurrence rate or probability over time. Verification / Alternative check: These form the axes and curves of IDF relationships used in drainage sizing and flood estimation. Why Other Options Are Wrong: Options a–c are individually correct; thus option d (“All of the above”) is the comprehensive correct answer. Common Pitfalls: Confusing frequency with duration or interpreting return period deterministically rather than probabilistically. Final Answer: All of the above

Question 10

Non-recording rain gauges — key characteristic Non-recording (storage) rain gauges primarily do which of the following?

Accepted Answer

Introduction / Context: Rain gauges are broadly divided into non-recording (storage) and recording types. Knowing their basic operation helps in choosing instruments for field monitoring. Given Data / Assumptions: Non-recording gauges such as Symon’s gauge collect rainfall in a container. Readings are taken manually at set intervals to compute depth (mm). Concept / Approach: A non-recording gauge has a funnel and receiver. Rainfall is stored and then measured with a graduated cylinder (calibrated typically in mm of rainfall over the gauge’s catch area). No timing mechanism records intensity automatically. Step-by-Step Solution: Recognize the absence of an automatic recording device.Identify manual measurement of collected volume as depth.Conclude that option (a) captures the primary characteristic. Verification / Alternative check: In contrast, recording gauges (tipping bucket, weighing, float) generate time-series data of rainfall intensity and totals without manual measurement after each event. Why Other Options Are Wrong: (b) describes recording gauges. (c) wind speed is measured by anemometers, not rain gauges. (d) telemetry is possible only with recording/IoT add-ons, not inherent to non-recording gauges. (e) is the purpose of evaporimeters (pan A, Class-A pans), not rain gauges. Common Pitfalls: Assuming any cylindrical device is recording; shape does not determine functionality. Final Answer: Collect rainfall which is later measured manually (as depth) using a graduated measuring cylinder

Question 11

In hydrology and meteorology, isohyets are imaginary contour lines drawn on a map that join locations having equal total rainfall (precipitation) over a specified period. What do isohyets represent?

Accepted Answer

Introduction / Context: Iso-lines are foundational tools in environmental mapping. Just as contours depict equal elevation and isobars show equal pressure, isohyets specifically map equal rainfall. Understanding isohyets is essential for catchment delineation, water resources planning, flood studies, and agricultural scheduling. Given Data / Assumptions: The map displays accumulated precipitation over a chosen duration (storm event, month, season, or year). Point rainfall data from gauges are spatially interpolated to create continuous lines of equal precipitation. The question asks what an isohyet represents. Concept / Approach: An isohyet is defined as a line joining points with the same precipitation depth. These lines allow quick visual assessment of spatial variability in rainfall, aiding estimation of areal average rainfall for basins (e.g., by Thiessen polygons, isohyetal method, or arithmetic mean when appropriate). Step-by-Step Solution: Identify the iso-term: isohyet → hyetos (rain) + iso (equal).Relate to other iso-lines: isobar (pressure), isotherm (temperature), isohume (humidity), contour (height).Therefore, isohyets represent equal rainfall totals across a mapped region. Verification / Alternative check: In areal rainfall estimation, the isohyetal method calculates the weighted average based on areas between successive isohyets, confirming that each line represents a constant precipitation value. Why Other Options Are Wrong: Equal atmospheric pressure: That is an isobar, not an isohyet. Equal elevation: Those are topographic contours. Equal humidity: Those are isohumes or iso-humidity lines. Equal wind speed: These are isotachs. Common Pitfalls: Confusing isohyets with contours or isobars due to similar line maps. Assuming isohyets are valid for instantaneous intensity; they are typically drawn for totals over a period. Final Answer: Equal rainfall (precipitation depth)

Question 12

Design flood estimation (regional formula): In Dickens’ formula, the peak flood discharge Q (in m^3/s) is taken proportional to the catchment area A raised to which power (Q ∝ A^n)?

Accepted Answer

Introduction / Context: Before wide availability of long hydrologic records, Indian practice often used regional empirical formulae for preliminary design flood estimates. Dickens’ formula is one such classical relation linking peak discharge to drainage area with an exponent less than 1, reflecting non-linear growth of floods with catchment size. Given Data / Assumptions: Dickens’ formula: Q = C * A^(3/4), where Q is in m^3/s, A is catchment area in km^2, and C is a regional coefficient (typically 6 to 30 depending on rainfall and region). Question asks only the exponent n in Q ∝ A^n. Concept / Approach: Empirical flood formulae (e.g., Ryve’s, Dickens’, Inglis’) use a power-law Q–A relationship. The exponent encapsulates hydrologic scaling: as basin area increases, unit response attenuates, so n is less than 1. Dickens’ established n = 3/4 for Indian conditions historically. Step-by-Step Solution: Recall Dickens’ relation: Q = C * A^(3/4).Hence the exponent n = 3/4. Verification / Alternative check: Comparison with other formulae shows similar exponents (generally between 0.6 and 0.8) capturing the diminishing incremental increase in floods with larger areas. This cross-check supports n = 3/4 for Dickens’ formula. Why Other Options Are Wrong: 2/3, 4/5, 1/2: Not the classic Dickens exponent. 5/4: Greater than 1, which would imply super-linear growth and is inconsistent with attenuation observed in larger basins. Common Pitfalls: Mixing up Dickens with other regional formulae and their exponents. Using empirical formulae outside their calibrated regions or without local coefficient C. Final Answer: n = 3/4

Question 13

Boston Society of Civil Engineers (B.S.C.E.) runoff relation (expressed in cumecs per square kilometre): This empirical relation primarily depends on which factors?

Accepted Answer

Introduction / Context: Regional empirical relations for design discharge often appear in the form of unit discharge (cumecs per km^2). The Boston Society of Civil Engineers (B.S.C.E.) relation is a classical example linking peak runoff to catchment size and rainfall. Recognizing the governing variables helps engineers apply such formulae appropriately and understand their limitations. Given Data / Assumptions: Discharge is expressed as cumecs per square kilometre (unit discharge). Empirical form generally uses rainfall intensity or an index storm parameter and drainage area. Question asks which factors it is based upon. Concept / Approach: Unit discharge normalizes by area, so remaining primary drivers are rainfall (intensity/duration appropriate to the basin response) and catchment properties. In these simplified relations, area is always present; rainfall is the forcing input. Basin shape or total runoff volume are not typically the core variables in the simple B.S.C.E. peak relation stated per unit area. Step-by-Step Reasoning: Identify required variables for peak runoff generation: rainfall forcing + catchment size.Other details (shape, slope, infiltration) are often embedded in empirical coefficients; they are not the explicit paired factors in the simplest expression. Verification / Alternative check: Many historical US and regional formulae express Q = C * P^m * A^n; when cast per unit area, Q/A primarily scales with P and weakly with A through the exponent or coefficient. This supports option (a). Why Other Options Are Wrong: Total runoff volume is not the input for an instantaneous peak relation. Drainage area alone omits the meteorological driver. Wind speed and humidity are not primary controls for short-term peak runoff estimates. Common Pitfalls: Assuming unit discharge depends only on area after normalization. Applying empirical formulae beyond their validated hydro-climatic region. Final Answer: Rainfall characteristics and drainage area

Question 14

Minor losses in a well pipeline system: If the head loss due to pipe friction is 4 m, what approximate allowance should be made for additional losses in the strainer and bends (rule-of-thumb for preliminary estimates)?

Accepted Answer

Introduction / Context: In preliminary groundwater and small water-supply designs, engineers often approximate minor losses (entries, strainers, bends, fittings) as an additive allowance to the computed pipe-friction head loss. While detailed coefficients can be used, a practical rule-of-thumb speeds early sizing and pump selection. Given Data / Assumptions: Pipe friction head loss h_f = 4 m (from Darcy–Weisbach or Hazen–Williams). System includes a well strainer and several bends/fittings. Preliminary estimate stage where a reasonable allowance is customary. Concept / Approach: Minor losses h_m are summed as K * V^2/(2g) over fittings and inlets. In many small well systems, the aggregate of these is often taken as about 10% to 20% of h_f or a nominal fraction of a metre. A commonly used preliminary allowance is near half a metre when friction is a few metres, pending refinement in final design. Step-by-Step Solution: Recognize that exact K-values depend on geometry and Reynolds number.For h_f = 4 m, choose a practical allowance h_m in the range 0.3 to 0.8 m for typical small systems.Adopt 0.50 m as a balanced preliminary allowance for strainer + bends. Verification / Alternative check: Detailed checks would compute K for each component: entrance to screen, screen head loss, bend coefficients, and any valves. Such detailed summation often lands near a fraction of a metre to about a metre, depending on layout—supporting the 0.50 m preliminary choice. Why Other Options Are Wrong: 0.10 m and 0.20 m: Often too low for combined strainer and multiple bends. 1.00 m and 1.50 m: Possible in complex layouts, but generally conservative for a small installation with 4 m of friction loss. Common Pitfalls: Ignoring minor losses entirely, leading to undersized pumps. Applying a blanket percentage without checking unusual layouts (many fittings, valves, sharp entries). Final Answer: 0.50 m

Question 15

Pan evaporation computation with rainfall and withdrawals: A Class-A type pan (diameter 1.5 m) initially had 8.0 cm of water. During the period, rainfall = 5.0 cm, water withdrawn to maintain operation = 3.0 cm, and final water depth = 9.0 cm. If th…

Accepted Answer

Introduction / Context: Pan evaporation is routinely converted to reservoir evaporation using a pan coefficient. Correct bookkeeping of pan water balance with rainfall inputs and operational withdrawals is essential for accurate estimates used in reservoir yield studies and irrigation planning. Given Data / Assumptions: Initial pan depth = 8.0 cm. Rainfall during the period = 5.0 cm (falls into the pan). Water withdrawn (to maintain operation) = 3.0 cm. Final pan depth = 9.0 cm. Pan coefficient Kp = 0.6 (assume negligible seepage/leakage). Concept / Approach: Pan evaporation E_pan is obtained from the pan water balance: Initial + Rainfall − Withdrawals − Final = Evaporation from panThen convert to reservoir evaporation using E_reservoir = Kp * E_pan. Step-by-Step Solution: Compute E_pan = 8.0 + 5.0 − 3.0 − 9.0 = 1.0 cm.Convert to reservoir evaporation: E_res = Kp * E_pan = 0.6 * 1.0 cm = 0.6 cm.Therefore, E_res = 0.6 cm = 6 mm. Verification / Alternative check: Consistency check: Despite rainfall, the final depth increased by only 1 cm after withdrawing 3 cm. The remainder must have evaporated: 1 cm pan evaporation. Applying a pan coefficient less than 1 scales to the reservoir surface, yielding 6 mm. Why Other Options Are Wrong: 2 mm, 4 mm, 8 mm, 10 mm: Do not match the calculated 6 mm after applying Kp = 0.6 to the 1.0 cm pan evaporation. Common Pitfalls: Forgetting to subtract final depth and withdrawals correctly in the water balance. Applying the inverse of Kp instead of Kp itself. Final Answer: 6 mm

Question 16

Groundwater hydraulics: In well hydraulics, the “radius of influence” is defined as which of the following?

Accepted Answer

Introduction / Context: During pumping, a cone of depression develops in the potentiometric surface (or water table). The lateral extent of measurable drawdown defines the effective impact zone of the well, a key parameter in spacing wells and in analyzing interference between wells. Given Data / Assumptions: Confined or unconfined aquifer with a pumped well at steady or quasi-steady conditions. Drawdown s(r) decreases radially and tends to zero at some large distance. Definition focuses purely on geometry of drawdown. Concept / Approach: The radius of influence R is the radial distance from the well centre to the point in the aquifer where drawdown becomes negligible (approaches zero). Analytical solutions (e.g., Thiem, Theis) and field tests (pumping tests) are used to estimate R for design and management decisions. Step-by-Step Clarification: Consider drawdown as a function of radius r from well centre.Find r where s(r) → 0 (practical cut-off based on field sensitivity).Define that r as the radius of influence. Verification / Alternative check: In design, R helps determine whether neighbouring wells will hydraulically interfere during simultaneous pumping and informs the size of the capture zone. Why Other Options Are Wrong: Well wall distance (option b) misses the centre reference used in radial flow solutions. Casing radius and gravel-pack radius are construction dimensions, not drawdown reach. Saturated thickness is a vertical dimension, not lateral reach of drawdown. Common Pitfalls: Confusing construction geometry with hydraulic influence radius. Assuming a fixed R; in reality, it varies with transmissivity, storativity, and pumping rate/time. Final Answer: The distance from the well centre to the point of zero drawdown

Question 17

Normal-ratio method (precipitation estimation at an ungauged station): Storm totals at three surrounding stations are A = 110 mm, B = 90 mm, and C = 70 mm. Normal annual precipitation values are: X = 1000 mm (target), A = 1100 mm, B = 1200 mm, C = 1…

Accepted Answer

Introduction / Context: When a storm total is missing at a station, the normal-ratio method adjusts neighbouring station storm totals by the ratios of normal annual precipitation to account for climatic gradients. It is widely used for hydrologic data reconstruction when normals differ by more than about 10% between stations. Given Data / Assumptions: Normal annual precipitation: N_X = 1000 mm, N_A = 1100 mm, N_B = 1200 mm, N_C = 1250 mm. Observed storm totals: P_A = 110 mm, P_B = 90 mm, P_C = 70 mm. Normals differ sufficiently to warrant the normal-ratio adjustment. Concept / Approach: Normal-ratio method formula: P_X = (1/n) * Σ [ (N_X / N_i) * P_i ] , for i = A, B, C and n = 3.This scales each neighbour’s storm by the relative wetness/dryness indicated by normals. Step-by-Step Solution: Term A: (1000/1100) * 110 = 0.9091 * 110 ≈ 100.0Term B: (1000/1200) * 90 = 0.8333 * 90 = 75.0Term C: (1000/1250) * 70 = 0.8 * 70 = 56.0Sum = 231.0; Average = 231 / 3 = 77.0 mm Verification / Alternative check: If normals were similar (within ~10%), the simple arithmetic mean (110 + 90 + 70)/3 = 90 mm might be used. Because normals differ notably, the adjusted result (77 mm) is more appropriate. Why Other Options Are Wrong: 75, 79, 81, 83 mm: Do not match the computed normal-ratio average of 77 mm. Common Pitfalls: Forgetting to scale each neighbour by N_X/N_i. Using a weighted average without proper justification from normals or distances. Final Answer: 77 mm

Question 18

Sharp-crested (thin-plate) weirs in flow measurement: For which situation are sharp-crested weirs generally preferred as the primary measuring device?

Accepted Answer

Introduction / Context: Sharp-crested weirs are classical hydraulic structures for measuring flow. Their head–discharge relations are well established under free-flow conditions and are widely used in labs, calibration channels, and small canals where flow is relatively clean and submergence is avoidable. Given Data / Assumptions: Thin, sharp crest with nappe aeration and adequate approach conditions. Flow ranges small to moderate, with limited sediment and debris. Free (not submerged) outflow ideally maintained. Concept / Approach: Accurate head measurement over a sharp crest requires well-defined approach flow and cleanliness at the crest. For large rivers or sediment-laden streams, broad-crested weirs, flumes, or other structures are more robust. Debris and sediment can distort the nappe and calibration, reducing accuracy in sharp-crested devices. Step-by-Step Reasoning: Assess flow magnitude and quality (sediment/debris).Match device to conditions: sharp-crested → small, clean flows with good control.Hence, best suited for small laboratory or minor channel discharges. Verification / Alternative check: Standard hydraulics texts list V-notch and rectangular sharp-crested weirs for precise measurement of small discharges; for large or sediment-laden flows, robust structures are recommended. Why Other Options Are Wrong: Large river flows and tidal estuaries: Structural and backwater issues make sharp-crested weirs unsuitable. Sediment-laden streams or debris-rich rivers: Crest blockage and nappe distortion degrade accuracy. Common Pitfalls: Ignoring submergence effects which invalidate free-flow calibration. Not providing adequate upstream stilling and approach conditions for head measurement. Final Answer: For small flows in laboratory channels, flumes, or minor irrigation outlets

Question 19

Drainage-basin form factor – compute from given area and axial length A drainage basin has an axial length of 100 km and a plan area of 2,500 sq. km. Using the standard definition (form factor = basin area / axial length^2), determine the basin’s fo…

Accepted Answer

Introduction / Context: The form factor is a popular basin-shape index in hydrology used to infer hydrograph characteristics. A higher form factor suggests a “rounder” basin that tends to yield a sharper flood peak, while a lower value indicates an elongated basin with a flatter hydrograph. Given Data / Assumptions: Basin area A = 2,500 sq. km. Axial length L = 100 km. Standard definition: form factor Ff = A / L^2. Planar (map) measurements; no need for relief corrections. Concept / Approach: The form factor is a dimensionless ratio comparing area to the square of the maximum basin length. It normalizes area by length scale so basins of different size can be compared by shape alone. No unit conversion is needed because the km units cancel in the ratio. Step-by-Step Solution: Compute L^2: L^2 = 100^2 = 10,000 km^2.Apply formula: Ff = A / L^2 = 2,500 / 10,000.Evaluate: Ff = 0.25.Match with the options provided → 0.25. Verification / Alternative check: If the basin were a perfect square of side 50 km, its axial length would be 50 km and A/L^2 would be 2,500/2,500 = 1. Here, L is twice that (100 km), so A/L^2 naturally becomes 1/4 = 0.25, consistent with an elongated plan shape. Why Other Options Are Wrong: 0.10, 0.20, 0.30, 0.35 are not equal to A / L^2 with the given numbers. Common Pitfalls: Confusing axial length with mean width; or squaring the area instead of length. Always use the exact definition Ff = A / L^2 with consistent units. Final Answer: 0.25

Question 20

Unit hydrograph applications – choose the comprehensive correct statements Regarding the use of unit hydrographs (UH) for design and analysis, identify the correct statements about duration tolerance, scaling with runoff depth, and transferability t…

Accepted Answer

Introduction / Context: The unit hydrograph (UH) is a cornerstone of catchment hydrology. It represents the direct runoff hydrograph (DRH) produced by a unit depth of effective rainfall uniformly distributed over the catchment for a specified duration. Once derived, it enables synthesis of storm hydrographs for other runoff depths. Given Data / Assumptions: Linear, time-invariant catchment response within practical limits. UH duration matches (or is close to) the effective rainfall duration. Runoff depth is in consistent units (e.g., mm or cm). Catchment conditions between events are broadly similar. Concept / Approach: Under linear superposition, a storm DRH equals the UH ordinates scaled by the runoff depth (in the same units as the unit depth defining the UH). If the UH is a 1 cm UH, multiplying factor = runoff depth (mm) / 10; if a 25 mm UH, factor = runoff (mm) / 25. Limited duration mismatch (≈ ±25%) is commonly tolerable, though using a matching-duration UH is best practice. Step-by-Step Solution: Transferability: The UH of a given duration can generate future storm hydrographs of like duration → true.Scaling: Multiply UH ordinates by runoff depth in UH-units → true.Factor example: For a 25 mm UH, factor = runoff(mm)/25 → true.Duration tolerance: ±25% is a commonly cited acceptable range → true. Verification / Alternative check: Hydrology texts illustrate convolution of a UH with an excess-rain hyetograph to reproduce storm DRHs, confirming the linear scaling of ordinates and practical duration tolerance. Why Other Options Are Wrong: Each of (a)–(d) is correct individually; hence the comprehensive answer is (e) All the above. Common Pitfalls: Confusing UH depth (1 cm or 25 mm) with total rainfall depth; or applying UH outside its climatic/catchment context without adjustment. Final Answer: All the above.

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