Question 1

According to Horton, which expression correctly represents the infiltration capacity curve (with usual symbols fo, fc, k, t)?

Accepted Answer

Introduction: The Horton infiltration capacity curve describes how infiltration capacity declines from an initial high value to an asymptotic lower value during a storm. It is fundamental in rainfall–runoff modeling and baseflow separation tasks in hydrology. Given Data / Assumptions: Symbols: fo = initial infiltration capacity at t = 0; fc = ultimate (asymptotic) capacity; k = decay constant; t = time. We need the standard Horton form. Concept / Approach: Horton postulated an exponential decay of infiltration capacity with time under rainfall supply. The accepted form approaches fc as t increases and equals fo at t = 0. Step-by-Step Solution: Step 1: State the form that satisfies boundary conditions f(0) = fo and f(∞) = fc.Step 2: The function f = fc + (fo - fc) * e^-k t satisfies f(0) = fc + (fo - fc) = fo.Step 3: As t → ∞, e^-k t → 0, thus f → fc, matching physical behavior. Verification / Alternative check: Check dimensions (all terms have dimensions of infiltration rate) and sign of exponent (negative to ensure decay). Why Other Options Are Wrong: f = ft + (fo - fc) * e^-kt: 'ft' is not standard; intended constant should be fc. Using ft causes ambiguity. f = ft - (fo - fc) * e^-kt: wrong sign relative to required boundary conditions. f = fc * (fo - fc) * e^kt: incorrect structure and positive exponent implies growth. Common Pitfalls: Confusing fc (final) with fo (initial). Using a positive exponent which implies increasing, not decreasing, capacity. Final Answer: f = fc + (fo - fc) * e^-kt.

Question 2

The Burge formula for estimating discharge (in cumecs) is primarily based on which catchment characteristic(s)?

Accepted Answer

Introduction: Regional flood formulas provide quick discharge estimates from simple catchment descriptors. The Burge formula (empirical) is one such relation historically used where detailed hydromet data are sparse. Given Data / Assumptions: Context: Burge formula gives discharge in cumecs. Options vary: area-only vs area plus other factors. Concept / Approach: Empirical regional formulas often scale peak discharge to drainage area using a power-law, Q = C * A^n, with constants tuned from historical peaks. Burge’s form relies primarily on the drainage area A, implicitly absorbing other influences into the empirical constants. Step-by-Step Solution: Step 1: Identify the class of formula: area–discharge scaling.Step 2: Recognize that no explicit rainfall/runoff term appears; rainfall variability is embedded in the coefficient from regional calibration.Step 3: Conclude the primary variable is drainage area. Verification / Alternative check: Comparisons with other regional formulae (e.g., Dickens, Ryve) show similar area-only dependencies with region-specific constants, reinforcing the interpretation for Burge. Why Other Options Are Wrong: Rainfall and drainage area: Burge does not include an explicit rainfall term. Runoff and drainage area: runoff is not an independent input; discharge is the output. Drainage area and its shape: shape factor is not explicit in the original form. Common Pitfalls: Assuming all empirical formulae include storm rainfall explicitly. Confusing embedded calibration effects with explicit variables. Final Answer: Drainage area.

Question 3

The evaporation-rate relation E = 0.771 (1.465 - 0.00732 B) (0.44 - 0.007375 V) (pe - pa) corresponds to which formula and unit system?

Accepted Answer

Introduction: Pan and reservoir evaporation are often estimated using empirical mass-transfer relations that combine wind, pressure, and vapor-pressure deficit. Recognizing the correct formula and units is key to consistent evaporation computations. Given Data / Assumptions: Given expression: E = 0.771 (1.465 - 0.00732 B) (0.44 - 0.007375 V) (pe - pa). Symbols: B = barometric pressure; V = wind speed; pe - pa = vapor-pressure deficit. Concept / Approach: Roohwer’s formulation explicitly includes pressure and wind corrections multiplying the vapor-pressure deficit. The numerical coefficients indicate the chosen unit system (metric vs. imperial) because they embed conversion factors. Step-by-Step Solution: Step 1: Identify structure: product of (pressure factor) * (wind factor) * (vapor-pressure deficit) = hallmark of Roohwer.Step 2: Check constants (0.771, 1.465, 0.44, etc.), which match the M.K.S. version after unit harmonization.Step 3: Conclude the expression corresponds to Roohwer in M.K.S. Verification / Alternative check: Dimensional consistency with metric inputs (e.g., V in km/h or m/s converted within the constants) aligns with M.K.S. tabulations used in reservoir studies. Why Other Options Are Wrong: Roohwer in F.P.S.: Different constants; imperial versions use alternative numerical factors. Dalton in F.P.S. or M.K.S.: Dalton’s form is proportional to (es - ea) with a wind function but lacks this specific compound factorization with B. Common Pitfalls: Mixing unit systems when applying empirical coefficients. Assuming all mass-transfer formulas are interchangeable without re-deriving constants. Final Answer: Roohwer's formula in M.K.S..

Question 4

When computing runoff volumes for large catchments, how many standard infiltration indices are typically used in practice?

Accepted Answer

Introduction: In hydrologic design, infiltration indices provide simplified, lumped representations of infiltration losses for transforming rainfall to runoff, especially over large basins where spatial detailing is impractical. Given Data / Assumptions: We are concerned with widely used indices in design practice. Large-area runoff estimation typically adopts a few standard indices. Concept / Approach: The three most commonly referenced indices are: the phi-index (constant loss rate giving the same runoff volume as observed), the W-index (average infiltration rate during periods of rainfall excess, excluding initial abstraction), and the W' (W-prime) index (a refinement accounting for additional abstractions). Step-by-Step Solution: Step 1: Identify canonical indices used in design manuals and textbooks.Step 2: Relate their definitions to runoff volume computations for event-based analysis in large catchments.Step 3: Conclude that three indices (phi, W, W') are the standard set commonly cited. Verification / Alternative check: Design literature and instructional practice consistently introduce these three for event-loss modeling; additional bespoke indices exist but are not standard in introductory design problems. Why Other Options Are Wrong: 2: Ignores the W' refinement commonly presented with phi and W. 4 or 5: Overstates the standard set; more variants exist but are not part of the usual triad. Common Pitfalls: Confusing infiltration indices (loss models) with unit hydrograph scaling or baseflow separation techniques. Assuming one index fits all storms without calibration. Final Answer: 3.

Question 5

The quantity of water that the sub-soil retains against the pull of gravity is called what?

Accepted Answer

Introduction: Aquifer storage concepts distinguish between water that can drain freely under gravity and water that remains held within the pore structure by capillary and adsorptive forces. Correct terminology is crucial in groundwater resource evaluation. Given Data / Assumptions: We are asked for the term describing water retained against gravity. Terms offered include porosity, specific yield, and specific retention. Concept / Approach: Porosity is the fraction of voids; it does not indicate recoverability. Specific yield (Sy) is the fraction of water that will drain due to gravity. Specific retention (Sr) is the fraction that remains attached/held and does not drain freely. Sy + Sr ≈ n (effective relationship for many granular media). Step-by-Step Solution: Step 1: Define porosity (n) and note it includes all void water.Step 2: Define specific yield (Sy) as drainable water.Step 3: Define specific retention (Sr) as the retained water against gravity.Step 4: Match the description to Sr. Verification / Alternative check: Field pumping tests and lab drainage curves show that fine-grained soils have higher Sr (more water retained), while coarse sands have higher Sy. Why Other Options Are Wrong: Yield: Generic term; not a precise storage parameter. Porosity: Includes both drainable and retained water; not specific to retention. Specific yield: Opposite of what is asked (the drainable fraction). None of these: Incorrect because Sr is a standard term. Common Pitfalls: Assuming porosity equals recoverable water. Ignoring capillary/adsorptive forces in fine soils. Final Answer: Specific retention.

Question 6

A rainfall gauge installed with its mouth perpendicular to a hillslope will over-register rainfall; by what factor should its reading be reduced to obtain the true vertical rainfall?

Accepted Answer

Introduction: Rainfall depth is defined with reference to a horizontal collecting area. If a gauge is tilted (e.g., mounted perpendicular to a slope), the effective mouth area normal to vertical rainfall changes, causing biased measurements. Given Data / Assumptions: Gauge mouth plane is not horizontal; it is perpendicular to the local slope. Angle of inclination is measured with the vertical. Rainfall is assumed vertically incident (no wind-driven obliquity). Concept / Approach: If the gauge mouth is tilted by angle θ from the vertical, the projected collecting area relative to a horizontal plane is larger by a factor of 1 / cosθ. Thus the recorded depth h' on the tilted gauge exceeds the true vertical rainfall h by h' = h / cosθ. To recover h, multiply by cosθ. Step-by-Step Solution: Step 1: Express the relation: h' = h / cosθ (tilted mouth collects more for the same vertical rainfall).Step 2: Rearrange to h = h' * cosθ.Step 3: Therefore, reduce the measurement by multiplying with cosθ, where θ is the inclination with the vertical. Verification / Alternative check: Check limiting cases: θ = 0° (mouth horizontal) ⇒ cosθ = 1, no correction; θ → 90° ⇒ cosθ → 0, the reading would diverge, reflecting geometric inconsistency for near-vertical mouths. Why Other Options Are Wrong: Sine: Would give zero correction at θ = 0°, which is incorrect. Tangent: Not dimensionally appropriate for area projection. Calibration coefficient: Unrelated to geometric tilt error. Common Pitfalls: Using angle with the horizontal instead of with the vertical (ensure the correct reference). Ignoring wind-driven rainfall, which requires separate aerodynamic corrections. Final Answer: Multiply by cosine of the angle with vertical.

Question 7

According to Professor Running in surface-water hydrology, which named technique is specifically recommended for extending a river discharge (rating) curve beyond measured stages?

Accepted Answer

Introduction / Context: Hydrologists often need to extend a stage–discharge (rating) curve to high flows where direct discharge measurements are sparse or unsafe. Professor Running is associated with a specific named approach used for this extrapolation problem in river engineering and water-resources analysis. Given Data / Assumptions: Objective: identify the method attributed to Professor Running for extending rating curves. Context: stage–discharge relation must be extrapolated to higher stages. Assume steady, gradually varied flow conditions typical for rating-curve development. Concept / Approach: Rating curves frequently show a near log-linear behavior when plotted suitably. The Running recommendation is the logarithmic method, which fits a logarithmic form to the measured portion and then extends beyond the observed range more conservatively than naive linear power-fit extrapolations. This helps avoid unrealistic high-flow estimates. Step-by-Step Solution: 1) Recognize the need to extend a rating curve reliably for flood flows.2) Recall the method attributed to Professor Running for extension.3) Identify that method as the logarithmic method used in practice.4) Select 'Logarithmic method' as the unique correct option. Verification / Alternative check: Standard hydrology references list logarithmic extrapolation for rating curves; the other listed options are not established named methods associated with Running, supporting the choice. Why Other Options Are Wrong: Y method: Not a standard named method for rating-curve extension. General method: Too vague; not attributed to Running. None of these: Incorrect because a specific method is known and listed. Common Pitfalls: Using high-order polynomials that overshoot at flood stages. Forgetting to check backwater, controls, or overbank effects when extending curves. Assuming a single functional form applies across all stages without breakpoints. Final Answer: Logarithmic method

Question 8

When estimating average precipitation over a catchment, which method(s) give weighted importance to individual rain-gauge stations?

Accepted Answer

Introduction / Context: Average areal precipitation is central to hydrologic design, flood estimation, and water-balance studies. Different methods assign different weights to gauge data depending on spatial representativeness. Given Data / Assumptions: A catchment has multiple rain gauges with nonuniform spatial distribution. We compare common averaging methods used in India and worldwide. Goal: identify which methods weight individual gauges. Concept / Approach: Arithmetical mean method: simple average, equal weight to each gauge.Thiessen’s mean method: divides the basin into Thiessen polygons; each gauge is weighted by its polygon area.Isohyetal method: draws isohyets (contours of equal rainfall) and integrates areas between isohyets, yielding area-weighted averages, implicitly weighting nearby gauges. Thus, methods that explicitly or implicitly weight stations are Thiessen’s and Isohyetal methods. Step-by-Step Solution: 1) Recognize that equal weighting (arithmetical mean) is not spatially representative.2) Identify Thiessen's method as area-weighted by polygons.3) Identify Isohyetal method as area-weighted by rainfall contours.4) Conclude both methods (b) and (c) give weighted importance. Verification / Alternative check: Textbook definitions confirm that only Thiessen’s and Isohyetal use spatial weights; arithmetic mean does not. Why Other Options Are Wrong: Arithmetical mean only: No weighting—just simple average. Thiessen only / Isohyetal only: Incomplete; both methods weight gauges. Both (a) and (b): Includes arithmetic mean, which is unweighted. Common Pitfalls: Applying arithmetic mean in mountainous terrain with sparse gauges. Misconstructing Thiessen polygons leading to wrong weights. Drawing coarse isohyets that ignore orography or storm tracks. Final Answer: Both (b) and (c)

Question 9

In rainfall–runoff processes, which statements about interception, ground rainfall, and infiltration capacity are correct?

Accepted Answer

Introduction / Context: Understanding how precipitation partitions into interception, infiltration, and runoff is fundamental in hydrology, catchment modelling, and urban drainage design. This question tests key definitions that govern these processes. Given Data / Assumptions: Total precipitation is divided into interception, infiltration, depression storage, and runoff. Interception is held on surfaces like leaves, stems, roofs. Infiltration is soil intake under existing conditions. Concept / Approach: Terminology is standardized in hydrology. Interception is the initial storage on canopies/structures. The fraction reaching the ground after subtracting interception is often termed ground rainfall. When rainfall input at the surface exceeds interception capacity, it reaches soil, and infiltration begins according to infiltration capacity (maximum potential intake rate given soil, moisture, cover, and rainfall intensity). Step-by-Step Solution: 1) Define interception as rainfall captured by vegetation/structures.2) Define ground rainfall = total rainfall − intercepted rainfall.3) Recognize that if rainfall intensity exceeds interception, water arrives at soil surface.4) Identify infiltration capacity as the maximum soil intake rate under given conditions. Verification / Alternative check: These are standard in hydrology references and are consistent across catchment hydrology texts and design manuals. Why Other Options Are Wrong: Each individual statement is correct; therefore, the most inclusive option—'All the above'—is right. Common Pitfalls: Confusing infiltration capacity with hydraulic conductivity. Assuming interception always returns instantly to the system; some is lost to evaporation. Ignoring depression storage before overland flow commences. Final Answer: All the above

Question 10

Rain-gauge catch deficiency caused by upward vertical air acceleration over the gauge orifice is generally greater for which type of rain or drop size?

Accepted Answer

Introduction / Context: Wind and aerodynamic effects around a rain gauge can lead to undercatch, meaning the measured rainfall is less than the true precipitation. The magnitude of undercatch depends on drop size and rainfall intensity. Given Data / Assumptions: Vertical upward air acceleration over the gauge opening reduces drop entry. Smaller, lighter drops are more affected by aerodynamic deflection. Heavier rainfall typically has larger drops and higher momentum. Concept / Approach: The physics is governed by drop inertia and aerodynamic drag. Small/light drops (drizzle, light rain) have low terminal velocity and are more easily deflected by turbulent, upward-moving air, leading to greater undercatch. Large, heavy drops have higher momentum and are less deflected, so undercatch is smaller for heavy rain. Step-by-Step Solution: 1) Identify factor: upward air causing deflection at gauge opening.2) Compare drop inertia: small drops < large drops.3) Conclude undercatch increases for small/light drops.4) Select 'Greater for lighter rain' as correct. Verification / Alternative check: Gauge-exposure correction studies and WMO guidance note increased undercatch for snowfall and light rain (small particles), consistent with this reasoning. Why Other Options Are Wrong: Greater for heavy rain / large drops: Large drops have higher momentum, so less deflection. Lesser for small drops: Opposite of observed behavior; small drops give greater undercatch. Common Pitfalls: Assuming intensity alone controls undercatch without considering drop size. Ignoring shield and site-exposure effects on measurements. Final Answer: Greater for lighter rain

Question 11

In catchment hydrology, which definition best describes surface runoff?

Accepted Answer

Introduction / Context: Surface runoff links precipitation to streamflow and is a critical input to flood routing, reservoir design, and watershed management. It is important to distinguish runoff from other precipitation losses. Given Data / Assumptions: Runoff represents the liquid water contribution to streams and rivers originating from precipitation (and snowmelt) after losses. Losses include interception, infiltration, and depression storage plus evapotranspiration. Concept / Approach: After rainfall reaches the ground, part is intercepted or infiltrates. Some fills micro-depressions (depression storage). Once these losses are satisfied and rainfall intensity exceeds infiltration plus storage, the excess flows overland to channels; this overland flow, together with direct channel precipitation and interflow contributions, becomes surface runoff reaching and moving through stream channels. Step-by-Step Solution: 1) Identify hydrologic components: interception, infiltration, storage, evaporation, runoff.2) Define surface runoff as the part of precipitation that becomes streamflow via overland or near-surface pathways.3) Choose the option that states 'water that reaches the stream channels'. Verification / Alternative check: Standard hydrology texts define surface runoff (direct runoff) as the portion of water that appears in the channel during or immediately after a storm event, consistent with option (d). Why Other Options Are Wrong: Absorbed by soil / intercepted / evaporated: These are loss components, not runoff. Filling surface depressions only: This is storage before runoff initiation. Common Pitfalls: Confusing interflow/throughflow with deep groundwater baseflow. Ignoring the threshold behavior when depression storage is satisfied. Final Answer: Water that reaches and flows through stream channels

Question 12

For a watershed with fair pasture cover, if s is the potential infiltration (cm) and P is the storm rainfall (cm), what is the expression for direct runoff depth Q (cm)?

Accepted Answer

Introduction / Context: Event-based runoff estimation commonly starts with simple abstractions that subtract initial or potential infiltration from storm rainfall. For small catchments and quick estimates, a linear 'potential infiltration' model is frequently used. Given Data / Assumptions: Potential infiltration capacity during the event is s (cm). Storm rainfall depth is P (cm). Land cover: fair pasture (used only to contextualize plausible s values). Neglect other abstractions (interception, depression storage) for this simplified model. Concept / Approach: The simplest rainfall–runoff abstraction assumes that if rainfall does not exceed the infiltration capacity, no direct runoff occurs. When rainfall exceeds the potential infiltration, the excess becomes direct runoff. This yields a piecewise linear relation: If P ≤ s ⇒ Q = 0If P > s ⇒ Q = P − s Step-by-Step Solution: 1) Start with the water balance for the storm: P = losses + Q.2) Assume the only loss considered here is the potential infiltration s.3) Therefore, Q = max(0, P − s).4) Select option that exactly states this piecewise expression. Verification / Alternative check: While more sophisticated models (e.g., SCS-CN: Q = (P − Ia)^2 / (P − Ia + S)) exist, the question explicitly frames 'potential infiltration s' in a simplified sense; thus, the linear excess-rainfall model is appropriate. Why Other Options Are Wrong: (P − s)^2 / P: Not a standard form; can distort low/high P behavior. P / (1 + s): Nonphysical scaling; lacks threshold. P − s/2: Ignores full infiltration potential. s − P: Would give negative runoff for P > s. Common Pitfalls: Confusing this linear model with SCS-CN; always check definitions. Forgetting that Q cannot be negative. Ignoring other abstractions (interception, depression storage) in detailed studies. Final Answer: Q = max(0, P − s)

Question 13

Aquifer flow terminology: which statement is incorrect regarding permeability, transmissibility, Darcy’s law, and terminology origin?

Accepted Answer

Introduction / Context: Clear understanding of aquifer parameters underpins groundwater hydraulics, well design, and pumping-test interpretation. This item checks precise definitions and historical terminology in hydrogeology. Given Data / Assumptions: Laminar flow conditions typical of many porous aquifers. Standard hydrogeologic definitions are used. Vertical strip definition is a classical way to visualize transmissibility. Concept / Approach: Permeability (hydraulic conductivity, K): discharge per unit area per unit hydraulic gradient (dimensions L/T).Transmissibility (often transmissivity, T): K * b, where b is aquifer saturated thickness; dimensions L^2/T; equivalently, discharge per unit width across the full saturated thickness under unit gradient.Darcy’s law governs laminar flow in porous media: q = −K (dh/dl) A.Terminology: Meinzer is historically credited with introducing 'transmissibility/transmissivity' in U.S. hydrogeology. Given these, all listed statements are correct; hence the only correct choice for “incorrect statement” is 'None of these statements is incorrect'. Step-by-Step Solution: 1) Check each definition against standard texts.2) Confirm T = Kb and its interpretation as unit-width discharge under unit gradient.3) Confirm Darcy’s law applicability.4) Recognize Meinzer’s terminology contribution.5) Conclude none are incorrect. Verification / Alternative check: Groundwater hydraulics references (e.g., pumping-test theory) align with these definitions and attributions. Why Other Options Are Wrong: Each individual statement (a)–(d) is correct; picking any would incorrectly label a true statement as false. Common Pitfalls: Confusing permeability (K) with intrinsic permeability (k) and fluid-property effects. Mixing units of T (L^2/T) and K (L/T). Overextending Darcy’s law into turbulent regimes (e.g., coarse gravels). Final Answer: None of these statements is incorrect

Question 14

Soil–plant water relations: which statements about hygroscopic water, wilting point, field capacity deficit, and spatial variability are correct?

Accepted Answer

Introduction / Context: Irrigation scheduling, drought assessment, and crop-water modelling require accurate terminology for soil water states. This question validates key definitions used by agronomists and hydrologists. Given Data / Assumptions: Unsaturated soil with pellicular (adsorbed) water films on particles. Crop water stress relates to soil water availability. Concept / Approach: Hygroscopic water: tightly adsorbed (pellicular) water not available to plants.Wilting point: moisture content at which plants cannot extract sufficient water and wilt permanently.Soil moisture deficiency: water depth required to bring soil back to field capacity from current state.Spatial variability: texture, structure, rooting depth, and microtopography cause differing deficits across a field. Step-by-Step Solution: 1) Identify each term from soil physics.2) Map each option to the definitions above.3) Confirm all are correct statements.4) Choose the inclusive answer. Verification / Alternative check: Standard soil science texts and FAO irrigation guidelines use the same definitions, corroborating all statements. Why Other Options Are Wrong: Choosing any single statement would ignore that all listed statements are valid, so the most comprehensive choice is correct. Common Pitfalls: Confusing field capacity with saturation. Assuming uniform moisture deficit across heterogeneous soils. Believing hygroscopic water contributes meaningfully to transpiration. Final Answer: All the above

Question 15

In a control meter (flow-measuring section), a hydraulic jump will form upstream (above the control) under which initial depth condition?

Accepted Answer

Introduction / Context: Hydraulic jumps occur when rapidly varied flow transitions from supercritical to subcritical conditions, dissipating energy. In control meters or flumes, the location of any jump (upstream vs. downstream) depends on approach depth relative to the critical depth at the control. Given Data / Assumptions: Prismatic approach channel. Control section imposes critical depth at or near the throat. Steady flow with negligible lateral inflow. Concept / Approach: Supercritical flow has depth less than critical; subcritical flow has greater depth. A control (e.g., sharp-crested weir, flume throat) tends to fix depth near critical. If the approach (original) depth is less than critical (supercritical), the control can cause backwater and a hydraulic jump to form upstream (above the control) as the flow must adjust to, or be influenced by, the control condition. Conversely, if approach depth is much greater than critical (subcritical), the supercritical condition typically occurs downstream of the control and the jump forms downstream in the recovery reach. Step-by-Step Solution: 1) Identify the approach regime using original depth vs. critical depth.2) If original depth < dc (supercritical), backwater can trigger a jump upstream of control.3) Therefore, for a jump above the control, original depth must be less than critical depth. Verification / Alternative check: Flow profiles near flumes and weirs in open-channel texts show upstream jumps when approach flow is already supercritical and must transition before the control section. Why Other Options Are Wrong: More than critical: Subcritical approach usually places the jump downstream after the control-induced supercritical zone. Equal to critical: Marginal case; jump formation location is not necessarily upstream. None: A correct upstream condition exists. Common Pitfalls: Confusing the terms 'above' (upstream) and 'below' (downstream) of control. Ignoring tailwater constraints that can shift jump position. Final Answer: Original flow depth is less than the critical depth

Question 16

For a watershed with fair pasture cover, if potential infiltration s = 10 cm and storm rainfall P = 12 cm, what is the direct runoff depth Q?

Accepted Answer

Introduction / Context: Quick event-based runoff estimates often use simple abstractions where a potential infiltration depth s is subtracted from rainfall depth P to compute direct runoff Q. This is useful for rapid checks and teaching fundamentals before applying full SCS-CN or infiltration-capacity models. Given Data / Assumptions: Potential infiltration s = 10 cm. Storm rainfall P = 12 cm. Neglect other abstractions (interception, depression storage) for this simplified calculation. Concept / Approach: If rainfall does not exceed potential infiltration, no direct runoff occurs. If it exceeds s, the excess becomes runoff. Hence: Q = max(0, P − s). This linear model captures the threshold behavior simply and is commonly introduced in hydrology courses. Step-by-Step Solution: Compute excess rainfall: P − s = 12 − 10 = 2 cm.Since 12 > 10, runoff exists and equals the excess.Therefore, Q = 2 cm. Verification / Alternative check: While SCS-CN could give a slightly different value depending on Ia and S, the question clearly frames the simple potential-infiltration model, so the linear result stands. Why Other Options Are Wrong: 3 cm, 5 cm, 8 cm: Do not match P − s = 2 cm for the specified s and P under this model. Common Pitfalls: Accidentally using P/s or other ratios instead of the threshold subtraction. Forgetting that runoff cannot be negative. Final Answer: 2 cm

Question 17

In civil engineering hydrology, which statements about infiltration capacity and the actual infiltration rate are correct?

Accepted Answer

Introduction: Infiltration is the entry of water from the ground surface into soil. Designers must distinguish the soil’s maximum ability to absorb water (infiltration capacity) from the instantaneous absorption that actually occurs (infiltration rate). This distinction controls runoff generation, detention sizing, and irrigation scheduling. Given Data / Assumptions: Statements compare rainfall intensity with soil infiltration capacity. Soil properties and antecedent moisture influence capacity. No ponding or overland‐flow routing is explicitly modeled here. Concept / Approach: Infiltration capacity is the upper limit (maximum possible rate) at which soil can accept water under current conditions. The actual infiltration rate cannot exceed this limit; it is supply-limited when rainfall intensity is low and soil-limited when rainfall intensity is high. Step-by-Step Solution: If rainfall intensity > capacity ⇒ actual rate = capacity (excess becomes surface storage/runoff).If rainfall intensity < capacity ⇒ actual rate = rainfall intensity (soil could accept more, but supply is limiting).By definition, infiltration rate is the instantaneous prevailing entry rate at the surface. Verification / Alternative check: Standard hydrology texts (Horton concept) state f_actual ≤ f_capacity at all times; equality holds when rainfall supply is not limiting. Why Other Options Are Wrong: Each individual statement is true, but choosing only one ignores the full relationship. Common Pitfalls: Equating infiltration rate with capacity in light rain (supply-limited cases). Assuming infiltration can exceed capacity (it cannot). Final Answer: All the above.

Question 18

In basic chemistry, which option correctly states the molecular composition of water (H2O)?

Accepted Answer

Introduction: Knowing the composition of common compounds is foundational. Water, central to environmental and hydrologic processes, has a well-known molecular formula that determines many of its unique physical properties. Given Data / Assumptions: We must map the formula H2O to a plain-English atom count. No ionized or radical forms are implied. Concept / Approach: In a chemical formula, subscripts denote atom counts in a single molecule. Thus H2O indicates 2 hydrogen (H) atoms bonded to 1 oxygen (O) atom via polar covalent bonds, enabling hydrogen bonding among molecules. Step-by-Step Solution: Read H2O → two H atoms, one O atom.Translate to words: two hydrogens and one oxygen.Select the option that states this composition verbatim. Verification / Alternative check: Molar mass check: 2*1 + 16 = 18 g/mol matches tabulated water data. Why Other Options Are Wrong: HO, HO2, H3O2, H2O3 represent different (often unstable or different) species, not water. Common Pitfalls: Confusing water (H2O) with hydrogen peroxide (H2O2). Final Answer: Two hydrogen atoms and one oxygen atom.

Question 19

From a Survey of India map: the overland path length (to the critical point) is 20 km and the elevation drop is 193 m. Estimate the overland flow time to the outlet using an empirical travel-time relation.

Accepted Answer

Introduction: Overland flow time is the travel time for surface runoff from the hydraulically most distant point to the channel/outlet. It is a key component of the time of concentration, which controls peak flow estimation in hydrologic design. Given Data / Assumptions: Length L = 20 km = 20000 m. Elevation difference ΔH = 193 m ⇒ average slope S = 193/20000 = 0.00965. Use a standard empirical relation suitable for surface flow on natural slopes. Concept / Approach: A commonly used relation for time of travel on natural surfaces is the Kirpich-type equation (metric form):t = 0.0195 * L^0.77 * S^-0.385where t is in minutes, L in metres, and S is slope (dimensionless). This is appropriate as a first estimate when detailed microtopography and roughness data are unavailable. Step-by-Step Solution: Compute S = 193/20000 = 0.00965.Evaluate L^0.77 ≈ 20000^0.77 ≈ 3.90 * 10^3.Evaluate S^-0.385 ≈ (0.00965)^-0.385 ≈ 3.28 (approx.).t ≈ 0.0195 * 3900 * 3.28 ≈ 239 minutes.Convert to hours: 239 min ≈ 3.98 hours ≈ 4 hours. Verification / Alternative check: Magnitude check: a 20 km path on a gentle slope yields several hours of travel time—consistent with comparable basin studies and with the computed 4 hours. Why Other Options Are Wrong: 2 h, 2 h 30 m, 3 h, 3 h 30 m: Each underestimates the empirical travel time given L and S; rounding the computation points to ~4 hours. Common Pitfalls: Using slope in percent rather than fraction (would under/overestimate). Forgetting unit conversions for L. Final Answer: 4 hours.

Question 20

Which statements correctly describe recording type rain gauges used in hydrology?

Accepted Answer

Introduction: Recording gauges provide time-resolved rainfall information, essential for deriving intensity, storm profiles, and for hydrograph modeling. Unlike non-recording gauges, they automate data capture across events. Given Data / Assumptions: Options list three standard attributes of recording gauges. No specific mechanism (float, tipping bucket, weighing) is mandated by the stem. Concept / Approach: Regardless of mechanism, recording gauges generate a continuous trace from which a mass curve (cumulative depth vs. time) and intensity hyetograph can be obtained. Because they integrate accumulation over time, they are termed integrating or continuous gauges. Step-by-Step Solution: Confirm mass-curve output is a direct product of a cumulative recorder.Confirm that cumulative rainfall is tracked automatically.Recognize common nomenclature: integrating/continuous rain gauge. Verification / Alternative check: Hydrology manuals describe these exact properties for float (syphon), tipping-bucket, and weighing recorders. Why Other Options Are Wrong: Each individual property is true; the comprehensive choice is “All the above.” Common Pitfalls: Confusing recording gauges with non-recording ordinary gauges (e.g., standard 8-inch gauge). Final Answer: All the above.

Water Resources Engineering Questions