Question 1

For preliminary hydraulic design, what is a reasonable assumed efficiency for a typical pump when exact performance curves are unavailable?

Accepted Answer

Introduction:Pump efficiency (η) = water horsepower / shaft horsepower. In early‐stage design and quick checks, engineers adopt representative values to size motors and estimate energy use before selecting a specific pump curve. Given Data / Assumptions:Pump type is not specified; assume a conventional centrifugal pump.Field installation and off-BEP operation often reduce ideal efficiencies. Concept / Approach:Typical field efficiencies for medium-size centrifugal pumps cluster around 60–70%. A mid-range assumption balances conservatism with realism for screening studies and preliminary power estimates. Step-by-Step Solution:Select a representative mid-range η.0.65 is widely used as a planning-level default.Refine later with manufacturer's BEP curves and system curve intersection. Verification / Alternative check:Design handbooks and codes of practice list similar ranges; detailed selections may exceed 70% under ideal conditions but should not be assumed globally. Why Other Options Are Wrong:0.55: Too conservative for a general assumption.0.60: Acceptable but typically not the default mid-range pick.0.70: Feasible for selected units near BEP; optimistic for general assumptions. Common Pitfalls:Using best-case lab efficiency in field energy estimates. Final Answer:0.65.

Question 2

In meteorology, how does absolute humidity (mass of water vapour per unit volume of air) vary with altitude?

Accepted Answer

Introduction:Absolute humidity (AH) quantifies vapour mass per air volume and, unlike relative humidity, is not normalized by saturation capacity. Understanding AH variation with height is important for flight operations, cloud physics, and rainfall formation. Given Data / Assumptions:Altitude increase implies lower pressure and temperature on average.Most atmospheric water vapour resides in the lower troposphere. Concept / Approach:As pressure falls with height, a fixed vapour mass occupies a larger volume, and cooler temperatures reduce evaporation sources. Consequently, the actual vapour mass within a given volume typically declines with altitude, so AH decreases. Step-by-Step Solution:Recognize vertical gradients: pressure ↓, temperature ↓ with altitude.Lower temperatures curb evaporation and support condensation.Therefore, average AH diminishes upward. Verification / Alternative check:Profiles from radiosondes show sharp AH decreases above ~3–5 km; most precipitable water is concentrated near the surface. Why Other Options Are Wrong:Increase/Constant: Contradict observed vertical humidity profiles.None of these: Not applicable, as decrease is the correct trend. Common Pitfalls:Confusing absolute humidity (per volume) with specific humidity (per mass of air), which varies differently. Final Answer:It decreases at higher altitudes.

Question 3

Which factors significantly influence runoff from a catchment?

Accepted Answer

Introduction:Runoff is controlled by how, when, and where precipitation falls, and by how much the soil can absorb. Recognizing the multi-factor nature of runoff is central to flood estimation and watershed management. Given Data / Assumptions:Options list storm form, intensity/duration, spatial distribution, antecedent moisture, and storm track. Concept / Approach:Type (rain/snow) affects losses and melt. High intensity and long duration raise excess rainfall. Non-uniform distribution alters hydrograph shape. Dry/wet antecedent soil changes infiltration. Storm movement relative to the basin can synchronize tributary peaks and amplify the outlet peak. Step-by-Step Solution:Assess each factor’s physical influence on rainfall excess and timing.Recognize that all listed factors act together in real storms.Select the comprehensive option. Verification / Alternative check:Design manuals (unit hydrograph and SCS methods) explicitly include these factors via loss models and areal reduction/time-of-concentration methods. Why Other Options Are Wrong:Any single factor alone omits important controls; “All the above” captures the full set. Common Pitfalls:Considering only intensity and ignoring storm track/areal variability. Final Answer:All the above.

Question 4

On what soil characteristics does infiltration capacity primarily depend?

Accepted Answer

Introduction:Infiltration capacity is the maximum rate at which soil can accept water under given conditions. It is a function of soil structure and texture, which control pore connectivity and capillary forces. Given Data / Assumptions:Options list void ratio/porosity, particle geometry, arrangement, and compaction. Concept / Approach:Large, well-connected pores (macropores) raise capacity; fine, poorly connected pores reduce it. Particle size/shape and fabric influence pore geometry. Compaction reduces voids and connectivity, lowering infiltration capacity. Therefore, all listed properties matter. Step-by-Step Solution:Relate each property to pore size distribution and connectivity.Infer impact on gravitational and capillary flow paths.Conclude that all properties jointly control capacity. Verification / Alternative check:Soil physics shows infiltration increases from clays → loams → sands (texture/structure effects) and declines with compaction and crusting. Why Other Options Are Wrong:Any single factor explanation is incomplete; multiple soil attributes interact. Common Pitfalls:Ignoring soil structure (aggregation, biopores) and focusing only on texture. Final Answer:All the above.

Question 5

In open-channel flow, if y is the flow depth at a section, an accepted empirical relation for the mean velocity V (m/s) is V = k * y, where k is approximately which of the following?

Accepted Answer

Introduction:Velocity distribution in open channels is non-uniform: it is zero at the boundary and maximum near the surface, with the depth-averaged (mean) velocity occurring below the surface. Field practice often uses empirical depth–velocity relations for quick checks. Given Data / Assumptions:y is the flow depth.An empirical proportionality V ≈ k * y is referenced. Concept / Approach:For many steady, uniform flows in rough channels, the depth of mean velocity occurs at about 0.6 y below the free surface; equivalently, the velocity measured at 0.6 y is close to the depth-average. Hence a shorthand rule uses the 0.6-depth velocity (or coefficient 0.6 in simple proportionalities). Step-by-Step Solution:Recall common field rule: V_mean ≈ V at 0.6 y below surface.Translate to proportional form: coefficient closest to 0.6.Select option “0.6 y”. Verification / Alternative check:USGS and standard hydraulics references use the two-point (0.2 y and 0.8 y) or single-point (0.6 y) method; the 0.6-point is the accepted single-point estimate for mean velocity. Why Other Options Are Wrong:0.1–0.5 y: Do not match the observed location/estimate of mean velocity in typical natural channels. Common Pitfalls:Confusing the depth where mean velocity occurs (0.6 y below surface) with the location of maximum velocity (near surface). Final Answer:0.6 y.

Question 6

Hydrology is applied to which of the following planning and design tasks in water resources engineering?

Accepted Answer

Introduction:Hydrology provides the quantitative basis for designing and operating water resources systems. From flood control to water supply, hydrologic analyses inform risk, reliability, and economic decisions. Given Data / Assumptions:Tasks include flood prediction, reservoir sizing, yield estimation, and backwater/level impacts of dams. Concept / Approach:Design floods are obtained from frequency analysis or event modeling; dependable yields and storage come from flow duration and mass curve methods; expected river level changes are forecast using inflow hydrographs combined with routing and stage–discharge relations. Step-by-Step Solution:Relate each listed task to a standard hydrologic method (flood frequency, reservoir planning curves, yield analysis, hydraulic backwater modeling with hydrologic inputs).All tasks rely on hydrologic inputs and analysis.Therefore, the comprehensive option is correct. Verification / Alternative check:Codes of practice and planning manuals explicitly require hydrologic studies for these decisions before hydraulic/structural detailing. Why Other Options Are Wrong:Any single task alone is incomplete—hydrology underpins all the listed tasks. Common Pitfalls:Confusing hydrology (catchment response) with pure hydraulics (reach flow profiles) and ignoring their linkage in design. Final Answer:All the above.

Question 7

In surface-water hydrology, who first proposed the widely used 'infiltration capacity' concept for describing the maximum possible rate at which soil can absorb rainfall under given conditions?

Accepted Answer

Introduction / Context:The concept of infiltration capacity is fundamental in hydrology and storm-runoff modeling. It represents the upper limit of soil intake rate under prevailing conditions (soil type, moisture, cover, and rainfall intensity). This question checks your recall of the originator of the theory used in many infiltration and excess-rainfall computations.Given Data / Assumptions:The term 'infiltration capacity' refers to a rate (depth per unit time).We focus on the origin of the theory, not its formulae or parameter estimation.Names listed are historical hydrology figures; only one is directly credited with this theory.Concept / Approach:Horton advanced a physically inspired, empirically supported description in which infiltration capacity typically decreases exponentially during a storm toward an asymptotic minimum. His framework underpins Hortons infiltration model and many textbook treatments, making 'Horton' the definitive answer.Step-by-Step Solution:Identify the hydrologist associated with infiltration capacity → Robert E. Horton.Recognize that alternatives are not credited with originating the theory.Select the correct option that matches the historical attribution.Verification / Alternative check:Standard hydrology texts (storm-runoff analysis, unit hydrograph derivations, infiltration modeling) consistently cite Horton for the infiltration capacity concept and its exponential decay form.Why Other Options Are Wrong:Merrill Bernard / W. W. Horner / LeRoy K. Sherman: Important contributors in hydrology, but not originators of the infiltration capacity theory.None of these: Incorrect because the correct name appears in the list.Common Pitfalls:Confusing Horton's infiltration concept with SCS Curve Number abstraction; they are distinct methods.Attributing the theory to general 'infiltration tests' or empirical charts rather than the original theoretical framework.Final Answer:Robert E. Horton

Question 8

Using Slichter’s velocity relation for groundwater flow, compute the seepage velocity given: viscosity factor = 1.00, Slichter’s constant C = 400, effective size d = 0.5 mm, and hydraulic gradient i = 1/80. Choose the closest value (m/day).

Accepted Answer

Introduction / Context:Empirical groundwater-flow relations like Slichter’s are useful for quick estimates of seepage velocity in granular aquifers when effective size and gradient are known. This problem applies the standard form to obtain a daily velocity.Given Data / Assumptions:Viscosity factor (temperature corrected) = 1.00 (typical reference condition).Slichter’s constant C = 400.Effective grain size d = 0.5 mm.Hydraulic gradient i = 1/80 = 0.0125.Use the common form v = C * i * d^2 with d in mm and v in m/day.Concept / Approach:Slichter’s empirical velocity relation is expressed as:v = C * i * d^2where v is the apparent seepage velocity (m/day), i is hydraulic gradient, d is effective size (mm), and C is a temperature-dependent constant (dimension-adjusted). With the viscosity factor set to 1.00, we use the supplied C directly.Step-by-Step Solution:Compute d^2: d = 0.5 mm → d^2 = 0.5^2 = 0.25.Compute i: i = 1/80 = 0.0125.Compute v: v = 400 * 0.0125 * 0.25.First multiply 400 * 0.0125 = 5.0, then 5.0 * 0.25 = 1.25 m/day.Verification / Alternative check:Order of magnitude is reasonable for coarse sand under a mild gradient; daily seepage of about a meter is credible in such materials.Why Other Options Are Wrong:Values 1.00 m/day is close but still below the exact calculation.Common Pitfalls:Using d in meters rather than millimeters in the Slichter form provided with C.Forgetting to square the grain size.Final Answer:1.25 m/day

Question 9

Perched aquifers: identify the correct statements regarding occurrence, perched water table, and formation within an unconfined setting.

Accepted Answer

Introduction / Context:Perched aquifers are local saturated zones situated above the regional water table due to intervening low-permeability layers. Recognizing perched conditions is essential for siting wells, assessing seepage, and interpreting soil moisture behavior.Given Data / Assumptions:Unconfined setting with heterogeneous stratigraphy.Presence of a discontinuous impervious or semi-impervious lens (e.g., clay).Local recharge that can temporarily saturate the overlying zone.Concept / Approach:When infiltrating water encounters an intervening impervious lens above the regional phreatic surface, it can pond, creating a saturated zone with its own upper surface—the perched water table. This perched zone is hydraulically separated from the main water table and can be transient, depending on recharge and leakage.Step-by-Step Solution:Confirm that perched conditions are a phenomenon within unconfined systems (true).Define the upper saturated surface as the 'perched water table' (true).State the cause: a local impervious lens causing temporary saturation above the main water table (true).Hence, the inclusive answer is 'All the above'.Verification / Alternative check:Hydrogeology references consistently define perched water tables and their formation over discontinuous lenses.Why Other Options Are Wrong:Each individual statement (a)–(c) is correct on its own; selecting any single one would be incomplete.'None of these' contradicts standard definitions.Common Pitfalls:Confusing perched aquifers with truly confined aquifers.Assuming perched conditions are permanent; they are often seasonal or event-based.Final Answer:All the above

Question 10

In hydrology: During a rainstorm, which of the following factors significantly reduce the infiltration capacity of a soil surface?

Accepted Answer

Introduction / Context:Infiltration capacity is the maximum rate at which soil can absorb rainfall. Understanding what reduces this capacity is critical for flood estimation, urban drainage design, and soil conservation. Multiple surface and subsurface processes can act together during a storm to lower infiltration.Given Data / Assumptions:Rainstorm conditions with sustained rainfall intensity.Typical natural or compacted soil profiles.No special soil amendments or artificial drainage in place.Concept / Approach:Factors that reduce infiltration generally decrease the soil’s ability to transmit water by closing macropores, sealing the surface, or saturating the profile. Each option addresses a known mechanism: surface storage, antecedent moisture, mechanical sealing, and fines migration.Step-by-Step Explanation:Surface detention creates temporary ponding; when storage fills, excess becomes runoff, while infiltration rate may decline as hydraulic gradient changes.High antecedent moisture means pores are already occupied by water; less suction and fewer empty pores remain to accept new rainfall.Compaction from raindrop impact or traffic reduces pore volume and permeability, lowering infiltration capacity.Fine particles wash into surface pores, forming a seal or crust that dramatically reduces near-surface conductivity.Verification / Alternative check:Field infiltration tests (double-ring infiltrometers) commonly show declining infiltration rates during storms due to sealing and saturation effects, corroborating the combined influence listed above.Why Other Options Are Wrong:Any single factor alone can reduce capacity, but the best answer recognizes that all listed mechanisms contribute during storms.Common Pitfalls:Assuming initial infiltration capacity remains constant throughout a storm.Ignoring the role of fine sediment in forming surface crusts.Final Answer:All of the above

Question 11

Unit hydrograph selection: The best (most informative) unit period for a unit hydrograph is approximately equal to the basin lag divided by which number?

Accepted Answer

Introduction / Context:Hydrologists select a unit hydrograph duration to represent the catchment response to a unit depth of effective rainfall. Choosing an appropriate unit period is important for accurate runoff modeling and design flood estimation.Given Data / Assumptions:Basin lag (time from centroid of excess rainfall to peak of direct runoff hydrograph) is known or can be estimated.Objective is to choose a practical unit duration that captures catchment dynamics without over-smoothing.Concept / Approach:A commonly used guideline is to select the unit duration as a fraction of the basin lag so that the hydrograph adequately represents the catchment response while remaining manageable for computations and convolution. A typical recommendation is about one-third of the basin lag.Step-by-Step Reasoning:Let T_lag be the basin lag.Choose unit duration T_u such that T_u ≈ T_lag / 3.This balances temporal resolution with the need to avoid an excessively short unit that introduces noise and numerical instabilities.Verification / Alternative check:Hydrology texts and practice often suggest a range (about one-third to one-fifth of lag); adopting one-third is a standard rule-of-thumb for routine problems and objective-type questions.Why Other Options Are Wrong:2: Too short; may cause spiky unit hydrographs.4 or 5 (or 6): Unit duration becomes too long, smoothing the response and losing definition of peak timing.Common Pitfalls:Selecting unit duration equal to storm duration rather than basin dynamics.Using an excessively long duration leading to underestimation of peak discharge.Final Answer:3

Question 12

In surface hydrology, the time taken by runoff from the most remote point of a watershed to reach the outlet is called the __________.

Accepted Answer

Introduction / Context:Design of storm drains, culverts, and small basins frequently relies on the concept of time of concentration, which defines when the entire watershed contributes simultaneously to flow at the outlet. Correct terminology is essential for selecting rainfall intensities from IDF curves and applying rational or unit hydrograph methods.Given Data / Assumptions:Watershed with a definable hydraulic path from the hydraulically most remote point (in time) to the outlet.Uniform rainfall intensity considered for a design duration equal to or exceeding this time.Concept / Approach:The time of concentration, Tc, is the travel time of runoff from the critical remote point to the outlet. When storm duration equals or exceeds Tc, peak discharge is expected because the whole area contributes. Terms like “overland flow time” are components of Tc only for the sheet flow segment.Step-by-Step Clarification:Identify the hydraulically longest (in time) path including sheet flow, shallow concentrated flow, and channel flow segments.Compute travel times for each segment using appropriate empirical or kinematic formulas.Sum to obtain Tc, which dictates the critical design rainfall duration.Verification / Alternative check:Using Tc to select rainfall intensity from IDF curves is standard. Peak flow from the Rational method Q = C * I * A is taken with I corresponding to duration ≈ Tc.Why Other Options Are Wrong:Time of overland flow / Concentration time of overland flow: Only part of the total concentration time; excludes channel segments.Duration of rainfall: Independent storm characteristic, not a watershed property.Baseflow recession time: Describes groundwater discharge decay, not travel time of surface runoff.Common Pitfalls:Equating Tc to just channel travel time or just overland flow time.Using Tc from one method outside its valid watershed size/land use range.Final Answer:Time of concentration

Question 13

Flood-frequency prediction: For estimating floods of a specified return period at a site using annual-maximum series, which of the following methods is generally regarded as the most reliable among the choices given?

Accepted Answer

Introduction / Context:Design discharges for bridges, culverts, and spillways often require flood-frequency analysis rather than just runoff transformation methods. Statistical methods based on extreme value theory are preferred to estimate discharges for specified return periods (e.g., 50-year, 100-year floods).Given Data / Assumptions:Annual maximum (or peak-over-threshold) discharge records are available.Objective is to estimate a quantile associated with a chosen return period.Concept / Approach:Gumbel’s method (Type I Extreme Value) fits a theoretical distribution to annual maxima, enabling extrapolation to desired return periods. In contrast, a unit hydrograph converts rainfall to runoff (not a frequency tool), and empirical regional formulae (e.g., California method) are rough and site-limited.Step-by-Step Reasoning:Compile annual maximum flood series.Estimate Gumbel parameters (mean and standard deviation or method of moments).Compute reduced variate and derive design flood for the target return period.Verification / Alternative check:For many basins, regional frequency analysis (e.g., Log-Pearson Type III) may be adopted by standards. Among the listed options, however, Gumbel’s method is the most directly intended for frequency estimation of extremes.Why Other Options Are Wrong:Unit hydrograph: A runoff transformation tool; not a statistical frequency method.California method: Empirical and regional; less reliable than analytical fitting of extremes.Rational method without frequency: Lacks statistical basis for return period selection.None of these: Incorrect because Gumbel is suitable.Common Pitfalls:Confusing hydrologic transformation tools with statistical frequency tools.Applying empirical regional equations outside their calibration region.Final Answer:Gumbel’s analytical method (extreme value distribution)

Question 14

Rainfall terminology and indices (hydrology): Pick the correct statement(s) about annual rainfall, normal rainfall and index of wetness.

Accepted Answer

Introduction / Context:Basic rainfall definitions are foundational for hydrologic design and water resources planning. Examinations frequently test precise terminology such as annual rainfall, normal rainfall, and indices used to describe inter-annual variability.Given Data / Assumptions:Station rainfall records exist over multiple years.Normal (long-term mean) is computed over a suitably long reference period (often several decades).Index of wetness compares a year’s rainfall with the normal value.Concept / Approach:Annual rainfall is the total depth accumulated over the defined year. Normal annual rainfall is a long-term mean over many years (commonly a few decades). Index of wetness (or related indices) indicates relative wetness/dryness compared to the normal, aiding quick classification of years.Step-by-Step Clarification:Define the annual total at a station by summing daily/monthly totals across the year.Compute the normal by averaging annual totals over a baseline period (e.g., a few decades, often cited around 30–35 years in many curricula).Compute an index (e.g., ratio to normal or percentage departure) to judge wetness of the year.Verification / Alternative check:Hydrologic services commonly publish normals and departures. A year whose total is close to the normal is classified as an “average” year, supporting option (d).Why Other Options Are Wrong (if taken individually):Each individual statement (a–d) is correct in standard textbook usage; hence the composite best answer is “All the above”.Common Pitfalls:Confusing “normal” with median or a short-term mean.Using inconsistent reference periods across stations, leading to biased comparisons.Final Answer:All the above

Question 15

Precipitation mechanisms in hydrology – choose the most comprehensive correct statement In applied hydrology and meteorology, different precipitation types arise from different atmospheric processes. Which of the following statements correctly descr…

Accepted Answer

Introduction / Context:Engineers and hydrologists must understand how different atmospheric mechanisms generate rainfall because flood estimation, drainage design, and reservoir operation depend on the expected intensity, duration, and areal spread of precipitation. Three dominant mechanisms are cyclonic (frontal), convective, and orographic rainfall, and each has characteristic spatial and temporal patterns.Given Data / Assumptions:Mid-latitude synoptic systems with warm and cold fronts are considered.Convective storms arise from buoyant instability in warm, moist air.Orographic effects refer to air forced to ascend mountain ranges or elevated terrain.Statements refer to typical behavior under mature, well-developed conditions.Concept / Approach:Warm-front precipitation involves gentle uplift of warm air over cooler air, producing steady stratiform rain over large areas. Convective storms feature strong localized updrafts, leading to intense, short-lived showers. Orographic ascent cools air adiabatically on the windward side, maximizing condensation and precipitation there; the leeward side often lies in a rain shadow. In cyclones, organized vertical motions include a central updraft region that enhances cloud growth and precipitation.Step-by-Step Solution:Review (a): Mature cyclones exhibit organized ascent near the center → statement aligns with updraft “chimney”.Review (b): Warm fronts yield continuous, widespread rainfall → correct for cyclonic/frontal precipitation.Review (c): Convective rainfall is intense and of short duration → characteristic of thunderstorms.Review (d): Orographic lifting causes maximum rainfall on windward slopes → classical result.Therefore, all given statements are simultaneously true.Verification / Alternative check:Intensity-Duration-Frequency (IDF) data commonly show convective storms with high short-duration intensities. Synoptic rainfall exhibits lower intensities but longer durations and larger aerial coverage. Orographic climatologies display windward maxima and leeward minima.Why Other Options Are Wrong:(a) alone is true but incomplete.(b) alone omits convective/orographic characteristics.(c) alone neglects frontal and orographic patterns.(d) alone omits the other two mechanisms.Common Pitfalls:Assuming all heavy rain is convective; warm-front events can produce large volumes over time. Also, forgetting that topography can strongly modulate frontal and convective precipitation totals.Final Answer:All the above.

Question 16

Pettis empirical flood formula – variables on which it is based The Pettis formula for peak flood discharge, Q = C (P · B)^(5/4) cumecs, relates discharge to which catchment descriptors?

Accepted Answer

Introduction / Context:For preliminary hydrologic design in ungauged basins, engineers sometimes employ empirical regional flood formulas. The Pettis formula is one such relation that predicts peak discharge based on rainfall potential and basin scale.Given Data / Assumptions:P denotes a characteristic storm rainfall depth (or index) for the basin and design frequency.B denotes the catchment (drainage basin) area.C is an empirical coefficient calibrated for region and units.Formula is intended for quick estimation, not a substitute for detailed hydrologic modeling.Concept / Approach:Empirical flood formulae typically link peak discharge to a measure of storm input and basin size. In Q = C (P · B)^(5/4), the product P · B captures the magnitude of potential runoff volume scaled by area, while the exponent 5/4 reflects observed nonlinearity in peak response with respect to rainfall and basin area.Step-by-Step Solution:Identify symbols: P = rainfall metric; B = basin area.Check options: Only option (a) explicitly lists rainfall and drainage area.Therefore, (a) is correct.Verification / Alternative check:Comparable regional relations (e.g., Dicken’s, Ryve’s, Inglis) also combine rainfall or a regional constant with basin area to estimate peak flows, lending conceptual support.Why Other Options Are Wrong:(b) “Runoff and area” is not what the stated variables represent; P is rainfall, not measured runoff.(c) Shape factor is not present explicitly in the formula.(d) Area alone cannot capture storm input variability.Common Pitfalls:Using empirical formulas outside their calibrated region; ensure consistent units for P and B so that discharge units emerge correctly with the chosen C.Final Answer:Rainfall (P) and drainage basin area (B)

Question 17

IMD standard non-recording rain gauges – collector aperture sizes According to the Indian Meteorological Department (IMD), the standard non-recording rain gauges are manufactured with collector apertures of which areas?

Accepted Answer

Introduction / Context:Rainfall measurement is foundational for hydrologic analysis. In India, the IMD specifies standardized non-recording gauges (often called Symon’s type) with specific collector areas so that measured depths are comparable across networks and long-term records.Given Data / Assumptions:Non-recording gauges require manual observation of collected rainfall depth.Standardization ensures uniform calibration and acceptable accuracy under field conditions.Two commonly used sizes are specified for different exposure and rainfall regimes.Concept / Approach:Collector aperture area determines the volume-to-depth conversion factor: volume collected / area = rainfall depth. IMD standardizes this area so that observers can use standard graduated measuring cylinders and comparison tables without site-specific recalibration.Step-by-Step Solution:Recall typical IMD specifications: standard apertures of 100 cm² and 200 cm².Match with provided options → option (c).Verification / Alternative check:Field manuals and hydrology textbooks summarizing IMD practices list 100 cm² and 200 cm² collectors for non-recording gauges deployed across networks.Why Other Options Are Wrong:(a) and (b) include an incorrect size (50 or 150 cm²).(d) 250 or 500 cm² are atypically large for standard non-recording gauges.(e) is unnecessary because a correct pair exists.Common Pitfalls:Confusing non-recording (manual) gauges with recording gauges, which use tipping buckets or weighing mechanisms and are specified differently (e.g., orifice diameter, bucket volume).Final Answer:100 cm² or 200 cm²

Question 18

Symon’s rain gauge – identify its operating type In hydrometeorological practice, a Symon’s rain gauge is classified as which type of rain gauge?

Accepted Answer

Introduction / Context:Rain gauges are broadly divided into recording and non-recording types. The Symon’s gauge is a classic, robust instrument widely used for long-term rainfall observations in India and other countries, especially where daily manual readings are standard practice.Given Data / Assumptions:Symon’s gauge comprises a collector funnel, a receiving bottle, and a calibrated measuring cylinder.It does not contain an automatic recording mechanism.Observations are typically read once or twice per day at fixed times.Concept / Approach:Non-recording gauges collect rainfall for subsequent manual measurement of depth. Recording gauges (tipping-bucket, weighing, or float types) produce continuous charts or digital records of rainfall versus time. Symon’s gauge specifically belongs to the non-recording category.Step-by-Step Solution:Identify function: Symon’s lacks moving recording components → non-recording.Select the matching option → (d) Non-recording (manual) gauge.Verification / Alternative check:Standard hydrology references describe Symon’s as the IMD standard non-recording gauge used for daily totals.Why Other Options Are Wrong:(a), (b), and (c) are recording gauges with mechanisms that log rainfall continuously.(e) is invalid because the correct classification is present.Common Pitfalls:Assuming “standard gauge” implies “recording”; in many networks, the standard reference is a non-recording gauge used alongside a subset of recording gauges for intensity studies.Final Answer:Non-recording (manual) gauge

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