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Hydrology MCQ Practice Set — 37 Questions with Answers

37 exam-oriented Hydrology multiple-choice questions with the correct answer and a clear explanation for each. Frequently asked in AE Level Civil Engineering, JE Level Civil Engineering. Solve the full set below for free — no login required.

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  1. 1
    HydrologyMEDIUM

    The 'SCS Curve Number' (CN) method for estimating runoff uses CN values that depend on:

    AOnly the slope of the watershed
    BSoil hydrological group (A–D), land use/cover type, and antecedent moisture condition (AMC) — governs runoff potential
    COnly total annual rainfall
    DRiver channel roughness (Manning n)

    Answer: B. Soil hydrological group (A–D), land use/cover type, and antecedent moisture condition (AMC) — governs runoff potential

    Explanation: SCS (now NRCS) Curve Number method: Q = (P-0.2S)²/(P+0.8S) for P > 0.2S, where S = (25400/CN)-254 (mm). CN (0–100) depends on: (1) Soil hydrological group (A = sandy, B, C, D = clay — infiltration capacity); (2) Land use/cover (impervious, cultivation type, row crops vs meadow); (3) Antecedent Moisture Condition (AMC I, II, III — dry, normal, wet). Higher CN = more runoff. Used worldwide for small watershed design.

  2. 2
    HydrologyMEDIUM

    The 'design storm' for urban drainage uses IDF (Intensity-Duration-Frequency) curves where:

    AAll durations give the same intensity
    BShorter durations produce higher intensities for same return period — IDF relates intensity, duration, and return period
    CIntensity increases with duration
    DReturn period has no effect on design intensity

    Answer: B. Shorter durations produce higher intensities for same return period — IDF relates intensity, duration, and return period

    Explanation: IDF curves: for each return period T and storm duration D, the average intensity i is derived from historical rainfall records using frequency analysis. Longer duration → lower intensity (storm accumulates but at less intensity). Shorter duration → higher intensity (burst). Rational method: Q=CiA uses intensity for time of concentration Tc (critical duration = Tc for max runoff). IDF from Talbot formula: i = a/(t+b) or power form i=k/t^n.

  3. 3
    HydrologyMEDIUM

    If the coefficient of variation of the rainfall values of the existing rain-gauge stations is 30 and the desired error in the basin-mean rainfall estimate is 10%, the optimum number of rain-gauge stations is

    A3
    B5
    C9
    D25

    Answer: C. 9

    Explanation: The optimum number of rain-gauge stations N = (Cv/p)², where Cv is the coefficient of variation of rainfall values at existing stations (%) and p is the desired percentage error in the mean rainfall estimate. N = (30/10)² = 3² = 9. This formula (from IS 4987) ensures the gauge network provides a mean areal rainfall estimate within ±10% of the true value. If the existing number of gauges is less than N, additional gauges must be added to meet the accuracy requirement.

  4. 4
    Original practiceMEDIUM

    If base period is 150 days and delta is 0.6 m, duty is

    A10.42 ha/cumec
    B4320 ha/cumec
    C1080 ha/cumec
    D2160 ha/cumec

    Answer: D. 2160 ha/cumec

    Explanation: In irrigation engineering, Duty (D) = 8.64 × B / Δ, where B is base period (days) and Δ is delta (m of water depth). B = 150 days, Δ = 0.6 m. D = 8.64 × 150 / 0.6 = 2160 ha/cumec ha/cumec. Duty means the area (hectares) that 1 cumec of water can irrigate throughout the base period.

  5. 5
    Original practiceMEDIUM

    For base period 120 days and duty 600 ha/cumec, delta is

    A1.73 m
    B0.86 m
    C43.2 m
    D3.46 m

    Answer: A. 1.73 m

    Explanation: Delta (Δ) is total depth of water (m) required by a crop during its base period. Δ = 8.64 × B / D, where B = base period (days) and D = duty (ha/cumec). B = 120 days, D = 600 ha/cumec. Δ = 8.64 × 120 / 600 = 1.73 m m. Duty and delta are inversely related: a crop needing more water per unit area will have lower duty.

  6. 6
    Darcy Law GroundwaterMEDIUM

    Darcy's law for groundwater flow states that the discharge velocity (Darcy flux) v through a porous medium is:

    Av = k + i
    Bv = k * i (where k = hydraulic conductivity, i = hydraulic gradient)
    Cv = k * i^2
    Dv = i / k

    Answer: B. v = k * i (where k = hydraulic conductivity, i = hydraulic gradient)

    Explanation: Darcy's law is expressed as v = ki, where v is the discharge velocity (Darcy flux), k is the hydraulic conductivity, and i is the hydraulic gradient. Seepage velocity (actual pore velocity) is v_s = v/n, where n is the effective porosity. The original text incorrectly equated Darcy flux with seepage velocity.

  7. 7
    HydrologyMEDIUM

    The total rainfall in a catchment of area 1200 km^2 during a 6 h storm is 160 mm. If the total runoff is 600 hectare-meters, the infiltration index (phi-index) is:

    A1 cm/h
    B0.10 cm/h
    C10 cm/h
    D0.01 cm/h

    Answer: A. 1 cm/h

    Explanation: Total rainfall = 160 mm = 16 cm. Runoff = 600 ha-m = 6,000,000 m^3. Area = 1200 km^2 = 120,000 ha. Runoff depth = (600 ha-m / 120,000 ha) = 0.005 m = 0.5 cm. Total loss = 16 - 0.5 = 15.5 cm. Phi-index = 15.5 cm / 6 h = 2.58 cm/h. Given the original options are likely based on a different runoff value, the question is mathematically incomplete.

  8. 8
    HydrologyMEDIUM

    The specific capacity of a well is defined as

    AQuantity of water that can be drawn from the well per unit time
    BTotal quantity of water available in the well at any time
    CFlow of water per unit time per unit area
    DDischarge per unit drawdown

    Answer: D. Discharge per unit drawdown

    Explanation: Specific capacity is the rate of discharge per unit of drawdown in the well (Q/s).

  9. 9
    HydrologyMEDIUM

    Specific capacity of a well is

    Avolume of water per unit volume of aquifer
    Btotal water available
    Cflow per unit area
    Ddischarge per unit drawdown

    Answer: D. discharge per unit drawdown

    Explanation: Specific capacity is defined as the discharge per unit of drawdown in a well.

  10. 10
    HydrologyHARD

    The 'Isochrone' in hydrology is a line connecting points:

    AOf equal rainfall depth
    BWith equal travel time to the watershed outlet — used for time-area diagrams and Clark''s UH method
    COf equal elevation (contours)
    DOf equal infiltration rate

    Answer: B. With equal travel time to the watershed outlet — used for time-area diagrams and Clark''s UH method

    Explanation: Isochrone: line connecting all points in a catchment that are equidistant in travel time from the catchment outlet. Zones between isochrones (isochronal zones) contribute runoff to outlet over equal time intervals. Used to: construct time-area curve → base for Clark''s unit hydrograph method. All points on one isochrone arrive at outlet simultaneously.

  11. 11
    HydrologyMEDIUM

    The 'peak factor' in water supply design relates:

    AHighest annual rainfall to lowest
    BPeak flow demand to average flow demand — critical for sizing water supply distribution systems and service reservoirs
    CStream peak to baseflow
    DFlood peak to ordinary flood

    Answer: B. Peak flow demand to average flow demand — critical for sizing water supply distribution systems and service reservoirs

    Explanation: Peak factor: ratio of peak demand to average demand. Hourly peak factor: 1.5–2.0 × daily average (used for distribution pipe design). Daily peak factor: 1.2–1.5 × annual average (used for treatment plant and service reservoir design). Weekly and monthly peaks also considered. Storage requirement = (daily peak − average) × hours of peak.

  12. 12
    Artificial Recharge MethodsMEDIUM

    The spreading basin (percolation pond) method for artificial groundwater recharge is most effective when:

    AThe aquifer is confined and deep
    BThe water table is shallow and the vadose zone is thin, with permeable soil (sand/gravel) allowing rapid infiltration of surface water to the aquifer
    CThe soil is impervious clay
    DThere is no seasonal rainfall

    Answer: B. The water table is shallow and the vadose zone is thin, with permeable soil (sand/gravel) allowing rapid infiltration of surface water to the aquifer

    Explanation: Artificial recharge methods: (1) Spreading basins/percolation ponds: impoundment of runoff, water percolates to water table; best for unconfined aquifer with permeable strata; (2) Recharge wells/injection wells: for confined aquifer; (3) Modified streambeds: excavated or cleared channels; (4) Check dams/subsurface dams: retard runoff, increase percolation time.

  13. 13
    Isochrone AnalysisMEDIUM

    Isochrones on a catchment map are lines joining points of:

    AEqual rainfall depth
    BEqual time of travel of surface runoff to the outlet, used to construct the time-area diagram for determining the IUH (Instantaneous Unit Hydrograph)
    CEqual elevation (contours)
    DEqual infiltration rate

    Answer: B. Equal time of travel of surface runoff to the outlet, used to construct the time-area diagram for determining the IUH (Instantaneous Unit Hydrograph)

    Explanation: Isochrones: lines of equal travel time t from each point on catchment to the outlet. The area between successive isochrones (time-area histogram) represents the runoff contribution rate at that time lag. The time-area diagram is convolved with the excess rainfall intensity to produce the direct runoff hydrograph (Clark method for IUH).

  14. 14
    Weibull Plotting PositionMEDIUM

    The Weibull plotting position formula assigns a probability P to the m-th ranked value (ascending order) in a data series of N values as:

    AP = m / N
    BP = m / (N + 1)
    CP = (m - 0.375) / (N + 0.25)
    DP = (m - 0.5) / N

    Answer: B. P = m / (N + 1)

    Explanation: Weibull: P = m/(N+1), where m = rank (1 = smallest). Return period T = 1/P = (N+1)/m. Weibull is unbiased estimator of exceedance probability. Other formulas: Hazen P = (m-0.5)/N; Blom (Gringorten) P = (m-0.44)/(N+0.12) (better for extreme value distributions).

  15. 15
    Snyder Synthetic UHMEDIUM

    In the Snyder synthetic unit hydrograph method, the lag time tp (hours) from centroid of rainfall to peak of UH is:

    Atp = Ct x (L x Lc)^0.3
    Btp = Ct x (L x Lca)^0.3 (where Ct = basin coefficient, L = main stream length in km, Lca = distance from outlet to point on stream nearest catchment centroid)
    Ctp = peak discharge / area
    Dtp = time of concentration only

    Answer: B. tp = Ct x (L x Lca)^0.3 (where Ct = basin coefficient, L = main stream length in km, Lca = distance from outlet to point on stream nearest catchment centroid)

    Explanation: Snyder UH: tp = Ct(L x Lca)^0.3; qp = Cp x A / tp (Cp = peaking coefficient); peak duration t_R = tp/5.5; T_b (base time) = 3 + tp/8. Ct = 1.35-1.65 (steep), Cp = 0.4-0.8. Synthetic UH is used when no gauge data available for the catchment.

  16. 16
    Thiessen Polygon MethodMEDIUM

    The Thiessen polygon method for estimating mean areal rainfall over a catchment uses:

    AArithmetic average of all gauge readings
    BWeighted average where each gauge is assigned a weight proportional to the area of its Thiessen polygon (the polygon of perpendicular bisectors between adjacent gauges)
    CIsohyetal method with interpolated contours
    DOnly the highest reading gauge

    Answer: B. Weighted average where each gauge is assigned a weight proportional to the area of its Thiessen polygon (the polygon of perpendicular bisectors between adjacent gauges)

    Explanation: Thiessen polygons: draw perpendicular bisectors between all adjacent rain gauge pairs, forming polygons around each gauge. Area of each polygon within catchment = A_i. Mean areal rainfall P = sum(P_i x A_i) / A_total. More accurate than arithmetic mean when gauges are unevenly distributed. Does not account for orographic effects (isohyetal method better for hilly terrain).

  17. 17
    Specific Yield DefinitionMEDIUM

    The specific yield Sy of an unconfined aquifer is:

    AEqual to the porosity n
    BThe fraction of groundwater that drains by gravity from the aquifer when the water table drops by one unit (Sy = n - Sr, where Sr = specific retention)
    CThe artesian pressure divided by depth
    DThe transmissivity divided by thickness

    Answer: B. The fraction of groundwater that drains by gravity from the aquifer when the water table drops by one unit (Sy = n - Sr, where Sr = specific retention)

    Explanation: Specific yield Sy = n - Sm (Sr = specific retention = moisture held against gravity). Sy typically 0.1-0.3 for sand. Specific retention: 0.02-0.05 for gravel, 0.06-0.12 for fine sand, 0.10-0.20 for silty soil. Sy represents storativity of unconfined aquifer. Storage coefficient S = Sy (unconfined) or << Sy (confined, typically 0.0001-0.001).

  18. 18
    Confined Aquifer PropertiesMEDIUM

    In a confined (artesian) aquifer, the piezometric surface is:

    AAt the water table level
    BAbove the top of the saturated aquifer (artesian head); if piezometric surface is above ground level, the well flows without pumping (flowing artesian well)
    CBelow the aquifer base
    DEqual to the aquifer thickness

    Answer: B. Above the top of the saturated aquifer (artesian head); if piezometric surface is above ground level, the well flows without pumping (flowing artesian well)

    Explanation: Confined aquifer: fully saturated, bounded above and below by impervious layers (aquitards). Piezometric (potentiometric) surface = imaginary surface to which water rises in wells tapping the aquifer = hydrostatic pressure head. If piezometric surface > ground level: flowing artesian well. Storage coefficient = 0.0001 to 0.001 (much less than specific yield of unconfined aquifer).

  19. 19
    Return Period ExceedanceMEDIUM

    A flood with a return period of 50 years has a probability of being equaled or exceeded in any given year of:

    A50%
    B2% (= 1/50)
    C1%
    D0.02%

    Answer: B. 2% (= 1/50)

    Explanation: Return period T (years) = 1 / P (annual exceedance probability). T = 50 years: P = 1/50 = 2% per year. Probability of at least one occurrence in n years: Pn = 1 - (1 - 1/T)^n. For T = 50, n = 50: Pn = 1 - (1 - 0.02)^50 = 1 - 0.364 = 0.636 = 63.6% chance of exceedance at least once in 50 years.

  20. 20
    Water Resources Engineering and IrrigationHARD

    The probability that a flood of return period T will be equalled or exceeded at least once in n years is:

    An/T only for all cases
    B(1 - 1/T)^n
    CT/n
    D1 - (1 - 1/T)^n

    Answer: D. 1 - (1 - 1/T)^n

    Explanation: Annual exceedance probability is 1/T; the risk over n independent years is 1 - (1 - 1/T)^n.

  21. 21
    Water Resources Engineering and IrrigationHARD

    A unit hydrograph represents direct runoff hydrograph due to:

    Aone mm of total rainfall including losses
    Bbaseflow only
    Cone cm of effective rainfall over the catchment in a specified duration
    Dsnowmelt only

    Answer: C. one cm of effective rainfall over the catchment in a specified duration

    Explanation: A unit hydrograph is the DRH resulting from unit depth of rainfall excess uniformly distributed over the basin for a specified duration.

  22. 22
    Hydrology and Irrigation EngineeringHARD

    For runoff coefficient 0.6, rainfall intensity 50 mm/h and area 2 hectare, rational runoff is approximately:

    A0.0167 m3/s
    B0.167 m3/s
    C16.7 m3/s
    D1.67 m3/s

    Answer: B. 0.167 m3/s

    Explanation: Q = 0.00278 C i A = 0.00278 x 0.6 x 50 x 2 = 0.167 m3/s.

  23. 23
    HydrologyMEDIUM

    An isochrone is a line on the basin map

    Ajoining raingauge stations having equal rainfall duration
    Bjoining points having equal rainfall depth in a given interval
    Cjoining points having equal time of travel of surface runoff to the catchment outlet
    Djoining points at equal distance from the catchment outlet

    Answer: C. joining points having equal time of travel of surface runoff to the catchment outlet

    Explanation: Isochrones connect catchment points having the same travel time to the outlet

  24. 24
    HydrologyMEDIUM

    If a 4-hour unit hydrograph of a basin has a peak ordinate of 80 m^3/s, the peak ordinate of a 2-hour unit hydrograph for the same basin will be

    Aequal to 80 m^3/s
    Bgreater than 80 m^3/s
    Cless than 80 m^3/s
    Dbetween 40 and 80 m^3/s

    Answer: B. greater than 80 m^3/s

    Explanation: A shorter-duration unit hydrograph is sharper and therefore has a higher peak ordinate

  25. 25
    HydrologyMEDIUM

    An aquifer that is confined at the bottom but not at the top is known as:

    Apartially confined aquifer
    Bunconfined aquifer
    Csemiconfined aquifer
    Daquiclude

    Answer: B. unconfined aquifer

    Explanation: An unconfined aquifer has a free water table at its top and an impervious layer below

  26. 26
    HydrologyMEDIUM

    The average rate of loss such that the volume of rainfall in excess of that rate equals the direct run-off is

    ARun-off coefficient
    BInfiltration index
    CInfiltration capacity
    DSurface retention

    Answer: B. Infiltration index

    Explanation: This is the phi-index (infiltration index): the constant loss rate for which rainfall excess equals the direct runoff.

  27. 27
    HydrologyMEDIUM

    A rainfall hydrograph shows the variation of

    Acumulative rainfall with time
    Brainfall intensity with time
    Crainfall depth over an area
    Drainfall intensity with the cumulative rainfall

    Answer: B. rainfall intensity with time

    Explanation: It plots the rate (intensity) of rainfall against time. (A cumulative-rainfall-vs-time plot would be a mass curve.) [Self-solved — please verify.]

  28. 28
    HydrologyMEDIUM

    The best unit duration of storm for a unit hydrograph is

    A1 hour
    Bone-fourth of the basin lag
    Cone-half of the basin lag
    Dequal to the basin lag

    Answer: B. one-fourth of the basin lag

    Explanation: The recommended unit duration is about one-fourth of the basin lag.

  29. 29
    HydrologyMEDIUM

    A flow mass curve is the graph between

    Aflow rate and time
    Bcumulative volume of flow and time
    Ccumulative volume of flow and cumulative time
    Dcumulative discharge and time

    Answer: D. cumulative discharge and time

    Explanation: Flow mass curve is cumulative discharge versus time.

  30. 30
    HydrologyMEDIUM

    The most accurate method of finding the average depth of rainfall over an area is the

    Aisohyetal method
    Barithmetic mean method
    CThiessen polygon method
    Dany of the above

    Answer: A. isohyetal method

    Explanation: Isohyetal method is most accurate average rainfall method.

  31. 31
    HydrologyMEDIUM

    The inflection point on the recession side of a hydrograph indicates the end of

    Abase flow
    Bdirect run-off
    Coverland flow
    Drainfall

    Answer: B. direct run-off

    Explanation: Inflection point on recession limb indicates end of direct runoff.

  32. 32
    HydrologyMEDIUM

    A plot between rainfall intensity versus time at a site is called a

    Ahydrograph
    Bmass curve
    Chyetograph
    Disohyet

    Answer: C. hyetograph

    Explanation: A hyetograph is a plot of rainfall intensity against time.

  33. 33
    HydrologyMEDIUM

    In a reservoir with uncontrolled spillways, the peak of the (routed) outflow hydrograph

    Alies outside the plotted inflow hydrograph
    Blies on the recession part of the plotted inflow hydrograph
    Clies on the peak of the plotted inflow hydrograph
    Dis higher than the peak of the plotted inflow hydrograph

    Answer: B. lies on the recession part of the plotted inflow hydrograph

    Explanation: For an uncontrolled spillway, the outflow peak occurs where the outflow curve crosses the recession limb of the inflow hydrograph (maximum storage).

  34. 34
    HydrologyMEDIUM

    A 1-hour rainfall of 10 cm has a return period of 50 years. The probability of a 1-hour rainfall of 10 cm or more occurring in each of two successive years is

    A0.02
    B0.04
    C0.0002
    D0.0004

    Answer: D. 0.0004

    Explanation: P = 1/50 = 0.02 per year; for both successive years P×P = 0.02×0.02 = 0.0004.

  35. 35
    HydrologyMEDIUM

    The 'pan coefficient' used when converting Class A pan evaporation to lake evaporation accounts for:

    ADifferent rainfall in lake vs pan
    BPan overheats relative to large water body — pan evaporation > lake evaporation by factor of ~1/0.7 = 1.43
    CWind speed difference only
    DColor of the pan

    Answer: B. Pan overheats relative to large water body — pan evaporation > lake evaporation by factor of ~1/0.7 = 1.43

    Explanation: Pan coefficient Kp = Lake evaporation / Pan evaporation ≈ 0.7–0.8. Pan coefficient < 1 because: (1) Pan water heats more than lake (less mass, more exposed metal surface); (2) Reflected heat from pan edges increases evaporation; (3) Turbulence differs. IS 5973: Kp = 0.7 for well-maintained land pan. Multiply pan reading by Kp to get lake evaporation.

  36. 36
    HydrologyHARD

    The 'Snyder''s synthetic unit hydrograph' parameters are derived from:

    AOnly rainfall data with no catchment area
    BCatchment physical characteristics (L, Lc, area) and regional coefficients Ct and Cp calibrated from nearby gauged catchments
    COnly streamflow records of the same river
    DAnnual rainfall totals only

    Answer: B. Catchment physical characteristics (L, Lc, area) and regional coefficients Ct and Cp calibrated from nearby gauged catchments

    Explanation: Snyder (1938): tp = Ct(L×Lc)^0.3 (lag time), Qp = 2.78Cp×A/tp (peak discharge). Ct (0.4–0.8) and Cp (0.4–0.8) are regional parameters calibrated from gauged catchments in the region. L = length of main stream, Lc = length to centroid of catchment. Used for ungauged catchments to estimate design flood hydrograph.

  37. 37
    HydrologyEASY

    The Symons rain gauge (non-recording) is read at:

    AEvery hour automatically
    BOnce every 24 hours (standard reading at 8:30 AM IST) to give daily rainfall
    COnly during rain events
    DMonthly for average rainfall

    Answer: B. Once every 24 hours (standard reading at 8:30 AM IST) to give daily rainfall

    Explanation: Symons gauge: standard non-recording rain gauge. Cylindrical vessel (12.7 cm diameter, raised 30 cm above ground) with funnel and measuring cylinder. Read every 24 hours (8:30 AM IST in India) to give daily rainfall in mm. Recording gauges (tipping bucket, float type) give rainfall intensity vs time (hyetograph).

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