GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersRain to runoff: measuring precipitation, subtracting losses, estimating runoff and routing floods.
In the exams: GATE AG Section 4 Hydrology (SWCE is the heaviest GATE section, ~16 marks a year 2019–26); ICAR AIEEA-PG Unit VII; ICAR AICE subjects 40 and 43; NCEES PE hydrology; state PSCs
GATE — Agricultural Engineering (AG)
Numerical questions in the Soil and Water Conservation section: unit hydrograph and S-curve ordinates, rain-gauge network size and missing data, rational method and SCS curve number runoff, φ-index and Horton infiltration, Muskingum coefficients and reservoir routing, return period and risk. Answers are often numerical-answer-type, so units and rounding must be right.
How our questions for GATE spread across the units of Watershed Hydrology. Difficulty: 8 easy, 34 medium, 15 hard.
| Hydrographs and the unit hydrograph | 8 | |
| Infiltration: Horton, Philip and the phi-index | 6 | |
| S-curve, IUH and synthetic unit hydrographs | 6 | |
| Frequency analysis, return period and IDF curves | 5 | |
| Flood routing: Muskingum and reservoir (Puls) methods | 5 | |
| The hydrologic cycle | 4 | |
| Precipitation measurement and rain-gauge networks | 4 | |
| Missing rainfall data and double-mass analysis | 4 | |
| Runoff estimation: rational method | 4 | |
| Runoff estimation: SCS curve number | 4 | |
| Interception and evaporation (pan) | 3 | |
| Streamflow measurement | 3 | |
| Drought classification | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A 200 km² catchment has mean annual rainfall 900 mm and a mean annual flow of 1.9 m³/s at its outlet. Taking long-term storage change and groundwater exchange as zero, estimate mean annual evapotranspiration.
D. About 600 mm
Runoff depth = 1.9 × 365 × 86,400 / (200 × 10⁶) = 0.2996 m = 300 mm. ET = P − R = 900 − 300 ≈ 600 mm. 300 mm is the runoff itself, not ET; 899 mm comes from converting only one day of flow; 1,200 mm adds runoff to rainfall.
Further reading: Subramanya, Engineering Hydrology
In hydrograph analysis, 'effective rainfall' means:
C. rainfall that becomes direct runoff
Effective rainfall (rainfall excess) = total rainfall − abstractions, and its depth equals the direct runoff depth. Rain that recharges groundwater is an abstraction from storm runoff, not effective rainfall.
Further reading: Subramanya, Engineering Hydrology
Isohyets of 20, 30, 40 and 50 mm are drawn over a 200 km² catchment. Areas between successive isohyets are 60 km² (20–30), 90 km² (30–40) and 50 km² (40–50). What is the mean rainfall by the isohyetal method?
A. 34.5 mm
Each band takes the mean of its bounding isohyets: (25×60 + 35×90 + 45×50)/200 = 34.5 mm. Using the lower isohyet of each band gives 29.5 mm and the upper gives 39.5 mm; 35.0 mm is an unweighted average of the band means.
Further reading: Subramanya, Engineering Hydrology
The probability that a 50-year flood is equalled or exceeded in any one year is:
B. 0.02
p = 1/T = 1/50 = 0.02. 0.98 is the probability of non-exceedance; a 50-year event is not guaranteed once in 50 years. 0.64 is the chance of at least one exceedance in 50 years (1 − 0.98⁵⁰), and 0.36 is the chance of none (0.98⁵⁰).
Further reading: Subramanya, Engineering Hydrology
A spillway must have no more than a 10% risk of its design flood being exceeded during a 50-year life. The minimum design return period is about:
D. 475 years
T = 1/[1 − (1 − R)^(1/n)] = 1/[1 − 0.9^(1/50)] = 475 years. 500 years comes from the approximation T = n/R, which is close but not exact. 10 years is simply 1/R, ignoring the 50-year life.
Further reading: Subramanya, Engineering Hydrology
Using Meyer's formula E = C(ew − ea)(1 + u₉/16) with C = 0.36 for a large deep reservoir, estimate daily evaporation when ew = 23.76 mm Hg, ea = 11.88 mm Hg and u₉ = 16 km/h.
B. 8.55 mm/day
E = 0.36 × (23.76 − 11.88) × (1 + 16/16) = 0.36 × 11.88 × 2 = 8.55 mm/day. C = 0.50 is for small shallow waters; omitting the wind term gives 4.28; using ew instead of the difference gives 17.11.
Further reading: Subramanya, Engineering Hydrology
Urbanisation of a small catchment doubles its runoff coefficient and, by shortening tc, raises the design intensity from 50 to 70 mm/h. By the rational method, the peak increases by a factor of:
D. 2.8
Q ∝ C × i, so the factor is 2 × (70/50) = 2 × 1.4 = 2.8. 2.0 ignores the change in tc; 1.4 ignores the change in C; 3.4 adds the factors.
Further reading: Subramanya, Engineering Hydrology
For a curve number of 80, the potential maximum retention S in SI units is:
A. 63.5 mm
S = 25400/CN − 254 = 317.5 − 254 = 63.5 mm. 317.5 forgets to subtract 254; 2.5 is the inch value (1000/80 − 10) mis-labelled in mm; 12.7 mm is Ia = 0.2S.
Further reading: USDA NRCS National Engineering Handbook Part 630
A 6-hour unit hydrograph can be obtained from a 3-hour unit hydrograph by:
B. adding two 3-hour unit hydrographs lagged by 3 hours and dividing the ordinates by 2
The lagged sum represents 2 cm in 6 hours, so dividing by 2 returns it to 1 cm. Simply doubling ordinates gives 2 cm in 3 hours, a different storm.
Further reading: Subramanya, Engineering Hydrology
By Snyder's method a basin has lag tp = 11 h, so the standard duration is tr = tp/5.5 = 2 h. For a 4-hour unit hydrograph, the adjusted lag is:
B. 11.5 h
tpR = tp + (tR − tr)/4 = 11 + (4 − 2)/4 = 11.5 h. 13.0 adds the whole difference in duration; 11.0 ignores the adjustment. 12.0 h divides the duration difference by 2 instead of 4.
Further reading: Subramanya, Engineering Hydrology
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