GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersPlanning a whole watershed: delineating it, classifying its land, budgeting its water and choosing harvesting structures with the people who live there.
In the exams: GATE AG Section 4 Watershed Management; ICAR AIEEA-PG Unit VII; ICAR AICE subject 40 Unit 6; TNPSC Unit III; RPSC AAE Unit 2
GATE — Agricultural Engineering (AG)
Section 4 Watershed Management: numerical-answer questions on morphometric parameters (form factor, circularity and elongation ratios, drainage density, bifurcation ratio), water budget arithmetic and harvesting volumes; short conceptual questions on land capability classes and prioritisation. Unit slips between mm, ha and m³ cost the most marks.
How our questions for GATE spread across the units of Watershed Planning & Management. Difficulty: 6 easy, 20 medium, 10 hard.
| Watershed characterisation and morphometry | 12 | |
| Watershed delineation | 5 | |
| Rainwater harvesting: check dams, percolation tanks and farm ponds | 5 | |
| Cost-benefit analysis of watershed projects | 5 | |
| Land capability classification | 3 | |
| Water budgeting | 3 | |
| Roof-top rainwater harvesting | 2 | |
| Land irrigability classification | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
The highest point on the divide of a watershed is at 1240 m and the outlet at 580 m. The basin length is 16.5 km. The relief ratio (dimensionless) is:
B. 0.040
Basin relief H = 1240 − 580 = 660 m. Rh = H / Lb = 660 / 16,500 m = 0.040. 40.0 is H in metres divided by Lb in kilometres, a unit slip. Using the highest elevation (0.075) or the outlet elevation alone (0.035) instead of the difference is wrong.
Further reading: Horton (1945); Strahler (1957); Schumm (1956); Miller (1953) — as summarised in Suresh, Soil and Water Conservation Engineering
In land capability classification, the subclass symbols e, w, s and c stand for limitations due to:
A. erosion, excess water, soil root zone, and climate
The four subclasses are e (erosion hazard), w (wetness, drainage or overflow), s (soil root zone limitations such as shallowness, stones, low water capacity, salinity or sodicity) and c (climatic limitation). Salinity is included under s, and slope is expressed through erosion hazard.
Further reading: Klingebiel and Montgomery, USDA Agriculture Handbook 210 (1961)
A farmer's field is on a gentle slope with deep, slowly permeable clay soil in a region of fairly high rainfall. Which bund type suits it better, and why?
D. Graded bund, because it leads excess runoff safely to an outlet instead of ponding it on the clay
Contour bunds hold water behind them and suit low-rainfall areas with permeable soils, where the ponded water infiltrates. On slowly permeable clays with high rainfall, ponding causes waterlogging and bund overtopping, so graded bunds with a slight longitudinal grade lead excess water to a grassed waterway. Clay soils can erode, especially when wet.
For a watershed over one year: rainfall 850 mm, runoff at the outlet 120 mm, deep percolation 95 mm and an increase in soil moisture storage of 15 mm. There is no inflow across the divide. Actual evapotranspiration is:
B. 620 mm
P = ET + Q + R + ΔS, so ET = 850 − 120 − 95 − 15 = 620 mm. Adding the storage increase (650 mm) gets its sign wrong: water that went into storage did not evaporate. 635 mm ignores storage altogether. 730 mm subtracts only the runoff (850 − 120).
A farm pond has a catchment of 8 ha. Seasonal rainfall is 650 mm and the seasonal runoff coefficient is 0.25. The runoff available to fill the pond is:
B. 13,000 m³
Runoff = C × P × A = 0.25 × 0.65 m × 80,000 m² = 13,000 m³. 52,000 m³ is total rainfall on the catchment (runoff coefficient omitted). The other two are factor-of-ten slips in converting hectares to m².
Further reading: Critchley and Siegert, FAO Water Harvesting manual (1991)
A percolation tank has an average water-spread area of 2.0 ha and an average stored depth of 1.5 m. It fills 3 times in the rainy season, and evaporation takes 20% of the water captured. Assuming all remaining water percolates, the seasonal recharge is:
C. 72,000 m³
One filling = 20,000 m² × 1.5 m = 30,000 m³. 3 fillings = 90,000 m³. Recharge = 90,000 × (1 − 0.2) = 72,000 m³. 90,000 m³ ignores evaporation; 24,000 m³ counts only one filling; 18,000 m³ is the evaporation loss.
When computing the volume of a farm pond with sloping sides, the average end-area formula compared with the prismoidal formula generally:
C. overestimates the volume
For a frustum of a pyramid the average end-area method, h(A1 + A2)/2, weights the large top area too heavily and gives a larger value than the exact prismoidal result, (h/6)(A1 + 4Am + A2). For a 30 m square top, 18 m square bottom and 3 m depth the two give 1,836 m³ and 1,764 m³.
Ten here, 36 in the app. The Laisha app has every practice question we have written to GATE in this subject, marked as you go, with timed papers, a jump-to-unanswered list and a certificate when you pass at 70 per cent.
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