GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersRemoving excess water and salt: surface and subsurface drain design, and reclaiming saline and sodic land.
In the exams: GATE AG Section 5 (drainage coefficient in 5 of 8, SAR and leaching in 3 of 8 papers 2019–26); ICAR AIEEA-PG Unit VII; ICAR AICE subjects 40 and 43; NCEES PE drainage; RPSC AAE
GATE — Agricultural Engineering (AG)
Numerical items in the Irrigation and Drainage section: drainage coefficient and discharge from an area, drainable porosity and water-table rise, drain pipe size, SAR from ion concentrations (including mg/L to meq/L), and leaching requirement and applied water depth. Hooghoudt spacing is in the syllabus but has been asked rarely; learn it, but weight the coefficient, porosity, SAR and LR items higher.
How our questions for GATE spread across the units of Agricultural Drainage Engineering. Difficulty: 14 easy, 24 medium, 13 hard.
| Drainage coefficient | 7 | |
| Subsurface drain spacing: Hooghoudt and Ernst | 6 | |
| Salinity and the leaching requirement | 6 | |
| Irrigation water quality: EC, SAR and RSC | 6 | |
| Kirkham and Glover-Dumm equations | 5 | |
| Reclamation of saline and sodic soils and gypsum requirement | 5 | |
| Causes and effects of waterlogging | 4 | |
| Surface drainage design | 4 | |
| Drain pipes and envelopes | 3 | |
| Vertical drainage and bio-drainage | 3 | |
| Interceptor and relief drains | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Before a canal was built the water table stood at 8 m depth. After irrigation began, net recharge to the groundwater increased by 60 mm/year. If the drainable porosity is 0.12, how many years until the water table reaches 2 m depth (assume no extra outflow)?
A. 12 years
Rise per year = recharge / drainable porosity = 0.060 / 0.12 = 0.50 m/year. The water table must rise 8 - 2 = 6 m, so t = 6 / 0.50 = 12 years. 100 years ignores the drainable porosity; 16 years uses 8 m instead of 6 m; 833 years multiplies instead of dividing by the porosity.
Further reading: Ritzema (ed.), Drainage Principles and Applications, ILRI Publication 16
A rainfall of 30 mm percolates to a shallow water table in a soil of drainable porosity 0.06. By how much does the water table rise?
B. 0.50 m
Rise = recharge / drainable porosity = 0.030 m / 0.06 = 0.50 m. Multiplying (30 x 0.06 = 1.8 mm) is the common error; 0.05 m and 5 m are decimal slips. Small drainable porosity means a small recharge causes a large rise.
Further reading: Ritzema (ed.), Drainage Principles and Applications, ILRI Publication 16
Using the rational method, find the peak runoff from a 20 ha field with runoff coefficient 0.5 and design rainfall intensity 36 mm/h.
D. 1.0 m3/s
Q = CIA/360 = 0.5 x 36 x 20 / 360 = 1.0 m3/s with I in mm/h and A in ha. 360 forgets the unit-conversion constant; 10 and 0.1 are decimal slips.
Further reading: Michael, Irrigation: Theory and Practice
For the trapezoidal drain with bed width 1.0 m, depth 0.8 m, side slopes 1.5:1 carrying 0.69 m3/s, what is the mean velocity?
C. 0.39 m/s
V = Q/A = 0.69 / 1.76 = 0.39 m/s. Using the rectangle b x y = 0.8 m2 ignores the side triangles (0.69/0.8 = 0.86); dividing by the wetted perimeter of 3.88 m (0.18) or multiplying by the area (1.21) is dimensionally wrong. The velocity is then checked against the permissible non-erosive and non-silting limits for the soil.
Further reading: USDA-NRCS NEH Part 650, Chapter 14
A surface drainage design module is 4 L/s per hectare. What drainage coefficient in mm/day is this?
D. 34.6 mm/day
4 L/s x 86,400 s = 345.6 m3/day per ha. Over 10,000 m2 that is 0.03456 m/day = 34.6 mm/day. 14.4 is the flow in m3/h per hectare (4 x 3,600 / 1,000), not a depth per day; 4.0 reads L/s per ha directly as mm/day; 3.46 is a decimal slip. A useful identity: 1 L/s/ha = 8.64 mm/day.
Further reading: Ritzema (ed.), Drainage Principles and Applications, ILRI Publication 16
Why did Hooghoudt replace the depth to the impermeable layer D by an equivalent depth d?
B. To account for the extra head loss from radial convergence of flow near a pipe drain
Flow converges radially into a small pipe, which the horizontal-flow (Dupuit-Forchheimer) assumption ignores. Using a smaller d adds the right amount of resistance. It has nothing to do with pipe roughness or time.
Further reading: Ritzema (ed.), Drainage Principles and Applications, ILRI Publication 16
In the USBR dynamic-equilibrium method, each irrigation adds 40 mm of deep percolation to a soil with drainable porosity 0.08. How much does each irrigation raise the water table?
A. 0.50 m
Rise = R / mu = 0.040 / 0.08 = 0.50 m. 3.2 mm multiplies; 0.05 m is a decimal slip; 2.0 m divides porosity by recharge. At dynamic equilibrium this rise equals the fall achieved between irrigations.
Further reading: US Bureau of Reclamation Drainage Manual; ILRI Publication 16
In the Glover-Dumm spacing equation, with all else unchanged, the drainable porosity of a soil is found to be twice the design value. The required spacing becomes:
C. The original spacing divided by about 1.41
L^2 is inversely proportional to mu, so L scales with 1/sqrt(mu): doubling mu divides L by sqrt 2 = 1.41. More water must drain per metre of fall, so drains must be closer, not wider.
Further reading: Ritzema (ed.), Drainage Principles and Applications, ILRI Publication 16
Irrigation water has EC 1.2 dS/m and the drainage water salinity must not exceed 8 dS/m. Using the US Salinity Laboratory formula, what is the leaching requirement?
A. 15 %
LR = ECiw / ECdw = 1.2 / 8 = 0.15 = 15 %. 85 % is 1 - LR; 6.67 is the inverse ratio; 1.5 % is a decimal slip.
Further reading: Richards (ed.), USDA Agriculture Handbook 60
Using TDS (mg/L) = 640 x EC (dS/m), what is the approximate dissolved-salt content of water with EC 2.5 dS/m?
D. 1,600 mg/L
640 x 2.5 = 1,600 mg/L. 25 is roughly the total cations in meq/L (10 x EC), a different quantity. The factor 640 varies roughly 550-700 with composition. 16,000 and 160 mg/L are factor-of-ten slips.
Further reading: Ayers and Westcot, FAO Irrigation and Drainage Paper 29 Rev. 1
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