ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR
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
ICAR AIEEA-PG — Agricultural Engineering & Technology (Unit VII)
Single-answer MCQs on causes and effects of waterlogging, drainage coefficient, surface and subsurface system types, envelope functions, saline-sodic classification limits, amendments and water-quality indices, with short calculations.
How our questions for ICAR AIEEA-PG spread across the units of Agricultural Drainage Engineering. Difficulty: 26 easy, 30 medium, 1 hard.
| Causes and effects of waterlogging | 7 | |
| Salinity and the leaching requirement | 7 | |
| Reclamation of saline and sodic soils and gypsum requirement | 7 | |
| Drainage coefficient | 6 | |
| Drain pipes and envelopes | 6 | |
| Surface drainage design | 5 | |
| Vertical drainage and bio-drainage | 5 | |
| Irrigation water quality: EC, SAR and RSC | 5 | |
| Subsurface drain spacing: Hooghoudt and Ernst | 4 | |
| Interceptor and relief drains | 3 | |
| Kirkham and Glover-Dumm equations | 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
Heavy weed growth in an open drain has which effect?
A. It raises Manning's n and so lowers the drain's capacity
Vegetation adds roughness; with a higher n, Q = (1/n) A R^(2/3) S^(1/2) falls, and water levels rise. That is why maintenance is part of design.
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
Mole drainage is best suited to which situation?
A. A stable clay subsoil in which an unlined channel keeps its shape
A mole plough forms an unlined channel that survives only in cohesive, stable clay. It collapses in sand, cannot be drawn through stones, and deforms in peat.
Further reading: Smedema, Vlotman and Rycroft, Modern Land Drainage
Vertical drainage by tubewells is most attractive where:
D. A transmissive aquifer lies beneath and the pumped water is fresh enough to reuse
The pumped water then becomes an irrigation supply, giving conjunctive use. Thin clay over rock and perched tables have no aquifer to pump; saline water without an outlet creates a disposal problem.
Further reading: Ritzema (ed.), Drainage Principles and Applications, ILRI Publication 16
An interceptor drain is used to:
B. Cut off groundwater seeping laterally into an area from higher land or a canal
Interceptors deal with lateral inflow from outside the area. Water arriving over the area itself is handled by relief (parallel) drains; artesian pressure by relief wells.
Further reading: Smedema, Vlotman and Rycroft, Modern Land Drainage
Which description fits a sodic (alkali) soil?
C. ESP above 15, ECe below 4 dS/m, pH 8.5-10, dispersed clay and poor permeability
Sodic soils have dispersed clay and organic matter ('black alkali'), high pH and very low permeability. White crusts mark saline soils; acid sulphate soils are a different problem.
Further reading: Richards (ed.), USDA Agriculture Handbook 60
A saline-sodic soil is to be reclaimed. What is the correct sequence?
B. Apply a calcium amendment such as gypsum first, then leach
If the salts are leached first, the clay disperses and the soil becomes sodic and nearly impermeable. Calcium must replace exchangeable sodium first. Lime is insoluble at high pH and raising pH is the wrong direction.
Further reading: Richards (ed.), USDA Agriculture Handbook 60
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