ASRB NET — Farm Machinery & Power / Land & Water Management Engineering / Agricultural Structures & Process Engineering · India · Agricultural Scientists Recruitment Board (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
ASRB NET — Land & Water Management Engineering
Like the ICAR PhD paper: standard formulas, classification limits and concepts across drainage design, reclamation and water quality.
How our questions for ASRB NET spread across the units of Agricultural Drainage Engineering. Difficulty: 0 easy, 25 medium, 19 hard.
| Kirkham and Glover-Dumm equations | 8 | |
| Vertical drainage and bio-drainage | 6 | |
| Subsurface drain spacing: Hooghoudt and Ernst | 5 | |
| Drain pipes and envelopes | 5 | |
| Reclamation of saline and sodic soils and gypsum requirement | 5 | |
| Drainage coefficient | 4 | |
| Interceptor and relief drains | 4 | |
| Irrigation water quality: EC, SAR and RSC | 3 | |
| Causes and effects of waterlogging | 2 | |
| Salinity and the leaching requirement | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
In a piezometer nest, the piezometer open at 2 m depth has its water level 0.8 m below the ground surface and the one open at 6 m depth has its water level 0.3 m below the surface. What is the direction and magnitude of the vertical hydraulic gradient between them?
A. Upward, gradient 0.125
Taking the ground surface as datum, head at 2 m = -0.8 m and at 6 m = -0.3 m. The deeper point has the higher head, so water moves upward. Gradient = 0.5 m / (6 - 2) m = 0.125. Reporting 0.5 forgets to divide by the 4 m distance; 'downward' reads the levels as depths instead of heads. Upward seepage means relief wells or drains designed for extra inflow.
Further reading: Ritzema (ed.), Drainage Principles and Applications, ILRI Publication 16
Over six consecutive days the water-table depths were 5, 15, 25, 35, 20 and 40 cm. What is SEW30 for this period?
C. 55 cm-days
SEW30 sums (30 - depth) only for days when the water table is shallower than 30 cm: 25 + 15 + 5 + 10 = 55 cm-days; the 35 and 40 cm days add nothing. 40 wrongly subtracts the deep days; 65 adds depths instead of excesses; 4 counts days.
Further reading: Sieben's SEW30; Skaggs and van Schilfgaarde, Agricultural Drainage
In Hiler's stress day index (SDI) for drainage design, what are the two factors multiplied for each growth stage?
A. A stress-day factor such as SEW30 and a crop susceptibility factor for that stage
SDI = sum of (SD x CS): the stress-day factor measures how wet the soil was, and the crop susceptibility factor measures how sensitive the crop is at that stage. The other pairs are design inputs, not stress indices.
Further reading: Skaggs and van Schilfgaarde, Agricultural Drainage
Ernst: q = 6 mm/day, L = 40 m, Kv = 0.5 m/day over Dv = 0.3 m above drains, transmissivity Sum(KD) = 4.0 m2/day, radial layer Kr = 0.8 m/day and Dr = 5 m, wetted perimeter u = 0.3 m, a = 1. What total midpoint head h results?
D. 0.57 m
hv = qDv/Kv = 0.006 x 0.3 / 0.5 = 0.0036 m. hh = qL^2 / (8 Sum KD) = 0.006 x 1600 / 32 = 0.300 m. hr = (qL / pi Kr) ln(Dr/u) = (0.24 / 2.513) x ln(16.67) = 0.269 m. Total = 0.57 m. 0.30 m keeps only the horizontal term; 0.42 m uses log10 in the radial term; 0.27 m is the radial term alone.
Further reading: Ritzema (ed.), Drainage Principles and Applications, ILRI Publication 16
Glover-Dumm: K = 1.0 m/day, average flow depth D = 4.5 m, drainable porosity 0.05, initial midpoint head 1.0 m above drains, required head 0.7 m after 3 days (initial shape a fourth-degree parabola). What drain spacing is needed?
C. 73 m
L^2 = pi^2 K D t / [mu ln(1.16 h0/ht)] = 9.870 x 1.0 x 4.5 x 3 / [0.05 x ln(1.657)] = 133.2 / (0.05 x 0.5051) = 5276 m2; L = 73 m. 23 m omits pi^2; 67 m uses 1.27 (flat initial water table); 7.3 m enters porosity as 5 instead of 0.05.
Further reading: Ritzema (ed.), Drainage Principles and Applications, ILRI Publication 16
What distinguishes Kirkham's drain-flow solutions from Hooghoudt's?
D. They solve Laplace's equation without the Dupuit-Forchheimer assumption of horizontal flow
Kirkham used potential theory, so vertical and converging flow are handled exactly for the stated geometry. Hooghoudt uses horizontal flow with an equivalent-depth correction. Kirkham's solutions are analytical, not empirical.
Further reading: Ritzema (ed.), Drainage Principles and Applications, 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
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
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
Using the leaching relation (C/C0)(Dw/Ds) = 0.3 for continuous ponding, how much water must infiltrate to reduce salinity in the top 0.6 m to 20 % of its initial value?
C. 0.9 m
Dw = k x Ds / (C/C0) = 0.3 x 0.6 / 0.2 = 0.9 m. 0.3 m uses k = 0.1, which applies to intermittent ponding; 0.036 m multiplies by C/C0; 3.0 m omits k. Constants are soil-specific; confirm in a trial.
Further reading: Hoffman, in FAO Irrigation and Drainage Paper 29
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