GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersThe soil an engineer tills, irrigates, drains and builds on: its physical, water and chemical properties.
In the exams: ICAR AIEEA-PG Unit VII; ICAR AICE-JRF/SRF subjects 40 and 43; NCEES PE soil and water
GATE — Agricultural Engineering (AG)
Numerical answer items on soil-water-plant relations: gravimetric to volumetric water content, bulk density and porosity, available water and irrigation depth, infiltration equations (Kostiakov, Horton, Philip), Darcy's law and layered-soil conductivity, and SAR or gypsum requirement in the drainage and salinity part. Watch units and the dry-mass basis.
How our questions for GATE spread across the units of Soil Science for Engineers. Difficulty: 12 easy, 14 medium, 15 hard.
| Soil water: field capacity, wilting point and available water | 8 | |
| Infiltration | 8 | |
| Hydraulic conductivity and Darcy's law | 7 | |
| Soil texture and structure | 6 | |
| Bulk density, particle density and porosity | 6 | |
| Sodicity: SAR and ESP | 4 | |
| Problem soils: saline, sodic and acid | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Using Stokes' law at 20 °C (viscosity 1.002 × 10⁻³ Pa s, Δρ = 1650 kg/m³), what is the settling velocity of a 0.02 mm silt particle?
D. 0.359 mm/s
v = 9.81 × (2 × 10⁻⁵)² × 1650/(18 × 1.002 × 10⁻³) = 3.590 × 10⁻⁴ m/s = 0.359 mm/s. 0.090 puts the 0.01 mm radius in place of the diameter; 0.577 uses particle density 2650 instead of the density difference; 0.718 uses 9η in place of 18η.
Further reading: Stokes' law
A cylindrical core 5.0 cm in diameter and 5.0 cm long contains 135 g of oven-dry soil. What is the bulk density?
B. 1.38 g/cm³
Volume = π × 2.5² × 5 = 98.2 cm³; ρb = 135/98.2 = 1.375 g/cm³. Using the diameter as the radius gives 0.34; treating the core as a 5 cm cube gives 1.08; 2.65 is the default particle density, not bulk density.
Further reading: Brady and Weil, The Nature and Properties of Soils
A soil sample has a gravimetric water content of 22% and a bulk density of 1.45 g/cm³. What is its volumetric water content?
A. 31.9%
θ = w × ρb/ρw = 0.22 × 1.45 = 0.319, or 31.9%. Dividing by bulk density gives 15.2%, leaving it unconverted gives 22%, and multiplying by particle density instead of bulk density gives 58.3%.
Further reading: Hillel, Introduction to Environmental Soil Physics
A soil layer 60 cm thick has a volumetric water content of 0.25. What is the equivalent depth of water stored in it?
B. 15 cm
Equivalent depth = θ × depth = 0.25 × 60 cm = 15 cm. 25 cm treats the 0.25 as a depth; 2.4 cm divides 60 by 25; 240 cm divides the depth by θ instead of multiplying.
Further reading: Hillel, Introduction to Environmental Soil Physics
In a constant-head permeameter test, a soil column 15 cm long and 10 cm in diameter under a head difference of 20 cm passes 120 cm³ of water in 30 minutes. What is the saturated hydraulic conductivity?
C. 2.29 cm/h
A = π × 10²/4 = 78.5 cm²; K = V L/(A h t) = 120 × 15/(78.5 × 20 × 1,800) = 6.37 × 10⁻⁴ cm/s = 2.29 cm/h. Using the diameter as the radius gives 0.57; swapping L and h gives 4.07; converting s to h with 60 instead of 3,600 gives 0.038.
Further reading: Darcy's law
Darcy's law is valid for flow in soils as long as the flow is:
D. Laminar, at low Reynolds number
Darcy's law assumes viscous, laminar flow, which holds in nearly all soils (Reynolds number below roughly 1–10). It fails in turbulent flow through very coarse gravels. Froude number and supercritical flow are open-channel concepts.
Further reading: Hillel, Introduction to Environmental Soil Physics
For the Kostiakov equation I = 2.5 t^0.6 (I in cm, t in hours), what is the infiltration rate at t = 2 hours?
D. 1.14 cm/h
Rate = dI/dt = a b t^(b−1) = 2.5 × 0.6 × 2^(−0.4) = 1.14 cm/h. 3.79 is the cumulative depth; 1.89 is the average rate over the first 2 hours, which is higher than the instantaneous rate; 1.50 is the rate at 1 hour.
Further reading: Kostiakov (1932)
How many tonnes of gypsum dihydrate (equivalent weight 86.09 g/eq) are equivalent in reclamation value to 1 t of elemental sulphur (equivalent weight 16 g/eq)?
D. 5.38 t
Sulphur oxidises to H₂SO₄, which supplies 2 equivalents per mole of S (32 g), so its equivalent weight is 16 g/eq. One tonne of S therefore carries 10⁶/16 equivalents, and the same equivalents of gypsum weigh 10⁶/16 × 86.09 g: 1 t S ≈ 86.09/16 = 5.38 t gypsum. 0.19 t is the inverse (tonnes of S per tonne of gypsum); 2.69 t uses the molar mass of S with the equivalent weight of gypsum; 10.76 t uses the molar mass of gypsum with the equivalent weight of S.
Further reading: USDA Agriculture Handbook 60 (1954)
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