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 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
ICAR AIEEA-PG — Agricultural Engineering & Technology (Unit VII)
Single-answer MCQs on soil water constants, infiltration measurement, hydraulic conductivity methods, salinity and sodicity class limits, amendments and short calculations of SAR, ESP, porosity and available water.
How our questions for ICAR AIEEA-PG spread across the units of Soil Science for Engineers. Difficulty: 36 easy, 46 medium, 10 hard.
| Problem soils: saline, sodic and acid | 15 | |
| Soil chemistry: pH and electrical conductivity | 14 | |
| Bulk density, particle density and porosity | 11 | |
| Soil formation and profile | 10 | |
| Sodicity: SAR and ESP | 10 | |
| Soil water: field capacity, wilting point and available water | 9 | |
| Soil texture and structure | 8 | |
| Infiltration | 8 | |
| Hydraulic conductivity and Darcy's law | 7 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Which master horizon designation is used for hard, consolidated bedrock beneath the soil?
A. R
R is hard bedrock. C is unconsolidated, little-altered parent material that can be dug with a spade, O is an organic layer at the surface, and B is the subsoil horizon of illuviation or alteration.
Further reading: Soil Survey Manual, USDA Handbook 18
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
What is the purpose of the outer ring in a double-ring infiltrometer?
B. It buffers lateral spread so the inner ring measures nearly vertical infiltration
Water from the outer ring wets the soil around the inner ring, so water from the inner ring is not drawn sideways and its fall represents one-dimensional vertical infiltration. Readings are taken from the inner ring only. The outer ring is not an evaporation pan, a reservoir or an anchor.
Further reading: ASTM D3385
Using the common approximation TDS (mg/L) ≈ 640 × EC (dS/m), what is the total dissolved solids of an irrigation water with EC 2.5 dS/m?
C. 1600 mg/L
TDS ≈ 640 × 2.5 = 1600 mg/L. Dividing gives 256; 25 is the approximate total cations in meq/L (10 × EC), not mg/L; 16000 is a decimal slip. The factor varies with salt type (about 550–700).
Further reading: FAO Irrigation and Drainage Paper 29 Rev. 1
A crop has a salinity threshold ECe of 2.0 dS/m and a yield decline of 7% per dS/m above it (Maas-Hoffman model). What is the expected relative yield at ECe 5.0 dS/m?
D. 79%
Relative yield = 100 − b(ECe − a) = 100 − 7 × (5.0 − 2.0) = 79%. Ignoring the threshold gives 65%; 21% is the yield loss, not the yield; adding the threshold gives 51%.
Further reading: Maas and Hoffman (1977); FAO Paper 29
A sodic soil with GR 7.75 t/ha of pure gypsum is to be reclaimed with agricultural gypsum of 80% purity. How much product is needed?
B. 9.69 t/ha
Product = 7.75/0.80 = 9.69 t/ha. Multiplying by purity gives 6.20, the common slip; 7.75 ignores purity; multiplying by 1.2 treats 80% purity as a 20% surcharge.
Further reading: USDA Agriculture Handbook 60 (1954)
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)
An acid soil has 2 cmol(+)/kg of exchangeable acidity to be neutralised in the top 15 cm, bulk density 1.3 g/cm³. How much pure CaCO₃ (equivalent weight 50 g/eq) is required per hectare?
D. 1.95 t/ha
2 cmol(+)/kg = 0.02 eq/kg; CaCO₃ = 0.02 × 50 = 1.0 g/kg. Soil mass = 10,000 × 0.15 × 1,300 = 1.95 × 10⁶ kg/ha, so lime = 1.95 t/ha. Using the molar mass 100 doubles it; using bulk density 1.5 gives 2.25; treating 2 cmol as 0.002 eq gives the tenfold-low answer.
Further reading: Brady and Weil, The Nature and Properties of Soils
A laboratory reports a sodic soil as calcareous (free CaCO₃ present). Which amendment option becomes available that would not work in a non-calcareous sodic soil?
A. Elemental sulphur or sulphuric acid, which dissolve native CaCO₃ to release calcium
Acid formers work only where CaCO₃ is present to be dissolved. Calcium chloride, gypsum and organic manures work in both calcareous and non-calcareous sodic soils, so none of them is the option that calcareousness newly makes possible.
Further reading: USDA Agriculture Handbook 60 (1954)
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