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ASRB NET Soil Science for Engineers: practice questions

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 papers
10worked here, free
57written to this exam in this subject
9syllabus units

What this subject covers

The 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

What ASRB NET asks from it

ASRB NET — Land & Water Management Engineering

Same ground as AICE at a similar depth, with more recall of definitions, classification systems and measurement methods.

Where the 57 questions fall

How our questions for ASRB NET spread across the units of Soil Science for Engineers. Difficulty: 3 easy, 37 medium, 17 hard.

Problem soils: saline, sodic and acid11
Soil formation and profile10
Soil chemistry: pH and electrical conductivity10
Bulk density, particle density and porosity5
Soil water: field capacity, wilting point and available water5
Hydraulic conductivity and Darcy's law5
Infiltration4
Sodicity: SAR and ESP4
Soil texture and structure3

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Soil formation and profile · medium

Which master horizon designation is used for hard, consolidated bedrock beneath the soil?

  1. R
  2. O
  3. B
  4. C
Answer and working

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

2 · Hydraulic conductivity and Darcy's law · medium

As a soil dries from saturation, its unsaturated hydraulic conductivity:

  1. Falls steeply, often by several orders of magnitude, as the larger pores empty
  2. Stays equal to the saturated value until the soil reaches the wilting point
  3. Rises steadily, because the matric suction gradients become much steeper
  4. Falls, usually in direct proportion to the volumetric water content of the soil
Answer and working

A. Falls steeply, often by several orders of magnitude, as the larger pores empty

Flow capacity depends roughly on the fourth power of pore radius, so emptying the large pores first cuts K(θ) drastically; it can fall by orders of magnitude between saturation and field capacity. Steeper gradients can raise flux but not conductivity; K is not constant to wilting point, and the decline is strongly non-linear.

Further reading: Hillel, Introduction to Environmental Soil Physics

3 · Hydraulic conductivity and Darcy's law · medium

For designing subsurface drain spacing, which field method is standard for measuring saturated hydraulic conductivity below a shallow water table?

  1. Ring infiltrometer
  2. Falling-head permeameter on a small core
  3. Auger-hole method
  4. Tension disc infiltrometer
Answer and working

C. Auger-hole method

The auger-hole method measures the rate of rise of water in a hole bored below the water table and gives K of a large soil volume at the depth that matters for drains. Ring and tension infiltrometers measure surface infiltration above the water table, and small cores in a laboratory permeameter often miss the macropores that dominate field flow.

Further reading: ILRI Publication 16

4 · Soil chemistry: pH and electrical conductivity · medium

The permanent negative charge of 2:1 clay minerals such as montmorillonite arises mainly from:

  1. Protonation of hydroxyl groups at broken edges when the soil pH falls
  2. Isomorphous substitution, such as Mg²⁺ for Al³⁺ in the octahedral sheet
  3. Adsorption of nitrate and chloride anions onto the planar clay surfaces
  4. Dissociation of hydrogen from carboxyl groups on the edges of the clay
Answer and working

B. Isomorphous substitution, such as Mg²⁺ for Al³⁺ in the octahedral sheet

Substitution of a lower-valence cation during crystal formation (Al³⁺ for Si⁴⁺, Mg²⁺ for Al³⁺) leaves a permanent negative charge independent of pH. Carboxyl dissociation is the pH-dependent charge of humus; protonation at low pH creates positive, not negative, charge; and anion adsorption is not the source.

Further reading: Brady and Weil, The Nature and Properties of Soils

5 · Soil chemistry: pH and electrical conductivity · easy

Which of the following is a micronutrient essential for higher plants?

  1. Sulphur
  2. Calcium
  3. Magnesium
  4. Boron
Answer and working

D. Boron

Boron is one of the micronutrients (Fe, Mn, Zn, Cu, B, Mo, Cl, Ni). Sulphur, magnesium and calcium are secondary macronutrients, required in larger amounts.

Further reading: Brady and Weil, The Nature and Properties of Soils

6 · Soil chemistry: pH and electrical conductivity · hard

Why does a soil's lime requirement depend on its CEC and organic matter, and not just on its measured pH?

  1. Reserve acidity on the colloids far exceeds active acidity and buffers pH change
  2. Organic matter reacts with lime to form insoluble salts, so more is needed to reach the same pH
  3. The pH electrode misreads in high-CEC soils, so readings need correcting
  4. High-CEC soils contain calcium carbonate that dissolves and neutralises part of the applied lime
Answer and working

A. Reserve acidity on the colloids far exceeds active acidity and buffers pH change

Most soil acidity is held on the exchange complex and pH-dependent sites (reserve acidity), far exceeding the H⁺ in solution; a soil with more clay and humus has more reserve acidity and needs more lime for the same pH change. Electrode error, carbonate in acid soils and insoluble organic salts are not the reason.

Further reading: Brady and Weil, The Nature and Properties of Soils

7 · Sodicity: SAR and ESP · hard

Under the USSL Handbook 60 convention, a soil is classed as sodic when its ESP exceeds 15. Which alternative criterion is used in Australian practice?

  1. ESP above 6
  2. ESP above 25
  3. SAR above 30
  4. ECe above 15 dS/m
Answer and working

A. ESP above 6

Australian classification (following Northcote and Skene) treats soils with ESP above 6 as sodic, recognising that dispersion problems begin well below ESP 15 in low-salinity conditions; some authorities use SAR 13 in place of ESP 15. ESP 25, SAR 30 and ECe 15 are not the standard alternative thresholds.

Further reading: Northcote and Skene (1972)

8 · Problem soils: saline, sodic and acid · hard

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)?

  1. 2.69 t
  2. 0.19 t
  3. 10.76 t
  4. 5.38 t
Answer and working

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)

9 · Problem soils: saline, sodic and acid · hard

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?

  1. 2.25 t/ha
  2. 3.90 t/ha
  3. 0.195 t/ha
  4. 1.95 t/ha
Answer and working

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

10 · Problem soils: saline, sodic and acid · medium

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?

  1. Elemental sulphur or sulphuric acid, which dissolve native CaCO₃ to release calcium
  2. Farmyard manure, which improves structure as it decomposes in the surface soil
  3. Calcium chloride, which supplies soluble calcium directly to the exchange complex
  4. Gypsum, which supplies soluble calcium directly to the exchange complex of the soil
Answer and working

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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Soil Science for EngineersThe full study manual, free ASRB NETFormat, sections and how it has changed

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