ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersSoil as an engineering material: its phases, flow of water through it, compaction, consolidation, shear strength and stability.
In the exams: GATE AG Section 4 (soil mechanics asked in 6 of 8 papers 2019–26); ICAR AIEEA-PG Unit IV; NCEES PE soils and foundations
ICAR AICE-JRF/SRF (PhD) and ASRB NET — Soil and Water Conservation / Land and Water Management Engineering
Deeper treatment of seepage through and under earth dams (phreatic line, filters, piping), consolidation theory and its assumptions, drained versus undrained analysis, stability cases for earth dams including rapid drawdown, and methods of slices.
How our questions for ICAR AICE-JRF/SRF spread across the units of Soil Mechanics. Difficulty: 3 easy, 30 medium, 28 hard.
| Seepage and flow nets | 8 | |
| Terzaghi one-dimensional consolidation | 8 | |
| Shear strength and the Mohr-Coulomb criterion | 8 | |
| Slope stability, including infinite slopes | 8 | |
| Bearing capacity basics | 7 | |
| Earth pressure: active and passive (Rankine) | 6 | |
| Permeability | 5 | |
| Effective stress | 4 | |
| Index properties and soil classification | 3 | |
| Phase relations | 2 | |
| Compaction and the Proctor test | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A soil has a bulk unit weight of 19.0 kN/m³, a water content of 12% and G = 2.65. Taking γw = 9.81 kN/m³, what is its degree of saturation?
B. 0.597
γd = γ/(1 + w) = 19.0/1.12 = 16.96 kN/m³. e = Gγw/γd − 1 = 2.65 × 9.81/16.96 − 1 = 0.532. S = wG/e = 0.12 × 2.65/0.532 = 0.597. Using the bulk instead of the dry unit weight to find e gives e = 0.368 and S = 0.864; 0.318 is just wG with no division by e.
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
In a hydrometer analysis, how long does a spherical particle of diameter 0.01 mm and specific gravity 2.65 take to settle 100 mm in water of viscosity 1.0 × 10⁻³ Pa·s (γw = 9.81 kN/m³)?
B. 18.5 min
Stokes' law: v = (γs − γw)D²/(18μ) = (2.65 − 1) × 9810 × (1.0 × 10⁻⁵)²/(18 × 10⁻³) = 8.993e-05 m/s. Time = 0.1/v = 1112 s = 18.5 min. Using γs instead of (γs − γw) ignores buoyancy and gives 11.5 min; the other options are factor-of-ten unit slips.
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
A soil pat dried below its shrinkage limit has a shrinkage ratio of 1.85. The specific gravity of solids is 2.70. What is the shrinkage limit?
B. 17.0%
ws = 1/R − 1/G = 1/1.85 − 1/2.70 = 0.5405 − 0.3704 = 17.0%. The shrinkage ratio R is the dry density of the pat divided by the density of water, so 1/R is the volume of dry pat per unit mass of solids and 1/G the volume of solids; their difference is the void volume, filled with water at the shrinkage limit. 54.1% is 1/R alone, 37.0% is 1/G alone, and 31.5% is (G − R)/G = 0.85/2.70.
Further reading: Punmia, Jain and Jain, Soil Mechanics and Foundations
Downstream of a weir, the last square of a flow net has a length of 1.2 m along the flow line. The total head loss of 6 m is spread over 12 equal equipotential drops. The foundation sand has G = 2.68 and e = 0.72. What is the factor of safety against boiling at the exit?
A. 2.34
Head drop per equipotential = 6/12 = 0.5 m. Exit gradient = 0.5/1.2 = 0.417. ic = (G − 1)/(1 + e) = 0.977. FS = ic/iexit = 2.34. Using the whole head over one square (5.0) gives 0.20; 0.43 is the ratio upside down. A factor near 2.3 is below the 3 to 4 usually wanted, so the exit needs a loaded filter. 3.74 computes ic as G/(1 + e) = 1.558, forgetting buoyancy (1.558/0.417).
Further reading: Craig's Soil Mechanics
Laplace's equation for two-dimensional seepage does NOT assume which of the following?
A. Turbulent flow
Laplace's equation follows from continuity plus Darcy's law, which is valid for laminar flow. It assumes a saturated, homogeneous, isotropic soil with incompressible soil and water. Turbulent flow would violate Darcy's law.
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
A soil (G = 2.70) has a standard Proctor maximum dry unit weight of 17.5 kN/m³ at an optimum moisture content of 16%. With γw = 9.81 kN/m³, what is the percentage air voids at optimum?
C. 5.4%
e = Gγw/γd − 1 = 2.70 × 9.81/17.5 − 1 = 0.5135. S = wG/e = 0.16 × 2.70/0.5135 = 0.841. n = e/(1 + e) = 0.3393. na = n(1 − S) = 5.4%, within the few-per-cent range typical at optimum. 15.9% is air content (fraction of voids), not air voids (fraction of total volume). 8.2% uses e instead of n (0.5135 × 0.159); 84.1% is the degree of saturation.
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
A clay has k = 1 × 10⁻⁹ m/s and mv = 0.25 m²/MN. With γw = 9.81 kN/m³, what is its coefficient of consolidation in m²/year?
C. 12.9 m²/year
mv = 0.25 m²/MN = 2.5 × 10⁻⁴ m²/kN. cv = k/(mv·γw) = 10⁻⁹/(2.5 × 10⁻⁴ × 9.81) = 4.077e-07 m²/s × 3.156 × 10⁷ s/year = 12.9 m²/year. Using 0.25 as if in m²/kN gives 0.0129; omitting γw gives 126. 1.29 is a factor-of-ten slip in the unit conversion.
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
A clay layer 3 m thick (e0 = 0.90) has σ′0 = 80 kPa and a preconsolidation pressure of 150 kPa at mid-depth. Cc = 0.30 and Cr = 0.05. What settlement results from a stress increase of 50 kPa at mid-depth?
B. 16.6 mm
Final stress 80 + 50 = 130 kPa is below σ′p = 150 kPa, so only recompression occurs: S = [Cr·H/(1 + e0)]·log(130/80) = (0.05 × 3/1.90) × 0.2109 = 16.6 mm. Using Cc treats the clay as normally consolidated and gives 99.9 mm, about six times too much; leaving out (1 + e0) gives 31.6 mm. OCR = 150/80 = 1.88. 38.3 mm uses the natural logarithm instead of log10 (0.0789 × ln 1.625).
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
A soil has c′ = 10 kPa and φ′ = 25°. In a drained triaxial test at a cell pressure of 150 kPa, what major principal effective stress will cause failure?
B. 401.0 kPa
σ′1 = σ′3·tan²(45° + φ′/2) + 2c′·tan(45° + φ′/2). tan 57.5° = 1.5697, tan² = 2.4639. σ′1 = 150 × 2.4639 + 2 × 10 × 1.5697 = 369.6 + 31.4 = 401.0 kPa. Leaving out cohesion gives 369.6; the cohesion term uses tan(45° + φ′/2), not its square. 418.9 kPa squares the tangent in the cohesion term (369.6 + 2 × 10 × 2.4639); 385.3 kPa leaves out the 2 (369.6 + 10 × 1.5697).
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
A vertical cut is made in a saturated clay with cu = 25 kPa, φu = 0 and γ = 18 kN/m³. By Rankine's theory, what is the depth of tension cracks and the critical unsupported height?
D. Crack depth 2.78 m; critical height 5.56 m
With φu = 0, Ka = 1. zc = 2cu/γ = 50/18 = 2.78 m. The net active thrust is zero at depth 2zc, so Hc = 4cu/γ = 5.56 m. In practice cracks fill with water and cuts are supported well below this height; Taylor's circular analysis gives a lower value near 3.85cu/γ.
Further reading: Das and Sobhan, Principles of Geotechnical Engineering
Ten here, 61 in the app. The Laisha app has every practice question we have written to ICAR AICE-JRF/SRF in this subject, marked as you go, with timed papers, a jump-to-unanswered list and a certificate when you pass at 70 per cent.
Free for ten days, then one subscription of US$10 a month. We would rather say so here than have you find it later.
An AI operating system for agriculture: free guides, manuals and tools written for every country.