Soil Mechanics MCQ With Answers for JKSSB Junior Engineer (Civil)

Technical · 2026-10-27 · 6 min read

Soil mechanics carried 10 marks in the board's earlier Junior Engineer (Civil) syllabus, which lists soil classification, the three-phase system, permeability and seepage, effective stress, consolidation, compaction, shear strength, field density tests, soil exploration and bearing capacity. The board's 2025 paper had 120 one-mark questions in 120 minutes, so expect quick definition and formula questions with a few short calculations. Our guide to the 2026 notice shows where this section fits. This page gives a revision list, worked problems and solved MCQs.

Quick revision list

  • Phase relations: void ratio e, porosity n = e ÷ (1 + e), degree of saturation S, and the relation S × e = w × G
  • Index properties: liquid limit, plastic limit, plasticity index PI = LL − PL, and the flow and consistency indices
  • Classification: in the Indian system, clay of low plasticity has LL below 35, intermediate between 35 and 50, and high above 50
  • Permeability: Darcy's law v = k × i, the constant head test for coarse soils and the falling head test for fine soils
  • Stress: total, neutral and effective stress, and the effect of a rising water table
  • Consolidation: Terzaghi's theory, and the time factor for the rate of settlement
  • Compaction: the Proctor test, optimum moisture content and maximum dry density, and field control
  • Strength and bearing capacity: Mohr–Coulomb failure, Terzaghi's equation, the plate load test and the standard penetration test (SPT)
  • Earth pressure: the Rankine coefficient Ka = (1 − sin φ) ÷ (1 + sin φ)

Worked numerical problems

  • Porosity: a soil with a void ratio of 0.6 has a porosity of 0.6 ÷ 1.6 = 0.375
  • Dry density: with G = 2.65 and e = 0.6, γd = G γw ÷ (1 + e) = 2.65 × 9.81 ÷ 1.6 = 16.25 kN/m³
  • Plasticity index: LL = 45% and PL = 25% give PI = 20%
  • Saturated clay: w = 20% and G = 2.7 with S = 1 give e = wG ÷ S = 0.54
  • Darcy's law: k = 0.01 cm/s and a hydraulic gradient of 0.5 give a discharge velocity of 0.005 cm/s
  • Effective stress: at 5 m in a saturated soil of unit weight 19 kN/m³ with the water table at the surface, σ = 95 kPa, u = 5 × 9.81 = 49.05 kPa, so σ' = 45.95 kPa
  • Earth pressure: for φ = 30° the Rankine coefficient is Ka = 1/3. For a wall 4 m high retaining soil of 18 kN/m³, the thrust is ½ × (1/3) × 18 × 4² = 48 kN per metre

Compaction in brief

Compaction squeezes air out of the soil to raise its dry density and strength. In the Proctor test, soil is compacted at several moisture contents, and the plot of dry density against moisture content has a peak. The moisture content at the peak is the optimum moisture content and the density there is the maximum dry density. Too little water leaves friction between the grains, and too much water leaves the pores filled with water that cannot be squeezed out. The zero air voids line lies above the curve and can never be crossed.

Common mistakes

  • Using the total stress when the question asks for the effective stress
  • Writing PI = PL − LL, when the plasticity index is the liquid limit minus the plastic limit
  • Using the core cutter method for gravelly soil, where sand replacement is needed
  • Forgetting that the effective stress falls when the water table rises

Shear strength and bearing capacity in brief

The shear strength of a soil is found in the laboratory by the direct shear, triaxial or unconfined compression tests. The unconfined test suits saturated clays, where the shear strength is half the unconfined compressive strength. Terzaghi's bearing capacity equation adds three terms, one for cohesion, one for the surcharge from the soil above the foundation level and one for the width of the footing. A deeper or wider footing raises the capacity, and a rising water table lowers it.

Practise

Work the solved questions below, then read the Junior Engineer (Civil) guide for the dates and the marks for every subject. The other subject pages in this series cover surveying, fluid mechanics, RCC and construction management.

By Shivam · Written with AI assistance and fact-checked · Questions on this page are original · Editorial policy · Report an error

Practice questions

Q1. The porosity of a soil with a void ratio of 0.6 is:

  1. 0.6
  2. 0.375
  3. 0.4
  4. 1.6
Show answer

0.375 — n = e ÷ (1 + e) = 0.6 ÷ 1.6 = 0.375.

Q2. The relation between the degree of saturation, void ratio, water content and specific gravity is:

  1. S × e = w × G
  2. S × w = e × G
  3. S = e × w × G
  4. e = S × w × G
Show answer

S × e = w × G — For a fully saturated soil, S = 1 and e = wG.

Q3. The dry unit weight of a soil with G = 2.65 and e = 0.6 is about:

  1. 19.6 kN/m³
  2. 26.0 kN/m³
  3. 9.81 kN/m³
  4. 16.25 kN/m³
Show answer

16.25 kN/m³ — γd = Gγw ÷ (1 + e) = 2.65 × 9.81 ÷ 1.6 = 16.25 kN/m³.

Q4. The plasticity index of a soil is:

  1. Plastic limit minus liquid limit
  2. Liquid limit plus plastic limit
  3. Liquid limit minus plastic limit
  4. Liquid limit divided by plastic limit
Show answer

Liquid limit minus plastic limit — PI = LL − PL, the range of water content over which the soil is plastic.

Q5. A soil with a liquid limit of 45% and a plastic limit of 25% has a plasticity index of:

  1. 70%
  2. 20%
  3. 35%
  4. 1.8%
Show answer

20% — 45 − 25 = 20%.

Q6. In the Indian soil classification, a clay with a liquid limit above 50% is classified as having:

  1. High plasticity
  2. Low plasticity
  3. Intermediate plasticity
  4. No plasticity
Show answer

High plasticity — Below 35 is low, 35 to 50 intermediate and above 50 high plasticity.

Q7. Darcy's law states that the discharge velocity is proportional to:

  1. The square of the gradient
  2. The void ratio only
  3. The unit weight of soil
  4. The hydraulic gradient
Show answer

The hydraulic gradient — v = k × i, where k is the coefficient of permeability.

Q8. The falling head permeability test is used for:

  1. Coarse-grained soils
  2. Rocks
  3. Fine-grained soils
  4. Peat only
Show answer

Fine-grained soils — The constant head test is for coarse soils with a high permeability.

Q9. The effective stress at a point in a soil mass is:

  1. Total stress plus pore water pressure
  2. Total stress minus pore water pressure
  3. Pore water pressure minus total stress
  4. Total stress divided by the void ratio
Show answer

Total stress minus pore water pressure — Terzaghi's principle: σ' = σ − u.

Q10. When the water table rises in a soil deposit, the effective stress at a given depth:

  1. Decreases
  2. Increases
  3. Remains the same
  4. Becomes infinite
Show answer

Decreases — The pore pressure rises and the soil is buoyed up.

Q11. The Proctor compaction test gives:

  1. The bearing capacity of the soil
  2. The permeability of the soil
  3. The liquid limit of the soil
  4. The optimum moisture content and the maximum dry density
Show answer

The optimum moisture content and the maximum dry density — The peak of the dry density versus moisture content curve gives both.

Q12. The core cutter method for the field density is suited to:

  1. Gravelly soils
  2. Rock
  3. Fine-grained cohesive soils free of gravel
  4. Boulders
Show answer

Fine-grained cohesive soils free of gravel — The sand replacement method is used where gravel or hard soil prevents driving the cutter.

Q13. The shear strength of a soil according to the Mohr–Coulomb theory is:

  1. τ = c − σ' tan φ
  2. τ = c + σ' tan φ
  3. τ = σ' + c tan φ
  4. τ = c × σ' × tan φ
Show answer

τ = c + σ' tan φ — c is the cohesion and φ is the angle of internal friction.

Q14. A clean dry sand has a cohesion of:

  1. Zero
  2. High
  3. Equal to its unit weight
  4. Equal to the friction angle
Show answer

Zero — Sands are cohesionless, and their strength comes from friction alone.

Q15. Terzaghi's theory of consolidation deals with:

  1. The compaction of dry sand
  2. The bearing capacity of rock
  3. The shear strength of gravel
  4. The time-dependent settlement of saturated clays
Show answer

The time-dependent settlement of saturated clays — Water is squeezed out of the pores slowly, so settlement takes time.

Q16. The Rankine active earth pressure coefficient for φ = 30° is:

  1. 3
  2. 1/2
  3. 1/3
  4. 2/3
Show answer

1/3 — Ka = (1 − sin 30°) ÷ (1 + sin 30°) = 0.5 ÷ 1.5 = 1/3.

Q17. The active thrust on a 4 m high smooth wall retaining dry cohesionless soil of unit weight 18 kN/m³ and φ = 30° is:

  1. 96 kN per metre
  2. 48 kN per metre
  3. 24 kN per metre
  4. 144 kN per metre
Show answer

48 kN per metre — Pa = ½ Ka γ H² = ½ × (1/3) × 18 × 16 = 48 kN/m.

Q18. The effective stress at 5 m depth in a saturated clay of unit weight 19 kN/m³ with the water table at the surface is about:

  1. 46 kPa
  2. 95 kPa
  3. 49 kPa
  4. 144 kPa
Show answer

46 kPa — σ = 95 kPa, u = 49.05 kPa, so σ' = 95 − 49.05 = 45.95 kPa.

Q19. The standard penetration test (SPT) is used to find:

  1. The liquid limit
  2. The specific gravity
  3. The grain size distribution
  4. The relative density or consistency of the soil in the field
Show answer

The relative density or consistency of the soil in the field — The N value is the number of blows for 300 mm of penetration.

Q20. The plate load test is used to estimate the:

  1. Liquid limit of the soil
  2. Permeability of the soil
  3. Bearing capacity and settlement of the soil
  4. Specific gravity of the soil
Show answer

Bearing capacity and settlement of the soil — A plate is loaded in steps and the load-settlement curve is plotted.

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