Soil Mechanics Chapter-wise MCQs with PYQs & MEQs | Civil Engineering

Soil Mechanics – Chapter-wise MCQs

PYQs • MEQs • Conceptual & Numerical Questions | Civil Engineering

Note: Har question ke niche “Answer Dekho” button dabao. Questions mein variety hai – theoretical, numerical, PYQ-style aur expected (MEQ). Major chapters cover kiye gaye hain. Practice ke liye best!

1. Phase Relations 2. Index Properties 3. Classification 4. Compaction 5. Permeability 6. Effective Stress 7. Consolidation 8. Shear Strength 9. Earth Pressure 10. Slope Stability

Chapter 1: Basic Definitions & Phase Relationships

1Soil mass in its natural state is a ________ system. MEQ

a) One-phase b) Two-phase c) Three-phase d) Four-phase

Answer: c) Three-phase (solids + water + air). Fully saturated soil behaves as two-phase.

2Void ratio (e) is defined as the ratio of: PYQ

a) Volume of voids to total volume b) Volume of voids to volume of solids c) Volume of solids to volume of voids d) Volume of water to volume of voids

Answer: b) e = V_v / V_s

3Porosity (n) of a soil with void ratio e = 0.35 is approximately: Numerical

a) 19.5% b) 25.9% c) 35% d) 40.6%

Answer: b) n = e/(1+e) = 0.35/1.35 ≈ 0.259 → 25.9%

4Relationship between porosity n and void ratio e is: MEQ

a) n = e/(1+e) b) e = n/(1+n) c) Both a and b d) n = 1 + e

Answer: c) Both relations are valid and interchangeable.

5Degree of saturation S for a fully saturated soil is: PYQ

a) 0 b) 0.5 c) 1.0 d) > 1.0

Answer: c) S = 1 (or 100%)

6A soil sample has G = 2.7, e = 0.66, w = 20%. Degree of saturation is: Numerical

a) 70% b) 81.8% c) 90% d) 100%

Answer: b) S = (wG)/e = (0.2 × 2.7)/0.66 ≈ 0.818 → 81.8%

7Bulk unit weight γ is related to dry unit weight γ_d by: MEQ

a) γ = γ_d (1 + w) b) γ = γ_d / (1 + w) c) γ = γ_d × w d) γ = γ_d + w

Answer: a) γ = γ_d (1 + w)

8If volume of voids equals volume of solids, then: PYQ

a) e = 1, n = 0.5 b) e = 0.5, n = 1 c) e = 1, n = 1 d) e = 0, n = 0

Answer: a) e = V_v/V_s = 1; n = e/(1+e) = 0.5

9Air content a_c is the ratio of: MEQ

a) Volume of air to total volume b) Volume of air to volume of voids c) Volume of air to volume of water d) Volume of voids to total volume

Answer: b) a_c = V_a / V_v = 1 − S

10Specific gravity G of soil solids is usually determined by: PYQ

a) Oven drying method b) Pycnometer method c) Sand bath method d) Core cutter method

Answer: b) Pycnometer (or density bottle) method is standard for G.

11A soil weighs 190 kN. After oven drying it weighs 150 kN. Weight of water is: Numerical

a) 30 kN b) 40 kN c) 45 kN d) 50 kN

Answer: b) W_w = 190 − 150 = 40 kN

12Relative density is used for: MEQ

a) Cohesive soils only b) Cohesionless soils c) Both d) Organic soils

Answer: b) Relative density (Density Index) is primarily for sands/gravels.

Chapter 2: Index Properties & Consistency Limits

1Liquid Limit is the water content at which soil: PYQ

a) Starts to shrink b) Passes from liquid to plastic state c) Passes from plastic to semi-solid state d) Has maximum dry density

Answer: b) LL is the boundary between liquid and plastic states.

2Plasticity Index (PI) is equal to: MEQ

a) LL − PL b) LL − SL c) PL − SL d) LL + PL

Answer: a) PI = Liquid Limit − Plastic Limit

3Shrinkage Limit is the water content below which: PYQ

a) Soil starts cracking b) Further reduction in water content does not cause volume reduction c) Soil becomes plastic d) Soil loses strength completely

Answer: b) Below SL, volume remains constant even if water content decreases.

4Atterberg limits are determined for: MEQ

a) Coarse-grained soils b) Fine-grained soils c) Gravels only d) All soils

Answer: b) Atterberg limits are meaningful for fine-grained (cohesive) soils.

5Liquidity Index (LI) is given by: Numerical

a) (w − PL)/PI b) (LL − w)/PI c) (w − LL)/PI d) PI/(w − PL)

Answer: a) LI = (Natural water content − PL) / PI

6Casagrande’s liquid limit device uses how many blows for LL determination? PYQ

a) 10 b) 25 c) 30 d) 50

Answer: b) Standard is 25 blows for groove closure of 12.5 mm.

7Activity of clay is defined as: MEQ

a) PI / % clay fraction b) LL / % clay fraction c) PI × % clay d) % clay / PI

Answer: a) Activity = Plasticity Index / (% of clay-sized particles)

8Sensitivity of a clay is the ratio of: PYQ

a) Undisturbed strength to remoulded strength b) Remoulded strength to undisturbed strength c) Peak strength to residual strength d) Cohesion to friction

Answer: a) S_t = q_u (undisturbed) / q_u (remoulded)

9Which method is most accurate for water content determination? MEQ

a) Sand bath method b) Oven drying method c) Calcium carbide method d) Alcohol method

Answer: b) Oven drying at 105–110°C is the standard accurate method.

10A soil has LL = 50%, PL = 25%. Its plasticity index is: Numerical

a) 25% b) 50% c) 75% d) 12.5%

Answer: a) PI = 50 − 25 = 25%

Chapter 3: Soil Classification

1Unified Soil Classification System (USCS) was developed by: PYQ

a) Rankine b) Casagrande c) Terzaghi d) Coulomb

Answer: b) Arthur Casagrande developed USCS.

2According to IS classification, a soil is coarse-grained if more than ____ % is retained on 75 micron sieve. MEQ

a) 30% b) 50% c) 70% d) 90%

Answer: b) > 50% retained on 75 μ sieve → coarse-grained.

3In USCS, symbol ‘M’ stands for: PYQ

a) Clay b) Silt c) Sand d) Organic

Answer: b) M = Silt (from Swedish word “Mo”)

4A well-graded soil has: MEQ

a) Excess of one size particles b) Good representation of all particle sizes c) Gap in particle sizes d) Only fine particles

Answer: b) Well-graded soils have particles of all sizes in good proportion.

5Coefficient of uniformity C_u = D₆₀/D₁₀. For well-graded gravel, C_u should be: Numerical

a) > 4 b) > 6 c) < 4 d) = 1

Answer: a) For gravel C_u > 4; for sand C_u > 6 (along with C_c between 1–3).

6Plasticity chart is used for classification of: PYQ

a) Coarse-grained soils b) Fine-grained soils c) Organic soils only d) Gravels

Answer: b) A-line on plasticity chart separates clays from silts.

7IS classification system is based on: MEQ

a) AASHTO b) USCS with some modifications c) MIT system only d) USDA system

Answer: b) IS 1498 is largely based on USCS with Indian modifications.

8Group symbol ‘CH’ means: PYQ

a) Clay of high plasticity b) Clay of low plasticity c) Silt of high plasticity d) Organic clay

Answer: a) C = Clay, H = High plasticity (LL > 50%)

Chapter 4: Soil Compaction

1Compaction is a process of: MEQ

a) Expelling air by mechanical means b) Expelling water by drainage c) Reducing volume by consolidation d) Increasing water content

Answer: a) Compaction densifies soil by expelling air (not water).

2Standard Proctor test uses: PYQ

a) 2.5 kg rammer, 30 cm drop, 3 layers b) 4.5 kg rammer, 45 cm drop, 5 layers c) 2.5 kg rammer, 45 cm drop, 5 layers d) 4.5 kg rammer, 30 cm drop, 3 layers

Answer: a) Standard Proctor: 2.5 kg, 305 mm drop, 3 layers, 25 blows each.

3Optimum Moisture Content (OMC) corresponds to: MEQ

a) Maximum dry density b) Minimum dry density c) Zero air voids d) Maximum bulk density

Answer: a) At OMC, dry density is maximum for given compactive effort.

4Modified Proctor test has higher energy than Standard Proctor because: PYQ

a) Heavier rammer and more layers b) Lighter rammer c) Less number of blows d) Smaller mould

Answer: a) 4.54 kg rammer, 457 mm drop, 5 layers → about 4.5 times energy.

5Zero air voids line represents: MEQ

a) Fully saturated condition b) Dry soil c) Optimum condition d) Partially saturated

Answer: a) Theoretical maximum dry density for given water content (S=100%).

6Relative compaction is: Numerical

a) (γ_{d field} / γ_{d max}) × 100% b) (γ_{d max} / γ_{d field}) × 100% c) e_max − e / e_max − e_min d) Both a and c

Answer: a) Relative compaction = field dry density / maximum dry density × 100%

7Which of the following increases with increase in compactive effort? PYQ

a) OMC b) MDD c) Both d) Neither

Answer: b) Higher energy → higher MDD and lower OMC.

Chapter 5: Permeability & Seepage

1Darcy’s law is valid for: PYQ

a) Laminar flow through soils b) Turbulent flow c) Both laminar and turbulent d) Only coarse gravels

Answer: a) Darcy’s law: v = ki holds for laminar flow (Reynolds number < 1 approximately).

2Coefficient of permeability k has units of: MEQ

a) m/s or cm/s b) m²/s c) N/m² d) Dimensionless

Answer: a) Velocity units (length/time).

3Which soil has the highest permeability? PYQ

a) Clay b) Silt c) Sand d) Gravel

Answer: d) Gravel > Sand > Silt > Clay

4Constant head permeability test is suitable for: MEQ

a) Fine-grained soils b) Coarse-grained soils c) Both d) Organic soils

Answer: b) Constant head for permeable soils (sands/gravels); falling head for clays/silts.

5Seepage velocity is related to discharge velocity by: Numerical

a) v_s = v / n b) v_s = v × n c) v_s = v × e d) v_s = v

Answer: a) Seepage velocity = discharge velocity / porosity

6Quick sand condition occurs when effective stress becomes: PYQ

a) Zero b) Maximum c) Equal to total stress d) Negative

Answer: a) When seepage force makes σ′ = 0, soil loses strength (boiling/quick condition).

7Flow net is used to determine: MEQ

a) Seepage quantity and exit gradient b) Only permeability c) Only shear strength d) Compaction characteristics

Answer: a) Flow nets give seepage discharge, uplift pressure and exit gradient.

Chapter 6: Effective Stress Principle

1Effective stress principle was given by: PYQ

a) Rankine b) Terzaghi c) Coulomb d) Darcy

Answer: b) Karl Terzaghi (σ′ = σ − u)

2Effective stress σ′ is equal to: MEQ

a) Total stress − pore water pressure b) Total stress + pore water pressure c) Pore water pressure only d) Total stress only

Answer: a) σ′ = σ − u

3Increase in pore water pressure causes: PYQ

a) Increase in effective stress b) Decrease in effective stress c) No change d) Increase in shear strength

Answer: b) Higher u → lower σ′ → lower strength.

4In a submerged soil, effective unit weight is: Numerical

a) γ_sat − γ_w b) γ_sat + γ_w c) γ_d d) γ_sat

Answer: a) γ′ = γ_sat − γ_w (submerged/buoyant unit weight)

5Capillary rise in soil causes: MEQ

a) Increase in effective stress above water table b) Decrease in effective stress c) Zero effective stress d) Negative pore pressure only without stress change

Answer: a) Negative pore pressure (suction) increases effective stress in capillary zone.

Chapter 7: Consolidation

1Consolidation is a process of: PYQ

a) Expelling air b) Expelling water under sustained load c) Mechanical densification d) Increasing void ratio

Answer: b) Time-dependent volume reduction due to expulsion of water.

2Terzaghi’s one-dimensional consolidation theory assumes: MEQ

a) Soil is fully saturated b) Flow is one-dimensional c) Darcy’s law is valid d) All of the above

Answer: d) All are assumptions of Terzaghi’s theory.

3Coefficient of consolidation C_v has units of: Numerical

a) m²/year or cm²/s b) m/s c) kN/m² d) Dimensionless

Answer: a) C_v = k / (m_v γ_w) → length²/time

4Primary consolidation is due to: PYQ

a) Expulsion of air b) Expulsion of water c) Plastic readjustment of particles d) Creep

Answer: b) Primary = hydrodynamic (pore pressure dissipation). Secondary = creep.

5Compression index C_c is the slope of: MEQ

a) e–log σ′ curve in virgin compression b) e–σ′ curve c) Void ratio vs time d) Settlement vs time

Answer: a) C_c = Δe / Δlog σ′ (virgin compression line)

6Time factor T_v for 50% consolidation (U=50%) is approximately: Numerical

a) 0.197 b) 0.848 c) 1.0 d) 0.05

Answer: a) T_v ≈ 0.197 for U = 50%

Chapter 8: Shear Strength of Soil

1Mohr-Coulomb failure criterion is: PYQ

a) τ = c + σ tan φ b) τ = c − σ tan φ c) τ = σ tan φ only d) τ = c only

Answer: a) τ_f = c + σ′ tan φ

2For saturated clay under undrained condition, φ_u is approximately: MEQ

a) 0° b) 30° c) 45° d) Equal to φ′

Answer: a) φ_u ≈ 0 for saturated clay in undrained loading (c_u only).

3Direct shear test is suitable for: PYQ

a) Determining both c and φ of sands and clays b) Only clays c) Only residual strength d) Pore pressure measurement

Answer: a) Widely used for both, though has some limitations (failure plane forced).

4Unconfined compression test is a special case of: MEQ

a) Triaxial UU test with σ₃ = 0 b) Direct shear c) Consolidated drained test d) Vane shear

Answer: a) UC test → q_u = 2c_u (σ₃ = 0)

5In triaxial test, the major principal stress at failure is: Numerical

a) Cell pressure + deviator stress b) Cell pressure only c) Deviator stress only d) Pore pressure

Answer: a) σ₁ = σ₃ + (σ₁ − σ₃)

6Shear strength of soil increases with: PYQ

a) Increase in effective normal stress b) Decrease in effective normal stress c) Increase in pore pressure d) None

Answer: a) Higher σ′ → higher shear strength (Coulomb).

Chapter 9: Earth Pressure

1Rankine’s theory of earth pressure assumes: PYQ

a) Smooth vertical wall and horizontal backfill b) Rough wall c) Inclined wall only d) Cohesive soil only

Answer: a) Rankine assumes no wall friction, vertical wall, horizontal backfill.

2Active earth pressure coefficient K_a for φ = 30° is: Numerical

a) 1/3 b) 1/2 c) 1 d) 3

Answer: a) K_a = (1−sinφ)/(1+sinφ) = 1/3 for φ=30°

3Passive earth pressure is greater than active because: MEQ

a) Wall moves towards the soil b) Wall moves away from soil c) Soil is denser d) Water pressure is higher

Answer: a) Passive state develops when wall pushes into the soil (higher resistance).

4Earth pressure at rest K₀ is approximately: PYQ

a) 1 − sin φ b) (1 − sin φ)/(1 + sin φ) c) tan²(45 − φ/2) d) 1

Answer: a) Jaky’s formula: K₀ ≈ 1 − sin φ

5Coulomb’s theory considers: MEQ

a) Wall friction and inclined backfill b) Only horizontal backfill c) No wall friction d) Only cohesive soils

Answer: a) Coulomb’s wedge theory includes wall friction (δ) and backfill inclination.

Chapter 10: Stability of Slopes

1Factor of safety against sliding for infinite slope in cohesionless soil is: PYQ

a) tan φ / tan β b) tan β / tan φ c) c / (γ H) d) 1

Answer: a) FOS = tan φ / tan i (where i = slope angle)

2Swedish circle method is used for: MEQ

a) Finite slopes (mainly cohesive) b) Infinite slopes only c) Retaining walls d) Bearing capacity

Answer: a) Method of slices / Swedish circle for finite slopes.

3Critical height of a vertical cut in pure clay (φ=0) is: Numerical

a) 4c / γ b) 2c / γ c) c / γ d) 8c / γ

Answer: a) H_c = 4c_u / γ (for φ_u = 0)

4Taylor’s stability number is: PYQ

a) c / (γ H F) b) γ H / c c) c / γ d) F / (γ H)

Answer: a) S_n = c / (γ H F_c)

5For a slope, the most critical failure surface is the one with: MEQ

a) Minimum factor of safety b) Maximum factor of safety c) Maximum length d) Steepest angle

Answer: a) The surface giving the lowest FOS is critical.

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Soil Mechanics MCQ Blog • Prepared for Civil Engineering students (GATE / ESE / SSC-JE / University exams)
Questions based on standard concepts & previous year patterns