All Free Civil Engineering MCQs with Answers

• A. Liquid limit plus plastic limit
• B. Plastic limit minus shrinkage limit
• C. Liquid limit minus plastic limit
• D. Liquid limit minus shrinkage limit

Explanation: Plasticity index is PI = LL minus PL, and it represents the range of water content over which the soil remains plastic.

• A. Degree of saturation only
• B. Specific gravity of solids
• C. Plasticity index
• D. Hydraulic gradient

Explanation: Darcy's law gives discharge per unit area as q = ki, where k is the coefficient of permeability and i is the hydraulic gradient.

• A. Effective stress becomes nearly zero
• B. Total stress becomes zero
• C. Pore pressure becomes zero
• D. The soil reaches maximum dry density

Explanation: Quicksand occurs when upward seepage reduces the effective stress to nearly zero, so the soil loses its shear strength.

• A. Soil minerals
• B. Pore air
• C. Pore water
• D. Organic matter

Explanation: Compaction rapidly densifies soil mainly by expelling air from its voids through mechanical energy.

• A. Present pore water pressure
• B. Current total overburden pressure
• C. Current effective overburden pressure
• D. Shear strength at failure

Explanation: In normally consolidated clay, the current effective overburden pressure is the greatest effective stress the soil has experienced.

• A. Cohesive resistance
• B. Volumetric strain
• C. Pore water pressure
• D. Frictional resistance

Explanation: The Mohr-Coulomb equation is τ = c + σ tan φ, where c is cohesion and σ tan φ is frictional resistance.

• A. 0.25
• B. 0.33
• C. 0.50
• D. 1.00

Explanation: Rankine's active coefficient is Ka = (1 − sin φ)/(1 + sin φ). For φ = 30 degrees, Ka = 0.5/1.5, which is approximately 0.33.

• A. A strong bearing layer lies near ground level
• B. The upper soil is weak and a deeper bearing layer is available
• C. The applied load is negligible
• D. The groundwater table is absent

Explanation: Piles are useful when weak or highly compressible near-surface soil cannot safely support the structure.

• A. Volume of solids to total volume
• B. Volume of air to volume of solids
• C. Volume of water to volume of voids
• D. Volume of water to total volume

Explanation: Degree of saturation is the percentage of the void space occupied by water, so it is Vw divided by Vv.

• A. Solid state to semisolid state
• B. Semisolid state to plastic state
• C. Liquid state to gaseous state
• D. Plastic state to liquid state

Explanation: At the liquid limit, soil changes from a plastic consistency to a liquid consistency.

• A. 1.0 and 3.0
• B. 0.5 and 1.0
• C. 3.0 and 5.0
• D. 5.0 and 10.0

Explanation: For well-graded sand, Cu must exceed 6 and Cc must lie from 1 to 3. These limits indicate a broad and suitably distributed range of…

• A. Increases substantially
• B. Decreases substantially
• C. Remains constant
• D. Becomes uniformly graded only

Explanation: Capillary rise is greater in smaller pores, which are commonly associated with finer soil particles.

• A. q = kH Nd/Nf
• B. q = HNf/(kNd)
• C. q = kH Nf/Nd
• D. q = kNd/(HNf)

Explanation: The standard flow-net expression is q = kH(Nf/Nd), where Nf is the number of flow channels and Nd is the number of potential drops.

• A. Shrinkage limit
• B. Liquid limit
• C. Saturation moisture content
• D. Optimum moisture content

Explanation: The optimum moisture content is the water content at which a specified compactive effort produces maximum dry unit weight.

• A. Rate at which consolidation settlement occurs
• B. Ultimate consolidation settlement
• C. Shear strength at failure
• D. Initial specific gravity of solids

Explanation: The coefficient of consolidation combines the effects of permeability and compressibility and controls the time rate of excess pore-water…

• A. 0 degrees
• B. 15 degrees
• C. 30 degrees
• D. 45 degrees

Explanation: In the total-stress analysis of saturated clay under undrained conditions, the apparent cohesion carries the shear resistance and phi is…

• A. 0.33
• B. 1.00
• C. 2.00
• D. 3.00

Explanation: The Rankine passive coefficient is Kp = (1 + sin phi)/(1 - sin phi). For phi equal to 30 degrees, this gives 1.5 divided by 0.5, or…

• A. Columns are widely spaced on strong rock
• B. The soil has high bearing capacity near the surface
• C. Only a single light column carries the structure
• D. Individual footings would cover most of the building area

Explanation: A raft supports many columns over a large area and is useful when individual footings would occupy most of the plan area.

• A. Gs γw/(1 + e)
• B. Gs γw(1 + e)
• C. (Gs + e)γw
• D. eγw/Gs

Explanation: Dry unit weight equals the weight of soil solids divided by the total soil volume.

• A. Liquid state to plastic state
• B. Semisolid state to solid state
• C. Plastic state to semisolid state
• D. Solid state to liquid state

Explanation: At the shrinkage limit, further loss of water does not cause a further reduction in soil volume.