Foundation Engineering MCQ Practice Set — 19 Questions with Answers
19 exam-oriented Foundation Engineering multiple-choice questions with the correct answer and a clear explanation for each. Frequently asked in AE Level Civil Engineering, JE Level Civil Engineering. Solve the full set below for free — no login required.
- 1Original practiceMEDIUM
If water volume is 0.2 m³ and void volume is 0.5 m³, degree of saturation is
A20.0%B25.0%C60.0%D40.0%Answer: D. 40.0%
Explanation: Degree of saturation S = (Vw / Vv) × 100%, where Vw is volume of water and Vv is volume of voids. S = (0.2 / 0.5) × 100 = 40.0%. S = 0% means dry soil; S = 100% means fully saturated (all voids filled with water).
- 2Original practiceMEDIUM
For a soil mass, volume of voids is 0.5 m³ and total volume is 1.0 m³. Porosity is
A75.0%B25.0%C100.0%D50.0%Answer: D. 50.0%
Explanation: Porosity n = (Vv / V) × 100%, where Vv is volume of voids and V is total volume of soil mass. n = (0.5 / 1.0) × 100 = 50.0%. Porosity indicates the fraction of total soil volume occupied by voids (air + water). Typical values: gravel 25–40%, sand 30–45%, clay 40–60%.
- 3Original practiceMEDIUM
A soil has water weight 10 N and dry solid weight 100 N. Water content is
A10.0%B9.1%C5.0%D20.0%Answer: A. 10.0%
Explanation: Gravimetric water content w = (Ww / Ws) × 100%, where Ww is weight of water and Ws is weight of dry soil. w = (10 / 100) × 100 = 10.0%. Water content is always expressed as a percentage of dry weight. Natural soils range from ~5% (arid) to >100% (soft clays).
- 4Original practiceMEDIUM
A soil sample weighs 30 kN and occupies 1.0 m³. Bulk unit weight is
A30 kN/m³B31 kN/m³C15 kN/m³D60 kN/m³Answer: A. 30 kN/m³
Explanation: Bulk (wet) unit weight γ = W / V, where W is total weight of the soil mass and V is total volume. γ = 30 / 1.0 = 30 kN/m³ kN/m³. Bulk unit weight includes weight of solids plus pore water. Typical values: loose sand ~16 kN/m³, dense sand ~20 kN/m³, saturated clay ~18–22 kN/m³.
- 5Original practiceMEDIUM
Using Darcy law, k=0.0001 m/s, hydraulic gradient=0.5 and area=2.0 m². Discharge is
A0.0002 m³/sB1e-05 m³/sC0.001 m³/sD0.0001 m³/sAnswer: D. 0.0001 m³/s
Explanation: Darcy's law: Q = k * i * A, where k = coefficient of permeability (m/s), i = hydraulic gradient (= head loss / flow length, dimensionless), A = cross-sectional area of flow (m^2). Valid for laminar flow through saturated porous media (Re_porous < 1). Not valid for gravels or at high gradients (turbulent seepage). Q=k i A=0.0001×0.5×2.0=0.0001 m³/s.
- 6PYQ/PYQ-PatternMEDIUM
A soil sample has volume of voids 0.6 m³ and volume of solids 1.0 m³. Void ratio is
A1.60B1.67C0.37D0.60Answer: D. 0.60
Explanation: Void ratio e = V_v / V_s (dimensionless), where V_v = volume of voids (air + water) and V_s = volume of soil solids. Related to porosity: n = e/(1+e). Typical e values: dense gravel 0.25–0.45; loose sand 0.55–0.90; soft clay 1.5–3.0; peat > 3.0. Void ratio influences permeability, compressibility, and shear strength. Void ratio e=Vv/Vs=0.6/1.0=0.60.
- 7Foundation EngineeringMEDIUM
The additional pile driven to increase the capacity of supporting loads on vertical pile is known as:
Asinking pileBbatter pileCeccentric pileDsimplex pileAnswer: B. batter pile
Explanation: A batter pile (or raking pile) is driven at an inclination to the vertical to resist horizontal or inclined forces, thereby increasing the lateral load-carrying capacity of the pile group.
- 8Foundation EngineeringMEDIUM
The maximum settlement of raft foundation on sand should be limited to the following values:
A20 to 40 mmB40 to 65 mmC65 to 80 mmD80 to 100 mmAnswer: B. 40 to 65 mm
Explanation: According to IS 1904, the permissible settlement for raft foundations on sand is typically 40-65 mm, whereas 20-40 mm is often cited for isolated footings on sand.
- 9Eccentric FootingsHARD
For a rectangular footing of width B subjected to a vertical load with one-way eccentricity e, the soil pressure remains compressive over the whole base only if e does not exceed:
AB/3BB/6CB/2DB/12Answer: B. B/6
Explanation: The resultant must lie within the middle third of the base to avoid tension at the soil-footing contact. For one-way eccentricity, e <= B/6.
- 10Soil Mechanics and Foundation EngineeringEASY
Well foundation is commonly used for:
APavement shouldersBRoof slabsCLight partition wallsDBridge piers in riversAnswer: D. Bridge piers in rivers
Explanation: Wells provide deep foundations resistant to scour and lateral forces in river bridges.
- 11Well Foundation SteiningMEDIUM
The minimum thickness of steining (well wall) for a well foundation is governed by:
ADepth of well below scour level onlyBStructural requirements (hoop stress during sinking and service loads) and the need to provide adequate self-weight for sinking through soilCNumber of compartmentsDRiver widthAnswer: B. Structural requirements (hoop stress during sinking and service loads) and the need to provide adequate self-weight for sinking through soil
Explanation: Well foundation steining thickness: minimum t = KD (D = external diameter, K = 0.08 for stone, 0.05 for RCC). Must be checked for: (1) bending during sinking (steining acts as a cylinder shell under lateral soil/water pressure); (2) hoop tension; (3) dead weight needed to sink without kentledge. IS 3955 gives design guidelines.
- 12SPT N Value CorrectionMEDIUM
The Standard Penetration Test (SPT) N value is corrected for overburden pressure using the relation (Peck, 1974):
AN_c = N (no correction needed)BN_c = 0.77 log(2000/sigma_v) x N where sigma_v is effective overburden (kPa), applicable for sigma_v > 25 kPaCN_c = N x depthDN_c = N / 2Answer: B. N_c = 0.77 log(2000/sigma_v) x N where sigma_v is effective overburden (kPa), applicable for sigma_v > 25 kPa
Explanation: SPT N correction for overburden: Peck correction: N_c = 0.77 log(2000/sigma_v) x N for sigma_v > 25 kPa (no correction for sigma_v = 1 ton/ft2 = 100 kPa). Also water table correction: if water table is within B above footing: N_corrected = 0.5 N + 15 (Terzaghi). Corrected N used for bearing capacity and settlement.
- 13Pile Group EfficiencyMEDIUM
The Converse-Labarre formula for efficiency of a pile group with n rows x m columns, diameter d, and center-to-center spacing s is:
AEg = 1 - (theta/90) x [(n-1)m + (m-1)n] / (mn)BEg = 1.0 alwaysCEg = n x m x individual pile capacityDEg = s/d ratioAnswer: A. Eg = 1 - (theta/90) x [(n-1)m + (m-1)n] / (mn)
Explanation: Converse-Labarre: Eg = 1 - (theta/90) x [(n-1)m + (m-1)n] / (mn), where theta = arctan(d/s) in degrees, m = number of columns, n = number of rows. Typical Eg = 0.6-0.8. Group capacity = Eg x n x m x individual pile capacity (or lesser of group block failure).
- 14Negative Skin FrictionMEDIUM
Negative skin friction (NSF) on a pile occurs when:
AThe pile is in tensionBSurrounding soil settles more than the pile, causing downward drag on the pile shaft, which ADDS to the axial load on the pileCThe pile is too longDGroundwater table is very highAnswer: B. Surrounding soil settles more than the pile, causing downward drag on the pile shaft, which ADDS to the axial load on the pile
Explanation: NSF (drag-down force): occurs in piles passing through consolidating soft clay overlying firm bearing stratum. The settling soil drags the pile down, adding to the structural load. NSF = K_s x sigma_v x tan(delta) x perimeter x length of settling layer. Must be subtracted from pile capacity in design.
- 15GeneralMEDIUM
A soil sample has liquid limit 45% and plastic limit 25%. Its plasticity index is
A20%B1.80%C70%D15%Answer: A. 20%
Explanation: Plasticity index PI = LL - PL = 45 - 25 = 20%.
- 16Soil Mechanics and Foundation EngineeringHARD
Terzaghi's ultimate bearing capacity for a strip footing includes the surcharge term:
A0.5 gamma B N_gammaBqNqCcNcDgamma Df/NqAnswer: B. qNq
Explanation: The three main terms for strip footing are cNc + qNq + 0.5 gamma B N_gamma; qNq is the surcharge contribution.
- 17Foundation EngineeringMEDIUM
According to Terzaghi's theory, the ultimate bearing capacity at ground surface for a purely cohesive soil and smooth base of strip footing is
A2.57 CB5.14 CC5.7 CD6.2 CAnswer: B. 5.14 C
Explanation: For phi = 0 and smooth strip footing, Nc = 5.14, hence qu = 5.14C
- 18Foundation EngineeringMEDIUM
The minimum centre-to-centre distance between piles of a pile group in clay should be equal to
Adiameter of the pileBtwice the diameter of pileC3 times the diameter of pileDnone of the aboveAnswer: C. 3 times the diameter of pile
Explanation: Usual minimum spacing for piles in clay is about 3 times the pile diameter
- 19Foundation EngineeringMEDIUM
The ultimate bearing capacity of a strip footing resting on clay compared to a square footing of same size is:
AmoreBlessCequalDcannot be predictedAnswer: B. less
Explanation: For clay under undrained condition, the square footing has a higher shape factor; hence strip footing capacity is lower