Q: Moisture content determination — which method gives the most accurate routine measurement of soil water content in laboratory practice?

Introduction / Context: Moisture content w strongly affects compaction, strength, and compressibility. Several techniques exist to determine w, trading speed for accuracy. For quality control and index testing, the industry relies on a reference method that provides repeatable, unbiased measurements across soil types (with noted exceptions for organics and gypsum). Given Data / Assumptions: Standard laboratory conditions available with thermostatically controlled oven. Typical mineral soils; special handling may be needed for organic/gypsiferous soils. Representative samples and airtight tins are used. Concept / Approach: The oven-drying method determines water content by removing pore water through evaporation at 105 to 110 degC until constant mass. The calculation is straightforward: w (%) = (M_wet - M_dry) / M_dry * 100. This direct gravimetric approach minimizes systematic error and is adopted in IS/ASTM standards as the reference for calibration of faster field methods. Step-by-Step Solution: 1) Record mass of empty, dry tin (M_tin).2) Add wet soil and record M_wet_total.3) Dry at 105-110 degC to constant mass and record M_dry_total.4) Compute M_wet = M_wet_total - M_tin; M_dry = M_dry_total - M_tin; then w = (M_wet - M_dry)/M_dry * 100. Verification / Alternative check: Replicate tests converge to the same w within small tolerance. Field rapid methods (calcium carbide) can be correlated but typically show greater scatter due to reaction completeness and sample representativeness. Why Other Options Are Wrong: Calcium carbide — fast but sensitive to fines and sealing; gives approximate values. Sand bath and alcohol methods can under/over-dry, leading to bias. Infrared lamp is quicker but less consistent without careful control. Common Pitfalls: Not drying to constant mass; leaving tins open to ambient humidity between weighings; overheating organic soils which decomposes solids and biases w high. Final Answer: oven-drying method

Q: Soil phase relationships — the ratio of the volume of voids to the volume of soil solids in a given soil mass is called:

Introduction / Context: Two foundational volumetric indices in soil mechanics are porosity (n) and void ratio (e). They quantify how much of the total space is empty (air and/or water) relative to solids and are indispensable in consolidation, permeability, and compaction analyses. Correctly distinguishing e from n is vital when converting between laboratory measurements and design parameters. Given Data / Assumptions: Total volume V consists of Vv (voids) and Vs (solids): V = Vv + Vs. Void space may be partially or fully saturated with water. Representative elementary volume is considered. Concept / Approach: The void ratio is defined as e = Vv / Vs. Porosity is defined as n = Vv / V. The two are related by n = e / (1 + e) and e = n / (1 - n). Engineers use e to link volumes to masses through Gs and unit weights, for example in the equation gamma_d = (Gs * gamma_w) / (1 + e) for dry unit weight (symbol meanings as usual). Step-by-Step Solution: 1) Measure or compute Vs and Vv from test data (e.g., core cutter and pycnometer).2) Compute e = Vv / Vs.3) If porosity is needed, compute n = e / (1 + e).4) Apply e in consolidation and permeability correlations as required. Verification / Alternative check: Cross-validate using mass-based measurements: if gamma_d and Gs are known, rearrange gamma_d = (Gs * gamma_w) / (1 + e) to solve for e and compare with volumetric determination. Why Other Options Are Wrong: Porosity uses total volume V, not solids volume. Specific gravity compares weights of equal volumes of solids and water. Water content is mass of water over mass of dry soil. Degree of saturation is volume of water over volume of voids. Common Pitfalls: Interchanging e and n; mixing inconsistent units; using bulk volumes that include container voids. Final Answer: void ratio

Q: Atterberg limits interpretation — the liquidity index (LI) is defined as which percentage ratio of natural water content w, plastic limit PL, and plasticity index PI?

Introduction / Context: Liquidity index LI situates a cohesive soil’s in-situ water content within its Atterberg limits. It is widely used for qualitative assessment of consistency, workability, and expected undrained strength. Values below zero indicate stiff to hard soils; values near one indicate very soft behavior approaching liquid limit. Given Data / Assumptions: w is the natural water content in percent. PL is the plastic limit determined by roll thread test. PI is the plasticity index defined as PI = LL - PL, where LL is the liquid limit. Concept / Approach: The standard definition is LI (%) = (w - PL) / PI * 100. This normalizes the natural water content relative to the plastic range. If LI = 0, w equals PL; if LI = 1, w equals LL. Practitioners correlate LI with field behavior (e.g., hand vane shear) and with settlement potential under loading. Step-by-Step Solution: 1) Compute PI = LL - PL from Atterberg tests.2) Subtract PL from the natural water content: w - PL.3) Divide by PI to obtain LI (fractional), then multiply by 100 for percent.4) Interpret LI: negative (semi-solid), 0-1 (plastic to soft), greater than 1 (very soft/liquid). Verification / Alternative check: The consistency index Ic is complementary: Ic = (LL - w) / PI. Note that LI + Ic = 1 when expressed as fractions, providing a cross-check between reported values. Why Other Options Are Wrong: Options reversing signs or adding PL do not reflect the definition. Using LL - w corresponds to Ic, not LI. Common Pitfalls: Mixing percentage and decimal forms; using disturbed samples with moisture loss before testing; misreporting PI due to inconsistent LL/PL determinations. Final Answer: natural water content minus plastic limit, divided by plasticity index

Q: Rankine active condition — the active earth pressure of a cohesionless soil is proportional to which expression in terms of the friction angle phi?

Introduction / Context: For retaining walls with granular backfill and negligible wall friction, Rankine’s theory provides a simple coefficient relating horizontal and vertical stresses at failure. The active earth pressure coefficient Ka determines the magnitude of lateral pressure when the wall yields sufficiently away from the soil to mobilize active conditions. Given Data / Assumptions: Backfill is level and cohesionless (c = 0). Wall friction and wall batter are neglected (pure Rankine state). phi is the soil’s angle of shearing resistance. Concept / Approach: Rankine’s expression for the active coefficient is Ka = tan^2(45 - phi/2) which is algebraically equal to (1 - sin phi)/(1 + sin phi). As phi increases, Ka decreases, reflecting the greater shear resistance and lower lateral thrust exerted by denser, more interlocked grains. The active pressure at depth z is then sigma_h = Ka * gamma * z (plus any surcharge term Ka * q). Step-by-Step Solution: 1) Recall Ka = tan^2(45 - phi/2).2) For checks, convert to the sine form Ka = (1 - sin phi)/(1 + sin phi).3) Compute sigma_h = Ka * sigma_v for a given depth if needed.4) Compare with passive coefficient Kp = tan^2(45 + phi/2) for design envelopes. Verification / Alternative check: Check limiting cases: phi = 0 gives Ka = 1 (fluid-like); higher phi gives progressively smaller Ka, consistent with observed behavior of dense sands. Why Other Options Are Wrong: tan^2(45 + phi/2) corresponds to Kp (passive). Linear tan forms lack the square and are dimensionally inconsistent for the Rankine coefficients. cot^2(45 - phi/2) is the inverse, not Ka. Common Pitfalls: Confusing active and passive states; forgetting that surcharge contributes Ka * q uniformly; applying Rankine where wall friction or sloping backfills require Coulomb theory. Final Answer: tan^2(45 - phi/2)

Q: Soil type comparisons for bearing and strength — identify the set of correct statements about composition, shear strength, and safe bearing among listed ground materials:

Introduction / Context: Preliminary foundation decisions often rely on broad soil descriptions and expected engineering behavior. Knowing which loam has more sand or silt, and recognizing relative bearing capacities of common natural strata, helps in site reconnaissance and in evaluating desk-study data before detailed testing. Given Data / Assumptions: “Sandy clay loam” indicates a clay loam with higher sand fraction. “Silty clay loam” indicates a clay loam with higher silt fraction. “Boulder clay” is a stiff glacial till with gravel/boulder inclusions and strong matrix. “Soft chalk” is weak, porous carbonate rock, often low in safe bearing without improvement. Concept / Approach: Textural names directly reference dominant fractions: sandy variants have relatively more sand; silty variants more silt. Stiff boulder clay typically has high undrained shear strength due to dense, overconsolidated matrix and gravel/boulder reinforcement. Soft chalk, due to its low density, fissuring, and crushability, usually provides lower safe bearing pressure unless improved or confined. Step-by-Step Solution: Relate names to dominant fractions: “sandy” → more sand; “silty” → more silt.Recall qualitative strengths: stiff overconsolidated tills > soft weak rocks like chalk for bearing.Synthesize: each statement reflects standard qualitative behavior.Therefore, select “All of the above”. Verification / Alternative check: Typical bearing-capacity tables list higher allowable pressures for stiff clays/tills and lower for soft chalk; agricultural soil texture triangles also support the fraction-based naming. Why Other Options Are Wrong: Choosing any single statement would omit other true facts; the question asserts multiple correct facts. Common Pitfalls: Confusing “clay loam” with “loamy clay”; assuming chalk is always strong rock (its softness and microstructure matter). Final Answer: All of the above

Question 1

A soil has bulk density 2.30 g/cm3 and water content 15 per cent, the dry density of the sample, is

Accepted Answer

2.0 g/cm3

Question 2

A coarse-grained soil has a voids ratio 0.75, and specific gravity as 2.75. The critical gradient at which quick sand condition occurs, is

Accepted Answer

1.00

Question 3

Rankine's theory of active earth pressure assumes

Accepted Answer

all the above.

Question 4

Back fill with a sloping surface exerts a total active pressure Pa on the wall of height H and acts at

Accepted Answer

H/3 above the base parallel to base

Question 5

Degree of saturation from compaction data: A compacted soil has moisture content w = 10% and bulk density rho = 2.0 g/cm^3. Given Gs = 2.7 and rho_w = 1.0 g/cm^3, determine the degree of saturation S.

Accepted Answer

Introduction / Context: The degree of saturation S indicates how much of a soil’s void space is filled with water. It is a critical parameter for compaction control, permeability assessment, and predicting volume change behavior. From routine compaction data (bulk density and water content), along with specific gravity, S can be calculated using standard index relationships. Given Data / Assumptions: w = 10% = 0.10. rho = 2.0 g/cm^3 (bulk density). Gs = 2.7, rho_w = 1.0 g/cm^3. Representative sample with no unusual chemical effects. Concept / Approach: Use two key relations. First, rho_d = rho / (1 + w). Second, rho_d = (Gs / (1 + e)) * rho_w, which gives 1 + e = (Gs * rho_w) / rho_d. With e known, use w = (S * e) / Gs, so S = (w * Gs) / e. Step-by-Step Solution: Compute dry density: rho_d = rho / (1 + w) = 2.0 / 1.10 = 1.818 g/cm^3.Find void ratio: 1 + e = (Gs * rho_w) / rho_d = (2.7 * 1.0) / 1.818 = 1.485 -> e = 0.485.Compute saturation: S = (w * Gs) / e = (0.10 * 2.7) / 0.485 = 0.556 -> 55.6%. Verification / Alternative check: Back-calculate w using w = (S * e) / Gs = (0.556 * 0.485) / 2.7 = 0.10 = 10%, confirming consistency. Why Other Options Are Wrong: 11.1% and 33.3% are too low given the computed e and w. 69.6% is higher than the value indicated by the mass-volume relationships. None of these is incorrect because 55.6% matches the calculation. Common Pitfalls: Confusing porosity n with void ratio e; here formulas use e. Forgetting to convert percentage to decimal form before substitution. Final Answer: 55.6%

Question 6

Under-reamed piles are generally

Accepted Answer

bored piles

Question 7

Pick up the correct statement from the following:

Accepted Answer

When water table is located somewhat below the base of a footing, the elastic wedge is partly of moist soil and partly of submerged soil, and a suitable reduction factor is used

Question 8

Shallow foundations in geotechnical engineering — the minimum depth D of a footing carrying a heavy load is commonly estimated from bearing considerations using which formula (use q for allowable bearing pressure, gamma for unit weight of soil, and …

Accepted Answer

Introduction / Context: Determining a safe minimum depth for a shallow footing is a classic task in foundation engineering. The depth must be sufficient so that the overburden confinement provides the necessary shear resistance in the underlying soil and the bearing pressure at the base can be carried without shear failure. Empirical design aids often recast Rankine/Prandtl concepts into practical formulas that relate depth to soil strength and allowable bearing pressure. Given Data / Assumptions: Footing is on a cohesionless or effectively drained granular soil where phi governs shear strength. q represents the net allowable bearing pressure at foundation level. gamma is the effective unit weight of soil above the base of the footing. Groundwater effects, surcharge variations, and local factors are either included in q or treated separately. Concept / Approach: For granular soils, a convenient expression for minimum embedment is obtained by equating the available passive confinement from the overburden to the mobilized active tendency beneath the footing edge. Using Rankine relationships, the ratio (1 - sin phi)/(1 + sin phi) appears naturally, and squaring it reflects the conversion to a depth term when expressed against unit weight. This yields a compact depth estimate that increases if q is larger and decreases for stronger soils (higher phi) or denser soils (higher gamma). Step-by-Step Solution: 1) Recognize that deeper embedment increases vertical effective stress sigma_v = gamma * D, improving confinement.2) Use the Rankine form for active/passive ratios: K_a = (1 - sin phi)/(1 + sin phi); K_p = 1/K_a.3) Rearrange customary design relations to isolate D proportional to (q / gamma) * K_a^2.4) Substitute K_a and simplify to D = (q / gamma) * ((1 - sin phi)/(1 + sin phi))^2. Verification / Alternative check: Check limits: if phi increases, the fraction (1 - sin phi)/(1 + sin phi) decreases, so D reduces, matching intuition. If q becomes larger, D increases to supply more confinement. Designers should still verify against bearing capacity (e.g., Terzaghi's equation) and settlement criteria. Why Other Options Are Wrong: tan^2(45 + phi/2) and cot^2(45 - phi/2) correspond to passive trends, not minimum embedment for active control. The inverted ratio ((1 + sin phi)/(1 - sin phi))^2 grows with phi, contradicting expected behavior. sec^2(phi) is not a recognized embedment relation. Common Pitfalls: Using total instead of effective unit weight when groundwater reduces gamma; ignoring surcharge q_s that should be combined with overburden; treating the result as a final design without checking settlement. Final Answer: D = (q / gamma) * ((1 - sin phi) / (1 + sin phi))^2

Question 9

Soil mechanics terminology — the ratio of the weight of a given volume of soil solids to the weight of an equal volume of distilled water at the same temperature is known as:

Accepted Answer

Introduction / Context: Specific gravity of soil solids (Gs) is a cornerstone property used in almost every phase-relationship calculation in geotechnical engineering. It allows conversion between mass/weight and volume for the solids phase and is essential in determinations of unit weights, void ratio, and saturation when combined with water content data. Understanding what Gs represents avoids confusion with other volumetric ratios like porosity and void ratio. Given Data / Assumptions: The comparison is between equal volumes: one of soil solids and one of distilled water. Temperature is the same for both, typically 20 °C, because water density varies with temperature. Only soil solids are considered (voids and moisture are excluded from the solids volume). Concept / Approach: Specific gravity of soil solids is defined by Gs = Ws / Ww for equal volumes, where Ws is the weight of soil solids occupying a reference volume and Ww is the weight of water occupying the same volume. For most mineral soils, Gs ranges from about 2.60 to 2.75, with quartz-rich soils near 2.65. Higher Gs values may indicate heavy minerals; lower values can indicate organics or porous lightweight constituents. Step-by-Step Solution: 1) Fix a reference volume V.2) Measure or infer Ws for V of soil solids and Ww for V of distilled water at the same temperature.3) Compute Gs = Ws / Ww.4) Use Gs with other measurements (e.g., bulk unit weight) to back-calculate void ratio or degree of saturation as needed. Verification / Alternative check: Laboratory methods such as the pycnometer test determine Gs by replacing the unknown solid volume with water and applying mass balances. Cross-checks with known mineralogy (e.g., silica content) can validate expected ranges. Why Other Options Are Wrong: Porosity is Vv/V (voids to total). Void ratio is Vv/Vs (voids to solids). Water content is mass of water to mass of dry solids. Degree of saturation is volume of water to volume of voids. None of these compare equal volumes of solids and water by weight. Common Pitfalls: Including pore water or air in the solids volume; ignoring temperature control for water density; confusing Gs with unit weight ratios. Final Answer: specific gravity

Question 10

Particle-size distribution indices — the coefficient of curvature (Cc, also called coefficient of gradation) for a soil gradation curve is defined as:

Accepted Answer

Introduction / Context: Gradation characteristics govern engineering behavior of sands and gravels, affecting permeability, shear strength, and compaction. Two widely used indices are the coefficient of uniformity (Cu) and the coefficient of curvature (Cc). Correctly defining Cc ensures proper soil classification (e.g., well-graded vs. poorly graded) under systems like IS, ASTM, or USCS. Given Data / Assumptions: D10, D30, and D60 are particle sizes corresponding to 10%, 30%, and 60% finer by weight from the semi-logarithmic gradation curve. Standard sieve analysis has been performed with consistent units. Soil is primarily coarse-grained where gradation indices are most informative. Concept / Approach: The formal definition is Cc = (D30)^2 / (D10 * D60). This index reflects the curvature of the gradation curve around the 30% passing ordinate. For well-graded sands, typical acceptance criteria include Cu >= 6 and 1 <= Cc <= 3; for gravels, Cu >= 4 and 1 <= Cc <= 3. Meeting both Cu and Cc thresholds helps avoid pronounced gaps in particle sizes, promoting denser packing and better drainage control. Step-by-Step Solution: 1) Read D10, D30, D60 accurately from the semi-log plot using straight-line interpolation across log-scale abscissa.2) Compute numerator (D30)^2.3) Compute denominator D10 * D60.4) Evaluate Cc = (D30)^2 / (D10 * D60) and compare with grading criteria to assess well-graded status. Verification / Alternative check: Cross-check Cu = D60 / D10 simultaneously. If both Cu and Cc meet specified ranges, the soil is typically well-graded; otherwise it is poorly graded (uniform) or gap-graded. Why Other Options Are Wrong: Options expressing D10/D60 or D60/D10 represent Cu or its inverse, not Cc. Other algebraic forms lack the squared D30 term central to the curvature definition. Common Pitfalls: Reading D-values off a linear x-axis; mixing units (e.g., mm and microns) between D-values; rounding too aggressively leading to borderline misclassification. Final Answer: Cc = (D30)^2 / (D10 * D60)

Question 11

Seepage force in soils — identify the complete and correct description of seepage force with respect to flow-net geometry and hydraulic parameters.

Accepted Answer

Introduction / Context: Seepage force is the drag exerted by flowing water on the soil skeleton. It influences piping, heave beneath excavations, and stability of filters and drains. Recognizing its direction and how its magnitude scales with hydraulic quantities is central to safe geotechnical design near exits and interfaces. Given Data / Assumptions: Flow is steady and laminar (Darcy’s law applies). A valid flow net exists with orthogonal flow lines and equipotential lines. Unit weight of water is gamma_w; hydraulic gradient is i. Concept / Approach: The seepage force per unit volume is f = i * gamma_w, acting in the direction of flow. Flow lines are by definition perpendicular to equipotential lines; hence the force is perpendicular to equipotentials. The gradient i can be written as delta_h / delta_l for a given flow path length, so for a fixed path the force is proportional to head loss delta_h. Near the exit boundary, i approximates the exit gradient, making the force particularly sensitive there and explaining why critical gradients govern piping risk. Step-by-Step Solution: 1) Note direction: force follows flow lines and is perpendicular to equipotentials.2) Note magnitude: f/volume = i * gamma_w, so it scales with gradient.3) Express gradient: i = delta_h / delta_l; for fixed delta_l, f is proportional to delta_h.4) Recognize local maxima: at the exit, i is high, hence force is high and must be checked against critical gradient. Verification / Alternative check: Compute effective stress under upward seepage: sigma_prime = sigma_total - u - i * gamma_w * z_effective (conceptually). When i approaches the critical gradient i_c = (Gs - 1)/(1 + e), quick conditions occur, validating the link between gradient and destabilizing force. Why Other Options Are Wrong: Each of A, B, C is individually true but incomplete; only “all the above” covers direction and both proportionalities. Common Pitfalls: Confusing flow direction with equipotential orientation; ignoring exit gradients in design; assuming seepage force is negligible in coarse filters without checking i. Final Answer: all the above

Question 12

Refraction of flow lines at soil interfaces — a flow line meets the interface between soils of permeability k1 and k2 making angles theta1 and theta2 with the normal before and after crossing. According to the law of deflection at a boundary between…

Accepted Answer

Introduction / Context: When seepage transitions from one soil layer to another with a different coefficient of permeability, the flow lines “refract” at the interface, similar in spirit to optical refraction. The correct relationship is essential for sketching accurate flownets across stratified deposits and for estimating exit gradients and seepage quantities. Given Data / Assumptions: Steady-state, 2D, Darcian flow in isotropic layers. Angles theta1 and theta2 are measured from the normal to the interface. Continuity of head and discharge holds across the boundary. Concept / Approach: Equate tangential components of specific discharge across the boundary using Darcy’s law: q_t = k * i_t. For a given head drop across an elemental strip, the tangential hydraulic gradient i_t is proportional to tan(theta), leading to k1 * tan(theta1) = k2 * tan(theta2). This is sometimes rearranged as tan(theta1) / tan(theta2) = k2 / k1. The result preserves orthogonality between flow lines and equipotentials while changing the “mesh aspect ratio” as k varies. Step-by-Step Solution: 1) Resolve flow components parallel to the interface on each side.2) Apply Darcy’s law to tangential components: k1 * i_t1 = k2 * i_t2.3) Use flow-net geometry where i_t is proportional to tan(theta) when angles are measured from the normal.4) Obtain k1 * tan(theta1) = k2 * tan(theta2). Verification / Alternative check: Construct a flownet with equal potential drops; the rectangles distort in proportion to permeability contrast, and checking the ratio of tangents against k values confirms the law numerically. Why Other Options Are Wrong: sin/cos forms do not arise from tangential Darcy continuity with angles referenced to the normal; equality of angles ignores permeability contrast; the inverted ratio in option D is simply a rearrangement but is not the standard compact expression. Common Pitfalls: Measuring angles from the tangent rather than the normal; applying the relation to anisotropic media without coordinate transformation. Final Answer: k1 * tan(theta1) = k2 * tan(theta2)

Question 13

Moisture content determination — which method gives the most accurate routine measurement of soil water content in laboratory practice?

Accepted Answer

Introduction / Context: Moisture content w strongly affects compaction, strength, and compressibility. Several techniques exist to determine w, trading speed for accuracy. For quality control and index testing, the industry relies on a reference method that provides repeatable, unbiased measurements across soil types (with noted exceptions for organics and gypsum). Given Data / Assumptions: Standard laboratory conditions available with thermostatically controlled oven. Typical mineral soils; special handling may be needed for organic/gypsiferous soils. Representative samples and airtight tins are used. Concept / Approach: The oven-drying method determines water content by removing pore water through evaporation at 105 to 110 degC until constant mass. The calculation is straightforward: w (%) = (M_wet - M_dry) / M_dry * 100. This direct gravimetric approach minimizes systematic error and is adopted in IS/ASTM standards as the reference for calibration of faster field methods. Step-by-Step Solution: 1) Record mass of empty, dry tin (M_tin).2) Add wet soil and record M_wet_total.3) Dry at 105-110 degC to constant mass and record M_dry_total.4) Compute M_wet = M_wet_total - M_tin; M_dry = M_dry_total - M_tin; then w = (M_wet - M_dry)/M_dry * 100. Verification / Alternative check: Replicate tests converge to the same w within small tolerance. Field rapid methods (calcium carbide) can be correlated but typically show greater scatter due to reaction completeness and sample representativeness. Why Other Options Are Wrong: Calcium carbide — fast but sensitive to fines and sealing; gives approximate values. Sand bath and alcohol methods can under/over-dry, leading to bias. Infrared lamp is quicker but less consistent without careful control. Common Pitfalls: Not drying to constant mass; leaving tins open to ambient humidity between weighings; overheating organic soils which decomposes solids and biases w high. Final Answer: oven-drying method

Question 14

Soil phase relationships — the ratio of the volume of voids to the volume of soil solids in a given soil mass is called:

Accepted Answer

Introduction / Context: Two foundational volumetric indices in soil mechanics are porosity (n) and void ratio (e). They quantify how much of the total space is empty (air and/or water) relative to solids and are indispensable in consolidation, permeability, and compaction analyses. Correctly distinguishing e from n is vital when converting between laboratory measurements and design parameters. Given Data / Assumptions: Total volume V consists of Vv (voids) and Vs (solids): V = Vv + Vs. Void space may be partially or fully saturated with water. Representative elementary volume is considered. Concept / Approach: The void ratio is defined as e = Vv / Vs. Porosity is defined as n = Vv / V. The two are related by n = e / (1 + e) and e = n / (1 - n). Engineers use e to link volumes to masses through Gs and unit weights, for example in the equation gamma_d = (Gs * gamma_w) / (1 + e) for dry unit weight (symbol meanings as usual). Step-by-Step Solution: 1) Measure or compute Vs and Vv from test data (e.g., core cutter and pycnometer).2) Compute e = Vv / Vs.3) If porosity is needed, compute n = e / (1 + e).4) Apply e in consolidation and permeability correlations as required. Verification / Alternative check: Cross-validate using mass-based measurements: if gamma_d and Gs are known, rearrange gamma_d = (Gs * gamma_w) / (1 + e) to solve for e and compare with volumetric determination. Why Other Options Are Wrong: Porosity uses total volume V, not solids volume. Specific gravity compares weights of equal volumes of solids and water. Water content is mass of water over mass of dry soil. Degree of saturation is volume of water over volume of voids. Common Pitfalls: Interchanging e and n; mixing inconsistent units; using bulk volumes that include container voids. Final Answer: void ratio

Question 15

Atterberg limits interpretation — the liquidity index (LI) is defined as which percentage ratio of natural water content w, plastic limit PL, and plasticity index PI?

Accepted Answer

Introduction / Context: Liquidity index LI situates a cohesive soil’s in-situ water content within its Atterberg limits. It is widely used for qualitative assessment of consistency, workability, and expected undrained strength. Values below zero indicate stiff to hard soils; values near one indicate very soft behavior approaching liquid limit. Given Data / Assumptions: w is the natural water content in percent. PL is the plastic limit determined by roll thread test. PI is the plasticity index defined as PI = LL - PL, where LL is the liquid limit. Concept / Approach: The standard definition is LI (%) = (w - PL) / PI * 100. This normalizes the natural water content relative to the plastic range. If LI = 0, w equals PL; if LI = 1, w equals LL. Practitioners correlate LI with field behavior (e.g., hand vane shear) and with settlement potential under loading. Step-by-Step Solution: 1) Compute PI = LL - PL from Atterberg tests.2) Subtract PL from the natural water content: w - PL.3) Divide by PI to obtain LI (fractional), then multiply by 100 for percent.4) Interpret LI: negative (semi-solid), 0-1 (plastic to soft), greater than 1 (very soft/liquid). Verification / Alternative check: The consistency index Ic is complementary: Ic = (LL - w) / PI. Note that LI + Ic = 1 when expressed as fractions, providing a cross-check between reported values. Why Other Options Are Wrong: Options reversing signs or adding PL do not reflect the definition. Using LL - w corresponds to Ic, not LI. Common Pitfalls: Mixing percentage and decimal forms; using disturbed samples with moisture loss before testing; misreporting PI due to inconsistent LL/PL determinations. Final Answer: natural water content minus plastic limit, divided by plasticity index

Question 16

Rankine active condition — the active earth pressure of a cohesionless soil is proportional to which expression in terms of the friction angle phi?

Accepted Answer

Introduction / Context: For retaining walls with granular backfill and negligible wall friction, Rankine’s theory provides a simple coefficient relating horizontal and vertical stresses at failure. The active earth pressure coefficient Ka determines the magnitude of lateral pressure when the wall yields sufficiently away from the soil to mobilize active conditions. Given Data / Assumptions: Backfill is level and cohesionless (c = 0). Wall friction and wall batter are neglected (pure Rankine state). phi is the soil’s angle of shearing resistance. Concept / Approach: Rankine’s expression for the active coefficient is Ka = tan^2(45 - phi/2) which is algebraically equal to (1 - sin phi)/(1 + sin phi). As phi increases, Ka decreases, reflecting the greater shear resistance and lower lateral thrust exerted by denser, more interlocked grains. The active pressure at depth z is then sigma_h = Ka * gamma * z (plus any surcharge term Ka * q). Step-by-Step Solution: 1) Recall Ka = tan^2(45 - phi/2).2) For checks, convert to the sine form Ka = (1 - sin phi)/(1 + sin phi).3) Compute sigma_h = Ka * sigma_v for a given depth if needed.4) Compare with passive coefficient Kp = tan^2(45 + phi/2) for design envelopes. Verification / Alternative check: Check limiting cases: phi = 0 gives Ka = 1 (fluid-like); higher phi gives progressively smaller Ka, consistent with observed behavior of dense sands. Why Other Options Are Wrong: tan^2(45 + phi/2) corresponds to Kp (passive). Linear tan forms lack the square and are dimensionally inconsistent for the Rankine coefficients. cot^2(45 - phi/2) is the inverse, not Ka. Common Pitfalls: Confusing active and passive states; forgetting that surcharge contributes Ka * q uniformly; applying Rankine where wall friction or sloping backfills require Coulomb theory. Final Answer: tan^2(45 - phi/2)

Question 17

Pycnometer calculation — a pycnometer containing 400 g of sand and filled to the top with water weighs 2150 g. The same pycnometer filled only with clean water weighs 1950 g. If the specific gravity of soil solids Gs is 2.5, estimate the water conte…

Accepted Answer

Introduction / Context: The pycnometer test is often used to determine specific gravity and, with appropriate mass balances, to back-calculate moisture content of a granular sample. Here, Gs is given, and two weighed conditions allow us to solve for the mass of water within the sand before dilution, hence its water content w. Given Data / Assumptions: Mass of dry sand solids Ms is unknown; total initial sand mass is 400 g (includes water). Mass of pycnometer + sand + water to top = 2150 g. Mass of pycnometer + water to top (no sand) = 1950 g. Specific gravity of solids Gs = 2.5; density of water rho_w is the reference. Concept / Approach: Let M_p be the mass of empty pycnometer. In the water-only case, mass of water is 1950 - M_p. In the sand case, the solids displace an equivalent volume of water equal to Vs * rho_w, where Vs = Ms / (Gs * rho_w). Balancing masses in the filled-to-top condition yields a simple relation independent of M_p and rho_w, allowing direct solution for Ms and then w. Step-by-Step Solution: 1) Let Ms = mass of dry solids; Mw = mass of water originally in moist sand. Given Ms + Mw = 400.2) In the filled condition with sand, the water mass in the pycnometer equals (1950 - M_p) - displaced_water + Mw_added, but a standard shortcut is: (pyc with sand to top) - (pyc water only) = Ms - (mass of water displaced by solids).3) Therefore: 2150 - 1950 = Ms - displaced_water. So 200 = Ms - displaced_water.4) displaced_water = volume of solids * rho_w = Ms / Gs (since rho_w cancels in consistent units). Thus 200 = Ms - Ms/Gs = Ms * (1 - 1/Gs).5) With Gs = 2.5, 1 - 1/Gs = 1 - 0.4 = 0.6. Hence Ms = 200 / 0.6 = 333.33 g.6) Moisture in the original 400 g = Mw = 400 - Ms = 66.67 g. Water content w (%) = Mw / Ms * 100 = 66.67 / 333.33 * 100 = 20%. Verification / Alternative check: Cross-check by computing solid volume Vs = Ms / (Gs * rho_w) and confirming that the difference in water mass between the two pycnometer fillings equals Ms - Ms/Gs. The numbers are internally consistent. Why Other Options Are Wrong: 5%, 10%, 15% are inconsistent with the derived Ms and Mw values from the displacement relation. 25% would require a larger mass difference than observed. Common Pitfalls: Forgetting that mass of displaced water equals Ms/Gs when using grams with water density as 1 g/cc; mixing up wet mass (400 g) with dry mass Ms; attempting to use pycnometer volume explicitly (not needed here). Final Answer: 20%

Question 18

Atterberg limits in soil mechanics — the minimum water content at which a soil still behaves as a liquid yet just develops a small shear resistance to flow is called:

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Introduction / Context: The consistency of fine-grained soils changes with water content. Atterberg limits define key transition points between the liquid, plastic, semi-solid, and solid states. The threshold at which a soil stops flowing like a viscous liquid and just begins to sustain a minute shear resistance is fundamental for classification and for predicting field behavior such as workability and settlement. Given Data / Assumptions: Soil is fine-grained (silt/clay) and responds to changes in water content. Standard definitions per Atterberg limits are used (liquid limit LL, plastic limit PL, shrinkage limit SL). “Small shearing strength” refers to the barely measurable resistance that differentiates liquid from plastic behavior. Concept / Approach: The liquid limit (LL) is the water content at which soil changes from a liquid-like state to a plastic state. In the lab, LL is determined by standardized methods (e.g., Casagrande cup or cone penetrometer) that correspond to a very low shear strength. This boundary is used with the plastic limit to compute plasticity index PI = LL - PL, a key classifier in systems like USCS and BIS/IS codes. Step-by-Step Solution: Identify the state described: “retains liquid state yet has small shear resistance”.Match with the Atterberg definition: this is precisely the liquid limit.Recall that plastic limit refers to the transition from plastic to semi-solid, not liquid.Conclude the correct term is liquid limit. Verification / Alternative check (if short method exists): In practice, the cone penetrometer LL test targets a specific penetration at which the soil’s undrained shear strength is about 1.7–2.0 kPa, aligning with the “small shear strength” description. Why Other Options Are Wrong: Plastic limit — boundary between plastic and semi-solid, not liquid behavior. Shrinkage limit — boundary below which further drying does not reduce volume. Permeability limit/consistency limit — not standard Atterberg terms defining this transition. Common Pitfalls (misconceptions, mistakes): Confusing LL with PL; assuming LL is a fixed property independent of test method; ignoring sensitivity of LL to sample remolding and salts. Final Answer: liquid limit

Question 19

Coulomb's wedge theory — the direction of movement of the active earth-pressure wedge relative to the retaining wall is:

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Introduction / Context: Retaining walls supporting backfill are subjected to earth pressures that depend on the wall movement. Under active conditions, the wall yields sufficiently away from the soil to mobilize a lower lateral pressure. Coulomb's wedge theory models the backfill behind the wall as a potential sliding wedge bounded by a planar failure surface and evaluates equilibrium to find the extreme (active/passive) pressures. Given Data / Assumptions: Backfill approximated as a rigid wedge with planar failure surface. Active condition: wall moves away from backfill. Frictional soil with angle of shearing resistance phi. Resultant of soil weight and reactions drives motion along the failure plane. Concept / Approach: Under active conditions, the soil mass seeks to expand laterally, and the failure wedge tends to slide downwards and outwards along a slip plane inclined to the horizontal. The retaining wall moves away from the backfill, and the wedge moves toward the wall’s back face while descending due to gravity, consistent with Coulomb’s free-body analysis. Step-by-Step Solution: Visualize the wedge bounded by the wall back face and a planar slip surface in the backfill.For active conditions, reduce lateral confinement by moving the wall outward; the wedge then seeks a lower-energy configuration.Gravity components resolve along the plane, causing sliding downward and outward.Hence, the motion direction is down and outwards on the slip surface. Verification / Alternative check: Compare with passive state: for passive pressure, the wall pushes into the soil and the wedge tends to move upward and inward, opposite to the active case — reinforcing the active-direction conclusion. Why Other Options Are Wrong: Up and inwards — characteristic of passive, not active conditions. Horizontal movements parallel to base neglect gravitational component along the slip plane. Pure rotation about the heel is not Coulomb’s assumed wedge mechanism. Common Pitfalls: Confusing active with passive wedge kinematics; ignoring wall friction and backfill slope effects when applying Coulomb formulas. Final Answer: down and outwards on a slip surface

Question 20

Soil type comparisons for bearing and strength — identify the set of correct statements about composition, shear strength, and safe bearing among listed ground materials:

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Introduction / Context: Preliminary foundation decisions often rely on broad soil descriptions and expected engineering behavior. Knowing which loam has more sand or silt, and recognizing relative bearing capacities of common natural strata, helps in site reconnaissance and in evaluating desk-study data before detailed testing. Given Data / Assumptions: “Sandy clay loam” indicates a clay loam with higher sand fraction. “Silty clay loam” indicates a clay loam with higher silt fraction. “Boulder clay” is a stiff glacial till with gravel/boulder inclusions and strong matrix. “Soft chalk” is weak, porous carbonate rock, often low in safe bearing without improvement. Concept / Approach: Textural names directly reference dominant fractions: sandy variants have relatively more sand; silty variants more silt. Stiff boulder clay typically has high undrained shear strength due to dense, overconsolidated matrix and gravel/boulder reinforcement. Soft chalk, due to its low density, fissuring, and crushability, usually provides lower safe bearing pressure unless improved or confined. Step-by-Step Solution: Relate names to dominant fractions: “sandy” → more sand; “silty” → more silt.Recall qualitative strengths: stiff overconsolidated tills > soft weak rocks like chalk for bearing.Synthesize: each statement reflects standard qualitative behavior.Therefore, select “All of the above”. Verification / Alternative check: Typical bearing-capacity tables list higher allowable pressures for stiff clays/tills and lower for soft chalk; agricultural soil texture triangles also support the fraction-based naming. Why Other Options Are Wrong: Choosing any single statement would omit other true facts; the question asserts multiple correct facts. Common Pitfalls: Confusing “clay loam” with “loamy clay”; assuming chalk is always strong rock (its softness and microstructure matter). Final Answer: All of the above

Soil Mechanics and Foundation Engineering Questions