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NICET Geotech practice questions

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Question 1Civil Engineering Technology - Soils (CMT-Soils)

A technician obtains a moist soil sample and oven-dries it. The mass of the moist sample is 250.0 g and the dry mass is 210.0 g. What is the water content?

  1. 19.0 percent Correct answer.
  2. 8.4 percent
  3. 16.0 percent
  4. 84.0 percent

Explanation

Water content is the mass of water divided by the mass of solids, expressed as a percentage. The water mass is 250.0 minus 210.0, or 40.0 g, and 40.0 divided by 210.0 equals 0.190, or 19.0 percent. Dividing by the total moist mass instead of the dry solids mass is the common error that produces 16 percent.

Question 2Civil Engineering Technology - Soils (CMT-Soils)

In the Unified Soil Classification System, which group symbol is assigned to a clean well-graded gravel?

  1. GM
  2. GC
  3. GP
  4. GW Correct answer.

Explanation

The USCS uses the letter W for well graded and G for gravel, so GW denotes a well-graded gravel with little or no fines. GP denotes a poorly graded gravel, while the M and C letters indicate significant silt and clay content respectively.

Question 3Civil Engineering Technology - Soils (CMT-Soils)

A sample of saturated soil has a natural water content of 40 percent and a specific gravity of solids of 2.70. What is its void ratio?

  1. 0.74
  2. 2.70
  3. 1.08 Correct answer.
  4. 0.40

Explanation

For a saturated soil, the phase relationship reduces to e equals w times Gs, because the volume of voids is filled entirely with water. Multiplying 0.40 by 2.70 gives 1.08, which is the void ratio.

Question 4Civil Engineering Technology - Soils (CMT-Soils)

A standard Proctor compaction test is being performed on a silt. As molding water content increases past the optimum, what happens to the compacted dry unit weight?

  1. It remains constant once the soil is fully saturated.
  2. It decreases because the increasing water occupies space that would otherwise hold soil solids. Correct answer.
  3. It increases sharply because the soil becomes impermeable.
  4. It continues to increase because water lubricates the particles.

Explanation

On the wet side of optimum, additional water fills voids and displaces soil solids per unit volume, so the dry unit weight falls. The lubrication benefit that raises dry unit weight applies on the dry side of optimum only, and the curve is not flat beyond saturation.

Question 5Civil Engineering Technology - Soils (CMT-Soils)

A soil sample is tested in a direct shear box. The normal stress is 100 kPa and the measured shear stress at failure is 60 kPa. Using a cohesion intercept of zero, what is the angle of internal friction?

  1. 53 degrees
  2. 37 degrees
  3. 31 degrees Correct answer.
  4. 60 degrees

Explanation

For a cohesionless soil the Mohr-Coulomb envelope gives shear stress equal to normal stress times the tangent of the friction angle. The tangent is 60 divided by 100, or 0.60, and the arctangent of 0.60 is about 31 degrees.

Question 6Civil Engineering Technology - Soils (CMT-Soils)

A sieve analysis reports that 12 percent of a sand passes the No. 200 sieve. How is this material described under the USCS regarding fines?

  1. Clean, since less than 5 percent passes the No. 200 sieve is required for that descriptor
  2. With fines, since more than 12 percent passes the No. 200 sieve
  3. Clean, since less than 50 percent passes the No. 200 sieve
  4. With fines, since more than 5 percent passes the No. 200 sieve Correct answer.

Explanation

The USCS treats a coarse-grained soil as clean when less than 5 percent passes the No. 200 sieve and as containing fines when 5 percent or more passes. At 12 percent passing, the material is classified as a sand with fines.

Question 7Civil Engineering Technology - Soils (CMT-Soils)

A field density test using the sand cone method yields a dry unit weight of 108 pcf. A laboratory standard Proctor test on the same soil gives a maximum dry unit weight of 120 pcf. What is the relative compaction?

  1. 111 percent
  2. 10 percent
  3. 12 percent
  4. 90 percent Correct answer.

Explanation

Relative compaction is the field dry unit weight divided by the laboratory maximum dry unit weight, expressed as a percentage. Dividing 108 by 120 gives 0.90, or 90 percent.

Question 8Civil Engineering Technology - Soils (CMT-Soils)

Which Atterberg limit is defined as the water content at which a soil, when rolled into a thread, crumbles at a diameter of about 3 mm?

  1. Liquid limit
  2. Liquidity index
  3. Shrinkage limit
  4. Plastic limit Correct answer.

Explanation

The plastic limit is the water content at which a soil thread crumbles when rolled to about 3 mm diameter, marking the boundary between plastic and semisolid behavior. The liquid limit is measured with the Casagrande device or fall cone, and the shrinkage limit marks the boundary where further drying stops reducing volume.

Question 9Civil Engineering Technology - Soils (CMT-Soils)

A clay has a liquid limit of 55 and a plastic limit of 20. What is the plasticity index, and how is the soil described?

  1. PI of 2.75, a low-plasticity silt
  2. PI of 35, a high-plasticity clay Correct answer.
  3. PI of 35, a low-plasticity clay
  4. PI of 75, a high-plasticity clay

Explanation

The plasticity index is the liquid limit minus the plastic limit, so 55 minus 20 gives 35. Under the USCS, a PI of 35 combined with a liquid limit of 55 places the soil in the high-plasticity range, so it is described as a high-plasticity clay.

Question 10Civil Engineering Technology - Soils (CMT-Soils)

In a constant head permeability test, which change would double the measured coefficient of permeability if all other conditions remained the same?

  1. Doubling the length of the specimen
  2. Halving the volume of water collected
  3. Doubling the head difference across the specimen Correct answer.
  4. Doubling the cross-sectional area of the specimen

Explanation

Darcy's law in the constant head form gives k proportional to the head difference and specimen length and inversely to area. Doubling the head difference doubles the flow rate through the same specimen, which doubles the computed permeability. Changing the specimen area or length alters the flow but not the material property in the way the formula implies for a fixed head.

Question 11Civil Engineering Technology - Soils (CMT-Soils)

A stratum of soil has a hydraulic conductivity of 0.002 cm/s and a hydraulic gradient of 0.5. What is the seepage velocity through the voids if the porosity is 0.40?

  1. 0.001 cm/s
  2. 0.0025 cm/s Correct answer.
  3. 0.25 cm/s
  4. 0.10 cm/s

Explanation

Darcy velocity equals conductivity times gradient, giving 0.001 cm/s, and seepage velocity equals Darcy velocity divided by porosity. Dividing 0.001 by 0.40 gives 0.0025 cm/s, which is the average velocity through the pore space.

Question 12Civil Engineering Technology - Soils (CMT-Soils)

A saturated normally consolidated clay is loaded in a consolidation test. Which quantity is used to describe the rate of consolidation?

  1. Compression index
  2. Recompression index
  3. Preconsolidation pressure
  4. Coefficient of consolidation Correct answer.

Explanation

The coefficient of consolidation governs how quickly excess pore pressure dissipates and thus the rate of settlement with time. The compression and recompression indices describe the magnitude of volume change, and preconsolidation pressure describes stress history rather than rate.

Question 13Civil Engineering Technology - Soils (CMT-Soils)

A clay layer is 4 m thick and drains at both the top and bottom. What drainage path length should be used in the time factor calculation?

  1. 2 m Correct answer.
  2. 1 m
  3. 4 m
  4. 8 m

Explanation

For double drainage, the longest path a water particle travels to reach a drainage boundary is half the layer thickness. Half of 4 m is 2 m, which is the drainage path used in the time factor relationship.

Question 14Civil Engineering Technology - Soils (CMT-Soils)

The effective stress at a point in a saturated soil is best described by which relationship?

  1. Total stress minus pore water pressure Correct answer.
  2. Pore water pressure minus total stress
  3. Total stress multiplied by porosity
  4. Total stress plus pore water pressure

Explanation

Terzaghi's principle states that effective stress equals total stress minus pore water pressure, representing the stress carried by the soil skeleton. Pore pressure reduces the intergranular contact stress and therefore reduces shear strength.

Question 15Civil Engineering Technology - Soils (CMT-Soils)

A soil profile has 3 m of saturated sand with a total unit weight of 19 kN per cubic meter overlying a clay. The water table is at the ground surface. What is the vertical effective stress at a depth of 3 m?

  1. 27 kPa Correct answer.
  2. 57 kPa
  3. 30 kPa
  4. 0 kPa

Explanation

Total stress at 3 m is 19 times 3, or 57 kPa, and pore pressure is 9.81 times 3, or about 29.4 kPa. Subtracting pore pressure from total stress gives an effective stress of about 27 kPa.

Question 16Civil Engineering Technology - Soils (CMT-Soils)

A retaining wall holds a dry cohesionless backfill with a unit weight of 18 kN per cubic meter and a friction angle of 30 degrees. Using Rankine theory, what is the active lateral earth pressure coefficient?

  1. 0.50
  2. 1.0
  3. 0.33 Correct answer.
  4. 3.0

Explanation

The Rankine active earth pressure coefficient is the tangent squared of 45 degrees minus half the friction angle. For a friction angle of 30 degrees this is the tangent squared of 30 degrees, or one third.

Question 17Civil Engineering Technology - Soils (CMT-Soils)

A slope in a stiff clay fails along a deep rotational surface. Which factor of safety definition is most appropriate for this analysis?

  1. Cohesion divided by the unit weight of the soil
  2. Effective stress divided by total stress
  3. Sum of vertical forces divided by sum of horizontal forces
  4. Resisting moment divided by driving moment about the center of rotation Correct answer.

Explanation

For a rotational slip surface, the factor of safety is the ratio of the resisting moment provided by shear strength along the arc to the driving moment from the soil weight. Force ratios apply to translational or wedge analyses rather than rotational ones.

Question 18Civil Engineering Technology - Soils (CMT-Soils)

A standard penetration test records 18 blows for the second and third 6 inch increments in a medium sand. What is the reported N-value?

  1. 12
  2. 36
  3. 6
  4. 18 Correct answer.

Explanation

The standard penetration resistance is the sum of the blow counts over the second and third 6 inch increments, excluding the seating increment. Those two increments total 18 blows, so the reported N-value is 18.

Question 19Civil Engineering Technology - Soils (CMT-Soils)

A sample is recovered from a borehole with a split-spoon sampler driven by the standard drop hammer. Which sample quality does this represent?

  1. Disturbed sample Correct answer.
  2. Block sample
  3. Core sample
  4. Undisturbed sample

Explanation

The split-spoon sampler disturbs the soil significantly and yields a disturbed sample suitable for classification and index testing, not for strength or consolidation testing. Undisturbed samples require thin-walled tube or block sampling methods.

Question 20Civil Engineering Technology - Soils (CMT-Soils)

A fine-grained soil has a liquid limit of 40 and a plastic limit of 22. Its natural water content is 30. What is the liquidity index?

  1. 0.53
  2. 1.22
  3. 0.44 Correct answer.
  4. 0.80

Explanation

The liquidity index is the natural water content minus the plastic limit, divided by the plasticity index. Here the plasticity index is 18 and the difference is 8, so the ratio is about 0.44, indicating a stiff consistency.

Question 21Civil Engineering Technology - Soils (CMT-Soils)

Which laboratory test is used to determine the consolidation characteristics of a saturated clay?

  1. California bearing ratio test
  2. One-dimensional oedometer test Correct answer.
  3. Direct shear test
  4. Los Angeles abrasion test

Explanation

The oedometer, or one-dimensional consolidation test, measures compression of a laterally confined saturated specimen under incrementally applied loads. Direct shear measures shear strength, CBR measures bearing capacity for pavement design, and the abrasion test measures aggregate durability.

Question 22Civil Engineering Technology - Soils (CMT-Soils)

A soil has a dry unit weight of 105 pcf and a specific gravity of solids of 2.65. What is the void ratio?

  1. 1.65
  2. 0.61 Correct answer.
  3. 0.63
  4. 0.38

Explanation

The relationship between dry unit weight and void ratio is dry unit weight equal to the unit weight of water times specific gravity divided by one plus the void ratio. Solving with 62.4 pcf for water gives one plus e equal to about 1.61, so the void ratio is about 0.61.

Question 23Civil Engineering Technology - Soils (CMT-Soils)

A soil is classified as CL under the USCS. Which description matches this group symbol?

  1. Clayey sand with high plasticity
  2. Silty gravel with low plasticity
  3. Lean clay with low to medium plasticity Correct answer.
  4. Organic silt with high plasticity

Explanation

In the USCS, C stands for clay and L stands for low to medium liquid limit, so CL denotes a lean clay. The symbol applies to fine-grained soils with a liquid limit below 50 and a plasticity index above the A-line.

Question 24Civil Engineering Technology - Soils (CMT-Soils)

A triaxial compression test is performed on a saturated sand with a confining pressure of 100 kPa. At failure the axial stress is 300 kPa. What is the deviator stress at failure?

  1. 100 kPa
  2. 200 kPa Correct answer.
  3. 300 kPa
  4. 400 kPa

Explanation

The deviator stress is the difference between the axial stress at failure and the confining pressure. Subtracting 100 kPa from 300 kPa gives 200 kPa, which is the additional axial stress applied by the loading ram.

Question 25Civil Engineering Technology - Soils (CMT-Soils)

During a falling head permeability test, which observation indicates that the test setup is leaking?

  1. The collected water volume matches the computed volume
  2. The head drops at exactly the predicted rate
  3. The head remains constant throughout the test
  4. The head drops more slowly than predicted from the calibration curve Correct answer.

Explanation

A leak in the apparatus allows water to escape or enter outside the soil specimen, so the measured head change no longer reflects flow through the soil alone. A slower drop than predicted suggests loss of water from the system rather than flow through the specimen.

Question 26Civil Engineering Technology - Soils (CMT-Soils)

A technician receives a cohesive soil sample for classification testing. The liquid limit is 48 and the plastic limit is 19. What is the plasticity index, and how is the soil described in terms of plasticity?

  1. PI = 19, low plasticity
  2. PI = 29, medium plasticity Correct answer.
  3. PI = 29, high plasticity
  4. PI = 67, high plasticity

Explanation

The plasticity index is the liquid limit minus the plastic limit, so 48 minus 19 equals 29. Under the standard plasticity ranges, a PI between 16 and 30 corresponds to a medium plasticity soil. A PI above 30 would indicate high plasticity.

Question 27Civil Engineering Technology - Soils (CMT-Soils)

A fine-grained soil is classified under the Unified Soil Classification System. Its liquid limit is 60 and its plasticity index is 30, plotting above the A-line. Which group symbol applies?

  1. CL
  2. OH
  3. CH Correct answer.
  4. ML

Explanation

A fine-grained soil with a liquid limit of 50 or greater that plots above the A-line is classified as CH, an inorganic clay of high plasticity. The high liquid limit of 60 places it in the H group, and the position above the A-line confirms clay rather than silt.

Question 28Civil Engineering Technology - Soils (CMT-Soils)

A sample of dry sand has a void ratio of 0.62 and a specific gravity of solids of 2.65. What is the dry unit weight of this soil, using a unit weight of water of 62.4 pcf?

  1. 63 pcf
  2. 166 pcf
  3. 102 pcf Correct answer.
  4. 40 pcf

Explanation

The dry unit weight is computed as the specific gravity of solids times the unit weight of water divided by one plus the void ratio. That gives 2.65 times 62.4 divided by 1.62, which equals about 102 pcf.

Question 29Civil Engineering Technology - Soils (CMT-Soils)

A soil sample is weighed moist and then oven-dried. The moist mass is 486 g and the dry mass is 412 g. What is the water content of the sample?

  1. 17.9 percent
  2. 18.0 percent Correct answer.
  3. 84.8 percent
  4. 15.2 percent

Explanation

Water content is the mass of water divided by the mass of solids, expressed as a percentage. The mass of water is 486 minus 412, or 74 g, and 74 divided by 412 equals about 0.180, or 18.0 percent.

Question 30Civil Engineering Technology - Soils (CMT-Soils)

A compaction test on a silty clay produces a moisture-density curve. Which statement correctly describes the typical effect of increasing compaction effort on that curve?

  1. Neither the maximum dry unit weight nor the optimum water content changes.
  2. The maximum dry unit weight decreases and the optimum water content increases.
  3. Both the maximum dry unit weight and the optimum water content increase.
  4. The maximum dry unit weight increases and the optimum water content decreases. Correct answer.

Explanation

Greater compaction effort packs the soil particles more closely at a given water content, raising the peak dry unit weight. Because the soil becomes more workable at lower water content with more energy, the optimum water content shifts to a lower value.

Question 31Civil Engineering Technology - Soils (CMT-Soils)

A field density test reports a dry unit weight of 112 pcf. Laboratory compaction of the same soil gives a maximum dry unit weight of 120 pcf at optimum water content. What is the relative compaction of the field soil?

  1. 96.0 percent
  2. 85.7 percent
  3. 107 percent
  4. 93.3 percent Correct answer.

Explanation

Relative compaction is the field dry unit weight divided by the laboratory maximum dry unit weight, expressed as a percentage. Dividing 112 by 120 gives 0.933, or 93.3 percent.

Question 32Civil Engineering Technology - Soils (CMT-Soils)

A specimen of saturated clay is subjected to a consolidated-undrained triaxial test. Which quantity is measured directly as the difference between the major and minor principal stresses at failure?

  1. The effective cohesion
  2. The pore pressure parameter
  3. The deviator stress Correct answer.
  4. The consolidation settlement

Explanation

The deviator stress is the additional axial stress applied beyond the confining pressure, equal to the major principal stress minus the minor principal stress at failure. It is the directly controlled and measured quantity in a triaxial test.

Question 33Civil Engineering Technology - Soils (CMT-Soils)

A direct shear test is run on a dry sand at a normal stress of 100 kPa, and the soil fails at a shear stress of 58 kPa. What is the friction angle of this sand?

  1. 35 degrees
  2. 45 degrees
  3. 25 degrees
  4. 30 degrees Correct answer.

Explanation

For a cohesionless soil the Mohr-Coulomb envelope passes through the origin, so the tangent of the friction angle equals the shear stress divided by the normal stress. That ratio is 58 divided by 100, or 0.58, and the arctangent of 0.58 is about 30 degrees.

Question 34Civil Engineering Technology - Soils (CMT-Soils)

A sample of stiff clay is loaded in a consolidation test and the void ratio is plotted against the logarithm of pressure. Which feature of the resulting curve indicates that the soil is overconsolidated?

  1. A recompression curve that is flatter than the virgin line Correct answer.
  2. A horizontal line across all pressures
  3. A curve that rises steadily with pressure
  4. A straight virgin compression line at high pressure

Explanation

An overconsolidated clay first follows a flatter recompression curve until the preconsolidation pressure is reached, after which it follows the steeper virgin compression line. The presence of that flatter initial portion is the signature of prior loading.

Question 35Civil Engineering Technology - Soils (CMT-Soils)

A saturated clay stratum is 6 m thick and drains only at its top surface because the base is impermeable rock. What drainage path length should be used in the time factor for consolidation?

  1. 1.5 m
  2. 6 m Correct answer.
  3. 12 m
  4. 3 m

Explanation

For single drainage the longest drainage path equals the full thickness of the layer, because water must travel the entire thickness to reach the draining boundary. For double drainage the path would be half the thickness.

Question 36Civil Engineering Technology - Soils (CMT-Soils)

A soil sample is tested in the laboratory and the results show a liquid limit of 35, a plastic limit of 18, and a natural water content of 30. What is the liquidity index, and what does it indicate about consistency?

  1. LI = 0.71, soft consistency
  2. LI = 0.71, stiff consistency Correct answer.
  3. LI = 0.35, medium consistency
  4. LI = 1.41, soft consistency

Explanation

The liquidity index is the natural water content minus the plastic limit, divided by the plasticity index. Here the plasticity index is 17, and (30 minus 18) divided by 17 equals about 0.71. A liquidity index below 1 with the water content closer to the plastic limit indicates a stiffer, more workable consistency.

Question 37Civil Engineering Technology - Soils (CMT-Soils)

A falling head permeability test is performed on a compacted clay specimen in a permeameter. Which observation most reliably indicates that water is leaking around the sides of the specimen rather than flowing through the soil?

  1. The head drops more rapidly than predicted and the measured permeability is too high. Correct answer.
  2. The head drops more slowly than predicted from the soil's known permeability.
  3. The head remains constant throughout the test.
  4. The water temperature rises during the test.

Explanation

Side leakage provides an additional flow path that bypasses the soil, so water drains faster than it should and the computed permeability is erroneously high. Comparing the measured rate with an expected value exposes the leak.

Question 38Civil Engineering Technology - Soils (CMT-Soils)

A soil sample is classified by the AASHTO system. Its sieve analysis shows 60 percent passing the No. 10 sieve, 45 percent passing the No. 40 sieve, and 20 percent passing the No. 200 sieve. Which AASHTO group does this material most closely fit?

  1. A-1-a Correct answer.
  2. A-2-4
  3. A-6
  4. A-3

Explanation

A-1-a requires a high percentage passing the No. 10 sieve with limited fines, which matches the 60 percent passing No. 10 and 20 percent passing No. 200. The gradation and low fines content fit this granular group better than the silty or clayey groups.

Question 39Civil Engineering Technology - Soils (CMT-Soils)

A standard Proctor compaction test is being performed on a soil. Which procedure correctly prepares the specimen for the test?

  1. Compact the soil in three layers using a 5.5 lb hammer dropped 12 inches. Correct answer.
  2. Compact the soil in one layer using a 5.5 lb hammer dropped 12 inches.
  3. Compact the soil in five layers using a 10 lb hammer dropped 18 inches.
  4. Compact the soil in three layers using a 10 lb hammer dropped 18 inches.

Explanation

The standard Proctor test uses a 5.5 lb rammer dropped 12 inches, with the soil compacted in three equal layers in the standard mold. The heavier 10 lb rammer with an 18 inch drop belongs to the modified Proctor test.

Question 40Civil Engineering Technology - Soils (CMT-Soils)

A soil profile consists of 2 m of dry sand with a total unit weight of 17 kN per cubic meter overlying a saturated clay. The water table is at the top of the clay. What is the total vertical stress at a depth of 5 m below the ground surface?

  1. 85 kPa Correct answer.
  2. 51 kPa
  3. 68 kPa
  4. 102 kPa

Explanation

Total stress is the sum of unit weight times thickness for each layer. The dry sand contributes 2 m times 17 kN per cubic meter, or 34 kPa, and the saturated clay contributes 3 m times about 17 kN per cubic meter, or 51 kPa, for a total of 85 kPa.

Question 41Civil Engineering Technology - Soils (CMT-Soils)

A technician is asked to determine the coefficient of consolidation of a saturated clay. Which laboratory test is appropriate for this purpose?

  1. The one-dimensional consolidation test Correct answer.
  2. The constant head permeability test
  3. The direct shear test
  4. The sieve analysis

Explanation

The one-dimensional consolidation test, commonly the oedometer test, produces the time-settlement data needed to back-calculate the coefficient of consolidation. Permeability, shear strength, and gradation tests do not provide the rate of consolidation behavior.

Question 42Civil Engineering Technology - Soils (CMT-Soils)

A sample of clean sand is placed in a graduated cylinder and the minimum and maximum void ratios are determined. Which statement correctly describes the relative density of a sand with a void ratio close to its maximum value?

  1. The sand is dense and well compacted.
  2. The sand is loose and in a loose state. Correct answer.
  3. The sand is at its densest possible state.
  4. The sand has zero relative density and no voids.

Explanation

Relative density compares the natural void ratio with the maximum and minimum void ratios. When the void ratio approaches the maximum, the soil is in its loosest condition, so the relative density is low and the sand is described as loose.

Question 43Civil Engineering Technology - Soils (CMT-Soils)

A soil sample is classified as GW under the Unified Soil Classification System. Which description matches this group symbol?

  1. An organic silt
  2. A well-graded gravel with little or no fines Correct answer.
  3. A poorly graded sand with silt
  4. A lean clay with sand

Explanation

The G indicates gravel and the W indicates well-graded, with little or no fines present. This group is assigned to clean gravels that have a broad, smooth gradation curve.

Question 44Civil Engineering Technology - Soils (CMT-Soils)

A retaining wall is backfilled with a clean dry sand having a unit weight of 18 kN per cubic meter and a friction angle of 32 degrees. Using Rankine theory, what is the active earth pressure coefficient?

  1. 0.47
  2. 0.69
  3. 0.55
  4. 0.31 Correct answer.

Explanation

The Rankine active earth pressure coefficient is the tangent squared of 45 degrees minus half the friction angle. For 32 degrees this is the tangent squared of 29 degrees, which equals about 0.31.

Question 45Civil Engineering Technology - Soils (CMT-Soils)

A slope stability analysis is performed on a slope in a stiff overconsolidated clay. Which condition most appropriately defines the factor of safety for this analysis?

  1. The driving moment divided by the resisting moment along the critical slip surface
  2. The resisting moment divided by the driving moment along the critical slip surface Correct answer.
  3. The pore pressure divided by the effective stress
  4. The total weight of the slope divided by the cohesion

Explanation

The factor of safety is the ratio of resisting forces or moments to driving forces or moments, so it must exceed one for a stable slope. For a rotational failure in clay, moments are taken about the center of rotation.

Question 46Civil Engineering Technology - Soils (CMT-Soils)

A standard penetration test is performed in a borehole. Which factor most directly affects the blow count recorded for a given soil?

  1. The water content of the drill fluid only
  2. The color of the soil sample
  3. The size of the borehole only
  4. The energy delivered by the hammer and the depth of the test Correct answer.

Explanation

The SPT blow count depends on the energy delivered by the hammer and on the effective overburden stress, which changes with depth. Corrections for energy ratio and overburden are applied to standardize the value.

Question 47Civil Engineering Technology - Soils (CMT-Soils)

A soil sample is recovered from a borehole using a thin-walled Shelby tube pushed into a soft clay. How is this sample quality best described?

  1. An undisturbed sample suitable for strength and consolidation testing Correct answer.
  2. A disturbed sample for sieve analysis
  3. A remolded sample for compaction testing
  4. A disturbed sample suitable only for classification

Explanation

Thin-walled Shelby tube sampling causes minimal disturbance to soft cohesive soils, so the sample can be treated as undisturbed for strength and consolidation testing. Disturbed or remolded samples are obtained by other methods and are not suitable for those tests.

Question 48Civil Engineering Technology - Soils (CMT-Soils)

A soil specimen is tested in a falling head permeameter and the coefficient of permeability is calculated. Which change would increase the measured coefficient of permeability if all other conditions are held constant?

  1. Decreasing the head difference across the specimen
  2. Increasing the length of the soil specimen
  3. Decreasing the cross-sectional area of the standpipe Correct answer.
  4. Increasing the cross-sectional area of the soil specimen

Explanation

In the falling head equation the coefficient of permeability is proportional to the standpipe cross-sectional area and inversely proportional to the specimen area. A smaller standpipe area causes the head to drop more slowly, which corresponds to a higher computed permeability for the same soil.

Question 49Civil Engineering Technology - Soils (CMT-Soils)

A soil is described as having a plasticity index of 8 and a liquid limit of 28. Which description best fits this material?

  1. A silt of low plasticity Correct answer.
  2. A well-graded gravel
  3. A highly plastic clay
  4. An organic peat

Explanation

A low plasticity index combined with a low liquid limit is characteristic of a silt of low plasticity, often classified as ML. Highly plastic clays have much higher liquid limits and plasticity indices.

Question 50Civil Engineering Technology - Soils (CMT-Soils)

A compaction test is performed on a soil and the maximum dry unit weight is found at a water content of 14 percent. A field sample of the same soil has a water content of 20 percent. What is the most appropriate field action before compaction?

  1. Compact the soil as is, since water content does not matter.
  2. Add cement to stabilize the soil immediately.
  3. Allow the soil to dry to near the optimum water content. Correct answer.
  4. Add water to raise the water content further.

Explanation

The soil is wet of optimum, so it should be allowed to dry toward the optimum water content to achieve the required dry density. Compacting soil that is too wet tends to produce lower dry unit weight and higher pore pressures.