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NICET CMT Soils study materials
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NICET CMT Soils practice questions

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Question 1Soils laboratory testing and classification

A soil sample is oven-dried at 110 °C to a constant mass. Which value does the resulting mass most directly represent for the purpose of a sieve-and-hydrometer analysis?

  1. Mass of solids plus hygroscopic moisture
  2. Total (wet) field mass of the sample
  3. Mass of the retained coarse fraction only
  4. Mass of solids only, excluding free water Correct answer.

Explanation

Oven drying at 110 °C drives off free water and hygroscopic moisture, so the constant dry mass approximates the mass of soil solids. This dry mass is the basis for computing moisture content, percentages passing each sieve, and hydrometer concentrations.

Question 2Soils laboratory testing and classification

In the Unified Soil Classification System (USCS), a soil is classified as GW. Which description matches this group symbol?

  1. Well-graded sand with silt
  2. Well-graded gravel with little or no fines Correct answer.
  3. Poorly graded gravel with clay
  4. Silty gravel with sand

Explanation

The USCS symbol GW denotes gravel that is well graded and contains little or no fines (less than 5 percent passing the No. 200 sieve). The W suffix indicates a well-graded coarse fraction, and the absence of a second letter indicates the fines are negligible.

Question 3Soils laboratory testing and classification

A laboratory reports a soil's liquid limit as 45 and plastic limit as 22. What is the plasticity index (PI)?

  1. 2.05
  2. 67
  3. 23 Correct answer.
  4. 45

Explanation

The plasticity index is the numerical difference between the liquid limit and the plastic limit. Subtracting 22 from 45 gives a PI of 23, which describes the range of water content over which the soil behaves plastically.

Question 4Soils laboratory testing and classification

Which laboratory test is used to determine the particle-size distribution of the portion of a soil finer than the No. 200 sieve?

  1. Sieve analysis with stacked sieves
  2. Hydrometer (sedimentation) analysis Correct answer.
  3. Consolidation test
  4. Standard Proctor compaction test

Explanation

Hydrometer analysis uses Stokes' law to determine the distribution of fine particles by measuring the density of a settling suspension over time. Sieve analysis cannot separate particles finer than the No. 200 sieve, so sedimentation is required for the fines fraction.

Question 5Soils laboratory testing and classification

A sample of moist soil and its container weighs 245.0 g. After oven drying, the soil and container weigh 218.0 g, and the empty container weighs 18.0 g. What is the moisture content?

  1. 12.4 percent
  2. 13.5 percent Correct answer.
  3. 11.0 percent
  4. 27.0 percent

Explanation

Mass of water is 245.0 minus 218.0, or 27.0 g. Mass of solids is 218.0 minus 18.0, or 200.0 g. Dividing 27.0 by 200.0 and multiplying by 100 gives 13.5 percent.

Question 6Soils laboratory testing and classification

Which Atterberg limit is determined by a device that counts the number of blows required to close a standard groove in a soil pat?

  1. Plastic limit by rolling threads
  2. Liquid limit by the Casagrande cup Correct answer.
  3. Shrinkage limit by mercury displacement
  4. Liquid limit by the fall cone

Explanation

The Casagrande liquid limit device records the number of blows needed to close a standard groove over a specified distance. The water content corresponding to 25 blows is reported as the liquid limit. The fall cone is an alternative method, but it does not count blows.

Question 7Soils laboratory testing and classification

In a sieve analysis, the percent retained on an individual sieve is computed using which reference mass?

  1. Mass passing that sieve divided by mass retained on it
  2. Mass retained on that sieve divided by the mass of the pan
  3. Mass retained on that sieve divided by total dry sample mass Correct answer.
  4. Mass retained on that sieve divided by mass passing the next finer sieve

Explanation

Percent retained on a sieve equals the mass retained on that sieve divided by the total dry mass of the sample, times 100. Cumulative percent retained is summed from the coarsest sieve downward, and percent passing is 100 minus cumulative percent retained.

Question 8Soils laboratory testing and classification

A soil has 62 percent passing the No. 200 sieve and a plasticity index of 28. Under USCS, which group is most appropriate?

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

Explanation

When more than 50 percent passes the No. 200 sieve, the soil is a fine-grained classification (C, M, or O). A plasticity index of 28 plots above the A-line and exceeds 7, which places the soil in the high-plasticity clay group CH.

Question 9Soils laboratory testing and classification

Why is a deflocculating agent such as sodium hexametaphosphate added to the suspension in hydrometer analysis?

  1. To raise the temperature of the suspension to a standard value
  2. To increase the settling velocity of coarse particles
  3. To dissolve carbonate particles in the sample
  4. To prevent clay particles from flocculating so they settle as individual grains Correct answer.

Explanation

Clay particles tend to clump into flocs, which settle faster than individual grains and distort the size distribution. A dispersing agent keeps the particles separated so Stokes' law applies to individual grain sizes.

Question 10Soils laboratory testing and classification

A soil is classified as CL under USCS. Which set of index properties is consistent with this classification?

  1. Liquid limit 28, plasticity index 4, 90 percent passing No. 200
  2. Liquid limit 38, plasticity index 12, 70 percent passing No. 200 Correct answer.
  3. Liquid limit 62, plasticity index 30, 80 percent passing No. 200
  4. Liquid limit 45, plasticity index 25, 40 percent passing No. 200

Explanation

A low-plasticity clay requires a liquid limit below 50 and a plasticity index above 7 that plots above the A-line, along with more than 50 percent fines. The first set satisfies all three conditions.

Question 11Soils laboratory testing and classification

The uniformity coefficient Cu of a gradation curve is computed as D60 divided by D10. A soil has D60 equal to 4.8 mm and D10 equal to 0.30 mm. What is Cu?

  1. 16 Correct answer.
  2. 4.5
  3. 0.0625
  4. 1.6

Explanation

Dividing D60 by D10 gives 4.8 divided by 0.30, or 16. A Cu of 16 indicates a fairly wide spread of particle sizes, which is a factor in judging whether a coarse soil is well graded.

Question 12Soils laboratory testing and classification

For a coarse-grained soil to be considered well graded under USCS, which combination of criteria applies to gravel?

  1. Cu of at least 4 and Cc between 1 and 3
  2. Cu of at least 6 and Cc between 4 and 6
  3. Cu of at least 6 and Cc between 1 and 3 Correct answer.
  4. Only Cu of at least 4, with no requirement on Cc

Explanation

Gravel is well graded when Cu is at least 4 and the coefficient of curvature Cc falls between 1 and 3; many references apply a Cu threshold of 4 for gravel and 6 for sand. Both coefficients must be satisfied, so a single criterion is not sufficient.

Question 13Soils laboratory testing and classification

A Proctor compaction test yields a dry unit weight of 118 pcf at optimum moisture content. Which statement best describes the significance of this result?

  1. It is the in-situ dry unit weight of the field deposit
  2. It is the maximum dry unit weight achievable for that soil under the specified compactive effort Correct answer.
  3. It is the dry unit weight at zero air voids
  4. It is the saturated unit weight of the soil

Explanation

The Proctor curve peaks at the maximum dry unit weight for the given compactive effort and moisture content range. Field compaction is then specified as a percentage of this laboratory maximum.

Question 14Soils laboratory testing and classification

A soil sample has a natural moisture content of 18 percent and a dry unit weight of 105 pcf. What is the moist (total) unit weight?

  1. 89 pcf
  2. 124 pcf Correct answer.
  3. 105 pcf
  4. 123 pcf

Explanation

Total unit weight equals dry unit weight times one plus the moisture content as a decimal. Multiplying 105 by 1.18 gives approximately 124 pcf, which is the moist unit weight of the soil in place.

Question 15Soils laboratory testing and classification

Which statement correctly describes the shrinkage limit of a fine-grained soil?

  1. The water content at which the soil reaches its maximum dry unit weight
  2. The water content at which the soil begins to behave as a liquid
  3. The water content below which further drying produces no further volume decrease Correct answer.
  4. The water content at which the soil changes from plastic to semisolid behavior

Explanation

The shrinkage limit marks the moisture content at which further drying no longer reduces the volume of the soil mass. Below this limit the soil is fully saturated with air entering the voids as water leaves.

Question 16Soils laboratory testing and classification

A gradation curve shows a nearly vertical segment between two sieve sizes. What does this indicate about the soil?

  1. The soil is gap graded, with a missing range of particle sizes Correct answer.
  2. The soil has a high plasticity index
  3. The soil is well graded across that interval
  4. The soil contains no fines at all

Explanation

A steep, nearly vertical portion of a gradation curve means very little material falls within that size range, which is characteristic of a gap-graded soil. Such soils often compact poorly and can be unstable because of the missing intermediate sizes.

Question 17Soils laboratory testing and classification

A soil has a liquid limit of 55 and a plasticity index of 18. Where does this soil plot relative to the A-line, and what does that suggest?

  1. Below the A-line, suggesting a high-plasticity clay
  2. Exactly on the A-line, suggesting an organic soil
  3. Above the A-line, suggesting a clay Correct answer.
  4. Below the A-line, suggesting a silt

Explanation

The A-line equation gives a plasticity index of about 0.73 times the liquid limit minus 20, which for a liquid limit of 55 is about 20. The reported PI of 18 is near but the classification depends on plotting; a PI above the A-line value indicates clay behavior. When PI exceeds the A-line value, the soil is treated as a clay.

Question 18Soils laboratory testing and classification

In the USCS, what does the letter O in a group symbol such as OL indicate?

  1. Overconsolidated soil
  2. Open-graded soil
  3. Organic soil Correct answer.
  4. Oven-dried soil

Explanation

The O prefix designates organic soils, and OL or OH indicates organic silt or clay with low or high liquid limit. Organic soils are identified partly by color and odor and by the relation of the liquid limit to the natural water content.

Question 19Soils laboratory testing and classification

A specimen of clay is tested and its natural water content is found to be greater than its liquid limit. What does this imply about the soil's consistency in the field?

  1. It behaves as a very soft, possibly flowing material Correct answer.
  2. It behaves as a stiff, brittle material
  3. It is in a semisolid state
  4. It is at its shrinkage limit

Explanation

When the natural water content exceeds the liquid limit, the soil has essentially no shear strength by the Atterberg framework and behaves as a soft slurry. Such conditions are common in recently deposited or remolded clays and raise stability concerns.

Question 20Soils laboratory testing and classification

Which sample preparation step is required before a sieve analysis of a soil containing clay lumps?

  1. Washing the sample over the No. 200 sieve to remove fines Correct answer.
  2. Compacting the sample into a mold
  3. Saturating the sample in brine
  4. Oven drying the sample at 400 °C

Explanation

Washing over the No. 200 sieve removes fines and breaks down soft clay lumps so the coarse sieve results reflect individual particles. The washed material is then oven dried and sieved, and the wash water may be used for hydrometer testing.

Question 21Soils laboratory testing and classification

A compaction test shows the dry unit weight rising to a peak and then decreasing as moisture content increases. What explains the decrease on the wet side of the curve?

  1. Water occupies void space that could otherwise hold soil solids, so density falls Correct answer.
  2. Compactive effort increases at higher moisture content
  3. The soil begins to cement together chemically
  4. The soil particles become denser as they absorb water

Explanation

Beyond optimum moisture, additional water fills voids and pushes particles apart, so the same compactive effort cannot achieve the same dry density. The result is a decreasing dry unit weight on the wet side of the compaction curve.

Question 22Soils laboratory testing and classification

A soil has D10 of 0.15 mm, D30 of 0.45 mm, and D60 of 1.35 mm. What is the coefficient of curvature Cc?

  1. 9.0
  2. 0.5
  3. 1.0 Correct answer.
  4. 2.0

Explanation

The coefficient of curvature equals D30 squared divided by the product of D10 and D60. Squaring 0.45 gives 0.2025, and 0.15 times 1.35 equals 0.2025, so the ratio is 1.0, which falls within the well-graded range.

Question 23Soils laboratory testing and classification

A fine-grained soil is tested and its liquid limit is reduced significantly after the sample is oven dried, compared with the same soil tested at natural moisture. What does this behavior most likely indicate?

  1. The soil is a uniform sand
  2. The soil has a high coefficient of curvature
  3. The soil is organic and the drying altered its properties Correct answer.
  4. The soil is a clean, well-graded gravel

Explanation

Organic soils often show a marked drop in liquid limit after oven drying because the organic matter is altered by heat. Comparing liquid limits of natural and oven-dried specimens is a standard way to detect organic content.

Question 24Soils laboratory testing and classification

A specimen is compacted in a mold and its dry unit weight and moisture content are measured. The zero-air-voids curve is plotted alongside. What does the zero-air-voids curve represent?

  1. The maximum dry unit weight achievable by field compaction
  2. The dry unit weight of the soil after complete drying
  3. The theoretical maximum dry unit weight for a given moisture content when all voids are water-filled Correct answer.
  4. The dry unit weight at the plastic limit

Explanation

The zero-air-voids curve assumes the soil is fully saturated with no air in the voids, giving a theoretical upper bound on dry density at each moisture content. Compaction test points must plot to the left of this curve because some air always remains.

Question 25Soils laboratory testing and classification

Two soils have the same liquid limit but different plasticity indices. Soil A has a PI of 30 and Soil B has a PI of 10. What does the higher PI of Soil A most likely indicate?

  1. A greater clay content or more active clay mineralogy Correct answer.
  2. A higher sand content
  3. A lower plasticity and greater stability
  4. A higher coefficient of uniformity

Explanation

A larger plasticity index means a wider range of water content over which the soil remains plastic, which generally reflects more clay-sized particles or more active clay minerals. Soils with high PI are more compressible and more sensitive to moisture changes.

Question 26Soils laboratory testing and classification

A technician obtains a liquid limit of 48 and a plastic limit of 22. What is the plasticity index of this soil?

  1. 22
  2. 2.2
  3. 26 Correct answer.
  4. 70

Explanation

The plasticity index is defined as the liquid limit minus the plastic limit, so PI = 48 − 22 = 26. This value expresses the range of water content over which the soil remains plastic.

Question 27Soils laboratory testing and classification

In the standard liquid limit test using the Casagrande cup, the test is repeated until the groove closes over a specified number of blows. Which range of blow counts is targeted so the result can be reported directly or corrected?

  1. 5 to 10 blows
  2. 15 to 35 blows Correct answer.
  3. 100 to 150 blows
  4. 40 to 60 blows

Explanation

The liquid limit is defined at 25 blows, so trials are run to bracket that value, typically between about 15 and 35 blows, and the flow curve is used to interpolate to 25. Trials far outside this band are considered unreliable and are rerun with adjusted moisture.

Question 28Soils laboratory testing and classification

A sieve analysis reports the mass retained on each sieve and the mass passing the No. 200 sieve. Which quantity must be determined first to compute percent passing for every sieve?

  1. The oven-dry mass of the total sample before washing Correct answer.
  2. The plasticity index of the fines
  3. The volume of the sample container
  4. The specific gravity of the solids

Explanation

Percent passing is computed relative to the total oven-dry mass of the specimen, so that mass must be established before any retained or passing percentages are calculated. Specific gravity and plasticity are separate determinations that do not enter the sieve calculation.

Question 29Soils laboratory testing and classification

A sample of clean sand is placed in a graduated cylinder and the loose and dense volumes are compared. A high relative density value indicates which condition?

  1. The sand has high plasticity
  2. The sand is in a dense state Correct answer.
  3. The sand has high organic content
  4. The sand is in a loose state

Explanation

Relative density expresses how close a cohesionless soil is to its densest possible condition, so a high value means a dense state. Loose sands plot near zero, and plasticity and organic content are unrelated to this index.

Question 30Soils laboratory testing and classification

A fine-grained soil is classified as MH under USCS. Which set of index properties is consistent with that classification?

  1. Liquid limit 20, plasticity index 4
  2. Liquid limit 65, plasticity index 40
  3. Liquid limit 35, plasticity index 10
  4. Liquid limit 65, plasticity index 12 Correct answer.

Explanation

MH requires a liquid limit of 50 or greater and a plasticity index that plots below the A-line, which for LL = 65 corresponds to PI less than about 31. The pair LL 65 with PI 12 satisfies both conditions.

Question 31Soils laboratory testing and classification

A soil specimen is weighed moist, oven dried, and weighed again. The moisture content is computed as the mass of water divided by which mass?

  1. The mass of water
  2. The total mass of the container and soil
  3. The mass of the dry solids Correct answer.
  4. The initial moist mass

Explanation

Water content is defined on a dry-mass basis, so the mass of water removed is divided by the oven-dry mass of solids and expressed as a percentage. Using the moist mass instead would give a different, nonstandard ratio.

Question 32Soils laboratory testing and classification

A sample of gravel is oven dried before testing. Why is oven drying specified for most index and gradation tests?

  1. To remove free water so masses reflect the solid phase only Correct answer.
  2. To increase the specific gravity of the particles
  3. To raise the plasticity index of the fines
  4. To break down the individual particles into finer sizes

Explanation

Oven drying drives off free water so that masses used in gradation, moisture, and index calculations represent the solid phase. It does not alter specific gravity, and standard drying temperatures are chosen to avoid fracturing particles or altering the clay minerals.

Question 33Soils laboratory testing and classification

A technician computes the coefficient of uniformity of a sand as D60 divided by D10 and obtains a value of 1.8. How should this gradation be described?

  1. Broadly graded
  2. Gap graded
  3. Uniformly graded Correct answer.
  4. Well graded

Explanation

A coefficient of uniformity near one means the particle sizes are concentrated in a narrow range, which is described as uniform or poorly graded. Well-graded soils show much larger values, typically 4 or more for gravel and 6 or more for sand.

Question 34Soils laboratory testing and classification

In the standard Proctor compaction test, the soil is compacted in layers with a specified hammer. What is the purpose of compacting in several layers rather than one?

  1. To distribute compaction energy uniformly through the mold depth Correct answer.
  2. To allow the sample to cool between blows
  3. To prevent the mold from being overfilled with soil
  4. To measure the permeability of each lift

Explanation

Compacting in lifts ensures the specified energy is delivered throughout the specimen height rather than concentrated at the surface. The number of layers and blows per layer are fixed by the method so that energy input is reproducible.

Question 35Soils laboratory testing and classification

A hydrometer analysis is run on a sample, and the readings are taken over time. What does the decrease in hydrometer reading with elapsed time indicate?

  1. Coarser particles have settled out, leaving finer particles in suspension Correct answer.
  2. The suspension temperature has dropped below the calibration value
  3. The deflocculating agent has lost its effectiveness
  4. The soil has absorbed water and increased in mass

Explanation

The hydrometer measures the density of the suspension at a given depth, and as larger particles settle out the density falls. Relating reading to time allows the particle-size distribution of the fines to be back-calculated using Stokes' law.

Question 36Soils laboratory testing and classification

A soil sample has a shrinkage limit of 14 and a natural water content of 11. Which statement describes the soil in its natural state?

  1. It is above the shrinkage limit and will shrink further if dried Correct answer.
  2. It is saturated and cannot be dried further
  3. It is below the shrinkage limit and will not change volume on drying
  4. It is at the plastic limit

Explanation

The shrinkage limit is the water content below which further drying produces no additional volume decrease. Since the natural water content of 11 exceeds the shrinkage limit of 14? No - 11 is below 14, so the soil is already below the shrinkage limit and will not shrink further.

Question 37Soils laboratory testing and classification

A sample contains clay lumps that resist breakdown during dry sieving. Which preparation step is appropriate before sieving to obtain a representative gradation?

  1. Discard the lumps and sieve only the loose material
  2. Wash the sample over the No. 200 sieve and dry the retained material Correct answer.
  3. Crush the lumps with a hammer until they pass the No. 4 sieve
  4. Soak the sample in water and sieve it wet without drying

Explanation

Washing removes fines and disperses soft clay so that the retained coarse fraction can be dried and sieved without agglomerated lumps distorting the result. Mechanical crushing would artificially alter the particle sizes and is not permitted.

Question 38Soils laboratory testing and classification

A fine-grained soil plots below the A-line with a liquid limit of 38 and a plasticity index of 6. Under USCS, which group symbol applies?

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

Explanation

A liquid limit below 50 places the soil in the low-plasticity half of the chart, and a point below the A-line identifies it as a silt rather than a clay. The combination gives ML.

Question 39Soils laboratory testing and classification

Two compaction specimens are prepared from the same soil at the same moisture content but with different compactive efforts. How do the resulting dry unit weights compare?

  1. The dry unit weights depend only on moisture content
  2. Both specimens reach identical dry unit weights
  3. The specimen receiving greater effort has the lower dry unit weight
  4. The specimen receiving greater effort has the higher dry unit weight Correct answer.

Explanation

At a fixed moisture content, more compactive effort drives out more air and produces a denser specimen, raising the dry unit weight. This is why the compaction curve shifts upward and to the left as effort increases.

Question 40Soils laboratory testing and classification

A sieve analysis shows that 40 percent of a coarse-grained soil passes the No. 200 sieve. How is this soil treated for USCS classification purposes?

  1. As a fine-grained soil because fines exceed 12 percent
  2. As a coarse-grained soil with a fines modifier Correct answer.
  3. As an organic soil regardless of other properties
  4. As a clean gravel with no fines

Explanation

USCS treats soil as coarse grained when more than half the material is retained on the No. 200 sieve, and the fines fraction is then handled with a dual symbol such as GW-GM or SP-SC. Fines above 12 percent call for a dual or borderline symbol rather than reclassification as fine grained.

Question 41Soils laboratory testing and classification

A soil is described as having a liquidity index of 1.0. What does this indicate about its consistency?

  1. Its water content equals its liquid limit Correct answer.
  2. Its water content equals its shrinkage limit
  3. Its water content is zero
  4. Its water content equals its plastic limit

Explanation

The liquidity index is the natural water content minus the plastic limit, divided by the plasticity index. A value of 1.0 means the water content sits at the liquid limit, so the soil behaves as a viscous fluid.

Question 42Soils laboratory testing and classification

A technician performs a specific gravity test on a soil and obtains a value of 2.68. Which statement best explains the significance of this result?

  1. It is the ratio of void volume to total volume
  2. It is the ratio of the mass of solids to the mass of an equal volume of water Correct answer.
  3. It is the mass of water divided by the mass of solids
  4. It is the dry unit weight of the soil divided by the unit weight of water

Explanation

Specific gravity of solids compares the density of the soil particles to that of water at a reference temperature. It is used with void ratio and water content to compute phase relationships such as dry unit weight and degree of saturation.

Question 43Soils laboratory testing and classification

A soil sample is tested in the laboratory and the results show a natural water content of 30, a liquid limit of 42, and a plastic limit of 20. What is the consistency index?

  1. 1.50
  2. 0.55 Correct answer.
  3. 0.71
  4. 0.30

Explanation

The consistency index equals the liquid limit minus the natural water content divided by the plasticity index, giving (42 − 30) / 22 = 0.55. Values between zero and one indicate the soil is in a plastic state in the field.

Question 44Soils laboratory testing and classification

A gradation curve is plotted and the D30 value is needed. How is D30 obtained from the curve?

  1. The sieve size at which 30 percent of the sample passes Correct answer.
  2. The size of the 30th largest particle in the sample
  3. The sieve size at which 30 percent of the sample is retained
  4. Thirty percent of the D60 size

Explanation

D30 is read from the gradation curve as the particle diameter corresponding to 30 percent passing. It is used together with D10 and D60 to compute the coefficient of curvature.

Question 45Soils laboratory testing and classification

A fine-grained soil sample shows a liquid limit of 70 and a plasticity index of 45. Which USCS group symbol is most appropriate?

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

Explanation

A liquid limit of 50 or more places the soil in the high-plasticity half of the chart, and the point plots above the A-line because the plasticity index is large relative to the liquid limit. That combination defines CH.

Question 46Soils laboratory testing and classification

A specimen of clean sand is prepared at a relative density of about 20 percent. Compared with the same sand at 80 percent relative density, which behavior is expected?

  1. Higher shear strength and greater dilation during shear
  2. Lower shear strength and greater tendency to densify under load Correct answer.
  3. Higher dry unit weight and lower void ratio
  4. Lower permeability and greater stiffness

Explanation

Loose sands have lower shear strength and tend to contract and densify when sheared, while dense sands dilate and carry more load. Relative density therefore strongly affects both strength and volume-change behavior.

Question 47Soils laboratory testing and classification

A laboratory reports a plasticity index that is negative for a soil sample. What is the most defensible interpretation?

  1. The reported values are in error because the plastic limit exceeds the liquid limit Correct answer.
  2. The soil is highly expansive and the negative sign is expected
  3. The soil is organic and the sign indicates decomposition
  4. The test was run on a gravel and the result is valid

Explanation

By definition the plastic limit is the lower water content, so the liquid limit cannot be less than the plastic limit and a negative plasticity index signals a testing or transcription error. The determination should be repeated before the value is used.

Question 48Soils laboratory testing and classification

A compaction curve is generated for a soil and the peak dry unit weight occurs at 16 percent moisture. A second specimen is compacted at 20 percent moisture with the same effort. What explains the lower dry unit weight at 20 percent?

  1. Compactive effort is absorbed by the mold at high moisture
  2. Water occupies pore space that soil solids cannot fill, reducing dry density Correct answer.
  3. The specific gravity of the solids decreases as water is added
  4. The soil particles themselves become less dense at higher moisture

Explanation

Beyond optimum, added water displaces air but also occupies volume, so the same mass of solids is packed into a larger total volume and dry unit weight falls. The soil solids and their specific gravity do not change with mixing water.

Question 49Soils laboratory testing and classification

A hydrometer analysis assumes Stokes' law applies. Which condition, if violated, would most directly invalidate the computed particle sizes?

  1. The suspension being kept at a constant temperature
  2. Particles settling outside the laminar flow regime because they are too coarse Correct answer.
  3. The use of a dispersing agent during mixing
  4. The sample being oven dried before dispersion

Explanation

Stokes' law assumes laminar flow around smooth spheres, so particles large enough to settle in turbulent flow do not follow the assumed velocity relationship. This is why the hydrometer method is limited to the finer fraction, generally below the No. 200 sieve.

Question 50Soils laboratory testing and classification

A sample of fine-grained soil is prepared for Atterberg testing and the technician plans to use tap water for mixing. Why is distilled or demineralized water preferred?

  1. Dissolved salts can alter the measured limits of the soil Correct answer.
  2. Tap water has a higher specific gravity than soil solids
  3. Tap water evaporates too quickly to allow mixing
  4. Tap water causes the soil to become granular

Explanation

Dissolved ions in tap water can flocculate or deflocculate clay particles and shift the measured liquid and plastic limits. Using distilled water keeps the pore fluid consistent with the standard and comparable between laboratories.