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Question 1
ప్రశ్న 1
Which mapping of Indian Standards to soil compaction and in-place density tests is correct?
Explanation:
IS 2720 (Part 7) covers the laboratory water-content and dry-density relationship using light compaction. IS 2720 (Part 8) covers the corresponding relationship using heavy compaction. IS 2720 (Part 28) addresses in-place dry density determined by sand replacement. IS 2720 (Part 29) addresses in-place dry density determined by the core-cutter method. Identifying the correct test method is essential before interpreting a field compaction result.
వివరణ:
Question 2
ప్రశ్న 2
A project requires at least 95% of the heavy-compaction laboratory MDD. Light and heavy laboratory MDDs for the same soil are 1.80 and 1.95 Mg/m³, respectively, and the field dry density is 1.86 Mg/m³. Which conclusion is correct?
Explanation:
The specification expressly identifies heavy-compaction MDD as the reference. The relevant relative compaction is therefore 1.86/1.95 multiplied by 100. This equals approximately 95.38%, which exceeds the specified 95% minimum. Using 1.80 Mg/m³ from the light test would apply the wrong reference and could give a misleading result. The layer passes the stated density criterion, subject to any additional project requirements.
వివరణ:
Question 3
ప్రశ్న 3
For a 1000 cm³ mould, a light-compaction test uses a 2.6 kg rammer falling 0.310 m, 25 blows per layer and three layers. Taking g = 9.81 m/s², what is the approximate compactive energy per unit volume?
Explanation:
Compactive energy per unit volume equals rammer mass times gravitational acceleration times drop height times total blows divided by mould volume. The number of blows is 25 times 3, giving 75. The mould volume is 1000 cm³, or 0.001 m³. Substitution gives E = 2.6 × 9.81 × 0.310 × 75 / 0.001 = 593014.5 J/m³. Thus the energy is approximately 593 kJ/m³.
వివరణ:
Question 4
ప్రశ్న 4
For a 1000 cm³ mould, a heavy-compaction test uses a 4.9 kg rammer falling 0.450 m, 25 blows per layer and five layers. Taking g = 9.81 m/s², what is the approximate compactive energy per unit volume?
Explanation:
The mechanical compactive energy is calculated from mass, gravitational acceleration, drop height and the total number of blows. Five layers with 25 blows each give 125 blows. The 1000 cm³ mould has a volume of 0.001 m³. Hence E = 4.9 × 9.81 × 0.450 × 125 / 0.001 = 2703881.25 J/m³. The heavy-compaction energy is therefore approximately 2704 kJ/m³.
వివరణ:
Question 5
ప్రశ్న 5
An engineer compares laboratory MDD and OMC values obtained by light and heavy compaction for the same soil. Which interpretation is technically appropriate?
Explanation:
Laboratory compaction effort affects the water-content and dry-density relationship. For the same soil, heavy compaction commonly produces a higher MDD and a lower OMC than light compaction. The OMC and MDD values from different procedures should not be mixed into an artificial reference pair. A project acceptance specification must identify which laboratory compaction procedure governs. Correct comparison prevents an incorrect pass or fail decision in field quality control.
వివరణ:
Question 6
ప్రశ్న 6
A laboratory compaction mould has volume 1000 cm³ and weighs 3.05 kg empty. The mould plus compacted moist soil weighs 4.85 kg, and the measured water content is 12.5%. What is the dry density?
Explanation:
The moist soil mass is the filled-mould mass minus the empty-mould mass. It equals 4.85 - 3.05 = 1.80 kg. Since 1000 cm³ equals 0.001 m³, the bulk density is 1.80 Mg/m³. Dry density is obtained by dividing bulk density by 1 + w, with w expressed as a decimal. Thus the dry density is 1.80/1.125 = 1.60 Mg/m³.
వివరణ:
Question 7
ప్రశ్న 7
Which laboratory procedure most reliably establishes OMC and MDD from a conventional compaction curve?
Explanation:
A compaction curve is developed from dry-density results at different water contents under the selected compactive effort. Measurements should adequately cover both the rising and falling sides of the curve. Bulk density alone does not identify the maximum dry density because moisture contributes to bulk mass. A smooth curve through reliable results helps estimate the peak dry density and corresponding water content. These are reported as MDD and OMC for the specified laboratory procedure.
వివరణ:
Question 8
ప్రశ్న 8
Two field locations both have water contents within the project tolerance about OMC, but one has field dry density below the specified minimum. What is the correct quality-control decision?
Explanation:
A water-content tolerance controls an important condition during field compaction. It is not a substitute for achieving the specified field dry density. Soil may be near OMC yet remain under-compacted because of inadequate passes, unsuitable equipment or excessive lift thickness. Therefore a location failing the density criterion cannot be accepted solely on moisture compliance. The contractor should correct the compaction deficiency and verify compliance by retesting.
వివరణ:
Question 9
ప్రశ్న 9
In a moisture-content test associated with field density, the container plus wet soil weighs 42 g, the container plus oven-dry soil weighs 36 g, and the empty container weighs 12 g. What is the gravimetric water content?
Explanation:
Water content is the mass of water divided by the mass of dry soil solids. The mass of water lost on drying is 42 - 36 = 6 g. The dry soil mass is 36 - 12 = 24 g. Therefore w = 6/24 × 100 = 25%. The container mass must be subtracted before calculating the dry soil mass.
వివరణ:
Question 10
ప్రశ్న 10
Which field condition most strongly favours using the core-cutter method under IS 2720 (Part 29)?
Explanation:
The core-cutter method measures the mass of a cylindrical volume of in-place soil. It is well suited to appropriate fine-grained soils that allow the cutter to enter and retain a representative specimen. Gravel, cobbles or hard inclusions can prevent correct penetration and disturb the sampled soil. Such disturbance makes the nominal cutter volume an unreliable representation of the soil actually sampled. A suitable alternative such as sand replacement should then be considered.
వివరణ:
Question 11
ప్రశ్న 11
A core cutter of internal volume 1000 cm³ has empty mass 1.10 kg and mass 3.06 kg when filled with trimmed moist soil. The soil water content is 12%. What is its field dry density?
Explanation:
The moist soil mass is 3.06 - 1.10 = 1.96 kg. A 1000 cm³ core cutter has internal volume 0.001 m³. Therefore the bulk density is 1.96 Mg/m³. Converting to dry density gives 1.96/(1 + 0.12) = 1.75 Mg/m³. This field dry density can then be compared with the specified laboratory MDD.
వివరణ:
Question 12
ప్రశ్న 12
During a core-cutter test, the cutter is forced through a compacted layer containing oversized angular stones and the cutting edge visibly disturbs the soil. What is the best response?
Explanation:
Core-cutter testing assumes that the recovered soil occupies a known and representative internal volume. Angular stones can obstruct the cutter and distort or disturb the material being sampled. The resulting measured density may no longer represent the undisturbed field condition. An arbitrary numerical correction does not restore test validity. The engineer should use a suitable field-density method and obtain a fresh, properly executed result.
వివరణ:
Question 13
ప్రశ్న 13
In the sand-replacement method under IS 2720 (Part 28), what is the primary purpose of the calibrated dry sand?
Explanation:
The sand-replacement method uses a hole excavated in the field soil. Soil excavated from the hole is collected and weighed. Calibrated dry sand is poured into the hole to determine its volume. After making the required apparatus correction, the sand mass occupying the hole is divided by its calibrated bulk density. This volume, together with the excavated soil mass and water content, gives the field dry density.
వివరణ:
Question 14
ప్రశ్న 14
In a sand-replacement test, the sand-pouring apparatus contains 5.40 kg of sand before discharge and 2.82 kg afterward. The previously calibrated sand mass occupying the cone and base-plate region is 0.18 kg, and the dry sand bulk density is 1.60 Mg/m³. What is the excavated hole volume?
Explanation:
The mass of sand discharged from the apparatus is 5.40 - 2.82 = 2.58 kg. Part of this sand remains in the cone and base-plate region and does not occupy the hole. Subtracting the calibrated correction of 0.18 kg leaves 2.40 kg in the hole. The calibrated sand density of 1.60 Mg/m³ is equal to 1.60 g/cm³. Therefore the hole volume is 2400/1.60 = 1500 cm³.
వివరణ:
Question 15
ప్రశ్న 15
A sand-replacement test yields an excavated hole volume of 1500 cm³. The collected moist soil has mass 2.70 kg and water content 12.5%. What is the in-place dry density?
Explanation:
The volume of the excavated hole is 1500 cm³, or 0.0015 m³. Dividing 2.70 kg of moist soil by this volume gives a bulk density of 1800 kg/m³. Thus the bulk density equals 1.80 Mg/m³. Dry density is bulk density divided by 1 + w, with w = 0.125. The field dry density is therefore 1.80/1.125 = 1.60 Mg/m³.
వివరణ:
Question 16
ప్రశ్న 16
In a sand-replacement test, an operator forgets to subtract the sand held in the cone and base-plate region but uses the correct calibrated sand density. If all other measurements are correct, how is the reported field dry density affected?
Explanation:
Omitting the cone correction counts sand outside the hole as if it filled the hole. The calculated sand mass assigned to the hole is therefore too large. Dividing this excessive mass by the correct sand density gives an overestimated hole volume. The collected soil mass is then divided by a volume larger than the true volume. Consequently the calculated bulk and dry densities are underestimated.
వివరణ:
Question 17
ప్రశ్న 17
During calibration of sand used for a sand-replacement test, 2.55 kg of sand is found to occupy a calibrated container of 1500 cm³ after all apparatus corrections. What is the calibrated bulk density of the sand?
Explanation:
The calibrated sand bulk density is sand mass divided by the known occupied volume. The corrected sand mass is given as 2.55 kg. The calibration container volume is 1500 cm³, or 0.0015 m³. Dividing 2.55 by 0.0015 gives 1700 kg/m³. The bulk density is therefore 1.70 Mg/m³, equivalent to 1.70 g/cm³.
వివరణ:
Question 18
ప్రశ్న 18
A sand-replacement hole excavated in a loose, wet, nearly cohesionless soil suffers sidewall slumping before the sand is poured. The fallen soil remains in the hole and is not collected or weighed. Which systematic error is likely?
Explanation:
The original excavated soil mass should correspond to the full volume of the excavated cavity. Sidewall slumping introduces soil into the hole after excavation without adding that soil to the collected mass. Consequently the sand fills a smaller effective volume than the volume associated with the excavated mass. Dividing the excavated mass by this underestimated volume makes the computed density too high. Such unstable holes require careful method selection and testing practice to prevent erroneous results.
వివరణ:
Question 19
ప్రశ్న 19
The laboratory MDD for an earthfill is 1.90 Mg/m³ and the measured field dry density is 1.805 Mg/m³. What is the achieved relative compaction?
Explanation:
Relative compaction compares field dry density with the specified laboratory MDD. Both densities are supplied in the same units. The ratio is 1.805/1.90 = 0.95. Multiplying this ratio by 100 gives 95.0%. Any decision to accept this result must use the minimum relative compaction specified for the project.
వివరణ:
Question 20
ప్రశ్న 20
A project specification requires at least 98% relative compaction using a laboratory MDD of 1.90 Mg/m³. What is the minimum acceptable field dry density?
Explanation:
The minimum field dry density is the required relative-compaction fraction multiplied by the reference MDD. The specified fraction is 0.98. Therefore the minimum is 0.98 × 1.90 = 1.862 Mg/m³. A measured field value below this level fails the stated density criterion. The actual contract may also prescribe separate moisture-content and testing requirements.
వివరణ:
Question 21
ప్రశ్న 21
An earthwork specification requires both at least 95% of laboratory MDD and water content within 2 percentage points of OMC. If MDD is 1.84 Mg/m³, OMC is 12%, field dry density is 1.76 Mg/m³ and field water content is 14.5%, how should the layer be assessed?
Explanation:
The minimum permitted dry density is 0.95 × 1.84 = 1.748 Mg/m³. The field dry density of 1.76 Mg/m³ exceeds that minimum. A water-content tolerance of 2 percentage points about 12% allows values from 10% to 14%. The measured water content of 14.5% is outside that permissible interval. Accordingly the layer meets the density condition but fails the separate moisture condition.
వివరణ:
Question 22
ప్రశ్న 22
Four field density tests on the same layer give dry densities A = 1.82, B = 1.79, C = 1.83 and D = 1.81 Mg/m³. The specified minimum is 95% of laboratory MDD = 1.90 Mg/m³, and every individual test point must pass. Which location fails?
Explanation:
The specified minimum dry density is 0.95 × 1.90 = 1.805 Mg/m³. Location A has 1.82 Mg/m³, which satisfies the criterion. Location B has 1.79 Mg/m³, which is less than 1.805 Mg/m³. Locations C and D have 1.83 and 1.81 Mg/m³, respectively, both exceeding the minimum. Since every point must pass, only location B fails the stated criterion.
వివరణ:
Question 23
ప్రశ్న 23
A field sample has measured bulk density 2.10 Mg/m³ and water content 20%. Laboratory MDD is 1.90 Mg/m³, and at least 95% relative compaction is required. Which assessment is correct?
Explanation:
Bulk density includes the mass contribution of pore water and is not directly comparable with a dry-density criterion. The dry density equals 2.10/(1 + 0.20) = 1.75 Mg/m³. Relative compaction is 1.75/1.90 × 100 = approximately 92.1%. This is below the specified minimum of 95%. The layer therefore fails the stated density acceptance requirement despite its apparently high bulk density.
వివరణ:
Question 24
ప్రశ్న 24
A compacted clay lift shows low dry density and water content materially above laboratory OMC. Which remedial plan is most technically sound before final acceptance?
Explanation:
Soil moisture substantially above optimum can hinder achievement of the specified dry density. Repeated roller passes alone may be ineffective if moisture and lift conditions are unsuitable. Moisture conditioning, including controlled drying or remixing where appropriate, can improve the compaction response. Lift thickness and compaction equipment must also be matched to the material. The reworked layer should be accepted only after the required field tests demonstrate compliance.
వివరణ:
Question 25
ప్రశ్న 25
Quality-control tests show that the upper portion of a thick granular fill lift is well compacted but its lower portion remains insufficiently dense. Which engineering action best addresses this pattern?
Explanation:
Compactive energy delivered at the surface does not necessarily densify a thick soil layer uniformly through its depth. Granular soils commonly respond well to vibration when the equipment and moisture conditions are suitable. An excessively thick lift can therefore remain loose near its base despite a dense upper zone. Reducing loose lift thickness is a direct way to improve through-depth compaction for the available equipment. The corrected layer should then be checked by suitable field quality-control tests.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. IS 2720 (Part 7) - light compaction; (Part 8) - heavy compaction; (Part 28) - sand replacement; (Part 29) - core cutter. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: C. The field passes because relative compaction against heavy MDD is approximately 95.4%. సమాధానం: C.
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Question 3 ప్రశ్న 3Answer: A. 593 kJ/m³. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: D. 2704 kJ/m³. సమాధానం: D.
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Question 5 ప్రశ్న 5Answer: C. Each OMC-MDD pair belongs to its specified compactive effort, and the reference test for acceptance must match the project specification. సమాధానం: C.
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Question 6 ప్రశ్న 6Answer: B. 1.60 Mg/m³. సమాధానం: B.
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Question 7 ప్రశ్న 7Answer: B. Test several water contents on both sides of the peak, calculate dry density at each point, and determine the peak of a smooth moisture-density curve. సమాధానం: B.
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Question 8 ప్రశ్న 8Answer: D. Reject or rework the deficient location according to the specification because moisture compliance does not replace the dry-density requirement. సమాధానం: D.
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Question 9 ప్రశ్న 9Answer: C. 25%. సమాధానం: C.
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Question 10 ప్రశ్న 10Answer: A. A soft, fine-grained cohesive soil without significant gravel or hard inclusions, into which the cutter can be driven without substantial disturbance. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: B. 1.75 Mg/m³. సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: C. Select a more suitable applicable field-density method, such as sand replacement, and repeat the measurement. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: D. To determine the volume of an excavated test hole from the mass of sand filling it and the calibrated bulk density of that sand. సమాధానం: D.
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Question 14 ప్రశ్న 14Answer: A. 1,500 cm³. సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: C. 1.60 Mg/m³. సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: B. It is underestimated because the calculated hole volume is too large. సమాధానం: B.
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Question 17 ప్రశ్న 17Answer: C. 1.70 Mg/m³. సమాధానం: C.
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Question 18 ప్రశ్న 18Answer: A. The reported soil dry density will be overestimated because the sand-measured hole volume is reduced. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: D. 95.0%. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: B. 1.862 Mg/m³. సమాధానం: B.
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Question 21 ప్రశ్న 21Answer: C. It satisfies the density criterion but fails the specified moisture-content tolerance. సమాధానం: C.
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Question 22 ప్రశ్న 22Answer: B. B only. సమాధానం: B.
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Question 23 ప్రశ్న 23Answer: C. The layer fails, because its dry density is 1.75 Mg/m³ and relative compaction is approximately 92.1%. సమాధానం: C.
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Question 24 ప్రశ్న 24Answer: D. Condition the soil moisture appropriately, adjust lift thickness or equipment effort as needed, recompact, and retest dry density and water content. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: B. Reduce the loose lift thickness to a depth that the selected vibratory compactor can effectively densify, then recompact and verify the layer. సమాధానం: B.
