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Question 1
ప్రశ్న 1
According to IS 1498, which criterion distinguishes a coarse-grained soil from a fine-grained soil in laboratory classification?
Explanation:
IS 1498 distinguishes coarse-grained and fine-grained soils principally using the 75-micron IS sieve boundary. A soil is coarse-grained when more than half its material by mass is coarser than this boundary. A fine-grained soil has more than half its material finer than the same sieve size. The 4.75 mm sieve is subsequently used to distinguish gravel-dominated and sand-dominated coarse fractions. Gradation coefficients are used for further classification and do not determine the initial coarse-grained versus fine-grained division.
వివరణ:
Question 2
ప్రశ్న 2
Which particle-size boundaries are consistent with IS 1498 for the gravel and sand components of soil?
Explanation:
The applicable IS 1498 classification identifies gravel particles between the 4.75 mm and 75 mm sieve-size boundaries. Sand particles fall between 75 microns and 4.75 mm. Particles finer than 75 microns constitute the fines fraction. The upper gravel boundary of 75 mm must be used rather than the superseded 80 mm value corrected by the standard's amendment. Further subdivisions identify coarse and fine gravel and coarse, medium and fine sand.
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Question 3
ప్రశ్న 3
An engineer finds that most particles in a fine soil are smaller than 2 microns. Which statement is most accurate under the IS 1498 engineering-classification framework?
Explanation:
Particle sizes smaller than approximately 2 microns can be described as clay-size particles. This size designation is not identical to the engineering classification of a clay soil. IS 1498 classifies fine-grained soils principally using plasticity characteristics and liquid-limit ranges. Mineralogical clays can sometimes exhibit engineering behaviour associated with silts. Therefore particle-size data alone are insufficient to assign a particular fine-grained soil group symbol.
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Question 4
ప్రశ్న 4
A soil sample has 62% of its dry mass passing the 75-micron IS sieve. What is its primary classification division under IS 1498?
Explanation:
The dry mass percentage passing the 75-micron sieve is 62%. Consequently only 38% is retained above that size. More than half the soil is therefore in the fines fraction. IS 1498 places the sample in the fine-grained soil division. Additional liquid-limit, plasticity and organic-character information would be needed to determine its final group symbol.
వివరణ:
Question 5
ప్రశ్న 5
A soil contains 38% gravel-size particles, 30% sand-size particles and 32% fines by dry mass. Under IS 1498, how should its primary division and coarse-grained subdivision be identified?
Explanation:
The coarse fraction is the sum of gravel and sand percentages. This gives 38 + 30 = 68% of the total soil mass. Since the coarse fraction exceeds 50%, the soil is coarse-grained. Gravel represents 38 of the 68 percentage points in the coarse fraction and therefore exceeds half that fraction. The primary subdivision is gravel, although further information about the fines is needed for a complete group symbol.
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Question 6
ప్రశ్న 6
A soil contains 12% gravel, 56% sand and 32% fines by dry mass. According to IS 1498, which statement correctly identifies its primary coarse-grained subdivision?
Explanation:
The percentage coarser than 75 microns is 12 + 56 = 68%. This establishes the coarse-grained division. Within the coarse fraction, sand accounts for 56 of the 68 percentage points. The soil is therefore classified in the sand subdivision. Because the fines content is appreciable, a complete classification must consider the plasticity of those fines rather than automatically assigning the clean-sand symbol SW.
వివరణ:
Question 7
ప్రశ్న 7
Which treatment of fines content in coarse-grained soils agrees with the laboratory-classification rules of IS 1498?
Explanation:
IS 1498 distinguishes clean coarse-grained soils from soils containing appreciable fines. Less than 5% fines generally permits clean gravel or sand symbols. More than 12% fines requires the silty or clayey classification criteria to be considered. The interval from 5% to 12% is treated as a boundary classification using an appropriate dual symbol. This rule prevents an intermediate-fines soil from being assigned a misleading single clean-soil symbol.
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Question 8
ప్రశ్న 8
A 1000 g dry soil sample contains 150 g gravel, 650 g sand and 200 g fines. The fines are non-plastic, and their behaviour is consistent with silt. What is the correct IS 1498 group symbol?
Explanation:
The total coarse fraction is 150 + 650 = 800 g, representing 80% of the sample. The soil is therefore coarse-grained. Sand dominates the coarse fraction because 650 g is greater than 150 g. The fines constitute 20% of total mass, which exceeds the 12% threshold for appreciable fines. Since those fines are non-plastic and silt-like, the appropriate group symbol is SM, meaning silty sand.
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Question 9
ప్రశ్న 9
For a soil grain-size distribution, D10 = 0.15 mm and D60 = 1.50 mm. What is the coefficient of uniformity Cu?
Explanation:
The coefficient of uniformity is calculated as Cu = D60/D10. D60 represents the particle size at which 60% of the sample is finer by mass. D10 similarly represents the size at which 10% is finer. Substituting the supplied values gives Cu = 1.50/0.15 = 10.0. The coefficient is dimensionless because both particle sizes are expressed in the same unit.
వివరణ:
Question 10
ప్రశ్న 10
A soil has D10 = 0.20 mm, D30 = 0.60 mm and D60 = 2.00 mm. What is its coefficient of curvature Cc?
Explanation:
The coefficient of curvature is Cc = D30 squared divided by D10 times D60. The numerator is 0.60 squared = 0.36 square millimetres. The denominator is 0.20 times 2.00 = 0.40 square millimetres. Hence Cc = 0.36/0.40 = 0.90. This value falls below the usual 1 to 3 range required for well-graded clean gravels and sands.
వివరణ:
Question 11
ప్రశ్న 11
A clean gravel containing 3% fines has Cu = 5.0 and Cc = 2.0. Assume gravel predominates over sand in the coarse fraction. What is its group symbol under IS 1498?
Explanation:
The soil is identified as a clean gravel because gravel predominates and fines content is below 5%. IS 1498 requires Cu greater than 4 for the well-graded gravel group. It also requires Cc to lie between 1 and 3. The supplied values of 5.0 and 2.0 satisfy both conditions. Therefore GW, or well-graded gravel, is the correct group symbol.
వివరణ:
Question 12
ప్రశ్న 12
A clean sand contains 2% fines and has Cu = 8.0 and Cc = 0.60. Which IS 1498 group symbol is appropriate?
Explanation:
The fines percentage is below 5%, so the clean-sand gradation criteria are applicable. Well-graded sand requires Cu greater than 6 and Cc between 1 and 3. The given Cu of 8.0 satisfies the uniformity criterion. However, the coefficient of curvature of 0.60 falls outside the required range. The sand therefore fails the combined well-graded criteria and is classified as SP, or poorly graded sand.
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Question 13
ప్రశ్న 13
A sand-dominated coarse-grained soil contains 20% fines. The fines have plasticity index 3 and show silt-like behaviour. Which group symbol is most appropriate under IS 1498?
Explanation:
A fines content of 20% exceeds the 12% boundary for appreciable fines. The classification must therefore consider the nature of the finer particles. A plasticity index below 4 is consistent with the silty-fines criterion in the relevant coarse-soil classification table. The coarse skeleton is stated to be sand-dominated. Consequently the appropriate group symbol is SM, or silty sand.
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Question 14
ప్రశ్న 14
A soil contains 60% gravel, 25% sand and 15% fines. Its fines have liquid limit 42% and plasticity index 22. Using the A-line Ip = 0.73(wL - 20), which group symbol is appropriate?
Explanation:
The coarse fraction is 60 + 25 = 85%, and gravel is dominant within it. The fines content is 15%, which exceeds the 12% limit for clean or borderline gradation classification. At liquid limit 42%, the A-line plasticity index is 0.73 times 22 = 16.06. The actual plasticity index of 22 is above that line and also exceeds 7. Therefore the fines are clayey for engineering classification and the gravel is designated GC.
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Question 15
ప్రశ్న 15
A sand-dominated coarse-grained soil contains 8% non-plastic fines. Its gradation parameters are Cu = 8.0 and Cc = 1.5. Which IS 1498 classification is most suitable?
Explanation:
The soil is sand-dominated and its fines percentage lies between 5% and 12%. Therefore a dual group symbol is required. Its Cu exceeds 6 and its Cc lies between 1 and 3, satisfying the well-graded sand criteria. The fines are non-plastic and are treated as silty rather than clayey. The appropriate designation is SW-SM, indicating well-graded sand with silt-like fines.
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Question 16
ప్రశ్న 16
A soil has 65% by mass passing the 75-micron sieve. It is inorganic, with liquid limit 42% and plasticity index 19. What is its IS 1498 group symbol?
Explanation:
Since 65% passes the 75-micron sieve, the soil belongs to the fine-grained division. Its liquid limit of 42% lies in the Indian intermediate-compressibility range between 35% and 50%. The A-line value at this liquid limit is 0.73 times 22 = 16.06. The actual plasticity index of 19 lies above the A-line and is outside the low-plasticity boundary zone. The soil is therefore classified as CI, an inorganic clay of intermediate compressibility.
వివరణ:
Question 17
ప్రశ్న 17
An inorganic fine-grained soil has liquid limit 30% and plasticity index 2. What is its IS 1498 classification, assuming it is not a borderline organic soil?
Explanation:
The liquid limit of 30% is below 35%, placing the soil in the low-compressibility category under IS 1498. The A-line value is 0.73 times (30 - 20), or 7.3. Its actual plasticity index of 2 lies well below this boundary. The soil is inorganic and behaves as a low-plasticity silt rather than a clay. The appropriate group symbol is ML.
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Question 18
ప్రశ్న 18
An inorganic fine-grained soil has liquid limit 60% and plasticity index 35. Which IS 1498 group symbol applies?
Explanation:
The liquid limit of 60% exceeds the Indian high-compressibility boundary of 50%. The A-line ordinate is 0.73 times (60 - 20) = 29.2. The measured plasticity index of 35 lies above that line. The soil is specified to be inorganic, excluding organic group symbols. It is consequently classified as CH, an inorganic clay of high compressibility.
వివరణ:
Question 19
ప్రశ్న 19
A fine-grained inorganic soil has liquid limit 65% and plasticity index 10. Which classification is appropriate under IS 1498?
Explanation:
The liquid limit is greater than 50%, placing the soil in the high-compressibility category. The A-line ordinate at liquid limit 65% is 0.73 times 45 = 32.85. The measured plasticity index of 10 is substantially below the A-line. The soil is inorganic, so an organic soil symbol is not justified. Its classification is MH, representing inorganic silt of high compressibility.
వివరణ:
Question 20
ప్రశ్న 20
For a soil with liquid limit 45%, the A-line is defined as Ip = 0.73(wL - 20). If its measured plasticity index is 21, which statement is correct?
Explanation:
The specified liquid limit is 45%. Substitution in the A-line equation gives 0.73 times (45 - 20). This equals 18.25. The measured plasticity index is 21, which exceeds the A-line ordinate. Therefore the soil plots above the A-line, indicating clay-like plasticity behaviour under the relevant classification rules.
వివరణ:
Question 21
ప్రశ్న 21
A fine-grained soil has a liquid limit of 40% before oven drying and 24% after oven drying and remixing with water. Under the organic-soil identification criterion of IS 1498, what is the appropriate interpretation?
Explanation:
The ratio of liquid limit after drying to liquid limit before drying is 24/40 = 0.60. IS 1498 identifies a substantial reduction to less than three-fourths of the original value as evidence of organic soil. The calculated ratio is below 0.75. The result therefore supports organic-soil identification under the stated method. Additional classification information is required to establish the appropriate organic group symbol and detailed soil description.
వివరణ:
Question 22
ప్రశ్న 22
During field identification, a moist fine-grained soil pat shows rapid dilatancy when shaken, little dry strength and negligible toughness near its plastic limit. Which interpretation is most defensible?
Explanation:
Dilatancy describes the response of a moist fine soil to shaking and subsequent squeezing. Rapid dilatancy is commonly associated with inorganic silts or similar low-plasticity soils. Low dry strength and negligible toughness reinforce that interpretation. Highly plastic clays generally exhibit greater toughness and a weak or absent dilatancy response. Field observations should still be supplemented by laboratory grain-size and Atterberg-limit measurements before assigning a definitive group symbol.
వివరణ:
Question 23
ప్రశ్న 23
Which statement correctly describes the purpose and physical basis of hydrometer analysis in soil particle-size determination?
Explanation:
Hydrometer analysis is used to investigate the particle-size distribution of fine soil fractions. The method observes how suspension density changes as particles settle with time. Stokes' law relates settling velocity to an equivalent particle diameter under appropriate laminar-settling assumptions. Suitable dispersion and corrections are necessary for a reliable test. The resulting distribution complements sieve analysis, but it does not determine mineral identity or plasticity classification by itself.
వివరణ:
Question 24
ప్రశ్న 24
A dry soil sample weighs 500 g and contains 45 g gravel, 280 g sand and 175 g fines. The fines are non-plastic and exhibit silt-like behaviour. Which group symbol follows from the IS 1498 coarse/fine division and fines criteria?
Explanation:
The coarse fraction is 45 + 280 = 325 g out of 500 g, which equals 65% of the sample. The soil is therefore coarse-grained. Sand dominates the coarse fraction because its mass of 280 g is much larger than the 45 g gravel component. Fines account for 175/500 times 100 = 35%, exceeding the 12% fines threshold. Because the fines are non-plastic and silt-like, the group symbol is SM rather than a clean-sand or clayey-sand symbol.
వివరణ:
Question 25
ప్రశ్న 25
A soil investigation report provides an IS 1498 group symbol obtained from sieve analysis and Atterberg limits. Which additional information is most important for a reliable engineering description of the actual foundation stratum?
Explanation:
IS 1498 provides a systematic classification based on grain-size and consistency characteristics. A group symbol is useful for communication but does not fully describe the natural soil deposit. In-situ structure, density, moisture condition, stratification and cementation can strongly affect engineering behaviour. Groundwater and drainage conditions also influence foundation performance. Therefore a geotechnical report should combine the classification symbol with relevant field observations and engineering test results.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. More than 50% of the total soil material by dry mass must be coarser than the 75-micron IS sieve size for the soil to be coarse-grained. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. Gravel: 4.75–75 mm; sand: 75 microns–4.75 mm. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. The particles may be described as clay-size, but the engineering classification as clay or silt also requires consideration of plasticity and related behaviour. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. Fine-grained soil because more than half its mass is finer than 75 microns. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: B. Coarse-grained gravel because 68% is coarser than 75 microns and gravel predominates within that coarse fraction. సమాధానం: B.
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Question 6 ప్రశ్న 6Answer: A. It is coarse-grained sand because the coarse fraction is 68% and sand is its dominant component. సమాధానం: A.
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Question 7 ప్రశ్న 7Answer: D. Less than 5% fines generally permits clean-soil symbols, more than 12% requires classification according to the fines, and 5–12% is a borderline range requiring dual symbols. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: C. SM. సమాధానం: C.
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Question 9 ప్రశ్న 9Answer: B. 10.0. సమాధానం: B.
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Question 10 ప్రశ్న 10Answer: A. 0.90. సమాధానం: A.
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Question 11 ప్రశ్న 11Answer: D. GW. సమాధానం: D.
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Question 12 ప్రశ్న 12Answer: C. SP. సమాధానం: C.
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Question 13 ప్రశ్న 13Answer: B. SM. సమాధానం: B.
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Question 14 ప్రశ్న 14Answer: A. GC. సమాధానం: A.
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Question 15 ప్రశ్న 15Answer: D. SW-SM. సమాధానం: D.
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Question 16 ప్రశ్న 16Answer: C. CI. సమాధానం: C.
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Question 17 ప్రశ్న 17Answer: B. ML. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: A. CH. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: D. MH. సమాధానం: D.
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Question 20 ప్రశ్న 20Answer: C. The A-line ordinate is 18.25, and the soil plots above it. సమాధానం: C.
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Question 21 ప్రశ్న 21Answer: B. The ratio is 0.60, below 0.75, which provides positive identification of organic soil under the stated criterion. సమాధానం: B.
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Question 22 ప్రశ్న 22Answer: A. The observations are characteristic of silt-like, low-plasticity behaviour, subject to confirmation by laboratory classification tests. సమాధానం: A.
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Question 23 ప్రశ్న 23Answer: D. It estimates the distribution of fine particle sizes from changes in suspension density during settling, using principles based on Stokes' law under suitable assumptions. సమాధానం: D.
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Question 24 ప్రశ్న 24Answer: C. SM. సమాధానం: C.
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Question 25 ప్రశ్న 25Answer: A. In-situ density or consistency, moisture, stratification, cementation, drainage conditions and other relevant field observations in addition to the laboratory group symbol. సమాధానం: A.
