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Question 1
ప్రశ్న 1
Which statement correctly identifies the liquid limit, plastic limit and shrinkage limit of a fine-grained soil?
Explanation:
Atterberg limits describe changes in the consistency of fine-grained soils as water content varies. The liquid limit is the conventional boundary between plastic and liquid-like behaviour. The plastic limit separates plastic behaviour from the semi-solid state. The shrinkage limit identifies the water content below which further drying produces essentially no additional volume reduction. These boundaries are determined through standardized index tests rather than from the degree of saturation alone.
వివరణ:
Question 2
ప్రశ్న 2
Under the mechanical liquid-limit procedure in IS 2720 (Part 5), which statement is correct?
Explanation:
The mechanical liquid-limit method uses a standardized cup and grooving arrangement. The soil groove is closed over the specified distance by repeated drops of the cup. IS 2720 (Part 5) uses approximately 12 mm closure in the mechanical procedure. The liquid limit is obtained from the flow curve at 25 blows. This procedure differs from the cone-penetration method, which uses a specified penetration depth instead of a blow count.
వివరణ:
Question 3
ప్రశ్న 3
Which observation correctly represents the endpoint used to determine the plastic limit of a soil by the rolling-thread method?
Explanation:
The plastic-limit test evaluates the water content at which a soil ceases to be satisfactorily rolled into a thin thread. The conventional procedure uses a thread diameter of approximately 3 mm. Rolling and kneading are continued until the prescribed crumbling condition is reached. The crumbled material is then used for determining water content. This endpoint is distinct from the liquid-limit and shrinkage-limit test conditions.
వివరణ:
Question 4
ప్రశ్న 4
As the water content of a typical plastic fine-grained soil decreases from above its liquid limit to below its shrinkage limit, which sequence of consistency states is correct?
Explanation:
A highly wet fine-grained soil may exhibit liquid-like behaviour. As water content falls below the liquid limit, the soil enters the plastic consistency range. Further drying below the plastic limit produces semi-solid behaviour. The shrinkage limit marks the transition toward the solid consistency state. These states are idealized index-property descriptions, and the behaviour of natural undisturbed soil can also depend on its structure and stress history.
వివరణ:
Question 5
ప్రశ్న 5
A fine-grained soil has liquid limit 62% and plastic limit 27%. What is its plasticity index?
Explanation:
Plasticity index is the numerical difference between liquid limit and plastic limit. The specified liquid limit is 62%, while the plastic limit is 27%. Subtraction gives I_P = 62 - 27 = 35 percentage points. This represents the width of the water-content interval over which the soil exhibits plastic consistency. Plasticity index is an index property and does not by itself establish a unique shear strength or compressibility.
వివరణ:
Question 6
ప్రశ్న 6
A laboratory reports a liquid limit of 24% and a plastic limit of 29% for the same tested soil. According to the reporting convention in IS 2720 (Part 5), what plasticity index should be reported when the plastic limit is equal to or greater than the liquid limit?
Explanation:
The usual plasticity-index equation is the liquid limit minus the plastic limit. Direct subtraction of the reported values would give 24 - 29 = -5. IS 2720 (Part 5) provides a reporting convention under which the plasticity index is reported as zero when the plastic limit equals or exceeds the liquid limit. A negative plasticity index is therefore not reported as the test result under this convention. The laboratory should nevertheless review the measurements if their consistency or reliability is in doubt.
వివరణ:
Question 7
ప్రశ్న 7
A soil has liquid limit 60%, plastic limit 25% and natural water content 39%. What is its liquidity index?
Explanation:
Liquidity index is calculated using I_L = (w - w_P)/(w_L - w_P). The plasticity index is 60 - 25 = 35 percentage points. The natural water content exceeds the plastic limit by 39 - 25 = 14 percentage points. Hence I_L = 14/35 = 0.40. This places the natural water content within the plastic consistency range.
వివరణ:
Question 8
ప్రశ్న 8
For a soil having liquid limit 60%, plastic limit 25% and natural water content 39%, what is its consistency index?
Explanation:
Consistency index is given by I_C = (w_L - w)/(w_L - w_P). The plasticity index is 60 - 25 = 35. The difference between liquid limit and natural water content is 60 - 39 = 21. Therefore I_C = 21/35 = 0.60. This value is consistent with the corresponding liquidity index of 0.40 because the two indices sum to one.
వివరణ:
Question 9
ప్రశ్న 9
For a soil with a positive plasticity index, which mathematical relationship between liquidity index I_L and consistency index I_C is necessarily correct when both use the same liquid limit, plastic limit and natural water content?
Explanation:
Liquidity index and consistency index use the same plasticity index as their denominator. Their numerators are w - w_P and w_L - w, respectively. Adding these numerators gives w_L - w_P, which equals the plasticity index. Dividing by the same positive plasticity index gives I_L + I_C = 1. The relationship remains valid even when the natural water content lies outside the plastic consistency range.
వివరణ:
Question 10
ప్రశ్న 10
A soil has liquid limit 48%, plastic limit 18% and natural water content 12%. What are its liquidity index and the appropriate interpretation?
Explanation:
The plasticity index is 48 - 18 = 30. The liquidity-index numerator is 12 - 18 = -6. Dividing gives I_L = -6/30 = -0.20. A negative liquidity index indicates that natural water content is below the plastic limit. Whether the soil is specifically semi-solid or solid requires the shrinkage limit, which has not been supplied.
వివరణ:
Question 11
ప్రశ్న 11
A fine-grained soil has liquid limit 55%, plastic limit 30% and natural water content 65%. What is its liquidity index, and what does it indicate?
Explanation:
The soil has plasticity index I_P = 55 - 30 = 25. Natural water content exceeds the plastic limit by 65 - 30 = 35 percentage points. Therefore I_L = 35/25 = 1.40. A liquidity index greater than one means the natural water content exceeds the liquid limit. It suggests liquid-like remoulded consistency, although actual in-situ strength also depends on soil fabric and stress conditions.
వివరణ:
Question 12
ప్రశ్న 12
A soil has liquid limit 65% and plastic limit 25%. If its liquidity index is 0.75, what is its natural water content?
Explanation:
The plasticity index is 65 - 25 = 40. Liquidity index is given by I_L = (w - w_P)/I_P. Rearranging gives w = w_P + I_L I_P. Therefore w = 25 + 0.75 times 40 = 55%. This lies between the plastic limit and liquid limit, which agrees with the given liquidity index.
వివరణ:
Question 13
ప్రశ్న 13
A soil has liquid limit 70%, plastic limit 30% and consistency index 0.25. Determine its natural water content.
Explanation:
The plasticity index is 70 - 30 = 40. Consistency index is I_C = (w_L - w)/I_P. Rearranging gives w = w_L - I_C I_P. The substitution is w = 70 - 0.25 times 40 = 60%. As an independent check, the liquidity index is 0.75, giving the same water content from the plastic limit.
వివరణ:
Question 14
ప్రశ్న 14
Soil A has liquid limit 45% and plastic limit 20%, while Soil B has liquid limit 60% and plastic limit 40%. Which comparison is correct?
Explanation:
For Soil A, plasticity index is 45 - 20 = 25. For Soil B, plasticity index is 60 - 40 = 20. Soil A therefore has the wider plastic consistency range by 5 percentage points. A comparison of plasticity indices is justified from the supplied limits. However, plasticity index alone does not guarantee a particular ranking of field compressibility or swelling potential.
వివరణ:
Question 15
ప్రశ్న 15
In a Casagrande flow-curve investigation, water content is 50% at 10 blows and 30% at 100 blows. Assuming a straight line on the plot of water content against log10(number of blows), what is the flow index?
Explanation:
The flow index is the magnitude of the slope of the water-content versus logarithm-of-blows relationship. The water-content change is 50 - 30 = 20 percentage points. The corresponding logarithmic interval is log10(100) - log10(10) = 2 - 1 = 1. Therefore the flow index is 20/1 = 20. The sign is conventionally reported as a positive magnitude because water content decreases as the number of blows increases.
వివరణ:
Question 16
ప్రశ్న 16
A soil gives water content 45% at 20 blows and 35% at 80 blows in a mechanical liquid-limit test. Assuming an exactly linear relationship between water content and log10(blow count), what is the interpolated water content at 25 blows, before rounding to the reporting precision?
Explanation:
The water content varies linearly with the logarithm of the blow count under the specified assumption. The water-content difference is 10 percentage points between 20 and 80 blows. The fraction of the logarithmic interval from 20 to 25 blows is log10(25/20)/log10(80/20). Subtracting the corresponding decrease from 45% gives approximately 43.39%. This is the interpolated value before applying the reporting precision of the specified laboratory method.
వివరణ:
Question 17
ప్రశ్న 17
A soil has plasticity index 30 and flow index 12. What is its toughness index?
Explanation:
Toughness index is defined as the ratio of plasticity index to flow index. The plasticity index is 30 and the flow index is 12. Therefore I_T = 30/12 = 2.50. The index provides a comparative measure associated with the soil's plastic consistency behaviour. It is not a direct measurement of undrained shear strength or a substitute for a mechanical strength test.
వివరణ:
Question 18
ప్రశ్న 18
Two soils have equal plasticity indices, but Soil P has a greater flow index than Soil Q. What relationship between their toughness indices necessarily follows?
Explanation:
Toughness index equals plasticity index divided by flow index. The two soils are stated to have equal plasticity indices. A greater flow-index denominator therefore produces a smaller toughness index. Consequently Soil P has a lower toughness index than Soil Q. This mathematical comparison does not independently establish which soil will have greater in-situ bearing capacity.
వివరణ:
Question 19
ప్రశ్న 19
A remoulded soil has plastic limit 20% and shrinkage limit 12%. What is its shrinkage index?
Explanation:
Shrinkage index is the difference between plastic limit and shrinkage limit for the relevant remoulded-soil determination. The supplied plastic limit is 20%. The supplied shrinkage limit is 12%. Therefore shrinkage index is 20 - 12 = 8 percentage points. This describes the water-content interval between the plastic and shrinkage limits, not the percentage change in specimen volume.
వివరణ:
Question 20
ప్రశ్న 20
A remoulded soil pat initially has wet mass 50 g and wet volume 30 cm³. Its oven-dry mass is 35 g and dry volume is 22 cm³. Taking water density as 1 g/cm³ and using w_s = [(M_wet - M_dry) - rho_w(V_wet - V_dry)]/M_dry times 100, what is its shrinkage limit?
Explanation:
The initial water mass is 50 - 35 = 15 g. The reduction in specimen volume during drying is 30 - 22 = 8 cm³. At the stated water density, this volume change corresponds to 8 g of water. The water mass remaining at the shrinkage limit is therefore represented by 15 - 8 = 7 g. Dividing 7 by the dry mass of 35 g and multiplying by 100 gives a shrinkage limit of 20.00%.
వివరణ:
Question 21
ప్రశ్న 21
A soil pat has initial volume 40 cm³ and oven-dry volume 32 cm³. Using volumetric shrinkage defined as the reduction in volume expressed as a percentage of the oven-dry volume, determine its volumetric shrinkage.
Explanation:
The specimen undergoes a volume reduction of 40 - 32 = 8 cm³. The question defines the reference volume as the oven-dry volume. Therefore volumetric shrinkage is 8/32 times 100. This gives 25%. Dividing by the initial volume instead would produce 20%, which is a different percentage measure.
వివరణ:
Question 22
ప్రశ్న 22
A soil pat has oven-dry mass 48 g and oven-dry volume 30 cm³. Take water density as 1 g/cm³, initial water content as 27% and shrinkage limit as 12%. Using shrinkage ratio R = M_dry/(rho_w V_dry) and volumetric shrinkage V_S = R(w_initial - w_s), with water contents entered as percentages, what is the volumetric shrinkage?
Explanation:
The shrinkage ratio is calculated from dry mass divided by the mass of water occupying the dry specimen volume. With the stated quantities, R = 48/(1 times 30) = 1.60. The water-content difference above the shrinkage limit is 27 - 12 = 15 percentage points. Hence volumetric shrinkage is 1.60 times 15 = 24%. The calculation applies the supplied shrinkage relationship and uses the dry specimen volume as the reference.
వివరణ:
Question 23
ప్రశ్న 23
A soil has shrinkage limit 10%, plastic limit 22% and liquid limit 48%. Its current water content is 16%. Which consistency description and drying behaviour are appropriate under the conventional Atterberg-state model?
Explanation:
The supplied water content of 16% is below the plastic limit of 22%. It is nevertheless above the shrinkage limit of 10%. Under the conventional Atterberg-state description, this places the soil in the semi-solid range. Drying through this range may continue to reduce the specimen volume. Once the shrinkage limit is reached, further drying produces essentially no additional shrinkage under the idealized model.
వివరణ:
Question 24
ప్రశ్న 24
A foundation engineer observes that Soil P has a substantially higher plasticity index than Soil Q. Which conclusion is most defensible without additional testing?
Explanation:
Plasticity index measures the difference between liquid limit and plastic limit. A higher value indicates a wider interval of water content over which remoulded soil exhibits plastic consistency. Plasticity can be useful for anticipating potential engineering concerns and selecting further tests. However, mineralogy, soil structure, density, stress history and water conditions also affect behaviour. It is therefore unsafe to calculate a specific strength, settlement or swelling magnitude from plasticity index alone.
వివరణ:
Question 25
ప్రశ్న 25
A soil has liquid limit 64%, plastic limit 28%, natural water content 40% and flow index 18. Which combination correctly gives plasticity index, liquidity index, consistency index and toughness index, respectively?
Explanation:
Plasticity index is 64 - 28 = 36. Liquidity index is (40 - 28)/36 = 12/36, or approximately 0.333. Consistency index is (64 - 40)/36 = 24/36, or approximately 0.667. Toughness index is 36/18 = 2.00. As an independent verification, liquidity index plus consistency index equals one, and the natural water content lies within the plastic range.
వివరణ:
Answer Key సమాధానాల పట్టిక
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Question 1 ప్రశ్న 1Answer: B. Liquid limit separates plastic and liquid consistency states, plastic limit separates semi-solid and plastic states, and shrinkage limit marks the boundary below which further drying produces essentially no additional volume reduction. సమాధానం: B.
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Question 2 ప్రశ్న 2Answer: D. Liquid limit is determined from the water content corresponding to 25 blows on the flow curve, with groove closure measured over approximately 12 mm in the specified mechanical test. సమాధానం: D.
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Question 3 ప్రశ్న 3Answer: A. A soil thread repeatedly rolled to approximately 3 mm diameter reaches a condition at which it crumbles under the specified rolling procedure. సమాధానం: A.
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Question 4 ప్రశ్న 4Answer: C. Liquid, plastic, semi-solid, solid. సమాధానం: C.
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Question 5 ప్రశ్న 5Answer: A. 35 percentage points. సమాధానం: A.
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Question 6 ప్రశ్న 6Answer: C. 0. సమాధానం: C.
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Question 7 ప్రశ్న 7Answer: D. 0.40. సమాధానం: D.
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Question 8 ప్రశ్న 8Answer: B. 0.60. సమాధానం: B.
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Question 9 ప్రశ్న 9Answer: A. I_L + I_C = 1. సమాధానం: A.
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Question 10 ప్రశ్న 10Answer: C. -0.20; the natural water content is below the plastic limit. సమాధానం: C.
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Question 11 ప్రశ్న 11Answer: B. 1.40; its natural water content exceeds the liquid limit. సమాధానం: B.
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Question 12 ప్రశ్న 12Answer: D. 55%. సమాధానం: D.
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Question 13 ప్రశ్న 13Answer: A. 60%. సమాధానం: A.
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Question 14 ప్రశ్న 14Answer: C. Soil A has plasticity index 25, while Soil B has plasticity index 20, so Soil A has the wider plastic water-content range. సమాధానం: C.
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Question 15 ప్రశ్న 15Answer: C. 20. సమాధానం: C.
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Question 16 ప్రశ్న 16Answer: D. 43.39%. సమాధానం: D.
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Question 17 ప్రశ్న 17Answer: B. 2.50. సమాధానం: B.
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Question 18 ప్రశ్న 18Answer: A. Soil P has the lower toughness index because toughness index is inversely proportional to flow index when plasticity index is fixed. సమాధానం: A.
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Question 19 ప్రశ్న 19Answer: C. 8. సమాధానం: C.
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Question 20 ప్రశ్న 20Answer: D. 20.00%. సమాధానం: D.
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Question 21 ప్రశ్న 21Answer: A. 25%. సమాధానం: A.
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Question 22 ప్రశ్న 22Answer: C. 24%. సమాధానం: C.
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Question 23 ప్రశ్న 23Answer: B. Semi-solid state; further drying toward the shrinkage limit can still reduce its volume. సమాధానం: B.
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Question 24 ప్రశ్న 24Answer: D. Soil P has a wider plastic water-content range, but swelling, compressibility and strength require additional soil-characterization and engineering assessment. సమాధానం: D.
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Question 25 ప్రశ్న 25Answer: A. 36, 0.333, 0.667 and 2.00. సమాధానం: A.
