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In a shrinkage limit test, the volume and mass of a dry soil pat are found to be 50 cm3 and 88 g. respectively. The specific gravity of the soil solids is 2.71 and the density of water is 1 g/cc. The shrinkage limit (in % up to two decimal places) is___________.
Correct answer is '(19.90)'. Can you explain this answer?
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Verified Answer
In a shrinkage limit test, the volume and mass of a dry soil pat are f...
Dry soil mass = 88 gm
Volume of dry soil = 50 cc
Dry density of soil mass,
∴ Shrinkage limit,
Volume of dry soil = 50 cc
Dry density of soil mass,
∴ Shrinkage limit,
Most Upvoted Answer
In a shrinkage limit test, the volume and mass of a dry soil pat are f...
Solution:
Given data:
Volume of dry soil pat (V) = 50 cm³
Mass of dry soil pat (M) = 88 g
Specific gravity of soil solids (G) = 2.71
Density of water (d) = 1 g/cc
Shrinkage limit:
The shrinkage limit of soil is defined as the minimum water content at which further reduction in water content does not cause any volume change in the soil mass.
Formula used:
Shrinkage limit (SL) = (M/V) × (G/d) × 100
Calculation:
Substituting the given values in the above formula, we get:
SL = (88/50) × (2.71/1) × 100
SL = 19.90% (approx)
Therefore, the shrinkage limit of soil is 19.90%.
HTML Code:
Given data:
Volume of dry soil pat (V) = 50 cm³
Mass of dry soil pat (M) = 88 g
Specific gravity of soil solids (G) = 2.71
Density of water (d) = 1 g/cc
Shrinkage limit:
The shrinkage limit of soil is defined as the minimum water content at which further reduction in water content does not cause any volume change in the soil mass.
Formula used:
Shrinkage limit (SL) = (M/V) × (G/d) × 100
Calculation:
Substituting the given values in the above formula, we get:
SL = (88/50) × (2.71/1) × 100
SL = 19.90% (approx)
Therefore, the shrinkage limit of soil is 19.90%.
HTML Code:
Given data:
- Volume of dry soil pat (V) = 50 cm³
- Mass of dry soil pat (M) = 88 g
- Specific gravity of soil solids (G) = 2.71
- Density of water (d) = 1 g/cc
Formula used:
Shrinkage limit (SL) = (M/V) × (G/d) × 100
Calculation:
Substituting the given values in the above formula, we get:
- SL = (88/50) × (2.71/1) × 100
- SL = 19.90% (approx)
Therefore, the shrinkage limit of soil is 19.90%.
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In a shrinkage limit test, the volume and mass of a dry soil pat are found to be 50 cm3 and 88 g. respectively. The specific gravity of the soil solids is 2.71 and the density of water is 1 g/cc. The shrinkage limit (in % up to two decimal places) is___________.Correct answer is '(19.90)'. Can you explain this answer?
Question Description
In a shrinkage limit test, the volume and mass of a dry soil pat are found to be 50 cm3 and 88 g. respectively. The specific gravity of the soil solids is 2.71 and the density of water is 1 g/cc. The shrinkage limit (in % up to two decimal places) is___________.Correct answer is '(19.90)'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about In a shrinkage limit test, the volume and mass of a dry soil pat are found to be 50 cm3 and 88 g. respectively. The specific gravity of the soil solids is 2.71 and the density of water is 1 g/cc. The shrinkage limit (in % up to two decimal places) is___________.Correct answer is '(19.90)'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a shrinkage limit test, the volume and mass of a dry soil pat are found to be 50 cm3 and 88 g. respectively. The specific gravity of the soil solids is 2.71 and the density of water is 1 g/cc. The shrinkage limit (in % up to two decimal places) is___________.Correct answer is '(19.90)'. Can you explain this answer?.
In a shrinkage limit test, the volume and mass of a dry soil pat are found to be 50 cm3 and 88 g. respectively. The specific gravity of the soil solids is 2.71 and the density of water is 1 g/cc. The shrinkage limit (in % up to two decimal places) is___________.Correct answer is '(19.90)'. Can you explain this answer? for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about In a shrinkage limit test, the volume and mass of a dry soil pat are found to be 50 cm3 and 88 g. respectively. The specific gravity of the soil solids is 2.71 and the density of water is 1 g/cc. The shrinkage limit (in % up to two decimal places) is___________.Correct answer is '(19.90)'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In a shrinkage limit test, the volume and mass of a dry soil pat are found to be 50 cm3 and 88 g. respectively. The specific gravity of the soil solids is 2.71 and the density of water is 1 g/cc. The shrinkage limit (in % up to two decimal places) is___________.Correct answer is '(19.90)'. Can you explain this answer?.
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