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The laboratory tests on a soil sample gave the following results:
1). Natural water content, wn = 25%
2). Liquid limit, wL = 60%
3). Plastic limit, wp = 35%
4). Percentage of particle less than 2μ = 20%
Q. The value of activity number is
  • a)
    1.25
  • b)
    1.20
  • c)
    1.0
  • d)
    0.8
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The laboratory tests on a soil sample gave the following results:1). N...
Plasticity index,
Ip = WL-WP = 60-35 = 25%
Soil activity (A) is a number equal to IP/CF, where IP is the plasticity index of soil, and the CF is percentage of clay particles in soil (particles smaller than 2 microns)
Activity number
IP/c = 0.25/0.20 = 1.25
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Most Upvoted Answer
The laboratory tests on a soil sample gave the following results:1). N...
Mm, D2 = 40%
Based on these results, we can determine the following parameters:

1) Plasticity Index (PI): The plasticity index is the difference between the liquid limit and the plastic limit. PI = wL - wp = 60% - 35% = 25%.

2) Liquidity Index (LI): The liquidity index is the ratio of the plasticity index to the difference between the natural water content and the plastic limit. LI = PI / (wn - wp) = 25% / (25% - 35%) = -2.5.

3) Activity Index (AI): The activity index is the ratio of the plasticity index to the difference between the liquid limit and the particle size index. AI = PI / (wL - D2) = 25% / (60% - 40%) = 1.

These parameters can provide insight into the behavior and engineering properties of the soil sample.
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The laboratory tests on a soil sample gave the following results:1). Natural water content, wn = 25%2). Liquid limit, wL = 60%3). Plastic limit, wp = 35%4). Percentage of particle less than 2μ = 20%Q. The value of activity number isa)1.25b)1.20c)1.0d)0.8Correct answer is option 'A'. Can you explain this answer?
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