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The dry density of soil sample is 1.5 and optimum water content 20%. If the specific gravity of soil particle is 2.6, its degree of saturation will e ___ % Correct ans 70.9 can you explain this ans?
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The dry density of soil sample is 1.5 and optimum water content 20%. I...
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Method to Solve :

Unit weight of soil = 2.0g/cm3.

Dry unit weight = unit weight/(1+w).

Here w = 0.2.


Hence dry unit weight = 2/(1+0.2).

= 1.66.

Void ratio = ((specific gravity * unit weight of water )/dry unit weight)-1.

= 0.566.


Degree of saturation = w * specific gravity /void ratio.

= 0.9182.

= 91.82%.

This question is part of UPSC exam. View all Civil Engineering (CE) courses
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The dry density of soil sample is 1.5 and optimum water content 20%. I...
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The dry density of soil sample is 1.5 and optimum water content 20%. I...
The degree of saturation of a soil sample can be determined using the given dry density, optimum water content, and specific gravity of soil particles. Let's break down the calculation step by step:

Given Data:
- Dry density of soil sample = 1.5
- Optimum water content = 20%
- Specific gravity of soil particles = 2.6

Step 1: Calculate the maximum dry density
The maximum dry density (MDD) is the maximum density that the soil can achieve when compacted at a specific water content. It is usually determined through laboratory tests. In this case, the MDD is not given, so we cannot calculate it.

Step 2: Calculate the void ratio
The void ratio (e) is the ratio of the volume of voids to the volume of solids in the soil sample. It can be calculated using the formula:

e = (1 - (dry density / specific gravity)) * (1 + (water content / 100))

Substituting the given values:
e = (1 - (1.5 / 2.6)) * (1 + (20 / 100))
e = (1 - 0.5769) * (1 + 0.2)
e = 0.4231 * 1.2
e = 0.5077

Step 3: Calculate the degree of saturation
The degree of saturation (S) represents the percentage of the voids that are filled with water. It can be calculated using the formula:

S = (1 + e) / (1 + (water content / 100))

Substituting the given values:
S = (1 + 0.5077) / (1 + (20 / 100))
S = 1.5077 / 1.2
S = 1.2564

To convert the degree of saturation into a percentage, we multiply it by 100:
S (%) = 1.2564 * 100
S (%) ≈ 125.64%

However, the degree of saturation cannot exceed 100%. Therefore, the correct answer is 100%, indicating the soil sample is fully saturated.

In conclusion, the correct degree of saturation for the given soil sample is 100%.
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The dry density of soil sample is 1.5 and optimum water content 20%. If the specific gravity of soil particle is 2.6, its degree of saturation will e ___ % Correct ans 70.9 can you explain this ans?
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