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The porosity of a soil sample is 30% and specific gravity of its particles is 2.75. Then Saturated unit weight is equal to
- a)1.69gm/cm3
- b)1.94 gm/cm3
- c)2.23 gm/cm3
- d)2.46 gm/cm3
Correct answer is option 'C'. Can you explain this answer?
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The porosity of a soil sample is 30% and specific gravity of its parti...
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The porosity of a soil sample is 30% and specific gravity of its parti...
Preliminary Information:
Porosity is the ratio of the volume of voids in a soil sample to the total volume of the soil sample. It is expressed as a percentage. Specific gravity is the ratio of the density of a substance to the density of a reference substance (usually water for solids and liquids). It is a dimensionless quantity.
Given data:
Porosity (n) = 30% = 0.3
Specific gravity (G) = 2.75
Calculation:
Step 1: Calculate the dry unit weight of the soil sample (γd).
The dry unit weight is the weight of the solid particles in a soil sample divided by the total volume of the sample. It is calculated using the formula:
γd = (1 - n) * G * γw
Where:
γd = Dry unit weight of soil
n = Porosity
G = Specific gravity of soil particles
γw = Unit weight of water (assumed to be 9.81 kN/m3)
Substituting the given values into the formula:
γd = (1 - 0.3) * 2.75 * 9.81
γd = 0.7 * 2.75 * 9.81
γd = 19.13 kN/m3
Step 2: Calculate the saturated unit weight (γsat).
The saturated unit weight is the weight of the saturated soil sample divided by its total volume. It can be calculated using the formula:
γsat = γd + γw
Where:
γsat = Saturated unit weight of soil
γd = Dry unit weight of soil
γw = Unit weight of water (assumed to be 9.81 kN/m3)
Substituting the calculated value of γd and the assumed value of γw into the formula:
γsat = 19.13 + 9.81
γsat = 28.94 kN/m3
Conversion:
To convert the unit weight from kN/m3 to gm/cm3, divide the value by 9.81.
1 kN/m3 = 1 gm/cm3
Therefore, the saturated unit weight is:
γsat = 28.94 / 9.81
γsat ≈ 2.95 gm/cm3
Conclusion:
The saturated unit weight of the soil sample is approximately 2.95 gm/cm3. However, none of the given options match this value. Therefore, it seems there is an error in the provided answer options. The correct answer cannot be determined based on the given options.
Porosity is the ratio of the volume of voids in a soil sample to the total volume of the soil sample. It is expressed as a percentage. Specific gravity is the ratio of the density of a substance to the density of a reference substance (usually water for solids and liquids). It is a dimensionless quantity.
Given data:
Porosity (n) = 30% = 0.3
Specific gravity (G) = 2.75
Calculation:
Step 1: Calculate the dry unit weight of the soil sample (γd).
The dry unit weight is the weight of the solid particles in a soil sample divided by the total volume of the sample. It is calculated using the formula:
γd = (1 - n) * G * γw
Where:
γd = Dry unit weight of soil
n = Porosity
G = Specific gravity of soil particles
γw = Unit weight of water (assumed to be 9.81 kN/m3)
Substituting the given values into the formula:
γd = (1 - 0.3) * 2.75 * 9.81
γd = 0.7 * 2.75 * 9.81
γd = 19.13 kN/m3
Step 2: Calculate the saturated unit weight (γsat).
The saturated unit weight is the weight of the saturated soil sample divided by its total volume. It can be calculated using the formula:
γsat = γd + γw
Where:
γsat = Saturated unit weight of soil
γd = Dry unit weight of soil
γw = Unit weight of water (assumed to be 9.81 kN/m3)
Substituting the calculated value of γd and the assumed value of γw into the formula:
γsat = 19.13 + 9.81
γsat = 28.94 kN/m3
Conversion:
To convert the unit weight from kN/m3 to gm/cm3, divide the value by 9.81.
1 kN/m3 = 1 gm/cm3
Therefore, the saturated unit weight is:
γsat = 28.94 / 9.81
γsat ≈ 2.95 gm/cm3
Conclusion:
The saturated unit weight of the soil sample is approximately 2.95 gm/cm3. However, none of the given options match this value. Therefore, it seems there is an error in the provided answer options. The correct answer cannot be determined based on the given options.
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Question Description
The porosity of a soil sample is 30% and specific gravity of its particles is 2.75. Then Saturated unit weight is equal toa)1.69gm/cm3b)1.94 gm/cm3c)2.23 gm/cm3d)2.46 gm/cm3Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The porosity of a soil sample is 30% and specific gravity of its particles is 2.75. Then Saturated unit weight is equal toa)1.69gm/cm3b)1.94 gm/cm3c)2.23 gm/cm3d)2.46 gm/cm3Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The porosity of a soil sample is 30% and specific gravity of its particles is 2.75. Then Saturated unit weight is equal toa)1.69gm/cm3b)1.94 gm/cm3c)2.23 gm/cm3d)2.46 gm/cm3Correct answer is option 'C'. Can you explain this answer?.
The porosity of a soil sample is 30% and specific gravity of its particles is 2.75. Then Saturated unit weight is equal toa)1.69gm/cm3b)1.94 gm/cm3c)2.23 gm/cm3d)2.46 gm/cm3Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The porosity of a soil sample is 30% and specific gravity of its particles is 2.75. Then Saturated unit weight is equal toa)1.69gm/cm3b)1.94 gm/cm3c)2.23 gm/cm3d)2.46 gm/cm3Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The porosity of a soil sample is 30% and specific gravity of its particles is 2.75. Then Saturated unit weight is equal toa)1.69gm/cm3b)1.94 gm/cm3c)2.23 gm/cm3d)2.46 gm/cm3Correct answer is option 'C'. Can you explain this answer?.
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