Problem Statement (Civil Services Exam 1990)
A soil sample coated with paraffin wax has:
- Mass of coated soil: \( 690.6 \, \text{g} \)
- Mass of soil alone: \( 683 \, \text{g} \)
- Volume of displaced water: \( 350 \, \text{ml} \)
- Specific gravity of soil (\( G \)): \( 2.73 \)
- Specific gravity of wax (\( G_p \)): \( 0.89 \)
- Water content (\( w \)): \( 17\% \)
Find the void ratio (\( e \)) and degree of saturation (\( S \)).
Solution
1. Paraffin Wax Calculations
\( M_p = 690.6 \, \text{g} – 683 \, \text{g} = 7.6 \, \text{g} \)
\( V_p = \frac{M_p}{G_p} = \frac{7.6}{0.89} \approx 8.539 \, \text{cm}^3 \)
2. Soil Volume and Bulk Density
\( V_{\text{soil}} = 350 \, \text{cm}^3 – 8.539 \, \text{cm}^3 = 341.461 \, \text{cm}^3 \)
\( \rho_{\text{bulk}} = \frac{683}{341.461} \approx 2 \, \text{g/cm}^3 \)
3. Dry Density and Void Ratio
\( \rho_d = \frac{2}{1 + 0.17} \approx 1.709 \, \text{g/cm}^3 \)
\( e = \frac{2.73 \times 1}{1.709} – 1 \approx 0.597 \)
4. Degree of Saturation
\( S = \frac{0.17 \times 2.73}{0.597} \approx 77.73\% \)
Results:
- Void ratio: \( e \approx 0.597 \)
- Degree of saturation: \( S \approx 77.73\% \)
Explanation
Key Steps:
- The paraffin coating’s mass and volume are subtracted to isolate the soil’s properties.
- Bulk density accounts for total mass and volume, while dry density removes water’s effect.
- Void ratio and saturation link porosity and water content to soil behavior.
Physical Meaning
1. Void Ratio (0.597):
- Moderate porosity, typical of sandy soils, allows for drainage and root penetration.
2. Degree of Saturation (77.73%):
- Soil is mostly saturated, reducing air voids and increasing stability for construction.
Exam Context: Tests practical understanding of soil mechanics for geotechnical engineering applications.
