Degree of saturation from bulk density, water content, and Gs A soil has mass specific gravity (bulk density relative to water) ρ/ρw = 1.92 and moisture content w = 30%. If the specific gravity of solids Gs = 2.75, compute the degree of saturation S (in %).

Introduction / Context:Phase relations in soils connect densities, water content, void ratio, and degree of saturation. Given bulk density, w, and Gs, we can find S by combining standard formulas—useful in compaction control and assessing near-saturation field conditions.

Given Data / Assumptions:

• Bulk density ratio: ρ/ρw = 1.92 (so ρ = 1.92 g/cm³).
• Moisture content w = 30% = 0.30.
• Specific gravity Gs = 2.75.
• ρw = 1.0 g/cm³.

Concept / Approach:

Use the relationships: ρd = ρ / (1 + w) ρd = (Gs * ρw) / (1 + e) ⇒ e = (Gs * ρw / ρd) − 1 S = (w * Gs) / e These follow from definitions of water content and degree of saturation (no LaTeX used).

Step-by-Step Solution:

Verification / Alternative check:

Cross-check using bulk formula ρ = ρw * (Gs + S*e) / (1 + e) with S ≈ 0.957 gives ρ ≈ 1.92 g/cm³, confirming consistency.

Why Other Options Are Wrong:

95.4–95.6% are near-misses from rounding; 94.0% is too low for the given high density and w.

Common Pitfalls:

Confusing mass and unit weight; using w in percent rather than fraction; arithmetic errors when computing e.

Final Answer:

95.7%