Void ratio from mass specific gravity and water content: A soil has mass specific gravity G_m = 1.92 and water content w = 30%. If the specific gravity of solids G_s = 2.75, determine the void ratio e.

Difficulty: Medium

Correct Answer: 0.862

Explanation:


Introduction / Context:
Relating mass specific gravity G_m, water content w, and specific gravity of solids G_s to the void ratio e is a common task in soil phase-relationship problems. Using two standard equations allows elimination of degree of saturation S and solution for e directly.


Given Data / Assumptions:

  • G_m = γ/γ_w = 1.92.
  • w = 0.30 (30%).
  • G_s = 2.75.
  • Water unit weight/ density normalization is implied.


Concept / Approach:

Key relations:
1) G_m = (G_s + S e) / (1 + e).
2) w = (S e) / G_s.
Use (2) to substitute S e = w G_s into (1), then solve for e.


Step-by-Step Solution:

Compute S e from w: S e = w * G_s = 0.30 * 2.75 = 0.825.Substitute in G_m: (G_s + S e) / (1 + e) = 1.92 → (2.75 + 0.825) / (1 + e) = 1.92.Sum numerator: 3.575 / (1 + e) = 1.92.Solve: 1 + e = 3.575 / 1.92 = 1.86198 → e ≈ 0.86198.Round suitably: e ≈ 0.862.


Verification / Alternative check:

Back-calculate S from w = S e / G_s → S = (w G_s) / e ≈ 0.825 / 0.862 ≈ 0.957 (reasonable near full saturation), confirming consistency.


Why Other Options Are Wrong:

0.858, 0.860, 0.864 are close but reflect rounding beyond justified precision; 0.880 deviates from the derived e.


Common Pitfalls:

Using bulk unit weight formula for saturated soil; forgetting to eliminate S correctly; rounding too early.


Final Answer:

0.862

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