A laboratory compaction test conducted on a soil sample having specific gravity equal to 2.65 gave a maximum dry density of 1.8 g/cm³ and a degree of saturation of 96%. Take density of water ρw = 1 g/cc. Find the optimum moisture content and air content.

Asked in SSC JE 06/11/2024 (Morning)

Detailed Explanation

At the peak of the compaction curve, the measured dry density is the maximum dry density and the water content is the OMC. Using the phase relationships of soil, we can back-calculate the void ratio from the dry density, and then the water content from the degree of saturation.

Given: G = 2.65, ρ_d,max = 1.8 g/cm³, S = 0.96, ρ_w = 1 g/cm³

Step 1 — Void ratio: \( \rho_d = \dfrac{G\rho_w}{1+e} \Rightarrow 1 + e = \dfrac{2.65 \times 1}{1.8} = 1.472 \Rightarrow e = 0.472 \)

Step 2 — Water content (OMC): \( Se = wG \Rightarrow w = \dfrac{0.96 \times 0.472}{2.65} = 0.171 = 17.10\% \)

Step 3 — Air content: \( a_c = \dfrac{V_a}{V_v} = 1 - S = 1 - 0.96 = 0.04 = 4\% \)

Why “17.10%; 4%” Is Correct

17.10% and 4% come straight from the two standard relationships ρ_d = Gρ_w/(1+e) and Se = wG, with air content defined as the fraction of voids filled with air.

Why the Other Options Are Wrong

• B. 16.75%; 2%: 16.75% and 2% would need a different void ratio or saturation; they do not satisfy Se = wG with the given data.
• C. 15.58%; 5.6%: 15.58% and 5.6% mix up air content with percentage air voids or use a wrong void ratio.
• D. 17.98%; 6%: 17.98% and 6% overestimate both; 6% air content would mean S = 94%, not 96%.

Key Concepts for Students

• Always find e first from the dry density; everything else follows.
• Se = wG is the most-used phase relation in compaction numericals.
• Read the question carefully: ‘air content’ and ‘percentage air voids’ are not the same.

Practice more: this is question 10 of Compaction of Soil MCQs – Part 1. Try the whole set and check every answer instantly.