The maximum dry density and optimum moisture content of a soil are 1.65 g/cc and 20.5% respectively. What is the percentage of air content of the soil at OMC, if the specific gravity of particles is 2.65?
Try it first: pick an option, then read the explanation.
Asked in SSC JE 24/01/2018 (Evening)
Detailed Explanation
This is the reverse of a typical phase-relation problem: we know dry density, water content and specific gravity, and need the air content at the peak of the compaction curve.
Step 1 — Void ratio: \( 1 + e = \dfrac{G\gamma_w}{\gamma_d} = \dfrac{2.65}{1.65} = 1.606 \Rightarrow e = 0.606 \)
Step 2 — Degree of saturation: \( S = \dfrac{wG}{e} = \dfrac{0.205 \times 2.65}{0.606} = 0.896 \)
Step 3 — Air content: \( a_c = 1 - S = 1 - 0.896 = 0.104 = 10.4\% \)
(Equivalently, \( \gamma_d = \dfrac{G\gamma_w}{1 + wG/S} \Rightarrow 1.65 = \dfrac{2.65}{1 + 0.205\times 2.65/S} \Rightarrow S = 0.896 \).)
Why “10.4” Is Correct
Air content is the fraction of the voids filled with air, ac = Va/Vv = 1 − S. With S ≈ 0.896, ac ≈ 10.4%.
Why the Other Options Are Wrong
- B. 15.5: 15.5 does not follow from the given data with either air content or air voids.
- C. 26.8: 26.8 does not follow from 1 − S with the given data; it comes from mixing up air content with another ratio.
- D. 35.7: 35.7 is close to the porosity of this soil (n = e/(1 + e) ≈ 37.7%), not its air content.
Key Point
Percentage air voids would be na = n × ac = (0.606/1.606) × 0.104 ≈ 3.9%; make sure you answer the quantity asked.
Key Concepts for Students
- γd = Gγw/(1 + e) and Se = wG solve almost every compaction numerical.
- At OMC a soil typically holds a few percent air voids; it is not fully saturated.
- Check: S must be less than 1; if you get S > 1, recheck G or w.
Practice more: this is question 5 of Compaction of Soil MCQs – Part 2. Try the whole set and check every answer instantly.