Question

Proctor's compaction test for the maximum dry density of a certain soil gave the results as $1.8 \text{ gm/cc}$ and Optimum Moisture Content (OMC) is $15%$. The specific gravity of the clay soil grain was 2.7. What was the saturation degree for this soil?

• A. 61%
• B. 71%
• C. 81%
• D. 51%

Hint:

When $\gamma_d$ is given in gm/cc, always use $\gamma_w = 1 \text{ gm/cc}$ to simplify the math.

Answer 1

The Correct Option is C

Concept: Dry density ($\gamma_d$) is related to specific gravity ($G$), water content ($w$), and degree of saturation ($S$). The goal is to find $S$ using the given dry density at OMC.

Step 1: Calculate the void ratio ($e$).
The formula for dry density is: \[ \gamma_d = \frac{G \gamma_w}{1+e} \] Taking $\gamma_w = 1 \text{ gm/cc}$: \[ 1.8 = \frac{2.7 \times 1}{1+e} \] \[ 1+e = \frac{2.7}{1.8} = 1.5 \] \[ e = 0.5 \]

Step 2: Calculate the degree of saturation ($S$).
Using $Se = wG$: \[ S \times 0.5 = 0.15 \times 2.7 \] \[ S = \frac{0.405}{0.5} = 0.81 \] \[ S = 81% \]