Question

The natural dry density of a soil deposit was found to be $18.0 \, \text{kN/m}^3$. A sample of soil was brought to the laboratory, and the minimum and maximum dry density were found to be $15.0 \, \text{kN/m}^3$ and $20.0 \, \text{kN/m}^3$, respectively. The density index for the soil deposit will be:

• A. $66.66\%$
• B. $54.3\%$
• C. $33.33\%$
• D. $27.2\%$

Hint:

Always ensure correct substitution into the Id formula. Pay attention to units when calculating density indices for accurate results.

Answer 1

The Correct Option is A

The density index ($I_D$) is calculated as:
\[I_D = \frac{\gamma_d - \gamma_{\text{dmin}}}{\gamma_{\text{dmax}} - \gamma_{\text{dmin}}} \times 100\]
Substitute the given values:
\[I_D = \frac{18.0 - 15.0}{20.0 - 15.0} \times 100 = \frac{3.0}{5.0} \times 100 = 60\%\]
Answer: 66.66%