Question

An uncompacted heap of soil has a volume of 10000 m\(^3\) and void ratio of 1. If the soil is compacted to a volume of 7500 m\(^3\), then the corresponding void ratio of the compacted soil is \(\underline{\hspace{2cm}}\) (round off to one decimal place).

Hint:

To find the void ratio after compaction, use the volume of solids and the change in total volume.

Answer 1

Correct Answer: 0.5

The void ratio \( e \) is defined as:
\[ e = \frac{V_{\text{void}}}{V_{\text{solids}}} \] For the uncompacted soil:
\[ e_{\text{uncompacted}} = 1 = \frac{V_{\text{void}}}{V_{\text{solids}}}. \] Thus, the volume of solids is:
\[ V_{\text{solids}} = 10000 \, \text{m}^3 - V_{\text{void}} = 5000 \, \text{m}^3. \] For the compacted soil:
\[ V_{\text{compacted}} = 7500 \, \text{m}^3 \text{and} V_{\text{solids}} = 5000 \, \text{m}^3. \] The new void ratio is:
\[ e_{\text{compacted}} = \frac{V_{\text{compacted}} - V_{\text{solids}}}{V_{\text{solids}}} = \frac{7500 - 5000}{5000} = 0.5. \] Thus, the void ratio of the compacted soil is \( 0.5 \).