Question

A soil which was compacted at 10% moisture content has bulk unit weight of \( 22 \, \text{kN/m}^3 \). Then its dry unit weight is

• A. \( 22 \, \text{kN/m}^3 \)
• B. \( 20 \, \text{kN/m}^3 \)
• C. \( 24.2 \, \text{kN/m}^3 \)
• D. \( 11 \, \text{kN/m}^3 \)

Hint:

The dry unit weight is a crucial parameter in soil mechanics as it represents the weight of the solid particles per unit volume, excluding the weight of water. It is always less than or equal to the bulk unit weight, with equality occurring when the soil is completely dry ($w=0$). This concept is fundamental for compaction control and settlement analysis.

Answer 1

The Correct Option is B

Step 1: Identify the given data.
Let \( \gamma_b \) be the bulk unit weight of the soil.
Given \( \gamma_b = 22 \, \text{kN/m}^3 \).
Let \( w \) be the moisture content.
Given \( w = 10% = 0.10 \).
We need to find the dry unit weight, denoted as \( \gamma_d \).
Step 2: Recall the relationship between bulk unit weight, dry unit weight, and moisture content.
The relationship between bulk unit weight, dry unit weight, and moisture content is given by the formula:
$$\gamma_b = \gamma_d (1 + w)$$
Where:
\( \gamma_b \) = Bulk unit weight
\( \gamma_d \) = Dry unit weight
\( w \) = Moisture content (as a decimal)
Step 3: Rearrange the formula to solve for dry unit weight.
From the formula, we can express the dry unit weight as:
$$\gamma_d = \frac{\gamma_b}{1 + w}$$
Step 4: Substitute the given values into the formula and calculate \( \gamma_d \).
$$\gamma_d = \frac{22 \, \text{kN/m}^3}{1 + 0.10}$$
$$\gamma_d = \frac{22 \, \text{kN/m}^3}{1.10}$$
$$\gamma_d = 20 \, \text{kN/m}^3$$
Step 5: Select the correct option.
Based on the calculation, the dry unit weight is \( 20 \, \text{kN/m}^3 \). $$\boxed{20 \, \text{kN/m}^3}$$