Question

For a partially saturated soil deposit at a construction site, water content (\(w\)) is 15%, degree of saturation (\(S\)) is 67%, void ratio (\(e\)) is 0.6 and specific gravity of solids in the soil (\(G_s\)) is 2.67. Consider unit weight of water as 9.81 kN/m\(^3\).
To fully saturate 5 m\(^3\) of this soil, the required weight of water (in kN) will be __________ (round off to the nearest integer).
 

• A. 5
• B. 6
• C. 7
• D. 8

Hint:

To calculate the weight of water required for full saturation, use the void ratio, degree of saturation, and the specific gravity of solids in the soil.

Answer 1

The Correct Option is B

Given:
Initial water content: \( w_1 = 0.15 \)
Degree of saturation: \( S = 67\% \)
Void ratio: \( e = 0.6 \)
Specific gravity of solids: \( G_s = 2.67 \)
Unit weight of water: \( \gamma_w = 9.81 \, \text{kN/m}^3 \)

Let water content after full saturation be \( w_2 \). The weight of water after full saturation can be calculated using the following steps.

Step 1: Water content at full saturation (\( w_2 \)) \[ w_2 = \frac{e}{G_s} = \frac{0.6}{2.67} = 0.2247 \]
Step 2: Change in weight of water (\( w_2 - w_1 \)) \[ w_2 - w_1 = \frac{\text{Weight of water}}{\text{Weight of solid}} = \frac{w}{w_s} \] where \( w_s \) is the weight of the solid.

Step 3: Weight of solid (\( w_s \)) \[ w_s = V_s \cdot G_s \cdot \gamma_w = \frac{V_t}{1+e} \cdot G_s \cdot \gamma_w \] where: - \( V_t = 5 \, \text{m}^3 \) is the total volume. - \( \gamma_w = 9.81 \, \text{kN/m}^3 \) is the unit weight of water.

Step 4: Substituting the values \[ 0.2247 - 0.15 = \frac{w}{\frac{5}{1.6} \times 2.67 \times 9.81} \] Solving for \( w \): \[ w = 0.0747 \times 3.125 \times 2.67 \times 9.81 = 6 \, \text{kN} \]
Thus, the required weight of water is: \[ \boxed{6} \, \text{kN} \]