Question

What will be the unit weight of a fully saturated soil sample having water content of 38% and grain specific gravity of 2.65?

• A. 19.88 kN/m\(^3\)
• B. 17.88 kN/m\(^3\)
• C. 16.52 kN/m\(^3\)
• D. 14.65 kN/m\(^3\)

Hint:

- The unit weight of fully saturated soil depends on specific gravity, water content, and saturation. - Use the formula: \[ \gamma_{{saturated}} = \frac{G_s + S w}{1 + w} \times \gamma_{{w}} \] for accurate calculations.

Answer 1

The Correct Option is B

Step 1: Using the formula for unit weight of fully saturated soil: \[ \gamma_{{saturated}} = \frac{G_s + S w}{1 + w} \times \gamma_{{w}} \] where, \( G_s = 2.65 \) (specific gravity), \( w = 38\% = 0.38 \) (water content), \( S = 1 \) (fully saturated), \( \gamma_{{w}} = 9.81 \) kN/m\(^3\) (unit weight of water). 
Step 2: Substituting values: \[ \gamma_{{saturated}} = \frac{2.65 + (1 \times 0.38)}{1 + 0.38} \times 9.81 \] 
Step 3: Calculating: \[ \gamma_{{saturated}} = \frac{2.65 + 0.38}{1.38} \times 9.81 = \frac{3.03}{1.38} \times 9.81 \] \[ \gamma_{{saturated}} = 2.196 \times 9.81 = 17.88 { kN/m}^3 \] Thus, the correct answer is (B) 17.88 kN/m\(^3\).