The moist unit weight of the soil taken from the field is \( 1900 \mathrm{~kg} / \mathrm{m}^{3} \) and the moisture content is \( 11.5 \% \). The unit weight of solid soil is 2600 \( \mathrm{~kg} / \mathrm{m}^{3} \). 47. Compute the dry unit weight. \( \begin{array}{ll}\text { a. } 1804 \mathrm{~kg} / \mathrm{m}^{3} & \text { b. } 1904 \mathrm{~kg} / \mathrm{m}^{3} \\ \text { c. } 1704 \mathrm{~kg} / \mathrm{m}^{3} & \text { d. } 1604 \mathrm{~kg} / \mathrm{m}^{3} \\ \text { 48. Compute the void ratio. } \\ \begin{array}{ll}\text { a. } 0.626 & \text { b. } 0.526 \\ \text { c. } 0.426 & \text { d. } 0.326 \\ \text { 49. Compute the degree of saturation. } \\ \begin{array}{ll}\text { a. } 0.968 & \text { b. } 0.568\end{array} \\ \text { c. } 0.468 & \text { d. } 0.868\end{array}\end{array} \).

To find the dry unit weight, we can use the formula: \[ \text{Dry Unit Weight} = \frac{\text{Moist Unit Weight}}{1 + w} \] where \( w \) is the moisture content in decimal form (i.e., \( 11.5\% = 0.115 \)). Substituting the values, we get: \[ \text{Dry Unit Weight} = \frac{1900 \, \text{kg/m}^3}{1 + 0.115} \approx 1704 \, \text{kg/m}^3 \] Thus, the answer is \( c. \, 1704 \mathrm{~kg/m}^{3} \). For the void ratio, we can calculate it using: \[ e = \frac{w \cdot \text{Unit Weight of Solid}}{\text{Dry Unit Weight}} \] Substituting the values, we find: \[ e = \frac{0.115 \cdot 2600}{1704} \approx 0.426 \] So, the answer for the void ratio is \( c. \, 0.426 \). For the degree of saturation, we can use the formula: \[ S = \frac{w \cdot \text{Dry Unit Weight}}{e \cdot \text{Unit Weight of Water}} \] Assuming the unit weight of water is \( 9.81 \, \text{kg/m}^3 \), we plug in the values to calculate the degree of saturation. \[ S \approx \frac{0.115 \cdot 1704}{0.426 \cdot 9.81} \approx 0.868 \] Thus, the answer for the degree of saturation is \( d. \, 0.868 \).