Q5. A construction site has a soil sample with the following details: - Wet weight of the sample (Wwet): 6.2 kg - Dry weight (Wdry): 5.3 kg - Total volume \( (\mathrm{V}): 0.004 \mathrm{~m}^{3} \) - Volume of solids \( (\mathrm{Vs}): 0.003 \mathrm{~m}^{3} \) Tasks: 1. Calculate the density of the wet sample. 2. Calculate the unit weight of the wet sample. 3. Calculate the porosity \( (\mathrm{n}) \).

To solve the tasks related to the construction site's soil sample, let's break them down step by step. 1. **Density of the Wet Sample**: The formula for density \( (\rho) \) is given by the wet weight divided by the total volume. \[ \rho_{\text{wet}} = \frac{W_{\text{wet}}}{V} = \frac{6.2 \, \text{kg}}{0.004 \, \text{m}^3} = 1550 \, \text{kg/m}^3 \] 2. **Unit Weight of the Wet Sample**: The unit weight \( (\gamma) \) is equal to the density multiplied by the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). \[ \gamma_{\text{wet}} = \rho_{\text{wet}} \times g \approx 1550 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \approx 15280.5 \, \text{N/m}^3 \] 3. **Porosity \( (n) \)**: Porosity can be calculated using the formula: \[ n = \frac{V - V_s}{V} = \frac{0.004 \, \text{m}^3 - 0.003 \, \text{m}^3}{0.004 \, \text{m}^3} = \frac{0.001 \, \text{m}^3}{0.004 \, \text{m}^3} = 0.25 \text{ or } 25\% \] Hence, the answers to the tasks are: - Density of the wet sample: \( 1550 \, \text{kg/m}^3 \) - Unit weight of the wet sample: \( 15280.5 \, \text{N/m}^3 \) - Porosity \( (n) \): \( 25\% \) Understanding the behavior of soil samples is essential in construction. For example, knowing the soil density helps engineers determine how much load a foundation can safely bear. On the other hand, porosity provides insights into how well water can drain through the soil, crucial for managing groundwater levels and preventing erosion at a construction site.