Let's analyze and compute the required parameters step-by-step. --- ## Given Data: - Soil A: - Effective grain size \( D_{10A} = .005\, \text{mm} \) - Soil B: - Effective grain size \( D_{10B} = .5\, \text{mm} \) - Effective pore size: \[ D_{eff} = .2 \times D_{10} \] - The pore pressure distribution is hydrostatic. --- ## Step 1: Calculate the pore sizes ### Soil A: \[ D_{effA} = .2 \times .005\, \text{mm} = .001\, \text{mm} \] ### Soil B: \[ D_{effB} = .2 \times .5\, \text{mm} = .1\, \text{mm} \] --- ## Step 2: Calculate the potential height of capillary rise (\( h_{max} \)) The capillary rise height \( h \) is given by Jurin's law: \[ h = \frac{2 \sigma \cos \theta}{\rho g D_{eff}} \] Assuming: - Surface tension of water \( \sigma \approx .0728\, \text{N/m} \), - Contact angle \( \theta \approx ^\circ \Rightarrow \cos \theta \approx 1 \), - Density of water \( \rho \approx 100\, \text{kg/m}^3 \), - Acceleration due to gravity \( g = 9.81\, \text{m/s}^2 \), - Convert \( D_{eff} \) to meters. ### Convert \( D_{eff} \) to meters: - For Soil A: \[ D_{effA} = .001\, \text{mm} = 1 \times 10^{-6}\, \text{m} \] - For Soil B: \[ D_{effB} = .1\, \text{mm} = 1 \times 10^{-4}\, \text{m} \] ### Calculate \( h_{max} \): \[ h_{max} = \frac{2 \times .0728 \times 1}{100 \times 9.81 \times D_{eff}} \] ### For Soil A: \[ h_{maxA} = \frac{2 \times .0728}{100 \times 9.81 \times 1 \times 10^{-6}} = \frac{.1456}{9.81 \times 10^{-3}} \] \[ h_{maxA} \approx \frac{.1456}{.00981} \approx 14.85\, \text{meters} \] ### For Soil B: \[ h_{maxB} = \frac{.1456}{100 \times 9.81 \times 1 \times 10^{-4}} = \frac{.1456}{.981} \approx .148\, \text{meters} \] --- ## **Final Answer for potential capillary height:** | Soil | \( D_{10} \) (mm) | \( D_{eff} \) (mm) | \( D_{eff} \) (m) | \( h_{max} \) (m) | |--------|-------------------|--------------------|------------------|----------------| | A | .005 | .001 | \(1 \times 10^{-6}\) | **14.85 m** | | B | .5 | .1 | \(1 \times 10^{-4}\) | **.148 m** | --- ## Step 3: Water pressure at the meniscus Since the pressure distribution is hydrostatic: \[ P = \rho g h \] At the top of the capillary (water table): - For Soil A: \[ P_A = 100 \times 9.81 \times 14.85 \approx 145,628\, \text{Pa} \approx 146\, \text{kPa} \] - For Soil B: \[ P_B = 100 \times 9.81 \times .148 \approx 1,450\, \text{Pa} \approx 1.45\, \text{kPa} \] --- ## **Summary:** ### 1. **Potential height of capillary rise:** - Soil A: **14.85 meters** - Soil B: **.148 meters** ### 2. **Water pressure at the meniscus:** - Soil A: approximately **146 kPa** - Soil B: approximately **1.45 kPa** --- ## **Impact of Capillarity on Effective Stresses:** Capillarity increases the effective stress in unsaturated soils by creating a negative pore water pressure (suction). The magnitude depends on the height of the capillary rise and the corresponding pressure at the meniscus, as calculated above. --- **Note:** These calculations assume ideal conditions with perfect wetting and hydrostatic pressure distribution.