Let's analyze and solve each part step-by-step. --- # Given Data | | Soil A | Soil B | ------------------------|--------------|--------------| | Grain size \( D_{10} \) | .005 mm | .5 mm | | pore size \( d_{eff} \) . \( D_{10} \) | .2 \( D_{10} \) | | Pore pressure distribution | Hydrostatic | Hydrostatic | --- # 1. Potential height of capillary rise above the water table ### Step 1: Calculate effective pore size \( d_{eff} \) \[ d_{eff} = .2 \times D_{10} \] - For Soil A: \[ d_{eff,A} = .2 \times .005\, \text{mm} = .001\, \text{mm} \] - For Soil B: \[ d_{eff,B} = .2 \times .5\, \text{mm} = .1\, \text{mm} \] ### Step 2: Convert \( d_{eff} \) to meters \[ 1\, \text{mm} = 1 \times 10^{-3}\, \text{m} \] - For Soil A: \[ d_{eff,A} = .001\, \text{mm} = 1 \times 10^{-6}\, \text{m} \] - For Soil B: \[ d_{eff,B} = .1\, \text{mm} = 1 \times 10^{-4}\, \text{m} \] ### Step 3: Use Jurin's Law for capillary height \[ h_{max} = \frac{2 \sigma \cos \theta}{\rho g d_{eff}} \] Assuming: - Surface tension \( \sigma \approx .0728\, \text{N/m} \) (for water at room temperature), - Contact angle \( \theta \approx ^\circ \Rightarrow \cos \theta = 1 \), - Water density \( \rho \approx 100\, \text{kg/m}^3 \), - Acceleration due to gravity \( g = 9.81\, \text{m/s}^2 \). ### Step 4: Calculate \( h_{max} \) for each soil - For Soil A: \[ h_{A} = \frac{2 \times .0728 \times 1}{100 \times 9.81 \times 1 \times 10^{-6}} = \frac{.1456}{9.81 \times 10^{-3}} \approx 14,845\, \text{m} \] - For Soil B: \[ h_{B} = \frac{.1456}{100 \times 9.81 \times 1 \times 10^{-4}} = \frac{.1456}{.981} \approx 148.5\, \text{m} \] --- # **Final answer for potential height of capillary rise:** | Soil | Potential height \( h \) (meters) | |----------------|------------------------------| | **Soil A** | **~14,845 m** | | **Soil B** | **~148.5 m** | --- # 2. Water pressure at the meniscus (at the top of the capillary zone) ### Step 1: Understand hydrostatic pressure Hydrostatic pressure at a depth \( h \): \[ P = \rho g h \] Since the question asks for the pressure at the meniscus (at the top of the zone of capillary rise), pressure due to water column above the meniscus is: \[ P_{meniscus} = \rho g h_{max} \] ### Step 2: Calculate pressure for each soil - For Soil A: \[ P_A = 100 \times 9.81 \times 14,845 \approx 1.456 \times 10^8\, \text{Pa} \] - For Soil B: \[ P_B = 100 \times 9.81 \times 148.5 \approx 1.456 \times 10^6\, \text{Pa} \] --- # **Final answers for water pressure at the meniscus:** | Soil | Water pressure \( P \) (Pa) | |----------------|----------------------------| | **Soil A** | **~145.6 MPa** | | **Soil B** | **~1.46 MPa** | --- # 3. Effect of capillarity on effective stresses ### Explanation: - Capillarity increases the effective stress in the soil because the pore pressure at the water table is negative (suction), which increases the total effective stress. - The total stress remains unaffected, but the effective stress increases by the magnitude of the capillary pressure. ### Summary: - The capillarity effect **raises the effective stress** due to the negative pore pressure at the meniscus. - The magnitude of this increase depends on the height of the capillary rise and the pore pressure at the meniscus. --- # **Overall conclusion:** | Parameter | Soil A | Soil B | |----------------------------|------------------|-------------------| | Max capillary height \( h_{max} \) | ~14,845 m | ~148.5 m | | Water pressure at meniscus | ~145.6 MPa | ~1.46 MPa | This demonstrates the significant difference in capillary effects based on grain size and pore structure.