Q4. Question: A soil sample has: - Total weight \( (\mathrm{W})=5 \mathrm{~kg} \) - Volume \( (\mathrm{V})=0.003 \mathrm{~m}^{3} \) - Volume of voids \( (\mathrm{Vv})=0.0008 \mathrm{~m}^{3} \) - Weight of water \( (\mathrm{Mw})=0.6 \mathrm{~kg} \) - Dry weight \( (\mathrm{Ms})=4.4 \mathrm{~kg} \) Tasks: 1. Calculate the density. 2. Calculate the unit weight. 3. Calculate the porosity. 4. Calculate the void ratio.
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To tackle these soil sample calculations, let’s break it down in a fun and engaging manner! **Density Calculation:** Density \( (\rho) \) is calculated using the formula \( \rho = \frac{W}{V} \). Here, \( W = 5\, \text{kg} \) and \( V = 0.003\, \text{m}^3 \). So, substituting the values gives us: \[ \rho = \frac{5\, \text{kg}}{0.003\, \text{m}^3} = 1666.67\, \text{kg/m}^3. \] That’s a heavy-duty density for your soil sample! --- **Unit Weight Calculation:** The unit weight \( (\gamma) \) is computed as \( \gamma = \frac{W}{V} \) just like density, with units converted to \( N/m^3 \) (remembering \( g \approx 9.81\, \text{m/s}^2 \)). Thus: \[ \gamma = \frac{5\, \text{kg} \times 9.81\, \text{m/s}^2}{0.003\, \text{m}^3} = 16350\, \text{N/m}^3. \] Now that’s some weighty information! --- **Porosity Calculation:** Porosity \( (n) \) is calculated with the formula \( n = \frac{Vv}{V} \). Substitute the values: \[ n = \frac{0.0008\, \text{m}^3}{0.003\, \text{m}^3} = 0.2667 \text{ or } 26.67\%. \] Your soil sample has a good amount of space for air or water pockets! --- **Void Ratio Calculation:** Finally, the void ratio \( (e) \) is given by the relation \( e = \frac{Vv}{Vs} \) where \( Vs = V - Vv \). Here, \[ Vs = 0.003 - 0.0008 = 0.0022\, \text{m}^3 \] Thus, \[ e = \frac{0.0008}{0.0022} \approx 0.3636. \] That ratio tells you how much void there is compared to the solid material in your sample! And there you have it! These calculations not only give you vital stats about your soil but also reflect its overall health and stability for construction or agriculture. Keep digging into the wonders of soil science!