Question

When 220g of core shaped fresh soil with 5 cm length and 6 cm diameter lost 20g on drying, what is its bulk density?

• A. \(1.31\text{ g}\cdot\text{cm}^{-3}\)
• B. \(1.41\text{ g}\cdot\text{cm}^{-3}\)
• C. \(1.51\text{ g}\cdot\text{cm}^{-3}\)
• D. \(1.61\text{ g}\cdot\text{cm}^{-3}\)

Hint:

Dry mass $= 220 - 20 = 200\text{ g}$. Core volume $= \frac{\pi}{4} (6^2) (5) = 141.37\text{ cm}^3$. $\rho_b = \frac{200}{141.37} = 1.41\text{ g/cm}^3$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Dry bulk density of soil is defined as the mass of oven-dry soil solids divided by the total undisturbed bulk volume of the soil core.
Key Formula or Approach:
\[ \rho_b = \frac{M_{\text{dry}}}{V_{\text{core}}} \]
\[ V_{\text{core}} = \frac{\pi}{4} D^2 L \]

Step 2: Detailed Explanation:
Given parameters:
- Fresh soil mass: \(M_{\text{fresh}} = 220\text{ g}\)
- Moisture lost on drying: \(\Delta M = 20\text{ g}\)
- Oven-dry soil mass: \(M_{\text{dry}} = 220 - 20 = 200\text{ g}\)
- Core diameter: \(D = 6\text{ cm} \implies R = 3\text{ cm}\)
- Core length (height): \(L = 5\text{ cm}\)

Step 1: Calculate cylindrical core volume:
\[ V = \pi R^2 L = \pi \times (3\text{ cm})^2 \times 5\text{ cm} = 45\pi\text{ cm}^3 \]
\[ V = 45 \times 3.14159 = 141.37\text{ cm}^3 \]
Calculate dry bulk density:
\[ \rho_b = \frac{M_{\text{dry}}}{V} = \frac{200\text{ g}}{141.37\text{ cm}^3} = 1.4147\text{ g/cm}^3 \approx 1.41\text{ g}\cdot\text{cm}^{-3} \]

Step 3: Final Answer:
Therefore, the bulk density is \(1.41\text{ g}\cdot\text{cm}^{-3}\), corresponding to option (B).