Question

If the soil in a core of diameter 4 cm, height 7 cm, weighs 170 g after drying and the weight of core without soil is 50 g, then the bulk density of the soil is :

• A. $0.60 \text{ Mg m}^{-3}$
• B. $1.15 \text{ Mg m}^{-3}$
• C. $1.36 \text{ Mg m}^{-3}$
• D. $1.85 \text{ Mg m}^{-3}$

Hint:

The units $\text{g/cm}^3$ and $\text{Mg/m}^3$ (Megagrams per cubic meter) are numerically identical. Typical soil bulk densities range from $1.0$ to $1.6 \text{ g/cm}^3$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Bulk density ($\rho_b$) is defined as the mass of dry soil divided by its total bulk volume (volume of soil solids plus pore space).
Key Formula or Approach:
The formulas required are:
- Volume of a cylindrical core: \[ V = \pi r^2 h = \pi \left(\frac{d}{2}\right)^2 h \] - Bulk density: \[ \rho_b = \frac{M_{\text{dry soil}}}{V} \]

Step 2: Detailed Explanation:
Given values from the problem:
- Core diameter ($d$) = $4\text{ cm} \implies$ radius ($r$) = $2\text{ cm}$
- Core height ($h$) = $7\text{ cm}$
- Total dry weight = $170\text{ g}$
- Core weight empty = $50\text{ g}$
First, calculate the volume of the cylindrical core: \[ V = \pi \times r^2 \times h \] \[ V = \frac{22}{7} \times 2^2 \times 7 = 22 \times 4 = 88\text{ cm}^3 \] Second, find the mass of the oven-dry soil: \[ M_{\text{dry soil}} = \text{Total weight} - \text{Core weight} = 170 - 50 = 120\text{ g} \] Third, calculate the bulk density: \[ \rho_b = \frac{120}{88} \approx 1.364\text{ g/cm}^3 \] Since $1\text{ g/cm}^3 = 1\text{ Mg/m}^3$: \[ \rho_b \approx 1.36\text{ Mg m}^{-3} \]

Step 3: Final Answer:
The bulk density of the soil is $1.36\text{ Mg m}^{-3}$, which corresponds to Option (C).