Question

An undisturbed soil sample is collected from a field with core cylinder having an internal diameter of \(60\) mm and length of \(100\) mm. The weights of moist soil and oven dried soil are \(0.48\) kg and \(0.44\) kg, respectively. Assuming density of water as \(1000\) kg/m\(^3\), the water depth (in metres per metre depth of soil) is ________ (Rounded off to two decimal places)

Hint:

Find the mass of water lost on drying and convert its volume into an equivalent depth.

Answer 1

Correct Answer: 0.14

Step 1: Find the volume of the core sample.
\[ V = \frac{\pi}{4}d^2L = \frac{\pi}{4}(0.06)^2(0.1) \approx 2.827\times10^{-4}\ \text{m}^3 \]
Step 2: Find the mass of water lost on oven drying.
\[ m_w = 0.48-0.44 = 0.04\ \text{kg} \]
Step 3: Convert this mass to a volume of water.
\[ V_w = \frac{m_w}{\rho_w} = \frac{0.04}{1000} = 4\times10^{-5}\ \text{m}^3 \]
Step 4: Spread this water volume over the sample's cross-section to get a depth.
Cross-sectional area \(= \dfrac{\pi}{4}d^2 \approx 2.827\times10^{-3}\) m\(^2\).
\[ d_w = \frac{V_w}{\text{area}} = \frac{4\times10^{-5}}{2.827\times10^{-3}} \approx 0.01415\ \text{m (over the 100 mm sample)} \]
Step 5: Express this per metre depth of soil.
\[ \frac{d_w}{L} = \frac{0.01415}{0.1} \approx 0.1415\ \text{m of water per m of soil} \]
Final Answer:
Each metre depth of this soil holds close to \(0.14\) m depth of water at the sampled moisture level. \[ \boxed{\approx 0.14} \]