Question

The capillary rise in soil (h, cm) with average pore radius (r, cm) is related as

• A. $hr= 0.15$
• B. $hr= 1.5$
• C. $hr= 0.30$
• D. $hr= 3.0$

Hint:

This relationship shows that capillary rise is inversely proportional to the pore radius ($h \propto \frac{1}{r}$):
- Clay soils have very small pore radii, resulting in a high capillary rise (up to several meters) but a slow rate of rise.
- Sandy soils have large pore radii, resulting in a low capillary rise but a rapid rate of rise.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Capillary rise describes the upward movement of water in soil pores against gravity.
This process is driven by the surface tension of water and adhesive forces between water molecules and soil particle walls.
Soil pores act like a network of narrow capillary tubes.

Step 2: Key Formula or Approach:
The height of capillary rise ($h$) in a cylindrical tube of radius $r$ is given by the Jurin's Law equation:
\[ h = \frac{2\gamma \cos \theta}{\rho \cdot g \cdot r} \]
where $\gamma$ is the surface tension of water ($72.8\text{ dynes/cm}$ at $20\text{ }^\circ\text{C}$), $\theta$ is the contact angle (assumed to be $0$ for pure water on clean silica, so $\cos \theta = 1$), $\rho$ is the density of water ($1\text{ g/cm}^3$), $g$ is the acceleration due to gravity ($981\text{ cm/s}^2$), and $r$ is the pore radius.

Step 3: Detailed Explanation:
Let us substitute the physical values into Jurin's Law, keeping all parameters in CGS units (cm, g, s):
\[ h = \frac{2 \times 72.8 \times 1}{1 \times 981 \times r} \]
\[ h = \frac{16}{981 \times r} \]
\[ h \approx \frac{0.1484}{r} \approx \frac{0.15}{r}\text{ cm} \]
Rearranging this relationship:
\[ h \cdot r \approx 0.15\text{ cm}^2 \]
Thus, the product of the capillary rise height ($h$ in cm) and the average pore radius ($r$ in cm) is a constant value of approximately $0.15$.

Step 4: Final Answer:
The relationship between capillary rise ($h$, cm) and average pore radius ($r$, cm) is $hr = 0.15$.
Therefore, the correct option is (A).