Question

The water table in an unconfined aquifer at a place near the coast is 1 m above the Mean Sea Level. Given the densities of fresh and saline water as 1.001 and 1.025 g/cc, respectively, the fresh-saline water interface at the same location should be at a depth of \(\underline{\hspace{1cm}}\) m from the water table.

Hint:

The depth of the fresh-saline water interface can be calculated by using the densities of both types of water and the height above the mean sea level.

Answer 1

Correct Answer: 42.6

The depth of the fresh-saline water interface is calculated using the formula:
\[ d = \frac{\rho_f}{\rho_s - \rho_f} \cdot h \] where:
- \( \rho_f \) is the density of fresh water,
- \( \rho_s \) is the density of saline water,
- \( h \) is the height above Mean Sea Level.
Given:
- \( \rho_f = 1.001 \, \text{g/cc} \),
- \( \rho_s = 1.025 \, \text{g/cc} \),
- \( h = 1 \, \text{m} \).
Substituting the values: \[ d = \frac{1.001}{1.025 - 1.001} = \frac{1.001}{0.024} \approx 42.7 \, \text{m} \] Thus, the depth is approximately \( 42.7 \, \text{m} \).