Aquifer storage change due to water-table decline An unconfined aquifer under an area of 200 hectares experiences a water-table drop of 4 m. Porosity n = 0.35 and specific retention Sr = 0.15. Compute the change in volume of water stored (m^3).

Introduction / Context:
In an unconfined aquifer, the change in stored water due to a water-table movement equals the product of the area, the change in head, and the specific yield. Specific yield Sy is the drainable portion of porosity after accounting for water retained by capillarity (specific retention).

Given Data / Assumptions:

• Area A = 200 ha = 200 × 10^4 m^2 = 2.0 × 10^6 m^2.
• Water-table decline Δh = 4 m.
• Porosity n = 0.35; specific retention Sr = 0.15 → specific yield Sy = n − Sr = 0.20.
• Gravity drainage dominates; compressibility effects neglected.

Concept / Approach:
For unconfined storage change: ΔV = A * Δh * SySpecific yield represents the fraction of pore water that actually drains when the water table falls.

Step-by-Step Solution:
1) Compute Sy = 0.35 − 0.15 = 0.20.2) Convert area: A = 200 ha = 2.0 × 10^6 m^2.3) Multiply: ΔV = 2.0 × 10^6 * 4 * 0.20.4) ΔV = 1.6 × 10^6 m^3.

Verification / Alternative check:
Order-of-magnitude: A * Δh = 8 × 10^6 m^3 of aquifer volume; with 20% drainable yield → 1.6 × 10^6 m^3, consistent.

Why Other Options Are Wrong:
8 × 10^6 m^3 ignores Sy; 160 m^3 and 1.6 × 10^3 m^3 are far too small for the stated area and drawdown.

Common Pitfalls:
Using porosity instead of specific yield; mis-converting hectares to square metres; forgetting to apply the water-table change.

Final Answer:
1.6 × 10^6 m^3