Unconfined aquifer storage change — convert to hectare–metre Over an irrigable land of 50 hectares, the water table drops by 3 m. Given porosity n = 0.30 and specific retention Sr = 0.10, compute the change in storage in hectare–metres (ha–m).

Introduction / Context:
For unconfined aquifers, a fall in the water table releases water equal to the product of the area, the decline in head, and the specific yield. Reporting storage change in hectare–metres is common in irrigation hydrology.

Given Data / Assumptions:

• Area A = 50 hectares.
• Water-table fall Δh = 3 m.
• Porosity n = 0.30; specific retention Sr = 0.10.
• Specific yield Sy = n − Sr (drainable portion).

Concept / Approach:
Specific yield represents the fraction of the saturated volume that actually drains under gravity. The volumetric change is ΔV = A * Δh * Sy. Because 1 ha–m equals the volume under 1 hectare area with 1 m depth, i.e., 10,000 m^3, multiplying A (in hectares) by Δh (in metres) by Sy directly gives ha–m.

Step-by-Step Solution:
1) Compute Sy = n − Sr = 0.30 − 0.10 = 0.20.2) Compute gross depth–area product: A * Δh = 50 * 3 = 150 ha–m.3) Apply Sy: ΔStorage = 150 * 0.20 = 30 ha–m.4) Select 30 ha–m.

Verification / Alternative check:
In cubic metres: 30 ha–m = 30 * 10,000 = 300,000 m^3, which is 20% of the gross aquifer volume change (50 ha * 3 m = 1,500,000 m^3).

Why Other Options Are Wrong:
15 and 45 ha–m correspond to Sy of 0.10 and 0.30 (wrong bases). 60 ha–m exceeds the maximum drainable volume for the given Sy.

Common Pitfalls:
Using porosity instead of specific yield; converting to m^3 unnecessarily; misinterpreting ha–m units.

Final Answer:
30