Question

Water table drops by 4 m in a field having 180 ha area. If the porosity of soil in the field is 0.25 and the specific retention is 0.15, then the change in volume of storage (in $M{m}^3)$ in the aquifer is:

• A. 0.72
• B. 1.80
• C. 7.20
• D. 18.0

Answer 1

The correct option is A 0.72

We know that
$n=S_y+S_r$

Where n = porosity

$S_y=\text{specific yield}$

$S_r=\text{specific retention}$

$\Rightarrow 0.25=S_y+0.15$

$\Rightarrow S_y=0.10$

$\text{Specific yield}=\left[\frac{\text{Volume of water extracted}}{\text{Volume of aquifer}}\right]$

$\Rightarrow 0.10=\left[\frac{\text{Change in storage volume of aquifer}}{180\times {10}^4\times 4}\right]$

$\Rightarrow \text{Change in storage}=0.72\times {10}^6{m}^3$

$=0.72M{m}^3$