Question

Irrigation scheduling for a 1 ha rice field is planned in such a way that the amount of irrigation to be provided per day is equal to the amount of ET/day. On a day if 1 cm of ET was observed, how much of water is to be used for irrigation on that day?

• A. \(1\text{ m}^3\)
• B. \(10\text{ m}^3\)
• C. \(100\text{ m}^3\)
• D. \(0.1\text{ m}^3\)

Hint:

$1\text{ ha-cm} = 10,000\text{ m}^2 \times 0.01\text{ m} = 100\text{ m}^3 = 100,000\text{ litres}$. $1\text{ ha-mm} = 10\text{ m}^3$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Volumetric irrigation application is the product of field surface area and the replacement water depth.
Key Formula or Approach:
\[ V = A \times d \]
where \(A\) is field area (\(\text{m}^2\)) and \(d\) is water depth (m).

Step 2: Detailed Explanation:
Given parameters:
- Field area: \(A = 1\text{ ha} = 10,000\text{ m}^2\)
- Observed daily evapotranspiration depth: \(d = 1\text{ cm} = 0.01\text{ m}\)
Computing the total volume of irrigation water required:
\[ V = A \times d = 10,000\text{ m}^2 \times 0.01\text{ m} = 100\text{ m}^3 \]
\[ (100\text{ m}^3 = 100,000\text{ litres}) \]

Step 3: Final Answer:
Thus, the amount of water to be used is \(100\text{ m}^3\), corresponding to option (C).