Question

A crop requires 1000 mm of water for a base period of 100 days. The duty of water in hectares per cumec is

• A. \( 1000 \)
• B. \( 864 \)
• C. \( 100 \)
• D. \( 1728 \)

Hint:

Duty, Delta, and Base Period are fundamental concepts in irrigation engineering. Duty refers to the area of land that can be irrigated by a unit discharge of water for the entire base period of a crop. Delta is the total depth of water required by the crop during its base period. The constant 8.64 in the formula arises from unit conversions (days to seconds, cubic meters to cubic millimeters, etc.) to yield hectares per cumec.

Answer 1

The Correct Option is B

To determine the duty of water, we need to calculate the area irrigated per unit volume of water flow over a specific period. Given that the base period is 100 days and the crop requires 1000 mm (or 1 meter) of water, we proceed with the following steps:

Step 1: Convert water requirement into meters per day
Since the crop requires 1000 mm over 100 days, this is equivalent to \( \frac{1000 \text{ mm}}{1000} = 1 \text{ m over 100 days} \). Thus, the daily requirement is \( \frac{1 \text{ m}}{100 \text{ days}} = 0.01 \text{ m/day} \).

Step 2: Understand the concept of duty
Duty (D) is defined as the area irrigated (in hectares) by a unit discharge (in cubic meters per second or cumec) during the base period. We know that 1 cumec = 1 m³/s, which over one day (86400 seconds) will convey:

\( \text{Volume per day} = 1 \times 86400 = 86400 \text{ m}^3 \).

Step 3: Calculate the area (in hectares) that can be irrigated
A hectare is 10000 m². The volume required to irrigate 1 hectare to a depth of 0.01 m is:

\( \text{Volume for 1 hectare} = 10000 \text{ m}^2 \times 0.01 \text{ m} = 100 \text{ m}^3 \).

Thus, for 86400 m³ of water:

\( \text{Area irrigated} = \frac{86400}{100} = 864 \text{ hectares} \).

Conclusion:
The duty of water is 864 hectares per cumec. Therefore, the correct answer is \( 864 \).