A mustard crop is irrigated in 30 hours, with a discharge rate of 200 litres per minute, workout the average depth of irrigation
Show Hint
If an exam question asks for irrigation depth without specifying the area, always test an area of \(1 \text{ ha } (10,000 \text{ m}^2)\) first. This is a very common omission in competitive agricultural exams.
Step 1: Understanding the Concept:
To find the average depth of irrigation applied to a crop, we must calculate the total volume of water discharged and divide it by the irrigated land area. Key Formula or Approach:
The total volume of water (\(V\)) applied is:
\[ V = \text{Discharge rate } (Q) \times \text{Time } (t) \]
The average depth of irrigation (\(d\)) is:
\[ d = \frac{\text{Volume of water } (V)}{\text{Area of land } (A)} \]
Note: In standard Indian agronomy and agricultural engineering exams, if the field area is not explicitly stated in a problem, it is conventionally assumed to be1 hectare (\(1\text{ ha} = 10,000\text{ m}^2\)). Let us perform the calculations based on this standard assumption. Step 2: Detailed Explanation:
Let us identify the given parameters and perform the calculations:
Calculate the total volume of water applied:
\[ V = 200 \text{ liters/min} \times 1800 \text{ min} = 360,000 \text{ liters} \] Convert the volume from liters to cubic meters (\(1 \text{ m}^3 = 1000 \text{ liters}\)):
\[ V = \frac{360,000}{1000} \text{ m}^3 = 360 \text{ m}^3 \] Assuming the standard field area of 1 hectare, \(A = 10,000 \text{ m}^2\):
\[ d = \frac{V}{A} = \frac{360 \text{ m}^3}{10,000 \text{ m}^2} = 0.036 \text{ m} \] Convert this depth to millimeters:
\[ d = 0.036 \text{ m} \times 1000 \text{ mm/m} = 36 \text{ mm} \]
This calculation confirms that the assumed area of 1 hectare is correct and yields an irrigation depth of 36 mm. Step 2: Final Answer:
The average depth of irrigation is 36 mm, which corresponds to Option (D).
