Irrigation Water Conveyance and Application Methods
Subtopic: Surface, sprinkler and micro irrigation methods
Question Statement:
A field is irrigated by constructing 100 m long furrows spaced at 0.75 m apart. The advance time to the end of furrows was 30 min with an inflow rate of 2 litre s⁻¹. After that the inflow rate was cutback to 0.5 litre s⁻¹ and continued for one hour. The average depth of irrigation is
Options:
(A) 2.4 cm
(B) 7.2 cm
(C) 9.0 cm
(D) 18.0 cm
Official Answer Key:
B
Step-by-Step Solution & Explanation
Total volume applied = (2 l/s × 30×60 s) + (0.5 l/s × 60×60 s) = 3600+1800 = 5400 litres = 5.4 m³. Irrigated area = furrow length × spacing = 100×0.75 = 75 m². Average depth = 5.4/75 = 0.072 m = 7.2 cm.
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