Subtopic: Leaching requirement and salinity control
Question Statement:
A crop grown over an area of 50 ha can tolerate 5 decisiemens m⁻¹ of electrical conductivity in the drainage water. The consumptive use of the crop is 1.0 m, 30% of which is obtained from the rainfall and the remainder is met from the irrigation water having electrical conductivity of 2 decisiemens m⁻¹. For efficient crop production, the leaching requirement and the quantity of water that must be drained from the area are
Options:
(A) 40% and 0.23 Mm³
(B) 60% and 0.23 Mm³
(C) 40% and 0.21 Mm³
(D) 60% and 0.21 Mm³
Official Answer Key:
A
Step-by-Step Solution & Explanation
Leaching requirement LR = EC(irrigation water)/EC(drainage water, tolerance limit) = 2/5 = 0.40 = 40%. Consumptive use met from irrigation = (1 − 0.30)×1.0 m = 0.70 m. Because a fraction LR of the applied irrigation water must pass through and drain below the root zone (over and above meeting consumptive use), the gross irrigation water applied Iw satisfies Iw(1−LR) = 0.70 m ⇒ Iw = 0.70/0.60 = 1.1667 m. Water draining out = Iw×LR = 1.1667×0.40 = 0.4667 m depth. Volume = 0.4667 m × 50 ha × 10,000 m²/ha = 233,333 m³ ≈ 0.23 Mm³, matching option (A).
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