Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
In a 9×20 cm fluted roller type seed drill, each fluted roller is discharging 4.25 g of seed per revolution of fluted roller shaft. The fluted roller shaft rotates once for two complete rotation of the ground drive wheel of the seed drill. The rolling diameter of the ground drive wheel is 0.35 m. Considering no skid of the ground drive wheel, the seed rate in kg ha⁻¹ is [Consider π = 3.14]
Options:
(A) 96.62
(B) 141.55
(C) 187.35
(D) 386.42
Official Answer Key:
A
Step-by-Step Solution & Explanation
“9×20 cm” denotes 9 furrow openers/fluted rollers spaced 20 cm apart, so working width = 9×0.20 = 1.8 m.
Distance covered per revolution of the ground wheel = πD = 3.14×0.35 = 1.099 m.
Since the fluted-roller shaft makes 1 revolution for every 2 ground-wheel revolutions, seed discharged per roller per ground-wheel revolution = 4.25×0.5 = 2.125 g. With 9 rollers: total seed per ground-wheel revolution = 9×2.125 = 19.125 g.
Area covered per ground-wheel revolution = width × distance = 1.8×1.099 = 1.978 m².
Seed rate = (19.125 g / 1.978 m²) × (10,000 m²/ha) ÷ (1000 g/kg) ≈ 96.6 kg/ha. Answer: (A) 96.62 (matches within normal rounding of πD).
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