Subtopic: Functional requirements, principles of working, construction and operation of manual, animal, tractor and renewable energy operated equipment for tillage, sowing, planting, fertilizer application, inter-cultivation, spraying, mowing, chaff cutting, harvesting and threshing
Question Statement:
An 8 × 20 cm seed drill has a 0.60 m diameter ground wheel. On a hard-surface calibration stand, the drill delivers 814 g of seeds in 30 wheel revolutions. When operated in a field having soft soil, the wheel’s effective rolling circumference reduces by 4%. Neglecting other losses, the actual-field seed application rate (in kg·ha⁻¹) is ________. (Rounded off to two decimal places) (Take π = 3.14)
Official Answer Key:
92.00 to 95.00
Step-by-Step Solution & Explanation
“8 × 20 cm” means 8 rows spaced 20 cm apart.
Step 1 — working width:
W = 8 × 0.20 = 1.60 m
Step 2 — distance covered in calibration (hard stand, no reduction):
Circumference = πD = 3.14 × 0.60 = 1.884 m
Distance in 30 revolutions = 30 × 1.884 = 56.52 m
Step 3 — calibration area:
A = 56.52 × 1.60 = 90.43 m²
Step 4 — calibration seed rate:
= (0.814 kg/90.43 m²) × 10,000 m²/ha = 90.01 kg·ha⁻¹
Step 5 — field correction. In soft soil the effective rolling circumference falls by 4%, so each wheel revolution advances 4% less ground while the metering mechanism (driven by wheel rotation) delivers the same seed per revolution. Seed per unit area therefore increases:
Field rate = 90.01/0.96 = 93.76 kg·ha⁻¹
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