Subtopic: Cost analysis of implements and tractors
Question Statement:
A level field of 1.2 ha (120 m × 100 m) is ploughed using a reversible mould board plough with total effective cutting width 0.64 m. Average field overlap is 80 mm between two consecutive laps. Average time taken for each turn is 30 seconds and mean operating speed is 5.0 km/h. The maximum effective field capacity in ha/h is
Options:
(A) 0.207
(B) 0.236
(C) 0.283
(D) 0.318
Official Answer Key:
A
Step-by-Step Solution & Explanation
Effective working width after overlap loss: W_eff = 0.64 − 0.080 = 0.56 m.
Working the field lengthwise (pass length L = 120 m, field width to be covered = 100 m):
Number of passes = 100/0.56 = 178.6 ≈ 179 passes, with 178 turns between them.
Working time per pass = L/S = 120 m ÷ (5.0×1000/3600 m/s) = 120/1.389 = 86.4 s.
Total working time = 179 × 86.4 s = 15,466 s. Total turn time = 178 × 30 s = 5,340 s.
Total field time = 15,466 + 5,340 = 20,806 s = 5.78 h.
Effective field capacity = Area/Total time = 1.2 ha / 5.78 h = 0.208 ≈ 0.207 ha/h (the small residual difference is due to rounding the pass count to a whole number). Answer: (A) 0.207.
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