A 2×0.3 m tractor drawn mould board plough operating at depth 0.15 m has draft 2.5 kN at forward speed 3 km h⁻¹, field efficiency 75%. Speed increased by 20%, draft increased by 10%. Assuming field efficiency, soil pulverization and soil inversion unchanged, the performance index of the plough increases by
Options:
(A) 0%
(B) 9%
(C) 20%
(D) 30%
Official Answer Key:
B
Step-by-Step Solution & Explanation
With width, depth, field efficiency and soil handling held constant, performance index (which measures effective tillage output per unit draft-power expended) scales as PI ∝ speed/draft (since output ∝ speed while power ∝ draft × speed, so PI ∝ 1/draft in the power sense — equivalently PI ∝ v/D). PI₁ ∝ 3/2.5 = 1.20. PI₂ ∝ (3×1.2)/(2.5×1.1) = 3.6/2.75 = 1.309. Increase = (1.309−1.20)/1.20 × 100 = 9.09% ≈ 9%.
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