Question Bank • QB_FM_001 NAT • 2 Marks

Section 2: Farm Machinery

Farm Machinery

Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses

Question Statement:

A 4-bottom tractor-mounted mouldboard plough has a cutting width of 35 cm per bottom. It operates at an average forward speed of $5.4\text{ km/h}$. If the field efficiency is $80\%$, the effective field capacity of the plough in $\text{ha/h}$ is ________ (round off to 2 decimal places).

Answer Key:

0.60

Step-by-Step Solution & Explanation

1. Total width of cut $W$: $W = 4 \times 0.35\text{ m} = 1.4\text{ m}$ 2. Forward speed $S$: $S = 5.4\text{ km/h}$ 3. Theoretical Field Capacity ($TFC$): $TFC = \frac{W \times S}{10} = \frac{1.4 \times 5.4}{10} = 0.756\text{ ha/h}$ 4. Effective Field Capacity ($EFC$): $EFC = TFC \times \eta = 0.756 \times 0.80 = 0.6048\text{ ha/h} \approx 0.60\text{ ha/h}$
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