Allied & ICAR Exam Bank • ALLIED_0231 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 tractor-drawn cultivator has width of cut $2.4\text{ m}$. Forward speed is $5.0\text{ km/h}$. Over an 8-hour workday, the machine covers $7.68\text{ ha}$. The field efficiency in percentage is:

Answer Key:

80

Step-by-Step Solution & Explanation

$TFC = (2.4 \times 5.0)/10 = 1.20\text{ ha/h}$. In 8 hours, theoretical area $= 1.20 \times 8 = 9.60\text{ ha}$. Efficiency $= (7.68 / 9.60) \times 100 = 80.0\%$.
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