Question Bank • QB_EXP_FM_004 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 seed-cum-fertilizer drill with 9 furrow openers spaced at 20 cm operates at a forward speed of 4.5 km/h. If the effective field capacity of the machine is 0.6075 ha/h, the field efficiency of the operation in percentage is ________ (in integer).

Answer Key:

75

Step-by-Step Solution & Explanation

1. Total width of operation $W = 9 \times 0.20\text{ m} = 1.8\text{ m}$. 2. Forward speed $v = 4.5\text{ km/h}$. 3. Theoretical Field Capacity $TFC = \frac{W \times v}{10} = \frac{1.8 \times 4.5}{10} = 0.81\text{ ha/h}$. 4. Field efficiency $\eta = \frac{EFC}{TFC} \times 100 = \frac{0.6075}{0.81} \times 100 = 75\%$.
← Previous (QB_FM_GATE_003) Next (QB_SUBEXP_S2_003) →

Practice In Authentic CBT Exam Interface

Test your skills under authentic timed conditions with live scoring and AIR percentile ranking.

Launch CBT Mock Test