Allied & ICAR Exam Bank • ALLIED_0176 NAT • 2 Marks

Section 2: Farm Machinery

Farm Machinery

Subtopic: Functional requirements, principles of working, construction and operation of manual, animal, tractor and renewable energy operated equipment for tillage, sowing, planting, fertilizer application, inter-cultivation, spraying, mowing, chaff cutting, harvesting and threshing

Question Statement:

A seed-cum-fertilizer drill has drive wheel circumference $2.0\text{ m}$. The drive wheel slips by $10\%$. Over a field distance of $100\text{ m}$, the actual number of revolutions of the drive wheel is:

Answer Key:

55.6

Step-by-Step Solution & Explanation

Zero slip revs $N_0 = 100 / 2.0 = 50\text{ revs}$. With $10\%$ slip ($S = (N - N_0)/N = 0.10$), $0.9 N = 50 \implies N = 50 / 0.9 = 55.56\text{ revs}$.
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