Question Bank • QB_SUB_FP_TRC_002 NAT • 2 Marks

Section 3: Farm Power

Tractors and Power Tillers

Subtopic: Traction theory

Question Statement:

A pneumatic drive wheel travels a nominal distance of $25.0\text{ m}$ under zero-load conditions for 5 revolutions. Under heavy load, the tractor travels only $20.0\text{ m}$ for the same 5 revolutions. The wheel travel reduction (wheel slip, in percent) is:

Answer Key:

20.0

Step-by-Step Solution & Explanation

Slip $S = \frac{V_0 - V_a}{V_0} \times 100 = \frac{25.0 - 20.0}{25.0} \times 100 = \frac{5.0}{25.0} \times 100 = 20.0\%$
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