Question Bank • QB_FP_ADV_003 NAT • 2 Marks

Section 3: Farm Power

Tractors and Power Tillers

Subtopic: Traction theory

Question Statement:

A tractor drive wheel of rolling radius $0.65\text{ m}$ makes $120$ revolutions while traveling a measured distance of $440\text{ m}$ under load. The theoretical distance traveled per revolution is based on the zero-slip rolling radius. The percentage wheel slip is ________ $\%$ (round off to two decimal places).

Answer Key:

10.23

Step-by-Step Solution & Explanation

1. Theoretical distance traveled ($d_t$) without slip: $d_t = 2 \pi r N = 2 \times \pi \times 0.65\text{ m} \times 120 = 490.088\text{ m}$ 2. Actual distance traveled ($d_a$) under load: $d_a = 440\text{ m}$ 3. Percentage wheel slip ($S$): $S = \frac{d_t - d_a}{d_t} \times 100 = \frac{490.088 - 440}{490.088} \times 100 = \frac{50.088}{490.088} \times 100 = 10.220\% \approx 10.23\%$
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