Question Bank • QB_FP_004 NAT • 2 Marks

Section 3: Farm Power

Tractors and Power Tillers

Subtopic: Traction theory

Question Statement:

A tractor drive wheel of rolling radius $0.75\text{ m}$ completes 100 revolutions while travelling a horizontal distance of $424\text{ m}$ in tilled soil. The wheel slip expressed in percentage is ________ (round off to 1 decimal place).

Answer Key:

10.0

Step-by-Step Solution & Explanation

1. Theoretical distance without slip $S_{th}$: $S_{th} = 2 \pi r \cdot N = 2 \times 3.14159 \times 0.75 \times 100 = 471.24\text{ m}$ 2. Actual distance travelled $S_a = 424\text{ m}$. 3. Wheel slip $s$: $s = \frac{S_{th} - S_a}{S_{th}} \times 100 = \frac{471.24 - 424}{471.24} \times 100 = \frac{47.24}{471.24} \times 100 = 10.02\% \approx 10.0\%$
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