Question Bank • QB_FP_GATE_003 NAT • 2 Marks

Section 3: Farm Power

Tractors and Power Tillers

Subtopic: Traction theory

Question Statement:

A tractor driving wheel of rolling radius $0.75\text{ m}$ rotates at $40\text{ rpm}$. The tractor travels a distance of $160\text{ m}$ in $1\text{ minute}$ under load. The travel reduction (wheel slip) of the tractor is ________ $\%$ (round off to 1 decimal place).

Answer Key:

15.1

Step-by-Step Solution & Explanation

Wheel Slip Formulation: $D_0 = 2\pi r N t = 2\pi (0.75)(40)(1) = 60\pi \approx 188.496\text{ m}$ $S = \frac{D_0 - D_a}{D_0} \times 100 = \frac{188.496 - 160}{188.496} \times 100 = 15.118\%$ Rounding to 1 decimal place: **15.1** $\%$.
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