GATE AG 2015 • Q64 MCQ • 2 Marks

Section 3: Farm Power

Farm Power

Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses

Question Statement:

The speed reductions in the 1st low gear of a tractor gearbox and differential with final drive are 5:1 and 40:1, respectively. For the tractor developing 24 kW power at an engine rpm of 2000 with an overall power transmission efficiency of 80%, the total torque in kN m available at the wheel axle will be

Options:

  • (A) 22.92
  • (B) 28.64
  • (C) 16.54
  • (D) 18.33

Official Answer Key:

D

Step-by-Step Solution & Explanation

Engine torque Tₑ = P/ω = 24000/(2π×2000/60) = 24000/209.44 = 114.6 N·m. Total speed reduction from engine to wheel axle = 5 × 40 = 200. Accounting for the 80% overall transmission efficiency, wheel-axle torque: T_wheel = Tₑ × (gear ratio) × (efficiency) = 114.6 × 200 × 0.80 = 18334.6 N·m = 18.33 kN·m. Matches option (D) exactly.

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