GATE AG 2008 • Q68 MCQ • 2 Marks

Section 3: Farm Power

Tractors and Power Tillers

Subtopic: Steering and hydraulic control systems used in tractors

Question Statement:

A hydraulic motor receives a flow rate of 72 L min⁻¹ at a pressure of 12 MPa. The motor speed is 800 rpm. If the motor has a power loss of 3 kW, the actual torque delivered by the motor is

Options:

  • (A) 136.08 N m
  • (B) 171.89 N m
  • (C) 204.62 N m
  • (D) 262.84 N m

Official Answer Key:

A

Step-by-Step Solution & Explanation

Hydraulic input power = pQ = 12×10⁶×(72/60000) = 12×10⁶×0.0012 = 14.4 kW. Output (shaft) power = 14.4−3 = 11.4 kW. ω = 2π×800/60 = 83.78 rad/s. Torque = P/ω = 11,400/83.78 = 136.08 N·m.

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