GATE AG 2007 • Q28 MCQ • 1 Mark

Section 2: Farm Machinery

Farm Machinery

Subtopic: Functional requirements, principles of working, construction and operation of manual, animal, tractor and renewable energy operated equipment for tillage, sowing, planting, fertilizer application, inter-cultivation, spraying, mowing, chaff cutting, harvesting and threshing

Question Statement:

A tractor drawn rotary cultivator in concurrent revolution mode is to be operated at a depth of 150 mm and at a forward speed of 3.6 km h⁻¹. The radius of working set is 280 mm. The number of blades, which would cut identical path is 3. The working width of the cultivator is 1.8 m. The cultivator is to be powered from the tractor PTO running at 540 rpm through a suitable gearbox. For getting a tilling pitch of 74.1 mm, the suitable gear ratio is

Options:

  • (A) 1:2
  • (B) 1:1.5
  • (C) 1.5:1
  • (D) 2:1

Official Answer Key:

D

Step-by-Step Solution & Explanation

Tilling pitch, ℓ = v_forward /(N_rotor × z), where z = 3 identical-path blades. v_forward = 3.6 km/h = 1.0 m/s. N_rotor(rev/s) = v/(ℓ×z) = 1.0/(0.0741×3) = 4.499 rev/s → N_rotor = 269.9 rpm. Gear ratio (PTO : rotor) = 540/269.9 = ≈2:1.

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