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GATE Tractor Systems & Controls: practice questions

GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)

Original practice questions written to the published syllabus · not past papers
10worked here, free
65written to this exam in this subject
15syllabus units

What this subject covers

Everything between the engine and the implement: clutch, gearbox, differential, PTO, brakes, steering, hydraulics, hitch and tyres.

In the exams: GATE AG Section 3 Tractors and Power Tillers; ICAR AIEEA-PG Unit V; ICAR AICE subject 39 Unit 4; ASRB NET FMP; PRC ABE

What GATE asks from it

GATE — Agricultural Engineering (AG)

Section 3 'Tractors and Power Tillers': numerical questions on overall gear ratio and travel speed, differential speeds, clutch torque capacity (uniform wear vs uniform pressure), PTO speed and power, Ackermann steering angles, and hydraulic force and power; conceptual MCQ/MSQ on gearbox types, draft vs position control, open vs closed centre hydraulics. NAT answers are sensitive to radius vs diameter and rev/min vs rev/s.

Where the 65 questions fall

How our questions for GATE spread across the units of Tractor Systems & Controls. Difficulty: 22 easy, 29 medium, 14 hard.

Power take-off: standards and speeds8
Hydraulic system: open and closed centre8
Steering geometry (Ackermann) and power steering6
Gearbox and transmission: sliding, constant and synchromesh5
Clutch: single and dual plate5
CVT and power-shift transmissions5
Differential and differential lock4
Final drive4
Brakes4
Tyres and ballasting4
Three-point hitch: categories, free and restrained links3
Tractor types and selection3
Power tillers2
Draft and position control2
Operator controls and ROPS2

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Gearbox and transmission: sliding, constant and synchromesh · easy

In a two-stage reduction, a 15-tooth pinion on the input shaft drives a 45-tooth gear on the countershaft, and a 12-tooth pinion on the countershaft drives a 48-tooth gear on the output shaft. If the input turns at 1,800 rev/min, the output speed is

  1. 257 rev/min
  2. 600 rev/min
  3. 21,600 rev/min
  4. 150 rev/min
Answer and working

D. 150 rev/min

Ratio = (45/15) × (48/12) = 3 × 4 = 12. Output = 1,800/12 = 150 rev/min. Adding the ratios (3 + 4 = 7) gives 257; using only the first stage gives 600; multiplying instead of dividing gives 21,600.

2 · Clutch: single and dual plate · medium

A single-plate clutch has linings of outer radius 150 mm and inner radius 100 mm, μ = 0.3, and spring force 4000 N. Assuming uniform wear, the torque capacity is

  1. 304 N·m
  2. 150 N·m
  3. 300 N·m
  4. 360 N·m
Answer and working

C. 300 N·m

Uniform wear: R_m = (R_o + R_i)/2 = 0.125 m. T = μ W n R_m = 0.3 × 4000 × 2 × 0.125 = 300 N·m (n = 2 because the single driven plate has two friction faces). Uniform pressure gives R_m = 0.1267 m and 304 N·m; n = 1 gives 150 N·m; using the outer radius gives 360 N·m.

3 · Clutch: single and dual plate · hard

A tractor engine develops a maximum torque of 250 N·m. The single-plate dry clutch must carry 1.4 times this torque. Linings: outer radius 140 mm, inner radius 90 mm, μ = 0.35. Using the uniform-wear assumption, the total spring force needed is closest to

  1. 4,348 N
  2. 8,696 N
  3. 3,106 N
  4. 3,571 N
Answer and working

A. 4,348 N

Design torque = 250 × 1.4 = 350 N·m. R_m = (0.14 + 0.09)/2 = 0.115 m. W = T/(μ n R_m) = 350/(0.35 × 2 × 0.115) = 4,348 N. Using one friction face doubles it to 8,696 N; ignoring the service factor gives 3,106 N, a clutch that slips at peak load; using the outer radius gives 3,571 N.

4 · Hydraulic system: open and closed centre · medium

An operator holds a tractor's open-centre hydraulic lever against the stop after the cylinder has reached the end of its stroke. The gear pump delivers 30 L/min and the relief valve is set at 17.5 MPa. The rate of heat generation in the oil is

  1. 0.875 kW
  2. zero, because the cylinder is not moving
  3. 525 kW
  4. 8.75 kW
Answer and working

D. 8.75 kW

All the flow passes over the relief valve and does no work, so heat = Δp × Q = 17.5 MPa × 30 L/min/60 = 8.75 kW. /600 applies to bar, not MPa. This is why levers must not be held against the relief valve and why open-centre systems overheat when misused.

5 · Tyres and ballasting · hard

A 2WD tractor must develop a net pull of 18 kN on a soil where the usable net traction ratio at acceptable slip is 0.45. The dynamic load on the rear wheels is at present 32 kN. The ballast to add over the rear wheels is about

  1. 4,077 kg
  2. 1,835 kg
  3. 8,000 kg
  4. 815 kg
Answer and working

D. 815 kg

Required rear load = pull/ratio = 18.0/0.45 = 40 kN. Extra load = 40 − 32 = 8 kN = 8,000 N ÷ 9.81 = 815 kg. 4,077 kg is the whole rear load, not the addition; 8,000 forgets to convert newtons to kilograms. Check slip in the field afterwards, because the traction ratio varies with soil. 1,835 kg converts the whole 18 kN pull to mass (18,000/9.81).

6 · Gearbox and transmission: sliding, constant and synchromesh · medium

The synchroniser in a synchromesh gearbox

  1. locks the differential when the gear lever is moved to low range
  2. uses a cone clutch to match speeds before the dogs engage
  3. slides the spur gear sideways along its splines into mesh with the layshaft gear
  4. reverses the output direction by inserting an idler between the shafts
Answer and working

B. uses a cone clutch to match speeds before the dogs engage

The synchro ring rubs on the gear cone to bring both to the same speed, then the collar's dogs engage without clash, allowing shifts on the move.

7 · Power take-off: standards and speeds · medium

A 35 mm PTO shaft with 21 splines is designed to run at

  1. 1,000 rev/min
  2. 540 rev/min (type 1)
  3. 540 rev/min economy mode
  4. Proportional to ground speed
Answer and working

A. 1,000 rev/min

Type 1 is 35 mm with 6 splines at 540 rev/min; type 2 is 35 mm with 21 splines at 1,000 rev/min; type 3 is 45 mm with 20 splines at 1,000 rev/min. Different splines prevent mismatching.

8 · Steering geometry (Ackermann) and power steering · easy

The recirculating-ball steering gear uses balls

  1. as the main bearing of the front wheel
  2. to lock the differential
  3. running between a worm and a nut to reduce friction
  4. in the power-steering pump
Answer and working

C. running between a worm and a nut to reduce friction

Steel balls roll in the grooves between the worm and nut and recirculate through a return tube, giving low friction and wear.

9 · Tyres and ballasting · medium

Rear drive tyres with chevron lugs should be mounted so that

  1. the open end of the chevron contacts the ground first, so mud is squeezed out
  2. the point of the chevron contacts the ground first as the wheel rolls forward
  3. direction does not matter
  4. the lugs point towards the tractor centre
Answer and working

B. the point of the chevron contacts the ground first as the wheel rolls forward

Mounted correctly the lugs dig in and self-clean; mounted backwards they pack with soil and lose traction.

10 · Tyres and ballasting · hard

If an MFWD tractor is fitted with front tyres of a smaller rolling circumference than specified, the likely result is

  1. better traction with no drawback, as smaller front tyres dig in harder under the drive torque
  2. higher road speed
  3. the front wheels lag the rear, causing tyre scrub, power circulation and drive-train wear
  4. lighter steering
Answer and working

C. the front wheels lag the rear, causing tyre scrub, power circulation and drive-train wear

The front axle ratio is set for a small lead with the specified tyres. Smaller front tyres reduce or reverse the lead, so the axles fight each other.

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Read the subject

Tractor Systems & ControlsThe full study manual, free GATEFormat, sections and how it has changed

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Tractor Systems & Controls for other exams

ICAR AICE-JRF/SRF · IndiaICAR AIEEA-PG · IndiaRajasthan PSC · IndiaFarm Machinery Training & Testing Institutes · India

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