GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersHow soil resists tools and supports wheels: draft, slip, tractive efficiency, weight transfer and stability.
In the exams: GATE AG Section 3 (traction asked in all 8 papers 2019–26; weight transfer and stability in 5 of 8); ICAR AICE subject 39 Unit 6; ASRB NET FMP
GATE — Agricultural Engineering (AG)
Numerical questions on wheel slip (from speeds, revolutions or distances), rolling radius, tractive efficiency, net traction and gross traction, maximum thrust on sand (W tan φ), weight transfer (P·h/L), front-axle reaction and CG location, overturning angles. Many are NAT questions: keep units and ratios straight.
How our questions for GATE spread across the units of Mechanics of Tillage & Traction (Soil Dynamics). Difficulty: 30 easy, 26 medium, 12 hard.
| Traction theory: slip, rolling resistance and net traction | 10 | |
| Prediction of draft | 6 | |
| Tractive efficiency | 6 | |
| Wheel and track mechanics | 6 | |
| Longitudinal and lateral stability | 6 | |
| Tractor chassis mechanics | 6 | |
| Weight transfer | 5 | |
| Determination of the centre of gravity | 5 | |
| Dynamic properties and strength of soil | 4 | |
| Soil-tool mechanics | 4 | |
| Soil compaction by vehicles | 4 | |
| Cone index and cone penetrometer | 3 | |
| Traction in wetland soils | 3 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A soil has cohesion 15 kPa and angle of internal friction 25°. What is its shear strength on a plane carrying a normal stress of 100 kPa?
C. 61.6 kPa
τ = c + σ tan φ = 15 + 100 × tan 25° = 15 + 46.6 = 61.6 kPa. Leaving out cohesion gives 46.6 kPa; using sin φ instead of tan φ gives 57.3 kPa. 105.6 kPa uses cos φ instead of tan φ.
For a cohesive soil, how does adhesion to a steel tool usually change as moisture content rises from dry to very wet?
A. It rises to a maximum at intermediate moisture and then falls as a water film lubricates the interface
Dry soil has little stickiness; adhesion rises with moisture as water bridges form, then falls when a free water film separates soil from steel. This is why sticky clays cause build-up at intermediate moisture.
In moist plastic soil, the draft of a tine recorded against time is smooth and continuous, while in dry friable soil it shows a repeating saw-tooth pattern. What does the saw-tooth indicate?
D. Successive shear failure of soil blocks ahead of the tool
In brittle soil, force builds as a block is compressed, drops when it shears away, and builds again: a saw-tooth. Flow failure in plastic soil gives a smoother trace.
Why does tractive efficiency pass through a maximum as slip rises?
D. Net traction rises quickly then levels off, while the speed loss grows in proportion to slip
At zero slip, net traction and TE are zero. NT/GT rises steeply then saturates as soil strength is mobilised, while (1 − s) falls steadily; their product peaks.
A standard cone of base area 323 mm² needs a force of 450 N to penetrate soil at the standard rate. What is the cone index?
A. 1393 kPa
CI = F/A = 450 N / 323 × 10⁻⁶ m² = 1393189 Pa ≈ 1393 kPa (1.39 MPa). Dividing by mm² directly gives MPa-scale numbers in the wrong unit; using the small cone's area gives 3462 kPa.
Further reading: ASABE S313
Weight transfer from the front to the rear axle of a pulling tractor increases when
D. the hitch height is raised
ΔW = P·h/L: higher hitch or higher pull increases transfer; a longer wheelbase reduces it. Front ballast raises the front load but does not change the amount transferred.
A tractor with a rigid front axle has track width 1.6 m and CG 0.9 m above ground, midway between the wheels. What is the static side slope at which it begins to overturn?
B. 41.6°
tan θ = (t/2)/h = 0.8/0.9 = 0.889 → θ = 41.6°. Using the full track gives 60.6°, which overstates stability. 48.4° inverts the ratio; 62.7° uses sin⁻¹ instead of tan⁻¹.
A tractor of weight 30 kN, wheelbase 2.2 m, has a level front reaction of 9.0 kN. With the front raised 20° (rear wheels on the ground) the front reads 7.5 kN. The axle centres are level on level ground at 0.7 m height. How high is the CG above the ground?
B. 1.00 m
y above the axle line = (Wf − Wf′)L/(W tan θ) = (9.0 − 7.5) × 2.2/(30 × tan 20°) = 3.30/10.919 = 0.302 m. Height above ground = 0.7 + 0.302 = 1.00 m. Stopping at y forgets the axle height; using sin θ or cos θ uses the wrong trigonometric function. 0.30 m stops at y; 1.02 m uses sin θ; 0.82 m uses cos θ.
When taking moments about the rear wheel contact point of a tractor on level ground, why do the front wheel rolling resistance and rear wheel thrust not appear?
D. They act horizontally at ground level, so their lines of action pass through the moment point's level and have zero arm
Both forces act at the tyre-soil interface, at ground level, so their moment about a point on the ground is zero. They do enter the horizontal force balance.
According to the widely used conclusion from Söhne's work, subsoil stress under a tyre is governed mainly by
D. the total axle load
Topsoil stress follows contact (roughly inflation) pressure; deeper stress follows total load. Lower pressure protects topsoil, but only lighter axle loads protect the subsoil.
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