GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersGATE — Agricultural Engineering (AG) is a postgraduate admission exam run by IITs and IISc on rotation (2026: IIT Guwahati), for candidates in India. Computer-based, 3 hours, 100 marks (15 General Aptitude + 85 technical: 25 one-mark and 30 two-mark questions). MCQ, MSQ and NAT; negative marking on MCQ only (−1/3 for one-mark, −2/3 for two-mark); no partial credit on MSQ. On-screen virtual calculator.
Source: gate2026.iitg.ac.in.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Numerical items on sprayers (swath = 2h tan(θ/2) minus overlap, application rate, field capacity with slip and field efficiency, pump power including bypass flow), reciprocating mower kinematics, reel index, threshing cylinder speed, and thresher or combine performance and losses.
Sprayers, weeders, mowers, reapers, combines and threshers: how they work, how to calibrate them and where losses come from.
Our 25 GATE questions in this subject: 3 easy, 14 medium, 8 hard, across 10 units.
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
These 10 were written to GATE and are not shown on any other exam's page.
A nozzle discharges 0.8 L/min at 3 bar. What pressure gives 1.0 L/min from the same nozzle?
D. 4.69 bar
Q is proportional to the square root of P, so P2 = P1 (Q2/Q1)^2 = 3 × (1.0/0.8)^2 = 3 × 1.5625 = 4.69 bar. Scaling pressure linearly with flow gives 3.75 bar, a common slip. 1.92 bar inverts the flow ratio (3 × 0.8²); 2.40 bar scales linearly with the inverted ratio (3 × 0.8).
A spray has Dv0.1 = 110 µm, Dv0.5 = 250 µm and Dv0.9 = 450 µm. Its relative span is
C. 1.36
Relative span = (Dv0.9 − Dv0.1)/Dv0.5 = (450 − 110)/250 = 340/250 = 1.36. Using Dv0.9/Dv0.5 gives 1.80; 110/250 gives 0.44 — both are wrong definitions. 2.43 also subtracts Dv0.1 in the denominator (340/140).
If droplet diameter is halved while the total spray volume is unchanged, the number of droplets becomes
A. 8 times as many
Droplet volume goes as d³; halving d reduces each droplet's volume to 1/8, so the same liquid forms 8 times as many droplets. Thinking in terms of area (d²) gives the wrong answer of 4.
An 80° flat-fan nozzle is 0.45 m above the crop, and adjacent patterns are to overlap by 0.25 m. The nozzle spacing (effective swath per nozzle) is about
A. 0.51 m
Pattern width = 2 × 0.45 × tan 40° = 0.755 m. Effective width = 0.755 − 0.25 = 0.505 m. Forgetting the overlap gives 0.76 m; adding it gives 1.01 m.
A boom has nozzles at 0.5 m spacing, each delivering 0.8 L/min, and travels at 6 km/h. The application rate is
C. 160 L/ha
q = 600 Q/(w v) = 600 × 0.8/(0.5 × 6) = 480/3 = 160 L/ha. Dropping or adding a factor of ten in the unit conversion gives 16 or 1600. 96 L/ha uses 360 in place of 600.
A boom has 12 nozzles each at 1.2 L/min, plus 10 L/min returned through the bypass for agitation. Operating pressure is 400 kPa and pump efficiency 60%. The shaft power is about
B. 271 W
Total flow = 12 × 1.2 + 10 = 24.4 L/min = 4.07 × 10^-4 m³/s. Hydraulic power = 400,000 × 4.07 × 10^-4 = 163 W; shaft power = 163/0.6 = 271 W. Ignoring the bypass gives 14.4 L/min and 160 W; forgetting efficiency gives 163 W. 98 W multiplies by the efficiency instead of dividing.
A disc mower's knife tips travel on a 0.5 m diameter circle at 3000 rev/min. The tip speed is about
D. 78.5 m/s
v = πDN/60 = π × 0.5 × 3000/60 = 78.5 m/s. Using radius in place of diameter in πDN gives 39.3 m/s. 157 m/s uses 2π with the diameter; 25.0 m/s leaves out π.
In the cleaning shoe, air velocity through the sieves must be
A. above the terminal velocity of chaff but below that of grain
Then chaff is lifted and blown out while grain falls through. Air faster than grain's terminal velocity blows grain out; slower than chaff's leaves chaff on the sieves.
In a wheat field, the crop yield is 5 t/ha and the loss is 20 kg/ha. Loss as a percentage of yield is
B. 0.4%
20/5000 × 100 = 0.4%. Using 500 kg/ha by mistake gives 4%. 0.04% is a factor-of-ten slip; 2.5% inverts the ratio (5000/20/100).
A power chaff cutter has 2 knives on a flywheel at 300 rev/min, and feed rolls move fodder at 0.1 m/s. The theoretical length of cut is
C. 10 mm
Cuts per second = 2 × 300/60 = 10. Length = 0.1/10 = 0.01 m = 10 mm. Forgetting the second knife gives 20 mm. 5 mm halves the length again; 100 mm leaves out the cuts per second.
This plan follows where our questions actually fall, so the time goes where the marks are likely to be. It is our reading of the syllabus, not the exam body's weighting.
Read the manual chapters for these first and do every worked question you can find on them.
Aim for one sitting per unit, then mixed questions across them.
Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.
Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.
Do these early in revision so they are fast on the day.
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