GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
Original practice questions written to the published syllabus · not past papersWhere the energy for farm work comes from, how much is available, and how mechanisation is measured.
In the exams: GATE AG Section 3 Sources of Power; ICAR AIEEA-PG Unit V; PRC ABE power and machinery (32%)
GATE — Agricultural Engineering (AG)
Section 3 'Sources of Power': short calculations and NAT questions on power in wind (½ρAV³, Betz limit), PV fill factor and module area or efficiency, biogas plant sizing, and power from draught and speed. Unit handling (kgf, km/h, metric hp) decides marks.
How our questions for GATE spread across the units of Farm Power Sources. Difficulty: 14 easy, 14 medium, 18 hard.
| Wind power as a farm power source | 8 | |
| Solar power as a farm power source | 7 | |
| Human power and its limits | 6 | |
| Mechanical power: engines and tractors | 6 | |
| Biomass and biofuels as power sources | 6 | |
| Animal draught power and harnessing | 5 | |
| Electrical power on the farm | 3 | |
| Farm power availability (kW/ha) | 3 | |
| Mechanisation indices | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Standard agricultural engineering textbooks commonly take the mechanical power an adult worker can sustain over a working day as roughly:
A. 0.075–0.1 kW
About 0.1 hp, i.e. roughly 75 W, is the usual textbook figure for continuous work; planning studies often use about 0.05 kW. 0.75–1 kW is only reachable for a few seconds in a sprint, and 0.3–0.4 kW is closer to a draught animal's output.
A worker lifts 2400 litres of water through 5 m by hand in one hour. What is the useful (water-lifting) power output?
C. 32.7 W
P = mgh/t = 2400 × 9.81 × 5 / 3600 = 32.7 W (1 litre of water = 1 kg). Forgetting g gives 3.33 W; dividing by 60 instead of 3600 (minutes for seconds) gives 1,962 W, which no person can sustain. 0.33 W is a factor-of-ten slip on the 3.33 W value.
Which statement about human power output and duration is correct?
A. The power a person can sustain falls steeply as the duration of effort increases
A person can produce several hundred watts or more for seconds but only about 75 W or less all day. Output is not constant, peak exceeds sustained power, and heat above body-comfort levels reduces output.
A pair of bullocks exerts a steady pull of 1.2 kN while walking at 3 km/h. What power do they develop?
A. 1.00 kW
v = 3/3.6 = 0.833 m/s; P = 1.2 kN × 0.833 m/s = 1.00 kW. Multiplying by km/h without converting gives 3.60 kW, the commonest slip. 12.96 kW multiplies by 3.6 instead of dividing (1.2 × 3 × 3.6); 10.0 kW is a factor-of-ten slip.
A diesel engine delivers 30 kW brake power while consuming 8 L/h of fuel (density 0.84 kg/L, calorific value 42 MJ/kg). What is its brake thermal efficiency?
B. 38.3%
Fuel energy rate = 8 × 0.84 × 42 = 282.24 MJ/h = 78.4 kW. Efficiency = 30/78.4 = 38.3%. Ignoring density gives 32.1%; dividing by density instead of multiplying gives 27.0%. 3.83% is a factor-of-ten slip.
The Betz limit states that the maximum fraction of the power in the wind that an ideal rotor can extract is:
B. 16/27, about 59.3%
Betz's analysis shows the maximum power coefficient is 16/27 ≈ 0.593, because air must keep flowing through the rotor. Real rotors achieve less.
If wind speed doubles, the power in the wind:
D. Increases eight times
Power ∝ V³, so 2³ = 8. Four times would be the effect of doubling rotor diameter (area ∝ D²).
A farm has 8 cattle each giving 10 kg of fresh dung per day. Taking a gas yield of 0.04 m³ per kg of fresh dung, what daily biogas production can be expected?
C. 3.2 m³/day
Dung = 8 × 10 = 80 kg/day; gas = 80 × 0.04 = 3.2 m³/day. 0.4 m³ is for one animal only. 0.32 and 32 m³/day are factor-of-ten slips.
The same 80 kg/day of dung is mixed 1:1 with water (slurry density about 1000 kg/m³). For a hydraulic retention time of 40 days, what digester volume is needed?
B. 6.4 m³
Slurry = 80 + 80 = 160 kg/day ≈ 0.16 m³/day; volume = 0.16 × 40 = 6.4 m³. Forgetting the water gives 3.2 m³; 0.16 m³ is one day's feed only. 64 m³ is a factor-of-ten slip.
A block has 400 ha net sown area, 25 tractors of 30 kW, 10 power tillers of 9 kW, 150 electric motors of 5 kW, 60 diesel engines of 4 kW, 300 draught animals (0.38 kW each) and 900 workers (0.05 kW each). What is the farm power availability?
B. 4.97 kW/ha
Total = 750 + 90 + 750 + 240 + 114 + 45 = 1,989 kW; ÷ 400 ha = 4.97 kW/ha. Excluding people and animals gives 4.58; counting tractors only gives 1.88; dividing by a gross cropped area of 700 ha gives 2.84.
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