Rajasthan PSC — Assistant Agriculture Engineer (Agriculture Engineering syllabus) · India · Rajasthan Public Service Commission
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%)
Rajasthan PSC — Assistant Agriculture Engineer and other state services
Farm power portion: sources of power, kW/ha, mechanisation schemes and custom hiring; direct formula questions.
How our questions for Rajasthan PSC spread across the units of Farm Power Sources. Difficulty: 21 easy, 7 medium, 4 hard.
| Farm power availability (kW/ha) | 6 | |
| Animal draught power and harnessing | 5 | |
| Mechanical power: engines and tractors | 4 | |
| Electrical power on the farm | 4 | |
| Biomass and biofuels as power sources | 4 | |
| Human power and its limits | 3 | |
| Wind power as a farm power source | 2 | |
| Solar power as a farm power source | 2 | |
| 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 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.
An animal-drawn mouldboard plough cuts a furrow 20 cm wide and 12 cm deep in a soil with a specific draught of 4 N/cm². What is the draught?
B. 960 N
Draught = specific draught × furrow area = 4 N/cm² × (20 × 12) cm² = 960 N. The other options are decimal slips or half the area; always multiply width by depth in the same units as the specific draught.
A 4-pole induction motor runs on a 50 Hz supply with 4% slip at full load. What is its rotor speed?
B. 1440 rpm
Ns = 120f/p = 120 × 50/4 = 1500 rpm; N = Ns(1 − s) = 1500 × 0.96 = 1440 rpm. 1500 rpm is synchronous speed; 2880 rpm uses pole pairs instead of poles; adding slip gives 1560 rpm.
Which motor is the standard choice for farm pumps, threshers and grinders where a three-phase supply is available?
D. Three-phase squirrel-cage induction motor
The squirrel-cage induction motor is robust, cheap and nearly maintenance-free. DC series and universal motors are for traction and small hand tools; synchronous motors are for special constant-speed or power-factor duties.
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 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.
A district of 10,000 ha net sown area has 2.0 kW/ha. The target is 4.0 kW/ha, to be met entirely with 35 kW tractors. About how many additional tractors are needed?
D. 572 tractors
Additional power = (4.0 − 2.0) × 10,000 = 20,000 kW; tractors = 20,000/35 = 571.4 → 572. Using the full 4 kW/ha ignores existing power (1143). In practice targets are met with a mix of sources. 58 uses 350 kW per tractor, a factor-of-ten slip (20,000/350 = 57.1); 1715 adds the existing 2.0 kW/ha to the target (60,000/35).
Custom hiring centres (machinery banks) mainly raise the farm power effectively available to smallholders by:
C. Letting many farmers use machines they could not justify owning, when they need them
Hiring spreads fixed costs over more farms and hours, improving access and timeliness without each farmer buying a machine. It does not change machine ratings.
Operation-wise mechanisation data show tillage and threshing almost fully mechanised, but rice transplanting and cotton picking largely manual. What use is this for policy?
C. It identifies the operations where research, machinery development and subsidy will do most to raise the overall level
An overall index hides gaps. Operation-wise figures show bottlenecks such as transplanting and harvesting where targeted interventions pay off.
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