ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 is a PhD, fellowship and lectureship exam run by National Testing Agency for ICAR, for candidates in India. LAN-based computer test, 2 hours, 120 MCQs: 20 Part A (general knowledge and reasoning), 50 Part B (core group), 50 Part C (specialisation); four options.
Source: icar.org.in.
Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.
Energy in agriculture at PG depth: energy audit coefficients and indices, mechanisation indices, farm power availability trends, and renewable energy conversion efficiencies.
Where the energy for farm work comes from, how much is available, and how mechanisation is measured.
Our 25 ICAR AICE-JRF/SRF questions in this subject: 5 easy, 13 medium, 7 hard, across 7 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 ICAR AICE-JRF/SRF and are not shown on any other exam's page.
A worker consumes oxygen at 1.5 L/min while working. Taking an energy equivalent of 20.1 kJ per litre of oxygen and a gross mechanical efficiency of 20%, estimate the mechanical power output.
B. 100.5 W
Metabolic power = 1.5 × 20.1 = 30.15 kJ/min = 30,150/60 = 502.5 W. Mechanical output = 0.20 × 502.5 = 100.5 W. 502.5 W is the metabolic rate before applying efficiency; dividing by 3600 treats the rate as per hour; dividing by the efficiency instead of multiplying gives 2,512.5 W.
A wind turbine rotor of 20 m radius turns at 30 rpm in a 10 m/s wind. What is its tip speed ratio?
C. 6.28
ω = 2π × 30/60 = 3.1416 rad/s; tip speed = ω R = 62.8 m/s; λ = 62.8/10 = 6.28. Using rpm directly gives 60; 0.628 is a decimal slip; 3.14 treats the 20 m radius as a diameter.
A 250 kW wind turbine on a farm produces 547.5 MWh in a year. What is its capacity factor?
B. 25.0%
Capacity factor = 547,500 kWh / (250 kW × 8760 h) = 0.250 = 25.0%. Dividing MWh by kW without converting gives 219% (impossible, above 100%); 75.0% assumes only 8 operating hours a day, but capacity factor is referred to all 8760 hours. 2.5% is a factor-of-ten slip.
At a site the wind blows at 4 m/s for half the time and 8 m/s for the other half. What is the ratio of the true mean wind power to the power calculated from the mean wind speed?
A. 1.33
Mean of V³ = (64 + 512)/2 = 288; (mean V)³ = 6³ = 216; ratio = 288/216 = 1.33. Because power depends on V³, using the mean speed underestimates energy; this is why wind speed distributions (Weibull, Rayleigh) are used. 1.00 assumes power is linear in speed; 0.75 is the inverse ratio, 216/288.
'Agrivoltaics' refers to:
C. Using the same land for crops or grazing and for PV electricity generation
Agrivoltaic systems raise or space PV arrays so that crops can grow or animals graze beneath them, sharing land between food and power.
Which statement comparing biodiesel with petroleum diesel is correct?
D. Biodiesel has a somewhat lower heating value but higher cetane number and better lubricity
Biodiesel's heating value is somewhat below diesel's; its cetane number, flash point and lubricity are higher; cold-flow behaviour is poorer. Transesterification greatly reduces viscosity compared with straight vegetable oil.
Across Indian states, farm power availability (kW/ha) is positively correlated with land productivity. What is the soundest interpretation?
D. The association is real, but it does not by itself show that power causes higher yield, because irrigation, soils and income also drive both
Power improves timeliness and quality of operations, so some causal effect is plausible, but cross-sectional correlation is confounded by irrigation, soil fertility and income. Causal claims need designs such as panel data or natural experiments.
Using the block data (total farm power 1989 kW, of which tractors, power tillers, electric motors and diesel engines give 1830 kW), what is the level of mechanisation expressed as the share of mechanical and electrical power?
A. 92.0%
Share = 1830/1989 = 92.0%. 8.0% is the animate share; 1151% divides mechanical by animate power instead of by the total; 37.7% counts tractors only.
An operation needs 150 man-hours per hectare. Using an energy equivalent of 1.96 MJ per man-hour, what human energy input is recorded?
D. 294 MJ/ha
Energy = 150 × 1.96 = 294 MJ/ha. Dividing gives 76.5; 81.7 is the kWh value mislabelled; multiplying by 8 treats man-hours as man-days.
With an energy output of 73,500 MJ/ha and input of 25,000 MJ/ha, what is the net energy?
A. 48,500 MJ/ha
Net energy = output − input = 73,500 − 25,000 = 48,500 MJ/ha. 2.94 is the energy ratio (dimensionless), not MJ/ha. 98,500 MJ/ha adds input to output.
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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