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.
Design of tillage and planting machinery: soil–tool mechanics, critical depth, rotary tiller kinematics, metering design and singulation indices, draft prediction and hitching.
Machines that prepare the seedbed and put seed and fertiliser in the ground, and how to calibrate and cost them.
Our 23 ICAR AICE-JRF/SRF questions in this subject: 0 easy, 5 medium, 18 hard, across 13 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.
On a disc plough, increasing the tilt angle (disc more nearly lying back from vertical) generally
B. improves inversion of the slice but reduces penetration
A larger tilt angle lays the disc back so it lifts and turns the slice more but presents a less penetrating edge; to improve penetration the tilt is reduced. Changing disc angle, not tilt, mainly changes width of cut.
According to the Godwin–Spoor analysis of narrow tines, what happens to soil below the critical depth?
D. It flows sideways around the tine and is compacted rather than loosened
Above critical depth soil fails in a crescent upward and forward; below it the soil cannot escape upward and moves laterally, causing compaction. That is why working depth should stay above critical depth.
Which change raises the critical depth of a subsoiler tine, allowing deeper loosening without compaction?
B. Fitting wings to the tine foot or adding shallow leading tines
Wings and shallow leading tines raise critical depth and increase loosened area per unit draft. Wetter soil lowers critical depth, and critical depth falls, not rises, as the tine gets narrower.
For a rotary tiller at fixed rotor speed and blade number, increasing forward speed will
B. lengthen bite and produce a coarser tilth
Bite length L = 60v/(nZ) rises in proportion to v, so each blade takes a longer slice and clods are larger. λ = ωR/v falls as v rises.
In the ASABE D497 draft equation D = Fᵢ[A + BS + CS²]WT, what does Fᵢ represent?
D. A dimensionless soil texture adjustment
Fᵢ adjusts draft for fine, medium or coarse soil texture; A, B and C are machine parameters, S speed in km/h, W width and T depth. Efficiencies are not part of the draft equation.
A vacuum planter must place maize at 20 cm spacing while travelling at 7 km/h. The disc has 30 holes. What disc speed is required?
B. 19.4 rev/min
Seeds per second = v/s = (7/3.6)/0.2 = 9.72. Disc rev/min = 9.72 × 60/30 = 19.4. Using 7 without converting gives 70; omitting 60 gives rev/s; forgetting 30 holes gives seeds per minute.
In a planter test, 200 seed spacings were measured. 12 were greater than 1.5 times the theoretical spacing and 8 were not greater than 0.5 times. What is the quality-of-feed index?
A. 90 %
Miss index = 12/200 = 6 %; multiples index = 8/200 = 4 %. Quality of feed = 100 − 6 − 4 = 90 %. Subtracting only misses gives 94 %. 10 % is the combined miss and multiple index, not the quality of feed.
A no-till double-disc opener in wet clay leaves a glazed slot that later opens as it dries. What is the underlying problem?
B. Sidewall smearing and compaction by the discs, giving poor seed–soil contact
Discs pushed through wet, plastic soil smear the slot walls; the slot then cracks open, exposing the seed. Sowing at better moisture or using closing wheels that break the walls helps.
A planter plate with 24 cells turns once per ground wheel revolution. The wheel is 60 cm in diameter with 10 % skid (travel per turn = πD × 0.9). What is the seed spacing, and how many seeds per hectare at 75 cm rows?
A. 7.07 cm; 188,628 seeds/ha
Spacing = π × 0.6 × 0.9/24 = 0.0707 m = 7.07 cm. Seeds/ha = 10 000/(0.75 × 0.0707) = 188,628. Ignoring skid gives 7.85 cm; 18.86 seeds is the count per m², not per hectare.
Using the capital recovery factor, what is the uniform annual ownership cost (depreciation plus interest) of a machine costing 300 000 with 30 000 salvage after 5 years at 12 % interest?
C. 78,501
CRF = i(1+i)ⁿ/((1+i)ⁿ − 1) = 0.2774. Annual cost = (P − S) × CRF + S × i = 270 000 × 0.2774 + 30 000 × 0.12 = 78,501. The average-investment approximation gives 73,800; ignoring salvage gives 83,223; depreciation alone gives 54,000.
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.
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