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 papersThe physical, frictional, aerodynamic, thermal, rheological and electrical properties of grains, fruits and foods that every machine and process is designed around.
In the exams: GATE AG Section 6 (first syllabus line; rheology in 4 of 8 papers 2019–26); ICAR AICE subject 41 Unit 1; NCEES PE properties of biological materials
ICAR AICE-JRF/SRF (PhD) — Processing and Food Engineering (41)
Unit 1 at PG depth: Hertz contact and the bioyield point, Burgers and generalised Maxwell models, Peleg relaxation, dynamic moduli and tan δ, Rabinowitsch-Mooney correction, Metzner-Reed Reynolds number, Choi-Okos and Sweat models, line heat source theory, GAB and BET fitting, isosteric heat, dielectric loss and penetration depth.
How our questions for ICAR AICE-JRF/SRF spread across the units of Engineering Properties of Agricultural Produce. Difficulty: 5 easy, 24 medium, 28 hard.
| Viscoelastic models | 12 | |
| Rheology: Newtonian and non-Newtonian flow | 8 | |
| Water activity | 8 | |
| Electrical and optical properties | 7 | |
| Thermal properties: specific heat, conductivity and diffusivity | 6 | |
| Aerodynamic properties: terminal velocity | 5 | |
| Size and shape: sphericity and roundness | 3 | |
| Porosity | 3 | |
| Frictional properties: angle of repose and coefficient of friction | 3 | |
| Volume and density: bulk and true | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Using Janssen's equation, the lateral pressure on the wall of a deep bin approaches L_max = ρ_b g R / μ', where R is the hydraulic radius (area/perimeter). For a 6.0 m diameter bin storing grain of bulk density 780 kg/m³ with a wall friction coefficient μ' = 0.40, L_max is
D. 28.7 kPa
For a circle R = A/P = (πD²/4)/(πD) = D/4 = 1.5 m, so L_max = 780 × 9.81 × 1.5 / 0.40 = 28694 Pa = 28.7 kPa. 57.4 kPa takes R as the radius D/2; 4.6 kPa multiplies by μ' instead of dividing; 114.8 kPa uses the diameter.
Further reading: Janssen's equation; ASABE S433.1
Why do bin design standards apply an overpressure factor to the static Janssen lateral pressure?
C. Because flow of grain during emptying produces wall pressures above the static values
Flow during discharge changes the stress state and produces lateral pressures higher than the static fill values, so codes multiply by an overpressure factor whose value differs between standards. The factor increases, not reduces, design pressure; bulk density rises with compaction rather than falling; and the vertical-to-lateral conversion uses k, not an overpressure factor.
Further reading: ASABE S433.1
A kernel is treated as a sphere of diameter 5 mm and density 1300 kg/m³ falling in air of density 1.2 kg/m³ with a constant drag coefficient of 0.44 (Newton regime). Its terminal velocity, V_t = [4 g d (ρ_p − ρ_f)/(3 C_d ρ_f)]^(1/2), is
D. 12.7 m/s
V_t² = 4 × 9.81 × 0.005 × 1298.8 / (3 × 0.44 × 1.2) = 160.9; V_t = 12.7 m/s. 9.0 m/s comes from using 2 instead of 4 in the numerator; 983 m/s applies Stokes' law (air viscosity 1.8 × 10⁻⁵ Pa·s) far outside its Reynolds-number range; 0.39 m/s puts particle density in the denominator instead of air density.
Further reading: Mohsenin, Physical Properties of Plant and Animal Materials
For spheres of the same density in air, doubling the diameter changes the terminal velocity by a factor of
C. √2 in the Newton regime and 4 in the Stokes regime
In the Newton regime V_t ∝ (d)^(1/2), so doubling d gives √2 = 1.41; in the Stokes regime V_t ∝ d², giving 4. Swapping the two is the common slip. Terminal velocity does depend on size, so neither 2 nor 1 is right.
Further reading: Mohsenin, Physical Properties of Plant and Animal Materials
A Kelvin-Voigt element (spring and dashpot in parallel) loaded with a constant stress shows
B. delayed creep towards σ0/E, with no instantaneous strain and no stress relaxation
In parallel, the dashpot prevents instantaneous deformation and the spring limits the final strain to σ0/E, so strain rises as 1 − e^(−t/λ). Instantaneous strain plus unlimited flow is Maxwell behaviour; stress relaxation to zero is also Maxwell; purely instantaneous strain is a Hookean spring.
Further reading: Mohsenin, Physical Properties of Plant and Animal Materials
A Burgers model has E1 = 400 kPa, η1 = 8000 kPa·s (Maxwell part) and E2 = 200 kPa, η2 = 2000 kPa·s (Kelvin part). Under a constant stress of 20 kPa, with ε(t) = σ0/E1 + σ0 t/η1 + (σ0/E2)[1 − exp(−E2 t/η2)], the strain at t = 10 s is
A. 0.1382
Instantaneous 20/400 = 0.0500; flow 20 × 10/8000 = 0.0250; delayed 20/200 × (1 − e^(−1)) = 0.1 × 0.6321 = 0.0632. Total = 0.1382. 0.1132 omits the viscous flow term; 0.1750 takes the delayed elastic strain as complete; 0.1500 keeps only the two springs.
Further reading: Mohsenin, Physical Properties of Plant and Animal Materials
A dough has a relaxation time of 100 s. During 1-second mixing strokes it behaves elastically, but over a 1-hour proof it flows. This is explained by
B. the Deborah number λ/t_obs, far above 1 during mixing and far below 1 during proofing
De = 100/1 = 100 during mixing (solid-like) and 100/3600 ≈ 0.03 over the proof (liquid-like). The Reynolds number concerns inertia against viscosity in flow, the Biot number concerns heat-transfer resistances, and a change in n is not what makes dough flow on resting.
Further reading: Steffe, Rheological Methods in Food Process Engineering
On the force-deformation curve of an apple compressed between plates, the bioyield point is where
D. deformation increases with no increase or a decrease in force, marking the start of cell rupture
Bioyield is the point of internal failure of cells, reached before the macroscopic rupture point, and it is the practical handling limit. Visible splitting is the rupture point; first contact is the start of the test; the steepest initial slope is used for modulus, not bioyield.
Further reading: Mohsenin, Physical Properties of Plant and Animal Materials
A ripening tomato reads a* = 24 and b* = 18 in CIE L*a*b*. Its hue angle h = arctan(b*/a*) and chroma C* = √(a*² + b*²) are
B. h = 36.9°, C* = 30.0
h = arctan(18/24) = 36.9° (first quadrant since a* and b* are positive) and C* = √(576 + 324) = 30.0. 53.1° inverts the ratio; C* = 42.0 adds the coordinates instead of combining them vectorially; 0.644 is the angle in radians labelled as degrees.
Further reading: Sahin and Sumnu, Physical Properties of Foods
A shelf-stable snack bar has a water activity of 0.83. Using the general minimum water activities for growth given in food microbiology texts, which group is the main spoilage risk?
A. Moulds, many growing down to about 0.80
Most spoilage bacteria need about 0.90–0.91, yeasts about 0.88, and most moulds about 0.80, so at 0.83 moulds are the chief threat. Bacteria do not grow at 0.70; xerophilic moulds and osmophilic yeasts can grow down to about 0.6, so 'no microbes below 0.90' is wrong; moulds tolerate lower a_w than yeasts, not higher. Limits shift with pH, temperature and preservatives.
Further reading: Fennema's Food Chemistry; Beuchat, minimum water activity for microbial growth
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