GATE — Agricultural Engineering (AG) · India · IITs and IISc on rotation (2026: IIT Guwahati)
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
GATE — Agricultural Engineering (AG)
Engineering properties is the first line of Section 6 (Agricultural Process Engineering). Expect numerical-answer questions on sphericity and geometric mean diameter, bulk density and porosity, moisture-basis conversion and water removed in drying, and power-law or Bingham flow (apparent viscosity, consistency index from two shear points). Rheology has appeared in about half of the recent papers; water activity and Raoult's law also recur. State the sphericity formula used and carry units through.
How our questions for GATE spread across the units of Engineering Properties of Agricultural Produce. Difficulty: 6 easy, 22 medium, 21 hard.
| Rheology: Newtonian and non-Newtonian flow | 8 | |
| Water activity | 8 | |
| Thermal properties: specific heat, conductivity and diffusivity | 6 | |
| Size and shape: sphericity and roundness | 5 | |
| Aerodynamic properties: terminal velocity | 5 | |
| Viscoelastic models | 5 | |
| Porosity | 4 | |
| Frictional properties: angle of repose and coefficient of friction | 4 | |
| Electrical and optical properties | 3 | |
| Volume and density: bulk and true | 1 |
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
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 tomato concentrate follows the power law τ = K γ̇ⁿ with K = 8.5 Pa·sⁿ and n = 0.35. Its apparent viscosity at a shear rate of 100 s⁻¹ is
C. 0.426 Pa·s
η_a = K γ̇^(n−1) = 8.5 × 100^(−0.65) = 0.426 Pa·s. 42.6 is the shear stress τ in Pa, not a viscosity; 1.696 uses exponent −n instead of n − 1; 0.085 divides K by γ̇, valid only for n = 0.
Further reading: Steffe, Rheological Methods in Food Process Engineering
A steel ball of radius 1.0 mm (density 7800 kg/m³) falls at a steady 8.0 mm/s through honey (density 1400 kg/m³). Using Stokes' law, μ = 2r²g(ρ_s − ρ_f)/(9v), the viscosity of the honey is
A. 1.74 Pa·s
μ = 2 × (10⁻³)² × 9.81 × 6400 / (9 × 0.008) = 0.12557/0.072 = 1.74 Pa·s. 6.98 uses the diameter in r²; 2.13 ignores the buoyancy of the honey; 0.436 treats 1.0 mm as the diameter and halves it again.
Further reading: Steffe, Rheological Methods in Food Process Engineering
A juice concentrate has a viscosity of 12.0 mPa·s at 20 °C and an Arrhenius flow activation energy of 25 kJ/mol, η = A exp(E_a/RT). Its viscosity at 60 °C is
B. 3.50 mPa·s
η2 = 12.0 × exp[(25000/8.314)(1/333.15 − 1/293.15)] = 12.0 × exp(−1.2316) = 3.50 mPa·s. Using Celsius temperatures sends the exponent to an absurd value and gives effectively zero (0.00); reversing the sign gives a rise to 41.1 mPa·s, the wrong direction; scaling by 20/60 assumes inverse proportionality to °C, which has no basis. That scaling gives 12.0 × 20/60 = 4.00 mPa·s.
Further reading: Steffe, Rheological Methods in Food Process Engineering
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 food has a dielectric constant ε' = 52 and a dielectric loss factor ε'' = 15 at 2450 MHz. Its loss tangent is
D. 0.288
tan δ = ε''/ε' = 15/52 = 0.288. 3.467 inverts the ratio; 0.224 divides by the sum; 0.780 multiplies the two, which has no meaning.
Further reading: Rao, Rizvi, Datta and Ahmed (eds.), Engineering Properties of Foods
At a fixed relative humidity of the drying air, raising the air temperature changes the equilibrium moisture content of grain in which direction?
C. It lowers the EMC, so warmer air at the same RH dries grain to a lower final moisture
Isotherms shift downward as temperature rises, so EMC at a given RH falls; ASABE equilibrium-moisture equations include temperature for this reason. The reverse is the common slip; RH alone does not fix EMC; and grain does not exceed 100% dry basis at high temperature.
Further reading: ASABE D245; Brooker, Bakker-Arkema and Hall
Moisture content on dry basis is preferred in drying calculations because
D. the mass of dry matter stays constant while water is removed, so changes are on a fixed base
Dry matter is conserved during drying, so dry-basis moisture changes linearly with water lost. Dry-basis values can exceed 100% (wet basis cannot); trade normally uses wet basis; the oven method gives both bases from the same weighings.
Further reading: Brooker, Bakker-Arkema and Hall, Drying and Storage of Grains and Oilseeds
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