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ICAR AIEEA-PG Engineering Properties of Agricultural Produce: practice questions

ICAR AIEEA-PG — Agricultural Engineering & Technology (Major Subject Group 10) · India · National Testing Agency for ICAR

Original practice questions written to the published syllabus · not past papers
10worked here, free
67written to this exam in this subject
10syllabus units

What this subject covers

The 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

What ICAR AIEEA-PG asks from it

ICAR AIEEA-PG — Agricultural Engineering & Technology

Unit VI (biological materials) asks one-step MCQs: definitions of sphericity, roundness, porosity, angle of repose and terminal velocity; the meaning of n in the power law; which model describes stress relaxation or creep; units of thermal diffusivity; and sorption-isotherm vocabulary such as hysteresis and monolayer moisture.

Where the 67 questions fall

How our questions for ICAR AIEEA-PG spread across the units of Engineering Properties of Agricultural Produce. Difficulty: 33 easy, 27 medium, 7 hard.

Size and shape: sphericity and roundness9
Water activity9
Rheology: Newtonian and non-Newtonian flow8
Volume and density: bulk and true7
Porosity7
Thermal properties: specific heat, conductivity and diffusivity7
Frictional properties: angle of repose and coefficient of friction6
Aerodynamic properties: terminal velocity6
Viscoelastic models4
Electrical and optical properties4

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Size and shape: sphericity and roundness · easy

Curray's roundness of a fruit resting in its natural position is calculated as

  1. geometric mean diameter divided by the major diameter, both measured with a vernier calliper on the fruit
  2. radius of the largest inscribed circle multiplied by the number of corners counted on the outline
  3. surface area of the fruit divided by its volume, expressed per millimetre of the fruit's length
  4. projected area of the fruit divided by the area of the smallest circle circumscribing the projection
Answer and working

D. projected area of the fruit divided by the area of the smallest circle circumscribing the projection

Curray's roundness = A_p / A_c. The geometric mean over major diameter is Mohsenin's sphericity, not roundness. Wadell's roundness divides the summed corner radii by N R, rather than multiplying. Surface-to-volume ratio is a different shape index used in heat and mass transfer.

Further reading: Mohsenin, Physical Properties of Plant and Animal Materials

2 · Aerodynamic properties: terminal velocity · easy

Pneumatic conveying of grain in dilute phase requires that the conveying air velocity be

  1. equal to half the terminal velocity of the grain, so that kernels are not damaged in bends
  2. less than the terminal velocity of the chaff, so that only clean grain is lifted
  3. independent of terminal velocity, since the blower pressure alone carries the grain
  4. greater than the terminal velocity of the grain, with a design margin to prevent settling
Answer and working

D. greater than the terminal velocity of the grain, with a design margin to prevent settling

Kernels stay in suspension only if the air moves faster than their terminal velocity; designers add a margin because bends and particle interactions cause settling. Half the terminal velocity would drop the grain; air slower than the chaff's terminal velocity lifts nothing; pressure without adequate velocity does not keep particles suspended.

Further reading: Mohsenin, Physical Properties of Plant and Animal Materials

3 · Rheology: Newtonian and non-Newtonian flow · hard

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

  1. 1.74 Pa·s
  2. 6.98 Pa·s
  3. 2.13 Pa·s
  4. 0.436 Pa·s
Answer and working

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

4 · Viscoelastic models · easy

A Kelvin-Voigt element (spring and dashpot in parallel) loaded with a constant stress shows

  1. instantaneous strain followed by unlimited steady flow at constant rate, as in a viscous liquid
  2. delayed creep towards σ0/E, with no instantaneous strain and no stress relaxation
  3. exponential stress relaxation to zero under a constant applied strain
  4. instantaneous elastic strain only, with no time-dependent deformation
Answer and working

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

5 · Viscoelastic models · easy

On the force-deformation curve of an apple compressed between plates, the bioyield point is where

  1. the skin splits visibly and the force drops sharply towards zero as the whole fruit fails in compression
  2. the curve first leaves the origin and the plates make full contact with the fruit
  3. the slope of the curve is steepest, which gives the apparent modulus of elasticity
  4. deformation increases with no increase or a decrease in force, marking the start of cell rupture
Answer and working

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

6 · Electrical and optical properties · medium

A food has a dielectric constant ε' = 52 and a dielectric loss factor ε'' = 15 at 2450 MHz. Its loss tangent is

  1. 3.467
  2. 0.224
  3. 0.780
  4. 0.288
Answer and working

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

7 · Electrical and optical properties · medium

Two batches of dried mango differ by ΔL* = −3, Δa* = 4 and Δb* = −12. The total colour difference ΔE = √(ΔL*² + Δa*² + Δb*²) is

  1. 19.0
  2. −11.0
  3. 13.0
  4. 169.0
Answer and working

C. 13.0

ΔE = √(9 + 16 + 144) = √169 = 13.0. 19.0 adds the absolute differences; −11.0 adds the signed differences; 169.0 forgets the square root.

Further reading: Sahin and Sumnu, Physical Properties of Foods

8 · Water activity · medium

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?

  1. Moulds, many growing down to about 0.80
  2. Most spoilage bacteria, which grow down to about 0.70
  3. No microbes, since none grow below a_w of 0.90
  4. Yeasts only, since moulds need a_w above 0.95
Answer and working

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

9 · Water activity · easy

Which saturated salt solution at 25 °C holds the relative humidity closest to 75% in a desiccator used for isotherm work?

  1. Lithium chloride
  2. Magnesium chloride
  3. Potassium sulfate
  4. Sodium chloride
Answer and working

D. Sodium chloride

Saturated NaCl gives about 75% RH at 25 °C. Lithium chloride gives about 11%, magnesium chloride about 33% and potassium sulfate about 97%; the values change slightly with temperature.

Further reading: Greenspan, Humidity fixed points of binary saturated aqueous solutions

10 · Water activity · easy

Moisture content on dry basis is preferred in drying calculations because

  1. dry-basis values never exceed 100%, which keeps every drying curve on a bounded and convenient scale
  2. grain trade standards worldwide quote moisture only on a dry basis
  3. wet-basis moisture cannot be measured by the air-oven method
  4. the mass of dry matter stays constant while water is removed, so changes are on a fixed base
Answer and working

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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Engineering Properties of Agricultural ProduceThe full study manual, free ICAR AIEEA-PGFormat, sections and how it has changed

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Engineering Properties of Agricultural Produce for other exams

ASRB NET · IndiaICAR AICE-JRF/SRF · IndiaGATE · IndiaNCEES PE Agricultural and Biological Engineering · USA

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