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Agricultural and Biosystems Engineers Licensure Examination Engineering Properties of Agricultural Produce: practice questions

Agricultural and Biosystems Engineers Licensure Examination (ABELE) · Philippines · Professional Regulatory Board of Agricultural and Biosystems Engineering, PRC

Original practice questions written to the published syllabus · not past papers
10worked here, free
19written to this exam in this subject
9syllabus units

About Agricultural and Biosystems Engineers Licensure Examination

Agricultural and Biosystems Engineers Licensure Examination (ABELE) is a licensing and registration exam run by Professional Regulatory Board of Agricultural and Biosystems Engineering, PRC, for candidates in Philippines. Paper-based, 2 days (Day 1 6 hours; Day 2 5 hours + 4 hours). Non-programmable calculator only; the Dec 2025 programme bans the Casio fx-991ES/ES Plus by name.

Source: prc.gov.ph.

Checked against the exam body's published information. Formats and fees change; confirm the current edition there before you book.

Engineering Properties of Agricultural Produce in Agricultural and Biosystems Engineers Licensure Examination

The physical, frictional, aerodynamic, thermal, rheological and electrical properties of grains, fruits and foods that every machine and process is designed around.

Our 19 Agricultural and Biosystems Engineers Licensure Examination questions in this subject: 5 easy, 13 medium, 1 hard, across 9 units.

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.

These 10 were written to Agricultural and Biosystems Engineers Licensure Examination and are not shown on any other exam's page.

Question 1 · medium

Two seed lots are to be metered on an inclined plate. Lot X has aspect ratio W/L = 0.92 and Lot Y has W/L = 0.45. What behaviour should the designer expect?

  1. Lot Y tends to roll down the plate while Lot X tends to slide, because elongated seeds present a smaller contact area
  2. Both lots behave identically on the plate because aspect ratio affects only volume and not motion on a surface
  3. Lot X tends to roll down the plate while Lot Y tends to slide, so the two need different plate angles and cell shapes
  4. Lot X will not move at all on an inclined plate since round seeds have a higher coefficient of static friction
Answer and working

C. Lot X tends to roll down the plate while Lot Y tends to slide, so the two need different plate angles and cell shapes

An aspect ratio near 1 indicates a rounded outline that rolls; elongated seeds with low W/L slide on their flat side. Reversing the two ignores that elongated seeds rest on their long flat face. Shape does affect motion, which is why aspect ratio is measured. Roundness does not raise static friction; rolling seeds move at lower plate angles.

Question 2 · easy

A 0.5-litre test-weight measure filled by the standard procedure holds 0.392 kg of wheat. The test weight in kg/hL is

  1. 784 kg/hL
  2. 7.84 kg/hL
  3. 78.4 kg/hL
  4. 39.2 kg/hL
Answer and working

C. 78.4 kg/hL

Bulk density = 0.392 kg / 0.5 L = 0.784 kg/L; one hectolitre is 100 L, so test weight = 78.4 kg/hL (equal to 784 kg/m³). 784 is the value in kg/m³, not kg/hL; 7.84 is a factor of ten too small; 39.2 forgets that the measure holds half a litre.

Question 3 · medium

A storage manager finds that the tonnage in a tall bin exceeds bin volume × bulk density measured with a standard poured-fill test. The best explanation is that

  1. the standard poured-fill test overstates bulk density because it vibrates and packs the measuring container
  2. grain in a tall bin is compacted under its own weight, so its in-bin bulk density exceeds the poured value
  3. kernel density falls with depth in the bin, so lower layers contain more grain per cubic metre of space
  4. the weighbridge records the moisture removed during storage as extra grain mass on the way in
Answer and working

B. grain in a tall bin is compacted under its own weight, so its in-bin bulk density exceeds the poured value

Overburden pressure packs kernels more closely, so in-bin bulk density rises with depth; storage capacity estimates allow for this compaction. The poured test is done without vibration, so it gives a loose value, not an overstatement. Lower kernel density would mean less mass, not more, and moisture loss reduces mass rather than adding to it.

Question 4 · easy

A sealed silo holds 100 m³ of grain bulk with a porosity of 40%. For a controlled-atmosphere treatment, what volume of intergranular air must be displaced to fill the voids once?

  1. 40 m³
  2. 60 m³
  3. 100 m³
  4. 140 m³
Answer and working

A. 40 m³

Void volume = ε × bulk volume = 0.40 × 100 = 40 m³. 60 m³ is the volume of the kernels themselves; 100 m³ is the whole bulk; 140 m³ adds void volume to bulk volume. Real treatments also allow for headspace and leakage, which this simple fill ignores.

Question 5 · medium

In a winnower, the terminal velocities measured are 1.5–2.5 m/s for chaff and 7.5–9.5 m/s for sound wheat kernels. Which air velocity through the cleaning zone best separates them?

  1. 1.5–2.0 m/s, equal to the lower end of the chaff range
  2. 8–9 m/s, inside the terminal-velocity range of the grain
  3. 12–14 m/s, above the terminal velocity of both fractions
  4. 4–5 m/s, above the chaff range and well below the grain range
Answer and working

D. 4–5 m/s, above the chaff range and well below the grain range

Air must lift chaff (speed above its terminal velocity) while letting grain fall (speed below the grain's terminal velocity). 1.5–2.0 m/s leaves much chaff behind; 8–9 m/s blows out a large part of the grain; 12–14 m/s carries everything out.

Question 6 · easy

Terminal velocity of grain kernels is measured in the laboratory by

  1. timing kernels as they slide down a smooth inclined plane set at the angle of repose
  2. raising an upward air stream until kernels float
  3. dropping kernels through a water column and measuring the time to reach the bottom
  4. measuring the pressure drop across a deep bed of kernels at known airflow
Answer and working

B. raising an upward air stream until kernels float

When the upward air speed equals the terminal velocity, drag plus buoyancy balance weight and the kernel hovers; this is the vertical wind-tunnel method. A sliding test measures friction, falling through water measures a hydrodynamic settling velocity, and bed pressure drop measures airflow resistance.

Question 7 · medium

Freshly harvested mangoes are placed in a water tank before grading, and some float. Which property is being used, and what do the floating fruit indicate?

  1. Terminal velocity in air; the floaters have lower aerodynamic drag and are therefore the most mature fruit
  2. Density relative to water; in many fruits the less dense floaters are immature or internally defective
  3. Coefficient of friction; floaters have a waxier skin that repels water more strongly
  4. Thermal diffusivity; floaters are warmer and less dense because they were picked in the sun
Answer and working

B. Density relative to water; in many fruits the less dense floaters are immature or internally defective

Flotation separates by density: an item less dense than the liquid floats. Immature, hollow or internally damaged fruit is often less dense, though the meaning of floating differs by crop and cultivar, so each grading rule is set for its own produce. Terminal velocity in air, skin friction and field heat are not what a flotation tank measures.

Question 8 · medium

A 200 g apple is dropped 0.30 m onto a hard surface. The energy available to cause bruising, E = m g h, is

  1. 0.060 J
  2. 0.589 J
  3. 588.6 J
  4. 5.89 J
Answer and working

B. 0.589 J

E = 0.20 × 9.81 × 0.30 = 0.589 J. 0.060 J omits g; 588.6 J uses the mass in grams; 5.89 J is a factor-of-ten slip from taking 0.30 m as 3.0 m.

Question 9 · medium

Under the United States regulation for thermally processed low-acid foods in hermetically sealed containers, a 'low-acid food' is one with

  1. finished equilibrium pH less than 4.6 and water activity less than 0.85
  2. finished equilibrium pH greater than 4.6 and water activity greater than 0.85
  3. finished equilibrium pH greater than 7.0 and water activity greater than 0.95
  4. finished equilibrium pH greater than 4.6 and water activity less than 0.60
Answer and working

B. finished equilibrium pH greater than 4.6 and water activity greater than 0.85

21 CFR 113 defines low-acid foods as those with pH > 4.6 and a_w > 0.85, other than alcoholic beverages; such foods need a scheduled thermal process for Clostridium botulinum. Other countries set their own categories. pH below 4.6 describes acid or acidified foods; pH 7.0 and a_w 0.95 are not the regulatory limits; a_w below 0.60 is far drier than the threshold.

Question 10 · medium

A storage manager uses the moisture content in equilibrium with about 65–70% relative humidity as the safe storage limit. Compared with wheat, the safe moisture content for groundnut kernels by this rule is

  1. higher, because oil-rich kernels hold water much more strongly than the starch-rich kernels of cereals
  2. the same, because safe storage moisture depends on relative humidity and not on composition
  3. lower, because oil does not bind water, so the same humidity is reached at lower moisture content
  4. higher, because groundnut kernels have a lower bulk density and so dry more slowly in store
Answer and working

C. lower, because oil does not bind water, so the same humidity is reached at lower moisture content

Oil is hydrophobic, so oilseeds reach a given equilibrium relative humidity at a much lower overall moisture content than cereals; their safe storage moisture is therefore lower. National guidelines give crop-specific values. Oil does not hold water strongly, composition does matter, and bulk density does not set the equilibrium.

A study plan for Engineering Properties of Agricultural Produce in Agricultural and Biosystems Engineers Licensure Examination

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.

  1. First, the heavy units (58% of the 19 questions): Volume and density: bulk and true (5, 26%); Aerodynamic properties: terminal velocity (3, 16%); Water activity (3, 16%).

    Read the manual chapters for these first and do every worked question you can find on them.

  2. Then the middle band: Frictional properties: angle of repose and coefficient of friction (2, 11%); Porosity (2, 11%); Electrical and optical properties (1, 5%).

    Aim for one sitting per unit, then mixed questions across them.

  3. Then the lighter units: Size and shape: sphericity and roundness (1, 5%); Thermal properties: specific heat, conductivity and diffusivity (1, 5%); Viscoelastic models (1, 5%).

    Learn the definitions, classifications and standard numbers; these units are more often recalled than worked.

  4. Leave time for the hard ones: Water activity (1 hard of 3).

    Come back to these after a first pass, untimed, and read every explanation, including for the options you rejected.

  5. Bank the easy marks: Electrical and optical properties (1 easy of 1); Porosity (1 easy of 2).

    Do these early in revision so they are fast on the day.

  6. Finish timed. Sit the subject as a timed set without notes, mark it, and send each wrong answer back to its unit above.

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