Properties of Agricultural Materials Questions and Answers

Answer: C

Correct answer: 31 degrees

The heap is a right circular cone, so the angle of repose is the base angle between the sloping surface and the horizontal:

\[ \tan\theta = \frac{h}{r} = \frac{0.90}{1.5} = 0.60 \]

\[ \theta = \arctan 0.60 = 30.96^\circ \approx 31^\circ \]

Use the base radius, not the diameter: taking 0.90/3.0 would give 17 degrees. A value near 30 degrees is typical of free-flowing dry cereal grain, and it rises as moisture and cohesion increase - which is why hopper walls are pitched well steeper than the measured repose angle.

Answer: D

Correct answer: 25.0 per cent

The 12 g lost on oven drying is the water. The dry matter is what remains:

\[ 60 - 12 = 48\ \mathrm{g} \]

On the dry basis:

\[ M_{db} = \frac{12}{48} \times 100 = 25.0\% \]

The distractor 20.0 per cent is the wet basis value, \( 12/60 \), obtained by dividing by the original mass instead of the dry matter. Whenever both figures appear among the options, check which denominator the question asks for - that is usually the whole of the difficulty.

Answer: A

The ratio is the rehydrated mass over the dried mass, \( 62/8 = 7.75 \). Over-drying damages the cell structure and pushes this figure down, so the ratio serves as a routine quality index for dried fruits and vegetables.

Answer: B

Porosity is the void fraction, so the air volume is \( 0.40 \times 2.5 = 1.0 \) m\(^3\). That air is both the path along which drying air travels and the reason a grain bed conducts heat so poorly that hot spots persist.

Answer: C

Correct answer: Both Statement I and Statement II are correct

Both statements are true, and the second is the reason for the first.

Bulk volume contains the kernels plus the air between them, while the same mass occupies only the solid volume when true density is computed. The larger denominator necessarily gives the smaller density, so \( \rho_b < \rho_t \) for any lot with non-zero porosity - which is every real one, since rigid particles cannot pack without voids. The gap between the two is exactly the porosity, typically around 40 per cent for cereals.

Answer: D

Correct answer: Both Statement I and Statement II are correct

Both are true, and again the second explains the first. Since the dry basis denominator is the dry matter alone, a sample holding more water than dry matter gives a ratio above one.

Fresh fruits and vegetables routinely do: a tomato at 95 per cent wet basis is \( 95/5 = 1900 \) per cent on the dry basis. The wet basis has no such freedom, being bounded at 100 per cent because its denominator includes the water. This is one reason the dry basis is preferred for high-moisture materials, where small wet basis differences correspond to large changes in the water actually present.

Answer: A

The reason is correct but density alone does not settle the matter; it is the combination of mass and projected area, expressed as terminal velocity, that the air stream sorts on. Two particles of equal density but different shape separate readily.

Answer: B

Correct answer: Both Statement I and Statement II are correct

Both statements are true and the second accounts for the first. Surface moisture forms liquid bridges between adjacent kernels, and the surface tension in those bridges resists relative sliding.

A heap of the more cohesive material stands at a steeper slope before it collapses, so the measured angle of repose rises - dry cereal near 30 degrees may reach the high thirties when damp. The same mechanism raises the coefficient of friction against bin walls and is the direct cause of bridging in hoppers, so the effect is felt in handling equipment long before it is measured in a test.

Answer: C

The porosity figure is right and it is exactly why the assertion fails: air is a poor conductor and the bed insulates rather than conducts. Hot spots in a store therefore persist for weeks unless aeration is used to shift them.

Answer: D

Correct answer: Statement I is incorrect but Statement II is correct

Statement II is sound: microorganisms need water that is thermodynamically available, and water activity \( a_w \) measures exactly that, being the ratio of the vapour pressure over the food to that over pure water.

Statement I contradicts it. Two products at the same moisture content can have quite different water activities, because binding to sugars, salts and the solid matrix removes water from availability. This is the whole basis of intermediate-moisture foods, which are microbiologically stable at moisture contents that would spoil an unformulated product. Most bacteria stop below \( a_w = 0.90 \) and moulds below about 0.70, thresholds expressed in activity because moisture content alone does not predict them.