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 papersMeasuring and assuring food quality, the food-safety systems regulators expect, and laying out a processing plant.
In the exams: ICAR AICE subject 41 Units 8 and 10; ICAR AIEEA-PG Unit VI (plant layout, feasibility reports); NCEES PE food safety
ICAR AIEEA-PG — Agricultural Engineering & Technology (Unit VI)
Definition-level MCQs and one-step calculations: layout types and SLP codes, location factors, contents of a feasibility report, break-even and payback, HACCP principles, FSSAI and Codex basics, quality attributes and their instruments.
How our questions for ICAR AIEEA-PG spread across the units of Food Quality, Safety & Process Plant Design. Difficulty: 33 easy, 8 medium, 2 hard.
| Plant location and layout | 6 | |
| Feasibility reports | 6 | |
| Equipment sizing | 6 | |
| Waste and by-product utilisation | 6 | |
| Quality attributes and their measurement | 5 | |
| Food law: FSSAI, FDA and Codex | 4 | |
| HACCP | 3 | |
| GMP and hygiene | 3 | |
| Hygienic design of equipment | 2 | |
| Process flow sheets | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
How many steps does the Codex logic sequence for applying HACCP contain, including the five preliminary steps?
A. 12
Five preliminary steps plus seven principles make twelve. Seven is the number of principles only. 5 counts only the preliminary steps, and 14 has no basis in the Codex sequence.
Further reading: Codex CXC 1-1969 (rev. 2020), General Principles of Food Hygiene
Why must a food-contact surface be cleaned before it is sanitised?
B. Soil shields microbes and inactivates many sanitisers
Organic soil protects microorganisms and reacts with sanitisers such as chlorine, so sanitising a dirty surface achieves little. The other reasons are false.
Further reading: Codex CXC 1-1969; Marriott, Principles of Food Sanitation
Which earlier Indian law was repealed and replaced by the Food Safety and Standards Act?
B. Prevention of Food Adulteration Act, 1954
The FSS Act consolidated the PFA Act 1954 and product orders such as the Fruit Products Order. The Essential Commodities Act, the grading (AGMARK) Act and the Legal Metrology Act remain separate laws.
Further reading: Food Safety and Standards Act, 2006
Which stainless steel grade is preferred over AISI 304 for equipment handling salty brines and acidic products?
D. AISI 316
316 contains molybdenum and resists chloride pitting and acids better than 304. 410 and 430 are martensitic and ferritic grades with lower corrosion resistance; painted mild steel is not acceptable on product contact.
Further reading: EHEDG Guideline Document 8
A tomato paste plant is best located near the growing area mainly because
C. Raw tomatoes are perishable and lose much weight in processing
Perishable, weight-losing raw materials pull plants towards the source. Paste is shelf-stable and lighter than the fruit, so it is cheaper to ship the product.
Further reading: Lopez-Gomez and Barbosa-Cánovas, Food Plant Design
A food plant has annual fixed costs of 1.2 million, a selling price of 60 per unit and variable cost of 36 per unit. Installed capacity is 80000 units per year. The break-even point is
C. 50000 units, 62.5 % of capacity
BEP = 1200000/(60 − 36) = 50000 units = 50000/80000 = 62.5 %. Dividing by the price alone gives 20000; dividing by the variable cost gives 33333; 37.5 % is the margin of safety, not the break-even utilisation.
Further reading: Peters, Timmerhaus and West, Plant Design and Economics for Chemical Engineers
2000 kg/h of juice (cp = 3.9 kJ/kg·K) is cooled from 30 °C to 4 °C. Ignoring other loads, the refrigeration capacity needed is
B. 16.0 TR
Q = 2000 × 3.9 × 26/3600 = 56.3 kW; 56.3/3.517 = 16.0 TR. 56.3 is the load in kW mislabelled as TR; 198.1 multiplies by 3.517 instead of dividing. 4.6 TR divides by 3.517 twice.
Further reading: Energy balance; refrigeration
A pump delivers 10 m³/h of water against a total head of 25 m at 60 % efficiency. The shaft power required is
A. 1.14 kW
P = ρgQH/η = 1000 × 9.81 × (10/3600) × 25/0.6 = 1135 W = 1.14 kW. 0.68 kW is the hydraulic power (efficiency not applied); multiplying by 0.6 instead of dividing gives 0.41 kW. 1.89 kW divides by the efficiency twice (0.681/0.6²).
Further reading: Pump fundamentals
A plant uses 4 m³ of water per tonne of product and produces 20 t/day. Storage for two days of supply interruption requires
B. 160 m³
Daily demand = 4 × 20 = 80 m³; two days = 160 m³. 80 m³ covers one day only. 8 m³ forgets the production rate (4 × 2).
Further reading: Peters, Timmerhaus and West, Plant Design and Economics for Chemical Engineers
BOD5 of a wastewater is measured by incubating the sample for
B. 5 days at 20 °C
Standard BOD5 is 5 days at 20 °C. The 2-hour, 150 °C digestion is the COD test.
Further reading: Metcalf & Eddy, Wastewater Engineering
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