ICAR AIEEA-PG / PGS — Code 18 Food Science Technology · India · ICAR through the National Testing Agency
Original practice questions written to the published syllabus · not past papersThe calculus, linear algebra, differential equations and numerical methods behind every food engineering calculation: heat penetration curves, drying rates, reaction kinetics and process optimisation. GATE XE makes this section compulsory for every candidate who takes the Food Technology section.
In the exams: GATE XE-A (compulsory section for XE candidates); foundation for every numerical question in GATE XE-G Section 4 and ICAR AIEEA-PG Code 18 Unit II
How our questions for ICAR AIEEA-PG / PGS spread across the units of Mathematics for Food Science & Engineering. Difficulty: 7 easy, 7 medium, 4 hard.
| Applications in food science: first-order kinetics, exponential decay and logarithmic survivor curves | 10 | |
| First-order and second-order ordinary differential equations | 2 | |
| Matrices, determinants, rank and systems of linear equations | 1 | |
| Vector calculus: gradient, divergence, curl and Green's theorem | 1 | |
| Complex numbers, analytic functions and Cauchy-Riemann equations | 1 | |
| Partial differential equations: one-dimensional heat equation and Laplace equation by separation of variables | 1 | |
| Numerical methods: Gauss elimination, LU decomposition, interpolation and Newton-Raphson | 1 | |
| Numerical integration and Euler's method for ODEs | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Fourier's law of heat conduction is written q = −k∇T. What does the negative sign express?
A. Heat flows in the direction of decreasing temperature
∇T points towards increasing temperature, and heat flows the opposite way, from hot to cold; the minus sign makes q point down the gradient. Conductivity is positive; the sign is not a measuring convention for surfaces; and it says nothing about how temperature depends on conductivity.
Further reading: Kreyszig, Advanced Engineering Mathematics; Grewal, Higher Engineering Mathematics
A sauce cools from 90 °C to 50 °C in 10 min in a room at 20 °C, following Newton's law of cooling. How long from the start does it take to reach 30 °C?
B. 23.0 min
T − 20 = 70e^(−kt). At 10 min: 30 = 70e^(−10k), so k = ln(70/30)/10 = 0.0847 per min. For T = 30: 10 = 70e^(−kt), so t = ln 7/k = 23.0 min. 20 min assumes a linear fall; 15 min and 30 min come from using temperatures rather than temperature differences above the room.
Further reading: Kreyszig, Advanced Engineering Mathematics; Grewal, Higher Engineering Mathematics
A slab of food takes 40 min to reach a target centre temperature. Under the same surface conditions and properties, how long will a slab of twice the thickness take?
A. 160 min
Time enters through the Fourier number Fo = αt/L², so equal Fo needs t ∝ L². Doubling L multiplies the time by 4: 4 × 40 = 160 min. 80 min assumes time proportional to thickness; 320 min uses L³. (With a finite Biot number the scaling is only approximate, because Bi also changes with L.)
Further reading: Kreyszig, Advanced Engineering Mathematics; Grewal, Higher Engineering Mathematics
Thermal conductivity of a product is 0.40 W/(m·K) at 20 °C and 0.50 W/(m·K) at 40 °C. By linear interpolation, what is it at 26 °C?
B. 0.43 W/(m·K)
k = 0.40 + (26 − 20)(0.50 − 0.40)/(40 − 20) = 0.40 + 6 × 0.005 = 0.43. 0.45 is the midpoint value; 0.46 uses 6/10 of the difference instead of 6/20; 0.41 uses 2/20.
Further reading: Chapra and Canale, Numerical Methods for Engineers; Kreyszig, Advanced Engineering Mathematics
A cold-point record gives 111.1 °C, 121.1 °C and 121.1 °C at 0, 1 and 2 min (z = 10 °C, reference 121.1 °C). What F₀ does the trapezoidal rule give for these 2 min?
A. 1.55 min
Lethal rates L = 10^((T − 121.1)/10): 0.1, 1 and 1. Trapezoid with h = 1: 0.5[0.1 + 2(1) + 1] = 1.55 min. 2.00 min treats every minute as fully lethal; 1.10 min adds the rates of only two points; 1.00 min uses only the last interval.
Further reading: Chapra and Canale, Numerical Methods for Engineers; Kreyszig, Advanced Engineering Mathematics
A first-order inactivation has rate constant k = 0.5 per minute. What is the D value?
A. 4.61 min
D = 2.303/k = 2.303/0.5 = 4.61 min, because a tenfold drop means ln 10 = 2.303 units of kt. 2.00 min is 1/k, the time for an e-fold (not tenfold) drop; 0.87 min divides by 2.303k instead of multiplying; 0.50 min confuses k with D.
Further reading: Singh and Heldman, Introduction to Food Engineering; Toledo, Fundamentals of Food Process Engineering
A can holds 10⁴ spores of a spoilage organism with D₁₂₁ = 1.5 min. What F₀ gives a probability of one surviving spore per 10⁵ cans?
B. 13.5 min
From 10⁴ to 10⁻⁵ per can is 9 log reductions: F = 9 × 1.5 = 13.5 min. 6.0 min is only the 4 logs to one spore per can; 7.5 min uses 5 logs; 15.0 min uses 10 logs.
Further reading: Singh and Heldman, Introduction to Food Engineering; Toledo, Fundamentals of Food Process Engineering
In thermal death kinetics, what does the z value mean?
A. The temperature change that changes D tenfold
z is the temperature interval for a one-log change in D. The time for a tenfold reduction is D; no standard parameter is defined as the temperature where D is one minute; the number of logs delivered is F/D.
Further reading: Singh and Heldman, Introduction to Food Engineering; Toledo, Fundamentals of Food Process Engineering
A population of 10⁶ cells per mL with D = 2 min is held for 10 min at the test temperature. How many cells per mL survive?
C. 10 cells per mL
Log reduction = t/D = 10/2 = 5, so N = 10⁶/10⁵ = 10 per mL. 100 corresponds to 4 logs, 1 to 6 logs, and 10⁵ to a single log.
Further reading: Singh and Heldman, Introduction to Food Engineering; Toledo, Fundamentals of Food Process Engineering
Non-enzymatic browning in a stored powder follows zero-order kinetics: the browning index rises from 0.10 to 0.30 in 20 days. When will it reach the rejection value of 0.50?
A. 40 days
Zero order: the index rises by a constant amount per day, (0.30 − 0.10)/20 = 0.01 per day. From 0.10 to 0.50 needs 0.40/0.01 = 40 days. 50 days ignores the starting value of 0.10; 20 days is only the further time after day 20; 30 days has no basis in a constant rate.
Further reading: Singh and Heldman, Introduction to Food Engineering; Toledo, Fundamentals of Food Process Engineering
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