ICAR AICE-JRF/SRF (PhD): Dairy Chemistry, Dairy Microbiology, Dairy Technology, Dairy Engineering · India · NTA for ICAR
Original practice questions written to the published syllabus · not past papersThe calculus, differential equations and statistics that dairy engineering problems, kinetics and quality control need.
In the exams: Supports numerical items across ICAR AICE subject 38 (Dairy Engineering) and GATE XL-U Section 4; statistics recurs in ASRB NET dairy disciplines and in quality-control units
How our questions for ICAR AICE-JRF/SRF spread across the units of Mathematics & Statistics for Dairy Science. Difficulty: 9 easy, 20 medium, 24 hard.
| Analysis of variance and basic experimental designs | 11 | |
| Tests of significance: t, chi-square and F | 9 | |
| Descriptive statistics and probability distributions | 8 | |
| Introduction to response surface methodology for product optimisation | 6 | |
| Sampling theory and confidence intervals | 5 | |
| Correlation and regression | 5 | |
| Matrices and linear equations | 3 | |
| Numerical methods for engineering problems | 3 | |
| Ordinary differential equations: first-order kinetics and cooling | 2 | |
| Differential and integral calculus | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
Rate constants for whey protein denaturation measured at several temperatures give a straight line when plotted as:
A. ln k against 1/T, with slope −Eₐ/R
Arrhenius: ln k = ln A − Eₐ/(RT), linear in 1/T with slope −Eₐ/R (T absolute). Plotting against T itself is not linear over a wide range.
Further reading: Fox and McSweeney, Dairy Chemistry and Biochemistry
Concentration data from a heating trial give a straight line when 1/C is plotted against time, but curves when ln C is plotted. The reaction order is:
A. Second order
1/C − 1/C₀ = kt is the integrated second-order law. A linear ln C plot would indicate first order; a linear C plot, zero order.
Further reading: Fox and McSweeney, Dairy Chemistry and Biochemistry
Using Simpson's one-third rule on the lethal rates 0.1, 0.6, 0.9, 0.5 and 0.1 logged at 2 min intervals, the F value is:
B. 4.27 min
Simpson: (h/3)(y₀ + 4y₁ + 2y₂ + 4y₃ + y₄) = (2/3)(0.1 + 2.4 + 1.8 + 2.0 + 0.1) = (2/3)(6.4) = 4.27 min. The trapezoidal value is 4.20 min; the bare sum 2.20 forgets the h/3 weighting.
Further reading: Chapra and Canale, Numerical Methods for Engineers
Five fat tests read 3.2, 3.4, 3.6, 3.8 and 4.0%. What is the sample standard deviation (divisor n − 1)?
A. 0.316%
Deviations −0.4, −0.2, 0, 0.2, 0.4; sum of squares 0.40; variance 0.40/4 = 0.10; s = √0.10 = 0.316%. Dividing by n gives 0.283%; 0.100 is the variance; 0.632 forgets to divide the sum of squares at all.
Further reading: Snedecor and Cochran, Statistical Methods
Spores are distributed at random (Poisson) at a mean of 0.1 per pack in an aseptically filled batch. What fraction of packs contain at least one spore?
C. 0.095
P(0) = e^(−0.1) = 0.905, so P(≥1) = 1 − e^(−0.1) = 0.095. Taking the mean (0.100) as the fraction slightly overestimates because some packs carry two or more spores.
Further reading: Snedecor and Cochran, Statistical Methods
If the sample size used for a confidence interval for a mean is increased from 25 to 100, the width of the interval becomes approximately:
B. Half as wide
Width is proportional to 1/√n; √(100/25) = 2, so the interval halves. One quarter would need a sixteenfold larger sample.
Further reading: Snedecor and Cochran, Statistical Methods
Of 300 farm milk samples, 18 fail an antibiotic screening test. What is the approximate 95% confidence interval for the failure rate?
C. 3.3% to 8.7%
p̂ = 0.06; SE = √(0.06 × 0.94/300) = 0.0137; half-width 1.96 × SE = 0.027, giving 3.3% to 8.7%. Forgetting to divide by n gives an impossible negative lower limit.
Further reading: Snedecor and Cochran, Statistical Methods
Fisher's three basic principles of experimental design are:
B. Replication, randomisation and local control
Replication estimates and reduces error, randomisation prevents bias, and local control (blocking) removes known variation.
Further reading: Montgomery, Design and Analysis of Experiments
In a CRD with 5 replicates per treatment, the error mean square is 2.0 on 16 df and t(0.025, 16) = 2.120. What is the least significant difference at 5%?
C. 1.90
LSD = t × √(2 × EMS/r) = 2.120 × √(4/5) = 2.120 × 0.894 = 1.90. Omitting the factor 2 gives 1.34; using z = 1.96 ignores the error df.
Further reading: Montgomery, Design and Analysis of Experiments
Replicated centre points in a response surface experiment are used to:
D. Estimate pure error and test for curvature and lack of fit
Replicates at one setting measure pure experimental error, enabling a lack-of-fit test; the centre mean versus factorial mean reveals curvature.
Further reading: Montgomery, Design and Analysis of Experiments
Ten here, 53 in the app. The Laisha app has every practice question we have written to ICAR AICE-JRF/SRF in this subject, marked as you go, with timed papers, a jump-to-unanswered list and a certificate when you pass at 70 per cent.
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