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GATE Physical Properties, Rheology & Texture of Foods: practice questions

GATE — Engineering Sciences (XE), Section G Food Technology · India · IITs and IISc on rotation (2026: IIT Guwahati)

Original practice questions written to the published syllabus · not past papers
10worked here, free
30written to this exam in this subject
7syllabus units

What this subject covers

Measuring and modelling how foods flow, deform, break and transfer heat: rheometry of fluids and gels, viscoelasticity, texture profile analysis, and thermal, electrical and optical properties. Agricultural Engineering's engineering-properties-biomaterials covers size, shape, density and friction of grains and produce; this subject goes deeper into processed foods and their measurement.

In the exams: ICAR AIEEA-PG Code 18 Units I–II (physical properties of foods; engineering properties of food materials); ICAR AICE 37 Unit 5 (rheology of dairy foods)

What GATE asks from it

GATE — Engineering Sciences (XE), Section G Food Technology

Section 4 names flow of Newtonian fluids and heat transfer; Section 1 names starch gelatinisation and retrogradation. Rheology and physical properties appear as the inputs to flow and heat-transfer calculations (viscosity, power-law parameters, thermal properties) and in reasoning about starch and gel behaviour.

GATE — Life Sciences (XL), Section U Food Technology

Same scope as XE-G; property values and flow-behaviour reasoning are tested through processing questions.

Where the 30 questions fall

How our questions for GATE spread across the units of Physical Properties, Rheology & Texture of Foods. Difficulty: 7 easy, 16 medium, 7 hard.

Fundamentals of rheology: stress, strain and viscosity6
Newtonian and non-Newtonian fluids: power-law, Herschel-Bulkley and Casson models6
Electrical and dielectric properties of foods6
Thermal properties: specific heat, thermal conductivity and diffusivity of foods4
Rheology of doughs and dairy gels4
Rotational and capillary viscometry3
Physical properties of foods: density, porosity and particle size distribution1

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.

1 · Physical properties of foods: density, porosity and particle size distribution · medium

Porosity strongly lowers the thermal conductivity of a dry food mainly because

  1. air in the pores conducts heat better than water
  2. porosity increases the specific heat of the solids
  3. pores always fill with ice during storage
  4. air in the pores conducts heat very poorly
Answer and working

D. air in the pores conducts heat very poorly

Air (about 0.025 W/m K) is a good insulator, so porous foods conduct poorly.

Further reading: Sahin and Sumnu, Physical Properties of Foods

2 · Thermal properties: specific heat, thermal conductivity and diffusivity of foods · medium

A food contains 75 % water, 20 % protein and 5 % fat. Taking component cp values of 4.18, 1.6 and 2.0 kJ/kg K, its specific heat by mass-fraction weighting is

  1. 7.78 kJ/kg K
  2. 3.56 kJ/kg K
  3. 2.59 kJ/kg K
  4. 4.18 kJ/kg K
Answer and working

B. 3.56 kJ/kg K

cp = 0.75 × 4.18 + 0.20 × 1.6 + 0.05 × 2.0 = 3.56. 7.78 adds the component values; 2.59 is their plain average; 4.18 is water alone.

Further reading: Singh and Heldman, Introduction to Food Engineering

3 · Thermal properties: specific heat, thermal conductivity and diffusivity of foods · hard

A food has k = 0.50 W/m K, ρ = 1,050 kg/m³ and cp = 3,600 J/kg K. Its thermal diffusivity is closest to

  1. 0.146 m²/s
  2. 1.32 × 10⁻⁴ m²/s
  3. 7.56 × 10⁶ m²/s
  4. 1.32 × 10⁻⁷ m²/s
Answer and working

D. 1.32 × 10⁻⁷ m²/s

α = k/(ρcp) = 0.50/(1,050 × 3,600) = 1.32 × 10⁻⁷ m²/s. Using cp in kJ gives a value 1,000 times too large; the other two rearrange the formula wrongly.

Further reading: Singh and Heldman, Introduction to Food Engineering

4 · Fundamentals of rheology: stress, strain and viscosity · medium

Milk has dynamic viscosity 2.0 mPa s and density 1,030 kg/m³. Its kinematic viscosity is closest to

  1. 2.06 mm²/s
  2. 515 mm²/s
  3. 1.94 mm²/s
  4. 0.002 mm²/s
Answer and working

C. 1.94 mm²/s

ν = μ/ρ = 0.002/1,030 = 1.94 × 10⁻⁶ m²/s = 1.94 mm²/s (cSt). 2.06 multiplies; 515 inverts; 0.002 confuses units.

Further reading: Steffe, Rheological Methods in Food Process Engineering

5 · Newtonian and non-Newtonian fluids: power-law, Herschel-Bulkley and Casson models · medium

A purée follows τ = Kγ̇ⁿ with K = 10 Pa sⁿ and n = 0.4. Its apparent viscosity at 50 s⁻¹ is closest to

  1. 47.8 Pa s
  2. 0.96 Pa s
  3. 2.09 Pa s
  4. 0.24 Pa s
Answer and working

B. 0.96 Pa s

η = Kγ̇^(n−1) = 10 × 50^(−0.6) = 0.96 Pa s. 47.8 Pa s is the shear stress; 2.09 uses exponent −n; 0.24 is the value at 500 s⁻¹.

Further reading: Steffe, Rheological Methods in Food Process Engineering

6 · Newtonian and non-Newtonian fluids: power-law, Herschel-Bulkley and Casson models · medium

On a log–log plot of shear stress against shear rate for a power-law fluid, the slope equals

  1. the flow behaviour index n
  2. the consistency coefficient K
  3. the yield stress
  4. the apparent viscosity
Answer and working

A. the flow behaviour index n

log τ = log K + n log γ̇: slope n, intercept log K.

Further reading: Steffe, Rheological Methods in Food Process Engineering

7 · Newtonian and non-Newtonian fluids: power-law, Herschel-Bulkley and Casson models · medium

The yield stress of a sauce is best measured without wall-slip errors using

  1. a glass capillary viscometer
  2. a vane geometry rotated slowly
  3. a falling-ball viscometer
  4. a refractometer
Answer and working

B. a vane geometry rotated slowly

The vane shears within the sample, avoiding slip at smooth walls and disturbing the structure little.

Further reading: Steffe, Rheological Methods in Food Process Engineering

8 · Newtonian and non-Newtonian fluids: power-law, Herschel-Bulkley and Casson models · medium

A shear-thinning salad dressing is desirable because it

  1. is thick at rest but thins when poured
  2. becomes thicker the harder it is shaken
  3. has the same viscosity at every shear rate
  4. flows only when frozen
Answer and working

A. is thick at rest but thins when poured

Apparent viscosity falls under the high shear of pouring and spreading but is high at rest, keeping herbs suspended.

Further reading: Rao, Rheology of Fluid, Semisolid, and Solid Foods

9 · Rotational and capillary viscometry · medium

Single-point readings from a spindle ('Brookfield-type') viscometer on a non-Newtonian sauce are

  1. true viscosities valid at any shear rate for any sample
  2. comparable only under identical test conditions
  3. independent of temperature
  4. identical to cone-and-plate flow curves
Answer and working

B. comparable only under identical test conditions

The shear rate is undefined for non-Newtonian samples, so readings are empirical apparent viscosities.

Further reading: Steffe, Rheological Methods in Food Process Engineering

10 · Electrical and dielectric properties of foods · hard

Using P = 2πfε0ε″E² with f = 2.45 GHz, ε0 = 8.854 × 10⁻¹² F/m, ε″ = 15 and E = 1,000 V/m, the power absorbed per unit volume is closest to

  1. 3.25 × 10⁵ W/m³
  2. 2.04 × 10³ W/m³
  3. 2.04 × 10⁶ W/m³
  4. 1.06 × 10⁷ W/m³
Answer and working

C. 2.04 × 10⁶ W/m³

P = 2π × 2.45 × 10⁹ × 8.854 × 10⁻¹² × 15 × 10⁶ = 2.04 × 10⁶ W/m³. Omitting 2π gives 3.25 × 10⁵; using E instead of E² gives a value 1,000 times smaller; 1.06 × 10⁷ uses the dielectric constant ε′ (about 78) instead of the loss factor ε″.

Further reading: Datta and Anantheswaran, Handbook of Microwave Technology for Food Applications

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Read the subject

Physical Properties, Rheology & Texture of FoodsThe full study manual, free GATEFormat, sections and how it has changed

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Physical Properties, Rheology & Texture of Foods for other exams

ICAR AICE-JRF/SRF · IndiaICAR AIEEA-PG / PGS · IndiaGATE · India

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