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Food Processing Technology [ Question No: 1 To 20 ] :
Direction for [ Question No: 1 To 1 ] :
650 g of a wet food containing 405 g water is dried in a tray dryer to a final moisture content of 6.8 % (dry basis). It is observed that the drying process occurs under constant rate period and it takes 8 h.
The rate of drying (in kg/h) is
The temperature for thermal processing of food materials depends upon the heat sensitivity of-
From the following, select the correct pair of ionizing radiations used in food processing.
Extrusion cooking is accomplished in four different stages, which are indicated as I, II, III and IV in the figure given below. Choose the correct option representing the name of each stage.
[The figure plots Temperature/Pressure against Extrusion cooking stages: it is low and flat in stage I, rises gently in stage II, rises steeply to a peak in stage III, and drops sharply in stage IV.]
Irradiation carried out to reduce viable non-spore forming pathogenic bacteria using a dose between 3 to 10 kGy is called
Match the following food products in Column I with their corresponding processes in Column II.
| Column I | Column II | ||
|---|---|---|---|
| P | Idli | 1 | Baking |
| Q | Parboiled rice | 2 | Fermentation |
| R | Soda beverage | 3 | Gelatinization |
| S | Cookies | 4 | Carbonation |
In canning and retorting of foods, which of the following is the correct expression of Ball process time (B)?
Assume: \(t_{p}\) = processor’s process time; \(t_{c}\) = come-up time
Processing of fluid milk at 72 °C for 15 seconds is termed as
Thermal death time (TDT) of Clostridium botulinum at 121 °C is 2.78 min with a z-value of 10 °C. The TDT of the microorganism at 116 °C (in min) is
650 g of a wet food containing 405 g water is dried in a tray dryer to a final moisture content of 6.8 % (dry basis). It is observed that the drying process occurs under constant rate period and it takes 8 h. The rate of drying (in kg/h) is
Most commonly used refrigerant in direct immersion freezing of food is
One thousand units of mixed fruit bar, each weighing 100 g with a surface area of 0.01 m2, are frozen from 70 °C molten mass condition to −20 °C frozen storage condition within 3 hours. The specific heat capacity values of the bar are 3.6 kJ kg−1 K−1 and 1.97 kJ kg−1 K−1 before and after freezing point (0 °C) respectively. If the latent heat of crystallization is 250 kJ kg−1, the cooling capacity of the refrigeration unit required in tons of refrigeration is
The total solids content in a milk sample is 18 %. It is desired to produce 1000 kg of sweetened condensed milk (SCM) having 40 % sugar, 25 % moisture and rest milk solids. What is the 'Sugar Ratio' (in percentage) in the SCM in terms of sugar and water content in the final product?
100 kg tomato juice containing 5% Total Solids (w/w) is concentrated to 25% Total Solids (w/w). The total amount of water removed from tomato juice in kg is
Apple juice of 10% total solids (TS) is being concentrated in a single effect evaporator working with a surface condenser to 40% TS under a vacuum of 20 kPa. After some time the vacuum pump stops but the evaporation process continued. Choose the combination of possible implications from the following.
(P) Product quality is affected
(Q) Substantial increase in thermal energy requirement
(R) Decrease in the rate of evaporation
Identify the CORRECT sequence of operations for dressing of poultry.
Make the correct pair of food packaging technology given in Column I with operating principle or description in Column II.
| Column I | Column II |
|---|---|
| P. Aseptic packaging | 1. Control of the concentration of O2 and CO2 inside the package |
| Q. Active packaging | 2. Create a skin tight package wall |
| R. Modified atmosphere packaging | 3. Independent sterilization of food and packaging material and packaging under sterile environment |
| S. Vacuum packaging | 4. Makes non-passive contribution to product development |
Match the edible oil refining stages given in Column I with their respective functions in Column II
| Column I | Column II |
|---|---|
| P. Degumming | 1. Separation of waxes |
| Q. Neutralization | 2. Removal of pigments |
| R. Bleaching | 3. Removal of phosphatides |
| S. Winterization | 4. Removal of free fatty acids |
Match the process operation in Group I with the separated constituent in Group II
| Group I | Group II |
|---|---|
| P. Extraction | 1. Phospholipids |
| Q. Degumming | 2. Free fatty acids |
| R. Neutralization | 3. Pigments |
| S. Bleaching | 4. Crude oil |
Which one of the following statements is true for a high-pressure homogenizer?
Total number of questions : 20
Number of answered questions : 0
Number of unanswered questions : 20
Food Processing Technology [ Question No: 1 To 20 ] :
Direction for [ Question No: 1 To 1 ] :
650 g of a wet food containing 405 g water is dried in a tray dryer to a final moisture content of 6.8 % (dry basis). It is observed that the drying process occurs under constant rate period and it takes 8 h.
The rate of drying (in kg/h) is
Your Answer : (Not Answered)
Correct Answer : D
The drying rate is about 0.0485 kg/h.
The dry solids are 650 - 405 = 245 g. At 6.8% moisture on a dry basis the final product holds
\[ \begin{aligned} m_{w,final} &= 0.068 \times 245 = 16.66\ \text{g} \\ \text{water removed} &= 405 - 16.66 = 388.3\ \text{g} \\ \text{rate} &= \frac{388.3}{8} = 48.5\ \text{g/h} = 0.0485\ \text{kg/h} \end{aligned} \]
Since drying is entirely in the constant-rate period, the average rate equals the constant rate. The large options, such as 128.79 and 126.35 kg/h, are far too high for a 650 g sample dried over 8 h, and 77.81 does not follow from the mass balance either.
The temperature for thermal processing of food materials depends upon the heat sensitivity of-
Your Answer : (Not Answered)
Correct Answer : B
It depends on the heat sensitivity of Bacteria.
Thermal processes such as pasteurization and canning are designed around the heat resistance of the target bacteria and their spores, because they are the most heat-resistant organisms of concern in food. Low-acid canned foods, for example, are processed to destroy spores of Clostridium botulinum, using its D and z values to set time and temperature.
Protozoa, fungi (yeasts and moulds) and most viruses are destroyed at lower temperatures, so a process that kills the target bacteria also takes care of them.
From the following, select the correct pair of ionizing radiations used in food processing.
Your Answer : (Not Answered)
Correct Answer : B
The correct pair is Electromagnetic waves- γ rays and X- rays.
Food may be irradiated with three kinds of ionizing radiation: gamma rays from cobalt-60 or caesium-137, X-rays from machine sources, and electron beams from accelerators. Gamma rays and X-rays are both high-energy electromagnetic waves, so they belong together under that heading.
Electron beams are streams of particles, not waves, so they cannot be paired with X-rays or gamma rays as in options A and C. Cathode rays are electron beams, and beta rays are electrons from radioactive decay, so neither is an electromagnetic wave as option D says.
Extrusion cooking is accomplished in four different stages, which are indicated as I, II, III and IV in the figure given below. Choose the correct option representing the name of each stage.
[The figure plots Temperature/Pressure against Extrusion cooking stages: it is low and flat in stage I, rises gently in stage II, rises steeply to a peak in stage III, and drops sharply in stage IV.]
Your Answer : (Not Answered)
Correct Answer : C
The stages are I - Feeding, II - Kneading, III - Cooking, IV - Expansion.
In an extruder the material first enters the feed section, where it is merely conveyed and temperature and pressure stay low and flat (I). In the kneading (compression) section it is compacted and worked into a dough, so temperature and pressure begin to rise (II). In the cooking (metering) section intense shear and heat drive both to a maximum, gelatinising starch and denaturing protein (III). At the die the melt exits to atmospheric pressure, and the sudden drop lets superheated water flash to steam and expand the product (IV).
The other options place cooking before kneading or kneading before feeding, which contradicts the rising temperature and pressure profile.
Irradiation carried out to reduce viable non-spore forming pathogenic bacteria using a dose between 3 to 10 kGy is called
Your Answer : (Not Answered)
Correct Answer : D
It is called radicidation.
Food irradiation treatments are named by purpose and dose. Radicidation uses about 2.5 to 10 kGy to reduce viable non-spore-forming pathogens such as Salmonella, a radiation counterpart of pasteurisation.
Radurization uses lower doses, roughly 0.75 to 2.5 kGy, to reduce spoilage organisms and extend shelf life. Radappertization uses high doses, about 25 to 45 kGy, to sterilise food, killing spores as well. Thermoradiation is the combination of heat with irradiation, not a dose band.
Match the following food products in Column I with their corresponding processes in Column II.
| Column I | Column II | ||
|---|---|---|---|
| P | Idli | 1 | Baking |
| Q | Parboiled rice | 2 | Fermentation |
| R | Soda beverage | 3 | Gelatinization |
| S | Cookies | 4 | Carbonation |
Your Answer : (Not Answered)
Correct Answer : A
The match is P-2; Q-3; R-4; S-1.
- Idli is made from a batter of rice and black gram that is left to ferment overnight, by lactic acid bacteria and yeasts that raise it: fermentation.
- Parboiled rice is soaked, steamed and dried in the husk; the steaming gelatinizes the starch, which hardens the grain and reduces breakage: gelatinization.
- A soda beverage is defined by its dissolved carbon dioxide: carbonation.
- Cookies are formed from dough and baked.
The other options swap these, for example giving idli gelatinization or cookies carbonation, which do not describe how the products are made.
In canning and retorting of foods, which of the following is the correct expression of Ball process time (B)?
Assume: \(t_{p}\) = processor’s process time; \(t_{c}\) = come-up time
Your Answer : (Not Answered)
Correct Answer : A
Ball's process time is \(B = t_{p} + 0.42\,t_{c}\).
During the come-up time the retort is still heating toward its operating temperature, so this period contributes only part of the lethality of a full period at retort temperature. Ball's method credits 42% of the come-up time as if it were at retort temperature. The process time B used in Ball's formula is therefore the operator's process time, timed from when the retort reaches temperature, plus 0.42 of the come-up time.
The factors 0.30, 0.50 and 0.25 are not Ball's correction. The 0.42 assumes a retort temperature that rises roughly linearly during come-up.
Processing of fluid milk at 72 °C for 15 seconds is termed as
Your Answer : (Not Answered)
Correct Answer : A
Heating milk to 72 °C for 15 seconds is High-temperature, short-time (HTST) pasteurization.
HTST pasteurization, 71.7-72 °C for at least 15 s in a plate heat exchanger with immediate cooling, destroys pathogens such as Coxiella burnetii and Mycobacterium tuberculosis while keeping flavour change small.
LTLT (batch) pasteurization uses about 63 °C for 30 minutes. UHT treatment uses about 135-150 °C for a few seconds and gives a shelf-stable product. Homogenization reduces fat globule size by pressure and is not a heat treatment.
Thermal death time (TDT) of Clostridium botulinum at 121 °C is 2.78 min with a z-value of 10 °C. The TDT of the microorganism at 116 °C (in min) is
Your Answer : (Not Answered)
Correct Answer : B
The TDT at 116°C is about 8.791 min.
The z-value is the temperature change that alters the thermal death time tenfold. A drop of 5°C is half of z, so the time rises by a factor of \(10^{0.5}\):
\[ \begin{aligned} TDT_{116} &= TDT_{121} \times 10^{(121 - 116)/z} \\ &= 2.78 \times 10^{0.5} = 2.78 \times 3.162 = 8.791\ \text{min} \end{aligned} \]
A lower temperature needs a longer time, so the options smaller than 2.78 min (1.390 and 0.712) have the direction wrong. 5.270 does not follow from the 100.5 factor.
650 g of a wet food containing 405 g water is dried in a tray dryer to a final moisture content of 6.8 % (dry basis). It is observed that the drying process occurs under constant rate period and it takes 8 h. The rate of drying (in kg/h) is
Your Answer : (Not Answered)
Correct Answer : D
The rate of drying is 0.0485 kg/h.
Dry solids = 650 - 405 = 245 g. A final moisture of 6.8% on a dry basis means water left = 0.068 x 245 g:
\[ \begin{aligned} \text{water left} &= 0.068 \times 245 = 16.66\ \text{g} \\ \text{removed} &= 405 - 16.66 = 388.34\ \text{g} \\ \text{rate} &= \frac{0.38834\ \text{kg}}{8\ \text{h}} = 0.0485\ \text{kg/h} \end{aligned} \]
A sample of 650 g cannot lose anywhere near 78 to 129 kg in an hour, so the large options cannot be right whatever the working. Because the whole drying is in the constant-rate period, the average rate equals the constant rate.
Most commonly used refrigerant in direct immersion freezing of food is
Your Answer : (Not Answered)
Correct Answer : C
The refrigerant most commonly used is liquid nitrogen.
In direct immersion (cryogenic) freezing the food contacts the refrigerant itself. Liquid nitrogen boils at about −196 °C, freezes the product very quickly with small ice crystals, and is inert, tasteless and leaves no residue on the food, which is why it is the usual immersion medium.
Monochlorodifluoromethane and dichlorodifluoromethane are halogenated refrigerants (freons); food-grade dichlorodifluoromethane was once used for immersion freezing, but these compounds deplete ozone and have been phased out, and 'Freon' is only their trade name.
One thousand units of mixed fruit bar, each weighing 100 g with a surface area of 0.01 m2, are frozen from 70 °C molten mass condition to −20 °C frozen storage condition within 3 hours. The specific heat capacity values of the bar are 3.6 kJ kg−1 K−1 and 1.97 kJ kg−1 K−1 before and after freezing point (0 °C) respectively. If the latent heat of crystallization is 250 kJ kg−1, the cooling capacity of the refrigeration unit required in tons of refrigeration is
Your Answer : (Not Answered)
Correct Answer : B
The refrigeration capacity is about 1.43 tons.
The batch is 1000 × 0.1 = 100 kg. It is cooled from 70 °C to the freezing point, frozen, then cooled to −20 °C:
\[ \begin{aligned} Q &= 100[3.6(70) + 250 + 1.97(20)] \\ &= 100(252 + 250 + 39.4) = 54140\ \text{kJ} \\ \dot{Q} &= 54140/10800 = 5.01\ \text{kW} \\ \text{TR} &= 5.01/3.517 = 1.43 \end{aligned} \]
One ton of refrigeration is 3.517 kW. The surface area is not needed for the load. Leaving out the latent heat of crystallization gives about 0.77 TR, which is the trap in the first option; the latent heat is nearly half the total load.
The total solids content in a milk sample is 18 %. It is desired to produce 1000 kg of sweetened condensed milk (SCM) having 40 % sugar, 25 % moisture and rest milk solids. What is the 'Sugar Ratio' (in percentage) in the SCM in terms of sugar and water content in the final product?
Your Answer : (Not Answered)
Correct Answer : B
The sugar ratio is 61.54%.
The sugar ratio of sweetened condensed milk is the sugar concentration in its water phase, which is what makes the product keep:
\[ \begin{aligned} \text{sugar ratio} &= \frac{\text{sugar}}{\text{sugar} + \text{water}} \times 100 \\ &= \frac{40}{40 + 25} \times 100 = 61.54\% \end{aligned} \]
The 18% total solids of the raw milk and the 1000 kg batch size are not needed for this ratio; they matter only for working out how much milk and sugar to use. The other options do not follow from the 40% sugar and 25% water.
100 kg tomato juice containing 5% Total Solids (w/w) is concentrated to 25% Total Solids (w/w). The total amount of water removed from tomato juice in kg is
Your Answer : (Not Answered)
Correct Answer : D
The water removed is 80 kg.
The solids are not evaporated, so they balance across the concentrator:
\[ \begin{aligned} \text{solids} &= 0.05 \times 100 = 5\ \text{kg} \\ \text{concentrate} &= \frac{5}{0.25} = 20\ \text{kg} \\ \text{water removed} &= 100 - 20 = 80\ \text{kg} \end{aligned} \]
The concentrate weighs only 20 kg, so 80 kg of the original 100 kg must have gone as water. Removing 65, 70 or 75 kg would leave 35, 30 or 25 kg holding 5 kg of solids, which is 14%, 17% or 20% solids, not the required 25%.
Apple juice of 10% total solids (TS) is being concentrated in a single effect evaporator working with a surface condenser to 40% TS under a vacuum of 20 kPa. After some time the vacuum pump stops but the evaporation process continued. Choose the combination of possible implications from the following.
(P) Product quality is affected
(Q) Substantial increase in thermal energy requirement
(R) Decrease in the rate of evaporation
Your Answer : (Not Answered)
Correct Answer : C
The implications are R & P: a decrease in the rate of evaporation and an effect on product quality.
The vacuum pump keeps the evaporator at low pressure, so the juice boils at a low temperature. When it stops, non-condensable gases accumulate and the pressure rises, and the boiling point rises with it. The temperature difference between the heating steam and the boiling juice shrinks, so the heat transfer rate and the rate of evaporation fall (R). The juice is also held at a higher temperature, which degrades heat-sensitive flavour, colour and nutrients, for example through browning (P).
The heat needed per kilogram of water evaporated is essentially the latent heat, which changes little with the boiling temperature (it is slightly smaller at higher temperature), so there is no substantial increase in thermal energy requirement (Q).
Identify the CORRECT sequence of operations for dressing of poultry.
Your Answer : (Not Answered)
Correct Answer : A
The correct sequence is slaughtering and bleeding → scalding → defeathering → eviscerating → chilling.
After the bird is killed and bled, it is dipped in hot water (scalding) to loosen the feather follicles. Only then can the feathers be removed cleanly by the rubber-fingered pickers (defeathering). The carcass is then opened and the viscera removed (eviscerating), and finally it is chilled quickly to limit bacterial growth.
Defeathering before scalding tears the skin because the feathers are still held firmly. Eviscerating before defeathering or scalding exposes the body cavity to dirt and scald water, and scalding after evisceration would contaminate the opened carcass.
Make the correct pair of food packaging technology given in Column I with operating principle or description in Column II.
| Column I | Column II |
|---|---|
| P. Aseptic packaging | 1. Control of the concentration of O2 and CO2 inside the package |
| Q. Active packaging | 2. Create a skin tight package wall |
| R. Modified atmosphere packaging | 3. Independent sterilization of food and packaging material and packaging under sterile environment |
| S. Vacuum packaging | 4. Makes non-passive contribution to product development |
Your Answer : (Not Answered)
Correct Answer : A
The match is P-3, Q-4, R-1, S-2.
- Aseptic packaging sterilizes the food and the packaging separately and fills them in a sterile environment.
- Active packaging contains components such as oxygen scavengers or antimicrobials that take a non-passive part in keeping the product.
- Modified atmosphere packaging sets the concentrations of O2 and CO2 inside the pack.
- Vacuum packaging removes air so the film pulls tight around the product, forming a skin-tight wall.
The other options confuse vacuum packaging with active packaging, or modified atmosphere with aseptic packaging.
Match the edible oil refining stages given in Column I with their respective functions in Column II
| Column I | Column II |
|---|---|
| P. Degumming | 1. Separation of waxes |
| Q. Neutralization | 2. Removal of pigments |
| R. Bleaching | 3. Removal of phosphatides |
| S. Winterization | 4. Removal of free fatty acids |
Your Answer : (Not Answered)
Correct Answer : C
The match is P-3, Q-4, R-2, S-1.
- Degumming removes phosphatides (gums) by hydration with water or acid.
- Neutralization (alkali refining) converts free fatty acids into soaps that are separated.
- Bleaching with adsorbent earths removes pigments such as chlorophyll and carotenoids.
- Winterization chills the oil to crystallize and filter off waxes and high-melting glycerides.
The other options misplace neutralization or bleaching, for example making neutralization remove pigments.
Match the process operation in Group I with the separated constituent in Group II
| Group I | Group II |
|---|---|
| P. Extraction | 1. Phospholipids |
| Q. Degumming | 2. Free fatty acids |
| R. Neutralization | 3. Pigments |
| S. Bleaching | 4. Crude oil |
Your Answer : (Not Answered)
Correct Answer : C
The match is P-4, Q-1, R-2, S-3.
These are the steps of edible oil processing, in order. Extraction (pressing or solvent extraction) separates the crude oil from the seed. Degumming with water or acid hydrates and removes the phospholipids (gums). Neutralization with alkali converts free fatty acids into soap, which is separated. Bleaching with adsorbent earth removes pigments such as chlorophyll and carotenoids.
The other options swap two of these, for example giving bleaching the free fatty acids or degumming the pigments.
Which one of the following statements is true for a high-pressure homogenizer?
Your Answer : (Not Answered)
Correct Answer : B
The true statement is Solid particles in the suspension are disintegrated due to the high shear rate exerted by the liquid.
A high-pressure homogenizer forces the liquid at pressures of tens to hundreds of MPa through a narrow valve gap. The fluid accelerates to very high velocity, and the intense shear, turbulence and cavitation in and just after the gap break fat globules and particles into much smaller ones. This is how milk is homogenized.
It is not mere mixing, it injects no steam, and it does not compress solids into a cake; that last describes filtration or pressing.
Marks
Total number of questions : 20
Number of answered questions : 0
Number of unanswered questions : 20