← MES Jr. Grade
Past Questions
A fruit juice processing plant receives 10,000 kg of oranges with an average juice content of 50%. How many liters of juice can be extracted from these oranges?
A)
5000 L
B)
4000 L
C)
4500 L
D)
5500L
Step-by-step calculation:
1. **Identify Total Mass**: The total mass of oranges received is 10,000 kg.
2. **Calculate Mass of Juice**: The average juice content is 50%. Therefore, the mass of the extracted juice is:
10,000 kg × 0.50 = 5,000 kg.
3. **Convert Mass to Volume**: In food processing calculations for fruit juices (which are primarily water-based), the density of the juice is typically assumed to be approximately 1 kg/L (similar to water) unless a specific gravity is provided.
Volume = Mass / Density = 5,000 kg / 1 kg/L = 5,000 L.
Comparing this result to the options:
- Option (a) 5000 L matches the calculation.
- Options (b), (c), and (d) are mathematically incorrect based on the 50% yield provided in the problem statement.
Determine the cross-sectional area (in cm?) of a circular steel rod with a diameter of 25 mm.
A)
392.7 cm?
B)
490.9 cm?
C)
589.0 cm?
D)
706.9 cm?
Option B is correct: 490.9 cm?.
What does the term “ground drive” refer to in farm machinery?
A)
Driving on uneven terrain
B)
Power transmission to wheels
C)
Mechanical operation without PTO
D)
Preventing soil erosion
In farm machinery, 'ground drive' refers to a mechanism where the movement of the implement is powered by its own contact with the ground. As the machine is pulled forward by a tractor or animal, the rotation of the drive wheels (ground wheels) is used to power the internal mechanical components (such as seed meters in a planter or beaters in a harvester) via chains, gears, or belts. This eliminates the need for a Power Take-Off (PTO) shaft from the tractor to provide rotational energy.
Analysis of options:
- Option (a) is incorrect as driving on uneven terrain refers to 'terrain capability' or 'traction'.
- Option (b) is incorrect as power transmission to wheels is generally referred to as 'traction' or 'drivetrain'.
- Option (c) is correct because ground-driven implements operate independently of the tractor's engine power (PTO), relying instead on the forward motion of the machine.
- Option (d) is incorrect as preventing soil erosion refers to 'conservation tillage' or 'soil management'.
The average firing interval for 6 cylinder, 4 cycle engine is -
A)
120°
B)
180°
C)
240°
D)
360°
The firing interval of an internal combustion engine is the number of crankshaft degrees that occur between the firing of one cylinder and the next.
1. For a 4-stroke (4-cycle) engine, a complete cycle takes 2 full revolutions of the crankshaft, which equals 720° (360° x 2).
2. To find the average firing interval, the total degrees of a full cycle are divided by the number of cylinders (n).
3. Formula: Firing Interval = 720° / n
4. For a 6-cylinder engine: 720° / 6 = 120°.
Therefore, in a 6-cylinder 4-stroke engine, a cylinder fires every 120° of crankshaft rotation.
Comparing with other options:
- 180° would be the interval for a 4-cylinder 4-stroke engine (720/4).
- 240° would be the interval for a 3-cylinder 4-stroke engine (720/3).
- 360° would be the interval for a 2-cylinder 4-stroke engine (720/2).
Constant rate of drying of agriculture produce is dependent of -
A)
Air velocity
B)
Thickness of bed
C)
Airhumidity
D)
Airtemperature
In the drying process of agricultural produce, the 'Constant Rate Period' occurs when the surface of the material is completely saturated with water. During this phase, the rate of evaporation is governed by the external conditions surrounding the product rather than the internal diffusion of moisture. The rate of moisture removal is determined by the mass transfer coefficient, which is directly influenced by the air velocity (the speed at which air moves over the surface). Higher air velocity removes the saturated boundary layer of air more quickly, thereby maintaining a higher constant rate of drying. While air temperature and humidity (options c and d) affect the driving force (vapor pressure deficit), the 'rate' of transfer in the constant period is specifically dependent on the convective mass transfer coefficient, which is a function of air velocity. Thickness of the bed (option b) primarily affects the falling rate period and overall drying time, but not the specific rate of evaporation from the surface during the constant rate phase.
In attrition mill, the size of the food grain is reduced by -
A)
impact
B)
impact and shear
C)
impact and cutting
D)
shear and crushing
An attrition mill is designed to reduce the size of materials through a combination of mechanisms. 'Attrition' refers to the scrubbing or rubbing action between particles or between a particle and a surface, which creates a 'shear' force. In most industrial attrition mills (such as hammer mills or fluid energy mills), the material is first hit by a high-speed rotating element (impact) and then rubbed against other particles or the mill walls (shear/attrition). Therefore, the size reduction occurs primarily through the combined action of impact and shear.
- Option (a) Impact alone is characteristic of impact mills.
- Option (c) Cutting is characteristic of slicing or chopping mechanisms (like slicers).
- Option (d) Crushing is characteristic of compression mills (like jaw crushers or roller mills).
Which law is also called as the elasticity law?
A)
Bernoulli’s law
B)
Stress law
C)
Hooke’s law
D)
Poisson’s law
Hooke's Law is the fundamental principle of elasticity. It states that within the elastic limit of a material, the stress induced is directly proportional to the strain produced (Stress ∝ Strain). This linear relationship is expressed as σ = Eε, where E is the Modulus of Elasticity (Young's Modulus).
Why other options are incorrect:
- Bernoulli’s law: Relates to fluid dynamics (conservation of energy in a flowing fluid).
- Stress law: This is not a standard named law in materials science; stress is a physical quantity, not a law.
- Poisson’s law: While related to elasticity, Poisson's ratio describes the ratio of lateral strain to longitudinal strain, rather than the general law of elasticity itself.
What is the primary advantage of using greenhouses in agriculture?
A)
Increased pest control
B)
Extended growing season
C)
Reduced water requirement
D)
Enhanced soil fertility
Step-by-step reasoning: 1. A greenhouse is a structure designed to trap solar radiation (the greenhouse effect), which keeps the internal temperature higher than the outside ambient temperature. 2. By controlling the internal climate (temperature, humidity, and light), farmers can protect plants from frost, extreme cold, or excessive heat. 3. This environmental control allows for the cultivation of crops during seasons when they would normally not survive outdoors, effectively extending the growing season or allowing year-round production. 4. Evaluating other options: (a) While greenhouses can help isolate plants from some pests, they can also create humid environments that encourage specific pests/fungi, so it is not the primary advantage. (c) Greenhouses often require precise irrigation and can actually increase water needs due to higher evaporation rates in heated environments. (d) Soil fertility depends on nutrient management and fertilization, not the physical structure of the greenhouse itself.
What is the primary purpose of designing farm structures using the working stress method?
A)
To minimize cost
B)
To increase safety factor
C)
To optimize material use
D)
To ensure structural stability
The Working Stress Method (WSM) is a traditional approach to structural design based on the elastic theory. In WSM, the structure is designed such that the stresses in the material under service loads do not exceed a specified 'allowable stress.' This allowable stress is calculated by dividing the yield stress or ultimate strength of the material by a Factor of Safety (FoS).
1. Why 'b' is correct: The primary characteristic of WSM is its conservative nature. By limiting the actual stress to a fraction of the material's capacity, it inherently increases the safety factor compared to Limit State Design (LSD), which allows for higher stresses by applying partial safety factors to both loads and materials.
2. Why 'a' and 'c' are incorrect: WSM is generally less economical than Limit State Design because it often results in larger sections and higher material consumption to keep stresses low. Therefore, it does not minimize cost or optimize material use.
3. Why 'd' is incorrect: While structural stability is a requirement for all design methods, it is not the 'primary purpose' specific to the choice of the Working Stress Method over other methods; stability is ensured regardless of whether WSM or LSD is used.
For a good building stone, how much is the required crushing strength?
A)
Less than 50 N/mm?
B)
Greater than 100 N/mm?
C)
155 N/mm?
D)
10N/mm?
In civil engineering and construction materials, the crushing strength (compressive strength) of a stone is a critical parameter to determine its suitability for load-bearing structures. For a stone to be classified as a 'good building stone' (especially for structural purposes), it should possess high compressive strength to withstand heavy loads without failure. Standard engineering guidelines generally specify that a good building stone should have a crushing strength greater than 100 N/mm² (or 100 MPa).
Analysis of options:
- Option (a) Less than 50 N/mm²: This would be considered weak and unsuitable for primary structural loads.
- Option (b) Greater than 100 N/mm²: This is the standard benchmark for high-quality building stones used in foundations and load-bearing walls.
- Option (c) 155 N/mm²: While a stone with this strength is excellent, the requirement is a minimum threshold (greater than), not a specific fixed value.
- Option (d) 10 N/mm²: This is extremely low and characteristic of very soft stones or soil-like materials, not building stones.
A centrifugal concentrator reduces the volume of orange juice from 1000 liters to 200 liters. What is the concentration ratio achieved?
A)
2:1
B)
4:1
C)
3:1
D)
5:1
The concentration ratio (CR) in food processing is defined as the ratio of the volume of the initial feed (dilute solution) to the volume of the final concentrate.
Formula: Concentration Ratio = Initial Volume / Final Volume
Given:
Initial Volume (V1) = 1000 liters
Final Volume (V2) = 200 liters
Calculation:
CR = 1000 / 200
CR = 5
This is expressed as a ratio of 5:1, meaning the original juice has been concentrated five times.
Comparing with options:
- a. 2:1 (Incorrect: 1000/500 = 2)
- b. 4:1 (Incorrect: 1000/250 = 4)
- c. 3:1 (Incorrect: 1000/333.3 = 3)
- d. 5:1 (Correct: 1000/200 = 5)
Operation and maintenance of farm machinery for primary tillage includes:
A)
Harvesting and threshing
B)
Land preparation and seedbed preparation
C)
Irrigation and drainage
D)
Crop storage and preservation
Step-by-step reasoning: 1. Definition of Primary Tillage: Primary tillage is the first soil tillage operation after the previous harvest. Its main objectives are to open up the soil, break the hardpan, incorporate crop residues, and loosen the soil to improve aeration and water infiltration. 2. Machinery Used: Common primary tillage implements include moldboard plows, disc plows, and subsoilers. 3. Purpose: These operations are the fundamental steps of 'Land preparation' and the initial phase of 'Seedbed preparation' (which is later refined by secondary tillage). 4. Analysis of Options: (a) Harvesting and threshing occur at the end of the crop cycle, not during tillage. (c) Irrigation and drainage are water management operations, not tillage. (d) Crop storage and preservation are post-harvest activities. Therefore, only option (b) aligns with the definition and purpose of primary tillage.
The highest stress that a material can withstand for a specified length of time without excessive deformation is called -
A)
Fatigue strength
B)
Endurance strength
C)
Creep strength
D)
Creep rupture strength
The question describes a specific material property related to time-dependent deformation under a constant load, which is the definition of 'Creep'.
1. **Creep Strength**: This is defined as the maximum stress that a material can withstand for a specified period of time at a given temperature without exceeding a specified amount of deformation (excessive deformation). It is a critical parameter in high-temperature engineering applications.
2. **Why other options are incorrect**:
- **Fatigue strength (a)**: This refers to the maximum stress a material can withstand for a specified number of loading cycles without failure. It is related to cyclic loading, not a 'specified length of time' under constant stress.
- **Endurance strength (b)**: Also known as the endurance limit, this is the stress level below which a material can theoretically withstand an infinite number of cycles without failure.
- **Creep rupture strength (d)**: This is the stress that causes a material to actually break (rupture) after a specified time. The question asks for the stress without 'excessive deformation', which refers to the strength limit before deformation becomes unacceptable, rather than the point of total failure/rupture.
What is the weight of rice bran obtained from 5000 kg of paddy with a bran recovery rate of 10%?
A)
500 kg
B)
100 kg
C)
50kg
D)
5kg
To find the weight of the rice bran, we apply the recovery rate to the total weight of the paddy.
1. **Given Data**:
- Total weight of paddy = 5000 kg
- Bran recovery rate = 10%
2. **Calculation**:
Weight of rice bran = (Recovery Rate / 100) × Total Weight of Paddy
Weight of rice bran = (10 / 100) × 5000 kg
Weight of rice bran = 0.10 × 5000 kg
Weight of rice bran = 500 kg
3. **Analysis of Options**:
- Option (a) 500 kg: Matches the calculated result.
- Option (b) 100 kg: Incorrect; this would be a 2% recovery rate.
- Option (c) 50 kg: Incorrect; this would be a 1% recovery rate.
- Option (d) 5 kg: Incorrect; this would be a 0.1% recovery rate.
The moisture content of paddy (at the time of milling) should be in the range of -
A)
9 to 10%
B)
11 to 12%
C)
13 to 14%
D)
16 to 18%
In post-harvest technology for paddy, the moisture content is a critical factor for both storage and milling efficiency.
1. **Why 13-14% is correct:** For milling, the ideal moisture content of paddy is generally recommended to be between 13% and 14%. At this level, the husk separates easily from the grain, and the bran layer is removed efficiently without causing excessive breakage of the rice kernels (head rice recovery is maximized).
2. **Why other options are incorrect:**
- **9 to 12% (Options a and b):** If the moisture content is too low (below 12%), the grain becomes too brittle. During the milling process, this leads to a high percentage of broken grains, reducing the commercial value of the rice.
- **16 to 18% (Option d):** If the moisture content is too high (above 14-15%), the grains are too soft and sticky. This leads to poor separation of the husk and bran, increased risk of fungal growth/mold during storage, and higher energy consumption during milling. It also increases the risk of 'heating' in silos.
The primary source of farm power in mechanized agriculture includes:
A)
Wind and solar energy
B)
Animal and electro-mechanical sources
C)
Geothermal and tidal energy
D)
Hydroelectric power
In the context of farm power and agricultural mechanization, power sources are generally categorized into human, animal, mechanical (internal combustion engines, tractors), and electrical sources.
1. Animal power has historically been a primary source for tillage and transport in traditional and semi-mechanized farming.
2. Electro-mechanical sources include the use of electric motors for irrigation pumps, milling, and processing, as well as internal combustion engines (mechanical power) which drive the majority of modern farm machinery.
Comparing the options:
- Option (a) Wind and solar are renewable energy sources but are not the 'primary' drivers of mechanized field operations (like plowing or harvesting).
- Option (c) Geothermal and tidal energy are not practically used as primary power sources for on-farm agricultural operations.
- Option (d) Hydroelectric power is a method of generating electricity, but the 'source' used on the farm is the resulting electrical power (electro-mechanical), not the hydroelectric plant itself.
Therefore, animal and electro-mechanical sources represent the fundamental pillars of power used in mechanized agriculture.
The theory of singly reinforced beams and slabs includes:
A)
Providing additional reinforcement at the bottom
B)
Using beams without any reinforcement
C)
Exceeding the load-carrying capacity
D)
Eliminating compression
In structural engineering, a singly reinforced beam or slab is one where reinforcement is provided only in the tension zone to resist tensile stresses, as concrete is strong in compression but very weak in tension. For a simply supported beam subjected to downward loads, the bottom fibers are in tension. Therefore, the theory of singly reinforced sections involves providing longitudinal steel reinforcement at the bottom (the tension zone) to carry the tensile load.
Analysis of other options:
- Option (b) is incorrect because a beam without reinforcement (plain concrete) would fail brittlely and immediately under tension.
- Option (c) is incorrect because the goal of design theory is to ensure the load-carrying capacity is sufficient for the applied loads, not to exceed it in a way that implies failure or inefficiency.
- Option (d) is incorrect because compression is a fundamental part of the internal couple (C=T) in a beam; the concrete in the top zone is specifically designed to handle the compression.
Calculate the total load (in KN) that can be supported by a wooden beam of 4 meters span, if the safe load carrying capacity is 10 kN/m.
A)
20 kN
B)
30 kN
C)
40 kN
D)
50 kN
To calculate the total load supported by the beam, we use the relationship between the distributed load (load per unit length) and the total span of the beam.
1. **Given Data:**
- Span of the beam (L) = 4 meters
- Safe load carrying capacity (w) = 10 kN/m (This is a uniformly distributed load, UDL).
2. **Formula:**
The total load (W) on a beam subjected to a uniformly distributed load is calculated by multiplying the load per unit length by the total length of the span:
Total Load (W) = Safe load carrying capacity (w) × Span (L)
3. **Calculation:**
W = 10 kN/m × 4 m
W = 40 kN
Therefore, the total load that can be supported by the wooden beam is 40 kN.
- Option (a) 20 kN is incorrect as it would imply a span of 2m or a load of 5 kN/m.
- Option (b) 30 kN is incorrect as it would imply a span of 3m or a load of 7.5 kN/m.
- Option (d) 50 kN is incorrect as it would imply a span of 5m or a load of 12.5 kN/m.
In deep litter system, how many birds can 2.4 x 2.4 m accommodate?
A)
12
B)
14
C)
16
D)
18
Step-by-step calculation:
1. Calculate the total floor area: Area = Length × Width = 2.4 m × 2.4 m = 5.76 square meters.
2. In a standard deep litter system for poultry (specifically for layers or growers), the recommended floor space per bird typically ranges from 0.3 to 0.4 square meters depending on the breed and management system.
3. Using the standard guideline of approximately 0.36 square meters per bird (which is a common benchmark for commercial deep litter housing for adult birds):
Number of birds = Total Area / Space per bird = 5.76 / 0.36 = 16 birds.
4. Alternatively, if we check the options provided, 16 is the most mathematically sound fit for standard agricultural engineering specifications for poultry housing density in a deep litter environment for this specific area size. Options 12, 14, and 18 do not align as precisely with the standard space requirements for this specific square footage.
Calculate the theoretical stress concentration factor for a steel plate with a circular hole, where the hole diameter is 20 mm and plate thickness is 10 mm.
A)
2.0
B)
2.5
C)
3.0
D)
3.5
The theoretical stress concentration factor (Kt) is a dimensionless ratio that describes how the stress increases near a geometric discontinuity (like a hole) compared to the nominal stress in the rest of the member. For a wide plate (where the width of the plate is significantly larger than the diameter of the hole) containing a small circular hole subjected to a uniaxial tensile load, the theoretical stress concentration factor is a well-established constant derived from the theory of elasticity (Kirsch's solution).
1. **Kirsch's Solution**: For an infinite plate with a circular hole, the maximum stress occurs at the edge of the hole on the axis perpendicular to the direction of the applied load. The formula for maximum stress is σ_max = 3 * σ_nominal.
2. **Calculation of Kt**: Kt = σ_max / σ_nominal = 3.
3. **Application to this problem**: The problem provides the hole diameter (20 mm) and plate thickness (10 mm), but it does not provide the total width of the plate. In standard engineering textbook problems of this type, if the plate width is not specified, it is assumed to be a 'wide plate' or 'infinite plate' relative to the hole size. For such a configuration, the theoretical value is exactly 3.0.
Therefore, option (c) is the correct theoretical value.
Which unit operation involves the removal of undesirable foreign materials from agricultural produce?
A)
Grading
B)
Cleaning
C)
Sizereduction
D)
Packaging
In Post Harvest Technology, 'Cleaning' is the primary unit operation dedicated to the removal of foreign materials (such as stones, dust, chaff, weed seeds, and other debris) from the agricultural produce to ensure purity and quality.
Here is why the other options are incorrect:
- a. Grading: This is the process of sorting produce into different categories based on physical properties like size, weight, color, or quality, rather than removing foreign contaminants.
- c. Size reduction: This involves breaking down larger particles into smaller ones (e.g., milling or grinding) to change the physical form of the product.
- d. Packaging: This is the final stage where the processed produce is enclosed in a container to protect it from contamination and damage during transport and storage.
Which type of stress is predominant in structural members subjected to bending loads?
A)
Torsional stress
B)
Compressive stress
C)
Shear stress
D)
Bending stress
When a structural member (such as a beam) is subjected to bending loads (transverse loads), the internal resistance developed to oppose the deformation is primarily bending stress. Bending stress is a combination of normal stresses: the fibers on one side of the neutral axis are subjected to tensile stress, while fibers on the opposite side are subjected to compressive stress. While shear stress is also present (due to the transverse load), the primary design consideration for members in bending is the bending stress (flexural stress), which is typically the predominant stress that determines the failure or deflection of the member.
Why other options are incorrect:
- a. Torsional stress: This occurs when a member is twisted about its longitudinal axis, not simply bent.
- b. Compressive stress: While bending induces compression in one part of the member, 'compressive stress' as a standalone term usually refers to axial loading (columns).
- c. Shear stress: Shear stress exists in bending members, but for most slender beams, the bending stress is significantly higher and more critical than the shear stress.
Aqueduct design focuses on:
A)
Water conservation
B)
Efficient water distribution
C)
Soil erosion control
D)
Crop rotation
An aqueduct is a hydraulic structure designed to convey water from a source to a destination, typically crossing over a valley, road, or another water body. Its primary engineering purpose is the efficient transport and distribution of water over distances or obstacles.
Analysis of options:
- (a) Water conservation: While aqueducts reduce seepage compared to open earth channels, their primary design focus is transport, not conservation techniques like rainwater harvesting or recycling.
- (b) Efficient water distribution: This is the correct answer. Aqueducts are fundamental components of water distribution systems, ensuring water reaches urban centers or agricultural fields efficiently.
- (c) Soil erosion control: This is the focus of structures like check dams, terraces, or retaining walls, not aqueducts.
- (d) Crop rotation: This is an agronomic practice related to soil health and farming, entirely unrelated to hydraulic engineering structures.
Why is site selection critical for constructing farmsteads?
A)
Tomaximize land use
B)
To minimize environmental impact
C)
To optimize resource availability
D)
To ensure accessibility
In farmstead construction, site selection is the most critical initial step because it determines the long-term efficiency and sustainability of the farm. 'Optimizing resource availability' is the comprehensive goal that encompasses several key factors: 1) Water availability (proximity to wells or streams for livestock and irrigation), 2) Soil quality (ensuring the farmstead is built on non-arable or well-drained land to save fertile soil for crops), 3) Topography (utilizing natural slopes for drainage), and 4) Energy/Wind patterns. While accessibility (d) and environmental impact (b) are important, they are subsets of the broader goal of resource optimization. Maximizing land use (a) is often counter-productive if it means building on the most fertile soil. Therefore, selecting a site that optimizes the available natural and infrastructural resources is the primary driver for farmstead planning.
Difference between a pasteurizer and a sterilizer is only in -
A)
heating agent used
B)
temperature attained
C)
design of equipment
D)
% of fat used
Pasteurization and sterilization are both thermal processing techniques used to eliminate microorganisms from food (most commonly milk), but they differ fundamentally in the intensity of heat applied.
1. Pasteurization: This is a mild heat treatment. For example, High-Temperature Short-Time (HTST) pasteurization typically heats milk to about 72°C for 15 seconds. The goal is to kill pathogenic bacteria and reduce spoilage organisms without significantly altering the nutritional or sensory properties of the food. It does not kill all spores.
2. Sterilization: This is a severe heat treatment. Commercial sterilization (such as UHT - Ultra High Temperature) typically heats the product to 135-150°C for a few seconds. The goal is to achieve 'commercial sterility,' meaning all pathogenic and spoilage microorganisms, including heat-resistant spores, are destroyed.
Analysis of options:
- (a) Heating agent: Both typically use steam or hot water/indirect heat exchangers.
- (b) Temperature attained: This is the primary difference (e.g., 72°C vs 135°C+).
- (c) Design of equipment: While there are differences in pressure ratings, the basic design (heat exchanger, holding tube, cooling section) is conceptually similar for both.
- (d) % of fat: The fat content of the food is a property of the raw material, not a difference between the two processing methods.