← MES Jr. Grade
Past Questions
The only irrigation module, which is not an orifice type but is of weir type is
A)
Kennedy’s gauge outlet
B)
open flume outlet
C)
pipe outlet
D)
adjustable proportional module.
In irrigation engineering, a module is a device used to regulate the discharge of water from a main canal into a distributary. Modules are generally classified into two types based on their hydraulic operation: Orifice type and Weir type.
1. **Orifice Type Modules**: These operate on the principle of flow through an opening (orifice). The discharge depends on the head of water above the opening. Examples include:
- **Kennedy's gauge outlet**: An orifice-type outlet.
- **Pipe outlet**: A standard orifice-type device where water flows through a pipe.
- **Adjustable proportional module**: These are typically designed as orifice-type devices to maintain a constant ratio of discharge.
2. **Weir Type Modules**: These operate on the principle of flow over a crest (weir). The discharge depends on the head of water above the crest of the weir.
- **Open flume outlet**: This is a weir-type module. It consists of a flume with a calibrated weir at the end, allowing the discharge to be measured and regulated by the height of the water flowing over the weir crest rather than through a submerged opening.
Therefore, the open flume outlet is the only one among the given options that is of the weir type.
The method of growing crops on ridges, running on the side of water ditches is known as
A)
Flood irrigation.
B)
Furrow irrigation.
C)
Check irrigation.
D)
Sprinkle irrigation.
Step-by-step reasoning: 1. Analysis of the question: The question describes a specific irrigation technique where crops are grown on 'ridges' and water flows in 'ditches' (small parallel channels) alongside these ridges. 2. Evaluating Option (b) Furrow Irrigation: In furrow irrigation, small parallel channels (furrows) are dug into the soil. The crops are planted on the raised strips of land (ridges) between these furrows. Water is applied to the furrows and reaches the root zone of the plants via lateral seepage. This matches the description perfectly. 3. Evaluating Option (a) Flood Irrigation: This involves flooding the entire field with water. There are no specific ridges and ditches for the water to flow through; the whole surface is submerged. 4. Evaluating Option (c) Check Irrigation: Also known as basin irrigation, this involves dividing the field into small plots (checks) surrounded by embankments. The entire plot is filled with water, rather than using narrow ditches. 5. Evaluating Option (d) Sprinkle Irrigation: This is a pressurized system where water is sprayed into the air through nozzles, mimicking rainfall. It does not involve ridges or ditches. Therefore, furrow irrigation is the correct method.
Water is flowing through a pipe of 5 cm diameter under a pressure of 29.43 N/cm² (gauge) and with mean velocity of 2.0 m/s. Find the total head or total energy per unit weight of the water at a cross-section, which is 5 m above the datum line.
A)
30.0m
B)
25.6m
C)
20.7m
D)
35.2m
Total head at a section is given by:
H = p/γ + v²/(2g) + z
Given:
- Pressure, p = 29.43 N/cm² = 29.43 × 10⁴ N/m²
- Unit weight of water, γ = 9810 N/m³
- Velocity, v = 2.0 m/s
- Elevation, z = 5 m
Pressure head:
p/γ = (29.43 × 10⁴) / 9810 = 30 m
Velocity head:
v²/(2g) = 2²/(2 × 9.81) = 4/19.62 ≈ 0.204 m
Therefore,
H = 30 + 0.204 + 5 = 35.204 m ≈ 35.2 m
So the correct option is d.
Why others are incorrect:
- a (30.0 m): this is only the pressure head, ignoring velocity and elevation heads.
- b (25.6 m): does not match any correct combination of the three heads.
- c (20.7 m): also incorrect; far below the required total head.
- d (35.2 m): matches the full Bernoulli total head calculation.
The head of water over an orifice of diameter 40 mm is 10 m. Find the actual velocity of the jet at vena-contracta. Take Cc = 0.6 and Cv = 0.98.
A)
0.010 m/s
B)
13.72 m/s
C)
4.49 m/s
D)
15.67 m/s
To find the actual velocity of the jet at the vena-contracta, we use the concept of the Coefficient of Velocity (Cv).
1. **Identify Given Data:**
- Head of water (H) = 10 m
- Diameter of orifice (d) = 40 mm (Note: This is extra information not needed for velocity calculation)
- Coefficient of Contraction (Cc) = 0.6
- Coefficient of Velocity (Cv) = 0.98
- Acceleration due to gravity (g) ≈ 9.81 m/s²
2. **Theoretical Velocity (Vt):**
According to Torricelli's theorem, the theoretical velocity of a jet of water flowing out of an orifice is given by:
Vt = √(2gh)
Vt = √(2 * 9.81 * 10)
Vt = √196.2 ≈ 14.007 m/s
3. **Actual Velocity (Va):**
The actual velocity is the product of the coefficient of velocity and the theoretical velocity:
Va = Cv * Vt
Va = 0.98 * 14.007
Va ≈ 13.727 m/s
Comparing this result to the given options, 13.72 m/s is the correct match.
**Why other options are incorrect:**
- Option (a) 0.010 m/s is far too low.
- Option (c) 4.49 m/s does not correspond to the physics of the given head.
- Option (d) 15.67 m/s is higher than the theoretical maximum velocity (14.01 m/s), which is physically impossible for a gravity-driven flow.
Which property of the fluid accounts for the major losses in pipes?
A)
pressure
B)
force
C)
specific gravity
D)
viscosity
In fluid mechanics, losses in pipes are categorized into 'major losses' and 'minor losses'. Major losses refer to the pressure drop caused by friction between the flowing fluid and the pipe wall, as well as internal friction between fluid layers. This internal resistance to flow is caused by the fluid's viscosity. According to the Darcy-Weisbach equation (h_f = f * (L/D) * (v^2/2g)), the friction factor 'f' is a function of the Reynolds number (Re = ρvD/μ), where 'μ' represents the dynamic viscosity. Therefore, viscosity is the fundamental property responsible for the shear stress that leads to major energy losses.
Why other options are incorrect:
- a. Pressure: Pressure is a state variable that drops as a result of losses; it is not the property causing the loss.
- b. Force: Force is a result of pressure gradients or shear stress, not an intrinsic fluid property causing friction.
- c. Specific gravity: This is the ratio of the density of a fluid to a reference fluid; while density affects the Reynolds number, the primary cause of frictional resistance is viscosity.
What will be the probability of a storm occurring in any one year if the return period of the storm is 20 years?
A)
0.05
B)
1.05
C)
2.05
D)
3.05
In hydrology and probability theory, the return period (T) is defined as the average time interval between occurrences of an event of a certain magnitude. The probability (P) of that event occurring in any single year is the reciprocal of the return period.
Formula: P = 1 / T
Given:
Return Period (T) = 20 years
Calculation:
P = 1 / 20
P = 0.05
Therefore, the probability of the storm occurring in any one year is 0.05 (or 5%).
Analysis of options:
- Option (a) 0.05: Correct, as it matches the calculation 1/20.
- Option (b) 1.05: Incorrect. Probability cannot exceed 1.0.
- Option (c) 2.05: Incorrect. Probability cannot exceed 1.0.
- Option (d) 3.05: Incorrect. Probability cannot exceed 1.0.
The settling velocity of a particle in a sedimentation tank is dependent on:
A)
Depth of the tank
B)
Surface area of the tank
C)
Both the depth and surface area of the tank
D)
None of the above
The settling velocity (v_s) of a particle, according to Stokes' Law for laminar flow, is given by the formula: v_s = [g(ρ_p - ρ_w)d²] / 18μ, where g is acceleration due to gravity, ρ_p is the density of the particle, ρ_w is the density of the fluid, d is the diameter of the particle, and μ is the dynamic viscosity of the fluid.
1. Analysis of variables: The settling velocity depends solely on the physical properties of the particle (size and density) and the properties of the fluid (viscosity and density), as well as gravity.
2. Analysis of Options:
- Depth of the tank (Option a): The depth affects the *time* it takes for a particle to reach the bottom, but it does not change the *velocity* at which the particle falls.
- Surface area of the tank (Option b): The surface area (specifically the surface overflow rate) determines the *efficiency* of the tank (which particles will be removed), but it does not influence the inherent settling velocity of an individual particle.
- Option c: Since neither depth nor surface area affects the velocity, this is incorrect.
Therefore, the settling velocity is independent of the tank dimensions.
Differential manometers are used to measure
A)
atmospheric pressure
B)
difference in pressure at two points
C)
air temperature
D)
atmospheric temperature
A differential manometer is a device specifically designed to measure the difference in pressure between two distinct points in a fluid system or between two different fluids. Unlike a simple manometer (which measures pressure relative to the atmosphere), a differential manometer is connected to two separate pressure taps. The difference in the height of the manometric fluid columns is then used to calculate the pressure difference (ΔP) using the hydrostatic equation: ΔP = (ρ_m - ρ_f)gh, where ρ_m is the density of the manometric fluid and ρ_f is the density of the flowing fluid.
Analysis of other options:
- Option (a) Atmospheric pressure is measured using a barometer.
- Option (c) Air temperature is measured using a thermometer or thermocouple.
- Option (d) Atmospheric temperature is also measured using a thermometer.
If the average daily consumption of Aizawl city is 10^6 m3, the maximum daily consumption on peak hourly demand will be
A)
1.7 10^5 m3
B)
9.7 x 10^5 m3
C)
3.7 x 10^6 m3
D)
2.7 x 10^6 m3
In Environmental Engineering (Water Supply Engineering), the design of water distribution systems requires accounting for fluctuations in demand. There are two primary peaking factors used:
1. Maximum Daily Demand: Typically taken as 1.8 times the Average Daily Demand.
2. Peak Hourly Demand: Typically taken as 1.5 times the Maximum Daily Demand, or 2.7 times the Average Daily Demand (since 1.8 * 1.5 = 2.7).
Given:
Average Daily Consumption (Avg) = 10^6 m³
Calculation for Peak Hourly Demand:
Peak Hourly Demand = 2.7 × Average Daily Consumption
Peak Hourly Demand = 2.7 × 10^6 m³
Comparing this to the options:
- Option (a) 1.7 x 10^5 is too low.
- Option (b) 9.7 x 10^5 is less than the average daily consumption, which is logically incorrect for a peak demand.
- Option (c) 3.7 x 10^6 exceeds the standard peaking factor of 2.7.
- Option (d) 2.7 x 10^6 matches the standard engineering calculation for peak hourly demand.
Which of the following wastewater treatment processes is an attached growth process?
A)
Oxidation Pond
B)
Imhoff tank
C)
Rotating Biological Contactor
D)
Activated Sludge process
In wastewater treatment, biological processes are categorized based on how the biomass (microorganisms) is maintained in the system:
1. **Attached Growth Processes (Fixed-Film Processes):** In these systems, microorganisms grow as a biofilm attached to a solid medium (support material). The wastewater flows over this biofilm, and the microbes absorb organic matter. The **Rotating Biological Contactor (RBC)** consists of rotating discs that provide the surface area for this biofilm to attach and grow, making it a classic attached growth process. Other examples include Trickling Filters.
2. **Suspended Growth Processes:** In these systems, microorganisms are maintained in suspension within the liquid through mixing or aeration.
Evaluating the options:
- **a. Oxidation Pond:** Primarily a suspended growth system (though some algae/bacteria may attach to the bottom, it is categorized as a lagoon/suspended system).
- **b. Imhoff tank:** This is a primary treatment unit used for sedimentation and anaerobic digestion of sludge; it is not a biological growth process in the context of aerobic attached/suspended growth.
- **c. Rotating Biological Contactor:** Correct. It uses a fixed medium (discs) for biofilm attachment.
- **d. Activated Sludge process:** The definitive example of a suspended growth process, where biomass is kept in suspension in an aeration tank and recycled.
Disinfection of drinking water, is done to remove
A)
color
B)
pH
C)
odour
D)
bacterias
Disinfection is the process of killing or inactivating pathogenic microorganisms (such as bacteria, viruses, and protozoa) in water to make it safe for human consumption. The most common method is chlorination.
Why other options are incorrect:
- a. Color: Color is typically removed through coagulation, flocculation, sedimentation, or activated carbon filtration.
- b. pH: pH is a measure of acidity or alkalinity; it is adjusted through pH correction/stabilization using chemicals like lime or soda ash, not disinfection.
- c. Odour: Odours are generally removed through aeration, activated carbon adsorption, or oxidation, although some disinfectants (like ozone) can help, the primary purpose of 'disinfection' as a water treatment stage is biological safety.
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).
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.
Freezing is particularly beneficial for which type of agricultural produce?
A)
Leafy vegetables
B)
Root crops
C)
Berries and fruits
D)
Nuts and seeds
Freezing is a preservation method that stops microbial growth and slows down enzymatic activity by turning water into ice.
1. Why 'Berries and fruits' is correct: Many berries and fruits are highly perishable and have high water content. Quick-freezing (especially Individual Quick Freezing or IQF) preserves their nutritional value, flavor, and color better than other methods, making it the industry standard for these produce types.
2. Why 'Leafy vegetables' is generally incorrect: While some leafy greens can be frozen (usually after blanching), freezing often destroys the cellular structure of fresh leafy vegetables due to ice crystal formation, leading to a loss of turgidity and a 'mushy' texture upon thawing, making it less beneficial compared to refrigeration or drying.
3. Why 'Root crops' is incorrect: Root crops (like potatoes) have high starch content. Freezing raw root crops often leads to 'cold injury' or starch conversion to sugar, which causes them to turn brown or develop an off-flavor when cooked.
4. Why 'Nuts and seeds' is incorrect: Nuts and seeds have very low moisture content. They are best preserved through drying and controlled atmosphere storage. Freezing is not the primary or most beneficial method for these, as they are naturally stable at room temperature if kept dry.
Which material property of timber makes it suitable for constructing farm buildings?
A)
Thermal conductivity
B)
Water absorption
C)
Fire resistance
D)
Termite resistance
Step-by-step reasoning: 1. Analysis of Timber Properties: Timber is a natural insulator. It has a very low thermal conductivity compared to materials like steel or concrete. 2. Application to Farm Buildings: Farm buildings often house livestock or store crops that require stable internal temperatures to prevent heat loss in winter and excessive heat gain in summer. The low thermal conductivity of timber helps in maintaining this thermal comfort and energy efficiency. 3. Evaluation of other options: (b) Water absorption is a disadvantage of timber, as it leads to swelling and decay. (c) Fire resistance is low for timber compared to masonry or steel, making it a liability rather than a reason for suitability. (d) Termite resistance is generally low for most timber species unless chemically treated, making it a weakness rather than a suitability factor. Therefore, the insulating property (low thermal conductivity) is the primary material advantage among the choices provided for agricultural structures.
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.
Size reduction of agricultural produce is essential for -
A)
increasing bulk density
B)
eeducing shelflife
C)
reducing transportation costs
D)
enhancing flavor
Size reduction (milling, grinding, or crushing) of agricultural produce involves breaking down large particles into smaller ones. This process reduces the void spaces (interstitial spaces) between the particles. By filling these gaps, more mass can be packed into a given volume, which directly increases the bulk density of the material.
Analysis of other options:
- Option b (reducing shelflife): Size reduction usually increases the surface area, which can lead to faster oxidation or spoilage; however, the 'essential' purpose of size reduction is not to reduce shelf life, but rather to facilitate processing or storage.
- Option c (reducing transportation costs): While increasing bulk density (Option a) can lead to lower transportation costs per unit mass, the immediate physical effect of size reduction is the increase in bulk density. In many contexts, size reduction is a prerequisite for efficient packing, but 'increasing bulk density' is the fundamental technical reason.
- Option d (enhancing flavor): While some size reduction (like grinding spices) releases volatile oils that enhance flavor, this is a specific application and not the general essential reason for size reduction across all agricultural produce.
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 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 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.
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.
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.
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)
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.
The process of detaching grains from the ear heads or from the plants is known as -
A)
Threshing
B)
Weeding
C)
Blowing
D)
Harvesting
The process of separating the edible grains from the straw, husks, or ear heads (the plant part) is technically defined as threshing.
Step-by-step analysis of the options:
1. Threshing (Correct): This is the specific post-harvest operation used to detach grains from the ear heads or the plant. It can be done manually (beating) or mechanically (using a thresher).
2. Weeding (Incorrect): This is a pre-harvest cultivation practice involving the removal of unwanted plants (weeds) from the field to reduce competition for nutrients.
3. Blowing (Incorrect): While air is used in 'winnowing' to separate grain from chaff, 'blowing' is not the term for detaching grains from the plant.
4. Harvesting (Incorrect): Harvesting is the broader process of cutting or gathering the mature crop from the field. Threshing is a subsequent step that occurs after harvesting to isolate the grain from the rest of the plant.