← MES Jr. Grade
Past Questions — 2025
Which of the following is not a primary air pollutant?
A)
Sulphur dioxide
B)
Nitrogen dioxide
C)
Ozone
D)
Carbon monoxide
Air pollutants are categorized into two main types: Primary and Secondary pollutants.
1. Primary Pollutants: These are emitted directly from a source into the atmosphere. Examples include:
- Sulphur dioxide (SO2): Emitted from volcanic eruptions and burning of fossil fuels (coal/oil).
- Nitrogen dioxide (NO2) and Nitric oxide (NO): Emitted from vehicle exhausts and industrial combustion.
- Carbon monoxide (CO): Produced by the incomplete combustion of carbon-based fuels.
2. Secondary Pollutants: These are not emitted directly but form in the atmosphere through chemical reactions between primary pollutants and other atmospheric components (like water vapor or sunlight).
- Ozone (O3): Ground-level ozone is a classic secondary pollutant. It is formed when Nitrogen Oxides (NOx) and Volatile Organic Compounds (VOCs) react in the presence of sunlight (photochemical reaction).
Therefore, while Sulphur dioxide, Nitrogen dioxide, and Carbon monoxide are primary pollutants, Ozone is a secondary pollutant.
Reynolds number is a ratio of:
A)
Inertial forces to viscous forces
B)
Pressure forces to surface tension forces
C)
Gravity forces to viscous forces
D)
Inertial forces to gravity forces
The Reynolds number (Re) is a dimensionless quantity used in fluid mechanics to predict flow patterns. It is defined as the ratio of inertial forces to viscous forces.
Mathematically, it is expressed as: Re = (ρvL) / μ, where:
- ρ is the density of the fluid (representing inertia),
- v is the flow velocity,
- L is the characteristic linear dimension (e.g., pipe diameter),
- μ is the dynamic viscosity of the fluid (representing the resistance to flow).
Analysis of options:
- Option (a) is correct: Re = Inertial Forces / Viscous Forces.
- Option (b) describes the Weber number (ratio of inertial forces to surface tension forces).
- Option (c) is not a standard named dimensionless number in this context, though gravity vs viscosity relates to certain stability problems.
- Option (d) describes the Froude number (ratio of inertial forces to gravitational forces).
A tank contains oil of specific gravity 0.8. Pressure at 4 m depth is:
A)
31.4 kN/m²
B)
39.2 kN/m²
C)
45.0 kN/m²
D)
50.0 kN/m²
To find the pressure at a certain depth in a fluid, we use the hydrostatic pressure formula: P = ρgh or P = (S × ρ_water) × g × h.
1. **Given Data:**
- Specific gravity of oil (S) = 0.8
- Depth (h) = 4 m
- Density of water (ρ_water) ≈ 1000 kg/m³
- Acceleration due to gravity (g) ≈ 9.81 m/s²
2. **Calculate Density of Oil (ρ_oil):**
ρ_oil = Specific Gravity × Density of Water
ρ_oil = 0.8 × 1000 kg/m³ = 800 kg/m³
3. **Calculate Pressure (P):**
P = ρ_oil × g × h
P = 800 kg/m³ × 9.81 m/s² × 4 m
P = 31,392 N/m²
4. **Convert to kN/m²:**
P = 31,392 / 1000 = 31.392 kN/m²
Rounding to one decimal place, we get 31.4 kN/m².
Comparing this result to the options:
- Option (a) 31.4 kN/m² matches our calculation.
- Options (b), (c), and (d) are incorrect as they do not align with the physical properties of oil with a specific gravity of 0.8 at the given depth.
Which of the following is a common method for treating wastewater to remove organic matter?
A)
Reverse osmosis
B)
Activated sludge process
C)
Ion exchange
D)
Ultrafiltration
The goal of removing organic matter from wastewater is typically achieved through biological treatment.
1. **Activated Sludge Process (Option b):** This is a widely used secondary treatment method. It employs a biological process where a mass of microorganisms (activated sludge) is mixed with wastewater in an aeration tank. These microbes consume and break down the biodegradable organic matter (measured as BOD - Biochemical Oxygen Demand), converting it into carbon dioxide, water, and new biomass. This is the standard method for organic matter removal.
2. **Reverse Osmosis (Option a):** This is a membrane filtration process used primarily for desalination or removing dissolved salts and very small molecules. While it can remove organic molecules, it is an advanced tertiary treatment and is not the 'common' method for bulk organic matter removal due to high energy costs and membrane fouling.
3. **Ion Exchange (Option c):** This process is used to remove specific dissolved ions (like hardness ions Ca2+ and Mg2+ or nitrates) by exchanging them with other ions. It is not used for the general removal of organic matter.
4. **Ultrafiltration (Option d):** This is a pressure-driven membrane process that removes suspended solids, bacteria, and high-molecular-weight organics. However, it cannot remove dissolved organic matter as effectively as biological processes and is generally used for polishing or pretreatment rather than primary organic removal.
Which of the following properties distinguishes a fluid from a solid?
A)
Ability to resist shear stress
B)
Fixed shape
C)
Ability to flow under shear stress
D)
High compressibility
The fundamental distinction between a solid and a fluid (liquids and gases) lies in their response to shear stress.
1. **Solid Behavior**: When a shear stress is applied to a solid, it undergoes a finite deformation (strain). If the stress is removed (within the elastic limit), it returns to its original shape. A solid can resist shear stress by maintaining a static equilibrium of deformation.
2. **Fluid Behavior**: A fluid is defined as a substance that deforms continuously (flows) when subjected to a shear stress, no matter how small that stress may be. In other words, a fluid cannot resist shear stress while remaining at rest; it will continue to deform as long as the stress is applied.
**Analysis of Options:**
- **Option a**: Incorrect. Solids resist shear stress by deforming to a fixed point; fluids cannot resist it statically.
- **Option b**: Incorrect. A fixed shape is a characteristic of a solid, not a fluid.
- **Option c**: Correct. This is the defining physical property of a fluid.
- **Option d**: Incorrect. While gases (fluids) are highly compressible, liquids (also fluids) are generally considered incompressible. Therefore, compressibility is not the universal property that distinguishes all fluids from solids.
The drainage coefficient is expressed in:
A)
T/day
B)
m3/day
C)
cm/day
D)
cm/km
The drainage coefficient is defined as the depth of water to be removed from a given area per unit of time to prevent crop damage or to maintain the water table at a desired depth. Mathematically, it is the volume of water removed per unit area per unit time. Since Volume/Area = Depth, the unit of the drainage coefficient is a unit of length (depth) per unit of time. In irrigation and drainage engineering, this is most commonly expressed as centimeters per day (cm/day) or millimeters per day (mm/day).
Analysis of options:
- Option (a) T/day: 'T' usually refers to tonnes, which is a unit of mass, not depth.
- Option (b) m3/day: This is a unit of discharge (volume per time), not a coefficient based on area.
- Option (c) cm/day: Correct. It represents the depth of water removed per day.
- Option (d) cm/km: This is a unit of slope or gradient, not a rate of water removal.
The direct runoff hydrograph of a storm obtained from a catchment is triangular in shape and has a base period of 80 hours. The peak flow rate is 30 m3/sec and catchment area is 86.4 km2. The rainfall excess that has resulted in the hydrograph is:
A)
0.5 cm
B)
5 cm
C)
15 cm
D)
25 cm
Step-by-step calculation:
1. **Identify Given Data**:
- Shape of Direct Runoff Hydrograph (DRH): Triangular
- Base period (T) = 80 hours
- Peak flow rate (Qp) = 30 m³/sec
- Catchment area (A) = 86.4 km² = 86.4 × 10⁶ m²
2. **Calculate Total Volume of Runoff (V)**:
For a triangular hydrograph, the volume is the area under the triangle:
Volume (V) = 1/2 × Base × Height
V = 1/2 × (80 hours × 3600 sec/hour) × 30 m³/sec
V = 0.5 × 288,000 sec × 30 m³/sec
V = 4,320,000 m³
3. **Calculate Rainfall Excess (Depth of Runoff)**:
The rainfall excess (d) is the total volume of runoff divided by the catchment area:
d = Volume / Area
d = 4,320,000 m³ / 86,400,000 m²
d = 0.05 meters
4. **Convert to Centimeters**:
d = 0.05 m × 100 cm/m = 5 cm
Conclusion: The rainfall excess that resulted in the hydrograph is 5 cm. Therefore, option (b) is correct.
What is the SI unit of dynamic viscosity?
A)
kg/m·s
B)
N·s/m²
C)
m²/s
D)
kg/m³
Dynamic viscosity (μ) is defined by Newton's Law of Viscosity as the ratio of shear stress (τ) to the velocity gradient (du/dy).
1. Formula: μ = τ / (du/dy)
2. Units of Shear Stress (τ): Force per unit area = N/m² = (kg·m/s²) / m² = kg/(m·s²)
3. Units of Velocity Gradient (du/dy): (m/s) / m = 1/s
4. Calculating the unit of μ: [kg/(m·s²)] / [1/s] = kg/(m·s).
Analysis of options:
- Option (a) kg/m·s: This is the derived SI unit for dynamic viscosity. It is also equivalent to Pascal-seconds (Pa·s).
- Option (b) N·s/m²: While N·s/m² is mathematically equivalent to Pa·s (since 1 Pa = 1 N/m²), the standard base SI unit representation is kg/(m·s). However, in many contexts, Pa·s is used. In this specific multiple-choice set, kg/m·s is the fundamental base unit representation.
- Option (c) m²/s: This is the SI unit for Kinematic Viscosity (ν = μ/ρ).
- Option (d) kg/m³: This is the SI unit for Density (ρ).
Turbidity of water is due to:
A)
Algae
B)
Fungi
C)
Organic salts
D)
Finely divided particles of clay, silt etc.
Turbidity is an optical property of water that describes the cloudiness or haziness of a fluid caused by individual particles (suspended solids) that are generally invisible to the naked eye.
1. **Why Option (d) is correct:** Turbidity is primarily caused by suspended inorganic matter such as finely divided particles of clay, silt, and other minerals. These particles scatter light, preventing it from passing through the water in a straight line, which results in the appearance of turbidity.
2. **Why other options are incorrect:**
- **Algae (a):** While algae can contribute to turbidity (biological turbidity), the primary and most common cause of turbidity in general water quality contexts is inorganic suspended solids. Algae are typically categorized under biological contaminants or specific 'algal blooms'.
- **Fungi (b):** Fungi are not a primary cause of water turbidity in standard environmental water sources.
- **Organic salts (c):** Dissolved salts (whether organic or inorganic) do not cause turbidity because they are dissolved at the molecular or ionic level and do not scatter light. Only suspended particles cause turbidity.
Bernoulli’s principle in hydraulics relates to:
A)
Conservation of mass
B)
Conservation of pressure
C)
Conservation of momentum
D)
Conservation of energy
Bernoulli's principle states that for an inviscid, incompressible fluid in steady flow, the sum of pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline. Mathematically, it is expressed as: P + ½ρv² + ρgh = constant. This is a direct application of the Law of Conservation of Energy.
Why other options are incorrect:
- Option (a) Conservation of mass is the basis for the 'Continuity Equation' (A1V1 = A2V2).
- Option (b) 'Conservation of pressure' is not a recognized fundamental law of physics; pressure changes as energy transforms between kinetic and potential forms.
- Option (c) Conservation of momentum is the basis for the 'Momentum Equation' or 'Euler's Equation' used to calculate forces exerted by flowing fluids.
What is the main advantage of drip irrigation over other methods?
A)
Low initial cost
B)
High water application efficiency
C)
Suitable for all types of crops
D)
Requires no maintenance
Drip irrigation, also known as trickle irrigation, delivers water slowly and precisely to the root zone of plants through a network of valves, pipes, and emitters.
1. Why Option (b) is correct: The primary advantage of drip irrigation is its extremely high water application efficiency (often exceeding 90%). It minimizes water loss due to evaporation, surface runoff, and deep percolation, making it the most efficient method for water conservation.
2. Why Option (a) is incorrect: Drip irrigation typically has a higher initial installation cost compared to surface or sprinkler irrigation due to the requirement for pumps, filters, and extensive piping networks.
3. Why Option (c) is incorrect: While versatile, it is not suitable for all crops. For example, it is highly effective for row crops and orchards but impractical for close-growing crops like wheat, rice, or certain turf grasses.
4. Why Option (d) is incorrect: Drip systems require significant maintenance, particularly the cleaning of filters and the prevention of emitter clogging caused by mineral deposits or algae.
The permissible limit of arsenic in drinking water (WHO standard) is:
A)
0.01 ml/g
B)
0.001 mg/L
C)
0.0001 ml/L
D)
0.01 mg/L
The World Health Organization (WHO) guidelines for drinking-water quality specify the provisional guideline value for arsenic as 0.01 mg/L (which is equivalent to 10 micrograms per liter or 10 ppb). Arsenic is a toxic metalloid that can cause skin lesions, cancer, and cardiovascular disease upon long-term ingestion.
Analysis of options:
- Option (a) 0.01 ml/g: This is an incorrect unit for measuring dissolved chemical concentrations in water.
- Option (b) 0.001 mg/L: This is too low and does not match the WHO standard.
- Option (c) 0.0001 ml/L: This is an incorrect unit (volume per volume) and an incorrect value.
- Option (d) 0.01 mg/L: This is the correct WHO guideline value for arsenic in drinking water.
The area where water collects after rain is called:
A)
Watershed
B)
Aquifer
C)
Percolation area
D)
Catchment area
Step-by-step reasoning:
1. Definition of Catchment Area: A catchment area (also known as a drainage basin or drainage area) is the entire geographical area from which all precipitation (rain or snowmelt) collects and drains into a common outlet, such as a river, lake, or reservoir. This matches the question's description of the area where water collects after rain.
2. Analysis of other options:
- a. Watershed: While often used interchangeably with catchment area in common speech, technically, a watershed is the 'divide' or the boundary line (ridge) that separates one catchment area from another.
- b. Aquifer: An aquifer is an underground layer of water-bearing permeable rock, gravel, sand, or silt from which groundwater can be extracted. It is a storage medium, not the surface collection area.
- c. Percolation area: Percolation refers to the downward movement of water through soil and rock. A percolation area is a specific zone where water infiltrates the ground, rather than the total area collecting the runoff.
3. Conclusion: The most accurate term for the total area collecting rainwater is the Catchment area.
The most effective method to control waterlogging is:
A)
Increasing irrigation
B)
Deforestation
C)
Reducing crop rotation
D)
Installing tube wells
Waterlogging occurs when the water table rises to a level where the root zone of crops becomes saturated, displacing air and hindering plant growth. To control waterlogging, the excess groundwater must be removed to lower the water table.
1. **Installing tube wells (Correct):** This is a method of 'vertical drainage'. By pumping out groundwater through tube wells, the water table is artificially lowered, which drains the saturated root zone and mitigates waterlogging.
2. **Increasing irrigation (Incorrect):** Adding more water to an already waterlogged field will exacerbate the problem, raising the water table further.
3. **Deforestation (Incorrect):** Trees help in reducing waterlogging through transpiration (biological pumping). Removing trees reduces the amount of water removed from the soil, potentially worsening the condition.
4. **Reducing crop rotation (Incorrect):** Crop rotation, specifically including deep-rooted crops, can help lower the water table. Reducing this practice does not provide a mechanism to remove excess water.
Which device is used to measure the velocity of a fluid flow by utilizing the pressure difference?
A)
Manometer
B)
Hydrometer
C)
Barometer
D)
Venturimeter
The question asks for a device that measures fluid flow velocity (or discharge) based on pressure difference.
1. **Venturimeter (Correct):** A Venturimeter works on Bernoulli's principle. It consists of a converging section, a throat, and a diverging section. As fluid flows through the constricted throat, its velocity increases and its pressure decreases. By measuring the pressure difference between the inlet and the throat using a manometer, the flow velocity and discharge rate can be calculated.
2. **Manometer (Incorrect):** A manometer is a device used to measure the static pressure of a fluid or the pressure difference between two points, but it does not measure velocity on its own; it is often a component used *within* a Venturimeter.
3. **Hydrometer (Incorrect):** A hydrometer is used to measure the relative density (specific gravity) of a liquid based on the principle of buoyancy.
4. **Barometer (Incorrect):** A barometer is used to measure atmospheric pressure.
Intensity-duration-frequency (IDF) curves are used for:
A)
Irrigation duty
B)
Groundwater
C)
Rainfall analysis
D)
Duty
Intensity-Duration-Frequency (IDF) curves are fundamental tools in hydrology used for rainfall analysis.
1. **Definition**: An IDF curve represents the relationship between the rainfall intensity (amount of rain per unit time), the duration of the storm, and the frequency (or return period) with which such a storm is expected to occur.
2. **Application**: These curves are primarily used by civil and hydraulic engineers to design drainage systems, culverts, bridges, and storm sewers. By knowing the design return period (e.g., a 50-year storm) and the time of concentration for a catchment area, engineers can determine the peak rainfall intensity to calculate the peak runoff using the Rational Method (Q = CiA).
**Why other options are incorrect:**
- **a. Irrigation duty**: Duty refers to the area of land that can be irrigated by a unit discharge of water. It is a concept of irrigation engineering, not rainfall intensity.
- **b. Groundwater**: Groundwater analysis involves aquifer properties, permeability, and drawdown, which are unrelated to the surface rainfall intensity patterns described by IDF curves.
- **d. Duty**: As mentioned in option (a), duty is a measure of water efficiency in irrigation and is not related to storm frequency or intensity.
The hydraulic radius (R) is defined as:
A)
Length / Perimeter
B)
Area / Radius
C)
Area / Perimeter
D)
Area2 / Perimeter
In fluid mechanics and open channel flow, the hydraulic radius (R) is a geometric property of the channel cross-section. It is defined as the ratio of the cross-sectional area of the flow (A) to the wetted perimeter (P).
Formula: R = A / P
- Area (A): The total cross-sectional area through which the fluid flows.
- Wetted Perimeter (P): The total length of the channel boundary that is in contact with the flowing liquid.
Analysis of options:
- Option (a) Length / Perimeter: This is not a standard hydraulic parameter.
- Option (b) Area / Radius: This does not define hydraulic radius; radius is a property of a circle, whereas hydraulic radius applies to any channel shape.
- Option (c) Area / Perimeter: This matches the standard definition of Hydraulic Radius.
- Option (d) Area^2 / Perimeter: This is mathematically incorrect and does not represent any standard hydraulic property.
In cold working:
A)
strength, hardness and ductility increase
B)
strength and hardness increase, but ductility decreases
C)
strength, hardness and ductility decrease
D)
strength and hardness decrease, but ductility increases
Cold working (also known as strain hardening or work hardening) is the process of plastically deforming a metal below its recrystallization temperature. During this process, the dislocation density within the crystal lattice increases significantly. As these dislocations move and interact, they become entangled and impede each other's motion, making it more difficult for further deformation to occur. This results in an increase in the material's yield strength and hardness. However, because the capacity for further plastic deformation is reduced as the material becomes 'saturated' with dislocations, the ductility (the ability of the material to deform plastically before fracture) decreases. Therefore, strength and hardness increase while ductility decreases.
Extrusion force does not depend upon the:
A)
extrusion ratio
B)
type of extrusion process
C)
material of the die
D)
working temperature
Extrusion force is the force required to push a metal billet through a die to create a desired cross-section. It is governed by the formula: F = A₀ * σ_f * ln(R), where A₀ is the initial area, σ_f is the flow stress of the material, and R is the extrusion ratio.
1. Extrusion Ratio (a): The force is directly proportional to the natural log of the extrusion ratio (R = A₀/A_final). A higher ratio requires more force.
2. Type of Extrusion Process (b): Direct extrusion requires more force than indirect extrusion because in direct extrusion, there is additional friction between the billet and the container wall, whereas in indirect extrusion, there is no relative motion between the billet and the container.
3. Working Temperature (d): The flow stress (σ_f) of the metal is highly dependent on temperature. Increasing the temperature reduces the yield strength of the material, thereby reducing the required extrusion force.
4. Material of the Die (c): While the die material must be strong and wear-resistant enough to withstand the process, the *force required to deform the workpiece* depends on the workpiece's properties and the geometry of the process, not the chemical composition or material of the die itself (assuming the die does not deform).
Explain the TTT diagram with a neat and labelled schematic.
In a transportation problem, North-West corner rule would yield:
A)
An optimum solution
B)
An initial feasible solution
C)
A Vogel’s approximate solution
D)
A minimum cost solution
In Operations Research, the Transportation Problem is solved in two main phases: finding an Initial Basic Feasible Solution (IBFS) and then optimizing that solution to find the minimum cost.
1. **North-West Corner Rule (NWCR)**: This is a heuristic method used solely to find an Initial Basic Feasible Solution. It starts at the top-left (North-West) cell of the transportation matrix and allocates as much as possible based on supply and demand, moving right or down. It completely ignores the transportation costs associated with the cells.
2. **Why other options are incorrect**:
- **Option (a) An optimum solution**: The NWCR does not consider costs, so it rarely yields the cheapest (optimum) solution. Optimization requires further tests like the MODI (Modified Distribution) method or the Stepping Stone method.
- **Option (c) A Vogel’s approximate solution**: Vogel's Approximation Method (VAM) is a *different* method for finding an IBFS. While VAM also provides an initial solution, it uses 'penalties' to get closer to the optimum than NWCR does.
- **Option (d) A minimum cost solution**: Similar to the optimum solution, the NWCR does not minimize costs; it only ensures that supply and demand constraints are met (feasibility).
The main purpose of spheroidizing treatment is to improve:
A)
hardenability of low carbon steels
B)
machinability of low carbon steels
C)
hardenability of high carbon steels
D)
machinability of high carbon steels
Spheroidizing is a heat treatment process used primarily for high-carbon steels (typically those with carbon content above 0.6%). In these steels, cementite usually exists as hard, brittle plates (lamellar pearlite), which makes the material extremely difficult to machine and prone to tool wear. Spheroidizing involves heating the steel to just below the lower critical temperature (A1) for a prolonged period, causing the cementite lamellae to coalesce into spherical particles (spheroids) embedded in a soft ferrite matrix. This microstructure is the softest possible state for steel, significantly reducing hardness and increasing ductility, which directly improves the machinability of high-carbon steels.
Why other options are incorrect:
- Options (a) and (c): Spheroidizing actually decreases hardness and does not aim to improve 'hardenability' (which is the ease with which a steel forms martensite upon quenching).
- Option (b): Low carbon steels are already relatively soft and easy to machine; they do not possess the coarse lamellar carbide structures that necessitate spheroidizing treatment.
How does the distribution of heat between the tool, chip, and workpiece affect machining performance?
What is the iron-carbon diagram? Draw a neat and labelled schematic iron-carbon diagram (need not be in scale).
Explain the different types of crystal imperfections.