MES Jr. Grade • 2025 • Paper-II
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).
MES Jr. Grade • 2025 • Paper-II
The Isohyetal method is used for:
A)
Measuring evaporation
B)
Calculating runoff coefficient
C)
Determining groundwater level
D)
Estimating average rainfall over an area
The Isohyetal method is one of the primary techniques used in hydrology to calculate the average depth of precipitation over a specific catchment area. An 'isohyet' is a line connecting points of equal rainfall depth (similar to isotherms for temperature or contours for elevation). By mapping these lines and calculating the area between successive isohyets, the weighted average rainfall for the entire basin can be determined. This method is generally considered the most accurate because it accounts for the spatial distribution of rainfall and topographic influences.
Why other options are incorrect:
- Option (a): Evaporation is measured using evaporation pans or lysimeters.
- Option (b): The runoff coefficient is typically determined using the Rational Method (Q=CiA) or based on land-use characteristics, not isohyetal maps.
- Option (c): Groundwater levels are determined using observation wells or piezometers.
MES Jr. Grade • 2025 • Paper-II
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.
MES Jr. Grade • 2025 • Paper-II
Which process is used to remove hardness from water?
A)
Filtration
B)
Ion exchange
C)
Sedimentation
D)
Condensation
Hardness in water is primarily caused by the presence of multivalent cations, most commonly Calcium (Ca2+) and Magnesium (Mg2+).
1. **Ion Exchange (Correct):** This process involves passing hard water through a resin bed (typically a strong acid cation exchange resin). The resin contains sodium ions (Na+) or hydrogen ions (H+) which are exchanged for the Ca2+ and Mg2+ ions in the water. This effectively removes the hardness-causing minerals, making it a standard industrial and domestic water softening method.
2. **Filtration (Incorrect):** Filtration is used to remove suspended solids, turbidity, and some microorganisms from water. It cannot remove dissolved ions like calcium and magnesium.
3. **Sedimentation (Incorrect):** Sedimentation is a physical process where heavier particles settle to the bottom of a tank due to gravity. Dissolved minerals causing hardness do not settle out through sedimentation.
4. **Condensation (Incorrect):** Condensation is the phase change from gas to liquid. While distillation (which involves evaporation and condensation) can remove hardness, condensation alone is not a water treatment process for hardness removal.
MES Jr. Grade • 2025 • Paper-II
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.
MES Jr. Grade • 2025 • Paper-II
Which type of turbine is most suitable for a high head and low discharge hydropower plant?
A)
Kaplan turbine
B)
Francis turbine
C)
Pelton turbine
D)
Propeller turbine
The selection of a hydraulic turbine depends primarily on the available head (the vertical distance the water falls) and the discharge (the volume of water flowing per unit time).
1. **Pelton Turbine (Impulse Turbine):** These are designed specifically for **high head** and **low discharge** conditions. They use high-velocity jets of water that strike buckets on the periphery of the wheel to transfer kinetic energy into mechanical energy.
2. **Francis Turbine (Reaction Turbine):** These are versatile and are typically used for **medium head** and **medium discharge** applications.
3. **Kaplan and Propeller Turbines (Reaction Turbines):** These are axial-flow turbines designed for **low head** and **high discharge** conditions. Kaplan turbines have adjustable blades to maintain efficiency over a range of flows, while propeller turbines have fixed blades.
Since the question specifies 'high head and low discharge', the Pelton turbine is the most suitable choice.
MES Jr. Grade • 2025 • Paper-II
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.
MES Jr. Grade • 2025 • Paper-II
The term ‘aquifer’ refers to:
A)
A body of surface water
B)
A layer of impermeable rock
C)
A saturated underground layer that can yield water
D)
A type of dam structure
Step-by-step reasoning:
1. Definition: In hydrology and geology, an aquifer is a body of permeable rock or sediment (such as sand, gravel, or limestone) that is saturated with groundwater.
2. Key Characteristics: To be classified as an aquifer, the geological formation must possess two critical properties: porosity (the ability to store water in voids) and permeability (the ability to allow water to flow through those voids). This allows the layer to 'yield' water to a well or spring at a useful rate.
3. Evaluating Options:
- Option (a) is incorrect because an aquifer is a subsurface (underground) formation, not surface water.
- Option (b) is incorrect because a layer of impermeable rock is called an 'aquiclude' or 'aquitard'; it prevents the flow of water rather than yielding it.
- Option (c) is correct as it accurately describes the saturated nature and the capacity to yield water.
- Option (d) is incorrect because a dam is a man-made surface structure used to impound water, not a geological layer.
MES Jr. Grade • 2025 • Paper-II
What is the primary pollutant responsible for acid rain?
A)
Sulfur dioxide (SO2)
B)
Carbon monoxide (CO)
C)
Ozone (O3)
D)
Particulate matter (PM)
Acid rain is caused by a chemical reaction that begins when compounds like sulfur dioxide (SO2) and nitrogen oxides (NOx) are released into the air. These substances rise high into the atmosphere, where they mix and react with water, oxygen, and other chemicals to form mild solutions of sulfuric and nitric acid. Sulfur dioxide, primarily emitted from fossil fuel combustion (like coal-fired power plants), is considered a primary driver of this phenomenon.
Why other options are incorrect:
- b. Carbon monoxide (CO): This is a colorless, odorless toxic gas that interferes with oxygen transport in the blood, but it does not contribute significantly to the acidification of precipitation.
- c. Ozone (O3): While ground-level ozone is a pollutant and a component of smog, it is not the primary cause of acid rain.
- d. Particulate matter (PM): These are microscopic solids or liquid droplets that can cause respiratory issues, but they are not the chemical precursors that create acid rain.
MES Jr. Grade • 2025 • Paper-II
Which of the following is a positive displacement pump?
A)
Centrifugal pump
B)
Axial flow pump
C)
Reciprocating pump
D)
Jet pump
Pumps are broadly classified into two main categories based on how they move fluid: Dynamic (Kinetic) pumps and Positive Displacement pumps.
1. **Positive Displacement Pumps**: These pumps move fluid by trapping a fixed volume of it and then forcing (displacing) that volume into the discharge pipe. They provide a constant flow rate regardless of the pressure head. Examples include Reciprocating pumps (piston/plunger) and Rotary pumps (gear, screw, vane).
2. **Dynamic (Kinetic) Pumps**: These pumps add kinetic energy to the fluid using a rotating impeller, which is then converted into pressure energy. The flow rate varies significantly with the pressure head. Examples include Centrifugal pumps and Axial flow pumps.
Evaluating the options:
- **a. Centrifugal pump**: A dynamic pump that uses centrifugal force to move fluid.
- **b. Axial flow pump**: A dynamic pump where the fluid is pushed parallel to the axis of rotation.
- **c. Reciprocating pump**: A positive displacement pump that uses a piston or plunger to displace a specific volume of fluid per stroke.
- **d. Jet pump**: An ejector pump that uses the Venturi effect (high-velocity fluid jet) to move another fluid; it is not a positive displacement pump.
MES Jr. Grade • 2025 • Paper-II
Double mass curve method is used for:
A)
Runoff estimation
B)
Evaporation estimation
C)
Rainfall data consistency
D)
Temperature estimation
The Double Mass Curve technique is a graphical method used in hydrology to check the consistency and homogeneity of precipitation (rainfall) data from a particular gauging station.
1. **How it works**: The cumulative rainfall of the station under test is plotted on the y-axis against the cumulative average rainfall of a group of surrounding reliable stations (the 'base' stations) on the x-axis.
2. **Interpretation**: If the data is consistent, the plot will be a straight line. A break or change in the slope of the line indicates a change in the conditions at the station (e.g., relocation of the rain gauge, change in surroundings like the growth of a tree, or change in the instrument), suggesting that the data is inconsistent.
3. **Why other options are incorrect**:
- **Runoff estimation**: This is typically done using the Rational Method, SCS Curve Number method, or Unit Hydrographs.
- **Evaporation estimation**: This is measured using evaporation pans or estimated via Penman-Monteith equations.
- **Temperature estimation**: This is measured using thermometers and validated through spatial interpolation or regional climate models, not double mass curves.
MES Jr. Grade • 2025 • Paper-II
The most common disinfectant used in water treatment is:
A)
Chlorine
B)
Ozone
C)
Ultraviolet rays
D)
RO
Step-by-step reasoning:
1. **Definition of Disinfection**: Disinfection is the process of killing pathogenic microorganisms in water to make it safe for human consumption.
2. **Evaluating Option (a) Chlorine**: Chlorine is the most widely used disinfectant globally due to its effectiveness, low cost, ease of application, and most importantly, its ability to provide 'residual disinfection.' This means chlorine remains in the water as it travels through the distribution pipes, preventing re-contamination.
3. **Evaluating Option (b) Ozone**: While ozone is a very powerful oxidant and disinfectant (more effective than chlorine against certain cysts), it is expensive to produce on-site and provides no residual protection once the water leaves the treatment plant.
4. **Evaluating Option (c) Ultraviolet (UV) rays**: UV radiation is a physical process that disrupts the DNA of microorganisms. Like ozone, it provides no residual effect and requires the water to be very clear (low turbidity) to be effective.
5. **Evaluating Option (d) RO (Reverse Osmosis)**: RO is a membrane filtration process used for desalination and removing dissolved solids. While it removes microbes physically, it is a filtration method, not a chemical disinfectant.
Conclusion: Because of its cost-effectiveness and residual protection, chlorine is the most common disinfectant used in municipal water treatment.
MES Jr. Grade • 2025 • Paper-II
The term ‘return period’ in hydrology refers to:
A)
Average interval between two events
B)
Frequency of rainfall
C)
Duration of flood
D)
Base period of crop
In hydrology, the 'Return Period' (also known as the recurrence interval) is defined as the average time interval between the occurrence of an event (such as a flood or a heavy rainfall) of a given magnitude or greater. Mathematically, it is the reciprocal of the probability (P) of occurrence in any given year: T = 1/P. For example, a '100-year flood' has a 1% probability of occurring in any single year, meaning on average, such an event occurs once every 100 years.
Why other options are incorrect:
- Option b (Frequency of rainfall): Frequency refers to how often an event occurs (the probability), whereas the return period is the time interval associated with that frequency.
- Option c (Duration of flood): Duration refers to the length of time a flood event lasts from start to finish, not the interval between separate events.
- Option d (Base period of crop): This is an irrigation engineering term referring to the period from the first watering at the time of sowing to the last watering before harvest; it is unrelated to hydrological return periods.
MES Jr. Grade • 2025 • Paper-II
The intensity of irrigation means:
A)
Percentage of cultivable area irrigated annually
B)
Depth of water supplied
C)
Number of waterings given to a crop
D)
Duration of irrigation season
In irrigation engineering, 'Intensity of Irrigation' is defined as the percentage of the total cultivable area of a region that is irrigated annually. It is calculated as: (Area Irrigated / Total Cultivable Area) × 100.
Analysis of other options:
- Option (b) 'Depth of water supplied' refers to the 'Duty' or 'Delta' concepts (specifically Delta is the total depth of water required by a crop).
- Option (c) 'Number of waterings given to a crop' refers to the 'Frequency of Irrigation'.
- Option (d) 'Duration of irrigation season' refers to the 'Base Period' (the time between the first watering at sowing and the last watering before harvest).
MES Jr. Grade • 2025 • Paper-II
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.
MES Jr. Grade • 2025 • Paper-II
The term ‘hydraulic efficiency’ of a turbine refers to:
A)
the ratio of mechanical power output to electrical power output
B)
the ratio of power delivered by the turbine to the power available in the water
C)
the ratio of actual head to theoretical head
D)
the ratio of pump input power to turbine output power
Hydraulic efficiency of a turbine is defined as the ratio of the power actually transferred to the runner to the power available at the turbine inlet from the water. In terms of head, it is written as η_h = actual head utilized / theoretical head available, so option (c) is correct.
Why the others are incorrect:
- (b) describes overall turbine efficiency more generally as power output to water power available, but the standard definition of hydraulic efficiency specifically uses the head ratio (or equivalent water-power-to-runner-power relation), not just the broad delivered-power wording.
- (a) is incorrect because it compares mechanical power output to electrical power output, which is not a turbine efficiency definition; that would relate to generator/electrical conversion.
- (d) is incorrect because it mixes pump and turbine terms and does not define turbine efficiency.
MES Jr. Grade • 2025 • Paper-II
Which of the following is a surface irrigation method?
A)
Drip irrigation
B)
Sprinkler irrigation
C)
Furrow irrigation
D)
Subsurface irrigation
Irrigation methods are broadly classified based on how water is applied to the soil.
1. Surface Irrigation: Water is applied and distributed over the soil surface by gravity. Examples include Border irrigation, Basin irrigation, and Furrow irrigation. In Furrow irrigation, water is applied in small parallel channels (furrows) between crop rows.
2. Sub-surface Irrigation: Water is applied below the root zone through perforated pipes or tiles. (Option d is incorrect).
3. Sprinkler Irrigation: Water is sprayed into the air and falls like rain, which is a form of overhead irrigation, not surface irrigation. (Option b is incorrect).
4. Drip (Trickle) Irrigation: Water is applied slowly and precisely to the root zone through emitters. This is a localized irrigation method. (Option a is incorrect).
Therefore, Furrow irrigation is the only surface irrigation method listed.
MES Jr. Grade • 2025 • Paper-II
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.
MES Jr. Grade • 2025 • Paper-II
NPSH stands for:
A)
Net Positive Suction Head
B)
Net Pressure Suction Head
C)
Nominal Pressure Suction Head
D)
Net Pump Suction Head
NPSH stands for Net Positive Suction Head. In hydraulic machinery, specifically centrifugal pumps, NPSH is a critical parameter used to determine if a pump will operate without cavitation. Cavitation occurs when the local pressure of the liquid drops below its vapor pressure, causing bubbles to form and collapse, which can damage the impeller.
1. 'Net' refers to the total head available after accounting for losses.
2. 'Positive' indicates that the pressure must be above the vapor pressure of the liquid.
3. 'Suction Head' refers to the head at the inlet (suction side) of the pump.
Options b, c, and d are incorrect because 'Pressure', 'Nominal', and 'Pump' are not the standard technical terms used in the definition of this specific hydraulic parameter.
MES Jr. Grade • 2025 • Paper-II
The permissible limit of turbidity in drinking water as per WHO guidelines is:
A)
1 NTU
B)
5 NTU
C)
10 NTU
D)
15 NTU
Turbidity is a measure of the cloudiness or haziness of a fluid caused by individual particles (suspended solids). According to the World Health Organization (WHO) Guidelines for Drinking-water Quality, for an effective disinfection process and for aesthetic acceptability, turbidity should ideally be less than 1 NTU (Nephelometric Turbidity Unit). While WHO notes that turbidity levels up to 5 NTU may be acceptable in some contexts, the gold standard for treated drinking water to ensure the removal of pathogens (like Cryptosporidium and Giardia) and to ensure the efficiency of chlorination is 1 NTU. In many competitive engineering exams, 1 NTU is recognized as the permissible limit for treated drinking water to ensure safety and clarity.
MES Jr. Grade • 2025 • Paper-II
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.
MES Jr. Grade • 2025 • Paper-II
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.
MES Jr. Grade • 2025 • Paper-II
A hydraulic jump occurs when:
A)
Flow changes from critical to supercritical
B)
Flow changes from laminar to turbulent
C)
Flow changes from supercritical to subcritical
D)
Flow changes from turbulent to laminar
A hydraulic jump is a phenomenon in open channel flow where a high-velocity, shallow flow (supercritical flow) transitions abruptly to a low-velocity, deep flow (subcritical flow). This transition is characterized by a sudden rise in the water surface level and significant energy dissipation through turbulence.
1. Supercritical flow occurs when the Froude number (Fr) is greater than 1 (Fr > 1).
2. Subcritical flow occurs when the Froude number (Fr) is less than 1 (Fr < 1).
3. During a hydraulic jump, the flow must move from a state of high kinetic energy (supercritical) to a state of higher potential energy (subcritical) to dissipate energy, often used in dam spillways to prevent erosion of the riverbed.
Analysis of other options:
- Option (a) is incorrect because a change from critical to supercritical is typically a gradual acceleration (e.g., down a steep slope), not a jump.
- Options (b) and (d) are incorrect because hydraulic jumps are related to the flow regime (Froude number) and depth, not the viscosity-driven transition between laminar and turbulent flow.
MES Jr. Grade • 2025 • Paper-II
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.
MES Jr. Grade • 2025 • Paper-II
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.