← MES Jr. Grade
Past Questions
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 product of duty and delta is equal to __________ times the base period (B).
A)
81.64
B)
9.46
C)
8.64
D)
8.12
In irrigation engineering, the relationship between Duty (D), Delta (Δ), and Base Period (B) is defined by the formula:
Δ = 8.64 B / D
Where:
- Δ (Delta) is the total depth of water required by a crop during the entire base period, usually measured in centimeters (cm).
- D (Duty) is the area of land irrigated per unit discharge of water, usually measured in hectares per cumec (ha/cumec).
- B (Base Period) is the time between the first watering and the last watering, measured in days.
Rearranging the formula to find the product of Duty and Delta:
Duty × Delta = 8.64 × Base Period
Derivation of the constant 8.64:
1 cumec = 1 m³/s = 60 × 60 × 24 m³/day = 86,400 m³/day.
1 hectare = 10,000 m².
Therefore, 1 cumec flowing for 1 day covers 86,400 / 10,000 = 8.64 hectares to a depth of 1 meter (or 100 cm).
Thus, the constant is 8.64.
Comparing this to the options:
- Option (a) 81.64 is incorrect.
- Option (b) 9.46 is incorrect.
- Option (c) 8.64 is the correct constant.
- Option (d) 8.12 is incorrect.
The term ‘priming’ in pumps refers to:
A)
Removing air from the pump casing
B)
Lubricating moving parts
C)
Starting the motor
D)
Cleaning filters
Priming is a critical process in centrifugal pumps. Centrifugal pumps are designed to move liquids, not gases. If air is trapped in the pump casing and suction pipe, the pump cannot create the necessary vacuum (pressure differential) to lift the liquid from the source (sump) because air is highly compressible compared to water. Therefore, 'priming' is the process of filling the suction pipe and the pump casing with the liquid to be pumped and removing all air/gas pockets before starting the pump.
Why other options are incorrect:
- Option b (Lubricating moving parts) refers to maintenance or lubrication, not priming.
- Option c (Starting the motor) is the act of powering the pump, which must happen AFTER priming is complete.
- Option d (Cleaning filters) is a maintenance task to prevent clogging, unrelated to the removal of air from the casing.
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.
In a hydraulic system, the continuity equation implies:
A)
Pressure remains constant throughout the system
B)
Flow rate is constant in a closed system
C)
Velocity increases with increasing pipe diameter
D)
Energy is always conserved
The continuity equation is a mathematical expression of the Law of Conservation of Mass. For an incompressible fluid (like most liquids in hydraulic systems) flowing through a closed conduit, the mass flow rate must remain constant. This is expressed as Q = A1V1 = A2V2, where Q is the volumetric flow rate, A is the cross-sectional area, and V is the velocity.
Analysis of options:
- Option (a) is incorrect: Pressure typically changes along a system due to friction (head loss) or changes in velocity (Bernoulli's principle).
- Option (b) is correct: It directly states the principle of conservation of mass/flow rate for a closed system.
- Option (c) is incorrect: According to Q = AV, if the pipe diameter (and thus area A) increases, the velocity V must decrease to keep Q constant. Velocity decreases with increasing diameter.
- Option (d) is incorrect: While energy is conserved in an ideal system (Bernoulli's equation), the 'continuity equation' specifically refers to the conservation of mass/flow, not energy.
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.
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.
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.
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.
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.
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.
The main function of a spillway in a hydropower dam is:
A)
Generate electricity
B)
Discharge excess water safely
C)
Store water for irrigation
D)
Prevent soil erosion
A spillway is a critical safety structure designed to provide a controlled release of water from a dam or reservoir. Its primary purpose is to prevent the reservoir from overflowing the dam crest (overtopping), which could lead to catastrophic structural failure, especially in embankment dams.
Step-by-step analysis of options:
- Option (a) 'Generate electricity' is the function of the powerhouse and turbines, not the spillway.
- Option (b) 'Discharge excess water safely' is the correct definition of a spillway's function, acting as a 'safety valve' for the dam during floods.
- Option (c) 'Store water for irrigation' is the function of the reservoir created by the dam body itself.
- Option (d) 'Prevent soil erosion' is generally handled by riprap, vegetation, or energy dissipators at the toe of the spillway, but the spillway's main function is water discharge, not erosion prevention (in fact, uncontrolled spillways can cause erosion).
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).
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.
The method of irrigation which uses narrow channels between crop rows:
A)
Sprinkler irrigation
B)
Drip irrigation
C)
Check basin irrigation
D)
Furrow irrigation
Step-by-step reasoning: 1. Analysis of the question: The key phrase is 'narrow channels between crop rows'. 2. Evaluating Option (d) Furrow Irrigation: In this method, small parallel channels (furrows) are constructed between the rows of crops. Water is applied to these furrows and infiltrates the soil laterally to reach the root zone. This matches the description perfectly. 3. Evaluating Option (a) Sprinkler Irrigation: This is an overhead irrigation method where water is sprayed into the air to mimic rainfall; it does not use channels. 4. Evaluating Option (b) Drip Irrigation: This is a micro-irrigation system that delivers water directly to the root zone via emitters/drippers through pipes, not open channels. 5. Evaluating Option (c) Check Basin Irrigation: In this method, the field is divided into rectangular plots (basins) surrounded by embankments, and the entire basin is flooded. It does not involve narrow channels between rows. Therefore, Furrow irrigation is the correct method.
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.
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.
A repair shop is manned by a single worker. Customers arrive at the rate of 30 per hour. Time required to provide service is exponentially distributed with mean of 100 seconds. Find the mean waiting time of a customer, needing repair facility in the queue.
In shell moulding, the resin-coated sand is hardened by:
A)
Cooling
B)
Chemical catalyst
C)
Heating to cure the resin
D)
Applying pressure
Shell moulding is a precision casting process that uses a mixture of sand and a thermosetting resin binder. The process works as follows: 1. A metal pattern is heated to a specific temperature (typically 200-300°C). 2. Resin-coated sand is dropped onto the heated pattern. 3. The heat from the pattern causes the thermosetting resin to melt and then cure (polymerize), forming a hard, thin 'shell' around the pattern. 4. Once the shell is hardened, the excess uncured sand is dumped, and the shell is removed from the pattern.
Why other options are incorrect:
- a. Cooling: Cooling would not trigger the polymerization of a thermosetting resin; heat is required for the chemical reaction.
- b. Chemical catalyst: While some resins use catalysts, shell moulding specifically relies on thermal energy (heat) from the pattern to initiate curing.
- d. Applying pressure: Pressure is not the mechanism used to harden the resin in this process; the chemical bond is formed via thermal curing.
For a ductile material, toughness is the measure of:
A)
resistance to scratching
B)
ability to absorb energy up to fracture
C)
ability to absorb energy till elastic limit
D)
resistance to indentation
Toughness is a fundamental material property defined as the ability of a material to absorb energy and deform plastically before it eventually fractures. In a stress-strain curve, toughness is represented by the total area under the curve from the origin to the point of fracture.
Analysis of options:
- Option (a) 'resistance to scratching' and Option (d) 'resistance to indentation' both describe 'Hardness', not toughness.
- Option (c) 'ability to absorb energy till elastic limit' describes 'Resilience' (the capacity of a material to absorb energy when it is deformed elastically and then release that energy upon unloading).
- Option (b) correctly identifies toughness as the total energy absorption capacity up to the point of failure (fracture), which is especially critical for materials subjected to impact loading.
For preventing shrinkage defects, the riser should:
A)
Solidify before the casting
B)
Solidify after the casting
C)
Solidify at the same time as the casting
D)
Have no relation to casting solidification
In casting, shrinkage occurs because most metals contract as they transition from a liquid to a solid state. To prevent 'shrinkage cavities' (voids) within the main casting, a riser (or feeder) is used. The riser acts as a reservoir of molten metal. For the riser to effectively feed the casting, it must remain liquid longer than the casting itself. This ensures that as the casting shrinks and creates a void, molten metal from the still-liquid riser flows into that void via gravity or pressure. If the riser solidified before or at the same time as the casting, it would be unable to supply the necessary liquid metal, leading to internal shrinkage defects in the final part. Therefore, the riser must have a higher volume-to-surface area ratio (according to Chvorinov's Rule) to ensure it solidifies last.
A spherical drop of molten metal of radius 5 mm was found to solidify in 12 s. In how much time will a similar drop of radius 10 mm would solidify?
The maximum heat in resistance welding is at the:
A)
tip of the positive electrode
B)
tip of the negative electrode
C)
top surface of the plate at the time of electric contact with the electrode
D)
interface between the two plates being joined
In resistance welding (such as spot welding), heat is generated based on Joule's Law: H = I²Rt, where 'H' is the heat generated, 'I' is the current, 'R' is the electrical resistance, and 't' is the time. The heat generated is directly proportional to the electrical resistance of the material. In a resistance welding setup, the electrodes are typically made of high-conductivity materials (like copper alloys) to minimize their own resistance and prevent them from overheating. The highest electrical resistance in the entire circuit occurs at the interface between the two workpieces (the plates) due to the presence of surface oxides, contaminants, and the microscopic gaps between the two surfaces before they fuse. Because the resistance (R) is maximum at this interface, the maximum heat is generated there, allowing the metal to melt and form a weld nugget while the electrodes remain relatively cool.
Plastic deformation is caused by:
A)
stored elastic energy
B)
dislocation motion
C)
low value of modulus
D)
complex elastic stress
Plastic deformation refers to the permanent, non-reversible deformation of a material that occurs when the applied stress exceeds the yield strength. In crystalline materials (like metals), this process occurs primarily through a mechanism called 'slip'. Slip is the process by which planes of atoms slide over one another. This sliding is not a simultaneous movement of an entire plane of atoms, but rather the movement of linear defects known as 'dislocations' through the crystal lattice. As dislocations move, they break and reform bonds, allowing the material to deform permanently without fracturing.
Why other options are incorrect:
- a. Stored elastic energy: This is associated with elastic deformation (temporary), where the material returns to its original shape once the load is removed.
- c. Low value of modulus: The modulus of elasticity (Young's modulus) describes the stiffness of a material in the elastic region; it does not cause plastic deformation.
- d. Complex elastic stress: Elastic stress, by definition, results in elastic (reversible) deformation, not plastic (permanent) deformation.
In thread cutting operation the saddle is moved by:
A)
Head stoke
B)
Tail stoke
C)
Tool post
D)
Lead screw
In a lathe machine, thread cutting requires a precise relationship between the rotation of the spindle (workpiece) and the longitudinal movement of the cutting tool. This is achieved using the lead screw. The lead screw is a long, threaded shaft that is engaged with the saddle via a half-nut mechanism. As the lead screw rotates, it translates the rotary motion into a precise linear motion of the saddle, ensuring that the tool moves a specific distance (the lead) for every single revolution of the spindle, thereby creating a consistent thread pitch.
Why other options are incorrect:
- a. Headstock: This is the stationary part of the lathe that houses the main spindle and gear mechanism; it does not move the saddle.
- b. Tailstock: This is used to support long workpieces or hold drilling/reaming tools; it does not drive the saddle for threading.
- c. Tool post: This is the component that holds the cutting tool securely in place on top of the compound rest; it is a passive holder and not a driving mechanism.