← MES Jr. Grade
Past Questions
The percentage of Chlorine in fresh bleaching powder is about
A)
10-15%
B)
30-35%
C)
20-25%
D)
40-45%
Bleaching powder, chemically known as Calcium Hypochlorite [Ca(OCl)₂], is widely used in water treatment for disinfection. Freshly manufactured bleaching powder typically contains about 30% to 35% 'available chlorine'. Available chlorine is the amount of chlorine that can be released from the compound to act as an oxidizing agent. Over time, bleaching powder decomposes due to exposure to air, moisture, and light, leading to a decrease in this percentage. Therefore, for 'fresh' bleaching powder, the standard range is 30-35%. Options a, c, and d are incorrect as they either represent degraded powder (lower percentages) or values that exceed the chemical composition of standard commercial bleaching powder.
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.
Capacity inflow ratio for a storage reservoir is defined as
A)
reservoir capacity / average annual flood inflow
B)
reservoir capacity / average annual sediment inflow
C)
dead storage capacity of the reservoir / average annual flood inflow
D)
dead storage capacity of the reservoir / average annual sediment deposited
The Capacity-Inflow (C-I) ratio is a critical parameter used in reservoir sedimentation studies to estimate the lifespan of a reservoir. It is defined as the ratio of the reservoir's dead storage capacity (the volume of the reservoir below the lowest outlet level, intended to collect sediment) to the average annual volume of sediment entering the reservoir.
1. **Dead Storage Capacity**: This is the portion of the reservoir that is not used for active water storage but is specifically allocated to accumulate sediment over the design life of the dam.
2. **Average Annual Sediment Inflow**: This is the volume of sediment deposited in the reservoir per year.
Mathematically: C-I Ratio = (Dead Storage Capacity) / (Average Annual Sediment Inflow).
Why other options are incorrect:
- Option (a) and (c) refer to 'flood inflow' (water volume), which relates to flood routing or storage capacity, not the C-I ratio used for sedimentation.
- Option (b) refers to total reservoir capacity, whereas the C-I ratio specifically focuses on the 'dead storage' allocated for sediment to determine how many years it will take for the sediment to fill that specific volume.
The fluid coming into the centrifugal pump is accelerated by
A)
nozzle
B)
impeller
C)
throat
D)
valve
In a centrifugal pump, the primary function of the impeller is to transfer energy from the motor to the fluid. As the impeller rotates, the fluid enters the eye of the impeller and is flung radially outward by centrifugal force. This process increases the kinetic energy of the fluid, thereby accelerating it. The high-velocity fluid then enters the volute casing (or diffuser), where the kinetic energy is converted into pressure energy.
Why other options are incorrect:
- a. Nozzle: A nozzle is used to increase velocity by decreasing the cross-sectional area, typically used in turbines or jets, not for the primary acceleration of fluid entering a pump.
- c. Throat: The throat refers to the narrowest part of a nozzle or venturi; it is a geometric feature, not the component responsible for accelerating fluid in a pump.
- d. Valve: A valve is used to control the flow rate or stop the flow of fluid, not to accelerate it.
Multiple arch dams is an example of
A)
arch dams.
B)
buttress dams.
C)
shell-arch dams.
D)
earth dam.
A multiple arch dam is a specific variation of a buttress dam. In a standard buttress dam, the upstream face is supported by a series of triangular supports (buttresses). In a multiple arch dam, the upstream face consists of a series of small arches that transfer the water pressure to the buttresses. Because the primary structural support system consists of buttresses that stabilize the arches, it is classified under the broader category of buttress dams.
Why other options are incorrect:
- a. Arch dams: While it uses arches, a true 'arch dam' is a single, monolithic curved structure that transfers the load primarily to the canyon walls (abutments), not to a series of supporting buttresses.
- c. Shell-arch dams: These are a hybrid of gravity and arch dams, where a thin arch is supported by a thicker gravity section (shell).
- d. Earth dam: These are embankment dams made of compacted soil and rock, which are fundamentally different from the concrete structural design of a multiple arch dam.
On an indicator diagram of a reciprocating pump,
A)
Pressure head is taken as abscissa and stroke length as ordinate.
B)
Pressure head is taken as ordinate and stroke length as abscissa.
C)
Both pressure head and stroke length are taken as abscissa.
D)
Both pressure head and abscissa are taken as ordinate.
In the context of reciprocating pumps, an indicator diagram is a plot used to visualize the pressure variations inside the cylinder throughout one complete cycle of the piston.
1. **Abscissa (X-axis):** The horizontal axis represents the position of the piston, which is the stroke length (or displacement). As the piston moves from the Top Dead Center (TDC) to the Bottom Dead Center (BDC) and back, the distance traveled is plotted on this axis.
2. **Ordinate (Y-axis):** The vertical axis represents the pressure (or pressure head) of the fluid inside the cylinder at each specific position of the piston.
By plotting pressure against stroke length, the area enclosed by the loop of the indicator diagram represents the work done per cycle.
Comparing this to the options:
- Option (a) is incorrect because it swaps the axes.
- Option (b) is correct: Pressure head (Y-axis/Ordinate) and Stroke length (X-axis/Abscissa).
- Options (c) and (d) are logically incorrect as they suggest both variables are on the same axis.
The intensity of irrigation means
A)
percentage of culturable commanded area to be irrigated annually
B)
percentage of gross commanded area to be irrigated annually
C)
percentage of the mean of culturable commanded area and the gross commanded area to be irrigated annually
D)
total depth of water supplied by the number of waterings.
In irrigation engineering, the following terms are used to define the area of land:
1. Gross Command Area (GCA): The total area which can be physically reached by a canal system, including forests, roads, villages, and non-culturable land.
2. Culturable Command Area (CCA): The portion of the GCA which is fit for cultivation.
3. Intensity of Irrigation: This is defined as the ratio of the area actually irrigated to the Culturable Command Area (CCA), expressed as a percentage.
Formula: Intensity of Irrigation = (Area to be irrigated / CCA) × 100.
Analysis of options:
- Option (a) is correct as it matches the standard definition.
- Option (b) is incorrect because the intensity is measured against the CCA, not the GCA (which includes non-farmable land).
- Option (c) is incorrect as there is no such 'mean' calculation in irrigation terminology.
- Option (d) describes the 'Duty' or 'Delta' concepts (specifically Delta refers to the total depth of water), not the intensity of irrigation.
Presence of tail water in a gravity dam
A)
increases the principal stress and decreases the shear stress.
B)
increases the principal stress and increases the shear stress.
C)
decreases the principal stress and increases the shear stress.
D)
decreases the principal stress and decreases the shear stress.
Tail water acts on the downstream face of a gravity dam and provides an opposing hydrostatic pressure to the upstream water pressure. This reduces the net horizontal thrust on the dam, so the compressive principal stress at the base decreases. Since shear stress is produced mainly by the unbalanced horizontal force, the reduced net thrust also decreases the shear stress. Therefore, the correct option is (d) decreases the principal stress and decreases the shear stress.
Why others are wrong:
- (a) and (b): These say principal stress increases, but tail water actually counteracts the upstream water pressure, so the resultant stress effect is reduced, not increased.
- (c): It correctly says principal stress decreases, but shear stress does not increase; it also decreases because the net horizontal force is smaller.
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 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.
Open channel flow takes place
A)
In the pump
B)
Within a cylindrical depth
C)
On a free surface
D)
In the pipe
Open channel flow is defined as the flow of a liquid (usually water) in a conduit where the liquid surface is exposed to the atmosphere. This interface between the liquid and the air is known as the 'free surface.' In such flows, the driving force is typically gravity.
Analysis of options:
- Option (a) 'In the pump': Pumps are closed systems where fluid is pressurized; this is pressure flow, not open channel flow.
- Option (b) 'Within a cylindrical depth': This is a vague description and does not define the nature of the flow boundary.
- Option (c) 'On a free surface': This is the defining characteristic of open channel flow (e.g., rivers, canals, or partially filled pipes).
- Option (d) 'In the pipe': While a partially filled pipe can exhibit open channel flow, a pipe generally refers to 'pipe flow' (closed conduit flow) where the fluid fills the entire cross-section and is driven by a pressure gradient rather than gravity alone.
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.
When a body is totally or partially immersed in a fluid, it is buoyed up by a force equal to
A)
weight of the body and fluid displaced by the body
B)
difference of weights of the fluid displaced and that of the body
C)
sum of weights of the fluid displaced and that of the body
D)
weight of the fluid displaced by the body
This question is a direct application of Archimedes' Principle. Archimedes' Principle states that any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.
1. **Buoyant Force (Upthrust):** When a body is immersed in a fluid, it experiences an upward force called the buoyant force. This force arises because the pressure at the bottom of the object is greater than the pressure at the top (due to the increase in pressure with depth).
2. **Mathematical Relation:** Buoyant Force (F_b) = Volume of displaced fluid (V) × Density of fluid (ρ) × Acceleration due to gravity (g). Since (V × ρ × g) is the formula for the weight of the displaced fluid, the buoyant force is exactly equal to that weight.
Evaluating the options:
- Option (a) is incorrect because the buoyant force does not include the weight of the body itself.
- Option (b) describes the 'apparent weight' or the net force acting on the body, not the buoyant force itself.
- Option (c) is incorrect as it sums the weights, which has no physical basis in buoyancy laws.
- Option (d) is the correct statement of Archimedes' Principle.
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.
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 total hardness of water is greater than its total alkalinity, the carbonate hardness will be equal to
A)
Total Hardness
B)
Total Alkalinity
C)
Non-carbonate hardness
D)
Bicarbonate hardness
In environmental engineering and water chemistry, hardness is categorized into Carbonate Hardness (CH) and Non-Carbonate Hardness (NCH).
1. **Definitions**:
- Total Hardness (TH) is the sum of multivalent metallic cations (primarily Ca²⁺ and Mg²⁺).
- Total Alkalinity (TA) is the measure of the water's capacity to neutralize acids, primarily due to bicarbonate (HCO₃⁻), carbonate (CO₃²⁻), and hydroxide (OH⁻) ions.
2. **The Relationship**:
Carbonate hardness is the portion of total hardness that is associated with alkalinity. Mathematically, Carbonate Hardness is defined as the minimum of Total Hardness and Total Alkalinity:
- If Total Hardness > Total Alkalinity, then Carbonate Hardness = Total Alkalinity.
- If Total Alkalinity > Total Hardness, then Carbonate Hardness = Total Hardness.
3. **Application to the Question**:
The question specifies that Total Hardness > Total Alkalinity. Therefore, the carbonate hardness is limited by the available alkalinity and is equal to the Total Alkalinity. The remaining hardness (TH - TA) is classified as Non-Carbonate Hardness (NCH).
4. **Analysis of Options**:
- (a) Total Hardness: Incorrect, as this would imply NCH is zero, which contradicts the condition TH > TA.
- (b) Total Alkalinity: Correct, based on the chemical relationship defined above.
- (c) Non-carbonate hardness: Incorrect; NCH is the difference between TH and TA in this scenario.
- (d) Bicarbonate hardness: Incorrect; while bicarbonate contributes to alkalinity, the standard definition for the calculation of CH uses Total Alkalinity.
Orthotolidine test is used for the determination of
A)
Residual Chlorine
B)
Dissolved Oxygen
C)
Biochemical Oxygen Demand
D)
Chloride Ion
The Orthotolidine (OT) test is a standard chemical method used in water treatment to determine the concentration of residual chlorine in water. When orthotolidine reagent is added to water containing chlorine, it reacts to produce a yellow color; the intensity of this yellow color is proportional to the amount of chlorine present, which is then measured using a colorimeter or compared against standard color discs.
Why other options are incorrect:
- b. Dissolved Oxygen (DO) is typically measured using the Winkler Method or an electronic DO probe.
- c. Biochemical Oxygen Demand (BOD) is determined by measuring the difference in dissolved oxygen of a sample before and after a 5-day incubation period at 20°C.
- d. Chloride Ion is commonly determined using the Argentometric method (Mohr's method) involving titration with silver nitrate (AgNO3).
If the duration of a storm is 30 minutes and the depth of rainfall is 30 cm then what will be the rainfall intensity in cm /hr?
A)
10
B)
20
C)
60
D)
100
Rainfall intensity (I) is defined as the depth of rainfall (P) divided by the duration of the storm (t).
Given:
- Depth of rainfall (P) = 30 cm
- Duration of storm (t) = 30 minutes
To find the intensity in cm/hr, we must first convert the duration from minutes to hours:
- Duration (t) = 30 minutes / 60 minutes per hour = 0.5 hours
Now, apply the formula for intensity:
- Intensity (I) = Depth (P) / Duration (t)
- I = 30 cm / 0.5 hr
- I = 60 cm/hr
Therefore, the rainfall intensity is 60 cm/hr. Option (c) is correct. Options (a), (b), and (d) are mathematically incorrect based on the provided data.
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.
The phenomenon occuring in an open channel when a rapidly flowing stream abruptly changes to a slowly flowing stream causing a distinct rise of liquid surface, is
A)
hydraulic jump
B)
critical discharge
C)
water hammer
D)
critical velocity
A hydraulic jump occurs in an open channel when a flow transitions rapidly from a supercritical state (high velocity, low depth) to a subcritical state (low velocity, high depth). This abrupt change causes a sudden rise in the liquid surface and is characterized by significant turbulence and energy dissipation.
Why other options are incorrect:
- b. Critical discharge: This refers to the discharge that occurs when the flow is at critical depth; it is a state of flow, not a phenomenon of abrupt surface rise.
- c. Water hammer: This is a pressure surge or wave caused when a fluid in motion is forced to stop or change direction suddenly (typically in a closed conduit/pipe), not an open channel surface phenomenon.
- d. Critical velocity: This is the velocity at which the specific energy is minimum for a given discharge; it is a threshold value, not the phenomenon of the surface rise itself.
A pipe section has water flowing through it as shown below in figure. The diameter of a pipe at section 1 and 2 are 10 cm and 15 cm respectively. The velocity of the water flowing through the pipe at section 1 is 5 m/s. Determine the velocity at section 2.
A)
5m/s
B)
3.3 m/s
C)
2.2m/s
D)
0.8 m/s
Option C is correct: 2.2m/s.
Buckets and blades used in a turbine are used to
A)
Switch off the turbine
B)
To regulate the wind speed
C)
Alter the direction of water
D)
To regulate the temperature
In a hydraulic turbine, the primary purpose of the buckets (used in Pelton wheels) or blades (used in Francis or Kaplan turbines) is to extract kinetic energy from the flowing fluid. This is achieved by changing the momentum of the water. According to Newton's Second Law of Motion, a force is exerted on the turbine runner when the direction of the water flow is altered (deflected) by the blades/buckets. The change in the velocity vector (direction and magnitude) creates a torque on the shaft, which rotates the turbine to produce mechanical work.
Analysis of other options:
- Option (a) is incorrect because switching off the turbine is handled by guide vanes, needles, or intake valves, not the buckets/blades themselves.
- Option (b) is incorrect because the question refers to a turbine (generally implying hydraulic in this context), and blades do not regulate the speed of the wind; they capture energy from it.
- Option (d) is incorrect because turbines are designed for energy conversion, not for temperature regulation.
Determine the specific gravity of one litre (1 L) of a liquid which weighs 7 N.
A)
1
B)
0.7
C)
1.2
D)
0.5
Specific gravity = (weight of liquid) / (weight of equal volume of water). For 1 L of water, weight ≈ 9.81 N (about 10 N). Given liquid weight = 7 N for the same volume, specific gravity = 7/9.81 ≈ 0.71 ≈ 0.7. So option b is correct. Option a (1) would mean the liquid has the same density as water, which it does not. Option c (1.2) would mean it is denser than water, which is impossible here since 7 N < 9.81 N for the same volume. Option d (0.5) is too low and does not match the given weight ratio.
Immediately after a rain, when all the gravity water has drained down to the water table, a certain amount of water is retained on the surface of the soil grains by molecular attraction and is not readily available to the plants, is known as
A)
Capillary water
B)
Gravitational water
C)
Pellicular water
D)
Hygroscopic water
In soil physics and hydrology, soil water is classified based on how it is held within the soil matrix.
1. **Hygroscopic water (Option d):** This is the water held tightly as a thin film around soil particles by molecular attraction (adhesion). It is held with such high tension that it cannot be removed by gravity or extracted by plant roots. This matches the description in the question perfectly.
2. **Capillary water (Option a):** This is water held in the small pores (capillaries) of the soil by surface tension. Unlike hygroscopic water, capillary water is available for plant uptake.
3. **Gravitational water (Option b):** This is the water that occupies the larger pores and drains away under the influence of gravity shortly after a rain event. The question specifies the state 'after all the gravity water has drained down,' meaning this is not the answer.
4. **Pellicular water (Option c):** This is another term sometimes used for the thin film of water surrounding particles, but in the context of standard engineering and hydrological terminology for water unavailable to plants due to molecular attraction, 'Hygroscopic water' is the precise technical term.
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.