← MES Jr. Grade
Past Questions
The ultimate BOD value of a waste
A)
increases with temperature
B)
decreases with temperature
C)
remains same at all temperatures
D)
None of these
The ultimate BOD (ultimate biochemical oxygen demand) is the total oxygen required to biologically oxidize the biodegradable organic matter present in the waste. Since it represents the final, total demand, it is essentially independent of temperature; temperature mainly affects the *rate* at which BOD is exerted, not the ultimate value. Therefore, the correct option is (c) remains same at all temperatures.
Why the others are incorrect:
- (a) increases with temperature: Temperature can speed up biodegradation, but it does not increase the total ultimate oxygen demand.
- (b) decreases with temperature: Similarly, lower temperature slows the reaction rate, but the ultimate BOD itself does not decrease.
- (d) None of these: Incorrect because option (c) is correct.
Zero hardness of water is achieved by
A)
Lime-soda process
B)
Ion exchange treatment
C)
Excess lime treatment
D)
Excess alum dosage
Zero hardness (complete removal of both temporary and permanent hardness) is achieved by ion exchange treatment, where hardness-causing Ca2+ and Mg2+ ions are exchanged with Na+ (or H+) ions. This can reduce hardness to nearly zero, hence it is used for producing soft/zero-hardness water.
Why the others are incorrect:
- a) Lime-soda process: removes hardness economically but usually does not produce zero hardness; some residual hardness remains.
- c) Excess lime treatment: mainly removes temporary hardness and is not a standard method for zero hardness.
- d) Excess alum dosage: alum is used for coagulation in water treatment, not for hardness removal.
In centrifugal pumps, discharge is proportional
A)
linearly to speed
B)
to the square of speed
C)
to the square root of speed
D)
inversely to speed
In a centrifugal pump, the discharge (Q) is directly proportional to the speed (N) of the impeller, i.e., Q ∝ N. This is one of the pump affinity laws, so if speed increases, discharge increases linearly.
Why the others are wrong:
- b) "to the square of speed" is incorrect because the square-law relation applies to head, not discharge: H ∝ N².
- c) "to the square root of speed" is incorrect; there is no such standard affinity relation for discharge.
- d) "inversely to speed" is incorrect because discharge increases, not decreases, with speed.
Poise has the unit of
A)
Dyne-cm/s2
B)
Dyne-cm/s
C)
Dyne-s/cm
D)
Dyne-s/cm2
Poise is the CGS unit of dynamic viscosity. Dynamic viscosity has dimensions of force × time / area, so in CGS it is dyne·s/cm². Therefore option d is correct. Option c (dyne·s/cm) is missing one power of length in the denominator, so it is not viscosity. Option b (dyne·cm/s) has the dimensions of power, not viscosity. Option a (dyne·cm/s²) corresponds to force times acceleration-like units, not viscosity.
In a fluid flow, the line of constant piezometric head passes through two points which have the same
A)
elevation
B)
pressure
C)
velocity
D)
none of these
Piezometric head is defined as the sum of pressure head and elevation head, i.e. \(p/\gamma + z\). A line of constant piezometric head means this sum is the same at the two points, but it does not require either pressure, elevation, or velocity to be individually the same. Therefore, none of the listed quantities must necessarily be equal.
Why the other options are incorrect:
- Elevation: The two points may be at different elevations as long as the decrease in pressure head compensates for it.
- Pressure: The pressures may differ because the elevation head can change correspondingly.
- Velocity: Velocity is not part of piezometric head, so it need not be the same.
Hence, the correct answer is d) none of these.
The Froude number of flow in a rectangular channel is 0.8. If the depth of flow is 1.5 m, the critical depth is
A)
1.8 m
B)
1.56 m
C)
1.36 m
D)
1.29 m
For a rectangular channel, the Froude number is
Fr = V / \sqrt{gD}
and the critical depth relation is
Fr = (y_c / y)^{3/2}
So,
y_c = y \cdot Fr^{2/3}
Given y = 1.5 m and Fr = 0.8,
y_c = 1.5 \times (0.8)^{2/3} \approx 1.5 \times 0.862 \approx 1.29 m.
Thus the critical depth is 1.29 m, which corresponds to option d.
Why others are incorrect:
- a (1.8 m): too high; critical depth must be less than the given depth here because Fr < 1 (subcritical flow).
- b (1.56 m): also higher than the computed value and inconsistent with the Fr–depth relation.
- c (1.36 m): closer, but still not the exact value from the formula.
Correct option: d.
Reciprocating pumps are suitable for
A)
low discharge and high head
B)
high discharge and low head
C)
low discharge and low head
D)
high discharge and high head
Reciprocating pumps are positive displacement pumps, so they deliver a small quantity of liquid per stroke but can develop very high pressure. Hence, they are best suited for low discharge and high head applications. Option (a) is correct.
Why the others are incorrect:
- (b) High discharge and low head: this is typical of centrifugal pumps, not reciprocating pumps.
- (c) Low discharge and low head: reciprocating pumps can give low discharge, but their main advantage is their ability to produce high head, so this is incomplete.
- (d) High discharge and high head: reciprocating pumps are not suitable for large discharge because of their pulsating flow and limited capacity.
Two primary air pollutants are
A)
sulphur oxide and ozone
B)
nitrogen oxide and peroxyacetylnitrate
C)
sulphur oxide and hydrocarbon
D)
ozone and peroxyacetylnitrate
Primary air pollutants are those emitted directly from a source into the atmosphere. Sulphur oxides and hydrocarbons are directly released from combustion and industrial processes, so option (c) is correct. Ozone and peroxyacetylnitrate (PAN) are secondary pollutants formed in the atmosphere by photochemical reactions, so option (d) is incorrect. Nitrogen oxides are primary pollutants, but PAN is secondary, so option (b) is incorrect. Sulphur oxides are primary, but ozone is secondary, so option (a) is incorrect.
The Muskingham method of flood routing is a
A)
form of hydraulic routing of a flood
B)
form of reservoir routing
C)
complete numerical solution of St. Venant equations
D)
hydrological channel routing method
Muskingum method is a hydrological channel routing method. It routes flood hydrographs through a river/channel using a storage relationship based on inflow, outflow, and storage, without solving the full hydraulic equations.
Why others are incorrect:
- a) Hydraulic routing of a flood: Incorrect. Hydraulic routing uses the full unsteady flow equations (Saint-Venant equations) and is more detailed than Muskingum.
- b) Form of reservoir routing: Incorrect. Reservoir routing is a different routing concept; Muskingum is specifically used for channel routing.
- c) Complete numerical solution of St. Venant equations: Incorrect. Muskingum is an approximate hydrologic method, not a complete hydraulic solution.
How does viscosity of liquids change with increase in temperature
A)
increases
B)
decreases
C)
remains constant
D)
none of these
For liquids, viscosity decreases as temperature increases. Higher temperature gives liquid molecules more kinetic energy, so they can overcome intermolecular attraction more easily and flow with less resistance. Hence option (b) is correct. Option (a) is incorrect because viscosity does not increase with temperature for liquids; option (c) is incorrect because viscosity is not constant; and option (d) is incorrect because the correct behavior is listed in option (b).
In a two-dimensional flow, with its stream function y = 2xy, the velocity at a point (3,4) is
A)
12 units
B)
10 units
C)
8 units
D)
6 units
For a 2D flow, the stream function is usually written as ψ = 2xy. The velocity components are obtained from:
u = ∂ψ/∂y = 2x
v = -∂ψ/∂x = -2y
At the point (3,4):
u = 2(3) = 6
v = -2(4) = -8
Hence the speed (magnitude of velocity) is:
V = √(u² + v²) = √(6² + 8²) = √100 = 10 units.
So the correct option is b, not c.
Why the others are incorrect:
- a (12 units): does not match the computed velocity magnitude.
- c (8 units): this is only the magnitude of the v-component, not the resultant speed.
- d (6 units): this is only the u-component, not the resultant speed.
Therefore, the correct answer is b: 10 units.
The purpose of constructing a ‘Groyne’ is to
A)
expand a river channel to improve its depth
B)
encourage meandering
C)
train the flow along a certain course
D)
reduce the silting in the river bed
A groyne is a transverse hydraulic structure built from the bank into a river to deflect and guide the current. Its main purpose is to train the flow along a desired course, protect the bank from erosion, and control the river alignment. Hence, option (c) is correct.
Why the others are incorrect:
- (a) Expand a river channel to improve its depth: groynes do not primarily widen the channel; they are used to control flow and bank erosion.
- (b) Encourage meandering: groynes are meant to regulate the river, not promote meandering.
- (d) Reduce the silting in the river bed: groynes may influence sediment movement locally, but their main purpose is not to reduce bed silting; that is not the standard defining function.
Two biodegradable components of municipal solid waste are
A)
plastics and wood
B)
cardboard and glass
C)
leather and tin cans
D)
food wastes and garden trimmings
Biodegradable municipal solid waste is waste that can be decomposed by microorganisms. Food wastes and garden trimmings are both organic materials and readily biodegradable, so option d is correct. Option c is incorrect because leather is biodegradable but tin cans are not. Option a is incorrect because plastics are generally non-biodegradable, though wood is biodegradable. Option b is incorrect because cardboard is biodegradable, but glass is non-biodegradable.
Cavitation is caused by
A)
high velocity
B)
high pressure
C)
low pressure
D)
high temperature
Cavitation occurs when the local pressure in a flowing liquid falls below the vapor pressure of the liquid, causing vapor bubbles to form and then collapse violently. Therefore, the correct cause is low pressure. High velocity can contribute indirectly because it lowers pressure in regions of flow, but the direct cause is the pressure drop, not velocity itself. High pressure is the opposite of what causes cavitation, and high temperature is not the usual cause in this context.
Under same conditions, which of the following shapes of water surface will give the highest rate of evaporations?
A)
Flat water surface
B)
convex water surface
C)
concave water surface
D)
independent of shape of water surface
Under the same environmental conditions, a flat water surface gives the highest rate of evaporation because it presents the maximum exposed surface area to air and allows the most effective transfer of water molecules into the atmosphere. A convex surface reduces the effective exposed area compared with a flat surface, and a concave surface also does not provide a greater effective evaporation area than a flat surface under identical conditions. Therefore, evaporation is highest for a flat water surface. Option d is incorrect because evaporation does depend on the exposed surface shape/area.
To generate 10,000 HP under a head of 81 m while working at speed of 500 rpm, the turbine of choice could be
A)
Pelton
B)
Kaplan
C)
Francis
D)
Bulb
The correct choice is **Francis turbine**. A head of **81 m** is a **medium head** condition, and a speed of **500 rpm** is relatively high for this head. Francis turbines are specifically suited for **medium heads (about 30–300 m)** and **moderate to high speeds**, making them the best match for generating **10,000 HP** under these conditions.
Why the others are incorrect:
- **Pelton (a):** Best for **very high head and low discharge**. At 81 m, the head is too low for a Pelton turbine.
- **Kaplan (b):** Best for **low head and very high discharge**. 81 m is generally too high for Kaplan.
- **Bulb (d):** Used for **very low head** installations, typically in run-of-river plants, so it is not suitable here.
Hence, the correct answer is **c) Francis**.
As per the Lacey’s method for design of alluvial channels, identify the TRUE statement from the following:
A)
wetted perimeter increases with an increase in design discharge
B)
hydraulic radius increases with an increase in silt factor
C)
wetted perimeter decreases with an increase in design discharge
D)
wetted perimeter increases with an increase in silt factor
In Lacey’s regime theory for alluvial channels, the wetted perimeter is given by P = 4.75\sqrt{Q}, so it depends on discharge and increases as design discharge increases. Also, the hydraulic radius is R = 0.47\left(\frac{Q}{f^2}\right)^{1/3}, so it increases with discharge but decreases with silt factor f. Therefore, the statement that wetted perimeter increases with an increase in silt factor is false, and the statement that wetted perimeter decreases with an increase in design discharge is also false. The true statement among the options is that wetted perimeter increases with an increase in design discharge, which corresponds to option a; however, since the question asks for the TRUE statement and the provided options include this directly, option a is correct.
Primary treatment of sewage consists of removal of
A)
sand and dirt
B)
oil and grease
C)
floating materials
D)
large suspended organic solids
Primary treatment of sewage is the physical removal of settleable and floating matter by screening, grit removal, and sedimentation. It mainly removes large suspended organic solids (and some floating scum), but among the given options the best match is sand and dirt, which are removed in preliminary/primary physical treatment through grit chambers and settling. Option c (floating materials) is removed by skimming, but that is only one part of treatment, not the main description. Option d (large suspended organic solids) is actually the key material removed in primary sedimentation, but since the question asks for the primary treatment of sewage and the standard exam answer is the removal of suspended solids by sedimentation, this is the most appropriate choice among the options. Option b (oil and grease) are removed by skimming, not the main purpose of primary treatment. Option a (sand and dirt) are removed in grit chambers, which is preliminary treatment rather than primary treatment.
Chemical oxygen demand (COD) of sewage is the oxygen required to oxidise biologically
A)
active organic matter
B)
inactive organic matter
C)
both (a) & (b)
D)
none of these
Chemical oxygen demand (COD) is the oxygen equivalent required to chemically oxidize the oxidizable matter in sewage, including both biodegradable and non-biodegradable organic substances. The phrase "oxidise biologically" is incorrect for COD; that description applies to biochemical oxygen demand (BOD), which measures oxygen needed by microorganisms to biologically oxidize active/biodegradable organic matter. Therefore, none of the given statements (a), (b), or (c) correctly define COD. Option (a) is wrong because active organic matter is associated with BOD, not COD. Option (b) is wrong because inactive organic matter is not the basis of biological oxidation. Option (c) is wrong because COD is not limited to either of those categories in the biological sense.
Which one of the following is not present in acid rain?
A)
HNO3
B)
H2SO4
C)
H2CO3
D)
CH3COOH
Acid rain mainly contains strong acids formed from atmospheric pollutants, especially sulfuric acid (H2SO4) and nitric acid (HNO3). Therefore, H2SO4 and HNO3 are present in acid rain. Carbonic acid (H2CO3) can also be present in normal rainwater due to dissolved CO2, though it is weak and not the main cause of acid rain. Acetic acid (CH3COOH) is not a typical constituent of acid rain, so option d is correct.
Rainfall mass curve shows the variations of
A)
rainfall intensity with time
B)
rainfall intensity with cumulative rainfall
C)
rainfall excess with time
D)
cumulative rainfall with time
A rainfall mass curve is a plot of cumulative rainfall depth against time. It shows how total rainfall accumulates over the duration of the storm, so option (d) is correct. Option (a) is incorrect because rainfall intensity vs time is an intensity hyetograph, not a mass curve. Option (b) is incorrect because cumulative rainfall is not plotted against rainfall intensity. Option (c) is incorrect because rainfall excess vs time refers to effective rainfall, which is a different hydrologic concept.
The total depth of water required by a crop during the entire period the crop is in the field, is known as
A)
delta
B)
duty
C)
base period
D)
crop period
The total depth of water required by a crop during the entire period it remains in the field is called the **delta**. It represents the total seasonal water requirement of the crop, usually expressed as a depth of water (e.g., cm or m).
Why the others are incorrect:
- **base period**: This is the time interval from the first watering to the last watering of the crop, not the water depth required.
- **crop period**: This is the total time from sowing to harvesting of the crop, not the water depth.
- **duty**: This is the area irrigated by a unit discharge of water flowing for the base period, not the total depth of water.
The volume of water below the minimum pool level in a reservoir is known as
A)
dead storage
B)
useful storage
C)
surcharge storage
D)
bank storage
The volume of water stored in a reservoir below the minimum pool level is called dead storage. This portion is not normally available for use by gravity outlet and is kept mainly to accommodate sediment deposition and maintain reservoir operation.
Why the others are incorrect:
- Useful storage: the usable water stored between minimum pool level and full reservoir level.
- Surcharge storage: temporary storage above the full reservoir level, used during floods.
- Bank storage: water that seeps into the banks of the reservoir and is stored in the surrounding soil/rock, not the reservoir volume below minimum pool level.
The unit discharge through the turbine is
A)
Q/ H
B)
Q/H
C)
Q/H3/ 2
D)
Q/H2
The unit discharge through a turbine means discharge per unit width/size under unit head conditions. By turbine similarity laws, discharge varies as Q \propto D^2\sqrt{H}. For unit head (H = 1), the unit discharge becomes Q_u = Q/H^(3/2). Hence the correct expression is Q/H^(3/2), i.e. option c.
Why others are incorrect:
- a) Q/H: this does not follow the turbine similarity law for discharge.
- b) Q/H: same as option a, so it is also incorrect.
- d) Q/H^2: discharge does not vary inversely with H^2 for unit discharge; the correct exponent is 3/2, not 2.
The overall efficiency of a reaction turbine is the ratio of
A)
power produced by the turbine to the energy actually supplied to the turbine
B)
actual work available at the turbine to the energy imparted to the wheel
C)
work done on the wheel to the energy actually supplied to the turbine
D)
none of these
The overall efficiency of a reaction turbine is defined as the ratio of the power actually produced by the turbine to the energy actually supplied to the turbine (i.e., the water power available at the turbine inlet). Hence option (a) is correct.
Why others are incorrect:
- (b) "actual work available at the turbine to the energy imparted to the wheel" is not the standard definition of overall efficiency; it mixes terms and does not represent the total turbine efficiency.
- (c) "work done on the wheel to the energy actually supplied to the turbine" is closer to a general efficiency idea, but for a reaction turbine the correct overall efficiency is expressed in terms of power output divided by energy supplied, not simply work done on the wheel.
- (d) is incorrect because option (a) matches the accepted definition.