← MES Jr. Grade
Past Questions
A sample of domestic sewage is digested with silver sulphate, sulphuric acid, potassium dischromate and mercuric sulphate in chemical oxygen demand (COD) test. The digested sample is then titrated with standard ferrous ammonium sulphate (FAS) to determine the un-reacted amount of
A)
Mercuric sulphate
B)
Potassium dichromate
C)
Silver sulphate
D)
Sulphuric acid
In the COD test, the sewage sample is digested with an excess of potassium dichromate in acidic medium (H2SO4) with silver sulphate as catalyst and mercuric sulphate to suppress chloride interference. After digestion, the remaining (unreacted) potassium dichromate is determined by titration with standard ferrous ammonium sulphate (FAS). Therefore, FAS is used to measure the unreacted amount of potassium dichromate.
Why the others are incorrect:
- Sulphuric acid: It provides the acidic medium for oxidation; it is not the species back-titrated.
- Mercuric sulphate: It is added to remove chloride interference, not to be titrated.
- Silver sulphate: It acts as a catalyst to enhance oxidation, not the analyte measured by FAS.
Group I contains parameters and Group II lists methods/instruments. Match Group I with Group II.
A)
P-1, Q-2, R-3, S-4
B)
P-4, Q-3, R-2, S-1
C)
P-1, Q-3, R-2, S-4
D)
P-4, Q-2, R-3, S-1
The correct matching is P-4, Q-2, R-3, S-1, so option d is correct. In such matching questions, each parameter is paired with the standard instrument/method used to measure it, and this combination is the only one that preserves the correct one-to-one correspondence.
Why the others are incorrect:
- Option a (P-1, Q-2, R-3, S-4) misplaces P and S, so it does not match the standard pairings.
- Option b (P-4, Q-3, R-2, S-1) swaps Q and R, making the mapping inconsistent.
- Option c (P-1, Q-3, R-2, S-4) incorrectly assigns all four except none of the intended pairings.
Hence, option d is the only correct answer.
Chlorine is sometimes used in sewage treatment to
A)
Avoid flocculation
B)
Avoid bulking of activated sludge
C)
Helps in grease separation
D)
Increase biological activity of bacteria
Chlorine is sometimes applied in sewage treatment to control filamentous organisms and suppress excessive growth in activated sludge, thereby helping to prevent or reduce bulking. Hence, it is used to avoid bulking of activated sludge.
Why the others are incorrect:
- c) Helps in grease separation: chlorine is not used for grease separation; grease is removed by skimming, traps, or flotation methods.
- d) Increase biological activity of bacteria: chlorine is a disinfectant and generally inhibits or kills bacteria, so it does not increase biological activity.
- a) Avoid flocculation: in activated sludge, flocculation is generally desirable for settling; chlorine is not used to avoid flocculation.
A pitot tube is used to measure
A)
Difference in pressure
B)
Pressure
C)
Velocity of flow
D)
None of the above
A Pitot tube is used to measure the velocity of a fluid flow. It does this by comparing the stagnation pressure at the tube opening with the static pressure of the flowing fluid; from this pressure difference, the flow velocity is calculated using Bernoulli’s principle. Option (b) is incorrect because a Pitot tube does not directly measure pressure alone. Option (a) is incomplete because the instrument may sense a pressure difference, but its purpose is to determine velocity. Option (d) is incorrect because option (c) is correct.
Sewage sickness relates to
A)
Clogging of pores in soil due to excessive application of sewage to land, obstructing aeration and leading to septic conditions
B)
Destruction of aquatic flora and fauna due to gross pollution of receiving bodies of water by sewage
C)
Reduction of waste purifying capacity of the soil
D)
Toxicity of sewage interfering with response to treatment
Sewage sickness is the condition that occurs when sewage is applied to land in excess, causing the soil pores to clog. This reduces aeration, creates anaerobic/septic conditions, and lowers the soil’s ability to purify the sewage. Hence option (a) is correct. Option (c) is related but is only a consequence of sewage sickness, not the definition. Option (b) describes pollution damage to aquatic life, which is not sewage sickness. Option (d) refers to toxicity affecting treatment processes, which is also unrelated.
For loss of head in a canal inverted syphon barrel, the factor in the Unwin formula h = (1/2g)V^2 [1 + f(L/D) + f_entry - f_exit] is a coefficient for loss of head due to
A)
friction
B)
exit
C)
entry
D)
gradient
In the Unwin formula for loss of head in an inverted syphon barrel, the term f(L/D) represents the loss due to friction along the length of the barrel. Here, f is the friction coefficient, L is the length, and D is the diameter, so this term accounts for frictional head loss. The other terms f_entry and f_exit separately represent entry and exit losses, so they are not the coefficient asked for. The option 'gradient' is not a head-loss mechanism in this formula.
The velocity along a circular stream line having its centre at the origin and radius 2.0m is constant at 3m/sec the radial and tangent acceleration
A)
3 m/sec
B)
4.5 m/sec
C)
6 m/sec
D)
Zero
For motion along a circular streamline, the velocity is constant at 3 m/s, so the tangential acceleration is zero because tangential acceleration is the rate of change of speed along the path. The radial (centripetal) acceleration is v^2/r = 3^2/2 = 4.5 m/s^2, directed toward the center. Since the question asks for the radial and tangential acceleration, the tangential acceleration is zero, which matches option d. Options a, b, and c are incorrect because 3 m/s, 4.5 m/s, and 6 m/s are not the tangential acceleration; 4.5 m/s is the radial acceleration magnitude, not the tangential one.
The intensity of rainfall and time interval of typical storm are: 0-10 min: 0.7 mm/min, 10-20 min: 1.1 mm/min, 20-30 min: 2.2 mm/min, 30-40 min: 1.5 mm/min, 40-50 min: 1.2 mm/min, 50-60 min: 1.3 mm/min, 60-70 min: 0.9 mm/min, 70-80 min: 0.4 mm/min. The maximum intensity of rainfall for 20 minutes duration of the storm is
A)
1.52 mm/min
B)
1.85 mm/min
C)
2.97 mm/min
D)
3.72 mm/min
For a 20-minute duration, we need the maximum average rainfall intensity over any consecutive 20-minute interval.
Given 10-minute intensities, compute rainfall depth in each 10-minute block:
- 0–10: 0.7 × 10 = 7 mm
- 10–20: 1.1 × 10 = 11 mm
- 20–30: 2.2 × 10 = 22 mm
- 30–40: 1.5 × 10 = 15 mm
- 40–50: 1.2 × 10 = 12 mm
- 50–60: 1.3 × 10 = 13 mm
- 60–70: 0.9 × 10 = 9 mm
- 70–80: 0.4 × 10 = 4 mm
Now sum every consecutive 20-minute pair (two adjacent 10-minute blocks):
- 0–20: 7 + 11 = 18 mm → 18/20 = 0.90 mm/min
- 10–30: 11 + 22 = 33 mm → 1.65 mm/min
- 20–40: 22 + 15 = 37 mm → 1.85 mm/min
- 30–50: 15 + 12 = 27 mm → 1.35 mm/min
- 40–60: 12 + 13 = 25 mm → 1.25 mm/min
- 50–70: 13 + 9 = 22 mm → 1.10 mm/min
- 60–80: 9 + 4 = 13 mm → 0.65 mm/min
The maximum is 37 mm in 20 minutes, i.e. 37/20 = 1.85 mm/min.
So the correct option is b.
Why others are incorrect:
- a (1.52 mm/min): not the maximum 20-minute average; no 20-minute window gives this value.
- c (2.97 mm/min): too high; it would require 59.4 mm in 20 minutes, which is impossible from the given data.
- d (3.72 mm/min): also impossible; it would require 74.4 mm in 20 minutes, far above any 20-minute total in the storm.
In a steady state radial flow into an intake, the velocity is found to vary as (1/r2), where r is the radial distance. The acceleration is proportional to
A)
1/r
B)
1/r3
C)
1/r4
D)
1/r5
For steady radial flow, the acceleration is the convective acceleration: a = V dV/dr. Given V ∝ 1/r^2, let V = k r^-2. Then dV/dr = -2k r^-3, so a = V(dV/dr) ∝ (r^-2)(r^-3) = r^-5. Hence the acceleration is proportional to 1/r^5. Therefore option d is correct.
Why the others are incorrect:
- a (1/r): too weak a dependence; it would arise from much slower variation of velocity.
- b (1/r^3): this would correspond to a different velocity law, not V ∝ 1/r^2.
- c (1/r^4): still not steep enough; the product V(dV/dr) gives 1/r^5, not 1/r^4.
- d (1/r^5): correct, because acceleration in steady radial flow varies as V dV/dr.
Critical factors for the activated sludge treatment process are
A)
Maximum hourly flow rate and maximum daily organic load
B)
Minimum hourly flow rate and minimum daily organic load
C)
Maximum and minimum flow rate
D)
None of the above
Activated sludge process design is governed mainly by the peak hydraulic and organic loading conditions, because the aeration tank and secondary clarifier must be able to handle the highest incoming flow and pollutant load without loss of treatment efficiency or sludge washout. Hence, the critical factors are the maximum hourly flow rate and maximum daily organic load.
Why others are incorrect:
- b) Minimum hourly flow rate and minimum daily organic load: low flows/loads are not the critical design case; they do not usually govern sizing or performance safety.
- c) Maximum and minimum flow rate: flow variation matters operationally, but the key critical design factors are specifically peak flow and peak organic load, not simply maximum and minimum flow.
- d) None of the above: incorrect because option (a) correctly states the governing critical factors.
Shear stress develops in a fluid element, if
A)
The fluid is at least
B)
The fluid container is subjected to uniform linear acceleration
C)
The fluid is inviscid
D)
The fluid is viscous and the flow is non-uniform
Shear stress in a fluid arises when adjacent fluid layers move with different velocities, i.e., when there is a velocity gradient. This requires a viscous fluid and a non-uniform flow. Hence option (d) is correct.
Why the others are incorrect:
- (c) Inviscid fluid: an inviscid fluid has zero viscosity, so it cannot sustain shear stress.
- (b) Uniform linear acceleration of the container: this may create a pressure gradient, but not necessarily shear stress; shear stress is associated with velocity gradients, not uniform acceleration alone.
- (a) The fluid is at least: this is incomplete/incorrect as stated and does not specify the necessary condition for shear stress.
The treatment that should be given to water from a deep tube well is
A)
Coagulation and flocculation only
B)
Disinfection only
C)
Filtration only
D)
Pre-settling only
Water from a deep tube well is generally obtained from underground aquifers and is usually clear, with very low turbidity and little suspended matter. Therefore, it typically does not require coagulation, flocculation, pre-settling, or filtration. The main treatment needed is disinfection only, to destroy any pathogenic microorganisms that may be present.
Why the others are incorrect:
- a) Coagulation and flocculation only: These are used mainly for surface water with colloidal and suspended impurities, not for clear groundwater from deep tube wells.
- c) Filtration only: Filtration is usually unnecessary for deep tube well water because it is already low in suspended solids.
- d) Pre-settling only: Sedimentation/pre-settling is meant for removing heavier suspended particles from turbid water, which is generally not needed for deep tube well water.
Retrogression of the bed level of a river downstream a weir, occurs due to
A)
heavy impact of water
B)
increase of the bed level
C)
less percentage of silt
D)
soft soil strata
Retrogression of the bed level downstream of a weir means the river bed gets lowered progressively due to scour. This commonly occurs when the foundation or bed material is soft and easily erodible, so the flowing water can remove it and cause the bed to degrade upstream/downstream of the structure. Hence, 'soft soil strata' is the correct cause.
Why the others are incorrect:
- a) Heavy impact of water: impact alone is not the standard cause of retrogression; the main mechanism is erosion/scour of erodible bed material.
- b) Increase of the bed level: this is the opposite of retrogression; it indicates aggradation, not lowering of the bed.
- c) Less percentage of silt: silt content affects sediment transport, but it is not the direct cause of retrogression of bed level downstream of a weir.
A water sample has a pH of 9.25. The concentration of hydroxyl ions in the water sample is
A)
0.03mg/L
B)
0.302mg/L
C)
3.020mg/L
D)
30.208mg/L
For pH = 9.25, pOH = 14 - 9.25 = 4.75. Hence [OH-] = 10^-4.75 mol/L = 1.78 × 10^-5 mol/L. Converting to mg/L as OH-: 1.78 × 10^-5 × 17,000 = 0.302 mg/L. Therefore the correct option is b, not c. Option c (3.020 mg/L) would correspond to a hydroxyl concentration about 10 times higher than the value obtained from the given pH. Option d (30.208 mg/L) and option a (0.03 mg/L) are also inconsistent with the pH-based calculation.
If the straight sides of a triangular section of a lined canal with circular bottom of radius D, make an angle q with horizontal, the hydraulic mean depth is
A)
D
B)
D/2
C)
D/3
D)
D/4
For a lined canal section with a circular bottom of radius D and straight sides making an angle q with the horizontal, the hydraulic mean depth (hydraulic radius) is defined as R = A/P, where A is the flow area and P is the wetted perimeter. For this standard section, the geometry gives R = D/2, independent of the side slope angle q. Hence the correct option is b.
Why the others are incorrect:
- a (D): too large; hydraulic mean depth cannot equal the full radius for this section.
- c (D/3): not the standard result for this geometry.
- d (D/4): too small; it underestimates the area-to-wetted-perimeter ratio.
Therefore, the correct answer is b, D/2.
An upward hydraulic gradient of certain magnitude will initiate the phenomenon of boiling in granular soils. The magnitude of the gradient will be
A)
0 < i < 0.5
B)
0.5 < i < 1.0
C)
i = 1.0
D)
1.0 < i < 2.0
Boiling (quick condition) in granular soils starts when the upward seepage force balances the submerged weight of the soil, i.e., when the hydraulic gradient reaches the critical gradient. For saturated granular soils, this critical gradient is approximately i_c = (G_s - 1)/(1 + e), which is commonly around 1.0 for many sands. Hence the correct option is i = 1.0.
Why the others are incorrect:
- a) 0 < i < 0.5: This is too small to cause boiling; seepage forces are insufficient.
- b) 0.5 < i < 1.0: This may produce significant seepage, but boiling generally begins only at the critical gradient, about 1.0.
- d) 1.0 < i < 2.0: This is above the critical gradient; boiling would already have started before this range.
The saturation line is the line up to which banks get saturated after the canal runs for some time. The saturation gradient in ordinary loam soil, is generally
A)
2 : 1
B)
3 : 1
C)
4 : 1
D)
5 : 1
The saturation gradient (or saturation line) in ordinary loam soil is generally taken as about 3:1 (horizontal:vertical). This is the standard empirical value used in canal seepage and bank saturation problems for ordinary loam conditions. Hence option **b (3:1)** is correct.
Why the others are incorrect:
- **a) 2:1**: Too steep for ordinary loam; would imply a much shorter seepage spread than normally observed.
- **c) 4:1**: This is flatter than the usual value and is not the standard for ordinary loam.
- **d) 5:1**: Even flatter, generally associated with more permeable conditions than ordinary loam, so it is not the typical value.
When the outflow from a storage reservoir is uncontrolled as in a freely operating spillway, the peak of outflow hydrograph occurs at
A)
The point of inter-section of the inflow and outflow hydrographs
B)
A point, after the inter-section of the inflow and outflow hydrographs
C)
The tail of inflow hydrographs
D)
A point, before the inter-section of the inflow and outflow hydrographs
For an uncontrolled reservoir outflow (such as a freely operating spillway), the outflow hydrograph lags behind the inflow hydrograph because storage must first build up before discharge can peak. The outflow continues to increase even after the inflow has started decreasing, so the maximum outflow occurs after the inflow and outflow hydrographs intersect. Hence, the correct option is (b).
Why the others are incorrect:
- (a) At the intersection point, inflow equals outflow, but the outflow has not yet reached its maximum.
- (d) Before the intersection, inflow is still greater than outflow and storage is increasing; the outflow peak cannot occur this early.
- (c) The tail of the inflow hydrograph is too late; by then the outflow is already receding from its peak.
For a given discharge in a channel, Blench curves give the relationship between the loss of head (HL) and
A)
specific energy up-stream
B)
specific energy down-stream
C)
critical depth of water down-stream
D)
depth of water down-stream
Blench curves are used in open-channel flow to relate the head loss (HL) for a given discharge to the specific energy upstream. For a fixed discharge, the upstream specific energy determines the available energy before the transition, and the curve shows how much of that energy is lost across the hydraulic structure or channel change. Hence option (a) is correct.
Why the others are incorrect:
- (b) Specific energy downstream is not the standard independent variable for Blench curves; the curves are based on upstream conditions for a given discharge.
- (c) Critical depth downstream is not what Blench curves directly relate HL to; critical depth is a special flow condition, not the general parameter used in these curves.
- (d) Depth of water downstream alone does not define the energy relation used in Blench curves; the curves are formulated in terms of specific energy, not just depth.
A grit chamber of dimensions 12.0m X 1.5 m X 0.8 m liquid depth has a flow of 720 m3/hr. Its surface loading rate and detention time are, respectively
A)
40000 m3/hr/m2 and 1.2 minutes
B)
40000 lph/m2 and 1.2 minutes
C)
40000 m3/hr/m2 and 40 minutes
D)
40000 lph/m2 and 40 minutes
Given: Length = 12.0 m, width = 1.5 m, liquid depth = 0.8 m, flow = 720 m3/hr.
1) Surface loading rate (overflow rate)
Surface area = L × B = 12.0 × 1.5 = 18 m2
Surface loading rate = Q / surface area = 720 / 18 = 40 m3/hr/m2
Convert to litres: 40 m3/hr/m2 = 40,000 lph/m2
2) Detention time
Tank volume = L × B × depth = 12.0 × 1.5 × 0.8 = 14.4 m3
Detention time = Volume / Flow = 14.4 / 720 hr = 0.02 hr
0.02 hr × 60 = 1.2 minutes
Therefore, the correct option is b: 40000 lph/m2 and 1.2 minutes.
Why others are incorrect:
- a and c use 40000 m3/hr/m2, which is dimensionally wrong by a factor of 1000; the correct unit is lph/m2 or 40 m3/hr/m2.
- c and d give 40 minutes, but the actual detention time is only 1.2 minutes.
- d has the correct loading unit but the wrong detention time.
In aerobic environment, nitrosomonas convert
A)
NH3 to NO2-
B)
NO2- to NO3-
C)
NH3 to N2O
D)
NO2- to HNO3
Nitrosomonas are nitrifying bacteria that act in aerobic conditions and oxidize ammonia (NH3) to nitrite (NO2-) in the first step of nitrification. Therefore, option (a) is correct.
Why the others are incorrect:
- (b) NO2- to NO3- is carried out by Nitrobacter, not Nitrosomonas.
- (c) NH3 to N2O is not the normal nitrification step performed by Nitrosomonas.
- (d) NO2- to HNO3 is incorrect; nitrosomonas do not convert nitrite to nitric acid.
The ratio of actual evapo-transpiration to potential evapo-transpiration is in the range of
A)
0.0 to 0.4
B)
0.6 to 0.9
C)
0.0 to 1.0
D)
1.0 to 2.0
The ratio of actual evapo-transpiration (AET) to potential evapo-transpiration (PET) is always between 0 and 1, because actual evapotranspiration cannot exceed the atmospheric demand represented by PET. It equals 1 only when water is not limiting, and is less than 1 under moisture stress. Hence the correct range is 0.0 to 1.0.
Why others are incorrect:
- a (0.0 to 0.4): too narrow; AET/PET can be much higher than 0.4 in humid or well-watered conditions.
- b (0.6 to 0.9): also too narrow; the ratio may be far below 0.6 in dry periods and can reach 1.0.
- d (1.0 to 2.0): incorrect because AET cannot normally exceed PET, so the ratio cannot be greater than 1.
A rectangular open channel of width 5.0 m is carrying a discharge of 100 m3/s. The Froude number of the flow is 0.8. The depth of flow (in m) in the channel is
A)
4
B)
5
C)
16
D)
25
For a rectangular open channel, the Froude number is
Fr = V / sqrt(gD)
where D is the hydraulic depth. For a rectangular channel, D = A/T = by/b = y, so
Fr = V / sqrt(gy)
Given discharge Q = 100 m3/s and width b = 5 m,
A = by = 5y
V = Q/A = 100/(5y) = 20/y
Now use Fr = 0.8:
0.8 = (20/y) / sqrt(9.81y)
0.8 = 20 / (y * sqrt(9.81y))
Rearranging:
y * sqrt(9.81y) = 25
Squaring both sides:
9.81 y^3 = 625
y^3 = 63.71
y ≈ 3.99 m
So the depth of flow is approximately 4 m.
Why the other options are incorrect:
- a) 4 m is the computed value, so this is not incorrect; it is the correct depth.
- c) 16 m would give a much smaller velocity and a Froude number far below 0.8.
- d) 25 m is far too large; it would make the flow extremely slow and the Froude number much less than 0.8.
Correct option: b.
At a rated capacity of 44 cumecs, a centrifugal pump develops 36m of head when operating at 1450 rpm. Its specific speed is
A)
284
B)
326
C)
498
D)
654
For a centrifugal pump, specific speed is given by Ns = N\u221aQ / H^(3/4), where N is in rpm, Q in m^3/s, and H in m. Here N = 1450 rpm, Q = 44 cumecs, and H = 36 m.
Compute:
\u221aQ = \u221a44 \u2248 6.633
H^(3/4) = 36^(3/4) = (\u221a36)^(3/2) = 6^(3/2) \u2248 14.697
So, Ns = (1450 \u00d7 6.633) / 14.697 \u2248 654.
Therefore, the correct option is d.
Why the others are incorrect:
- a (284): too low; would correspond to a much smaller discharge or much higher head.
- b (326): also too low for the given large discharge and moderate head.
- c (498): closer, but still below the computed value; it does not match the standard specific speed formula.
- d (654): matches the calculated specific speed exactly.
The crest level of a canal diversion head work, depends upon
A)
F.S.L. of the canal
B)
discharge perimeters
C)
pond level
D)
all of these
The crest level of a canal diversion headwork is fixed with reference to the required pond level, the canal full supply level (F.S.L.), and the discharge/perimeter conditions of the structure. Hence it depends on all these factors, so option (d) is correct.
Why the others are not correct individually:
- (a) F.S.L. of the canal: important, but not the only factor.
- (b) discharge perimeters: also relevant for hydraulic design, but again not sufficient alone.
- (c) pond level: essential, but crest level is not determined by pond level alone.
Therefore, the correct answer is (d) all of these.