← MES Jr. Grade
Past Questions — 2015
According to World Health Organization (WHO), what is the standard permissible limit of pH in water?
A)
7-8.5
B)
Less than 6.5
C)
6.5-9.2
D)
7
According to WHO drinking-water guidelines, the acceptable pH range is generally around 6.5 to 8.5, and many exam keys in this context use the range 6.5–9.2 as the standard permissible limit mentioned in such questions. Among the given options, (c) best matches the expected WHO-based permissible range. Option (d) is only a single value, (b) is too narrow/incorrect, and (a) does not match the standard cited in exam references.
Dynamic similarity between model and prototype means
A)
The similarity of forces
B)
The similarity of shape
C)
The similarity of motion
D)
None of these
Dynamic similarity means that the ratios of all corresponding forces acting on the model and prototype are the same. This ensures that the flow behavior is similar in both. Geometric similarity refers to similarity of shape, and kinematic similarity refers to similarity of motion. Therefore, the correct answer is the similarity of forces.
As per the middle third rule, for concrete gravity dams with base width b, eccentricity e should be less than or equal to (b/6) for
A)
No sliding to occur
B)
No overturning to occur
C)
No tension to develop
D)
None of these
For a concrete gravity dam, the middle third rule states that the resultant force at the base should lie within the middle third of the base. This means the eccentricity must satisfy e \u2264 b/6. When this condition is met, the stress distribution at the base remains compressive throughout, so no tensile stress develops. Sliding and overturning are checked by separate stability criteria, not by the middle third rule.
In a homogenous embankment type of earth dam, the phreatic line is kept well within the body of the dam by
A)
Providing proper w/s slope protection
B)
Providing proper d/s slope protection
C)
Suitably increasing the top width
D)
Providing horizontal drainage filter at the d/s face
In a homogeneous embankment earth dam, the phreatic line (seepage line) is controlled by providing drainage on the downstream side. A horizontal drainage filter at the downstream face lowers the seepage line and keeps it well within the body of the dam, improving stability and preventing sloughing. Water/slope protection mainly prevents erosion, and increasing top width does not control the phreatic line effectively.
A pelton turbine has following specifications: Shaft power = 9560 kw, Head = 350m, Overall efficiency = 85%, wheel dia = 15.8 cm, Co-eff of velocity = 0.985. No of jet is
A)
2
B)
1
C)
0
D)
3
For a Pelton wheel, the speed ratio is usually taken as u/v1 ≈ 0.46, where v1 = Cv√(2gH).
Given:
- H = 350 m
- Cv = 0.985
- Wheel diameter D = 15.8 cm = 0.158 m
First find jet velocity:
v1 = 0.985 × √(2 × 9.81 × 350) ≈ 0.985 × 82.87 ≈ 81.6 m/s
Wheel peripheral speed:
u = 0.46 × 81.6 ≈ 37.5 m/s
Now speed relation:
u = πDN/60
So N ≈ 60u/(πD) ≈ 60 × 37.5 /(π × 0.158) ≈ 4530 rpm
This is a very high speed for a Pelton turbine, so to keep the runner diameter as given, the turbine would operate with a single jet. Also, for high-head Pelton turbines with small wheel diameter, the usual arrangement is one jet.
Hence, the number of jets is 1.
Option (b) is correct.
The following operations are generally employed for tunnelling in hard rock: 1. Removing ground water, 2. Loading holes and firing the explosive 3. Setting up and drilling 4. Grouting and lining 5. Removing muck 6. Ventilation and removing explosion dust. The correct sequence is
A)
412365
B)
346152
C)
542136
D)
326514
In hard-rock tunnelling, the usual cycle is: first set up and drill the holes (3), then load the holes and fire the explosive (2), after blasting provide ventilation and remove explosion dust (6), then remove the muck (5), followed by removing ground water if present (1), and finally grouting and lining (4). Hence the correct sequence is 326514.
In urban air pollution, the most poisonous gas is supposed to be carbon monoxide. It is hazardous because it
A)
Affects our sense of smell
B)
Is carcinogenic in nature
C)
Combines with haemoglobin
D)
Causes blindness
Carbon monoxide is highly poisonous because it binds strongly with haemoglobin in blood to form carboxyhaemoglobin. This reduces the blood’s ability to carry oxygen, leading to oxygen starvation in body tissues. It does not mainly affect smell, is not known as a carcinogen, and does not cause blindness.
In a homogenous earth dam without d/s horizontal filter, portion of the d/s face through which seepage flow comes out is
A)
Stream line
B)
Equipotential line
C)
Neither a stream line nor an equipotential line
D)
Both (a) and (b)
In a homogeneous earth dam without a downstream horizontal filter, the seepage emerges through the downstream face at the phreatic surface (seepage line). This boundary is not a stream line, because flow crosses it as water exits the dam, and it is not an equipotential line either, because hydraulic head is not constant along it. Hence the correct answer is that it is neither a stream line nor an equipotential line.
For effective control of silt entry into the canal, the sill of the head regulator should be
A)
Below the sill of the under sluices
B)
Above the sill of the under sluices
C)
At the same level as the sill of under sluices
D)
At the maximum flood level
For effective control of silt entry into the canal, the head regulator sill is kept higher than the sill of the under sluices. This helps the heavier silt-laden bottom layers of water to pass through the under sluices, while relatively clearer water is admitted into the canal through the head regulator. If the head regulator sill were below or at the same level, more silt would enter the canal. Hence, the correct answer is (b).
The dimensions of dynamic viscosity μ are
A)
ML-1T-1
B)
ML-2T-1
C)
ML0T-1
D)
MLT-1
Dynamic viscosity μ is defined by Newton’s law of viscosity: τ = μ (du/dy), where shear stress τ has dimensions of force/area = M L^-1 T^-2 and velocity gradient du/dy has dimensions T^-1. Therefore, μ = τ / (du/dy) = (M L^-1 T^-2) / (T^-1) = M L^-1 T^-1. Hence the correct option is (a).
An irrotational flow is one in which
A)
The streamlines of flow are curved and closely spaced
B)
The fluid does not rotate as it moves along
C)
The net rotation of fluid particles about their mass centres remains zero
D)
None of the above
An irrotational flow is defined as a flow in which the vorticity is zero, meaning fluid particles do not have net angular rotation about their own mass centres. Option (c) states this correctly. Option (b) is close in meaning, but it is less precise; irrotational flow specifically refers to zero rotation of fluid elements, not merely that the fluid does not rotate in a general sense. Options (a) and (d) are incorrect.
The critical shear stress τc, at which incipient motion of sediment takes place, is proportional to grain size d as
A)
y^d
B)
d
C)
d^1/2
D)
d^3/2
For incipient motion of sediment, the critical shear stress is given by the Shields criterion: τc = θc(ρs − ρ)gd. Since θc is approximately constant for a given flow condition, τc is directly proportional to grain size d. Hence the correct relation is τc ∝ d. The other powers (d^1/2, d^3/2) are not the standard dependence.
The foundation of a weir consists of a horizontal floor of length 30m, a w/s pile of depth 8m and a d/s pile of depth 12m. the creep length according to Bligh’s theory is
A)
50m
B)
70m
C)
90m
D)
110m
According to Bligh’s creep theory, the creep length is the total length of the seepage path along the contact surface of the floor and piles. Here, it is the sum of the horizontal floor length and the depths of the upstream and downstream piles: 30 + 8 + 12 = 50 m. However, in Bligh’s theory, the vertical portions are counted twice when considering the full creep path around the impervious floor profile, so the effective creep length becomes 30 + 2(8) + 2(12) = 30 + 16 + 24 = 70 m? Wait—this still does not match the options. For a weir with upstream and downstream piles, the standard creep length is taken as floor length plus twice the depth of both piles: 30 + 2×8 + 2×12 = 70 m. Since 70 m is among the options, it is the correct value by Bligh’s theory.
Water logging is the state of the soil where
A)
The water table is brought very near to the ground surface
B)
The water table is at deep depth
C)
The moisture in soil is beyond the reach of plant roots
D)
None of the above
Water logging means the soil becomes saturated due to the water table rising close to the ground surface. In this condition, air spaces in the soil are reduced, harming root respiration and crop growth. Option (b) is the opposite condition, and (c) describes moisture being too deep for roots, not water logging.
The purpose of the end sill in the silting basin of a hydraulic jump type energy dissipater is
A)
To increase the tail water depth
B)
To reduce the length of the jump and control scour
C)
To counteract the uplift on the floor
D)
To dissipate the energy by impact action
In a hydraulic jump type energy dissipater, the end sill helps stabilize the jump by creating a backwater effect near the downstream end. This shortens the jump length and reduces the tendency of high-velocity flow to scour the downstream bed. Hence, its purpose is to reduce the length of the jump and control scour. Option (a) is not the main purpose, (c) relates to floor safety against uplift, and (d) describes energy dissipation by impact, which is not the specific function of the end sill.
Cross drainage works are not required when the canal is completely
A)
A ridge canal
B)
A contour canal
C)
Side slope canal
D)
Carrier canal
Cross drainage works are needed when a canal intersects natural drains or streams. A ridge canal runs along the watershed or ridge line, so it does not cross any natural drainage channels. Hence, cross drainage works are not required. A contour canal and side slope canal may still encounter drainage crossings, and a carrier canal is not defined by complete freedom from such crossings.
If the depletion of oxygen is found to be 2.5 mg/lit after incubating 2.5 ml of sewage diluted to 250 ml for 5 days at 20°C, BOD of the sewage is
A)
100 mg/l
B)
150mg/l
C)
200 mg/l
D)
250 mg/l
BOD is calculated as BOD = DO depletion × dilution factor. Here, 2.5 mL sewage is diluted to 250 mL, so dilution factor = 250/2.5 = 100. Given oxygen depletion = 2.5 mg/L, BOD = 2.5 × 100 = 250 mg/L. Hence, the correct answer is (d) 250 mg/l. The other options are lower values and do not match the dilution-based calculation.
A 12-h unit hydrograph of a catchment is triangular in shape with a base width of 144 hours and a peak discharge value of 23 m3/s. This unit hydrograph refers to a catchment of area
A)
756 km2
B)
596 km2
C)
1000 km2
D)
None of these
For a unit hydrograph, the volume of direct runoff under the hydrograph equals 1 cm over the catchment area.
Given a triangular 12-h unit hydrograph:
- Base width = 144 h
- Peak discharge = 23 m3/s
Area under triangular hydrograph = (1/2) × base × height.
Convert time to seconds:
- 144 h = 144 × 3600 s
So runoff volume = 1/2 × 144 × 3600 × 23
= 5,961,600 m3
This equals 1 cm = 0.01 m runoff over area A:
- 0.01 × A = 5,961,600
- A = 596,160,000 m2
- A = 596.16 km2
So the catchment area is approximately 596 km2.
Note: The computed area matches option (b).
After which of the following treatment units is the turbidity maximum?
A)
Chlorination
B)
Primary sedimentation
C)
Flocculation basin
D)
Secondary sedimentation
In conventional water treatment, turbidity is highest before settling begins and is reduced progressively by coagulation and flocculation. The flocculation basin is the unit where fine particles aggregate into larger flocs, so the turbidity is still relatively high there compared with later units. After primary sedimentation and secondary sedimentation, turbidity decreases further, and chlorination does not remove turbidity. Hence, the maximum turbidity among the given units is after the flocculation basin.
If the moisture content of a sludge is reduced from 98% to 96%, the volume of sludge will be decreased by
A)
2%
B)
20%
C)
25%
D)
50%
Assume the mass of dry solids in sludge remains constant.
At 98% moisture content, solids = 2% of total mass. Let initial total mass be 100 kg, so dry solids = 2 kg.
At 96% moisture content, solids = 4% of total mass. To keep dry solids = 2 kg, the new total mass must be 2/0.04 = 50 kg.
So the sludge volume (proportional to total mass for same density) decreases from 100 to 50, i.e. a decrease of 50%.
Why others are wrong:
- 2% and 20% are too small for this change in moisture content.
- 25% would correspond to a smaller reduction in moisture content.
Hence, the correct answer is 50%.
As per Lacey’s theory, the silt factor is
A)
Directly proportional to average particle size
B)
Directly proportional to square root of average particle size
C)
Inversely proportional to average particle size
D)
Not related to average particle size
In Lacey’s regime theory, the silt factor is given by the empirical relation f = 1.76 \u221a d, where d is the mean particle size in mm. Hence, the silt factor is directly proportional to the square root of the average particle size. Therefore, option (b) is correct. Options (a), (c), and (d) are incorrect because they do not match Lacey’s formula.
Surge tanks are used
A)
For storage of water
B)
To increase the velocity in pipeline
C)
As overflow valves
D)
To guard against water hammer
A surge tank is provided in a pipeline or penstock system to absorb sudden pressure changes caused by rapid changes in flow, such as valve closure or turbine load variation. This helps protect the pipeline from water hammer. It is not mainly for water storage, increasing velocity, or acting as an overflow valve.
The continuity equation ∂u/∂x + ∂v/∂y + ∂w/∂z = 0
A)
Is not valid for unsteady, incompressible fluids
B)
Is valid for incompressible fluids whether the flow is steady or unsteady
C)
Is valid for steady flow whether flow is compressible or incompressible
D)
Is valid for ideal fluid only
For an incompressible fluid, density remains constant, so the continuity equation reduces to ∂u/∂x + ∂v/∂y + ∂w/∂z = 0. This form does not depend on whether the flow is steady or unsteady. Option (a) is wrong because the equation is valid for unsteady incompressible flow too. Option (c) is wrong because compressible flow generally does not satisfy this simplified form. Option (d) is wrong because the equation is not limited to ideal fluids.
In pipe lines, a surge tank is provided
A)
To relieve the pressure due to water hammer
B)
To provide additional water head
C)
To overflow the pipeline when suddenly closed
D)
All of these
A surge tank is installed in pipelines to absorb sudden changes in pressure caused by rapid valve closure or pump failure. It reduces the effect of water hammer by temporarily storing excess water and releasing it back gradually. It is not meant to provide additional head, and it does not overflow the pipeline when suddenly closed. Therefore, the correct option is (a).
In slow sand filters, the turbidity of raw water can be removed only up to
A)
60 mg/lit
B)
75 mg/lit
C)
100 mg/lit
D)
150 mg/lit
Slow sand filters are suitable for relatively clear water and can effectively treat raw water with turbidity up to about 100 mg/lit. Beyond this, the filter clogs rapidly and needs very frequent cleaning. Hence, 100 mg/lit is the standard upper limit among the given options.