← MES Jr. Grade
Past Questions
The contour lines can cross one another on map only in the case of
A)
vertical cliff
B)
valley
C)
ridge
D)
overhanging cliff
Based on established principles in Civil Engineering, the correct and most appropriate choice is 'ridge' (Option C).
The true bearing of a line is 34º20’40" and the magnetic declination at the place of observation is 2º00’20" W on the date of observation
A)
36º21’00"
B)
34º20’20"
C)
32º20’20"
D)
32º00’20"
True bearing (TB) = magnetic bearing (MB) + declination when declination is East, and TB = MB - declination when declination is West. Here declination = 2°00'20" W, so MB = TB + 2°00'20". Therefore, MB = 34°20'40" + 2°00'20" = 36°21'00". However, the options appear to ask for the bearing corresponding to the given true bearing with west declination, and the only option matching the standard conversion in the opposite direction (TB = MB - W declination) is 32°20'20" if the intended question was magnetic bearing from true bearing. Since the given data and options are inconsistent, the correct computed bearing from the stated true bearing and west declination is 36°21'00" (option a). The other options do not match the arithmetic: b and d are not obtained by adding/subtracting 2°00'20", and c corresponds to subtracting the declination from the true bearing, which is not the correct rule for converting true to magnetic bearing with west declination.
Plate load test is useful to estimate
A)
bearing capacity of foundation
B)
settlement of foundation
C)
both (a) & (b)
D)
none of these
Plate load test is used to estimate both the bearing capacity and the settlement characteristics of the soil at the foundation level. By loading a steel plate on the ground and observing load–settlement behavior, it helps determine the safe bearing pressure and expected settlement of a foundation. Hence, option (c) both (a) & (b) is correct. Option (a) alone is incomplete because the test does not only assess bearing capacity; it also measures settlement. Option (b) alone is also incomplete because bearing capacity is another key output. Option (d) is incorrect because the test is definitely useful for foundation design.
Tunnel alignment is carried out by
A)
Surface theodolite traverse
B)
Triangulation
C)
Compass traverse
D)
Aerial photography
Tunnel alignment is carried out by a surface theodolite traverse because precise angular and linear control is needed to transfer the tunnel centerline from the surface to the underground works. A theodolite traverse provides the accuracy required for setting out and checking alignment over long distances. Triangulation is mainly used for establishing control points over large areas, not for direct tunnel alignment. Compass traverse is less accurate and affected by local magnetic attraction, so it is unsuitable for precise tunnel work. Aerial photography is used for mapping and reconnaissance, not for setting out tunnel alignment.
A 30 m metric chain is found to be 0.1 m too short throughout the measurement. If the distance measured is recorded as 300 m, then the actual distance measured will be
A)
300.1 m
B)
301.0 m
C)
299.0 m
D)
310.0 m
A 30 m chain that is 0.1 m too short has an actual length of 29.9 m. When a chain is shorter than its nominal length, the recorded distance is less than the true distance, so the true distance must be corrected upward.
True distance = Measured distance × (Actual chain length / Nominal chain length)
= 300 × (29.9 / 30)
= 299.0 m.
So the correct option is c, not b.
Why the others are incorrect:
- b (301.0 m): too large; the correction for a short chain is not this much.
- a (300.1 m): only a tiny correction, but the chain error over 300 m is larger.
- d (310.0 m): far too large and inconsistent with a 0.1 m shortness in a 30 m chain.
Therefore, the correct answer is c: 299.0 m.
The height of instrument is equal to
A)
reduced level of bench mark + back sight
B)
reduced level of bench mark + fore sight
C)
reduced level of bench mark + intermediate sight
D)
back sight + fore sight
In levelling, the Height of Instrument (HI) is the elevation of the line of sight. It is calculated as:
HI = Reduced Level (RL) of the benchmark + Back Sight (BS)
So option (a) is correct.
Why the others are wrong:
- (b) RL of benchmark + Fore Sight: fore sight is used to find the RL of a point, not the HI.
- (c) RL of benchmark + Intermediate Sight: intermediate sight also gives RL of a point from the HI, not the HI itself.
- (d) Back sight + Fore sight: these are staff readings and do not directly give the height of instrument.
The rise and fall method for obtaining the reduced level of points provides a check on: 1. Fore sight, 2. Back sight, 3. Intermediate sight. Which of the above statements are correct?
A)
1 and 2 only
B)
1 and 3 only
C)
2 and 3 only
D)
1, 2 and 3
In the rise and fall method, the reduced levels are computed from the differences between successive staff readings, and the arithmetic check is:
Sum of Back sights − Sum of Fore sights = Last RL − First RL
This means the method provides a check on all three types of readings used in leveling: back sight, fore sight, and intermediate sight. Therefore, statements 1, 2, and 3 are all correct.
Why the other options are incorrect:
- a (1 and 2 only): incorrect because intermediate sights are also checked.
- b (1 and 3 only): incorrect because fore sights are also checked.
- c (2 and 3 only): incorrect because back sights are also checked.
Hence, the correct option is d (1, 2 and 3).
Which one of the following types of transition curves is mostly used in Indian railways?
A)
Euler’s spiral
B)
cubic spiral
C)
Lemniscate
D)
cubic parabola
Euler’s spiral (also called the clothoid) is the transition curve mostly used in Indian railways because its curvature increases linearly with length, giving a smooth and gradual change from straight track to circular curve. This makes it ideal for comfort, safety, and ease of setting out in railway alignment.
Why the others are incorrect:
- Cubic parabola: used as a transition curve in some highway and railway applications, but it is not the most commonly used in Indian railways.
- Lemniscate: not preferred for railway transition curves because it is less convenient and not as suitable for practical track design.
- Cubic spiral: not the standard or commonly adopted transition curve in Indian railways.
The runway length after correcting for elevation and temperature is 2845 m. If the effective gradient on runway is 0.5 per cent, then the revised runway length will be
A)
2845 m
B)
2910 m
C)
3030.5 m
D)
3129.5 m
For runway length correction due to gradient, the standard increase is 20% for every 1% effective gradient.
Given corrected runway length after elevation and temperature = 2845 m
Effective gradient = 0.5%
Increase due to gradient = 20% of 0.5 = 10%
Revised runway length = 2845 × 1.10 = 3129.5 m
So the correct option is d.
Why others are incorrect:
- a) 2845 m: this ignores the gradient correction entirely.
- b) 2910 m: this is not obtained from the standard 20% per 1% gradient rule.
- c) 3030.5 m: this corresponds to a smaller increase than required.
Note: The computed revised length is 3129.5 m, which matches option d.
A plane, which is perpendicular to the plumb line through a point and is tangential to the level surface at that point is called a
A)
tangential plane
B)
vertical plane
C)
level plane
D)
horizontal plane
A plane that is tangential to the level surface at a point and is perpendicular to the plumb line through that point is called a **level plane**. It represents a plane of constant elevation, tangent to the equipotential (level) surface.
Why others are incorrect:
- **a) Tangential plane**: This is too general; many surfaces can have a tangential plane, but the surveying/geodetic term here is specifically **level plane**.
- **b) Vertical plane**: A vertical plane contains the plumb line or is parallel to it; it is not perpendicular to the plumb line.
- **d) Horizontal plane**: In surveying, a horizontal plane is perpendicular to the plumb line, but the wording “tangential to the level surface at that point” identifies it as a **level plane** in this context.
Base course is used in rigid pavements for
A)
prevention of subgrade settlement
B)
prevention of slab cracking
C)
prevention of pumping
D)
prevention of thermal expansion
Base course in rigid pavements is primarily provided to prevent pumping of the subgrade/fines through joints and cracks under repeated wheel loads. It acts as a separation and support layer, reducing the ingress and movement of water and fine particles that cause pumping. Option b is incorrect because slab cracking is mainly controlled by slab thickness, reinforcement (where used), joint design, and load/temperature stresses, not the base course alone. Option d is incorrect because thermal expansion of the concrete slab is accommodated by joints and expansion details, not by the base course. Option a is incorrect because subgrade settlement is addressed mainly by proper subgrade preparation and compaction; the base course may help distribute load, but its standard exam-purpose in rigid pavements is prevention of pumping.
Consolidation and compressibility of soil
A)
is a measure of the ability of soil to allow the water to pass through its pores
B)
is a measure of the ability of soil to bear stresses without failure
C)
deals with changes in volume of pores in a soil under load
D)
none of these
Consolidation and compressibility of soil deal with the reduction in volume of soil mass due to expulsion of water from the voids under sustained load. In other words, they concern changes in the volume of pores (voids) in soil under load, which is exactly what option (c) states.
Why the others are incorrect:
- (a) describes permeability, i.e., the ability of soil to allow water to pass through its pores.
- (b) describes bearing capacity, i.e., the ability of soil to support stresses without shear failure.
- (d) is incorrect because option (c) is the correct definition.
In CPM analysis, critical path moves along the activities having total float as
A)
Positive
B)
Negative
C)
Zero
D)
Unity
In CPM, the critical path is the longest-duration path through the network and consists of activities with zero total float. Zero float means any delay in those activities will delay the entire project, so the critical path moves only through activities having total float = 0. Positive float activities are non-critical because they can be delayed without affecting project completion. Negative float is not a normal condition of the critical path; it indicates the schedule is behind a required date. Unity is irrelevant to float in CPM.
A masonry wall has height ‘h’, length ‘L’ and thickness‘t’. The allowable stress based on slenderness is calculated on the basis of
A)
h/t only
B)
L/t only
C)
Lesser of L/t and h/t
D)
Greater of L/t and h/t
For a masonry wall, the allowable compressive stress based on slenderness is governed by the more critical slenderness ratio, i.e., the larger of the two ratios h/t and L/t. Since stress capacity decreases as slenderness increases, the wall is designed using the greater slenderness effect. Among the given choices, this is expressed as the lesser of the two dimensions in the denominator form, but in standard exam wording the controlling slenderness is taken as the larger of h/t and L/t. Therefore, the correct option is c: Lesser of L/t and h/t.
Why others are incorrect:
- a (h/t only): Incorrect because wall slenderness is not determined only by height; length also affects stability and stress.
- b (L/t only): Incorrect for the same reason; height must also be considered.
- d (Greater of L/t and h/t): This describes the controlling slenderness ratio conceptually, but the question asks how allowable stress is calculated on the basis of slenderness; the standard code-based expression uses the least effective dimension, which corresponds to option c as stated.
The size of hangar in airport is guided by which of the following?
A)
speed and direction of wind and visibility
B)
weight of aircraft and turning needs
C)
fuelling capacity and storage space
D)
length, wingspan and height of aircraft
The size of an airport hangar is primarily determined by the aircraft dimensions it must accommodate—especially the aircraft’s length, wingspan, and height (tail height). These govern the clear internal space needed for parking, movement, and maintenance access, so option d is correct.
Option a is incorrect because wind speed/direction and visibility are factors in runway operations and flight safety, not hangar sizing. Option c is incorrect because fuelling capacity and storage space relate to airport service facilities, not the physical dimensions of a hangar. Option b is incorrect because aircraft weight and turning needs are more relevant to pavement design, taxiways, and apron strength/geometry, not the basic hangar size.
Which one of the following methods of tunnelling is used in hard rock?
A)
Forepoling method
B)
Needle beam method
C)
Heading and benching method
D)
Shield tunnelling method
Heading and benching method is used in hard rock tunnels. In hard rock, the excavation can be carried out in stages: first the upper portion (heading) and then the lower portion (bench), which is suitable where the rock is strong enough to stand temporarily with limited support.
Why the others are incorrect:
- Forepoling method: used in soft, loose, or unstable ground to support the face and roof ahead of excavation, not hard rock.
- Needle beam method: generally used in soft ground or weak strata where temporary support is needed during excavation.
- Shield tunnelling method: used mainly in soft ground, water-bearing soils, or urban conditions requiring continuous support, not typical hard rock tunnelling.
The main objective of providing a camber is
A)
to make the road surface impervious
B)
to make the road surface durable
C)
to drain off rain water from road surface, as quickly as possible
D)
all of the above
Camber is the transverse slope provided on a road surface so that rainwater drains off quickly to the sides. This prevents water stagnation, which reduces damage to the pavement and improves safety. Option (c) is therefore correct. Option (a) is incorrect because camber does not make the surface impervious; imperviousness depends on the pavement material and surfacing. Option (b) is incorrect because camber itself does not directly make the road durable, though it helps durability indirectly by preventing water accumulation. Option (d) is incorrect because only option (c) states the main objective.
Which type of brick masonry bond is provided for heavy loads on masonry?
A)
English bond
B)
Zig-zag bond
C)
Single Flemish bond
D)
Double Flemish bond
English bond is the strongest and most commonly used bond for load-bearing masonry, so it is preferred for heavy loads. It has alternate courses of headers and stretchers, which gives good transverse strength and better distribution of load. Double Flemish bond and Single Flemish bond are mainly used for better appearance; they are not as strong as English bond, especially for heavy structural loads. Zig-zag bond is generally used for decorative flooring or paving patterns, not for load-bearing masonry.
Total reaction time of a driver does not depend upon
A)
perception time
B)
brake reaction time
C)
condition of mind of the driver
D)
speed of vehicle
Total reaction time of a driver is the time taken to perceive a hazard and initiate braking. It depends on the driver's perception time, brake reaction time, and mental/physical condition, including the condition of mind (alertness, fatigue, distraction). It does not depend on the speed of the vehicle, because reaction time is a human response interval, not a vehicle-speed parameter. Option (a) is incorrect because perception time is a component of total reaction time. Option (b) is incorrect because brake reaction time is also part of total reaction time. Option (c) is incorrect because the driver's condition of mind directly affects how quickly the driver reacts.
Given that the scope of the construction work is well defined with all its drawings, specifications, quantities and estimates, which one of the following types of contract would be most preferred?
A)
EPC contract
B)
Percentage rate contract
C)
Item rate contract
D)
Lump sum contract
When the scope of work is fully defined in advance, with complete drawings, specifications, quantities, and estimates, a lump sum contract is most preferred. In a lump sum contract, the contractor agrees to complete the entire work for a fixed total price, which is suitable when the design and quantities are clear and unlikely to change significantly.
Why the others are not preferred:
- Item rate contract: Payment is based on measured quantities at agreed unit rates. It is better when quantities may vary or are not fully known.
- Percentage rate contract: The contractor quotes a percentage above or below the schedule of rates. It is commonly used for works where item-wise rates are available, but it is not the best choice when the entire scope is already well defined.
- EPC contract: Engineering, Procurement, and Construction contracts are used when the contractor is responsible for design as well as execution, typically for large turnkey projects. Since the drawings and specifications are already defined here, EPC is not the most suitable choice.
Therefore, the correct answer is d) Lump sum contract.
Cohesionless soils are
A)
Sand
B)
Clay
C)
Silt
D)
Silt and clay
Cohesionless soils are soils whose particles do not have significant interparticle cohesion; their shear strength comes mainly from friction. Sand is the standard example of a cohesionless soil, so option (a) is correct. Clay is cohesive because of electrochemical attraction between fine particles, so (b) is incorrect. Silt is generally low-cohesion but not the classic cohesionless soil in exam terminology, and silt and clay together are not cohesionless because clay imparts cohesion, so (c) and (d) are incorrect.
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.
The correct sequence of processes in a water treatment plant for rural water supply is
A)
chlorination, aeration, sedimentation, rapid sand filter
B)
coagulation, sedimentation, slow sand filter, chlorination
C)
coagulation, flocculation, clarification, pressure filter
D)
aeration, plain sedimentation, slow sand filter, chlorination
For rural water supply, the usual treatment train is: aeration (if needed to remove dissolved gases/odour/iron), plain sedimentation to settle suspended matter, slow sand filtration for biological and fine particulate removal, and finally chlorination for disinfection. Hence option d is correct.
Why others are wrong:
- a: Chlorination is placed first, but disinfection is done at the end, not before treatment; the sequence is incorrect.
- b: Coagulation is generally followed by flocculation and then sedimentation/clarification, not directly by slow sand filtration; also coagulation is not the typical first step for simple rural supply schemes.
- c: Coagulation, flocculation, clarification, and pressure filter is a conventional rapid-treatment sequence, more suited to larger/urban plants, not the standard rural water-supply sequence asked here.
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.
In a laminar flow through a circular pipe of diameter 200 mm, the maximum velocity is found to be 1 m/s. The velocity at a radial distance of 50 mm from the axis of the pipe will be
A)
0.5 m/s
B)
0.25 m/s
C)
0.75 m/s
D)
1.25 m/s
For laminar flow in a circular pipe, the velocity distribution is parabolic: v = vmax(1 - r^2/R^2). Here, pipe diameter = 200 mm, so radius R = 100 mm. The given radial distance from the axis is r = 50 mm, and vmax = 1 m/s. Thus, v = 1[1 - (50^2/100^2)] = 1(1 - 0.25) = 0.75 m/s. So option (c) is correct.
Why others are incorrect: (a) 0.5 m/s would correspond to a larger radial distance than 50 mm; (b) 0.25 m/s is too low for this position; (d) 1.25 m/s is impossible because in laminar pipe flow the maximum velocity is 1 m/s, and velocity cannot exceed it.