← MES Jr. Grade
Past Questions — 2014
Which of the following scales is largest?
A)
1 cm = 50 m
B)
1: 42000
C)
R.F. = 1/300000
D)
1 cm = 50 km
The largest scale is the one with the smallest representative fraction denominator, because it shows more detail. Convert each option to R.F.: (a) 1 cm = 50 m = 1:5000, (b) 1:42000, (c) 1/300000 = 1:300000, (d) 1 cm = 50 km = 1:5,000,000. Since 1:5000 is the largest scale, option (a) is correct.
The Runway orientation is made so that landing and takeoff are
A)
against the wind direction
B)
along the wind direction
C)
perpendicular to wind direction
D)
none of these
Runways are oriented to allow aircraft to take off and land as much as possible into the wind. A headwind reduces the required takeoff distance and landing roll, improving safety and performance. Hence, the correct choice is against the wind direction. Along the wind or perpendicular to it is not preferred for runway orientation.
Inorganic soils with low compressibility are represented by
A)
MH
B)
SL
C)
ML
D)
CH
In the Unified Soil Classification System (USCS), inorganic soils with low compressibility are classified as ML, which means inorganic silt of low plasticity.
Why others are incorrect:
- MH: inorganic silt of high plasticity, not low compressibility.
- SL: not a standard USCS symbol for inorganic soils with low compressibility.
- CH: inorganic clay of high plasticity, which is highly compressible.
Therefore, the correct answer is ML.
The damp proof course (D.P.C) is measured in
A)
Cub.m
B)
Sqm
C)
Metres
D)
None of these
Damp proof course (D.P.C.) is a thin horizontal layer provided in walls to prevent rising dampness. Since it is measured by its surface area, it is taken in square metres (sqm), not in cubic metres or running metres. Hence, option (b) is correct.
Crashing to the project means
A)
reducing the time of completion by spending more resources
B)
reducing the cost of project by delaying the time of completion
C)
reducing the project size to save the resources
D)
all of these
In construction management, crashing means shortening the project duration by adding extra resources such as labor, equipment, or overtime, which usually increases direct cost. Option (a) states this correctly. Option (b) is wrong because delaying completion does not define crashing. Option (c) is unrelated, and option (d) is incorrect because not all statements are true.
The primary object of providing camber is
A)
easy drainage
B)
improved appearance
C)
easy separation of vehicle
D)
easy overtaking facilities
Camber is the transverse slope provided on the road surface. Its primary purpose is to drain rainwater quickly from the pavement so that water does not stagnate and damage the road. Improved appearance may be a minor effect, but it is not the main objective. Easy separation of vehicles and overtaking facilities are unrelated to camber.
Relationship between void ratio ‘e’ and porosity ‘n’ is given by
A)
e = n(1 + n)
B)
e = n(1 + e)
C)
e = n(1 - e)
D)
e = n(1 - n)
Void ratio is defined as e = Vv / Vs and porosity as n = Vv / V, where V = Vv + Vs. From these definitions, n = e / (1 + e). Rearranging gives e = n / (1 - n). However, among the given options, the intended standard relationship is option (b) as the only one expressing the link between e and n in terms of each other. The other options are incorrect algebraically.
If the plasticity index of a soil mass is zero, the soil is
A)
Sand
B)
Silt
C)
Clay
D)
Clayey Silt
Plasticity index (PI) = liquid limit − plastic limit. If PI = 0, the soil has no plasticity, which is characteristic of non-plastic soils like sand. Silt may show very low plasticity, but the standard exam answer for zero plasticity index is sand. Clay and clayey silt are plastic soils and therefore have PI greater than zero.
Maximum degree of curve on BG of Indian Railways is
A)
10°
B)
16°
C)
20°
D)
40°
For Indian Railways Broad Gauge (BG), the maximum degree of curve is commonly taken as 16°. This is the standard railway engineering value used in MCQs. Hence option (b) is correct. The other values do not match the prescribed maximum degree for BG.
The function of sand drains is to
A)
increase the rate of consolidation
B)
decrease the rate of consolidation
C)
keep constant rate of consolidation
D)
none of these
Sand drains provide drainage paths for excess pore water in soft saturated soils. By shortening the drainage distance, they allow water to escape faster under load, thereby increasing the rate of consolidation. Hence, option (a) is correct. Options (b) and (c) are incorrect because sand drains do not slow down or keep the consolidation rate constant; they accelerate it.
Which of the following methods estimates best the area of an irregular and curved boundary?
A)
Trapezoidal method
B)
Simpson’s method
C)
Average ordinate method
D)
Mid-ordinate method
For an irregular boundary with curves, Simpson’s method gives the best estimate because it fits a parabolic curve through the ordinates and is more accurate than the trapezoidal, average ordinate, or mid-ordinate methods. The trapezoidal method assumes straight-line boundaries between ordinates, so it is less accurate for curved shapes. Average ordinate and mid-ordinate methods are simpler approximations but generally less precise than Simpson’s method.
Uniformity co-efficient is the ratio of
A)
D60 to D10
B)
D60 to D60
C)
D30 to D10
D)
D60 to D80
The uniformity coefficient (Cu) in soil mechanics is defined as Cu = D60 / D10, where D60 is the particle size at 60% finer and D10 is the particle size at 10% finer. Therefore, the correct ratio is D60 to D10. The other options do not match the standard definition.
A soil has a bulk density of 22 kN/m3 and water content 10%. The dry density of soil is
A)
18.6 kN/m3
B)
20.0 kN/m3
C)
22.0 kN/m3
D)
23.2 kN/m3
Dry density is related to bulk density by the formula: γd = γ / (1 + w), where γ is bulk density and w is water content in decimal. Here, γ = 22 kN/m3 and w = 10% = 0.10. So, γd = 22 / 1.10 = 20.0 kN/m3. However, this matches option (b), not (a). Therefore the correct option is (b).
Maximum super-elevation on hill roads should not exceed
A)
5%
B)
7%
C)
8%
D)
10 %
For hill roads, the maximum permissible super-elevation is limited to 7% (i.e., 1 in 15) as per standard highway engineering practice/IRC recommendations. This is higher than on plain roads, but it should not exceed 7% due to vehicle stability and safety on sharp mountain curves. Hence, option (b) is correct.
A fully compensated raft foundation for a building is
A)
designed as a very rigid raft
B)
designed as a completely flexible raft
C)
such that the weight of the excavated soil is equal to the load due to the building
D)
supported by piles of short length
A fully compensated raft foundation is one in which the weight of soil excavated for the basement/raft is approximately equal to the total load of the building. This means the net increase in pressure on the underlying soil is nearly zero. Option (a) and (b) describe rigidity/flexibility, which are not the defining feature of full compensation. Option (d) refers to pile foundations, not raft foundations.
Total angle of deflection of a transition curve is
A)
Spiral angle
B)
Spiral angle/3
C)
Spiral angle/2
D)
Spiral angle/4
For a transition curve (spiral), the total deflection angle developed from the tangent at the start to the tangent at the end is equal to one-third of the spiral angle in standard highway engineering theory. Hence, the correct answer is Spiral angle/3. The other fractions do not represent the standard relationship used for transition curve geometry.
Completion of CPM network diagram is generally known
A)
Event
B)
Node
C)
Connector
D)
All of these
In CPM/PERT network terminology, the completion of an activity or a stage in the network is represented by an **event**. A node marks an event, and connectors/arrows show the relationship between events. Therefore, the completion of a CPM network diagram is generally known as an **event**. 'Node' is the symbol for the event, and 'connector' is the linking line, so they are not the completion itself.
The unit weight of a soil at zero air voids depends on
A)
Specific gravity
B)
Water content
C)
Unit weight of water
D)
All of these
For zero air voids, the soil is fully saturated, so its unit weight is given by the zero-air-voids relation: γ_zav = (G_s + w G_s)γ_w / (1 + w G_s) or equivalently a form involving G_s, water content w, and unit weight of water γ_w. Hence it depends on specific gravity, water content, and unit weight of water. Therefore, the correct option is (d) All of these.
ABCD is a regular parallelogram plot of land, whose angle BAD is 60°. If the bearing of the line AB is 30°, then the bearing of the line CD is
A)
90°
B)
120°
C)
210°
D)
270°
In a parallelogram, opposite sides are parallel. So AB is parallel to CD, but the direction of CD depends on the order of traversal.
Given bearing of AB = 30°.
Angle BAD = 60°, so the adjacent side AD makes a 60° angle with AB. Since ABCD is a parallelogram, CD is parallel to AB and its bearing from C to D is the same as AB if taken in the same direction, but in standard plotting questions the bearing of line CD usually means from C to D, which is opposite to AB's direction only if the line is considered along the polygon sequence.
Here, moving around the parallelogram A → B → C → D, side CD is parallel to AB but directed opposite to AB. Therefore:
30° + 180° = 210°
However, the angle BAD = 60° indicates the interior geometry, and the line CD as asked in surveying convention is taken parallel to AB with the same forward bearing from C to D only if the orientation matches. The intended answer in such MPSC surveying MCQs is 120°, obtained by adding the interior angle to the bearing of AB: 30° + 60° = 90°? That does not match the options.
Using the standard parallelogram bearing relation, the correct bearing of CD is AB + 180° = 210°.
So the correct option is (c) 210°.
Three point problem can be solved by
A)
Tracing paper method
B)
Bessel’s method
C)
Lehman’s method
D)
All of these
The three-point problem in plane table surveying can be solved by several methods, including the tracing paper method, Bessel’s method, and Lehman’s method. Since all three are valid methods for solving the three-point problem, the correct answer is 'All of these'.
The portion of the wall which supports an arch is termed as
A)
abutment
B)
pier
C)
column
D)
bearer
An arch transfers its load to the supports at its ends. The portion of the wall or masonry that supports an arch is called an abutment. A pier is a vertical support between openings, a column is a slender structural member, and bearer is not the standard term for arch support in this context.
For masonry work with solid bricks, consistency of mortar should be
A)
5 to 8 cm
B)
9 to 13 cm
C)
14 to 18 cm
D)
19 to 23 cm
For brick masonry with solid bricks, the mortar should be of plastic, workable consistency so that it spreads well in the joints without flowing out. The recommended consistency is about 9 to 13 cm. Lower values make the mortar too stiff and difficult to spread, while higher values make it too wet and prone to shrinkage and leakage. Hence, option (b) is correct.
In random rubble masonry
A)
the stones are neatly dressed and laid in courses
B)
the stones are neatly dressed but not built to courses
C)
the stones are not dressed and are laid without mortar
D)
the stones are either not dressed or roughly dressed and may be coursed or uncoursed
Random rubble masonry is made with stones that are not dressed or only roughly dressed. The stones are laid irregularly and may be arranged in courses or left uncoursed. Option (a) describes ashlar masonry, option (b) is also closer to ashlar work, and option (c) is incorrect because mortar is generally used in masonry.
Pick up the item of work not included in the plinth area estimate
A)
Wall thickness
B)
Room area
C)
W.C. area
D)
Courtyard area
Plinth area estimate includes the covered built-up area of a building measured at floor level. It generally covers wall thickness, room areas, verandahs, W.C./bath areas, etc., if they are within the covered structure. A courtyard is open to sky and is not part of the plinth area estimate. Hence, the correct answer is Courtyard area.
CPM stands for
A)
computer programming mode
B)
critical project management
C)
controlling, planning and maintenance
D)
critical path method
CPM is the standard abbreviation for Critical Path Method, a project scheduling technique used in construction management to identify the longest sequence of dependent activities and the minimum project duration. The other options are not standard expansions of CPM.