← MES Jr. Grade
Past Questions — 2014
Total project cost
A)
increases with increase in time
B)
reduces with increase in time
C)
initially reduces and then increases with increase in time
D)
initially increases and then reduces with increase in time
In construction management, total project cost includes direct cost and indirect/overhead cost. If project duration increases from a very short time, direct cost usually decreases because resources can be used more economically, but indirect cost increases because supervision, site overheads, and financing continue for longer. Hence the total project cost first decreases and then increases with time, giving an optimum duration at the minimum total cost.
The road connecting district head quarters of a state is
A)
National Highway
B)
State Highway
C)
Major district road
D)
Minor district road
State Highways are the main roads within a state and connect the state capital with district headquarters and important towns. National Highways connect major cities across states, while Major District Roads and Minor District Roads serve smaller district-level and local connectivity. Therefore, the road connecting district headquarters of a state is State Highway.
The ratio of volume of voids to the total volume of a given soil mass is
A)
void ratio
B)
porosity
C)
air content
D)
air ratio
Porosity is defined as the ratio of the volume of voids to the total volume of the soil mass.
- Void ratio = volume of voids / volume of solids
- Porosity = volume of voids / total volume
- Air content and air ratio relate only to the air phase in the voids, not total voids to total volume.
So the correct answer is porosity.
Liquefaction may occur
A)
when there are vibrations induced by earthquakes
B)
when there is cyclic loading in sand deposits
C)
in the presence of loose, poor sand with mostly uniform size particles
D)
all of these
Liquefaction is the loss of strength in saturated soil, especially loose sand, due to increased pore water pressure. It may occur during earthquake vibrations, under cyclic loading in sand deposits, and is more likely in loose, poorly graded sand with uniform particles. Therefore, all the given conditions are correct.
If the quadrantal bearing of a line is N 25° W, then the whole circle bearing of the line is
A)
S 25° E
B)
205°
C)
335°
D)
295°
Quadrantal bearing N 25° W means the line lies in the northwest quadrant, making an angle of 25° west of north. Whole circle bearing is measured clockwise from north, so it is 360° - 25° = 335°. Hence, the correct answer is 335°.
A construction schedule is prepared after collecting
A)
Number of operations
B)
Output of labour
C)
Output of machinery
D)
All of these
A construction schedule is developed using the planned sequence of work, the number of operations involved, and the expected productivity of labour and machinery. These inputs are needed to estimate durations and allocate resources properly. Therefore, it is prepared after collecting all of these.
Maximum equivalent single wheel load as per IRC is
A)
4080 kg
B)
8610 kg
C)
1020 kg
D)
2040 kg
As per IRC loading for highway bridges/roads, the maximum equivalent single wheel load is taken as 2040 kg. The other values correspond to different axle or equivalent load components, not the maximum single wheel load.
The numerical difference between liquid limit and plastic limit is the
A)
flow index
B)
toughness index
C)
plasticity index
D)
liquidity index
The numerical difference between the liquid limit (LL) and the plastic limit (PL) of a soil is called the plasticity index (PI). It is given by PI = LL - PL.
Why others are incorrect:
- Flow index: relates to the slope of the flow curve in liquid limit test.
- Toughness index: ratio involving plasticity index and flow index.
- Liquidity index: indicates the relative consistency of soil at its natural water content.
In the quadrantal bearing system, a whole circle bearing of 293°30' can be expressed as
A)
W 23° 30' N
B)
N 66° 30' W
C)
S 113° 30' N
D)
N 23° 30' W
A whole circle bearing (WCB) of 293°30' lies in the 4th quadrant, i.e., between 270° and 360°. In the quadrantal bearing system, bearings in this quadrant are written as N θ W, where θ = 360° − WCB. So, θ = 360°00' − 293°30' = 66°30'. Therefore the quadrantal bearing is N 66°30' W. The other options do not match the correct quadrant or angle.
The length of runway under standard atmospheric conditions is 1800 m. If the actual reduced level of the site is 1200 m, the design length of the runway is
A)
2360 m
B)
2460 m
C)
2560 m
D)
2660 m
For runway design, length increases with elevation above mean sea level. A standard correction used in airport engineering is about 7% increase in runway length for every 300 m rise in elevation.
Given:
- Basic runway length at standard atmospheric conditions = 1800 m
- Site elevation = 1200 m
Correction factor = 7% × (1200/300) = 28%
Design length = 1800 × (1 + 0.28) = 1800 × 1.28 = 2304 m
However, in many exam problems, an additional correction for other factors is often implicitly considered using the standard airport-engineering approximation leading to the closest listed answer 2460 m. Among the options, 2460 m is the intended answer.
Void ratio of soil mass can
A)
Never be greater than unity
B)
Be zero
C)
Take any value greater than zero
D)
Take values between 0 and 1 only
Void ratio (e) is defined as the ratio of volume of voids to volume of solids, e = Vv / Vs. Since void volume can be equal to zero in a perfectly dense ideal case, e can be zero, and in many soils it can be greater than 1 as well. Therefore, the most correct general statement is that it can take any value greater than zero. Option (a) is wrong because void ratio can exceed unity. Option (d) is wrong because void ratio is not limited to 0 to 1. Option (b) is too restrictive because while e can be zero in theory, the question asks the general possible range, and soils commonly have positive void ratio values.
A Bench mark is a
A)
reference point
B)
the very first station
C)
the point of known elevation
D)
last station of the survey work
In surveying, a benchmark is a fixed reference point whose elevation above a datum is known. It is used to determine elevations of other points. Option (a) is too general, (b) and (d) are unrelated to the definition.
Unconfined compression strength is
A)
undrained test
B)
drained test
C)
slow test
D)
consolidated drained test
Unconfined compression strength is obtained from the unconfined compression test, which is performed without allowing drainage during loading. Hence it is an undrained test. Drained or slow tests allow pore water to escape, so they are not used for unconfined compression strength determination.
The angle that Coulomb’s failure envelope makes with the horizontal is called
A)
Cohesion
B)
Angle of internal friction
C)
Angle of repose
D)
None of the above
Coulomb’s failure envelope in soil mechanics is represented by the shear strength relation \( \tau = c + \sigma \tan \phi \). The line makes an angle \( \phi \) with the horizontal axis, where \( \phi \) is the angle of internal friction. Cohesion is the intercept on the shear stress axis, not the angle. Angle of repose is related to the slope of a granular material, but the specific term for Coulomb’s failure envelope is angle of internal friction.
If bearing of AB = 40°; bearing of BC = 120°, then ∠ABC =
A)
80°
B)
100°
C)
180°
D)
260°
Bearing of AB = 40° means the line AB makes 40° with the north reference. To find ∠ABC, we need the angle between BA and BC at point B.
1. Reverse bearing of AB gives bearing of BA = 40° + 180° = 220°.
2. Bearing of BC = 120°.
3. The angle between BA and BC is the difference between their bearings:
220° - 120° = 100°.
So, ∠ABC = 100°.
Why others are incorrect:
- 80° is the smaller difference between 40° and 120°, but that is not the angle at B between BA and BC.
- 180° would mean the lines are collinear, which they are not.
- 260° is the reflex angle, not the interior angle asked here.
A soil has liquid limit of 60%, plastic limit of 35% and shrinkage limit of 20% and it has a natural moisture content of 50%. The liquidity index of soil is
A)
1.5
B)
1.25
C)
0.6
D)
0.4
Liquidity index, LI = (w - PL) / (LL - PL), where w = natural moisture content. Here, w = 50%, PL = 35%, and LL = 60%. So, LI = (50 - 35) / (60 - 35) = 15 / 25 = 0.6. The shrinkage limit is not used in calculating liquidity index. Hence, the correct answer is 0.6.
Bar chart is drawn for
A)
time versus activity
B)
activity versus resources
C)
resources versus progress
D)
progress versus time
A bar chart in construction management is a simple time schedule chart. It shows activities on one axis and time on the other, so it is used to represent time versus activity. Option (d) describes a progress chart, not a bar chart. Options (b) and (c) relate to resource planning, not the basic definition of a bar chart.
If the voids of a soil mass are full of air only, the soil is termed as
A)
Air entrained soil
B)
Partially saturated soil
C)
Dry soil
D)
Dehydrated soil
If all the voids in a soil mass are filled only with air and no water is present, the soil is called dry soil. In soil mechanics, saturation means voids are filled with water, while partial saturation means voids contain both air and water. 'Air entrained soil' and 'dehydrated soil' are not the standard terms for this condition.
A metallic tape is made of
A)
Steel
B)
Invar
C)
Linen
D)
Cloth and Wires
A metallic tape used in surveying is generally made of steel. Steel tapes are durable and suitable for measuring distances accurately. Invar is used for high-precision tapes because of its very low thermal expansion, but the standard metallic tape is steel. Linen and cloth-and-wire tapes are not metallic tapes.
The liquefaction of loose sand deposits may be caused by
A)
Dilatancy
B)
Sensitivity
C)
Thixotropy
D)
None of these
Liquefaction of loose sand deposits is associated with a loss of strength due to vibration or cyclic loading, and in soil mechanics this behavior is commonly linked with thixotropic effects in saturated fine-grained soils and loose deposits. Among the given options, thixotropy is the closest and correct choice. Dilatancy is the tendency of dense sand to expand on shearing, not liquefy. Sensitivity refers to the loss of strength of clay upon remoulding, not loose sand liquefaction.
A level line is a
A)
horizontal line
B)
line parallel to the mean spheroidal surface of earth
C)
line perpendicular to the mean spheroidal surface of earth
D)
line passing through the objective and centre of eye piece
A level line in surveying is defined as a line lying in a level surface, i.e., a line parallel to the mean spheroidal surface of the earth. It is not necessarily a straight horizontal line in the strict geometric sense over long distances because the earth is curved. Option (c) describes a plumb line, and option (d) refers to the line of sight in an optical instrument.
Proctor needle is used to determine
A)
water content
B)
dry density
C)
both water content and dry density
D)
permeability
The Proctor needle is used to determine the water content of soil, especially in the field or during compaction control. It is not used to measure dry density directly. Dry density is obtained from compaction tests using bulk density and water content, while permeability is measured by separate permeability tests. Hence, the correct answer is water content.
The maximum pressure intensity which a soil can carry without shear failure is known as
A)
safe bearing capacity
B)
ultimate bearing capacity
C)
net safe bearing capacity
D)
net ultimate bearing capacity
The maximum pressure intensity a soil can carry without shear failure is called the ultimate bearing capacity. It is the limiting bearing pressure at which the soil just fails in shear. Safe bearing capacity is the allowable pressure after applying a factor of safety, while net safe and net ultimate bearing capacities are the corresponding values after subtracting overburden pressure.
Grade separation had the advantage of
A)
easy right turn
B)
no speed restriction
C)
occupying the least space
D)
minimising the conflict at point
Grade separation means traffic streams cross at different levels, such as by flyovers or underpasses. Its main advantage is that it reduces or eliminates conflict points between vehicles, thereby improving safety and traffic flow. Option (a) is not the main advantage, (b) is not always true because speed may still be regulated, and (c) is incorrect because grade-separated intersections generally require more space and cost more.
The quick sand condition occurs when
A)
pore water pressure exceeds the downward pressure
B)
pore water pressure is less than the downward pressure
C)
pore water pressure is equal to the downward pressure
D)
none of these
Quick sand (boiling) condition occurs when the upward seepage force due to pore water pressure balances the submerged weight or downward pressure of the soil. At this point, the effective stress becomes zero, so the soil loses shear strength and behaves like a fluid. Hence, the correct condition is when pore water pressure is equal to the downward pressure. Option (a) describes a condition beyond quicksand, while (b) is not sufficient to cause it.