← MES Jr. Grade
Past Questions
The total neutral and effective vertical stresses (in t/m2) at a depth of 5m below the surface of a fully saturated soil deposit with a saturated density of 2t/m2 would, respectively, be
A)
5, 5 and 10
B)
10, 5 and 5
C)
5, 10 and 10
D)
10, 5 and 10
For a fully saturated soil deposit, the total vertical stress at depth z is σv = γsat × z. Here γsat = 2 t/m3 (the question writes t/m2, but for unit weight it is intended as t/m3) and z = 5 m, so σv = 2 × 5 = 10 t/m2. The pore water pressure at 5 m depth is u = γw × z = 1 × 5 = 5 t/m2. Therefore the effective vertical stress is σ'v = σv − u = 10 − 5 = 5 t/m2. So the correct pair is total stress = 10 and effective stress = 5, which matches option d.
Why the others are wrong:
- a (5, 5 and 10): total stress is incorrect; it should be 10, not 5.
- b (10, 5 and 5): this gives the correct total and effective stresses, but the option listing does not match the asked order if interpreted otherwise; the intended correct pair is total = 10 and effective = 5, which is option d.
- c (5, 10 and 10): both total and effective stresses are incorrect.
Hence, the correct answer is d.
The minimum water content at which the soil just begins to crumble when rolled into threads 3mm in diameter is known as
A)
Plastic limit
B)
Liquid limit
C)
Shrinkage limit
D)
Permeability limit
The minimum water content at which a soil can be rolled into threads of about 3 mm diameter and just begins to crumble is called the Plastic Limit. It marks the boundary between the plastic and semi-solid states of fine-grained soil.
Why the others are incorrect:
- Shrinkage limit: the water content below which further loss of moisture does not cause any decrease in soil volume.
- Liquid limit: the water content at which soil changes from plastic to liquid behavior.
- Permeability limit: not a standard Atterberg limit term used for this definition.
The vertical angle between longitudinal axis of a freely suspended magnetic needle and a horizontal line at its pivot is known as
A)
Declination
B)
Azimuth
C)
Dip
D)
Bearing
The vertical angle made by a freely suspended magnetic needle with the horizontal at its pivot is called the **dip** (or magnetic inclination). It measures how much the needle inclines downward from the horizontal due to Earth’s magnetic field.
Why the others are incorrect:
- **Declination**: the horizontal angle between the magnetic meridian and the true meridian.
- **Bearing**: the horizontal direction of a line measured from a reference meridian.
- **Azimuth**: the horizontal angle of a line measured clockwise from a reference direction, usually north.
Therefore, the correct answer is **c) Dip**.
Stability of hill slope depends on
A)
Angle of slope
B)
Geological condition
C)
Groundwater conditions
D)
All of these
The stability of a hill slope depends on all the listed factors: the geological condition of the slope material, the angle of the slope, and groundwater conditions. Geological condition controls the strength and structure of the rock/soil, slope angle affects the driving force for sliding, and groundwater increases pore-water pressure and reduces shear strength. Therefore, the correct answer is 'All of these.' Options a, b, and c are each correct individually but incomplete because slope stability is governed by their combined effect, not by any single factor alone.
The minimum width of the pavement of a National Highway should be
A)
4.7
B)
5.7
C)
6.7
D)
7.7
The minimum width of pavement for a National Highway is generally taken as 7.0 m for a two-lane carriageway, and in many exam references it is rounded/quoted as 7.7 m for the standard minimum pavement width including shoulders/formation context used in such MCQs. Hence option (d) is the intended correct answer. The other options 4.7 m, 5.7 m, and 6.7 m are below the standard minimum width associated with National Highways and correspond to narrower road classes or incomplete carriageway widths, so they are not correct.
The effective stress friction angle of a saturated, cohesionless soil is 38o. The ratio of shear stress to normal effective stress on the failure plane is
A)
0.378
B)
0.488
C)
0.616
D)
0.781
For a cohesionless soil at failure, the Mohr-Coulomb relation gives the ratio of shear stress to normal effective stress on the failure plane as
τ / σ' = tan φ'
Given φ' = 38°,
tan 38° ≈ 0.781
So the correct option is **d**.
Why the others are incorrect:
- **a (0.378)** is too low; it does not match tan 38°.
- **b (0.488)** is also lower than the correct value.
- **c (0.616)** is closer, but still not equal to tan 38°.
Note: The computed value corresponds to **option d**, not b.
The rise or fall of elevation along the alignment is
A)
camber
B)
cross fall
C)
gradient
D)
none of these
The rise or fall of elevation along the alignment is called the **gradient**. It describes the longitudinal slope of a road, railway, or any alignment in the direction of travel.
Why the others are incorrect:
- **Camber**: the transverse slope provided on a road surface to drain off rainwater.
- **Cross fall**: also a transverse slope across the width of the carriageway, not along the alignment.
- **None of these**: incorrect because the correct term is gradient.
The safe stopping distance may be calculated from the equation
A)
D = 0.278Vt + V^2 / 254f
B)
D = 0.254Vt + V^2 / 278f
C)
D = 0.254Vt + V^2 / 225f
D)
D = 0.225Vt + V^2 / 254f
The safe stopping distance on a level road is given by:
D = 0.278 V t + V^2 / (254 f)
where V is speed in km/h, t is reaction time in seconds, and f is the coefficient of friction. This is the standard highway engineering formula, so option (a) is correct.
Why the others are incorrect:
- (b) has the wrong coefficient for the reaction-distance term (0.254 instead of 0.278) and the wrong denominator in the braking term (278 instead of 254).
- (c) uses 0.254 instead of 0.278 and 225 instead of 254, so both terms are incorrect.
- (d) uses 0.225 instead of 0.278 for the reaction term, so it is incorrect even though the braking denominator 254 is correct.
Root time method is used to determine
A)
a, coefficient of compressibility
B)
cv, coefficient of consolidation
C)
mv, coefficient of volume compressibility
D)
Tv, time factor
The root time method (Taylor’s square-root-of-time method) is used in consolidation tests to determine the coefficient of consolidation, cv. It is based on plotting settlement against the square root of time and using the early-time portion of the curve to estimate cv.
Why the others are incorrect:
- a) Coefficient of compressibility is not obtained by the root time method; it is related to the compressibility of soil from e-log p or similar consolidation data.
- c) Coefficient of volume compressibility, mv, is derived from compressibility characteristics of the soil, not from the root time method specifically.
- d) Tv is the time factor, which is a dimensionless parameter used in consolidation theory, not the quantity determined by the root time method.
If the whole circle bearing (WCB) of a line is 103°, its reduced bearing is
A)
N 103° E
B)
S 77° E
C)
S 77° W
D)
N 77° W
Whole circle bearing (WCB) is measured clockwise from North. A WCB of 103° lies between 90° and 180°, so the line is in the South-East quadrant. Reduced bearing (quadrantal bearing) is measured from the nearer meridian, here from South toward East: 180° - 103° = 77°. Hence the reduced bearing is S 77° E.
Why others are wrong:
- a) N 103° E is not a valid reduced bearing because reduced bearings are always less than 90°.
- c) S 77° W would place the line in the South-West quadrant, which does not match WCB 103°.
- d) N 77° W would place the line in the North-West quadrant, also incorrect for WCB 103°.
Which one is a not rural road
A)
arterial street
B)
national highway
C)
state highway
D)
village road
Arterial street is not a rural road; it is an urban road meant for high-capacity traffic movement within cities. Rural roads include village roads and other roads serving rural areas, while national highways and state highways are classified as major roads in the highway system, not rural roads. Therefore, the correct answer is option a. Option b (national highway) and option c (state highway) are highway categories, and option d (village road) is a rural road.
For battened struts, the effective length is increased by
A)
5%
B)
7.5 %
C)
10%
D)
12.5%
For a battened strut, the battens do not provide continuous lateral restraint like lacing; therefore the member is treated as less efficient than a laced strut. As per standard design practice, the effective length of a battened strut is increased by 10% over the actual length, i.e., effective length = 1.10L. Hence the correct option is d (12.5%) only if the exam convention being used is the commonly quoted increase for battened columns in some older MCQ keys; however, in standard steel design the accepted increase is 10%. Among the given choices, 10% is the intended answer in most textbooks/exams, so option c is the correct one.
Why others are incorrect: 5% and 7.5% are too small and do not match standard code-based treatment; 12.5% is not the usual design value for battened struts. The commonly used increase is 10%.
The correct prismoidal formula for volume is
A)
D [first area + last area + Σ Even area + 2 Σ odd areas]
B)
D/3 [first area + last area + 4 Σ Even area + 2 Σ odd areas]
C)
D/3 [first area + last area + 2 Σ Even area + 4 Σ odd areas]
D)
D/6 [first area + last area + 2 Σ Even area + 4 Σ odd areas]
The prismoidal formula for volume is V = D/6 [A1 + An + 4(odd intermediate areas) + 2(even intermediate areas)], where D is the common interval and A1, An are the first and last areas. This matches option d exactly: D/6 [first area + last area + 2 Σ even areas + 4 Σ odd areas].
Why others are wrong:
- b uses D/3 instead of D/6, so the coefficient is incorrect.
- c also uses D/3, so it is incorrect despite having the 2 and 4 multipliers.
- a uses D instead of D/6 and has the multipliers reversed/incorrect for the prismoidal rule.
Falling head permeability is used for
A)
Fine grained soil
B)
Coarsed grained soil
C)
None of the above
D)
Both (a) & (b)
Falling head permeability test is used for fine-grained soils such as silts and clays, because their permeability is low and the discharge rate is too small for the constant head test to be convenient. In the falling head test, the drop in water head over time is measured to determine permeability.
Option (b) is incorrect because coarse-grained soils like sands and gravels have high permeability, so the constant head test is generally used for them. Option (c) is incorrect because the test is definitely used for a specific soil type. Option (d) is incorrect because falling head is not typically used for both fine- and coarse-grained soils; it is mainly for fine-grained soils.
Horizontal curves on highways are provided
A)
To break the monotony of driving
B)
To discourage the tendency to increase speed
C)
To decrease the mental strain on drivers
D)
All of these
Horizontal curves are provided on highways for several purposes: they break the monotony of straight driving, reduce the mental strain on drivers by making the route more natural and less tiring, and help discourage excessive speed by requiring drivers to negotiate the alignment. Therefore, all the listed reasons are correct. Options a, b, and c are each individually true, so the combined option d ('All of these') is the correct answer.
The GIVE WAY control
A)
Requires the driver in the minor road to slow down to a minimum speed and allow the vehicle on the major road to proceed.
B)
Requires the driver in the major road to slow down to a minimum speed and allow the vehicle on the minor road to proceed.
C)
Requires the drivers on both minor and major roads to stop.
D)
Is similar to one way control.
‘Give Way’ means the driver on the minor road must slow down and yield priority to traffic on the major road, proceeding only when it is safe. Hence option (a) is correct. Option (b) is wrong because the major road traffic has priority, not the minor road. Option (c) is wrong because ‘Give Way’ does not require both roads to stop; that is closer to a stop control. Option (d) is wrong because ‘Give Way’ is not the same as one-way control; one-way control regulates direction of travel, not right-of-way.
A pavement designer has arrived at a design traffic of 100 million standard axles for a newly developing national highway as per IRC:37 guidelines using the following data: design life = 15 years, commercial vehicle count before pavement construction = 4500 vehicles/day, annual traffic growth rate = 8%. The vehicle damage factor used in the calculation was
A)
1.56
B)
2.24
C)
3.20
D)
4.42
As per IRC:37, design traffic in msa is computed by N = 365 × A × [(1+r)^n − 1]/r × D × F, where A is initial commercial vehicles/day, r is annual growth rate, n is design life, D is lane distribution factor, and F is vehicle damage factor (VDF). For a newly developing national highway, take D = 0.75 for a single lane of a two-lane carriageway (or equivalent design lane factor used in IRC practice). Given N = 100 msa, A = 4500, r = 0.08, n = 15: the growth factor [(1.08)^15 − 1]/0.08 ≈ 24.31. Hence 100 × 10^6 = 365 × 4500 × 24.31 × 0.75 × F. Solving gives F ≈ 2.24. Therefore option b is correct.
Why others are incorrect: 1.56 is too low and would give a design traffic well below 100 msa for the given traffic and growth. 3.20 and 4.42 are too high; they would overestimate the axle repetitions and produce a design traffic much greater than 100 msa under the same conditions.
The Terzaghi’s general bearing capacity equation for a continuous footing is given by
A)
q = cNc + γDNq + 0.5γBNγ
B)
q = cNc - γDNq + 0.5γBNγ
C)
q = cNc + γDNq - 0.5γBNγ
D)
None of these
For a continuous (strip) footing, Terzaghi’s general bearing capacity equation is q_u = cNc + γDNq + 0.5γBNγ, where c is cohesion, γ is unit weight of soil, D is depth of footing, B is footing width, and Nc, Nq, Nγ are bearing capacity factors. Option (a) matches this standard form exactly.
Option (c) is incorrect because the Nγ term has the wrong sign; it should be positive, not negative. Option (b) is incorrect because both the Nq term and the Nc term/sign pattern are wrong; the depth surcharge term γDNq must be positive, and cohesion contributes positively as cNc. Option (d) is incorrect because the correct equation is present in option (a).
The curve composed of two arcs of different radii having their centres on the opposite side of the curve, is known as
A)
A simple curve
B)
A compound curve
C)
A reverse curve
D)
A vertical curve
A curve made up of two arcs of different radii with their centres on opposite sides of the curve is called a reverse curve. This arrangement means the direction of curvature changes from one arc to the next.
Why the others are incorrect:
- a) Simple curve: consists of a single arc of constant radius, not two arcs.
- b) Compound curve: consists of two or more arcs of different radii, but their centres lie on the same side of the curve.
- d) Vertical curve: is used in the vertical alignment of roads/railways to connect different gradients, not described by centres on opposite sides.
Planimeter is used for measuring
A)
Volume
B)
Area
C)
Contour gradient
D)
Slope angle
A planimeter is a surveying instrument used to measure the area of a plane figure on a map, drawing, or plan. Hence, option **b (Area)** is correct. The other options are incorrect because a planimeter does not measure **volume** (a), **contour gradient** (c), or **slope angle** (d); those require different instruments or calculations.
Stability of hill slope depends on
A)
Angle of slope
B)
Geological condition
C)
Groundwater conditions
D)
All of these
The stability of a hill slope depends on all the listed factors: the angle of slope, groundwater conditions, and geological condition. A steeper slope increases the driving force for sliding, groundwater raises pore-water pressure and reduces shear strength, and the geological nature of the material (rock/soil type, bedding, joints, faults, weathering) controls resistance to failure. Therefore, "All of these" is correct. The individual options a, b, and c are not complete by themselves because each is only one contributing factor, not the full set of controls on slope stability.
Free float for any activity is defined as the difference between
A)
Its earliest finish time and earliest start time for its successive activity
B)
Its latest start time and earliest start time
C)
Its latest finish time and earliest start time for its successive activity
D)
Its earliest finish time and latest start time for its successive activity
Free float is the amount of time an activity can be delayed without delaying the earliest start of its immediate successor. It is defined as: Free Float = Earliest Start of successor (ESj) - Earliest Finish of activity (EFi). This matches option (a): the difference between its earliest finish time and the earliest start time of its successive activity.
Why the others are incorrect:
- (b) Latest start time minus earliest start time is Total Float, not Free Float.
- (c) Latest finish time minus earliest start time of successor is not the standard definition of any float.
- (d) Earliest finish time minus latest start time of successor is also not the definition of free float and does not represent the allowable delay of the activity.
In construction management PERT technique of network analysis is mainly useful for
A)
Small project
B)
Large and complex project
C)
Deterministic activity
D)
Research and Development project
PERT (Program Evaluation and Review Technique) is mainly used for projects with uncertain activity durations, especially research and development projects, where time estimates are probabilistic rather than fixed. It helps plan, schedule, and control such non-repetitive, uncertain work.
Why others are incorrect:
- a) Small project: PERT is generally not needed for small projects; simpler scheduling methods are usually sufficient.
- b) Large and complex project: This is more commonly associated with CPM/PERT network analysis in general, but PERT is specifically most useful when activity times are uncertain, as in R&D.
- c) Deterministic activity: PERT is not meant for deterministic activities; CPM is better suited when activity durations are known with certainty.
Site order book is used for recording
A)
Instruction of the executive engineer
B)
Construction measurements
C)
Requisition of plants and equipment
D)
Indents for materials to be ordered.
The site order book is maintained at the work site to record instructions, directions, and orders issued by the executive engineer or other supervising officers. It serves as an official communication record for site execution.
Why others are incorrect:
- b) Construction measurements are recorded in the measurement book, not the site order book.
- c) Requisition of plants and equipment is handled through requisition/procurement records, not the site order book.
- d) Indents for materials to be ordered are entered in indent or material requisition registers, not in the site order book.
The type of bond in a brick masonry containing alternate courses of strechers and headers is called
A)
Flemish bond
B)
English bond
C)
Strecher bond
D)
Header bond
The correct answer is **English bond**. In English bond, the brick masonry is laid in **alternate courses of headers and stretchers**. This arrangement gives good strength and is commonly used in load-bearing walls.
Why the others are incorrect:
- **Flemish bond (a):** Each course contains **alternate headers and stretchers in the same course**, not alternate courses.
- **Stretcher bond (c):** All bricks are laid as **stretchers only** in every course.
- **Header bond (d):** All bricks are laid as **headers only** in every course.
Hence, the bond with alternate courses of stretchers and headers is **English bond**.