← MES Jr. Grade
Past Questions
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°.
Rankine theory of active earth pressure assumes
A)
Soil mass is homogeneous, dry and cohesionless
B)
Ground surface is a plane which may be horizontal or inclined
C)
Back of the wall is vertical and smooth
D)
All of these
Rankine’s active earth pressure theory assumes: (1) the backfill soil is homogeneous, dry, and cohesionless; (2) the ground surface is a plane, either horizontal or uniformly inclined; and (3) the back of the retaining wall is vertical and smooth (so wall friction is neglected). Since options a, b, and c are all correct assumptions, the correct answer is d, All of these.
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).
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.
When drainage is permitted under initially applied normal stress only and full primarily consolidation is allowed to take place, the test is known as
A)
Consolidated drained test
B)
Drained test
C)
Quick test
D)
None of these
The correct answer is **Consolidated drained test**. In this test, drainage is allowed during the application of the initial normal stress, and the specimen is then allowed to undergo full primary consolidation before further loading/shearing. This matches the condition described in the question.
Why the others are incorrect:
- **b) Drained test**: This is too general; it only indicates drainage is permitted, but the question specifically describes drainage under the initial normal stress followed by full consolidation, which is the consolidated drained condition.
- **c) Quick test**: In a quick test, drainage is not permitted and consolidation does not occur before shearing.
- **d) None of these**: Incorrect because the described test is a standard **consolidated drained test**.
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.
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.
CBR is used for
A)
Shear test
B)
Penetration test
C)
Bearing test
D)
Flexural test
CBR stands for California Bearing Ratio, which is a penetration test used to assess the strength of subgrade soil and base materials for pavement design. In the test, a standard plunger is driven into the soil and the resistance is measured, so the correct option is penetration test. Shear test measures shear strength, bearing test is a broader term not specific to CBR, and flexural test is used for bending strength of materials like concrete or beams.
The following data are related to a horizontal curved of a two lane highway: length of curve = 200 m, radius of curve = 300 m and width of pavement = 7.5 m. Inorder to provide a stopping sight distance (SSD) of 80 m, the set back distance (in m) required from the centre line of the inner lane of the pavement is
A)
2.64
B)
4.55
C)
7.10
D)
7.96
For a horizontal curve, the required set back distance (middle ordinate) to provide sight distance S is:
m = R \left[1 - \cos\left(\frac{S}{2R}\right)\right]
Given:
R = 300 m, S = 80 m
m = 300 \left[1 - \cos\left(\frac{80}{2 \times 300}\right)\right]
= 300 \left[1 - \cos\left(\frac{80}{600}\right)\right]
= 300 \left[1 - \cos(0.1333)\right]
Using cos(0.1333) \approx 0.9911,
m \approx 300(1 - 0.9911) = 300(0.0089) = 2.67 m
This is the set back distance from the centre line of the inner lane. The nearest option is 2.64 m, so option a is correct.
Why the other options are incorrect:
- 4.55 m, 7.10 m, and 7.96 m are too large for the given radius and sight distance.
- They would correspond to either a much smaller radius or a much larger required sight distance than 80 m.
Note: The curve length and pavement width are not needed for this basic sight-distance set back calculation when the radius of the inner lane centerline is already given.
Which one of the following statements is true?
A)
Clays are more porous than sands
B)
Pressure of organic matter in a soil decreases the bearing capacity of the soil
C)
Aluminous cement is used for foundations in soils with chemical deposits
D)
All the above
All the listed statements are true. (a) Clays generally have higher porosity than sands because clay particles are very fine and create many voids. (b) Organic matter in soil reduces bearing capacity since it is compressible and weak, lowering soil strength. (c) Aluminous cement is used in foundations exposed to chemically aggressive soils because it has better resistance to chemical attack than ordinary Portland cement. Hence, the correct answer is 'All the above'.
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.
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 minimum water content at which the soil retains its liquid state and also possesses a small shearing strength against flowing is known as
A)
Plastic limit
B)
Liquid limit
C)
Shrinkage limit
D)
Permeability limit
The correct answer is **Liquid limit (b)**. It is defined as the **minimum water content at which a soil remains in the liquid state and has only a small shearing strength, just enough to resist flowing**. This matches the statement exactly.
Why the others are incorrect:
- **Plastic limit (a):** The water content at which soil changes from **plastic to semi-solid** state; it is not the liquid-state boundary.
- **Shrinkage limit (c):** The water content below which **further drying does not cause any decrease in volume**; it relates to shrinkage, not liquid behavior.
- **Permeability limit (d):** This is **not a standard Atterberg limit** in soil mechanics and is not related to the consistency state described.
Two series of compaction tests were performed in the laboratory on an inorganic clayey soil employing two different levels of compaction energy per unit volume of soil. With regard to the above tests, the following two statements were made: (i) The optimum moisture content is expected to be more for the tests with higher energy; (ii) The maximum dry density is expected to be more for the tests with higher energy.
A)
Only (i) is true
B)
Only (ii) is true
C)
Both are true
D)
Both are false
For an inorganic clayey soil, increasing the compaction energy generally shifts the compaction curve upward and to the left. This means:
(i) The optimum moisture content (OMC) decreases, not increases, with higher compaction energy. Higher energy helps soil particles rearrange more efficiently at lower water content, so statement (i) is false.
(ii) The maximum dry density (MDD) increases with higher compaction energy because the soil can be packed more tightly. So statement (ii) is true.
Therefore, only statement (ii) is correct.
Why the other options are wrong:
- (a) Only (i) is true: incorrect because (i) is false.
- (c) Both are true: incorrect because (i) is false.
- (d) Both are false: incorrect because (ii) is true.
Brake is applied on a vehicle which then skids a distance of 16m before coming to stop. If the developed average coefficient of friction between the tyres and the pavement is 0.4, then, the speed of the vehicle before skidding would have been nearly
A)
50 kmph
B)
20 kmph
C)
30 kmph
D)
40 kmph
For a skidding vehicle, the braking work done by friction equals the initial kinetic energy:
v^2 = 2 μ g s
Given μ = 0.4, g ≈ 9.81 m/s², s = 16 m:
v^2 = 2 × 0.4 × 9.81 × 16 = 125.57
v ≈ 11.2 m/s
Convert to km/h:
11.2 × 3.6 ≈ 40.3 km/h
So the speed before skidding was nearly 40 km/h, which corresponds to option d.
Why others are incorrect:
- 20 km/h and 30 km/h are too low; they would produce much shorter skid distances for the same friction.
- 50 km/h is too high; with μ = 0.4, a vehicle at 50 km/h would skid farther than 16 m before stopping.
Correct option: d
The coefficient of compressibility of soil is the ratio of
A)
Stress to strain
B)
Strain to stress
C)
Stress to settlement
D)
Rate of loading to that of settlement
The coefficient of compressibility of soil is defined as the ratio of strain to stress (more precisely, change in volume or void ratio per unit increase in effective stress, depending on the context). Hence option **b** is correct.
Why others are incorrect:
- **a. Stress to strain** is the inverse of compressibility; it represents stiffness or modulus, not compressibility.
- **c. Stress to settlement** is not the definition of coefficient of compressibility; settlement is a total deformation measure, not the standard ratio used.
- **d. Rate of loading to that of settlement** is unrelated to the soil compressibility definition.
The length of Summit Curve on a two lane two way highway depends
A)
upon Allowable rate of change of centrifugal acceleration
B)
Coefficient of lateral friction
C)
Required Stopping Sight Distance
D)
Required Overtaking Sight Distance
For a summit (crest) curve on a two-lane, two-way highway, the controlling design criterion is usually the **required overtaking sight distance (OSD)**, because the curve must provide enough visibility for safe passing on a two-way road. Hence the length of the summit curve depends on OSD.
Why the others are incorrect:
- **a) Allowable rate of change of centrifugal acceleration**: This is relevant to **transition curves** and comfort in horizontal alignment, not summit curves.
- **b) Coefficient of lateral friction**: This is used in **horizontal curve design** to relate speed, radius, and superelevation, not for summit curve length.
- **c) Required Stopping Sight Distance**: SSD is also a visibility criterion for crest curves, but for a **two-lane two-way highway**, the usual governing requirement for summit curve length is **overtaking sight distance**, which is more demanding than SSD.
In a liquid limit test, the moisture content at 10 blows was 70% and that at 100 blows was 20%. The liquid limit of the soils, is
A)
35%
B)
50%
C)
65%
D)
none of these
In the liquid limit test, the water content is plotted against the logarithm of number of blows, and the liquid limit is the moisture content corresponding to 25 blows. Using the two given points (10 blows, 70%) and (100 blows, 20%), the flow line is assumed linear on a semi-log plot. Since 25 blows lies between 10 and 100, interpolate on the log scale:
w at 25 blows = 70 + (20 - 70) * [log10(25) - log10(10)] / [log10(100) - log10(10)]
= 70 - 50 * [log10(2.5)] / 1
≈ 70 - 50 * 0.398
≈ 50.1%
So the liquid limit is approximately 50%, hence option (b).
Why others are incorrect:
- (a) 35% is too low; it would correspond to a much drier soil than indicated by the flow line.
- (c) 65% is too high; it is close to the 10-blow moisture content, not the 25-blow value.
- (d) none of these is incorrect because the computed liquid limit is about 50%.
A saturated clay layer with double drainage takes 5 years to attain 90% degree of consolidation under a structure. If the same layer were to be single drained, what would be the time (in years) required to attain the same consolidation under the same loading conditions?
A)
10
B)
15
C)
20
D)
25
For 1-D consolidation, the time factor is t = Tv Hdr^2 / Cv. For the same clay and same degree of consolidation (90%), Tv and Cv remain the same, so time is proportional to the square of the drainage path length Hdr^2. In double drainage, the maximum drainage path is H/2; in single drainage, it is H. Therefore, the drainage path doubles, and the time becomes 2^2 = 4 times. Given 5 years for double drainage, the single-drainage time = 5 × 4 = 20 years. Hence the correct answer is 20 years. Among the options, 10, 15, and 25 years are incorrect because they do not reflect the square-law increase in consolidation time when the drainage path doubles.
A saturated soil mass has a total density 22kN /m3 and water content of 10%. The bulk density and dry density of this soil are
A)
12kN /m3 and 20kN/m3 respectively
B)
22kN /m3 and 20kN/m3 respectively
C)
19.8kN /m3 and 19.8kN/m3 respectively
D)
23.2kN /m3 and 19.8kN/m3 respectively
For a saturated soil mass, the given total density (unit weight) is the bulk/saturated density: \(\gamma = 22\,\text{kN/m}^3\). Water content \(w = 10\% = 0.10\). Using the relation \(\gamma = \gamma_d(1+w)\), the dry density is \(\gamma_d = \frac{\gamma}{1+w} = \frac{22}{1.10} = 20\,\text{kN/m}^3\). Hence the bulk density and dry density are \(22\,\text{kN/m}^3\) and \(20\,\text{kN/m}^3\), respectively.
Why others are incorrect:
- a: bulk density 12 is not given and does not match the saturated unit weight.
- c: 19.8 and 19.8 are incorrect because dry density must be less than bulk density when water content is positive.
- d: 23.2 is higher than the given saturated bulk density, so it cannot be correct; dry density 19.8 is also not the exact value from the formula.
Degree of saturation of a natural soil deposit having water content 15%, specific gravity 2.50 and void ratio 0.5 is
A)
60%
B)
50%
C)
75%
D)
80%
Use the relation between water content (w), specific gravity (G), void ratio (e), and degree of saturation (S): w = (S·e)/G. Hence S = (w·G)/e = (0.15 × 2.50)/0.5 = 0.75 = 75%. Therefore the correct answer is option c, not b. The other options are incorrect because 50%, 60%, and 80% do not satisfy the standard soil phase relation for the given data.
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**.
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 lateral earth pressure on the retaining wall is
A)
Equal to the mass of the soil retained
B)
Proportional to the square of the depth of the soil
C)
Proportional to the internal friction of the soil
D)
Proportional to the depth of the soil
The lateral earth pressure on a retaining wall increases with depth, and the total resultant earth pressure on the wall is proportional to the square of the height/depth of the retained soil (for a uniform backfill). This is because the pressure intensity varies linearly with depth, so integrating it over the wall height gives a triangular distribution whose resultant is proportional to h^2. Option (b) is therefore correct.
Why the others are incorrect:
- (a) Equal to the mass of the soil retained: earth pressure is a force due to soil stress, not simply equal to the soil mass.
- (d) Proportional to the depth of the soil: this is true for pressure intensity at a given depth, not for the total resultant force on the wall.
- (c) Proportional to the internal friction of the soil: internal friction affects the magnitude of earth pressure coefficient, but the pressure is not directly proportional to friction alone.
The boundary of water of a still lake represents
A)
Level surface
B)
Horizontal surface
C)
Contour line
D)
Concave surface
The boundary of water in a still lake is a **level surface** because the free surface of a liquid at rest is everywhere perpendicular to the direction of gravity and has the same elevation throughout. Hence it represents a level surface.
Why others are incorrect:
- **b. Horizontal surface**: A horizontal surface is a plane surface; the water surface of a lake is not necessarily a plane, though it is level.
- **c. Contour line**: A contour line is a line joining points of equal elevation on a map, not the actual water surface.
- **d. Concave surface**: The water surface of a still lake is not concave; it is level due to gravity.