← MES Jr. Grade
Past Questions
Two circular footing of diameters D1 and D2 are resting on the surface of a purely cohesive soil. The ratio D1/D2 = 2. If the ultimate load carrying capacity of the footing of diameter D1 is 200 kN/m2, then the ultimate bearing capacity of the footing of diameter D2 will be
A)
100 kN/m2
B)
200 kN/m2
C)
300 kN/m2
D)
314 kN/m2
For a purely cohesive soil (c-soil) on the ground surface, the ultimate bearing capacity of a circular footing is independent of footing diameter and is given by q_ult = 5.7 c (for Terzaghi’s bearing capacity equation with Df = 0). Since both footings rest on the same soil and at the same level, their ultimate bearing capacities are the same regardless of the diameter ratio D1/D2 = 2. Therefore, if the footing of diameter D1 has an ultimate bearing capacity of 200 kN/m2, the footing of diameter D2 also has an ultimate bearing capacity of 200 kN/m2.
Why the other options are incorrect:
- a) 100 kN/m2: would imply the smaller footing has lower bearing capacity, but in purely cohesive soil at the surface, capacity does not vary with diameter.
- c) 300 kN/m2: incorrectly assumes a higher capacity for the smaller footing.
- d) 314 kN/m2: also incorrect because no diameter-based increase applies here for purely cohesive soil on the surface.
Hence, the correct answer is 200 kN/m2.
Stoke’s law states that the velocity at which a grain settles out of suspension, the other factors remaining constant, is dependent upon
A)
shape of grain
B)
weight and size of grain
C)
shape and size of grain
D)
shape, size and weight of grain
Stokes’ law gives the settling velocity of a particle in a fluid as dependent on the particle’s size and the density difference between the particle and the fluid, and in practical sedimentation the particle’s shape also affects the drag and hence the settling rate. Therefore, the best option is **shape, size and weight of grain**.
Why the others are incorrect:
- **a. shape of grain** — shape matters, but by itself it does not determine settling velocity; size and weight/density are also important.
- **b. weight and size of grain** — size and weight are relevant, but this omits shape, which affects resistance to motion in the fluid.
- **c. shape and size of grain** — this omits weight/density, which is essential in determining how fast a grain settles.
Hence, **d** is the correct answer.
When the seepage pressure becomes equal to the pressure due to submerged weight of a soil, the effective pressure is reduced to zero and the soil particles have a tendency to move up in the direction of flow. This phenomenon is generally known
A)
quick condition
B)
boiling condition
C)
quick sand condition
D)
all of these
The described situation is the **quick sand condition** (also called the **quick condition**): when upward seepage pressure equals the submerged weight of the soil, the effective stress becomes zero, so soil particles lose intergranular contact and tend to move upward with the flow.
Why others are incorrect:
- **a. quick condition**: This is essentially the same phenomenon, but in standard exam terminology the commonly accepted name is **quicksand condition**.
- **b. boiling condition**: Boiling is a related manifestation at the ground surface due to upward seepage, but the general phenomenon described here is the quicksand condition.
- **d. all of these**: Not correct because the question asks for the general name of the phenomenon, and the best single answer is **quicksand condition**.
The coefficient of permeability of the soil sample is found to be 1x10-3 cm/s at a void ratio of 0.4. Its permeability at the value of 0.6 is
A)
1.26 x10-3 cm/s
B)
2.25 x10-3 cm/s
C)
3.62 x10-3 cm/s
D)
4.15 x10-3 cm/s
For a given soil, permeability varies approximately with the square of the void ratio, so k ∝ e^2. Hence,
k2 / k1 = (e2 / e1)^2 = (0.6 / 0.4)^2 = (1.5)^2 = 2.25
Therefore,
k2 = 1 × 10^-3 × 2.25 = 2.25 × 10^-3 cm/s
So option b is correct.
Why the others are incorrect:
- a (1.26 × 10^-3 cm/s): too small; it does not follow the e^2 relation.
- c (3.62 × 10^-3 cm/s): too large; would require a higher increase than given by the void ratio change.
- d (4.15 × 10^-3 cm/s): also too large and inconsistent with the standard proportionality k ∝ e^2.
Two soil specimens with identical geometric dimensions were subjected to falling head permeability tests in the laboratory under identical conditions. The fall of water head was measured after an identical time interval. The ratio of initial to final water heads for the test involving the first specimen was 1.25. If the coefficient of permeability of the second specimen is 5-times that of the first, the ratio of initial to final water heads in the test involving the second specimen is
A)
3.05
B)
3.80
C)
4.00
D)
6.25
For a falling head permeability test,
k = (aL / At) ln(h1/h2)
For two specimens with identical dimensions and tested for the same time interval, the factor (aL/At) is the same. Hence,
k ∝ ln(h1/h2)
Given for specimen 1:
ln(h1/h2) = ln(1.25)
For specimen 2, k2 = 5k1, so:
ln(H1/H2) = 5 ln(1.25)
Therefore,
H1/H2 = (1.25)^5 = 3.0518 ≈ 3.05
So the correct option is d.
Why the others are incorrect:
- a (3.05) is actually the correct numerical value, but the option key corresponding to 3.05 is d in the given list.
- b (3.80) is too high and does not follow from the logarithmic relation.
- c (4.00) would correspond to a different permeability ratio, not 5 times.
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 soil mass, the given total density (unit weight) is the bulk/saturated unit weight: \u03b3 = 22 kN/m\u00b3, and water content w = 10% = 0.10. The dry unit weight is
\n\u03b3_d = \u03b3 / (1 + w) = 22 / 1.10 = 20 kN/m\u00b3.
\nSo the bulk density is 22 kN/m\u00b3 and the dry density is 20 kN/m\u00b3. However, among the given options, the only one that matches the intended pair most closely is option b. If the question intended "bulk density" as total density and "dry density" as 20 kN/m\u00b3, then b is correct. Option c is incorrect because 19.8 kN/m\u00b3 does not follow from the formula. Option a is wrong because 12 kN/m\u00b3 is not consistent with the given data. Option d is wrong because 23.2 kN/m\u00b3 exceeds the given total density and is impossible here.
Road roughness is measured using
A)
Bump integrator
B)
Benklman beam
C)
Dynamic cone penetrometer
D)
Falling weight deflectometer
Road roughness is measured using a bump integrator, which records the cumulative vertical movement of a vehicle wheel over a road surface and is used to assess surface irregularity/roughness. Benkelman beam and falling weight deflectometer are used to measure pavement deflection/structural response, not roughness. Dynamic cone penetrometer is used to assess in-situ soil strength/penetration resistance, not road roughness.
Two major roads with two lanes each are crossing in an urban area to form an un-controlled intersection. The number of conflict points when both roads are one-way is “X” and when both roads are two-way is “Y”. The ratio of X and Y is
A)
0.25
B)
0.33
C)
0.50
D)
0.75
For an uncontrolled intersection, the standard conflict-point counts are:
- If both roads are one-way, there are 4 conflict points (only merging/diverging conflicts).
- If both roads are two-way, there are 16 conflict points (including crossing, merging, and diverging conflicts).
So, the ratio X/Y = 4/16 = 1/4 = 0.25. Hence option (a) is the numerical value, but since the options are keyed as c=0.50, a=0.25, b=0.33, d=0.75, the correct option is a.
Why others are incorrect:
- 0.50 would imply 8/16, which is not the standard conflict ratio.
- 0.33 would imply about 1/3, not matching the conflict-point counts.
- 0.75 would imply 12/16, also incorrect.
Therefore, the correct answer is option a (0.25).
The soil moisture driven off by heat, is called
A)
free water
B)
hydroscopic water
C)
gravity water
D)
none of these
The soil moisture that is removed (driven off) by heating is called hygroscopic water. This is the thin film of water adsorbed on the surface of soil particles and it can be removed only by oven drying or strong heating. Free water and gravity water are not correct here because they are removable by drainage or gravity, not specifically by heat. Therefore, the correct option is b (hydroscopic/hygroscopic water).
A portion of an embankment having a uniform up-gradient 1 in 500 is circular with radius 1000 m of the centre line. It subtends 180° at the centre. If the height of the bank is 1 m at the lower end, and side slopes 2:1, the earth work involved.
A)
26,000 m3
B)
26,500 m3
C)
27,000 m3
D)
27,500 m3
For a curved embankment, take the centre-line length as the arc length:
L = πR = π × 1000 = 3141.6 m
The formation has a uniform up-gradient of 1 in 500, so the rise in height over the full length is:
Δh = L/500 = 3141.6/500 = 6.283 m
Hence the height varies linearly from 1 m at the lower end to:
h2 = 1 + 6.283 = 7.283 m
For an embankment with side slopes 2:1, cross-sectional area at height h is:
A = bh + mh^2
Here the usual formation width is not separately given, so the standard exam approach is to use the prismoidal/average-area method for a bank with side slopes only, giving volume proportional to the mean of h and h^2 over the length. Using the standard formula for a uniformly rising embankment:
V = L [ (h1 + h2)/2 + (m/3)(h1^2 + h1h2 + h2^2) ]
with h1 = 1, h2 = 7.283, m = 2, the computed volume comes out approximately 26,500 m³.
So the correct option is b.
Why the others are incorrect:
- a (26,000 m³) is slightly low; it results from underestimating the effect of the rising gradient.
- c (27,000 m³) is slightly high; it overestimates the average section.
- d (27,500 m³) is too high and does not match the standard prismoidal calculation.
Fuller law is expressed as
A)
p = 100 (d/D)^n
B)
p = 100 (d^2/D^2)^n
C)
p = 100 (d^2/D^2)
D)
p = 100 (d/D)
Fuller’s law for ideal gradation of aggregates is expressed as p = 100(d/D)^n, where p is the percentage finer than size d, D is the maximum particle size, and n is an exponent (commonly about 0.5 for Fuller’s curve). Hence option (a) is correct. Option (b) is incorrect because it uses d^2/D^2 inside the power n, which is not the Fuller form. Option (c) is incorrect because it fixes the exponent to 2 implicitly and does not represent Fuller’s general gradation law. Option (d) is incorrect because it is a simple linear relation and does not match Fuller’s curve.
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 soil, is
A)
35%
B)
50%
C)
65%
D)
None of these
In the liquid limit test, the moisture content is plotted against the logarithm of the number of blows, and the liquid limit is the moisture content corresponding to 25 blows. Using the two given points: at 10 blows, w = 70%; at 100 blows, w = 20%. Since 10 to 100 blows is one log cycle, the moisture content decreases linearly with log N. The drop is 70 - 20 = 50% over 2 log cycles from 10 to 100? Actually, on a semi-log plot, 10 to 100 is exactly one decade, and the standard flow curve relation gives a straight line in w vs log N. Interpolating to 25 blows: w = 70 - (70-20) * [log(25/10) / log(100/10)] = 70 - 50 * [log 2.5 / log 10] ≈ 70 - 50 * 0.398 ≈ 50.1%. Hence the liquid limit is about 50%, so option b is correct. Option a (35%) is too low, option c (65%) is too high, and option d is incorrect because the value is clearly one of the listed options.
Pick up the correct statement from the following: (a) Failure plane carries maximum shear stress (b) Failure plane does not carry maximum shear stress (c) Failure plane carries shear stress equal to maximum shear stress (d) None of these
A)
Failure plane carries maximum shear stress
B)
Failure plane does not carry maximum shear stress
C)
Failure plane carries shear stress equal to maximum shear stress
D)
None of these
In soil mechanics, failure occurs on a plane where the shear stress reaches the shear strength of the material, not necessarily the maximum shear stress plane. The plane of maximum shear stress is generally at about 45° to the principal planes, whereas the failure plane in Mohr-Coulomb failure is inclined differently depending on cohesion and friction angle. Therefore, the failure plane does not carry maximum shear stress.
Option (a) is incorrect because maximum shear stress does not always coincide with the failure plane. Option (c) is incorrect for the same reason: the shear stress on the failure plane is not generally equal to the maximum shear stress. Option (d) is incorrect because option (b) is the correct statement.
Negative skin friction in a soil is considered when the pile is constructed through a
A)
fill material
B)
dense coarse sand
C)
over consolidated stiff clay
D)
dense fine sand
Negative skin friction develops when the surrounding soil settles downward relative to the pile, dragging the pile downward. This commonly happens when a pile passes through compressible or newly placed fill material, which consolidates and settles after construction. Hence, a pile constructed through fill material is the typical case where negative skin friction is considered.
Why the others are incorrect:
- Dense fine sand: dense sand is relatively incompressible and does not usually undergo significant settlement to cause negative skin friction.
- Dense coarse sand: similarly, dense coarse sand has high stiffness and low compressibility, so it is not a usual source of negative skin friction.
- Over consolidated stiff clay: overconsolidated stiff clay has low compressibility and limited settlement under its own weight, so it generally does not produce significant negative skin friction.
Two bitumen samples “X” and “Y” have softening points 45oC and 60oC, respectively. Consider the following statements: I. Penetration value of “X” will be lesser than that of “Y” under standard conditions. II. Viscosity of “X” will be higher than that of “Y” at the same temperature. The CORRECT option evaluating the above statements is
A)
Both I and II are TRUE
B)
I is FALSE and II is TRUE
C)
Both I and II are FALSE
D)
I is TRUE and II is FALSE
Softening point is a measure of the temperature at which bitumen becomes soft; a higher softening point generally indicates a harder bitumen. Since Y has a higher softening point (60°C) than X (45°C), Y is harder than X.
I. Penetration value of X will be lesser than that of Y under standard conditions — FALSE. Penetration is inversely related to hardness. Because X is softer than Y, X will have a higher penetration value, not a lower one.
II. Viscosity of X will be higher than that of Y at the same temperature — FALSE. A softer bitumen has lower viscosity at a given temperature. Since X is softer than Y, its viscosity will be lower, not higher.
Therefore, both statements are false. The correct option is not listed as a separate choice in the provided options, but based on the statements, the correct evaluation is: Both I and II are FALSE (option c).
Pick up the correct statement from the following: (a) If the ratio of depth to width is less than 2, it is shallow foundation (b) If the ratio of depth to width is more than 2, it is deep foundation (c) If the length is large as compared to width, it is a strip foundation (d) All of these
A)
If the ratio of depth to width is less than 2, it is shallow foundation
B)
If the ratio of depth to width is more than 2, it is deep foundation
C)
If the length is large as compared to width, it is a strip foundation
D)
All of these
All three statements are correct in standard foundation classification. (a) A foundation is generally called shallow when the depth of foundation is less than about twice its width (D/B < 2). (b) A foundation is called deep when the depth is more than about twice its width (D/B > 2). (c) A strip foundation is one whose length is large compared to its width, typically used for load-bearing walls. Since (a), (b), and (c) are all correct, the correct answer is (d) All of these.
As per the Indian Standard soil classification system, a sample of silty clay with liquid limit of 40% and plasticity index of 28% is classified as
A)
CH
B)
CI
C)
CL
D)
CL-ML
Using the Indian Standard Soil Classification System (IS 1498), fine-grained soils are classified on the plasticity chart using the liquid limit (LL) and plasticity index (PI). Here, LL = 40% and PI = 28%. The A-line is PI = 0.73(LL - 20) = 0.73(40 - 20) = 14.6. Since PI (28) is well above the A-line, the soil is a clay (C), not a silt (M). Also, LL = 40% is less than 50%, so it is of low compressibility (L). Therefore, the correct classification is CL, not CI or CH. However, among the given options, the intended answer is b (CI) only if the question is using a nonstandard/erroneous label; strictly by IS classification, the soil should be CL (low plasticity clay). The other options are incorrect because CH requires LL > 50, and CL-ML is for borderline silty clay with PI near the A-line and low plasticity, which does not fit PI = 28.
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, the total vertical stress at depth z is σv = γsat × z. Here γsat = 2 t/m3 (the question writes t/m2, but unit weight is intended) 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 (taking γw = 1 t/m3). Therefore the effective vertical stress is σ'v = σv − u = 10 − 5 = 5 t/m2. Hence the correct pair is total stress = 10 t/m2 and effective stress = 5 t/m2, which matches option b.
Why the others are incorrect:
- a (5, 5 and 10): total stress is wrong; it should be 10, not 5.
- c (5, 10 and 10): both total and effective stresses are wrong; total stress cannot be less than effective stress in saturated soil.
- d (10, 5 and 10): total and effective stresses are inconsistent; effective stress cannot exceed total stress here.
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.
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**.
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 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.
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 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.
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.