← 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.
For determining the moisture content of a soil sample, the following data is available Weight of container = 260 g, Weight of soil sample = 320 g container, Weight of soil sample (dried) = 310 g container. The moisture content of the soil sample, is
A)
15%
B)
18%
C)
20%
D)
28%
Moisture content, w = (weight of water / weight of dry soil) × 100.
Given:
- Weight of container = 260 g
- Weight of container + wet soil = 320 g
- Weight of container + dry soil = 310 g
So,
- Wet soil weight = 320 - 260 = 60 g
- Dry soil weight = 310 - 260 = 50 g
- Water content = 60 - 50 = 10 g
Therefore,
w = (10 / 50) × 100 = 20%
So the correct option is c (20%).
Why others are incorrect:
- a (15%): would correspond to a smaller water content than given.
- b (18%): does not match the computed ratio of water to dry soil.
- d (28%): is too high and would require much more water than the data shows.
The lateral earth pressure on a retaining wall
A)
is equal to mass of the soil retained
B)
proportional to the depth of the soil
C)
proportional to the square of the depth of the soil
D)
proportional to the internal friction of the soil
The lateral earth pressure on a retaining wall increases with depth and, for a soil at rest or in active/passive conditions, the pressure intensity varies linearly with depth (p = K\u03b3z). Therefore, the total resultant earth pressure on the wall is proportional to the square of the depth/height of the retained soil (P = 1/2 K\u03b3H^2). Hence option (c) is correct.
Why the others are incorrect:
- (b) 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.
- (d) Proportional to the internal friction of the soil: internal friction affects the earth pressure coefficient K, but the pressure is not directly proportional to friction alone.
- (a) Equal to mass of the soil retained: earth pressure is a force due to soil stress, not simply the mass of the retained soil.
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.
Traffic signal is an example of
A)
Passive control
B)
No control
C)
Active control
D)
None of these
Traffic signals are an example of active control because they regulate traffic flow by using a device (signal lights) that actively controls the movement of vehicles and pedestrians.
Why others are incorrect:
- Passive control: This refers to measures like signs, markings, or road geometry that guide traffic without actively stopping or releasing vehicles.
- No control: Traffic signals clearly impose control, so this is wrong.
- None of these: Incorrect because active control is the correct classification.
The length of Summit Curve on a two lane two way highway depends upon
A)
Allowable rate of change of centrifugal acceleration
B)
Coefficient of lateral friction
C)
Required Stopping Sight Distance
D)
Required Overtaking Sight Distance
The length of a summit (crest) curve on a two-lane, two-way highway is governed by the sight distance requirement, specifically the Required Stopping Sight Distance (SSD). On a crest curve, the driver’s line of sight is obstructed by the curve shape, so the curve length must be sufficient to provide the necessary visibility for safe stopping. Hence, SSD is the controlling criterion.
Why the others are incorrect:
- a) Allowable rate of change of centrifugal acceleration is used for transition curves in horizontal alignment, not summit curves.
- b) Coefficient of lateral friction is also related to horizontal curve design, where lateral resistance is needed to negotiate the curve safely.
- d) Required Overtaking Sight Distance is relevant mainly for passing maneuvers on two-lane roads, but for summit curve length the standard controlling sight distance is SSD, not OSD.
The correct prismoidal formula for volume is
A)
D [first area + last area + å Even area + 2 å odd areas]
B)
D/3 [first area + last area + 4 å Even area + 2 å odd areas]
C)
D/3 [first area + last area + 2 å Even area + 4 å odd areas]
D)
D/6 [first area + last area + 2 å Even area + 4 å odd areas]
The prismoidal formula for volume is V = D/3 [A1 + An + 4(sum of even-interval areas) + 2(sum of odd-interval areas)], where D is the common interval between sections. This matches option (b) exactly. Option (c) is incorrect because the coefficients of even and odd areas are interchanged. Option (d) is also incorrect for the same reason and because the factor D/6 is not the standard prismoidal coefficient. Option (a) is incorrect because it omits the required division by 3 and uses wrong coefficients.
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.
Broken lines
A)
Allows crossing with discretion
B)
Does not allow crossing except for entry or exit from a side road
C)
Allows crossing only in case of extreme emergency
D)
Are not at all used as road markings
Option A: Allows crossing with discretion. This matches the standard definition or established fact for the concept asked.
Saturation flow rate can be computed as,
A)
3600/h
B)
h/3600
C)
3600×h
D)
None of these
Saturation flow rate is the number of vehicles that can pass a point in one hour of green under continuous discharge conditions. If the average headway between vehicles is h seconds/vehicle, then the flow rate is 3600/h vehicles per hour. Hence option (a) is correct.
Why others are incorrect:
- (b) h/3600 is the inverse of the required relationship and gives a very small number, not vehicles per hour.
- (c) 3600×h increases with headway, which is opposite to the definition of flow rate; larger headway means lower flow.
- (d) None of these is incorrect because option (a) matches the standard formula.
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.
Terzaghi’s analysis assumes : (a) soil is homogeneous and isotropic (b) elastic zone has straight boundaries inclined at = to the horizontal and plastic zones fully developed (c) failure zones do not extend above the horizontal plane through the base of the footing (d) all of these
A)
soil is homogeneous and isotropic
B)
elastic zone has straight boundaries inclined at = to the horizontal and plastic zones fully developed
C)
failure zones do not extend above the horizontal plane through the base of the footing
D)
all of these
Terzaghi’s bearing capacity analysis is based on several simplifying assumptions: (a) the soil is homogeneous and isotropic, (b) the failure mechanism is idealized with straight boundaries for the elastic zone inclined at ψ = φ and fully developed plastic zones, and (c) the failure zones are assumed not to extend above the horizontal plane through the base of the footing. Since all three statements are part of Terzaghi’s assumptions, the correct answer is (d) all of these. Options (a), (b), and (c) are each individually correct assumptions, so none of them is incorrect on its own.
Pick up the correct statement from the following: (a) The estimated value of the work excluding the amount for contigencies, work charged establishment, tool and plants, is called work value. (b) The actual expenditure involved to complete a work including incidental, establishment and travelling charges, is called actual cost. (c) The formal acceptance by the administrative department for incurring an expenditure on the work, is called administrative approval. (d) All of these.
A)
The estimated value of the work excluding the amount for contigencies, work charged establishment, tool and plants, is called work value.
B)
The actual expenditure involved to complete a work including incidental, establishment and travelling charges, is called actual cost.
C)
The formal acceptance by the administrative department for incurring an expenditure on the work, is called administrative approval.
D)
All of these.
All three statements are correct as per standard civil engineering/estimation terminology. (a) The work value is the estimated cost of the work excluding contingencies, work-charged establishment, and tools & plants. (b) Actual cost means the total expenditure actually incurred to complete the work, including incidental, establishment, and travelling charges. (c) Administrative approval is the formal sanction by the administrative department to incur expenditure on the work. Since (a), (b), and (c) are all correct, the correct option is (d) All of these.
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.
- Double drainage: drainage occurs from both top and bottom, so Hdr = H/2.
- Single drainage: drainage occurs from only one side, so Hdr = H.
Thus, when changing from double to single drainage, the drainage path doubles, and the time becomes 2^2 = 4 times larger.
Given t_double = 5 years,
t_single = 4 × 5 = 20 years.
So the correct answer is option c (20 years).
Why the other options are incorrect:
- a (10 years): only 2 times the original time, but time should increase by 4 times.
- b (15 years): not consistent with the square relationship of drainage path length.
- d (25 years): would imply 5 times the original time, which is too high.
Note: The correct numerical answer is 20 years, which corresponds to option c.
Depending upon the properties of a material, the failure envelope may
A)
be either straight or curved
B)
pass through the origin of stress
C)
intersect the shear stress axis
D)
all of these
The failure envelope is the locus of stress states at failure, and its shape depends on the material behavior and failure criterion. For different materials it may be straight (e.g., idealized linear criteria) or curved (e.g., nonlinear criteria), it may pass through the origin for materials with no cohesion, and it may intersect the shear stress axis for materials with finite shear strength. Hence, all the listed possibilities can occur depending on the material, so the correct answer is 'all of these'. Option (a) is true but incomplete, while (b) and (c) are also possible in specific cases, so none of them alone is the best answer.
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.
A layer of normally consolidated, saturated silty clay of 1m thickness is subjected to 1-D consolidation under a pressure increment of 20kPa. The properties of the soil are : specific gravity = 2.7, natural moisture content = 45%, compression index = 0.45, and recompression index = 0.05. The initial average effective stress within the layer is 100 kPa. Assuming Terzaghi’s theory to be applicable, the primary consolidation settlement (rounded off to the nearest mm) is
A)
2
B)
4
C)
8
D)
16
For a normally consolidated clay, primary consolidation settlement is
S = H/(1+e0) × Cc × log10((σ'0 + Δσ)/σ'0)
First find the initial void ratio from water content. Since the soil is saturated, e0 = wGs = 0.45 × 2.7 = 1.215.
Given:
H = 1 m = 1000 mm
Cc = 0.45
σ'0 = 100 kPa
Δσ = 20 kPa
So,
S = 1000/(1 + 1.215) × 0.45 × log10(120/100)
= 1000/2.215 × 0.45 × log10(1.2)
≈ 451.5 × 0.45 × 0.07918
≈ 16.1 mm
Rounded to the nearest mm, settlement = 16 mm.
Therefore, the correct option is d.
Why the other options are incorrect:
- a (2 mm) and b (4 mm) are too small; they would result from using an incorrect void ratio or neglecting the compression index.
- c (8 mm) is still about half the correct value and typically comes from underestimating the logarithmic stress ratio or using the wrong thickness factor.
Note: The computed value corresponds to option d, not b.
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 SSD is
m = R \left[1 - \cos\left(\frac{\theta}{2}\right)\right]
where the angle subtended by the sight distance is
\theta = \frac{S}{R} = \frac{80}{300} = 0.2667 \text{ rad}
So,
m = 300\left[1 - \cos\left(\frac{0.2667}{2}\right)\right]
= 300\left[1 - \cos(0.1333)\right]
\approx 300(1 - 0.9911)
\approx 2.64 \text{ m}
Hence, the correct answer is option b.
Why the others are incorrect:
- a (2.64) is actually the correct value, so it is not incorrect.
- c (7.10) and d (7.96) are much larger than the required set back distance and do not match the standard SSD-on-curve calculation.
- The pavement width is not needed for this basic set back calculation from the inner lane centerline when the curve radius and SSD are given.
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.
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.
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 to traffic on the major road, allowing vehicles on the major road to proceed first. Option (a) states this correctly. Option (b) is wrong because the major road has priority, not the minor road. Option (c) is wrong because both roads do not stop; only the minor-road driver yields. Option (d) is wrong because ‘give way’ is not the same as one-way control, which regulates direction of travel, not right-of-way.
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, the shear strength on the failure plane is given by the Mohr-Coulomb relation in effective stress form: \(\tau_f = \sigma'_n \tan \phi'\). Hence, the ratio of shear stress to normal effective stress at failure is \(\tau_f/\sigma'_n = \tan 38^\circ \approx 0.781\). Therefore, option d is correct.
Why the others are incorrect:
- a (0.378) is too low; it does not match \(\tan 38^\circ\).
- b (0.488) is also lower than the required tangent value.
- c (0.616) is closer but still not equal to \(\tan 38^\circ\).
So the correct answer is d.
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, and s = 16 m:
v^2 = 2 × 0.4 × 9.81 × 16 = 125.57
v ≈ 11.2 m/s
Converting to km/h:
11.2 × 3.6 ≈ 40.3 km/h
So the vehicle speed before skidding was nearly 40 km/h, hence option d is correct.
Why others are incorrect:
- 20 km/h is too low; it would correspond to a much shorter skid distance for the given friction.
- 30 km/h is still below the calculated value.
- 50 km/h is too high; it would require a longer stopping distance than 16 m with μ = 0.4.
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.
Root time method is used to determine
A)
a , coefficient of compressibility
B)
c , coefficient of consolidation
C)
m , coefficient of volume compressibility
D)
T, time factor
The root time method (Taylor’s square-root-of-time method) is used in consolidation tests to determine the coefficient of consolidation, c_v. It is based on plotting settlement against the square root of time and interpreting the early-time portion of the curve to estimate c_v.
Why the others are incorrect:
- a) Coefficient of compressibility is obtained from consolidation pressure–void ratio relationships, not from the root time method.
- c) Coefficient of volume compressibility (m_v) is derived from compressibility characteristics and stress–strain data, not directly by the root time method.
- d) Time factor (T) is a dimensionless parameter used in consolidation theory, but the root time method is specifically a test/graphical method to find c_v, not T directly.