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Civil Engineering : Soil Mechanics - Quiz(MCQ)
A)
Weathering → upheaval → transportation → deposition
B)
Weathering → transportation → deposition → upheaval
C)
Upheavel → transportation → deposition → weathering
D)
Transportation → upheaval → weathering → deposition

Correct Answer :   Weathering → transportation → deposition → upheaval

A)
Illites do not swell when wet
B)
Illite bond is weaker than Kaolinite bond
C)
Illite bond is stronger than montmorillonite bond
D)
All of the above

Correct Answer :   All of the above


Explanation : * Illites do not swell when wet
* Illite bond is weaker than Kaolinite bond
* Illite bond is stronger than montmorillonite bond
* Illites are composed of two silica tetrahedralsheets with a central octahedral sheet

A)
Clay
B)
Red earth
C)
Black cotton soil
D)
Compacted ground

Correct Answer :   Black cotton soil

A)
sand, gravel and air
B)
solids, water and air
C)
solids and air only
D)
solids and water only

Correct Answer :   solids, water and air


Explanation : Solids and water, solids and air are two-phase systems. Sand, gravel and air are not the phases. Solids, water and air are considered to be a three-phase system of the soil mass.

A)
20.4 kN/m³
B)
19.3 kN/m³
C)
24.4 kN/m³
D)
22.6 kN/m³

Correct Answer :   20.4 kN/m³


Explanation : Dry unit weight γd = 17 kN/m³
Water content w = 20% = 0.2
Bulk unit weight γ = γd *(1+w)
γ = 17*(1+0.2)
γ = 20.4 kN/m³.

A)
are of spongy nature
B)
swell with decrease of moisture
C)
shrink with increase of moisture content
D)
None of the above

Correct Answer :   are of spongy nature

A)
True
B)
False
C)
Can not say
D)
--

Correct Answer :   True


Explanation : The density index is used to express the relative compactness or degree of compaction of a natural cohesion-less soil deposit. It is also known as relative density or degree of density.

A)
4.0 to 5.0
B)
3.0 to 4.0
C)
1.0 to 3.0
D)
0.5 to 1.0

Correct Answer :   1.0 to 3.0

A)
18.350 kN/m³
B)
18.545 kN/m³
C)
18.535 kN/m³
D)
18.640 kN/m³

Correct Answer :   18.535 kN/m³


Explanation : Porosity n=30%=0.3
Specific gravity G=2.7
Voids ratio e = n / (1-n)
e=0.3/(1-0.3)=0.429
Dry unit weight γd = Gγw/(1-e)
γd=(2.7*9.81)/(11+0.429)
γd=18.535 kN/m³

A)
saturated soils
B)
homogenous soils
C)
stratified soils
D)
non-uniform slopes

Correct Answer :   homogenous soils

A)
rectangular
B)
hexagonal
C)
triangular
D)
rhombohedral

Correct Answer :   rectangular


Explanation : When the sphere centres form a rectangular space lattice, then they form a packing known as cubic packing, wherein each sphere is in contact with six surrounding neighbouring spheres.

A)
Illites
B)
Kaolinites
C)
Montomorillonite
D)
All of the above

Correct Answer :   All of the above


Explanation : * Illites
* Kaolinites .
* Montomorillonite
* Vermiculite

A)
runways
B)
bridge foundations
C)
sky scrapper buildings
D)
residential buildings

Correct Answer :   sky scrapper buildings

A)
the volume of air and volume of solids
B)
the volume of air and volume of water
C)
the volume of water and volume of solids
D)
the volume of water and the weight of water

Correct Answer :   the volume of air and volume of water


Explanation : The voids in soil mass are occupied by air and water. Hence, the volume of voids equals to the volume of air and volume of water.

A)
block diagram
B)
element diagram
C)
soil grain diagram
D)
constituents diagram

Correct Answer :   block diagram


Explanation : It is always convenient to show the constituents occupying separate spaces as blocks. In the phase diagram, the soil occupies the bottom position. Water and air occupy the middle and top positions.

A)
top
B)
bottom
C)
left side
D)
right side

Correct Answer :   left side


Explanation : Volumes are represented on the left side whereas weights are represented on the right side of the block diagram.

A)
under-consolidated
B)
pre-consolidated
C)
normally consolidated
D)
None of the above

Correct Answer :   under-consolidated

A)
0
B)
0.5
C)
1
D)
1.5

Correct Answer :   1

A)
Cohesion and adhesion between the surfaces of the soil particles
B)
To the structural relation to displacement of the soil because of the interlocking of the particles
C)
The frictional resistance to translocation between the individual soil particles at their contact point
D)
All of the above

Correct Answer :   All of the above

A)
loess
B)
marine
C)
alluvial
D)
lacustrine

Correct Answer :   alluvial

A)
0-15
B)
15-35
C)
35-65
D)
85-100

Correct Answer :   15-35


Explanation :

Relative Density (%) Density Description
0-15 Very loose
15-35 Loose
35-65 Medium
65-85 Dense
85-100 Very Dense

A)
5° to 20°
B)
20° to 30°
C)
30° to 45°
D)
40° to 55°

Correct Answer :   5° to 20°

A)
distribution of pores
B)
compaction of stratum
C)
shape and size of particles
D)
All of the above

Correct Answer :   All of the above

A)
1
B)
0.326
C)
0.247
D)
0.543

Correct Answer :   0.247


Explanation : Degree of saturation S=1
Specific gravity G=2.7
Voids ratio e = (W*G)/S = Wsat*G
Wsat = e/G
Wsat = 0.667/2.7
Wsat = 0.247.

A)
angular-grained loose sand
B)
angular-grained dense sand
C)
round-grained dense sand
D)
round-grained loose sand

Correct Answer :   angular-grained dense sand

A)
is treated as spread foundation
B)
is provided for heavily loaded isolated columns
C)
consists of two sets of perpendicularly placed steel beams
D)
All of the above

Correct Answer :   All of the above

A)
47.64%
B)
42.66%
C)
30.21%
D)
25.95%

Correct Answer :   47.64%


Explanation : Consider a unit cube of soil having spherical particles of diameter d.
Volume of each spherical particle = (π/6)d³
Total volume of container = 1*1*1=1
No. of solids in the container = (1/d)*(1/d)*(1/d)=(1/d³)
Volume of the solids Vs = (π/6)*d³*(1/d³)=(π/6)
Volume of the voids Vv = 1-(π/6)
Voids ratio e = (1-(π/6))/(π/6)
Porosity n = e/(1+e) = 0.9099/(+0.9099) = 0.4764
In percentage, n = 47.64%.

A)
120 years
B)
140 years
C)
160 years
D)
180 years

Correct Answer :   160 years

A)
safe bearing capacity
B)
net ultimate bearing capacity
C)
ultimate bearing capacity
D)
net safe bearing capacity

Correct Answer :   net ultimate bearing capacity

A)
Wd
B)
Ww
C)
Wa
D)
W

Correct Answer :   Wd


Explanation : The weights of Water, air and total weight are represented as Ww, Wa and W. The weight of solids is represented as Wd or Ws in the phase diagram.

A)
Wv
B)
Wa+Wd
C)
Wa+Ww
D)
Ww+Wd

Correct Answer :   Ww+Wd


Explanation : The sum of the weight of voids Wv and weight of solid is the total weight. The weight of total voids is equal to the weight of Water and weight of air. Since the weight of air is considered to be negligible, the weight of total voids is equal to the weight of Water. Therefore, the total weight of the moist sample is the sum of Ww+Wd.

A)
natural water content plus its plastic limit to its plasticity index
B)
liquid limit minus the natural water content to the plasticity index
C)
natural water content minus its plastic limit to its plasticity index
D)
plastic limit minus the natural water content, to its plasticity index

Correct Answer :   natural water content minus its plastic limit to its plasticity index

A)
plastic state
B)
liquid state
C)
semi solid state
D)
All of the above

Correct Answer :   All of the above

A)
bending of particles as elastic sheets
B)
sliping of particles to new positions of greater density
C)
expulsion of double layer water from in between the grains
D)
All of the above

Correct Answer :   All of the above

A)
In hydrometer method, weight Wd per ml of suspension is found directly
B)
In pipette analysis, weight Wd per ml of suspension is found directly
C)
In pipette analysis, weight Wd per ml of suspension is found indirectly
D)
None of the above

Correct Answer :   In pipette analysis, weight Wd per ml of suspension is found directly

A)
circular
B)
elliptical
C)
parabolic
D)
a straight line

Correct Answer :   parabolic

A)
dry density
B)
rate of loading
C)
nature of loading
D)
confining pressure

Correct Answer :   confining pressure

A)
sodium-hexametaphosphate 33 g, sodium carbonate 7 g and distilled water one litre
B)
sodium-hexametaphosphate 7 g, sodium carbonate 33 g and distilled water one litre
C)
sodium-hexametaphosphate 23 g, sodium carbonate 17 g and distilled water one litre
D)
None of the above

Correct Answer :   sodium-hexametaphosphate 33 g, sodium carbonate 7 g and distilled water one litre

A)
plastic but incompressible
B)
plastic and also compressible
C)
elastic and also compressible
D)
None of the above

Correct Answer :   plastic and also compressible

A)
20%
B)
30%
C)
50%
D)
60%

Correct Answer :   20%


Explanation : Porosity n = 40% = 0.4
Air content ac = 50% = 0.5
Percentage air voids na= n*ac
na = 0..4*0.5
na = 20%.

A)
24.5 kN/m³
B)
25 kN/m³
C)
19.5 kN/m³
D)
25.5 kN/m³

Correct Answer :   19.5 kN/m³


Explanation : Voids ratio e = 0.6
Specific gravity G = 2.65
Saturated unit weight γsat = (G+e) γw/ (1+e)
γsat = (2.65+0.6)*9.81/(1+0.6)
γsat= 19.927 N/m3.

A)
dilatancy test
B)
toughness test
C)
dry strength test
D)
All of the above

Correct Answer :   All of the above

A)
by expelling water from pores
B)
by elastic compression of soil grains
C)
by reducing the space occupied by air
D)
All of the above

Correct Answer :   All of the above

A)
1
B)
1.5
C)
2
D)
2.5

Correct Answer :   2

A)
90°
B)
75°
C)
60°
D)
40°

Correct Answer :   90°


Explanation : The soil will have maximum possible voids when the soil grains are arranged in a cubical array of spheres. In a cubical array, the angle of orientation α = 90°.

A)
n = (1-cosα)*( √(1+2cosα) )
B)
n = 1-[π/(6*(1-cosα)*(√(1-2cosα))]
C)
n = 6*(1-cosα)*(√(1-2cosα))
D)
n = 1-[1/(6*(1-cosα)*( √(1-2cosα) )]

Correct Answer :   n = 1-[Ï€/(6*(1-cosα)*(√(1-2cosα))]


Explanation : The formula was given by Slichter. In an intermediate packing, the centres of any 8 spheres, originally arranged in a cubic packing form the corners of the rhombohedron, with an acute face angle α.

A)
Ï€/6
B)
Ï€/4
C)
2Ï€/3
D)
None of the above

Correct Answer :   Ï€/6


Explanation : Consider a unit cell,
Volume of each spherical particle = (π/6)*d³
Total volume of container=1*1*1=1
No. of solids in the container=(1/d)*(1/d)*(1/d)=(1/d³)
Volume of the solids Vs=(π/6)*d³*(1/d³)=(π/6).

A)
type of soil
B)
moisture content
C)
compacting content
D)
All of the above

Correct Answer :   All of the above

A)
swelling
B)
compaction
C)
consolidation
D)
None of the above

Correct Answer :   consolidation

A)
particle classification
B)
textural classification
C)
unified soil classification
D)
High Way Research Board classification

Correct Answer :   textural classification

A)
2 m
B)
2 m to 3 m
C)
3 m to 4 m
D)
5 m

Correct Answer :   3 m to 4 m

A)
40 kN
B)
35 kN
C)
30 kN
D)
25 kN

Correct Answer :   40 kN


Explanation : Weight of soil sample W=190kN
Weight after oven drying Wd=150kN
Weight of Water Ww=W-Wd
Ww=190-150
Ww=40kN.

A)
0.286
B)
0.279
C)
0.329
D)
0.432

Correct Answer :   0.286


Explanation : Volume of Water Va= (1/7) total volume V
Volume of Water VW= (1/8) total volume V
Volume of voids Vv= Va + Vw
Vv=(1/7)V+(1/8)V
Vv=(15/56)V=0.268V.

A)
sodium silicate
B)
clay suspension
C)
cement and water mix
D)
All of the above

Correct Answer :   All of the above

A)
H/2 above the base parallel to base
B)
H/3 above the base parallel to base
C)
H/4 above the base parallel to base
D)
H/5 above the base parallel to base.

Correct Answer :   H/3 above the base parallel to base

A)
negative equipotential lines
B)
positive equipotential lines
C)
positive hydrostatic pressure
D)
negative hydrostatic pressure

Correct Answer :   positive hydrostatic pressure

A)
wind
B)
water
C)
glacier
D)
gravitational force

Correct Answer :   wind

A)
dry density
B)
zero air voids
C)
saturation dry density
D)
All of the above

Correct Answer :   All of the above

A)
equal to total shearing strength
B)
less than total shearing strength
C)
greater than total shearing strength
D)
None of the above

Correct Answer :   greater than total shearing strength

A)
A soil having pH value more than 7 is an acidic soil
B)
A soil having pH value less than 7 is an acidic soil
C)
A soil having pH value more than 7 is an alkaline soil
D)
A soil containing chemicals for the manufacture of portland cement is preferred.

Correct Answer :   A soil having pH value less than 7 is an acidic soil

A)
silt
B)
sand
C)
clay
D)
clay and silt

Correct Answer :   sand

A)
3
B)
3/2
C)
1
D)
1/3

Correct Answer :   3

A)
Kray
B)
Ralph Brazelton Peck
C)
Arthur Casagrande
D)
Karl Terzaghi

Correct Answer :   Karl Terzaghi


Explanation : If civil engineering were a game, Karl Terzaghi had a right to lay down the rules—he had invented and established much of the groundwork.

Terzhaghi (1883-1963) is one of the leading civil engineers of the 20th century and is widely known as the father of soil mechanics.

A)
air
B)
water
C)
partly with air and partly with water
D)
None of the above

Correct Answer :   air

A)
9.256
B)
9.678
C)
10.423
D)
11.502

Correct Answer :   10.423


Explanation : Voids ratio e=0.6
Specific gravity G=2.7
Submerged unit weight Y’ = (G -1) Yw/ (1+e)
Y’ = (2.7-1)*9.81/(1+0.6)
Y’ = 10.423 kN/m³.

A)
size of the particles
B)
shape of the particles
C)
cohesive properties of particles
D)
All of the above

Correct Answer :   All of the above


Explanation : * size of the particles
* shape of the particles
* cohesive properties of particles
* internal frictional resistance of particles

A)
A relatively thin thickness of sample permits quick drainage
B)
As the test progresses the area under shear, gradually changes
C)
A relatively thin thickness of sample permits quick dissipation of pore pressure developed during the test
D)
None of the above

Correct Answer :   As the test progresses the area under shear, gradually changes

A)
12
B)
8
C)
6
D)
4

Correct Answer :   12


Explanation : In the densest state of packing, sphere centres form a rhombohedral array with face angle α=60°. Thus, each sphere is in contact with 12 neighbouring spheres.

A)
0.5
B)
0.7071
C)
0.9099
D)
1

Correct Answer :   0.7071


Explanation : The total volume of rhombohedron is given by,
V = (1-cosα)*(√(1+2cosα))
V = (1-cos60°)*(√(1+2 cos60°))
V = 0.7071.

A)
0.20
B)
0.25
C)
0.30
D)
0.35

Correct Answer :   0.35


Explanation : Consider a unit cube of soil having spherical particles of diameter d.
Volume of the solids Vs= (π/6)=0.5236
The total volume V=(1-cosα)*(√(1+2cosα))
Minimum void ratio is possible in densest state with α=60°
∴ V=(1-cos60°)*(√(1+2cos60°))
V=0.7071
Volume of the voids Vv=V-Vs=0.7071-0.5236
Vv=0.1835
Voids ratio e= Vv/ Vs=0.1835/0.5236
∴ e=0.35.

A)
0.3595
B)
0.3895
C)
0.4266
D)
0.7439

Correct Answer :   0.3895


Explanation : α=75°
The total volume V=(1-cosα)*(√(1+2cosα))
∴ V=(1-cos75°)*(√(1+2cos75°))
V=0.9131
Volume of the solids Vs=(π/6)=0.5236
Volume of the voids Vv=V-Vs=0.9131-0.5236
Vv=0.3895.

A)
excess pore pressure
B)
hydrodynamic pressure
C)
excess hydrostatic pressure
D)
All of the above

Correct Answer :   All of the above

81 .
The ratio of emax and emin of silty sand, is :
A)
2.0
B)
3.0
C)
3.5
D)
5.0

Correct Answer :   3.0

A)
be either straight or curved
B)
intersect the shear stress axis
C)
pass through the origin of stress
D)
All of the above

Correct Answer :   All of the above

A)
shrinkage limit
B)
plastic limit
C)
liquid limit
D)
permeability limit

Correct Answer :   shrinkage limit

A)
2.2
B)
2.4
C)
2.6
D)
2.9

Correct Answer :   2.6

A)
pipette
B)
alcohol
C)
oven drying
D)
calcium carbide

Correct Answer :   pipette


Explanation : The pipette method is the standard sedimentation method used in laboratory for wet mechanical or sedimentation analysis.

A)
1
B)
3
C)
24
D)
48

Correct Answer :   24


Explanation : For complete drying, sandy soil takes about 4 hours and fat clays take about 14 to 16 hours. To assure complete drying of sample, it is kept for about 24 hours.

A)
14 cm
B)
10 cm
C)
9 cm
D)
8 cm

Correct Answer :   10 cm


Explanation : The sampling depth (h) in hydrometer method is taken as 10 cm.

A)
RL
B)
RR
C)
RS
D)
RH

Correct Answer :   RH


Explanation : The reducing reading is designated as RH, which is used for calibration of hydrometer.

A)
porosity
B)
void ratio
C)
specific gravity
D)
water content

Correct Answer :   specific gravity

A)
void ratio
B)
compressive index
C)
depth of the compressible soil strata
D)
Both (B) and (C)

Correct Answer :   Both (B) and (C)

A)
Plasticity
B)
Tenacity
C)
Elasticity
D)
None of the above

Correct Answer :   Plasticity


Explanation : Plasticity is the property of a soil which allows it to be deformed rapidly, without rupture, without elastic rebound and without volume change.

A)
Consistency index
B)
Plasticity index
C)
Both (A) and (B)
D)
None of the above

Correct Answer :   Consistency index


Explanation : consistency index or relative consistency is defined as the ratio of liquid limit, minus the natural water content to the plasticity index of the soil.

A)
Silts where cementing agent is soluble
B)
Loose bulky grains hold together by capillary stresses
C)
Both (A) and (B)
D)
None of the above

Correct Answer :   Loose bulky grains hold together by capillary stresses


Explanation : A common feature of collapsible soil is the loose bulky grain held together by capillary stresses.

A)
Macveren
B)
Holtz and Gibbs
C)
Both (A) and (B)
D)
None of the above

Correct Answer :   Holtz and Gibbs


Explanation : In 1967, Holtz and Gibbs provide guidance for identifying the potential collapsible soil in arid and semi-arid regions.

A)
Severe problem
B)
Moderate problem
C)
No problem
D)
None of the above

Correct Answer :   No problem


Explanation : The severity of the problem will be zero, if the CP value lies between 0-1 percent.

A)
17°C
B)
23°C
C)
27°C
D)
30°C

Correct Answer :   27°C

A)
ultimate bearing capacity
B)
ultimate bearing resistant
C)
ultimate load carrying capacity
D)
All of the above

Correct Answer :   All of the above

A)
clay
B)
silt
C)
fine sand
D)
coarse sand

Correct Answer :   clay

A)
90° + effective angle of shearing resistance
B)
90° + half of the angle of shearing resistance
C)
45° - half of the angle of shearing resistance
D)
45° + half of the angle of shearing resistance

Correct Answer :   45° + half of the angle of shearing resistance

A)
Total area of soil particles
B)
Total surface area in a unit mass
C)
Total volume of soil particles
D)
Ratio of total surface to the total volume

Correct Answer :   Total surface area in a unit mass


Explanation : Specific surface is usually defined as the total surface area of the soil particle in a unit mass.

A)
Decreases
B)
Bulges
C)
Increases
D)
None of the above

Correct Answer :   Decreases


Explanation : The subdivision of a mass yields greater surface area than the same mass itself. The specific surface of a mass increases as the grain size decreases.

A)
Grain size
B)
Grain molecular structure
C)
Grain shape
D)
None of the above

Correct Answer :   Grain shape


Explanation : Specific surface varies for different shapes of particle having the same mass. Hence grain shape is important factor for determining specific surface.

A)
IS classification
B)
Unified soil classification
C)
Textural classification
D)
Particle size classification

Correct Answer :   Textural classification


Explanation : Soil classification of composite soils exclusively based on the particle size distribution is known as textural classification.

A)
Indian classification system
B)
International classification
C)
M.I.T. classification system
D)
Triangular classification of U.S. public road administration

Correct Answer :   Triangular classification of U.S. public road administration


Explanation : Triangular classification of U.S. public road administration is a commonly used system for textural classification.

A)
down at the toe
B)
down at the centre
C)
upward at the centre
D)
None of the above

Correct Answer :   down at the centre

A)
head loss to the length of the seepage
B)
flow line to slope
C)
total head to the length of seepage
D)
head upstream to that at downstream

Correct Answer :   head loss to the length of the seepage

A)
soil particles are spherical
B)
walls of the jar do not affect the settlement
C)
particles settle independent of other particles
D)
All of the above

Correct Answer :   All of the above

A)
Kozney
B)
Rankine
C)
Casagrande
D)
Atterberg

Correct Answer :   Casagrande


Explanation : In 1940, casagrande originally developed the unified soil classification system (USCS).

A)
Foundation
B)
Earth slopes
C)
Dam construction
D)
Air field construction

Correct Answer :   Air field construction


Explanation : In the earlier stages of World War 2, the USCS system was used for air field construction.

A)
Plasticity
B)
Group index
C)
Grain size distribution
D)
Particle size composition

Correct Answer :   Plasticity


Explanation : According to USCS the fine grained soils, whose behavior is controlled by plasticity, are classified on the basis of their plasticity.

A)
1951
B)
1956
C)
1959
D)
1963

Correct Answer :   1959


Explanation : The Indian standard classification system (ISCS) was first developed in 1959 and revised in 1970.

A)
5
B)
4
C)
3
D)
2

Correct Answer :   3


Explanation : Fine grained soil are subdivided into 3 types :

* Inorganic silts and very fine sands
* Inorganic clays
* Organic silts and clay and organic matter.

A)
18 groups
B)
16 groups
C)
15 groups
D)
12 groups

Correct Answer :   18 groups


Explanation : The ISCS classifies the soil into 18 groups as against 15 groups of USCS.

A)
seepage
B)
exit gradient
C)
hydrostatic pressure
D)
All of the above

Correct Answer :   All of the above


Explanation : * seepage
* exit gradient
* hydrostatic pressure
* seepage pressure

A)
force per neutral area
B)
stess taken up by the pore water
C)
force per effective area
D)
stress taken up by solid particles

Correct Answer :   stess taken up by the pore water

A)
70%
B)
75%
C)
80%
D)
85%

Correct Answer :   80%

122 .
If the coefficients of volume change and compressibility of a soil sample are respectively 6.75 x 10⁻² and 3 x 10⁻², the void ratio of the soil sample, is
A)
1.10
B)
1.15
C)
1.20
D)
1.25

Correct Answer :   1.25


Explaination : Coeff.compressibility = Coeff.volume change/(1 + e).

(1 + e) = Coeff.volume change/Coeff.compressibility.

e = (Coeff.volume change/Coeff.compressibility) - 1.

e = ((6.75 x 10⁻²)/(3 x 10⁻²)) - 1.

e = 2.25 - 1.

e = 1.25.

A)
tan (45° - φ)
B)
tan (45° + φ)
C)
tan² (45° - φ
D)
tan² (45° + φ)

Correct Answer :   tan² (45° + φ)

A)
Resilience of soils
B)
Elasticity of soils
C)
Compressibility of soils
D)
None of the above

Correct Answer :   Elasticity of soils

A)
60°
B)
80°
C)
105°
D)
110°

Correct Answer :   60°


Explanation : For organic soils like peat, a lower temperature of about 60° is preferable to prevent the oxidation of the organic matter.

A)
Alcohol
B)
Sand bath
C)
Calcium carbide
D)
Oven drying

Correct Answer :   Calcium carbide


Explanation : Calcium carbide method is very quick as the result can be obtained in 5 – 10 minutes. The field kit contains the moisture tester, a small single – pan weighing balance, a bottle containing calcium carbide and a brush.

A)
oxygen
B)
methane
C)
acetylene
D)
carbon dioxide

Correct Answer :   acetylene


Explanation : The acetylene gas is produced by the reaction of moisture of soil and the calcium carbide. The gas exerts pressure on a sensitive diaphragm attached to a dial gauge which reads the water content.

A)
Equal
B)
Negative
C)
Positive
D)
Both negative and positive

Correct Answer :   Negative


Explanation : The addition of dispersing agent in water increases its water level, therefore the dispersing correction is always negative.

A)
Taking the observation
B)
Principle of test
C)
Method of procedure
D)
None of the above

Correct Answer :   Taking the observation


Explanation : In sedimentation analysis the hydrometer method differs from the pipette method in the method of taking the observation.

A)
Negative sometime
B)
Positive sometime
C)
Always negative
D)
Always positive

Correct Answer :   Always positive


Explanation : The actual reading to be taken at water level will be more; hence the meniscus correction will be always positive.

A)
R = Cm ± Rm – Cd + CL
B)
R = Rh’ + Cm ± CL – Cd
C)
R = Rh + Cm ± Cd + CL
D)
R = Rh’ + cm + CL – Cd

Correct Answer :   R = Rh’ + Cm ± CL – Cd


Explanation : According to the derived equation of hydrometer the equation is
R = Rh’ + Cm ± CL – Cd.

A)
R=Rh’ + C
B)
R=Rh’ + Cm
C)
R=Rh’+ Cm ± CL – Cd
D)
None of the above

Correct Answer :   R=Rh’ + C


Explanation : The three corrections Cm, CL, Cd can be combined into one corrections Cm and the equation can be written as R = Rh’ + Cm.

A)
proportional to the head loss
B)
proportional to the exit graident
C)
perpendicular to the equipotential lines
D)
All of the above

Correct Answer :   All of the above

A)
1.0 g/cm²
B)
1.5 g/cm³
C)
2.0 g/cm³
D)
2.5 g/cm³

Correct Answer :   2.0 g/cm³

A)
shape of grain
B)
weight of grain
C)
shape and size of grain
D)
shape, size and weight of grain

Correct Answer :   shape, size and weight of grain

A)
clay soils
B)
silty soils
C)
sandy soils
D)
gravel soils

Correct Answer :   clay soils

A)
stress to strain
B)
strain to stress
C)
stress to settlement
D)
rate of loading to that of settlement

Correct Answer :   strain to stress

A)
IC = W-WP/IP
B)
IC = W-WL/IP
C)
IC = WL-W/IP
D)
IC = W-WL/IP

Correct Answer :   IC = WL-W/IP


Explanation : The consistency or relative density is defined as ratio of liquid limit, minus the natural water content to the plasticity index of the soil IC = WL-W/IP.

A)
IC
B)
IP
C)
WS
D)
WP

Correct Answer :   WS


Explanation : Shrinkage limit is represented as WS.

A)
Atterberg
B)
Laplace
C)
Dupuit’s
D)
Boussinesq

Correct Answer :   Atterberg


Explanation : In 1911, Atterberg who was a Swedish Agriculturist divided the entire range of liquid to solid state in to four stages.

A)
Dry regions
B)
Semi-arid regions and Arid regions
C)
Both (A) and (B)
D)
None of the above

Correct Answer :   Semi-arid regions and Arid regions


Explanation : Since collapsible soil exhibit loose, weakly cemented sands they are usually present in arid and semi- arid regions.

A)
Coulomb
B)
Mohr
C)
Rankine
D)
Culaman

Correct Answer :   Coulomb

A)
alcohol method
B)
oven-drying method
C)
calcium carbide method
D)
pycnometer

Correct Answer :   pycnometer

A)
60%
B)
62%
C)
64%
D)
66%

Correct Answer :   64%

A)
1
B)
2
C)
3
D)
4

Correct Answer :   3

A)
Proportional to the compressive force
B)
Directly proportional to electric force
C)
Proportional to inter-molecular surface
D)
Inversely proportional to electric force

Correct Answer :   Directly proportional to electric force


Explanation : The electrical force responsible for inter-particle effect is proportional to the specific surface.

A)
kozney’s method
B)
Atterberg method
C)
Skempton’s method
D)
casagrande method

Correct Answer :   kozney’s method


Explanation : The harmonic diameter DH can be found by using kozney’s method using the formula : F=6/ρs×1/Dn.

A)
Wet soil
B)
Clay soil
C)
Soil with moist
D)
Coarse grained soil

Correct Answer :   Coarse grained soil


Explanation : Since the textural classification is based on the percentages of sand, silt, and clay size making up the soil. Such a classification is more suitable for describing coarse-grained soil.

A)
Parallel to the three sides of the triangle
B)
Adjacent to the one sides of the triangle
C)
Adjacent to the three sides of the triangle
D)
Parallel to the only one side of the triangle

Correct Answer :   Parallel to the three sides of the triangle


Explanation : To use the textural classification chart, for a given percentage of three constituents forming a soil, lines are drawn parallel to three sides of the equilateral triangle.

A)
net safe bearing capacity
B)
ultimate bearing capacity
C)
safe bearing capacity
D)
net ultimate bearing capacity

Correct Answer :   safe bearing capacity

A)
unified soil classification system
B)
textural classification system
C)
particle size classification system
D)
High Way Research Board (HRB), classification system

Correct Answer :   unified soil classification system

A)
a parabola
B)
a semi-circle
C)
a straight line
D)
a semi-ellipse

Correct Answer :   a semi-ellipse

A)
value of the angle of repose
B)
cohesive strength of soil
C)
amount of compaction required
D)
All of the above

Correct Answer :   All of the above

A)
sinks downwards
B)
moves away from the backfill
C)
moves towards the backfill
D)
stresses equally by vertical and horizontal forces

Correct Answer :   moves away from the backfill

A)
80% of the soil
B)
70% of the soil
C)
50% of the soil
D)
20% of the soil

Correct Answer :   50% of the soil


Explanation : If more than 50 % of the soil is retained on a NO.200 US sieve (0.075), it is designated as coarse grained soil.

A)
CL-ML
B)
GW-GM and SP-SM
C)
Both (A) and (B)
D)
None of the above

Correct Answer :   GW-GM and SP-SM


Explanation : If the percentage of fines lies between 5 to 12 %, coarse grained soils are designated by dual symbols GW-GM or SP-SM.

A)
Silt and clay
B)
Sand and clay
C)
Gravel and peat
D)
Organic and silt

Correct Answer :   Silt and clay

A)
a circular arc
B)
a straight line
C)
a parabolic arc
D)
an elliptical arc

Correct Answer :   a circular arc

159 .
For a base failure of a slope, depth factor :
A)
Df = 1
B)
Df < 1
C)
Df > 1
D)
None of the above

Correct Answer :   Df > 1

A)
flow is in upward direction
B)
effective pressure is zero
C)
seepage pressure is in upward direction
D)
All of the above

Correct Answer :   All of the above

A)
rocky soils is 5 cm to 50 cm
B)
sandy soils is 80 cm to 100 cm
C)
clay soils is 90 cm to 160 cm
D)
All of the above

Correct Answer :   All of the above

A)
IP=WL-20
B)
IP=20-WL
C)
IP=0.73(WL-20)
D)
IP=0.73(20-wL)

Correct Answer :   IP=0.73(WL-20)


Explanation : The A-line, dividing the inorganic clay from silt and organic soil has the equation: IP=0.73(WL-20).

A)
Sand gravel
B)
Clayey gravel
C)
Silt gravel
D)
Well graded gravel

Correct Answer :   Clayey gravel


Explanation : As per Indian standards of group symbols and typical names, G-refers to gravel and C-refers to clay.

A)
Bulky hard
B)
Round to angular shape
C)
Diameter more than 30cm
D)
All of the above

Correct Answer :   All of the above


Explanation : For a soil component to be a boulder, size range must be of average diameter more than 30cm and shape of round to angular and bulky hard.

A)
loose elasticity
B)
loose permeability
C)
decrease their shear strength
D)
gain shear strength

Correct Answer :   decrease their shear strength

A)
wind
B)
water
C)
glacier
D)
gravitational force

Correct Answer :   gravitational force

A)
is caused in soft clays
B)
decreases the pile capacity
C)
is caused due to relative settlement of the soil
D)
All of the above

Correct Answer :   All of the above

A)
load at which soil fails
B)
safe load on the bearing area
C)
total load on the bearing area
D)
load at which soil consolidates

Correct Answer :   load at which soil fails

A)
minor principal plane
B)
major principal plane
C)
intermediate principal plane
D)
None of the above

Correct Answer :   major principal plane

A)
0.08 mm
B)
0.06 mm
C)
0.04 mm
D)
0.002 mm

Correct Answer :   0.002 mm

A)
radiation
B)
pycnometer
C)
oven – dry
D)
calcium carbide

Correct Answer :   radiation


Explanation : The radiation method consists of 2 steel casings. A device containing some radio–active isotope is lowered in one caning and a detector in another. The neutrons emitted by the device are detected by the detector and the loss of energy of neutrons is equated to the water content.

A)
Opaque
B)
Transparent
C)
Translucent
D)
None of the above

Correct Answer :   Opaque


Explanation : Since the density of the soil suspension is high, the visibility will be opaque.

A)
100
B)
120
C)
500
D)
1000

Correct Answer :   1000


Explanation : The quantity of dry soil and dispersing agent is taken as double of pipette which is 500 ml.

A)
Liters
B)
Millimeters
C)
Grams
D)
All of the above

Correct Answer :   Grams


Explanation : since the soil is weighed in terms of mass, the volume is taken in terms of grams.

A)
base failure
B)
toe failure
C)
face failure
D)
slope failure

Correct Answer :   toe failure

A)
volume of water to volume of given soil
B)
volume of water to volume of voids in soil
C)
weight of water to weight of air in voids
D)
weight of water to weight of solids of given mass of soil

Correct Answer :   weight of water to weight of solids of given mass of soil

A)
oven-drying method
B)
alcohol method
C)
sand bath method
D)
calcium carbide method

Correct Answer :   oven-drying method

A)
Liquid limit
B)
Plastic limit
C)
Solid limit
D)
Shrinkage limit

Correct Answer :   Solid limit


Explanation : Liquid limit, plastic limit, shrinkage limit are useful for engineering purpose.

A)
Smooth surface
B)
Electro-magnetic charges
C)
Both (A) and (B)
D)
None of the above

Correct Answer :   Electro-magnetic charges

A)
Oil
B)
Kerosene
C)
Soap solution
D)
None of the above

Correct Answer :   Kerosene


Explanation : According to Goldschmidt theory, the clay does not become plastic when mixed with liquids of non-polarizing agents like kerosene.

A)
Capillary stress
B)
Compressive stress
C)
Inter-molecular force between the soil
D)
All of the above

Correct Answer :   Capillary stress


Explanation : As the common Feature exhibited by collapsible soil is capillary stress, the loose bulky grains in the soil are held together by it.

A)
Pressure
B)
Moisture content
C)
Temperature coefficient
D)
Degree of saturation

Correct Answer :   Degree of saturation


Explanation : Collapsible potential varies with degree of saturation, at an overburden pressure of Po.

A)
free water
B)
gravity water
C)
hydroscopic water
D)
None of the above

Correct Answer :   hydroscopic water

A)
0.2200 cm/sec
B)
0.2250 cm/sec
C)
0.2275 cm/sec
D)
0.2350 cm/sec

Correct Answer :   0.2275 cm/sec

A)
is not related with plastic limit
B)
increases with an increase in the liquid limit
C)
decreases with an increase in the liquid limit
D)
decreases with an increase in the plastic limit

Correct Answer :   increases with an increase in the liquid limit

A)
Physical properties of the soil
B)
Chemical properties of the soil
C)
General characteristic of the soil
D)
All of the above

Correct Answer :   Physical properties of the soil


Explanation : As specific surface is associated with surface characteristics of a particle, therefore it is related with physical properties.

A)
Larger diameter
B)
A larger volume
C)
Small specific surface
D)
Larger specific surface

Correct Answer :   Larger specific surface


Explanation : As the specific surface increases as the size of the grain decreases. Therefore Microscopic particles have a large specific surface.

A)
Moisture content of the soil
B)
Physical properties of the soil
C)
Surface area of the particle per unit volume
D)
None of the above

Correct Answer :   Surface area of the particle per unit volume


Explanation : From the expression, specific surface = surface area of particle/volume of the particle, it can be seen that specific surface is the surface area of a particle.

A)
Right angled triangle
B)
Equilateral triangle
C)
Perpendicular triangle
D)
None of the above

Correct Answer :   Equilateral triangle


Explanation : A triangle, having equal length of sides (i.e. Equilateral triangle) is used in textural classification chart.

A)
Particle size distribution
B)
Particle-size composition and Plasticity characteristics
C)
Both (A) and (B)
D)
None of the above

Correct Answer :   Particle-size composition and Plasticity characteristics


Explanation : HRB classification system, also known as public road administration (PRA) is based on both the particle-size composition and plasticity characteristics.

A)
Indian classification system
B)
M.I.T. classification system
C)
International classification system
D)
Public road administration (PRA) system

Correct Answer :   Public road administration (PRA) system


Explanation : Highway research board classification system is also known as public road administration (PRA) system.

A)
Shrinkage limit
B)
liquid limit
C)
Plastic limit
D)
semi-solid limit

Correct Answer :   Shrinkage limit

A)
shear strength
B)
bearing capacity
C)
Both (A) and (B)
D)
None of the above

Correct Answer :   Both (A) and (B)

A)
mean sea level
B)
water table
C)
phreatic surface
D)
Both (B) and (C)

Correct Answer :   Both (B) and (C)

A)
flow lines and equipotential lines cross each other at right angles
B)
fields are rectangles whose length is twice the breadth
C)
for homogeneous soil, the curves are smooth and circular.
D)
smaller the dimensions of the field, smaller will be the hydraulic gradient and velocity of flow through it

Correct Answer :   flow lines and equipotential lines cross each other at right angles

A)
porosity
B)
void ratio
C)
percentage voids
D)
degree of saturation

Correct Answer :   degree of saturation

A)
Group Index
B)
Liquid limit and Plasticity index
C)
Both (A) and (B)
D)
None of the above

Correct Answer :   Liquid limit and Plasticity index


Explanation : Based on the liquid limit and plasticity index, fine grained soil is subdivided.

A)
Sand and silt
B)
Silt and clay
C)
Gravel and sand
D)
Gravel and clay

Correct Answer :   Gravel and sand


Explanation : Based on the soil size, coarse grained soil is divided in to gravel and sand.

A)
Grade of the soil
B)
Division of the group
C)
Relative suitability
D)
None of the above

Correct Answer :   Relative suitability


Explanation : In Casagrande’s plasticity chart, the numbers denote relative suitability with No.1 as the best and No.10 as the worst.

A)
it can be performed under all three driange conditions with complete control
B)
stress distribution on the failure plane, is non uniform
C)
precise measurement of pore pressure and change in volume during test, is not posible
D)
None of the above

Correct Answer :   it can be performed under all three driange conditions with complete control

A)
hydration
B)
oxidation
C)
carbonation
D)
All of the above

Correct Answer :   All of the above

A)
meniscus correction
B)
dispersing agent correction
C)
meniscus and dispersing agent corrections
D)
temperature, meniscus and dispersing agent corrections

Correct Answer :   temperature, meniscus and dispersing agent corrections

A)
is less in wet soil than in dry soil
B)
depends upon the force responsible
C)
increases as the size of the soil particles increases
D)
decreases as the size of the soil particles decreases

Correct Answer :   depends upon the force responsible

A)
back fill is dry, cohesionless, homogeneous and isotropic
B)
slip surface is the plane which passes through the heel of the wall
C)
position and direction of the resultant earth pressure, are known
D)
All of the above

Correct Answer :   All of the above


Explanation : * back fill is dry, cohesionless, homogeneous and isotropic
* slip surface is the plane which passes through the heel of the wall
* position and direction of the resultant earth pressure, are known
* sliding wedge itself acts as a rigid body and the value of earth pressure is obtained by considering the limiting equilibrium of the wedge

A)
q times the lateral pressure within the surface
B)
1/q times the lateral pressure within the surface
C)
equal to a fill of height Z equal to q/r, where r is the density of the backfill
D)
All of the above

Correct Answer :   equal to a fill of height Z equal to q/r, where r is the density of the backfill

A)
Isotropic consolidation of clay can be obtained in the triaxial apparatus under equal all-round pressure.
B)
If the present effective stress in the past was more than present effective stress, it is called over-consolidated clay
C)
If the present effective stress is the maximum to which the clay has ever been subjected, it is called normally consolidated clay
D)
All of the above

Correct Answer :   All of the above

A)
35°c
B)
27°c
C)
-35°c
D)
-27°c

Correct Answer :   27°c


Explanation : The hydrometer are generally calibrated at 27°c if the temperature is more than 27°c, temperature correction will be taken negative. If less than 27°c then the temperature correction will be positive.

A)
down and outwards on a slip surface
B)
up and inwards on a slip surface
C)
horizontal upward and parallel to base
D)
horizontal inward and parallel to base

Correct Answer :   down and outwards on a slip surface

A)
D₁₀
B)
Dâ‚‚â‚€
C)
D₃₀
D)
D₆₀

Correct Answer :   D₁₀

A)
Illite group
B)
Kaolinite group
C)
Vermiculite group
D)
Montrorillonite group

Correct Answer :   Illite group

A)
shear strength to shear stress along the surface
B)
actual cohesion to that required to maintain stability of slope
C)
Both (A) and (B)
D)
Neither (A) nor (B)

Correct Answer :   Both (A) and (B)

A)
porosity
B)
void ratio
C)
water content
D)
specific gravity

Correct Answer :   void ratio

A)
Area
B)
Volume
C)
Per cent water content
D)
All of the above

Correct Answer :   Per cent water content

A)
Gaseous state
B)
Solid state
C)
Liquid state
D)
Semi-solid state

Correct Answer :   Gaseous state

A)
7.24
B)
4.44
C)
3.85
D)
2.04

Correct Answer :   4.44


Explanation : Sensitivity,
St = QU(undistributed)/QU(distributed)
=240/54
=4.44.

A)
Last
B)
Does not occur
C)
First saturation
D)
At an intermediate

Correct Answer :   First saturation


Explanation : As there is a sudden compression of the soil under pressure, the structural collapse occurs upon first saturation.

229 .
For a homogeneous earth dam 50 m high having 2 m free broad, a flow net was constructed and the results were : Number of potential drops = 2.4 Number of flow channels = 0.4. If coefficiency of permeability of the dam mateiral is 3 x 10⁻³ cm³/sec, the discharge per metre length of dam, is
A)
6 x 10⁻⁵ m³/sec
B)
12 x 10⁻⁵ m³/sec
C)
24 x 10⁻³ m³/sec
D)
24 x 10⁻⁵ m³/sec

Correct Answer :   24 x 10⁻⁵ m³/sec

A)
stress distribution on the failure plane, is uniform
B)
it can be performed under all three drainage conditions
C)
precise measurement of the pore pressure and volume change during the test is possible
D)
All of the above

Correct Answer :   All of the above


Explanation : * stress distribution on the failure plane, is uniform
* it can be performed under all three drainage conditions
* precise measurement of the pore pressure and volume change during the test is possible
* state of stress within the specimen during any stage of the test as well as at failure, is completely determined

A)
105° to 110°C
B)
103° to 105°C
C)
95° to 100°C
D)
90° to 95°C

Correct Answer :   105° to 110°C

A)
consolidation index
B)
degree of consolidation
C)
co-efficient of consolidation
D)
consolidation of undisturbed soil

Correct Answer :   degree of consolidation

A)
sandy soils
B)
clayey soils
C)
stratified soils
D)
cohesive soils

Correct Answer :   stratified soils

234 .
260 g of wet soil was taken in a pycnometer jar of weight 400 g in order to find the moisture content in the soil, with specific gravity of soil particles 2.75. The weight of soil and remaining water filled in pycnometer without air bubbles was 1415 g and the weight of pycnometer filled with water alone was 1275 g. The moisture content in the soil is____.
A)
24.2%
B)
18.2%
C)
53.8%
D)
None of the above

Correct Answer :   18.2%

A)
Density of the soil
B)
Moisture present in soil
C)
Size of the soil particle
D)
Specific surface of the soil particle

Correct Answer :   Specific surface of the soil particle


Explanation : Permeability of soil greatly depends on the specific surface of the soil because of the viscous effects associated with surface characteristics of the grains.

A)
Surface area
B)
Unit mass
C)
Compactness
D)
Electric charge

Correct Answer :   Surface area


Explanation : The chemical activity of the clay mineral depends upon the surface area, because of the electrical charge of clay minerals is high per unit mass.

A)
6 primary groups
B)
7 primary groups
C)
2 primary groups
D)
None of the above

Correct Answer :   7 primary groups


Explanation :  Based on HRB system, the soil is divided into 7 primary groups, designated as A-1, A-2…A-7.

A)
Quality test
B)
Material test
C)
Group index
D)
None of the above

Correct Answer :   Group index


Explanation : As group index is a mean of rating the value of soil as a sub-grade material. It can be used for finding the performance and quality of the soil.

A)
GI=0.2a+0.5ac+0.001bd
B)
GI=0.2b+0.005ac+0.01bd
C)
GI=0.2a+0.005ac+0.001bd
D)
GI=0.2a+0.005bd+0.001ac

Correct Answer :   GI=0.2a+0.005ac+0.001bd


Explanation : Group index (GI) is given by the following equation :

GI = 0.2a+0.005ac+0.01bd.

A)
0.2 mm
B)
0.02 mm
C)
0.002 mm
D)
0.0002 mm

Correct Answer :   0.0002 mm

A)
Directional angles
B)
perpendicular bisector of slope and the locus of the centre
C)
perpendicular drawn at l/3rd slope from toe and the locus of the centre
D)
perpendicular drawn at 2/3rd slope from toe and the locus of the centre

Correct Answer :   directional angles

A)
major principal stress is vertical
B)
minor principal stress is vertical
C)
major principal stress is horizontal
D)
minor and major principal stresses are equally inclined to the horizontal

Correct Answer :   minor principal stress is vertical

A)
major principal stress is vertical
B)
minor principal stress is vertical
C)
major principal stress is horizontal
D)
minor and major principal stresses are equally inclined to horizontal

Correct Answer :   major principal stress is vertical

A)
2.72
B)
2.85
C)
2.65
D)
2.90

Correct Answer :   2.65

A)
0.04 mm
B)
0.06 mm
C)
0.08 mm
D)
0.002 mm

Correct Answer :   0.002 mm

A)
driven piles
B)
bored piles
C)
precast piles
D)
All of the above

Correct Answer :   bored piles

A)
proportional to the internal friction of the soil
B)
proportional to the depth of the soil
C)
is equal to mass of the soil retained
D)
proportional to the square of the depth of the soil

Correct Answer :   proportional to the internal friction of the soil

A)
is more in well compacted clays
B)
depends upon the external applied load
C)
increases as the moisture content decreases
D)
decreases as the moisture content increases

Correct Answer :   is more in well compacted clays

A)
quick test
B)
drained test
C)
Both (A) and (B)
D)
consolidated undrained test

Correct Answer :   consolidated undrained test

A)
plastic limit
B)
shrinkage limit
C)
liquid limit
D)
permeability limit

Correct Answer :   liquid limit

A)
swell more when wet
B)
shrink more when dry
C)
possess high plasticity
D)
possess high coefficient of internal coefficient

Correct Answer :   possess high coefficient of internal coefficient

A)
primary compression
B)
primary time effect
C)
primary consolidation
D)
All of the above

Correct Answer :   All of the above

A)
wind
B)
glaciated water
C)
gravitational force
D)
All of the above

Correct Answer :   glaciated water

A)
liquid limit minus the natural water content to the plasticity index of the soil
B)
liquid limit plus the natural water content to the plasticity index of the soil
C)
natural water content of a soil minus plastic limit to the plasticity index of the soil
D)
natural water content of a soil plus its plastic limit to the plasticity index of the soil

Correct Answer :   liquid limit minus the natural water content to the plasticity index of the soil

A)
cohesion
B)
internal friction
C)
inter-granular friction
D)
None of the above

Correct Answer :   cohesion

A)
pre-compressed
B)
pre-consolidated
C)
over-consolidated
D)
All of the above

Correct Answer :   All of the above

A)
H
B)
H²
C)
H³
D)
None of the above

Correct Answer :   H²

A)
0.4
B)
0.52
C)
1.08
D)
None of the above

Correct Answer :   1.08

A)
direct shear test
B)
triaxial compression test
C)
unconfined compression test
D)
None of the above

Correct Answer :   unconfined compression test

A)
0.188
B)
0.288
C)
0.388
D)
0.488

Correct Answer :   0.188

A)
2.93
B)
2.90
C)
2.85
D)
2.72

Correct Answer :   2.72

A)
the φ line
B)
repose line
C)
natural slope line
D)
All of the above

Correct Answer :   All of the above

A)
it starts flowing
B)
it is on the verge of failure
C)
it is in plastic stage
D)
it is stressed to maximum

Correct Answer :   it is on the verge of failure

A)
65%
B)
50%
C)
35%
D)
None of the above

Correct Answer :   65%

A)
30°
B)
45°
C)
60°
D)
90°

Correct Answer :   45°

A)
tan (45° + φ)
B)
tan (45° - φ)
C)
tan² (45° + φ/2)
D)
tan² (45° - φ/2)

Correct Answer :   tan² (45° - φ/2)

A)
minus unit weight of water
B)
plus unit weight of water
C)
divided by unit weight of water
D)
multiplied by unit weight of water

Correct Answer :   minus unit weight of water

A)
one phase soil
B)
two phase soil
C)
three phase soil
D)
four phase soil

Correct Answer :   three phase soil

A)
is proportional to the cohesion of the soil
B)
is generally known as the strength of the soil
C)
increases with an increase in the normal stress
D)
All of the above

Correct Answer :   All of the above


Explanation : * is proportional to the cohesion of the soil
* is generally known as the strength of the soil
* increases with an increase in the normal stress
* is proportional to the tangent of the angle of internal friction

A)
capillarity
B)
permeability
C)
moisture content
D)
None of the above

Correct Answer :   permeability

A)
water content
B)
dry density
C)
voids ratio
D)
density index

Correct Answer :   water content

A)
porosity
B)
water content
C)
void ratio
D)
specific gravity

Correct Answer :   water content

A)
perpendicular to stream lines
B)
parallel to equipotential lines
C)
perpendicular to equipotential lines
D)
None of the above

Correct Answer :   perpendicular to equipotential lines

A)
ductility
B)
porosity
C)
permeability
D)
plasticity

Correct Answer :   plasticity