Soil Mechanics and Foundation Engineering MCQ Quiz - Objective Question with Answer for Soil Mechanics and Foundation Engineering - Download Free PDF
Last updated on May 15, 2025
Latest Soil Mechanics and Foundation Engineering MCQ Objective Questions
Soil Mechanics and Foundation Engineering Question 1:
The Atterberg limits of of a clay soil are liquid limit = 75 %, Plastic limit = 45 % and shrinkage limit = 25 %. If a sample of this soil has a volume of 30 cm3 at the liquid limit and a volume 16.6 cm3 at the shrinkage limit. Determine the specific gravity of solids.
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 1 Detailed Solution
Concept:
Shrinkage ratio (S.R.):
It is defined as the ratio of volume change expressed as the percentage of dry volume to the corresponding change in water content above the shrinkage limit.
Here,
Vo - Initial volume of saturated sample at water content wl
Vd - Final volume of the soil sample at water content ws
Md - Mass of dry volume 'Vd' of the sample
γd, γw - The unit weight of the dry volume of soil and water respectively.
Gm - Mass-specific gravity of soil in its dry state.
Shrinkage limit:
Where, Ws = Shrinkage limit and Gs = Specific gravity of solids
Calculation:
Given, Wl = 75 %, Ws = 25 %, Wp = 45 %
Volume of soil sample(V0) = 30 cm3, Vd = 16.6 cm3
S.R. = 1.614
Shrinkage limit:
Gs = 2.71
Soil Mechanics and Foundation Engineering Question 2:
In a 1250 cm3 soil sample, 10 cm3 is volume of voids in which 3 cm3 is volume of air voids. The degree of saturation will be:
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 2 Detailed Solution
Explanation:
The degree of saturation (S) is the ratio of the volume of water to the volume of voids in a soil sample and mathematically, it is given as:
Volume of Voids (Vv) = Volume of Water (Vw) + Volume of air (Va)
Given that Vv = 10 cm3 and Va = 3 cm3
⇒ 10 = Vw + 3
Or
∴ The volume of water in a soil sample, Vw = 7 cm3
Substituting the values of Vw and Vv in the above formula, we get
⇒ Degree of Saturation, S = 70 %
Soil Mechanics and Foundation Engineering Question 3:
Which soil type has sensitivity less than 1?
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 3 Detailed Solution
Concept:
Sensitivity(St):
- Sensitivity is a measure of loss in the strength of soils as a result of remolding.
- In general, it denotes the degree of disturbance of undisturbed samples of the soil upon remolding.
- Sensitivity is defined as the ratio of the unconfined compressive strength of the clay in the undisturbed state to the unconfined compressive strength of the clay in the remolded state.
- Sensitivity is generally observed in cohesive soils and these soils generally have a sensitivity value greater than 1.
The sensitivity of clay soil is as follows:
Classification |
Sensitivity |
Insensitive clay |
|
Slightly sensitive clay |
1-2 |
Medium sensitive clay |
2-4 |
Very sensitive clay |
4-8 |
Slightly quick clay |
8-16 |
Medium quick clay |
16-32 |
Very quick clay |
32-64 |
Extra quick clay |
>64 |
Soil Mechanics and Foundation Engineering Question 4:
The average permeability perpendicular to the bedding planes in case of stratified soil deposits is determined using the formula:
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 4 Detailed Solution
Explanation:
The average permeability for flow perpendicular to the bedding planes is given by,
The average permeability for flow parallel to the bedding planes is given by,
∴ Average permeability for flow perpendicular to the bedding plane is harmonic mean whereas, average permeability for flow parallel to the bedding plane is the arithmetic mean.
Important PointsAs arithmetic mean is always greater than harmonic mean therefore, the average permeability for flow perpendicular to the bedding plane is always lesser than the average permeability for flow parallel to the bedding plane.
Soil Mechanics and Foundation Engineering Question 5:
As per IS classification (IS1498 : 1970), the grain size pertaining to fine sand type soils is:
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 5 Detailed Solution
Explanation:
Particle Size classification:
This classification of soils is done on the basis of particle size composition. According to which soil may be termed as clay, silt, sand, gravel, cobbles and boulders.
I.S. Classification (Grain size distribution):
(i) Clay = d
(ii) Silt = 2μ
- Fine silt = 2μ
- Medium silt = 10μ
- Coarse silt = 20μ
(iii) Sand = 75μ
- Fine sand = 0.075 mm
- Medium sand = 0.425 mm
- Coarse sand = 2 mm
(iv) Gravel = 4.75 mm
- Fine gravel = 4.75 mm
- Coarse gravel = 20 mm
(v) Cobbles = 80 mm
(vi) Boulders = d > 300 mm
Top Soil Mechanics and Foundation Engineering MCQ Objective Questions
Which of the following represents the percentage limit of porosity of the compacted sand?
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 6 Detailed Solution
Download Solution PDFExplanation:
Porosity (η) is defined as the ratio of volume of voids to the total volume of the soil sample in a given soil mass.
The porosity of soils can vary widely.
The porosity of loose soils can be about η = 50 to 60%.
The porosity of compact soils is about η = 30 to 40%.
A value of toughness index less than unity indicates that the soil is friable at the ______
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 7 Detailed Solution
Download Solution PDFConcept:
Toughness Index:
- Toughness index is defined as the ratio of plasticity index (IP) of the soil to the flow index (IF) of the soil.
- This gives us an idea of the shear strength of soil at its plastic limit.
- The toughness index varies between 0 to 3. If If is more, the rate of loss of shear strength is more. Hence, the toughness index is less.
- When the toughness index is less than 1, the soil is said to be friable, which means it can be easily crushed at the plastic limit.
As per IS 4651(Part 2) : 1989, if the value of coefficient of earth pressure 'Ko' at rest is 0.4, then the soil type is __________.
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 8 Detailed Solution
Download Solution PDFExplanation:
Coefficient of Earth Pressure at rest:
Where,
μ is the coefficient is the Poisson ratio.
As per IS 4651 (Part 2): 1989, Table 1
Soil | Ko |
Loose sand | 0.4 |
Dense sand | 0.5-0.6 |
Sand, well tamped | 0.8 |
Soft clay | 0.6 |
Hard clay | 0.5 |
A tracer takes 100 days to travel from Well-1 to Well-2 which are 100 m apart. The elevation of water surface in Well-2 is 3 m below that in Well-1. Assuming porosity equal to 15%, the coefficient of permeability (expressed in m/day) is
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 9 Detailed Solution
Download Solution PDFConcept:
According to Darcy’s law,
For laminar flow Vα i
V = K × i
Where,
V = velocity of water flowing in soil, K = coefficient of permeability
hL = head difference, L = seepage length
Seepage velocity is given by,
VS = V/n
Where,
n = porosity of the soil
Calculation:
T = Time = 100 days
D = distance = 100 m
hL = head difference = 3 m
n = 15 % = 0.15
i = hL/L = 3/100
VS = D/T = 100/100 = 1 m/day
V = K × i
V = VS × n
∴ VS × n = K × i
According to I.S. classification system, the soil can be classified into:-
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 10 Detailed Solution
Download Solution PDFSoil classification as per Indian standards:
Coarse-grained soils (More than half of the material is larger than 75 μ sieve) |
||||||||
Gravels |
Sands |
|||||||
More than half of the coarse fraction is larger than the 4.75 mm sieve |
More than half of the coarse fraction is smaller than the 4.75 mm sieve |
|||||||
Clean gravel |
Gravel with appreciable fines |
Clean sands |
Sand with appreciable fines |
|||||
GW |
GP |
GM |
GC |
SW |
SP |
SM |
SC |
|
Fine-grained soils (More than half of the material is smaller than 75 μ sieve) |
||||||||
Silts and clay |
||||||||
Low compressibility (wL |
Medium compressibility (50 L > 35) |
Higher compressibility (wL > 50) |
||||||
ML |
CL |
OL |
MI |
CI |
OI |
MH |
CH |
OH |
High organic peat soil → (Pt) |
∴ According to IS Classification, there are 18 groups of soils:
8 groups of coarse-grained, 9 groups of fine-grained, and 1 of peat
What is the optimum range of water content adopted for sand in the standard proctor test?
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 11 Detailed Solution
Download Solution PDFExplanation:
The range of optimum water content for different types of soil is as follows:
Soil |
Optimum Water Content (%) |
Sand |
6 – 10 |
Sandy Silt or Silty Sand |
8 – 12 |
Silt |
12 – 16 |
Clay |
14 – 20 |
For a clay, SL = 20%, PL = 40%. The total volume at PL & LL was 1.20 Vd and 1.50 Vd respectively wherein Vd = dry volume. Then, the plasticity Index Pl of the soil is
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 12 Detailed Solution
Download Solution PDFConcept:
slope =
⇒
where
wL = LL = liquid limit
wP = PL = plastic limit
ws = SL = shrinkage limit
VL = volume of soil mass at LL
VP = volume of soil mass at PL
Vs = volume of soil mass at SL
Calculation:
Given:
SL = 20%
PL = 40%
VP = 1.2 Vd
VL = 1.5 Vd
⇒
Now, putting values, we get
⇒
⇒ IP = 30%
Hence, the plasticity Index Pl of the soil is 30%.
For the clay with an OCR of greater than 4 in a CD test, the A-factor at failure will be.
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 13 Detailed Solution
Download Solution PDFConcept:
Over Consolidation Ratio:
Over Consolidation Ratio ( OCR ) is defined as the ratio of maximum pressure experienced in the past
( Preconsoildation pressure ) to the present overburden pressure.
OCR > 1 ⇒ It is Over consolidated soil
OCR = 1 It is Normally consolidated soil
OCR
Calculation:
OCR of clay = 4
Since the OCR of the given soil is greater than 1, it is overconsolidated clay.
A-factor:
A – factor is a function of the over-consolidation ratio in the case of overconsolidated clay.
For OC clays, A = f (OCR)
For heavily overconsolidated clay A –0.2 to –0.3
What is the effective stress at a depth of 10 m below the ground level, when water table is 3 m below ground level, saturated density is 20 kN/m3 and bulk density is 18 kN/m3?
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 14 Detailed Solution
Download Solution PDFConcept:
The effective stress concept was developed by "Terzaghi". A fully saturated soil relates three types of stress:
- Total Stress
- Pore Pressure( Neutral Stress)
- Effective Stress
Total Stress(
Pore pressure(u) is the pressure of the water filling the void space between solid particles.
Effective stress(
The relation between total stress, pore pressure, and effective stress is as follows:
where
Explanation:
Given:
γw = 10 kN/m3, γsat = 20 kN/m3, γb = 18 kN/m3
Effective stress at a depth of 10 m below the ground level:
γeff = γb × 3 + γsub × 7
γeff = 18 × 3 + (20 - 10) × 7
γeff = 124 kN/m3
As per IS: 1892 – 1979; what should be the maximum thickness of cutting edge of sampling tube of 70 mm external diameter which is required for sampling in undisturbed stiff clay soil?
Answer (Detailed Solution Below)
Soil Mechanics and Foundation Engineering Question 15 Detailed Solution
Download Solution PDFConcept:
Area ratio:
It can be defined as the ratio of the maximum cross-sectional area of the cutting edge to the area of the soil sample.
The area ratio can be expressed as
where
D1 = inner diameter of cutting edge
D2 = outer diameter of cutting edge
Note:
For stiff formation (Ar)max = 20%
Soft sensitive clay (Ar)max = 10%
Maximum thickness of cutting edge =
Calculation:
Given, D2 = 70 mm
Soil is stiff clay, So (Ar)max = 20%
D1 = 63.9 mm
Maximum thickness of cutting edge =
=