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Test: Soil Classification & Soil Structure - Civil Engineering (CE) MCQ


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10 Questions MCQ Test - Test: Soil Classification & Soil Structure

Test: Soil Classification & Soil Structure for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The Test: Soil Classification & Soil Structure questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Soil Classification & Soil Structure MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Soil Classification & Soil Structure below.
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Test: Soil Classification & Soil Structure - Question 1

Match List-I (Symbol) with List-ll (Soil) and select the correct answer using the codes given below the lists:
List-I
A. ML
B. SM
C. Pt
D. MH

List-ll
1. Silty sand
2 . Inorganic silt with large compressibility
3. Inorganic silt with small compressibility
4. Soil with high organic content with high compressibility

Test: Soil Classification & Soil Structure - Question 2

Match List-I (Soil classification symbol) with List-ll (Soil property) and select the correct answer using the codes given below the lists: 

List-I
A. GW
B. SW
C. ML
D. CL

List-ll
1. Soil having uniformity coefficient > 6
2. Soil having uniformity coefficient > 4
3. Soil have low plasticity
4. Soil have low compressibility

Detailed Solution for Test: Soil Classification & Soil Structure - Question 2

GW is well graded gravel for which coefficient of uniformity (Cu) > 4.
SW is well graded sand for which coefficient of uniformity (Cu) > 6
ML is silt with low plasticity (< 35%)
CL is clay with low plasticity (< 35%). It also possess low compressibility.

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Test: Soil Classification & Soil Structure - Question 3

Match List-I (Soils) with List-II (Group symbols) and select the correct answer using the codes given below the lists:
List-I
A. Clayey gravel
B. Clayey sand
C. Organicclay
D. Silty sand

List-ll
1. SM
2. OH
3. SC
4. GC

Test: Soil Classification & Soil Structure - Question 4

A soil mass contains 40% gravel, 50% sand and 10% silt. This soil can be classified as

Detailed Solution for Test: Soil Classification & Soil Structure - Question 4


∴ D60 = 4.75 mm
D10 = 0.075 mm

So correct answer is ‘c’.

Test: Soil Classification & Soil Structure - Question 5

Inorganic soil with low compressibility are represented by

Detailed Solution for Test: Soil Classification & Soil Structure - Question 5

In ML;
M represents inorganic soil,
L represents low compressibility, 
MH. Inorganic soil of high compressibility,
SL: Sand of low compressibility,
CH: Clay with high compressibility,

Test: Soil Classification & Soil Structure - Question 6

Sieve analysis on a dry soil sample of mass 1000g showed that 980g and 270g of soil pass through 4.75 mm and 0.075 mm sieve respectively. The liquid limit and plastic limit of the soil fraction passing through 425 μ sieves are 40% and 18%, respectively. The soil may be classified as

Detailed Solution for Test: Soil Classification & Soil Structure - Question 6

Total mass of dry sample = 1000 g
Soil passing through 4.75 mm sieve = 980 g
=9801000×100=98 %
Soil passing through 0.075 mm sieve
= 270 g = 27%
Liquid limit, wL=40%
Plastic limit, wp=18%
Plasticity Index, Ip=wL−wP
= 40 - 18 = 22%
Since less than 50% passes through 0.075 mm sieve, soil is coarse grained
More than 50% of coarser fraction passes through 4.75 mm sieve, so it is sand.
Since fines(< .075 mm) are more than 12%
so using A line chart
IPC=0.73(wL−20)
= 0.73 (40 - 20) = 14.6
Ip>IPC, So point lies above A line (clayey type)
Hence, soil is classified as SC.

Test: Soil Classification & Soil Structure - Question 7

In a particular soil sample, laboratory analysis has yielded the following result:
1. Sand - 20%
2. Silt - 30%
3. Clay - 50%
Without using the textural chart, the correct textural classification of the soil would be

Detailed Solution for Test: Soil Classification & Soil Structure - Question 7

As 50% of soil is clay. So it will be classified as clay.

Test: Soil Classification & Soil Structure - Question 8

The description of ‘sandy silty clay’ signifies that

Test: Soil Classification & Soil Structure - Question 9

Sieve analysis on a dry soil sample of mass 1000 g showed that 980 g and 270 g of soil pass through 4.75 mm and 0.075 mm sieve, respectively. The liquid limit and plastic limits of the soil fraction passing through 425 m sieves are 40% and 18%, respectively. The soil may be classified as

Detailed Solution for Test: Soil Classification & Soil Structure - Question 9

980 gm of soil passes through 4.75 mm sieve
270 gm of soil passes through 0.075 mm sieve
Liquid limit of the soil,
Plastic limit of the soil,
Formula of plasticity index is given by,


More than 50% of the material passes through 4.75 mm sieve therefore it is sand. Percentage finer is 27%. It is clayey sand, SC.

Test: Soil Classification & Soil Structure - Question 10

Consider the following statements:
1. Coarse-grained soil having fines (<75μ in size) between 5% and 12%, have a dual symbol according to IS code for soil classification
2, At liquid limit, all soils have the same shearing strength.
3. Lower the shrinkage limit, greater is the volume change in a soil with change in water content.
Of these statements:

Detailed Solution for Test: Soil Classification & Soil Structure - Question 10

Lower the shrinkage limit greater is the volume change.
For coarse grained soil with fines < 5% classification will be GP, GW, SP, and SW, for fines > 12% classification will be based on plasticity chart as GM, GC, SM and SC. For fines 5 -12% dual classification like GP - GM; GP - GC etc., will be used.
At liquid limit the soils possess a certain shear strength which is the smallest value that can be measured in a standard procedure. From direct shear tests on different types of clays it is found that liquid limit corresponds to a shearing strength of about 2.7 kN/m2.

If shrinkage limit Is less the volume change with change in water content will be more.

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