Test: Soil Classification & Soil Structure


10 Questions MCQ Test Topicwise Question Bank for Civil Engineering | Test: Soil Classification & Soil Structure


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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

Solution:
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

Solution:

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.

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

Solution:
QUESTION: 4

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

Solution:


∴ D60 = 4.75 mm
D10 = 0.075 mm

So correct answer is ‘c’.

QUESTION: 5

Inorganic soil with low compressibility are represented by

Solution:

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

QUESTION: 6

In a soil specimen, 70% of particles are passing through 4.75 mm IS sieve and 40% of particles are passing through 75 μ IS sieve. Its uniformity coefficient is 8 and coefficient of curvature is 2. AS per IS classification, this soil is classified as

Solution:

Since more than 50% of particles are passing through 4.75 mm sieve while less than 50% are passing through 75μ sieve, the soil is sand.

Therefore its well graded sand (SW).

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

Solution:

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

QUESTION: 8

The description of ‘sandy silty clay’ signifies that

Solution:
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

Solution:

Plasticity Index. PI = WC - WP.
= 40 - 18 = 22%
∵ PI is > 17. soil is highly plastic i.e. clayey soil.
Since. 98% of soil pass through 4.75 mm sieve and 27% through .075 mm sieve. majority of soil lies between 0.075 and 4.75 mm i.e., sand.
∴ Soil is clayey sand (SC) 

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:

Solution:

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.