This set of Pavement Design Interview Questions and Answers focuses on “Highway Materials – Subgrade Soil Properties – 2”.
1. Which of the below is the correct formula to find group index (GI) of soil?
a) GI = 0.1a+0.0025ac+0.005bd
b) GI = 0.2a+0.005ac+0.01bd
c) GI = 0.01a+0.025ac+0.05bd
d) GI = 0.02a+0.05ac+0.001bd
Explanation: GI = 0.2a+0.005ac+0.01bd is the right formula to find the group index of soil. In the formula a and b represent the portion of materials passing certain sieves; c and d represent ranged values of liquid and plastic limit respectively.
2. Soils having ______ values of liquid limit and plastic limit are generally considered to be poor engineering materials.
Explanation: Both the liquid limit and plastic limit depend on the type and amount of clay present in the soil. Higher value of both parameters indicates higher clay content and thus are not considered good materials for engineering purposes.
3. According to the BIS soil classification system, what does SM represent?
a) Sandy silt
b) Silty sand
c) Sandy clay
d) Clayey sand
Explanation: The letter S is used to represent sand and M for silt. So, SM would represent silty sand or poorly graded sand-silt mixture. For representing clay, the letter C is used.
4. Identify the right classification as per the unified soil classification system for the following data provided:
|SIEVE SIZE (mm)||PERCENT PASSING|
Cu = 2.45 and Cg = 1.1
Explanation: As per the unified soil classification system, if the percent of particles smaller than 4.75 mm sieve is more than 50%, then it is sand type soil. The question has 70%, so definitely it is under the major division sand. If the percentage of fines (particles passing 0.074 mm sieve) is less than 5%, it is either SW or SP. If Cu is < 4 and Cg is between 1 and 3, it is classified as SW. As per the question, there are no fines, so it either SW or SP. The values of Cu and Cg confirm that it is SW.
5. In the textural soil classification chart, the base of the triangle represents the percentage of ______
Explanation: In order to classify the soil according to the textural features, there is a chart available showing various percentages of sand, silt and clay. The chart is shaped like an equilateral triangle with percent sand as base, percent clay on the left side and percent silt on the right side.
6. The cohesion of a soil ______ with the increase in moisture content.
a) Doesn’t change
d) Decreases then increases
Explanation: Cohesion is a property of fine grained soil types and it is derived from the water films which bind the soil particles together. Cohesion is greater in soil types having higher moisture content, like clay.
7. Which of the below is the soil classification for the given data?
Passing 0.074 mm sieve = 60%
Liquid limit = 45%
Plastic limit = 39%
Use Highway research board classification system.
a) A – 4
b) A – 3
c) A – 5
d) A – 7
Explanation: According to the system, the soil is granular if percent passing 0.074 mm sieve is less than 35% and fine grained if it is more than 35%. Since the percent passing 0.074 mm sieve is more than 35% in question (60%), the soil is fine grained. The value of liquid limit must be a minimum of 41% for groups A – 5 and A – 7. Since the value in question is 45%, it may be A – 5 or A – 7. If the maximum plasticity index is 10, then it is A – 5, and if the minimum is 11, it is group A – 7. As per the question, plasticity index, PI = liquid limit – plastic limit = 45 – 39 = 6%. Therefore, it is classified as A – 5.
8. Which of the below soil group – value as subgrade pair has been paired correctly according to the unified soil classification system?
a) GC – good
b) SW – excellent
c) SC – poor
d) CH – good
Explanation: As per the unified soil classification system, the soil group GC is deemed to be good to be used as subgrade. The value of soil group SW is good, for SC it is fair to good and for CH it is poor.
9. Dry sieve analysis is carried out on cohesive soils.
Explanation: Sieve analysis is carried out to determine the grain size of soil components. There are two types of sieve analysis – dry and wet. Dry sieve analysis is carried out for non-cohesive soils and wet sieve analysis is carried out for cohesive soils. Cohesive soils have clay content and may have some small lumps which might be difficult to be separated if the dry sieve analysis is carried out.
10. Using the below given data, which of the options represent the group index of the soil?
Passing 0.074 mm sieve = 50%
Liquid limit = 46%
Plastic limit = 37%
Explanation: Using the equation for group index as
where “a” represents the portion of material passing 0.074 mm sieve, greater than 35 but not exceeding 75 percent (represented as whole number in the range 0-40),
“b” represents the portion of material passing 0.074 mm sieve, greater than 15 but not exceeding 35 percent (expressed as whole number in the range 0-40),
“c” represents the liquid limit in excess of 40 but less than 60 (expressed as whole number in the range 0-20),
“d” represents the plasticity index in excess of 10 but less than 30 (expressed as whole number in the range 0-20)
As per the question,
Plasticity index = liquid limit – plastic limit = 46 – 37 = 9
d=9-10=0 (whole number)
GI =0.2×15+0.005×15×6+0.01×35×0=3.45 \(\approx\) 4 (whole number).
11. The ISC system of soil classification consists of a total of 18 groups of soil.
Explanation: The ISC system is the Indian Standard Classification system. It classifies soil into three broad categories, namely coarse grained, fine grained and peat. There are 8 groups of coarse grained soil, 9 groups of fine grained soil and 1 group of peat. These add up to a total of 18 groups.
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