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Test: Index Properties - Civil Engineering (CE) MCQ


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10 Questions MCQ Test - Test: Index Properties

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

The three-phase diagram of the partially saturated soil sample is given in the figure. Determine the degree of saturation of the soil.

Detailed Solution for Test: Index Properties - Question 1

Concept:
Degree of saturation (Sr): 
Degree of saturation is defined as the ratio of the volume of water (Vw) to the volume of voids (Vv).

Calculation:
Given data:

Weight of soil solid in sample (Ws) =1.0 g
Weight of water in sample (Ww) = 0.15 g
Volume of soil solid in sample (Vs) = 0.3704 cm3
Volume of water in sample (Vw) = 0.15 cc or 0.15 cm3
Volume of voids in sample (Vv) = 0.2349 cm3
Volume of air  in sample (Va) = 0.2349 - 0.15 = 0.0849 cc
Volume of air  in sample (Va) = 0.0849 cc

Degree of saturation in a given sample (Sr) = 0.6385

Test: Index Properties - Question 2

In the geotechnical engineering lab, the pycnometer is used for the determination of

Detailed Solution for Test: Index Properties - Question 2

Concept:
Pycnometer method is used to determine specific gravity and water content both. This method is suitable for cohesionless soils.
Pycnometer is a glass jar of 1-liter capacity that is fitted at its top by a conical cap made of brass. It has a screw type cover and there is a small hole at its apex of 6 mm diameter.

W1 = Mass of empty volume of pycnometer
W2 = Mass of pycnometer + Mass of moist sample
W3 = Mass of pycnometer + soil + water
W4 = Mass of pycnometer full of water
G, = Specific gravity of soil solids
Water content of the soil sample is given by:

Specific gravity is usually reported at 27°C and it is given by:

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Test: Index Properties - Question 3

Based on Fuller's maximum density criterion, for 4 mm maximum size of soil particles, what is the percentage of particles between 4 mm and 2 mm, by weight?

Detailed Solution for Test: Index Properties - Question 3

Fuller's Maximum Density Criteria:

  • This method is economical to design for a minimum amount of cement-water paste in a concrete mix.
  • Fuller assumes that the greater the amount of solid particles that can be packed in a given volume of concrete, the higher will be its strength.
  • According to Fuller's maximum density criteria, Percentage of total material smaller than size d is given as:


Where P = % passing sieve size d , D = Maximum size of particle
Given, maximum size of particle = 4 mm 
∴ Percentage particles between 4 mm and 2 mm, by weight can be calculated by: P4mm - P2mm  

Test: Index Properties - Question 4

As per IS, the internal diameter and height of the steel dolly used in core cutter test are __________ respectively.

Detailed Solution for Test: Index Properties - Question 4

Concept:
Core cutter test:

  • This method is used to find in-situ unit weight in the case of fine-grained cohesive soils without stones.
  • This method cannot be used in the case of hard and gravelly soils.

Equipment used:

  1. Cylindrical core cutter, 100 mm internal diameter and 130 mm long.
  2. Steel rammer of mass 9 kg and overall length 900 mm
  3. Steel Dolley, 25 mm high and 100 mm diameter
  4. Weighing balance of accuracy up to 1 gm
Test: Index Properties - Question 5

A soil sample with specific gravity of solids 3 has a mass specific gravity of 2. Assuming the soil to be perfectly dry, determine the void ratio.

Detailed Solution for Test: Index Properties - Question 5

Concept:
From a fundamental relationship, the density of soil is given by:

Where
γ = Density of soil, 
γw = Density of water, G = Specific gravity of soil, S = Degree of saturation, w = Water content, and e = void ratio
For the dry density of soil: Degree of saturation (S) = 0
The dry density of soil(γd) is given by

Given data
The specific gravity of solids = 3
The mass-specific gravity = 2
The dry density of soil(γd) is given by

Where Gm = Mass specific gravity

e = 0.5
The void ratio is 0.5.

Test: Index Properties - Question 6

The void ratios at the densest, loosest, and natural states of a sand deposit are 0.2, 0.6, and 0.4, respectively. The relative density of the deposit is:

Detailed Solution for Test: Index Properties - Question 6

Concept:
Relative density: 

  • Relative density is the measure of the compactness of cohesion-less soil.
  • Relative density represents whether the soil is closest to its densest state or loosest state.
  • It is defined as the ratio of the difference between the voids ratio of the soil in its loosest state and its natural state to the difference between the voids ratio in the loosest and densest state.


The relationship between the Relative Density and Void ratio of the soil sample is depicted in the graph below:

∴ Relative density varies inversely to the void ratio.
Classification of soil as per relative density as follows:

Calculation:
emax = 0.6, emin = 0.2, and natural void ratio (e) = 0.4

Expressing it in percentage = 0.5 × 100 = 50%

Test: Index Properties - Question 7

A soil has a bulk density of 22 kN/m3, water content 10%, the dry density of soil is -

Detailed Solution for Test: Index Properties - Question 7

Concept:
The fundamental relation between dry density (γd), bulk density (γ) and water content (ω), is

Calculation:
Given, 
γ = 22 kN/m3,  w = 10 %

Test: Index Properties - Question 8

The total unit weight of the glacial outwash soil is 5 kN/m3. The water content in the soil is 17%. Find dry unit weight

Detailed Solution for Test: Index Properties - Question 8

Concept:
The formula for finding the dry unit weight of soil is given by:

where, 
γd = Dry unit weight of soil
γ = Total Unit weight of soil
w = Water content
Calculations:
Given Data:
γ = 5 kN/m3
w = 17 % = 0.17
So, 

Test: Index Properties - Question 9

When liquidity index is greater than 1:

Detailed Solution for Test: Index Properties - Question 9

Concept:
The liquidity index is defined as a ratio of the difference between the natural water content of a soil and plastic limit to the plasticity index.
The liquidity index is also known as the water plasticity ratio.
Liquidity Index of a soil is given by:


Hence, when the liquidity index is greater than 1, it indicates the soil is in a liquid state.

Test: Index Properties - Question 10

At which consistency limit, further reduction of the water content will not decrease the volume of soil mass?

Detailed Solution for Test: Index Properties - Question 10

The shrinkage limit is the water content of the soil when the water is just sufficient to fill all the pores of the soil and the soil is just saturated.
The volume of the soil does not decrease when the water content is reduced below the shrinkage limit.

Where,
wL = liquid limit
wp = plastic limit
ws = shrinkage limit
VL = Volume of soil mass corresponding to liquid limit
Vp = Volume of soil mass corresponding to plastic limit
Vd = Volume of soil mass corresponding to shrinkage limit
Vs = Volume of solids

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