Index Properties MCQ Quiz - Objective Question with Answer for Index Properties - Download Free PDF

Last updated on Jun 20, 2025

Latest Index Properties MCQ Objective Questions

Index Properties Question 1:

Select the correct statement.

  1. Unit weight of dry soil is greater than unit weight of wet soil.
  2. For dry soils, dry unit weight is less than total unit weight.
  3. Unit weight of soil increases due to submergence in water
  4. Unit weight of soil decreases due to submergence in water.

Answer (Detailed Solution Below)

Option 4 : Unit weight of soil decreases due to submergence in water.

Index Properties Question 1 Detailed Solution

Explanation:

  • Unit weight of dry soil (γd) → weight of soil solids only per unit volume.

  • Unit weight of wet soil (γ) → includes soil solids + water weight per unit volume.

  • Submerged unit weight (γsub) → effective weight of soil when immersed in water, calculated as: \(\gamma_{sub}=\gamma_{sat}-\gamma_{w}\), where γsat = saturated unit weight,  γw=unit weight of water

  • When soil is submerged, the buoyant force reduces its effective weight.

  • The apparent or submerged unit weight decreases compared to the saturated or total unit weight.

  • This is why submerged soils have lower unit weight (γsub) than saturated soils.

Additional Information Unit Weight of Dry Soil

  • Refers to the weight of soil solids only per unit volume.

  • It is the unit weight when no water is present in the soil pores.

  • Dry unit weight is an important property for understanding the compaction and strength of soils in geotechnical engineering.

  • It is always less than or equal to the wet (bulk) or saturated unit weight.

Unit Weight of Wet Soil

  • Represents the weight of both soil solids and the water present in the pores of the soil mass.

  • This is the natural condition of soil found in the field — where some amount of moisture is always present.

  • The presence of water increases the unit weight compared to dry unit weight. 

Index Properties Question 2:

The soil has a bulk density of \(44\text{kN}/m^3\) and water content 10%. The dry density of soil is

  1. \(4.4\text{kN}/m^3\)
  2. \(10\text{kN}/m^3\)
  3. \(20\text{kN}/m^3\)
  4. \(40\text{kN}/m^3\)

Answer (Detailed Solution Below)

Option 4 : \(40\text{kN}/m^3\)

Index Properties Question 2 Detailed Solution

Concept:

Dry density of soil is given by: \(\gamma_d=\frac{\gamma}{1+w} \)

where \(\gamma_d\) is dry density of soil; \(\gamma\) is bulk density of soil and w is water content of same soil.

Calculation:

Given: \(\gamma = 44kN/m^3; w=10%\)

So, dry density of soil is \(\gamma_d=\frac{44}{1+0.1}=40kN/m^3\)

Index Properties Question 3:

The ratio of unconfined compressive strength of an undisturbed sample of soil to that of a remoulded sample, at the same water content is known as

  1. Activity
  2. Sensitivity
  3. Damping
  4. Plasticity

Answer (Detailed Solution Below)

Option 2 : Sensitivity

Index Properties Question 3 Detailed Solution

Explanation

  • It is defined as the ratio of unconfined compressive strength (UCS) of an undisturbed soil sample to that of the remoulded sample at the same water content.
  • It indicates the loss in strength due to disturbance or remoulding of soil.

 Additional Information

Activity:

  • Defined as the ratio of plasticity index to the percentage of clay fraction (<2 micron size).

  • Indicates the expansive nature of clay soils.

Damping:

  • Refers to the energy dissipation property of soil during cyclic or dynamic loading.

  • Commonly discussed in earthquake and vibration analysis, not related to strength comparison.

Plasticity:

  • Refers to the ability of soil to undergo deformation without cracking or crumbling.

  • Measured using Atterberg limits: Liquid limit, Plastic limit, and Plasticity Index.

Index Properties Question 4:

When the plastic limit of a soil is greater than the liquid limit, then the plasticity index is reported as

  1. Zero
  2. 1
  3. Negative
  4. Positive

Answer (Detailed Solution Below)

Option 1 : Zero

Index Properties Question 4 Detailed Solution

Explanation:

  • Plasticity Index (PI) = Liquid Limit (LL) – Plastic Limit (PL)

  • Normally, LL > PL, so PI is positive.

  • However, if PL > LL, this is physically unrealistic, often due to testing errors or soil peculiarities.

  • As per IS 2720 (Part 5) and general practice, in such cases, the Plasticity Index is reported as zero, not negative.

Additional InformationLiquid Limit (LL):

  • The minimum water content at which a soil changes from a plastic state to a liquid state.

  • It represents the boundary between liquid and plastic behavior.

Plastic Limit (PL):

  • The minimum water content at which a soil can be rolled into 3 mm diameter threads without breaking.

  • It marks the boundary between plastic and semi-solid states.

Plasticity Index (PI):

  • The numerical difference between the liquid limit and the plastic limit:
    PI = LL − PL

  • It indicates the range of moisture content over which the soil remains plastic.

Index Properties Question 5:

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. 

  1. 2.71
  2. 2.77
  3. 2.67
  4. 2.76
  5. 2.50

Answer (Detailed Solution Below)

Option 1 : 2.71

Index Properties Question 5 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.

\(S.R. = \frac{\frac{(V_o-V_d)}{V_d}\times100}{w_l -w_s} = \frac{M_d}{\rho_wV_d}=\frac{γ_d}{γ_w} = G_m\)      ---(1)

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:

\({{\rm{W}}_{\rm{S}}} = \frac{1}{{\rm{S.R.}}} - \frac{1}{{\rm{G_s}}}\)

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

\(S.R. = \frac{\frac{(V_o-V_d)}{V_d}\times100}{w_l -w_s} \)

\(S.R. = \frac{\frac{(30-16.6)}{16.6}\times100}{75 \ -\ 25} \)

S.R. = 1.614

Shrinkage limit:

\({{\rm{W}}_{\rm{S}}} = \frac{1}{{\rm{S.R.}}} - \frac{1}{{\rm{G_s}}}\)

\({{0.25}} = \frac{1}{{\rm{1.614}}} - \frac{1}{{\rm{G_s}}}\)

Gs = 2.71

Top Index Properties MCQ Objective Questions

A value of toughness index less than unity indicates that the soil is friable at the ______

  1. plastic limit
  2. liquidity limit
  3. composite limit
  4. elastic limit

Answer (Detailed Solution Below)

Option 1 : plastic limit

Index Properties Question 6 Detailed Solution

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Concept:

Toughness Index: 

  • Toughness index is defined as the ratio of plasticity index (IP) of the soil to the flow index (IF) of the soil.

\(\begin{array}{l} Toughess\;index = \frac{{Plasticity\;index}}{{Flow\;index}}\\ \end{array}\)

  • 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.

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

  1. 10
  2. 20
  3. 30
  4. 40

Answer (Detailed Solution Below)

Option 3 : 30

Index Properties Question 7 Detailed Solution

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Concept:

F1 A.M Madhu 23.12.19 D10

slope = \(dy\over dx \) = constant

⇒ \({V_L-V_P}\over {w_L - w_p} \) = \(V_P-V_D\over {w_p - w_s}\)

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

⇒ \({V_L-V_P}\over {w_L - w_p} \) = \(V_P-V_D\over {w_p - w_s}\)

Now, putting values, we get 

⇒ \({1.5 V_d - 1.2 V_d}\over I_P\) = \(1.2V_d - V_d \over {40-20}\)

IP = 30%

Hence, the plasticity Index Pl of the soil is 30%.

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

  1. 10 cm and 2.73 cm
  2. 10 cm and 3 cm
  3. 10 cm and 2.5 cm
  4. 11.7 cm and 1.5 cm

Answer (Detailed Solution Below)

Option 3 : 10 cm and 2.5 cm

Index Properties Question 8 Detailed Solution

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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

F1 A.M Madhu 09.04.20 D11

Which of the following can be considered as quick clay?

  1. Sensitivity = 0
  2. Sensitivity = 1
  3. Sensitivity = 100
  4. Sensitivity = ∞

Answer (Detailed Solution Below)

Option 3 : Sensitivity = 100

Index Properties Question 9 Detailed Solution

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Explanation:

(i) A quick clay is defined as a clay where the undisturbed shear strength of the soil is at least 30 times greater than the remolded (or disturbed) shear strength. 

(ii) The ratio of undisturbed to disturbed strength is termed as sensitivity.

Classification 

Sensitivity 

Insensitive clay

< 1 

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 

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:

  1. 100%
  2. 75%
  3. 50%
  4. 25%

Answer (Detailed Solution Below)

Option 3 : 50%

Index Properties Question 10 Detailed Solution

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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.

\({I_D} = \frac{{{e_{max}} - {e_{natural}}}}{{{e_{max}} - {e_{min}}}}\)

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

RRB JE CE R46 17Q Geotechnical Engg(Hindi) 2

∴ Relative density varies inversely to the void ratio.

Classification of soil as per relative density as follows:

Relative Density

Density

Description

0-15

Very Loose

15-35

Loose

35-65

Medium

65-85

Dense

85-100

Very Dense

Calculation:

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

\({{\rm{I}}_{\rm{d}}} = \frac{{0.6 - 0.4}}{{0.6 - 0.2}} = \frac{{0.2}}{{0.4}} = 0.5\)

Expressing it in percentage = 0.5 × 100 = 50%

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.

  1. 1.0
  2. 1.5
  3. 0.5
  4. 0.8

Answer (Detailed Solution Below)

Option 3 : 0.5

Index Properties Question 11 Detailed Solution

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Concept:

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

\({{\rm{\gamma }}_{{\rm{bulk}}}} = \frac{{\left( {{\rm{G}} + S × e} \right) × {\gamma _w}}}{{1 + e}} = \frac{{\left( {{\rm{G}} + w × G} \right) × {\gamma _w}}}{{1 + e}} = \frac{{\left( {1 + {\rm{w}}} \right) × G × {\gamma _w}}}{{1 + e}}\)

Where

\(\gamma\) = Density of soil, \(\gamma_w\) = 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(\(\gamma_d\)) is given by

\({{\rm{\gamma }}_{\rm{d}}} = \frac{{\left( {\rm{G}} \right) × {\gamma _w}}}{{1 + e}}\)

Calculation:

Given data

The specific gravity of solids = 3

The mass-specific gravity = 2

The dry density of soil(\(\gamma_d\)) is given by

\({{\rm{\gamma }}_{\rm{d}}} = \frac{{\left( {\rm{G}} \right) × {\gamma _w}}}{{1 + e}}\)

\({\gamma_d \over \gamma_w} = {G \over 1+e}\)

\(G_m = {G \over 1+e}\)

Where Gm = Mass specific gravity

\(2 = \frac{{\left( {\rm{3}} \right)}}{{1 + e}}\)

\({{1 + e}} = 1.5\)

e = 0.5

The void ratio is 0.5.

Which of the following is one of the methods of specific gravity determination of solid particles?

  1. Oven drying method
  2. Alcohol method
  3. Shrinkage limit method
  4. Calcium carbide method

Answer (Detailed Solution Below)

Option 3 : Shrinkage limit method

Index Properties Question 12 Detailed Solution

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Concepts:

The shrinkage Limit Method is used to determine the specific gravity of solid particles. Based on Shrinkage Ratio (SR) and Shrinkage Limit (SL), the specific gravity of soil solids can be calculated using the following formula:

\(G = \frac{1}{\frac{1}{SR}-\frac{SL}{100}} \)

Important Points:

1. Oven dry method, Calcium Carbide Method, Alcohol Methods, Sand Bath Method, Torsion balance Methods are used to determine the water content of soil.

2. The other methods for determining the specific gravity of solid particles are – Core Cutter Method, Sand Replacement Method, Water displacement Method. These methods give the unit weight of soil, by dividing it with the unit weight of water, we can get specific gravity.

3. Pycnometer is used for determining the specific gravity of soil if we know the water content and also used for determining the water content if we know the specific gravity of soil. 

A soil sample is tested for its plasticity index and the obtained value of plasticity index is 21. In such case, the soil sample can be described as:

  1. non plastic
  2. low plastic
  3. highly plastic
  4. medium plastic

Answer (Detailed Solution Below)

Option 3 : highly plastic

Index Properties Question 13 Detailed Solution

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Explanation:

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Liquid limit (WL): The water content at which soil changes its consistency from plastic to a liquid state.

Plastic limit (WP): The water content at which soil changes its consistency from brittle/crumbly to a plastic state.

Shrinkage limit (WS): A state when the decrease in moisture content leads to solid-state, no change in volume of soil mass is observed, the consistency of soil changes from semi to solid-state. The boundary water content is called shrinkage limit.

Plasticity index (IP) = WL - WP

Where, WL = Liquid limit, WP = Plastic limit, WN = Natural water content

Based on the plasticity index, the soil may be classified as followed:

Plasticity Index

Soil Description

Soil Type

0

Non-plastic

Sand

1 – 7

Low plastic

Silt

7 - 17

Medium plastic

Silty clay

> 17

Highly plastic

Clay

The determination of plasticity index and the percentage by weight of clay mineral finer than 2 μ present in the soil were done for two soil samples A and B. The ratio of plasticity index to the percentage of clay size particle for the samples A and B are given in the table.

Soil Sample Ratio of plasticity index to percentage of clay size particle
Sample A Less than 0.75
Sample B Greater than 1.25

The two soil samples A and B are classified as:

  1. Sample A: Normal; Sample B: Inactive
  2. Sample A: Active; Sample B: Normal
  3. Sample A: Active; Sample B: Inactive
  4. Sample A: Inactive; Sample B: Active

Answer (Detailed Solution Below)

Option 4 : Sample A: Inactive; Sample B: Active

Index Properties Question 14 Detailed Solution

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Explanation:

The Activity of the soil (A)

The activity of the soil is given by the ratio of the plasticity index (Ip) and the percentage of clay fraction (C) in the soil.

It indicates water absorption capacity or indicates swelling and shrinkage characteristics.

Activity \(= \frac{{Plasticity\;Index}}{{Per\;cent\;of\;clay\;particles\;finer\;than\;2\;\mu m}}\)

Activity-based classification of clays:

Activity Classification
< 0.75 Inactive 
0.75 - 1.40 Normal
> 1.40 Active

Hence, The most appropriate answer is Sample A is inactive and Sample B is active.

Which statement is not true regarding specific gravity?

  1. 50 ml density bottle is useful for all types of soil
  2. Kerosene is used in density bottles
  3. 500ml flask is used only for fine grained soils
  4. Pycnometer is used only for coarse-grained soil

Answer (Detailed Solution Below)

Option 3 : 500ml flask is used only for fine grained soils

Index Properties Question 15 Detailed Solution

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Concept:

Specific Gravity:

Specific gravity can be defined as the ratio of the weight of a given volume of soil solids at a given temperature to the weight of the equal volume of distilled water at that temperature. It can be written as G = \(\dfrac{\gamma_s}{\gamma_w}\)

1. Density bottle method:

  • The specific gravity of solid particles can be determined in a laboratory using a density bottle fitted with a stopper having a hole.
  • The density bottle of 50 ml capacity is generally used [ IS:2720 (part III) 1980].

2. Pycnometer method:

  • Pycnometer method is used to determine specific gravity and water content both.
  • This method is suitable for cohesionless soils.
  • A 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 in diameter

Explanation:

  • Density bottle method can be used for every type of soil, and it is the most accurate one.
  • Flask and pycnometer are used for coarse-grained soil.
  • Distilled water is used in the pycnometer test and kerosene is used for the density bottle method. When kerosene is used, its specific gravity is determined accurately by a separate test. 
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