Index Properties MCQ Quiz in मराठी - Objective Question with Answer for Index Properties - मोफत PDF डाउनलोड करा

Last updated on Mar 13, 2025

पाईये Index Properties उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). हे मोफत डाउनलोड करा Index Properties एमसीक्यू क्विझ पीडीएफ आणि बँकिंग, एसएससी, रेल्वे, यूपीएससी, स्टेट पीएससी यासारख्या तुमच्या आगामी परीक्षांची तयारी करा.

Latest Index Properties MCQ Objective Questions

Top Index Properties MCQ Objective Questions

Index Properties Question 1:

The soil sample used for liquid limit, plastic limit and shrinkage limit tests should be finer than :

  1. 75 μm
  2. 150 μm
  3. 425 μm
  4. 200 μm

Answer (Detailed Solution Below)

Option 3 : 425 μm

Index Properties Question 1 Detailed Solution

Explanation:

The Atterberg limits are a basic measure of the critical water contents of a fine-grained soil: its shrinkage limit, plastic limit, and liquid limit. Depending on its water content, a soil may appear in one of four states: solid, semi-solid, plastic, and liquid as shown below:

SSC JE CE 23 SEPT 19 Morning Nitesh Upload 40Q 1

Shrinkage Limit: Water content below which, no further reduction in the volume of soil is possible, irrespective of change in the water content.

Plastic Limit: Water content just above which the soil just turns into a plastic state.

Liquid Limit: Water content at which soil changes from a liquid state to a plastic state.

Soil particles of size above 425 microns are not considered for the liquid limit, plastic limit, and shrinkage limit as the liquid flow on Casagrande's apparatus will be hindered by the particles of size greater than 425 microns resulting in a differed value of the liquid limit.

Index Properties Question 2:

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

  1. 0.29 kN/m3
  2. 6.27 kN/m3
  3. 3.40 kN/m3
  4. 4.27 kN/m3

Answer (Detailed Solution Below)

Option 4 : 4.27 kN/m3

Index Properties Question 2 Detailed Solution

Concept:

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

\({γ _d} = \frac{γ }{{1 + w}}\)

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, 

\({γ _d} = \frac{5 }{{1 + 0.17}}=4.27\;kN/m^3\)

Index Properties Question 3:

Water content of a soil sample can be determined by:

  1. the alcohol method
  2. the sand replacement method
  3. the jar test method
  4. the shrinkage limit method

Answer (Detailed Solution Below)

Option 1 : the alcohol method

Index Properties Question 3 Detailed Solution

Explanation:

Methods used for finding water content:

Method Properties
Oven drying method Most accurate method and is a standard laboratory method
Pycnometer method More suitable for cohesionless soil as removal of entrapped air from cohesive soil difficult.
Sand bath method It is a rapid method, hence not very accurate
Torsion balance method Drying and weighing done simultaneously, hence one of the accurate methods
Calcium carbide method Takes just 5-7 minutes and used as a field test
Alcohol test It is a quick field test and not used for soils containing calcium or organic compound
Radiation method Gives water content in an in-situ condition

Index Properties Question 4:

If the Consistency Index of soil is in the range of 50 - 75 % then the soil is said to be

  1. Soft
  2. Medium
  3. Stiff
  4. Hard

Answer (Detailed Solution Below)

Option 2 : Medium

Index Properties Question 4 Detailed Solution

Explanation:

Consistency Index: It is defined as a ratio of the difference between the liquid limit and the natural water content of the soil to the plasticity index.

\({I_c} = \frac{{{w_{L - W}}}}{{{I_p}}}\)

Where,

wL = water content at liquid limit

w = Natural water content

Ip = Platicity index

Consistency

Description

Ic

Liquid

Liquid

< 0

Plastic

Very soft

0-0.25

Soft

0.25-0.5

Medium stiff

0.5-0.75

Stiff

0.75-1.00

Semi-solid

Very stiff or hard

> 1

Solid

Hard or very hard

> 1

Index Properties Question 5:

The plastic limit and liquid limit of a soil are 30% and 42% respectively. The percentage volume change from liquid limit to dry state is 35% of the dry volume. Similarly the percentage volume change from plastic limit to dry state is 22% of the dry volume. The shrinkage ratio will be nearly

  1. 4.2
  2. 3.1
  3. 2.2
  4. 1.1

Answer (Detailed Solution Below)

Option 4 : 1.1

Index Properties Question 5 Detailed Solution

Concept:

Shrinkage ratio: It is the ratio of a given volume change expressed as a percentage of dry volume to the corresponding charge in water content above the shrinkage limit expressed as a percentage of the weight of the oven dried soil.

\(\text{SR}=\frac{\frac{\left( {{\text{V}}_{1}}-{{\text{V}}_{2}} \right)}{{{\text{V}}_{\text{d}}}}\times 100}{{{\text{w}}_{1}}-{{\text{w}}_{2}}}=\frac{{{\text{V}}_{1}}-{{\text{V}}_{2}}}{{{\text{V}}_{\text{d}}}\left( {{\text{w}}_{1}}-{{\text{w}}_{2}} \right)}\times 100\)

Where,

V1 = Volume of soil mass at water content (w1)

V2 = Volume of soil mass at water content (w2)

Vd = Volume of dry soil mass

Note: Shrinkage ratio is equal to mass specific gravity of soil in its dry state.

Calculation:

WL = 42%

WP = 30%

gate ft 5 images Q23

As per given data,

(VL – Vd )/Vd= 35

(VP – Vd )/Vd= 22

Also, \(\frac{{{V_L} - {V_d}}}{{{W_L} - {W_s}}} = \frac{{{V_p} - {V_d}}}{{{W_p} - {W_s}}}\)

\(\frac{{35}}{{0.42 - {W_s}}} = \frac{{22}}{{0.3 - {W_s}}}\)

Solving we get WS = 9.69%

Shrinkage Ratio, \(SR = {{{{{V_L} - {V_d}} \over {{V_d}}} \times 100} \over {{w_L} - {w_s}}}\)

∴ \(S.R = \frac{{35}}{{42 - 9.69}} = 1.1\)

Index Properties Question 6:

A specimen of clayey silt contains 70% silt size particles. Its liquid limit is 40% and its plastic limit is 20%. In the liquid limit test, at a moisture content of 30%, the required no. of blows was 50. Its plasticity index, activity, and consistency index will respectively be 

  1. 20, 0.67 and 0.5
  2. 20, 1.5 and 2.0
  3. 30, 1.5 and 0.72
  4. 20, 0.286 and 0.38

Answer (Detailed Solution Below)

Option 1 : 20, 0.67 and 0.5

Index Properties Question 6 Detailed Solution

Concept:

Plasticity Index:

  • Plasticity index (PI) is the range of water content over which the soil remains in the plastic state. Mathematically defined as,
  • Plasticity Index = Liquid Limit - Plastic Limit.
  • The plasticity of soil is its ability to undergo deformation without cracking.
  • It is an important index property of fine-grained soil, especially for clayey soils.

Activity:

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

Consistency Index or Relative consistency:

  • It is useful in the study of the field behavior of saturated fine-grained soil.
  • It is given as subtraction of natural water content from the liquid limit to the plasticity index of the soil.

\({{\rm{I}}_{\rm{c}}} = \frac{{{{\rm{w}}_{\rm{l}}} - {{\rm{w}}_{\rm{n}}}}}{I_P}\)

IP = Plasticity Index = wl - wP

\({{\rm{I}}_{\rm{c}}} = \frac{{{{\rm{w}}_{\rm{l}}} - {{\rm{w}}_{\rm{n}}}}}{{{{\rm{w}}_{\rm{l}}} - {{\rm{w}}_{\rm{p}}}}}\)

\({{\rm{w}}_{\rm{l}}}\) = liquid limit, \({{\rm{w}}_{\rm{P}}}\) = plastic limit, \({{\rm{w}}_{\rm{n}}}\) = water content of the soil at the natural state

Calculation:

Given, Wl = 40%, Wp = 20%, Clayey silt = 70%

Wn = 30%, Clay content = 100 - silt content = 100 - 70 = 30%

a) Plasticity Index

Plasticity Index = Liquid Limit – Plastic Limit

Plasticity Index = 40 - 20 = 20%

b) Activity

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

Activity \(= \frac{{20}}{{30}}\) = 0.67

Consistency Index:

\({{\rm{I}}_{\rm{c}}} = \frac{{{{\rm{w}}_{\rm{l}}} - {{\rm{w}}_{\rm{n}}}}}{{{{\rm{w}}_{\rm{l}}} - {{\rm{w}}_{\rm{p}}}}}\)

\({{\rm{I}}_{\rm{c}}} = \frac{40 \ - \ 30}{40 \ -\ 20}\) = 0.5

Index Properties Question 7:

The consistency index of soil is defined as the ratio of:

  1. Subtraction of natural water content from liquid limit to the plasticity index of the soil
  2. Subtraction of plastic limit from natural water content to the plasticity index of soil
  3. Summation of natural water content of a soil plus its plastic limit to the plasticity index of the soil
  4. Summation of liquid limit and natural water content to the plasticity index of soil

Answer (Detailed Solution Below)

Option 1 : Subtraction of natural water content from liquid limit to the plasticity index of the soil

Index Properties Question 7 Detailed Solution

Consistency Index or Relative consistency: 

It is useful in the study of the field behaviour of saturated fine-grained soil.

It is given as subtraction of natural water content from liquid limit to the plasticity index of the soil.

\({{\rm{I}}_{\rm{c}}} = \frac{{{{\rm{w}}_{\rm{l}}} - {{\rm{w}}_{\rm{n}}}}}{I_P}\)

IP = Plasticity Index = wl - wP

\({{\rm{I}}_{\rm{c}}} = \frac{{{{\rm{w}}_{\rm{l}}} - {{\rm{w}}_{\rm{n}}}}}{{{{\rm{w}}_{\rm{l}}} - {{\rm{w}}_{\rm{p}}}}}\)

\({{\rm{w}}_{\rm{l}}}\) = liquid limit

\({{\rm{w}}_{\rm{P}}}\) = plastic limit

\({{\rm{w}}_{\rm{n}}}\) = water content of the soil at the natural state

Range: It can be greater than or less than 1.

For

I= 1, the soil is at its plastic limit

I= 0, the soil is at its liquid limit

I> 1, the soil is in semi-solid state and is stiff

I< 0, it indicates that the natural water content is greater than the liquid limit and hence behaves like a liquid.

Index Properties Question 8:

A soil sample has a with liquid limit = 45 percent, plastic limit = 25 percent, and shrinkage limit = 15 percent, for natural water content of 30 percent, the consistency index will be:

  1. 40 percent
  2. 75 percent
  3. 85 percent
  4. 60 percent

Answer (Detailed Solution Below)

Option 2 : 75 percent

Index Properties Question 8 Detailed Solution

Concept:

Consistency index

 \({I_C} = \frac{{{w_L} - w}}{{{w_L} - {w_P}}}\)

So, consistency index is – ve means, w > wL i.e. soil is in the liquid state and behaves like a liquid. 

\({I_C} = 0;w = {w_L}\; \Rightarrow Soil\;is\;in\;liquid\;limit.\)

\({I_C} = 1;w = {w_P}\; \Rightarrow Soil\;is\;in\;plastic\;limit.\)

\({I_C} > 1\;;\left( {{w_L} - w} \right) > \left( {{w_L} - {w_P}} \right) \Rightarrow w < {w_P}\;i.e;soil\;is\;stiff.\)

Consistency Index

Behavior of Soil

Greater than 1

Soil behaves as solid

0-1

Soil behaves as a plastic material

Less than 0

Soil behaves as liquid and is unstable

Calculation:

Given data,

Liquid limit, wL = 45%, 

Plastic limit, wp = 25%,

Shrinkage limit, ws = 15% , and

Natural moisture content, w = 30%

Consistency index

\({I_C} = \frac{{{w_L} - w}}{{{w_L} - {w_P}}}\)

\(\therefore {{\rm{I}}_{\rm{C}}} = \frac{{45 - 30}}{{45 - 25}} = \frac{{15}}{{20}}\) = 0.75

Hence, the consistency index is 75%.

Index Properties Question 9:

Activity of soil is ratio of the plasticity index to:

  1. percentage by weight of clay fraction
  2. percentage of sensitivity
  3. liquidity index
  4. percentage of compressive strength

Answer (Detailed Solution Below)

Option 1 : percentage by weight of clay fraction

Index Properties Question 9 Detailed Solution

Concept:

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

So, C = Percentage of clay fraction finer than 2 μ

Index Properties Question 10:

Which of the following bonding is responsible to combine the silica-gibbsite sheet in kaolinite clay mineral?

  1. Covalent bond
  2. Hydrogen bond
  3. Ionic bond
  4. Polar covalent bond

Answer (Detailed Solution Below)

Option 2 : Hydrogen bond

Index Properties Question 10 Detailed Solution

Kaolinite:

SSC JE Civil 71 10Q FT 4 Part 2 images Q3

  • The basic Kaolinite mineral is a two-layer unit that is formed by stacking a gibbsite sheet on a silica sheet
  • These basic units are then stacked one on top of the other to form a lattice of the mineral
  • These units are held together by hydrogen bonds
  • The strong bonding does not permit water to enter the lattice
  • Thus, kaolinite minerals are stable and do not expand under saturation
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