Definitions and Relationships MCQ Quiz - Objective Question with Answer for Definitions and Relationships - Download Free PDF

Last updated on Apr 4, 2025

Latest Definitions and Relationships MCQ Objective Questions

Definitions and Relationships Question 1:

A soil has a bulk density of 2 gm cm3 and water content of 20%. What will be the water content if the soil partially dries to a density of 1.8 gm/cm3 and the void ratio remains unchanged? 

  1. 10% 
  2. 8% 
  3. 18% 
  4. 12% 

Answer (Detailed Solution Below)

Option 2 : 8% 

Definitions and Relationships Question 1 Detailed Solution

Concept:

The dry density of soil is given by:

\( \gamma_d = \frac{\gamma}{1 + w} \)

If the void ratio remains unchanged, the dry density of soil remains constant.

Given Data:

  • Initial bulk density (\( \gamma \)) = 2.0 g/cm³
  • Initial water content (\( w_1 \)) = 20% = 0.20
  • Final bulk density (\( \gamma' \)) = 1.8 g/cm³
  • Void ratio remains unchanged → dry density remains constant

Step 1: Calculate Initial Dry Density

\( \gamma_{d1} = \frac{2.0}{1 + 0.20} \)

\( \gamma_{d1} = \frac{2.0}{1.2} = 1.667 \) g/cm³

Since the void ratio remains unchanged, the dry density remains the same:

\( \gamma_{d2} = \gamma_{d1} = 1.667 \) g/cm³

Step 2: Calculate Final Water Content

\( \gamma_{d2} = \frac{1.8}{1 + w} \)

\(1+w = {1.8\over1.667}\)

w = 0.079 or 8%

 

 

 

Definitions and Relationships Question 2:

Which Atterberg limit re presents the water content at which clay changes from a plastic to a liquid state ?

  1. Shrinkage limit
  2. Plastic limit
  3. Liquid limit
  4. Permeability limit

Answer (Detailed Solution Below)

Option 3 : Liquid limit

Definitions and Relationships Question 2 Detailed Solution

Explanation:

Atterberg Limits

Atterberg limits are a basic measure of the critical water contents of a fine-grained soil. Depending on the water content, soil can exist in one of several states: solid, semi-solid, plastic, and liquid. These limits are used to distinguish the boundaries between these states. The three most common Atterberg limits are:

  • Shrinkage Limit (SL): The water content at which further loss of moisture will not result in any more volume reduction of the soil.

  • Plastic Limit (PL): The water content at which soil changes from a plastic state to a semi-solid state.

  • Liquid Limit (LL): The water content at which soil changes from a plastic state to a liquid state.

Analyzing the Given Options

  1. "Shrinkage limit" 

    • The shrinkage limit refers to the water content at which soil transitions from a semi-solid state to a solid state, not to a liquid state.

    • Therefore, this option is not the correct answer.

  2. "Plastic limit" 

    • The plastic limit is the water content at which soil changes from a plastic state to a semi-solid state, not to a liquid state.

    • Therefore, this option is not the correct answer.

  3. "Liquid limit" 

    • The liquid limit is the water content at which soil transitions from a plastic state to a liquid state.

    • This is the correct definition for the water content at which clay changes from a plastic to a liquid state.

    • Therefore, this is the correct answer.

  4. "Permeability limit" 

    • The permeability limit is not an Atterberg limit and does not refer to the transition of soil states based on water content.

    • Therefore, this option is not the correct answer.

Definitions and Relationships Question 3:

If the volume of voids is equal to the volume of the solids in a soil mass, then the values of porosity and voids ratio respectively are 

  1. 0.5 and 1.0
  2. 1.0 and 0.0
  3. 0.0 and 1.0
  4. 1.0 and 0.5

Answer (Detailed Solution Below)

Option 1 : 0.5 and 1.0

Definitions and Relationships Question 3 Detailed Solution

Concept:

Void ratio (e):

Void ratio (e) is the ratio of the volume of voids (Vv) to the volume of soil solids (Vs), and is expressed as a decimal.

e = Vv/Vs

Porosity (n):

Porosity (n) is the ratio of the volume of voids to the total volume of soil (V), and is expressed as a percentage.

n = (Vv/V) × 100

Relation between void ratio and porosity

n = e/(1+e)

Calculation:

Given,

V= Vs

∴ e = Vv/Vs

e = Vv/Vv = 1

∵ n = e/(1 + e)

n = 1/(1 + 1) = 1/2 = 0.5

Definitions and Relationships Question 4:

If the porosity of a soil sample is 40 percent and the degree of saturation is 85 percent, then its percentage of air voids is:

  1. 10%
  2. 8%
  3. 6%
  4. 5%

Answer (Detailed Solution Below)

Option 3 : 6%

Definitions and Relationships Question 4 Detailed Solution

CONCEPT: 

Air content(ac):

 Air content is defined as the ratio of the volume of air to the volume of voids. It is denoted by ac.

\({{\bf{a}}_{\bf{c}}} = \frac{{{{\bf{V}}_{\bf{a}}}}}{{{{\bf{V}}_{\bf{v}}}}} = \frac{{{\bf{Volume\ of\ air}}}}{{{\bf{Volume\ of\ voids}}}}\)

S + ac = 1

Porosity(n):

Porosity is defined as the ratio of the volume of voids to the total volume of soil. It is denoted by n. It varies between 0 and 1.

\({\bf{n}} = \frac{{{{\bf{V}}_{\bf{v}}}}}{{\bf{V}}} = \frac{{{\bf{Volume\ of\ voids}}}}{{{\bf{Total\ volume}}}}\)

Percentage air voids:

Percentage air voids are defined as the ratio of the volume of air to the total volume of soil. It is denoted by n­­a.

\({{\bf{n}}_{\bf{a}}} = \frac{{{{\bf{V}}_{\bf{a}}}}}{{\bf{V}}} × 100 = \frac{{{\bf{Volume\ of\ air}}}}{{{\bf{Total\ volume}}}} × 100\)

or, \({{\bf{n}}_{\bf{a}}} = \frac{{{{\bf{V}}_{\bf{a}}}}}{{\bf{V_v}}} × \frac{V_v}{V}\) = a× n 

Calculation:

n = 40% = 0.4, S = 85% = 0.85

For air content:

S + a= 1

0.85 + ac = 1

ac = 0.15

Percentage air voids:

na = ac × n

na = 0.15 × 0.4 = 0.06 or 6%

Definitions and Relationships Question 5:

A dry soil has a mass specific gravity of 1.35. If the specific gravity of solids is 2.7, then the void ratio will be:

  1. 0.5
  2. 1.0
  3. 0.9
  4. 0.75

Answer (Detailed Solution Below)

Option 2 : 1.0

Definitions and Relationships Question 5 Detailed Solution

Explanation:

Relation between Mass specific gravity (Gm), True specific gravity (Gs) and void ratio (e) will be:

\({G_m} = \frac{{{G_s}}}{{1 + e}}\)

Gm = 1.35

Gs = 2.7

\(1.35 = \frac{{2.7}}{{1 + e}}\)

e = 1.0

Top Definitions and Relationships MCQ Objective Questions

Which of the following represents the percentage limit of porosity of the compacted sand?

  1. 5% to 15%
  2. 15% to 30%
  3. 30% to 40%
  4. 40% to 50%

Answer (Detailed Solution Below)

Option 3 : 30% to 40%

Definitions and Relationships Question 6 Detailed Solution

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Explanation:

Porosity (η) is defined as the ratio of volume of voids to the total volume of the soil sample in a given soil mass.

\(\eta = \frac{{{V_v}}}{V}\)

The porosity of soils can vary widely.

The porosity of loose soils can be about η = 50 to 60%.

The porosity of compact soils is about η = 30 to 40%.

A fill having a volume of 15000 m3 is to be constructed at a void ratio of 0.5. The borrow pit solid has a void ratio of 1.5. The volume of soil required (in cubic meters) to be excavated from the borrow pit will be:

  1. 18750
  2. 20000
  3. 25000
  4. 75000

Answer (Detailed Solution Below)

Option 3 : 25000

Definitions and Relationships Question 7 Detailed Solution

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Concept:

Relation between void ratio and total volume of the pit is given by:

Let e = void ratio, n = Porosity, VS = Volume od solids, VV = Volume of voids, V = Total volume of soil

\(n=\frac{e}{1+e}\) ...... (1)

\(e = \frac{V_{V}}{V_{S}}\) ...... (2)

\(n = \frac{V_{V}}{V}\) ...... (3)

Putting the value of VV from 2 into 3 and value of n from 1 to 3

\(\frac{e}{1+e}= \frac{e× V_{s}}{V}\)

⇒ V = (1 + e) × VS

⇒ V ∝ (1 + e)   (As, VS is always constant)

Remember this relation as this is used to solve such problems

Calculation:

Given: 

Volume of fill = Vf = 15000 m3, Void ratio of fill = ef = 0.5, Void ratio of borrow pit = eb = 1.5, Volume of borrow pit = Vb

From above relation, We can write volume of fill as

Vf ∝ (1 + ef)

⇒ 15000 ∝ (1 + 0.5) ...... (1)

And similarly for volume of borrow pit

Vb ∝ (1 + eb)

⇒ Vb ∝ (1 + 1.5) ....... (2)

Dividing 1 by 2, we get

\(\frac{V_{f}}{V_{b}}=\frac{1+e_{f}}{1+e_{b}}\)

⇒ \(\frac{15000}{V_{b}}=\frac{1+0.5}{1+1.5}\)

⇒ \(\frac{15000}{V_{b}}=\frac{1+0.5}{1+1.5}\)

⇒ Vb = 25000 m3

∴ The volume of soil required to be excavated from the borrow pit is 25000 m3

A soil has bulk density of 20.1 kN/m3 and water content 15%. What is the water content if the soil partially dries to a density of 19.4 kN/m3 and the void ratio remains unchanged ?

  1. 10.86%
  2. 10.76%
  3. 10.68%
  4. 10.66%

Answer (Detailed Solution Below)

Option 1 : 10.86%

Definitions and Relationships Question 8 Detailed Solution

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Explanation:

Given,

Case - I: Bulk density = 20.1 kN/m3, water content = 15%

Case II: Dry Density = 19.4 kN/m⇒ water content = ??

No change in void ratio.

As, dry unit weight (γd) = γ/(1 + w)  

\(\begin{array}{l} {γ_d} = \frac{γ}{{1 + w}} = \frac{{20.1}}{{1 + 0.15}} = 17.478\; kN/{m^3}\\ {γ_d} = \frac{{{γ_w}.G}}{{1 + e}} \end{array}\)

 If void ratio 'e' remains unchanged during drying,  \(γ_d\)  also remains unchanged. 

\(\begin{array}{l} {\gamma_d} = \frac{\gamma}{{1 + w}}\\ 17.478 = \frac{{19.4}}{{1 + w}} \end{array}\)

w = 10.99 %

Hence, The most appropriate answer is option 1.

Water that forms a hydration shell around soil grains is known as:

  1. Structural water
  2. Solvate water
  3. Pore Water
  4. Infiltered water

Answer (Detailed Solution Below)

Option 2 : Solvate water

Definitions and Relationships Question 9 Detailed Solution

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1. Structural water

The water chemically combined in the crystal structure of the mineral of the soil is called structural water.it cannot be remove without breaking the structure of the mineral.

 

2. Solvate water

The solvate water forms a hydration shell around soil grain. It is influenced by ionic and polar forces.

 

3. Pore Water

free water present in a soil. Normally under hydrostatic pressure. The shear strength of adjacent soil depends on this pore pressure, which reduces frictional resistance and soil stability

4. Infiltered water

Infiltered water is the water which infiltrate into the ground surface through pores.

A saturated sand sample has a dry unit weight of 18 kN/m3 and a specific gravity of 2.65. If γw = 9.81 kN/m3, then, what is the water content of the soil?

  1. 0.166
  2. 0.3
  3. 0.189
  4. 0.277

Answer (Detailed Solution Below)

Option 1 : 0.166

Definitions and Relationships Question 10 Detailed Solution

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Concept:

Dry unit weight of soil,  \({γ _d} = \frac{{G{γ _w}}}{{1 + e}}\)

where, γd = dry unit weight of soil

γw = unit weight of water

e = void ratio of soil 

Relation between Degree of saturation (S), void ratio (e), water content (w) and specific gravity (G), Se = wG

Calculation:

Given,

γd = 18 kN/m3

γw = 9.81 kN/m3

G = 2.65 

S = 1 (saturation condition)

\({γ _d} = \frac{{G{γ _w}}}{{1 + e}}\)

\(18 = \frac{{2.65 × 9.81}}{{1 + e}}\)

e = 0.444

Se = wG

1 × 0.444 = w × 2.65

w = 0.167

If the unit weight (γ) of sand and clay samples are increased by 10% which of the following is true ?

  1. More volume reduction occurs in sand
  2. More volume reduction occurs in clay
  3. Equal volume reduction in sand and clay
  4. Volume reduction is independent of γ

Answer (Detailed Solution Below)

Option 2 : More volume reduction occurs in clay

Definitions and Relationships Question 11 Detailed Solution

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Explanation

 1) Bulk Unit weight γb: It is defined as the ratio of the total weight of soil to the total volume of the soil mass.

\({\gamma _b} = \frac{W}{V} = \frac{{{W_s} + {W_w}}}{{{V_s} + {V_w} + {V_a}}}\)

As it is given that if unit weight is increased by 10% (i.e.) it is signifying the weight of water is increased or the Volume of water or air or Volume of voids got decreased.

(Weight of solids remains unchanged)

As the option relates to Volume.

Definitely volume reduction takes place.

Comparatively to Sand, more Volume reduction takes place in clay.

Let's consider Montmorillonite clay (Expansive soil) It expands as well as shrinks more compared to sand.

Soil scientist collects unsaturated 200 cm3 sample of soil having weight 220 g. if the dried weight of soil is 180 g. then, find the water content available in the soil.

  1. 0.222
  2. 0.176
  3.  0.133
  4. 0.166

Answer (Detailed Solution Below)

Option 1 : 0.222

Definitions and Relationships Question 12 Detailed Solution

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Concept:

Water content formula is given by:

w = Ww/Ws

where, 

w = Water content

Ww = Weight of water

Ws = Weight of solids

Calculation:

Given data :

Weight of unsaturated sample = 220 g

Weight of dried sample = Weight of solids, Ws = 180 gm

Now, finding the weight of water, Ww

So, Ww = Weight of unsaturated sample - Weight of dried sample

= 220 - 180 = 40 g

So, Water content, w = \(40 \over 180\) = 0.222

Note: There is no use of sample volume in finding out water content

As per IS 2720 (Part II) for more than 90% of the particles passive through a 4.75 mm sieve, the minimum quantity of soil taken for water content determination is ______.

  1. 500 g
  2. 300 g
  3. 50 g
  4. 200 g

Answer (Detailed Solution Below)

Option 4 : 200 g

Definitions and Relationships Question 13 Detailed Solution

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Explanation:

Soil Specimen:

(i) As per CI 4.1 of IS 2720 (part II), The soil specimen taken shall be representative of the soil mass. The size of the specimen selected depends on the quantity required for good representation, which is influenced by the gradation and the maximum size of particles, and on the accuracy of weighing,

The following quantities are recommended for general laboratory use:

Size of particles more than 90% passing Minimum quantity of soil specimen to be taken for test mass in 'g'
425 micron IS Sieve 25
2 mm IS Sieve 50
4.75 mm IS Sieve 200
10 mm IS Sieve 300
20 mm IS Sieve 500
40 mm IS Sieve 1000

A saturated soil sample has a dry unit weight of 18000 N/m3 and specific gravity 2.65. If unit weight of water is 9810 N/m3, determine the water content of the soil sample?

  1. 0.25
  2. 0.41
  3. 0.34
  4. 0.17

Answer (Detailed Solution Below)

Option 4 : 0.17

Definitions and Relationships Question 14 Detailed Solution

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Concept:

\({{\rm{\gamma }}_{\rm{d}}} = \frac{{{\rm{G}}{{\rm{\gamma }}_{\rm{w}}}}}{{1 + {\rm{e}}}}{\rm{\;and\;e}} \times {{\rm{S}}_{\rm{r}}} = {\rm{w}} \times {\rm{G}}\)

Where,

𝝲d = Dry unit weight of soil, G = Specific gravity, 𝝲w = Unit weight of water, e = void ratio, and Sr = Degree of saturation

Calculation:

\({{\rm{\gamma }}_{\rm{d}}} = \frac{{{\rm{G}}{{\rm{\gamma }}_{\rm{w}}}}}{{1 + {\rm{e}}}}\)

18000 = (2.65 × 9810) / (1 + e)

e = 0.444

e × Sr = w × G

0.444 × 1 = w × 2.65

∴ w = 0.167 ≈ 0.17

An oven dried soil mass of 200 gm is placed is pycnometer and completely filled with water. Combined mass of bottle, soil and water is 1605 gm. Calculate specific gravity of soil if pycnometer with water alone has weight of 1480 gm:-

  1. 2.63
  2. 2.65
  3. 2.67
  4. 2.69

Answer (Detailed Solution Below)

Option 3 : 2.67

Definitions and Relationships Question 15 Detailed Solution

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Concepts:

The Specific Gravity of soil solids is given as:

\(G = \frac{W_s}{W_4 -W_3 + W_s}\)

Where,

Ws = Weight of soil solids in dry state

W4 = Weight of pycnometer and water filled in it completely

W3 = Weight of pycnometer plus soil solids and water filled in pycnometer

Calculation:

Given:

Oven dry weight of soil, Ws = 200 gm

Combined mass of bottle, soil and water, W3 = 1605 gm

Weight of pycnometer with water alone, W4 = 1480 gm

Using above formula, the specific gravity is calculated as:

\(G = \frac{200}{1480 - 1605 + 200}\)

Gs = 2.67

Important Points:

  1. Specific gravity of soil solids is determined by using 50 ml density bottle, 500 ml flask or by the use of pycnometer.
  2. Dry soil sample is used to determine the specific gravity.
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