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

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%

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.

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,

Vv = Vs

∴ e = Vv/Vs

e = Vv/Vv = 1

∵ n = e/(1 + e)

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

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

Calculation:

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

For air content:

S + ac = 1

0.85 + ac = 1

ac = 0.15

Percentage air voids:

na = ac × n

na = 0.15 × 0.4 = 0.06 or 6%

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

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%.

Concept:

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

Let e = void ratio, n = Porosity, V_S = Volume od solids, V_V = 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 V_V from 2 into 3 and value of n from 1 to 3

\(\frac{e}{1+e}= \frac{e× V_{s}}{V}\)

⇒ V = (1 + e) × V_S

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

Remember this relation as this is used to solve such problems

Calculation:

Given: 

Volume of fill = V_f = 15000 m³, Void ratio of fill = e_f = 0.5, Void ratio of borrow pit = e_b = 1.5, Volume of borrow pit = V_b

From above relation, We can write volume of fill as

V_f ∝ (1 + e_f)

⇒ 15000 ∝ (1 + 0.5) ...... (1)

And similarly for volume of borrow pit

Vb ∝ (1 + eb)

⇒ V_b ∝ (1 + 1.5) ....... (2)

Dividing 1 by 2, we get

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

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

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

⇒ V_b = 25000 m³

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

Explanation:

Given,

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

Case II: Dry Density = 19.4 kN/m³ ⇒ 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.

Concept:

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

where, γ_d = dry unit weight of soil

γ_w = 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,

γ_d = 18 kN/m³

γ_w = 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

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.

Concept:

Water content formula is given by:

w = W_w/W_s

where, 

w = Water content

W_w = Weight of water

W_s = Weight of solids

Calculation:

Given data :

Weight of unsaturated sample = 220 g

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

Now, finding the weight of water, W_w

So, W_w = 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

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:

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,

𝝲_d = Dry unit weight of soil, G = Specific gravity, 𝝲_w = Unit weight of water, e = void ratio, and S_r = 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 × S_r = w × G

0.444 × 1 = w × 2.65

Concepts:

The Specific Gravity of soil solids is given as:

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

Where,

W_s = 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, W_s = 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:

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

∴ G_s = 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.