A soil has bulk unit weight of 20 kN/m^{2} and water content of 17%. Calculate the water content if the soil particle dries to a unit weight of 19 kN/m^{2} and the void ratio remains constant. (Give the answer rounded to the nearest integer value.)

This question was previously asked in

SSC JE CE Previous Year Paper 10 (Held on 30 Oct 2020 Evening)

Option 4 : 11%

ST 1: Building Material and Concrete Technology

18752

20 Questions
20 Marks
12 Mins

**Concept:**

Bulk Unit weight (γ_{b}):

It is defined as the ratio of the total weight of soil to the total volume of the soil mass.

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

Water Content (w):

Water content or moisture content of a soil mass is defined as the ratio of the weight of water to the weight of solids (dry weight) of the soil mass.

\({\rm{w = }}\frac{{{{\rm{W}}_{\rm{w}}}}}{{{{\rm{W}}_{\rm{s}}}}};{\rm{w}} \ge 0\)

It is denoted by the w and is commonly expressed as a percentage. The minimum value for water content is 0. There is no upper limit for water content.

Dry Unit Weight (γd):-

Dry unit weight is defined as the weight of soil solids per unit volume of soil. It is denoted by the letter symbol γd it has the unit of kN/m3.

\({{\rm{γ }}_{\rm{d}}}{\rm{ = }}\frac{{{{\rm{W}}_{\rm{s}}}}}{{\rm{V}}}{\rm{ = }}\frac{{{{\rm{W}}_{\rm{d}}}}}{{\rm{V}}}\)

It is used as a measure of the denseness of soil. A high value of dry unit weight indicates that more solids are packed in a unit volume of soil hence a more compact soil.

Specific gravity of solids (G):-

The specific gravity of solids is defined as the **ratio of the unit weight of solids to the unit weight of water**. It is denoted by the letter G and is a unitless quantity.

\({\rm{G = }}\frac{{{{\rm{γ }}_{\rm{s}}}}}{{{{\rm{γ }}_{\rm{w}}}}}\)

__ Explanation:__

\(\begin{array}{l} {γ_d} = \frac{γ}{{1 + w}} = \frac{{20}}{{1 + 0.17}} = 17.094\; kN/{m^3}\\ {γ_d} = \frac{{{γ_w}.G}}{{1 + e}} \end{array}\)

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

\(\begin{array}{l} {γ_d} = \frac{γ}{{1 + w}}\\ 17.094 = \frac{{19}}{{1 + w}} \end{array}\)

w = 0.11 = 11 %

__Important Points__

Air content:

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

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

Porosity:

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.

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

**Degree of Saturation:**

Degree of Saturation of a soil mass is defined as the ratio of the volume of water in the voids to the volume of voids. It is denoted by S.

\(S = \frac{{{V_w}}}{{{V_v}}} \times 100\;\;;0 \le S \le 100\)

- For a fully saturated soil mass Vv = Vw, hence S = 100%
- For fully dry soil mass Vw = 0, hence S = 0%