# A soil has bulk unit weight of 20 kN/m2 and water content of 17%. Calculate the water content if the soil particle dries to a unit weight of 19 kN/m2

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A soil has bulk unit weight of 20 kN/m2 and water content of 17%. Calculate the water content if the soil particle dries to a unit weight of 19 kN/m2 and the void ratio remains constant. (Give the answer rounded to the nearest integer value.)

1. 20%
2. 15%
3. 8%
4. 11%

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Correct Answer - Option 4 : 11%

Concept:

Bulk Unit weight (γb):

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 %

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%