# A 588 cm3 volume of moist sand weighs 1010 gm. Its dry weight is 918 gm and specific gravity of solids, G is 2.67. Assuming density of water as 1 gm/c

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A 588 cm3 volume of moist sand weighs 1010 gm. Its dry weight is 918 gm and specific gravity of solids, G is 2.67. Assuming density of water as 1 gm/cm3, the void ratio is ___________.

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Concept:

According to the three-phase system of soil mass,

V = VV + V

Where,

VV = Total volume of voids = Va + Vw

Va = Volume of air, Vw = Volume of water

V = Total volume of soil, VS = Volume of soil solids

Unit weight of soil

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

${{\rm{γ }}_{\rm{s}}} = \frac{{\rm{W}}}{{\rm{V}}}$

The specific gravity of soil

It is defined as the ratio of unit weight of a given volume of solids to the unit weight of an equivalent volume of water.

${\rm{G}} = \frac{{{{\rm{γ }}_{\rm{s}}}}}{{{{\rm{γ }}_{\rm{w}}}}}$

Voids ratio:

It is defined as the ratio of total volume of voids to the volume of solids in given soil mass.

${\rm{e}} = \frac{{{{\rm{V}}_{\rm{v}}}}}{{{{\rm{V}}_{\rm{s}}}{\rm{\;}}}}$

e > 0, voids ratio has no upper limit

Calculation:

Given,

W = 1010 gm, Ws = 918 gm

V = 588 cm3, G= 2.67

γw = 1000 kg/m3 = 1 gm/cm3

$\begin{array}{l} {G_s} = \frac{{{W_s}}}{{{V_s}{Y_w}}}\\ 2.67 = \frac{{918}}{{{V_s} \times 1}} \end{array}$

Vs = 343.82 cm3

$\begin{array}{l} {V_v} = V - {V_s} = 588 - 343.82 = 244.18{m^3}\\ e = \frac{{{V_v}}}{{{V_s}}} = \frac{{244.18}}{{343.82}} = 0.71 \end{array}$