Consider a unit cube of soil having spherical particles of diameter d.
Volume of solids V_s= (?/6)=0.5236
The total volume V=(1-cos?)*(?(1+2cos?))
Minimum void ratio is possible in densest state with ?=60�
? V=(1-cos60�)*(?(1+2cos60�))
V=0.7071
Volume of voids V_v=V-V_s=0.7071-0.5236
V_v=0.1835
Voids ratio e= V_v/ V_s=0.1835/0.5236
? e=0.35.

Concepts:

Darcy’s Law states that velocity of flow through porous media is directly proportional to hydraulic gradient for any given saturated soil under steady laminar flow conditions.

 If rate of flow is Q through cross-sectional area (A) of soil mass, Darcy's Law can be expressed as

V ∝ i or V = K × i

Q = K × i × A

k  = permeability of soil
i = dh/dl

dh = difference in total heads
dl = length of small soil element

Darcy’s law assumes flow must be laminar i.e. flow of water through soil voids should have very less velocity. This only possible in case of soils having small but large no of voids. Therefore, it is applicable for soils such as fine saturated sands, clays, silts etc. It is not applicable for soils having large size voids like gravel, boulder etc because in se soils water is drained immediately due to action of gravity and hence, not possible to maintain laminar flow in se soils. 

It is always convenient to show constituents occupying separate spaces as blocks. In phase diagram, soil occupies bottom position. Water and air occupy middle and top positions.

Cohesion-less soil is in its loosest form will have its void ratio e=e_min.
Since I_D=\(\frac{(e_{max} ? e)}{(e_{max}-e_{min})}\)
I_D=\(\frac{(e_{max} ? e_{min})}{(e_{max}-e_{min})}\)
I_D=1.

Given,
?=75�
The total volume V=(1-cos?)*(?(1+2cos?))
? V=(1-cos75�)*(?(1+2cos75�))
V=0.9131
Volume of solids V_s=(?/6)=0.5236
Volume of voids V_v=V-V_s=0.9131-0.5236
V_v=0.3895.

The bulk unit weight is given by,
\(?=\frac{W}{V} = \frac{(?_s*V_s+ ?_w *V_w)}{V}\)
When V_s = 1, V_w = e_w and V = 1+e
? = \(\frac{(?_s*1 + ?_w * e_w)}{(1+e)}\)
since e_w = e*S and ?_s = G?_w
? = \(\frac{(G?_w*1 + ?_w * e*S)}{(1+e)}\)
?=\(\frac{(G+eS) ?_w}{(1+e)}\).

Given,
Porosity of 30%=0.3
voids ratio in loosest state e_min= 0.35
voids ratio in densest state e_max=0.92
e=\(\frac{n}{(1-n)}\)
e=\(\frac{0.3}{(1-0.3)}\)
e=0.429
I_D=\(\frac{(e_{max} ? e)}{(e_{max}-e_{min})}\)
I_D=\(\frac{(0.92 ?0.429)}{(0.92-0.35)}\)
I_D= 0.861.

Given,
Voids ratio e = 0.6
Specific gravity G = 2.65
Saturated unit weight ?_sat = (G+e) ?_w/ (1+e)
?_sat = (2.65+0.6)*9.81/(1+0.6)
?_sat= 19.927 N/m³.

Concept:

Dry unit weight of soil,  ${\gamma }_d=\frac{G{\gamma }_w}{1+e}$

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)

${\gamma }_d=\frac{G{\gamma }_w}{1+e}$

$18=\frac{2.65\times 9.81}{1+e}$

e = 0.444

Se = wG

1 × 0.444 = w × 2.65

w = 0.167

The density index is used to express relative compactness or degree of compaction of a natural cohesion-less soil deposit. It is also known as relative density or degree of density.