Q1: A compacted soil sample using 10% moisture content has a weight of 200 g and mass unit weight of 2.0 g/cm3. If the specific gravity of soil particles and water are 2.7 and 1.0, the degree of saturation of the soil is

ANS:B - 55.6%

g/cc= 9.81KN/m3.
2gm/cc=2*9.81=19.62KN/m3.
Dry density Yd=19.62/(1+w)=19.62/(1+0.1)=17.836KN/m3,
Yd = G.Yw/ (1+ e) , e=0.4855.

Yb = Yw*(G+e*Sr)/(1+e). Sr=0.5561=55.61%. This is very simple.

Yd = 2/1 + 0. 1 = 1.81.
Void raito e= (G. Yw/Yd) - 1.
E= (2. 7 X 1/1. 81) -1 = 0.49.

Now we know.
S = WG/e = 0. 1 x 2. 7/0. 49 = 0.55.
55% answer. Pt = (Gs(1+w) /1+e) pw g/cm^3.
2 = (2.7(1+0.1)/1+e)*1.
e = 0.485.
Se = Gs wc.
S 0.458 = 2.7 0.1
S = 0.589. Simply apply bulk mass unit wt formula

P= bulk mass unit wt
G= specific gravity of soil solids
e=void ratio
S=saturation
Pw= mass unit wt if water =1g/m3

Now,

P = {(G+Se)}/{1+e}
But,
Se = wG

Put values as a substitute for e, and solving

P = {(G+wG)}/{(1)+((wG)/S)}
Then, S = 0.55670 = 55.67%. Dry unit weight =bulk unit weight/1+w.
Dry unit weight =1.81.

Also,
Dry unit weight = G * unit weight of water/1+e.
1.81=2.7*1/1+e,
e=0.49,
Se=wG,
S * 0.49 = 0.10 * 2.7.
S = 55.10%. BULK UNIT WEIGHT = 9.81* Density = 9.81*2 = 19.62 kN/m^3.

DRY UNIT WEIGHT = (19.62/1+0.1) = 17.836 kN/m^3.

So, e = (G*Yw/Yd)-1 = (2.7*9.81/17.836) = 0.485.

Therefore, S = w*G/e = (0.1*2.7/0.485) = 55.67% Density (Sat)= Mass(sat)/Vol(sat)
2.0=200/ Vs
Vs=100.

Moisture Content = Wt of Water/Wt dry Soil.
0.1 = (200 - Wdry)/Wdry.
Wdry = 181.81 gm.

Solid Unit Wt = Wdry/Volume (dry).
2.7 = 181.81/V(dry),
V(dry)=67.33.

Void Ratio = Vv / Vs = {V(sat)-V(dry)}/V(dry)
e = (100-67.33) / 67.33
e = 0.485.

se = wGs.
s= {0.1(2.7)/0.485}*100.
s= 55.67. A natural soil deposit has a bulk density of 1.90 g/cm^3 and water content of 6 per cent assume G=2.67 assuming the voids ratio to remain constant what will be the degree of saturation at a water content of 16%.