Q1: A partially saturated sample of soil has a unit weight of 2.0 g/cm3 and specific gravity of soil particles is 2.6. If the moisture content in the soil is 20%, the degree of saturation is

ANS:C - 92%

Ws = Gs * 1 = 2.6
Wc = Ww/Ws.
0.2 = Ww/2.6.
Ww = 0.52.
Vw = 0.52.
Density = Wt/Vt.
2 = 2.6 + 0.52/Vt.
Vt = 1.56.

e = Vv/vs = 1.56-1/1 = 0.56,
S = Vw/Vv = 0.52/1.56-1 = 0.92. W=.2, bulk density = 2g/cc.
Dry density = 2/1 +.2 = 1.667g/cc.
e = [(2.6 * 1)/1.67-1] = .56.
S = (2.6 * 0.1)/.56 = .9265.
OR
S = 92.6%. The question is to find the degree saturation;

I am getting the answer as S=87%,
By using for:eS=wG,
e=.62. Unit weight of soil = 2.0g/cm3.
Dry unit weight = unit weight/(1+w).
Here w = 0.2.
Hence dry unit weight = 2/(1+0.2).
= 1.66.
Void ratio = ((specific gravity * unit weight of water )/dry unit weight)-1.
= 0.566.
Degree of saturation = w * specific gravity /void ratio.
= 0.9182.
= 91.82%.