ANS:D - 84%

Step1) bulk density(yb)= W/V=108/60 = 1.8.

Step2) dry density (yd)=Wd/V=86.4/60 = 1.44.

Step3) water content determination(w).
Yd=Yb/1+w.

So 1.44= 1.8/1+w,
So 1+w = 1.8/1.44,
So w = (1.8/1.44)-1,
So, w= 1.25-1= o.25,
w = 0.25.

Step4) void ratio determination;
Yd=G.Yw/1+e,
1.44= 2.52/1+e,
e= (2.52/1.44)-1,
e= 0.75.

Step5) degree of saturation determination(SR)
SR= WG/e.
Sr= 0.25 * 2.52/0.75,
Sr= 0.84,
Sr= 84%.

So, the answer is 84%. A partially saturated sample of soil has a volume of 60 cc and a mass of 92g. The sample is dried in an oven and its dried mass is 73.8g. If the specific gravity of solids be 2.62.

Sind the degree of saturation, water content, void ratio, porosity, bulk unit weight and dry unit weight. Mass of water=108-86.4 g =21.6 g.
Volume of water=21.6/1 cc=21.6cc.
Now, Mass of dried soil =86.4g and absolute density =2.52.
Therefore Volume of soil solids = 86.4/2.52=34.29cc.
Also, Total volume is given = 60cc.
Hence volume of air =60-(34.29+21.6) cc = 4.11 cc.
Volume of voids = 4.11 + 21.6 = 25.71 cc.
Thus Degree of Saturation =Vol. of water/Vo/. of voids *100=21.6/25.71*100=84%.