Answer:
degree of saturation at optimummoisture content is 78.1 %.
Step-by-step explanation:
The maximum dry unit weight of the soilYa is 16.8kN/mThe optimum moisture content of the soilwis 17 %.The specific gravity of soil G is 2.73.Calculation:Determine the degree of saturation (S)using the relation.Yd = Gs Yw/ 1+Gsw/sHere, Yw is the unit weight of water.Substitute 16.8 kN/m' for a, 2.73 for G,9.81 kN/m' for u, and 17 % for w.16.8= 2.73x9.811/2.73x1 .4641 26.78S-1.594S =-0.4641-0.594S= -0.4641S= 0.4641S = 0.781x 100S = 78.1 %16.80.594Therefore, the degree of saturation atoptimum moisture content is 78.1 %.