A chunk of moist soil has volume of 0.02135 meter saquare and mass of 41.54 kg spgravity of solids is 2.722 and moisture content is 23.2 determine void ratio and degree of saturation?

To determine the void ratio (e) and degree of saturation (S) for the given soil, we first need to find the dry density (ρ_d) and the specific gravity (G_s). The dry density can be calculated using the formula ( \rho_d = \frac{m}{V} ), where ( m ) is the mass of the dry soil and ( V ) is the volume. The mass of the dry soil can be calculated as ( m = \frac{mass}{1 + moisture\ content} = \frac{41.54}{1 + 0.232} \approx 33.69 ) kg, giving a dry density of ( \rho_d = \frac{33.69}{0.02135} \approx 1575.68 ) kg/m³.

Next, the void ratio ( e ) can be found using ( e = \frac{G_s \cdot \rho_w}{\rho_d} - 1 ), where ( \rho_w ) (density of water) is approximately 1000 kg/m³. Substituting ( G_s = 2.722 ) and ( \rho_d \approx 1575.68 ) kg/m³, we find ( e \approx 0.662 ).

To find the degree of saturation ( S ), we can use the formula ( S = \frac{w \cdot \rho_d}{e \cdot \rho_w} ), where ( w ) is the moisture content as a decimal (0.232). Plugging in the values, we get ( S \approx \frac{0.232 \cdot 1575.68}{0.662 \cdot 1000} \approx 0.223 ) or 22.3%.