A dry sample of sand is placed in a container having a volume of 0.3 ft3. The dry weight of the sample is 31 lb. Water is carefully added to the container so not to disturb the condition of the sand. When the container is filled, the combined weight of soil plus water is 38.2 lb. Calculate: a) the void ratio of the soil in the container, b) the specific gravity of the soil particles, c) the wet and the dry unit weights specifying their units.

Explanation:

Given:

Volume of dry soil = 0.3 ft^3

= 0.0085 m^3

Mass of dry soil = 31 lb

= 14.08 kg

Mass of dry soil + water = 38.2 lb

= 17.34 kg

Mass of water = 17.34 - 14.08

= 3.26 kg

A.

Density of water = 1000 kg/m^3

Volume of water = 0.0033 m^3

Porosity, e = Vv/Vs

Where,

Vv = volume of void-space (air and water)

Vs = volume of solids(dry soil)

= 0.0033/0.0085

= 0.39

B.

SG = density of dry soil/density of water

= (14.08/0.0085)/1000

= 1.656

answer:

sure does

explanation: