In non-cohesive soil in passive state of plastic equilibrium

Related MCQs

• 260 g of wet soil was taken in a pycnometer jar of weight 400 g in order to find the moisture content in the soil, with specific gravity of soil particles 2.75. The weight of soil and remaining water filled in pycnometer without air bubbles was 1415 g and the weight of pycnometer filled with water alone was 1275 g. The moisture content in the soil is
• In active state of plastic equilibrium in a non cohesive soil with horizontal ground surface
• For determine the specific gravity of soil solids, using a pycnometer of 500 cc., the following data is available : Weight of dry empty pycnometer = 125 g Weight of dried soil and pycnometer = 500 g Weight of dried soil and distilled = 850 g water filled in pycnometer up to top The specific gravity of soil solids, is
• For determining the moisture content of a soil sample, the following data is available Weight of container = 260 g, Weight of soil sample and = 320 g container, Weight of soil sample (dried) and = 310 g container. The moisture content of the soil sample, is
• A soil mass is said to be in plastic equilibrium if
• The water content of soil, which represents the boundary between plastic state and liquid state, is known as
• If the natural water content of soil mass lies between its liquid limit and plastic limit, the soil mass is said to be in
• A soil sample has passing 0.075 mm sieve = 60% liquid limit = 65% and plastic limit = 40%. The group index of the soil, is
• 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
• A saturated soil sample has water content of 40% and specific gravity of soil particles 2.7. The void ratio of the soil, is