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If the plasticity index of a soil is 45%, then the soil will be:
Plasticity index (PI) is the range of water content over which the soil remains in the plastic state. Mathematically defined as,
Plasticity Index = Liquid Limit (w_{L}) – Plastic Limit (w_{p})
Consider the following statements. The incorrect statement is:
The term ‘Loess’ indicates those soils which are
1. Uniformly graded
2. Poorly graded
3. Made up of more than 50% sand size particles
4. Made up of more than 50% of silt particles
5. These are transported by winds
Which of the above statements are correct:
Match List – I (Type of soil) with ListII (Mode of transportation and deposition) and select the correct answer using the codes given below the lists:
Lacustrine soil: Soil which is deposited from suspension in fresh still water of the lakes is known as Lacustrine soil. Lacustrine deposits are sedimentary rock formations which formed in the bottom of ancient lakes.
Alluvial Soil: Soil which is been deposited from suspension in running water is known as Alluvial soil. It is transported soil formed by physical wheathering.
Aeoline Soil: Soil which is formed due to transportation by wind is known as Aoeline soil. It is also a transported soil.
Marine soil: Soil which is deposited from the suspension in sea water is known as marine soil.
In wet soil mass, air occupies 1/6 of its volume and water occupies 1/3 of its volume. The void ratio of the soil is
Voids ratio: It is defined as the ratio of the total volume of voids to the volume of solids in a given soil mass.
Void ratio, e = V_{v }/ V_{s}
e > 0, voids ratio has no upper limit
V_{v} = air void (Va) + water filled voids (Vw)
Calculation:
Given,
V_{a} = 1/6 of V
V_{w} = 1/3 of V
V_{v} = Va + Vw = V/2
The given figure indicate the weights of different pycnometers:
The specific gravity of the solids is given by
Specific gravity of solids is the ratio of weight of given volume of solids to the weight of equivalent volume of water at 4° C.
Specific gravity is usually reported at 27°C and determined using pycnometer. It is given by
W_{1 }= Mass of empty volume of pycnometer
W_{2} = Mass of pycnometer + Mass of moist sample
W_{3} = Mass of pycnometer + soil + water
W_{4} = Mass of pycnometer full of water
G_{s} = Specific gravity of soil solids
The following data are given for soil:
• Porosity: n = 0.4
• The specific gravity of the soil solids: G_{s} = 2.68
• Moisture content: w = 12%
Determine the mass of water in kg to be added to 10 m^{3} of soil for full saturation.
Total volume of soil, V = 10 m^{3}
∴ V_{V} = 0.4 × 10 = 4m^{3}
Volume of solids, V_{s} = V – V_{V} = 10 – 4 = 6 m^{3}
Mass of solids, W_{s} = V_{s} G_{s} Y_{w} = 6 × 2.68 × 1000 = 16080 kg
Moisture content = 0.12
So, Mass of water = 0.12 × 16080 = 1929.6 kg
Volume of water = 1929.6/1000 = 1.929 m^{3}
So, phase diagram can be shown as
For full saturation, the amount of air has to be replaced by water.
So, water to be added = 2.071 × 1000 = 2071 kg
The water content of a soil sample is found to be 35%. If the percentage of air voids present in the soil sample is 5%. Then compute dry unit weight and air content of the soil sample when the void ratio of the soil sample is 0.65.
Take specific gravity of soil solids and unit weight of water to be 2.70 and 10 kN/m^{3}.
Dry unit weight is given by
η_{a} = 0.05
G = 2.70
ω = 0.35
η_{d} = 13.19 kN/m^{3}
η_{a} = a_{c}η
η → porosity
a_{c} → aircontent
a_{c} = 0.1269
a_{c} = 12.70%
A fully saturated clay has a water content of 40% and unit weight 19 kN/m^{3}. After oven drying the dry density of the soil becomes 18 kN/m^{3}. The shrinkage limit for the soil sample is ______%. (Take γw=γw=10 kN/m^{3})
And, e s_{r} = wG
⇒ e × 1 = 0.40 G
⇒ G = 2.97
Shrinkage limit, w_{s }=
Where, G_{m} = Mass specific gravity at dry state
The soil has to be excavated from a borrowpit which has a density of 1.12 gm/cc with a water content of 12% to fill a land. The soil is compacted to a density of 1.32 gm/cc with water content 20%. Find the volume of soil to be excavated in m^{3} for 1000 m^{3} volume of the fill?
Dry density of soil at excavation site,
∴ Volume of excavated soil =
= 1000 × 1.1/1
= 1100 m^{3}
27 docs296 tests

27 docs296 tests
