1. If a soil has a void ratio of 0.6 and a specific gravity of 2.7, then its submerged unit weight will be ______
a) 9.678
b) 11.502
c) 9.256
d) 10.423
Discussion
Explanation: Given,
Voids ratio e=0.6
Specific gravity G=2.7
Submerged unit weight γ’ = (G -1) γw/ (1+e)
γ’ = (2.7-1)*9.81/(1+0.6)
γ’ = 10.423 kN/m3.
2. A soil has a dry unit weight of 16.25 kN/m3 and saturated unit weight of 20 kN/m3. If the degree of saturation of soil is 50%, then its bulk unit weight will be ______
a) 20.748 kN/m3
b) 17.628 kN/m3
c) 18.123 kN/m3
d) 19.246 kN/m3
Discussion
Explanation: Given,
Dry unit weight γd = 16.25 kN/m3
Saturated unit weight γsat = 20 kN/m3
Degree of saturation S = 50% = 0.5
Bulk unit weight γ = γd + S[γsat– γd] γ = 16.25 + 0.5[20-16.25] γ =18.123 kN/m3.
3. The loosest stable arrangement of equal sized spheres is obtained when the sphere centres form a ______ space lattice.
a) triangular
b) rhombohedral
c) rectangular
d) hexagonal
Discussion
Explanation: When the sphere centres form a rectangular space lattice, then they form a packing known as cubic packing, wherein each sphere is in contact with six surrounding neighbouring spheres.
4. The porosity of cubic packing is ______
a) 42.66%
b) 47.64%
c) 25.95%
d) 30.21%
Discussion
Explanation: Consider a unit cube of soil having spherical particles of diameter d.
Volume of each spherical particle = (π/6)d3
Total volume of container = 1*1*1=1
No. of solids in the container = (1/d)*(1/d)*(1/d)=(1/d3)
Volume of the solids Vs = (π/6)*d3*(1/d3)=(π/6)
Volume of the voids Vv = 1-(π/6)
Voids ratio e = (1-(π/6))/(π/6)
Porosity n = e/(1+e) = 0.9099/(+0.9099) = 0.4764
In percentage, n = 47.64%.
5. The angle of orientation of the spheres in the maximum possible voids ratio is ______
a) 60°
b) 90°
c) 30°
d) 75°
Discussion
Explanation: The soil will have maximum possible voids when the soil grains are arranged in a cubical array of spheres. In a cubical array, the angle of orientation α = 90°.
6. The porosity of intermediate packing factor is given by ______
a) n = 1-[1/(6*(1-cosα)*( √(1-2cosα) )]
b) n = (1-cosα)*( √(1+2cosα) )
c) n = 1-[π/(6*(1-cosα)*(√(1-2cosα))]
d) n = 6*(1-cosα)*(√(1-2cosα))
Discussion
Explanation: The formula was given by Slichter. In an intermediate packing, the centres of any 8 spheres, originally arranged in a cubic packing form the corners of the rhombohedron, with an acute face angle α.
7. The volume of the solids in a unit cell is given by ______
a) π/6
b) π/4
c) π/6
d) 2π/3
Discussion
Explanation: Consider a unit cell,
Volume of each spherical particle = (π/6)*d3
Total volume of container=1*1*1=1
No. of solids in the container=(1/d)*(1/d)*(1/d)=(1/d3)
Volume of the solids Vs=(π/6)*d3*(1/d3)=(π/6).
8. The volume of solids in a unit cell is constant.
a) True
b) False
Discussion
Explanation: The volume of solids in a unit cell is (π/6) as it does not change with the acute face angle α. Whereas, the total volume of the unit cell varies with acute face angle α.
9. In the densest state of packing, each sphere is in contact with ______ neighbouring spheres.
a) 6
b) 12
c) 8
d) 18
Discussion
Explanation: In the densest state of packing, sphere centres form a rhombohedral array with face angle α=60°. Thus, each sphere is in contact with 12 neighbouring spheres.
10. The total volume of rhombohedron with face angle 60° is _______
a) 0.5
b) 0.7071
c) 0.9099
d) 1
Discussion
Explanation: The total volume of rhombohedron is given by,
V = (1-cosα)*(√(1+2cosα))
V = (1-cos60°)*(√(1+2 cos60°))
V = 0.7071.