1. A soil sample weighs 190kN. After oven drying it weighs 150kN. What will be the weight of Water before oven drying?
a) 35 kN
b) 40 kN
c) 30 kN
d) 45 kN
Discussion
Explanation: Given,
Weight of soil sample W=190kN
Weight after oven drying Wd=150kN
Weight of Water WW=W-Wd
WW=190-150
WW=40kN.
2. In wet soil mass, air occupies one-seventh of its volume and Water occupies one- eighth of its volume. What will be the ratio of volume of voids with respect to the total volume?
a) 0.329
b) 0.279
c) 0.432
d) 0.268
Discussion
Explanation: Given,
Volume of Water Va= (1/7) total volume V
Volume of Water VW= (1/8) total volume V
Volume of voids Vv= Va + VW
Vv=(1/7)V+(1/8)V
Vv=(15/56)V=0.268V.
3. What will be the porosity of a soil sample whose void ratio is 0.35?
a) 19.5%
b) 40.65%
c) 25.9%
d) 30%
Discussion
Explanation: Given,
Void ratio e=0.35
Porosity \(n = \frac{e}{1+e}\)
\(n = \frac{0.35}{1+0.35}\)
\(n = \frac{0.35}{1.35}\)
\(\)n=0.259
In percentage n=25.9%.
4. ______ is independent of the total volume of the soil mass.
a) Bulk unit weight
b) Dry unit weight
c) Submerged unit weight
d) Unit weight of solids
Discussion
Explanation: The unit weight of solids is a function of the volume of solids, which does not change with water or moisture content. Whereas, bulk, dry and submerged unit weights are functions of total volume.
5. Void ratio is the ratio of _____
a) volume of voids to total volume
b) volume of voids to volume of solids
c) volume of solids to volume of voids
d) volume of soil mass to volume of voids
Discussion
Explanation: The volume of voids to the volume of solids is void ratio. The volume of voids to total volume is porosity. The volume of voids includes the volume of air voids as well as the volume of water present.
6. If the porosity of a soil sample is 22%, then what will be its void ratio?
a) 0.3
b) 0.27
c) 0.28
d) 0.25
Discussion
Explanation: Given,
Porosity n= 22%= 0.22
Void ratio e = n/(1-n)
e= 0.22/(1-0.22)
e= 0.28
7. For a soil sample, the ratio of saturated unit weight to dry unit weight is equal to 1.2 and specific gravity of solids G is 2.65, then its void ratio will be _____
a) 0.58
b) 0.66
c) 0.53
d) 0.94
Discussion
Explanation: Given,
The ratio of saturated unit weight to dry unit weight is equal to 1.2
γsat/γd= 1.2
Specific gravity of solids G is 2.65
γsat = \(\frac{(G+e)}{1+e} \gamma_w \,and\, \gamma_d = \frac{G}{1+e}\gamma_w\)
→ γsat/γd= \(\frac{(G+e)}{1+e} \gamma_w /\frac{G}{1+e}\gamma_w\)
γsat/γd = (G+e) / e
→(G+e) / e = 1.2
e = 1.2*G – G
e = 1.2*2.65 – 2.65
e = 0.53.
8. For a fully saturated soil sample, the degree of saturation is ______
a) 0
b) 0.5
c) 0.8
d) 1
Discussion
Explanation: For a fully saturated soil sample, the volume of water is equal to the volume of voids and hence the degree of saturation is equal to one.
9. The percentage void ratio is defined as the ratio of _____
a) volume of air voids to total volume of the soil mass
b) volume of air voids to volume of total voids
c) volume of air voids to volume of water
d) volume of voids to total volume of the soil mass
Discussion
Explanation: The percentage void ratio is defined as the ratio of the volume of air voids to the total volume of the soil mass. It is expressed as percentage.
10. If air content and degree of saturation are represented as ac and S respectively, then the relationship between them is ________
a) ac= 1/S
b) ac=1-S
c) S=1+ac
d) S=ac
Discussion
Explanation: Air content is the ratio of the volume of air voids (Va) to the volume of voids (Vv). The degree of saturation (S) is the ratio of the volume of water (VW) to the volume of voids (Vv) in a given soil mass.
ac=Va/Vv
S=VW/Vv
Since Va=Vv-VW
ac=1-VW/Vv
ac=1-S.