Functional Relationships - Soil Mechanics Questions and Answers

This set of Soil Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Functional Relationships – 1”.

1. A soil has a dry unit weight of 17 kN/m³ and water content of 20%, then what will be its bulk unit weight?
a) 19.3 kN/m³
b) 20.4 kN/m³
c) 22.6 kN/m³
d) 24.4 kN/m³
View Answer

Answer: b
Explanation: Given,
Dry unit weight γ_d = 17 kN/m³
Water content w = 20% = 0.2
Bulk unit weight γ = γ_d *(1+w)
γ = 17*(1+0.2)
γ = 20.4 kN/m³.

2. The relationship between e, G, w and S is ______
a) \(e=\frac{wG}{S}\)
b) e = wGS
c) \(e=\frac{wS}{G}\)
d) \(e=\frac{GS}{w}\)
View Answer

Answer: a
Explanation: The degree of saturation is given by
S=\(\frac{V_w}{V_V}=\frac{e_w}{e}\) where e_w = water void ratio=eS ——-(1)
Water content w=\(\frac{W_w}{W_d}\)
w = \(\frac{e_wγ_w}{γ_s*1}\)
But γ_s = Gγ_w
∴ w = \(\frac{e_wγ_w}{Gγ_w}= \frac{e_w}{G}\)
e_w = wG ———–(2)
from equation (1) and (2)
\(e=\frac{wG}{S}\).

3. The relationship between e, S and n_a is ______
a) \(n_a=e*\frac{1+S}{1+e}\)
b) \(n_a=e*\frac{1-S}{1-e}\)
c) \(n_a=\frac{1-S}{1+e}\)
d) \(n_a=e*\frac{1-S}{1+e}\)
View Answer

Answer: d
Explanation: The percentage air voids is given by,
\(n_a=\frac{V_a}{V}\)
V_a = V_v – V_w = e – e_w
And V = V_s+V_v = 1+e
∴ n_a=\(\frac{e-e_w}{1+e}\)
Since e_w = eS
\(n_a=e*\frac{1-S}{1+e}\).

4. The relationship between n_a, a_c, and n is ____
a) n_a=n*a_c
b) n_a=n/a_c
c) n_a=n+a_c
d) n_a=n-a_c
View Answer

Answer: a
Explanation: The percentage air voids is given by,
a_c = \(\frac{V_a}{V_v}\) and n = \(\frac{V_v}{V}\)
n_a = \(\frac{V_a}{V}\) = n * a_c.

5. The relationship between γ_d, G and e is given by ______
a) γ_d = \(\frac{Gγ_w}{1-e}\)
b) γ_d = \(\frac{G+γ_w}{1+e}\)
c) γ_d = \(\frac{Gγ_w}{1+e}\)
d) γ_d = \(\frac{Gγ_w}{e}\)
View Answer

Answer: c
Explanation: The dry unit weight is given by,
γ_d = \(\frac{W_d}{V} = \frac{γ_s*V_s}{V}\)
when V_s = 1, we have V = (1+e)
γ_d = γ_s*1/(1+e)
But γ_s = Gγ_w
∴ γ_d = \(\frac{Gγ_w}{1+e}\).

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6. The relationship between γ_sat, G and e is _______
a) \(γ_{sat}=\frac{(G*e) γ_w}{(1+e)}\)
b) \(γ_{sat}=\frac{(G-e) γ_w}{(1+e)}\)
c) \(γ_{sat}=\frac{(G+e) γ_w}{(1+e)}\)
d) \(γ_{sat}=\frac{(G+e)}{(1+e)}\)
View Answer

Answer: c
Explanation: The saturated unit weight is given by,
\(γ_{sat}=\frac{W_{sat}}{V} = \frac{W_d+W_w}{V}\)
∴ \(γ_{sat}=\frac{(γ_s*V_s + γ_w *V_w)}{V}\)
When V_s = 1, V_w = e and V = 1+e
\(γ_{sat}=\frac{(γ_s*1 + γ_w *e)}{(1+e)} = \frac{(Gγ_w+ γ_w e)}{(1+e)}\)
\(γ_{sat}=\frac{(G+e) γ_w}{(1+e)}\).

7. The relationship between γ, S, G and e is ______
a) γ=\(\frac{(GS+e) γ_w}{(1+e)}\)
b) γ=\(\frac{(G+eS) γ_w}{(1+e)}\)
C) γ=\(\frac{(G+e)S γ_w}{(1+e)}\)
D) γ=\(\frac{(G+e) γ_w}{(1+e)S}\)
View Answer

Answer: b
Explanation: The bulk unit weight is given by,
\(γ=\frac{W}{V} = \frac{(γ_s*V_s+ γ_w *V_w)}{V}\)
When V_s = 1, V_w = e_w and V = 1+e
γ = \(\frac{(γ_s*1 + γ_w * e_w)}{(1+e)}\)
since e_w = e*S and γ_s = Gγ_w
γ = \(\frac{(Gγ_w*1 + γ_w * e*S)}{(1+e)}\)
γ=\(\frac{(G+eS) γ_w}{(1+e)}\).

8. The relationship between γ and γ_d can e derived when the degree of saturation is______
a) 0
b) 0.5
c) 0.8
d) 1
View Answer

Answer: a
Explanation: The bulk unit weight is given by,
γ = \(\frac{(G+e)S γ_w}{(1+e)}\)
When the degree of saturation S=0,
the unit weight is \(\frac{(G+e) γ_w}{(1+e)}\) which is equal to γ_d
∴ \(γ_d = \frac{Gγ_w}{(1+e)}\)
When the degree of saturation S=1,
the unit weight is \(\frac{(G+e) γ_w}{(1+e)}\) which is equal to γ_sat
∴ \(γ_{sat}=\frac{(G+e) γ_w}{(1+e)}\).

9. The relationship between γ’, G and e is ______
a) \(γ’=\frac{(G+1) γ_w}{(1+e)}\)
b) \(γ’=\frac{(G-1) γ_w}{(1+e)}\)
c) \(γ’=\frac{(G-1) γ_w}{(1-e)}\)
d) \(γ’=\frac{G γ_w}{(1+e)}\)
View Answer

Answer: b
Explanation: The submerged unit weight is given by,
γ’ = γ_sat – γ_w
γ’ = \(\frac{(G+e) γ_w}{(1+e)} – γ_w\)
∴ \(γ’=\frac{(G-1) γ_w}{(1+e)}\).

10. The relationship between γ_d, γ and w is given by _______
a) γ_d = γ(1+w)
b) γ_d = γ *(1+w)
c) γ_d = \(\frac{γ}{(1+w)}\)
d) γ_d = γ+1+w
View Answer

Answer: c
Explanation: The water content is given by,
\(w=\frac{W_w}{W_d}\)
hence \(1+w=\frac{(W_w + W_d)}{W_d} = \frac{W}{W_d}\)
\(W_d=\frac{W}{(1+w)}\)
\(γ_d=\frac{W_d}{V} = \frac{W}{(1+w)V}\)
∴ γ_d = \(\frac{γ}{(1+w)}\).

11. The relationship between γ_sat, γ_d, γ and S is ______
a) γ = γ_d*S[γ_sat-γ_d]
b) γ = \(\frac{γ_d}{S[γ_{sat}-γ_d]}\)
c) γ = γ_d+S[γ_sat-γ_d]
d) γ = γ_d-S[γ_sat-γ_d]
View Answer

Answer: c
Explanation: The bulk unit weight is given by,
\(γ = \frac{(G+eS) γ_w}{(1+e)} \,or\, γ = G\frac{γ_w}{(1+e)} + S\frac{eγ_w}{(1+e)}\)
\(γ = γ_d+S[\frac{(G+e) γ_w}{(1+e)} – \frac{G γ_w}{(1+e)}]\)
γ = γ_d+S[γ_sat-γ_d].

12. The relationship between γ_d, G, w and S is_______
a) γ_d = G γ_w*(1+(\(\frac{wG}{S}\)))
b) γ_d = G γ_w – (1+(\(\frac{wG}{S}\)))
c) γ_d = G γ_w/(1+(\(\frac{wG}{S}\)))
d) γ_d = G γ_w+(1+(\(\frac{wG}{S}\)))
View Answer

Answer: c
Explanation: The dry unit weight is given by,
γ_d = \(\frac{Gγ_w}{(1+e)}\)
but e=\(\frac{wG}{S}\)
substituting e in γ_d
∴ γ_d = G γ_w/(1+(\(\frac{wG}{S}\))).

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