Soil Mechanics Questions and Answers – Functional Relationships – 1

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 γd = 17 kN/m3
Water content w = 20% = 0.2
Bulk unit weight γ = γd *(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 ew = water void ratio=eS ——-(1)
Water content w=\(\frac{W_w}{W_d}\)
w = \(\frac{e_wγ_w}{γ_s*1}\)
But γs = Gγw
∴ w = \(\frac{e_wγ_w}{Gγ_w}= \frac{e_w}{G}\)
ew = wG ———–(2)
from equation (1) and (2)
\(e=\frac{wG}{S}\).

3. The relationship between e, S and na 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,
\(n_a=\frac{V_a}{V}\)
Va = Vv – Vw = e – ew
And V = Vs+Vv = 1+e
∴ na=\(\frac{e-e_w}{1+e}\)
Since ew = eS
\(n_a=e*\frac{1-S}{1+e}\).
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4. The relationship between na, ac, and n is ____
a) na=n*ac
b) na=n/ac
c) na=n+ac
d) na=n-ac
View Answer

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

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

Answer: c
Explanation: The dry unit weight is given by,
γd = \(\frac{W_d}{V} = \frac{γ_s*V_s}{V}\)
when Vs = 1, we have V = (1+e)
γd = γs*1/(1+e)
But γs = Gγw
∴ γd = \(\frac{Gγ_w}{1+e}\).
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6. The relationship between γsat, 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 Vs = 1, Vw = 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 Vs = 1, Vw = ew and V = 1+e
γ = \(\frac{(γ_s*1 + γ_w * e_w)}{(1+e)}\)
since ew = e*S and γs = Gγw
γ = \(\frac{(Gγ_w*1 + γ_w * e*S)}{(1+e)}\)
γ=\(\frac{(G+eS) γ_w}{(1+e)}\).
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8. The relationship between γ and γd 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 γd
∴ \(γ_d = \frac{Gγ_w}{(1+e)}\)
When the degree of saturation S=1,
the unit weight is \(\frac{(G+e) γ_w}{(1+e)}\) which is equal to γsat
∴ \(γ_{sat}=\frac{(G+e) γ_w}{(1+e)}\).

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’ = γsat – γw
γ’ = \(\frac{(G+e) γ_w}{(1+e)} – γ_w\)
∴ \(γ’=\frac{(G-1) γ_w}{(1+e)}\).
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10. The relationship between γd, γ and w is given by _______
a) γd = γ(1+w)
b) γd = γ *(1+w)
c) γd = \(\frac{γ}{(1+w)}\)
d) γd = γ+1+w
View Answer

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

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

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

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

Sanfoundry Global Education & Learning Series – Soil Mechanics.

To practice all areas of Soil Mechanics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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