Soil Mechanics Questions and Answers – Functional Relationships – 1

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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

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) $$e=\frac{GS}{w}$$

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) $$n_a=e*\frac{1-S}{1+e}$$

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

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 = $$\frac{Gγ_w}{e}$$

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 = $$\frac{Gγ_w}{1+e}$$.

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) $$γ_{sat}=\frac{(G+e)}{(1+e)}$$

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=$$\frac{(G+e) γ_w}{(1+e)S}$$

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=$$\frac{(G+eS) γ_w}{(1+e)}$$.

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

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) $$γ’=\frac{G γ_w}{(1+e)}$$

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

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

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]

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+($$\frac{wG}{S}$$))

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+($$\frac{wG}{S}$$)).

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