250+ TOP MCQs on Functional Relationships – 1 and Answers

Soil Mechanics Multiple Choice Questions 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
Answer: b
Clarification: 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})
Answer: a
Clarification: 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})
Answer: d
Clarification: 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
Answer: a
Clarification: 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})
Answer: c
Clarification: 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}).

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)})
Answer: c
Clarification: 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})
Answer: b
Clarification: 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
Answer: a
Clarification: 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)})
Answer: b
Clarification: 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
Answer: c
Clarification: 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]
Answer: c
Clarification: The bulk unit weight is given by,
(γ = frac{(G+eS) γ_w}{(1+e)} ,or, γ = Gfrac{γ_w}{(1+e)} + Sfrac{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})))
Answer: c
Clarification: 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}))).

Soil Mechanics,