Soil Mechanics Questions and Answers – Functional Relationships – 2

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This set of Soil Mechanics Interview Questions and Answers focuses on “Functional Relationships – 2”.

1. If the soil sample has porosity of 30% and specific gravity of solids G=2.7, then its dry unit weight will be ______
a) 18.350 kN/m3
b) 18.535 kN/m3
c) 18.640 kN/m3
d) 18.545 kN/m3
View Answer

Answer: b
Explanation: Given,
Porosity n=30%=0.3
Specific gravity G=2.7
Voids ratio e = n / (1-n)
e=0.3/(1-0.3)=0.429
Dry unit weight γd = Gγw/(1-e)
γd=(2.7*9.81)/(11+0.429)
γd=18.535 kN/m3.

2. A soil sample has a dry density of 8.5 kN/m3, specific gravity of solids G as 2.7 and voids ratio of 0.6. If the unit weight of soil is 50% saturated, then its bulk unit weight will be ______
a) 19.274 kN/m3
b) 23.034 kN/m3
c) 20.554 kN/m3
d) 21.428 kN/m3
View Answer

Answer: c
Explanation: Given,
Dry density γd = 18.5 kN/m3
Specific gravity G=2.7
Void ratio e= 0.6
Degree of saturation = 50% = 0.5
Since, e = (W*G)/S
W = (e*S)/G=(0.6*0.5)/2.7=0.111
Bulk unit weight γ=γd(1+W)
γ=18.5*1.111=20.554 kN/m3.

3. When the soil is fully saturated with its specific gravity of solids G=2.7 and voids ratio e=0.667, its water content is ______
a) 0.326
b) 0.247
c) 0.543
d) 1
View Answer

Answer: b
Explanation: Given,
Degree of saturation S=1
Specific gravity G=2.7
Voids ratio e = (W*G)/S = Wsat*G
Wsat = e/G
Wsat = 0.667/2.7
Wsat = 0.247.
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4. A soil has a dry unit weight of 17.5 kN/m3 and a water content of 20%. What will be its bulk unit weight?
a) 18 kN/m3
b) 19 kN/m3
c) 20 kN/m3
d) 21 kN/m3
View Answer

Answer: d
Explanation: Given,
Dry unit weight γd = 17.5 kN/m3
Water content W = 20% = 0.2
Bulk unit weight γ = γd(1+W)
γ = 17.5*(1+0.2)
γ=21 kN/m3.

5. A soil has a void ratio of 0.6 and a specific gravity of solids as 2.7. What will be its dry unit weight?
a) 15.5 kN/m3
b) 16 kN/m3
c) 16.5 kN/m3
d) 17 kN/m3
View Answer

Answer: c
Explanation: Given,
Voids ratio e=0.6
Specific gravity G=2.7
Dry unit weight γd = (G* γw/1+e)
Unit weight of water γw=9.8 kN/m3
γd = (2.7*9.81)/(1+0.6)
γd = 16.5 kN/m3.
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6. A soil has porosity of 40% and air content of 50%.What will be its percentage air voids?
a) 30%
b) 60%
c) 20%
d) 50%
View Answer

Answer: c
Explanation: Given,
Porosity n = 40% = 0.4
Air content ac = 50% = 0.5
Percentage air voids na= n*ac
na = 0..4*0.5
na = 20%.

7. If a solid has air content of 50%, then its degree of saturation will be ______
a) 40%
b) 50%
c) 30%
d) 20%
View Answer

Answer: b
Explanation: Given,
Air content ac = 0.5
Degree of saturation S= 1-ac
S = 1-0.5 = 0.5
In percentage S=50%.
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8. If a soil has a void ratio of 0.64, specific gravity of solids as 2.65, then its saturated unit weight will be ______
a) 24.5 kN/m3
b) 25 kN/m3
c) 25.5 kN/m3
d) 19.5 kN/m3
View Answer

Answer: d
Explanation: Given,
Voids ratio e = 0.6
Specific gravity G = 2.65
Saturated unit weight γsat = (G+e) γw/ (1+e)
γsat = (2.65+0.6)*9.81/(1+0.6)
γsat= 19.927 N/m3.

9. A soil has a void ratio of 0.6, specific gravity as 2.65 and degree of saturation as 50%. What will be its bulk unit weight?
a) 17 kN/m3
b) 18 kN/m3
c) 19 kN/m3
d) 20 kN/m3
View Answer

Answer: b
Explanation: Given,
Degree of saturation S = 50% = 0.5
Bulk unit weight γ = (G+eS)γw/ (1+e)
γ = (2.65+0.6*0.5)*9.81/(1+0.6)
γ = 18 kN/m3.
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10. If a soil has a saturated unit weight of 24.5 kN/m3, then its submerged unit weight will be ______
a) 15.69
b) 14.69
c) 9.81
d) 20.61
View Answer

Answer: c
Explanation: Given,
Saturated unit weight γsat = 24.5 kN/m3
Submerged unit weight γ’= γsat – γw
Unit weight of water γw = 9.8 kN/m3
Submerged unit weight γ’=2.45-9.81
γ’=14.69 kN/m3.

11. If a soil has a void ratio of 0.6 and a specific gravity of 2.7, then its submerged unit weight will be ______
a) 9.678
b) 11.502
c) 9.256
d) 10.423
View Answer

Answer: d
Explanation: Given,
Voids ratio e=0.6
Specific gravity G=2.7
Submerged unit weight γ’ = (G -1) γw/ (1+e)
γ’ = (2.7-1)*9.81/(1+0.6)
γ’ = 10.423 kN/m3.

12. A soil has a dry unit weight of 16.25 kN/m3 and saturated unit weight of 20 kN/m3. If the degree of saturation of soil is 50%, then its bulk unit weight will be ______
a) 20.748 kN/m3
b) 17.628 kN/m3
c) 18.123 kN/m3
d) 19.246 kN/m3
View Answer

Answer: c
Explanation: Given,
Dry unit weight γd = 16.25 kN/m3
Saturated unit weight γsat = 20 kN/m3
Degree of saturation S = 50% = 0.5
Bulk unit weight γ = γd + S[γsat– γd] γ = 16.25 + 0.5[20-16.25] γ =18.123 kN/m3.

Sanfoundry Global Education & Learning Series – Soil Mechanics.

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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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