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

1. A clay sample in laboratory test has 24mm thickness. It is tested with double drainage and consolidated 50%. The clay layer from which sample was obtained is 4m thick. Find the time to consolidate 50% with double drainage.

a) 365 days

b) 386 days

c) 750 days

d) 150 days

View Answer

Explanation: Given, double drainage for both specimen and field,

t

_{1}=20 min

d

_{1}=24 mm/2 = 2.4 cm/2 = 1.2 cm

d

_{2}=4/2 m 200cm

Since, for same degree of consolidation of 50%, T

_{v}is same,

\(t∝\frac{d^2}{C_r}\)

Therefore, \(\frac{t_2}{t_1} =(\frac{d_2}{d_1})^2\)

So, \(t_2=t_1 (\frac{d_2}{d_1} )^2=20(\frac{200}{1.2})^2=365 \,days.\)

2. A clay sample in laboratory test has 24mm thickness. It is tested with double drainage and consolidated 50%. The clay layer from which sample was obtained is 4m thick. Find the time to consolidate 50% with single drainage.

a) 1544 days

b) 1455 days

c) 1322 days

d) 1799 days

View Answer

Explanation: Given, double drainage for sample and single drainage for field layer,

t

_{1}=20 min

d

_{1}=24 mm/2 = 2.4 cm/2 = 1.2 cm

d

_{2}= 4 m = 400cm

Since, for same degree of consolidation of 50%, T

_{v}is same,

\(t∝\frac{d^2}{C_r}\)

Therefore, \(\frac{t_2}{t_1} =(\frac{d_2}{d_1} )^2\)

So, \(t_2=t_1 (\frac{d_2}{d_1} )^2=20(\frac{400}{1.2})^2=1544 \,days.\)

3. A clay sample in laboratory test has 24mm thickness. It is tested with single drainage and consolidated 50%. The clay layer from which sample was obtained is 4m thick. Find the time to consolidate 50% with double drainage.

a) 44 days

b) 55 days

c) 97 days

d) 107 days

View Answer

Explanation: Given, single drainage for sample and double drainage for field layer,

t

_{1}= 2 0 min

d

_{1}= 24 mm = 2.4 cm

d

_{2}= 2 m = 200cm

Since, for same degree of consolidation of 50%, T

_{v}is same,

\(t∝\frac{d^2}{C_r}\)

Therefore, \(\frac{t_2}{t_1} =(\frac{d_2}{d_1})^2\)

So, \(t_2=t_1 (\frac{d_2}{d_1} )^2=20(\frac{200}{2.4})^2=97 \,days.\)

4. A clay sample in laboratory test has 24mm thickness. It is tested with single drainage and consolidated 50%. The clay layer from which sample was obtained is 4m thick. Find the time to consolidate 50% with single drainage.

a) 386 days

b) 390 days

c) 404 days

d) 175 days

View Answer

Explanation: Given, single drainage for both the sample and for the field layer,

t

_{1}=20 min

d

_{1}=24 mm= 2.4 cm

d

_{2}=4 m = 400cm

Since, for same degree of consolidation of 50%, T

_{v}is same,

\(t∝\frac{d^2}{C_r}\)

Therefore, \(\frac{t_2}{t_1} =(\frac{d_2}{d_1} )^2\)

So, \(t_2=t_1 (\frac{d_2}{d_1})^2=20(\frac{400}{2.4})^2=386 \,days.\)

5. For a dry soil mass of 180.4 g, specific gravity 2.68 and cross-sectional area of specimen as 50 cm^{2}, find the height of solids H_{s}.

a) 13.45 mm

b) 14 mm

c) 17 mm

d) 19.5 mm

View Answer

Explanation: Given,

M

_{d}=180.4 g

G=2.68

A=50cm

^{2}

Since \(H_s=\frac{M_d}{GAρ_w}\)

\(H_s=\frac{180.4*10}{2.68*50*1} =13.45mm.\)

6. If the specimen height H is 24mm and height of solids H_{s} is 13.45mm, then find the voids ratio.

a) 0.786

b) 0.432

c) 2

d) 34

View Answer

Explanation: Given,

H=24mm

H

_{s}=13.45mm

\(e=\frac{H-H_s}{H_s}\)

\(e=\frac{24-13.45}{13.45}\)

e=0.786.

7. If the final water content is 30% and specific gravity is 2.65 for a soil sample, then its final void ratio is _________

a) 0.900

b) 0.795

c) 0.667

d) 0.549

View Answer

Explanation: W

_{f}=30%=0.30

G=2.65

e

_{f}=W

_{f}G

e

_{f}=0.3*2.65=0.795.

8. If the final voids ratio is 0.864 for a soil sample of specific gravity 2.7, then its final water content will be _______________

a) 32%

b) 48%

c) 56%

d) 62%

View Answer

Explanation: Given,

e

_{f}=0.864

G=2.7

e

_{f}=W

_{f}G

\(\frac{e_f}{G}=W_f\)

W

_{f}=32%.

9. If height of solids H_{s} is 6.725mm, mass is 180.4g with specific gravity G=2.68, then final the Cross-sectional area of specimen.

a) 220 cm^{2}

b) 342 cm^{2}

c) 100 cm^{2}

d) 659 cm^{2}

View Answer

Explanation: Given,

H

_{s}=6.725mm

G=2.68

M

_{d}=180.4g

Since \(H_s= \frac{M_d}{GH_s ρ_w}\)

\(A=\frac{M_d}{GH_s ρ_w}\)

\(A=\frac{180.4×10}{2.68×6.725×1}=100cm^2.\)

10. If the final height of specimen H_{f} is 13.45mm and final voids is 0.864, then the change of voids ratio ∆e with respect to ∆H is ____________

a) ∆e=0.139∆H

b) ∆e=0.123∆H

c) ∆e=0.178∆H

d) ∆e=0.148∆H

View Answer

Explanation: Given, H

_{f}=13.45

e

_{f}=0.864

\(∆e=\frac{1+e_f}{H_f} ∆H\)

\(∆e=\frac{1+0.864}{13.45} ∆H\)

∆e=0.139∆H.

11. The change in voids ratio is 0.18 and initial void ratio is 0.68 with the increase of pressure as 200KN/m^{2}. Find the coefficient of volume change.

a) 0.536 m^{2}/MN

b) 0.111 m^{2}/MN

c) 0.832 m^{2}/MN

d) 0.356 m^{2}/MN

View Answer

Explanation: Given,

∆e

_{f}=0.18

e

_{o}=0.68

∆σ’=200kN/m

^{2}

\(m_v=\frac{-∆e}{1+e_o}\frac{1}{∆σ’}\)

\(m_v=\frac{0.18}{1+0.68}+\frac{1}{200}=0.536m^2/MN.\)

12. The change in voids ratio is 0.18 and initial void ratio is 0.68 with the increase of pressure as 200KN/m^{2}. Find the coefficient of Compressibility.

a) 0.9 m^{2}/MN

b) 0.7 m^{2}/MN

c) 0.4 m^{2}/MN

d) 0.5 m^{2}/MN

View Answer

Explanation: Given,

∆e

_{f}=0.18

e

_{o}=0.68

∆σ’=200kN/m

^{2}

\(m_v=\frac{-∆e}{1+e_o}\frac{1}{∆σ’}\)

since, \(m_v=\frac{a_v}{1+e_0}\)

a

_{v}=m

_{v}(1+e

_{0})=0.536(1+0.68)

a

_{v}=0.9 m

^{2}/MN.

13. If coefficient of volume change is 0.291 m^{2}/MN, ∆e=0.1 and the initial voids ratio is 0.72, then find the change in effective pressure.

a) 200 KN/m^{2}

b) 50 KN/m^{2}

c) 300 KN/m^{2}

d) 00 KN/m^{2}

View Answer

Explanation: Given,

m

_{v}=0.291 m

^{2}/MN

∆e=0.10

e

_{o}=0.72

\(∆σ’=\frac{-∆e}{1+e_o}\frac{1}{m_v} \)

\(∆σ’=\frac{0.10}{1+0.72}×\frac{1×10^3}{0.291}=200KN/m^2.\)

14. If a soil A has coefficient of volume change as 0.536m^{2}/MN and the ratio of coefficient of volume change of A and B soils is 1.845, then find mv for soil B.

a) 0.255 m^{2}/MN

b) 0.291 m^{2}/MN

c) 0.234 m^{2}/MN

d) 0.345 m^{2}/MN

View Answer

Explanation: Given,

M

_{va}=0.536 m

^{2}/MN

\(\frac{m_{va}}{m_{vb}} =1.845\)

\(M_{vb}=\frac{0.536}{1.845}=0.291 m^2/MN.\)

15. A soil sample is having voids ratio 0.72 under a pressure of 200KN/m^{2}. If the void ratio is reduced to 0.5 under 450kn/m^{2} load, then find the change in effective pressure.

a) 4000 kN/m^{2}

b) 250 kN/m^{2}

c) 350 kN/m^{2}

d) 2000 kN/m^{2}

View Answer

Explanation: Given,

σ

_{0}’=200kN/m

^{2}

σ’=450kN/m

^{2}

∆σ’=σ’-σ

_{0}’=450-200

∆σ’=250kN/m

^{2}.

**Sanfoundry Global Education & Learning Series – Soil Mechanics.**

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