This set of Soil Mechanics Multiple Choice Questions & Answers (MCQs) focuses on “Calculation of Voids Ratio and Coefficient of Volume Change”.

1. From the consolidation test, the final voids ratio at end of pressure increment can be calculated by ________________

a) height of solids method only

b) change in voids ratio method only

c) both height of solids method and change in voids ratio method

d) no method are available

View Answer

Explanation: The equilibrium voids ratio or the final voids ratio can be calculated by the following methods:

- height of solids method
- change in voids ratio method.

2. The change in voids ratio method is applicable for _________________

a) only dry specimens

b) only fully saturated specimens

c) only partially saturated specimens

d) for all types of specimens

View Answer

Explanation: For the change in voids ratio method the final voids ratio is calculated by the relation of e

_{f}=w

_{f}G. Where the w

_{f}represents the final water content and the degree of saturation is one percent.

3. The height of solids method is applicable for ________

a) only dry specimens

b) only fully saturated specimens

c) both saturated and unsaturated samples

d) only partially saturated specimens

View Answer

Explanation: The height of solids method is applicable for both saturated and unsaturated samples but the change in voids ratio method is applicable only for fully saturated soil specimens.

4. The height H_{S} of the solids of the specimen is calculated from _______________

a) \(H_s=\frac{M_d}{GA}\)

b) \(H_s=\frac{M_d}{Gρ_w}\)

c) \(H_s=\frac{M_d}{Aρ_w}\)

d) \(H_s=\frac{M_d}{GAρ_w}\)

View Answer

Explanation: The height H

_{S}of the solids of the specimen is calculated from the expression,

\(H_s=\frac{M_d}{GAρ_w},\)

where, H

_{s}=height of solids

M

_{d}=mass of dried soil sample

G=specific gravity of solid particles

A=cross-sectional area of specimen

ρ

_{w}=mass density of water.

5. The height H_{s} of the solids of the specimen with respect to weight of dried specimen is _____________

a) \(H_s=\frac{W_d}{Gρ_w}\)

b) \(H_s=\frac{W_d}{GAρ_w}\)

c) \(H_s=\frac{W_d}{GA}\)

d) \(H_s=\frac{W_d}{Aρ_w}\)

View Answer

Explanation: Since the height H

_{S}of the solids of the specimen is calculated from the expression,

\(H_s=\frac{W_d}{GAρ_w}\) and the dry weight in relation to dry mass is given by,

\(W_d=\frac{M_d}{ρ_w}.\) Therefore, the The height H

_{s}of the solids of the specimen with respect to weight of dried specimen is,

\(H_s=\frac{W_d}{GA}.\)

6. The voids ratio with respect to height of solids is given by equation______________

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

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

c) \(e=\frac{H+H_s}{H_s}\)

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

View Answer

Explanation: The voids ratio is calculated from the equation of,

\(e=\frac{H-H_s}{H_s},\) where e=voids ratio

H=height of specimen at equilibrium

H

_{s}= height of solids.

7. The height of the specimen at various applied pressures is given by _____________

a) H=H_{0}*∑∆H

b) H=H_{0}-∑∆H

c) \(H=\frac{H_0}{∑∆H}\)

d) H=H_{0}+∑∆H

View Answer

Explanation: The height of specimen at equilibrium under various applied pressures is,

H=H

_{0}+∑∆H,

where, H

_{0}=initial height of the specimen

∆H= change in specimen thickness under any pressure increment. The sum of each pressure increment is ∑∆H.

8. From the consolidation test, by the change in voids ratio method, the final void is given by _______

a) e_{f}=w_{f} G

b) e_{f}=w_{f} AG

c) \(e_f=\frac{w_f}{G}\)

d) \(e_f=\frac{G}{w_f}\)

View Answer

Explanation: In the change in voids ratio method of consolidation test, the final voids ratio is given by, e

_{f}=w

_{f}G.

Where, w

_{f}=final water content

G=specific gravity of solid particles

The degree of saturation is one (S=1).

9. The change in voids ratio of the consolidation test is calculated from the relation of________

a) \(∆e=\frac{1+e_f}{2} ∆H\)

b) \(∆e=\frac{1+e_f}{H_f} ∆H_f\)

c) \(∆e=\frac{1+e_f}{H}\)

d) \(∆e=\frac{e_f}{H_f} ∆H\)

View Answer

Explanation: The change in voids ratio with the pressure increment is given by the relation,

\(\frac{∆e}{1+e_f}=\frac{∆H}{H_f},\) therefore on rearranging the equation, we get,

\(∆e=\frac{1+e_f}{2} ∆H.\)

10. From the consolidation test data using the ‘change in void ratio’ method, the coefficient of volume change can be calculated by _____________

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

b) \(m_v=-\frac{∆e}{1+e_o}\)

c) \(m_v=-\frac{∆e}{1+e_o} \frac{10}{∆σ’}\)

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

View Answer

Explanation: Using the change in void ratio method, the coefficient of volume change can be defined as the change in volume of a soil per unit of initial volume due to a given increase in unit pressure.

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

11. From the consolidation test data using the ‘Height of solids’ method, the coefficient of volume change can be calculated by _____________

a) \(m_v=-\frac{∆H_o}{H_o}\frac{1}{∆σ’}\)

b) \(m_v=\frac{∆H}{H_o} \frac{1}{∆σ’}\)

c) \(m_v=-\frac{∆H}{H_o} \frac{1}{∆σ’}\)

d) \(m_v=-\frac{∆H}{H}\frac{1}{∆σ’}\)

View Answer

Explanation: Since we know the relation of change in voids ratio with the pressure increment is given by the relation,

\(\frac{∆e}{1+e_0}=\frac{∆H}{H_0},\) and coefficient of volume change is given by \(m_v=-\frac{∆e}{1+e_o}\frac{1}{∆σ’}.\)

Therefore, from the two equations, we get,

\(m_v=-\frac{∆H}{H_o} \frac{1}{∆σ’}.\)

12. The coefficient of volume change is ______________

a) reciprocal of compressibility modulus

b) equal to compressibility modulus

c) double compressibility modulus

d) equal to three times the compressibility modulus

View Answer

Explanation: The compressibility modulus is a measure of the relative volume change of a fluid or solid as a response to a pressure change. The coefficient of volume change is the reciprocal of compressibility modulus and has units in the form of m

^{2}/kN.

13. The unit of coefficient of volume change is

a) m^{2}/kN

b) m^{3}/kN

c) m^{2}/N

d) m^{3}/N

View Answer

Explanation: The coefficient of volume change can be defined as the change in volume of a soil per unit of initial volume due to a given increase in unit pressure.

\(m_v=-\frac{∆H}{H_o}\frac{1}{∆σ’},\) Hence it has the units in the form of m

^{2}/kN (reciprocal of pressure).

14. From the calculations consideration of a consolidation test data, the change in thickness can be both positive and negative.

a) True

b) False

View Answer

Explanation: From the calculations consideration of a consolidation test data, the change in thickness can be both positive and negative. The negative sign denotes the decrease height due to loading and the positive sign denotes the increase in height due to unloading of the specimen.

15. The Rowe cell apparatus also measures the pore pressure.

a) False

b) True

View Answer

Explanation: The Rowe cell apparatus allows the consolidation test based on the theory of consolidation and measure pore water pressure and the rate of dissipation. The change in sample height can also be measured during the test.

**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__.

**Next Steps:**

- Get Free Certificate of Merit in Geotechnical Engineering
- Participate in Geotechnical Engineering Certification Contest
- Become a Top Ranker in Geotechnical Engineering
- Take Geotechnical Engineering Tests
- Chapterwise Practice Tests: Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
- Chapterwise Mock Tests: Chapter 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

**Related Posts:**

- Apply for Geotechnical Engineering I Internship
- Apply for Civil Engineering Internship
- Practice Geotechnical Engineering II MCQs
- Practice Civil Engineering MCQs
- Buy Civil Engineering Books