Foundation Engineering Questions and Answers – Geophysical Methods – 2

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This set of Foundation Engineering MCQs focuses on “Geophysical Methods – 2”.

1. The angle of internal friction for fines greater than 5% is ___________
a) φ=25+0.15 ID
b) φ=20+0.15 ID
c) φ=15+0.15 ID
d) φ=10+0.15 ID
View Answer

Answer: a
Explanation: Meyerhof in 1956 gave approximate values of φ with respect to density index ID. For fines greater than 5%, the general formula is,
φ=25+0.15 ID
where, φ=angle of internal friction
ID= density index.
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2. The angle of internal friction for fines lesser than 5% is ___________
a) φ=25+0.15 ID
b) φ=20+0.15 ID
c) φ=15+0.15 ID
d) φ=10+0.15 ID
View Answer

Answer: b
Explanation: Meyerhof in 1956 gave approximate values of φ with respect to density index ID. For fines lesser than 5%, the general formula is,
φ=20+0.15 ID
where, φ=angle of internal friction
ID= density index.

3. In Dutch cone test, the base of cone is ____________ cm in diameter.
a) 3.6
b) 4.6
c) 5.6
d) 6.6
View Answer

Answer: a
Explanation: In Dutch cone test, the base of cone has a cross-sectional area of 10 cm2. The base of cone in Dutch cone test is 3.6 cm in diameter. The cone is carried at the lower end of the steel driving rod.
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4. In Dutch cone test, the angle of vector of cone is ____
a) 30°
b) 60°
c) 20°
d) 45°
View Answer

Answer: b
Explanation: The Dutch cone test is used for getting a continuous record of resistance of soil by penetrating a cone with an angle of 60° at the vertex. The base of the cone has a diameter of 3.6 cm.

5. To know the cone resistance, it is first forced down to _________ depth.
a) 8 mm
b) 8 cm
c) 15 cm
d) 30 cm
View Answer

Answer: b
Explanation: The cone is carried at the lower end of the steel driving rod which passes through a mantle with external diameter equal to the base of cone. To know the cone resistance, it is first forced down to 8 cm depth.
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6. In Dutch cone test, the cone along with mantle is further penetrated after 8cm to a depth of ________
a) 20 cm
b) 8 cm
c) 15 cm
d) 30 cm
View Answer

Answer: a
Explanation: In Dutch cone test, the cone along with mantle is further penetrated after 8cm to a depth of 20 cm. This is done to know the cone resistance in penetration through the soil layers.

7. The cone test is very useful in __________
a) cohesion-less soils
b) cohesive soils
c) clays
d) silty clay
View Answer

Answer: a
Explanation: The cone test is considered very useful in determining the bearing capacity of pits or bores in cohesion-less soils. It is particularly useful in fine sands of varying density.
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8. The cone resistance is approximately equal to ______ times the penetration resistance N.
a) 10
b) 20
c) 30
d) 40
View Answer

Answer: a
Explanation: The cone resistance in kg/cm2 is approximately equal to 10 times the penetration resistance N from the standard penetration test. The Dutch cone test is very useful in cohesion-less soils.

9. In Seismic refraction method, the radiating shock waves are picked up by ______
a) geophone
b) wave buoy
c) thermometer
d) cohesive soils
View Answer

Answer: a
Explanation: In Seismic refraction method, the radiating shock waves are picked up by the vibration detector known as geophone or seismometer where the time of travel gets recorded. The wave buoy measures the height of waves.
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10. In a blind zone, the geophone cannot detect the seisdmic refraction.
a) False
b) True
View Answer

Answer: b
Explanation: In 1968, Goodman and Karol showed that in a blind zone, the geophone cannot detect the seisdmic refraction. The blind zone occurs when a layer having a velocity lower than that of the upper layer exists.

11. When deeper layers have increasingly greater density, the seismic refraction method is fast.
a) False
b) True
View Answer

Answer: b
Explanation: The seismic refraction method is fast and a reliable method in establishing profiles of different strata when the deeper layers have increasingly greater density and thus higher velocities.

12. The equivalent penetration resistance is given by ___________
a) Ne=15+\(\frac{1}{2}\)(N+15)
b) Ne=\(\frac{1}{2}\)(N-15)
c) Ne=15+\(\frac{1}{2}\)(N-15)
d) Ne=15+\(\frac{1}{2}\)(N)
View Answer

Answer: c
Explanation: Terzaghi and Peck have recommended the use of equivalent penetration resistance Ne instead of observed penetration values of N, When N is greater than 15.
Ne=15+\(\frac{1}{2}\)(N-15).

13. If σ’ is the effective overburden pressure, then the effective penetration resistance is given by _____________
a) Ne=\(N \frac{50}{1.42σ^{‘}}\)
b) Ne=\(N \frac{50}{σ^{‘}+10}\)
c) Ne=\(N \frac{1}{1.42σ^{‘}+10}\)
d) Ne=\(N \frac{50}{1.42σ^{‘}+10}\)
View Answer

Answer: d
Explanation: Gibbs and Holtz have experimentally studied the effect of overburden on values of N.
Ne=\(N \frac{50}{1.42σ^{‘}+10}\), where, Ne=corrected value for overburden effect
N=actual value of blows
σ’ =effective overburden pressure.

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