Soil Mechanics MCQ 1
The active earth pressure of a soil is proportional to (where φ is the angle of friction of the soil)
- tan(45°- φ)
- tan2(45°+φ/2)
- tan2(45°-φ/2)
- tan(45°+φ)
Answer: 3. tan2(45°-φ/2)
Solution
Active earth pressure,
pa=kaγH
where,
ka=(1- Sinφ)/(1+ Sinφ)=tan2(45°-φ/2)
Soil Mechanics MCQ 2
The quantity of seepage of water through soils is proportional to
- coefficient of permeability of soil
- total head loss through the soil
- neither (a) nor (b)
- both (a) and (b)
Answer: 4. both (a) and (b)
Solution
We know that
Seepage,
Q=kH(Nf/Nd)
where,
k = coefficient of permeability of the soil
H = Total head loss through the soil
Nf = number of flow channels
Nd = number of equipotential drops
Therefore, seepage quantity is proportional to both coefficients of permeability of the soil and total head loss through the soil.
Soil Mechanics MCQ 3
When drainage is permitted under initially applied normal stress only and full primarily consolidation is allowed to take place, the test is known as
- quick test
- drained test
- consolidated undrained test
- none of these
Answer: 3. Consolidated undrained test
Solution
Three different types of triaxial tests can be performed:
Unconsolidated Undrained Test (UU test) or quick test: Drainage is not permitted in both the consolidation and shearing stage.
Consolidated Drained Test (CD test): Drainage is permitted in both the consolidation and shearing stages.
Consolidated Undrained Test (CU test): Drainage is only permitted in the consolidation stage but not in the shearing stage.
Soil Mechanics MCQ 4
The maximum value of effective stress in the past divided by the present value is defined as the over consolidation ratio (OCR). The O.C.R. of an over consolidated clay is
- less than 1
- 1
- more than 1
- none of these
Answer: 3. more than 1
Solution
OCR=(Maximum applied effective stress in the past) /(effective stress applied in the present)
If OCR <= 1, soil is normally consolidated
If OCR >1, soil is over consolidated
Soil Mechanics MCQ 5
If Nf, Nd and H are total number flow channels, total number of potential drops and total hydraulic head differences respectively, the discharge q through the complete flow is given by (where K is a constant)
- q=√H(Nf/Nd)
- q=KH(Nd/Nf)
- q=KH(Nf/Nd)
- q=KH√(Nf/Nd)
Answer: 3. q=KH(Nf/Nd)
Solution
Seepage quantity is given by
q=KH(Nf/Nd)
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