Soil Mechanics and Foundation Sheet 7

1) Question.

The specific gravity of soil solids is determined by
Answer.
- 1
  
  Pycnometer method
- 2
  
  hydrometer analysis
- 3
  
  sieve analysis
- 4
  
  all of these
Answer: 1

Explanation:
2) Question.

The relation between the saturated unit weight $(γ_{s a t})$ , specific gravity (G) and porosity (n) is given by
Answer.
- 1
  
  $y_{s a t} = G . γ_{w} (1 + n)$
- 2
  
  $y_{s a t} = n γ_{w} (1 - n)$
- 3
  
  $y_{s a t} = G . γ_{w} (1 - n) + n γ_{w}$
- 4
  
  $y_{s a t} = G . γ_{w} (1 + n) + n γ_{w} (1 - n)$
Answer: 3

Explanation:
3) Question.

The submerged unit weight for completely saturated soil is given by
Answer.
- 1
  
  $\frac{(G + e) γ_{w}}{1 + e}$
- 2
  
  $\frac{(1 + e) γ_{w}}{G + e}$
- 3
  
  $\frac{(G - 1) γ_{w}}{1 + e}$
- 4
  
  $\frac{(1 - e) γ_{w}}{G + e}$
Answer: 3

Explanation:
4) Question.

The water content of soils can be accurately determined by
Answer.
- 1
  
  sand bath method
- 2
  
  calcium carbide method
- 3
  
  oven drying method
- 4
  
  pycnometer method
Answer: 3

Explanation:
5) Question.

The unit weight of a completely saturated soil $(γ_{s a t})$ is given by

[ G= specific gravity of soil, e= void ratio, $γ_{w}$ = unit weight of water]
Answer.
- 1
  
  (G+e)Yw / (1+e)
- 2
  
  (1+e)Yw / (G+e)
- 3
  
  (G-1)Yw / (1+e)
- 4
  
  (1-e)Yw / (G+e)
Answer: 1

Explanation:
6) Question.

The relation between dry unit weight $(γ_{d})$ , specific gravity (G), void ratio (e) or porosity (n) is
Answer.
- 1
  
  $γ_{d} = \frac{G . γ_{w}}{1 + e}$
- 2
  
  $γ_{d} = \frac{G . γ_{w}}{1 - e}$
- 3
  
  $γ_{d} = G . γ_{w} (1 - n)$
- 4
  
  $γ_{d} = G . γ_{w} (1 + n)$
Answer: 1

Explanation:
7) Question.

If w is the water content and $γ$ is the unit weight of soil mass, then the unit weight of dry soil $(γ_{d})$ is equal to
Answer.
- 1
  
  (w/ $γ$ ) +1
- 2
  
  ( $γ$ /w)+1
- 3
  
  ( $γ$ )/1+w
- 4
  
  (1+w)/ $γ$
Answer: 3

Explanation:
8) Question.

If the volume of voids is equal to the volume of soil solids, then the values of porosity and void ratio are respectively
Answer.
- 1
  
  0 and 0.5
- 2
  
  0 and 1
- 3
  
  0.5 and 1
- 4
  
  1 and 0.5
Answer: 3

Explanation:
9) Question.

The density of soil mass is expresssed in
Answer.
- 1
  
  $k g / m^{2}$
- 2
  
  $k g / m^{3}$
- 3
  
  $N / m^{2}$
- 4
  
  $N / m^{3}$
Answer: 2

Explanation:
10) Question.

Which of the following gives the correct decreasing order of the densities of a soil sample ?
Answer.
- 1
  
  Saturated, submerged, wet, dry
- 2
  
  Saturated, wet, submerged, dry
- 3
  
  Saturated, wet, dry, submerged
- 4
  
  wet, saturated, submerged, dry
Answer: 3

Explanation:
11) Question.

If a cohesive soil specimen is subjected to a vertical compressive load, the inclination of the cracks to the horizontal is
Answer.
- 1
  
  $90^{o}$
- 2
  
  $45^{o}$
- 3
  
  $22 . 5^{o}$
- 4
  
  $0^{o}$
Answer: 2

Explanation:
12) Question.

In a triaxial compression test on a soil speciment, the intermediate principal stress is equal to
Answer.
- 1
  
  major principal stress
- 2
  
  minor principal stress
- 3
  
  difference between major and minor principal stress
- 4
  
  none of the above
Answer: 2

Explanation:
13) Question.

For a loose sand sample and dense sand sample consolidated to the same effective stress
Answer.
- 1
  
  ultimate strength is same and also peak strength is same
- 2
  
  ultimate strength is different but peak strength is same
- 3
  
  ultimate strength is same but peak strength of dense sand is greater than that of loose sand
- 4
  
  ultimate strength is same but peak strength of loose sand is greater than that of dense sand
Answer: 3

Explanation:
14) Question.

Select the incorrect statement.
Answer.
- 1
  
  Effective cohesion of a soil can never have a negative value.
- 2
  
  Effective angle of internal fricion for coarse grained soils is rarely below $30^{o}$
- 3
  
  Effective angle of internal friction for a soil increases as state of compactness increases
- 4
  
  Effective angle of internal friction is a complicated function of mineralogy and clay size content
Answer: 1

Explanation:
15) Question.

For better strength and stability, the fine grained soils and coarse grained soils are compacted respectively as

where OMC is optimum moisture content
Answer.
- 1
  
  dry of OMC and wet of OMC
- 2
  
  wet of OMC and dry of OMC
- 3
  
  wet of OMC and wet of OMC
- 4
  
  dry of OMC and dry of OMC
Answer: 2

Explanation:
16) Question.

The maximum dry density upto which any soil can be compacted depends upon
Answer.
- 1
  
  moisture content only
- 2
  
  amount of compaction energy only
- 3
  
  both mosture content and amount of compaction energy
- 4
  
  none of the above
Answer: 3

Explanation:
17) Question.

With the increase in the amount of compaction energy
Answer.
- 1
  
  optimum water content increases but maximum dry density decreases
- 2
  
  optimum water content decreases but mamimum dry density increases
- 3
  
  both optimum water content and maximum dry density increase
- 4
  
  both optimum water content and maximum dry density decrease
Answer: 2

Explanation:
18) Question.

Coarse grained soils are best compacted by a
Answer.
- 1
  
  drum roller
- 2
  
  rubber tyred roller
- 3
  
  sheep's foot roller
- 4
  
  vibratory roller
Answer: 4

Explanation:
19) Question.

The ultimate consolidation settlement of a soil is
Answer.
- 1
  
  directly proportional to the voids ratio
- 2
  
  directly proportional to the compression index
- 3
  
  inversely proportional to the compression index
- 4
  
  none of the above
Answer: 2

Explanation:
20) Question.

The ultimate consolidation settlement of a structure resting on a soil
Answer.
- 1
  
  decreases with the increase in the initial voids ratio
- 2
  
  decreases with the decrease in the plastic limit
- 3
  
  increases iwth the increase in the initial voids ratio
- 4
  
  increases with the decrease in the porosity of the soil
Answer: 1

Explanation:

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