Soil Mechanics and Foundation Sheet 21

1) Question.

According to Coulomb's law, the shearing strength of soil is

[c= Cohesion (apparent or actual), p= Normal stress on the plane of shear, $ϕ$ = Angle of internal friction or angle of shearing resistance]
Answer.
- 1
  
  $c - p \tan ϕ$
- 2
  
  $p - c \tan ϕ$
- 3
  
  $c + p \tan ϕ$
- 4
  
  $p + c \tan ϕ$
Answer: 3

Explanation:
2) Question.

The angle of shearing resistance for dry loose sand is same as that of angle of internal friction.
Answer.
- 1
  
  Right
- 2
  
  Wrong.
Answer: 1

Explanation:
3) Question.

The angle of internal friction of round-grained dense sand is about
Answer.
- 1
  
  $5^{0} t o 25^{0}$
- 2
  
  $25^{0} t o 30^{0}$
- 3
  
  $30^{0} t o 35^{0}$
- 4
  
  $32^{0} t o 37^{0}$
Answer: 4

Explanation:
4) Question.

The angle of internal friction of round-grained loose sand is about
Answer.
- 1
  
  $5^{0} t o 25^{0}$
- 2
  
  $25^{0} t o 30^{0}$
- 3
  
  $30^{0} t o 35^{0}$
- 4
  
  $32^{0} t o 37^{0}$
Answer: 2

Explanation:
5) Question.

Which one of the following parameters can be used to estimate the angle of friction of a sandy soil ?
Answer.
- 1
  
  Particle size
- 2
  
  Roughness of particle
- 3
  
  Density index
- 4
  
  Particle size distribution
Answer: 3

Explanation:
6) Question.

The settlement of a footing in clays $(S_{F})$ is given by
Answer.
- 1
  
  $\frac{S_{P}}{B_{P}} \times B_{F}$
- 2
  
  $\frac{S_{P}}{B_{F}} \times B_{F}$
- 3
  
  $\frac{S_{P}}{{[\frac{B_{P} (B_{F +} 30.48)}{B_{F} (B_{P} + 30.48)}]}^{2}}$
- 4
  
  $\frac{S_{P} {(B_{P} + 30.48)}^{2}}{{(B_{F +} 30.48)}^{2}}$
Answer: 1

Explanation:
7) Question.

If the pressure on a clay layer increases slightly from p to p+ $δ$ p and the void ratio decreases from $e_{0}$ to e, then the coefficient of compressibility is given by
Answer.
- 1
  
  e/p
- 2
  
  $\frac{e_{0}}{δ p}$
- 3
  
  $\frac{e}{δ p}$
- 4
  
  $\frac{e_{0} - e}{δ p}$
Answer: 4

Explanation:
8) Question.

The unit of coefficient of compressibility is inverse of pressure.
Answer.
- 1
  
  Correct
- 2
  
  Incorrect.
Answer: 1

Explanation:
9) Question.

The coefficient of compressibility..... with increase in pressure.
Answer.
- 1
  
  increases
- 2
  
  decreases
- 3
  
  does not change
Answer: 2

Explanation:
10) Question.

The coefficient of volume of compressibility is given by

[ $a_{v}$ = Coefficient of compressibility, $e_{0}$ = Void ratio]
Answer.
- 1
  
  $\frac{a_{v}}{1 + e_{0}}$
- 2
  
  $\frac{1 + e_{0}}{a_{v}}$
- 3
  
  $\frac{e_{0}}{a_{v}}$
- 4
  
  $\frac{e_{0}}{a_{0}}$
Answer: 1

Explanation:
11) Question.

A 300 mm square bearing plate settles by 15 mm in a plate load test on a cohesive soil when the intensity of loading is 0.2 N/mm2. The settlement of a prototype shallow footing 1 m square under the same intensity of loading is
Answer.
- 1
  
  15 mm
- 2
  
  30 mm
- 3
  
  50 mm
- 4
  
  167 mm
Answer: 3

Explanation:
12) Question.

A 600 mm square bearing plate settles by 15 mm in plate load test on cohesionless soil under an intensity of loading of 0.2 N/mm2. The settlement of a prototype shallow footing 1 m square under the same intensity of loading is
Answer.
- 1
  
  15 mm
- 2
  
  between 15 mm and 25 mm
- 3
  
  25 mm
- 4
  
  greater than 25 mm
Answer: 2

Explanation:
13) Question.

The rise of water table below the foundation influences the bearing capacity of soil mainly by reducing
Answer.
- 1
  
  cohesion and effective angle of shearing resistance
- 2
  
  cohesion and effective unit weight of soil
- 3
  
  effective unit weight of soil and effective angle of shearing resistance
- 4
  
  effective angle of shearing resistance
Answer: 2
Explanation:
When the water table rises, the soil becomes partially or fully saturated. This mainly causes a reduction in the effective unit weight (γ'), which reduces the bearing capacity.

The effective unit weight below the water table is:

γ' = γsat − γw

where:
- γsat = saturated unit weight of soil
- γw = unit weight of water
- γ' = submerged/effective unit weight
Also, in the context of the given MCQ, the rise of the water table is considered to reduce the soil's effective cohesion contribution.

📌 Exam Point

Rise of water table → reduction in effective unit weight → reduction in bearing capacity.

✅ Answer: B. Cohesion and effective unit weight of soil
14) Question.

The net ultimate bearing capacity of a purely cohesive soil
Answer.
- 1
  
  depends on width of footing and is independent of depth of footing
- 2
  
  depends on depth of footing and is independent of width of footing
- 3
  
  depends on both depth and width of footing
- 4
  
  is independent of both depth and width of footing
Answer: 4

Explanation:
For purely cohesive soil:

φ = 0°

Terzaghi's ultimate bearing capacity equation:

qu = cNc + qNq + 0.5γBNγ

For φ = 0°:

Nc = 5.7, Nq = 1, Nγ = 0

Therefore,

qu = 5.7c + γD

Net ultimate bearing capacity:

qnu = qu − γD

So, qnu = 5.7c

Hence, net ultimate bearing capacity is independent of both the width (B) and depth (D) of the footing.
15) Question.

According to Terzaghi's theory, the ultimate bearing capacity at ground surface for a purely cohesive soil and for a smooth base of a strip footing is

where C = unit cohesion of soil
Answer.
- 1
  
  2.57 C
- 2
  
  5.14 C
- 3
  
  5.7 C
- 4
  
  6.2 C
Answer: 2

Explanation:
16) Question.

Base failure of a finite slope
Answer.
- 1
  
  occurs when soil below the level of toe is strong
- 2
  
  occurs when there is a relatively weak zone in upper part of the slope
- 3
  
  occurs when the soil below the toe is relatively soft and weak
- 4
  
  is a most common failure and occurs in relatively steep slopes
Answer: 3

Explanation:
17) Question.

A cylindrical specimen of saturated soil failed under an axial vertical stress of $100 k N^{2} / m^{2}$ when it was laterally unconfined. The failure plane was inclined to the horizontal plane at an angle of $45^{o}$ . The values of cohesion and angle of internal friction for the soil are respectively
Answer.
- 1
  
  $0.5 N / m m^{2} a n d 30^{o}$
- 2
  
  $0.05 N / m m^{2} a n d 0^{o}$
- 3
  
  $0.2 N / m m^{2} a n d 0^{o}$
- 4
  
  $0.05 N / m m^{2} a n d 45^{o}$
Answer: 2

Explanation:
18) Question.

In a deposit of normally consolidated clay
Answer.
- 1
  
  effective stress increases with depth but water content of soil and undrained strength decrease with depth
- 2
  
  effective stress and water content increase with depth but undrained strength decreases with depth
- 3
  
  effective stress and undrained strenth increase with depth but water content decreases with depth
- 4
  
  effective stress, water content and undrained strenth decrease with depth
Answer: 3

Explanation:
19) Question.

Shear strength of a soil is a unique function of
Answer.
- 1
  
  effective stress only
- 2
  
  total stress only
- 3
  
  both effective stress and total stress
- 4
  
  none of the above
Answer: 1

Explanation:
20) Question.

Skempton's pore pressure coefficient B for saturated soil is
Answer.
- 1
  
  1
- 2
  
  zero
- 3
  
  between 0 and 1
- 4
  
  greater than 1
Answer: 1

Explanation:

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