Previous Years GATE Questions - Soil Properties

1.   For sand of uniform spherical particles, the void ratios in the loosest and the densest states are _____ and _____ respectively.

[GATE 1991: 2 Marks]

Answer: 0.91, 0.35

Explanation:
In loosest state, the soil particles are arranged in a cubical array in which each soil particle is surrounded by 6 soil particles.
Volume of Spherical soil particle = (d/2)3 x

4Π / 3

Assume the cube side as   1 x 1 x 1 unit.
Total volume = 1 cubic unit.
The number of spherical particles in 1 cubic unit is   1/d x 1/d x 1/d =
1 / d3

Thus, Volume of soil solids = Volume of single soil particle x number of particles
⇒ Vsoil= (d/2)3 x
4Π / 3
x
1 / d3
= Π/6
⇒ Vsoil = 0.523
Total volume of voids, Vvoids = V - Vsoil
⇒ Vvoids = 1 - 0.523 = 0.477
Thus void ratio in loosest state,
e max =
Vvoids / Vsoil

⇒ e max =
0.477 / 0.523

⇒ e max = 0.91
Similarly, in densest state, the soil particles are arranged in a rhombohedral array in which each soil particle is surrounded by 12 soil particles.
The volume of soil solids remains constant.
Thus, Volume of soil solids,
Vsoil = Π/6 = 0.523
Total volume, V = (1 - cos α)√(1 + 2 cos α)
Here, α =
360° / 12
= 60°
⇒ V = 0.707 m3
Now, Volume of voids, Vvoids = V - Vsoil
⇒ Vvoids = 0.707 - 0.523
⇒ Vvoids = 0.184 m3
Thus void ratio in densest state,
e min =
Vvoids / Vsoil

⇒ e min =
0.184 / 0.523

⇒ e min = 0.35

2.  A saturated sand sample has dry unit weight of 18kN/m3 and a specific gravity of 2.65. If Ɣw = 10 kN/m3, the water content of the soil is _____ .

[GATE 1991: 2 Marks]

Answer: 0.17

Explanation:
Dry unit weight, Ɣd = 18 kN/m3
Specific Gravity of, G = 2.65
Unit weight of water, Ɣw = 10 kN/m3
From the existing relations,
Ɣd =

G Ɣw / 1 + e

⇒18 =
2.65 x 10 / 1 + e

⇒18 x (1 + e) = 2.65 x 10
⇒e =
26.5 – 18 / 18

⇒e= 0.47
Now,
For a saturated sand sample, Degree of saturation, S = 1
S.e = wG
⇒ 1 x 0.47 = w x 2.65
⇒ w =
1 x 0.47 / 2.65

⇒ w = 0.17

3.  The atterberg limits of a clay are 38%, 27% and 24.5%. It’s natural water content is 30%. The clay is in _____ state.

[GATE 1994: 2 Marks]

Answer: Plastic

Explanation:

Properties of soil - Graph pertaining to Consistency limits

The natural water content of soil is 30%.
It can be seen that natural water content of soil lies between Plastic Limit and Liquid Limit.
Thus, it can be concluded that the soil sample is in Plastic State.

4.  The void ratio of a soil sample is 1, the corresponding porosity of the sample is _____ .

[GATE 1994: 1 Mark]

Answer: 0.5

Explanation:
By using the existing relation,
n =

e / 1 + e

⇒n =
1 / 1 + 1

⇒n =
1 / 2

⇒n = 0.5

5.  The consistency of a saturated cohesive soil is affected by _____ .

[GATE 1995: 1 Mark]
  1. Water content
  2. Particle size distribution
  3. Density index
  4. Co-efficient of permeability

Answer: Option A

Explanation: Consistency is the term which is used to determine the degree of firmness of the soil. It is mostly influenced by water content present in the soil.

6.  The void ratio of soil can exceed unity. Is it true or false?

[GATE 1991: 1 Mark]

Answer: True

Explanation: Void Ratio, e =

Volume of voids / Volume of soil soilds

Hence, Void ratio can be greater than unity.

7.  Which one of the following relations is not correct?

[GATE 1996: 1 Mark]
  1. e =
    n / 1 – n
  2. Ɣsat =
    (G + e)Ɣw / 1 + e
  3. n =
    e / 1 – e
  4. e =
    w G / S

Answer: Option C

Explanation: Self Explanatory

8.  Consistency limit for a clayey soil is _____ (LL = Liquid Limit, PL = Plastic Limit, PI = Plasticity Index, w = Natural Moisture content).

[GATE 1997: 1 Mark]
  1. LL – w / PI
  2. w – LL / PI
  3. LL – PI
  4. 0.5 w

Answer: Option A

Explanation: Consistency limit is defined as the ratio of difference between Liquid Limit and Natural Water Content to its Plasticity Index

9.  If the porosity of soil sample is 20%, the void ratio is _____ .

[GATE 1997: 1 Mark]
  1. 0.20
  2. 0.80
  3. 1.00
  4. 0.25

Answer: Option D

Explanation:
Porosity, n = 20% = 0.20
Using, n =

e / 1 + e

⇒0.20 =
e / 1 + e

⇒0.20 + 0.20 e = e
⇒0.20 = ( 1 - 0.20) e
⇒e =
0.20 / 0.80

⇒e = 0.25

10.  The ratio of unconfined compressive strength of an undisturbed sample of soil to that of a remoulded sample at the same water content, is _____ .

[GATE 1997: 1 Mark]
  1. Activity
  2. Damping
  3. Plasticity
  4. Sensitivity

Answer: Option D

Explanation: Definition of Sensitivity

11.  If a soil is dried beyond its shrinkage limit, it will show _____ .

[GATE 1998: 1 Mark]
  1. Large volume change
  2. Moderate volume change
  3. Low volume change
  4. No volume change

Answer: Option D

Explanation: Volume change occurs only when the water content of soil is varied beyond its shrinkage limit. But, below shrinkage limit, the volume remains constant.

12.  Principle involved in the relationship between submerged unit weight and saturated unit weight of soil is based on _____.

[GATE 1999: 1 Mark]
  1. Equilibrium of floating bodies
  2. Archimedes principle
  3. Stoke’s law
  4. Darcy’s law

Answer: Option B

Explanation:
Archimedes Principle - When soil is immersed in water, the volume of water displaced will be equal to the volume of soil immersed in water.
Ɣsub = Ɣsat - Ɣw

13.  A soil sample in its natural state has mass of 2.290 kg and a volume of 1.15 x 10-3 m3. After being oven dried, the mass of the sample is 2.035 kg. Specific Gravity of soil solids is 2.68. The void ratio of natural soil is _____.

[GATE 1999: 1 Mark]
  1. 0.40
  2. 0.45
  3. 0.55
  4. 0.53

Answer: Option B

Explanation:
Specific Gravity of soil, G = 2.68
⇒ G =

Ɣ / Ɣw

⇒ 2.68 =
Ɣ / 9.81

⇒ Ɣ = 2.68 x 9.81 = 26.29 kN/m3
⇒ Ɣ = 26.29 x 103 =
Msoil x Ɣw / Vsoil

⇒ 26.29 x 103 =
2.035 x 9.81 / Vsoil

⇒ Vsoil = 0.75 x 10-3 m 3

Now,
Total Volume, V = Vsoil + Vvoids
⇒ Vvoids = 1.15x10-3 - 0.75x10-3
⇒ Vvoids = 0.40 x 10-3
Thus, Void ratio, e =
Vvoids / Vsoil

⇒ e =
0.40 x 10-3 / 0.75 x 10-3

⇒ e = 0.53
Thus, The void ratio of natural soil is 0.53

14.  The toughness index of a clayey soil is given by _____ .

[GATE 1999: 1 Mark]
  1. Plasticity Index / Flow Index
  2. Liquid Limit / Plastic Limit
  3. Liquidity Index / Plastic Limit
  4. Plastic Limit / Liquidity Index

Answer: Option A

Explanation: Toughness index is defined as the ratio of Plasticity index to Flow index.

15.  A soil sample has a void ratio of 0.5 and its porosity will be close to _____ .

[GATE 2000: 1 Mark]
  1. 50%
  2. 66%
  3. 100%
  4. 33%

Answer: Option D

Explanation:
From n =

e / 1 + e

⇒n =
0.5 / 1 + 0.5

⇒n =
0.5 / 1.5

⇒n = 0.33 (or) 33%
Thus, The porosity of soil sample is 33%.

16.  A burrow pit has a dry density of 17 kN/m3. How many cubic meters of this soil will be required to construct an embankment of 100 m3 volume with a dry density of 16 kN/m3?

[GATE 2000: 1 Mark]
  1. 94 m3
  2. 106 m3
  3. 100 m3
  4. 90 m3

Answer: Option A

Explanation:
Dry density of soil in embankment, Ɣdry_emb = 16 kN/m3
Also, Ɣdry_emb =

Wsoil / Vsoil

⇒ Wsoil = 16 x 100
⇒ Wsoil = 1600 kN
The weight of soil solids always remains constant.
Dry density of soil in burrow pit, Ɣdry_pit = 17 kN/m3
⇒ Ɣdry_pit =
Wsoil /Vsoil

⇒ Vsoil =
1600 /17

⇒ Vsoil = 94.11 m3
Therefore, the required volume of soil is 94.11 m3.

17.  The void ratio and specific gravity of a soil are 0.65 and 2.72 respectively. The degree of saturation (in percent) corresponding to water content of 20% is _____ .

[GATE 2000: 1 Mark]
  1. 65.3
  2. 20.9
  3. 83.7
  4. 54.4

Answer: Option C

Explanation:
Void ratio, e = 0.65
Specific gravity of soil, G = 2.72
water content, w = 20%
From, S e = w G
⇒ S =

w G /S

⇒ S =
0.2 x 2.72 /0.65

⇒ S = 0.837 (or) 83.7%
Thus, the degree of saturation of soil is 83.7%.

18.  The void ratios at the densest, loosest and the natural states of a sand deposit are 0.2, 0.6 and 0.4 respectively. The relative density of the deposit is _____ .

[GATE 2002: 1 Mark]
  1. 100%
  2. 75%
  3. 50%
  4. 25%

Answer: Option C

Explanation:
Relative Density, ID =

emax - e /emax - emin

⇒ ID =
0.6 - 0.4 /0.6 - 0.2

⇒ ID = 0.50 (or) 50%
Thus, the relative density of the deposit is 50%.

19.  The following data was obtained from a liquid test conducted on a soil sample.

Question data containing Number of blows and corresponding water content

The liquid limit of the soil is

[GATE 2002: 1 Mark]
  1. 63.1%
  2. 62.8%
  3. 61.9%
  4. 60.6%

Answer: Option C

Explanation:
Liquid limit is defined as the water content pertaining to 25 number of blows.
Thus, wL = 61.9%

20.  The undrained cohesion of a remoulded clay soil is 10 kN/m2. If the sensitivity of the clay is 20, the corresponding remoulded compressive strength is _____ .

[GATE 2004: 1 Mark]
  1. 5 kN/m2
  2. 10 kN/m2
  3. 20 kN/m2
  4. 200 kN/m2

Answer: Option C

Explanation:
Undrained Cohesion, Cu = 10kN/m2
Compressive strength, qu = 2Cu
⇒qu = 2 x 10
⇒qu = 20 kN/m2
Thus, Compressive strength of remoulded clay is 20 kN/m2

21.  The ratio of saturated unit weight to dry unit weight of a soil is 1.25. If the specific gravity of soilds (Gs) is 2.65, the void ratio of the soil is _____ .

[GATE 2004: 1 Mark]
  1. 0.625
  2. 0.663
  3. 0.944
  4. 1.325

Answer: Option B

Explanation:
Saturated unit weight,Ɣsat =

(G + e)Ɣw / 1 + e

Dry unit weight,Ɣdry =
G Ɣw / 1 + e

Now,
Ɣsat / Ɣdry
=
G + e / G

⇒1.25 =
2.65 + e / 2.65

⇒e = 0.663

22.  A saturated soil mass has a total density 22 kN/m3 and a water content of 10%. The bulk density and dry density of this soil are _____ .

[GATE 2005: 2 Marks]
  1. 12 kN/m3 and 20 kN/m3 respectively
  2. 22 kN/m3 and 20 kN/m3 respectively
  3. 19.8 kN/m3 and 19.8 kN/m3 respectively
  4. 23.2 kN/m3 and 19.8 kN/m3 respectively

Answer: Option B

Explanation:
Bulk density is also known as total density. Hence, Bulk density is 22 kN/m3.
Dry density, Ɣdry =

Ɣ / 1 + w

⇒ Ɣdry =
22 / 1 + 0.1

⇒ Ɣdry =
22 / 1.1

⇒ Ɣdry = 20 kN/m3

23.  The water content of a saturated soil and the specific gravity of soil solids were found to be 30% and 2.70 respectively. Assuming the unit weight of water to be 10 kN/m3, the saturated unit weight (in kN/m3) and the void ratio of the soil are _____.

[GATE 2007: 2 Marks]
  1. 19.4, 0.81
  2. 18.5, 0.30
  3. 19.4, 0.45
  4. 18.5, 0.45

Answer: Option A

Explanation:
Using S e = w G
⇒ 1 x e = 2.70 x 0.30
⇒ e = 0.81
Also, Ɣsat =

(G + e)Ɣw / 1 + e

⇒Ɣsat =
(2.7 + 0.81) x 10 / 1 + 0.81

⇒Ɣsat = 19.4 kN/m3

24.  The liquid limit (LL), plastic limit (PL) and shrinkage limit (SL) of a cohesive soil satisfy the relation.

[GATE 2008: 2 Marks]
  1. LL > PL < SL
  2. LL > PL > SL
  3. LL < PL < SL
  4. LL < PL > SL

Answer: Option B

Explanation: Self Explanatory

25.  A soil is composed of solid spherical grains of identical specific gravity and diameter between 0.075 mm and 0.0075 mm. If the terminal velocity of the largest particle falling through water without flocculation is 0.5 mm/s, that for the smallest particle would be _____ .

[GATE 2011: 1 Mark]
  1. 0.005 mm/s
  2. 0.05 mm/s
  3. 5 mm/s
  4. 50 mm/s

Answer: Option A

Explanation: According to Stoke's law,
Terminal velocity, V =

D2 x (Ɣsoil - Ɣw) / 18 η

⇒ V α D2
⇒
V1 / V2
=
0.0752 / 0.00752

⇒
0.5 / V2
=
0.0752 / 0.00752

⇒ V2 = 0.005 mm/s

26.  In its natural condition, a soil sample has a mass of 1.980 kg and a volume of 0.001 m3. After being completely dried in an oven, the mass of the sample is 1.800 kg. Specific gravity is 2.7. Unit weight of water is 10 kN/m3. The degree of saturation of soil is _____ .

[GATE 2013: 1 Mark]
  1. 0.65
  2. 0.70
  3. 0.54
  4. 0.61

Answer: Option C

Explanation:
Bulk density, ρ =

M / V
=
1.980 / 0.001
= 1980 kg/m3
Dry density, ρdry =
Mdry / V
=
1.800 / 0.001
= 1800 kg/m3
Also, ρdry =
ρ / 1 + w
= 1800 kg/m3
⇒ w = 0.1
Also, ρdry =
G ρw / 1 + e

⇒ 1800 =
2.7 x 1000 / 1 + e

⇒ e =0.5
Now, From the relation,
S e = w G
⇒ S x 0.5 = 0.1 x 2.7
⇒ S = 0.54

27.  A given cohesionless soil has emax = 0.85 and emin = 0.50. In the field, the soil is compacted to a mass density of 1800 kN/m3 at a water content of 8%. Take the mass density of water as 1000 kN/m3 and Gs as 2.7. The relative density (in %) of the soil is _____ .

[GATE 2014: 2 Marks, I Set]
  1. 56.43
  2. 60.25
  3. 62.87
  4. 65.71

Answer: Option D

Explanation:
Dry density, ρdry =

ρ / 1 + w

⇒ ρdry =
1800 / 1 + 0.08

⇒ ρdry = 1666.67 kg/m3
Also, ρdry = 1666.67 =
G ρw / 1 + e

⇒ 1666.67 =
2.7 x 1000 / 1 + e

⇒ e = 0.62
Thus, Relative Density ID =
emax - e / emax - emin

⇒ ID =
0.85 - 0.62 / 0.85 - 0.5

⇒ ID = 0.657

28.  A certain soil has the following properties: Gs = 2.71, n = 40% and w = 20%. The degree of saturation of the soil (rounded off to the nearest percent) is _____ .

[GATE 2014: 1 Mark, II Set]

Answer: 81%

Explanation:
From e=

n / 1 - n

⇒ e =
0.4 / 1 - 0.40

⇒ e = 0.67
Also, S e = w G
⇒ S x 0.67 = 0.20 x 2.71
⇒ S = 80.9% ≈ 81%

29.  A fine grained soil has 60% (by weight) silt content. The soil behaves fluid-like when the water content is more than 40%. The activity of the soil is _____ .

[GATE 2015: 1 Mark, I Set]
  1. 3.33
  2. 0.42
  3. 0.30
  4. 0.20

Answer: Option C

Explanation:
Activity, Ac =

Plasticity Index / % of particles finer than 2 micron

⇒Ac =
wL - wP / C

⇒Ac =
40 - 28 / 40

⇒Ac =
12 / 40

⇒Ac = 0.3

30.  An earth embankment is to be constructed with compacted cohesionless soil. The volume of the embankment is 5000 m3 and the target dry unit weight is 16.2 kN/m3. Three nearby sites (see figure below) have been identified from where the required soil can be transported to the construction site. The void ratio (e) of different sites are shown in the figure. Assume the specific gravity of soil to be 2.7 for all the three sites. If the cost of transportation per km is twice the cost of excavation per m3 of borrow pits, which site would you choose as the most economic solution? (use unit weight of water = 10 kN/m3).

Soil properties - Sample problem to depict dry density of soil is constant
[GATE 2015: 2 Marks, I Set]
  1. Site X
  2. Site Y
  3. Site Z
  4. Any of the sites

Answer: Option A

Explanation:
Dry density of soil in site, Ɣdry = 16.2 kN/m3
Also, Ɣdry =

Wsoil / Vsoil

⇒ Wsoil = 16.2 x 5000
⇒ Wsoil = 81000 kN
The weight of soil solids always remains constant.
Dry density of soil in Site X, Ɣdry_x =
G Ɣw /1 + e

⇒ Ɣdry_x =
2.7 x 10 /1 + 0.6

⇒ Ɣdry_x = 16.875 kN/m3
Also, Ɣdry_x =
Wsoil / Vsoil

⇒ 16.875 =
81000 / Vsoil

⇒ Vsoil = 4800 m3
Let 'z' be the cost of excavation of per m3 of soil.
Then, the cost of transportation of soil per km will be '2z'.
Total cost when site x is used = 4800z + (2z x 140) = 5080z
Similarly, total cost when site y is used = 5100z + (2z x 80) = 5260z
Similarly, total cost when site z is used = 4920z + (2z x 100) = 5120z
Thus, the most economic solution is to use Site X due to minimum total cost involved in excavation and transportation.

31.  If the water content of a fully saturated soil mass is 100%. The void ratio of the sample is _____ .

[GATE 2015: 1 Mark, II Set]
  1. Less than specific gravity of soil
  2. Equal to specific gravity of soil
  3. Greater than specific gravity of soil
  4. Independent of specific gravity of soil

Answer: Option B

Explanation:
For a saturated sample, Degree of saturation (S) = 1
Using S e = w G
⇒ 1 x e = 1 x G
⇒ e = G

32.  A 588 cm3 volume of moist sand weighs 1010 gm. Its dry weight is 918 gm and specific gravity of solids, Gs is 2.67. Assuming density of water as 1 gm/cm3, the void ratio is _____.

[GATE 2015: 2 Marks, II Set]

Answer: 0.71

Explanation:
Volume of soil, Vsoil = 588 cm3
Mass of dry soil, Msoil = 918 gm
Dry density of soil, ρdry =

Msoil / Vsoil

⇒ ρdry =
918 / 588

⇒ ρdry = 1.561 g/cc
Also, ρdry = 1.561 =
G Ɣw / 1 + e

⇒ 1.561 =
2.67 x 1 / 1 + e

⇒ 1.561 + 1.561e = 2.67
⇒ e = 0.71

33.  The porosity (n) and the degree of saturation (S) of a soil sample are 0.70 and 40% respectively. In a 100 m3 volume of the soil, the volume (expressed in m3) of air is _____.

[GATE 2016: 2 Marks, I Set]

Answer: 42

Explanation:
Porosity, n =

Volume of voids / Total volume

⇒ n =
Vv / V

⇒ 0.7 =
Vv / 100

⇒ Vv = 70 m3
Also, Degree of saturation, S =
Vw / Vv

⇒ S =
Vv - Vair / Vv

⇒ 0.40 =
70 - Vair / 70

⇒ 0.40 x 70 = 70 − Vair
⇒ Vair = 42 m3

34.  The laboratory tests on a soil sample yield the following results: natural water content = 18%, liquid limit = 60%, plastic limit = 25%, percentage of clay sized fraction = 25%. The liquidity index and activity (as per the expression proposed by Skempton) of the soil, respectively, are _____ .

[GATE 2017: 2 Marks, I Set]
  1. -0.2 and 1.4
  2. 0.2 and 1.4
  3. -1.2 and 0.714
  4. 1.2 and 0.714

Answer: Option A

Explanation:
Liquidity Index, IL =

wnat - wP / wL - wP

⇒ IL =
18 - 25 / 60 - 25

⇒ IL =
-7 / 35

⇒ IL = - 0.2
Also, Activity, Ac =
Plasticity Index / % of particles finer than 2 micron

⇒Ac =
wL - wP / C

⇒Ac =
60 - 25 / 25

⇒Ac =
35 / 25

⇒Ac = 1.4

35.  Let G be the specific gravity of soil solids, w is the water content in the soil sample, Ɣw is the unit weight of water and Ɣd is the dry unit weight of the soil. The equation for the zero air voids line in a compaction test plot is _____ .

[GATE 2017: 1 Mark, II Set]
  1. Ɣd =
    G Ɣw / 1 + wG
  2. Ɣd =
    G Ɣw / Gw
  3. Ɣd =
    G w / 1 + Ɣw
  4. Ɣd =
    G w / 1 – Ɣw

Answer: Option A

Explanation:
Equation of Air voids line, Ɣdry = (1 - na)

G Ɣw / 1 + wG

Incase of Zero air void line, na = 0
⇒ Ɣdry =
G Ɣw / 1 + wG

36.  In a shrinkage limit test, the volume and mass of a dry soil pat are found to be 50 cm3 and 88 gm respectively. The specific gravity of soil solids is 2.71 and the density of water is 1 g/cc. The shrinkage limit (in % upto two decimal places) is _____ .

[GATE 2018: 1 Mark, I Set]

Answer: 19.90%

Explanation:
Dry density of soil, ρdry =

Msoil / Vsoil

⇒ ρdry =
88 / 50

⇒ ρdry = 1.76 gm/cc
From the existing relation,
Specific Gravity, G =   
1 /
ρw / ρdry
-
wS / 100

⇒ 2.71 =   
1 /
1 / 1.76
-
wS / 100

⇒ wS = 0.1990 (or) 19.90%

37.  The clay mineral, whose structural units are held together by potassium bond is _____ .

[GATE 2018: 1 Mark, II Set]
  1. Halloysite
  2. Illite
  3. Kaolinite
  4. Smectite

Answer: Option B

Explanation:
Illite is the clay mineral in which the units are held together by means of Potassium bond.

38.  A soil has specific gravity of solids equal to 2.65. The mass density of water is 1000 kg/m3. Considering zero air voids and 10% moisture content of the soil sample, the dry density (in kg/m3, round off to 1 decimal place) would be _____ .

[GATE 2019: 1 Mark, I Set]

Answer: 2094.9

Explanation:
The equation of zero air void line is Ɣdry = (1 - na)

G Ɣw / 1 + wG

⇒ Ɣdry = (1 - 0)
2.65 x 1000 / 1 + (0.1 x 2.65)

⇒ Ɣdry =
2650 / 1.265

⇒ Ɣdry = 2094.86 kg/m3

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