# 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.

**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

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 =

^{3}

Thus, Volume of soil solids = Volume of single soil particle x number of particles

⇒ V

_{soil}= (d/2)

^{3}x

^{3}

⇒ V

_{soil}= 0.523

Total volume of voids, V

_{voids}= V - V

_{soil}

⇒ V

_{voids}= 1 - 0.523 = 0.477

Thus void ratio in loosest state,

e

_{max}=

_{voids}/ V

_{soil}

⇒ e

_{max}=

⇒ 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,

V

_{soil}= Π/6 = 0.523

Total volume, V = (1 - cos α)√(1 + 2 cos α)

Here, α =

⇒ V = 0.707 m

^{3}

Now, Volume of voids, V

_{voids}= V - V

_{soil}

⇒ V

_{voids}= 0.707 - 0.523

⇒ V

_{voids}= 0.184 m

^{3}

Thus void ratio in densest state,

e

_{min}=

_{voids}/ V

_{soil}

⇒ e

_{min}=

⇒ e

_{min}= 0.35

## 2. A saturated sand sample has dry unit weight of 18kN/m^{3} and a specific gravity of 2.65. If Æ”_{w} = 10 kN/m^{3}, the water content of the soil is _____ .

**Answer:** 0.17

**Explanation:**

Dry unit weight, Æ”_{d} = 18 kN/m^{3}

Specific Gravity of, G = 2.65

Unit weight of water, Æ”_{w} = 10 kN/m^{3}

From the existing relations,

Æ”_{d} =

_{w}/ 1 + e

⇒18 =

⇒18 x (1 + e) = 2.65 x 10

⇒e =

⇒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 =

⇒ w = 0.17

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

**Answer:** Plastic

**Explanation:**

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 _____ .

**Answer:** 0.5

**Explanation:**

By using the existing relation,

n =

⇒n =

⇒n =

⇒n = 0.5

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

- Water content
- Particle size distribution
- Density index
- 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?

**Answer:** True

**Explanation:** Void Ratio, e =

Hence, Void ratio can be greater than unity.

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

- e =n / 1 – n
- Æ”
_{sat}=(G + e)Æ”_{w}/ 1 + e - n =e / 1 – e
- 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).

- LL – w / PI
- w – LL / PI
- LL – PI
- 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 _____ .

- 0.20
- 0.80
- 1.00
- 0.25

**Answer:** Option D

**Explanation:**

Porosity, n = 20% = 0.20

Using, n =

⇒0.20 + 0.20 e = e

⇒0.20 = ( 1 - 0.20) e

⇒e =

⇒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 _____ .

- Activity
- Damping
- Plasticity
- Sensitivity

**Answer:** Option D

**Explanation:** Definition of Sensitivity

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

- Large volume change
- Moderate volume change
- Low volume change
- 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 _____.

- Equilibrium of floating bodies
- Archimedes principle
- Stokeâ€™s law
- Darcyâ€™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.

Æ”_{sub} = Æ”_{sat} - Æ”_{w}

## 13. A soil sample in its natural state has mass of 2.290 kg and a volume of 1.15 x 10^{-3} m^{3}. 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 _____.

- 0.40
- 0.45
- 0.55
- 0.53

**Answer:** Option B

**Explanation:**

Specific Gravity of soil, G = 2.68

⇒ G =

_{w}

⇒ 2.68 =

⇒ Æ” = 2.68 x 9.81 = 26.29 kN/m

^{3 ⇒ Æ” = 26.29 x 103 = Msoil x Æ”w / 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 _____ .

- Plasticity Index / Flow Index
- Liquid Limit / Plastic Limit
- Liquidity Index / Plastic Limit
- 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 _____ .

- 50%
- 66%
- 100%
- 33%

**Answer:** Option D

**Explanation:**

From n =

⇒n =

⇒n =

⇒n = 0.33 (or) 33%

Thus, The porosity of soil sample is 33%.

## 16. A burrow pit has a dry density of 17 kN/m^{3}. How many cubic meters of this soil will be required to construct an embankment of 100 m^{3} volume with a dry density of 16 kN/m^{3}?

- 94 m
^{3} - 106 m
^{3} - 100 m
^{3} - 90 m
^{3}

**Answer:** Option A

**Explanation:**

Dry density of soil in embankment, Æ”_{dry_emb} = 16 kN/m^{3}

Also, Æ”_{dry_emb} =

_{soil}/ V

_{soil}

⇒ W

_{soil}= 16 x 100

⇒ W

_{soil}= 1600 kN

The weight of soil solids always remains constant.

Dry density of soil in burrow pit, Æ”

_{dry_pit}= 17 kN/m

^{3}

⇒ Æ”

_{dry_pit}=

_{soil}/V

_{soil}

⇒ V

_{soil}=

⇒ V

_{soil}= 94.11 m

^{3}

Therefore, the required volume of soil is 94.11 m

^{3}.

## 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 _____ .

- 65.3
- 20.9
- 83.7
- 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 =

⇒ S =

⇒ 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 _____ .

- 100%
- 75%
- 50%
- 25%

**Answer:** Option C

**Explanation:**

Relative Density, I_{D} =

_{max}- e /e

_{max}- e

_{min}

⇒ I

_{D}=

⇒ I

_{D}= 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.

## The liquid limit of the soil is

- 63.1%
- 62.8%
- 61.9%
- 60.6%

**Answer:** Option C

**Explanation:**

Liquid limit is defined as the water content pertaining to 25 number of blows.

Thus, w_{L} = 61.9%

## 20. The undrained cohesion of a remoulded clay soil is 10 kN/m^{2}. If the sensitivity of the clay is 20, the corresponding remoulded compressive strength is _____ .

- 5 kN/m
^{2} - 10 kN/m
^{2} - 20 kN/m
^{2} - 200 kN/m
^{2}

**Answer:** Option C

**Explanation:**

Undrained Cohesion, C_{u} = 10kN/m^{2}

Compressive strength, q_{u} = 2C_{u}

⇒q_{u} = 2 x 10

⇒q_{u} = 20 kN/m^{2}

Thus, Compressive strength of remoulded clay is 20 kN/m^{2}

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

- 0.625
- 0.663
- 0.944
- 1.325

**Answer:** Option B

**Explanation:**

Saturated unit weight,Æ”_{sat} =

_{w}/ 1 + e

Dry unit weight,Æ”

_{dry}=

_{w}/ 1 + e

Now,

_{sat}/ Æ”

_{dry}

⇒1.25 =

⇒e = 0.663

## 22. A saturated soil mass has a total density 22 kN/m^{3} and a water content of 10%. The bulk density and dry density of this soil are _____ .

- 12 kN/m
^{3}and 20 kN/m^{3}respectively - 22 kN/m
^{3}and 20 kN/m^{3}respectively - 19.8 kN/m
^{3}and 19.8 kN/m^{3}respectively - 23.2 kN/m
^{3}and 19.8 kN/m^{3}respectively

**Answer:** Option B

**Explanation:**

Bulk density is also known as total density. Hence, Bulk density is 22 kN/m^{3}.

Dry density, Æ”_{dry} =

⇒ Æ”

_{dry}=

⇒ Æ”

_{dry}=

⇒ Æ”

_{dry}= 20 kN/m

^{3}

## 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/m^{3}, the saturated unit weight (in kN/m^{3}) and the void ratio of the soil are _____.

- 19.4, 0.81
- 18.5, 0.30
- 19.4, 0.45
- 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, Æ”_{sat} =

_{w}/ 1 + e

⇒Æ”

_{sat}=

⇒Æ”

_{sat}= 19.4 kN/m

^{3}

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

- LL > PL < SL
- LL > PL > SL
- LL < PL < SL
- 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 _____ .

- 0.005 mm/s
- 0.05 mm/s
- 5 mm/s
- 50 mm/s

**Answer:** Option A

**Explanation:** According to Stoke's law,

Terminal velocity, V =

^{2}x (Æ”

_{soil}- Æ”

_{w}) / 18 η

⇒ V α D

^{2}

⇒

_{1}/ V

_{2}

^{2}/ 0.0075

^{2}

⇒

_{2}

^{2}/ 0.0075

^{2}

⇒ V

_{2}= 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 m^{3}. 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/m^{3}. The degree of saturation of soil is _____ .

- 0.65
- 0.70
- 0.54
- 0.61

**Answer:** Option C

**Explanation:**

Bulk density, ρ =

^{3}

Dry density, ρ

_{dry}=

_{dry}/ V

^{3}

Also, ρ

_{dry}=

^{3}

⇒ w = 0.1

Also, ρ

_{dry}=

_{w}/ 1 + e

⇒ 1800 =

⇒ 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 e_{max} = 0.85 and e_{min} = 0.50. In the field, the soil is compacted to a mass density of 1800 kN/m^{3} at a water content of 8%. Take the mass density of water as 1000 kN/m^{3} and G_{s} as 2.7. The relative density (in %) of the soil is _____ .

- 56.43
- 60.25
- 62.87
- 65.71

**Answer:** Option D

**Explanation:**

Dry density, ρ_{dry} =

⇒ ρ

_{dry}=

⇒ ρ

_{dry}= 1666.67 kg/m

^{3}

Also, ρ

_{dry}= 1666.67 =

_{w}/ 1 + e

⇒ 1666.67 =

⇒ e = 0.62

Thus, Relative Density I

_{D}=

_{max}- e / e

_{max}- e

_{min}

⇒ I

_{D}=

⇒ I

_{D}= 0.657

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

**Answer:** 81%

**Explanation:**

From e=

⇒ e =

⇒ 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 _____ .

- 3.33
- 0.42
- 0.30
- 0.20

**Answer:** Option C

**Explanation:**

Activity, A_{c} =

⇒A

_{c}=

_{L}- w

_{P}/ C

⇒A

_{c}=

⇒A

_{c}=

⇒A

_{c}= 0.3

## 30. An earth embankment is to be constructed with compacted cohesionless soil. The volume of the embankment is 5000 m^{3} and the target dry unit weight is 16.2 kN/m^{3}. 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 m^{3} of borrow pits, which site would you choose as the most economic solution? (use unit weight of water = 10 kN/m^{3}).

- Site X
- Site Y
- Site Z
- Any of the sites

**Answer:** Option A

**Explanation:**

Dry density of soil in site, Æ”_{dry} = 16.2 kN/m^{3}

Also, Æ”_{dry} =

_{soil}/ V

_{soil}

⇒ W

_{soil}= 16.2 x 5000

⇒ W

_{soil}= 81000 kN

The weight of soil solids always remains constant.

Dry density of soil in Site X, Æ”

_{dry_x}=

_{w}/1 + e

⇒ Æ”

_{dry_x}=

⇒ Æ”

_{dry_x}= 16.875 kN/m

^{3}

Also, Æ”

_{dry_x}=

_{soil}/ V

_{soil}

⇒ 16.875 =

_{soil}

⇒ V

_{soil}= 4800 m

^{3}

Let 'z' be the cost of excavation of per m

^{3}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 _____ .

- Less than specific gravity of soil
- Equal to specific gravity of soil
- Greater than specific gravity of soil
- 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 cm^{3} volume of moist sand weighs 1010 gm. Its dry weight is 918 gm and specific gravity of solids, G_{s} is 2.67. Assuming density of water as 1 gm/cm^{3}, the void ratio is _____.

**Answer:** 0.71

**Explanation:**

Volume of soil, V_{soil} = 588 cm^{3}

Mass of dry soil, M_{soil} = 918 gm

Dry density of soil, ρ_{dry} =

_{soil}/ V

_{soil}

⇒ ρ

_{dry}=

⇒ ρ

_{dry}= 1.561 g/cc

Also, ρ

_{dry}= 1.561 =

_{w}/ 1 + e

⇒ 1.561 =

⇒ 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 m^{3} volume of the soil, the volume (expressed in m^{3}) of air is _____.

**Answer:** 42

**Explanation:**

Porosity, n =

⇒ n =

_{v}/ V

⇒ 0.7 =

_{v}/ 100

⇒ V

_{v}= 70 m

^{3}

Also, Degree of saturation, S =

_{w}/ V

_{v}

⇒ S =

_{v}- V

_{air}/ V

_{v}

⇒ 0.40 =

_{air}/ 70

⇒ 0.40 x 70 = 70 − V

_{air}

⇒ V

_{air}= 42 m

^{3}

## 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 _____ .

- -0.2 and 1.4
- 0.2 and 1.4
- -1.2 and 0.714
- 1.2 and 0.714

**Answer:** Option A

**Explanation:**

Liquidity Index, I_{L} =

_{nat}- w

_{P}/ w

_{L}- w

_{P}

⇒ I

_{L}=

⇒ I

_{L}=

⇒ I

_{L}= - 0.2

Also, Activity, A

_{c}=

⇒A

_{c}=

_{L}- w

_{P}/ C

⇒A

_{c}=

⇒A

_{c}=

⇒A

_{c}= 1.4

## 35. Let G be the specific gravity of soil solids, w is the water content in the soil sample, Æ”_{w} is the unit weight of water and Æ”_{d} is the dry unit weight of the soil. The equation for the zero air voids line in a compaction test plot is _____ .

- Æ”
_{d}=G Æ”_{w}/ 1 + wG - Æ”
_{d}=G Æ”_{w}/ Gw - Æ”
_{d}=G w / 1 + Æ”_{w} - Æ”
_{d}=G w / 1 – Æ”_{w}

**Answer:** Option A

**Explanation:**

Equation of Air voids line, Æ”_{dry} = (1 - n_{a})

_{w}/ 1 + wG

Incase of Zero air void line, n

_{a}= 0

⇒ Æ”

_{dry}=

_{w}/ 1 + wG

## 36. In a shrinkage limit test, the volume and mass of a dry soil pat are found to be 50 cm^{3} 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 _____ .

**Answer:** 19.90%

**Explanation:**

Dry density of soil, ρ_{dry} =

_{soil}/ V

_{soil}

⇒ ρ

_{dry}=

⇒ ρ

_{dry}= 1.76 gm/cc

From the existing relation,

Specific Gravity, G =

_{w}/ ρ

_{dry}

_{S}/ 100

⇒ 2.71 =

_{S}/ 100

⇒ w

_{S}= 0.1990 (or) 19.90%

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

- Halloysite
- Illite
- Kaolinite
- 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/m^{3}. Considering zero air voids and 10% moisture content of the soil sample, the dry density (in kg/m^{3}, round off to 1 decimal place) would be _____ .

**Answer:** 2094.9

**Explanation:**

The equation of zero air void line is Æ”_{dry} = (1 - n_{a})

_{w}/ 1 + wG

⇒ Æ”

_{dry}= (1 - 0)

⇒ Æ”

_{dry}=

⇒ Æ”

_{dry}= 2094.86 kg/m

^{3}