Previous Years GATE Questions - Soil Compaction

1.   The zero-air voids curve is non-linear owing to _____ .

[GATE 1992: 1 Mark]
  1. the standard proctor test data of dry density and corresponding water content plotting as a non-linear curve
  2. the dry density at 100% saturation being a non-linear function of the void ratio
  3. the water content altered during compaction
  4. the soil being compacted with an odd number of blows

Answer: Option B

Explanation:
As γdry is proportional to

1 / 1 + e
, the dry density of soil is a non-linear function of the void ratio.

2.  The unit weight of a soil at zero air voids depends on _____ .

[GATE 1995: 1 Mark]
  1. specific gravity
  2. water content
  3. unit weight of water
  4. All of the above

Answer: Option D

Explanation:
γdry =

G γw / 1 + e

γbulk =
γdry / 1 + w

Thus, γbulk depends on G, γw and w.

3.  The measure of soil compaction is its wet density. Is it True or False?

[GATE 1995: 1 Mark]

Answer: False

Explanation:
The measure of soil compaction is its dry density. Hence, the given statement is FALSE.

4.  In a compaction test, as the compaction effort is increased, the optimum moisture content _____ .

[GATE 1997: 1 Mark]
  1. decreases
  2. remains same
  3. increases
  4. increases first and then decreases

Answer: Option A

Explanation:

>Soil Compaction - Graph indicating the relationship between water content and dry density of soil

The increased compactive effort indicates that modified Proctor is used for compaction. As the compactive effort is increased, the Optimum Moisture Content decreases.

5.  Compaction of an embankment is carried out in 500 mm thick layers. The rammer used for compaction has a foot area of 0.05 m2 and the energy imparted in every drop of rammer is 400 Nm. Assuming 50% more energy in each pass over the compacted area due to overlap, the number of passes required to develop compactive energy equivalent to Indian Standard light compactin for each layer would be _____ .

[GATE 2003: 2 Marks]
  1. 10
  2. 16
  3. 20
  4. 26

Answer: Option D

Explanation:
Considering 50% more energy in each drop of rammer,
Total Energy imparted by rammer = 1.50 x Energy imparted in every drop of rammer = 1.5 x 400 = 600 Nm.
The volume of soil compacted by rammer = Foot area x layer thickness = 0.05 x 104 x 500 x 10-1 = 25000 cc
Now, For Indian Standard Light Compaction Test,
2.6 kg of hammer is allowed to fall from a height of 31 cm. Also, soil is compacted in 3 layers with each layer tamped 25 times. The volume of soil considered is 1000 cc.
Thus, Energy imparted in Light Compaction Test = (2.6 x 9.81) x (31 x 10-2) x 3 x 25 = 593.01 Nm.
Now, comparing energy from rammer and Light Compaction Test,
n x

600 / 25000
=
593.01 / 1000

⇒ n = 24.71 ≈ 25.
Thus, the number of blows required from rammer inorder to produce the same compactive energy as that of light compaction test is 25.
NOTE:
In any of the compaction tests, the energy imparted =
Weight of hammer x Height of fall x Number of layers x Number of blows / Volume of soil

6.  A clayey soil has a maximum dry density of 16 kN/m3 and optimum moisture content of 12%. A contractor during the construction of core of an earth dam obtained the dry density 15.2 kN/m3 and water content 11%. This construction is acceptable because _____ .

[GATE 2005: 1 Mark]
  1. the density is less than the maximum dry density and water content is on dry side of optimum
  2. the compaction density is very low and water content is less than 12%
  3. the compaction is done on the dry side of the optimum
  4. both the dry density and water content of the compacted soil are within the desirable limits

Answer: Option A

Explanation: Self Explanatory

7.  In a standard proctor test, 1.8 kg of moist soil was filling the mould (volume = 944 cc) after compaction. A soil sample weighing 23 g was taken from the mould and ovendried for 24 hours at a temperature of 110°C. Weight of the dry sample was found to be 20 g. Specific gravity of soil solids is G = 2.7. The theoretical maximum value of the dry unit weight of the soil at that water content is equal to _____ .

[GATE 2006: 2 Marks]
  1. 4.67 kN/m3
  2. 11.5 kN/m3
  3. 16.26 kN/m3
  4. 18.85 kN/m3

Answer: Option D

Explanation:
Bulk density of soil, γ =

Mass / Volume
=
18 x 1000 / 944
= 1.907 g/cc
water content of soil, w =
Mass of water / Mass of soil
=
23 - 20 / 20
= 0.15
For, theoretical maximum dry unit weight,
Degree of saturation, S = 100%
⇒ na = 0 (Since, s + na = 1)
From the existing relation,
S e = w G
⇒ 1 x e = 0.15 x 2.7
⇒ e = 0.405
Now, Theoretical Maximum Dry Unit Weight of soil, γd(max) =
(1 - na) G γw / 1 + e
=
2.7 x 9.81 / 1 + 0.405

⇒ γd(max) = 18.85 kN/m3

8.  Compaction by vibratory roller is the best method of compaction in case of _____ .

[GATE 2008: 1 Mark]
  1. moist silty sand
  2. well graded dry sand
  3. clay of medium compressibility
  4. silt of high compressibility

Answer: Option B

Explanation: Self Explanatory

9.  Deposit with flocculated structure is formed when _____ .

[GATE 2009: 1 Mark]
  1. clay particles settle on sea bed
  2. clay particles settle on fresh water lake bed
  3. sand particles settle on river bed
  4. sand particles settle on sea bed

Answer: Option A

Explanation:
Flocculated particles are found in clay minerals. Also, Sea bed is polar in nature.

10.  In a compaction test, G, w, S and e represent the specific gravity, water content, degree of saturation and void ratio of the soil sample respectively. If γw represents the unit weight of water and γd represents the dry unit weight of the soil, the equation for zero air voids line is _____ .

[GATE 2010: 1 Mark]
  1. γd =
    G γw / 1 + Se
  2. γd =
    G γw / 1 + Gw
  3. γd =
    Gw / 1 + γwS
  4. γd =
    Gw / 1 + Se

Answer: Option B

Explanation: Self Explanatory

11.  Two series of compaction tests were performed in the laboratory on an inorganic clayey soil employing two different level of compaction energy per unit volume of soil. With regard to the above tests, the following two statements are made
I. The optimum moisture content is expected to be more for the tests with higher energy.
II. The maximum dry density is expected to be more for the tests with higher energy.
The CORRECT option evaluating the above statement is

[GATE 2012: 1 Mark]
  1. Only I is TRUE
  2. Only II is TRUE
  3. Both I and II are TRUE
  4. Neither I nor II is TRUE

Answer: Option B

Explanation:

Soil Compaction - Graph indicating the relationship between water content and dry density of soil

From the graph, it can be seen that as compactive effort is increased,
i. Optimum Moisture Content is decreased.
ii. Maximum Dry Density is increased.

12.  Following statement are made on compacted soil, where DS stands for soil compaction on Dry side of OMC and WS stands for soil compacted on wet side of OMC. Identify incorrect statement.

[GATE 2013: 1 Mark]
  1. Soil structure is flocculated on DS and dispersed on WS
  2. Construction of pre water pressure is low on DS and high on WS
  3. Soil on drying, shrinkage is high on DS and low on WS
  4. On addition of water, swelling is high on DS and low on WS

Answer: Option C

Explanation: Self Explanatory

13.  OMC-SP and MDD-SP denote the optimum moisture content and maximum dry density obtained from Standard Proctor compaction test, respectively. OMC-MP and MDD-MP denote the optimum moisture content and maximum dry density obtained from the Modified Proctor compaction test, respectively. Which one of the following is correct?

[GATE 2016: 2 Marks, Set II]
  1. OMC-SP < OMC-MP and MDD-SP < MDD-MP
  2. OMC-SP > OMC-MP and MDD-SP < MDD-MP
  3. OMC-SP < OMC-MP and MDD-SP > MDD-MP
  4. OMC-SP > OMC-MP and MDD-SP > MDD-MP

Answer: Option B

Explanation:
Soil Compaction - Graph indicating the relationship between water content and dry density of soil
From the graph,
i. Incase of Optimum Moisture Content, SP > MP
ii. Incase of Maximum Dry Density, SP < MP.

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