Previous Years GATE Questions - Soil Compaction

1.   The maximum possible value of Group Index for a soil is _____ .

[GATE 1991: 2 Marks]

Answer: 20

Explanation: The Group Index values ranges from 0 to 20.

2.  The charge on kaolinite is due to one aluminium substitution for every four hundredth silicon ion. Is it True or False?

[GATE 1992: 1 Mark]

Answer: False

Explanation: Substitution of silica by aluminium does not occur in Kaolinite. Hence, the given statement is False.

3.  The description 'sandy silty clay' signifies that.

[GATE 1992: 1 Mark]
  1. the soil contains unequal proportions of the three constituents, in the order, sand > silt > clay.
  2. there is no information regarding the relative proportions of the three.
  3. the soil contains unequal proportions of the three constituents, in the order, clay > silt > sand.
  4. the soil contains equal proportions of sand, silt and clay.

Answer: Option C

Explanation: The constituent which is present in lesser proportion ends with '-y'. Thus, Here silt and sand will be present in lesser proportion in comparison with clay. Hence, it will be appropriate to indicate that the soil contains unequal proportions of the three constituents such that clay > silt > sand.

4.  The 'A' line in the plasticity chart separates organic clays from inorganic clays. Is it True or False?

[GATE 1992: 1 Mark]

Answer: True

Explanation: The soil present above A-line is inorganic and the soil present below the A-line is organic. Thus, the given statement is True.

5.  A soil having a uniformity coefficient smaller than 2 is considered as 'uniform'. Is it True or False?

[GATE 1992: 1 Mark]

Answer: True

Explanation: Self Explanatory

6.  The swelling nature of black cotton soil is primarily due to the presence of _____ .

[GATE 1993: 1 Mark]
  1. Kaolinite
  2. Illite
  3. Montmorillonite
  4. Vermiculite

Answer: Option C

Explanation: Montmorillonite has large affinity for water. Black cotton soil contains Montmorillonite which is responsible for its swelling action.

7.  A soil having particles of nearly the same size is known as _____ .

[GATE 1995: 1 Mark]
  1. well graded
  2. uniformly graded
  3. poorly graded
  4. gap graded

Answer: Option B

Explanation: Self Explanatory

8.  The equation of the 'A' line in the Plasticity chart is _____ .

[GATE 1995: 1 Mark]

Answer: IP = 0.73 (wL - 20)

Explanation: Self Explanatory

9.  Soils transported by wind are known as _____ soils .

[GATE 1995: 1 Mark]

Answer: Aeolian

Explanation: Self Explanatory

10.  The particle size distribution of curves are extremely useful for the classification of _____ .

[GATE 1996: 1 Mark]
  1. fine grained soils
  2. coarse grained soils
  3. both fine grained and coarse grained soils
  4. silts and clays

Answer: Option B

Explanation: The properties of coarse grained soils depends mainly on particle size and hence particle size distribution curve is useful for the classification of coarse grained soils.

11.  The shape of clay particle is usually _____ .

[GATE 1997: 1 Mark]
  1. angular
  2. flaky
  3. tubular
  4. rounded

Answer: Option B

Explanation: Self Explanatory

12.  Some of the structural strength of a clayey material that is lost by remoulding is slowly recovered with time. This property of soils to undergo an isothermal gel-to-sol-to-gel transformation upon agitation and subsequent rest is termed as _____ .

[GATE 1998: 1 Mark]
  1. Isotropy
  2. Anisotropy
  3. Thixotropy
  4. Allotropy

Answer: Option C

Explanation: When a clay is remoulded, its strength gets reduced even at the same water content. The phenomenon by which the strength of remoulded clay increases with passage of time without change in water content is called as Thixotropy. The increase in strength will be lesser than the original strength.

13.  The values of liquid limit and plasticity index for soils having common geological origin in a restricted locality usually define _____.

[GATE 1999: 1 Mark]
  1. a zone above A line
  2. a straight line parallel to A line
  3. a straight line perpendicular to A line
  4. points may be anywhere in the plasticity chart

Answer: Option B

Explanation: Self Explanatory

14.  Data from sieve analysis conducted on a given sample of soil showed that 67% of the particles passed through 75 micron IS sieve. The liquid limit and plastic limit of the finer fraction was found to be 45 and 33 percent respectively. The group symbol of the given soil as per IS 1498-1970 is _____ .

[GATE 2002: 1 Mark]
  1. SC
  2. MI
  3. CH
  4. MH

Answer: Option B

Explanation:
Here, more than 50% of the particles passes through 75 μ IS sieve.
⇒ Soil sample is either Silt (M) or clay (C).
Now, Equation of A-line, IP = 0.73(wL - 20)
⇒ IP = 0.73(45 - 20) = 18.25
Plasticity index of soil sample, IP = wL - wP = 45 - 33 = 12 < 18.25
⇒ Soil sample lies below A-line. Hence the soil sample is inorganic (Silt - M).
Now, Liquid limit of soil sample is, wL = 45.
⇒ 35 < wL < 50
⇒ The compressibility of soil sample is Medium (I).
Thus, the soil sample is MI.

15.  A soil mass contains 40% gravel, 50% sand and 10% silt. This soil can be classified as _____ .

[GATE 2005: 2 Marks]
  1. silty sandy gravel having coefficient of uniformity less than 60.
  2. silty gravelly sand having coefficient of uniformity equal to 10.
  3. gravelly silty sand having coefficient of uniformity greater than 60.
  4. gravelly silty sand and its coefficient of uniformity cannot be determined.

Answer: Option C

Explanation:
The constituent with major proportion is to be written at the end. Hence, Sand should be written at the end.
Among the minor proportions (Gravel and Silt), maximum proportion is to be written at first (with an '-y' at the end - Gravelly)
Since silt is also present in minor proportion, it is written with '-y' at the end.
Thus, Gravelly Silty Sand is the exact nomenclature. The option which supports this is Option C.

16.  Laboratory sieve analysis was carried out on a soil sample using a complete set of standard IS sieves. Out of 500 g of soil used in the test, 200 g was retained on IS 600 μ sieve, 250 g was retained on IS 500 μ sieve and the remaining 50 g was retained on IS 425 μ sieve. The coefficient of uniformity of the soil is _____ .

[GATE 2006: 2 Marks]
  1. 0.9
  2. 1.0
  3. 1.1
  4. 1.2

Answer: Option D

Explanation:

soil classification - Data pertaining to a soil sample subjected to sieve analysis

Here, D60 = 600 μ and D10 = 500 μ
Now, Uniformity Coefficient, CU =
D60 / D10
=
600 μ / 500 μ

⇒ CU = 1.2

17.  Laboratory sieve analysis was carried out on a soil sample using a complete set of standard IS sieves. Out pf 500 g of soil used in the test, 200 g was retained on IS 600 μ sieve, 250 g was retained on IS 500 μ sieve and the remaining 50 g was retained on IS 425 μ sieve. The classification of the soil is _____ .

[GATE 2006: 2 Marks]
  1. SP
  2. SW
  3. GP
  4. GW

Answer: Option A

Explanation:

soil classification - Data pertaining to a soil sample subjected to sieve analysis

Here, 60% of the soil sample passes through 600 μ(which is less than 4.75mm). Hence, the soil sample is Sand (S).
Also, D60 = 600 μ and D10 = 500 μ
Now, Uniformity Coefficient, CU =
D60 / D10
=
600 μ / 500 μ

⇒ CU = 1.2 < 6
⇒ Soil is poorly graded.
Thus, thus soil sample is Poorly Graded Sand.

18.  The sieve analysis on a dry soil sample of mass 1000 g showed that 980 g and 270 g of soil pass through 4.75 mm and 0.075 mm sieve, respectively. The liquid limit and plastic limit of the soil fraction passing through 425 μ sieves are 40% and 18%, respectively. The soil may be classified as _____ .

[GATE 2007: 2 Marks]
  1. SC
  2. MI
  3. CI
  4. SM

Answer: Option A

Explanation:
Percentage of particle passing through 75 μ sieve =

270 / 1000
= 27%

⇒ Percentage of particle retained through 75 μ sieve = 100 - 27 = 63%
⇒ Soil sample is Coarse grained (i.e) Either Sand (S) or Gravel (G).
Now, Percentage of particle passing through 4.75 mm sieve =
980 / 1000
= 98%

⇒ Percentage of particle retained through 4.75 mm sieve = 100 - 98 = 2%
⇒ Soil sample is Sand (S).
Also, Percentage finer = 18% > 12%
⇒ Soil sample is clay(C).
Thus, the soil sample is designated as SC.

19.  Group symbols assigned to silty sand and clayey sand are _____ respectively.

[GATE 2008: 1 Mark]
  1. SS and CS
  2. SM and CS
  3. SM and SC
  4. MS and CS

Answer: Option C

Explanation:
'S' denotes 'Sand'. 'M' denotes 'Silt'. 'C' denotes 'Clay'.
The constituent in minor proportion is to be mentioned at end.
Thus, Silty sand - SM and Clayey sand - SC.

20.  The laboratory tests results of a soil sample are given below:
Percentage finer than 4.75 mm = 60
Percentage finer than 0.075 mm = 30
Liquid limit = 35%
Plastic limit = 27%
The soil classification is _____ .

[GATE 2009: 2 Marks]
  1. GM
  2. SM
  3. GC
  4. ML – MI

Answer: Option B

Explanation:
Percentage of particle finer than 75 μ sieve = 30% < 50%
⇒ Soil sample is Coarse grained (i.e) either Sand (S) or gravel (G).
Now, Percentage of particle finer than 4.75 mm sieve = 60% > 50%
⇒ Soil sample is Sand (S).
Now, Equation of A-line, IP = 0.73(wL - 20)
⇒ IP = 0.73(35 - 20) = 10.95%
Plasticity index of soil sample, IP = wL - wP = 35 - 27 = 8% < 10.95%
⇒ Soil sample lies below A-line. Hence the soil sample is inorganic (Silt - M).
Thus, the soil sample is designated as SM.

21.  A fine grained soil has liquid limit of 60% and plastic limit of 20%. As per the plasticity chart, according to IS classification, the soil is represented by the letter symbol _____ .

[GATE 2010: 1 Mark]
  1. CL
  2. CI
  3. CH
  4. CL – ML

Answer: Option C

Explanation:
Equation of A-line, IP = 0.73(wL - 20)
⇒ IP = 0.73(60 - 20) = 29.2%
Plasticity index of soil sample, IP = wL - wP = 60 - 20 = 40% > 29.2%
⇒ Soil sample lies above A-line. Hence the soil sample is organic (Clay - C).
Now, Liquid limit of soil sample is, wL = 60%.
⇒ wL > 50%
⇒ The compressibility of soil sample is High (H).
Thus, the soil sample is CH.

22.  The results for sieve analysis carried out for three types of sand P, Q and R are given in the adjoining figure. If the fineness modulus values of the three sands are given as FMP, FMQ and FMR, it can be stated that

Graph showing the relation between Fineness modulus of three soil samples P, Q and R.
[GATE 2011: 1 Mark]
  1. FMQ = √(FMP x FMR)
  2. FMQ = 0.5 (FMP + FMR)
  3. FMP > FMQ > FMR
  4. FMP < FMQ < FMR

Answer: Option A

Explanation:

23.  As per the Indian Standard Soil classification system, a sample of silty clay with liquid limit of 40% and plasticity index of 28% is classified as _____ .

[GATE 2012: 1 Mark]
  1. CH
  2. CI
  3. CL
  4. CL – ML

Answer: Option B

Explanation:
Equation of A-line, IP = 0.73(wL - 20)
⇒ IP = 0.73(40 - 20) = 14.6%
Plasticity index of soil sample, IP = 28% > 14.6%
⇒ Soil sample lies above A-line. Hence the soil sample is organic (Clay - C).
Now, Liquid limit of soil sample is, wL = 40%.
⇒ 35% < wL < 50%
⇒ The compressibility of soil sample is Medium (I).
Thus, the soil sample is CI.

24.  As per the Indian Soil Classification System (IS 1498 - 1970), an expression for A-line is _____ .

[GATE 2014: 1 Mark, I Set]
  1. IP = 0.73(wL – 20)
  2. IP = 0.70(wL – 20)
  3. IP = 0.73(wL – 10)
  4. IP = 0.70(wL – 10)

Answer: Option A

Explanation: Self Explanatory

25.  A clay mineral primarily governing the swelling behaviour of black cotton soil is _____ .

[GATE 2014: 1 Mark, II Set]
  1. Halloysite
  2. Illite
  3. Kaolinite
  4. Montmorillonite

Answer: Option D

Explanation: Montmorillonite has large affinity for water. Black cotton soil contains Montmorillonite which is responsible for its swelling action.

26.  A fine grained soil is found to be plastic in the water content range of 26%-48%. As per Indian Standard Classification System, the soil is classified as _____ .

[GATE 2016: 1 Mark, I Set]
  1. CL
  2. CH
  3. Cl – ML
  4. CI

Answer: Option D

Explanation:
Plastic limit, wP = 26% and Liquid limit, wL = 48%.
Equation of A-line, IP = 0.73(wL - 20)
⇒ IP = 0.73(48 - 20) = 20.44%
Plasticity index of soil sample, IP = wL - wP28 > = 48 - 26 = 22% > 20.44%
⇒ Soil sample lies above A-line. Hence the soil sample is organic (Clay - C).
Also, 35% < wL < 50%
⇒ Soil sample is of Medium compressibility.
Thus, the soil sample is designated as CI.

27.  The notation 'SC' as per Indian Standard Soil Classification System refers to _____ .

[GATE 2019: 1 Mark, II Set]
  1. silty clay
  2. clayey sand
  3. sandy clay
  4. clayey silt

Answer: Option B

Explanation:
’S' denotes 'Sand'. 'C' denotes 'Clay'.
The constituent in minor proportion is to be mentioned at end.
Thus, Clayey sand - SC.

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