1. There are __________ fundamental building blocks for the clay mineral structures.
a) 3
b) 5
c) 4
d) 2
Answer: d
Explanation: Silica tetrahedral unit and octahedral unit are the 2 building blocks of clay mineral.

2. In silica tetrahedral unit, the silicon structure is enclosed by __________
a) Oxygen
b) Hydroxyls
c) Aluminum
d) None of the mentioned
Answer: a
Explanation: In silica tetrahedral unit, four oxygen or hydroxyls have the configuration of tetra hadron.

3. Octahedral unit, consist atoms of __________ element.
a) Aluminum atom
b) Oxygen atom
c) Magnesium atom
d) All of the mentioned
Answer: d
Explanation: In octahedral unit, aluminum, iron or magnesium atom are enclosed in six hydroxyls having the configuration of an octahedron.

4. The silicon sheet is represented by ___________ symbol.
a) trapezoidal
b) rectangle
c) triangle
d) none of the mentioned
Answer: a
Explanation: The silicon sheet is represented by trapezoidal symbol, representing the oxygen basal layer and hydroxyl apex layer.

5. The 15 clay minerals are mainly divided in to ________ groups.
a) 3 groups
b) 2 groups
c) 9 groups
d) 4 groups
Answer: d
Explanation: Clay minerals are classified into 3 groups of kaolin, montmorillonite, illite and palygorskite.

6. The kaolinite structural unit is made up of ____________ layer or sheet.
a) Gibbsite sheet
b) Silica sheet
c) Oxygen layer
d) Aluminum sheet
Answer: a
Explanation: The kaolinite structural unit is made up of gibbsite sheet with aluminum atoms at their centers.

7. In kaolinite crystal, the layers are held by _________ bond.
a) Ionic bond
b) Cationic linkage
c) Hydrogen bond
d) Electro static bond
Answer: c
Explanation: Since kaolinite is stable and water is unable to penetrate between the layers, they are held by a fairly strong bond of hydrogen.

8. Kaolinite exhibit the characteristic of ___________
a) Shrinkage limit
b) Plasticity
c) All of the mentioned
d) None of the mentioned
Answer: b
Explanation: The platelets in kaolinite carry negative charges which attract thick layers of adsorbed water thereby producing plasticity.

9. An Example of kaolinite crystal or mineral is ___________
a) China clay
b) Crystal
c) Clay mineral
d) None of the mentioned
Answer: a
Explanation: China clay show the property of plasticity, when mixed with water therefore it can be taken as an example for kaolinite crystal.

10. The illite crystal has a great tendency of _______________ properties.
a) Swelling
b) To split in to ultimate platelets
c) High shrinkage
d) None of the mentioned
Answer: b
Explanation: The cationic bond of illite is weaker than the hydrogen bond of kaolinite, but is stronger in water bond. Due to this, illite crystal has a great tendency to split in to ultimate platelets.

11. The structure of illite is similar to that of __________ crystal.
a) Kaolinite
b) Dickite
c) Montmorillonite
d) Attapulgite
Answer: c
Explanation: Both illite and montmorillonite are made up of sheet like unit.

12. The thickness of each sheet in montmorillonite is ______________
a) 5 Å
b) 7 Å
c) 10 Å
d) 15 Å
Answer: d
Explanation: Thickness of montmorillonite, T= 7×10-10mm
T=7 Å.

13. The property of a soil which allows it to be deformed rapidly, without rupture is _________
a) Elasticity
b) Plasticity
c) Tenacity
d) None of the mentioned
Answer: b
Explanation: Plasticity is the property of a soil which allows it to be deformed rapidly, without rupture, without elastic rebound and without volume change.

14. The ratio of liquid limit, minus the natural water content to the plasticity index of the soil is __________
a) Consistency index
b) Plasticity index
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: consistency index or relative consistency is defined as the ratio of liquid limit, minus the natural water content to the plasticity index of the soil.

15. The consistency (or) relative consistency (Ice) is given by the formula __________
a) IC = WL-W/IP
b) IC = W-WL/IP
c) IC = W-WL/IP
d) IC = W-WP/IP
Answer: a
Explanation: The consistency or relative density is defined as ratio of liquid limit, minus the natural water content to the plasticity index of the soil IC = WL-W/IP.

16. The shrinkage limit is represented by the term ___________
a) IP
b) WS
c) IC
d) WP
Answer: b
Explanation: Shrinkage limit is represented as WS.

17. The Swedish agriculturist who divided the entire range of consistency from liquid to solid states is __________
a) Dupuit’s
b) Laplace
c) Boussinesq
d) Atterberg
Answer: d
Explanation: In 1911, Atterberg who was a Swedish Agriculturist divided the entire range of liquid to solid state in to four stages.

18. Which of the following is not useful for engineer purpose, as proposed by Atterberg?
a) Plastic limit
b) Liquid limit
c) Solid limit
d) Shrinkage limit
Answer: c
Explanation: Liquid limit, plastic limit, shrinkage limit are useful for engineering purpose.

19. According to Goldschmidt theory, the plasticity in soil is due to __________
a) Electro-magnetic charges
b) Smooth surface
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: The plasticity in soil is due to the presence of thin scale like particles which carry on to their surfaces, electro-magnetic charges.

20. Clay does not become plastic, when mixed with ___________
a) Soap solution
b) Kerosene
c) Oil
d) None of the mentioned
Answer: b
Explanation: According to Goldschmidt theory, the clay does not become plastic when mixed with liquids of non-polarizing agents like kerosene.

21. In consistency of soil, the limits are expressed in terms of__________
a) Per cent water content
b) Area
c) Volume
d) All of the mentioned
Answer: a
Explanation: According to Atterberg, the consistency limits are expressed as percent water content.

22. Which of the following is not considered as one of the state, as divided by Atterberg?
a) Solid state
b) Gaseous state
c) Semi-solid state
d) Liquid state
Answer: b
Explanation: The four stages as divided by Atterberg are solid state, liquid state, plastic state, semi-solid state.

23. The grooving tool which is used for finding liquid limit is _________
a) ASTM tools
b) Grooving tools
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: ASTM tool and casagrande tools are two types of grooving tools are used in determining liquid and plastic limit.

24. The depth of the groove cut by Casagrande tool for determining the liquid limit is ______
a) 10 mm
b) 11.0 mm
c) 2 mm
d) 8 mm
Answer: b
Explanation: The casagrande tools cut a groove of size 11.0 mm wide at the bottom.

25. The type of tools which is preferred for sandy soil for the purpose of grooving is _______
a) ASTM tool
b) Casagrande tool
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: ASTM tool is used only for more sandy soil, where the Casagrande tool tends to tears the side of the groove.

26. What is the diameter of the sieve that is used for finding the liquid limit?
a) 275 microns
b) 700 microns
c) 425 microns
d) 200 microns
Answer: c
Explanation: 425 micron sieve is used as a standard, for filtering the given specimen.

27. The soviet liquid limit device is based on the principle of ___________
a) Station penetration
b) Soil moisture
c) Soil water content
d) None of the mentioned
Answer: a
Explanation: Static cone penetration method is used in operation of soviet liquid limit device.

28. The toughness index (It) is defined by the ratio of __________
a) It=WP/IP
b) It=IP/If
c) It=IF/IP
d) It=WL/If
Answer: b
Explanation: Toughness index (It) is determined by the ratio of plasticity index to the flow index: It=IP/If.

29. The number of revolutions per second, at which the handle is rotated in process of finding the liquid limit is _________
a) 2
b) 7
c) 4
d) 9
Answer: a
Explanation: In liquid limit apparatus, the handle is rotated at a rate of about 2 revolutions per second.

30. The cone which is used to penetrate the soil pat has a central angle of ___________
a) 26 degree
b) 30 degree
c) 31 degree
d) 40 degree
Answer: c
Explanation: The cone has a central angle of 31° since the total sliding mass is 80 g.

31. The plastic index is calculated from the relation ____________
a) IP = WP-WL
b) IP = WL-WP
c) IP= IL-IS
d) IP=IW-IS
Answer: b
Explanation: The plasticity index is given by the formula, IP=WL-WP.

32. One-point method, used for determining the liquid limit is majorly applicable for finding ________
a) Accurate value
b) Precise value
c) Rough value
d) None of the mentioned
Answer: c
Explanation: Since the liquid limit is found out by only one reading of water content and its corresponding number of blows. One point method is applicable for finding rough value.

33. The volume shrinkage (VS) is defined by the formula _________
a) VS = (VL-VD/VD)×100
b) VS = (VD-VL/VD)×100
c) VS = (WL-WS)SR
d) None of the mentioned
Answer: c
Explanation: Since VS = (VL-Vd) × 100/Vd
But, (VL-Vd)×100/Vd = (WL-WS) SR
Therefore, VS = (WL-WS) SR.

34. The shrinkage ratio of soil is equal__________ the soil in its dry state.
a) Mass specific gravity
b) Mass density
c) Water content
d) Specific gravity
Answer: a
Explanation: The shrinkage ratio is equal to mass specific gravity in a dry state.

35. Shrinking limit can be found out using alternate method, if __________
a) The specific gravity of soil is known
b) Dry volume of soil is known
c) Water content in the soil is known
d) Dry density of soil is known
Answer: a
Explanation: Alternatively, shrinking limit can be found out if the specific gravity G of the soil grains is known.

36. Which of the following apparatus does not include, in a determination of shrinkage limit?
a) Porcelain evaporating dish
b) Two glass plates
c) Brass cup
d) Stainless steel shrinking dish
Answer: c
Explanation: The equipment for determination of shrinkage limit consist of Porcelain evaporating dish, two glass plates, stainless steel shrinking dish.

38. The inside of the shrinkage dish is coated with a thin layer of _________
a) Oil
b) Vaseline
c) Acid
d) None of the mentioned
Answer: b
Explanation: Vaseline prevents the formation of air-bubbles. Hence it is used as for inner coating in a shrinking dish.

39. The volume of the wet soil, present in shrinkage dish_________ of volume of dish.
a) One –fourth
b) Two-third
c) One-third
d) Same
Answer: c
Explanation: One –third of shrinkage dish is filled with wet soil to prevent overflow.

40. What are the ways of preventing of inclusion of air bubbles in shrinkage dish?
a) Mixing the soil with sufficient distilled water and Making the soil pasty enough
b) Coating the surface with a thin layer of oil
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: Mixing the soil with sufficient distilled water and making it pasty enough prevents forming air bubbles in shrinking dish.

1. The water content of a soil sample cannot be determined by the _____ method.
a) oven drying
b) alcohol
c) calcium carbide
d) pipette
Answer: d
Explanation: The pipette method is the standard sedimentation method used in laboratory for wet mechanical or sedimentation analysis.

2. In the oven drying method, the soil sample is kept for about ______ hours in the oven.
a) 1
b) 2
c) 24
d) 48
Answer: c
Explanation: For complete drying, sandy soil takes about 4 hours and fat clays take about 14 to 16 hours. To assure complete drying of sample, it is kept for about 24 hours.

3. An organic soil sample is kept in an oven for its water content determination. The temperature preferred is ______
a) 60°
b) 80°
c) 105°
d) 110°
Answer: a
Explanation: For organic soils like peat, a lower temperature of about 60° is preferable to prevent the oxidation of the organic matter.

4. ______ method is especially suited to a circumstance where water content is to be quickly determined for the purpose of proper field control.
a) Oven drying
b) Sand bath
c) Alcohol
d) Calcium carbide
Answer: d
Explanation: The calcium carbide method is very quick as the result can be obtained in 5 – 10 minutes. The field kit contains the moisture tester, a small single – pan weighing balance, a bottle containing calcium carbide and a brush.

5. In the calcium carbide method, the gas produced is______
a) methane
b) carbon dioxide
c) acetylene
d) oxygen
Answer: c
Explanation: The acetylene gas is produced by the reaction of moisture of soil and the calcium carbide. The gas exerts pressure on a sensitive diaphragm attached to a dial gauge which reads the water content.

6. The water content of soil deposit in the in-situ condition is determinate by ______ method.
a) radiation
b) pycnometer
c) calcium carbide
d) oven – dry
Answer: a
Explanation: The radiation method consists of 2 steel casings. A device containing some radio–active isotope is lowered in one caning and a detector in another. The neutrons emitted by the device are detected by the detector and the loss of energy of neutrons is equated to the water content.

7. Sieve analysis is meant for______
a) coarse-grained soils
b) fine-grained soils
c) coarse-grained gravel
d) silt
Answer: a
Explanation: Sieve analysis is first stage of particle size analysis which is meant for coarse-grained soils only, while the second stage is the sedimentation analysis which is performed for fine-grained soils.

8. In Indian Standard (IS : 460-11962) the sieve sizes are given by_____
a) number of openings
b) number of openings per inch
c) size of aperture in mm
d) size of aperture in cm
Answer: c
Explanation: In the BS and ASTM standards, the sieve sizes are given in terms of number of openings per inch. In Indian Standard, the sieves are designed by the size of aperture in mm.

9. The portion retained on______ IS sieve is termed as gravel fraction.
a) 4.75mm
b) 2mm
c) 425micron
d) 75micron
Answer: a
Explanation: The portion retained on 4.75mm sieve is kept for coarse analysis, hence termed as gravel fraction. While the portion passing through 4.75mm sieve is subjected to fine sieve analysis.

10. The receiver at the bottom of the assembly in sieve shaking machine is________
a) 4.75mm sieve
b) 425micron
c) pan
d) 75micron
Answer: c
Explanation: A pan is kept at the bottom of the whole assembly to collect the finest particles of the soil sample that is used in the experiment.

11. Sieving is performed by arranging the various sieves one over the other in the order of their mech openings.
a) True
b) False
Answer: a
Explanation: The largest aperture sieve being kept at the top and the smallest aperture sieve is at the bottom of the assembly filled on a sieve shaking machine.

12. ______ minutes of shaking is done for soil with small particles.
a) 2
b) 10
c) 15
d) 60
Answer: b
Explanation: The amount of shaking depends upon the shape and number of particles. At least 10 minutes of shaking is desirable for soil particles.

13. The percentage of soil retained on each sieve is calculated on the basis of ______
a) total mass
b) total weight
c) volume of sample
d) density of soil
Answer: a
Explanation: The portion of the soil sample retained on each sieve is weighed and the percentage of soil retained in each sieve is calculated. Mass is measured in terms of mass, and weight in terms of Newton.

14. The soil portion passing through 4.75 mm sieve is washed for further sieve analysis.
a) True
b) False
Answer: a
Explanation: It is advisable to wash the soil portion passing through 4.75 mm sieve so that silt and cay particles sticking to the sand particles may be dislodged.

15. ______ is used for washing the soil portion passing through 4.75 mm sieve.
a) distilled water
b) 2g of sodium hexametaphosphate per litre of water
c) 10% of brine solution
d) kerosene
Answer: b
Explanation: Sodium hexametaphosphate acts as a dispersing agent that prevents flocculation or the combining of suspended matter into aggregates that settle due to gravity.

16. In sedimentation analysis, the soil fraction should be of what micron size, so as to be kept in a liquid medium (water).
a) 75
b) 83
c) 57
d) 70
Answer: a
Explanation: According to Indian standards the soil fraction is set as 75 microns to be used in a liquid medium.

17. The sedimentation analysis is done with the help of________
a) Hydrometer and Pipette
b) Sieve
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: Both the hydrometer and pipette method can be used for taking sedimentation analysis.

18. The time for a particle to settle, of diameter 0.06 through a height of 10 cm is_________
a) 25.8 s
b) 42 m 59 s
c) 3 m 52 s
d) 42.8 s
Answer: a
Explanation: Based on the time of settlement of particles of various diameters it is found that minimum time of 25.8 s is needed for a particle to settle of diameter 0.006 mm.

19. Which of the following assumptions is wrong, with respect to sedimentation analysis?
a) The soil particles are spherical
b) Particle settle independent of other particles do not have any effect on its velocity of settlements
c) Soil particles has different specific gravity
d) The walls of jar, in which the suspension is kept do not affect the settlement
Answer: c
Explanation: In sedimentation analysis, soil particle has average specific gravity.

20. The sedimentation analysis is based on _________ law.
a) Stoke
b) Cullman’s
c) Rankine’s
d) None of the mentioned
Answer: a
Explanation: The sedimentation analysis is based on stokes law according to which the velocity at which the grains settles down depend on shape, size, weight of grain.

21. In the formula, f=105√(300 ɳ/(g-1)Vw), the factor ɳ depends on__________
a) Pressure
b) Velocity
c) Temperature
d) Specific gravity
Answer: c
Explanation: Since ɳ=viscosity of water/liquid, it can be affected by temperature.

22. The lower limit of particle size allowed in sedimentation analysis is ________
a) 0.0002 mm
b) 0.04 mm
c) 0.074 mm
d) 0.0004 mm
Answer: a
Explanation: The particle smaller than 0.0002 mm is affected by Brownian movement and stokes law remains no longer valid.

23. 1 poise is equivalent to _________
a) 0.5 NS/m2
b) 0.1 NS/ m2
c) 10-4 K NS/ m2
d) 0.8 NS/ m2
Answer: b
Explanation: As,1 poise =1/10 NS/ m2
1 poise =0.1 NS/ m2.

24. The unit weight of water in v=(2/g)r2(Vs-Vw/ɳ) is taken as ________
a) 9.81 kn/m2
b) 98.1 n/m2
c) 0.981 kn/m2
d) 9.81 n/m3
Answer: a
Explanation:The constant value of unit weight of water/liquid is taken as γw=9.81 kn/m2.

25. Viscosity of water/liquid (ɳ) is expressed in ________
a) KN-s/m3
b) Ns/m2
c) KN-s/m2
d) Ns/m
Answer: c
Explanation: The viscosity of distilled water in sedimentation analysis is taken as 0.00855 KN-s/m2.

26. In hydrometer method, the sampling depth h is kept at a constant of __________
a) 14 cm
b) 9 cm
c) 8 cm
d) 10 cm
Answer: d
Explanation: The sampling depth (h) in hydrometer method is taken as 10 cm.

27. In the calibration of hydrometer reading, the reduced reading are designated as __________
a) RS
b) RH
c) RR
d) RL
Answer: b
Explanation: The reducing reading is designated as RH, which is used for calibration of hydrometer.

28. The dispersing agent correction in hydrometer reading is always__________
a) Positive
b) Both negative and positive
c) Negative
d) Equal
Answer: c
Explanation: The addition of dispersing agent in water increases its water level, therefore the dispersing correction is always negative.

29. The hydrometer method differs from the pipette method on the basis of which of the following?
a) Principle of test
b) Taking the observation
c) Method of procedure
d) None of the mentioned
Answer: b
Explanation: In sedimentation analysis the hydrometer method differs from the pipette method in the method of taking the observation.

30. The corrections which are applied to the hydrometer reading, in hydrometer method is _______
a) Meniscus correctionc and Dispersing agent correction
b) Magnitude correction
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: Both meniscus and dispersing agent correction are applied to a hydrometer reading.

31. The meniscus correction in hydrometer reading is ___________
a) Always positive
b) Always negative
c) Negative sometime
d) Positive sometime
Answer: a
Explanation: The actual reading to be taken at water level will be more; hence the meniscus correction will be always positive.

32. The visibility of the soil suspension in hydrometer method is __________
a) Transparent
b) Translucent
c) Opaque
d) None of the mentioned
Answer: c
Explanation: Since the density of the soil suspension is high, the visibility will be opaque.

33. The volume of suspension taken in the hydrometer test is __________
a) 500
b) 120
c) 100
d) 1000
Answer: d
Explanation: The quantity of dry soil and dispersing agent is taken as double of pipette which is 500 ml.

34. Initially, the volume of hydrometer is taken in terms of ___________
a) Grams
b) Millimeters
c) Liters
d) All of the mentioned
Answer: a
Explanation: since the soil is weighed in terms of mass, the volume is taken in terms of grams.

35. The hydrometer are generally calibrated at ____________
a) 35°c
b) 27°c
c) -35°c
d) -27°c
Answer: b
Explanation: The hydrometer are generally calibrated at 27°c if the temperature is more than 27°c, temperature correction will be taken negative. If less than 27°c then the temperature correction will be positive.

36. The shape of the particle size curve is represented by _________
a) Effective size
b) Effective diameter
c) Uniform coefficient
d) Co-efficient of curvature
Answer: d
Explanation: Coefficient of the curvature Cc represent the shape of the particle size curve given by CC = (D40)2/D10×D30.

37. A particle-size distribution curve gives us an idea about __________
a) Type of soil
b) Properties of soil
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: A particle-size curve shows both type and gradation of the soil.

38. A soil sample may be well-graded if __________
a) If it has most number of particles of same size
b) Excess of certain particles
c) Good representation of particles of all size
d) None of the mentioned
Answer: c
Explanation: A soil is said to be well graded when it has a good representation of particle of all size.

39. For coarse grained soil, the particle size D10 is sometimes called as __________
a) Effective size and effective diameter
b) Uniform diameter
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: The size D10 is sometimes called as the effective size or effective diameter.

40. The shape of particle size curve, which is represented by the coefficient of curvature (Cc) is given by __________
a) Cc = (D30)2/D10×D40
b) Cc = (D40)2/D10×D30
c) Cc = (D30)2/D10×D60
d) Cc = D60/D10
Answer: c
Explanation: The coefficient of curvature (Cc) is given by the equation
Cc = (D30)2/D10×D60.

41. The coefficient of uniformity (CV) is the ratio of _________
a) D60 and D10
b) D30 and D10
c) D10 and D30
d) D10 and D30
Answer: a
Explanation: The coefficient of uniformity CV, is defined as the ratio of D60 and D10
i.e.CV = D60/ D10.

42. The curve situated at the right side of the particle size distribution curve is _________
a) Coarse-grained soil
b) Fine-grained soil
c) coarse-grained soil
d) None of the mentioned
Answer: a
Explanation: The right side the particle size distribution curve shows the coarse-grained while the left shows the relatively fine-grained soil.

43. What is the time of settlement of coarse particle of a soil sample, of diameter 0.5?
Take γ=0.905D2 and height of water tank as 5 m.
a) 11.6 seconds
b) 72.8 seconds
c) 14 seconds
d) 22.1 seconds
Answer: d
Explanation: γ=0.905(0.5)2=0.2263 m/sec
Time of settlement, T=h/v =5/0.2263=22.1 seconds.

44. A curve with a flat portion, in particle size distribution curve represent __________
a) Intermediate size particle are missing
b) Intermediate size particles are present
c) Smaller size particle are present
d) Large size particles are present
Answer: a
Explanation: A curve with a flat portion represents oil in which intermediate size particle are missing.

45. The D10 represents a size, such that _________ of the particles are finer than this size.
a) 20%
b) 60%
c) 10%
d) 100%
Answer: c
Explanation: The D10 represents a size, in mm such that 10% of the particle is finer than D10 size

46. The Indian standard soil classification system, (ISCS) was first developed in ___________
a) 1947
b) 1950
c) 1959
d) 1960
Answer: c
Explanation: The Indian standard classification system (ISCS) was first developed in 1959 and revised in 1970.

47. According to ISCS, fine grained soils are subdivided in to ___________
a) 2
b) 4
c) 5
d) 3
Answer: d
Explanation: Fine grained soil are subdivided into 3 types
i. Inorganic silts and very fine sands
ii. Inorganic clays
iii. Organic silts and clay and organic matter.

48. The ISCS classifies the soil in to __________
a) 12 groups
b) 15 groups
c) 18 groups
d) 16 groups
Answer: c
Explanation: The ISCS classifies the soil into 18 groups as against 15 groups of USCS.

49. Laboratory classification of fined grained soil is done with the help of ________
a) Plasticity chart
b) Textural classification chart
c) Kozney’s graphical method
d) None of the mentioned
Answer: a
Explanation: Laboratory classification criteria of fine grained soil is done by plasticity chart.

50. The A-line, in unified classification system table has the equation of __________
a) IP=WL-20
b) IP=0.73(WL-20)
c) IP=0.73(20-wL)
d) IP=20-WL
Answer: b
Explanation: The A-line, dividing the inorganic clay from silt and organic soil has the equation: IP=0.73(WL-20).

51. According to IS classification, the symbol GC means ____________
a) Clayey gravel
b) Silt gravel
c) Sand gravel
d) Well graded gravel
Answer: a
Explanation: As per Indian standards of group symbols and typical names, G-refers to gravel and C-refers to clay.

52. The symbol ‘L’ represents which of the following soil types?
a) Silt and Clay
b) Gravel
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: By IS classification, silts and clays of low compressibility having a liquid less than 35, are represented by the ‘L’.

53. What are the features required, for classifying a soil component as boulder?
a) Bulky hard
b) Diameter more than 30cm
c) Round to angular shape
d) All of the mentioned
Answer: d
Explanation: For a soil component to be a boulder, size range must be of average diameter more than 30cm and shape of round to angular and bulky hard.

1. What are the types of water flow in the soil?
a) Turbulent flow and Laminar flow
b) Linear flow
c) None of the mentioned
d) All of the mentioned
Answer: a
Explanation: The flow of water through soils may be either a laminar flow or turbulent flow.

2. In what way, do the fluid particles travel in a turbulent flow?
a) Twisting
b) Crossing
c) Re-crossing
d) Definite path
Answer: d
Explanation: In laminar flow, each fluid particles travels along a definite path which never crosses the path of any other particles.

3. Gravels are __________ permeable.
a) Highly
b) Least
c) Partially
d) All of the mentioned
Answer: a
Explanation: The pore size in gravel is larger, hence the gravels are highly permeable than sand which is least permeable.

4. The law of flow of water through the soil was first studied by ___________
a) Taylor
b) Darcy
c) Lambe
d) Khosla
Answer: b
Explanation: In 1856, Darcy studied the law of flow of water through soil and demonstrated experimentally the laminar flow conditions.

5. Which of the following equation defines Darcy’s law?
a) q=KA
b) q=K/A
c) q=K i A
d) q=K i/A
Answer: c
Explanation: According to Darcy’s law, the rate of flow or discharge per unit area is proportional to the hydraulic gradient.
q α i
q = K i A, where K =Darcy’s coefficient of permeability
A = total cross-sectional area.

6. Coefficient of permeability or simple permeability is expressed in terms of ____________
a) cm/sec
b) m/day
c) feet/day
d) all of the mentioned
Answer: d
Explanation: As the dimension of the coefficient of permeability k are same as those of velocity. It is expressed in terms of cm/sec or m/day or feet/day.

7. Darcy’s law is valid for only ___________
a) Laminar flow
b) Turbulent flow
c) Hydraulic flow
d) All of the mentioned
Answer: a
Explanation: Darcy’s law of linear dependency between the velocity of flow and hydraulic agent is valid for only laminar flow conditions in the soil.

8. Darcy’s law is valid as long as it is equal to or less than ____________
a) 1
b) 0
c) 0.1
d) 2
Answer: a
Explanation: In 1933, Lewis and Barnes demonstrated experimentally that Darcy’s law is valid if it is ≤ 1.

9. Stiff clays are ______________ permeable.
a) Highly
b) Least
c) Partially
d) None of the mentioned
Answer: b
Explanation: As stiff clay does not contain any porous materials, it may be termed as least permeable or impermeable.

10. The study of seepage of water through soil is important for, which of the following purpose?
a) Drainage of soils
b) Stability of slopes
c) Ground water flow towards well
d) All of the mentioned
Answer: d
Explanation: The study of seepage of water through soil is important for the following engineering problem
i. Ground water towards soil and drainage of soil
ii. Calculation of seepage through the body of earth dams, and stability of slopes.

11. Based on Allen Hazen experiments, permeability can be expressed as ___________
a) K=CD102
b) K=CD210
c) K=DC210
d) K=DC102
Answer: a
Explanation: In 1892, Allen Hazen based on his experiment on filter of sand particles between 0.1 and 3 mm, found that the permeability can be expressed as K=CD102.

12. Which of the following factors affects the permeability of soil?
a) Grain size
b) Properties of pore fluid
c) Void ratio of soils
d) All of the mentioned
Answer: d
Explanation: Grain size, properties of pore fluid, structural soil arrangement of soil particles, entrapped air and foreign matter and adsorbed water.

13. Physical permeability of a soil KP is related to the coefficient of permeability by the equation ___________
a) KP=kɳ
b) KP=kɳ/γW
c) KP=k/ɳ
d) KP=k γW
Answer: b
Explanation: In 1937, Muskat pointed out a relationship between physical permeability and Darcy’s coefficient of permeability as KP=kɳ/γW.

14. Loudon’s experiments demonstrated the relationship between _____________
a) Permeability and specific surface
b) Permeability and grain size
c) Permeability and adsorbed water
d) Permeability and volume
Answer: a
Explanation: Loudon’s experiment demonstrated that the permeability of coarse grained soils is inversely proportional to the specific surface at a given porosity.

15. What is the constant value of a and b in Loudon’s empirical formula “log10(KSs2) = a + bn”?
a) 1 and 0
b) 1.515 and 1.365
c) 1.365 and 1.515
d) 0 and 1
Answer: c
Explanation: Based on his experiments, Loudon found out the value of a and b s 1.365 and 1.515 for permeability of 10°c.

16. What is the relationship between permeability and viscosity of water?
a) Directly proportional
b) Inversely proportional
c) Both are equal
d) None of the mentioned
Answer: b
Explanation: From Poiseuille’s law, K=Ds2 γw /ɳ
It is found that permeability is directly proportional to the unit weight of water and inversely proportional to its viscosity.

17. What is the effect of adsorbed water on the permeability of soil?
a) Structural arrangement is varied
b) Reduced degree of saturation
c) Size of the particles is diminished
d) Reduces the pore size
Answer: d
Explanation: The adsorbed water surrounding the fine soil particles is not free to move, and reduces the effective pore space available for the passage of water.

18. What is the approximate value, which can be taken as void ratio occupied by adsorbed water?
a) 1
b) 0
c) 0.1
d) 10
Answer: c
Explanation: According to approximation value after casagrande, 0.1 may be taken as the void ratio occupied by adsorbed water.

19. The effect of structural disturbance is on permeability is more in _____________
a) Fine-grained soil
b) Coarse grained soil
c) Clay soil
d) All of the mentioned
Answer: a
Explanation: The effect of structural disturbance on permeability is much pronounced in fine-grained soil than any other.

20. The structural arrangement of soil Particle vary depending upon ______________
a) Method of deposition and Compacting the soil mass
b) Degree of saturation
c) None of the mentioned
d) All of the mentioned
Answer: a
Explanation: The structural arrangement of the particle may vary, at the same void ratio, depending upon the method of deposition or compacting the soil mass.

21. How do degree of saturation effect permeability of soil?
a) By reducing the pore size
b) By entrapping air in the voids
c) Not allowing soil particles to move freely
d) By changing the void ratio
Answer: b
Explanation: By entrapping the air in the voids present in the soil, the permeability is greatly reduced, thus reducing the degree of saturation.

22. Permeability can be determined by direct measurement with the help of ___________
a) Permeameter
b) Consolidation test
c) Horizontal capillary test
d) Pumping-out test
Answer: a
Explanation: Permeability of soil can be determined by permeameters, by allowing the water to flow through soil sample under either constant head or under variable head.

23. Which of the following formula, cannot be used for determining the permeability of soil?
a) Jacky’s formula
b) Allen Hazen’s formula
c) Kozney’s formula
d) Darcy’s formula
Answer: d
Explanation: In Darcy’s equation or Darcy’s law, there is no permeability constant.
q = k i A
Hence it cannot be used for finding permeability of soil.

24. Coefficient of permeability of soil can be determined by which of the following method?
a) Laboratory methods
b) Field methods
c) Indirect methods
d) All of the mentioned
Answer: d
Explanation: The coefficient of permeability can be determined by the following methods:
i. Laboratory method – constant head permeability and falling head permeability test
ii. Field method –pumping-out and pumping-in test
iii. Indirect methods- horizontal capillary test and consolidation test data.

25. Allen Hazen’s formula is given by which of the following equations?
a) K = 100m2
b) K = C D102
c) K = 200De2e2
d) None of the mentioned
Answer: b
Explanation: Allen Hazen’s formula is K=CD102, where c is constant, which is taken as approximately equal to 100 when D10 is expressed in cm.

26. The unit of coefficient of absolute permeability(K) is ____________
a) Kg/cm
b) m/s2
c) m2
d) All of the mentioned
Answer: c
Explanation: Since K has the dimension of area (i.e. [k] = [L2]), the unit of K is m2.

27. Coefficient of absolute permeability (K) depends on ____________
a) Permeant
b) Properties of soil mass
c) Degree of saturation
d) All of the mentioned
Answer: b
Explanation: From the equation of coefficient of absolute permeability,
K = C (e3/1+e) D3
We can find that coefficient of permeability is independent of the properties of permeant (i.e. water) and it depends solely on the properties of soil mass.

28. In Terzaghi’s formula K = 200De2e2, De2 represents ____________
a) Effective grain shape
b) Void ratio
c) Effective grain size
d) Permeability
Answer: c
Explanation: De=effective grain size, the diameter of the sphere for which the ratio of its volume to its surface area is the same as the similar ratio for a given assemblage of soil particles.

29. K = 100Dm2, is given by which of the following formula?
a) Terzaghi’s formula
b) Kozney’s formula
c) Jacky’s formula
d) Allen Hazen’s formula
Answer: c
Explanation: In 1944, Jacky found out the order of magnitude of k can be obtained from all soils from the formula K= 100Dm2, where Dm denotes grain size.

30. Terzhagi’s formula can be applied for_________
a) Uniform sand
b) Clay soil
c) Coarse-grained soil
d) Fine-grained soil
Answer: a
Explanation: Terzaghi’s formula can be successfully applied to fairly uniform sand, which reflects the effect of grain size and void ratio.

31. When water flows through a saturated soil mass, piezometric head is the sum of velocity head and position head.
a) True
b) False
Answer: b
Explanation: When water flows through a saturated soil mass, total head is the sum of piezometric head, velocity head and position head.

32. A piezometric surface is the line joining ________
a) water level in piezometers
b) soil stratum
c) equal voids ratio in soil mass
d) equal velocity of flow
Answer: a
Explanation: A piezometric surface is the surface formed by the line joining water level in piezometers. The piezometric head is equal to the height of water rise in piezometric tube.

33. The piezometric head is also called ________
a) position head
b) elevation head
c) velocity head
d) pressure head
Answer: d
Explanation: A piezometric head is a measurement of liquid pressure above a vertical datum. Hence piezometric head is also known as pressure head.

34. The difference between the elevation of water surfaces in piezometers is _______
a) hydraulic gradient
b) velocity
c) head loss
d) depth or length of sample
Answer: c
Explanation: Head loss refers to the measurement of energy dissipated in a system due to friction. This energy dissipation is measured by the difference in the elevation of water surfaces in piezometers.

35. The position head is also called ________
a) total head
b) elevation head
c) velocity head
d) pressure head
Answer: b
Explanation: The position or elevation head is the elevation of a point with respect to any arbitrary datum. The elevation head can be taken at any point with respect to a datum.

36. The elevation head is taken as negative if the point is situated above the datum and positive if below the datum.
a) True
b) False
Answer: b
Explanation: The datum for elevation head is considered as abscissa of a graph for calculation purpose, where the region above it positive and the region below it is negative.

37. The total head at any point may be regarded as _________ measure with respect to the datum.
a) potential energy
b) potential energy per unit weight of water
c) unit weight of water
d) volume of water
Answer: b
Explanation: The total head consists of pressure head, velocity head and position head. The velocity head is considered negligible due to small flow of water.

38. Total head is the sum of ________
a) piezometric head and velocity head
b) piezometric head and position head
c) position head and velocity head
d) velocity head and pressure head
Answer: b
Explanation: The total head is the sum of pressure head, velocity head and position head. The velocity head is considered negligible due to small flow of water. Therefore, total head is the sum of pressure head and position head.

39. Flow occurs between two points only if _______
a) there is difference in total heads
b) there is no difference in total heads
c) the total heads are equal
d) the total heads are equal to zero
Answer: a
Explanation: Flow occurs between two points when there is a difference in elevation of the points which implies to the difference in total heads or potential energies. When there is no difference in total heads, no flow occurs.

40. In the ______ direction of flow, the effective pressure is increased.
a) perpendicular
b) tangential
c) upward
d) downward
Answer: d
Explanation: The effective pressure subjected to seepage pressure is given by,
σ’=zγ’± ps = zγ’±izγw. If the flow occurs in downward direction, + sign is used. If the flow occurs in upward direction – sign is used.

41. The seepage pressure always acts _______
a) in the direction of flow
b) opposite to direction of flow
c) perpendicular to direction of flow
d) tangential to direction of flow
Answer: a
Explanation: When the water flows through a soil mass, due to its viscous friction, an energy transfer is effected between water and soil. This energy transfer is seepage pressure.

42. Seepage force is the energy transfer between __________
a) air and water
b) water and soil
c) air and soil
d) earth and air
Answer: b
Explanation: The energy transfer is affected between water and soil by virtue of viscous friction exerted on water flowing through the soil pores. The force corresponding to this energy transfer is called seepage force.

43. Seepage pressure is responsible for quicksand conditions.
a) True
b) False
Answer: a
Explanation: When the water flows or percolates through the soil mass, then it produces stresses in the soil particles by virtue of seepage pressure. When this pressure is equal to submerged unit weight γ’ of soil, quick sand condition is obtained.

44. The first rational approach to the problem of seepage through soils was ________
a) Archimedes
b) Poiseuille
c) Darcy
d) Terzaghi
Answer: d
Explanation: The first rational approach to the problem was presented by Terzaghi in 1922 and it forms the basis of subsequent studies.

45. A flow net can be used for which of the following purpose?
a) Determination of seepage
b) Determination of seepage pressure
c) Determination of hydrostatic pressure
d) All of the mentioned
Answer: d
Explanation: A flow net can be utilized for determination of seepage, seepage pressure, hydrostatic pressure, and exit gradient.

46. The portion between two successive flow lines is known as ___________
a) Field channel
b) Flow channel
c) Open channel
d) All of the mentioned
Answer: b
Explanation: The portion between any two successive flow lines is called a flow channel and the portion enclosed between two successive equipotential lines and successive flow lines are known as field.

47. Who was the first to give a graphical method of flow net construction?
a) Casagrande
b) Darcy
c) Forchheimer
d) Kozney
Answer: c
Explanation: The graphical method of flow net construction first given by Forchheimer in 1930, based on trial sketching.

48. The Darcy’s law governing the flow of water through is related to which of the following law?
a) Ohm’s law
b) Stokes law
c) Faraday’s law
d) None of the mentioned
Answer: a
Explanation: The electric models suggested by ohm, have the same geometric shape as the soil through which the water flows. And both Darcy and ohm’s law have corresponding analogous quantities.

49. The flow lines and equipotential lines are ____________
a) Parallel
b) Perpendicular
c) Elliptical
d) All of the mentioned
Answer: b
Explanation: The flow lines and equipotential lines always meet at right angles to one another.

50. The hydrostatic pressure in terms of piezometric head can be calculated from which of the following equation?
a) hW = h – Z
b) hW = h + Z
c) hW = u/γW
d) hW = h/z
Answer: a
Explanation: The equation hW=h – Z, can be used to plot pressure net representing lines of equal water pressure without the saturated soil mass since all the three quantities in the equation can be expressed a the percentage of total hydraulic head H.

51. The seepage medium can be replaced by ____________electric model having the same geometric shape.
a) Potential divider
b) Insulator
c) Electric conductor
d) Potentiometer
Answer: c
Explanation: The seepage medium is replaced by an electric conductor consisting of water with some salt or dilutes hydrochloric acid.

52. What will be the hydrostatic pressure if, hW = 30 % and Z = 10 %?
a) 30%
b) 10%
c) 40%
d) 50%
Answer: c
Explanation: Using the formula, hW = h -Z
30 = h -10
h=30+10 = 40
Hydrostatic pressure, h=40 %.

53. Inhomogeneous soil, every transition in the shape of curves drawn in flow net must be ____________
a) Smooth
b) Sharp
c) Rough
d) All of the mentioned
Answer: a
Explanation: According to practical suggestion given by A.Casagrande, every transition in the shape of curve is smooth, being either elliptical or parabolic in shape.

54. The exit gradient can be expressed by which of the following expression?
a) ie = Δ h/i
b) ie = Δ h.i
c) ie = l/h
d) ie = h/i
Answer: a
Explanation: ie =Δ h /l, represent gradient formula where Δ h=potential drop and l is the average of last field in the flow net at the exit end.

1. Groundwater may be also called as ___________
a) Capillary water
b) Gravitational water and Free water
c) None of the mentioned
d) All of the mentioned
Answer: a
Explanation: Since the ground water is subjected to no forces other than gravity. Hence the water is also known as gravitational water or free water.

2. Water present in the voids of soil mass is called __________
a) Soil water
b) Free water
c) Ground water
d) Pore water
Answer: a
Explanation: Soil water is the water present in the voids of the soil mass.

3. Hygroscopic water is affected by which of the following factor?
a) Gravity
b) Capillary forces
c) All of the mentioned
d) None of the mentioned
Answer: d
Explanation: Since hygroscopic water has greater density and viscosity than ordinary water. It is neither affected by gravity or capillary forces or by any other force.

4. What are the forces involved in hygroscopic water or contact moisture?
a) Adhesion force
b) Capillary force
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: In hygroscopic water, soil particle freely adsorbs water from the atmosphere by the physical force of attraction, and is held by adhesion force.

5. The average hygroscopicity of clay is ___________
a) 6 %
b) 16 %
c) 1 %
d) 4 %
Answer: b
Explanation: The average hygroscopicity of sands, silts and clay is 1 %, 6 % and 16 %.

6. The water which soaks into ground by moving downward, subjected to capillary force is ___________
a) Ground water
b) Pore water
c) Infiltrated water
d) Capillary water
Answer: c
Explanation: Infiltrated water is the portion of surface precipitation which soaks into ground moving downward through air-containing zones.

7. Based on inter-particle forces, soil water can be classified in to __________
a) Adsorbed water
b) Pore water
c) All of the mentioned
d) None of the mentioned
Answer: a
Explanation: Adsorbed water which is attracted by forces within the soil and pore water are the two types of soil water divided based on inter-particle forces.

8. Solvate water is subjected to __________ forces.
a) Polar
b) Electrostatic
c) Binding
d) All of the mentioned
Answer: d
Explanation: Since the solvate water forms a hydration shell around soil grain. It is influenced by ionic and polar forces.

9. The soil water which is impossible to remove from the soil is ___________
a) Structural water
b) Capillary water
c) Solvate water
d) Pore water
Answer: a
Explanation: Under loading encountered in soil engineering, it is found out that structural water cannot be separated or removed and therefore it is considered as a part and parcel of the soil particle.

10. Ground water is influenced by ____________ force.
a) Vander Waals force
b) Surface force
c) Hydrodynamic force
d) Electro static force
Answer: c
Explanation: As the ground water obeys laws of hydraulic force, it is capable of moving under hydrodynamic forces.

11. Water can be classified in to __________ types based on structural aspect.
a) 3
b) 5
c) 2
d) 4
Answer: d
Explanation: Structural classifications of water in a soil are 4 types, which are: pore water, solvate water, adsorbed water, structural water.

12. The capillary tension or capillary potential can also be called as ___________
a) Pressure deficiency
b) Pressure reduction
c) Negative pressure
d) All of the mentioned
Answer: d
Explanation: The capillary potential is the pressure deficiency, pressure reduction or negative pressure in the pore water.

13. The maximum tensile stress of water in capillary tube is ___________
a) Proportional to the diameter of tube
b) Inversely proportional to height of capillary rise
c) Inversely proportional to the radius of meniscus
d) All of the mentioned
Answer: c
Explanation: From the formula, (UC) max = 2Ts /R
Where, (UC) max = maximum tensile stress
R = radius of meniscus
TS = surface tension.

14. The pF value of 2 in soil mass represent ____________
a) Soil suction of 200 cm of water
b) Suction pressure of 100 g/cm2
c) Soil suction of 100 cm of water
d) Capillarity of 100 g/cm2
Answer: c
Explanation: From the relationship,
pF=log10 (hc)
2 = log10 (hc)
hc = 2/log10
hc = 100 cm.

15. The pressure deficiency of held water in a capillary tube is termed as ____________
a) Soil suction and suction pressure
b) Negative pressure
c) None of the mentioned
d) All of the mentioned
Answer: a
Explanation: As the water is decreased linearly in the tube due to the suction of soil mass. The pressure deficiency is also called as soil suction or suction pressure.

16. Which one of the following factor does not affect soil suction?
a) Water content
b) Temperature
c) Plasticity index of a soil
d) Atmospheric pressure
Answer: d
Explanation: As the atmospheric pressure, does not have any effect on water in the soil mass. It does not contribute to the affecting factor.

17. The soil suction is maximum, when angle of contact(α) is _____________
a) α=0
b) α=90
c) α=60
d) α=1
Answer: a
Explanation: As soil suction depend on capillary height (hC)
Using the formula hc=4Tscos α/γw d
When α=0
hc = 4Ts/γw d, which is maximum.

18. Rise in temperature___________soil suction.
a) Decreases
b) Increases
c) Removed
d) None of the mentioned
Answer: a
Explanation: Rise in temperature results in a decrease of surface tension (TC) and hence decrease in soil suction.

19. Capillary tension is developed in saturated soil, when ___________
a) Soil is dry
b) Water content is reduced
c) Water content is increased
d) Soil particle is large
Answer: b
Explanation: On decreasing of water content, the menisci recede, resulting in a reduction of curvature and corresponding increase in capillary tension.

20. When do the soil suction is reduced to zero value?
a) Water content is reduced
b) Soil is oven dry
c) Dry soil is submerged
d) All of the mentioned
Answer: c
Explanation: When a dry soil is submerged under water, meniscus is destroyed resulting in a reduction of soil suction to zero value.

21. The size of interstices in a soil depends on ____________
a) Particle size
b) Plasticity index
c) Soil structure
d) Angle of contact
Answer: c
Explanation: Change in structure of the soil, changes the size of interstices in soil, resulting in soil suction variation.

22. The capillary pressure, transferred from grain to grain in soil may be also called as ____________
a) Inter granular pressure
b) Contact pressure
c) Effective pressure
d) All of the mentioned
Answer: d
Explanation: Inter granular or contact or effective pressure is the capillary pressure, transferred from grain to grain in soil.

23. Total stress or unit pressure on a soil mass is ___________
a) Total load
b) Total surface area
c) Total volume
d) Total weight
Answer: a
Explanation: At any plane in a soil mass, the total stress or unit pressure is the total load per unit area.

24. At any plane, pore pressure is equal to ___________
a) Ratio of Piezometric head to weight of water
b) Equal to piezometric head times the unit weight of water
c) Ratio of weight of water to the piezometric head
d) None of the mentioned
Answer: b
Explanation: u = hw × γw
Pore pressure=piezometric head × weight of water.

25. Pressure transmitted from particles to the soil mass is called ___________
a) Neutral pressure
b) Effective pressure
c) Pore pressure
d) Capillary pressure
Answer: b
Explanation: Effective pressure ‘σ’ or intergranular pressure is the pressure transmitted from particle through their point of contact through the soil mass above the plane.

26. The neutral pressure does not have any effect on ___________
a) Shearing resistance
b) Shearing strength
c) Shearing stress
d) All of the mentioned
Answer: a
Explanation: The neutral pressure does not have measurable influence on the mechanical property of the soils, such as shearing resistance.

27. The total pressure in a soil mass consists of _____________ distinct components.
a) 3
b) 4
c) 2
d) 5
Answer: c
Explanation: i. Effective pressure or intergranular pressure.
ii. Neutral pressure or pore pressure, are the two components of total pressure.

28. The neutral pressure is transmitted through ___________
a) Soil particle
b) Pore fluid
c) Air particle
d) Atmosphere
Answer: b
Explanation: As the neutral pressure should not have any effect on void ratio or property of soil, it is transmitted through pore fluid.

29. Total vertical pressure at any plane is equal to __________
a) σ = σ’ + u
b) σ’ = σ + u
c) σ = σ’ + v
d) None of the mentioned
Answer: a
Explanation: Total vertical pressure at any plane is equal to the sum of effective pressure and the total pressure i.e. σ = σ’ + u.

30. Factor of unit cross-section χ, depends on ____________
a) degree of saturation
b) soil structure
c) stress change
d) all of the mentioned
Answer: d
Explanation: Factor of unit cross-section varies with the degree of saturation, soil structure, process by which the soil present degree of saturation and stress change.

31. For degree of saturation, it is recommended to take χ as _____________
a) 0
b) 1
c) 2
d) ∞
Answer: b
Explanation: For degree of saturation of S ≥90%, it is recommended to take χ as unity (i.e., 1).

32. The capillary siphoning will take place if the top of the core is situated at height y _______ which is above H.F.L.
a) greater than capillary rise hc
b) lesser than capillary rise hc
c) equal to capillary rise hc
d) zero
Answer: b
Explanation: When the height of impervious core of the earth dam is less than that of the capillary rise, then due to the capillary action, the water from one side of the dam will move to the other side over the impervious core by a phenomenon known as capillary siphoning.

33. To check the seepage of water through the body of dam, the core provided is ________
a) pervious
b) semi-pervious
c) impervious
d) hollow
Answer: c
Explanation: The core of the dam is provided with an impervious material to prevent the seepage of the water through the core. If a pervious material is used, then the whole structure would easily collapse.

34. To check the seepage of water through the body of dam, the outer shells of dam provided is ________
a) pervious
b) semi-pervious
c) impervious
d) hollow
Answer: a
Explanation: The outer shells of the dam are made of highly pervious granular soil so as to check the seepage through it. The water can rise through the capillary action through the voids of the dry soil.

35. The capillary forces can rise the water in the capillary tube against the gravity force.
a) True
b) False
Answer: a
Explanation: Capillarity is the phenomenon of movement of water through narrow spaces due to capillary forces. In capillary rise of water, it will hang in tension in capillary tube.

36. The maximum capillary rise at 20°C is given by ________ cm.
a) 0.1234/d
b) 0.2975/d
c) 0.3857/d
d) 0.3084/d
Answer: b
Explanation: The maximum capillary rise is given by,
(hc)max=4Ts/(γwd)
At 20°C, Ts=72.8 dynes/cm and γw=9.78kN/m3
∴ (hc)max=4*72.8 *10-8/(9.78*10-6* d)
∴ (hc)max=0.2975/d cm.

37. With increase in temperature, the capillary rise _______
a) decreases
b) increases
c) has no change
d) does not depend on temperature
Answer: a
Explanation: As the surface tension decreases with the temperature increase, the capillary rise which is directly proportional to the surface tension, would decrease with the increase in the temperature.

38. Decrease in water content causes _______
a) shrinkage
b) swelling
c) frost heave
d) frost boil
Answer: a
Explanation: Decrease in water content of soil results in decrease in the volume of the soil thus causing the shrinkage of the soil.

39. Increase in water content causes _______
a) shrinkage
b) swelling
c) frost heave
d) frost boil
Answer: b
Explanation: Increase in water content of soil results in increase in the volume of the soil thus causing the swelling of the soil.

40. Arrange the following to explain the shrinkage of the soils.
I. Compression in soil
II. Formation of meniscus
III. Reduction in volume
a) (II), (I), (III)
b) (I), (II), (III)
c) (III), (I), (II)
d) (III), (II), (I)
Answer: a
Explanation: When saturated soil is dried, a meniscus develops in the voids at the solid surface. Formation of meniscus leads to tension in the soil water that leads to the compression of the soil structure and consequent reduction in the volume.

41. When the meniscus attains a minimum value, shrinkage is ________
a) maximum
b) minimum
c) no effect
d) zero
Answer: a
Explanation: When the meniscus attains a minimum value, compressive forces which are developed are maximum causing the shrinkage to also be maximum.

42. The degree of shrinkage does not depend upon _________
a) initial water content
b) type of clay
c) cation exchange capacity
d) amount of clay
Answer: c
Explanation: Cation exchange capacity is factor influencing the swelling of soil. The degree of shrinkage depends upon type and amount of clay minerals, mode and environment of deposition.

43. Arrange the following to explain the swelling of the soils.
I. Increase in tension in the pore water
II. Reduction in compressive stress in the soil
III. Menisci are destroyed
a) (II), (I), (III)
b) (I), (II), (III)
c) (III), (I), (II)
d) (III), (II), (I)
Answer: c
Explanation: When water is added to a soil menisci are destroyed resulting in tension in the pore water and the consequent reduction in the compressive stress in the soil.

44. Free swell ceases when the water content reaches ________
a) liquid limit
b) plastic limit
c) shrinkage limit
d) plasticity index
Answer: c
Explanation: Free swell of a soil is the increase in the volume of soil without any constrains on submergence in water and it ceases when the water content reaches the plastic limit.

45. The slaking of the clay occurs when clayey soil is dried ________
a) below the shrinkage limit
b) above the shrinkage limit
c) below the plastic limit
d) above the plastic limit
Answer: a
Explanation: When the clayey soil is dried below its shrinkage limit, it will attain its minimum volume. When this is suddenly immersed in water it causes slaking.

46. Slaking is due to the ______ into the drying of the soil below the shrinkage limit.
a) entry of water
b) exit of water
c) entry of air
d) exit of air
Answer: c
Explanation: When the water enters the voids (containing air during drying process), due to immersion, menisci is formed, causing high pressures and subsequent explosion of the voids leading to the disintegration soil structure.

47. The bulking of sand is maximum when the water content is _________
a) 0
b) 5 to 6%
c) 15 to 17%
d) 20 to 21%
Answer: b
Explanation: The bulking of sand is maximum when the water content is 5 to 6%. Saturating the soil above 5 or 6% will result in destruction of the menisci and decrease in the volume.

1. Compaction decreases the porosity.
a) True
b) False
Answer: a
Explanation: Compaction is a process by which the soil particles are artificially rearranged into a closer state. This arrangement reduces the voids ratio and hence there is decrease in the porosity.

2. Dry density of soil is increased by _________
a) compaction
b) swelling
c) bulking
d) addition of excess of water
Answer: a
Explanation: Compaction is a process by which the soil particles are artificially rearranged into a closer state which reduces the voids ratio and hence there is decrease in the porosity. Due to this, the dry density of the soil sample is increased.

3. Which of the following does not cause compaction?
a) vibration
b) tamping
c) rolling
d) adding excess water
Answer: d
Explanation: When the water is added to the soil sample, it first occupies the voids present in between the soil particles. When excess of water is added to the soil sample, then swelling of the sand takes place, hence reducing its compaction.

4. In 1933, ____________ showed the existence of a relationship between soil water content and degree of dry density.
a) Terzaghi
b) Skempton
c) Darcy
d) Proctor
Answer: d
Explanation: In 1933, Proctor showed the existence of relationship between soil water content and degree of dry density to which a soil might be compacted. Darcy is recognised for his woks in the field of permeability of the soil.

5. At Optimum water content, the soil has _____
a) minimum density
b) maximum density
c) no weight
d) no density
Answer: b
Explanation: In 1933, Proctor showed that for a specific amount of compaction energy applied on the soil, there was a water content termed as Optimum water content at which the soil attains the maximum density.

6. To test the compaction characteristics, the method are based on ______
a) dynamic loading only
b) kneading only
c) vibration loads only
d) dynamic, static, kneading
Answer: d
Explanation: There are many methods available for the test of compaction and these tests are based on the types of compaction and these are:
dynamic or impact
kneading
static and vibration.

7. The Standard Proctor test was developed by ________
a) R.R. Proctor
b) Skempton
c) Darcy
d) Terzaghi
Answer: a
Explanation: In 1933, Proctor showed the existence of relationship between soil water content and degree of dry density to which a soil might be compacted. He developed the Standard Proctor test for the construction of the earth fill dams in the state of California.

8. In Standard Proctor test, the inside diameter of the mould is ________ inches.
a) 1
b) 4
c) 3
d) 6
Answer: b
Explanation: In Standard Proctor test, the test equipment is a cylindrical metal mould that has an internal diameter of 4 inches or 10.15 cm.

9. In Standard Proctor test internal effective height is ________ inches.
a) 2.3
b) 4.5
c) 4.6
d) 45
Answer: c
Explanation: In Standard Proctor test, the test equipment is a cylindrical metal mould that has an internal effective height of 4.6 inches or 11.7 cm.

10. In Standard Proctor test, the capacity of the mould is ________ cu.ft.
a) 1/10
b) 1/20
c) 1/30
d) 1/40
Answer: c
Explanation: In Standard Proctor test, the test equipment is a cylindrical metal mould that has an internal effective height of 4.6 inches and an internal diameter of 4 inches. Therefore the volume of the cylinder is, πr2 h=π*42*4.6 cu.in=130 cu.ft.

11. In Standard Proctor test, the weight of rammer is ________
a) 4.5 kg
b) 18 lb
c) 2.5kg
d) 20 lb
Answer: c
Explanation: In Standard Proctor test, the layers of the specimen are tested with a free fall of rammer of weight 2.5 kg or 5.5 lb. The test consists in compacting at various water contents.

12. In Standard Proctor test, the free fall height is ________ inches.
a) 5
b) 12
c) 16
d) 18
Answer: b
Explanation: In Standard Proctor test, the layers of the specimen are tested with a free fall of rammer of weight 2.5 kg or 5.5 lb with the height of free fall being 12 inches or 310 mm.

13. The void ratio is increased by compaction.
a) True
b) False
Answer: b
Explanation: Compaction is a process by which the soil particles are artificially rearranged and packed together in a closer state of contact by mechanical means. This arrangement reduces the porosity of the specimen and hence the voids ratio.

14. The compaction reduces _________
a) strength
b) stability
c) air voids
d) dry density
Answer: c
Explanation: When the soil is compacted, it reduces the voids ratio and also its porosity. Hence the sir voids are reduced by the compaction process. The compaction increases the strength, stability and the dry density of the specimen.

15. _______ is a process in which soil particles are artificially rearrange into a close state.
a) Tension
b) Compaction
c) Sway mechanism
d) Consolidation
Answer: b
Explanation: Compaction is a process by which the soil particles are artificially rearranged and packed together in a closer state of contact by mechanical means. While, consolidation is the natural process of reduction in the voids ratio of the specimen.

16. Which of the following factors affects compacted density?
a) Water content and Type of compaction
b) Degree of saturation
c) None of the mentioned
d) All of the mentioned
Answer: a
Explanation: Water content, type of soil, addition of admixtures, amount and type of compaction are the various factors which affect the compacted density.

17. The compacted density is increased when the water content of the soil is _________
a) Increased
b) Decreased
c) Constant
d) None of the mentioned
Answer: a
Explanation: It has been seen from a laboratory experiment that, as the water content is increased the compact density goes on increasing, until a maximum dry density is achieved after which further addition of water decreases the density.

18. The force which is responsible for withholding of soil particles, of lower content is __________
a) Vander Waals force
b) Electric force
c) Frictional force
d) Cationic linkage
Answer: b
Explanation: When only a relatively small amount of water is present in the soil, it is firmly held by the electrical forces at the surface of soil particles with a high concentration of the electrolyte.

19. The amount of compaction greatly affects ___________
a) Water content and Maximum dry density
b) Saturation of soil
c) None of the mentioned
d) All of the mentioned
Answer: a
Explanation: The amount of compaction affects maximum dry density and optimum water content of a given soil. The effect of increasing the compactive energy results in an increase in the maximum dry density and decrease in the optimum water content.

20. Higher density and lower optimum water content is easily achieved by _________
a) Coarse grained soil
b) Fine grained soil
c) Cohesion less soil
d) Saturated soil
Answer: a
Explanation: Well graded coarse-grained soil attains a much higher density and lower water optimum water content then fine grained soil which requires more water.

21. The dry density decreases in cohesionless soil with an increase in water content due to which of the following reasons?
a) Capillary rise
b) Bulking of sand
c) Degree of saturation
d) Water content
Answer: b
Explanation: In case of cohesion less soil the dry density decreases with an increase in water content under a low compactive effect, this is due to bulking of sands where in the capillary tension resists the tendency of soil particles to take a density state.

22. The maximum density is reached in cohesion less soil when the soil is _________
a) Zero water content
b) Partially saturated
c) Fully saturated
d) Maximum specific surface
Answer: c
Explanation: The density reaches the maximum value when the cohesion less soil is fully saturated, on further addition of water, the dry density again increases.

23. The maximum bulking of sand occurs at a water content between ___________
a) 4 to 5 %
b) 2 to 6 %
c) 4 to 8 %
d) 1 to 5 %
Answer: a
Explanation: The maximum bulking occurs at water content between 4 to 5 %. On further additions of water, the meniscus is destroyed and the soil particles are able to shift to a closer packing.

24. The initial decrease of dry density at lower water content is exhibited in ____________ type of soil.
a) Fine grained soil
b) Black cotton soil
c) Alluvial soil
d) Cohesion soil
Answer: b
Explanation: The initial decrease of dry density at lower water content is a characteristic feature of black cotton soils, high swelling clays and fat clays. The optimum water content for such soils ranges between 20 to 25%.

25. The attainment of maximum density of soil at full saturation is due to __________
a) Lubrication action
b) Hydrostatic pressure
c) Bulking of sand
d) None of the mentioned
Answer: b
Explanation: The attainment of maximum density at full saturation is due to the reduction of effective pressure between soil particles by hydrostatic pressure.

26. The number of layers of soil compaction depends on ___________
a) Type of soil and Amount of compaction required
b) Water content of soil
c) None of the mentioned
d) All of the mentioned
Answer: a
Explanation: The number of soil layers and number of tamps per layers depends upon the type of soil and the amount of compaction required.

27. The rolling equipment’s are of __________ types.
a) 5
b) 4
c) 2
d) 3
Answer: a
Explanation: The rolling equipment’s are of five types:
i. Smooth level rollers
ii. Pneumatic tyred rollers
iii. Sheep foot rollers
iv. Pneumatic tyred construction plant
v. Track laying vehicles.

28. Sheep-foot-rollers is most suitable for compacting ____________
a) Fine-grained soil
b) Cohesive soil
c) Cohesion-less soil
d) Clay soil
Answer: b
Explanation: Sheet-foot-rollers are recommended for compacting cohesive soils, but are not considered effective on coarse grained cohesion-less soil.

29. The tyre pressures in the smaller rollers are in the order of ____________
a) 250 kN/m2
b) 400 kN/m2
c) 500 kN/m2
d) 100 kN/m2
Answer: a
Explanation: Smaller rollers are having a tyre load of about 7.5kN and pressure in the order of 250 kN/m2.

30. Which of the following rollers have wheels, mounted at slight angle with respect to axle?
a) Sheep foot roller
b) Smooth wheel roller
c) Wobble wheel roller
d) Tandem roller
Answer: c
Explanation: To provide a kneading action, the wobble wheel rollers have wheels mounted at a slight angle with respect to the axle.

31. The performance of a compaction equipment depends on _____________
a) Soil type
b) Water type
c) Particle size distribution
d) All of the mentioned
Answer: d
Explanation: The performance of compaction equipment depends on the soil type, its particle distribution and its water content.

32. Which of the following type of vibrating unit used in a vibrator?
a) Out-of balance weight type and Pulsating hydraulic type
b) Piston hydraulic type
c) None of the mentioned
d) All of the mentioned
Answer: a
Explanation: The vibrators consist of vibrating unit of either the out-balance weight type or a pulsating hydraulic type mounted on a screed, plate or roller.

33. Jumping rammers that are used for compacting the soil is also known as ____________
a) Frog rammers
b) Combustion rammers
c) Pneumatic rammers
d) None of the mentioned
Answer: a
Explanation: Internal combustion type jumping rammers are also known as frog rammers. They weigh up to 10 tons.

34. Pneumatic tyred rollers are best suited for ____________
a) Cohesion-less sand
b) Cohesive soils
c) Cohesion-less gravels
d) All of the mentioned
Answer: d
Explanation: Due to the combined action of pressure and kneading makes pneumatic rollers best suitable for cohesion-less sand and gravels and, and on cohesive soils.

35. The foot pressure in sheep foot rollers ranges from _____________
a) 800-350 kN/m2
b) 800-3500 kN/m2
c) 300-1500 kN/m2
d) 350 kN/m2
Answer: b
Explanation: For sheep foot rollers, the foot pressure ranges from 800 to 3500 kN/m2.

36. Which of the following property of soil is improved by compacting the soil?
a) Reduction of compressibility
b) Water absorption
c) Permeability
d) All of the mentioned
Answer: d
Explanation: The main aim of compacting soil is to improve some desirable properties of the soil, such as reduction of compressibility, water absorption and permeability, increase in soil strength, bearing capacity.

37. When compacted dry of optimum, the structure of clay is always ___________
a) Flocculated
b) Dispersed
c) Disoriented
d) Honey comb shaped
Answer: a
Explanation: As the structure of compacted clay is complicated and increasingly oriented, the structure of clay is always flocculated.

38. The flocculated structure of compacted dry soil is broken due to _____________
a) High strains
b) Low water content
c) Low strains
d) Dry density
Answer: a
Explanation: Due to higher strains, the flocculated structure of the compacted on the dry side is broken, giving rise to ultimate strength.

39. The shear strength of compacted clays depends upon ____________
a) Dry density
b) Water content
c) Degree of saturation
d) Addition of admixtures
Answer: a
Explanation: The shear strength of compacted clays depends upon dry density, moulding water content, soil structure, method of compaction, drainage condition and type of soil.

40. The structure of composite soil after compacted will be in the form of ______________
a) Coarse grained skeleton structure and Cohesive matrix structure
b) Single grained structure
c) None of the mentioned
d) All of the mentioned
Answer: a
Explanation: The structure of composite soils, after being compacted depend upon the relative proportion of coarse particles their structure can either be coarse grained skeleton structure or cohesive matrix structure.

41. Strength of soil sample compacted wet of optimum is influenced by ____________
a) Compressibility
b) Permeability
c) Manner of compaction
d) Amount of compaction
Answer: c
Explanation: The manner of compaction influences the strength of soil sample compacted wet of optimum to a certain extent.

42. The standard proctor test was developed by ___________
a) Darcy
b) Terzaghi
c) Proctor
d) Rendulic
Answer: c
Explanation: The standard proctor test was developed by R.R.Proctor in 1933 for the construction of earth fill dams in the state of California.

43. The compaction process can be accomplished by ___________ process.
a) Rolling
b) Tampering
c) Vibration
d) All of the mentioned
Answer: d
Explanation: Compaction can be done by rolling, tamping or vibration by a steel-tyred or rubber-tyred roller.

44. The compaction energy used for standard proctor test is ___________
a) 595 kJ/m3
b) 300 kJ/m3
c) 6065 kJ/m3
d) 1000 kJ/m3
Answer: a
Explanation: The compaction energy used in the proctor test is 6065 kg cm per 1000 ml of soil which is equal to 595kJ/m3.

45. Which of the following test are used in the laboratory, for compaction?
a) Vibration test
b) Standard proctor test and Jodhpur-mini compactor test
c) None of the mentioned
d) All of the mentioned
Answer:b
Explanation: Some of the compaction tests used in the laboratory is: standard and modified proctor test, Harvard Miniature compaction test, Abbot Compaction test and Jodhpur-mini compactor test.

46. A line showing the water content dry density relation for the compacted soil is ___________
a) Zero air voids lines
b) Air-voids line
c) Density line
d) All of the mentioned
Answer: b
Explanation: Air-voids line is a line which shows the water content dry density relation for the compacted soil containing a constant percentage of air voids.

47. Which of the following equipment is not used in standard compactor test?
a) Cylindrical metal mould
b) Rammer
c) Circular face plate
d) Collar
Answer: c
Explanation: The compactor test equipment consists of cylindrical metal mould, detachable base plate, collar, rammer of 2.5kg.

48. The initial percentage of water content taken for coarse-grained soil in proctor test is __________
a) 4
b) 10
c) 25
d) 50
Answer: a
Explanation: The initial water content may be taken as 4 % for coarse-grained soils and 10 % for fine-grained soils as the quantity of water to be added for the first test depends on the probable optimum water content for the soil.

49. The water content corresponding to the maximum density in compaction curve is called ____________
a) Water content of compacted soil
b) Optimum water content
c) Air void water content
d) None of the mentioned
Answer: b
Explanation: The water content goes on increasing till the maximum density is reached. The water content corresponding to the maximum density is called optimum water content W0.

50. The modified compactor test is also known as __________
a) Standard compactor test
b) AASHO test
c) Dietert test
d) Compaction test
Answer: b
Explanation: The modified compactor test was standardized by the American Associates of State highway officials and is known as the modified AASHO test.

51. In standard compactor test, soil is compacted into _____________ layers.
a) 2
b) 4
c) 3
d) 5
Answer: c
Explanation: The standard compactor test consists of compacting the soil at various water contents in the mould, in three equal layers, each layer being given 25 blows from the rammer.

52. The air voids line contains ___________
a) constant percentage air voids
b) water voids
c) varying percentage air voids
d) specific gravity of solids
Answer: a
Explanation: A line that shows the water content – dry density relation for the compacted soil containing a constant percentage air voids is known as air voids line.

53. The zero air voids line is also known as ___________
a) density line
b) saturation line
c) specific gravity line
d) unit weight line
Answer: c
Explanation: When the voids present in between the soil particles are completely filled by the moisture or the water, then it has no air entrapped in it or has no air voids. Therefore, the zero air voids line is also known as saturation line.

54. The theoretical maximum compaction for any given water content corresponds to ______
a) least specific gravity
b) maximum specific gravity
c) zero air void condition
d) only air voids present without the presence of water
Answer: c
Explanation: Compaction is a process by which the soil particles are artificially rearranged into a closer state. This arrangement reduces the voids ratio. When the voids present in between the soil particles are completely filled by the water, then it has no air voids. This zero air void condition, corresponds to the theoretical maximum compaction for any given water content.

55. The zero air void line shows the _________
a) dry density as a function of water content for soil containing no air voids
b) saturated density as a function of water content
c) partially saturated density as a function of water content
d) dry density as a function of specific gravity
Answer: a
Explanation: When the dry density is plotted as a function of water content, then for soil containing no air voids is represented by a line that is known as the zero air void line.

56. The Modified Proctor test was standardised by ____________
a) IS
b) AASHO
c) ASCE
d) CBR
Answer: b
Explanation: The Modified Proctor test was standardised by the American Association of State Highway Officials and the Modified Proctor test is also known as the AASHO test.

57. In Modified Proctor test the number of layers of the soil is compacted is ___________
a) 1
b) 2
c) 5
d) 3
Answer: c
Explanation: In Modified Proctor test the number of layers of the soil is compacted is five. While in Standard Proctor test the number of layers of the soil is compacted is three.

58. The weight of rammer in Modified proctor test is ______________
a) 4.5 kg
b) 2.5 kg
c) 18 lb
d) 22 lb
Answer: a
Explanation: In Standard Proctor test, the layers of the specimen are tested with a free fall of rammer of weight 2.5 kg or 5.5 lb. The test consists in compacting at various water contents. Whereas in Modified proctor test, the layers of the specimen are tested with a free fall of rammer of weight 4.5 kg or 10 lb.

59. The free fall height of rammer in Modified Proctor test is __________ inches.
a) 5
b) 12
c) 16
d) 18
Answer: b
Explanation: In Modified Proctor test, the layers of the specimen are tested with a free fall of rammer of weight 4.5 kg or 10 lb with the height of free fall being 18 inches. The free fall height of rammer in Standard Proctor test is 12 inches.

60. The Modified Proctor test curve of the water content – dry density curve lies __________ the SPT curve.
a) above
b) below
c) left side
d) right side
Answer: a
Explanation: The compactive energy given to the soil is more in the Modified Proctor test than the Standard Proctor test due to the weight of the rammer and the height of the free fall. The Modified Proctor test curve of the water content – dry density curve lies above the SPT curve.

61. The Modified Proctor test has its peak relatively towards ____________ in water content – dry density curve.
a) right
b) left
c) centre
d) near origin
Answer: b
Explanation: The compactive energy given to the soil is 27260 kg-cm per 1000 cm3 of soil for which the Modified Proctor test has its peak relatively towards left in water content – dry density curve.

62. For same soil, the effect of heavier compaction is in ________
a) increase in the maximum dry density
b) decrease in the maximum dry density
c) decrease in stability
d) decrease in strength
Answer: a
Explanation: For same soil, the effect of heavier compaction is in increase in the maximum dry density. Compaction reduces the voids ratio and hence there is decrease in the porosity. Due to this, the dry density of the soil sample is increased.

63. For same soil, the effect of heavier compaction is to ________
a) decrease the optimum water content
b) decrease in the maximum dry density
c) decrease in stability
d) decrease in strength
Answer: a
Explanation: Compaction reduces the voids ratio and hence there is decrease in the porosity. Hence, for same soil, the effect of heavier compaction is to decrease the optimum water content.

64. In Harvard miniature compaction test, the soil is compacted by _______
a) vibration
b) kneading action
c) impact
d) dynamic
Answer: b
Explanation: In Harvard miniature compaction test, the soil is compacted by kneading action of a cylindrical tamping foot that has a diameter of 0.5 inch or 12.7 mm. The soil is compacted in small cylindrical mould.