Multiple choice question for engineering
Set 1
1. 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
Answer: d [Reason:] 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.
2. When compacted dry of optimum, the structure of clay is always___________
a) Flocculated
b) Dispersed
c) Disoriented
d) Honey comb shaped
Answer
Answer: a [Reason:] As the structure of compacted clay is complicated and increasingly oriented, the structure of clay is always flocculated.
3. 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
Answer: a [Reason:] Due to higher strains, the flocculated structure of the compacted on the dry side is broken, giving rise to ultimate strength.
4. The shear strength of compacted clays depends upon____________
a) Dry density
b) Water content
c) Degree of saturation
d) Addition of admixtures
Answer
Answer: a [Reason:] The shear strength of compacted clays depends upon dry density, moulding water content, soil structure, method of compaction, drainage condition and type of soil.
5. 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
Answer: a [Reason:] 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.
6. What is the maximum dry density for a soil sample having sp. gr. of 2.7 and OMC=16 %?
a) 3.0 g/cm3
b) 1.88 g/cm3
c) 0.562 g/cm3
d) 1.00 g/cm3
Answer
Answer: b [Reason:] γd, max occurs when S is maximum, i.e. when S=1
γd, max = G γW/1+ (w.G/S)
= G γW/1+w.G (since S=1)
Hence, ρd max = G. ρW/1+w.G
= 2.7×1/1 + (0.16×2.7)
Maximum dry density, γd max =1.885 g/cm3.
7. Strength of soil sample compacted wet of optimum is influenced by____________
a) Compressibility
b) Permeability
c) Manner of compaction
d) Amount of compaction
Answer
Answer: c [Reason:] The manner of compaction influences the strength of soil sample compacted wet of optimum to a certain extent.
8. A cohesive soil yields a maximum dry density of 1.8 g/cc at an OMC of 16 % during a standard proctor test. What will be its degree of saturation?Take G=2.65
a) 100 %
b) 60.45 %
c) 43.27 %
d) 89.79 %
Answer
Answer: d [Reason:] Given ρd = 1.8 g/cm3; w=0.1; G=2.65
e=G ρw/ρd -1 = (2.65×1/1.8) -1 =0.4722
S =w G/e = 0.16×2.65/0.4722=0.8979=89.79 %
Degree of saturation, S=89.79 %.
Set 2
1. 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
Answer: a [Reason:] According to Indian standards the soil fraction is set as 75 micron to be used in a liquid medium.
2. 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
Answer: a [Reason:] Both hydrometer and pipette method can be used for taking sedimentation analysis.
3. 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
Answer: a [Reason:] 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.
4. 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
Answer: c [Reason:] In sedimentation analysis, soil particle has average specific gravity.
5. The sedimentation analysis is based on _________ law.
a) Stoke
b) Cullman’s
c) Rankine’s
d) None of the mentioned
Answer
Answer: a [Reason:] 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.
6. 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
Answer: c [Reason:] Since ɳ=viscosity of water/liquid, it can be affected by temperature.
7. 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
Answer: a [Reason:] The particle smaller than 0.0002 mm is affected by Brownian movement and stokes law remains no longer valid.
8. 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
Answer: b [Reason:] As,1 poise =1/10 NS/ m2
1 poise =0.1 NS/ m2.
9. 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
Answer: a [Reason:] The constant value of unit weight of water/liquid is taken as γw=9.81 kn/m2.
10. Viscosity of water/liquid (ɳ) is expressed in________
a) KN-s/m3
b) Ns/m2
c) KN-s/m2
d) Ns/m
Answer
Answer: c [Reason:] The viscosity of distilled water in sedimentation analysis is taken as 0.00855 KN-s/m2.
Set 3
1. The seepage analysis is done on the assumption of____________
a) Flow is laminar and Darcy’s law is valid
b) Seepage of water
c) None of the mentioned
d) All of the mentioned
Answer
Answer: a [Reason:] The seepage analysis is done on the assumptions that that the flow is laminar and Darcy’s law is valid and further assumed that the soil to be dewatered is homogeneous and isotropic.
2. For unconfined flow, which of the following discharge slot is used?
a) Partially penetrating slot
b) Fully penetrating slot
c) None of the mentioned
d) All of the mentioned
Answer
Answer: b [Reason:] Fully penetrating slot is used for both confined and unconfined case of flow.
3. Fine grained cohesive soil can be drained or stabilized by__________
a) Vacuum method
b) Electro-osmosis method
c) Shallow well system
d) Well point system
Answer
Answer: b [Reason:] Dewatering of fine grained cohesive soil can be done only electro-osmosis method.
4. Discharge for a unconfined aquifer can be found from the expression developed by__________
a) Dupuit
b) Darcy
c) Chapman
d) Baron
Answer
Answer: c [Reason:] In 1956, chapmen developed an expression for finding the discharge qP from a partially penetrating slot in an unconfined aquifer from his model studies.
5. A complete design of dewatering system consist of___________
a) Spacing of wells
b) Penetration of wells
c) Pumping capacities
d) All of the mentioned
Answer
Answer: d [Reason:] Complete design of dewatering system consists of the diameter and spacing of well points or wells, penetration of wells and the pumping capacities.
6. When dewatering system consist of number of well-points___________can be considered for a simplified solution.
a) Linear drainage limit
b) Seepage analysis
c) Flow of water
d) All of the mentioned
Answer
Answer: a [Reason:] Where the dewatering system consists of a number of closely spaced well points or wells a simplified solutions can be obtained by considering the lines of wells equivalent to a linear drainage slot.
7. In discharge equation given by chapmen qp=kba(H-ho)/L+EA, EA refers to__________
a) Coefficient of permeability
b) Electric gradient
c) Extra length factor
d) Efficiency of permeability of soil
Answer
Answer: c [Reason:] EA= extra length factor, which is a function of the ratio of the ratio of slot penetration p to the thickness of pervious stratum b.
8. The application of electro-osmosis on dewatering of soil was developed by___________
a) L. casagrande
b) Darcy
c) Chapmen
d) Dupuit
Answer
Answer: a [Reason:] The application of electro-osmosis to dewatering of soil was largely developed by L. Casagrande.
9. For fine-grained soil what type of dewatering system can be used?
a) Electro-osmosis method
b) Shallow well system
c) Deep well system
d) Vacuum method
Answer
Answer: d [Reason:] For fine-grained soils, the well point system can be extended by the vacuum method.
10. The drawdown curve in fully penetrated slot is based on the assumptions of_____________
a) Dupuit and Forchheimer
b) Darcy
c) None of the mentioned
d) All of the mentioned
Answer
Answer: a [Reason:] Based on Dupuit-Forchheimer assumptions, the slope of the drawdown curve was assumed to be equal to hydraulic agent.
Set 4
1. The total settlement of a footing in clay is considered to be consisting of___________components.
a) One
b) Three
c) Two
d) Four
Answer
Answer: b [Reason:] According to Skempton and Bjerrum, the total settlement of a footing in clay may be considered to consist of three components
S = Si + Sc + Ss.
2. The component Sc, used in the total settlement of clay refers to which of the following?
a) Total settlement
b) Consolidation settlement
c) Immediate plastic settlement
d) Settlement due to secondary consolidation of clay
Answer
Answer: b [Reason:] The three components used in total settlement of clay are given below:
Sc = consolidation settlement
Si = immediate elastic settlement
Ss = settlement due to secondary consolidation of clay.
3. The immediate settlement can be computed from the expression, based on____________
a) Theory of plasticity
b) Theory of elasticity
c) Terzaghi’s analysis
d) Pressure distribution
Answer
Answer: b [Reason:] The immediate settlement is the elastic settlement and can be computed from the following expression based on the theory of elasticity,
Si =q B (1-μ2/E s) I w.
4. The influence factor for rigid square footing is____________
a) 0.88
b) 0.82
c) 1.06
d) 1.70
Answer
Answer: b [Reason:] As suggested by BIS Code IS: 8009 Part 1, 1976 the value of influence factor Iw for rigid footing is 0.82.
5. The equation for computing immediate settlement “Si = μ0μ1 q B ( 1-μ2/Es )” was proposed by___________
a) Janbu
b) Bjerrum
c) Kjaernsli
d) All of the mentioned
Answer
Answer: d [Reason:] In 1966, Janbu, Bjerrum, and Kjaernsli have proposed the following equation for computing the immediate settlement:
Si = μ0μ1 q B (1-μ2/Es).
6. The value of Es used in the immediate settlement equation, can be found out using_____________
a) Triaxial test
b) Compression test
c) Direct shear test
d) Rankine’s theory
Answer
Answer: a [Reason:] The modulus of elasticity Es used in the equation Si = μ0μ1 q B (1-μ2/Es) is computed from triaxial test data, where
Es = (σ1-σ3)/ΔL / L.
7. A combined footing may be rectangular in shape if both the columns carry___________
a) Unequal loads
b) Equal loads
c) No load
d) All of the mentioned
Answer
Answer: b [Reason:] The combined footing may be rectangular in shape if both the columns carry equal loads, or may be trapezoidal if they carry unequal loads.
8. The influence factor Iw for rigid rectangular footing with L / B = 1.5 is____________
a) 0.88
b) 0.82
c) 1.70
d) 1.06
Answer
Answer: d [Reason:] Iw = influence factor = 1.06 for rigid rectangular footing, with L / B =1.5
= 1.70 for rigid rectangular footing, with L / B =5.
Set 5
1. The common types of well shape are_________
a) Rectangular
b) Dumb bell
c) Single circular
d) All of the mentioned
Answer
Answer: d [Reason:] The common types of well shapes are: 1) single circular 2) Twin circular 3) Dumb-well 4) Double-D 5) Twin-hexagonal 6) Twin-octagonal 7) Rectangular.
2. Well are also called as__________
a) Open caissons
b) Shell sunk
c) Caisse
d) All of the mentioned
Answer
Answer: a [Reason:] Open caissons (“caisson” a French word) are called as wells.
3. ___________ forms the most common type of deep foundation for bridges.
a) Pile foundation
b) Well foundation
c) Shallow foundation
d) Pier foundation
Answer
Answer: b [Reason:] In India, well foundation forms the most common type of deep foundation for bridges.
4. The choice of a particular shape for a well depends on________
a) Dimension of the base and Cost of sinking
b) Type of soil condition
c) None of the mentioned
d) All of the mentioned
Answer
Answer: a [Reason:] The choice of a particular shape of well depends upon the dimension of the pier or abutment, the care and cost of sinking, the considerations of the tilt and shift during sinking.
5. What are the types of caissons that can be used as a foundation?
a) Box caissons and Open caissons
b) Closed caissons
c) None of the mentioned
d) All of the mentioned
Answer
Answer: a [Reason:] The caissons are of three types: i) Box caissons ii) open caissons iii) pneumatic caissons.
6. Which of the following caissons, permits excavation in dry?
a) Pneumatic caissons
b) Box caissons
c) Open caissons
d) All of the mentioned
Answer
Answer: a [Reason:] A pneumatic caisson has its lower end designed as a working chamber in which compressed air is forced to prevent the entry of water and thus permit excavation in dry.
7. The sinking of the dredge is uniform in which of the following, shapes of well?
a) Double-D
b) Rectangular
c) Circular
d) Dumb-well
Answer
Answer: c [Reason:] As the perimeter of the circle is equidistant at all points from the center of the dredge hole, the sinking is more uniform than other shapes.
8. The components that are needed to be considered in designing of a well foundations is_______
a) Shape of the well
b) Sand filing
c) Bottom plug
d) Dredge hole
Answer
Answer: c [Reason:] The following components of a well have to be considered in the design of a well foundation: well curb cutting edge, steining, Bottom plug and well cap.
9. A disadvantage in using circular shape of well foundation is___________
a) Minimum perimeter
b) Skin friction is maximum
c) Diameter of well is more than required
d) Sinking is more uniform
Answer
Answer: c [Reason:] The disadvantage of a circular well is that in the direction parallel to the span of bridge, the diameter of the well is much more than the minimum size required.
10. The most economical shape, of a well for construction of large pier is_________
a) Twin circular
b) Double-D
c) Rectangular
d) Single circular
Answer
Answer: b [Reason:] The dredge area for a double-D shape is smaller when compared to others; hence it can be more economical than a singular circular well.
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