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# Multiple choice question for engineering

## Set 1

1. The curve obtained by plotting the normal and shear stress is called as___________
a) Mohr’s envelope
b) Coulomb envelope
c) Strength envelope
d) Stress envelope

Answer: c [Reason:] If the normal and shear stress corresponding to failure are plotted, then a curve is obtained .The plot or the curve is called the strength envelope.

2. Which of the following is coulomb’s strength equation?
a) S = c + tan φ
b) C = s + c tan φ
c) S = c + σ tan φ
d) S = tan φ

Answer: c [Reason:] Coulomb defined the function F (σ) as a linear function of σ and gave the following strength equation: S = c + σ tan φ.

3. The critical shear stress causing failure of material depends upon____________
a) Properties of the material and normal stress on the plane
b) Intermediate principal stress
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] According to Mohr’s strength theory, the critical shear stress causing failure depends upon the properties of the materials as well as on normal stress on the failure plane.

4. Theory of failure, was first proposed by____________
a) Coulomb
b) Mohr
c) Casagrande
d) Darcy

Answer: a [Reason:] The theory of failures was first expressed by coulomb in 1776 and later generalized by Mohr.

5. The Mohr-Coulomb theory can be expressed algebraically by, which of the following equation.
a) S = c + σ tan φ
b) τf = s = F(σ)
c) s = F(σ)
d) τf = F(σ)

Answer: b [Reason:] The Mohr-Coulomb failure theory can be expressed algebraically by the equation:τf = s = F (σ) Where, τf = s=shear stress on failure plane, at failure=shear resistance of material F (σ) = function of normal stress.

6. According to Coulomb, the relationship between shear strength and normal stress could be represented by_________
a) Linear curve
b) Parabolic curve
c) Straight line
d) None of the mentioned

Answer: b [Reason:] Coulomb considered that the relationship between shear strength and normal stress could be adequately represented by the straight line.

7. Mohr envelope can be considered to be straight if the angle of internal friction φ is assumed to be__________
a) 90°
b) >90°
c) <90°
d) None of the mentioned

Answer: d [Reason:] Mohr envelope can be considered straight if the angle of internal friction φ is assumed to be a constant.

8. Which of the following stresses does not have any influence on strength of a material?
a) Major principal stress
b) Minor principal stress
c) Intermediate principal stress
d) Shearing stress

Answer: c [Reason:] When a material is subjected to three dimensional stresses, the intermediate principal stress does not have any influence on the strength of material.

9. The parameter φ in coulomb’s equation “S = c + σ tan φ”, represents___________
a) Shearing resistance and Angle of internal friction
b) Angle of slope
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] The empirical constant φ in coulomb’s equation represents angle of internal friction or shearing resistance respectively.

## Set 2

1. The shearing resistance of a soil is constituted by___________
a) Structural resistance and Frictional resistance
b) Shearing strength
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] The shearing resistance of soil is constituted by 1) The structural resistance to displacement of the soil because of the interlocking of the particles 2) The frictional resistance to translocation between the individual soil particles at their contact points.

2. The shear strength in cohesion less soil is due to___________
a) Internal friction
b) Cohesion
c) Inter granular friction
d) Inter particle force

Answer: c [Reason:] The shear strength in cohesion less soils from inter granular friction, while in other soils it results both from internal friction as well as cohesion.

3. The planes that exist in soil mass is______________
a) Principal plane
b) Principal stress
c) Stress plane
d) None of the mentioned

Answer: a [Reason:] In a loaded soil mass, there exist three typical planes, mutually orthogonal to each other. These planes are called the principal planes.

4. The failure condition for a soil cam be expressed in terms of limiting shear stress, called__________
a) Principal stresses and Shear strength
b) Shearing resistances
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] The failure conditions for a soil may be expressed in terms of limiting stress, called shear strength or as a function of the principal stresses.

5. The shear deformation of soil in a building can cause___________of the following?
a) Sinking of footing
b) Slide in an earth embankment
c) Movement of wedge
d) All of the mentioned

Answer: d [Reason:] Shearing deformation of soil may be in the form of sinking of a footing, or movement of a wedge of soil particles or on masses upon the action of a shear stress.

6. Major principal stress in a soil is represented by the symbol________________
a) σ1
b) σ2
c) σ3
d) σ4

Answer: a [Reason:] Major, intermediate and minor principal stresses are represented by the corresponding symbols σ1, σ2, and σ3.

7. The circle obtained from two dimensional stress system is known as
a) Principal stress circle
b) Mohr circle
c) Shearing stress circle
d) None of the mentioned

Answer: b [Reason:] The circle drawn from two dimensional stress systems is known as Mohr’s circle of stress (Mohr, 1870).

8. The maximum shear stress τmax, for a soil mass is equal to___________
a) (σ1 – σ3 )/2
b) (σ1 + σ3 )/2
c) (σ1×σ3 )/2
d) (σ3 – σ1 )/2

Answer: a [Reason:] The maximum shear stress τmax is equal to (σ1 – σ3)/2 which occurs on planes with α= 45°.

9. The normal stresses acting on planes of the soil are known as_____________
a) Major principal stresses
b) Principal stresses
c) Minor principal stresses
d) Principal planes

Answer: a [Reason:] The normal stresses acting on principal planes are called the principal stresses.

10. Stress component on planes of a loaded soil mass depends upon____________
a) Stress acting on plane
b) Direction of plane
c) Shearing resistance
d) All of the mentioned

Answer: b [Reason:] Through a point in a loaded soil mass, innumerable planes passes and stress component on each plane depend upon the direction of the plane.

## Set 3

1. For engineering purpose, soil can be classified in terms of___________
a) Particle size
b) Textural
c) Unified soil
d) All of the mentioned

Answer: d [Reason:] For general engineering purposes, soil can be classified in terms of particle size, textural and unified soil.

2. Systems which are used for classification of soil based on particle size are__________
a) PRA system of united states
b) Indian standard classification system
c) International soil classification
d) All of the mentioned

Answer: d [Reason:] Based on particle size, soil can be classified as 1) U.S. Bureau of soil and public road administration system 2) The M.I.T. classification system 3) Indian standard classification system 4) International soil classification system.

3. In particle size classification system ,the soils are classified according to__________
a) Grain size
b) Properties
c) Shape
d) Solubility

Answer: a [Reason:] In the particle soil classification system, soils are arranged to the grain size.

4. Which of the following terms are used to indicate grain sizes?
a) Gravel
b) Sand
c) Silt
d) All of the mentioned

Answer: d [Reason:] Terms such as gravel, sand, silt, and clay are used to indicate grain sizes .these terms are used only as designation of particle size.

5. The term ‘silt’ in particle size classification system ,can be replaced by___________
a) Silt size
b) Clay size
c) Silt type
d) None of the mentioned

Answer: a [Reason:] As the system is classified based on size, it is preferable to use the word ‘silt size’ and ‘clay size’ in place of simply ‘silt’ or ‘clay’.

6. Indian standard classification of soil is based on_________
a) PRA system
b) International soil classification
c) M.I.T system
d) Indian system

Answer: c [Reason:] Indian standard classification (IS: 1948-1970) is based on the M.I.T. system.

7. Soil occurring in nature, is composed of__________
a) Sand
b) Decomposed substance
c) All of the mentioned
d) None of the mentioned

Answer: a [Reason:] Soil occurring in nature is composed of different percentage of sand, silt, and clay size particles.

8. The purpose of soil classification is to___________
a) To arrange various soils types in to groups
b) To use it for various purpose
c) For finding its properties
d) For investigating the soil

Answer: a [Reason:] The purpose of soil classification is to arrange various types of soils in to groups according to their engineering or agricultural properties.

9. Particle size classification is best suited for_________
a) Coarse grained soil
b) Clay soil
c) Dry soil
d) All of the mentioned

Answer: b [Reason:] Since properties of clay soils is less dependent on particle size distrubution(textural classification) and more dependent on size it is best suited for clay soil.

10. Particle size classification system does not signify__________
a) Soil types
b) Soil Particle size
c) All of the mentioned
d) None of the mentioned

Answer: a [Reason:] Particle size classification used only as designation of particle sizes, and do not signify the naturally occurring types which are mixtures of particle of different sizes.

## Set 4

1. 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 [Reason:] A soil is said to be well graded when it has a good representation of particle of all size.

2. 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 [Reason:] The size D10 is sometimes called as the effective size or effective diameter.

3. 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: b [Reason:] The coefficient of curvature (Cc) is given by the equation Cc = (D40)2/D10×D30.

4. 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 [Reason:] The coefficient of uniformity CV, is defined as the ratio of D60 and D10 i.e.CV = D60/ D10.

5. 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 [Reason:] The right side the particle size distribution curve shows the coarse-grained while the left shows the relatively fine-grained soil.

6. 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 [Reason:] γ=0.905(0.5)2=0.2263 m/sec Time of settlement, T=h/v =5/0.2263=22.1 seconds.

7. 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 [Reason:] A curve with a flat portion represents oil in which intermediate size particle are missing.

8. The D10 represent a size, such that _________ of the particles are finer than this size.
a) 20%
b) 60%
c) 10%
d) 100%

Answer: c [Reason:] The D10 represent a size, in mm such that 10% of the particle is finer than D10 size.

9. The time of settlement of finest particle of soil sample, is 15hr20min49sec .calculate the height of the water tank .Take D=0.01 mm,γ=0.905D2.
a) 5 m
b) 7 m
c) 12 m
d) 4 m

Answer: a [Reason:] γ=0.905(0.01) γ=9.05×10-5 t = 15hr 20min 49sec =55429 sec t=h/v 55429=h/9.05×10-5 h=55429×9.05×10-5 h = 5.000034 m.

10. 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 [Reason:] Coefficient of the curvature Cc represent the shape of the particle size curve given by CC= (D40)2/D10×D30.

11. 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 [Reason:] A particle-size curve shows both type and gradation of the soil.

## Set 5

1. What are the essentials, required to draw a flow net?
a) Top Flow and Phreatic line
b) Stream line
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] In order to draw the flow net, it is first essential to find out the location and shape of phreatic line and top flow line separating the saturated and unsaturated zones.

2. The phreatic line can be located by which of the following method?
a) Graphical method
b) Experimental method
c) Analytical method
d) All of the mentioned

Answer: d [Reason:] As suggested by casagrande, the phreatic line can be located by graphical, experimental and analytical methods.

3. Kozney’s top flow lines is called as____________
a) Basic and Base parabola
b) Simple parabola
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] kozney’s solution for flow of water consists of a family of confocal parabola s representing the flow lines. Hence kozney’s top flow line is called the basic or base parabola.

4. For the water flowing above an impervious ,infinite ,horizontal plane .the net flow is given by___________
a) Casagrande
b) Kozney
c) Forchheimer
d) Darcy

Answer: a [Reason:] In 1938, kozney analytically derived a flow net for the case of water flowing above an impervious, infinite, horizontal plane which at a certain place becomes permeable.

5. What is the line within a dam section, below which there are positive hydrostatic pressures?
a) Phreatic and Seepage line
b) Equipotential line
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] the phreatic line or seepage line is the line within a dam section below which there are hydrostatic pressures in the dam .the hydrostatic pressure below on the phreatic line itself is atmospheric.

6. The flow lines and equipotential proposed by Kozney is in the shape of____________
a) Hyperbola
b) Ellipse
c) Parabola
d) Circle

Answer: c [Reason:] According to flow net derived by Kozney, the flow lines and equipotential consist in the shape of parabola.

7. The analytical solution of schaffernak and Van Iterson gave result only for slopes, having__________
a) α>30°
b) α<30°
c) α>60°
d) α<60°