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# Multiple choice question for engineering

## Set 1

1. 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 [Reason:] ASTM tool and casagrande tools are two types of grooving tools are used in determining liquid and plastic limit.

2. 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 [Reason:] The casagrande tools cut a groove of size 11.0 mm wide at the bottom.

3. 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 [Reason:] ASTM tool is used only for more sandy soil, where the Casagrande tool tends to tears the side of the groove.

4. 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 [Reason:] 425 micron sieve is used as a standard, for filtering the given specimen.

5. 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 [Reason:] Static cone penetration method is used in operation of soviet liquid limit device.

6. 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 [Reason:] Toughness index (It) is determined by the ratio of plasticity index to the flow index: It=IP/If.

7. 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 [Reason:] In liquid limit apparatus, the handle is rotated at a rate of about 2 revolutions per second.

8. 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 [Reason:] The cone has a central angle of 31°, since the total sliding mass is 80 g.

9. 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 [Reason:] The plasticity index is given by the formula, IP=WL-WP.

10. 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 [Reason:] 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.

11. The shear strength of the soil, at the liquid limit is________
a) 1.75 kN/m2
b) 1.89 kN/m2
c) 17.9 kN/m2
d) 17.6 g/cm2

Answer: d [Reason:] The shear strength of the soil is about 17.6 g/cm2 which the penetrometer gives for a depth of 25 mm under a total sliding mass of 148 g.

12. The apparatus used for finding, WL value by static cone penetrometer method is_________
a) Liquid limit apparatus
b) Pycnometer
c) Cone penetration
d) Rubber balloon cylinder

Answer: c [Reason:] Cone penetration apparatus is used for finding WL value, according to IS: 2720 (part 5) -1965.

## Set 2

1. Originally, Rankine’s theory of lateral earth pressure can be applied to only_______
a) Cohesion less soil
b) Cohesive soil
c) Fine grained soil
d) Coarse grained soil

Answer: a [Reason:] As originally proposed, Rankine’ s theory of lateral pressure is applied to uniform cohesion less soils only. Later, it was extended to include cohesive soil.

2. Rankine’s theory of lateral pressure was extended to other soil by__________
a) Resal and Bell
b) Mohr
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] The theory of lateral pressure was extended to cohesive, stratified, partially immersed and submerged soil was by Resal in 1910 and by Bell in 1915.

3. Based on the assumptions of Rankine’s theory, the soil mass is__________
a) Stratified
b) Submerged
c) Homogeneous
d) All of the mentioned

Answer: c [Reason:] According to the assumption of Rankine’s theory, the soil mass is semi-infinite, homogeneous, dry and cohesion less.

4. Which of the following cases for cohesion less backfill in Rankine’s theory is considered?
a) Submerged backfill
b) Moist backfill with no surcharge
c) Backfill with sloping surface
d) All of the mentioned

Answer: d [Reason:] In Rankine’s theory the following cases of cohesionless backfill are considered 1) Dry or moist backfill with no surcharge 2) Submerged backfill 3) Backfill with sloping surface and inclined back and surcharge.

5. The factor that is responsible for inclination of resultant pressure to the retaining wall is__________
a) Frictional force
b) Surcharge
c) Earth pressure
d) Weight of the wall

Answer: a [Reason:] The retaining walls are constructed of masonry or concrete, due to this the frictional force develops. The existence of the friction makes the resultant pressure inclined to the wall at an angle that approaches the frictional angle between the soil and the wall.

6. If the sand filled behind the retaining wall with saturated water with water, then the possible lateral pressure is__________
a) Lateral pressure due to submerged weight and Lateral pressure due to water
b) Lateral pressure due to retaining wall
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] The lateral pressure is made up of two components for wetted soil in back of the retaining wall: 1) Lateral pressure due to submerged weight ‘γ’ of the soil, and 2) Lateral pressure due to water.

7. The earth pressure at rest, exerted on a retaining structure can be calculated using__________
a) Theory of plasticity
b) Theory of elasticity
c) Mohr’s theory of rupture
d) None of the mentioned

Answer: b [Reason:] The earth pressure at rest, exerted on the back of the rigid, unyielding retaining structure, can be calculated using theory of elasticity.

8. The value of K0(coefficient of earth pressure at rest) for loose sand is__________
a) 0.6
b) 0.5
c) 0.4
d) 0.8

Answer: c [Reason:] The value of K0 for different soils is: a) Loose sand – 0.4 b) Dense sand – 0.6 c) Soft clay -0.6 d) Hard clay – 0.5

9. The expression for K0 as given by Jacky is___________
a) K0=1 – sin φ
b) K0= sin φ
c) K0=1 – cos φ
d) K0=1 + sin φ

Answer: a [Reason:] K0 can be calculated by using the following equation as computed by Jacky in 1944: K0=1 – sin φ.

10. What will be the coefficient of earth pressure at rest for a rigid retaining wall, If the backfill consist of cohesion less soil having φ = 26° ?
a) 0.1295
b) 0.6552
c) 0.5616
d) 0.7383

Answer: c [Reason:] Coefficient of earth pressure, K0 can be calculated using Jacky’s formula: Given: φ = 26° Formula: K0=1 – sin φ K0 = 1 – sin 26° K0 = 1 – 0.4383 K0 = 0.561628.

## Set 3

1. The plasticity of clay depends on which of the following?
a) The nature of clay minerals present
b) Volume of clay present
c) All of the mentioned
d) None of the mentioned

Answer: a [Reason:] The plasticity depends on i) The nature of clay minerals present ii) amounts of clay minerals present.

2. The typical activity value of illite, present in clay is________
a) 0.4-0.5
b) 0.5-1.0
c) 1.0-7.0
d) 2.7-5.3

Answer: b [Reason:] Typical active values are kaolinite : 0.4-0.5 Illite: 0.5-1.0 Montmorillonite : 1.0-7.0.

3. Based on activity number, clays can be classified in to_______ categories.
a) Five
b) Two
c) Three
d) More than five

Answer: c [Reason:] Clays are classified into 3 types as kaolinite, illite, and montmorillonite.

4. A clay is said to be inactive, if the activity is_________
a) <0.75
b) 0.75-1.40
c) >1.40
d) None of the mentioned

Answer: a [Reason:] Clay is classified as inactive if the activity is less than 0.75.

5. Clay containing kaolinite, will relatively have a_________ activity.
a) High
b) Intermediate
c) Low
d) Moderate

Answer: c [Reason:] Clay containing will have relatively low activity, while montorillonite have high activity.

6. Activity of a clay can be determined by laboratory test like__________
a) Wet analysis
b) Liquid limit
c) Plastic limit
d) All of the mentioned

Answer: d [Reason:] Activity can be determined by from the result of usual laboratory tests such as wet analysis, liquid limit, and plastic limit.

7. Based on the observation made by Skempton, for a given soil__________
a) Plastic index is proportional to percent of clay-size fraction
b) Plastic index is inversely proportional to percent of clay-size fraction
c) Plastic index is equal to percent of clay-size fraction
d) None of the mentioned

Answer: a [Reason:] In 1953 skempton, observed that for a given soil the plasticity index is proportional to the percent of clay-size fraction.

8. For a clay to be active, the activity of it should be__________
a) 0.75-1.40
b) >1.40
c) <0.75
d) <0.75

Answer: b [Reason:] Based on activity number, the activity should be greater than 1.4 for clay to be active.

9. The sensitivity of most clays usually falls in a range of_________
a) 2 to 8
b) 1 to 6
c) 1 to 8
d) 1 to 5

Answer: c [Reason:] From referring sensitivity classification table, it can be seen that clays have a common sensitivity between: 1 to 8.

10. Highly over consolidating clays tends to be_________
a) sensitive
b) normal sensitive
c) less sensitive
d) extra sensitive

Answer: c [Reason:] This type of clay tends to have sensitive of 1(unity) .Hence it is classified as in sensitive.

## Set 4

1. The selection of the depth of a well is based on which of the following criteria?
a) Adequate embedded length should be provided and The well should rest on strata of adequate bearing capacity
b) The perimeter of the well should be minimum
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] The selection of the depth of a well is based on the following two criteria: 1) There should be adequate embedded length of well, called the grip length below the lowest scour level. 2) The well should be taken deep enough to rest on strata of adequate bearing capacity in a relation to the load transmitted.

2. For alluvial soil, the normal scour depth can be calculated by___________ formula.
a) Rankine’s formula
b) Lacey’s formula
c) Converse Labarre formula
d) Terzaghi’s formula

Answer: b [Reason:] For alluvial soil mostly met with in North Indian Rivers, the normal scour depth can be calculated by Lacey’s formula: RL =1.35(q2/f)1/3.

3. The grip length of well is taken as_________ below the scour level in roads.
a) ½ R
b) 1½ R
c) 1/3 R
d) 1 R

Answer: c [Reason:] According to code of practice of the Indian Road Congress, the grip length is taken as 1/3 R below the scour level.

4. For piers and abutment, the minimum depth of embedment below the scour level should be taken as _________m.
a) 2.0
b) 1.2
c) 3
d) 1.5

Answer: a [Reason:] According to IRC code, it is recommended that the minimum depth of embankment below the scour level should not be less than 2.0 m for piers and abutment with arches and 1.2 m for other type of structures.

5. What are the horizontal forces that are prone to act on a well foundation?
a) Wind force
b) Earth pressure
c) Seismic force
d) All of the mentioned

Answer: d [Reason:] A well foundation is liable to the following horizontal forces: 1) force on account of water current 2) wind forces 3) seismic forces 3) earth pressure 4) centrifugal forces.

6. The analysis of well foundation is done based on the assumptions of_________
b) Lacey
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] Based on the assumptions done by Banerjee and Gangopadhyay in 1960, the analysis of well foundation can be done.

7. The maximum depth of scour at the nose of pier is_______ the Lacey’s value.
a) Twice
b) Thrice
c) Half
d) None of the mentioned

Answer: a [Reason:] The maximum depth of scour at the nose of pier, is found to be twice the Lacey’s value of normal scour depth: R = 2RL.

8. Coefficient of horizontal soil reaction m, depends on_________
a) Shape of the area carrying load and Nature of soil
b) Skin friction
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] Factor m, called the coefficient of horizontal soil reaction depends on the nature of soil, size and shape of the area which carries the load.

9. When a rigid well starts moving parallel, it transforms the soil in to__________
a) Active and passive states
b) Rigid state
c) Loose sand
d) All of the mentioned

Answer: a [Reason:] When a rigid soil, embedded in sand, starts moving parallel to its original position under the action of a horizontal force, it transforms the soil on one side to passive state and the other side in to active state.

10. According to Terzaghi and Peck, the ultimate bearing capacity can be determined by which of the following expression?
a) Q f = Q p + 2πRfs Df
b) Q p = πR2(Q p + 2πRfs Df)
c) Q p = 2πRfs Df
d) All of the mentioned

Answer: a [Reason:] According to Terzaghi and Peck, the ultimate bearing capacity can be determined from the following expression: Q f = Q p + 2 π R fs Df Where R = radius of well, Df = depth of well, fs =average skin friction.

## Set 5

1. 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 [Reason:] A flow net can be utilized for determination of seepage, seepage pressure .hydrostatic pressure, and exit gradient.

2. 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 [Reason:] The portion between any two successive flow lines is called as flow channel and the portion enclosed between two successive equipotential lines and successive flow lines is known as field.

3. Who was the first to give graphical method of flow net construction?
a) Casagrande
b) Darcy
c) Forchheimer
d) Kozney

Answer: c [Reason:] The graphical method of flow net construction first given by Forchheimer in 1930, based on trial sketching.

4. 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
d) None of the mentioned

Answer: a [Reason:] 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.

5. The flow lines and equipotential lines are____________
a) Parallel
b) Perpendicular
c) Elliptical
d) All of the mentioned

Answer: b [Reason:] The flow lines and equipotential lines always meet at right angles to one another.

6. 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 [Reason:] : 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.

7. 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 [Reason:] The seepage medium is replaced by an electric conductor consisting of water with some salt or dilute hydrochloric acid.

8. What will be the hydrostatic pressure if, hW =30 % and Z =10 %?
a) 30%
b) 10%
c) 40%
d) 50%

Answer: c [Reason:] Using the formula, hW= h -Z 30 = h -10 h=30+10 =40 Hydrostatic pressure, h=40 %.

9. In homogeneous 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