Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

# Multiple choice question for engineering

## Set 1

1. 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 [Reason:] Plasticity is the property of a soil which allows it to be deformed rapidly, without rupture, without elastic rebound and without volume change.

2. 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 [Reason:] 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.

3. 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 [Reason:] 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.

4. The shrinkage limit is represented by the term___________
a) IP
b) WS
c) IC
d) WP

Answer: b [Reason:] Shrinkage limit is represented as WS.

5. 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 [Reason:] In 1911, Atterberg who was a Swedish Agriculturist divided the entire range of liquid to solid state in to four stages.

6. 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 [Reason:] Liquid limit, plastic limit, shrinkage limit are useful for engineering purpose.

7. 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 [Reason:] The plasticity in soil is due to presence of thin scale like particles which carry on to their surfaces, electro-magnetic charges.

8. Clay does not become plastic, when mixed with___________
a) Soap solution
b) Kerosene
c) Oil
d) None of the mentioned

Answer: b [Reason:] According to Goldschmidt theory, the clay does not become plastic when mixed with liquids of non-polarizing agents like kerosene.

9. 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 [Reason:] According to Atterberg, the consistency limits are expressed as per cent water content.

10. 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 [Reason:] The four stages as divided by Atterberg are solid state, liquid state, plastic state, semi-solid state.

## Set 2

1. Falling head permeability test can be used for___________
a) Coarse-grained soil
b) Less permeable soil
c) Clayey soil
d) All of the mentioned

Answer: b [Reason:] Falling head test is used for relatively less permeable soils where the discharge is small.

2. Which of the following formula is used in Falling head permeability test?
a) Jacky’s formula
b) Louden’s formula
c) Darcy’s law
d) Kozney’s formula

Answer: c [Reason:] In falling head permeability test, Darcy’s law is used from which the rate of flow q, can be used q = K i A.

3. In falling head permeability test apparatus, the water head at any time instant is equal to____________
a) Difference in the water level
b) Cross-sectional area of pipe
c) Height of permeameter
d) None of the mentioned

Answer: b [Reason:] The head h1 and h2 at any time instant t is equal to the difference in the water level in the stand pipe and the bottom tank.

4. In constant permeability test, the length of specimen is measured by___________
a) Calculating the difference in water level
b) Using Piezometric tube
c) Using measuring scale
d) None of the mentioned

Answer: b [Reason:] In constant head test, the length of the sample is measured by inserting piezometric tubes on top and bottom.

5. The constant head permeability test is used for___________
a) Fine-grained soil
b) Coarse-grained soil
c) Clay soil
d) Saturated soil

Answer: b [Reason:] Coarse-grained soil, for which a reasonable discharge can be collected in a given time, is used in constant test.

6. The laboratory observation falling head test consist of measurements of____________
a) h1 at an interval of t1 and h2 at an interval of t2
b) only t1 and t2
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] The laboratory measurement of falling head permeability test consists of measurement of the heads h1 and h2 at two chosen time intervals t1 and t2.

7. Which of the following tubes are used in constant head test?
a) Inlet tube
b) Overflow tube
c) Outlet tube
d) All of the mentioned

Answer: d [Reason:] The constant head test apparatus consist of three tubes: the inlet tube, the overflow tube and outlet tube.

8. In falling head permeability test, change in the head at a time interval dt is denoted as___________
a) dh
b) –dh
c) h
d) –h

Answer: b [Reason:] -dh is used to denote change in a smaller time interval dt, the minus sign has been used since h decreases as t increases.

## Set 3

1. The wedge theory of earth pressure is based on the concept of_________
a) Active earth pressure
b) Sliding wedge
c) Wall friction
d) All of the mentioned

Answer: b [Reason:] The wedge theory of earth pressure is based on the concept of a sliding wedge which is torn off from the rest of the backfill on movement of the wall.

2. Which of the following is a basic assumption of the wedge theory?
a) The slip surface is plane
b) The backfill is dry
c) The backfill is homogeneous
d) All of the mentioned

Answer: d [Reason:] Based on the assumptions of the wedge theory: 1 The backfill is dry, cohesion less, homogeneous, isotropic, and elastically undeformable. 2 The slip surface is plane which passes through the heel of the wall 3 The sliding wedge itself acts as a rigid body.

3. The force acting on a wedge of soil are____________
a) Frictional force
b) Weight of the wedge and Active thrust
c) None of the mentioned
d) All of the mentioned

Answer: b [Reason:] According to wedge theory, the forces acting on a wedge of soil are: its weight W, the reaction R along the plane of sliding and the active thrust against the retaining wall.

4. The active lateral pressure of intact saturated clays is calculated by assuming__________
a) φ=0 and φc=0
b) φ=90
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] The active lateral pressure of intact saturated clays for temporary works or immediately after construction of retaining wall is calculated by assuming φ=φc=0.

5. The forces acting on the trial wedge which is used for finding Rankine’s active earth pressure is__________
a) Weight W of the wedge
b) Resultant force
c) Resultant reaction between wedge and sol
d) All of the mentioned

Answer: d [Reason:] The three forces acting on a trial wedge are: i) the resultant force that exist between the free body and the wall ii) the weight W of the wedge iii) the resultant reaction between the wedges and the rest of the soil along the surface.

6. The Rankine’s active earth pressure can also be found out by method of_____________
a) Trial wedges
b) Graphical method
c) Sliding wedge
d) Theoretical calculation

Answer: a [Reason:] The Rankine’s active earth pressure (assuming the wall to be smooth) can be found out by method of trial wedges.

7. Based on the assumptions of the wedge theory, pressure distribution is assumed to be__________
a) Planar
b) Hydrostatic
c) Equal
d) Distributed

Answer: b [Reason:] The assumption is based on that, the total pressure distribution is hydrostatic,i.e., triangular

8. In Coulomb’s wedge theory, the angle λ is referred as___________
a) Angle of wall friction
b) Surcharge angle
c) Critical slip angle
d) None of the mentioned

Answer: c [Reason:] In wedge theory the angle between the slip plane and the ground is called the critical angle λ.

9. The φ-line in wedge theory can also be called as__________
a) Surcharge line
b) Natural slope line and Repose line
c) None of the mentioned
d) All of the mentioned

Answer: b [Reason:] The φ-line shows a plane inclined at an angle φ to the horizontal at which the soil is expected to stay in the absence of any lateral support. The line therefore is called the natural slope line or repose line.

10. In coulomb’s wedge theory, the criterion for maximum active pressure is present at___________
a) Slip-plane
b) Repose line
c) Surcharge line
d) Ground line

Answer: a [Reason:] In 1971, Professor Rebhann presented that the criterion for maximum active pressure is present at the slip –plane of the assumed triangle.

## Set 4

1. What are the types of water flow in soil?
a) Turbulent flow and Laminar flow
b) Linear flow
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] 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 [Reason:] 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 [Reason:] 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 soil was first studied by___________
a) Taylor
b) Darcy
c) Lambe
d) Khosla

Answer: b [Reason:] 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 [Reason:] 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 [Reason:] 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. The co-efficient of permeability of fine sand is____________
a) 1.0 and greater
b) 1×10-6 and smaller
c) 1×10-2 to 5×10 -3
d) 5×10-2 to 1×10-3

Answer: d [Reason:] For fine sand, the coefficient of permeability is 0.05 to 0.001 cm /sec or 5×10-2 to 1×10-3.

8. Darcy’s law is valid for only___________
a) Laminar flow
b) Turbulent flow
c) Hydraulic flow
d) All of the mentioned

Answer: a [Reason:] Darcy’s la of linear dependency between the velocity of flow and hydraulic agent is valid for only laminar flow conditions in the soil.

9. 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 [Reason:] In 1933, Lewis and Barnes demonstrated experimentally that Darcy’s law is valid if it is ≤ 1.

10. Stiff clays are______________permeable.
a) Highly
b) Least
c) Partially
d) None of the mentioned

Answer: b [Reason:] As stiff clay does not contain any porous materials, it may be termed as least permeable or impermeable.

11. 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 [Reason:] The study of seepage of water through soil is important for the following engineering problem 1. Ground water towards soil and drainage of soil 2. Calculation of seepage through the body of earth dams ,and stability of slopes.

## Set 5

1. Shearing resistance can be determined in the laboratory by_________methods.
a) 2
b) 6
c) 4
d) 8

Answer: c [Reason:] shearing resistance can be determined in the laboratory by following four methods 1) Direct shear test 2) Tri axial shear test 3) Unconfined shear test 4) Vane shear test.

2. Which of the following shear test is developed based on drainage conditions?
a) Quick test and Consolidated un drained test
b) Direct shear test
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] Depending upon drainage conditions, three types of shear tests have been developed: a) Un-drained test or quick test b) Consolidated un-drained test c) Drained test.

3. The direct shear test can also be called as___________
a) Simple shear test
b) Stress test
c) Strain controlled shear box test
d) All of the mentioned

Answer: b [Reason:] Since the shearing strain is made to increase at a constant rate in direct shear test, and hence the test is called the Strain controlled shear box test.

4. The commonly used apparatus used for performing shear box test is___________
a) Shear-box apparatus
b) Bishop’s pore pressure apparatus
c) Tri axial shear test apparatus
d) None of the mentioned

Answer: a [Reason:] Shear-box test is most commonly used apparatus for direct shear test.

5. To conduct un-drained test, which of the following is used?
a) Slope grids
b) Perforated grids
c) Plain grids
d) All of the mentioned

Answer: c [Reason:] To conduct un-drained test, plane grids are used and for the drained test, perforated grids are used.

6. The drained test is also known as___________
a) Direct shear test
b) Slow test
c) Vane shear test
d) Quick test

Answer: b [Reason:] In drained test, perforated grids are sheared sufficiently slowly so that complete dissipation of pore pressure takes place.

7. Which of the following is a disadvantage of shear box test?
a) Stress condition of soil is complex
b) The test cannot be used for coarse grained soil
c) No control on the drainage of soil
d) The shear box test is more complex test

Answer: a [Reason:] The stress condition across the soil sample is very complex. The distribution of normal stresses and shearing stresses over the potential surface of sliding is not uniform. The entire strength of the soil is not mobilized simultaneously.

8. The shearing of cohesive soil in drained test requires___________days.
a) 2
b) 1 to 2
c) 2 to 5
d) 1

Answer: c [Reason:] As the soil in drained test is sheared sufficiently slowly so that complete dissipation of pore pressure takes place, it takes 2 to 5 days long for shearing cohesive type of soil.

9. A major difference between the direct shear test and tri axial shear test is_____________
a) Control on the drainage level
b) Stress condition
c) None of the mentioned
d) All of the mentioned

Answer: a [Reason:] As compared to the tri axial system, there is a little control on the drainage of soil in direct shear test.

10. In direct shear test, the soil load is subjected to more stress at_____________
a) Centre
b) Edges
c) Top and bottom
d) All of the mentioned