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Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  APSC JE CE Paper 2 Mock Test - 2 - Civil Engineering (CE) MCQ

APSC JE CE Paper 2 Mock Test - 2 - Civil Engineering (CE) MCQ


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30 Questions MCQ Test - APSC JE CE Paper 2 Mock Test - 2

APSC JE CE Paper 2 Mock Test - 2 for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The APSC JE CE Paper 2 Mock Test - 2 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The APSC JE CE Paper 2 Mock Test - 2 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for APSC JE CE Paper 2 Mock Test - 2 below.
Solutions of APSC JE CE Paper 2 Mock Test - 2 questions in English are available as part of our course for Civil Engineering (CE) & APSC JE CE Paper 2 Mock Test - 2 solutions in Hindi for Civil Engineering (CE) course. Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free. Attempt APSC JE CE Paper 2 Mock Test - 2 | 100 questions in 120 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study for Civil Engineering (CE) Exam | Download free PDF with solutions
APSC JE CE Paper 2 Mock Test - 2 - Question 1
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Velocity distribution in a turbulent boundary layer follows:

Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 1
In turbulent boundary layer velocity distribution follows ((1/7)th) power-law i.e., logarithmic law.
APSC JE CE Paper 2 Mock Test - 2 - Question 2
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Dry rot -

Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 2
Dry rot is the decomposition of felled timber. It is caused by the action of various fungi. The fungus reduces fibre to a fine powder, and the timber losses its strength.
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APSC JE CE Paper 2 Mock Test - 2 - Question 3
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The shearing strength of a cohesion less coil depends on -

Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 3
Factor Affecting Shear Strength of Cohesionless Soil-

  • Shear strength of cohesionless soils is only developed with the presence of effective stress.

  • Shear strength increases with increasing effective stress.

  • The increase of shear strength of cohesionless soil depends upon the internal friction angles of the soil.

  • The type of soil most susceptible to liquefaction is one in which the resistance to deformation is mobilized by friction between particles. If other factors such as grain shape, uniformity coefficient, and relative density are equal, the frictional resistance of cohesion less soil decreases as the grain size of soils becomes smaller.

APSC JE CE Paper 2 Mock Test - 2 - Question 4
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Physical properties which influence permeability are -
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 4
Physical properties which influence permeability are Both viscosity and unit weight.

Hence the correct answer is option C.

APSC JE CE Paper 2 Mock Test - 2 - Question 5
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In a pipe when the valve is closed suddenly, then due to high pressure
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 5

During sudden closure of valve, a wave of high pressure intensity will be produced in the water, due to this high pressure circumferential and longitudinal stresses will be produced.

APSC JE CE Paper 2 Mock Test - 2 - Question 6
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Directions: A pair of number is given. Select the pair which shows the similar relationship that is shown by the given pair. 1984 : 4891
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 6

The pattern is as follows,
1984 is the reverse of 4891
Similarly, 2874 is the reverse of 4782.
Hence the required pair is 2874 : 4782.

APSC JE CE Paper 2 Mock Test - 2 - Question 7
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Bitumen is generally obtained from;
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 7
Destructive distillation
APSC JE CE Paper 2 Mock Test - 2 - Question 8
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For a continuous floor slab supported on beam, the ratio of the end span of length and intermediate span length is-
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 8
For a continuous floor slab supported on a beam, the ratio of the end span of length and intermediate span length is 0.9.

Hence the option D is correct.

APSC JE CE Paper 2 Mock Test - 2 - Question 9
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Section modulus for a rectangular section is given as-
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 9
Let rectangular section of width b and depth d then.

Section modulus, (Z=1/Ymax)

where (I=bd3/12)&(ymax=d/2)

Now

(Z=(bd3/12)/(d/2)=bd2/6)

Hence the option B is correct.

APSC JE CE Paper 2 Mock Test - 2 - Question 10
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Standard loads are given in-
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 10
IS 875-1964: Code of practice for Structural Safety of Buildings; Loading Standards.

Hence the option D is correct.

APSC JE CE Paper 2 Mock Test - 2 - Question 11
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The effective span of the simply supported slab is-
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 11
The effective span of a simply supported slab is the distance between the centre of the bearing and the clear distance between the inner face of the wall, thicker bar diameter plus twice the thickness of the wall

Hence the option D is correct.

APSC JE CE Paper 2 Mock Test - 2 - Question 12
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In an activated sludge plant F/M (Food to Microorganisms), the ratio is equal to.
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 12

and Sludge Age

Hence the correct answer is option B.

APSC JE CE Paper 2 Mock Test - 2 - Question 13
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The ratio of the rate of change of discharge of the outlet to the rate of change of discharge of the distributor channel is known as -
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 13
Flexibility: It provides criteria for judging the behaviour of modules and semi-modules. It is generally denoted by the letter 'F.,’ The ratio of the rate of change of outlet discharge to the rate of change of discharge of the parent channel.

Hence the correct answer is option C

APSC JE CE Paper 2 Mock Test - 2 - Question 14
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The principal rafter is designed as a:
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 14
A simply supported beam is a type of beam that has pinned support at one end and roller support at the other end. Depending on the load applied, it undergoes shearing and bending. It is one of the simplest structural elements in existence. The following image illustrates a simply supported beam.

A diagonal member of a roof principal, usually forming part of a truss and supporting the purlins on which the common rafters rest.

Hence the correct answer is option C.

APSC JE CE Paper 2 Mock Test - 2 - Question 15
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In-plane table, the operation which must be carried out is-
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 15
It is an operation for proper orientation of a plane table setup at any station needed to perform any plane tabbing measurement by the method.

Hence the correct answer is option B.

APSC JE CE Paper 2 Mock Test - 2 - Question 16
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The relationship between Young's modulus, E, shear modulus, G, and Poisson's ratio, v, is given by
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 16

The relationship between Young's modulus, E, shear modulus, G, and Poisson's ratio, v, is given by

Hence the correct answer is option A.

APSC JE CE Paper 2 Mock Test - 2 - Question 17
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Identify which grade of cement is not available in the Indian market.
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 17
23 grade of cement is not available in the Indian market.

Hence the correct answer is option A.

APSC JE CE Paper 2 Mock Test - 2 - Question 18
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Steel beam theory is a method of designing-
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 18
Steel beam theory is used to find the MR of the doubly reinforced beams. Steel beam theory is used to find the approximate value of the moment of resistance of a doubly reinforced beam, especially when the area of compression steel is equal to or more than the area of the tensile steel.
APSC JE CE Paper 2 Mock Test - 2 - Question 19
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Force method in structural analysis always ensures
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 19
The force method of analysis, also known as the method of consistent deformation, uses equilibrium equations and compatibility conditions to determine the unknowns in statically indeterminate structures. In this method, the unknowns are the redundant forces.
APSC JE CE Paper 2 Mock Test - 2 - Question 20
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For determining the ultimate bearing capacity of the soil, the recommended size of the square bearing plate used in the Plate load test is 30-75 cm with a minimum thickness of :
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 20
To determine the ultimate bearing capacity of the soil, the recommended size of the square bearing plate used in the Plate load test is 30-35 cm with a minimum thickness of 25mm.
APSC JE CE Paper 2 Mock Test - 2 - Question 21
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Age of a tree may be ascertained by :
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 21
If you know when the tree was planted, you can easily and accurately determine its age. The second most accurate way to estimate tree age is to count the annual rings of wood growth. Annual rings can be counted using two different methods. You can extract an increment core from the tree using an increment borer.
APSC JE CE Paper 2 Mock Test - 2 - Question 22
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The concrete mix which causes difficulty in obtaining a smooth finish is known to possess-
APSC JE CE Paper 2 Mock Test - 2 - Question 23
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When the deposits of efflorescence are more than 10 percent and less than 50 percent of the exposed areas of brick, the presence of efflorescence is classified as :
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 23
The degree of efflorescence is given as

(a)NIL – When there is no perceptible deposit of efflorescence.

(b)SLIGHT- Not more than 10% area of the brick covered with a thin deposit of salt.

(c)MODERATE- Covering up to 50% area of the brick.

(d)HEAVY- Covering 50% or more but unaccompanied by powdering or flaking of the brick surface.

(e)SERIOUS- When there is a heavy deposit of salts accompanied by powdering and/or flacking of the exposed surfaces.

Hence, the correct option is (B)

APSC JE CE Paper 2 Mock Test - 2 - Question 24
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Clumann's method assumes, the plane failure surface passing through-
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 24
Culmann's method assumes, the plane failure surface passing through the toe

A technique for the calculation of slope stability based upon the assumption of a plane

surface of failure through the toe of the slope has been proposed by Culmann. This slope stability analysis method is not widely used since it has been found that plane surfaces of sliding are observed only with very steep slopes, and for relatively flat slopes, the surfaces of sliding are almost always curved.

Hence, the correct option is (C)

APSC JE CE Paper 2 Mock Test - 2 - Question 25
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Which of the following defect arise due to seasoning of timber?
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 25
Defects that arises in timber during its seasoning are

Bow: Curvature of wood in the direction of its length is known as Bow.

Cup: Curvature in the transverse direction of timber is known as a cup.

Twist: It is the spiral distortion in the direction of timber.

Spring: Curvature of the timber in its plane

Note: While Upsets and Rind galls are the defects that arise due to natural forces.

APSC JE CE Paper 2 Mock Test - 2 - Question 26
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When the deposits of efflorescence are more than 10 percent and less than 50 percent of the exposed areas of brick, the presence of efflorescence is classified as:
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 26
The degree of efflorescence is given as

(a) NIL – When there is no perceptible deposit of efflorescence.

(b) SLIGHT- Not more than 10% area of the brick covered with a thin deposit of salt.

(c) MODERATE- Covering up to 50% area of the brick.

(d) HEAVY- Covering 50% or more but unaccompanied by powdering or flaking the brick surface.

(e) SERIOUS- When there is a heavy deposit of salts accompanied by powdering and flaking of the exposed surfaces.

APSC JE CE Paper 2 Mock Test - 2 - Question 27
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Scrap value of a property may be __________.
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 27
Scrap value is the value of dismantled material. Let's take an ln example of building, for a building when the life is over at the end of utility period, the dismantled materials such as steel, bricks, timber, etc. will fetch a certain amount which is the scrap value of the building, the amount of scrap value maybe about 10% of its total cost of construction. If the cost of dismantling the property is more than the value of scrap obtained from it, scrap value is negative.
APSC JE CE Paper 2 Mock Test - 2 - Question 28
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For controlling the growth of algae, the chemical generally used is:
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 28
Algae are microscopic free-floating plants that play an important role in aquatic ecosystems and directly affect the health of these systems. Through their photosynthetic processes, algae use carbon dioxide, converting it to organic matter and oxygen.

  • Algae are important sources of dissolved oxygen in a water impoundment and constitute the basic building blocks of the aquatic food chain

  • They cause water quality problems only when they proliferate so much that they upset the ecological balance

  • Copper sulfate is frequently used for algal control, and sometimes it is applied on a routine basis during summer months, whether or not it is actually needed

Note: Alum is used as a coagulant in the coagulation process.

APSC JE CE Paper 2 Mock Test - 2 - Question 29
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A structure has two degrees of indeterminacy. The number of plastic hinges that would be formed at complete collapse is
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 29
A number of plastic hinges for the following situations are given below

Therefore,

Number of plastic hinges = Degree of indeterminacy + 1

Number of plastic hinges = 2 + 1 = 3

APSC JE CE Paper 2 Mock Test - 2 - Question 30
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The durability of concrete is directly proportional to:
Detailed Solution for APSC JE CE Paper 2 Mock Test - 2 - Question 30
Durability of concrete may be defined as the ability of concrete to resist weathering action, chemical attack, and abrasion while maintaining its desired engineering properties.

  • More will be the cement-aggregate ratio, more will be the durability of concrete

  • The durability of concrete is inversely proportional to the water-cement ratio

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