Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  JKSSB JE Civil Mock Test Series 2024  >  JKSSB JE Civil Mock Test - 4 - Civil Engineering (CE) MCQ

JKSSB JE Civil Mock Test - 4 - Civil Engineering (CE) MCQ


Test Description

30 Questions MCQ Test JKSSB JE Civil Mock Test Series 2024 - JKSSB JE Civil Mock Test - 4

JKSSB JE Civil Mock Test - 4 for Civil Engineering (CE) 2024 is part of JKSSB JE Civil Mock Test Series 2024 preparation. The JKSSB JE Civil Mock Test - 4 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The JKSSB JE Civil Mock Test - 4 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JKSSB JE Civil Mock Test - 4 below.
Solutions of JKSSB JE Civil Mock Test - 4 questions in English are available as part of our JKSSB JE Civil Mock Test Series 2024 for Civil Engineering (CE) & JKSSB JE Civil Mock Test - 4 solutions in Hindi for JKSSB JE Civil Mock Test Series 2024 course. Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free. Attempt JKSSB JE Civil Mock Test - 4 | 120 questions in 120 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study JKSSB JE Civil Mock Test Series 2024 for Civil Engineering (CE) Exam | Download free PDF with solutions
JKSSB JE Civil Mock Test - 4 - Question 1

When Froude’s number is equal to unity, the flow in an open channel is called

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 1

Critical Flow:
It is defined as the flow for which the specific energy is minimum and Froude’s number is equal to unity.

The characteristics of critical flow are:
(i) The specific energy and specific force are minimum for the given discharge.
(ii) The Froude number is equal to unity.
(iii) For given specific energy the discharge is maximum at the critical flow.
(iv) The velocity head is equal to half the hydraulic depth in a channel of a small slope.
(v) Flow at the critical state is unstable.

Sub-critical flow
If
 the depth is greater and the velocity lower, therefore the Froude number is always less than 1.0.

Supercritical flow
If the depth is smaller and the velocity greater, the Froude number is always greater than 1.0.

JKSSB JE Civil Mock Test - 4 - Question 2

Consider a thin-walled cylindrical vessel having diameter 'x'. What would be the value of shear stress given that it is constrained to internal pressure 'I'? (where t is thickness)

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 2

Concept:
Given diameter = x, internal pressure = I
At any point in the material of the thin cylinder shell, there are two principal stresses.

Longitudinal stress: 

Hoop stress: 

As this is the case of a thin cylinder:
σr = 0
All these above stress can be considered principal stresses as no shear stress is acting.
Maximum shear stress,

1 Crore+ students have signed up on EduRev. Have you? Download the App
JKSSB JE Civil Mock Test - 4 - Question 3

What will be the optimum depth of ballast cushion required for a BG railway track below the sleepers with sleeper density of (M + 5) and bottom width of 22.22 cm?

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 3

Concept:
Sleeper density is denoted as L + m.
Where L=Length of one rail; m= any integer
Sleeper density signifies the number of sleepers per one rail length.
Optimum depth of ballast cushion = (S-W)/2
Where S = Spacing of sleepers
W=Bottom width of sleeper

Calculation:
Sleeper density = (M + 5) and bottom width = 22.22 cm
Length of one rail = 12.8 m
No of sleepers = 12.8 + 5 = 17.8
Spacing of sleepers = (12.8x100)/17.8 = 71.91 cm
Optimum depth of ballast cushion=(S-W)/2 = (71.91 - 22.22)/2 = 24.84 cm = 25 cm

JKSSB JE Civil Mock Test - 4 - Question 4

In Terazaghi's theory of bearing capacity of shallow foundation, which of the following zones is assumed to act as if it were part of the footing

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 4

Mechanism of Failure:
The shapes of the failure surface under ultimate loading conditions.
The zones of plastic equilibrium represented in this figure by the area may be subdivided into three zones:
1 . Zone I of elastic equilibrium
2. Zones II of radial shear state
3. Zones III of Rankine passive state

When load qu per unit area acting on the base of the footing of width B with a rough base is transmitted into the soil, the tendency of the soil located within zone I is to spread but this is counteracted by friction and adhesion between the soil and the base of the footing.
Due to the existence of this resistance against lateral spreading, the soil located immediately beneath the base remains permanently in a state of elastic equilibrium, and the soil located within this central Zone I behaves as if it were a part of the footing and sinks with the footing under the superimposed load.

JKSSB JE Civil Mock Test - 4 - Question 5

The recommended alum dosage for coagulation is in the range of?

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 5

Concept:
Coagulation:

The term coagulation describes the effect produced when certain chemicals are Added to Raw water containing slowly setting or Nonsettleable particles. The chemicals hydrolyse And  Neutralise the electrical charges in the form of colloidal particles, which began to form  Agglomerations Termed floc which Will be Removed by clarification and filtration.
According to IS code is provide this range 5 – 85 mg/l.
Different types of Coagulants added in the treatment of  water  are

  • Alum (Al2(SO4)3.18H2O –Hydrated aluminium sulphate
  • Copper (FUSO4. 7H2O)-Hydrated Ferrous Sulphate.
  • Chlorinated Copperos (Fe2(SO4)3.FeCl3)
  • Sodium  Aluminate (Na2 Al2O4)
JKSSB JE Civil Mock Test - 4 - Question 6

In a laminar flow between two parallel plates with a separation distance of 6 mm, the centre line Velocity is 1.8 m/s. The velocity at a distance of 1 mm from the boundary is:

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 6

Concept:
Laminar flow between the parallel fixed plate

Velocity distribution equation is given by:

Where,
u = velocity of the fluid at any distance y from the boundary
H = Distance between two parallel fixed plates.

Calculation:
H = 6 mm, ucentre = 1.8 m/sec

Where,
u = velocity of the fluid at any distance y from the boundary
H = Distance between two parallel fixed plates.

Calculation:
H = 6 mm, ucentre = 1.8 m/sec

JKSSB JE Civil Mock Test - 4 - Question 7

Formation of successive bends of reverse order may lead to the formation of a complete S curve called

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 7

Scouring:

  • Scouring can be defined as a process due to which the particles of the soil or rock around the periphery of the abutment or pier of the highway bridge spanning over a water body, get eroded and removed over a certain depth called scour depth.
  • Scouring usually occurs when the velocity of the flowing water increases or crosses the limiting value that the soil particles can easily handle.

Meander:

  • The formation of successive bends of reverse order in a river may lead to the formation of a complete S curve called meander.
  • When consecutive curves of reverse order connected with short straight reaches called crossings are developed in a river reach, the river is stated to be a meandering river. 

Causes:

  • Extra turbulence generated by an excess of river sediment during floods. When the silt charge is in excess of the quantity required for stability, the river starts building up its slope by depositing the silt on its bed.
  • The increase in bed slope tends to increase the width of channel if the banks are not resistant.
  • The banks are attacked by river water and in the process one bank is likely to be attacked slightly more than the other causing a slight deviation in flow. This slight deviation from the uniform axial flow helps in improving more and more flow towards one bank than towards the other.
  • This process continues causing one bank convex and other concave.
JKSSB JE Civil Mock Test - 4 - Question 8

A gully trap is provided at the junction of:

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 8

Gully Trap:

  • A gully trap or gully is provided at a junction of a roof drain, and other drains coming from the kitchen or bathroom.
  • It collects wastewater from the kitchen sink, wash basins, bathroom, and other wash areas.
JKSSB JE Civil Mock Test - 4 - Question 9

In a lemniscate curve, the angle between the tangent at the end of the polar ray and the tangent at the commencement of the curve (i.e. straight) is

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 9


Let P be a point on Laminscate with coordinates (μ, α).
When a tangent is drawn and it meets at P' to initial tangent, the angle between tangent and initial tangent is 'ϕ'
Where, β = The angle of polar ray op with tangent pp', and ∝ = polar deflection angle, ⇒ β = 2 α 
The equation of laminscate is given by: 
Differentiating w.r.t α. on both the sides,
We get,

∵ β = 2 α
We know ϕ = α + β = 2 α + α = 3 α 
⇒ ϕ = 3 α

JKSSB JE Civil Mock Test - 4 - Question 10

If stream function ψ = 2xy, then the velocity at a point (1, 2) is equal to ________.

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 10

Concept: 
The stream function in a two-dimensional flow automatically satisfies the continuity equation.
Stream Function: It is the scalar function of space and time.
The partial derivative of stream function with respect to any direction gives the velocity component perpendicular to that direction. Hence it remains constant for a streamline

Stream function defines only for the two-dimensional flow which is steady and incompressible..

Properties of stream function:

  • If ψ exists, it follows continuity equation and the flow may be rotational or irrotational.
  • If ψ satisfies the Laplace equation, then the flow is irrotational.

Continuity equation:

Calculation:

then the velocity at a point (1, 2),
V = √[(-2)2 + (- 4)2]
⇒ V = √20

JKSSB JE Civil Mock Test - 4 - Question 11

If the sewer is to be designed for the non-scouring velocity of 5 m/sec, which among the following material would you recommend?

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 11

Concept:
1. Brick Sewers:

  • Brick Sewers are made at the site and used for Construction large size sewer and carrying high-velocity sewage.
  • Brick Sewers are very useful for the Construction of storm Sewer or Combined sewers.
  • It can use the non-scouring velocity of 5 m/sec.
  • Nowadays brick sewers are replaced by Concrete sewers. Brick Sewers I get deformed and leakage may take place.

2. Cast iron Sewers:

  • These types of sewers are high strength and durability watertight.
  • Cast iron sewers can withstand high internal pressure and can bear an external load.
  • When there is a considerable difference in temperature.

3. Steel Sewers:

  • The Sewage is carried under pressure
  • They are generally used for outfall and trunk sewers.

4. Asbestos Cement Sewer:

  • Asbestos cement Sewers are manufacturing from a mixture of cement and asbestos fiber. Asbestos Cement Sewers are suitable for carrying domestic sanitary sewage 
  • Asbestos cement sewer is best as a vertical pipe for carrying sullage from the upper floors of the multistory building, but cannot be used for high sewage load or where velocity is high, as it is brittle in nature.

Advantages of Asbestos cement Sewer:

  • Smooth
  • Light in weight 
  • Can easily be cut, fitted, and drilled
  • Durable against soil corrosion.

Disadvantages of Asbestos Cement Sewer:

  • Brittle cannot withstand heavy loads.
  • They are easily broken in handling and transport.

5. Cement Concrete:

  • They may be cast in situ or precast, resistant to heavy loads, corrosion, and high pressure.
  • But these are very heavy and difficult to transport.
JKSSB JE Civil Mock Test - 4 - Question 12

A light house is visible just above the horizon at a certain station at the sea level. The distance between the station and the light house is 50km. The height of light house is

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 12

Concept:
Distance of visible horizon is given by:
D = 3.855√H
H = height of visible horizon (meters)
D= Distance of visible horizon (km)

Calculation:
Given, D = 50 km
50=3.855√H
H = 168.35 m

JKSSB JE Civil Mock Test - 4 - Question 13

Match the items under List I (Terms associated with placing of concrete) with those under List 2 (Definition of the phenomena). Use codes in lists for matching.

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 13
  • Mobility: The mobility is the property of fresh concrete to flow into the formwork around the steel reinforcement without the occurrence of segregation or bleeding. It is the ability of the concrete to be molded.
  • Bleeding: If excess water in the mix comes up at the surface causing small pores through the mass of concrete, it is called bleeding.
  • Segregation: It is caused when coarse aggregate is separated out from the finer materials resulting in large voids, less durability, and less strength.
  • Consistency: It refers to the fineness of the concrete or the ease with which it flows. It refers to the mean degree of wetness. It is an indication of workability.
JKSSB JE Civil Mock Test - 4 - Question 14

Find the delta for a crop when its duty is 1728 hectares/cumec on the field, the base period for the crop is 240 days.

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 14

Concept: 
The relation between Duty (D), Delta (Δ), and Base period (B) is given as:

Calculation:
Base period (B) = 240 days
Duty (D) = 1728 hec/cumec

Δ = 1.2 m = 120 cm

JKSSB JE Civil Mock Test - 4 - Question 15

The Garret diagram is applicable only for channel sections having side slopes of -

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 15

Concepts:
Garret's diagram

  • It is graphical representation of design of canal dimensions based on Kennedy theory.
  • The diagram is discharge plotted on abscissa (X-axis), and slope of the channel on the left side of ordinate (Primary Y-axis) and depth of the channel and critical velocity on the right side of the ordinate. (Secondary Y-axis).
  • Garret's diagram is applicable only for side slope of the channel of 0.5: 1.
  • The diagrams are specified for manning's coefficient (N) of 0.0225. However, these charts can be extended to any value of N using a nomogram at the top of each diagram.
JKSSB JE Civil Mock Test - 4 - Question 16

The hydrostatic law

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 16
  • According to Hydrostatic Law, the rate of increase of pressure in a vertical direction is equal to the weight density of the fluid at that point when the fluid is stationary.
  • The pressure at any point in a fluid at rest is obtained by the Hydrostatic Law which states that the rate of increase of pressure in a vertically downward direction must be equal to the specific weight of the fluid at that point.

    dP/dh = w for downward direction
    dP/dh = -w in upward direction

Thus, The rate of increase of pressure in a vertically downward direction is equal to Specific weight.
The hydrostatic law holds unaltered to for a liquid in a vessel subjected to constant rotation:

  • A liquid, contained in a vessel, may be rotated at a constant rotational velocity without any relative movement being created between different elements of the liquid in the vessel.
  • The liquid reorients itself once and for all to stay in that position with respect to the axis of rotation. The liquid mass is said to undergo a solid-body type of rotation, also known as a forced vortex motion.
  • This state of relative rest is also characterized by the absence of rates of strain and shear stresses in the liquid and the hydrostatic law may be used to estimate the pressure distribution, hydrostatic forces etc., due to the liquid.
JKSSB JE Civil Mock Test - 4 - Question 17

The optimistic, most likely and pessimistic time estimates of activity are 8, 15 and 18 weeks respectively. The Probable range of completion time of the activity will be

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 17

Concept:
The expected completion time (tE)

Where,
t0 = optimistic time, tL = most likely time
tP = pessimistic time
The probable range is from (te - 3σ) to (te + 3σ)   
σ is Standard Deviation

Solution:
The expected completion time (tE)​

The Probable range of completion time of the activity will be from (t− 3σ) t o(t+ 3σ)
i.e. (14.33 - 3 × 1.67) to (14.33 + 3 × 1.67)
9.32 weeks to 19.34 weeks

JKSSB JE Civil Mock Test - 4 - Question 18

Arrange the order of permanent adjustments of a theodolite
I. plate level test
II. cross-hair ring test
III. bubble tube adjustment test
IV. spire test
V. collimation in azimuth test
VI. vertical are test

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 18

The permanent adjustment is made to establish the fixed relationships between the fundamental lines of the instrument, and once made, they last for a long time.
The following are the permanent adjustments in a sequential manner in transit Theodolites:

  • Adjustment of the Horizontal Plate Levels
  • Cross-Hair ring test: This test is done to make the vertical crosshair lie in a plane perpendicular to the horizontal axis.
  • Collimation Adjustment (or Azimuth test): Azimuth test is done It is done to make the line of sight perpendicular to the horizontal axis.
  • Horizontal Axis Adjustment (Spire test) – The spire test is carried out to remove errors due to non-perpendicular of the horizontal axis and the vertical axis in the surveying Instrument.             
  • Adjustment of the Telescope Level or the Altitude level/bubble tube level.
  • Vertical Circle Index Adjustment (Vertical arc test): Vertical arc test is done to make the vertical circle indicate zero when the line of sight is perpendicular to the vertical axis.
JKSSB JE Civil Mock Test - 4 - Question 19

The sag in the dissolved oxygen curve results because of DO is a function of

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 19

Dissolved oxygen Curve:


Hence, The sag in the dissolved oxygen curve results because DO is a function of both addition and depletion of oxygen from the stream.

JKSSB JE Civil Mock Test - 4 - Question 20

Which one of the following is a gap developed in the canal bank due to erosion of some portion of the bank?

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 20

Different Definitions:
Remodeling

  • Remodeling of the canal aims at correcting the distribution system as per existing demand. It includes silt clearing operations, remodeling of outlets, the rectification of channels, modification of faulty hydraulic structures, etc.

Berm:

  • It is a narrow horizontal strip of natural ground left between the canal section and the inner toe of the bank. This is provided in such a way that the bed line and the bank line remain parallel.

Breaches:

  • They are the gaps created in the canal banks due to the breaking up of the banks. For bigger canals, this is reduced first followed by cutting the sides of the gap and then driving the double pile in the opening of the breach.

Internal Erosion:

  • It is also possible that a large slip surface could intersect the canal’s prism, but deformations are not large enough to release the water. Seepage along the slip surface and crack network may then develop as a scour-type internal erosion failure.
JKSSB JE Civil Mock Test - 4 - Question 21

Which method of contouring is most suitable for hilly terrains?

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 21

Different methods of contouring are:

Tachometric method of contouring

  • This method is used for angular surveying.
  • This method is best suited for hilly terrains as the number of stations which can be commanded by a tachometer is far more than those by a level and thus the number of instrument settings is considerably reduced.

Direct Method

  • The reduced level of different selected points is located and joined to form a contour.
  • This method is accurate but very tedious and slow.

Square method or grid method

  • In this method, we draw a grid point on the ground and measure the reduce level of all these points.
  • It is generally used in plain terrain.

Cross Section method

  • This method is generally used for hilly terrain to produce contour lines.
JKSSB JE Civil Mock Test - 4 - Question 22

Which of the following square slab will behave as one-way slab ?

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 22

Concept:
One-way slab:

(i) If a slab is supported only on two opposite supports, it is called a one-way slab.
(ii) If the slab is opposite on all four sides and span ratio 'ly/lx > 2', it is called a one-way slab.
(iii) Main reinforcement is provided along a shorter span.
(iv) Distribution reinforcement is provided along a longer span.
(v) maximum moment in the slab is along a shorter span direction.
(vi) When simply supported along two opposite edges square slab will behave as the one-way slab.
The maximum moment for the simply supported slab is given by,

Where lx = shorter span length 

JKSSB JE Civil Mock Test - 4 - Question 23

Vehicle damage factor (VDF) as given by I.R.C., is used in

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 23

Concept:

Vehicle Damage Factor (VDF): Different commercial vehicles have different weights and the conversion of the design number of commercial vehicles during the design life of the road into a number of standard axle repetitions. This is done using VDF.

The vehicle damage factor (VDF) is a multiplier for converting the number of commercial vehicles of different axle loads and axle configurations to the number of standard axle-load repetitions.

Mathematically,

This is used in the estimation of cumulative standard axles during the design life of the road which in turn is used in CBR method of design of flexible pavements.

JKSSB JE Civil Mock Test - 4 - Question 24

Which of the following represents the correct conditions for maximum discharge through a rectangular section?(Symbols and notations carry their usual meaning)

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 24

Concept:
Most economical section is the one whose wetted perimeter is minimum for the given value of discharge.

For minimum wetted perimeter,
d = b/2
We know, Hydraulic Radius (m) = wetted Area/Wetted perimeter
Wetted area = d × b = 2d2 
Wetted Perimeter = b + 2d = 4d
m = 2d2/4d
m = d/2 
∴ m = d/2 i.e. Half the depth of flow.

JKSSB JE Civil Mock Test - 4 - Question 25

How many CSR theories are there?

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 25

Concept:

Corporate social responsibility(CSR):

The title corporate social responsibility has two meanings. First, it’s a general name for any theory of the corporation that emphasizes both the responsibility to make money and the responsibility to interact ethically with the surrounding community. Second, corporate social responsibility is also a specific conception of that responsibility to profit while playing a role in broader questions of community welfare. Corporate social responsibility is composed of four obligations:

  • Economic responsibility
  • Legal responsibility
  • Ethical responsibility
  • Philanthropic responsibility

Theories of CSR:
The present practice of corporate social responsibility has been depicted and informed by three CSR theories:

  • The stakeholder theory of CSR
  • The business ethics theory of CSR
  • The shareholder value theory of CSR
JKSSB JE Civil Mock Test - 4 - Question 26

Which roller is most suitable for the compaction of a gravelly sand mixture with 25% fines? 

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 26

Smooth-Wheeled Roller: 

  • A smooth wheel roller generally consists of three wheels one small wheel at the front and two large wheels at the rear.
  • Smooth wheel rollers are useful for finishing operations after the compaction of fills and for compacting the granular base course of the highway.
  • These are not effective for compacting deep layers of soils as the resulting compaction pressure is low and also cause stratification in deep layers due to nonuniform compaction.
  • It is used to compact soil having gravel, sand, crushed rock, etc.

  • These roller compact the soil by the Pressure action.

Sheepfoot Roller: 

  • It consists of hollow drums with large numbers of small projections (known as feet) on its surface.
  • The projections penetrate the soil layers during rolling operation and cause compaction.
  • The sheep-foot rollers are ideally suited for the compaction of cohesive soils and these roller compact the soil by the combination of temping and kneading action.

Vibrating Roller:

  • In the case of vibratory rollers, vibrators are mounted on the drum and these rollers are available both as pneumatic type and smooth wheel type.
  • These rollers are suitable for granular soil with no fines in layers up to 1 m thick, however, if there is an appreciable percentage of fines, the thickness has to be reduced.

Heavy Roller:

  • Heavy ​rollers are suitable for compacting soil or clay, hardcore and gravel on smaller sites where a high level of compaction is still required.
  • Rubber-tired wheels at the rear allow these machines to climb steep gradients and provide traction in clay and soil conditions.
  • These heavy rollers provide considerably more compaction than smaller tandem rollers and their width allows large areas to be covered quickly.
JKSSB JE Civil Mock Test - 4 - Question 27

The value of Poisson's ratio for Brass material is

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 27

Poisson's Ratio:
Poisson's ratio is the ratio of transverse strain to longitudinal strain.

The value of Poisson's ratio for different materials is as follows:

JKSSB JE Civil Mock Test - 4 - Question 28

The minimum width recommended for kerbed urban road, so as to give allowance for stalled vehicle is:

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 28

Concept:
As per IRC 86-1983 the width of different type carriageways are given below.

Notes:

  • For access roads to residential areas, a lower lane width of 3 m is permissible.
  • Minimum width of a kerbed urban road is 5.5 m including allowance for a stalled vehicle.
JKSSB JE Civil Mock Test - 4 - Question 29

In Construction Management, the Critical Path Network helps an engineer

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 29

Concept:

  • The critical path is the longest path time-wise which can be determined from the forwarding path method only.
  • The critical path is the longest path through the network and time along this path gives the project duration. The critical path joins those activities which are critical.
  • The activities on the critical path are those activities that have total float equal to zero.
  • The activities that control the project duration are the ones that have zero total floats and form a continuous chain or path starting at the first node and ending with the last node.

So the Critical Path Network helps an engineer:

  • To concentrate his attention on critical activities.
  • To divert the resources from non-critical advanced activities to critical activities.
  • To be cautious for avoiding any delay in the critical activities to avoid delay of the whole project.

So here all of the above options are correct.

JKSSB JE Civil Mock Test - 4 - Question 30

Assumptions made in Euler's theory is/are -
(i) The column is initially straight
(ii) The cross-section is uniform throughout
(iii) The line of thrust coincides exactly with the axis of the column
(iv) The material is homogeneous but not isotropic
(v) The shortening of the column due to axial compression is negligible

Detailed Solution for JKSSB JE Civil Mock Test - 4 - Question 30

Euler’s theory:

  • It gives the buckling load for the column (compression member).

The following are the assumptions made in this theory:

  • The column is very long in proportion to its cross-sectional dimensions.
  • The column is initially straight and the compressive load is applied axially.
  • The material of the column is elastic, homogeneous, and isotropic.
  • The effect of direct stress is very small in comparison with bending stress.
  • Column shall fail by buckling alone.
  • The effect of the self-weight of the column is negligible.
  • The cross-section is uniform.
  • The shortening of the column due to axial compression is negligible.
  • The line of thrust coincides exactly with the axis of the column.
View more questions
8 tests
Information about JKSSB JE Civil Mock Test - 4 Page
In this test you can find the Exam questions for JKSSB JE Civil Mock Test - 4 solved & explained in the simplest way possible. Besides giving Questions and answers for JKSSB JE Civil Mock Test - 4, EduRev gives you an ample number of Online tests for practice

Top Courses for Civil Engineering (CE)

Download as PDF

Top Courses for Civil Engineering (CE)