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JKSSB JE Civil Mock Test - 6 - Civil Engineering (CE) MCQ


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30 Questions MCQ Test JKSSB JE Civil Mock Test Series 2024 - JKSSB JE Civil Mock Test - 6

JKSSB JE Civil Mock Test - 6 for Civil Engineering (CE) 2024 is part of JKSSB JE Civil Mock Test Series 2024 preparation. The JKSSB JE Civil Mock Test - 6 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The JKSSB JE Civil Mock Test - 6 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 - 6 below.
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JKSSB JE Civil Mock Test - 6 - Question 1

As per IS : 10500-2012, for drinking water, in the absence of alternate source of water, the permissible limit for fluorides is

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 1

Water quality for different compounds As per l.S.10500 are as follows:

JKSSB JE Civil Mock Test - 6 - Question 2

In triangulation, the best shape of the triangle is______ triangle with base angle equal to _____.

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 2

Well Conditioned Triangles of a Triangulation System

  • The arrangement of triangles in the layout and the magnitude of the angles in individual triangles, affect the accuracy of a triangulation system.
  • The shape of the triangle in which an error in angular measurements, has a minimum effect upon the lengths of the computed sides, is known as a well-conditioned triangle.
  • Hence, the best shape of a triangle is an isosceles triangle whose base angles are 56° 14' each.
  • But, for all practical purposes, an equilateral triangle may be treated as a well-conditioned triangle. The triangles, whose angles are less than 30° or more than 120°, should be avoided in the chain of triangles.
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JKSSB JE Civil Mock Test - 6 - Question 3

If there are n number of events in the networks diagram then the number of dual role event will be

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 3

In any network, there can be only one initial and one final event then if there are n number of events in the networks diagram then the number of dual role event = n – initial event – final event
= n – 2
A few important definitions which are used in Project Management Techniques like CPM or PERT are specified below:

  1. Event:  It is a stage that refers to a particular instant of time that indicates the starting or completion of an activity. The occurrence of an event neither requires time nor any other resources. Example: excavation started, beam cast is some events.
  2. Tail and End event: An event that marks the starting of an activity called tail event and an event that marks the completion of an activity called end event.
  3. Dual Role event: An event that acts as both tail and end event is called a dual role event.
  4. Activity: It is the actual performance of the task which requires resources like time and cost for its completion.
JKSSB JE Civil Mock Test - 6 - Question 4

Fish plates are used in rail joints to :

  1. Maintain the continuity of rails
  2. Provide for any expansion or contraction
  3. Transfer the load to the ballast 
  4. Maintain correct alignment
Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 4

Fish plate is a metal bar that is bolted to the ends of two rails to join them together in a track.
The name is derived from fish, a wooden bar with a curved profile used to strengthen a ship's mast. The top and bottom edges are tapered inwards so the device wedges itself between the top and bottom of the rail when it is bolted into place.
Functions:

  1. Provide for any expansion or contraction
  2. Maintain correct alignment
  3. Provide for any expansion or contraction
JKSSB JE Civil Mock Test - 6 - Question 5

If ‘ϕ’ is the angle of blade tip at outlet, then maximum hydraulic efficiency of an impulse turbine is

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 5

The velocity diagram of a Pelton wheel is given below -

V1, V2 = Velocity of the jet at inlet and outlet.
u1, u2 = Velocity of the vane at inlet and outlet.
Vr1, Vr2 = Relative velocity of jet at inlet and outlet.
Vw1, Vw2 = Whirl velocity i.e. component of velocity parallel to the direction of motion.
Vf1, Vf2 = Flow velocity i.e component of velocity perpendicular to the direction of motion.
α = ∠ between the direction of jet and direction of motion.
θ = ∠ between Vr1 and the direction of motion.
β = ∠ between V2 with the direction of motion.
ϕ = ∠ between Vr2 with the direction of motion.
Hydraulic efficiency is given by - 

Given:
For Pelton wheel 
V1 = Vw1, u1 = u2 = u, Vr1 = Vr2.
If Nozzle is 100 % efficient -

Hydraulic efficiency is:

For ηh to be maximum; 

∴ 2u = 2V1 - 2u
⇒ 4u = 2V1
∴  V1 = 2u
Putting this in eq (1)

JKSSB JE Civil Mock Test - 6 - Question 6

The minimum depth of foundation for the load bearing wall of a building is restricted to

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

As per Cl. 7.2, IS:1904,
All foundations shall extend to a depth of at least 50 cm below the natural ground level.
The minimum depth of footing on sandy soil is 0.8 - 1 m, for rocky soil, it is 0.05 - 0.5 m and for clayey soil, it is 0.9 - 1.6 m.
The minimum depth of foundation for the load-bearing wall is 900 mm.

JKSSB JE Civil Mock Test - 6 - Question 7

The Nagpur Road Plan which is the First 20-year Road Development Plan assumes which of the following patterns of the road network? 

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 7

The first 20-year road plan is called the Nagpur Road plan (1943 - 1963)
Feature: This plan was a major attempt in planning for road development in a scientific manner. A road length of 5,32,700 km with a density of 16 km of road length per 100 km area would be available by 1963.
Note:
They recommended the construction of a Radial or Star and grid pattern of roads throughout the country.

All roads were classified into 5 categories
(i) National Highway
(ii) State Highway
(iii) Major District Roads
(iv) Other district roads
(v) Village roads

JKSSB JE Civil Mock Test - 6 - Question 8

If the sequent depth ratio of a hydraulic jump in a rectangular channel is 16.48, the Froude number at the beginning of the jump is

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 8

Relation between pre jump, post-jump and Froude number is given by,

Where,
y2 = Post jump depth
y1 = Pre-jump depth
F1 = Froude number before the jump
Calculation:
Given,
y2/y1 = 16.48

JKSSB JE Civil Mock Test - 6 - Question 9

The observed N value from a standard penetration test conducted on a saturated sandy soil stratum is 35. The corrected N value for dilatancy can be estimated as:

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 9

Dilatancy correction:
It is to be applied when N’ obtained after overburden correction, exceeds 15 in saturated fine sands and silts. IS: 2131-1981 incorporates the Terzaghi and Peck recommended dilatancy correction (when N’ > 15) using the equation.

Given,
N value = 35
Corrected  value for dilatancy is 

N'' = 15 + 10 = 25
∴ Corrected value N'' = 25 

JKSSB JE Civil Mock Test - 6 - Question 10

The road length of National Highway by Third Road Plan Formulae, in a certain district in India having its area as 13,400 sq.km will be

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 10

Length of different types of road according to Lucknow Road Plan:
1. Length of NH (in km) = 
2. Length of SH (in km) = 
or Length (in km) = 62.5 x Number of towns with population above 5000− 
3. Length of MDR (in km) = 
Length (in km) = 90 x Number of towns with a population above 5,000
4. Total road length (in km) = 4.74 x Number of villages and town
5. Rural Road Length (in km) ⇒  This can be calculated by finding the total road length and subtracting the other categories.
Calculation:
Given, Area = 13,400 m2
Length of NH (in km) = 

JKSSB JE Civil Mock Test - 6 - Question 11

The following Four statements (S1, S2, S3, S4) in connection with soil behaviour are given.
S1: Consolidation is a function of total stress
S2: Permeability of Coarse grained soil is inversely proportional to the specific surface at a given porosity.
S3: For a fully saturated soil the pore pressure parameter is equal to zero.
S4 : Terzaghi's theory of consolidation considers only primary consolidation.
Which of the above statement(s) is /are incorrect?

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 11

S1: False
The expulsion of pore water by applying a steady load for a long time from a fully saturated soil sample is called Primary Consolidation or simply consolidation.  It is completed when all the excess pore water gets expelled from the soil and it is time taking phenomenon.
As per Terzaghi's one-dimensional theory of consolidation, primary consolidation is a function of effective stress only and it is independent of total stress.
S2: True
The higher the specific surface area, the higher will be the friction so water will not flow easily, and hence, low will be permeability. Therefore, the Permeability of Coarse-grained or even fine-grained soil is inversely proportional to the specific surface area at a given porosity.
S3: False
Skempton's Pore Pressure coefficients are A and B, Where A depends upon the degree of saturation and over consolidation ratio and B depends upon the degree of saturation of the soil.
In general, A < 1 and B < 1. However, for over-consolidated soil the value of A < 0 and for fully saturated soil, the value of B is 1.
S4: True
Terzaghi's one-dimensional theory of consolidation considered only primary consolidation. Even after complete dissipation of excess pore pressure, the soil undergoes a little more consolidation due to plastic rearrangement of soil solids which is called secondary consolidation.

JKSSB JE Civil Mock Test - 6 - Question 12

If the bar shown in the figure is heated, _______ stresses would be developed in it.

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 12

Thermal stress:

  • When a bar or beam is subjected to a change of temperature and its deformation is prevented, the stress induced in the bar is called thermal stress.
  • In other words if the material is constrained (i.e. body is not allowed to expand or contract freely), change in length due to rise or fall of temperature is prevented, stresses are developed in the body which is known as thermal stress.
  • Example in case of fixed beam thermal stress will develop on heating or cooling as its deformation is prevented.
  • When the temperature of a material changes there will be corresponding change in dimension.
  • When a member is free to expand or contract due to rise or fall of the temperature, no stress will be induced in the member.
  • For example in free bar or beam, cantilever and simply supported beam, no thermal stress will be developed because they are free to expand along the length or line of deformation.
JKSSB JE Civil Mock Test - 6 - Question 13

In a pipe of diameter 5 cm, water is flowing at a rate of 8 cm/sec. If the dynamic viscosity of water is 1.6 × 10-2 Pa-s, what type of flow is present?

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 13

Reynold Number:

  • It is the ratio of inertia force to viscous force.
  • It determines the type of fluid flow.

The expression for Reynold's number is given as:

Where ρ = density of the fluid, V = mean velocity, D = diameter of the pipe, and μ = dynamic viscosity of the fluid.

  • The flow is laminar if Reynold number < 2000.
  • The flow is turbulent if Reynold number > 4000.

Given:
D = 5 cm = 5 × 10-2 m, V = 8 cm/sec = 0.08 m/sec, μ = 1.6 × 10-2 Pa-sec, ρ = 1000 kg/m3 

The flow is laminar as Reynold number < 2000.

JKSSB JE Civil Mock Test - 6 - Question 14

As per IS-456: 2000, Side face reinforcemnet shall be provided along the two faces of a reinforced concrete beam, when the depth of the web exceeds

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 14

Side face reinforcement:
Side face reinforcement shall be provided as per CI. 26.5.1.3 and 26.5.17(6) of IS 456:2000

  • Beam depth > 450 mm (if beam subjected to torsion)
  • Beam depth > 750 mm (if beam not subjected to torsion)
  • Provide @ 0.1% of web area and distribute it equally on both side faces
JKSSB JE Civil Mock Test - 6 - Question 15

Hazardous substances have the following attributes: 

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 15

Hazardous wastes have the following four characteristics:
1. Ignitability: 
Any waste that is flammable can create fires. Examples of this include liquids with flashpoints below 140 °F, non-liquids with the potential to ignite via specific conditions, and compressed gases.
2. Corrosivity:
Any waste (typically acids and bases) that can rust and decompose has the ability to melt through steel materials. Examples of this include aqueous wastes with an acidity level of equal to or less than 2 pH or equal to or greater than 12.5 pH.
3. Reactivity:
Potential for explosive mixture or violent reaction when combined with water.
4. Toxicity: 
Any waste that is fatally poisonous when ingested or absorbed. Examples of this include lithium-Sulphur batteries.

JKSSB JE Civil Mock Test - 6 - Question 16

Cant deficiency is

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 16

The maximum cant deficiency permitted for BG track for Speed ≤ 100 km/hr is 7.6 cm or 76 mm.
The actual cant provided on a railway track is based on the average speed of trains. However, in the case of high-speed trains, cant requirement will be more than the actual value of cant provided so that train will be forced to move on a lower value of cant then maximum cant is required for high-speed train. This deficiency in cant for high speed trains is called the cant deficiency, by this we can design the track for average speed between maximum speed of fast trains and other trains
The maximum permissible limit of cant deficiency is given below in tabulated form different type of railway gauges:

JKSSB JE Civil Mock Test - 6 - Question 17

If 'L' is the length of the cantilever slab measured parallel to the fixed edge, the effective width of the cantilever slab shall not exceed: (For slabs carrying the concentrated load)

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 17

If a solid slab supported on 2 opposite edges, carries concentrated loads the maximum bending moment caused by the concentrated loads shall be assumed to be resisted by an effective width of the slab as follows:
(1) For the cantilever solid slab, the effective width shall be calculated in accordance with the following equation
beff = 1.2 a1 + a
Where, beff = effective width, a1 = distance of the concentrated load from the face of the cantilever support, and a = width of the contact area of the concentrated load measured parallel to the supporting edge
Provided that the effective width of the cantilever slab shall not exceed one-third of the length of the cantilever slab measured parallel to the fixed edge.

JKSSB JE Civil Mock Test - 6 - Question 18

Consider the following statements regarding Bulk modulus of Elasticity:
I. It is a constant.
II. It varies with temperature.
III. It varies with pressure.
Which of the following is correct?

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 18

Bulk modulus of Elasticity:

  1. The bulk modulus of elasticity (K) of a fluid is not constant, but it increases with an increase in pressure. This is so because when a fluid mass is compressed its molecules become close together and its resistance to further compression increases i.e., K increases.
  2. The bulk modulus of elasticity (K) of the fluid is affected by the temperature of the fluid. In the case of liquids, there is a decrease of K with an increase in temperature. However, for gases since pressure and temperature are inter-related and as temperature increases, pressure also increases, an increase in temperature results in an increase in the value of K.

Thus, statement II and III are correct and statement I is incorrect.

JKSSB JE Civil Mock Test - 6 - Question 19

The ratio of lateral strain to linear strain is known as

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 19

Poisson's ratio (μ):

  • When a homogeneous material is loaded within its elastic limit (or limit of proportionality) the ratio of lateral strain to linear strain is constant and is known as Poisson's ratio.

Strain:

  • The ratio of change in dimension to original dimension is called strain
  • It is a dimensionless quantity.

Linear strain:

  • The ratio of axial deformation to the original length of the body is known as longitudinal or linear strain.

Lateral strain:

  • The strain at right angles to the direction of the applied load is known as lateral strain.
  • The length of the bar will increase while the breadth and depth will decrease.
JKSSB JE Civil Mock Test - 6 - Question 20

Calculate the number of sleepers required for constructing a railway track of 640 m long broad gauge section when the sleeper density is M + 5.

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 20

Given:
For BG track, length of rail = 12.8 m [General information]
∴ n = 13 [next positive integer of 12.8]
Number of sleepers per rail length = M + 5
So,
Number of sleepers per rail length = 13 + 5 = 18
Number of rails = 640/12.8 = 50
So number of sleepers needed = 50 × 18 = 900

JKSSB JE Civil Mock Test - 6 - Question 21

The minimum 28-day cube compressive strength prescribed in the Indian standard code IS 1343 for pre-tensioned member is_____.

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 21

A minimum grade of concrete to be used in the design of the prestressed concrete structure as per IS 1343 is as below:

  • For Post-tensioning minimum grade of concrete used is M-30.
  • For Pre-tensioning minimum grade of concrete used is M-40.

Hence, The minimum 28 days compressive strength for the pre-tensioned member is: M 40 or 40 N/mm2

JKSSB JE Civil Mock Test - 6 - Question 22

Match the items in List 1 (Name of field exploration) with those in List 2 (soil properties) and select the correct option.
Use codes for matching

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 22
  • Cyclic Pile Load test:
    • The cyclic load test helps to estimate how much load is taken up by the end bearing of the pile and how much by skin friction. The main objective of the cyclic load test is to separately determine the point bearing and friction bearing capacities of a pile.
  • Plate bearing or Load test:
    • The plate load test is a field test, which is performed to determine the ultimate bearing capacity of the soil and settlement under a given load.  
    • The modulus of subgrade reaction is calculated by using the relation K = P/Δ; where K is the modulus of subgrade reaction, p applied pressure, Δ is the settlement estimated using the Plate load test.
  • Pressure meter test:
    • It is used to determine the stress-strain relations of in-situ soil from which elastic constants are calculated.  It is mostly used for hard clays and dense sands.
  • Standard penetration test
    • It is used to determine the relative density, bearing capacity, and settlement of granular soil. 
JKSSB JE Civil Mock Test - 6 - Question 23

A scaled drawing of the proposed construction site showing all the relevant features such as entry and exit points to the site, storage area for materials, toilets, workers quarters, etc. is called

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 23

Job Layout:
A job layout is prepared so that the work proceeds smoothly without any obstruction. The various construction resources such as materials, machinery, man power should be arranged in such a way as to achieve optimal utilization of space. To place everything properly job layout is prepared.
Relevant features shown in a job layout are:

  • Entry point
  • Exit point
  • Storage place of materials
  • Temporary services [washing and toilet area]
  • Contractor's office
  • Area to keep equipment
  • Bar bending area
  • Labour housing [if any]
JKSSB JE Civil Mock Test - 6 - Question 24

Which type of levelling is used to measure across large gaps such as rivers or valleys?

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 24

​Reciprocal Leveling:

  • ​This technique of leveling work is used to find the exact height difference or to find the exact RL(Reduce level) of the point by equalizing the distance when there is a large obstruction like a river, ponds, lakes, etc. in direction of the survey.
  • It eliminates the following errors:
    (i) error in instrument adjustments i.e error due to collimation
    (ii) the combined effect of Earth's curvature and the refraction of the atmosphere
    (iii) variation in the average refraction.

JKSSB JE Civil Mock Test - 6 - Question 25

If a point in a strained material is subjected to two mutually perpendicular stresses, σx = 100 MPa (T) and σy = 50 MPa (C), then what will be the magnitude of maximum shear stress?

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 25

Given:
σx = 100 MPa (Tensile) and σy = - 50 MPa (Compressive)
Since it is a bi-axial state of stress ⇒ σ3 = 0 
Now, maximum shear stress is given by

∴ τ2 = 75 MPa.

JKSSB JE Civil Mock Test - 6 - Question 26

If D is the depth of scour below original bed, then the width of launching apron is generally taken as

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 26

Launching Apron:

  • To protect the face of the guide bank at the river bed level a thick stone cover is laid on the bed. It is called an apron. When the scour undermines the river bed the apron comes down or launches to cover the face of the scour. Hence it is called Launching apron also.
  • Heavy scour of the river bed at curved heads and shanks of guide banks can cause undermining of stone pitching thereby resulting in failure of guide banks. Such a failure can be prevented by providing a launching apron beyond the toe of guide banks.
  • The width of the launching apron is generally taken as 1.5D.

JKSSB JE Civil Mock Test - 6 - Question 27

PERT is

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 27

CPM does not directly model uncertainty.
PERT was developed to address the needs of projects which are being done for the first time – a challenge to estimate activity duration.
PERT considers the uncertainty in execution timings.
PERT is event oriented.
PERT (Program Evaluation and Review Technique) uses 3 cases:

  • Most Optimistic
  • Most Pessimistic
  • Most likely durations

PERT determines the probability for each duration, whereas CPM considers the most likely duration.
In the standard PERT analysis, the distribution assumed for the activity times is a Beta distribution.

JKSSB JE Civil Mock Test - 6 - Question 28

Sleeper spacing is the width of the sleeper + ______.

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 28

The minimum depth of ballast cushion is given by 


Where,
S is center to center distance between two sleepers
W is the width of sleeper
Db is minimum depth of ballistic cushion

JKSSB JE Civil Mock Test - 6 - Question 29

Match the following units with their respective conversions. 

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 29

Different unit's conversion of length are as follows:

Different unit's conversion of area are as follows:

JKSSB JE Civil Mock Test - 6 - Question 30

A detailed estimate that is prepared when there is a requirement of additional work during the progress of the original work is known as:

Detailed Solution for JKSSB JE Civil Mock Test - 6 - Question 30

Supplementary estimate
(i) It is a detailed estimate of additional work and is prepared when additional works or changes are required to supplement the original works, during the execution of work.
(ii) Then a fresh detailed estimate of additional works is prepared in addition to the original works.
(iii) The abstract should show the amount of the original estimate and the total amount including the Supplementary amount, for which sanction is required.  
Revised estimate
It is a detailed estimate and is required to be prepared under any one of the following circumstances:
(i) When the original sanctioned estimate is likely to exceed by more than 5 %.
(ii) When the expenditure on work exceeds or likely to exceeds the amount of administrative sanctioned by more than 10 %.
(iii) If there is a change of rate or quantity of materials.
(iv) Major additions or alterations are introduced in the original work.

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