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JKSSB JE Civil Mock Test - 3 - Civil Engineering (CE) MCQ


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30 Questions MCQ Test - JKSSB JE Civil Mock Test - 3

JKSSB JE Civil Mock Test - 3 for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The JKSSB JE Civil Mock Test - 3 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The JKSSB JE Civil Mock Test - 3 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 - 3 below.
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JKSSB JE Civil Mock Test - 3 - Question 1

Schmutze decks will be formed in

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 1

In slow sand filter biological metabolism occurs which helps in filtering process.
Biological metabolism:
Certain micro-organisms and bacteria are generally present in the voids of the filters. They may either reside initial process of filtration. Nevertheless these organisms require organic impurities as their food for their survival.
These organisms therefore utilise such organic impurities and convert them into harmless compounds by the process of biological metabolism. The harmless compounds so formed generally form a layer on the top, which is called schmutzdecke or dirty skin. This layer further helps in absorbing and straining out the impurities.

JKSSB JE Civil Mock Test - 3 - Question 2

The maximum spacing of shear reinforcement (as vertical stirrups) measured along the axis of a reinforced concrete beam of width 300 mm and effective depth 500 mm, with a span 6 m is:

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 2

Maximum spacing of shear reinforcement(According to cl. 26.5.1.5 of IS 456:2000)
The maximum spacing of shear reinforcement measured along the axis of the member shall not exceed 0.75d for vertical stirrups and d for inclined stirrups at 450 where d is the effective depth of the section under consideration.
In no case shall the spacing exceed 300 mm.
​Calculation:
Given data:
Width = 300 mm
Effective depth = 500 mm
Span = 6 m
The maximum spacing of shear reinforcement (as vertical stirrups)
= 0.75 × 500 = 375 mm(In no case shall the spacing exceed 300 mm)
Hence, The maximum spacing of shear reinforcement (as vertical stirrups) = 300 mm

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JKSSB JE Civil Mock Test - 3 - Question 3

Consider the following statements:

  1. Bitumen is more resistant to water than tar.
  2. Bitumen is a petroleum product and tar is produced by destructive distillation of coal or wood.

Which of the above statements is(are) correct?

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

Difference between Bitumen and Tar:

JKSSB JE Civil Mock Test - 3 - Question 4

For a 5° Non-transitioned curve on narrow gauge track, the safe speed as per Martin formula is

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

Safe speed as per Martin’s formula is:
 [For transitioned curve]
R → Radius of curve in metres
For non-transitioned curves, the safe speed is reduced by 20%

For Non-transition curve
Safe speed = 0.80 × 67.69 = 54.16 kmph

JKSSB JE Civil Mock Test - 3 - Question 5

Which one of the following is not a self-reading staff?

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 5

A levelling staff is a straight rectangular rod having graduations, the foot of the staff representing zero reading. The purpose of a level is to establish a horizontal line of sight.
The purpose of the levelling staff is to determine the amount by which the station is above or below the line of sight. Types of Levelling staves may be divided into two classes

  • Self-reading staff
  • Target staff

A self-reading staff is one that can be read directly by the instrument man through the telescope. A target staff, on the other hand, contains a moving target against which reading is taken by staff man.
Self-reading staff:

  • This staff reading is directly read by the instrument man through a telescope. In a metric system staff, one-meter length is divided into 200 subdivisions, each of uniform thickness of 5 mm.
  • All divisions are marked with black on a white background. Meters and decimetres are written in red colour. The following three types of self-reading staff are available:
    • Solid staff: It is a single piece of 3 m.
    • Folding staff: A staff of two pieces each of 2 m which can be folded one over the other.
    • Telescopic staff: A staff of 3 pieces with upper one solid and lower two hollow. The upper part can slide into the central one and the central part can go into the lower part.

​Target staff:

  • If the sighting distance is more, instrument man finds it difficult to read self-reading staff.
  • Target staff is similar to self-reading staff but provided with a movable target. Target is a circular or oval shape, painted red and white in alternate quadrants. It is fitted with an a vernier at the centre.
  • The instrument man directs the person holding the target staff to move the target, till its centre is in the horizontal line of sight. Then target man reads the target and is recorded.
JKSSB JE Civil Mock Test - 3 - Question 6

A circular plate 1.5 m diameter is submerged in water with its greatest and least depths below the surface being 2 m and 0.75 m respectively. What is the approximate total pressure on one face of the plate?

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


Here,

Now,
Total pressure force = 
Total pressure force = 
∴ Total pressure force= 23.84 kN ≈ 24 kN

JKSSB JE Civil Mock Test - 3 - Question 7

In a flow field, the streamlines and equipotential lines

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 7

Streamline is an imaginary curve drawn through a flowing fluid in such a way so that the tangents to it at any point gives the direction of the instantaneous velocity of flow at that point.
Equipotential lines are formed by joining the different points having the same value of velocity potential function.
Properties:

Slope of equipotential Line × Slope of stream function = -1
i.e. streamlines and equipotential lines always meet orthogonally.
Meshes formed by them may be in square, rectangular or curvilinear.
At stagnation point streamlines and equipotential lines have no meaning, So they are not orthogonal at the stagnation point.

JKSSB JE Civil Mock Test - 3 - Question 8

The length of transition curve is dependent on

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 8

Length of transition curve (Maximum value is to be considered among the below options)
a) Based on the rate of change of centrifugal acceleration (c)
 
where  c value should be between 0.5 to 0.8
(b) Based on the rate of change of superelevation (e)
1. Superelevation attained by rotation about the inner edge or outer edge

2. Superelevation attained by rotation about the centre

c) Based on IRC Formula
 for Hilly terrain and  for Plain and Rolling terrain.
Where
Ls = length of the transition curve (m)
v = speed of the vehicle (m/s)
V = speed of the vehicle (km/hour)
e = required superelevation
N = rate of change of superelevation to be attained
R = radius of the transition (m)
w = width of road/pavement

JKSSB JE Civil Mock Test - 3 - Question 9

You are provided with a sample of soil. You have to determine the water content of the given sample. Which of the following method cannot be help in the given scenario?

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 9

Among the given options, the calcium carbide, oven dry method, pycnometer method are used to determine the water content of soil but pipette method is used to particle analysis of fine grained soils
Note:
Calcium carbide method: (Rapid Moisture Meter)

  • The calcium carbide method is the quickest and reasonably accurate method of determination of water content of soil using a portable moisture content kit. It is usually requires 5 to 7 minutes for determining the water content.
  • The method is based on the principle that when the water in the soil reacts with calcium carbide, acetylene gas is produced and the pressure exerted by the acetylene gas on a diaphragm gives a measure of the water content.
  • The water content obtained from the calcium carbide method is based on the initial weight of wet soil. It should be converted to water content based on the dry weight of soil.​

Pycnometer method:

  • It is quick method gives result in 10 – 20 minutes.
  • This method is used only when specific gravity of soil solids is known.
  • It is used only for cohesionless soils because removal of entrapped gasses is very difficult in case of cohesive soils.

Oven drying method:

  • Oven dry method is the most accurate and simplest method for water content determination.
  • In this method complete drying of soil sample occurs and water content in the sample is calculated accurately by a maintained temperature in the oven (105°C to 110°C) for 24 hours.
JKSSB JE Civil Mock Test - 3 - Question 10

What will be the cant deficiency (cm), if maximum safe speed on a 5 Degree curve of a broad gauge track is 110 km/h and average speed of train is 85 km/h?

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 10

If the speed of the train is equal to or greater than 100 kmph the actual cant deficiency (Ca) is 10 cm. If the speed of the train is less than 100 kmph, the actual cant is 7.5 cm
Since the maximum speed of the train is 110 kmph > 100 kmph, the cant deficiency is equal to 10 cm.
Important point: The radius of curve is given by
R = 1750/D
Where, D is in degrees.

JKSSB JE Civil Mock Test - 3 - Question 11

Match List – I with List – II and select the correct answer using the code given below the Lists:

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 11


Nitrate is converted into nitrite due to presence of acids in the intestine of infants which has very affinity for haemoglobin. Hence it replaces oxygen from it, leading to deficiency of oxygen in the body. The disease caused is blue baby disease and Methemoglobinemia.
Up to 1 mg/l of fluorine is required for the growth of permanent teeth and to prevent dental cavities. If it is more than 1.5 mg/l, it causes the decolourisation of teeth and name of this disease is Fluorosis.
If it is more than 5 mg/l, it causes the deformation of bones and called as Bone Fluorosis.

JKSSB JE Civil Mock Test - 3 - Question 12

Ultimate limit states deal with the,

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 12

Limit states:

  • Ultimate limit states (or limit state of collapse)- deals with safety of a structural member in terms of the strength, overturning, sliding, buckling etc.
  • Serviceability limit states- deals with serviceability of a structural member in terms of deflection, cracking, vibration, durability etc.
JKSSB JE Civil Mock Test - 3 - Question 13

In the figure given below, survey line EB passes through an obstacle. The ∠BAC and ∠BAD are 30o and 45o respectively. If AB is 90 m, what will be the length of AC, AD and obstructed portion BC respectively, such that CD is in line with EB produced?

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 13

AC = AB sec 30 = 90 x 2/√3 =180/√3 = (180 x√3) / (√3 x√3) = 180√3/3 = 60√3
AD = AB sec 45 = 90√2
BC = AB tan 30 = 90/√3 = (90 x√3)/ (√3 x√3) = 90√3/3= 30√3

JKSSB JE Civil Mock Test - 3 - Question 14

Which of the following statements is correct about corrections applied to length measured by a tape?

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 14

Tape corrections-

  • Correction for Absolute Length
  • Slope Correction
  • Correction for Temperature
  • Correction for Tension
  • Correction for Sag
  • Correction for Incorrect Alignment
  • Reduction to Mean Sea Level

Slope correction is always subtractive
Sag correction is always negative
Temperature correction is positive when applied pull is more than the standard pull and negative when applied pull is less than the standard pull
Correction due to mean sea level is always subtractive

JKSSB JE Civil Mock Test - 3 - Question 15

If the diameter of the most economical circular section is 5 m, then maximum velocity occurs when the depth of flow is,

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 15

For the most economical circular section, maximum velocity occurs when the depth of flow is 0.81 times the diameter of the channel
Depth of flow = 0.81 X 5 m = 4.05 m

JKSSB JE Civil Mock Test - 3 - Question 16

What will be the minimum required area of longitudinal reinforcement in a short axially loaded square column (Fe 415 steel and M20 concrete), if its gross cross-sectional area is 250000 mm2?

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 16

As per IS 456:2000, Cl.26.5.3.1(a)- The minimum cross-sectional area of longitudinal reinforcement in a column = 0.8% of the gross cross-sectional area of the column.
As per IS 456:2000, Cl.39.3-
Gross cross-sectional area (Ag) = 250000 mm2
Area of longitudinal reinforcement (Asc) = 

JKSSB JE Civil Mock Test - 3 - Question 17

If event '1' is the initial event, earliest expected time for event '2' in the given network will be,

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 17

Expected time (tE) for activity 1-2 = (tO+4tL+tP)/6 = 8.5
Since event 1 is the initial event which occurs at zero time,
Earliest expected time for event '2' = 0 + 8.5 = 8.5

JKSSB JE Civil Mock Test - 3 - Question 18

Reinforcement in beams should be such that,

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 18

As per IS 456:2000, Cl. 26.1:
Reinforcing steel of same type and grade shall be used as main reinforcement in a structural member. However, simultaneous use of two different types or grades of steel for main and secondary reinforcement respectively is permissible. Bars may be arranged singly, or in pairs in contact, or in groups of three or four bars bundled in contact. Bundled bars shall be enclosed within stirrups or ties. Bundled bars shall be tied together to ensure the bars remaining together. Bars larger than 32 mm diameter shall not be bundled, except in columns.

JKSSB JE Civil Mock Test - 3 - Question 19

Maximum surface width of cracks allowed by IS 456:2000 for a structural member located in tidal zone, will be,

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 19

As per IS 456:2000, Cl. 35.3.2:
For particularly aggressive environment, such as severe/very severe/extreme category in Table 3 of code, the assessed surface width of cracks should not in general, exceed 0.1 mm.
Extreme Environment- Surface of members in tidal zone, Members in direct contact with liquid/solid aggressive chemicals.

JKSSB JE Civil Mock Test - 3 - Question 20

What will be the minimum stress in a section (20 cm x 25 cm) of a column carrying a load of 30 kN in a plane bisecting 25 cm side, at an eccentricity of 10 cm from the geometric axis of the section?

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 20

Minimum stress in the section = direct stress - bending stress
Direct stress = 30 / (0.2 x 0.25) = 600 kN/m2

M = 30 x 0.1 = 3 kN-m
d = 0.25 m, b = 0.2 m
Z = db2/6 = 0.01/6
Bending stress = 3 x 6/0.01 = 1800 kN/m2
Minimum stress in the section = direct stress - bending stress = 600 - 1800 = -1200 kN/m2 (tensile) = 1.2 MN/m2(tensile)

JKSSB JE Civil Mock Test - 3 - Question 21

Which of the following is/are relevant to theodolite traversing?

  1. Centring
  2. Face left
  3. Face right
Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 21
  •  Centring: in this, setting of a Theodolite is done exactly over a station mark by means of a plumb-bob
  • Face left: It means that the vertical circle of the Theodolite is on the left of the observer while the readings are being taken.
  • Face right: It means that the vertical circle of the Theodolite is on the right of the observer while the readings are being taken.
JKSSB JE Civil Mock Test - 3 - Question 22

Seepage losses in a canal does not depend upon,

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 22

Seepage losses through a canal depend upon the following factors:

  • Type of seepage
  • Soil permeability
  • Condition of canal (seepage through a silted canal is less than that of a new canal)
  • Velocity of canal water
  • Cross-section of the canal
JKSSB JE Civil Mock Test - 3 - Question 23

50 g of soil sample was oven dried at 106oC. What will be the water content of the sample if the weight of the sample after oven drying is 36 g?

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 23

Water content (%) = weight of water x 100/ weight of soil solids
Water content = (50-36) x 100/36 = 39 %

JKSSB JE Civil Mock Test - 3 - Question 24

Water that remains in the soil after permanent wilting point is reached, is called,

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 24

Field capacity = capillary water + hygroscopic water
Capillary water- water available to plants
Hygroscopic water or unavailable moisture- water attached to soil molecules by loose chemical bond and is not available to plants. It is also called the water that remains in the soil after permanent wilting point is reached.

JKSSB JE Civil Mock Test - 3 - Question 25

What will be the pressure exerted by a penguin of weight 30kg walking on snow with a total foot implant area of 200 cm2?

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 25

Pressure P = force/area
Force = mass x g = 30 X 9.81 = 294.3 N
Area = 200 X 10-4m2

JKSSB JE Civil Mock Test - 3 - Question 26

Base period of a crop is,

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 26

Base period or crop period- The time period between the first watering of a crop at the time of its sowing to its last watering before harvesting. It is represented by 'B' (in days).

JKSSB JE Civil Mock Test - 3 - Question 27

There is no need of cross-drainage work for a canal aligned along a natural watershed because,

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 27

Aligning a canal along a natural watershed or ridge line ensure gravity irrigation on both the sides of the canal. No drainage crosses the canal since drainage flows away from the ridge. Therefore, no cross-drainage work is needed in such cases.

JKSSB JE Civil Mock Test - 3 - Question 28

In which of the following methods of calculating depreciation, a property loses its value at a fixed percentage of its value at the beginning of every year.

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 28

Constant percentage method/declining balance method- property loses its value at a fixed percentage of its value at the beginning of every year.
Annual depreciation D = 1-(S / C)1/n
Where, S = scrap value
C= original cost
n= life of property in years

JKSSB JE Civil Mock Test - 3 - Question 29

If perfect conditions prevail, the shortest possible time required to complete an activity, is called,

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 29

Optimistic time estimate- shortest possible time required to complete an activity under ideal conditions.

JKSSB JE Civil Mock Test - 3 - Question 30

What will be the additional length required for a bar (of diameter 'd') which is cranked or bent up at both ends at 30º?

Detailed Solution for JKSSB JE Civil Mock Test - 3 - Question 30

For 30o cranked or bent up at 30º-
Inclined length of crank = 2d
Horizontal length of the crank = 1.73d
Extra length required for one crank = 2d-1.73d=0.27d
Extra length required for two cranks (cranked at both ends) = 0.27 x 2 x d= 0.54d

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