Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  Mock Test Series for IES/ESE (CE)  >  ESE (CE) Paper II Mock Test - 5 - Civil Engineering (CE) MCQ

ESE (CE) Paper II Mock Test - 5 - Civil Engineering (CE) MCQ


Test Description

30 Questions MCQ Test Mock Test Series for IES/ESE (CE) - ESE (CE) Paper II Mock Test - 5

ESE (CE) Paper II Mock Test - 5 for Civil Engineering (CE) 2024 is part of Mock Test Series for IES/ESE (CE) preparation. The ESE (CE) Paper II Mock Test - 5 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The ESE (CE) Paper II Mock Test - 5 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for ESE (CE) Paper II Mock Test - 5 below.
Solutions of ESE (CE) Paper II Mock Test - 5 questions in English are available as part of our Mock Test Series for IES/ESE (CE) for Civil Engineering (CE) & ESE (CE) Paper II Mock Test - 5 solutions in Hindi for Mock Test Series for IES/ESE (CE) course. Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free. Attempt ESE (CE) Paper II Mock Test - 5 | 150 questions in 180 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study Mock Test Series for IES/ESE (CE) for Civil Engineering (CE) Exam | Download free PDF with solutions
ESE (CE) Paper II Mock Test - 5 - Question 1

Consider the following structure:
 

The horizontal shear in column ‘NQ’ is [Use portal method]

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 1

P + 2P + P = 20 + 30 + 40 - 30
4P = 60
P = 15 kN
∴ Horizontal shear in column NQ = 2P = 2 × 15 = 30 kN
Hence option (A) is correct.

ESE (CE) Paper II Mock Test - 5 - Question 2

In PERT, slack is computed as the difference between which are of the following?

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 2

The slack of an activity can be calculated as the difference between its latest start and earliest start time, or alternatively, as the difference between its latest and earliest finishing time.

Hence option 'B' is correct i.e Earliest expected time and latest allowable occurrence time

1 Crore+ students have signed up on EduRev. Have you? Download the App
ESE (CE) Paper II Mock Test - 5 - Question 3

What is the anchorage value of a standard hook of a reinforcement bar of diameter D?

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 3
16D
ESE (CE) Paper II Mock Test - 5 - Question 4

Two bars one of material A and the other of material B of same length are tightly secured between two unyielding walls. Coefficient of the thermal expansion of bar A is more than that of B. When temperature rises then the stresses induced are

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 4

In the question it is no where said that two bars are connected. Since they are placed at different places, both bars will experience compression corresponding to their respective values of thermal expansion.

ESE (CE) Paper II Mock Test - 5 - Question 5

Which of the following losses occur in pre-tensioned prestressed concrete?

1. Elastic deformation of concrete

2. Relaxation of steel

3. Friction

4. Anchorage slip

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 5

Friction and anchorage slip occur only in post-tensioned members.

ESE (CE) Paper II Mock Test - 5 - Question 6

Contact pressure beneath a rigid footing resting on a cohesive soil is

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 6

Settlement is uniform throughout beneath a rigid footing on a cohesive as well as cohesionless soil.

The contact pressure under a rigid footing is more at edges than at middle for cohesive soils and it is less at edge than middle for cohesionless soils. But for general practical cases, the contact pressure can be taken uniform under the rigid footing.

ESE (CE) Paper II Mock Test - 5 - Question 7

A square footing of 2.5 m × 2.5 m size has been founded at 1.2 m below the ground level in a cohesive soil having a bulk density of 1.8 t /m3 and an unconfined compressive strength of 5.5 t /m2. Determine the safe bearing capacity of the footing for a factor of safety of 2.5. Use Skempton’s theory.

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 7

ESE (CE) Paper II Mock Test - 5 - Question 8

In the following questions two statements are given. Read them carefully and choose from the options given below:

a) Both (A) and (R) are true and (R) is the correct explanation of (A).

b) Both (A) and (R) are true and (R) is NOT the correct explanation of (A).

c) (A) is true but (R) is false.

d) (A) is false but (R) is true.

Assertion (A): Mohr’s circle of an element subjected to hydrostatic loading, reduces into a point.

Reason (R): In hydrostatic loading, shear stress is zero and principal stresses are equal and alike.

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 8

ESE (CE) Paper II Mock Test - 5 - Question 9

Improved stream lining produces 25% reduction in the drag coefficient of a torpedo. When it is travelling fully submerged and assuming the driving power to remain the same, the increase in speed will be

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 9

ESE (CE) Paper II Mock Test - 5 - Question 10

In the boundary layer, the flow is

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 10

viscous and rotational

ESE (CE) Paper II Mock Test - 5 - Question 11

If the hinged end of a propped cantilever beam of span L settles by an amount  then the rotation of the hinged end will be

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 11

For beam with both ends fixed,

for cantilever beam,

By moment area theorem

ESE (CE) Paper II Mock Test - 5 - Question 12

Consider the following statements:

  1. The moisture content in a properly seasoned timber will be in the rage of 10 - 12%.
  2. The sapwood contains most of the moisture and dries more rapidly than the heartwood.

Which of the above statements is(are) CORRECT?

ESE (CE) Paper II Mock Test - 5 - Question 13

Traffic flow equation for a section of road is U = 100 – 0.8 K, where ‘U’ is speed (in kmph) and ‘K’ is density (in vehicles per km). The maximum expected flow is

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 13

Traffic flow equation: U = 100 - 0.8 K

When K = 0:

∴ U = 100 km/h

When U = 0:

K = 100 / 0.8 = 125 vehicles per km

ESE (CE) Paper II Mock Test - 5 - Question 14

The amount of irrigation water required to meet the evapo-transpiration needs of a crop for its full growth is called

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 14

consumptive irrigation requirement

ESE (CE) Paper II Mock Test - 5 - Question 15

Two different granular soils are placed in a permeameter tube and flow is allowed to take place under a constant total head. The total head and pressure head at point A in centimeters, are respectively

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 15

Total head at a point is the sum of pressure head and elevation head. It depends upon the datum from which the elevation is measured. Normally, the datum is fixed at the bottom, i.e., at the bottom layer of upper soil at point A.

Elevation head = 0
Pressure head = 75 cm
Total head = 0 + 75 = 75 cm

ESE (CE) Paper II Mock Test - 5 - Question 16

The arithmetical check for the computation of RL by “Rise and Fall” method is given by

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 16

ΣBS – ΣFS = RL of last point – RL of first point = ΣRise – ΣFall

ESE (CE) Paper II Mock Test - 5 - Question 17

A beaker filled with water is falling freely under the influence of gravity. Point B is on the free surface of water and point C is vertically below B near the bottom of the beaker. If PB is the pressure at the point B and PC the pressure of point C, then which one of the following is correct?

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 17

When beaker is falling freely under gravity then pressure

Here,
a  = -g 
P  = P0
∴ The pressure at any point in the beaker is equal to atmospheric pressure i.e. PB.

ESE (CE) Paper II Mock Test - 5 - Question 18

A column ABCD of rectangular section is subjected to an eccentric load at Z as shown in the given figure.

Under the compressive load, the direct stress is 400 t/m2 and maximum bending stresses are 

σxx = +1200 t/m2 and σyy = +800 t/m2.

The stress at corner A will be

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 18

For corner A, y is positive while x is negative,

∴ σA = 400 + 1200 - 800
= 800 t/m2 (compressive)

ESE (CE) Paper II Mock Test - 5 - Question 19

When subjected to a torque, a circular shaft undergoes a twist of 1° in a length of 1200 mm, and the maximum shear stress induced is 80 N/mm2. The modulus of rigidity of material of shaft is 0.8 × 105 N/mm2. The radius of shaft is

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 19

ESE (CE) Paper II Mock Test - 5 - Question 20

In the portal frame shown below the ratio of moments MCD  to MBA induced in the column due to side sway is

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 20

ESE (CE) Paper II Mock Test - 5 - Question 21

Consider the following statements:

  1. One year strength of continuously moist cured concrete is 40% higher than that of 28 days strength.
  2. Moist curing for the first seven to fourteen days results in a compressive strength of 70 to 80% of that of 28 days moisture curing.
  3. Curing of concrete by steam under pressure results in early gain of compressive strength of concrete.

Which of the above statements is(are) CORRECT?

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 21

IS 456 : 2000 stipulates a minimum of 7 day moist curing while IS : 7861 (Part I) 1975 stipulates a minimum of 10 days curing under hot weather condition. On an average, the one-year strength of continuously moist cured concrete is 40% higher than that of 28 days moist cured concrete, while non moist curing can lower the strength to about 40%. Moist curing for the first 7 to 14 days may result in a compressive strength of 70% - 85% of that of 28 days moist curing.

ESE (CE) Paper II Mock Test - 5 - Question 22

If a beam of cross–section 300 mm × 600 mm is subjected to a torsional moment of 50 kNm, then the equivalent shear force is:

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 22

ESE (CE) Paper II Mock Test - 5 - Question 23

If the atomic radius of aluminum is 0.143 nm; then the volume of its unit cell in cubic meter is given by

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 23

The volume of a unit cell of aluminium. Aluminium has an FCC crystal structure. The FCC unit cell volume:

VC = 16R3√2
= 16 x (0.143 x 10-9 m)3 x (√2) = 6.62 c 10-29 m3

ESE (CE) Paper II Mock Test - 5 - Question 24

The two cars are approaching from the opposite directions at a speed of 90 km/hr and 60 km/hr. If the coefficient of friction is 0.7 and brake efficiency is 50% then the minimum sight distance required to avoid head on collision of two vehicles is ___________. [Take reaction time as 2.5 sec]

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 24

ESE (CE) Paper II Mock Test - 5 - Question 25

If the actual thickness of brick masonry wall is 20 cm, its effective length is 3.50 m, its effective height is 3 m and its code specified stiffening coefficient is 1.2, then, for design considerations, the slenderness ratio of the wall will be taken as

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 25

ESE (CE) Paper II Mock Test - 5 - Question 26

The natural frequency of the structural system as shown in figure will be :

Where, L = 3.0 m, E = 24000 MPa, I = 1.2 x 10-4 m4, K = 40 kN/m, m = 10 kN. [Take g = 10 m/s2]

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 26

Stiffness of beam is given by,

Since both the springs will have the same displacement if the mass is displaced, and this the springs are in parallel. Therefore,

So,
Natural frequerucy,

ESE (CE) Paper II Mock Test - 5 - Question 27

A beam of uniform strength refers to which one of the following?

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 27

A beam in which extreme fibre stresses are same at all cross-sections along the length of the beam.

ESE (CE) Paper II Mock Test - 5 - Question 28

In any project network, the execution of various activities can be expedited if necessary by

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 28

Crashing

ESE (CE) Paper II Mock Test - 5 - Question 29

The coefficient of friction in the longitudinal direction of highway is estimated as 0.4. The braking distance for a car moving at a speed of 50 km/hr is

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 29
ESE (CE) Paper II Mock Test - 5 - Question 30

The maximum bending moment in purlin of span ‘L’ and total load ‘W’ is

Detailed Solution for ESE (CE) Paper II Mock Test - 5 - Question 30

WL/10

View more questions
9 tests
Information about ESE (CE) Paper II Mock Test - 5 Page
In this test you can find the Exam questions for ESE (CE) Paper II Mock Test - 5 solved & explained in the simplest way possible. Besides giving Questions and answers for ESE (CE) Paper II Mock Test - 5, 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)