# Civil Engineering Objective Questions

By BYJU'S Exam Prep

Updated on: September 25th, 2023

Civil Engineering Objective Questions are provided here for candidates who are actively preparing for the GATE Civil Engineering exam 2023 and other Civil Engineering competitive exams. Civil Engineering Objective Questions can provide a quick overview of the types of questions to be expected in the exams.

Subjects will be better understood by using objective-based MCQ questions and answers. 5 specially picked questions are mentioned here based on the recent pattern in Civil Engineering competitive exams. Attempt and get a detailed answer to each question mentioned in this article.

## Civil Engineering Objective Question 1

Two simply supported beams B1 and B2 have spans, l and 2l, respectively. Beam B1 has a cross-section of 1 x 1 units and beam B2 has a cross-section of 2 x 2 units. These beams are subjected to concentrated loads W each at the centre of their spans. The ratio of the maximum flexural stress in these beams is

1. 4
2. 2
3. 1/2
4. 1/4

Solution

For beam B1

Maximum bending moment (M1) =Wl/4

Section modulus (Z1) =bd2/6=12/6=1/6

Maximum flexular stress (f1) =M1/Z1=Wl×6/4=6Wl/4

For beam B2

Maximum bending moment (M2) =2l/4=2Wl/4

Section modulus (Z2) =bd2/6=22/6=8/6

Maximum flexular stress (f2) =M2/Z2=2Wl×6/8=12Wl/32

f1/f6Wl/12Wl/32 = 4

## Civil Engineering Objective Question 2

A particle of mass 3 kg moving in a straight line decelerates uniformly from a speed of 40 m/s to 20 m/s in a distance of 300 m. Before it comes to rest, it will travel a further distance of

1. 10 m
2. 20 m
3. 50 m
4. 100 m

Solution

Given,

Mass = 3 kg

For decrease in speed from 40 m/s to 20 m/s

Initial speed = 40 m/s

Final speed = 20 m/s

Distance travelled = 300 m

v2=u2+2aS

202=402+2×300

400=1600+600a

a=-2 m/s2

For body coming to rest from speed 20 m/s

Initial speed = 20 m/s

Final speed = 0 m/s

acceleration = – 2 m/s2

v2=u2+2aS

02=202+2×-2×S

4S=400

S=100 m

## Civil Engineering Objective Question 3

A 40° slope is excavated to a depth of 10 m in a deep layer of saturated clay of unit weight 20 kN/m3; the relevant shear strength parameters are cu = 70 kN/m2 and ɸu = 0. The rock ledge is at a great depth. The Taylor’s stability coefficient for ɸu = 0 and 40° slope angle is 0.18. The factor of safety of the slope is:

1. 2.0
2. 2.1
3. 2.2
4. 2.3

Solution

Given,

H=10 m

cu=70 kN/m2

Φu=0

Unit weight, =20 kN/m3

For 40° slope, Taylor’s stability coefficient =0.18

We know that,

Taylor’s stability coefficient,

Ns=cuF.O.S

0.18=70/20×10×F.O.S

F.O.S=70/20×10×0.18

F.O.S=1.94≅2

## Civil Engineering Objective Question 4

The ruling minimum radius of horizontal curve of a national highway in plain terrain for ruling design speed of 100 km/hour with e = 0.07 and f = 0.15 is close to

1. 250 m
2. 360 m
3. 36 m
4. 300 m

Solution

Rruling=(Vdesign)2/127 (e+f)

Rruling=1002/127 (0.07+0.15)

Rruling=357.91 m ≅ 360 m

## Civil Engineering Objective Question 5

The kinetic energy correction factor for a full developed laminar flow through a circular pipe is

1. 1.00
2. 1.33
3. 2.00
4. 1.50

Solution

The kinetic energy correction factor is the ratio of the rate of kinetic energy based on the fluid’s actual velocity to the rate of kinetic energy based on the fluid’s average velocity.

Kinetic energy correction factors for different cases are as follows:

• Laminar flow in circular pipe = 2
• Laminar flow between parallel plates = 1.543
• Turbulent flow in circular pipe = 1.03 – 1.06

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