Engineering Jobs   »   Civil Engineering quizs   »   DFCCIL QUIZ

DFCCIL’21 CE: Daily Practice Quiz. 09-September-2021

DFCCIL (Dedicated Freight Corridor Corporation of India Limited) Exam Date 2021 on September and October 2021 so boost your preparation by attempt this Civil Engineering MISCELLANEOUS quiz for DFCCIL recruitment 2021.

Quiz: Civil Engineering
Exam: DFCCIL-JE
Topic: MISCELLANEOUS

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 12 Minutes

Q1. The camber of road should be approximately equal to
(a) Longitudinal gradient
(b) Two times the longitudinal gradient
(c) Three times of longitudinal gradient
(d) Half of longitudinal gradient

Q2. The off- tracking of a vehicle having a wheelbase of 6.0 m and negotiating a curve path of mean radius 50 m
(a)0.36
(b)0.72
(c)0.24
(d)0.33

Q3. Which of the following equation is correct?
(a) Q=KV
(b) K=QV
(c) K=QV2
(d) V=KQ

Q4. The minimum radius for rotary recommended by IRC is
(a) 1.2 times of entry radius
(b) 1.33 times of entry radius
(c) 1.5 times of entry radius
(d) 1.7 times of entry radius

Q5. Radius of Mohr’s Circle represents
(a) Maximum shear stress
(b) Maximum tensile stress
(c) Minimum tensile stress
(d) None of these

Q6. A beam fixed at both ends carries a uniformly distributed load on entire length. The ratio of bending moment at the support to the bending moment at mid span is given by
(a) 0.5
(b) 1.0
(c) 1.5
(d) 2.0

SOLUTION

S1. Ans.(d)
Sol. For straight line camber

DFCCIL'21 CE: Daily Practice Quiz. 09-September-2021 |_30.1
Therefore, camber of road should be approximately equal to half of longitudinal gradient.

S2. Ans.(a)
Sol. Given ,
l=60 m
R = 50 m
Off – tracking =l^2/2R
=(6)^2/(2×50)
=0.36 m

S3. Ans.(a)
Sol. The relation between traffic volume (Q), traffic density (K) and traffic speed (V) is –
▭(Q(veh\/hr.)=K(veh\/km)×V(km\/hr.) )

S4. Ans.(b)
Sol. As per IRC, the minimum radius of rotary is 1.33 times of entry radius.

S5. Ans.(a)
Sol.DFCCIL'21 CE: Daily Practice Quiz. 09-September-2021 |_40.1

Radius of Mohr circle represent maximum shear stress
▭(τ_max=Radius of Mohr Circle=(σ_max-σ_min)/2)

S6. Ans.(d)
Sol. A fixed beam carries UDL on entire length, then the bending moment –
DFCCIL'21 CE: Daily Practice Quiz. 09-September-2021 |_50.1
→ At Supports =(WL^2)/12
→ At Center = (WL^2)/24
Hence ratio of bending moment at support to the bending moment at mid span is = ((WL^2)/12)/((WL^2)/24)
=24/12
=2.0

Sharing is caring!

Thank You, Your details have been submitted we will get back to you.

Leave a comment

Your email address will not be published.