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# DFCCIL’21 CE: Daily Practice Quiz. 23-July-2021

Quiz: Civil Engineering
Exam: DFCCIL-JE
Topic: HIGHWAY ENGINEERING

Each question carries 1 mark
Negative marking: 1/4 mark
Time: 8 Minutes

Q1. Whose theory is primarily applied for designing rigid pavements?
(a) Bossinessq
(b) Westergaard
(c) Skempton
(d) Mc Cormec

Q2. In rigid pavements, the contraction joints spacing is normally provided as-
(a) 2.5 m
(b) 3.5 m
(c) 4.5 m
(d) 5.5 m

Q3. The ballast packed below and around sleeper to transfer the load from sleeper to sub-base, generally, consists of
(a) broken stones
(b) gravel
(c) moorum
(d) all of these

Q4. Find the number of sleeper required for 1280 m long B.G. track. Take sleeper density (M +5) and length of rail as 12.8 m. (terms have their usual meaning)
(a) 1800
(b) 1000
(c) 1200
(d) 1500

Q5. The minimum depth of ballast prescribed for B.G. trunk lines on Indian railways is: –
(a) 28 cm
(b) 10 cm
(c) 25 cm
(d) 14 cm

Q6. If α is the angle of crossing, and then the number of crossing ‘N’ according to centre line method is given by:
(a) 1/2 cot α/2
(b) cot α/2
(c) cot⁡α
(d) 1/2 cosec α/2
SOLUTION

S1. Ans.(b)
Sol. Westergaard theory is primarily applied for designing rigid pavements.

S2. Ans.(c)
Sol. Contraction joints are used to prevent the contraction in rigid pavements. The contraction joints spacing is normally provided as 4.5. m.

S3. Ans.(d)
Sol. Ballast provide proper drainage and distribute the load from sleeper to sub-base and generally consist of broken stones, gravel and moorum.

S4. Ans.(a)
Sol. Number of sleepers per rail length generally varies from (M+2) to (M+7).
M = Length of a rail (in m)
for 12.8m rail length, the sleeper density is given = (M+5)
= (12.8 + 5)
= 17.8 ≃ 18
sleeper No. of sleeper required for 1280 m. long B.G. track = (1280×18)/128
= 1800 sleeper.

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