HIGHWAY NUMERICALS II

Quiz
•
Engineering
•
Professional Development
•
Hard
WINCENTRE CLASSES
FREE Resource
15 questions
Show all answers
1.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
Calculate the theoretical capacity of a traffic with one way traffic flow for a stream speed of 40 kmph and average centre to centre spacing 12.8 m
312.5 vehicles/hour/lane
31.25 vehicles/hour/lane
3125 vehicles/hour/lane
3.125 vehicles/hour/lane
2.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
Calculate the equivalent radius of the resisting section of 20 cm slab, if the ratio of radius of wheel load distribution to the thickness of the slab is 2
20
35.6
40
40.9
3.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
The spot speed observations in kmph are 50,40,60,54,45,31,72,58,43,52,46,56,60,65,33. then the time mean speed in kmph is:
50
51
51.5
52
4.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
It was noted that on a section of road , the free speed was 60 kmph and the jam density was 80 vpkm. The maximum flow in vph that could be expected on this road is ?
800
1400
2800
1200
5.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
The free mean speed on a road wing is found to be 60 kmph under stopped condition the average spacing between vehicles is 6 m. The capacity of flow, assuming linear speed density relations
2333 veh/hr
3333 veh/hr
2500 veh/hr
3838 veh/hr
6.
MULTIPLE CHOICE QUESTION
45 sec • 1 pt
A moto vlogger travels 300km. If he starts at 6 am in morning and reaches 10am that time, he takes half an hour for rest, the journey speed and running speed respectively are
85.71kmph, 75 kmph
75 kmph, 85.71 kmph
50 kmph, 75 kmph
300 kmph, 200 kmph
Answer explanation
Journey speed= 300/4 =75 kmph,
Running speed = 300/ 3.5 = 85.71 kmph
7.
MULTIPLE CHOICE QUESTION
30 sec • 1 pt
While aligning a hill road with a ruling gradient of 1 in 20, a horizontal curve of radius 75 m is encountered.
1 in 15
1 in 17
1 in 25
1 in 27
Answer explanation
Since 1 in 20 is higher than 4%
So as per IRC, Compensation is required.
Hence (30+75)/75 = 1.4
and 75/75=1.0%.
Ruling gradient = 1 in 20 = 5 %.
Hence Compensated gradient on the curve is 4% = 1 in 25
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