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HIGHWAY NUMERICALS II

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HIGHWAY NUMERICALS II
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15 questions

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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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