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Question

A 7.5 wide two-lane road on a plain terrain is to be laid along a horizontal curve of radius 500 m. For a design speed of 100kmph, super elevation is provided as per IRC:731980. Consider acceleration due to gravity as 9.81m/s2. The level difference between the inner and outer edges of the road will be

A
0.52 m
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B
0.66 m
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C
0.62 m
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D
0.58 m
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Solution

The correct option is A 0.52 m
e=V2225 R

e=1002225×500=0.088>0.07

So,
e=0.07
Δh=eB=0.07×7.5=0.525m

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