SSC JE Civil Engineering 30-Oct-2020 Shift 2 Paper

Question : 173 of 200

Marks: +1, -0

A vehicle has wheel base of 5.5 m. What is the off tracking negotiation of a curved path with a mean radius of 31.5 m?
(take width of pavement as 3.5 m)

0.48 m
0.96 m
0.17 m
0.087 m

Validate

Solution:

Concept:
The off-tracking on the road is given by:

\E_{{\off }}={\n \l^{2}}/{2 \R}

where
n = number of lanes
L = wheel-base
R = Radius of the curve

\E_\off

= Off-tracking
Calculations:
Pavement width is 2.5 m, So Number of lanes(n) = 1

\E_{{\off }}={1 5.5^{2}}/{2 31.5}=0.48 {\m}

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