On a railway track of radius of curvature 1600 m. If the distance between two trackes is 1.8 m, then the elevation of the outer track above the inner
On a railway track of radius of curvature 1600 m. If the distance between two trackes is 1.8 m, then the elevation of the outer track above the inner track will be `(g=10m//s^(2))`
A. `0.450 m`
B. `0.0450 m`
C. `4.50 m`
D. `4.0 m`
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Correct Answer - B
` tan theta =(v^(2))/(rg) therefore (h)/(l)=(v^(2))/(rg)`
Where l is distance between two tracks.
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