The radius of curvature of railway track at a place where the train is moving at a speed of `72km h^(-1)` is `625m` The distance between the rails is `1.5m` Find the angle and the elevation of the outer rail so that there may be no side pressure on the rails Take `g = 9.8ms^(-2)` .

Answer

`3.74^(@) ; 9.78 cm` .

Explanation

Here, `upsilon = 72km//h^(-1) =(72xx100)/(60 xx60) = 20ms^(-1)`
`r = 625 m, b = 1.5 m, 0 = ? H =? `
As `tan theta =(upsilon^(2))/(rg) :. tan theta = (20 xx20)/(625 xx 9.8) = 0.06531, `
`theta = tan^(-1) (0.06531) = 3.74^(@)`
Again , `tan theta = (h)/sqrt(b^(2) -h^(2))`
`h = tan theta sqrt(b^(2)-h^(2)) = 0.06531 sqrt(b^(2) - h^2)`
`h^(2) = 0.004265 (b^(2) -h^(2))`
`1.004265 h^(2) = 0.004265 (1.5)^(2) = 0.009596`
`h^(2) =( 0.009596)/(1.004265) = 0.009555`
`h = 0.09775 m=9.78 cm`.