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What is the angle of banking of a railway track of radius of curvature 250 m, if the maximum velocity of train is 90 km/hr (g = 10 m/s²)

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$θ = \tan^{- 1} (\frac{1}{3})$

$θ = \tan^{- 1} (\frac{1}{4})$

$θ = \tan^{- 1} (\frac{1}{2})$

$θ = \tan^{- 1} (\frac{1}{5})$

answer is C.

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Detailed Solution

$\begin{array}{l} v = 90 \times \frac{5}{18} = 25 m / s \\ \tan θ = \frac{v^{2}}{rg} \end{array}$

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