If G is gauge in metres, V is speed of trains in km/hour and R is radius of a curve in metres, the equilibrium super elevation is:

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OSSC JE Civil Mains Official Paper: (Held On: 16th July 2023)
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  1. \( \frac{G V^2}{R} \)
  2. \(\frac{G V^2}{17 R}\)
  3. \(\frac{G V^2}{127 R} \)
  4. \( \frac{G V^2}{130 R}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{G V^2}{127 R} \)
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Detailed Solution

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Concept:

Equilibrium cant \(= \frac{{{\rm{G}}{{\rm{V}}^2}}}{{127{\rm{R}}}}\)

V - Design speed (kmph)

R - Radius of curvature (m)

For B.G track G = 1.676

\(\begin{array}{l} {\rm{Cant}} = \frac{{1.676{\rm{\;}}{{\rm{V}}^2}}}{{127{\rm{\;R}}}}{\rm{m}}\\ {\rm{e}} = \frac{{1.315{\rm{\;}}{{\rm{V}}^2}}}{{\rm{R}}}{\rm{cm}} \end{array}\)

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