Radius of circular bend, r = 30 m, speed of train $=v=54\,\mathrm{km}\,\mathrm{h}^{-1}=54 \times \frac{5}{18}$ $=15\,\mathrm{m}\,\mathrm{s}^{-1}$ mass of train, $m=10^{6}\,\mathrm{kg}$ angle of banking $=\theta=?$ The centripetal force is provided by the lateral thrust by the rails on the wheels. According to Newton’s third law of motion, the train exerts an equal and opposite thrust on the rails causing its wear and tear. The angle of banking is given by $\tan \theta = \frac{v^{2}}{r g} = \frac{(15)^{2}}{30 \times 10}$ $= \frac{225}{30 \times 10} = 0.75$$\Rightarrow \theta = \tan^{-1}(0.75) \approx 37^{\circ}$