A train is moving with a speed of 12m s ^ - 1 on rails which are 1...

For the train, moving along the curved rails, it can be written, with reference to the figure below, that

$tan\theta = \dfrac{{v}^{2}}{Rg} . . . \left(1\right)$

Also, from the above figure, it follows that

$tan\theta = \dfrac{h}{1 . 5} . . . \left(2\right)$

Equations (1) and (2) imply that

$\dfrac{h}{1 . 5} = \dfrac{{v}^{2}}{Rg} ⇒ h = 1 . 5 \times \dfrac{{v}^{2}}{Rg} = 1 . 5 \times \dfrac{{\left(12m{s}^{- 1}\right)}^{2}}{400m \times 10m{s}^{- 2}} = 0 . 054m = 5 . 4cm$