A train is moving with a speed of $$12 \text{ m s}^{-1}$$ on rails which are $$1.5$$ m apart. To negotiate a curve radius $$400$$ m, the height by which the outer rail should be raised with respect to the inner rail is (Given, $$g = 10 \text{ m s}^{-2}$$):

For banking of tracks, the formula for the height of the outer rail is:

$$\tan\theta = \frac{v^2}{Rg}$$

For small angles, $$\tan\theta \approx \sin\theta \approx \frac{h}{l}$$, where $$h$$ is the height difference and $$l$$ is the distance between rails.

$$\frac{h}{l} = \frac{v^2}{Rg}$$

$$h = \frac{v^2 l}{Rg} = \frac{(12)^2 \times 1.5}{400 \times 10} = \frac{144 \times 1.5}{4000} = \frac{216}{4000} = 0.054 \text{ m} = 5.4 \text{ cm}$$

The answer is $$5.4$$ cm, which corresponds to Option (2).