Q 11-04-083JEE MainJEE Main 2024 (27 Jan, Shift 1)Medium
A train is moving with a speed of $12\ \text{m s}^{-1}$ on rails which are $1.5\ \text{m}$ apart. To negotiate a curve of radius $400\ \text{m}$, the height by which the outer rail should be raised with respect to the inner rail is (Given, $g = 10\ \text{m s}^{-2}$):
Answer: (B) $5.4\ \text{cm}$
For a banked track with no reliance on friction,
$$\tan\theta = \frac{v^2}{rg} = \frac{144}{400\times10} = 0.036$$
For a small angle, $\sin\theta \approx \tan\theta$, so the height of the outer rail is
$$h = l\sin\theta \approx 1.5\times0.036 = 0.054\ \text{m} = 5.4\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics