Question

A railway track is banked for a speed v, by making the height of the outer rail h higher than that of the inner rail. The distance between the rails is d. The radius of curvature of the track is r. Then,

• A. $\frac{h}{d}=\frac{v^2}{rg}$
• B. $tan\left(si{n}^{-1}\frac{h}{d}\right)=\frac{v^2}{rg}$
• C. $ta{n}^{-1}\left(\frac{h}{d}\right)=\frac{v^2}{rg}$
• D. $\frac{h}{r}=\frac{v^2}{dg}$

Answer 1

The correct option is B $tan\left(si{n}^{-1}\frac{h}{d}\right)=\frac{v^2}{rg}$

$Ncos\theta =mg$

$Nsin\theta =m{v}^2/r$

$tan\theta =v^2/rg$

for train, AB= d and BB = h

$sin\theta =h/d$

$\theta =si{n}^{-1}\left(h/d\right)$

$tan\theta =tan\left(si{n}^{-1}h/d\right)=v^2/rg$

So, option B is correct.