Question

For traffic moving at $60km/hr$ along a smooth circular track of radius $0.1km$, the correct angle of banking should be:

• A. $ta{n}^{-1}\left[\frac{(100\times 9.8)}{{\left(\frac{50}{3}\right)}^2}\right]$
• B. $ta{n}^{-1}\left[\frac{{\left(\frac{50}{3}\right)}^2}{(100\times 9.8)}\right]$
• C. $ta{n}^{-1}\left(\frac{{60}^2}{0.1}\right)$
• D. $ta{n}^{-1}\sqrt{(60\times 0.1\times 9.8)}$

Answer 1

The correct option is B $ta{n}^{-1}\left[\frac{{\left(\frac{50}{3}\right)}^2}{(100\times 9.8)}\right]$

Relation between banking angle, speed and radius of circular track is given by
$\theta ={\tan }^{-1}\left[\frac{v^2}{rg}\right]$
here
$v=60km/hr=60\times \frac{5}{18}m/s=\frac{50}{3}m/s$
and $r=0.1km=100m$
$\Rightarrow \theta ={\tan }^{-1}\left[\frac{{\left(\frac{50}{3}\right)}^2}{(100\times 9.8)}\right]$