A circular track of radius 600 m is to be designed for cars at an average speed of 180 km/hr. What should be the angle of banking of the track?

The correct option is A : $ta{n}^{-1}\left(0.42\right)$

Let's convert speed in m/s first: speed in m/s = 5/18 * speed in km/h

speed, v = 5/18 * 180 = 50 m/s

Let the angle of banking be θ. The forces on the car are
(a) weight of the car Mg downward and
(b) normal force N

For proper banking, static frictional force is not needed.
For vertical direction the acceleration is zero. So,
N cos θ = Mg ..........(i)
For horizontal direction, the acceleration is $\frac{v^2}{r}$ towards the centre, so
N sin θ = $\frac{M{v}^2}{r}$ .......... (ii)
From (i) and (ii),
tan θ = $\frac{v^2}{rg}$
Putting the values, tan θ = $\frac{(50m/s{)}^2}{\left(600m\right)\left(10m/s^2\right)}$ = 0.4167

$\theta$ = $ta{n}^{-1}\left(0.42\right)$