$\begin{array}{1 1} 12.2^{\circ} \\ 20.1^{\circ} \\ 4.8^{\circ}\\ 2.9^{\circ}\end{array} $

Answer :$ 20.1^{\circ}$

On a level (unbanked) road, the force that causes the centripetal acceleration is the force of static friction between car and road.

However, if the road is banked at an angle $\theta$, Only the component $n= n \sin \theta$ causes the centripetal acceleration.

Hence , Newton's second law for the radical direction gives

$\sum F_r = n \sin \theta= \large\frac{mv^2}{r}$-----(1)

The car is in equilibrium in the vertical direction.

Thus, from $\sum F_y =0$ we have

$n \cos \theta =mg$ ----------(2)

Dividing (1) and (2) gives

(3) $\tan \theta =\large\frac{v^2}{rg}$

$\theta = \tan ^{-1} \bigg( \large\frac{(13.4\;m/s)^2}{(50.0\;m)(9.80\;m/s^2)} \bigg) $$=20.1^{\circ}$

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