A car travels on a circular racetrack of radius 50 m, which is...

For minimum wear and tear on the tyres, the frictional force should be zero. The required centripetal force is provided entirely by the horizontal component of the normal reaction.

The optimum speed $v$ on a banked road is given by:

$v = \sqrt{R g \tan \theta}$

Squaring both sides:

$v^2 = R g \tan \theta$

Substituting the given values $v = 10$ ms$^{-1}$, $R = 50$ m, and $g = 10$ ms$^{-2}$:

$(10)^2 = 50 \times 10 \times \tan \theta$

$100 = 500 \tan \theta$

$\tan \theta = \dfrac{100}{500} = \dfrac{1}{5}$

$\theta = \tan^{-1}\left(\dfrac{1}{5}\right)$

Answer: $\tan^{-1}\left(\dfrac{1}{5}\right)