For a car moving on a banked circular track, the wear and tear on the tyres is minimum when the frictional force is zero. This occurs when the centripetal force is provided entirely by the horizontal component of the normal reaction.
The condition for safe turning without friction is given by the formula:
$$ \tan(\theta) = \frac{v^2}{rg} $$Where $v$ is the velocity, $r$ is the radius of the track, and $g$ is the acceleration due to gravity.
Given: $v = 10 \, ms^{-1}$, $r = 50 \, m$, and $g = 10 \, ms^{-2}$.
$$ \tan(\theta) = \frac{10^2}{50 \times 10} $$ $$ \tan(\theta) = \frac{100}{500} $$ $$ \tan(\theta) = \frac{1}{5} $$Taking the inverse tangent on both sides:
$$ \theta = \tan^{-1}\left(\frac{1}{5}\right) $$Final Answer: \tan^{-1}(\frac{1}{5})
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