A car turns a corner on a slippery road at a constant speed of $10\,m/s$. If the coefficient of friction is $0.5$, the minimum radius of the arc in meter in which the car turns is
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(a)$v = \sqrt {\mu \;g\;r} \Rightarrow r = \frac{{{v^2}}}{{\mu g}} = \frac{{100}}{{0.5 \times 10}} = 20$
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