(c) Here, $v_{\max} = ?,r = 18\ m,g = 10\ ms^{- 2}$ $\mu = 0.2$ $\frac{mv_{\max}^{2}}{r} = F = \mu R = \mu mg$ $v_{\max} = \sqrt{\mu rg} = \sqrt{0.2 \times 18 \times 10} = 6\ ms^{- 1}$ $= 6 \times \frac{18}{5}km{\ h}^{- 1} = 21.6\ km\ h^{- 1}$
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