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A motorcyclist of mass m is to negotiate a curve of radius r with a speed v. The minimum value of the coefficient of friction so that this negotiation may take place safely, is

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$\frac{v^{2}}{gr}$

v²rg

$\frac{gr}{v^{2}}$

$\frac{g}{v^{2} r}$

answer is B.

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Detailed Solution

$v = \sqrt{μgr} \Rightarrow μ = \frac{v^{2}}{rg}$

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