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A coin is present in horizontal disc. Disc is rotating uniformly such that there is no slipping between coin and disc. The coefficient of friction between the coin and the disc varies with the radial distance from axis as $μ = μ_{0} (1 - \frac{r}{2}) .$ The radial distance for which speed of the coin is maximum

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$\frac{1}{2}$

$\frac{1}{4}$

answer is C.

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Detailed Solution

$\frac{m V^{2}}{r} = μ m g V = \sqrt{μ r g} V = \sqrt{μ_{0} (1 - \frac{r}{2}) r g}$

For maximum $v \frac{d v}{d r} = 0 \Rightarrow r = 1$

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