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A circular disc is made to rotate in horizontal plane about its centre at the rate of $2 r p s$ . The greatest distance of a coin placed on the disc from its centre so that it does not skid is ( $μ$ is coefficient of friction)

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$\frac{μ}{6.2} m$

$6.2 μ cm$

$\frac{μ}{6.2} cm$

$\frac{μ g}{6.2} cm$

answer is B.

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Detailed Solution

$μ mg = mr ω^{2} \Rightarrow r = \frac{μ g}{ω^{2}} = \frac{μ (10)}{4 π^{2} (2)^{2}} = 6.2 μ cm$

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