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A coin, placed on a rotating turn-table slips, when it is placed at a distance of $9 cm$ from the centre. If the angular velocity of the turn-table is tripled, it will just slip, if its distance from the centre is

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$27 cm$

$9 cm$

$3 cm$

$1 cm$

answer is D.

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Detailed Solution

Necessary centripetal force to the coin is provided by friction. Thus,

${mrω}_{\max}^{2} = μmg$

$r = \frac{μg}{ω_{\max}^{2}}$

$r_{2} = 1 c m$

since $ω^{2} 9 = {(3 ω)}^{2} r_{2}$

$ω_{\max}$ is made three times. Therefore, distance from centre r will remain $\frac{1}{9}$ times.

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