The angular speed of seconds needle in a mechanical watch is

$\frac{π}{30} r a d / s$

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$2 π r a d / s$

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$π r a d / s$

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$\frac{60}{π} r a d / s$

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Solution

The correct option is A
$\frac{π}{30} r a d / s$

$ω = \frac{2 π}{T} = \frac{2 π}{60} = \frac{π}{30} r a d / s$

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