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A tuning fork arrangement (pair) produces $4 b e a t s / s$ with one fork of frequency $288 c p s$ . A little wax is placed on the unknown fork and it then produces $2 b e a t s / s$ . The frequency of the unknown fork is

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$286 c p s$

$292 c p s$

$294 c p s$

$288 c p s$

answer is B.

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Detailed Solution

$f = 288 + 4 = 292 H z o r 284 H z$ . Suppose, $f = 292 H z$ , On waxing, let its frequency becomes $290$ , which when sounded with known tuning fork gives $2 b e a t s / s$ . So $f$ must be $292 H z$ .

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