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Question

A tuning fork arrangement (pair) produces 4 beats/s with one fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is:

A
286 Hz
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B
292 Hz
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C
294 Hz
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D
288 Hz
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Solution

The correct option is B 292 Hz
Let the frequency of the unknown fork be v.
As It produces, 4 beats with 288 Hz,
|v288|=4
v288=±4v=292 or 284 Hz
waxing of fork decreases its frequency.
Since, beat frequency also drops, the frequency of unknown fork has to be, v=292 Hz.

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