Two tuning forks give 4 beats per second when sounded simultaneously. The frequency of one of the fork is 384 Hz. When the other fork is loaded with a little wax 6 beats per second are produced. The frequency of the second fork is
A
380 Hz
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B
370 Hz
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C
388 Hz
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D
350 Hz
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Solution
The correct option is A 380 Hz
Frequency of fork one is 384Hz. When other fork is waxed beats is increased. So, frequency of other fork is less than frequency of fork one. So, beat = frequency of fork one − frequency of second fork 4=384− Frequency of second fork Frequency of second fork =380Hz