The natural frequency of a tuning fork P is 432 Hz. 3 beats/s are produced when tuning fork P and another tuning fork Q are sounded together. If P is loaded with wax, the number of beats increases to 5 beats/ s. The frequency of Q is
A
429 Hz
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B
435 Hz
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C
437 Hz
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D
427 Hz
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Solution
The correct option is A 435 Hz Given: fP=432Hz As P is loaded with wax, frequency of P decreases which led to increase in beat.
So, frequency of P is less than frequency of Q. So, beat =fQ−fP 3=fQ−432 ⇒fQ=435Hz