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The fork A of frequency 100 is sounded with another tuning fork B. The number of beats produced is 2. On
putting some wax on the prong of B, the number of beats reduces to 1. The frequency of the fork B is

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101

102

answer is C.

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Detailed Solution

the possible frequency = $= 100 \pm 2 = 102 or 98$ After loading, the frequency is decreased. Now number
of beats is reduced to one. This is possible only with frequency 102. Hence frequency of fork B =102

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