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The frequency of a fork A is 3% more than the frequency of a standard fork whereas the frequency of fork B is 3 % less, The fork A and .B produce 6 beats per second. The frequency of standard fork will be

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110Hz

103Hz

100Hz

106 Hz

answer is A.

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Detailed Solution

Let n be the frequency of standard fork. Then
$n_{A} = n + \frac{3 n}{100} and n_{B} = n - \frac{3 n}{10} Given that n_{A} - n_{B} = 6 (n + \frac{3 n}{100}) - (n - \frac{3 n}{100}) = 6 Solving, we get n = 100 Hz$

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