Explanation:
There are 4 beats between \(A\) and \(B\), therefore the possible frequencies of \(A\) are 316 or 324 , \(i.\,e.\) \((320 \pm 4)Hz\).
When the prong of \(A\) is filled, its frequency becomes greater than the original frequency. If we assume that original frequency of \(A\) is 324 , then on filling its frequency will be greater than 324. The beats between \(A\) and \(B\) will be more than 4.
But it is given that, the beats are again 4, therefore, 324 is not possible.
Therefore, required frequency must be \(316\;Hz.\) This is true because on filing the frequency may increase so as to give 4 beats with \(B\) of frequency \(320\;Hz.\)