A) \[328\,\,Hz\]
B) \[316\,\,Hz\]
C) \[324\,\,Hz\]
D) \[320\,\,Hz\]
Correct Answer: D
Solution :
There are 4 beats between \[A\] and \[B\] therefore, the possible frequencies of \[A\] are \[316\] or \[324\] that is\[(320\pm 4)Hz\]. 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 filing its frequency will be greater than\[324\]. The beats between \[A\] and \[B\] will 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 6 of frequency\[320\,\,Hz\].You need to login to perform this action.
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