A) \[328\,\,Hz\]
B) \[316\,\,Hz\]
C) \[324\,\,Hz\]
D) \[320\,\,Hz\]
Correct Answer: B
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 then\[324\]. 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\[320\,\,Hz).\]You need to login to perform this action.
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