frequency of tuning fork A is 256 hertz it produces 4 beats per second with tuning fork B when wax was applied at tuning fork B then 6 beats per second are heard .By reducing little amount of wax 4 beats per second are heard .Frequency of B is?
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Answer:
256 - nB = 4 eq 1
nB - 256 = 4. eq 2
nB is Decrease
number of Beat increase
eq 1 is right
256- nB = 4
nB = 252Hz
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