Q.

Two tuning forks A and B vibrating simultaneously produce 5 beats/s. Frequency of B is 512 Hz. If tuning fork B is now loaded with wax, when it vibrated with A the number of beats become 6 beats per second. Frequency of A is

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a

507 Hz

b

517 Hz

c

502 Hz

d

522 Hz

answer is C.

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Detailed Solution

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nB = 512Hz  ; number of beats n = 5 b/s 
nB = (increased) number of beats m = 4   

So nA>nB.
nA = nB + n = 512 + 5 = 517 Hz

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