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A tuning fork A of frequency 512Hz produces 5 beats per second when sounded with another tuning fork B of unknown frequency. If B is located with wax the number of beats is again 5 per second. The frequency of the tuning fork B before it was loaded is
(A) 502Hz
(B) 507Hz
(C) 517Hz
(D) 522Hz

Answer
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HintThe frequency of B is equal to the sum of the frequency of the A and the number of beats produced from the frequency of A. And then the condition is given, the B is located in the wax and again 5 per second. By sing this condition, the frequency of B is determined.

Complete step by step solution
Given that,
The tuning fork A of frequency is, n=512Hz,
The tuning fork of A produces the beats of, m=5.
Then the tuning fork B is placed in the wax and again 5 beats per second.
Now, the frequency of B is given by,
n=n±m.....................(1)
Where, n is the frequency of the tuning fork B.
By substituting the frequency of A and the number of beats produced by the frequency of A in the equation (1), then
n=512±5
By adding and subtracting the terms in the above equation, then the above equation is written as,
n=517 or n=507
On loading the tuning fork B, the frequency decreases from 517 to 507, so that the number of beats per second remains 5.
Thus, the frequency of the tuning fork B before it was loaded is 517Hz.

Hence, the option (C) is the correct answer.

Note A tuning fork A of frequency 512Hz produces 5 beats per second when sounded with another tuning fork B of unknown frequency. If B is located with wax the number of beats is again 5 per second. The frequency of the tuning fork B before it was loaded is 517Hz.
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