Two tuning forks A and B produced 10 beats per second when sounded together. On slightly loading fork A with a little wax, it was observed that 15 beats are heard per second. If the frequency of fork B is 480 Hz, then the frequency of A before it was loaded would be

Options

A. 465 Hz
B. 470 Hz
C. 490 Hz
D. 495 Hz

Answer

B

Explanation

Let `n_(A) and n_(B)` be the frequencies of fork A and B respectively. There are two possibilities
`n_(A) gt n_(B) or n_(A) lt n_(B)`
If `n_(A) gt n_(B)`
i.e. `n_(A) - n_(B) = 10`
`n_(A) = n_(B) + 10 = 480 + 10 = 490 Hz`
On loading fork A its frequency decreases in other words `n_(A)` decreases.
`:. n_(A) - n_(B) = 10` also decreases.
But here the beat frequency increases to 15 beats.
Hence `n_(A)` is not greater than `n_(2)`.
i.e. `n_(A) lt n_(B)` in this case `n_(B) - n_(A) = 10`
`n_(A) = n_(B) - 10 = 480 - 10 = 470 Hz`