A tunning fork A has a frequency of 3 % more than that of a standard force. A second fork B has a frequency 2% less than that of the standard fork. When A and B are sounded together, the number of beats produced per sound is 8. the frequency off fork B is

Options

A. 154.5 Hz
B. 250 Hz
C. 157.8 Hz
D. 300 Hz

Answer

C

Explanation

Let the frequency of standard fork be n. then frequencies of A+B are [n+3n/100] and [n-2n/100]
respectively.
Now, according to questions.
(n+(3n)/(100))-(n-(2n)/(100))=8`
`implies(5n)/(100)=8impliesn=160`
`becausen_(B)=n-(2n)/(100)`
`=160-(320)/(100)=157.8Hz`