The frequency of tuning forks `A` and `B` are respectively `3%` more and `2%` less than the frequency of tuning fork `C`. When `A` and `B` are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork `A` (in Hz) Is

Options

A. 98
B. 100
C. 103
D. 105

Answer

C

Explanation

`f_(A)=f+(3f)/(100)`
`f_(B)=f-(2f)/(100)=(98f)/(100)`
`=(103f)/(100)`
`f_(A)-f_(B)=5implies(5f)/(100)=5impliesf=100Hz`
`f_(A)=103Hz`.