The frequencies of two tuning forks `A` and `B` are respectively `1.5%` more `2.5%` less than that of tuning fork `C`. When `A` and `B` sounded together, 12 beats are produced in `1` second. The frequency of tuning fork `C` is

Options

A. `200Hz`
B. `240Hz`
C. `400Hz`
D. `300Hz`

Answer

D

Explanation

Let `v` be the frequency of fork `C`. The frequency of fork `A`, `v_(A)=v+(1.5)/(100)v=v+(3)/(200)v`
Frequency of fork `B`, `v_(B)=v-(2.5)/(100)v=v-(5)/(200)v`
As per question, `v_(A)-v_(B)=12` ,brgt `:. (v+(3)/(200)v)-(v-(5)/(200)v)=12`
or `(8)/(200)v=12` or `v=(200xx12)/(8)=300Hz`