Two forks A and B when sounded together produce 4 beats per second. The fork A is in unison with 35 cm length of a sonometer wire, and B is in unison with 30 cm length of the same wire at the same tension. Then frequency of A is 

As in case of vibrations of string under specific tension, we know that frequency  $n\infty (1/L)$  or  $n=(K/L)$. As forks A and B are in unison (equal frequencies) with 30 cm and 25 cm l ength of a given sonometer wire respectively. If $n_A$ and $n_B$  be the frequencies of the two  tuning forks, we have,
$n_A=\frac{K}{30}and{n}_B=\frac{K}{25}$  
  Or              $\frac{n_A}{n_B}=\frac{25}{30}$                       …..(6.248)
   Further as the forks A and B produce 4 beats when sounded together,
                   $n_A~n_B=4$
  But from equation-(6.248) it is clear that $n_A<n_B$
    Thus we have                  $n_B-n_A=4$     ….(6.249)
   Solving equation-(6.248) and (6.249) for $n_A$  and $n_B$ , we get
       $n_A=20Hzand{n}_B=24Hz$