A number of tuning forks are arranged in the order increasing frequency and any two successive tuning forks produce 4 beats per senond, when sounded together. If the last tuning fork has a frequency octave higher than that of the first tuning, fork and the frequency of the first tuning fork is 256 Hz, then the number of tuning forks is

 Let the frequency of first fork be n. Then, the frequency of last fork is 2n. Since the successive turning fork gives four beats then frequency of first fork = n
Frequency of second fork = n + 4
Frequency of third fork = n + 8 = n + 2 4
Frequency of Nth fork = n + ( N – 1)4
As Nth fork is the last fork, so
  $2n=n+\left(N-1\right)4$or 
$n=n\left(N-1\right)4$   or  $256=\left(N-1\right)4$
 $\therefore$N = 65
Hence the correct answer is option (c).