MCQ
$25$ tunning forks are arranged in series in the order of decreasing frequency. Any two successive forks produce $3\, beats/sec.$ If the frequency of the first turning fork is the octave of the last fork, then the frequency of the $21^{st}$ fork is .... $Hz$
  • A
    $72$
  • B
    $288$
  • $84 $
  • D
    $87 $

Answer

Correct option: C.
$84 $
c
(c) According to the question frequencies of first and last tuning forks are $2n$ and $n$ respectively.
Hence frequency in given arrangement are as follows
$==> 2n -24 × 3 = n ==> n= 72 Hz$
So, frequency of $21^{st}$ tuning fork
${n_{21}} = (2 \times 72 - 20 \times 3) = 84Hz$

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