MCQ
Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce $4$ beats/sec. The highest frequency is twice of the lowest. Possible highest and the lowest frequencies are
  • A
    $80$ and $40$
  • B
    $100$ and $50$
  • C
    $44$ and $22$
  • $72$ and $36$

Answer

Correct option: D.
$72$ and $36$
d
(d) Using $n_{Last} = n_{First} + (N -1)x$

where $N =$ Number of tuning fork in series 

$x =$ beat frequency between two successive forks 

$==> 2n = n + (10 -1) × 4 n = 36 Hz$

$\therefore$ $ n_{First} = 36 Hz$ and $n_{Last} = 2 × n_{First} = 72 Hz$

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