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25 tuning 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 tuning fork is the octave of the last fork, then the frequency of the 21^st fork is

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288 Hz

87 Hz

84 Hz

72 Hz

answer is C.

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Detailed Solution

According to the question frequencies of first and last tuning forks are 2n and n respectively.
Hence frequency in given arrangement are as follows

Þ 2n – 24 ´ 3 = n Þ n = 72 Hz
So, frequency of 21^st tuning fork
$n_{21} = (2 \times 72 - 20 \times 3) = 84 Hz$

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