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

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80 and 40

100 and 50

44 and 22

72 and 36

answer is D.

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Detailed Solution

Using n_Last = n_First + (N – 1)x
where N = Number of tuning fork in series
x = beat frequency between two successive forks
$\Rightarrow$ 2n = n + (10 – 1) ´ 4 $\Rightarrow$ n = 36 Hz
$∴$ n_First = 36 Hz and n_Last = 2 ´ n_First = 72 Hz

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