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  4. 10 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 (in Hz ) are

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Q. $10$ 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 (in $Hz$ ) are

BITSATBITSAT 2018

Solution:

Using $n_{\text {Last }}=n_{\text {First }}+(N-1) x$
where $N =$ Number of tuning forks in series
$x=$ beat frequency between two successive forks
$\Rightarrow 2 n=n+(10-1) \times 4$
$ \Rightarrow n=36 \,H z$

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