$50$ tuning forks are arranged in increasing order of their frequencies such that each gives $4 \,beats/sec$ with its previous tuning fork. If the frequency of the last fork is octave of the first, then the frequency of the first tuning fork is ... $Hz$
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(c) Frequencies of tuning forks is given by
${n_{{\rm{last}}}} = {n_{{\rm{first}}}} + (N - 1)x$
$2n = n + (50 - 1) \times 4$ ==> $n = 196Hz.$
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