The frequency of tuning forks $A$ and $B$ are respectively $3\%$ more and $2\%$ less than the frequency of tuning fork $C.$ When $A$ and $B$ are simultaneously excited, $5$ beats per second are produced. Then the frequency of the tuning fork $'A'$ (in $Hz$) is
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(c) Let $n$ be the frequency of fork $C$
then ${n_A} = n + \frac{{3n}}{{100}} = \frac{{103n}}{{100}}$
and ${n_B} = n - \frac{{2n}}{{100}} = \frac{{98}}{{100}}$
but ${n_A} - {n_B} = 5$
==> $\frac{{5n}}{{100}} = 5$
==> $n = 100\,Hz$
$ \therefore$ ${n_A} = \frac{{(103)(100)}}{{100}} = 103\,Hz$
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