Two closed pipe produce $10$ beats per second when emitting their fundamental nodes. If their length are in ratio of $25 : 26$. Then their fundamental frequency in $Hz$, are
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(c) ${n_1} - {n_2} = 10$..…$(i)$
Using ${n_1}, = \frac{v}{{4{l_1}}}$ and ${n_2} = \frac{v}{{4{l_2}}}$
==> $\frac{{{n_1}}}{{{n_2}}} = \frac{{{l_2}}}{{{l_1}}} = \frac{{26}}{{25}}$…..$(ii)$
After solving these equation ${n_1} = 260Hz$,${n_2} = 250$$Hz$
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