\(v=\frac{v}{4 L}\)

Here, \(v=340 \mathrm{m} \mathrm{s}^{-1}, L=85 \mathrm{cm}=0.85 \mathrm{m}\)

\(\therefore \quad v=\frac{340 \mathrm{ms}^{-1}}{4 \times 0.85 \mathrm{m}}=100 \mathrm{Hz}\)

The natural frequencies of the closed organ pipe will be

\(v_{n} =(2 n-1) v=v, 3 v, 5 v, 7 v, 9 v, 11 v, 13 v, \ldots\)

\(=100 \mathrm{Hz}, 300 \mathrm{Hz}, 500 \mathrm{Hz}, 700 \mathrm{Hz}, 900 \mathrm{Hz}\)

\(1100 \mathrm{Hz}, 1300 \mathrm{Hz}, \ldots\) and so on

Thus, the natural frequencies lies below the \(1250 \mathrm{Hz}\) is \(6.\)