The number of possible natural oscillations of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are (Velocity of sound = 340 m $s^{-1}$) 

Fundamental frequency of the closed organ pipe is

$v=\frac{v}{4L}$

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

$\therefore v=\frac{340 \mathrm{m}{ \mathrm{s}}^{-1}}{4\times 0.85 \mathrm{m}}=100 \mathrm{Hz}$

The natural frequencies of the closed organ pipe will be

$v_n=(2n-1)v=v,3v,5v,7v,9v,11v,13v,\dots$

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

                                                 $1100 \mathrm{Hz},1300 \mathrm{Hz},\dots \text{ and so on. }$

Thus, the natural frequencies lies below the 1250 Hz is 6.