The number of possible natural oscillation of air column in a pipe closed at one end of length 85 cm whose frequencies lie below 1250 Hz are : (velocity of sound $=340{\mathrm{ms}}^{-1}\text{ ) }$

In case of closed organ pipe frequency, 

${\mathrm{f}}_{\mathrm{n}}=(2\mathrm{n}+1)\frac{\mathrm{v}}{4\mathrm{l}}$

for $\mathrm{n}=0, {\mathrm{f}}_{\mathrm{n}}=100\mathrm{Hz}$

$\begin{array}{l}\mathrm{n}=1, {\mathrm{f}}_1=300\mathrm{Hz}\\ \mathrm{n}=2, {\mathrm{f}}_2=500\mathrm{Hz}\\ \mathrm{n}=3, {\mathrm{f}}_3=700\mathrm{Hz}\\ \mathrm{n}=4, {\mathrm{f}}_4=900\mathrm{Hz}\\ \mathrm{n}=5, {\mathrm{f}}_5=1100\mathrm{Hz}\\ \mathrm{n}=6, {\mathrm{f}}_6=1300\mathrm{Hz}\end{array}$

Hence, number of possible natural oscillation whose frequencies $<1250\mathrm{Hz}=6(\mathrm{n}=0,1,2,3,4,5)$.