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Q.
A pipe closed at one end has length $83 \,cm$. The number of possible natural oscillations of air column whose frequencies lie below $1000\, Hz$ are (velocity of sound in air $= 332\, m/s$)
Fundamental frequency of one end closed pipe,
$v=\frac{n v}{4 L}$
Here, $n=1$
or $\,\,\,\,\,v_{1}=\frac{332}{4 \times 83 \times 10^{-2}}$
or $\,\,\,\,\,\,\,v_{1}=100$
As odd harmonics alone are produced in closed organ pipe, therefore possible frequencies are
$3 v_{1}=300 Hz , 5 v_{2}=500\, Hz $
$7 v_{2}=700 Hz , 9 v_{2}=900 \,Hz$
Hence, the number of possible natural oscillation of air column is $5 .$