Correct Answer - `L_(C) = 0.75 m` ; `L_(0) = 0.99 m` or `1.006 m`
for closed pipe, fundamental frequency
`(v)/(4l) = 110 Hz`
`rArr l = (v)/(4 xx 110) = (330)/(4 xx 110) = 0.75 m`
frequency of `1st` overtone `= 330 Hz`.
for open pipe :
frequency of `1st` overtone `= 2.(v)/(2l) = (v)/(l)`
Now `(330)/(l) = 330 +- 2.2 rArr l = 0.99 m` or `1.006 m`