Correct Answer - C
Funkdamental frequency of closed organ pipe,
`v_(o)=v/(4l_(o))rArrL_(o)=v/(4xx110)=0.75m`
Frequency of first overtone of open organ pipe,
`v_(1)=2xxv/(2L_(o))=v/(L_(o))`
Frequency of first overtone of closed prgan pipe, `v_(2)=3v_(o)`
`=3xxv/(4L_(o))=(3xx330)/(4xx0.75)=330 Hz`
`therefore" "v_(1)-v_(2)=2.2,v/(L_(o))-330=2.2`
`rArr" "v/(L_(o))=332.2`
`rArr" " L_(o)=(330)/(332.2)=0.993m`