Correct Answer - D
For closed organ pipe possible frequency,
`f_n=(2m+1)(v)/(4l)`
for `n=0,f_0=100Hz`
`n=1,f_1=300Hz`
`n=2,f_2=500Hz`
`n=3,f_3=700Hz`
`n=4,f_4=900Hz`
`n=5,f_5=1100Hz`
`n=6,f_6=1300Hz`
Hence possible natural oscillation whose frequencies
are less than `1250 hz` will be `6(n=0,1,2,3,4,5)`