Correct Answer - c
In closed organ pipe. First resonance occurs at `lambda//4.`
So, in fundamental mode of vibration of organ pipe
`lambda/4=(l+0.3d)`
where, `0.3` d is necessary end correction.
Frequency of vibration, `n=v/lambda = v/(4(l+0.3d))`
As / is same, wide pipe A will resonate at a lower
Frequency, I.e. `n_(A) lt n_(B).`