Correct Answer - B
Here,
Speed of sound,v=`330ms^(-1)`
Length of pipe,L=30cm=`30xx10^(-2)m`
In a open pipe (open at both ends), the frequency of its `n^(th)` harmonic is
`upsilon_(n)=(nv)/(2L)` where n=1,2,3,,……..
`thereforen=(2Lupsilon_(n))v`
Let `n^(th)` harmonic of open pipe resonate with 1.1 kHz source.
`thereforeupsilon_(n)=1.1kHz=1.1xx10^(3)Hz`
`thereforen=(2xx30xx10^(-2)xx1.1xx10^(3))/330=2`