JEE Main · 2012 · 29 Apr 2012 · Offline · Waves & Sound

Open at both ends, the tube has fundamental frequency $f=\frac{v}{2L}$, where $L$ is its length. Dipping it so that half is submerged leaves an air column of length $L/2$ that is now closed at the water surface and open at the top. A pipe closed at one end has fundamental $f^{\prime }=\frac{v}{4L^{\prime }}$ with $L^{\prime }=\frac{L}{2}$, which gives $f^{\prime }=\frac{v}{4(L/2)}=\frac{v}{2L}$. The fundamental frequency is therefore unchanged and still equals $f$.