Two open organ pipes of fundamental frequencies $n_1$ and $n_2$ are joined in series. The fundamental frequency of the new pipe so obtained will be

An open air pipe of length $80 \,cm$ has the second harmonic frequency equal to the fundamental frequency of a closed organ air pipe. The length of the closed pipe is (in $\,cm$)

If the length of a closed organ pipe is $1 \ m$ and the velocity of sound is $330 \ m/s$,then the frequency for the second note (first overtone) is:

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is $60 \,cm$, the length of the closed pipe will be: (in $\,cm$)

$A$ pipe open at both ends has a fundamental frequency '$f$' in air. The pipe is dipped in water, so that $\frac{2}{3}$ of its length is in water. Now, the fundamental frequency of the air column is:

$A$ tuning fork of frequency $340 \ Hz$ is vibrated just above a tube of $120 \ cm$ height. Water is poured slowly into the tube. What is the minimum height of water necessary for resonance (in $cm$)? (Speed of sound in air $= 340 \ m/s$)