$A$ cylindrical tube,open at both ends,has a vibrating air column of fundamental frequency '$f$' in air. The tube is dipped vertically in water so that half of its length is in water. The fundamental frequency of the vibrating air column is now:

An organ pipe has a fundamental frequency of $80 \ Hz$. If one end of the pipe is closed,what are the frequencies produced? (Neglect end correction)

$A$ closed pipe is in resonance with a tuning fork at $27^{\circ} C$ when its length is $20 \,cm$. If the pipe is to be in resonance with the same tuning fork at $7^{\circ} C$, then the change in the length of the pipe required is nearly: (in $\,mm$)

$A$ string fixed at both ends is in resonance in its $2^{nd}$ harmonic with a tuning fork of frequency $f_1$. Now,one end becomes free. If the frequency of the tuning fork is increased slowly from $f_1$,then again a resonance is obtained when the frequency is $f_2$. If in this case the string vibrates in the $n^{th}$ harmonic,then:

$A$ cylindrical tube,open at both ends,has a fundamental frequency $f_0$ in air. The tube is dipped vertically into water such that half of its length is inside water. The fundamental frequency of the air column now is

When an open pipe is closed from one end, the third overtone of the closed pipe is higher in frequency by $150 \,Hz$ than the second overtone of the open pipe. The fundamental frequency of the open pipe will be: (in $\,Hz$)