The frequency of the fifth harmonic of a closed pipe is equal to the frequency of the third harmonic of an open pipe. If the length of the open pipe is $72 \ cm$,then the length of the closed pipe is: (in $cm$)

Two pipes of lengths $L_1$ and $L_2$,open at both ends,are joined in series. If $f_1$ and $f_2$ are the fundamental frequencies of the two pipes,then the fundamental frequency of the series combination will be (neglect end correction).

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

What should be the length of a closed pipe to produce resonance with a sound wave of wavelength $62 \,cm$,in fundamental mode (in $\,cm$)? [Neglect end correction]

$A$ glass tube is open at both ends. $A$ tuning fork of frequency $f$ resonates with the air column inside the tube. Now,the tube is placed vertically inside water so that half the length of the tube is filled with water. Now,the air column inside the tube is in unison with another fork of frequency $f^{\prime}$. Then,

The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is $5 / x$. The value of $x$ is . . . . . . .