In fundamental mode,the time required for the sound wave to reach up to the closed end of a pipe filled with air is $t$ seconds. The frequency of vibration of the air column is

In a physics lab,a student is performing an experiment with a resonance tube to find the speed of sound and its end correction. For this,he used a resonance tube of length $120 \ cm$. When the length of the air column in the tube is $16 \ cm$ and $50 \ cm$,he obtains the $I$ and $II$ resonance respectively,while a tuning fork of frequency $500 \ Hz$ is sounded just above the tube. Match the parameters in List-$I$ with their suitable values in List-$II$.

List-$I$	List-$II$
$A$. Wavelength of sound $(cm)$	$p$. $1$
$B$. Height of liquid column at $II$ resonance $(cm)$	$q$. $2$
$C$. Speed of sound $(m/s)$	$r$. $340$
$D$. End correction $(cm)$	$s$. $68$
$E$. Minimum level of liquid column at resonance $(cm)$	$t$. $70$

The end correction of a resonance tube is $1 \ cm$. If the shortest length resonating with a tuning fork is $15 \ cm$,the next resonating length will be: (in $cm$)

$A$ string is stretched between fixed points separated by $75.0\, cm$. It is observed to have resonant frequencies of $420\, Hz$ and $315\, Hz$. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is .... $Hz$.

$A$ pipe open at both ends of length $1.5 \,m$ is dipped in water such that the second overtone of the vibrating air column resonates with a tuning fork of frequency $330 \,Hz$. If the speed of sound in air is $330 \,m/s$, then the length of the pipe immersed in water is (Neglect end correction). (in $\,m$)

The second overtone of an open organ pipe has the same frequency as the first overtone of a closed organ pipe $L$ metre long. The length of the open pipe will be: