$A$ 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 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 $cm$)

An open tube is in resonance with a string (frequency of vibration of the tube is $n_0$). If the tube is dipped in water so that $75\%$ of the length of the tube is inside the water,then the ratio of the frequency of the tube to the string now will be

$A$ tuning fork of frequency $500 \ Hz$ is sounded on a resonance tube. The first and second resonance are obtained at $17 \ cm$ and $52 \ cm$. The velocity of sound is $........ \ m/s$.

In the experiment for the determination of the speed of sound in air using the resonance column method,the length of the air column that resonates in the fundamental mode with a tuning fork is $0.1 \ m$. When this length is changed to $0.35 \ m$,the same tuning fork resonates with the first overtone. Calculate the end correction in $m$.

$A$ uniform tube of length $60.5\,cm$ is held vertically with its lower end dipped in water. $A$ sound source of frequency $500\,Hz$ sends sound waves into the tube. When the length of tube above water is $16\,cm$ and again when it is $50\,cm,$ the tube resonates with the source of sound. Two lowest frequencies (in $Hz$),to which the tube will resonate when it is taken out of water,are (approximately).