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

In a resonance tube,the first resonance with a tuning fork occurs at $16 \ cm$ and the second at $49 \ cm$. If the velocity of sound is $330 \ m/s$,the frequency of the tuning fork is: (in $Hz$)

$A$ closed organ pipe of length $1.2 \, m$ vibrates in its first overtone mode. The pressure variation is maximum at:

An open pipe is in resonance in its $2^{nd}$ harmonic with a tuning fork of frequency ${f_1}$. Now,it is closed at one end. If the frequency of the tuning fork is increased slowly from ${f_1}$,then again a resonance is obtained with a frequency ${f_2}$. If in this case the pipe vibrates in its $n^{th}$ harmonic,then:

An open pipe of length $33 \ cm$ resonates with a frequency of $1000 \ Hz$. If the speed of sound is $330 \ m/s$,then this frequency is:

$A$ pipe,$30.0 \; cm$ long,is open at both ends. Which harmonic mode of the pipe resonates a $1.1 \; kHz$ source? Will resonance with the same source be observed if one end of the pipe is closed? Take the speed of sound in air as $330 \; m s^{-1}$.