The first overtone frequency of a closed pipe of length $l_{1}$ is equal to the $2^{\text{nd}}$ harmonic frequency of an open pipe of length $l_{2}$. The ratio $\frac{l_{1}}{l_{2}}$ is:

The air column in an organ pipe closed at one end is made to vibrate so that there are $2$ nodes and $2$ antinodes each. The mode of vibration is called:

If $l_1$ and $l_2$ are the lengths of the air column for the first and second resonance when a tuning fork of frequency $n$ is sounded on a resonance tube,then the distance of the displacement antinode from the top end of the resonance tube is:

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

An organ pipe $P_1$ closed at one end vibrating in its first overtone and another pipe $P_2$ open at both ends vibrating in its third overtone are in resonance with a given tuning fork. The ratio of lengths of $P_1$ and $P_2$ is

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is $60 \,cm$, the length of the closed pipe will be: (in $\,cm$)