Two closed organ pipes,when sounded simultaneously,give $4$ beats per second. If the longer pipe has a length of $1 \ m$,then the length of the shorter pipe will be,... $cm$ $(v = 300 \ m/s)$.

An air column in a pipe,which is closed at one end,will be in resonance with a vibrating body of frequency $166 \,Hz$,if the length of the air column is .... $m$. (Assume the speed of sound in air is $332 \,m/s$)

Consider the following statements about stationary waves:
$A$. The distance between two adjacent nodes or antinodes is equal to $\frac{\lambda}{2}$ $(\lambda = \text{wavelength of the wave})$.
$B$. $A$ pressure node is always formed at the open end of the open organ pipe.
Choose the correct option from the following:

The fundamental frequency of a closed pipe is equal to the frequency of the second harmonic of an open pipe. The ratio of their lengths is

Sound waves of frequency $660 \, Hz$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particle has maximum amplitude of vibration is .... $m$ (velocity of sound in air is $330 \, m/s$)

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]