Two open organ pipes of length $25 \ cm$ and $25.5 \ cm$ produce $10 \ \text{beats/sec}$. The velocity of sound will be ..... $m/s$.

The vibrations of four air columns under identical conditions are represented in the figure below. The ratio of frequencies $n_p: n_q: n_r: n_s$ will be

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

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,... $cm$ $(v = 300 \ m/s)$.

$A$ cylindrical tube open at both ends has a fundamental frequency of $390 \,Hz$ in air. If $\frac{1}{4}$th of the tube is immersed vertically in water, what is the fundamental frequency of the air column (in $\,Hz$)?

$A$ closed organ pipe and an open organ pipe of the same length produce $2$ beats per second when they are set into vibrations together in their fundamental mode. The length of the open pipe is now halved and that of the closed pipe is doubled. The number of beats produced per second will be: