$A$ wire of length $0.4 \,m$ stretched at both ends vibrates $250$ times per second. If the length of the wire is increased by $0.1 \,m$ and the stretching force is reduced to $1/4$ of its original value, then the new frequency is (in $\,Hz$)

$A$ sonometer wire of length $114\, cm$ is fixed at both the ends. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio $1 : 3 : 4$?

If we add $3 \ kg$ load to the hanger of a sonometer,the fundamental frequency becomes two times its initial value. The initial load must be (in $kg$)

$A$ sonometer wire is in unison with a tuning fork of frequency '$n$' when it is stretched by a weight of specific gravity '$d$'. When the weight is completely immersed in water,'$x$' beats are produced per second,then

$A$ thin wire of length $99 \ cm$ is fixed at both ends as shown in the figure. The wire is kept under a tension and is divided into three segments of lengths $l_1, l_2$ and $l_3$ as shown in the figure. When the wire is made to vibrate, the segments vibrate respectively with their fundamental frequencies in the ratio $1: 2: 3$. Then, the lengths $l_1, l_2$ and $l_3$ of the segments respectively are (in $cm$):

$A$ sonometer wire is vibrating in the third overtone. There are: