$A$ stretched wire of length $110 \, cm$ is divided into three segments whose frequencies are in the ratio $1 : 2 : 3$. Their lengths must be:

$A$ uniform wire of length $L$,diameter $D$,and density $\rho$ is stretched under a tension $T$. The correct relation between its fundamental frequency $f$,the length $L$,and the diameter $D$ is:

$A$ sonometer wire stretched by weight '$w$' is in unison with a tuning fork. The corresponding resonating length is '$L_1$'. If the weight is increased by '$3w$',the corresponding resonating length of the sonometer in unison with the tuning fork becomes '$L_2$'. The ratio $\left(\frac{L_1}{L_2}\right)$ is:

$A$ wire of length $1 \,m$ under a certain initial tension emits a sound of fundamental frequency $256 \,Hz$. When the tension is increased by $1 \,kg \,wt$,the frequency of the fundamental note increases to $320 \,Hz$. The initial tension is ........... $kg \,wt$.

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 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 $1 : 3 : 4$?