The fundamental frequency of a sonometer wire is $n$. If the tension is increased $3$ times,the length is increased $3$ times,and the diameter is increased $2$ times,what will be the new frequency?

$A$ wire of length $L$,diameter $d$,and density of material $\rho$ is under tension $T$,having a fundamental frequency of vibration $n_A$. Another wire of length $2L$,tension $2T$,density $2\rho$,and diameter $3d$ has a fundamental frequency of vibration $n_B$. The ratio $n_B : n_A$ is

Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the lengths of the strings is :

If the length of a stretched string is shortened by $x \%$ and the tension is increased by $44 \%$,then the ratio of the initial and final fundamental frequencies is $1:2$. The value of '$x$' is:

The total length of a sonometer wire between fixed ends is $110 \ cm$. Two bridges are placed to divide the length of the wire in the ratio $6 : 3 : 2$. The tension in the wire is $400 \ N$ and the mass per unit length is $0.01 \ kg/m$. What is the minimum common frequency with which the three parts can vibrate 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 $1 : 3 : 4$?