$A$ string in a musical instrument is $50 \ cm$ long and its fundamental frequency is $800 \ Hz$. If a frequency of $1000 \ Hz$ is to be produced,then the required length of the string is ..... $cm$.

The length of a sonometer wire $AB$ is $110\; cm$. Where should the two bridges be placed from $A$ to divide the wire into $3$ segments whose fundamental frequencies are in the ratio of $1:2:3$?

$A$ violin emits sound waves of frequency $n_1$ under tension $T$. When tension is increased by $44\%$,keeping the length and mass per unit length constant,the frequency of sound waves becomes $n_2$. The ratio of frequency $n_2$ to frequency $n_1$ is:

$A$ string vibrates in its fundamental mode when a tension $T_1$ is applied to it. If the length of the string is decreased by $25 \%$ and the tension applied is changed to $T_2$,the fundamental frequency of the string increases by $100 \%$,then $\frac{T_2}{T_1} =$ (Linear density of the string is constant)

If the length of a stretched string is shortened by $40 \%$ and the tension is increased by $44 \%$,then the ratio of the final and initial fundamental frequencies is :