Two uniform stretched steel strings $A$ and $B$ are vibrating under the same tension. The first overtone of $A$ is equal to the second overtone of $B$. If the radius of $A$ is twice that of $B$,then the ratio of the lengths of the strings $l_A/l_B$ is:

$A$ string in a musical instrument is $50 \,cm$ long and its fundamental frequency is $800 \,Hz$. Keeping the tension applied to the string same, the change in the length to produce a sound note of fundamental frequency $1000 \,Hz$ will be: (in $\,cm$)

When a vibrating tuning fork is placed on a sound box of a sonometer, $8$ beats per second are heard when the length of the sonometer wire is kept at $101 \,cm$ or $100 \,cm$. Then the frequency of the tuning fork is (Consider that the tension in the wire is kept constant). (in $\,Hz$)

$A$ tuning fork gives $4$ beats with $50 \ cm$ length of a sonometer wire. If the length of the wire is shortened by $1 \ cm$,the number of beats is still the same. The frequency of the fork is (in $Hz$)

Two vibrating strings of the same material but lengths $L$ and $2L$ have radii $2r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes,the one of length $L$ with frequency $f_1$ and the other with frequency $f_2$. The ratio $\frac{f_1}{f_2}$ is given by

In order to double the frequency of the fundamental note emitted by a stretched string,the length is reduced to $\frac{3}{4}$ of the original length and the tension is changed. The factor by which the tension is to be changed is: