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:

$A$ wire of length $L$ and mass per unit length $6.0 \times 10^{-3} \; kg/m$ is put under a tension of $540 \; N$. Two consecutive frequencies at which it resonates are $420 \; Hz$ and $490 \; Hz$. Then $L$ in meters is: (in $; m$)

An iron load of $2 \ kg$ is suspended in air from the free end of a sonometer wire of length $1 \ m$. $A$ tuning fork of frequency $256 \ Hz$ is in resonance with $\frac{1}{\sqrt{7}}$ times the length of the sonometer wire. If the load is immersed in water, the length of the wire in metres that will be in resonance with the same tuning fork is (Specific gravity of iron $= 8$).

Two uniform wires of the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the $2^{\text{nd}}$ overtone of the $2^{\text{nd}}$ wire and the radius of the first wire is twice the radius of the $2^{\text{nd}}$ wire,then the ratio of the length of the first wire to the $2^{\text{nd}}$ wire is:

The fundamental frequency of a sonometer wire of length $l$ is $n_0$. $A$ bridge is now introduced at a distance of $\Delta l$ (where $\Delta l \ll l$) from the center of the wire. The lengths of wire on the two sides of the bridge are now vibrated in their fundamental modes. Then,the beat frequency is nearly

The frequency of a sonometer wire is $f$. When the weights producing the tension are completely immersed in water,the frequency becomes $f/2$. When the weights are immersed in a certain liquid,the frequency becomes $f/3$. The specific gravity of the liquid is: