Two strings A and B, made of the same material, are stretched by the same tension. The radius of string A is double of the radius of B. A transverse wave travels on A with speed \[{v_A}\] and on B with speed \[{v_B}\] ​. The ratio \[\dfrac{{{v_A}}}{{{v_B}}}\] ​ is
A. \[\dfrac{1}{2} \\ \]
B. 2
C. \[\dfrac{1}{4} \\ \]
D. 4

Hint:Using the relation given for the radius of strings, we find the relation for the mass per unit length of both the strings. As both are made of the same material, the density should be the same.

Formula used:
\[v = \sqrt {\dfrac{T}{\mu }} \]
where v is the speed of the transverse wave on a string with \[\mu \] as the mass per unit length of the string, i.e. \[\mu = \dfrac{m}{l}\] and T is the applied tension.

Complete step by step solution:
If the length of the string is l , the mass of the string is m and the applied tension is T. Then the speed of the transverse wave on the string is,
\[v = \sqrt {\dfrac{T}{\mu }} \]
here \[\mu \] is the mass per unit length of the string, i.e. \[\Rightarrow \mu = \dfrac{m}{l}\]

If the density of the material of string is \[\rho \] then the mass per unit length of the string will be,
\[\mu = \dfrac{{\rho Al}}{l} = \rho A\]
So, the speed of the transverse wave on the string is,
\[v = \sqrt {\dfrac{T}{{\rho A}}} = \sqrt {\dfrac{T}{{\pi \rho {r^2}}}} \]
The wave speed on the string A will be,
\[{v_A} = \dfrac{1}{{{r_A}}}\sqrt {\dfrac{{{T_A}}}{{\pi {\rho _A}}}} \]
The wave speed on the string B will be,
\[{v_B} = \dfrac{1}{{{r_B}}}\sqrt {\dfrac{{{T_B}}}{{\pi {\rho _B}}}} \]

It is given that the radius of A is double of the radius of B,
\[{r_A} = 2{r_B}\]
And both the strings are made of the same material,
\[{\rho _A} = {\rho _B}\]
The tension in both the strings are the same,
\[{T_A} = {T_B}\]

On dividing the wave speed of string A by the wave speed of string B, we get
\[\dfrac{{{v_A}}}{{{v_B}}} = \dfrac{{{r_B}}}{{{r_A}}}\sqrt {\dfrac{{{T_A}}}{{{T_B}}} \times \dfrac{{{\rho _B}}}{{{\rho _A}}}} \\ \]
\[\Rightarrow \dfrac{{{v_A}}}{{{v_B}}} = \dfrac{{{r_B}}}{{2{r_B}}}\sqrt {1 \times 1} \\ \]
\[\therefore \dfrac{{{v_A}}}{{{v_B}}} = \dfrac{1}{2}\]
Hence, the ratio of the speed of the transverse wave on the string A to the speed of the transverse wave on string B is equal to \[\dfrac{1}{2}\].

Therefore, the correct option is A.

Note: We can say from the obtained relation for the transverse wave and the density of the material that if we increase the density of the material the speed of the transverse wave decreases but the speed of longitudinal wave increases.