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(C) The fundamental frequency $f$ of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,a

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$2 : 1$

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(C) The fundamental frequency $f$ of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $L$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.Since $\mu = ho 	imes A = ho 	imes \pi r^2$,where $ho$ is the density of the material and $r$ is the radius,the formula becomes $f = \frac{1}{2L} \sqrt{\frac{T}{ho \pi r^2}} = \frac{1}{2Lr} \sqrt{\frac{T}{ho \pi}}$.Given that $L$,$T$,and $ho$ are the same for both wires,we have $f \propto \frac{1}{r}$.Therefore,the ratio of the fundamental frequencies is $\frac{f_1}{f_2} = \frac{r_2}{r_1}$.Given the ratio of radii $r_1 : r_2 = 1 : 2$,we have $\frac{r_2}{r_1} = \frac{2}{1}$.Thus,the ratio of the fundamental frequencies is $2 : 1$.

Two wires made up of the same material are of equal lengths but their radii are in the ratio $1 : 2$. On stretching each of these two strings by the same tension,the ratio between the fundamental frequencies is

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