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(C) The fundamental frequency $n$ of a sonometer wire is given by the formula: $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$ is t

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$n$

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(C) The fundamental frequency $n$ of a sonometer wire is given by the formula: $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$ is the linear mass density.Thus,$n \propto \sqrt{\frac{T}{r^2 \rho}}$.Let the initial values be $T_1 = T$,$r_1 = r$,and $\rho_1 = \rho$. The new values are $T_2 = 2T$,$r_2 = 2r$,and $\rho_2 = \frac{\rho}{2}$.The ratio of the frequencies is given by:$\frac{n_2}{n_1} = \sqrt{\frac{T_2}{T_1} \cdot \left(\frac{r_1}{r_2}\right)^2 \cdot \frac{\rho_1}{\rho_2}}$Substituting the values:$\frac{n_2}{n_1} = \sqrt{\frac{2T}{T} \cdot \left(\frac{r}{2r}\right)^2 \cdot \frac{\rho}{\rho/2}} = \sqrt{2 \cdot \frac{1}{4} \cdot 2} = \sqrt{1} = 1$.Therefore,$n_2 = n_1 = n$.

If the tension and diameter of a sonometer wire of fundamental frequency $n$ are doubled and density is halved,then its fundamental frequency will become

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