The fundamental frequency of sonometer wire is ' $n$ '. If the tension and length are increased 3 times and diameter is increased twice, the new frequency will be

The fundamental frequency of a sonometer wire is given by $n = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the linear mass density. Expressing $\mu$ in terms of material density $\rho$ and diameter $d$, we have $\mu = \rho \left(\frac{\pi d^2}{4}\right)$.

Substituting yields $n = \frac{1}{2L} \sqrt{\frac{4T}{\rho \pi d^2}} = \frac{1}{Ld} \sqrt{\frac{T}{\rho \pi}}$, establishing the proportionality $n \propto \frac{1}{Ld} \sqrt{T}$.

With the new parameters $L' = 3L$, $T' = 3T$, and $d' = 2d$, the frequency ratio becomes:

$\frac{n'}{n} = \frac{Ld}{L'd'} \sqrt{\frac{T'}{T}} = \frac{Ld}{(3L)(2d)} \sqrt{\frac{3T}{T}} = \frac{1}{6} \sqrt{3}$

Thus $n' = \frac{\sqrt{3}}{6} n = \frac{1}{2\sqrt{3}} n$, which corresponds to option C.