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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$\frac{n}{4}$

$\sqrt{2} n$

$\frac{n}{\sqrt{2}}$

answer is C.

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Detailed Solution

$n = \frac{1}{2 l} \sqrt{\frac{T}{{πr}^{2} ρ}} \propto \sqrt{\frac{T}{r^{2} ρ}}$

$\Rightarrow \frac{n_{1}}{n_{2}} = \sqrt{(\frac{T_{1}}{T_{2}}) {(\frac{r_{2}}{r_{1}})}^{2} (\frac{ρ_{2}}{ρ_{1}})} = \sqrt{(\frac{1}{2}) {(\frac{2}{1})}^{2} (\frac{1}{2})} = 1$

$∴ n_{1} = n_{2}$

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