Two wires whose lengths are in the ratio $3 : 2$ are stretched with equal tension and give equal frequency. If the radii of the wires are in the ratio $3 : 2$ , then the ratio of densities of the wires is:

81 : 16

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4 : 9

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16 : 81

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9 : 4

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Solution

The correct option is A $16 : 81$
$\begin{matrix} l_{1} : l_{2} = 3 : 2 r_{1} : r_{2} = 3 : 2 equal tension, equal frequency of the 2 wises To fund: ρ_{1} : ρ_{2} (density) \end{matrix}$

$for n = 1$
$\begin{matrix} f_{0} & = \frac{v}{2 L} \Rightarrow {(f_{0})}_{1} & = \frac{\sqrt{T / d_{1}}}{2 L_{1}} \Rightarrow {(f_{0})}_{2} & = \frac{\sqrt{T / d_{2}}}{2 L_{2}} \frac{{(f_{0})}_{1}}{{(f_{0})}_{2}} & = \frac{\sqrt{T / d_{1}}}{2 L_{1}} \times \frac{2 L_{2}}{\sqrt{T / d_{2}}} \end{matrix}$

$\begin{matrix} \Rightarrow 1 = \frac{\sqrt{d_{2}}}{\sqrt{d_{1}}} \frac{L_{2}}{L_{1}} \Rightarrow \frac{\sqrt{d_{1}}}{\sqrt{d_{2}}} = \frac{L_{2}}{L_{1}} = \frac{2}{3} \Rightarrow \frac{d_{1}}{d_{2}} = \frac{4}{9} \end{matrix}$

$\begin{matrix} d = \frac{m}{l} ρ = \frac{m}{A l} = \frac{m}{m r^{2} l} \Rightarrow ρ = \frac{d}{π r^{2}} \end{matrix}$

$\frac{ρ_{1}}{ρ_{2}} = \frac{d_{1} r_{2}^{2}}{r_{1}^{2} d_{2}} = \frac{4}{9} \times \frac{4}{9} = \frac{16}{81}$

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