The specific resistance of the material of a wire is ’ $ρ$ ’ and its volume is 3m³ and its resistance is 3 $Ω$ . The length of the wire will be

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$\frac{ρ}{\sqrt{3}}$

$\frac{3}{\sqrt{ρ}}$

$\frac{1}{\sqrt{ρ}}$

$\frac{\sqrt{3}}{ρ}$

answer is B.

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Detailed Solution

$\begin{aligned} R = \frac{ρl}{A}; V = Axl; A = \frac{V}{l}; l = \frac{RA}{ρ} = \frac{RV}{ρl} \\ l^{2} = \frac{RV}{ρ}; l = \sqrt{\frac{RV}{ρ}} = \sqrt{\frac{3 \times 3}{ρ}} = \frac{3}{\sqrt{ρ}} \end{aligned}$

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