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If the masses of three wires of same material are in the ratio of $1: 2: 3$ and their lengths are in the ratio of $3: 2: 1$, then electrical resistances of these wires are in the ratio
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$27: 6: 1$
$\begin{aligned} & \text { } \mathrm{R}=\frac{\rho \mathrm{L}}{\mathrm{A}} \quad(\because \mathrm{V}=\mathrm{AL}) \\ & =\frac{\rho \mathrm{L}^2}{\mathrm{~V}} \quad \because \mathrm{d}=\frac{\mathrm{m}}{\mathrm{v}}=\frac{\rho \mathrm{L}^2}{\mathrm{~m}} \\ & \mathrm{R} \propto \frac{\mathrm{L}^2}{\mathrm{~m}} \\ & \mathrm{~m}_1: \mathrm{m}_2: \mathrm{m}_3=1: 2: 3 \text { and } \mathrm{L}_1 \mathrm{~L}_2 \mathrm{~L}_3=3: 2: 1 \\ & \mathrm{R}_1: \mathrm{R}_2: \mathrm{R}_3=\frac{\mathrm{L}_1^2}{\mathrm{~m}_1}: \frac{\mathrm{L}_2^2}{\mathrm{~m}_2}: \frac{\mathrm{L}_3^2}{\mathrm{~m}_3} \\ & =\frac{3^2}{1}: \frac{2^2}{2}: \frac{1^2}{3} \\ & =9: 2: \frac{1}{3} \\ & =27: 6: 1\end{aligned}$
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