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(C) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional a

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$125:15:1$

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(C) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.Since $A = \frac{V}{l} = \frac{m}{l \cdot d}$ (where $m$ is mass and $d$ is density),we can write:$R = \rho \frac{l^2 \cdot d}{m}$Since $\rho$ and $d$ are constant for copper wires,$R \propto \frac{l^2}{m}$.Given ratios: $m_1:m_2:m_3 = 1:3:5$ and $l_1:l_2:l_3 = 5:3:1$.Calculating the ratio of resistances:$R_1:R_2:R_3 = \frac{l_1^2}{m_1} : \frac{l_2^2}{m_2} : \frac{l_3^2}{m_3}$$R_1:R_2:R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5}$$R_1:R_2:R_3 = 25 : 3 : \frac{1}{5}$Multiplying by $5$ to clear the fraction:$R_1:R_2:R_3 = 125 : 15 : 1$.

Masses of three wires of copper are in the ratio $1:3:5$ and their lengths are in the ratio $5:3:1$. The ratio of their electrical resistances is:

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