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(D) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$.Since the volume $V = A \cdot l$ and mass $m = d \cdot V = d \cdot A \cdot l$ (whe

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$125 : 15 : 1$

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(D) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$.Since the volume $V = A \cdot l$ and mass $m = d \cdot V = d \cdot A \cdot l$ (where $d$ is density),we have $A = \frac{m}{d \cdot l}$.Substituting $A$ into the resistance formula: $R = \rho \frac{l}{(m / (d \cdot l))} = \rho \cdot d \cdot \frac{l^2}{m}$.Since $\rho$ and $d$ are constant for copper,$R \propto \frac{l^2}{m}$.Given the ratios $m_1 : m_2 : m_3 = 1 : 3 : 5$ and $l_1 : l_2 : l_3 = 5 : 3 : 1$,the ratio of resistances is:$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} = 25 : 3 : 0.2$Multiplying by $5$ to clear the fraction: $125 : 15 : 1$.

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

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