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(D) Given,the ratio of masses is $m_{1}: m_{2}: m_{3} = 1: 3: 5$ and the ratio of lengths is $l_{1}: l_{2}: l_{3} = 5: 3: 1$.We know that the electric

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$125: 15: 1$

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(D) Given,the ratio of masses is $m_{1}: m_{2}: m_{3} = 1: 3: 5$ and the ratio of lengths is $l_{1}: l_{2}: l_{3} = 5: 3: 1$.We know that the electrical resistance $R$ is given by $R = \rho \frac{l}{A}$.Since density $d = \frac{m}{V} = \frac{m}{Al}$,we have $A = \frac{m}{dl}$.Substituting $A$ in the resistance formula,we get $R = \rho \frac{l}{(m/dl)} = \rho d \frac{l^{2}}{m}$.Since $\rho$ and $d$ are constant for copper wires,$R \propto \frac{l^{2}}{m}$.Therefore,the ratio of resistances is $R_{1}: R_{2}: R_{3} = \frac{l_{1}^{2}}{m_{1}}: \frac{l_{2}^{2}}{m_{2}}: \frac{l_{3}^{2}}{m_{3}}$.Substituting the given values: $R_{1}: R_{2}: R_{3} = \frac{5^{2}}{1}: \frac{3^{2}}{3}: \frac{1^{2}}{5} = \frac{25}{1}: \frac{9}{3}: \frac{1}{5} = 25: 3: 0.2$.To simplify the ratio,multiply by $5$: $125: 15: 1$.

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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