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(D) The resistance $R$ of a wire is given by $R = ho \frac{l}{A}$.Since the volume $V = A 	imes l$ and density $\sigma = \frac{m}{V}$,we have $A =

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$125 : 15 : 1$

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(D) The resistance $R$ of a wire is given by $R = ho \frac{l}{A}$.Since the volume $V = A 	imes l$ and density $\sigma = \frac{m}{V}$,we have $A = \frac{m}{\sigma l}$.Substituting this into the resistance formula: $R = ho \frac{l^2 \sigma}{m}$.Since the material is the same,$ho$ and $\sigma$ are constant. Thus,$R \propto \frac{l^2}{m}$.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}$$R_1 : R_2 : R_3 = \frac{5^2}{1} : \frac{3^2}{3} : \frac{1^2}{5}$$R_1 : R_2 : R_3 = 25 : 3 : 0.2$Multiplying by $5$ to clear the fraction: $125 : 15 : 1$.

If the ratio of the masses of wires of the same material is $1 : 3 : 5$ and the ratio of their lengths is $5 : 3 : 1$,what is the ratio of their resistances?

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