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(B) The resistance $R$ of a wire is given by $R = \frac{ho l}{A}$.Since volume $V = A 	imes l$,we can write $A = \frac{V}{l}$.Substituting this int

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$\sqrt{125}: 15: 1$

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(B) The resistance $R$ of a wire is given by $R = \frac{ho l}{A}$.Since volume $V = A 	imes l$,we can write $A = \frac{V}{l}$.Substituting this into the resistance formula,we get $R = \frac{ho l^2}{V}$.Since density $d = \frac{m}{V}$,we have $V = \frac{m}{d}$.Thus,$R = \frac{ho l^2 d}{m}$.For wires of the same material,$ho$ and $d$ are constant,so $R \propto \frac{l^2}{m}$.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:$R_1: R_2: R_3 = 125 : 15 : 1$.

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 resistance is . . . . . . .

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