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

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$27: 6: 1$

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(D) The resistance $R$ of a wire is given by $R = \frac{ho L}{A}$.Since the 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,$R = \frac{ho L^2 d}{m}$.Since the material is the same,$ho$ and $d$ are constant,so $R \propto \frac{L^2}{m}$.Given the ratios $m_1 : m_2 : m_3 = 1 : 2 : 3$ and $L_1 : L_2 : L_3 = 3 : 2 : 1$,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}$$R_1 : R_2 : R_3 = \frac{3^2}{1} : \frac{2^2}{2} : \frac{1^2}{3}$$R_1 : R_2 : R_3 = 9 : 2 : \frac{1}{3}$Multiplying by $3$ to clear the fraction,we get $27 : 6 : 1$.

If the masses of three wires of same material are in the ratio of $1: 2: 3$ and their lengths are in the ratio of $3: 2: 1$,then electrical resistances of these wires are in the ratio

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