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(D) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$.Since $Volume = V = A \cdot l$,we have $A = \frac{V}{l}$.Also,$Mass = m = d \cdot

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$27:6:1$

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(D) The resistance $R$ of a wire is given by $R = \rho \frac{l}{A}$.Since $Volume = V = A \cdot l$,we have $A = \frac{V}{l}$.Also,$Mass = m = d \cdot V$,where $d$ is the density. Thus,$V = \frac{m}{d}$.Substituting $A$ in the resistance formula: $R = \rho \frac{l}{(m/dl)} = \rho \cdot d \cdot \frac{l^2}{m}$.Since the material is the same,$\rho$ and $d$ are constant,so $R \propto \frac{l^2}{m}$.Given ratios: $m_1:m_2:m_3 = 1:2:3$ and $l_1:l_2:l_3 = 3:2:1$.Therefore,the ratio of resistances is $R_1:R_2:R_3 = \frac{l_1^2}{m_1} : \frac{l_2^2}{m_2} : \frac{l_3^2}{m_3}$.$R_1:R_2:R_3 = \frac{3^2}{1} : \frac{2^2}{2} : \frac{1^2}{3} = 9 : 2 : \frac{1}{3}$.Multiplying by $3$ to clear the fraction: $27 : 6 : 1$.

Masses of $3$ wires of same metal are in the ratio $1 : 2 : 3$ and their lengths are in the ratio $3 : 2 : 1$. The electrical resistances are in ratio

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