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(D) The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.For wires '$A$' and '$B$':$\frac{R_A}{R_B} = \frac{l_A}{l_B}

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$8 : 1$

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(D) The resistance of a wire is given by $R = ho \frac{l}{A} = ho \frac{l}{\pi r^2}$.For wires '$A$' and '$B$':$\frac{R_A}{R_B} = \frac{l_A}{l_B} 	imes \left( \frac{r_B}{r_A} ight)^2$.Given $\frac{l_A}{l_B} = \frac{1}{2}$ and $\frac{r_A}{r_B} = \frac{2}{1}$,so $\frac{r_B}{r_A} = \frac{1}{2}$.Substituting these values: $\frac{R_A}{R_B} = \frac{1}{2} 	imes \left( \frac{1}{2} ight)^2 = \frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}$.When connected in parallel,the potential difference $V$ across both wires is the same.The heat produced is given by $H = \frac{V^2}{R} t$.Therefore,the ratio of heat produced is $\frac{H_A}{H_B} = \frac{V^2/R_A}{V^2/R_B} = \frac{R_B}{R_A}$.Since $\frac{R_A}{R_B} = \frac{1}{8}$,then $\frac{R_B}{R_A} = \frac{8}{1}$.Thus,the ratio of heat produced is $8 : 1$.

Two wires '$A$' and '$B$' of the same material have lengths in the ratio $1 : 2$ and radii in the ratio $2 : 1$. If the two wires are connected in parallel to a battery,the ratio of heat produced in wire '$A$' to that in wire '$B$' will be:

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