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(D) The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi r^2}$.Since the material is the same,$\rho$ is constant. Thus,$R \p

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$8 : 1$

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(D) The resistance of a wire is given by $R = ho \frac{l}{A} = ho \frac{l}{\pi r^2}$.Since the material is the same,$ho$ is constant. Thus,$R \propto \frac{l}{r^2}$.Given $\frac{l_1}{l_2} = \frac{1}{2}$ and $\frac{r_1}{r_2} = \frac{2}{1}$.The ratio of resistances is $\frac{R_1}{R_2} = \frac{l_1}{l_2} 	imes (\frac{r_2}{r_1})^2 = \frac{1}{2} 	imes (\frac{1}{2})^2 = \frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}$.When connected in parallel,the potential difference $V$ across both wires is the same.The heat produced is given by $H = \frac{V^2 t}{R}$,so $H \propto \frac{1}{R}$.Therefore,the ratio of heat produced is $\frac{H_1}{H_2} = \frac{R_2}{R_1} = \frac{8}{1}$.

Two wires of the same material have lengths in the ratio $1 : 2$ and radii in the ratio $2 : 1$. If they are connected in parallel to a battery,what is the ratio of heat produced in them?

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