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(D) The power generated in a wire is given by $P = I^2 R$. Since the wires are in series,the current $I$ is the same for both.Resistance $R = \rho \fr

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$16 : 1$

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(D) The power generated in a wire is given by $P = I^2 R$. Since the wires are in series,the current $I$ is the same for both.Resistance $R = \rho \frac{l}{A}$.Since the mass $m$ is the same and the material is the same,the volume $V = A \cdot l$ is constant. Thus,$l = \frac{V}{A}$.Substituting this into the resistance formula: $R = \rho \frac{V/A}{A} = \rho \frac{V}{A^2}$.Since $A = \pi r^2$,we have $R \propto \frac{1}{r^4}$.Therefore,the ratio of power generated is $\frac{P_1}{P_2} = \frac{R_1}{R_2} = \frac{r_2^4}{r_1^4}$.Given $r_1 = 1 \, mm$ and $r_2 = 2 \, mm$,we get $\frac{P_1}{P_2} = \left( \frac{2}{1} \right)^4 = \frac{16}{1}$.

Two wires of thickness $1 \, mm$ and $2 \, mm$ have the same mass of copper. The two wires are connected in series and a current is passed through them. The ratio of the power generated in the wires will be ....

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