d
Let internal resistance of source \(=\mathrm{R}\)

Current in coil of resistance

\(\mathrm{R}_{1}=\mathrm{I}_{1}=\frac{\mathrm{V}}{\mathrm{R}+\mathrm{R}_{1}}\)

Current in coil of resistance

\(\mathrm{R}_{2}=\mathrm{I}_{2}=\frac{\mathrm{V}}{\mathrm{R}+\mathrm{R}_{2}}\)

Further, as heat generated is same, so

\(\mathrm{I}_{1}^{2} \mathrm{R}_{1} \mathrm{t}=\mathrm{I}_{2}^{2} \mathrm{R}_{2} \mathrm{t}\)

or \(\quad\left(\frac{\mathrm{V}}{\mathrm{R}+\mathrm{R}_{1}}\right)^{2} \mathrm{R}_{1}=\left(\frac{\mathrm{V}}{\mathrm{R}+\mathrm{R}_{2}}\right)^{2} \mathrm{R}_{2}\)

\(\Rightarrow \quad \mathrm{R}_{1}\left(\mathrm{R}+\mathrm{R}_{2}\right)^{2}=\mathrm{R}_{2}\left(\mathrm{R}+\mathrm{R}_{1}\right)^{2}\)

\(\Rightarrow \mathrm{R}^{2} \mathrm{R}_{1}+\mathrm{R}_{1} \mathrm{R}_{2}^{2}+2 \mathrm{RR}_{1} \mathrm{R}_{2}\)

\(\Rightarrow \quad \mathrm{R}^{2} \mathrm{R}_{2}+\mathrm{R}_{1}^{2} \mathrm{R}_{2}+2 \mathrm{RR}_{1} \mathrm{R}_{2} ?\)

\(\Rightarrow \quad \mathrm{R}^{2}\left(\mathrm{R}_{1}-\mathrm{R}_{2}\right)=\mathrm{R}_{1} \mathrm{R}_{2}\left(\mathrm{R}_{1}-\mathrm{R}_{2}\right)\)

\(\Rightarrow \quad \mathrm{R}=\sqrt{\mathrm{R}_{1} \mathrm{R}_{2}}\)