Correct answer is : 5
\(\mathrm{R}_{\text {heater }}=\frac{\mathrm{V}^{2}}{\mathrm{P}}=\frac{(100)^{2}}{1000}=10 \Omega\)
For heater \(\mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}} \Rightarrow \mathrm{V}=\sqrt{\mathrm{PR}}\)
\(\mathrm{V}=\sqrt{62.5 \times 10}\)
\(\mathrm{V}=25 \ \mathrm{v}\)
\(\mathrm{i}_{1}=\frac{75}{10}=7.5 \mathrm{~A}, \quad \mathrm{i}_{\mathrm{H}}=\frac{25}{10}=2.5 \mathrm{~A}\).
\(\mathrm{i}_{\mathrm{R}}=\mathrm{i}_{1}-\mathrm{i}_{\mathrm{H}}=5\)
\(\mathrm{V}=\mathrm{IR}\)
\(\mathrm{R}=\frac{25}{5}=5 \Omega\)