The resistance of the bulb filament is \( 100 \Omega \) at a temperature of \( 100^{\circ} \mathrm{C} \). If its temperature co-
efficient of resistance be \( 0.005 \) per \( { }^{\circ} \mathrm{C} \), its resistance will become \( 200 \Omega \) at a temperature

At \(T_{1}=100^{\circ} \mathrm{C}\), resistance, \(R_{1}=100 \Omega ;\) temperature coefficient, \(\alpha=0.005\) per \({ }^{\circ} \mathrm{C}\)
Let resistance \(R_{2}=200 \Omega\) at temperature \(T_{2}\), then
\(R_{2}=R_{1}\left(1+\alpha\left(T_{2}-T_{1}\right)\right) \Rightarrow R_{2}=R_{1}+\alpha R_{1}\left(T_{2}-T_{1}\right)\)
\(\Rightarrow R_{2}-R_{1}=\alpha R_{1}\left(T_{2}-T_{1}\right) \Rightarrow \frac{\left(R_{2}-R_{1}\right)}{\alpha R_{1}}=T_{2}-T_{1}\)
\(\Rightarrow T_{2}=T_{1}+\frac{\left(R_{2}-R_{1}\right)}{\alpha R_{1}}\)
Substituting the values, we get
\(T_{2}=100+\frac{(200-100)}{0.005 \times 100}=300^{\circ} \mathrm{C}\)