d
\(R(T)=R_{o}\left(1+\alpha\left(T-T_{o}\right)\right)\)

Applying boundary conditions, \(120=100(1+200 \alpha)\)

\(\alpha=10^{-3} \,K^{-1}\)

It is given that temperature increases at a constant rate from \(300\, \mathrm{K}\) to \(500 \,\mathrm{K}\) in \(30\, \mathrm{s}\). Hence, \(T(t)=300+20 t / 3\)

By Joule's Law, heat dissipated in a resistor is given by:

\(W=\int_{0}^{30} \frac{V^{2}}{R} d t\)

\(=\int_{0}^{30} \frac{V^{2}}{R_{0}\left(1+\alpha\left(T-T_{0}\right)\right)} d t\)

\(=\frac{V^{2}}{R_{o}} \int_{0}^{30} \frac{1}{(1+20 \alpha t / 3)} d t\)

Solving, \(W=400 \ln (6 / 5)\)

Work done on resistor \(=-W=400 \ln (5 / 6) \,J\)