1. Find the resistance as a function of temperature

Given $R(T)=R_0[1+\alpha (T-T_0)]$ with $R(300)=100\,\Omega,\qquad R(500)=120\,\Omega$

So, $120=100[1+\alpha(500-300)]$ $1.2=1+200\alpha$ $\alpha=\frac{0.2}{200}=0.001\,\text{K}^{-1}$

Hence, $R(T)=100[1+0.001(T-300)]$ $R(T)=100\left(0.7+0.001T\right)=70+0.1T$

Check:

• At $T=300$, $R=70+30=100\,\Omega$
• At $T=500$, $R=70+50=120\,\Omega$

Correct.

2. Find current and power at temperature $T$

The toaster is connected to a constant voltage source $V=200\,$V.

Instantaneous power: $P(T)=\frac{V^2}{R(T)}=\frac{200^2}{70+0.1T}$

So, $P(T)=\frac{40000}{70+0.1T}$

3. Relate temperature and time

Temperature rises uniformly from $300\,$K to $500\,$K in $30\,$s.

Thus, $\frac{dT}{dt}=\frac{500-300}{30}=\frac{200}{30}=\frac{20}{3}\,\text{K/s}$

Therefore, $dt=\frac{3}{20}\,dT$

4. Compute total work done

Total electrical work done is $W=\int P\,dt$

Substitute $P(T)$ and $dt$: $W=\int_{300}^{500} \frac{40000}{70+0.1T}\cdot \frac{3}{20}\,dT$

$W=6000\int_{300}^{500}\frac{dT}{70+0.1T}$

Now, $\int \frac{dT}{70+0.1T}=\frac{1}{0.1}\ln(70+0.1T)=10\ln(70+0.1T)$

Hence, $W=6000\cdot 10\left[\ln(70+0.1T)\right]_{300}^{500}$

$W=60000\left[\ln(120)-\ln(100)\right]$

$W=60000\ln\left(\frac{120}{100}\right)$

$W=60000\ln\left(\frac{6}{5}\right)\,\text{J}$

5. Match with options

This corresponds to: $\boxed{60000\ln\left(\frac{6}{5}\right)\,\text{J}}$

So the correct option is C.