The resistance of an electrical toaster has a temperature dependence given by $$R(T) = R_0[1 + \alpha(T - T_0)]$$ in its range of operation. At $$T_0 = 300$$ K, $$R = 100$$ $$\Omega$$ and at $$T = 500$$ K, $$R = 120$$ $$\Omega$$. The toaster is connected to a voltage source at 200 V and its temperature is raised at a constant rate from 300 to 500 K in 30 s. The total work done in raising the temperature is:
Note: This question was awarded as the bonus since all options were incorrect in the exam.

We are told that the resistance of the toaster varies with temperature according to

$$R(T)=R_0\left[1+\alpha\,(T-T_0)\right]$$

and that at the reference temperature $$T_0=300\ \text{K}$$ the resistance is $$R_0=100\ \Omega$$. We also know that at $$T=500\ \text{K}$$ the resistance is $$120\ \Omega$$. We first determine the temperature-coefficient $$\alpha$$.

Substituting $$T=500\ \text{K}$$ in the given relation, we have

$$120 = 100\left[1+\alpha\,(500-300)\right].$$

Hence

$$\frac{120}{100}=1+\alpha(200)\quad\Longrightarrow\quad1.2=1+200\alpha,$$

so

$$200\alpha = 0.2\quad\Longrightarrow\quad\alpha = 0.001\ \text{K}^{-1}.$$

Therefore the resistance as an explicit function of temperature is

$$R(T)=100\left[1+0.001\,(T-300)\right].$$

Expanding the bracket,

$$R(T)=100\bigl[1+0.001T-0.3\bigr]=100[0.7+0.001T]=70+0.1T.$$

We check: at $$T=300\ \text{K}$$, $$R=70+0.1(300)=100\ \Omega$$; at $$T=500\ \text{K}$$, $$R=70+0.1(500)=120\ \Omega$$, as required.

The toaster is connected to a fixed voltage source $$V=200\ \text{V}$$. The electrical power at any instant is given by Ohm’s law-power formula:

$$P(t)=\frac{V^2}{R(t)}.$$

The temperature is raised uniformly from 300 K to 500 K in 30 s; hence the temperature rises at the constant rate

$$\frac{\Delta T}{\Delta t} = \frac{500-300}{30}= \frac{200}{30}= \frac{20}{3}\ \text{K s}^{-1}.$$

If $$t$$ is the time (in seconds) measured from the start, the instantaneous temperature is

$$T(t)=300+\frac{20}{3}t.$$

Substituting this in our expression for resistance,

$$R(t)=70+0.1T(t)=70+0.1\left[300+\frac{20}{3}t\right].$$

Simplifying step by step,

$$R(t)=70+30+\frac{0.1\times20}{3}t=100+\frac{2}{3}t.$$

Thus the power as a function of time becomes

$$P(t)=\frac{200^2}{100+\frac{2}{3}t}=\frac{40000}{100+\frac{2}{3}t}.$$

The total electrical work done (energy supplied) while the temperature goes from 300 K to 500 K is the time integral of the power:

$$W=\int_{0}^{30}P(t)\,dt=\int_{0}^{30}\frac{40000}{100+\frac{2}{3}t}\,dt.$$

To evaluate the integral, we set

$$u = 100+\frac{2}{3}t \quad\Longrightarrow\quad du = \frac{2}{3}dt \quad\Longrightarrow\quad dt=\frac{3}{2}\,du.$$

When $$t=0$$, $$u=100$$; when $$t=30\ \text{s}$$, $$u=100+\frac{2}{3}\times30=100+20=120.$$

Hence

$$W = \int_{u=100}^{120}\frac{40000 \times \frac{3}{2}}{u}\,du = 40000\left(\frac{3}{2}\right)\int_{100}^{120}\frac{1}{u}\,du.$$

Multiplying the constants first,

$$40000\left(\frac{3}{2}\right)=60000.$$

Therefore

$$W = 60000\bigl[\ln u\bigr]_{100}^{120}=60000\ln\left(\frac{120}{100}\right).$$

Since $$\dfrac{120}{100}=\dfrac{6}{5},$$ we finally have

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

Comparing with the given options, this matches Option A.

Hence, the correct answer is Option A.