A heating element has a resistance of 100 $$\Omega$$ at room temperature. When it is connected to a supply of 220 V, a steady current of 2 A passes in it and temperature is 500$$^\circ$$C more than the room temperature. The temperature coefficient of resistance of the heating element is:

We are told that the resistance of the heating element at room temperature is $$R_0 = 100\ \Omega.$$

When this element is connected across a supply of $$V = 220\ \text{V},$$ a steady current of $$I = 2\ \text{A}$$ is observed. Ohm’s law, stated as $$V = IR,$$ allows us to find the resistance of the element at its elevated operating temperature.

Substituting the given values, we have $$R = \dfrac{V}{I} = \dfrac{220\ \text{V}}{2\ \text{A}} = 110\ \Omega.$$

This resistance $$R = 110\ \Omega$$ corresponds to a temperature that is $$\Delta T = 500^\circ\text{C}$$ higher than room temperature.

Now we recall the formula that relates resistance to temperature for a conductor:

$$R = R_0 \left(1 + \alpha\,\Delta T\right),$$

where $$\alpha$$ is the temperature coefficient of resistance, $$R_0$$ is the resistance at the reference temperature (room temperature here), and $$\Delta T$$ is the rise in temperature.

Substituting the known values, we get

$$110\ \Omega = 100\ \Omega \left(1 + \alpha \times 500^\circ\text{C}\right).$$

First divide both sides by $$100\ \Omega$$ to isolate the bracketed term:

$$\dfrac{110}{100} = 1 + 500\,\alpha.$$

Simplifying the left side gives

$$1.10 = 1 + 500\,\alpha.$$

Now subtract 1 from both sides:

$$1.10 - 1 = 500\,\alpha.$$

So we have

$$0.10 = 500\,\alpha.$$

Finally, divide both sides by 500 to solve for $$\alpha$$:

$$\alpha = \dfrac{0.10}{500} = 0.00020.$$

Expressing this in scientific notation,

$$\alpha = 2 \times 10^{-4}\ ^\circ\text{C}^{-1}.$$

Hence, the correct answer is Option B.