Resistance of the wire is measured as $$2 \Omega$$ and $$3 \Omega$$ at $$10°$$C and $$30°$$C respectively. Temperature coefficient of resistance of the material of the wire is

Given: $$R_1 = 2 \, \Omega$$ at $$T_1 = 10°$$C, and $$R_2 = 3 \, \Omega$$ at $$T_2 = 30°$$C.

Write the resistance-temperature relation.

$$R = R_0(1 + \alpha T)$$

where $$R_0$$ is the resistance at 0°C and $$\alpha$$ is the temperature coefficient.

Set up equations for the two temperatures.

$$2 = R_0(1 + 10\alpha)$$ ... (i)

$$3 = R_0(1 + 30\alpha)$$ ... (ii)

Divide equation (ii) by equation (i).

$$\frac{3}{2} = \frac{1 + 30\alpha}{1 + 10\alpha}$$

$$3(1 + 10\alpha) = 2(1 + 30\alpha)$$

$$3 + 30\alpha = 2 + 60\alpha$$

$$1 = 30\alpha$$

$$\alpha = \frac{1}{30} = 0.033°\text{C}^{-1}$$

The correct answer is Option A.