The resistance of a platinum resistance thermometer varies linearly with temperature. The relation can be given by:

$$R_t = R_0(1 + \alpha t)$$

where:

• $$R_t$$ is the resistance at temperature $$t$$,


• $$R_0$$ is the resistance at 0°C (ice point),


• $$\alpha$$ is the temperature coefficient of resistance, and


• $$t$$ is the temperature in degrees Celsius.

In this question, we are given:

• Resistance at the ice point ($$R_{0} = 8 \Omega$$),


• Resistance at the steam point ($$R_{100} = 10 \Omega$$).

To find the temperature coefficient of resistance ($$\alpha$$), we use the resistance values at the ice and steam points:

$$\alpha = \frac{R_{100} - R_0}{R_0 \times 100} = \frac{10\Omega - 8\Omega}{8\Omega \times 100} = \frac{2\Omega}{800\Omega} = \frac{1}{400} \text{ per } ^{\circ}C$$

Now, to find the resistance $$R_t$$ at $$400 ^{\circ}C$$, we substitute $$\alpha$$, $$R_0$$, and $$t = 400$$ into the formula:

$$R_t = R_0(1 + \alpha t) = 8\Omega(1 + \frac{1}{400} \times 400) = 8\Omega(1 + 1) = 8\Omega \times 2 = 16\Omega$$

Hence, the resistance of the platinum wire at $$400^{\circ}C$$ is $$16 \Omega$$. The correct option is:

Option B: 16 $$\Omega$$