The resistances of the platinum wire of a platinum resistance thermometer at the ice point and steam point are $8 \Omega$ and $10 \Omega$ respectively. After inserting in a hot bath of temperature $400^{\circ} \mathrm{C}$, the resistance of platinum wire is :

<p>The resistance of a platinum resistance thermometer varies linearly with temperature. The relation can be given by:</p> <p>$R_t = R_0(1 + \alpha t)$</p> <p>where: <ul> <li>$R_t$ is the resistance at temperature $t$,</li><br> <li>$R_0$ is the resistance at 0°C (ice point),</li><br> <li>$\alpha$ is the temperature coefficient of resistance, and</li><br> <li>$t$ is the temperature in degrees Celsius.</p></li> </ul> <p>In this question, we are given: <ul> <li>Resistance at the ice point ($R_{0} = 8 \Omega$),</li><br> <li>Resistance at the steam point ($R_{100} = 10 \Omega$).</p></li> </ul> <p>To find the temperature coefficient of resistance ($\alpha$), we use the resistance values at the ice and steam points:</p> <p>$$\alpha = \frac{R_{100} - R_0}{R_0 \times 100} = \frac{10\Omega - 8\Omega}{8\Omega \times 100} = \frac{2\Omega}{800\Omega} = \frac{1}{400} \text{ per } ^{\circ}C$$</p> <p>Now, to find the resistance $R_t$ at $400 ^{\circ}C$, we substitute $\alpha$, $R_0$, and $t = 400$ into the formula:</p> <p>$$R_t = R_0(1 + \alpha t) = 8\Omega(1 + \frac{1}{400} \times 400) = 8\Omega(1 + 1) = 8\Omega \times 2 = 16\Omega$$</p> <p>Hence, the resistance of the platinum wire at $400^{\circ}C$ is $16 \Omega$. The correct option is:</p> <p>Option B: 16 $\Omega$</p>