Resistance of a wire at $0^{\circ} \mathrm{C}, 100^{\circ} \mathrm{C}$ and $t^{\circ} \mathrm{C}$ is found to be $10 \Omega, 10.2 \Omega$ and $10.95 \Omega$ respectively. The temperature $t$ in Kelvin scale is _________.

<p>To determine the temperature $t$ in the Kelvin scale, we need to use the relationship between the resistance of a wire and temperature. The general formula for the resistance $R$ of a wire as a function of temperature is:</p> <p>$R_t = R_0 (1 + \alpha t)$</p> <p>where:</p> <ul> <li>$R_t$ is the resistance at temperature $t$</li> <li>$R_0$ is the resistance at the reference temperature (usually $0^{\circ} \mathrm{C}$)</li> <li>$\alpha$ is the temperature coefficient of resistance</li> <li>$t$ is the temperature</li> </ul> <p>We are given the following resistances:</p> <ul> <li>Resistance at $0^{\circ} \mathrm{C}$: $R_0 = 10 \Omega$</li> <li>Resistance at $100^{\circ} \mathrm{C}$: $R_{100} = 10.2 \Omega$</li> <li>Resistance at $t^{\circ} \mathrm{C}$: $R_t = 10.95 \Omega$</li> </ul> <p>First, we need to find the temperature coefficient of resistance $\alpha$. Using the resistance at $100^{\circ} \mathrm{C}$:</p> <p>$10.2 = 10 (1 + \alpha \cdot 100)$</p> <p>Solving for $\alpha$:</p> <p>$$ \frac{10.2}{10} = 1 + 100\alpha \implies 1.02 = 1 + 100\alpha \implies 100\alpha = 0.02 \implies \alpha = \frac{0.02}{100} = 0.0002 $$</p> <p>Now, we can find the temperature $t$ using the resistance at $t^{\circ} \mathrm{C}$:</p> <p>$10.95 = 10 (1 + 0.0002 \cdot t)$</p> <p>Solving for $t$:</p> <p>$$ \frac{10.95}{10} = 1 + 0.0002 t \implies 1.095 = 1 + 0.0002 t \implies 0.0002 t = 0.095 \implies t = \frac{0.095}{0.0002} = 475 $$</p> <p>The temperature $t$ in Celsius is $475^{\circ} \mathrm{C}$. To convert this to the Kelvin scale:</p> <p>$T_{K} = t_{C} + 273.15 = 475 + 273.15 = 748.15 \, \mathrm{K}$</p> <p>So, the temperature $t$ in Kelvin scale is approximately $748.15 \, \mathrm{K}$.</p>