Question

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

Answer 1

The temperature dependence of resistance is given by:
$R = R_0 (1 + \alpha \Delta T).$
From $0^\circ \text{C}$ to $100^\circ \text{C}$:
$\frac{\Delta R}{R_0} = \alpha \Delta T \implies \alpha = \frac{10.2 - 10}{10 \cdot 100} = 0.002.$
From $0^\circ \text{C}$ to $t^\circ \text{C}$:
$\frac{\Delta R}{R_0} = \alpha \Delta T \implies \Delta T = \frac{10.95 - 10}{10 \cdot 0.002}.$
$\Delta T = 475^\circ \text{C}.$
Convert to Kelvin:
T = 475 + 273 = 748 K
Final Answer: $748 \, \text{K}$.

Answer 2

The temperature dependence of resistance is given by:
$R = R_0 (1 + \alpha \Delta T).$
From $0^\circ \text{C}$ to $100^\circ \text{C}$:
$\frac{\Delta R}{R_0} = \alpha \Delta T \implies \alpha = \frac{10.2 - 10}{10 \cdot 100} = 0.002.$
From $0^\circ \text{C}$ to $t^\circ \text{C}$:
$\frac{\Delta R}{R_0} = \alpha \Delta T \implies \Delta T = \frac{10.95 - 10}{10 \cdot 0.002}.$
$\Delta T = 475^\circ \text{C}.$
Convert to Kelvin:
T = 475 + 273 = 748 K
Final Answer: $748 \, \text{K}$.