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

Correct Answer: 748

To solve this problem, we need to determine the temperature \(t\) at which the resistance of a wire is \(10.95 \, \Omega\). We are given the resistance at \(0^\circ C\) is \(10 \, \Omega\) and at \(100^\circ C\) is \(10.2 \, \Omega\). The resistance \(R(t)\) of a wire at temperature \(t\) is given by: \[ R(t)=R_0(1+\alpha t) \] where \( R_0 \) is the resistance at \(0^\circ C\), \(\alpha\) is the temperature coefficient of resistance. **Step 1: Calculate \(\alpha\)** At \(100^\circ C\), we have: \[ 10.2 = 10(1 + 100\alpha) \] \[ 10.2 = 10 + 1000\alpha \] \[ 0.2 = 1000\alpha \] \[ \alpha = 0.0002 \] **Step 2: Find temperature \(t\) for resistance \(10.95 \, \Omega\)** Using the equation: \[ 10.95 = 10(1 + 0.0002t) \] \[ 10.95 = 10 + 0.002t \] \[ 0.95 = 0.002t \] \[ t = \frac{0.95}{0.002} \] \[ t = 475 \] **Step 3: Convert temperature to Kelvin scale** \[ t = 475^\circ C \] \[ T(K) = 475 + 273.15 = 748.15 \] **Step 4: Verify within range** Given the range (748, 748), \(T\) calculated as 748.15 falls within this range when rounded appropriately. Thus, the temperature \(t\) in Kelvin scale is approximately 748 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: $748 \, \text{K}$.