Question

A wire has a resistance of \(2.5\ \Omega\) at \(28^\circ C\) and a resistance of \(2.9\ \Omega\) at \(100^\circ C\). The temperature coefficient of resistivity of the material of the wire is:

• A. \(1.06 \times 10^{-3}\ ^\circ C^{-1}\)
• B. \(3.5 \times 10^{-2}\ ^\circ C^{-1}\)
• C. \(2.22 \times 10^{-3}\ ^\circ C^{-1}\)
• D. \(3.95 \times 10^{-2}\ ^\circ C^{-1}\)

Hint:

Remember: \[ R=R_0(1+\alpha\Delta T) \]

• Resistance of metals increases with temperature
• \(\alpha\) is positive for conductors
• Always use: \[ \Delta T = T_2-T_1 \]

Answer 1

The Correct Option is C

Concept: Resistance varies with temperature as: \[ R_2=R_1\left[1+\alpha(T_2-T_1)\right] \] where: \[ \alpha = \text{temperature coefficient of resistivity} \]

Step 1: Write the given values. \[ R_1 = 2.5\ \Omega \] \[ T_1 = 28^\circ C \] \[ R_2 = 2.9\ \Omega \] \[ T_2 = 100^\circ C \]

Step 2: Substitute into resistance-temperature relation. \[ 2.9 = 2.5\left[1+\alpha(100-28)\right] \] \[ 2.9 = 2.5(1+72\alpha) \]

Step 3: Solve for \(\alpha\). \[ \frac{2.9}{2.5} = 1+72\alpha \] \[ 1.16 = 1+72\alpha \] \[ 0.16 = 72\alpha \] \[ \alpha = \frac{0.16}{72} \] \[ \alpha = 2.22\times10^{-3}\ ^\circ C^{-1} \] Therefore, the correct answer is: \[ \boxed{2.22\times10^{-3}\ ^\circ C^{-1}} \]