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

Step 1: Understanding the Concept:
The electrical resistance of metallic conductors increases linearly with temperature. This relationship occurs because rising temperatures cause metal ions in the lattice to vibrate more violently, increasing the collision frequency of traveling conduction electrons. The fractional change in resistance per degree change in temperature is defined by the temperature coefficient of resistivity ($\alpha$).
Step 2: Key Formula or Approach:
The linear temperature dependence formula for resistance is written as: \[ R_2 = R_1 [1 + \alpha(T_2 - T_1)] \] Rearranging this formula to isolate the temperature coefficient ($\alpha$) gives: \[ \alpha = \frac{R_2 - R_1}{R_1(T_2 - T_1)} \] Where: - $R_1 = 2.5 \ \Omega$ is the initial resistance at temperature $T_1 = 28^\circ\text{C}$. - $R_2 = 2.9 \ \Omega$ is the final resistance at temperature $T_2 = 100^\circ\text{C}$.
Step 3: Detailed Explanation:
Let's find the temperature change ($\Delta T$): \[ \Delta T = T_2 - T_1 = 100^\circ\text{C} - 28^\circ\text{C} = 72^\circ\text{C} \] Now calculate the absolute change in resistance ($\Delta R$): \[ \Delta R = R_2 - R_1 = 2.9 \ \Omega - 2.5 \ \Omega = 0.4 \ \Omega \] Substitute these values into the rearranged coefficient equation: \[ \alpha = \frac{0.4}{2.5 \times 72} \] \[ \alpha = \frac{0.4}{180} \] \[ \alpha = \frac{4}{1800} = \frac{1}{450} \approx 0.002222 \ ^\circ\text{C}^{-1} \] Converting this into standard scientific notation yields: \[ \alpha \approx 2.22 \times 10^{-3} \ ^\circ\text{C}^{-1} \] This perfectly matches option (C).
Step 4: Final Answer:
The temperature coefficient of resistivity of the wire material is $2.22 \times 10^{-3}$ °C$^{-1}$.