Question

The resistance of platinum wire at 0°C is 2Ω and 6.89Ω at 80°C. The temperature coefficient of resistance of the wire is:

• A. \(3 \times 10^{-3} \degree C^{-1}\)
• B. \(3 \times 10^{-2} \degree C^{-1}\)
• C. \(3 \times 10^{-1} \degree C^{-1}\)
• D. \(3 \times 10^{-4} \degree C^{-1}\)

Answer 1

The Correct Option is B

To determine the temperature coefficient of resistance for the platinum wire, we can use the formula:

\(R_t = R_0 (1 + \alpha \Delta T)\)

where:

• \(R_t\) is the resistance at temperature \(t\).
• \(R_0\) is the resistance at the reference temperature (usually 0°C).
• \(\alpha\) is the temperature coefficient of resistance.
• \(\Delta T\) is the change in temperature.

Given values:

• \(R_0 = 2 \, \Omega\) at \(0°C\)
• \(R_t = 6.89 \, \Omega\) at \(80°C\)
• \(\Delta T = 80°C - 0°C = 80°C\)

Rearranging the formula for \(\alpha\) gives:

\(\alpha = \frac{R_t - R_0}{R_0 \times \Delta T}\)

Substitute the known values:

\(\alpha = \frac{6.89 \, \Omega - 2 \, \Omega}{2 \, \Omega \times 80°C}\)

\(\alpha = \frac{4.89 \, \Omega}{160 \, \Omega \cdot °C}\)

\(\alpha = 0.0305625 \, ° C^{-1}\)

Given that the options are rounded to one significant figure, the closest value is:

\(\alpha \approx 3 \times 10^{-2} \, ° C^{-1}\)

Thus, the temperature coefficient of resistance is \(3 \times 10^{-2} \, ° C^{-1}\).

This confirms that the correct answer is:

\(3 \times 10^{-2} \degree C^{-1}\)