Question

The resistance of platinum wire at 0°C is 2$\Omega$ and 6$\Omega$ at 80°C. The temperature coefficient of resistance of the wire is

• A. 3$\times 10^{-3}$ °C$^{-1}$
• B. 0.25 $\times 10^{-1}$ °C$^{-1}$
• C. 0.25 $\times 10^{-2}$ °C$^{-1}$
• D. 0.25 °C$^{-1}$

Answer 1

Answer: (B)

0.25 $\times 10^{-1}$ °C$^{-1}$

Answer 2

The resistance of a wire at a given temperature $t$ is related to its resistance at 0°C by the formula:

$R_t = R_0 (1 + \alpha t)$

where:

$R_t$ is the resistance at temperature $t$

$R_0$ is the resistance at 0°C

$\alpha$ is the temperature coefficient of resistance

$t$ is the temperature in °C

Given values:

Resistance at 0°C, $R_0 = 2 \, \Omega$

Resistance at 80°C, $R_{80} = 6 \, \Omega$

Temperature, $t = 80 \, \text{°C}$

Substitute the given values into the formula:

$6 = 2 (1 + \alpha \times 80)$

Divide both sides by 2:

$\frac{6}{2} = 1 + 80\alpha$

$3 = 1 + 80\alpha$

Subtract 1 from both sides:

$3 - 1 = 80\alpha$

$2 = 80\alpha$

Solve for $\alpha$:

$\alpha = \frac{2}{80}$

$\alpha = \frac{1}{40}$

$\alpha = 0.025 \, \text{°C}^{-1}$

Therefore, $\alpha = 0.025 \, \text{°C}^{-1}$ which is equal to $0.25 \times 10^{-1} \, \text{°C}^{-1}$.