Question

A conducting wire has a resistance of $10\Omega$ at $0^{\circ }\text{C}$ and its coefficient of thermal resistance is $\alpha =\frac{1}{273}/{}^{\circ }\text{C}$. The resistance of wire at a temperature ${273}^{\circ }\text{C}$ will be
($\ln e=1$)

• A. $20\Omega$
• B. $5\Omega$
• C. $10e\Omega$
• D. $\frac{10}{e}\Omega$

Answer 1

The correct option is C $10e\Omega$

For conducting wire,
$T_i=0^{\circ }\text{C};T_f={273}^{\circ }\text{C}$

$\Rightarrow \Delta T={273}^{\circ }\text{C}$

Resistance is given by

$R=R_0(1+\alpha \Delta T)$

The formula is only valid when temperature change is very small. However, here term $\alpha \Delta T$ is significant thus it will contradict the truncation of series- expansion.

From basic definition of $\alpha$,

$dR=R\alpha dT$

$\Rightarrow \frac{dR}{R}=\alpha dT$

Integrating both side

$\underset{R_1}{\overset{R_2}{\int }}\frac{dR}{R}=\alpha \underset{T_1}{\overset{T_2}{\int }}dT=\alpha [T_2-T_1]$

$\Rightarrow [\ln R{]}_{10}^{R_2}=\alpha (273-0)=\frac{1}{273}\times 273=1$

$\Rightarrow \ln \frac{R_2}{10}=1\Rightarrow \frac{R_2}{10}=e$

$\therefore R_2=10e\Omega$

$\begin{array}{c}\text{Why this question?}\\ \text{The formula for obtaining resistance as}\\ \text{a function of temperature}R=R_0(1+\alpha \Delta T)\\ \text{must not be applied for large value of}\Delta T.\end{array}$