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 material of the wire is

• A. $1.06\times10^{-3}\,{}^{\circ}C^{-1}$
• B. $3.5\times10^{-2}\,{}^{\circ}C^{-1}$
• C. $2.22 \times 10^{-3}\,{}^{\circ}C^{-1}$
• D. $3.95\times10^{-2}\,{}^{\circ}C^{-1}$

Answer 1

Answer: (C)

Here, $R_{1}=2.5\,\Omega, T_{1} = 28\,{}^{\circ}C$
$R_{2} = 2.9\,\Omega$ and $T_{2} = 100\,{}^{\circ}C$
As $R_{2}= R_{1}\left[1+\alpha\left(T_{2}-T_{1}\right)\right]$
$\therefore 2.9 = 2.5\left[1+\alpha\left(100-28\right)\right]$
$\frac{2.9}{2.5}-1 = \alpha\left[72\right]$ or $\alpha = \frac{1}{72}\times\frac{2.9-2.5}{2.5}$
$ = \frac{1}{72}\times\frac{0.4}{2.5}$
$ = 2.22\times10^{-3}\,{}^{\circ}C^{-1}$