Question

A wire has a resistance of at $\sigma_{1}$ at $\sigma_{2}$ and a resistance of $\sigma_{1} + \sigma_{2}$ at 100 $\frac{\sigma_{1} + \sigma_{2}}{2}$. The temperature coefficient of resistivity of material of the wire is

• A. $9.4 \times 10^{18}$
• B. $\left( e = 1.6 \times 10^{- 19}C \right)$
• C. $4\Omega$
• D. $2 \times 10^{- 2}\Omega$

Answer 1

Answer: (C)

$4\Omega$

Answer 2

: Here,$R_{1} = 2.5\Omega,T_{1} = 28^{o}C$

$R_{2} = 2.9\Omega$ and $T_{2} = 100^{o}C$

As $R_{2} = R_{1}\lbrack 1 + \alpha(T_{2} - T_{1})\rbrack$

$\therefore 2.9 = 2.5\lbrack 1 + \alpha(100 - 28)\rbrack$

$\frac{2.9}{2.5} - 1 = \alpha\lbrack 72\rbrack$

or, $\alpha = \frac{1}{72} \times \frac{2.9 - 2.5}{2.5} = \frac{1}{72} \times \frac{0.4}{2.5} = 2.22 \times 1{0^{- 3}}^{o}C^{- 1}$