Explanation

(c) The resistance of a metallic wire at temperature $t{^\circ}C$ is given by
$\ \ \ \ \ \ \ \ \ \ \ R_{t} = R_{0}(1 + \alpha\ t)\ \ \ldots(i)$
Given, $\alpha = 0.00125K^{- 1}$
$\ \ \ \ \ \ \ \ R_{300} = 1\ \Omega$
From Eq. (i), we have
$1 = R_{o}(1 + 0.00125 \times 300)\ \ \ ..(ii)$
and $2 = R_{0}(1 + 0.00125 \times T)\ ..(iii)$
$\therefore\ \ \frac{2}{1} = \frac{1 + 0.00125 \times T}{1 + 0.00125 \times 300}$
or 2.75=1+0.00125$\times$T
or $T = \frac{1.75}{0.00125} = 1400K$