Explanation:
D Given, $\alpha=0.00125$ per $^{\circ} \mathrm{C}, \mathrm{R}_{300 \mathrm{~K}}=1 \Omega$,
$\Delta \mathrm{T}=300-273=27^{\circ} \mathrm{C}=27 \mathrm{~K}$
Applying relation between temperature and resistance,
$\mathrm{R}=\mathrm{R}_{\mathrm{o}}(1+\alpha \Delta \mathrm{T})$
Let resistance be $1 \Omega$ for temperature $27 \mathrm{~K}$.
$1=\mathrm{R}_{0}(1+27 \alpha)$
Let the resistance be $2 \Omega$ for temperature $\Delta T$.
$2=\mathrm{R}_{\mathrm{o}}(1+\alpha \Delta \mathrm{T})$
Dividing equation (i) by equation (ii), we get-
$1+\alpha \Delta \mathrm{T}=2(1+27 \alpha)$
$\alpha(\Delta \mathrm{T}-54)=1$
$\Delta \mathrm{T}-54=1 / 0.00125$
$\Delta \mathrm{T}=800+54=854^{\circ} \mathrm{C}$
$\mathrm{T}-273=854$
$\mathrm{~T}=854+273=1127 \mathrm{~K}$