In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line. One may conclude that:
JEE MAIN 2019, Diffcult
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$\frac{\frac{1}{T^{2}}}{\frac{1}{T_{0}^{2}}}+\frac{\ln R(T)}{\ln R\left(T_{0}\right)}=1$
$\Rightarrow \ln \mathrm{R}(\mathrm{T})=\ln \mathrm{R}\left(\mathrm{T}_{\mathrm{o}}\right)\left(1-\frac{\mathrm{T}_{\mathrm{o}}^{2}}{\mathrm{T}^{2}}\right)$
${\rm{R}}({\rm{T}}) = {{\rm{R}}_0}{{\rm{e}}^{ - \left( {\frac{{T_0^2}}{{{T^2}}}} \right)}}$
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