• # question_answer 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 10-4-2019 Morning] A) $R(T)=\frac{{{R}_{0}}}{{{T}^{2}}}$               B) $R(T)={{R}_{0}}{{e}^{-{{T}^{2}}/T_{0}^{2}}}$C) $R(T)={{R}_{0}}{{e}^{-T_{0}^{2}/T_{{}}^{2}}}$D) $R(T)={{R}_{0}}{{e}^{T_{{}}^{2}/T_{0}^{2}}}$

Correct Answer: C

Solution :

$\frac{\frac{1}{{{T}^{2}}}}{\frac{1}{T_{0}^{2}}}+\frac{\ell n(T)}{\ell nR({{T}_{0}})}=1$           $\Rightarrow \ell nR(T)=[\ell nR({{T}_{0}})]\left( 1-\frac{T_{0}^{2}}{{{T}^{2}}} \right)$          $\Rightarrow R(T)={{R}_{0}}{{e}^{\left( -\frac{T_{0}^{2}}{{{T}^{2}}} \right)}}$

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