According to Arrhenius equation rate constant (k) is equal to ${\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}$. Which of the following options represent the graph of ln k vs $\frac{1}{\mathrm{T}}$?

According to Arrhenius equation,$\mathrm{k}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}$
Taking log on both side ln $\mathrm{k}=\mathrm{In} \left(\mathrm{A}-{\mathrm{e}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}}\right)$
$\Rightarrow \ln \mathrm{k}=\ln \mathrm{A}-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}\Rightarrow \ln \mathrm{k}=-\frac{-{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\times \frac{1}{\mathrm{T}}+\ln \mathrm{A}$
                                          $\mathrm{y}=\mathrm{mx}+\mathrm{c}$
This equation can be related to equation of straight line as shown above. From the graph, it is very clear that slope of the plot $=\frac{-{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}$and intercept = ln A.