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:
A$R\left( T \right) = {R_o}{e^{{T^2}/T_0^2}}$
B$R\left( T \right) = \frac{{{R_o}}}{{{T^2}}}$
C$R\left( T \right) = {R_o}{e^{-{T^2}/T_0^2}}$
D$R\left( T \right) = {R_o}{e^{T_0^2/{T^2}}}$
JEE MAIN 2019, Diffcult
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C$R\left( T \right) = {R_o}{e^{-{T^2}/T_0^2}}$
c $\frac{\frac{1}{T^{2}}}{\frac{1}{T_{0}^{2}}}+\frac{\ln R(T)}{\ln R\left(T_{0}\right)}=1$
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