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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2019
  • Hard
  • MCQ

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:

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Answer & Solution

Correct Answer

(C) \(R\left( T \right) = {R_o}{e^{-{T^2}/T_0^2}}\)

Step-by-step Solution

Detailed explanation

\(\frac{\frac{1}{T^{2}}}{\frac{1}{T_{0}^{2}}}+\frac{\ln R(T)}{\ln R\left(T_{0}\right)}=1\)

\(\Rightarrow \ln \mathrm{R}(\mathrm{T})=\ln \mathrm{R}\left(\mathrm{T}_{\mathrm{o}}\right)\left(1-\frac{\mathrm{T}_{\mathrm{o}}^{2}}{\mathrm{T}^{2}}\right)\)

\({\rm{R}}({\rm{T}}) = {{\rm{R}}_0}{{\rm{e}}^{ - \left( {\frac{{T_0^2}}{{{T^2}}}} \right)}}\)

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