JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • 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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