Current Electricity

JEE Main level, single correct, medium

80 of 380

Question 80 of 380

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: (D) R(T)=R0e−T02/T2R\left(T\right)={R}_{0}{e}^{-{T}_{0}^{2}/{T}^{2}}


From the given graph,
lnR=−m(1T2)+ClnR=-m\left(\dfrac{1}{{T}^{2}}\right)+C
R=e−m(1T2)+C=e−m(1T2)eCR={e}^{-m\left(\dfrac{1}{{T}^{2}}\right)+C}={e}^{-m\left(\dfrac{1}{{T}^{2}}\right)}{e}^{C}
Further, it can be written as,
R=R0e−T02T2R={R}_{0}{e}^{-\dfrac{{T}_{0}^{2}}{{T}^{2}}}

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