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JEE Main 2019 (10 Apr, Shift 1) · Ohm's Law, Resistivity & Temperature

Ohm's Law, Resistivity & Temperature question from JEE Main 2019 (10 Apr, Shift 1)

Q48JEE Main 2019 · 10 APR · SHIFT 1Medium

Ohm's Law, Resistivity & Temperature

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)

Graph: $\ln R$ vs $1/T^2$ is a straight line with negative slope.
So $\ln R = \ln R_0 - T_0^2/T^2$ .
$R(T) = R_0 e^{-T_0^2/T^2}$ . The answer is (d).

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