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Q.

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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a

$R (T) = \frac{R_{0}}{T^{2}}$

b

$R (T) = R_{0} e^{T_{0}^{2} / T^{2}}$

c

$R (T) = R_{0} e^{T^{2} / T_{0}^{2}}$

answer is B.

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Detailed Solution

detailed_solution_thumbnail

Equation of straight line from graph

$y = - m x + c$

$\Rightarrow l n R = - m (\frac{1}{T^{2}}) + c$

Here, $m & c$ are constants

$R = e^{[- m (\frac{1}{T^{2}}) + c]} = e^{- m (\frac{1}{T^{2}})} \times e^{c} \Rightarrow R (T) = R_{0} e^{\frac{- T_{0}^{2}}{T^{2}}}$

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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