In an experiment,the resistance of a material | vedclass

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(D) The graph shows a straight line between $\ln R(T)$ and $1/T^2$ with a negative slope. The equation of a straight line is $y = mx + c$.Here,$y = \l

Vedclass · Accepted Answer

$R(T) = R_0 e^{T_0^2/T^2}$

Vedclass · Answer

(D) The graph shows a straight line between $\ln R(T)$ and $1/T^2$ with a negative slope. The equation of a straight line is $y = mx + c$.Here,$y = \ln R(T)$ and $x = 1/T^2$.So,$\ln R(T) = m(1/T^2) + c$,where $m$ is the negative slope and $c$ is the intercept.Let the intercept be $\ln R_0$ at $1/T^2 = 0$. Then $\ln R(T) = -k(1/T^2) + \ln R_0$,where $k$ is a positive constant.This can be written as $\ln R(T) = \ln R_0 - k/T^2$.Taking the exponential of both sides,we get $R(T) = R_0 e^{-k/T^2}$.Comparing this with the given options,if we set $k = T_0^2$,we get $R(T) = R_0 e^{-T_0^2/T^2}$.However,looking at the options provided,there seems to be a mismatch in the exponent sign or variable definition. Re-evaluating the slope: if the line passes through $(1/T_0^2, 0)$ and $(0, \ln R_0)$,the equation is $\frac{\ln R(T)}{\ln R_0} + \frac{1/T^2}{1/T_0^2} = 1$.$\ln R(T) = \ln R_0 (1 - T_0^2/T^2) = \ln R_0 - \ln R_0 (T_0^2/T^2)$.This implies $R(T) = R_0 e^{-\ln R_0 (T_0^2/T^2)}$.Given the standard form of such problems,the correct relation is $R(T) = R_0 e^{T_0^2/T^2}$ if the slope is positive,or $R(T) = R_0 e^{-T_0^2/T^2}$ if negative. Based on the provided options and the negative slope,option $D$ is the closest mathematical form if $T_0^2$ is defined appropriately.

In an experiment,the resistance of a material is plotted as a function of temperature. As shown in the figure,the plot of $\ln R(T)$ versus $1/T^2$ is a straight line. One may conclude that:

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