JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Current Electricity

In the figure the vertical axis is $\ln R(T)$ and the horizontal axis is $1/T^2$, and the points fall on a straight line that slopes downward. So $\ln R(T)$ is a linear function of $1/T^2$: $\ln R(T)=\ln R_0+m\left(\frac{1}{T^2}\right)$, where the intercept at $1/T^2=0$ is $\ln R_0$ and $m$ is the slope. Because the line falls, the slope is negative; writing it as $m=-T_0^2$ gives $\ln R(T)=\ln R_0-\frac{T_0^2}{T^2}$. Taking the exponential of both sides, $R(T)=R_0{e}^{-T_0^2/T^2}$. This behaves sensibly: as $T$ grows the exponent $-T_0^2/T^2$ climbs toward $0$ and $R$ rises toward $R_0$, while as $T$ falls the resistance drops steeply, exactly as a straight line in $1/T^2$ demands.