JEE PYQ: Current Electricity Question 50
Question 50 - 2019 (10 Apr 2019 Shift 1)
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.
[Graph: $\ln R(T)$ vs $1/T^2$ is a straight line with negative slope]
One may conclude that:
(1) $R(T) = \frac{R_0}{T^2}$
(2) $R(T) = R_0 e^{-T_0^2/T^2}$
(3) $R(T) = R_0 e^{-T^2/T_0^2}$
(4) $R(T) = R_0 e^{T^2/T_0^2}$
Show Answer
Answer: (2)
Solution
From graph: $\ln R = -m(1/T^2) + c$, where $m$ and $c$ are constants. $R = e^{-m/T^2 + c} = e^c \cdot e^{-m/T^2} = R_0 e^{-T_0^2/T^2}$.
BETA
