In an experiment,the resistance of a material | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

The resistivity of a wire:

Three copper wires of lengths and cross-sectional areas are $(l, A)$,$(2l, A/2)$,and $(l/2, 2A)$. Resistance is minimum in

$A$ and $B$ are two square plates of the same metal and same thickness,but the length of $B$ is twice that of $A$. The ratio of the resistances of $A$ and $B$ is

$A$ wire of resistance $R$ and radius $r$ is stretched until its radius becomes $r / 2$. If the new resistance of the stretched wire is $x R$,then the value of $x$ is $\qquad$

Vedclass Test Series

Exam Paper Generator

Online Exam Module